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[ "", null, "", null, "", null, "", null, "Wrox Programmer Forums", null, "Convert Currency To Words\n | Search | Today's Posts | Mark Forums Read\n Access Discussion of Microsoft Access database design and programming. See also the forums for Access ASP and Access VBA.\n Welcome to the p2p.wrox.com Forums. You are currently viewing the Access section of the Wrox Programmer to Programmer discussions. This is a community of software programmers and website developers including Wrox book authors and readers. New member registration was closed in 2019. New posts were shut off and the site was archived into this static format as of October 1, 2020. If you require technical support for a Wrox book please contact http://hub.wiley.com\n Kenny Alligood", null, "Authorized User Join Date: Jun 2003 Location: , FL, USA. Posts: 91 Thanks: 0 Thanked 0 Times in 0 Posts", null, "Convert Currency To Words\n\nHope someone can help me here.\n\nI am looking for a way to take a currency amount (say \\$1234.56) and convert it to words (like One Thousand Two Hundred Thirty Four Dollars & Fifty-Six Cents).\n\nWhat I am trying to do is create a form that is for check writing and when the user enters that amount of the check the equivalent words of the amount are appended to the appropriate line on the form.\n\nAny assistance would be greatly appreciated.\n\nKenny Alligood\n__________________\nKenny Alligood\n pgtips", null, "Friend of Wrox Join Date: Jun 2003 Location: , , United Kingdom. Posts: 1,212 Thanks: 0 Thanked 1 Time in 1 Post", null, "http://www.fabalou.com/Access/Modules/numbertowords.asp\n Kenny Alligood", null, "Authorized User Join Date: Jun 2003 Location: , FL, USA. Posts: 91 Thanks: 0 Thanked 0 Times in 0 Posts", null, "Never mind folks -- I found the answer. But for anyone that is interested the solution (http://www.ozgrid.com/VBA/CurrencyToWords.htm (from Microsoft)) is this:\n\nPrivate Sub txtAmount_LostFocus()\n\nDim strAmount As String\n\nstrAmount = txtAmount\n\nCall ConvertCurrencyToEnglish(strAmount)\n\nEnd Sub\n\nFunction ConvertCurrencyToEnglish(ByVal strAmount)\nDim Temp\nDim Dollars, Cents\nDim DecimalPlace, Count\n\nReDim Place(9) As String\nPlace(2) = \" Thousand \"\nPlace(3) = \" Million \"\nPlace(4) = \" Billion \"\nPlace(5) = \" Trillion \"\n\n' Convert strAmount to a string, trimming extra spaces.\nstrAmount = Trim(Str(strAmount))\n\n' Find decimal place.\nDecimalPlace = InStr(strAmount, \".\")\n\n' If we find decimal place...\nIf DecimalPlace > 0 Then\n' Convert cents\nTemp = Left(Mid(strAmount, DecimalPlace + 1) & \"00\", 2)\nCents = ConvertTens(Temp)\n\n' Strip off cents from remainder to convert.\nstrAmount = Trim(Left(strAmount, DecimalPlace - 1))\nEnd If\n\nCount = 1\nDo While strAmount <> \"\"\n' Convert last 3 digits of strAmount to English dollars.\nTemp = ConvertHundreds(Right(strAmount, 3))\nIf Temp <> \"\" Then Dollars = Temp & Place(Count) & Dollars\nIf Len(strAmount) > 3 Then\n' Remove last 3 converted digits from strAmount.\nstrAmount = Left(strAmount, Len(strAmount) - 3)\nElse\nstrAmount = \"\"\nEnd If\nCount = Count + 1\nLoop\n\n' Clean up dollars.\nSelect Case Dollars\nCase \"\"\nDollars = \"No Dollars\"\nCase \"One\"\nDollars = \"One Dollar\"\nCase Else\nDollars = Dollars & \" Dollars\"\nEnd Select\n\n' Clean up cents.\nSelect Case Cents\nCase \"\"\nCents = \" And No Cents\"\nCase \"One\"\nCents = \" And One Cent\"\nCase Else\nCents = \" And \" & Cents & \" Cents\"\nEnd Select\n\nlblAmount.Caption = Dollars & Cents\nEnd Function\n\nPrivate Function ConvertHundreds(ByVal strAmount)\nDim Result As String\n\n' Exit if there is nothing to convert.\nIf Val(strAmount) = 0 Then Exit Function\n\n' Append leading zeros to number.\nstrAmount = Right(\"000\" & strAmount, 3)\n\n' Do we have a hundreds place digit to convert?\nIf Left(strAmount, 1) <> \"0\" Then\nResult = ConvertDigit(Left(strAmount, 1)) & \" Hundred \"\nEnd If\n\n' Do we have a tens place digit to convert?\nIf Mid(strAmount, 2, 1) <> \"0\" Then\nResult = Result & ConvertTens(Mid(strAmount, 2))\nElse\n' If not, then convert the ones place digit.\nResult = Result & ConvertDigit(Mid(strAmount, 3))\nEnd If\n\nConvertHundreds = Trim(Result)\nEnd Function\n\nPrivate Function ConvertTens(ByVal MyTens)\nDim Result As String\n\n' Is value between 10 and 19?\nIf Val(Left(MyTens, 1)) = 1 Then\nSelect Case Val(MyTens)\nCase 10: Result = \"Ten\"\nCase 11: Result = \"Eleven\"\nCase 12: Result = \"Twelve\"\nCase 13: Result = \"Thirteen\"\nCase 14: Result = \"Fourteen\"\nCase 15: Result = \"Fifteen\"\nCase 16: Result = \"Sixteen\"\nCase 17: Result = \"Seventeen\"\nCase 18: Result = \"Eighteen\"\nCase 19: Result = \"Nineteen\"\nCase Else\nEnd Select\nElse\n' .. otherwise it's between 20 and 99.\nSelect Case Val(Left(MyTens, 1))\nCase 2: Result = \"Twenty \"\nCase 3: Result = \"Thirty \"\nCase 4: Result = \"Forty \"\nCase 5: Result = \"Fifty \"\nCase 6: Result = \"Sixty \"\nCase 7: Result = \"Seventy \"\nCase 8: Result = \"Eighty \"\nCase 9: Result = \"Ninety \"\nCase Else\nEnd Select\n\n' Convert ones place digit.\nResult = Result & ConvertDigit(Right(MyTens, 1))\nEnd If\n\nConvertTens = Result\nEnd Function\n\nPrivate Function ConvertDigit(ByVal MyDigit)\nSelect Case Val(MyDigit)\nCase 1: ConvertDigit = \"One\"\nCase 2: ConvertDigit = \"Two\"\nCase 3: ConvertDigit = \"Three\"\nCase 4: ConvertDigit = \"Four\"\nCase 5: ConvertDigit = \"Five\"\nCase 6: ConvertDigit = \"Six\"\nCase 7: ConvertDigit = \"Seven\"\nCase 8: ConvertDigit = \"Eight\"\nCase 9: ConvertDigit = \"Nine\"\nCase Else: ConvertDigit = \"\"\nEnd Select\nEnd Function\n\nKenny Alligood\njoeby", null, "Registered User\n Points: 3, Level: 1", null, "", null, "", null, "Activity: 0%", null, "", null, "", null, "Join Date: Oct 2017\nPosts: 1\nThanks: 0\nThanked 0 Times in 0 Posts", null, "Hello thank for the code.\nCan you help me get 131 to read One hundred and thirty one\n\nhow do I introduce the \"and\"\n\nQuote:\n Originally Posted by Kenny Alligood", null, "Never mind folks -- I found the answer. But for anyone that is interested the solution (http://www.ozgrid.com/VBA/CurrencyToWords.htm (from Microsoft)) is this: Private Sub txtAmount_LostFocus()      Dim strAmount As String      strAmount = txtAmount      Call ConvertCurrencyToEnglish(strAmount) End Sub Function ConvertCurrencyToEnglish(ByVal strAmount)          Dim Temp          Dim Dollars, Cents          Dim DecimalPlace, Count          ReDim Place(9) As String          Place(2) = \" Thousand \"          Place(3) = \" Million \"          Place(4) = \" Billion \"          Place(5) = \" Trillion \"          ' Convert strAmount to a string, trimming extra spaces.          strAmount = Trim(Str(strAmount))          ' Find decimal place.          DecimalPlace = InStr(strAmount, \".\")          ' If we find decimal place...          If DecimalPlace > 0 Then             ' Convert cents             Temp = Left(Mid(strAmount, DecimalPlace + 1) & \"00\", 2)             Cents = ConvertTens(Temp)             ' Strip off cents from remainder to convert.             strAmount = Trim(Left(strAmount, DecimalPlace - 1))          End If          Count = 1          Do While strAmount <> \"\"             ' Convert last 3 digits of strAmount to English dollars.             Temp = ConvertHundreds(Right(strAmount, 3))             If Temp <> \"\" Then Dollars = Temp & Place(Count) & Dollars             If Len(strAmount) > 3 Then                ' Remove last 3 converted digits from strAmount.                strAmount = Left(strAmount, Len(strAmount) - 3)             Else                strAmount = \"\"             End If             Count = Count + 1          Loop          ' Clean up dollars.          Select Case Dollars             Case \"\"                Dollars = \"No Dollars\"             Case \"One\"                Dollars = \"One Dollar\"             Case Else                Dollars = Dollars & \" Dollars\"          End Select          ' Clean up cents.          Select Case Cents             Case \"\"                Cents = \" And No Cents\"             Case \"One\"                Cents = \" And One Cent\"             Case Else                Cents = \" And \" & Cents & \" Cents\"          End Select           lblAmount.Caption = Dollars & Cents       End Function      Private Function ConvertHundreds(ByVal strAmount)          Dim Result As String          ' Exit if there is nothing to convert.          If Val(strAmount) = 0 Then Exit Function          ' Append leading zeros to number.          strAmount = Right(\"000\" & strAmount, 3)          ' Do we have a hundreds place digit to convert?          If Left(strAmount, 1) <> \"0\" Then             Result = ConvertDigit(Left(strAmount, 1)) & \" Hundred \"          End If          ' Do we have a tens place digit to convert?          If Mid(strAmount, 2, 1) <> \"0\" Then             Result = Result & ConvertTens(Mid(strAmount, 2))          Else             ' If not, then convert the ones place digit.             Result = Result & ConvertDigit(Mid(strAmount, 3))          End If          ConvertHundreds = Trim(Result)       End Function       Private Function ConvertTens(ByVal MyTens)          Dim Result As String          ' Is value between 10 and 19?          If Val(Left(MyTens, 1)) = 1 Then             Select Case Val(MyTens)                Case 10: Result = \"Ten\"                Case 11: Result = \"Eleven\"                Case 12: Result = \"Twelve\"                Case 13: Result = \"Thirteen\"                Case 14: Result = \"Fourteen\"                Case 15: Result = \"Fifteen\"                Case 16: Result = \"Sixteen\"                Case 17: Result = \"Seventeen\"                Case 18: Result = \"Eighteen\"                Case 19: Result = \"Nineteen\"                Case Else             End Select          Else             ' .. otherwise it's between 20 and 99.             Select Case Val(Left(MyTens, 1))                Case 2: Result = \"Twenty \"                Case 3: Result = \"Thirty \"                Case 4: Result = \"Forty \"                Case 5: Result = \"Fifty \"                Case 6: Result = \"Sixty \"                Case 7: Result = \"Seventy \"                Case 8: Result = \"Eighty \"                Case 9: Result = \"Ninety \"                Case Else             End Select             ' Convert ones place digit.             Result = Result & ConvertDigit(Right(MyTens, 1))          End If          ConvertTens = Result      End Function       Private Function ConvertDigit(ByVal MyDigit)          Select Case Val(MyDigit)             Case 1: ConvertDigit = \"One\"             Case 2: ConvertDigit = \"Two\"             Case 3: ConvertDigit = \"Three\"             Case 4: ConvertDigit = \"Four\"             Case 5: ConvertDigit = \"Five\"             Case 6: ConvertDigit = \"Six\"             Case 7: ConvertDigit = \"Seven\"             Case 8: ConvertDigit = \"Eight\"             Case 9: ConvertDigit = \"Nine\"             Case Else: ConvertDigit = \"\"          End Select       End Function Kenny Alligood", null, "Similar Threads Thread Thread Starter Forum Replies Last Post Can XSLT convert amount to words LeoMathew XSLT 4 January 10th, 2018 04:09 AM Convert Numbers to Words hasankelepir Classic ASP Basics 5 October 1st, 2007 07:15 AM Currency mjbkelly Beginning VB 6 1 March 22nd, 2007 07:14 AM System.Convert parameters for currency zoltac007 General .NET 0 September 27th, 2006 12:52 AM convert number into words on acces form superparim Wrox Book Feedback 0 September 19th, 2005 01:55 PM", null, "" ]

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[ "# Descriptive statistics\n\nA descriptive statistic (in the count noun sense) is a summary statistic that quantitatively describes or summarizes features from a collection of information, while descriptive statistics (in the mass noun sense) is the process of using and analysing those statistics. Descriptive statistics is distinguished from inferential statistics (or inductive statistics) by its aim to summarize a sample, rather than use the data to learn about the population that the sample of data is thought to represent. This generally means that descriptive statistics, unlike inferential statistics, is not developed on the basis of probability theory, and are frequently non-parametric statistics. Even when a data analysis draws its main conclusions using inferential statistics, descriptive statistics are generally also presented. For example, in papers reporting on human subjects, typically a table is included giving the overall sample size, sample sizes in important subgroups (e.g., for each treatment or exposure group), and demographic or clinical characteristics such as the average age, the proportion of subjects of each sex, the proportion of subjects with related co-morbidities, etc.\n\nSome measures that are commonly used to describe a data set are measures of central tendency and measures of variability or dispersion. Measures of central tendency include the mean, median and mode, while measures of variability include the standard deviation (or variance), the minimum and maximum values of the variables, kurtosis and skewness.\n\n## Use in statistical analysis\n\nDescriptive statistics provide simple summaries about the sample and about the observations that have been made. Such summaries may be either quantitative, i.e. summary statistics, or visual, i.e. simple-to-understand graphs. These summaries may either form the basis of the initial description of the data as part of a more extensive statistical analysis, or they may be sufficient in and of themselves for a particular investigation.\n\nFor example, the shooting percentage in basketball is a descriptive statistic that summarizes the performance of a player or a team. This number is the number of shots made divided by the number of shots taken. For example, a player who shoots 33% is making approximately one shot in every three. The percentage summarizes or describes multiple discrete events. Consider also the grade point average. This single number describes the general performance of a student across the range of their course experiences.\n\nThe use of descriptive and summary statistics has an extensive history and, indeed, the simple tabulation of populations and of economic data was the first way the topic of statistics appeared. More recently, a collection of summarisation techniques has been formulated under the heading of exploratory data analysis: an example of such a technique is the box plot.\n\nIn the business world, descriptive statistics provides a useful summary of many types of data. For example, investors and brokers may use a historical account of return behaviour by performing empirical and analytical analyses on their investments in order to make better investing decisions in the future.\n\n### Univariate analysis\n\nUnivariate analysis involves describing the distribution of a single variable, including its central tendency (including the mean, median, and mode) and dispersion (including the range and quartiles of the data-set, and measures of spread such as the variance and standard deviation). The shape of the distribution may also be described via indices such as skewness and kurtosis. Characteristics of a variable's distribution may also be depicted in graphical or tabular format, including histograms and stem-and-leaf display.\n\n### Bivariate and multivariate analysis\n\nWhen a sample consists of more than one variable, descriptive statistics may be used to describe the relationship between pairs of variables. In this case, descriptive statistics include:\n\nThe main reason for differentiating univariate and bivariate analysis is that bivariate analysis is not only simple descriptive analysis, but also it describes the relationship between two different variables. Quantitative measures of dependence include correlation (such as Pearson's r when both variables are continuous, or Spearman's rho if one or both are not) and covariance (which reflects the scale variables are measured on). The slope, in regression analysis, also reflects the relationship between variables. The unstandardised slope indicates the unit change in the criterion variable for a one unit change in the predictor. The standardised slope indicates this change in standardised (z-score) units. Highly skewed data are often transformed by taking logarithms. Use of logarithms makes graphs more symmetrical and look more similar to the normal distribution, making them easier to interpret intuitively.:47" ]

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[ "What is the number of moles in 20g of sodium hydroxide?\n\nContents\n\n⇒ Number of moles which make up 20 grams of NaOH = (20 grams × 1 Mole)/(40 grams) = 0.5 moles.\n\nHow many moles are in 20g of NaOH?\n\n20 gr of NaOH correspond to 0.5 moles which are contained in 0.5 liter of solution.\n\nHow many atoms are there in 20g of sodium hydroxide?\n\n1 mole of Na=6.022*10^23 atoms. therefore in 20g of sodium 5.22*10^23 atoms are present.\n\nHow many moles are in sodium hydroxide?\n\nThe molar mass tells you the mass of exactly one mole of a given substance. In this case, sodium hydroxide has a molar mass of 39.997 g mol−1 , which implies that one mole of sodium hydroxide has a mass of 39.997 g .\n\nHow many moles are in 1g of NaOH?\n\nDetermining the Molar Mass of a Compound\n\nIn a compound of NaOH, the molar mass of Na alone is 23 g/mol, the molar mass of O is 16 g/mol, and H is 1 g/mol. What is the molar mass of NaOH? The molar mass of the compound NaOH is 40 g/mol.\n\nIT IS INTERESTING:  Should I see a dermatologist for stretch marks?\n\nHow do I calculate moles?\n\nHow to find moles?\n\n1. Measure the weight of your substance.\n2. Use a periodic table to find its atomic or molecular mass.\n3. Divide the weight by the atomic or molecular mass.\n4. Check your results with Omni Calculator.\n\nHow many moles of NaOH are in 80.0 g of sodium hydroxide NaOH?\n\n80 gm of NaOH is two moles so you have two moles of sodium thus two times Avogadro’s constant of sodium atoms. 80 gm of NaOH is two moles so you have two moles of sodium thus two times Avogadro’s constant of sodium atoms. Sodium hydroxide is an ionic compound.\n\nWhat is the mass of 1 mole of NaOH?\n\nUsing periodic table you can find molar mass of HCl, which is 36.458 g/mol. This means 1 mol of HCl is 36.458 grams. So 20 g of HCl is 0.548 mol.\n\nHow many sodium atoms are in Naci?\n\nThere are 2 atoms in NaCl. This is because there is 1 atom of Na (sodium) and 1 atom of Cl (chlorine) in each NaCl molecule. Elements by themselves do not have a “number of atoms”- if you’re talking about the atomic number, it’s the number of protons (or electrons in a neutral atom) of an element." ]

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[ "Arduino Serial connection\n\nI've written a small program for Arduino, in which it receives a two digit number from serial, for example \"\"26\". Then it set's HIGH voltage to pin 2 with the duration of 6*100. That works fine, except that I also have a LED connected to pin 4, and it's voltage is set to HIGH at start-up, but when Arduino is receiving/sending through serial, it drops the voltage - is that normal?\n\nThe code is as follows:\n\nint pts = 4;\n\nString input;\n\nvoid flash(int pin, int duration){\nSerial.println(pin);\nSerial.println(duration);\npinMode(pin, OUTPUT);\ndigitalWrite(pin, HIGH);\ndelay(duration);\ndigitalWrite(pin, LOW);\n}\n\nvoid setup() {\ninput=\"\";\ndigitalWrite(pts, HIGH);\npinMode(pts,OUTPUT);\nSerial.begin(9600);\n}\n\nvoid parseInput(){\nif(Serial.available()>0){\n//digitalWrite(led, LOW);\ninput+=incomingByte;\ndelay(10);\n}\nelse{\n\nif(input!=\"\")\n{\nflash(input-'0', (input-'0') *100);\n}\ninput=\"\";\n}\n}\n\nvoid loop()\n{\nparseInput();\n}\n• Is it possible that PIN4 is used by the serial as txd or rxd? If it's so the led might dim but only during transmission or reception. And you are missing the main() function. – Vladimir Cravero Apr 19 '14 at 15:03\n• This is arduino, there's no main – Dzarda Apr 19 '14 at 15:08\n• @MustSeeMelons Just FYI. There is now a stack dedicated to Arduino arduino.stackexchange.com – Nick Alexeev Apr 19 '14 at 15:18\n• So what's the application entry point exactly? I really don't understand arduino :( – Vladimir Cravero Apr 19 '14 at 15:28\n• In Arduino, the main() is part of the library code. The key concept is that it calls setup() once and then calls loop() repeatedly (while also doing other internal functions). – Dave Tweed Apr 19 '14 at 16:01" ]

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[ "[cfe-dev] error: use of overloaded operator '==' is ambiguous\n\nDouglas Gregor dgregor at apple.com\nFri Jul 30 13:41:52 PDT 2010\n\n```On Jul 29, 2010, at 6:58 PM, John McCall wrote:\n\n> On Jul 29, 2010, at 6:03 PM, Shawn Erickson wrote:\n>> Sorry for the longish code example attached to this email but I am\n>> trying to understand if clang is being overly pedantic in this\n>> situation. Also please let me know if I should just file defects\n>> instead of posting to this mailing list (unsure if this is a clang\n>> problem or not).\n>\n> Filing bugs is generally nicer.\n>\n>> It seems like clang should be ignoring the private bool operators when\n>> searching for a candidate.\n>\n> Access doesn't affect visibility in C++: first you decide which operator you want,\n> then you decide whether you can access it.\n>\n>> I assume the built-in ones would be\n>> ignored/not listed if the only direct candidate was int32_t.\n>\n> If by 'direct candidate' you mean a member function candidate, you don't\n> consider member operators unless the left operand is of class type. In this\n> case, Status::eNo has type ResultCodeConst, i.e. int64_t. Neither ADL not\n> unqualified lookup find any user-defined operators, so the only possibilities\n> are the builtin candidates.\n>\n> That said, I think the builtin candidate at (int64_t, int32_t) should be viable;\n> Doug, thoughts?\n\nClang is doing the right thing here (and EDG agrees with Clang that this code is ill-formed due to an ambiguity), but it takes a bit of standards-reading to see why. Let's go with Eli's reduction of the issue, here, although the issue is the same:\n\nstruct A\n{\noperator int();\noperator bool();\n};\n\nint main(int argc, char** argv) {\nif (1000 == A()) {\n}\n}\n\nAnd we'll consider two built-in candidates (of the many required by C++ 13.6 [over.built]p12):\n\noperator==(int, int)\noperator==(int, float)\n\nFor both candidates, the first argument is trivially an exact match. It's the second argument, of type A, that is interesting.\n\nFor operator==(int, int), we perform overload resolution to find a conversion from A to int. Per C++ 13.3.1.5 [over.match.conv]p1, we consider all of the conversion functions in A as candidates. Both conversion functions work, so we perform overload resolution to see which conversion function is better. C++ 13.3.3 [over.match.best]p1 tells us that, since we're performing initialization by user-defined conversion, we can compare the last conversion sequences for the two conversion functions:\n\nfor operator int(), we have an exact match\nfor operator bool(), we have an integral promotion\n\nAn exact match is better than an integral promotion, so we pick the operator int() conversion sequence. Therefore, the conversion sequence for the second argument to operator==(int, int) is identity->user-defined operator int()->identity.\n\nFor operator==(int, float), we again perform overload resolution using the two conversion functions in A, but this time we're converting to float. The last conversion sequences for the two conversion functions are:\n\nfor operator int(), we have an integral-to-floating conversion\nfor operator bool(), we have an integral-to-floating conversion\n\nThe two integral-to-floating conversion functions are indistinguishable, so the conversion sequence for the second argument to operator==(int, float) is the ambiguous conversion sequence (C++ 13.3.3.1 [over.best.ics]p10).\n\nNow, we go to see which of operator==(int, int) or operator==(int, float) is better. The conversion sequences for the first argument are simple: both are exact matches, neither is better than the other. For the second argument, we're comparing a user-defined conversion sequence (using A::operator int()) against the ambiguous conversion sequence. Per C++ 13.3.3.1 [over.best.ics]p10, \"the ambiguous conversion sequence is treated as a user-defined sequence that is indistinguishable from any other user-defined conversion sequence\", meaning that we can't order these two. Hence, there is an ambiguity and the code is ill-formed.\n\nI don't know why GCC doesn't catch this ambiguity. My hunch is that it isn't including all of the built-in operator candidates required by the standard, since I've seen such problems with GCC before.\n\nTo fix the original code, you can add overloaded operator=='s such as\n\nbool operator==(int32_t, ResultCodeClass);\nbool operator==(ResultCodeClass, int32_t);\n\n- Doug\n\n```" ]

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[ "# 1742409\n\n## 1,742,409 is an odd composite number composed of two prime numbers multiplied together.\n\nWhat does the number 1742409 look like?\n\nThis visualization shows the relationship between its 2 prime factors (large circles) and 6 divisors.\n\n1742409 is an odd composite number. It is composed of two distinct prime numbers multiplied together. It has a total of six divisors.\n\n## Prime factorization of 1742409:\n\n### 32 × 193601\n\n(3 × 3 × 193601)\n\nSee below for interesting mathematical facts about the number 1742409 from the Numbermatics database.\n\n### Names of 1742409\n\n• Cardinal: 1742409 can be written as One million, seven hundred forty-two thousand, four hundred nine.\n\n### Scientific notation\n\n• Scientific notation: 1.742409 × 106\n\n### Factors of 1742409\n\n• Number of distinct prime factors ω(n): 2\n• Total number of prime factors Ω(n): 3\n• Sum of prime factors: 193604\n\n### Divisors of 1742409\n\n• Number of divisors d(n): 6\n• Complete list of divisors:\n• Sum of all divisors σ(n): 2516826\n• Sum of proper divisors (its aliquot sum) s(n): 774417\n• 1742409 is a deficient number, because the sum of its proper divisors (774417) is less than itself. Its deficiency is 967992\n\n### Bases of 1742409\n\n• Binary: 1101010010110010010012\n• Base-36: 11CG9\n\n### Squares and roots of 1742409\n\n• 1742409 squared (17424092) is 3035989123281\n• 1742409 cubed (17424093) is 5289934772306923929\n• The square root of 1742409 is 1320.0034090865\n• The cube root of 1742409 is 120.3326188539\n\n### Scales and comparisons\n\nHow big is 1742409?\n• 1,742,409 seconds is equal to 2 weeks, 6 days, 4 hours, 9 seconds.\n• To count from 1 to 1,742,409 would take you about four weeks!\n\nThis is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)\n\n• A cube with a volume of 1742409 cubic inches would be around 10 feet tall.\n\n### Recreational maths with 1742409\n\n• 1742409 backwards is 9042471\n• The number of decimal digits it has is: 7\n• The sum of 1742409's digits is 27\n• More coming soon!\n\nHTML: To link to this page, just copy and paste the link below into your blog, web page or email.\n\nBBCODE: To link to this page in a forum post or comment box, just copy and paste the link code below:\n\nMLA style:\n\"Number 1742409 - Facts about the integer\". Numbermatics.com. 2022. Web. 27 September 2022.\n\nAPA style:\nNumbermatics. (2022). Number 1742409 - Facts about the integer. Retrieved 27 September 2022, from https://numbermatics.com/n/1742409/\n\nChicago style:\nNumbermatics. 2022. \"Number 1742409 - Facts about the integer\". https://numbermatics.com/n/1742409/\n\nThe information we have on file for 1742409 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!\n\nKeywords: Divisors of 1742409, math, Factors of 1742409, curriculum, school, college, exams, university, Prime factorization of 1742409, STEM, science, technology, engineering, physics, economics, calculator, one million, seven hundred forty-two thousand, four hundred nine.\n\nOh no. Javascript is switched off in your browser.\nSome bits of this website may not work unless you switch it on." ]

[ null ]

[ "# Knudsen number\n\nJump to: navigation, search\n\nThe Knudsen number is defined as", null, "$Kn\\equiv \\frac{\\lambda}{L}$\n\nwhere\n\n•", null, "$\\lambda$ is the mean free path of the molecules\n•", null, "$L$ is the characteristic dimension\n\nKnudsen number is a measure of the rarefaction of the flow. The various Flow regimes can be classifed based on Knudsen number.\n\nThe Knudsen number is related to the Mach number and the Reynolds number by the following relation:", null, "$Kn = \\frac{Ma}{Re} \\; \\sqrt{ \\frac{\\pi \\gamma}{2}}$\n\nwhere", null, "$\\gamma$ is the ratio of specific heats." ]

[ null, "https://www.cfd-online.com/W/images/math/3/7/4/374b8de4de8bd80ec70b763c6f34b8df.png ", null, "https://www.cfd-online.com/W/images/math/e/0/5/e05a30d96800384dd38b22851322a6b5.png ", null, "https://www.cfd-online.com/W/images/math/d/2/0/d20caec3b48a1eef164cb4ca81ba2587.png ", null, "https://www.cfd-online.com/W/images/math/a/e/5/ae5787b262a10ca56495f20700da9d48.png ", null, "https://www.cfd-online.com/W/images/math/3/3/4/334de1ea38b615839e4ee6b65ee1b103.png ", null ]

[ "# Scanner nextShort() method in Java with Examples\n\nThe nextShort(radix) method of java.util.Scanner class scans the next token of the input as a short. If the translation is successful, the scanner advances past the input that matched. If the parameter radix is not passed, then it behaves similarly as nextShort(radix) where the radix is assumed to be the default radix.\n\nSyntax:\n\n`public int radix()`\n\nParameters: The function accepts a parameter radix which is used to interpret the token as a short value.\n\nReturn Value: This function returns the short scanned from the input.\n\nExceptions: The function throws three exceptions as described below:\n\n• InputMismatchException: if the next token does not matches the Integer regular expression, or is out of range\n• NoSuchElementException: throws if input is exhausted\n• IllegalStateException: throws if this scanner is closed\n\nBelow programs illustrate the above function:\n\nProgram 1:\n\n `// Java program to illustrate the ` `// nextShort() method of Scanner class in Java ` `// without parameter ` ` `  `import` `java.util.*; ` ` `  `public` `class` `GFG1 { ` `    ``public` `static` `void` `main(String[] argv) ` `        ``throws` `Exception ` `    ``{ ` ` `  `        ``String s = ``\"Gfg 9 + 6 = 12.0\"``; ` ` `  `        ``// create a new scanner ` `        ``// with the specified String Object ` `        ``Scanner scanner = ``new` `Scanner(s); ` ` `  `        ``while` `(scanner.hasNext()) { ` ` `  `            ``// if the next is a short, ` `            ``// print found and the short ` `            ``if` `(scanner.hasNextShort()) { ` `                ``System.out.println(``\"Found Short value :\"` `                                   ``+ scanner.nextShort()); ` `            ``} ` ` `  `            ``// if no short is found, ` `            ``// print \"Not Found:\" and the token ` `            ``else` `{ ` `                ``System.out.println(``\"Not found short value :\"` `                                   ``+ scanner.next()); ` `            ``} ` `        ``} ` `        ``scanner.close(); ` `    ``} ` `} `\n\nOutput:\n\n```Not found short value :Gfg\nFound Short value :9\nFound Short value :6\n```\n\nProgram 2:\n\n `// Java program to illustrate the ` `// nextShort() method of Scanner class in Java ` `// with parameter ` ` `  `import` `java.util.*; ` ` `  `public` `class` `GFG1 { ` `    ``public` `static` `void` `main(String[] argv) ` `        ``throws` `Exception ` `    ``{ ` ` `  `        ``String s = ``\"Gfg 9 + 6 = 12.0\"``; ` ` `  `        ``// create a new scanner ` `        ``// with the specified String Object ` `        ``Scanner scanner = ``new` `Scanner(s); ` ` `  `        ``while` `(scanner.hasNext()) { ` ` `  `            ``// if the next is a short, ` `            ``// print found and the short ` `            ``if` `(scanner.hasNextShort()) { ` `                ``System.out.println(``\"Found Short value :\"` `                                   ``+ scanner.nextShort(``12``)); ` `            ``} ` ` `  `            ``// if no short is found, ` `            ``// print \"Not Found:\" and the token ` `            ``else` `{ ` `                ``System.out.println(``\"Not found short value :\"` `                                   ``+ scanner.next()); ` `            ``} ` `        ``} ` `        ``scanner.close(); ` `    ``} ` `} `\n\nOutput:\n\n```Not found short value :Gfg\nFound Short value :9\nFound Short value :6\n```\n\nProgram 3: To demonstrate InputMismatchException\n\n `// Java program to illustrate the ` `// nextShort() method of Scanner class in Java ` `// InputMismatchException ` ` `  `import` `java.util.*; ` ` `  `public` `class` `GFG1 { ` `    ``public` `static` `void` `main(String[] argv) ` `        ``throws` `Exception ` `    ``{ ` ` `  `        ``try` `{ ` ` `  `            ``String s = ``\"Gfg 9 + 6 = 12.0\"``; ` ` `  `            ``// create a new scanner ` `            ``// with the specified String Object ` `            ``Scanner scanner = ``new` `Scanner(s); ` ` `  `            ``while` `(scanner.hasNext()) { ` ` `  `                ``// if the next is a short, ` `                ``// print found and the short ` `                ``// since the value 60 is out of range ` `                ``// it throws an exception ` ` `  `                ``System.out.println(``\"Next Short value :\"` `                                   ``+ scanner.nextShort()); ` `            ``} ` `            ``scanner.close(); ` `        ``} ` `        ``catch` `(Exception e) { ` `            ``System.out.println(``\"Exception thrown: \"` `+ e); ` `        ``} ` `    ``} ` `} `\n\nOutput:\n\n```Exception thrown: java.util.InputMismatchException\n```\n\nProgram 4: To demonstrate NoSuchElementException\n\n `// Java program to illustrate the ` `// nextShort() method of Scanner class in Java ` `// NoSuchElementException ` ` `  `import` `java.util.*; ` ` `  `public` `class` `GFG1 { ` `    ``public` `static` `void` `main(String[] argv) ` `        ``throws` `Exception ` `    ``{ ` ` `  `        ``try` `{ ` ` `  `            ``String s = ``\"Gfg\"``; ` ` `  `            ``// create a new scanner ` `            ``// with the specified String Object ` `            ``Scanner scanner = ``new` `Scanner(s); ` ` `  `            ``// Trying to get the next short value ` `            ``// more times than the scanner ` `            ``// Hence it will throw exception ` `            ``for` `(``int` `i = ``0``; i < ``5``; i++) { ` ` `  `                ``// if the next is a short, ` `                ``// print found and the short ` `                ``if` `(scanner.hasNextShort()) { ` `                    ``System.out.println(``\"Found Short value :\"` `                                       ``+ scanner.nextShort()); ` `                ``} ` ` `  `                ``// if no short is found, ` `                ``// print \"Not Found:\" and the token ` `                ``else` `{ ` `                    ``System.out.println(``\"Not found short value :\"` `                                       ``+ scanner.next()); ` `                ``} ` `            ``} ` `            ``scanner.close(); ` `        ``} ` `        ``catch` `(Exception e) { ` `            ``System.out.println(``\"Exception thrown: \"` `+ e); ` `        ``} ` `    ``} ` `} `\n\nOutput:\n\n```Not found short value :Gfg\nException thrown: java.util.NoSuchElementException\n```\n\nProgram 5: To demonstrate IllegalStateException\n\n `// Java program to illustrate the ` `// nextShort() method of Scanner class in Java ` `// IllegalStateException ` ` `  `import` `java.util.*; ` ` `  `public` `class` `GFG1 { ` `    ``public` `static` `void` `main(String[] argv) ` `        ``throws` `Exception ` `    ``{ ` ` `  `        ``try` `{ ` ` `  `            ``String s = ``\"Gfg 9 + 6 = 12.0\"``; ` ` `  `            ``// create a new scanner ` `            ``// with the specified String Object ` `            ``Scanner scanner = ``new` `Scanner(s); ` ` `  `            ``// close the scanner ` `            ``scanner.close(); ` `            ``System.out.println(``\"Scanner Closed\"``); ` ` `  `            ``System.out.println(``\"Trying to get \"` `                               ``+ ``\"next Short value\"``); ` ` `  `            ``while` `(scanner.hasNext()) { ` ` `  `                ``// if the next is a short, ` `                ``// print found and the short ` `                ``if` `(scanner.hasNextShort()) { ` `                    ``System.out.println(``\"Found Short value :\"` `                                       ``+ scanner.nextShort()); ` `                ``} ` ` `  `                ``// if no short is found, ` `                ``// print \"Not Found:\" and the token ` `                ``else` `{ ` `                    ``System.out.println(``\"Not found short value :\"` `                                       ``+ scanner.next()); ` `                ``} ` `            ``} ` `        ``} ` `        ``catch` `(Exception e) { ` `            ``System.out.println(``\"Exception thrown: \"` `+ e); ` `        ``} ` `    ``} ` `} `\n\nOutput:\n\n```Scanner Closed\nTrying to get next Short value\nException thrown: java.lang.IllegalStateException: Scanner closed\n```\n\nMy Personal Notes arrow_drop_up", null, "The function of education is to teach one to think intensively and to think critically\n\nIf you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to [email protected]. See your article appearing on the GeeksforGeeks main page and help other Geeks.\n\nPlease Improve this article if you find anything incorrect by clicking on the \"Improve Article\" button below.\n\nArticle Tags :\nPractice Tags :\n\nBe the First to upvote.\n\nPlease write to us at [email protected] to report any issue with the above content." ]

[ null, "https://media.geeksforgeeks.org/auth/profile/6m2zhcgue5xmoaktvc88", null ]

[ "# How can I add a signal to enter/exit trades on this indicator?", null, "1147\n\nHello everyone!\n\nHow can I add signal/buffer to this indicator to open and exit trades? I tried to add another buffer and read it, but I get numbers like \"2.3454682334345 e-308\" or just 0.", null, "405\n\nPedro Severin:\n\nHello everyone!\n\nHow can I add signal/buffer to this indicator to open and exit trades? I tried to add another buffer and read it, but I get numbers like \"2.3454682334345 e-308\" or just 0.\n\nPerhaps you could post the code where you've made the changes? That way people here will be better able to help...", null, "1147\n\nSeng Joo Thio:\n\nPerhaps you could post the code where you've made the changes? That way people here will be better able to help...\n\nHi Seng,\n\nSure...But I didn't change much of it. I will put it in yellow background for easier understanding. This is the full code, which can be found in the codebase.\n\nMy idea was to add a buffer that can give me a signal when the indicator generates signals. I also ran the indicator in backtest mode to see how many signals the indicator gives itself in 3 years (used the daily chart) and surprisingly it gave like 3-4 signals.\n\nAny idea why?\n\n```//+------------------------------------------------------------------+\n//|                                       Waddah_Attar_Explosion.mq4 |\n//|                              Copyright © 2006, Eng. Waddah Attar |\n//|                                          [email protected] |\n//+------------------------------------------------------------------+\n#property copyright \"Copyright © 2006, Eng. Waddah Attar\"\n#property link \"[email protected]\"\n//---- indicator version\n#property version   \"1.00\"\n//---- drawing the indicator in a separate window\n#property indicator_separate_window\n//---- number of indicator buffers 4\n#property indicator_buffers 5\n\n//---- only three plots are used\n#property indicator_plots   3\n//+-----------------------------------+\n//|  Indicator drawing parameters     |\n//+-----------------------------------+\n//---- drawing the indicator as a three-color histogram\n#property indicator_type1 DRAW_COLOR_HISTOGRAM\n//---- Gray, Lime and Magenta colors are used for the three-color histogram\n#property indicator_color1 clrGray,clrLime,clrRed\n//---- indicator line is a solid one\n#property indicator_style1 STYLE_SOLID\n//---- indicator line width is equal to 2\n#property indicator_width1 2\n//---- displaying the indicator label\n#property indicator_label1 \"MACD\"\n\n//---- drawing the indicator as a line\n#property indicator_type2 DRAW_LINE\n//---- use blue color for the line\n#property indicator_color2 Blue\n//---- indicator line is a solid curve\n#property indicator_style2 STYLE_SOLID\n//---- indicator line width is equal to 2\n#property indicator_width2 2\n//---- displaying the signal line label\n#property indicator_label2  \"Signal Line\"\n\n//---- drawing the indicator as a line\n#property indicator_type3 DRAW_LINE\n//---- use red color for the line\n#property indicator_color3 clrGold\n//---- the indicator line is a dash-dotted curve\n#property indicator_style3 STYLE_DASHDOTDOT\n//---- indicator line width is equal to 1\n#property indicator_width3 1\n//---- displaying the signal line\n#property indicator_label3  \"DeadZonePip Level\"\n\n#property  indicator_minimum 0.0\n//+------------------------------------+\n//|  Indicator input parameters        |\n//+------------------------------------+\ninput int Fast_MA = 20;       // Period of the fast MACD moving average\ninput int Slow_MA = 40;       // Period of the slow MACD moving average\ninput int BBPeriod=20;        // Bollinger period\ninput double BBDeviation=2.0; // Number of Bollinger deviations\ninput int  Sensetive=150;\ninput int  DeadZonePip=400;\ninput int  ExplosionPower=15;\ninput int  TrendPower=150;\ninput bool AlertWindow=false;\ninput int  AlertCount=2;\ninput bool AlertLong=false;\ninput bool AlertShort=false;\ninput bool AlertExitLong=false;\ninput bool AlertExitShort=false;\n//+-----------------------------------+\n//---- declaration of the integer variables for the start of data calculation\nint min_rates_total;\n//---- declaration of integer variables for the indicators handles\nint MACD_Handle,BB_Handle;\n//---- declaration of global variables\ndouble bask,bbid;\nint LastTime1,LastTime2,LastTime3,LastTime4,Status,PrevStatus;\n//---- declaration of dynamic arrays that\n//---- will be used as indicator buffers\ndouble IndBuffer1[],ColorIndBuffer1[],IndBuffer2[],IndBuffer3[];\ndouble signalTrades[];\n//+------------------------------------------------------------------+\n//| MACD indicator initialization function                           |\n//+------------------------------------------------------------------+\nvoid OnInit()\n{\n//---- initialization of variables of the start of data calculation\nmin_rates_total=BBPeriod+1;\n//---- initialization of variables\nLastTime1=1;\nLastTime2=1;\nLastTime3=1;\nLastTime4=1;\nStatus=0;\nPrevStatus=-1;\n//---- getting handle of the iMACD indicator\nMACD_Handle=iMACD(NULL,0,Fast_MA,Slow_MA,9,PRICE_CLOSE);\nif(MACD_Handle==INVALID_HANDLE)Print(\" Failed to get handle of the iMACD indicator\");\n//---- getting handle of the iBands indicator\nBB_Handle=iBands(NULL,0,BBPeriod,0,BBDeviation,PRICE_CLOSE);\nif(BB_Handle==INVALID_HANDLE)Print(\" Failed to get handle of the iBands indicator\");\n\n//---- set IndBuffer1[] dynamic array as an indicator buffer\nSetIndexBuffer(0,IndBuffer1,INDICATOR_DATA);\n//---- performing the shift of the beginning of the indicator drawing\nPlotIndexSetInteger(0,PLOT_DRAW_BEGIN,min_rates_total);\n//---- setting the indicator values that won't be visible on a chart\nPlotIndexSetDouble(0,PLOT_EMPTY_VALUE,0.0);\n//---- indexing the elements in buffers as timeseries\nArraySetAsSeries(IndBuffer1,true);\n\n//---- set ColorIndBuffer1[] as a colored index buffer\nSetIndexBuffer(1,ColorIndBuffer1,INDICATOR_COLOR_INDEX);\n//---- performing the shift of the beginning of the indicator drawing\nPlotIndexSetInteger(1,PLOT_DRAW_BEGIN,min_rates_total);\n//---- indexing the elements in buffers as timeseries\nArraySetAsSeries(ColorIndBuffer1,true);\n\n//---- set IndBuffer2[] as an indicator buffer\nSetIndexBuffer(2,IndBuffer2,INDICATOR_DATA);\n//---- performing the shift of the beginning of the indicator drawing\nPlotIndexSetInteger(2,PLOT_DRAW_BEGIN,min_rates_total);\n//---- setting the indicator values that won't be visible on a chart\nPlotIndexSetDouble(2,PLOT_EMPTY_VALUE,0.0);\n//---- indexing the elements in buffers as timeseries\nArraySetAsSeries(IndBuffer2,true);\n\n//---- set IndBuffer3[] as an indicator buffer\nSetIndexBuffer(3,IndBuffer3,INDICATOR_DATA);\n//---- performing the shift of the beginning of the indicator drawing\nPlotIndexSetInteger(3,PLOT_DRAW_BEGIN,min_rates_total);\n//---- setting the indicator values that won't be visible on a chart\nPlotIndexSetDouble(3,PLOT_EMPTY_VALUE,0.0);\n//---- indexing the elements in buffers as timeseries\nArraySetAsSeries(IndBuffer3,true);\n\nSetIndexBuffer(4,signalTrades);\nArraySetAsSeries(signalTrades,true);\n\n//---- creating a name for displaying in a separate sub-window and in a tooltip\nIndicatorSetString(INDICATOR_SHORTNAME,\"Waddah Attar Explosion\");\n//---- determination of accuracy of displaying the indicator values\nIndicatorSetInteger(INDICATOR_DIGITS,_Digits+1);\n//---- initialization end\n}\n//+------------------------------------------------------------------+\n//| MACD iteration function                                          |\n//+------------------------------------------------------------------+\nint OnCalculate(const int rates_total,     // number of bars in history at the current tick\nconst int prev_calculated, // number of bars calculated at previous call\nconst datetime &time[],\nconst double &open[],\nconst double &high[],\nconst double &low[],\nconst double &close[],\nconst long &tick_volume[],\nconst long &volume[],\nconst int &spread[])\n{\n//---- checking the number of bars to be enough for the calculation\nif(BarsCalculated(MACD_Handle)<rates_total\n|| BarsCalculated(BB_Handle)<rates_total\n|| rates_total<min_rates_total)\nreturn(0);\n//---- declarations of local variables\nint limit,to_copy,bar;\ndouble MACD[],BandsUp[],BandsDn[];\ndouble Trend1,Trend2,Explo1,Explo2,Dead;\ndouble pwrt,pwre,Ask,Bid;\nstring SirName;\n//---- indexing elements in arrays as timeseries\nArraySetAsSeries(spread,true);\nArraySetAsSeries(close,true);\nArraySetAsSeries(MACD,true);\nArraySetAsSeries(BandsUp,true);\nArraySetAsSeries(BandsDn,true);\n\n//---- calculation of the 'first' starting index for the bars recalculation loop\nif(prev_calculated>rates_total || prev_calculated<=0) // checking for the first start of the indicator calculation\n{\nlimit=rates_total-min_rates_total-1; // starting index for calculation of all bars\n}\nelse\n{\nlimit=rates_total-prev_calculated;   // starting index for calculation of new bars\n}\n\nto_copy=limit+2;\n//--- copy newly appeared data in the arrays\nif(CopyBuffer(BB_Handle,1,0,to_copy,BandsUp)<=0) return(0);\nif(CopyBuffer(BB_Handle,2,0,to_copy,BandsDn)<=0) return(0);\nto_copy+=2;\nif(CopyBuffer(MACD_Handle,0,0,to_copy,MACD)<=0) return(0);\n\n//---- main indicator calculation loop\nfor(bar=limit; bar>=0; bar--)\n{\nTrend1=(MACD[bar] - MACD[bar+1])*Sensetive;\nTrend2=(MACD[bar+2] - MACD[bar+3])*Sensetive;\nExplo1=BandsUp[bar] - BandsDn[bar];\nExplo2=BandsUp[bar+1] - BandsDn[bar+1];\n\nDead=_Point*DeadZonePip;\n\nIndBuffer1[bar]=MathAbs(Trend1);\n\nBid=close[bar];\nAsk=Bid+spread[bar];\n\nif(Trend1>0) ColorIndBuffer1[bar]=1;\nif(Trend1<0) ColorIndBuffer1[bar]=2;\nIndBuffer2[bar]=Explo1;\nIndBuffer3[bar]=Dead;\nif(bar==0)\n{\nif(Trend1>0 && Trend1>Explo1 && Trend1>Dead &&\nExplo1>Dead && Explo1>Explo2 && Trend1>Trend2 &&\nLastTime1<AlertCount && AlertLong==true && Ask!=bask)\n{\npwrt=100*(Trend1 - Trend2)/Trend1;\npwre=100*(Explo1 - Explo2)/Explo1;\nbask=Ask;\nif(pwre>=ExplosionPower && pwrt>=TrendPower)\n{\nsignalTrades=1.0;\n\nSirName=\"\";\nStringConcatenate(SirName,LastTime1,\"- \",Symbol(),\" - BUY \",\" (\",\nDoubleToString(bask,_Digits),\") Trend PWR \",\nDoubleToString(pwrt,0),\" - Exp PWR \",DoubleToString(pwre,0));\n\nif(AlertWindow==true) Alert(SirName);\nelse                  Print(SirName);\n\nLastTime1++;\n}\nStatus=1;\n}\nif(Trend1<0 && MathAbs(Trend1)>Explo1 && MathAbs(Trend1)>Dead &&\nExplo1>Dead && Explo1>Explo2 && MathAbs(Trend1)>MathAbs(Trend2) &&\nLastTime2<AlertCount && AlertShort==true && Bid!=bbid)\n{\npwrt=100*(MathAbs(Trend1) - MathAbs(Trend2))/MathAbs(Trend1);\npwre=100*(Explo1 - Explo2)/Explo1;\nbbid=Bid;\nif(pwre>=ExplosionPower && pwrt>=TrendPower)\n{\nsignalTrades=2.0;\nSirName=\"\";\nStringConcatenate(SirName,LastTime2,\"- \",Symbol(),\" - Sell \",\" (\",\nDoubleToString(bask,_Digits),\") Trend PWR \",\nDoubleToString(pwrt,0),\" - Exp PWR \",DoubleToString(pwre,0));\n\nif(AlertWindow==true) Alert(SirName);\nelse                  Print(SirName);\n\nLastTime2++;\n}\n\nStatus=2;\n}\nif(Trend1>0 && Trend1<Explo1 && Trend1<Trend2 && Trend2>Explo2 &&\nTrend1>Dead && Explo1>Dead && LastTime3<=AlertCount &&\nAlertExitLong==true && Bid!=bbid)\n{\nbbid=Bid;\nSirName=\"\";\nStringConcatenate(SirName,LastTime3,\"- \",Symbol(),\" - Exit BUY \",\" \",DoubleToString(bbid,_Digits));\nsignalTrades=-1.0;\n\nif(AlertWindow==true) Alert(SirName);\nelse                  Print(SirName);\n\nStatus=3;\nLastTime3++;\n}\nif(Trend1<0 && MathAbs(Trend1)<Explo1 &&\nMathAbs(Trend1)<MathAbs(Trend2) && MathAbs(Trend2)>Explo2 &&\nTrend1>Dead && Explo1>Dead && LastTime4<=AlertCount &&\nAlertExitShort==true && Ask!=bask)\n{\nbask=Ask;\nSirName=\"\";\nStringConcatenate(SirName,LastTime4,\"- \",Symbol(),\" - Exit SELL \",\" \",DoubleToString(bask,_Digits));\nsignalTrades=-2.0;\nif(AlertWindow==true) Alert(SirName);\nelse                  Print(SirName);\n\nStatus=4;\nLastTime4++;\n}\nPrevStatus=Status;\n}\nif(Status!=PrevStatus)\n{\nLastTime1=1;\nLastTime2=1;\nLastTime3=1;\nLastTime4=1;\n}\n}\n//----\nreturn(rates_total);\n}\n//+------------------------------------------------------------------+```", null, "405\n\nPedro Severin:\n\nHi Seng,\n\nSure...But I didn't change much of it. I will put it in yellow background for easier understanding. This is the full code, which can be found in the codebase.\n\nMy idea was to add a buffer that can give me a signal when the indicator generates signals. I also ran the indicator in backtest mode to see how many signals the indicator gives itself in 3 years (used the daily chart) and surprisingly it gave like 3-4 signals.\n\nAny idea why?\n\nOk, I did some tests, and found that\n\n```   SetIndexBuffer(4,signalTrades);\nArraySetAsSeries(signalTrades,true);\n```\n\nshould be changed to:\n\n```   SetIndexBuffer(4,signalTrades,INDICATOR_CALCULATIONS);\nPlotIndexSetInteger(4,PLOT_SHOW_DATA,false);\nArraySetAsSeries(signalTrades,true);\n```\n\nbefore I'm able to retrieve the values of signalTrades in my test EA.\n\nNext, I realised I need to set these flags to true for signalTrades to be filled:\n\n```input bool AlertLong=true;\ninput bool AlertShort=true;\ninput bool AlertExitLong=true;\ninput bool AlertExitShort=true;\n```\n\nLastly, you're right that signals are rare, but you can lower the values of DeadZonePip, ExplosionPower and TrendPower to get more signals.", null, "1147\n\nSeng Joo Thio:\n\nOk, I did some tests, and found that\n\nshould be changed to:\n\nbefore I'm able to retrieve the values of signalTrades in my test EA.\n\nNext, I realised I need to set these flags to true for signalTrades to be filled:\n\nLastly, you're right that signals are rare, but you can lower the values of DeadZonePip, ExplosionPower and TrendPower to get more signals.\n\nThank you Seng!\n\nIt worked in someway, but for some reason if I run a backtest from, lets say 2011 up to now, I get like 2-3 signals.\n\nThen if I run the same test but this time from 2019 up to now, I get only 2 signals.\n\nThis happens even if I try backtesting the pure indicator. I'm backtesting in OHLC mode to save some time.\n\nThis is how I'm getting the signal from the buffer.\n\n```int wae()\n{\n\nCopyBuffer(wHandle,4,0,1,wSignal2);     //signal buffer\n\nif(wSignal2==1) return 1; //buy\nif(wSignal2==2) return 2; //sell\nif(wSignal2==-1) return -1; //exit buy\nif(wSignal2==-2) return -2; //exit sell\n\nreturn 0;\n}```\n\nI also changed the default parameters to some lower ones that can give me more signals, but it does not seems to work in backtest.\n\nIts like in backtest, when it gives the first signal, it remains \"trapped\" in some point. I checked the code of the indicator, but couldn't figure it out.\n\nAny idea why this happens?", null, "1147\n\nIn fact, I don't know if the indicator is ment to give signals everytime the bars goes over the line, but I would like to use this like a volume indicator/exit indicator (I think its designed for volume, but honstly I think it shows entry points and exits pretty well)\n\nHow can I incorporate this into an EA? Maybe just putting the logic of the indicator as a function?", null, "405\n\nPedro Severin:\n\nThank you Seng!\n\nIt worked in someway, but for some reason if I run a backtest from, lets say 2011 up to now, I get like 2-3 signals.\n\nThen if I run the same test but this time from 2019 up to now, I get only 2 signals.\n\nThis happens even if I try backtesting the pure indicator. I'm backtesting in OHLC mode to save some time.\n\nThis is how I'm getting the signal from the buffer.\n\nI also changed the default parameters to some lower ones that can give me more signals, but it does not seems to work in backtest.\n\nIts like in backtest, when it gives the first signal, it remains \"trapped\" in some point. I checked the code of the indicator, but couldn't figure it out.\n\nAny idea why this happens?\n\nYou're right. The problem lies with these status flags:\n\n```int LastTime1,LastTime2,LastTime3,LastTime4,Status,PrevStatus;\n```\n\nLooking at the code, I seriously think that they were not being set/reset properly. E.g. \"PrevStatus=Status;\" must not be called right after the 4 big 'ifs' that checks the open/close conditions. So that is likely the direction you should head towards in checking through the execution.", null, "405\n\nPedro Severin:\n\nIn fact, I don't know if the indicator is ment to give signals everytime the bars goes over the line, but I would like to use this like a volume indicator/exit indicator (I think its designed for volume, but honstly I think it shows entry points and exits pretty well)\n\nHow can I incorporate this into an EA? Maybe just putting the logic of the indicator as a function?\n\nIt can be done, just remove all indicator-specific lines and group the remaining lines into a function in your EA. May not be necessary to debug the status flags issues if you're going to EA path, because in EA, you'll have access to other means to check order/trade status.", null, "1147\n\nSeng Joo Thio:\n\nIt can be done, just remove all indicator-specific lines and group the remaining lines into a function in your EA. May not be necessary to debug the status flags issues if you're going to EA path, because in EA, you'll have access to other means to check order/trade status.\n\nHmm I see...so the status thing is messing around.\n\nWhat do you think of this? All buffer arrays are set as series.\n\nI want to use this indicator as exit indicator, but it seems that it can be a good entry/volume indicator too. Thoughts?\n\n```int wae()\n{\nCopyBuffer(macdHandle,0,0,6,macd);\nCopyBuffer(bbHandle,1,0,6,bbUp);\nCopyBuffer(bbHandle,2,0,6,bbDown);\n\ndouble Trend1=(macd-macd)*Sensetive;\ndouble Trend2=(macd - macd)*Sensetive;\ndouble Explo1=bbUp - bbDown;\ndouble Explo2=bbUp - bbDown;\nAsk=SymbolInfoDouble(Symbol(),SYMBOL_ASK);\nBid=SymbolInfoDouble(Symbol(),SYMBOL_BID);\ndouble pwrt,pwre;\n\ndouble Dead=_Point*DeadZonePip;\n\nint retVal=0;\n\nif(Trend1>0 && Trend1>Explo1 && Trend1>Dead &&\nExplo1>Dead && Explo1>Explo2 && Trend1>Trend2 &&\nAsk!=bask)\n{\npwrt=100*(Trend1 - Trend2)/Trend1;\npwre=100*(Explo1 - Explo2)/Explo1;\nbask=Ask;\nif(pwre>=ExplosionPower && pwrt>=TrendPower)\n{\nretVal=1;//BUY\n\n}\n}\nif(Trend1<0 && MathAbs(Trend1)>Explo1 && MathAbs(Trend1)>Dead &&\nExplo1>Dead && Explo1>Explo2 && MathAbs(Trend1)>MathAbs(Trend2) &&\nBid!=bbid)\n{\npwrt=100*(MathAbs(Trend1) - MathAbs(Trend2))/MathAbs(Trend1);\npwre=100*(Explo1 - Explo2)/Explo1;\nbbid=Bid;\nif(pwre>=ExplosionPower && pwrt>=TrendPower)\n{\nretVal=2;//SELL\n\n}\n}\nif(Trend1>0 && Trend1<Explo1 && Trend1<Trend2 && Trend2>Explo2 &&\nTrend1>Dead && Explo1>Dead && Bid!=bbid)\n{\nbbid=Bid;\nretVal=-1;//Exit BUY\n\n}\nif(Trend1<0 && MathAbs(Trend1)<Explo1 &&\nMathAbs(Trend1)<MathAbs(Trend2) && MathAbs(Trend2)>Explo2 &&\nTrend1>Dead && Explo1>Dead && Ask!=bask)\n{\nbask=Ask;\nretVal=-2;//Exit SELL\n}\n\nreturn retVal;\n}```", null, "405\n\nPedro Severin:\n\nHmm I see...so the status thing is messing around.\n\nWhat do you think of this? All buffer arrays are set as series.\n\nI want to use this indicator as exit indicator, but it seems that it can be a good entry/volume indicator too. Thoughts?\n\nI think you're almost done :). Here're my thoughts:\n\n- Consider looking from bar 1 onwards, not 0... because 0 refers to the current bar, and will still change before the the next bar. So i would change all my references to macd, bbup, bbdown to use their indexes from 1 onwards.\n\n- I won't call wae every tick. instead, i'll keep track of the latest bar start time, to make sure that i call wae only when the bar changes.\n\n```static datetime latestbartime = 0;\n\nif (latestbartime<iTime(NULL,0,0))\n{\n// call wae\nlatestbartime = iTime(NULL,0,0);\n}```\n\n- There is probably no need for bbid and bask anymore.", null, "1147\n\nSeng Joo Thio:\n\nI think you're almost done :). Here're my thoughts:\n\n- Consider looking from bar 1 onwards, not 0... because 0 refers to the current bar, and will still change before the the next bar. So i would change all my references to macd, bbup, bbdown to use their indexes from 1 onwards.\n\n- I won't call wae every tick. instead, i'll keep track of the latest bar start time, to make sure that i call wae only when the bar changes.\n\n- There is probably no need for bbid and bask anymore.\n\nLike this?\n\n```int wae()\n{\nCopyBuffer(macdHandle,0,0,6,macd);\nCopyBuffer(bbHandle,1,0,6,bbUp);\nCopyBuffer(bbHandle,2,0,6,bbDown);\n\ndouble Trend1=(macd-macd)*Sensetive;\ndouble Trend2=(macd - macd)*Sensetive;\ndouble Explo1=bbUp - bbDown;\ndouble Explo2=bbUp - bbDown;\n//Ask=SymbolInfoDouble(Symbol(),SYMBOL_ASK);\n//Bid=SymbolInfoDouble(Symbol(),SYMBOL_BID);\ndouble pwrt,pwre;\n\ndouble Dead=_Point*DeadZonePip;\n\nint retVal=0;\n\nif(Trend1>0 && Trend1>Explo1 && Trend1>Dead &&\nExplo1>Dead && Explo1>Explo2 && Trend1>Trend2 &&\n/*Ask!=bask*/)\n{\npwrt=100*(Trend1 - Trend2)/Trend1;\npwre=100*(Explo1 - Explo2)/Explo1;\n//bask=Ask;\nif(pwre>=ExplosionPower && pwrt>=TrendPower)\n{\nretVal=1;//BUY\n\n}\n}\nif(Trend1<0 && MathAbs(Trend1)>Explo1 && MathAbs(Trend1)>Dead &&\nExplo1>Dead && Explo1>Explo2 && MathAbs(Trend1)>MathAbs(Trend2) /*&&\nBid!=bbid*/)\n{\npwrt=100*(MathAbs(Trend1) - MathAbs(Trend2))/MathAbs(Trend1);\npwre=100*(Explo1 - Explo2)/Explo1;\n//bbid=Bid;\nif(pwre>=ExplosionPower && pwrt>=TrendPower)\n{\nretVal=2;//SELL\n\n}\n}\nif(Trend1>0 && Trend1<Explo1 && Trend1<Trend2 && Trend2>Explo2 &&\nTrend1>Dead && Explo1>Dead /*&& Bid!=bbid*/)\n{\n//bbid=Bid;\nretVal=-1;//Exit BUY\n\n}\nif(Trend1<0 && MathAbs(Trend1)<Explo1 &&\nMathAbs(Trend1)<MathAbs(Trend2) && MathAbs(Trend2)>Explo2 &&\nTrend1>Dead && Explo1>Dead /*&& Ask!=bask*/)\n{\n//bask=Ask;\nretVal=-2;//Exit SELL\n}\n\nreturn retVal;\n}```\n\nI used the part that I posted in the last post and got this results, lol:", null, "It seems that, for some reason I have a valid setup to enter, but at the same time it gets wiped out because of wae. Maybe because I was using the current bar 0?\n\nTo add comments, please log in or register" ]

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[ "# Natural Frequency\n\n### Simply supported bridge\n\n$f \\left\\lbrack \\text{Hz} \\right\\rbrack = \\frac{17.75}{\\sqrt{\\delta\\ \\left\\lbrack \\text{mm} \\right\\rbrack}}$\n\nFor derivation see [natural-frequency-simply-supported-bridge].\n\nSymbol Description\n$$f$$ Natural frequency\n$$\\delta$$ Deflection of structure under dead-load\n\n### Two and three-span continuous\n\nSimplified methods for finding the natural frequency of other bridge arrangements can be found:\n\n• BS 5400-2:2006 or BD37/01 Clause B.2.3" ]

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[ "I'm trying to reverse engineer a key exchange protocol which is based on the Diffie-Hellman key exchange scheme and faced with the following problem:\n\nLet $g$ be the generator, $m$ the modulus.\n\nAlice has a private key $a$ and computes her public key $A = g^a\\pmod{m}$.\nBob has a private key $b$ and computes his public key $B = g^b\\pmod{m}$.\n\nDuring exchanges Bob generates password check signature:\n\n• $C = f(A,B,password)$ where $f$ is cryptographic hash function.\n• $D = f(m,g,password)$\n• $X = \\left(\\left(D+A\\right)\\pmod{m}\\right)^\\left(C+b\\right)\\pmod{m}$\n\nand then sends $f(X)$ to Alice.\n\nAlice calculates $X$ on her side without $b$ and I can't figure out how.\n\nCould somebody help me with the algorithm on Alice's side?\n\n• Is my interpretation of \"You don't know what $\\hspace{.02 in}f$ is and hope that something about $\\hspace{.02 in}f$ can be $\\hspace{.9 in}$ worked out from the fact that Alice can calculate X\" correct? $\\;$ – user991 Jun 10 '13 at 8:24\n• No, I know that $f$ is a cryptographic hash function e.g. sha1, and I'm sure that Alice can calculate X, but she doesn't know b and Bob doesn't send value of X he sends $f(X)$. I think $X = f_1(K,password)$ where K is the shared secret between Alice and Bob. They both know K and password, so there must be an expression equivalent to X on the Alice's side. – dr.g100k Jun 10 '13 at 12:36\n• it maybe that: $X \\equiv (DA)^{C+b} \\pmod{m}$, then it can be rewritten $X \\equiv g^{(d+a)(C+b)} \\pmod{m}$, so Alice can get $X$ by this way $X \\equiv (g^CB)^{d+a}\\pmod{m}$ , you can check that – T.B Sep 3 '13 at 1:17" ]

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[ "Search a number\nBaseRepresentation\nbin11110101010\n32200200\n4132222\n530322\n613030\n75502\noct3652\n92620\n101962\n111524\n121176\n13b7c\n14a02\n158ac\nhex7aa\n\n1962 has 12 divisors (see below), whose sum is σ = 4290. Its totient is φ = 648.\n\nThe previous prime is 1951. The next prime is 1973. The reversal of 1962 is 2691.\n\nSubtracting 1962 from its reverse (2691), we obtain a 6-th power (729 = 36).\n\nIt can be divided in two parts, 19 and 62, that added together give a 4-th power (81 = 34).\n\nIt is an interprime number because it is at equal distance from previous prime (1951) and next prime (1973).\n\nIt can be written as a sum of positive squares in only one way, i.e., 1521 + 441 = 39^2 + 21^2 .\n\nIt is a Smith number, since the sum of its digits (18) coincides with the sum of the digits of its prime factors.\n\nIt is a Harshad number since it is a multiple of its sum of digits (18), and also a Moran number because the ratio is a prime number: 109 = 1962 / (1 + 9 + 6 + 2).\n\nIt is a nude number because it is divisible by every one of its digits.\n\nIt is one of the 548 Lynch-Bell numbers.\n\nIt is a plaindrome in base 15 and base 16.\n\nIt is an unprimeable number.\n\n1962 is an untouchable number, because it is not equal to the sum of proper divisors of any number.\n\nIt is a pernicious number, because its binary representation contains a prime number (7) of ones.\n\nIt is a polite number, since it can be written in 5 ways as a sum of consecutive naturals, for example, 37 + ... + 72.\n\n1962 is an abundant number, since it is smaller than the sum of its proper divisors (2328).\n\nIt is a pseudoperfect number, because it is the sum of a subset of its proper divisors.\n\n1962 is a wasteful number, since it uses less digits than its factorization.\n\n1962 is an odious number, because the sum of its binary digits is odd.\n\nThe sum of its prime factors is 117 (or 114 counting only the distinct ones).\n\nThe product of its digits is 108, while the sum is 18.\n\nThe square root of 1962 is about 44.2944691807. The cubic root of 1962 is about 12.5189047278.\n\nThe spelling of 1962 in words is \"one thousand, nine hundred sixty-two\".\n\nDivisors: 1 2 3 6 9 18 109 218 327 654 981 1962" ]

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[ "# Convert 30.5 Celsius to Fahrenheit (30.5 c to f)\n\n What is 30.5 Celsius in Fahrenheit? Celsius: ℃ Fahrenheit: ℉\n\nYou may also interested in: Fahrenheit to Celsius Converter\n\nThe online Celsius to Fahrenheit (c to f) Converter is used to convert temperature from Degrees Celsius (℃) to Fahrenheit (℉).\n\n#### The Celsius to Fahrenheit Formula to convert 30.5 C to F\n\nHow to convert 30.5 Celsius to Fahrenheit? You can use the following formula to convert Celsius to Fahrenheit :\n\nX(℉)\n= Y(℃) ×\n9 / 5\n+ 32\n\nTo convert 30.5 Celsius to Fahrenheit: ?\n\nX(℉)\n= 30.5(℃) ×\n9 / 5\n+ 32\n\n#### Frequently asked questions to convert C to F\n\nHow to convert 103 celsius to fahrenheit ?\n\nHow to convert 167 celsius to fahrenheit ?\n\nHow to convert 105 celsius to fahrenheit ?\n\nHow to convert 184 celsius to fahrenheit ?\n\nHow to convert 137 celsius to fahrenheit ?\n\nHow to convert 194 celsius to fahrenheit ?\n\nTo convert from degrees Celsius to Fahrenheit instantly, please use our Celsius to Fahrenheit Converter for free.\n\n#### Best conversion unit for 30.5 ℃\n\nThe best conversion unit defined in our website is to convert a number as the unit that is the lowest without going lower than 1. For 30.5 ℃, the best unit to convert to is 30.5 ℃." ]

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[ "# Aptitude Percentage Test Paper 7\n\n31) If 40% of 60% of 3/5 of a number is 504, what will be the 25% of 2/5 of that number?\n\n1. 350\n2. 150\n3. 175\n4. 230\n\nExplanation:\nLet the number is x.\n(40/100) * (60/100) * (3/5) * x = 504\nx= 3500\nNow, (25/100)*(2/5)* 3500 = 350\n\n32) There are 35 students and 6 teachers in a class, each student gets toffees that are 20% of the total number of students, and each teacher gets toffees that are 40 % of the total number of students. How many toffees were there?\n\n1. 245\n2. 405\n3. 329\n4. 142\n\nExplanation:\nSolution 1:\nTotal number of toffees given to students= Number of students * Toffees given to one student\n= 35 * 20% of 35\n= 35* (20/100) * 35= 245\nTotal number of toffees given to teachers = Number of teachers * Toffees given to one teacher\n= 40 * 40% of 35 = 84\nSo, total number of toffees = 245+84 = 329\nSolution 2:\n20 % of 35 = 7 and 40% of 35 = 14\n35 students receive = 35*7=245 toffees\n6 teachers receive = 6*14=84 toffees\nSo, Total number of toffees = 245 + 84 = 329\n\n33) If 10% of A's income = 15% of B's income = 20% of C's income. If the sum of their income is 7800, find the income of B.\n\n1. 1800\n2. 2400\n3. 3000\n4. 3600\n\nExplanation:\nLet 1% =1\nThen, 10% = 10, 15% = 15, and 20% = 20\nLCM of 10, 15, and 20 is 60.\nThe ratio of the incomes of A: B: C = 6:4:3\nThe sum of the ratio is 13, but ATQ the sum=7800\nSo, multiply each one with 600\nHence, the income of B = 4*600 = 2400\n\n34) If 50% of (x-y) =30% of (x+y), what percent of x is y?\n\n1. 33.33%\n2. 40%\n3. 12.25%\n4. 25%\n\nExplanation:\n(50/100)*(x-y) = (30/100)*(x+y)\n(x-y)/2 = (3/10)(x+y)\n(x-y) = (3/5) * (x+y)\n5x-5y = 3x+3y\n2x=8y\ny= (1/4) of x\ni.e., y is 25% of x.\n\n35) If a number x is 10% less than another number y and y is 10% more than 125, find the value of x.\n\n1. 150\n2. 125\n3. 123.75\n4. None of these\n\nExplanation:\nWe can write 10% increase as 11/10 that indicates +1 increase per 10\nSimilarly, we can write 10% decrease as 9/10 that indicates -1 decrease per 10\nSo, the required number = 125 * (11/10) * (9/10) = 123.75\n\nPercentage Aptitude Test Paper 1\nPercentage Aptitude Test Paper 2\nPercentage Aptitude Test Paper 3\nPercentage Aptitude Test Paper 4\nPercentage Aptitude Test Paper 5\nPercentage Aptitude Test Paper 6\nPercentage Aptitude Test Paper 8\nPercentage Concepts\n\n### Feedback", null, "", null, "", null, "" ]

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[ "## ››Convert point [TeX] to centimetre\n\n point [TeX] centimeter\n\n Did you mean to convert point [Adobe] point [Britain, US] point [Didot] point [TeX] to centimeter\n\nHow many point [TeX] in 1 centimeter? The answer is 28.452755906694.\nWe assume you are converting between point [TeX] and centimetre.\nYou can view more details on each measurement unit:\npoint [TeX] or centimeter\nThe SI base unit for length is the metre.\n1 metre is equal to 2845.2755906694 point [TeX], or 100 centimeter.\nNote that rounding errors may occur, so always check the results.\nUse this page to learn how to convert between points and centimetres.\nType in your own numbers in the form to convert the units!\n\n## ››Quick conversion chart of point [TeX] to centimeter\n\n1 point [TeX] to centimeter = 0.03515 centimeter\n\n10 point [TeX] to centimeter = 0.35146 centimeter\n\n20 point [TeX] to centimeter = 0.70292 centimeter\n\n30 point [TeX] to centimeter = 1.05438 centimeter\n\n40 point [TeX] to centimeter = 1.40584 centimeter\n\n50 point [TeX] to centimeter = 1.7573 centimeter\n\n100 point [TeX] to centimeter = 3.5146 centimeter\n\n200 point [TeX] to centimeter = 7.0292 centimeter\n\n## ››Want other units?\n\nYou can do the reverse unit conversion from centimeter to point [TeX], or enter any two units below:\n\n## Enter two units to convert\n\n From: To:\n\n## ››Definition: Centimeter\n\nA centimetre (American spelling centimeter, symbol cm) is a unit of length that is equal to one hundreth of a metre, the current SI base unit of length. A centimetre is part of a metric system. It is the base unit in the centimetre-gram-second system of units. A corresponding unit of area is the square centimetre. A corresponding unit of volume is the cubic centimetre.\n\nThe centimetre is a now a non-standard factor, in that factors of 103 are often preferred. However, it is practical unit of length for many everyday measurements. A centimetre is approximately the width of the fingernail of an adult person.\n\n## ››Metric conversions and more\n\nConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3\", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!" ]

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[ "## Euler Archive - All Works\n\n#### Title\n\nDe mensura angulorum solidorum\n\n#### English Title\n\nOn the measure of solid angles\n\n514\n\n334\n\nLatin\n\nJournal article\n\n1781\n\n1775\n\n#### Content Summary\n\nEuler notes that plane angles may be measured by the circular arcs that they subtend. He also notes that a very clever geometer, Albert Girard (1595-1632), suggested measuring solid angles in the same way, by the part of a sphere that they subtend. He also defines the \"area of an angle\" of a sphere to be the area of the wedge-shaped region bounded by semicircles intersecting at the given angle: if the angle in radians measures a, then the area is 2ar2. If the radius is 1, as we will assume from now on, then the area is 2a. Euler then states and proves (with attribution) Girard's theorem: the area of a spherical triangle is always equal to the angle by which the sum of all three angles of the triangle exceeds two right angles. This gives the area of the triangle in terms of the angles. He then states his general problem: to give the area of the triangle in terms of the lengths of the sides. Euler manages to find a formula, which he follows with lots of examples. Then he derives a couple of rules for finding the measure of solid angles and wraps it up by measuring the solid angles of the regular solids.\n\n#### Original Source Citation\n\nActa Academiae Scientiarum Imperialis Petropolitanae, Volume 1778: II, pp. 31-54.\n\n#### Opera Omnia Citation\n\nSeries 1, Volume 26, pp.204-223.\n\n2018-09-25\n\nCOinS" ]

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[ "HDnyhudndunduundundndu\n\nWhat is 1,562 divided by 34 in partial quotients\n\n34  | 1562 |         34 won't divide into 1 or 15, but will divide into 156, say 4 times\n\n1360 |  40   34 * 4 = 136. Add on a zero (0) to make up 4 digits, and subtract\n\n202 |         34 won't divide into 2 or 20, but will divide into 202, say 5 times\n170 |    5   34 * 5 = 170. No need to add a zero (0), so just subtract\n\n32 |          32 is less than 34 so this is the remainder\n\nAdd the numbers on the rhs. 40 + 5 = 45\n\nAnswer: 1562 divided by 34 is 45 remainder 32\n\nby Level 11 User (80.9k points)" ]

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[ "# Find the 12th term from the end of the AP: –2, –4, –6,..., –100. - Mathematics\n\nSum\n\nFind the 12th term from the end of the AP: –2, –4, –6,..., –100.\n\n#### Solution\n\nGiven AP : –2, –4, –6,…., –100\n\nHere, first term (a) = –2\n\nCommon difference (d) = –4 – (–2) = –2\n\nAnd the last term (l) = –100\n\nWe know that, the nth term of an AP\n\nFrom the end an = l – (n – 1 )d\n\nWhere l is the last term\n\nAnd d is the common difference.\n\n∴ 12th term from the end\n\ni.e., a12 = –100 – (12 – 1)(–2)\n\n= –100 + (11)(2)\n\n= – 100 + 22\n\n= –78\n\nHence the 12th term from the end is –78.\n\nConcept: Arithmetic Progression\nIs there an error in this question or solution?\n\n#### APPEARS IN\n\nNCERT Mathematics Exemplar Class 10\nChapter 5 Arithematic Progressions\nExercise 5.3 | Q 16 | Page 53\nShare" ]

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[ "# Exploring Kaggle Titanic data with R Packages nanair and UpSetR\n\nExploring Kaggle Titanic data with R Packages nanair and UpSetR\n\nRecently (6/8/2018), I saw a post about a new R package “nanair”, which according to the package documentation, “provides data structures and functions that facilitate the plotting of missing values and examination of imputations. This allows missing data dependencies to be explored with minimal deviation from the common work patterns of ‘ggplot2’ and tidy data.”  nanair is authored by Nicholas Tierney, Di Cook, Miles McBain, Colin Fay, Jim Hester, and Luke Smith.\nIn addition to the user manual, there is a github repo with an excellent readme and links to really good vignettes and tutorials.  I decided to try nanair out on the Titanic dataset on Kaggle, as a way to look at missing values.  The code for this article is on github, and includes many other examples not detailed here.  The main feature of nanair is the creation of “shadow matrices” which generate columns with binary values describing if there are missing data in the original data.frame.  However, there are many other nice features, some of which I’ll highlight here.\nnanair’s handling of missing data can start right from reading a file.  Using the pipe notation, it looks like this:\ntrain_data <- read.csv(“train.csv”, header = TRUE, stringsAsFactors = FALSE) %>%     replace_with_na_all(condition = ~.x == “”)\nThe replace_with_na_all function allows you to replace any number of specific values, including blanks, with NA when you read the file.  Thus, if you have data that has blanks and NAs, you can clean that up in one operation.  Likewise, if there are values like -99 etc. you want to convert, you can do that there too.  Having had headaches where some data read in as blanks, and others NA, I really like this addition.  There are other variants of the replace_with_na available as well.\nLet’s start by looking at a simple, not prettified, ggplot of the age data in the dataset.  Note that in my code, I have put the test and train data into a list, hence the list notation below.\n\ni <- 1\nggplot(data = data_list[[i]],    aes(x = PassengerId, y = Age)) +    geom_point() +    labs(title = paste0(“ages for all passengersn”,                               “for “, names(data_list)[i], ” data”)) +    theme(plot.title = element_text(size = 12, hjust = 0.5)) +    theme(legend.title = element_blank())\nwhich produces this plot:\n\nggplot does not represent the missing values, so this chart is actually misleading.  Here’s where nanair can help.  nanair provides functions that mirror some ggplot functions, but explicitly handle missing values.  In this case, we’ll replace geom_point with geom_miss_point:\ni <- 1\nggplot(data = data_list[[i]],    aes(x = PassengerId, y = Age)) +    geom_miss_point() +    labs(title = paste0(“ages for all passengersn”,                               “for “, names(data_list)[i], ” data”)) +    scale_color_manual(labels = c(“Missing”, “Has Age”),    values = c(“coral2”, “aquamarine3”)) +    theme(plot.title = element_text(size = 12, hjust = 0.5)) +    theme(legend.title = element_blank())\nwhich produces this:\n\nMuch nicer!  What nanair geom_miss_point does in this case, is separate the missing values, then offset them from the other data (in this case, by putting them below 0).  The nanair functions work nicely with the ggplot enhancements, such as facets:\ni <- 1\nggplot(data = data_list[[i]],    aes(x = PassengerId, y = Age)) +     geom_miss_point() +     facet_wrap(~Pclass, ncol = 2) +     theme(legend.position = “bottom”) +    labs(title = paste0(“ages for all passengersn”,                               “for “, names(data_list)[i], ” datan”,                               “by passenger class”)) +    scale_color_manual(labels = c(“Missing”, “Has Age”),    values = c(“coral2”, “aquamarine3”)) +    theme(plot.title = element_text(size = 12, hjust = 0.5)) +    theme(legend.title = element_blank())\nproducing:\n\nSo we can quickly and easily see that there are more missing ages for the third class passengers.  Age is an important feature in the Titanic data set, so understanding its missingness informs what do do about it.\nA nice feature of the shador matrix is that it comes along for the ride in other operations.  Here, we impute the ages for the train data using the impute_lm function from package simputation (authored by Mark van der Loo).  Note that for the train we can leverage the Survived label which we cannot do for the test data.\ni <- 1\ndata_list[[i]] %>%    bind_shadow() %>%    impute_lm(Age ~ Fare + Survived + Pclass + Sex + Embarked) %>%    ggplot(aes(x = PassengerId,    y = Age,    color = Age_NA)) +     geom_point() +    labs(title = paste0(“ages for all passengersn”,                               “for “, names(data_list)[i], ” datan”,                               “with missing values imputed”)) +    scale_color_manual(labels = c(“Has Age”, “Missing”),    values = c(“aquamarine3”, “coral2”)) +    theme(plot.title = element_text(size = 12, hjust = 0.5)) +    theme(legend.title = element_blank())\ngenerating this:\n\nIn the above, the bind_shadow() function generates and attaches the shadow matrix to the original data.frame.\nAnother function in the nanair package is as_shadow_upset, which works with the UpSetR package to generate plots showing permutations of the missing values.\ni <- 1\ndata_list[[i]] %>%    as_shadow_upset() %>%    upset(main.bar.color = “aquamarine3”,    matrix.color = “red”,     text.scale = 1.5,    nsets = 3)\nWhich produces this great visualization:\n\nThe upset() function comes from UpSetR, and the combination of these two shows us missing values, combination of missing values, and the number in each case in two dimensions.  I had not used UpSetR before seeing it in the nanair vignettes, so props to its authors Jake Conway, and Nils Gehlenborg.\nAs a last example, nanair provides gg_mis_var() to generate basic plots of missingness:\ni <- 1\ngg_miss_var(data_list[[i]]) +    labs(title = paste0(“missing data for “,                                names(data_list)[i], ” data”)) +    theme(plot.title = element_text(size = 12, hjust = 0.5))\nproducing this:\n\nThere is much more you can do with nanair, and I would recommend you read the excellent tutorials and other information they have provided.  I think the value will really come through for more complex data sets and enhance may EDA workflows.\n\nLink: Exploring Kaggle Titanic data with R Packages nanair and UpSetR" ]

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[ "Courses\n\n# Test: Arun Sharma Based Level 2: Blood Relations\n\n## 25 Questions MCQ Test General Intelligence and Reasoning for SSC CHSL | Test: Arun Sharma Based Level 2: Blood Relations\n\nDescription\nThis mock test of Test: Arun Sharma Based Level 2: Blood Relations for SSC helps you for every SSC entrance exam. This contains 25 Multiple Choice Questions for SSC Test: Arun Sharma Based Level 2: Blood Relations (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Arun Sharma Based Level 2: Blood Relations quiz give you a good mix of easy questions and tough questions. SSC students definitely take this Test: Arun Sharma Based Level 2: Blood Relations exercise for a better result in the exam. You can find other Test: Arun Sharma Based Level 2: Blood Relations extra questions, long questions & short questions for SSC on EduRev as well by searching above.\nQUESTION: 1\n\n### Pointing to a photograph of a boy Suresh said, \"He is the son of the only son of my mother.\" How is Suresh related to that boy?\n\nSolution:\n\nThe boy in the photograph is the only son of the son of Suresh's mother i.e., the son of Suresh. Hence, Suresh is the father of boy.\n\nQUESTION: 2\n\n### If A + B means A is the mother of B; A - B means A is the brother B; A % B means A is the father of B and A x B means A is the sister of B, which of the following shows that P is the maternal uncle of Q?\n\nSolution:\n\nP - M → P is the brother of M\n\nM + N → M is the mother of N\n\nN x Q → N is the sister of Q\n\nTherefore, P is the maternal uncle of Q.\n\nQUESTION: 3\n\n### If A is the brother of B; B is the sister of C; and C is the father of D, how D is related to A?\n\nSolution:\n\nIf D is Male, the answer is Nephew.\n\nIf D is Female, the answer is Niece.\n\nAs the sex of D is not known, hence, the relation between D and A cannot be determined.\n\nNote: Niece - A daughter of one's brother or sister, or of one's brother-in-law or sister-in-law. Nephew - A son of one's brother or sister, or of one's brother-in-law or sister-in-law.\n\nQUESTION: 4\n\nIf A + B means A is the brother of B; A - B means A is the sister of B and A x B means A is the father of B. Which of the following means that C is the son of M?\n\nSolution:\n\nM x N → M is the father of N\n\nN - C → N is the sister of C\n\nand C + F → C is the brother of F.\n\nHence, M is the father of C or C is the son of M.\n\nQUESTION: 5\n\nIntroducing a boy, a girl said, \"He is the son of the daughter of the father of my uncle.\" How is the boy related to the girl?\n\nSolution:\n\nThe father of the boy's uncle → the grandfather of the boy and daughter of the grandfather → sister of father.\n\nQUESTION: 6\n\nPointing to a photograph Lata says, \"He is the son of the only son of my grandfather.\" How is the man in the photograph related to Lata?\n\nSolution:\n\nThe man in the photograph is the son of the only son of Lata's grandfather i.e., the man is the son of Lata's father. Hence, the man is the brother of Lata.\n\nQUESTION: 7\n\nIf A + B means A is the brother of B; A x B means A is the son of B; and A % B means B is the daughter of A then which of the following means M is the maternal uncle of N?\n\nSolution:\n\nBecause the sex of O is not known.\n\nQUESTION: 8\n\nIf D is the brother of B, how B is related to C? To answer this question which of the statements is/are necessary?\n\n1. The son of D is the grandson of C.\n2. B is the sister of D.\n\nSolution:\n\nGiven: D is the brother of B.\n\nFrom statement 1, we can detect that D is son of C (son of D is the grandson of C).\n\nFrom statement 2, we can detect that B is 'Female' (sister of D).\n\nTherefore, B is daughter of C.\n\nQUESTION: 9\n\nIf A + B means A is the father of B; A - B means A is the brother B; A % B means A is the wife of B and A x B means A is the mother of B, which of the following shows that M is the maternal grandmother of T?\n\nSolution:\n\nM x N → M is the mother of N\n\nN % S → N is the wife of S\n\nand S + T → is the father of T.\n\nHence, M is the maternal grandmother of T.\n\nQUESTION: 10\n\nPointing to a photograph. Bajpai said, \"He is the son of the only daughter of the father of my brother.\" How Bajpai is related to the man in the photograph?\n\nSolution:\n\nThe man in the photo is the son of the sister of Bajpai. Hence, Bajpai is the maternal uncle of the man in the photograph.\n\nQUESTION: 11\n\nDeepak said to Nitin, \"That boy playing with the football is the younger of the two brothers of the daughter of my father's wife.\" How is the boy playing football related to Deepak?\n\nSolution:\n\nFather's wife → mother. Hence, the daughter of the mother means sister and sister's younger brother means brother. Therefore, the boy is the brother of Deepak.\n\nQUESTION: 12\n\nPointing a photograph X said to his friend Y, \"She is the only daughter of the father of my mother.\" How X is related to the person of photograph?\n\nSolution:\n\n'The only daughter of the father of X's mother' means mother of X.\n\nHence X is the son of the lady in the photograph.\n\nNote: Still have doubt like \"How X is a male?\" - Let's see the discussions.\n\nQUESTION: 13\n\nVeena who is the sister-in-law of Ashok, is the daughter-in-law of Kalyani. Dheeraj is the father of Sudeep who is the only brother of Ashok. How Kalyani is related to Ashok?\n\nSolution:\n\nAshok is the only brother of Sudeep and Veena is the sister-in-law of Ashok. Hence Veena is the wife of Sudeep. Kalyani is the mother-in-law of Veena. Kalyani is the mother of Ashok.\n\nQUESTION: 14\n\nIf A + B means A is the sister of B; A x B means A is the wife of B, A % B means A is the father of B and A - B means A is the brother of B. Which of the following means T is the daughter of P?\n\nSolution:\n\nP x Q → P is the wife of Q\n\nQ % R → Q is the father of R\n\nR - T → R is the brother of T\n\nT + S → T is the sister of S.\n\nTherefore, T is the daughter of P.\n\nQUESTION: 15\n\nPointing to a woman, Abhijit said, \"Her granddaughter is the only daughter of my brother.\" How is the woman related to Abhijit?\n\nSolution:\n\nDaughter of Abhijit's brother → niece of Abhijit. Thus the granddaughter of the woman is Abhijit's niece.\n\nHence, the woman is the mother of Abhijit.\n\nQUESTION: 16\n\nAmit said - \"This girl is the wife of the grandson of my mother\". How is Amit related to the girl?\n\nSolution:\n\nGrandson of Amit’s brother means Son of Amit in this situation. Then the girl is the wife of Amit’s Son that makes Ami the Father-in-law of the Girl.\n\nQUESTION: 17\n\nA and B are children of D. Who is the father of A? To answer this question which of the statements (1) and (2) is necessary?\n\n1. C is the brother of A and the son of E.\n2. F is the mother B.\n\nSolution:\n\nA and B are children of D.\nFrom (1), C is the brother B and son of E.\nSince, the sex of D and E are not known. Hence (1) is not sufficient to answer the question.\nFrom (2). F is the mother of B. Hence, F is also the mother of A. Hence D is the father of A.\nThus, (2) is sufficient to answer the question.\n\nQUESTION: 18\n\nPointing towards a man, a woman said, \"His mother is the only daughter of my mother.\" How is the woman related to the man?\n\nSolution:\n\nOnly daughter of my mother → myself.\n\nHence, the woman is the mother of the man.\n\nQUESTION: 19\n\nIf P \\$ Q means P is the brother of Q; P # Q means P is the mother of Q; P * Q means P is the daughter of Q in A # B \\$ C * D, who is the father?\n\nSolution:\n\nA is the mother of B, B is the brother of C and C is the daughter of D. Hence, D is the father.", null, "QUESTION: 20\n\nIntroducing Sonia, Aamir says, \"She is the wife of only nephew of only brother of my mother.\" How Sonia is related to Aamir?\n\nSolution:\n\nBrother of mother means maternal uncle. Hence only nephew of Aamir's maternal uncle means Aamir himself. Therefore Sonia is the wife of Aamir.\n\nQUESTION: 21\n\nIf A + B means A is the brother of B; A % B means A is the father of B and A x B means A is the sister of B. Which of the following means M is the uncle of P?\n\nSolution:\n\nM + K → M is the brother of K\n\nK % T → K is the father of T\n\nT x P → T is the sister of P\n\nTherefore, K is the father of P and M is the uncle of P.\n\nQUESTION: 22\n\nPointing to Varman, Madhav said, \"I am the only son of one of the sons of his father.\" How is Varman related to Madhav?\n\nSolution:\n\nMadhav is the only son of one of the sons of Varman's father → Either Varman is the father or uncle of Madhav.\n\nQUESTION: 23\n\nIntroducing a woman, Shashank said, \"She is the mother of the only daughter of my son.\" How that woman is related to Shashank?\n\nSolution:\n\nThe woman is the mother of Shashank's granddaughter. Hence, the woman is the daughter-in-law of Shashank.\n\nQUESTION: 24\n\nIf A + B means B is the brother of A; A x B means B is the husband of A; A - B means A is the mother of B and A % B means A is the father of B, which of the following relations shows that Q is the grandmother of T?\n\nSolution:\n\nQ - P → Q is the mother of P\n\nP + R → R is the brother of P\n\nHence, → q is the mother of R\n\nR % T → R is the father of T.\n\nHence, Q is the grandmother of T.\n\nQUESTION: 25\n\n1. A3P means A is the mother of P\n2. A4P means A is the brother of P\n3. A9P means A is the husband of P\n4. A5P means A is the daughter of P\n\nWhich of the following means that K is the mother-in-law of M?\n\nSolution:\n\nM9N → M is the husband of N\n\nN5K → N is the daughter of K\n\nHence, → M is the son-in-law of K\n\nK3J → K is the mother of J\n\nHence, → K is a lady\n\nHence, → K is the mother-in-law of M." ]

[ null, "https://cdn3.edurev.in/ApplicationImages/Temp/f738dd29-5afe-47e3-809c-8fcb85a59ad7_lg.jpg", null ]

[ "# Difference between revisions of \"Dam break flow\"\n\nThis article discusses the often catastrophic flows that result from the failure of high dams that protect low-lying land. Dam break mechanisms are not dealt with; for this the reader is referred to the extensive literature existing on this subject, see for example Zhang et al. (2016) and Almog et al. (2011) .\n\n## Introduction\n\nFile:Dijkdoorbraak1953.png\nFig. 1. Sea dikes protecting low-lying polders in the Netherlands were breached during the extreme storm surge of 31 January 1953.\n\nDam failure can lead to disastrous situations. Most dam break tragedies are related to the collapse of reservoir dams in mountain rivers. The failure of sea dikes can also cause major disasters, although in this case the level difference is not as large. A dramatic example is the failure of more than fifty sea dikes in the Netherlands during the extreme storm surge of 1953, see figure 1. Many of these dikes protected land lying below the average sea level, while the storm surge level exceeded more than five meters. Rapid flooding killed more than two thousand people who were unable to flee in time to safe places. The most common dam failure mechanisms are related to overtopping and seepage (also called piping or internal erosion). In the case of the 1953 storm surge, overtopping and subsequent scour of the interior dike slope was the most important dike failure mechanism.", null, "Fig. 2. Left panel: Schematic representation of water retention behind a dam. Right panel: Positive downstream and negative upstream surges following instantaneous dam removal.\n\nThe consequences of dam failure have been studied for more than a century. It nevertheless remains a challenging topic due to the high non-linearity of the flood wave propagation. The problem can be tackled with numerical models, but rough estimates and insight into the tidal wave dynamics can be obtained with analytical solution methods.\n\n## Frictionless dam-break flow: analytical solution\n\nAnalytical solutions relate to idealized situations as depicted in Fig. 2. The initial situation consists of an infinite reservoir with a water level that is $h_0$ higher than the horizontal ground level downstream of the dam. Dam break is simulated by instantaneous removal of the dam. This causes a positive surge in the positive $x$-direction and a negative surge in the negative $x$-direction. The floor is dry in front of the positive surge and the body of water behind the negative surge is undisturbed: water level =$h_0$ and current speed $u = 0$.", null, "Fig. 3. The earth-filled South Fork Dam on Lake Conemaugh (Pennsylvania, US) collapsed in 1889, killing 2,209 people in downstream villages.\n\nThe study of the dam break-flow was triggered by several dam failures in the 19th century, in particular the breach of the South Fork Dam in Pennsylvania (USA) in 1889, Fig. 3. Three years later Ritter published an exact analytical solution for dam-break flow by assuming that frictional effects can be ignored. The derivation is given in box 1. According to this solution, the tip of the positive surge advances at high speed (supercritical flow) given by $u_f = 2 c_0$, where $c_0 = \\sqrt {gh_0}$ and $g$ is the gravitational acceleration. The shape of the positive and negative wave is a concave-up parabola,\n\n$h(x,t)= h_0, \\; x\\lt -c_0 t; \\quad h(x,t)= \\Large\\frac{h_0}{9}\\normalsize \\; (2 - \\Large\\frac{x}{c_0 t}\\normalsize)^2 , \\, -c_0 t\\lt x\\lt 2c_0 t; \\quad h(x,t)= 0, \\, x\\gt 2c_0 t , \\qquad (1)$\n\nsee Fig. 4. According to Eq. (1) and Box 1, the discharge through the breach per unit width is given by\n\n$q = \\frac{8}{27} h_0 \\sqrt{g h_0}. \\qquad (2)$\n\nFigure 4 also shows the shape of the dam break wave that was observed in laboratory experiments. For small values of $x / t$, the frictionless solution corresponds fairly well with the observations. However, the front zone is different: the shape is a bull nose, rather than a sharp edge. The front also advances more slowly; the speed of the front decreases with time and is closer to $u_f = c_0$ than to $u_f = 2 c_0$. The reason for this difference is the neglect of frictional effects that are important in the thin fluid layer near the front.\n\nFile:DamBreakFlowDerivation.png\nBox 1. Derivation of the solution for frictionless dam break flow.\n\n## Dam-break flow with friction", null, "Fig. 4. Wave profile after dam break. The red curve corresponds to the frictionless solution Eq. 1. The blue band represents data from different laboratory experiments performed at times $\\small t \\sim (40-80) \\, \\sqrt {h_0 / g}\\normalsize$ after removal of the dam . The frictionless solution only depends on $x/t$. Observed wave profiles also depend on time $t$ because of the decreasing wave tip speed.\n\nWhen friction terms are included in the flow equations, there is no exact analytical solution. It is generally assumed that the frictionless flow equations are approximately valid in a short period after dam break and for small values of $x / (c_0 t)$. In the front zone, where the water depth is small, the momentum balance is dominated by frictional momentum dissipation. If the inertia terms $\\partial u / \\partial t + u \\partial u / \\partial x$ in the momentum equation are ignored, then the current in the front region is mainly determined by the balance of gravitational acceleration $g \\partial h / \\partial x$ and shear stress $\\tau$, the latter term being proportional to the square of the flow velocity. With such models the shape of the wave front is given by $h(s) \\propto s^n, \\; n=1/2$, where $s$ is the distance measured from the wave front. The wave front speed $u_f$ decreases with time. At large times $t\\gt \\gt t_{\\infty}$, the wave front speed varies approximately as $u_f \\approx c_0 \\sqrt{t_{\\infty}/t}$ . Here, $t_{\\infty} = \\sqrt{h_0/g}/(3 c_D)$ and $c_D$ is the friction coefficient (value in the range $\\approx (1-5)\\, 10^{-3}$). A more elaborate model of the frictional boundary layer near the wave front was studied by Nielsen (2018) . According to this model, the wavefront is even more blunt, corresponding to $n = 1/4$, which agrees well with laboratory experiments. This model also reveals a strong downward velocity component at the wavefront, which can contribute to the dislodging of particles from the sediment bed . This may explain the high near-bed of suspended concentrations associated with propagating surges observed in the field, see also the article Tidal bore dynamics.\n\nData from numerous reservoir dam breaks that have occurred in the past have provided empirical formulas for the maximum discharge $Q_{max}$ through the breach . From this data it was deduced that the width of the breach is mainly related to the volume of water $V$ in the reservoir (volume above breach level). A reasonable estimate of the width is given by $B \\approx 0.3 V ^ {1/3}$. Empirical formulas for the maximum dam break discharge then yield \n\n$Q_{max} \\approx 0.04 \\sqrt{g} \\, V^{0.37} h_0^{1.4} \\approx 0.15 h_0 \\, B \\, \\sqrt{g h_0} \\, V^{0.04} h_0^{-0.1} , \\qquad (3)$\n\nor\n\n$q_{max} \\approx 0.22 h_0 \\, \\sqrt{g h_0} , \\qquad (4)$\n\nwhere in the approximation for the maximum discharge per unit width we have considered a large reservoir volume ($V \\approx 10^7 m^3$) and a water depth $h_0 \\approx 15 m$. The empirical estimate (4) is about 75% of the estimate (2) given by the frictionless flow solution.\n\n## Numerical models\n\nThe analytical methods for describing dam-break flow provide a qualitative picture and some first-order estimates for the wave profile and wave speed. They may only be used in situations that can be represented schematically by a simple prismatic channel. In most actual field situations, the dam-break flow must be modeled numerically. Detailed simulation of the forward and backward wave fronts requires 3D models of the Boussinesq type, taking into account vertical fluid accelerations. To correctly model the strong temporal and spatial gradients of the surging wave, a fine computational grid is required, with resulting long computer times. See also Tidal bore dynamics." ]

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[ "# #8c9fb0 Color Information\n\nIn a RGB color space, hex #8c9fb0 is composed of 54.9% red, 62.4% green and 69% blue. Whereas in a CMYK color space, it is composed of 20.5% cyan, 9.7% magenta, 0% yellow and 31% black. It has a hue angle of 208.3 degrees, a saturation of 18.6% and a lightness of 62%. #8c9fb0 color hex could be obtained by blending #ffffff with #193f61. Closest websafe color is: #999999.\n\n• R 55\n• G 62\n• B 69\nRGB color chart\n• C 20\n• M 10\n• Y 0\n• K 31\nCMYK color chart\n\n#8c9fb0 color description : Dark grayish blue.\n\n# #8c9fb0 Color Conversion\n\nThe hexadecimal color #8c9fb0 has RGB values of R:140, G:159, B:176 and CMYK values of C:0.2, M:0.1, Y:0, K:0.31. Its decimal value is 9215920.\n\nHex triplet RGB Decimal 8c9fb0 `#8c9fb0` 140, 159, 176 `rgb(140,159,176)` 54.9, 62.4, 69 `rgb(54.9%,62.4%,69%)` 20, 10, 0, 31 208.3°, 18.6, 62 `hsl(208.3,18.6%,62%)` 208.3°, 20.5, 69 999999 `#999999`\nCIE-LAB 64.568, -2.925, -11.055 31.048, 33.506, 45.903 0.281, 0.303, 33.506 64.568, 11.435, 255.178 64.568, -10.785, -16.083 57.884, -5.552, -6.499 10001100, 10011111, 10110000\n\n# Color Schemes with #8c9fb0\n\n• #8c9fb0\n``#8c9fb0` `rgb(140,159,176)``\n• #b09d8c\n``#b09d8c` `rgb(176,157,140)``\nComplementary Color\n• #8cb0af\n``#8cb0af` `rgb(140,176,175)``\n• #8c9fb0\n``#8c9fb0` `rgb(140,159,176)``\n• #8c8db0\n``#8c8db0` `rgb(140,141,176)``\nAnalogous Color\n• #b0af8c\n``#b0af8c` `rgb(176,175,140)``\n• #8c9fb0\n``#8c9fb0` `rgb(140,159,176)``\n• #b08c8d\n``#b08c8d` `rgb(176,140,141)``\nSplit Complementary Color\n• #9fb08c\n``#9fb08c` `rgb(159,176,140)``\n• #8c9fb0\n``#8c9fb0` `rgb(140,159,176)``\n• #b08c9f\n``#b08c9f` `rgb(176,140,159)``\n• #8cb09d\n``#8cb09d` `rgb(140,176,157)``\n• #8c9fb0\n``#8c9fb0` `rgb(140,159,176)``\n• #b08c9f\n``#b08c9f` `rgb(176,140,159)``\n• #b09d8c\n``#b09d8c` `rgb(176,157,140)``\n• #62798e\n``#62798e` `rgb(98,121,142)``\n• #6e869b\n``#6e869b` `rgb(110,134,155)``\n• #7d92a6\n``#7d92a6` `rgb(125,146,166)``\n• #8c9fb0\n``#8c9fb0` `rgb(140,159,176)``\n• #9bacba\n``#9bacba` `rgb(155,172,186)``\n• #aab8c5\n``#aab8c5` `rgb(170,184,197)``\n• #b9c5cf\n``#b9c5cf` `rgb(185,197,207)``\nMonochromatic Color\n\n# Alternatives to #8c9fb0\n\nBelow, you can see some colors close to #8c9fb0. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #8ca8b0\n``#8ca8b0` `rgb(140,168,176)``\n• #8ca5b0\n``#8ca5b0` `rgb(140,165,176)``\n• #8ca2b0\n``#8ca2b0` `rgb(140,162,176)``\n• #8c9fb0\n``#8c9fb0` `rgb(140,159,176)``\n• #8c9cb0\n``#8c9cb0` `rgb(140,156,176)``\n• #8c99b0\n``#8c99b0` `rgb(140,153,176)``\n• #8c96b0\n``#8c96b0` `rgb(140,150,176)``\nSimilar Colors\n\n# #8c9fb0 Preview\n\nThis text has a font color of #8c9fb0.\n\n``<span style=\"color:#8c9fb0;\">Text here</span>``\n#8c9fb0 background color\n\nThis paragraph has a background color of #8c9fb0.\n\n``<p style=\"background-color:#8c9fb0;\">Content here</p>``\n#8c9fb0 border color\n\nThis element has a border color of #8c9fb0.\n\n``<div style=\"border:1px solid #8c9fb0;\">Content here</div>``\nCSS codes\n``.text {color:#8c9fb0;}``\n``.background {background-color:#8c9fb0;}``\n``.border {border:1px solid #8c9fb0;}``\n\n# Shades and Tints of #8c9fb0\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #010101 is the darkest color, while #f5f6f8 is the lightest one.\n\n• #010101\n``#010101` `rgb(1,1,1)``\n• #090b0d\n``#090b0d` `rgb(9,11,13)``\n• #111519\n``#111519` `rgb(17,21,25)``\n• #191f24\n``#191f24` `rgb(25,31,36)``\n• #212930\n``#212930` `rgb(33,41,48)``\n• #29333b\n``#29333b` `rgb(41,51,59)``\n• #313d47\n``#313d47` `rgb(49,61,71)``\n• #394653\n``#394653` `rgb(57,70,83)``\n• #41505e\n``#41505e` `rgb(65,80,94)``\n• #495a6a\n``#495a6a` `rgb(73,90,106)``\n• #516476\n``#516476` `rgb(81,100,118)``\n• #596e81\n``#596e81` `rgb(89,110,129)``\n• #61788d\n``#61788d` `rgb(97,120,141)``\n• #698298\n``#698298` `rgb(105,130,152)``\n• #758ca0\n``#758ca0` `rgb(117,140,160)``\n• #8095a8\n``#8095a8` `rgb(128,149,168)``\n• #8c9fb0\n``#8c9fb0` `rgb(140,159,176)``\n• #98a9b8\n``#98a9b8` `rgb(152,169,184)``\n• #a3b2c0\n``#a3b2c0` `rgb(163,178,192)``\n• #afbcc8\n``#afbcc8` `rgb(175,188,200)``\n• #bbc6d0\n``#bbc6d0` `rgb(187,198,208)``\n• #c6d0d8\n``#c6d0d8` `rgb(198,208,216)``\n• #d2d9e0\n``#d2d9e0` `rgb(210,217,224)``\n• #dde3e8\n``#dde3e8` `rgb(221,227,232)``\n• #e9edf0\n``#e9edf0` `rgb(233,237,240)``\n• #f5f6f8\n``#f5f6f8` `rgb(245,246,248)``\nTint Color Variation\n\n# Tones of #8c9fb0\n\nA tone is produced by adding gray to any pure hue. In this case, #9b9ea1 is the less saturated color, while #41a3fb is the most saturated one.\n\n• #9b9ea1\n``#9b9ea1` `rgb(155,158,161)``\n• #939fa9\n``#939fa9` `rgb(147,159,169)``\n• #8c9fb0\n``#8c9fb0` `rgb(140,159,176)``\n• #859fb7\n``#859fb7` `rgb(133,159,183)``\n• #7da0bf\n``#7da0bf` `rgb(125,160,191)``\n• #76a0c6\n``#76a0c6` `rgb(118,160,198)``\n• #6ea1ce\n``#6ea1ce` `rgb(110,161,206)``\n• #67a1d5\n``#67a1d5` `rgb(103,161,213)``\n• #5fa1dd\n``#5fa1dd` `rgb(95,161,221)``\n• #58a2e4\n``#58a2e4` `rgb(88,162,228)``\n• #50a2ec\n``#50a2ec` `rgb(80,162,236)``\n• #49a3f3\n``#49a3f3` `rgb(73,163,243)``\n• #41a3fb\n``#41a3fb` `rgb(65,163,251)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #8c9fb0 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

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[ "## Monday, February 20, 2017\n\n### The Curry-Howard correspondence between programs and proofs\n\nThis post will explain the connection between programming languages and logical proofs, known as the Curry-Howard correspondence. I will provide several examples of this correspondence to help you build a working intuition for how these two fields relate to one another.\n\nThe Curry-Howard correspondence states that:\n\n• Logical propositions correspond to programming types\n• Logical proofs correspond to programming values\n• Proving a proposition corresponds to creating a value of a given type\n\nI'll use my Dhall programming language to illustrate the above connections since Dhall is an explicitly typed total programming language without any escape hatches. If you're not familiar with Dhall, you can read the Dhall tutorial and I will also make analogies to Haskell along the way, too.\n\n### Propositions\n\nLet's begin with the following logical proposition:\n\n``∀(a ∈ Prop): a ⇒ a``\n\nYou can read that as \"for any proposition (which we will denote using the letter `a`), if the proposition `a` is true then the proposition `a` is true\". This is true no matter what the proposition `a` is. For example, suppose that `a` were the following proposition:\n\n``a = {The sky is blue}``\n\nThen we could conclude that \"if the sky is blue then the sky is blue\":\n\n``{The sky is blue} ⇒ {The sky is blue}``\n\nHere the right double arrow (`⇒`) denotes logical implication. Anywhere you see (`A ⇒ B`) you should read that as \"if the proposition `A` is true then the proposition `B` is true\". You can also say \"the proposition `A` implies the proposition `B`\" or just \"`A` implies `B`\" for short.\n\nHowever, what if `a` were another proposition like:\n\n``a = {The sky is green}``\n\ngiving us:\n\n``{The sky is green} ⇒ {The sky is green}``\n\nThis is true, even though the sky might not be green. We only state that \"if the sky is green then the sky is green\", which is definitely a true statement, whether or not the sky is green.\n\nThe upside down A (i.e. `∀`) in our original proposition is mathematical shorthand that means \"for all\" (although I sometimes describe this as \"for any\"). The `∈` symbol is shorthand for \"in\" (i.e. set membership). So whenever you see this:\n\n``∀(a ∈ S): p``\n\n... you can read that as \"for any element `a` in the set `S` the proposition `p` is true\". Usually the set `S` is `Prop` (i.e. the set of all possible propositions) but we will see some examples below where `S` can be another set.\n\n### Types\n\nThis logical proposition:\n\n``∀(a ∈ Prop): a ⇒ a``\n\n... corresponds to the following Dhall type:\n\n``∀(a : Type) → a → a``\n\n... which you can read as \"for any type (which we will denote using the letter `a`), this is the type of a function that transforms an input of type `a` to an output of type `a`\". Here's the corresponding function that has this type:\n\n``λ(a : Type) → λ(x : a) → x``\n\nThis is an anonymous function of two arguments:\n\n• the first argument is named `a` and `a` is a `Type`\n• the second argument is named `x` and `x` has type `a`\n\nThe result is the function's second argument (i.e. `x`) which still has type `a`.\n\nThe equivalent Haskell function is `id`, the polymorphic identity function:\n\n``````id :: a -> a\nid x = x\n\n-- or using the `ExplicitForAll` language extension\nid :: forall a . a -> a\nid x = x``````\n\nThe main difference is that Haskell does not require you to explicitly bind the polymorphic type `a` as an additional argument. Dhall does require this because Dhall is an explicitly typed language.\n\n### Correspondence\n\nThe Curry-Howard correspondence says that we can use the type checker of a programming language as a proof checker. Any time we want to prove a logical proposition, we:\n\n• translate the logical proposition to the corresponding type in our programming language\n• create a value in our programming language that has the given type\n• use the type-checker to verify that our value has the specified type\n\nIf we find a value of the given type then that completes the proof of our original proposition.\n\nFor example, if we want to prove this proposition:\n\n``∀(a ∈ Prop): a ⇒ a``\n\n... then we translate the proposition to the corresponding type in the Dhall programming language:\n\n``∀(a : Type) → a → a``\n\n... then define a value in the Dhall language that has this type:\n\n``λ(a : Type) → λ(x : a) → x``\n\n... and then use Dhall's type checker to verify that this value has the given type:\n\n``````\\$ dhall <<< 'λ(a : Type) → λ(x : a) → x'\n∀(a : Type) → ∀(x : a) → a\n\nλ(a : Type) → λ(x : a) → x``````\n\nThe first line is the inferred type of our value and the second line is the value's normal form (which in this case is the same as the original value).\n\nNote that the inferred type slightly differs from the original type we expected:\n\n``````-- What we expected:\n∀(a : Type) → a → a\n\n-- What the compiler inferred:\n∀(a : Type) → ∀(x : a) → a``````\n\nThese are both still the same type. The difference is that the compiler's inferred type also includes the name of the second argument: `x`. If we were to translate this back to logical proposition notation, we might write:\n\n``∀(a ∈ Prop): ∀(x ∈ a): a``\n\n... which you could read as: \"for any proposition named `a`, given a proof of `a` named `x`, the proposition `a` is true\". This is equivalent to our original proposition but now we've given a name (i.e. `x`) to the proof of `a`.\n\nThe reason this works is that you can think of a proposition as a set, too, where the elements of the set are the proofs of that proposition. Some propositions are false and have no elements (i.e. no valid proofs), while other propositions can have multiple elements (i.e. multiple valid proofs).\n\nSimilarly, you can think of a type as a set, where the elements of the set are the values that have that type. Some types are empty and have no elements (i.e. no values of that type), while other types can have multiple elements (i.e. multiple values of that type).\n\nHere's an example of a proposition that has multiple valid proofs:\n\n``∀(a ∈ Prop): a ⇒ a ⇒ a``\n\nThe corresponding type is:\n\n``∀(a : Type) → a → a → a``\n\n... and there are two values that have the above type. The first value is:\n\n``λ(a : Type) → λ(x : a) → λ(y : a) → x``\n\n... and the second value is:\n\n``λ(a : Type) → λ(x : a) → λ(y : a) → y``\n\nWe can even translate these two values back into informal logical proofs. For example, this value:\n\n``λ(a : Type) → λ(x : a) → λ(y : a) → x``\n\n... corresponds to this informal proof:\n\n``````For each \"a\" in the set of all propositions:\n\nGiven a proof that \"a\" is true [Let's call this proof \"x\"]\n\nGiven a proof that \"a\" is true [Let's call this proof \"y\"]\n\nWe can conclude that \"a\" is true, according to the proof \"x\"\n\nQED``````\n\nSimilarly, this value:\n\n``λ(a : Type) → λ(x : a) → λ(y : a) → y``\n\n... corresponds to this informal proof:\n\n``````For each \"a\" in the set of all propositions:\n\nGiven a proof that \"a\" is true [Let's call this proof \"x\"]\n\nGiven a proof that \"a\" is true [Let's call this proof \"y\"]\n\nWe can conclude that \"a\" is true, according to the proof \"y\"\n\nQED``````\n\nWe can actually give these proofs a formal structure that parallels our code but that is outside of the scope of this post.\n\n### Function composition\n\nNow consider this more complex proposition:\n\n``````∀(a ∈ Prop):\n∀(b ∈ Prop):\n∀(c ∈ Prop):\n(b ⇒ c) ⇒ (a ⇒ b) ⇒ (a ⇒ c)``````\n\nYou can read that as saying: \"for any three propositions named `a`, `b`, and `c`, if `b` implies `c`, then if `a` implies `b`, then `a` implies `c`\".\n\nThe corresponding type is:\n\n`````` ∀(a : Type)\n→ ∀(b : Type)\n→ ∀(c : Type)\n→ (b → c) → (a → b) → (a → c)``````\n\n... and we can define a value that has this type in order to prove that the proposition is true:\n\n`````` λ(a : Type)\n→ λ(b : Type)\n→ λ(c : Type)\n→ λ(f : b → c)\n→ λ(g : a → b)\n→ λ(x : a)\n→ f (g x)``````\n\n... and the compiler will infer that our value has the correct type:\n\n``````\\$ dhall\nλ(a : Type)\n→ λ(b : Type)\n→ λ(c : Type)\n→ λ(f : b → c)\n→ λ(g : a → b)\n→ λ(x : a)\n→ f (g x)\n<Ctrl-D>\n∀(a : Type) → ∀(b : Type) → ∀(c : Type) → ∀(f : b → c) → ∀(g : a → b) → ∀(x : a) → c\n\nλ(a : Type) → λ(b : Type) → λ(c : Type) → λ(f : b → c) → λ(g : a → b) → λ(x : a) → f (g x)``````\n\nNote that this function is equivalent to Haskell's function composition operator (ie. `(.)`):\n\n``````(.) :: (b -> c) -> (a -> b) -> (a -> c)\n(f . g) x = f (g x)``````\n\nThis is because proofs and programs are equivalent to one another: whenever we prove a logical proposition we get a potentially useful function for free.\n\n### Circular reasoning\n\nThere are some propositions that we cannot prove, like this one:\n\n``∀(a ∈ Prop): a``\n\n... which you can read as saying \"all propositions are true\". This proposition is false, because if we translate the proposition to the corresponding type:\n\n``∀(a : Type) → a``\n\n... then there is no value in the Dhall language which has that type. If there were such a value then we could automatically create any value of any type.\n\nHowever, Haskell does have values which inhabit the above type, such as this one:\n\n``````example :: a\nexample = let x = x in x\n\n-- or using the `ExplicitForAll` language extension:\nexample :: forall a . a\nexample = let x = x in x``````\n\nThis value just cheats and endlessly loops, which satisfies Haskell's type checker but doesn't produce a useful program. Unrestricted recursion like this is the programming equivalent of \"circular reasoning\" (i.e. trying to claim that `a` is true because `a` is true).\n\nWe can't cheat like this in Dhall and if we try we just get this error:\n\n``````\\$ dhall <<< 'let x = x in x'\n\nUse \"dhall --explain\" for detailed errors\n\nError: Unbound variable\n\nx\n\n(stdin):1:9``````\n\nYou cannot define `x` in terms of itself because Dhall does not permit any recursion and therefore does not permit circular reasoning when used as a proof checker.\n\n### And\n\nWe will use the symbol `∧` to denote logical \"and\", so you can read the following proposition:\n\n``∀(a ∈ Prop): ∀(b ∈ Prop): (a ∧ b) ⇒ a``\n\n... as saying \"for any two propositions `a` and `b`, if `a` and `b` are both true then `a` is true\"\n\nThe type-level equivalent of logical \"and\" is a record with two fields:\n\n``∀(a : Type) → ∀(b : Type) → { fst : a, snd : b } → a``\n\n... which is equivalent to this Haskell type:\n\n``````(a, b) -> a\n\n-- or using `ExplicitForAll`:\nforall a b . (a, b) -> a``````\n\nThe programming value that has this type is:\n\n``λ(a : Type) → λ(b : Type) → λ(r : { fst : a, snd : b }) → r.fst``\n\n... which is equivalent to this Haskell value:\n\n``fst``\n\nSimilarly, we can conclude that:\n\n``∀(a ∈ Prop): ∀(b ∈ Prop): (a ∧ b) ⇒ b``\n\n... which corresponds to this type:\n\n``∀(a : Type) → ∀(b : Type) → { fst : a, snd : b } → b``\n\n... and this value of that type as the proof:\n\n``λ(a : Type) → λ(b : Type) → λ(r : { fst : a, snd : b }) → r.snd``\n\nNow let's try a slightly more complex proposition:\n\n``∀(a ∈ Prop): ∀(b ∈ Prop): (a ∧ (a ⇒ b)) ⇒ b``\n\nYou can read this as saying \"for any propositions `a` and `b`, if the proposition `a` is true, and `a` implies `b`, then the proposition `b` is true\".\n\nThat corresponds to this type:\n\n``∀(a : Type) → ∀(b : Type) → { fst : a, snd : a → b } → b``\n\n... and the following value of that type which proves that the proposition is true:\n\n`````` λ(a : Type)\n→ λ(b : Type)\n→ λ(r : { fst : a, snd : a → b })\n→ r.snd r.fst``````\n\nHere's a slightly more complex example:\n\n``∀(a ∈ Prop): ∀(b ∈ Prop): ∀(c ∈ Prop): ((a ∧ b) ⇒ c) ⇒ (a ⇒ b ⇒ c)``\n\nYou can read that as saying: \"for any three propositions named `a`, `b`, and `c`, if `a` and `b` imply `c`, then `a` implies that `b` implies `c`\".\n\nHere's the corresponding type:\n\n`````` ∀(a : Type)\n→ ∀(b : Type)\n→ ∀(c : Type)\n→ ({ fst : a, snd : b } → c) → (a → b → c)``````\n\n... and the corresponding value of that type:\n\n`````` λ(a : Type)\n→ λ(b : Type)\n→ λ(c : Type)\n→ λ(f : { fst : a, snd : b } → c)\n→ λ(x : a)\n→ λ(y : b)\n→ f { fst = x, snd = y }``````\n\nNote that this is equivalent to Haskell's `curry` function:\n\n``````curry :: ((a, b) -> c) -> (a -> b -> c)\ncurry f x y = f (x, y)``````\n\n### Or\n\nWe will use the symbol `∨` to denote logical \"or\", so you can read the following proposition:\n\n``∀(a ∈ Prop): ∀(b ∈ Prop): a ⇒ a ∨ b``\n\n... as saying \"for any propositions `a` and `b`, if `a` is true, then either `a` is true or `b` is true\".\n\nThe type-level equivalent of logical \"or\" is a sum type:\n\n``∀(a : Type) → ∀(b : Type) → a → < Left : a | Right : b >``\n\n... which is equivalent to this Haskell type:\n\n``````a -> Either a b\n\n-- or using `ExplicitForAll`\nforall a b . a -> Either a b``````\n\n... and the value that has this type is:\n\n``λ(a : Type) → λ(b : Type) → λ(x : a) → < Left = x | Right : b >``\n\n... which is equivalent to this Haskell value:\n\n``Left``\n\nSimilarly, for this proposition:\n\n``∀(a ∈ Prop): ∀(b ∈ Prop): b ⇒ a ∨ b``\n\n... the equivalent type is:\n\n``∀(a : Type) → ∀(b : Type) → b → < Left : a | Right : b >``\n\n... and the value that has this type is:\n\n``λ(a : Type) → λ(b : Type) → λ(x : b) → < Right = x | Left : a >``\n\nLet's consider a more complex example:\n\n``∀(a ∈ Prop): a ∨ a ⇒ a``\n\n... which says: \"for any proposition `a`, if `a` is true or `a` is true then `a` is true\".\n\nThe corresponding type is:\n\n``∀(a : Type) → < Left : a | Right : a > → a``\n\n... which is equivalent to this Haskell type:\n\n``````Either a a -> a\n\n-- or using `ExplicitForAll`\nforall a . Either a a -> a``````\n\n... and we can create a value of this type by pattern matching on the sum type:\n\n`````` λ(a : Type)\n→ λ(s : < Left : a | Right : a >)\n→ merge\n{ Left = λ(x : a) → x\n, Right = λ(x : a) → x\n}\ns\n: a``````\n\n... which is equivalent to this Haskell code:\n\n``````example :: Either a a -> a\nexample (Left x) = x\nexample (Right x) = x``````\n\nYou can read this \"proof\" as saying: \"There are two possibilities. The first possibility (named \"Left\") is that `a` is true, therefore `a` must be true in that case. The second possibility (named \"Right\") is that `a` is true, therefore `a` must be true in that case. Since `a` is true in both cases we can conclude that `a` is true, period.\"\n\n### True\n\nWe'll use `True` to denote a logical proposition that is always true:\n\n``True``\n\n... and the corresponding type is an empty record with no fields:\n\n``{}``\n\n... which is equivalent to Haskell's \"unit\" type:\n\n``()``\n\nAn empty record literal is the only value that has the type of an empty record:\n\n``{=}``\n\n... which is equivalent to Haskell's \"unit\" value:\n\n``()``\n\n### False\n\nWe'll use `False` to denote a logical proposition that is never true:\n\n``False``\n\n... and the corresponding type is an empty sum type (i.e. a type with no constructors/alternatives):\n\n``<>``\n\n... which is equivalent to Haskell's `Void` type.\n\nThere is no value that has the above type, so you cannot prove a `False` proposition.\n\nThe logical \"principle of explosion\" states that if you assume a `False` proposition then you can prove any other proposition. In other words:\n\n``False ⇒ ∀(a ∈ Prop): a``\n\n... which you can read as saying: \"if you assume a false proposition, then for any proposition named `a` you can prove that `a` is true\".\n\nThe corresponding type is:\n\n``<> → ∀(a : Type) → a``\n\n... and the value that inhabits the above type is:\n\n``λ(s : <>) → λ(a : Type) → merge {=} s : a``\n\nYou can read that \"proof\" as saying: \"There are zero possibilities. Since we handled all possibilities then the proposition `a` must be true.\" This line of reasoning is called a \"vacuous truth\", meaning that if there are no cases to consider then any statement you make is technically true for all cases.\n\nThe equivalent Haskell type would be:\n\n``Void -> a``\n\n... and the equivalent Haskell value of that type would be:\n\n``absurd``\n\n### Not\n\nWe'll use `not` to negate a logical proposition, meaning that:\n\n``````not(True) = False\n\nnot(False) = True``````\n\nWe can encode logical `not` in terms of logical implication and logical `False`, like this:\n\n``````not(a) = a ⇒ False\n\nnot(True) = True ⇒ False = False\n\nnot(False) = False ⇒ False = True``````\n\nIf logical `not` is a function from a proposition to another proposition then type-level `not` must be a function from a type to a type:\n\n``λ(a : Type) → (a → <>)``\n\nWe can save the above type-level function to a file to create a convenient `./not` utility we will reuse later on:\n\n``\\$ echo 'λ(a : Type) → a → <>' > ./not``\n\nNow let's try to prove a proposition like:\n\n``not(True) ⇒ False``\n\nThe corresponding type is:\n\n``./not {} → <>``\n\n... which expands out to:\n\n``({} → <>) → <>``\n\n... and the value that has this type is:\n\n``λ(f : {} → <>) → f {=}``\n\n### Double negation\n\nHowever, suppose we try to prove \"double negation\":\n\n``∀(a ∈ Prop): not(not(a)) ⇒ a``\n\n... which says that \"if `a` is not false, then `a` must be true\".\n\nThe corresponding type would be:\n\n``∀(a : Type) → ./not (./not a) → a``\n\n... which expands out to:\n\n``∀(a : Type) → ((a → <>) → <>) → a``\n\n... but there is no value in the Dhall language that has this type!\n\nThe reason why is that every type checker corresponds to a specific logic system and not all logic systems are the same. Each logic system has different rules and axioms about what you can prove within the system.\n\nDhall's type checker is based on System Fω which corresponds to an intuitionistic (or constructive) logic. This sort of logic system does not assume the law of excluded middle.\n\nThe law of excluded middle says that every proposition must be true or false, which we can write like this:\n\n``∀(a ∈ Prop): a ∨ not(a)``\n\nYou can read that as saying: \"any proposition `a` is either true or false\". This seems pretty reasonable until you try to translate the proposition to the corresponding type:\n\n``∀(a : Type) → < Left : a | Right : a → <> >``\n\n... which in Haskell would be:\n\n``example :: Either a (a -> Void)``\n\nWe cannot create such a value because if we could then that would imply that for any type we could either:\n\n• produce a value of that type, or:\n• produce a \"counter-example\" by creating `Void` from any value of that type\n\nWhile we can do this for some types, our type checker does not let us do this automatically for every type.\n\nThis is the same reason that we can't prove double negation, which implicitly assumes that there are only two choices (i.e. \"true or false\"). However, if we don't assume the law of the excluded middle then there might be other choices like: \"I don't know\".\n\nNot all hope is lost, though! Even though our type checker doesn't support the axiom of choice, we can still add new axioms freely to our logic system. All we have to do is assume them the same way we assume other propositions.\n\nFor example, we can modify our original proposition to now say:\n\n``(∀(b ∈ Prop): b ∨ not(b)) ⇒ (∀(a ∈ Prop): not(not(a)) ⇒ a)``\n\n... which you can read as saying: \"if we assume that all propositions are either true or false, then if a proposition is not false it must be true\".\n\nThat corresponds to this type:\n\n`````` (∀(b : Type) → < Left : b | Right : b → <> >)\n→ (∀(a : Type) → ((a → <>) → <>) → a)``````\n\n... and we can create a value of that type:\n\n`````` λ(noMiddle : ∀(b : Type) → < Left : b | Right : b → <> >)\n→ λ(a : Type)\n→ λ(doubleNegation : (a → <>) → <>)\n→ merge\n{ Left = λ(x : a) → x\n, Right = λ(x : a → <>) → merge {=} (doubleNegation x) : a\n}\n(noMiddle a)\n: a``````\n\n... which the type checker confirms has the correct type:\n\n``````\\$ dhall\nλ(noMiddle : ∀(b : Type) → < Left : b | Right : b → <> >)\n→ λ(a : Type)\n→ λ(doubleNegation : (a → <>) → <>)\n→ merge\n{ Left = λ(x : a) → x\n, Right = λ(x : a → <>) → merge {=} (doubleNegation x) : a\n}\n(noMiddle a)\n: a\n<Ctrl-D>\n∀(noMiddle : ∀(b : Type) → < Left : b | Right : b → <> >) → ∀(a : Type) → ∀(doubleNegation : (a → <>) → <>) → a\n\nλ(noMiddle : ∀(b : Type) → < Left : b | Right : b → <> >) → λ(a : Type) → λ(doubleNegation : (a → <>) → <>) → merge { Left = λ(x : a) → x, Right = λ(x : a → <>) → merge {=} (doubleNegation x) : a } (noMiddle a) : a``````\n\nYou can read that proof as saying:\n\n• The law of the excluded middle says that there are two possibilities: either `a` is true or `a` is false\n• If `a` is true then we're done: we trivially conclude `a` is true\n• If `a` is false then our assumption of `not(not(a))` is also false\n• we can conclude anything from a false assumption, therefore we conclude that `a` is true\n\nWhat's neat about this is that the compiler mechanically checks this reasoning process. You don't have to understand or trust my explanation of how the proof works because you can delegate your trust to the compiler, which does all the work for you.\n\n### Conclusion\n\nThis post gives a brief overview of how you can concretely translate logical propositions and proofs to types and programs. This can let you leverage your logical intuitions to reason about types or, vice versa, leverage your programming intuitions to reason about propositions. For example, if you can prove a false proposition in a logic system, then that's typically an escape hatch in the corresponding type system. Similarly, if you can create a value of the empty type in a programming language, that implies that the corresponding logic is not sound.\n\nThere are many kinds of type systems just like there are many kinds of logic systems. For every new logic system (like linear logic) you get a type system for free (like linear types). For example, Rust's type checker is an example of an affine type system which corresponds to an affine logic system.\n\nTo my knowledge, there are more logic systems in academic literature than there are type systems that have been implemented for programming languages. This in turn suggests that there are many awesome type systems waiting to be implemented.\n\n1.", null, "This comment has been removed by the author.\n\n1.", null, "This comment has been removed by the author.\n\n2.", null, "Excelent article, thank you so much.\n\n1.", null, "You're welcome!\n\n3.", null, "Vacuous truth is better written in today's Haskell as \\s -> case s of {}" ]

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[ "# Check your talent: are you ready for math or mathematical sciences or engineering\n\nAt the outset, let me put a little sweetener also: All I want to do is draw attention to the importance of symbolic manipulation. If you can solve this tutorial easily or with only a little bit of help, I would strongly feel that you can make a good career in math or applied math or mathematical sciences or engineering.\n\nOn the other hand, this tutorial can be useful as a “miscellaneous or logical type of problems” for the ensuing RMO 2019.\n\nI) Let S be a set having an operation * which assigns an element a*b of S for any", null, "$a,b \\in S$. Let us assume that the following two rules hold:\n\ni) If a, b are any objects in S, then", null, "$a*b=a$\n\nii) If a, b are any objects in S, then", null, "$a*b=b*a$\n\nShow that S can have at most one object.\n\nII) Let S be the set of all integers. For a, b in S define * by a*b=a-b. Verify the following:\n\na)", null, "$a*b \\neq b*a$ unless", null, "$a=b$.\n\nb)", null, "$(a*b)*c \\neq a*{b*c}$ in general. Under what conditions on a, b, c is", null, "$a*(b*c)=(a*b)*c$?\n\nc) The integer 0 has the property that", null, "$a*0=a$ for every a in S.\n\nd) For a in S,", null, "$a*a=0$\n\nIII) Let S consist of two objects", null, "$\\square$ and", null, "$\\triangle$. We define the operation * on S by subjecting", null, "$\\square$ and", null, "$\\triangle$ to the following condittions:\n\ni)", null, "$\\square * \\triangle=\\triangle = \\triangle * \\square$\n\nii)", null, "$\\square * \\square = \\square$\n\niii)", null, "$\\triangle * \\triangle = \\square$\n\nVerify by explicit calculation that if a, b, c are any elements of S (that is, a, b and c can be any of", null, "$\\square$ or", null, "$\\triangle$) then:\n\ni)", null, "$a*b \\in S$\n\nii)", null, "$(a*b)*c = a*(b*c)$\n\niii)", null, "$a*b=b*a$\n\niv) There is a particular a in S such that", null, "$a*b=b*a=b$ for all b in S\n\nv) Given", null, "$b \\in S$, then", null, "$b*b=a$, where a is the particular element in (iv) above.\n\nThis will be your own self-appraisal !!\n\nRegards,\n\nNalin Pithwa\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed." ]

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[ "# Divisors of 3802\n\n## Divisors of 3802\n\nThe list of all positive divisors (that is, the list of all integers that divide 22) is as follows :\n\nAccordingly:\n\n3802 is multiplo of 1\n\n3802 is multiplo of 2\n\n3802 is multiplo of 1901\n\n3802 has 3 positive divisors\n\n## Parity of 3802\n\nIn addition we can say of the number 3802 that it is even\n\n3802 is an even number, as it is divisible by 2 : 3802/2 = 1901\n\n## The factors for 3802\n\nThe factors for 3802 are all the numbers between -3802 and 3802 , which divide 3802 without leaving any remainder. Since 3802 divided by -3802 is an integer, -3802 is a factor of 3802 .\n\nSince 3802 divided by -3802 is a whole number, -3802 is a factor of 3802\n\nSince 3802 divided by -1901 is a whole number, -1901 is a factor of 3802\n\nSince 3802 divided by -2 is a whole number, -2 is a factor of 3802\n\nSince 3802 divided by -1 is a whole number, -1 is a factor of 3802\n\nSince 3802 divided by 1 is a whole number, 1 is a factor of 3802\n\nSince 3802 divided by 2 is a whole number, 2 is a factor of 3802\n\nSince 3802 divided by 1901 is a whole number, 1901 is a factor of 3802\n\n## What are the multiples of 3802?\n\nMultiples of 3802 are all integers divisible by 3802 , i.e. the remainder of the full division by 3802 is zero. There are infinite multiples of 3802. The smallest multiples of 3802 are:\n\n0 : in fact, 0 is divisible by any integer, so it is also a multiple of 3802 since 0 × 3802 = 0\n\n3802 : in fact, 3802 is a multiple of itself, since 3802 is divisible by 3802 (it was 3802 / 3802 = 1, so the rest of this division is zero)\n\n7604: in fact, 7604 = 3802 × 2\n\n11406: in fact, 11406 = 3802 × 3\n\n15208: in fact, 15208 = 3802 × 4\n\n19010: in fact, 19010 = 3802 × 5\n\netc.\n\n## Is 3802 a prime number?\n\nIt is possible to determine using mathematical techniques whether an integer is prime or not.\n\nfor 3802, the answer is: No, 3802 is not a prime number.\n\n## How do you determine if a number is prime?\n\nTo know the primality of an integer, we can use several algorithms. The most naive is to try all divisors below the number you want to know if it is prime (in our case 3802). We can already eliminate even numbers bigger than 2 (then 4 , 6 , 8 ...). Besides, we can stop at the square root of the number in question (here 61.66 ). Historically, the Eratosthenes screen (which dates back to Antiquity) uses this technique relatively effectively.\n\nMore modern techniques include the Atkin screen, probabilistic tests, or the cyclotomic test.\n\n## Numbers about 3802\n\nPrevious Numbers: ... 3800, 3801\n\nNext Numbers: 3803, 3804 ...\n\n## Prime numbers closer to 3802\n\nPrevious prime number: 3797\n\nNext prime number: 3803" ]

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[ "# Double Integral to compute conditional expectation\n\nLet $$V=X+Y$$ where $$X$$ and $$Y$$ are independent random variables, both normally distributed with $$\\mu=0$$ and $$\\sigma=1$$. I am interested in computing $$E[X \\mid V\\geq c]$$ where $$c$$ is an arbitrary constant. This is $$E[X \\mid V>c] = \\frac{1}{1-\\Phi_v(c)} \\int_c^\\infty \\int_{-\\infty}^{\\infty} x \\cdot f_X(x) \\cdot f_Y(v-x) \\quad dx ~dv$$ In Mathematica, with $$c=0.5$$, I am coding\n\n1/(1 - CDF[NormalDistribution[0, 1], 0.5])\nIntegrate[\nx PDF[NormalDistribution[0, 1], x] PDF[NormalDistribution[0, 1],\nv - x], {x, -Infinity, Infinity}, {v, 0.5, Infinity}]\n\n\nbut this seems a very difficult problem for Mathematica to solve, taking a long time and sometimes not solving at all. If I use NIntegrate it is done instantly instead. My question is whether this is to be expected, or whether there is a way to do it symbolically in a more efficient way.\n\n• Did you try replacing 0.5 with 1/2? – Tim Laska Jan 21 '19 at 23:57\n• @TimLaska Wow that makes a big difference! Can you explain me why? Also, in the symbolic answer I now get Erfc[-(1/(2 Sqrt))]. How should I deal with it since I am interested in the actual numerical value? – mariodrumblue Jan 21 '19 at 23:59\n• If you seek an analytical solution, sometimes it is better to provide rational fractions versus floating point numbers. It can change though, so you sometimes need to experiment. – Tim Laska Jan 22 '19 at 0:03\n• To convert the exact analytic result to an approximate numeric result use result // N – Bob Hanlon Jan 22 '19 at 0:57\n\nThanks to @AlexTourneau's slight modification to my approach exploiting the power of Mathematica:\n\nClear[x, y, c];\nExpectation[x \\[Conditioned] x + y > c,\n{x \\[Distributed] NormalDistribution[],\ny \\[Distributed] NormalDistribution[]}]\n\n\n$$\\frac{e^{-\\frac{c^2}{4}}}{\\sqrt{\\pi } \\text{erfc}\\left(\\frac{c}{2}\\right)}$$\n\n• Love the Expectation operator, which was unknown to me. This makes it a lot simple to write the code. It is also reasonably fast, however if I change the mean and standard deviations from a standard normal then it takes long or does not solve at all. For instance, this does not work Clear[x, y, c]; Expectation[ x \\[Conditioned] x + y > 0.5, {x \\[Distributed] NormalDistribution[1/2, 2], y \\[Distributed] NormalDistribution[1/2, 2]}] – mariodrumblue Jan 22 '19 at 1:19\n• Expectation[x \\[Conditioned] x + y > c, {x \\[Distributed] NormalDistribution[0, 1], y \\[Distributed] NormalDistribution[2, 2]}] took 1.78 seconds (v. 11.3.0.0, Mac). and Expectation[ x \\[Conditioned] x + y > 1/2, {x \\[Distributed] NormalDistribution[0, 2], y \\[Distributed] NormalDistribution[1/2, 2]}] to 0.5 seconds. Seems fast enough to me for a sophisticated calculation! – David G. Stork Jan 22 '19 at 1:25\n• I am also on (v. 11.3.0.0, Mac) and yes with your parametrisation it works, but not with mine (i.e. when both variables have mean 0.5 and standard deviation 2). In other words, it seems that the speed and overall ability to solve depends on the parametrisation. That's a bit of an issue for me since I was planning to simulate the model (which makes use of that expectation) under different values of the parameters. – mariodrumblue Jan 22 '19 at 1:33\n• @mariodrumblue: Experiment with exact and decimal numbers (e.g., $0.5$ versus $1/2$). – David G. Stork Jan 22 '19 at 1:38\n• thanks David. I did and it did not make any difference. Also, Expectation[ x \\[Conditioned] x + y > 1, {x \\[Distributed] NormalDistribution[0.5, 2], y \\[Distributed] NormalDistribution[0.5, 2]}] works, while changing the region over which the expectation is taken, such as Expectation[ x \\[Conditioned] x + y > 0, {x \\[Distributed] NormalDistribution[0.5, 2], y \\[Distributed] NormalDistribution[0.5, 2]}] does not work. Am I missing something or is this because it is a very difficult problem that Mathematica struggles with? – mariodrumblue Jan 22 '19 at 1:58" ]

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[ "# Answers\n\nSolutions by everydaycalculation.com\n\n## Divide 14/4 with 20/5\n\n1st number: 3 2/4, 2nd number: 4 0/5\n\n14/4 ÷ 20/5 is 7/8.\n\n#### Steps for dividing fractions\n\n1. Find the reciprocal of the divisor\nReciprocal of 20/5: 5/20\n2. Now, multiply it with the dividend\nSo, 14/4 ÷ 20/5 = 14/4 × 5/20\n3. = 14 × 5/4 × 20 = 70/80\n4. After reducing the fraction, the answer is 7/8\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:\nAndroid and iPhone/ iPad\n\n#### Divide Fractions Calculator\n\n÷\n\n© everydaycalculation.com" ]

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[ "## Arrhenius Graphs\n\nSanjay_Kumar_2J\nPosts: 15\nJoined: Fri Jul 22, 2016 3:00 am\n\n### Arrhenius Graphs\n\nFor an arrhenius graph, how do you calculate the Ea from the data given such as ln(K) and K-1?\n\nImani Johnson 1H\nPosts: 14\nJoined: Sat Jul 23, 2016 3:00 am\n\n### Re: Arrhenius Graphs\n\nIdk if this is what you're asking but, From the 'progress of reaction' energy graph, the activation energy is the difference between the energy of the transition state and the reactants,I believe.\n\nAzeel_Mohammed_1C\nPosts: 10\nJoined: Wed Nov 18, 2015 3:00 am\n\n### Re: Arrhenius Graphs\n\n\"The rate constant of a reaction can be expressed as\n\nk = Ae-Ea/RT\n\nwhich is called the Arrhenius equation. Taking the natural log of both sides of the Arrhenius equation gives\n\nln k = -Ea/R(1/T) + ln A\n\nThe equation above is of the form y = mx + b, where y = ln k, m = -Ea/RT, x = 1/T, and b = ln A. For a reaction whose rate constant obeys the Arrhenius equation, a plot of ln k vs 1/T gives a straight line and it's slope can be used to determine Ea.\"\n\nFrom there, it just depends the data you get and plug it into your equation.\n\nSarah_Kremer_1A\nPosts: 21\nJoined: Wed Sep 21, 2016 2:56 pm\n\n### Re: Arrhenius Graphs\n\nThe slope of the line on the graph is equal to -Ea/R. Since R is a constant and you can find the slope using the data points, you can then find Ea." ]

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[ "# Defining and Calling Simple Functions¶\n\nUse classiq to define a QMOD function by writing a Python function decorated with the @QFunc decorator:\n\nfrom classiq import QFunc, QParam, QBit, H, PHASE\n\n@QFunc\ndef foo(n: QParam[int], qv: QBit) -> None:\nH(qv)\nPHASE(n * pi, qv)\n\n\nThe @QFunc decorator relies on Python's type-hint mechanism to create a corresponding QMOD function definition from the specified Python function.\n\nUse the QParam type-hint to specify a classical parameter. Note it symbolically represents a classical parameter of a quantum function, and cannot be used in an arbitrary Python expression (but only as a parameter to other QParam arguments of QFunc objects). QParam inherits from the sympy.Symbol class, and thus can be used in a Sympy expression context.\n\nUse the QBit quantum type-hint to specify an inout port of the function of a single qubit. To specify an input-only or an output-only port, use Input[QBit] or Output[QBit] respectively. Alternatively, you can use QArray[QBit] to declare an array of qubits with an unspecified size (for more details on advanced quantum types, see Quantum Types and Expressions). The size of the array can also depend on previously defined classical parameters, for example:\n\nfrom classiq import QFunc, QParam, QArray, QBit\n\n@QFunc\ndef foo(n: QParam[int], qbv: QArray[QBit, \"n\"]) -> None:\npass\n\n\nInside the body of the @QFunc, you can call other @QFuncs, which results in quantum function calls in the body of the resulting QMOD definition. You can also introduce local quantum-variables, by using the declare_qvar function, specifying the variable's QMOD name and type (not specifying a type will default to QArray[QBit]). You need to allocate the quantum variable using builtin functions allocate or prepare_state before you pass it to a function with an input or an inout port. Many more builtin QMOD functions are available (see the full list of functions and their arguments under classiq/qmod/builtins.py).\n\nIn the next example, the local variable a_var must be allocated prior to applying X on it, since it is passed to an inout port. In contrast, the local variable b_var is passed without allocation, because prepare_state has an output-only port and allocates the variable inside.\n\nfrom classiq import (\nallocate,\nQFunc,\nQBit,\nQArray,\ncreate_model,\nprepare_state,\nCX,\n)\n\n@QFunc\ndef main() -> None:\na_var = QBit(\"a_var\")\nallocate(1, a_var)\nX(a_var)\nb_var = QArray(\"b_var\")\nprepare_state(probabilities=[0.25, 0.25, 0.25, 0.25], bound=0.01, out=b_var)\n\n\nWithin a function you can also use Python statements; e.g., a for loop to create multiple calls. (Do not use the n parameter as it is a declarative QParam, not a true Python integer.):\n\nfrom sympy import pi\nfrom classiq import QFunc, QParam, QBit, H, PHASE, allocate\nfrom classiq import create_model, synthesize, show\n\n@QFunc\ndef foo(n: QParam[int], qv: QBit) -> None:\nH(target=qv)\nfor i in range(4):\nPHASE(theta=(i / n) * pi, target=qv)\n\n@QFunc\ndef bar(m: QParam[int], qv: QBit) -> None:\nfoo(n=m * 2, qv=qv)\nfoo(n=m // 2, qv=qv)\n\n@QFunc\ndef main() -> None:\nqv = QBit(\"qv\")\nallocate(1, qv)\nbar(m=2, qv=qv)\n\nmodel = create_model(main)\nqprog = synthesize(model)\nshow(qprog)\n\n\nThis results in the following quantum program:", null, "## QVar Slicing¶\n\nWhen creating a model, you can use Python slicing on QArray[QBit] variables to connect to a port that accepts a smaller number of qubits:\n\nfrom sympy import pi\nfrom classiq import QFunc, QParam, QArray, QBit, H, PHASE, allocate\nfrom classiq import create_model, synthesize, show\n\n@QFunc\ndef foo(n: QParam[int], qv: QBit) -> None:\nH(target=qv)\nfor i in range(4):\nPHASE(theta=(i / n) * pi, target=qv)\n\n@QFunc\ndef bar(m: QParam[int], qbv: QArray[QBit, 2]) -> None:\nfoo(n=m * 2, qv=qbv)\nfoo(n=m // 2, qv=qbv)\n\n@QFunc\ndef main() -> None:\nqbv = QArray(\"qbv\")\nallocate(2, qbv)\nbar(m=2, qbv=qbv)\n\nmodel = create_model(main)\nqprog = synthesize(model)\nshow(qprog)\n\n\nThis results in the following quantum program:", null, "Note that the slice also accepts QParam objects and expressions involving them." ]

[ null, "https://docs.classiq.io/latest/user-guide/platform/qmod/python/resources/circuit_with_for.png", null, "https://docs.classiq.io/latest/user-guide/platform/qmod/python/resources/slicing.png", null ]

[ "Home - Systems Of Equations by Elimination Worksheet\n\n# Systems Of Equations by Elimination Worksheet\n\nHere is the Systems Of Equations By Elimination Worksheet section. Here you will find all we have for Systems Of Equations By Elimination Worksheet. For instance there are many worksheet that you can print here, and if you want to preview the Systems Of Equations By Elimination Worksheet simply click the link or image and you will take to save page section.\n\nSolving Systems Equations By Substitution Worksheet Methods Of Solving Systems Of Equations Free Math Worksheets Systems Equations Worksheet Systems Equations Elimination Worksheet 27 Best Systems Linear Solving Linear Systems In Three Variables Worksheet Lovely System Systems Of Equations Elimination Kuta Software Infinite Algebra 1 Solving Systems Of Equations By Substitution Kutasoftware Worksheet Energy Transformation Worksheet Answers Fronteirastral Worked Example Equivalent Systems Of Equations Video Solving Systems Equations By Elimination Worksheet Worksheet Part 3 Wizer Me Blended Worksheet \"solving Systems Of Linear Equations By Solving Systems Linear Equations By Elimination Worksheet Pdf Systems Of Equations With Elimination And Manipulation Video Age Word Problems Systems Of Linear Equations Practice." ]

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[ "", null, "", null, "高中化学知识点：氢氧化钠\n◎ 氢氧化钠的定义\n\n(1)与酸反应：NaOH+HCl==NaCl+H2O、2NaOH+H2SO4==Na2SO4+2H2O\n(2)与非金属氧化物反应：2NaOH+CO2==Na2CO3+H2O、2NaOH+SO2==Na2SO3+H2O 、2NaOH+SO3==Na2SO4+H2O、2NaOH+SiO2==Na2SiO3+H2O\n(3)与盐反应：2NaOH+CuCl2==Cu(OH)2+2NaCl\n\n◎ 氢氧化钠的知识扩展\n\n(1)与酸反应：NaOH+HCl==NaCl+H2O、2NaOH+H2SO4==Na2SO4+2H2O\n(2)与非金属氧化物反应：2NaOH+CO2==Na2CO3+H2O、2NaOH+SO2==Na2SO3+H2O 2NaOH+SO3==Na2SO4+H2O、2NaOH+SiO2==Na2SiO3+H2O\n(3)与盐反应：2NaOH+CuCl2==Cu(OH)2+2NaCl\n◎ 氢氧化钠的特性\n\n①使酸碱指示剂变色：能使石蕊溶液变蓝，能使酚酞溶液变红；\n②与酸发生中和反应生成盐和水；NaOH+HCl=NaCl+H2O\n③与某些盐反应生成新盐和新碱；2NaOH+CuSO4=Cu(OH)2↓+Na2SO4\n④与酸性氧化物反应生成盐和水。2NaOH+CO2====Na2CO3+H2O\n\n◎ 氢氧化钠的知识点拨", null, "◎ 氢氧化钠的考试要求\n\n◎ 氢氧化钠的所有试题" ]

[ null, "https://www.nduoti.cn/img/ad_holder.jpg", null, "https://www.nduoti.cn/img/ad_holder.jpg", null, "https://pic2.nduoti.cn/upload/papers/05/20131028/201310282025474005571.png", null ]

[ "# 7.14: Calculating pH of Strong Acid and Base Solutions\n\n$$\\newcommand{\\vecs}{\\overset { \\rightharpoonup} {\\mathbf{#1}} }$$ $$\\newcommand{\\vecd}{\\overset{-\\!-\\!\\rightharpoonup}{\\vphantom{a}\\smash {#1}}}$$$$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$$$\\newcommand{\\AA}{\\unicode[.8,0]{x212B}}$$\n\nLearning Objectives\n\n• Give the names and formulas of some strong acids and bases.\n• Explain the pH scale, and convert pH and concentration of hydronium ions.\n• Evaluate solution pH and pOH of strong acids or bases.\n\nAcids and bases that are completely ionized when dissolved in water are called strong acids and strong bases There are only a few strong acids and bases, and everyone should know their names and properties. These acids are often used in industry and everyday life. The concentrations of acids and bases are often expressed in terms of pH, and as an educated person, you should have the skill to convert concentrations into pH and pOH. The pH is an indication of the hydrogen ion concentration, $$\\ce{[H+]}$$.\n\n## Strong Acids\n\nStrong acids are acids that are completely or nearly 100% ionized in their solutions; Table $$\\PageIndex{1}$$ includes some common strong acids. Hence, the ionization in Equation $$\\ref{gen ion}$$ for a strong acid HA can be represented with a single arrow:\n\n$\\ce{HA(aq) + H2O(l) \\rightarrow H3O^{+}(aq) + A^{-}(aq)} \\label{gen ion}$\n\nWater is the base that reacts with the acid $$\\ce{HA}$$, $$\\ce{A^{−}}$$ is the conjugate base of the acid HA, and the hydronium ion is the conjugate acid of water. By definition, a strong acid yields 100% of $$\\ce{H3O+}$$ and $$\\ce{A^{−}}$$ when the acid ionizes in water. Table $$\\PageIndex{1}$$ lists several strong acids.\n\nTable $$\\PageIndex{1}$$: Some of the common strong acids and bases are listed here.\nStrong Acids Strong Bases\nperchloric acid ($$\\ce{HClO4}$$) lithium hydroxide ($$\\ce{LiOH}$$)\nhydrochloric acid ($$\\ce{HCl}$$) sodium hydroxide ($$\\ce{NaOH}$$)\nhydrobromic acid ($$\\ce{HBr}$$) potassium hydroxide ($$\\ce{KOH}$$)\nhydroiodic acid ($$\\ce{Hl}$$) calcium hydroxide ($$\\ce{Ca(OH)2}$$)\nnitric acid ($$\\ce{HNO3}$$) strontium hydroxide ($$\\ce{Sr(OH)2}$$)\nsulfuric acid ($$\\ce{H2SO4}$$) barium hydroxide ($$\\ce{Br(OH)2}$$)\n\nFor a strong acid, $$\\ce{[H+]}$$ = $$\\ce{[A^{-}]}$$ = concentration of acid if the concentration is much higher than $$1 \\times 10^{-7}\\, M$$. However, for a very dilute strong acid solution with concentration less than $$1 \\times 10^{-7}\\, M$$, the pH is dominated by the autoionization of water\n\n$\\ce{H2O \\rightleftharpoons H+ + OH-}$\n\nExample $$\\PageIndex{1}$$\n\nCalculate the pH of a solution with $$1.2345 \\times 10^{-4}\\; M \\ce{HCl}$$, a strong acid.\n\nSolution\n\nThe solution of a strong acid is completely ionized. That is, this equation goes to completion\n\n$\\ce{HCl(aq) -> H(aq) + Cl^{-}(aq)} \\nonumber$\n\nThus, $$\\ce{[H+]} = 1.2345 \\times 10^{-4}$$.\n\n$\\ce{pH} = -\\log(1.2345 \\times 10^{-4}) = 3.90851 \\nonumber$\n\nExercise $$\\PageIndex{1}$$\n\nWhat is the pH for a solution containing 0.234 M $$\\ce{[HCl]}$$?\n\npH = 0.63\n\n## Strong Bases\n\nStrong bases are completely ionized in solution. Table $$\\PageIndex{1}$$ includes some common strong bases. For example, $$\\ce{KOH}$$ dissolves in water in the reaction\n\n$\\ce{KOH \\rightarrow K+ + OH-} \\nonumber$\n\nRelative to the number of strong acids, there are fewer number of strong bases and most are alkali hydroxides. Calcium hydroxide is considered a strong base, because it is completely, almost completely, ionized. However, the solubility of calcium hydroxide is very low. When $$\\ce{Ca(OH)2}$$ dissolves in water, the ionization reaction is as follows:\n\n$\\ce{Ca(OH)2 \\rightarrow Ca^2+ + 2 OH-} \\nonumber$\n\nBecause of the stoichiometry of calcium hydroxide, upon dissociation, the concentration of $$\\ce{OH-}$$ will be twice the concentration of $$\\ce{Ca^2+}$$:\n\n$\\mathrm{[OH^-] = 2 [Ca^{2+}]} \\nonumber$\n\nExample $$\\PageIndex{4}$$\n\nCalculate the pH of a solution containing $$1.2345 \\times 10^{-4}\\; M \\; \\ce{Ca(OH)2}$$.\n\nSolution\n\nBased on complete ionization of\n\n$\\ce{Ca(OH)2 \\rightarrow Ca^{+2} + 2OH-} \\nonumber$\n\n\\begin{align*} \\ce{[OH^{-}]} &= 2 \\times 1.2345 \\times 10^{-4} \\\\[4pt] &= 2.4690 \\times 10^{-4}\\; M \\\\[4pt] \\ce{pOH} &= -\\log( 2.4690 \\times 10^{-4})\\\\[4pt] &= 3.6074 \\end{align*}\n\n$pH=14-pOH$\n\n$pH=14-3.6074$\n\n$pH=10.39$\n\nExercise $$\\PageIndex{4}$$\n\nThe molar solubility of calcium hydroxide is 0.013 M $$\\ce{Ca(OH)2}$$. Calculate the pH.\n\npOH = 12.42\n\n## Questions\n\n1. What is the pH of a solution containing 0.01 M $$\\ce{HNO3}$$?\n2. What is the pH of a solution containing 0.0220 M $$\\ce{Ba(OH)2}$$? Give 3 significant figures.\n3. Exactly 1.00 L solution was made by dissolving 0.80 g of $$\\ce{NaOH}$$ in water. What is $$\\ce{[H+]}$$? (Atomic mass: $$\\ce{Na}$$, 23.0; $$\\ce{O}$$, 16.0; $$\\ce{H}$$, 1.0)\n4. What is the pH for a solution which is 0.050 M $$\\ce{HCl}$$?\n5. Which of the following is usually referred to as strong acid in water solution?\n\n$$\\ce{HF}$$, $$\\ce{HNO2}$$, $$\\ce{H2CO3}$$, $$\\ce{H2S}$$, $$\\ce{HSO4-}$$, $$\\ce{Cl-}$$, $$\\ce{HNO3}$$, $$\\ce{HCN}$$\n\n## Solutions\n\nHint...\nYou do not need a calculator to evaluate $$-\\log (0.01) = 2$$\nHint...\n$$\\ce{Ba(OH)2 \\rightarrow Ba^2+ + 2 OH-}$$\n3. Answer $$5.0\\times 10^{-13}$$\n\nHint...\n$$\\mathrm{[OH^-] = \\dfrac{0.80}{40} = 0.020\\: M}$$; $$[H^+] = \\dfrac{1.0 \\times 10^{-14}}{0.020} = 5\\times 10^{-13} M$$. The pH is 12.30.\n\nThis solution contains 1.83 g of $$\\ce{HCl}$$ per liter. $$\\mathrm{[H^+] = 0.050}$$.\n5. Answer $$\\ce{HNO3}$$" ]

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[ "# What is 86% of 440?\n\n## 86 percent of 440 is equal to 378.4\n\n%\n\n86% of 440 equal to 378.4\n\nCalculation steps:\n\n( 86 ÷ 100 ) x 440 = 378.4\n\n### Calculate 86 Percent of 440?\n\n• F\n\nFormula\n\n(86 ÷ 100) x 440 = 378.4\n\n• 1\n\nPercent to decimal\n\n86% to decimal is 86 ÷ 100 = 0.86\n\n• 2\n\nMultiply decimal with the other number\n\n0.86 x 440 = 378.4\n\nExample" ]

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[ "# How Many Centimeters in 36.3 Inches?\n\n36.3 Inches to cm converter. How many centimeters in 36.3 inches?\n\n36.3 Inches equal to 92.202 cm or there are 92.202 centimeters in 36.3 inches.\n\n←→\nstep\nRound:\nEnter Inch\nEnter Centimeter\n\n## How to convert 36.3 inches to cm?\n\nThe conversion factor from inches to cm is 2.54. To convert any value of inches to cm, multiply the inch value by the conversion factor.\n\nTo convert 36.3 inches to cm, multiply 36.3 by 2.54, that makes 36.3 inches equal to 92.202 cm.\n\n36.3 inches to cm formula\n\ncm = inch value * 2.54\n\ncm = 36.3 * 2.54\n\ncm = 92.202\n\nCommon conversions from 36.3x inches to cm:\n(rounded to 3 decimals)\n\n• 36.3 inches = 92.202 cm\n• 36.31 inches = 92.227 cm\n• 36.32 inches = 92.253 cm\n• 36.33 inches = 92.278 cm\n• 36.34 inches = 92.304 cm\n• 36.35 inches = 92.329 cm\n• 36.36 inches = 92.354 cm\n• 36.37 inches = 92.38 cm\n• 36.38 inches = 92.405 cm\n• 36.39 inches = 92.431 cm\n• 36.4 inches = 92.456 cm\n\nWhat is a Centimeter?\n\nCentimeter (centimetre) is a metric system unit of length. The symbol is \"cm\".\n\nWhat is a Inch?\n\nInch is an imperial and United States Customary systems unit of length, equal to 1/12 of a foot. 1 inch = 2.54 cm. The symbol is \"in\".\n\nCreate Conversion Table\nClick \"Create Table\". Enter a \"Start\" value (5, 100 etc). Select an \"Increment\" value (0.01, 5 etc) and select \"Accuracy\" to round the result." ]

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[ "Java™ Platform\nStandard Ed. 7\njava.sql\n\n## Class Types\n\n• ```public class Types\nextends Object```\n\nThe class that defines the constants that are used to identify generic SQL types, called JDBC types.\n\nThis class is never instantiated.\n\n• ### Field Summary\n\nFields\nModifier and Type Field and Description\n`static int` `ARRAY`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `ARRAY`.\n`static int` `BIGINT`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `BIGINT`.\n`static int` `BINARY`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `BINARY`.\n`static int` `BIT`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `BIT`.\n`static int` `BLOB`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `BLOB`.\n`static int` `BOOLEAN`\nThe constant in the Java programming language, somtimes referred to as a type code, that identifies the generic SQL type `BOOLEAN`.\n`static int` `CHAR`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `CHAR`.\n`static int` `CLOB`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `CLOB`.\n`static int` `DATALINK`\nThe constant in the Java programming language, somtimes referred to as a type code, that identifies the generic SQL type `DATALINK`.\n`static int` `DATE`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `DATE`.\n`static int` `DECIMAL`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `DECIMAL`.\n`static int` `DISTINCT`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `DISTINCT`.\n`static int` `DOUBLE`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `DOUBLE`.\n`static int` `FLOAT`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `FLOAT`.\n`static int` `INTEGER`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `INTEGER`.\n`static int` `JAVA_OBJECT`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `JAVA_OBJECT`.\n`static int` `LONGNVARCHAR`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `LONGNVARCHAR`.\n`static int` `LONGVARBINARY`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `LONGVARBINARY`.\n`static int` `LONGVARCHAR`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `LONGVARCHAR`.\n`static int` `NCHAR`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `NCHAR`\n`static int` `NCLOB`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `NCLOB`.\n`static int` `NULL`\nThe constant in the Java programming language that identifies the generic SQL value `NULL`.\n`static int` `NUMERIC`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `NUMERIC`.\n`static int` `NVARCHAR`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `NVARCHAR`.\n`static int` `OTHER`\nThe constant in the Java programming language that indicates that the SQL type is database-specific and gets mapped to a Java object that can be accessed via the methods `getObject` and `setObject`.\n`static int` `REAL`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `REAL`.\n`static int` `REF`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `REF`.\n`static int` `ROWID`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `ROWID`\n`static int` `SMALLINT`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `SMALLINT`.\n`static int` `SQLXML`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `XML`.\n`static int` `STRUCT`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `STRUCT`.\n`static int` `TIME`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `TIME`.\n`static int` `TIMESTAMP`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `TIMESTAMP`.\n`static int` `TINYINT`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `TINYINT`.\n`static int` `VARBINARY`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `VARBINARY`.\n`static int` `VARCHAR`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `VARCHAR`.\n\n• ### Methods inherited from class java.lang.Object\n\n`clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait`\n• ### Field Detail\n\n• #### BIT\n\n`public static final int BIT`\n\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `BIT`.\n\nConstant Field Values\n• #### TINYINT\n\n`public static final int TINYINT`\n\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `TINYINT`.\n\nConstant Field Values\n• #### SMALLINT\n\n`public static final int SMALLINT`\n\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `SMALLINT`.\n\nConstant Field Values\n• #### INTEGER\n\n`public static final int INTEGER`\n\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `INTEGER`.\n\nConstant Field Values\n• #### BIGINT\n\n`public static final int BIGINT`\n\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `BIGINT`.\n\nConstant Field Values\n• #### FLOAT\n\n`public static final int FLOAT`\n\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `FLOAT`.\n\nConstant Field Values\n• #### REAL\n\n`public static final int REAL`\n\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `REAL`.\n\nConstant Field Values\n• #### DOUBLE\n\n`public static final int DOUBLE`\n\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `DOUBLE`.\n\nConstant Field Values\n• #### NUMERIC\n\n`public static final int NUMERIC`\n\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `NUMERIC`.\n\nConstant Field Values\n• #### DECIMAL\n\n`public static final int DECIMAL`\n\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `DECIMAL`.\n\nConstant Field Values\n• #### CHAR\n\n`public static final int CHAR`\n\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `CHAR`.\n\nConstant Field Values\n• #### VARCHAR\n\n`public static final int VARCHAR`\n\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `VARCHAR`.\n\nConstant Field Values\n• #### LONGVARCHAR\n\n`public static final int LONGVARCHAR`\n\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `LONGVARCHAR`.\n\nConstant Field Values\n• #### DATE\n\n`public static final int DATE`\n\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `DATE`.\n\nConstant Field Values\n• #### TIME\n\n`public static final int TIME`\n\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `TIME`.\n\nConstant Field Values\n• #### TIMESTAMP\n\n`public static final int TIMESTAMP`\n\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `TIMESTAMP`.\n\nConstant Field Values\n• #### BINARY\n\n`public static final int BINARY`\n\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `BINARY`.\n\nConstant Field Values\n• #### VARBINARY\n\n`public static final int VARBINARY`\n\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `VARBINARY`.\n\nConstant Field Values\n• #### LONGVARBINARY\n\n`public static final int LONGVARBINARY`\n\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `LONGVARBINARY`.\n\nConstant Field Values\n• #### NULL\n\n`public static final int NULL`\n\nThe constant in the Java programming language that identifies the generic SQL value `NULL`.\n\nConstant Field Values\n• #### OTHER\n\n`public static final int OTHER`\nThe constant in the Java programming language that indicates that the SQL type is database-specific and gets mapped to a Java object that can be accessed via the methods `getObject` and `setObject`.\nConstant Field Values\n• #### JAVA_OBJECT\n\n`public static final int JAVA_OBJECT`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `JAVA_OBJECT`.\nSince:\n1.2\nConstant Field Values\n• #### DISTINCT\n\n`public static final int DISTINCT`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `DISTINCT`.\nSince:\n1.2\nConstant Field Values\n• #### STRUCT\n\n`public static final int STRUCT`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `STRUCT`.\nSince:\n1.2\nConstant Field Values\n• #### ARRAY\n\n`public static final int ARRAY`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `ARRAY`.\nSince:\n1.2\nConstant Field Values\n• #### BLOB\n\n`public static final int BLOB`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `BLOB`.\nSince:\n1.2\nConstant Field Values\n• #### CLOB\n\n`public static final int CLOB`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `CLOB`.\nSince:\n1.2\nConstant Field Values\n• #### REF\n\n`public static final int REF`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `REF`.\nSince:\n1.2\nConstant Field Values\n\n`public static final int DATALINK`\nThe constant in the Java programming language, somtimes referred to as a type code, that identifies the generic SQL type `DATALINK`.\nSince:\n1.4\nConstant Field Values\n• #### BOOLEAN\n\n`public static final int BOOLEAN`\nThe constant in the Java programming language, somtimes referred to as a type code, that identifies the generic SQL type `BOOLEAN`.\nSince:\n1.4\nConstant Field Values\n• #### ROWID\n\n`public static final int ROWID`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `ROWID`\nSince:\n1.6\nConstant Field Values\n• #### NCHAR\n\n`public static final int NCHAR`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `NCHAR`\nSince:\n1.6\nConstant Field Values\n• #### NVARCHAR\n\n`public static final int NVARCHAR`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `NVARCHAR`.\nSince:\n1.6\nConstant Field Values\n• #### LONGNVARCHAR\n\n`public static final int LONGNVARCHAR`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `LONGNVARCHAR`.\nSince:\n1.6\nConstant Field Values\n• #### NCLOB\n\n`public static final int NCLOB`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `NCLOB`.\nSince:\n1.6\nConstant Field Values\n• #### SQLXML\n\n`public static final int SQLXML`\nThe constant in the Java programming language, sometimes referred to as a type code, that identifies the generic SQL type `XML`.\nSince:\n1.6" ]

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[ "# #cf3e08 Color Information\n\nIn a RGB color space, hex #cf3e08 is composed of 81.2% red, 24.3% green and 3.1% blue. Whereas in a CMYK color space, it is composed of 0% cyan, 70% magenta, 96.1% yellow and 18.8% black. It has a hue angle of 16.3 degrees, a saturation of 92.6% and a lightness of 42.2%. #cf3e08 color hex could be obtained by blending #ff7c10 with #9f0000. Closest websafe color is: #cc3300.\n\n• R 81\n• G 24\n• B 3\nRGB color chart\n• C 0\n• M 70\n• Y 96\n• K 19\nCMYK color chart\n\n#cf3e08 color description : Strong orange.\n\n# #cf3e08 Color Conversion\n\nThe hexadecimal color #cf3e08 has RGB values of R:207, G:62, B:8 and CMYK values of C:0, M:0.7, Y:0.96, K:0.19. Its decimal value is 13581832.\n\nHex triplet RGB Decimal cf3e08 `#cf3e08` 207, 62, 8 `rgb(207,62,8)` 81.2, 24.3, 3.1 `rgb(81.2%,24.3%,3.1%)` 0, 70, 96, 19 16.3°, 92.6, 42.2 `hsl(16.3,92.6%,42.2%)` 16.3°, 96.1, 81.2 cc3300 `#cc3300`\nCIE-LAB 47.92, 55.184, 57.338 27.5, 16.731, 2.011 0.595, 0.362, 16.731 47.92, 79.579, 46.097 47.92, 117.612, 37.966 40.904, 48.424, 25.718 11001111, 00111110, 00001000\n\n# Color Schemes with #cf3e08\n\n• #cf3e08\n``#cf3e08` `rgb(207,62,8)``\n• #0899cf\n``#0899cf` `rgb(8,153,207)``\nComplementary Color\n• #cf0835\n``#cf0835` `rgb(207,8,53)``\n• #cf3e08\n``#cf3e08` `rgb(207,62,8)``\n• #cfa208\n``#cfa208` `rgb(207,162,8)``\nAnalogous Color\n• #0836cf\n``#0836cf` `rgb(8,54,207)``\n• #cf3e08\n``#cf3e08` `rgb(207,62,8)``\n• #08cfa2\n``#08cfa2` `rgb(8,207,162)``\nSplit Complementary Color\n• #3e08cf\n``#3e08cf` `rgb(62,8,207)``\n• #cf3e08\n``#cf3e08` `rgb(207,62,8)``\n• #08cf3e\n``#08cf3e` `rgb(8,207,62)``\n• #cf0899\n``#cf0899` `rgb(207,8,153)``\n• #cf3e08\n``#cf3e08` `rgb(207,62,8)``\n• #08cf3e\n``#08cf3e` `rgb(8,207,62)``\n• #0899cf\n``#0899cf` `rgb(8,153,207)``\n• #852805\n``#852805` `rgb(133,40,5)``\n• #9e2f06\n``#9e2f06` `rgb(158,47,6)``\n• #b63707\n``#b63707` `rgb(182,55,7)``\n• #cf3e08\n``#cf3e08` `rgb(207,62,8)``\n• #e84509\n``#e84509` `rgb(232,69,9)``\n• #f65114\n``#f65114` `rgb(246,81,20)``\n• #f7642d\n``#f7642d` `rgb(247,100,45)``\nMonochromatic Color\n\n# Alternatives to #cf3e08\n\nBelow, you can see some colors close to #cf3e08. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #cf0c08\n``#cf0c08` `rgb(207,12,8)``\n• #cf1d08\n``#cf1d08` `rgb(207,29,8)``\n• #cf2d08\n``#cf2d08` `rgb(207,45,8)``\n• #cf3e08\n``#cf3e08` `rgb(207,62,8)``\n• #cf4f08\n``#cf4f08` `rgb(207,79,8)``\n• #cf5f08\n``#cf5f08` `rgb(207,95,8)``\n• #cf7008\n``#cf7008` `rgb(207,112,8)``\nSimilar Colors\n\n# #cf3e08 Preview\n\nThis text has a font color of #cf3e08.\n\n``<span style=\"color:#cf3e08;\">Text here</span>``\n#cf3e08 background color\n\nThis paragraph has a background color of #cf3e08.\n\n``<p style=\"background-color:#cf3e08;\">Content here</p>``\n#cf3e08 border color\n\nThis element has a border color of #cf3e08.\n\n``<div style=\"border:1px solid #cf3e08;\">Content here</div>``\nCSS codes\n``.text {color:#cf3e08;}``\n``.background {background-color:#cf3e08;}``\n``.border {border:1px solid #cf3e08;}``\n\n# Shades and Tints of #cf3e08\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #120501 is the darkest color, while #fffefe is the lightest one.\n\n• #120501\n``#120501` `rgb(18,5,1)``\n• #250b01\n``#250b01` `rgb(37,11,1)``\n• #381102\n``#381102` `rgb(56,17,2)``\n• #4b1603\n``#4b1603` `rgb(75,22,3)``\n• #5e1c04\n``#5e1c04` `rgb(94,28,4)``\n• #712204\n``#712204` `rgb(113,34,4)``\n• #832705\n``#832705` `rgb(131,39,5)``\n• #962d06\n``#962d06` `rgb(150,45,6)``\n• #a93307\n``#a93307` `rgb(169,51,7)``\n• #bc3807\n``#bc3807` `rgb(188,56,7)``\n• #cf3e08\n``#cf3e08` `rgb(207,62,8)``\n• #e24409\n``#e24409` `rgb(226,68,9)``\n• #f54909\n``#f54909` `rgb(245,73,9)``\n• #f6571c\n``#f6571c` `rgb(246,87,28)``\n• #f7652f\n``#f7652f` `rgb(247,101,47)``\n• #f87341\n``#f87341` `rgb(248,115,65)``\n• #f88154\n``#f88154` `rgb(248,129,84)``\n• #f98f67\n``#f98f67` `rgb(249,143,103)``\n• #fa9d7a\n``#fa9d7a` `rgb(250,157,122)``\n• #fbab8d\n``#fbab8d` `rgb(251,171,141)``\n• #fbb9a0\n``#fbb9a0` `rgb(251,185,160)``\n• #fcc7b3\n``#fcc7b3` `rgb(252,199,179)``\n• #fdd5c6\n``#fdd5c6` `rgb(253,213,198)``\n• #fee3d8\n``#fee3d8` `rgb(254,227,216)``\n• #fef0eb\n``#fef0eb` `rgb(254,240,235)``\n• #fffefe\n``#fffefe` `rgb(255,254,254)``\nTint Color Variation\n\n# Tones of #cf3e08\n\nA tone is produced by adding gray to any pure hue. In this case, #6c6b6b is the less saturated color, while #cf3e08 is the most saturated one.\n\n• #6c6b6b\n``#6c6b6b` `rgb(108,107,107)``\n• #746863\n``#746863` `rgb(116,104,99)``\n• #7c645b\n``#7c645b` `rgb(124,100,91)``\n• #856052\n``#856052` `rgb(133,96,82)``\n• #8d5c4a\n``#8d5c4a` `rgb(141,92,74)``\n• #955842\n``#955842` `rgb(149,88,66)``\n• #9d553a\n``#9d553a` `rgb(157,85,58)``\n• #a65131\n``#a65131` `rgb(166,81,49)``\n• #ae4d29\n``#ae4d29` `rgb(174,77,41)``\n• #b64921\n``#b64921` `rgb(182,73,33)``\n• #be4619\n``#be4619` `rgb(190,70,25)``\n• #c74210\n``#c74210` `rgb(199,66,16)``\n• #cf3e08\n``#cf3e08` `rgb(207,62,8)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #cf3e08 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

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[ " 【机器学习基础】熵、KL散度、交叉熵 - salon365手机版\n\n#", null, "", null, "13566568236\n\n### 来源：salon365手机版   发布时间：2019-06-24   点击量：306\n\n熵（entropy）、KL 散度（Kullback-Leibler (KL) divergence）和交叉熵（cross-entropy）在机器学习的很多地方会用到。比如在决策树模型使用信息增益来选择一个最佳的划分，使得熵下降最大；深度学习模型最后一层使用 softmax 激活函数后，我们也常使用交叉熵来计算两个分布的“距离”。KL散度和交叉熵很像，都可以衡量两个分布之间的差异，相互之间可以转化。\n\n1. 如何量化信息？\n\n信息论是应用数学的一个分支，主要研究的是对一个信号包含信息的多少进行量化。信息论的基本想法是一个不太可能的事件发生了，要比一个非常可能的事件发生，能提供更多的信息。\n\n在信息论中，我们认为：\n\n为了满足上面 3 个性质，定义了一事件 \\$mbox{x} = x\\$ 的自信息（self-information）为\n\n\begin{equation} I(x) = -log P(x)  end{equation}\n\n我们使用 \\$ext{x}\\$ 表示随机变量，使用 \\$x_1, x_2,...,x_i,..., x_N\\$ 或者 \\$x\\$ 表示随机变量 \\$ext{x}\\$ 可能的取值。当式（1）中 \\$log\\$ 以 2 为底数时，\\$I(x)\\$ 单位是比特（bit）或者香农（shannons）；当 \\$log\\$ 以自然常数 \\$e\\$ 为底数时，\\$I(x)\\$ 单位是奈特（nats）。这两个单位之间可以互相转换，通过比特度量的信息只是通过奈特度量信息的常数倍。（使用对数换底公式转化）\n\n自信息只能处理单个的输出。我们可以使用香农熵（Shannon entropy）来对整个概率分布中的不确定性总量进行量化：\n\n\begin{equation} H(ext{x}) = mathbb{E}_{ext{x} sim P}[I(x)] = sum_{i= 1}^{N} P(x_i)I(x_i) = - sum_{i= 1}^{N} P(x_i)log P(x_i) end{equation}\n\n在1948年，克劳德·艾尔伍德·香农将热力学的熵，引入到信息论，因此它又被称为香农熵。机器学习（ML）中熵的概念都是由信息论而来，所以在 ML 中能看到的熵都是香农熵，而不会是热力学的熵。\n\n熵的一些性质：\n\n2. KL 散度\n\nKL 散度全称 Kullback-Leibler (KL) divergence。\n\nKL 散度可以用来衡量两个分布的差异。\n\n在概率论与统计中，我们经常会将一个复杂的分布用一个简单的近似分布来代替。KL 散度可以帮助我们测量在选择一个近似分布时丢失的信息量。\n\n假设原概率分布为 \\$P(ext{x})\\$，近似概率分布为 \\$Q(ext{x})\\$，则使用 KL 散度衡量这两个分布的差异：\n\n\begin{equation} D_{KL}(P||Q) = mathbb{E}_{ext{x} sim P}[log frac{P(x)}{Q(x)}] = mathbb{E}_{ext{x} sim P}[log P(x) - log Q(x)]  end{equation}\n\n如果 \\$ext{x}\\$ 是离散型变量，式（3）还可以写成如下形式：\n\n\begin{equation}D_{KL}(P||Q) = sum_{i= 1}^{N} P(x_i) logfrac{P(x_i)}{Q(x_i)} = sum_{i= 1}^{N} P(x_i)[log P(x_i) - log Q(x_i)]   end{equation}\n\n对于连续型变量，则式（4）不能这么写，需要求积分。如果 \\$ext{x}\\$ 是连续型变量，则式（3）中概率分布最好用 \\$p(ext{x})\\$和\\$q(ext{x})\\$ 代替 \\$P(ext{x})\\$和\\$Q(ext{x})\\$。习惯上，用小写字母表示连续型变量的概率密度函数（probability density function，PDF），用大写字母表示离散型变量的概率质量函数（probability mass function，PMF）。（PDF和PMF都是用来描述概率分布）\n\nKL 散度的一些性质：\n\nKL 散度是非负的。KL 散度为 0，当且仅当 \\$P\\$ 和 \\$Q\\$ 在离散型变量的情况下是相同的分布，或者在连续型变量的情况下是“几乎处处”相同的。KL 散度不是真的距离，它不是对称的，即 \\$ D_{KL}(P||Q) e D_{KL}(Q||P)\\$。\n\n3. 交叉熵\n\n交叉熵（cross-entropy）和 KL 散度联系很密切。同样地，交叉熵也可以用来衡量两个分布的差异。以离散型变量 \\$ext{x}\\$ 为例：\n\n\begin{equation} H(P, Q) = - mathbb{E}_{ext{x} sim P}log Q(x) = - sum_{i= 1}^{N} P(x_i) log Q(x_i) end{equation}\n\n交叉熵 \\$H(P, Q) = H(P) +  D_{KL}(P||Q)\\$。其中 \\$H(P)\\$（即 \\$H(ext{x})\\$ ，其中 \\$ext{x} sim P\\$）为分布 \\$P\\$ 的熵，\\$D_{KL}(P||Q)\\$ 表示两个分布的 KL 散度。当概率分布 \\$P(ext{x})\\$ 确定了时，\\$H(P)\\$ 也将被确定，即 \\$H(P)\\$ 是一个常数。在这种情况下，交叉熵和 KL 散度就差一个大小为 \\$H(P)\\$ 的常数。下面给出一个简单的推导：\n\n我们将式（4）中 KL 散度的公式再进行展开：\n\n\begin{equation}label{equ:tpr}\begin{split}D_{KL}(P||Q) &= sum_{i= 1}^{N} P(x_i)[log P(x) - log Q(x)]  \\ &=  sum_{i= 1}^{N} P(x_i) log P(x_i)  - sum_{i= 1}^{N} P(x_i) log Q(x_i)  \\ &= -[- sum_{i= 1}^{N} P(x_i) log P(x_i)]+ [ - sum_{i= 1}^{N} P(x_i) log Q(x_i) ] \\ &= - H(P) + H(P, Q)end{split}end{equation}即 \\$H(P, Q) = H(P) +  D_{KL}(P||Q)\\$。\n\n交叉熵的一些性质：\n\n为什么既有 KL 散度又有交叉熵？在信息论中，熵的意义是对 \\$P\\$ 事件的随机变量编码所需的最小字节数，KL 散度的意义是“额外所需的编码长度”如果我们使用 \\$Q\\$ 的编码来表示 \\$P\\$，交叉熵指的是当你使用 \\$Q\\$ 作为密码来表示 \\$P\\$ 是所需要的 “平均的编码长度”。但是在机器学习评价两个分布之间的差异时，由于分布 \\$P\\$ 会是给定的，所以此时 KL 散度和交叉熵的作用其实是一样的，而且因为交叉熵少算一项，更加简单，所以选择交叉熵会更好。\n\nReferences\n\nGoodfellow, I., Bengio, Y., & Courville, A. (2016). Deep Learning.\n\nhttps://www.zhihu.com/question/65288314\n\nKullback-Leibler Divergence Explained\n\n, 1, 0, 9);" ]

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[ "Sidebilder PDF ePub\n .flow { margin: 0; font-size: 1em; } .flow .pagebreak { page-break-before: always; } .flow p { text-align: left; text-indent: 0; margin-top: 0; margin-bottom: 0.5em; } .flow .gstxt_sup { font-size: 75%; position: relative; bottom: 0.5em; } .flow .gstxt_sub { font-size: 75%; position: relative; top: 0.3em; } .flow .gstxt_hlt { background-color: yellow; } .flow div.gtxt_inset_box { padding: 0.5em 0.5em 0.5em 0.5em; margin: 1em 1em 1em 1em; border: 1px black solid; } .flow div.gtxt_footnote { padding: 0 0.5em 0 0.5em; border: 1px black dotted; } .flow .gstxt_underline { text-decoration: underline; } .flow .gtxt_heading { text-align: center; margin-bottom: 1em; font-size: 150%; font-weight: bold; font-variant: small-caps; } .flow .gtxt_h1_heading { text-align: center; font-size: 120%; font-weight: bold; } .flow .gtxt_h2_heading { font-size: 110%; font-weight: bold; } .flow .gtxt_h3_heading { font-weight: bold; } .flow .gtxt_lineated { margin-left: 2em; margin-top: 1em; margin-bottom: 1em; white-space: pre-wrap; } .flow .gtxt_lineated_code { margin-left: 2em; margin-top: 1em; margin-bottom: 1em; white-space: pre-wrap; font-family: monospace; } .flow .gtxt_quote { margin-left: 2em; margin-right: 2em; margin-top: 1em; margin-bottom: 1em; } .flow .gtxt_list_entry { margin-left: 2ex; text-indent: -2ex; } .flow .gimg_graphic { margin-top: 1em; margin-bottom: 1em; } .flow .gimg_table { margin-top: 1em; margin-bottom: 1em; } .flow { font-family: serif; } .flow span,p { font-family: inherit; } .flow-top-div {font-size:83%;} tre : Take the centre F, and through E the intersection of the straight lines AC, DB draw the diameter GEFH: And because the rectangle AE, EC is equal, as has been shewn to the rectangle GE, EH; and, for the same reason, the rectangle BE, ED is equal to the same rectangle GE, EH; therefore the rectangle AE, EC is equal to the rectangle BE, ED Wherefore, if two straight lines, &c. Q. E. D. Proposition XXXVI. Theorem. If from any point without a circle two straight lines be drawn, ono of which cuts the circle, and the other touches it; the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, shall be equal to the square of the line which touches it. Let D be any point without the circle ABC, and DCA, DB two straight lives drawn from it, of which DCA cuts the circle, and DB touches the same : The rectangle AD, DC is equal to the square of DB. Either DCA passes through the centre, or it does not ; first, let it pass through the centre E, and join EB : therefore the angle EBD is a right (18. 3.) angle : And because the straight line AC is bisected in E, and produced to the point D the rectangle AD, DC, together with the square o EC, is equal (6. 2.) to the square of ED, and CE i equal to EB : Therefore the rectangle AD, DC, to gether with the square of EB, is equal to the square of ED: But the square of ED is equal (47. 1.) to B E the squares of EB, BD, because EBD is a right angle : Therefore the rectangle AD, DC, together with the square of EB, is equal to the squares of EB, BD: Take away the coinmon square of EB : therefore the remaining rectangle AD, DC is equal to the square of the tangent DB. But if DCA does not pass through the centre of the circle ABC: take (1.3.) the centre E, and draw EF perpendicular (12. 1.) to AC, and join EB, EC, ED : And because the straight line EF, which passe. through the centre, cuts the straight line AC which does not pass through the centre, at right angles, it shall likewise bisect (3. 3.) it; therefore AF is equal to FC · And because the straight line AC is bisected in F, and produced to D, the rectangle AD, DC, to gether with the squarc of FC is equal (6. 2.) to the square of FD: To each of these equals add the square of FE; therefore the rectangle AD, DC, logether with the squares of CF, FE, is equal squares of DF, FE : But the square of ED is equal (47. 1.) to the squares of DF, FE, because EFD is a right angle : and the square of EC is equal to the squares of CF, FE; therefore the rectangle AD, DC, together with the square of EC, is equal to the square of ED, and CE is equal to the", null, "to EB ; therefore the rectargle AD, DC, together with the square of EB is equal to the square of ED : But the squares of EB, BD are equal to the square (47. 1.) of ED, because EBD is a right angle ; therefore the rectangle AD, DC, together with the square of EB, is equal to the squares of EB, BD: Take away the common square of EB ; therefore the remaining rectangle AD, DC is equal to the square of DB. Wherefore, if from any point, &c. Q. E. D. Cor. If from any point without a circle, there be drawn two straight lines cutting it, as AB, AC, the rectangles contained by the whole lines and the parts. of them without the circle, are equal to one another, viz. the rectangle BA, AE to the rectangle CA, AF: For each of them is equal to the square of the straight Ine AD which touches the circle. el Proposition XXXVII. Theorem. If from a point without a circle there be drawn two straight lines one of which cuts the circle, and the other meets it ; if the rectangle contained by the whole line which cuts the circle, and the part of it without the circle be equal to the square of the line which meets it, the line which meets shall touch the circle. Let any point D be taken without the circle ABC, and from it let two straight liues DCA and DB be drawn, of which DCA cuts the cir. cle, and DB meets it ; if the rectangle AD, DC be equal to the square of DB ; DB touches the circle. Draw (17.3.) the straight line De, touching the circle ABC, find its centre F, and join FE, FB, FD; then FED is a right (18.3) angle : And because DE touches the circle ABC, and DCĂ cuts it, the rectangle AD, DC is equal (36. 3.) to the square of DE : But the rectangle AD, DC is, by hypothesis, equal to the square of DB : Therefore the square of DE is equal to the square of DB; and the straight line DE equal to the straight line DB. And FE is equal to FB, wherefore DE, EF are equal to DB, BF; and the base FD is common to the two triangles DEF, DBF ; therefore the angle DEF, is equal (8. 1.) to the angle DBF; but DĒF is a right angle, therefore also DBF is a right angle : And FB, if produced, is a diameter, and the straight line which is drawn at right angles to a diameter, from the extremity of it touches (16. 3.) the circle : There fore DB touches the circle ABC. Wherefore, if from a point, &c. Q. E. D.", null, "BOOK IV.", null, "", null, "DEFINITIONS. I. scribed about a circle, when cach A IECTILINEAL figure is said to side of the circumscribed figure be inscribed in another rectilineal fie touches the circumference of the gure, when all the angles of the in- circle. acribed figure are upon the sides of the figure in which it is inscribed, In like manner, a circle cach upon each. is said to be inscribII. ed in a rectilineal fi. In like manner, a figure gure, when the cir. is said to be described cumference of the about another figure, circle touches each when all the sides of side of the figure. the circumscribed fi VI. gure pass through the angular A circle is said to be described about points of the figure about which a rectilineal figure, it is described, each through each. when the circumIII. ference of the circle A rectilineal figure is passes through all said to be inscribed the angular points in a circle, when all of the figure about the angles of the in which it is described. scribed figure are VII. upon the circumfer A straight line is said to be placed ence of the circle. in a circle, when the extremities IV. of it are in the circumference of A rectilineal figure is said to be de. the circle. PROP. I. PROB. In a given circle to place a straight line, equal to a given straight line not greater than the diameter of the circle. Let ABC be the given circle, and BC is greater D the given straight line, not greater than D; make than the diameter of the circle. CE equal (3.1.) Draw BC the diameter of the circle to D, aud from ABC; then, if BC is equal to D, the the centre C, thing required is done ; for in the at the distance circle ABC a straight' line BC is CE, describe placed equal to D: But, if it is not, the circle AEF, and join CA: Thereore, because C is the centre of the equal to the given straight line D, circle AEF, CA is equal to CE, but which is not greater than the diaD is equal to CE; therefore D is meter of the circle. Which was to equal to CA: Wherefore, in the be done. circle ABC, a straight line is placed", null, "PROP. II. PROB. In a given circle to inscribe a triangle equiangular to a given triangle. D Å Let ABC be the given circle, and DEF the giren triangle, it is required to inscribe in the circle ABC a triangle equiangular to the triangle DEF. Draw (17. 3.) the straight line GAH touching the circle in the point A, and at the point A, in the the alternate segment of the circle': straight line AH, make (23. 1.) the But HAC is equal to the angle DEF: angle HAC equal to the angle DEF; therefore also the angle ABC is equal and at the point A, in the straight to DEF: For the same reason, the line AG, make the angle GAB equal angle ACB is equal to the angle to the angle DFE, and join BC: DFE ; therefore the remaining angle Therefore, because HAG touches the BAC is equal (32. 1.) to the remain. circle ABC, and AC is drawn from ing angle EDF: Wherefore the triangle the point of contact, the angle HAC ABC is equiangular to the triangle is equal (32. 3.) to the angle ABC in DEF, and it is inscribed in the circle ABC. Which was to be done. PROP. III. PROB. About a given circle to describe a triangle equiangular to a given triangle. Let ABC be the given circle, and (17. 3.) the circle ABC; Therefore, DEF the given triangle ; it is require because LM, MN, NL touch the ed to describe a triangle about the circle ABC in the points A, B, C, to circle ABC equiangular to the tri- which from the centre are drawn KA, angle DEF. KB, KC, the angles at the points A, Produce EF both ways to the B, C, are right (18. 2.) angles : And points G, H, and find the centre K because the four angles of the quaof the circle ABC, and from it draw drilateral figure AMBK are equal to any straight line KB; at the point K, four right angles, for it can be dividin the straight line KB, make (23. 1.) ed into two triangles; and that two the angle BKA equal to the angle of them KAM, KBM are right angles, DEG, and the angle BKC equal to the other two AKB, AMB are equal the angle DFH; and through the to 'two right angles : But the angles points A, B, C, draw the straight DEG, DEF are likewise equal (13. 1.) lines LAM, MBN, NCL, touching to two right angles; therefore the A maining angle AMB is equal to the remaining angle DEF : In like manner, the angle LNM may be demonstrated to be equal to DFE ; and therefore the remaining angle MLN is equal (32. 1.) to the remaining angle EDÈ: Wherefore the triangle LMN is equiangular to the triangle angles AKB, AMB are equal to the DEF: And it is described about the angles DEG, DEF of which AKB is circle ABC. Which was to be done. eq nal to DEG; wherefore the re M B N PROP. IV. PROB. A To inscribe a circle in a given triangle. Let the given triangle be ABC, it equal angles in each, is common to is required to inscribe a circle in both; therefore their other sides shall ABC. be equal; (26. 1.) wherefore DE is Bisect (9. 1.) the angles ABC, BCA equal to DF: For the same reason by the straight lines BD, CD meet. DG is equal to DF; therefore the ing one another in the point D, from three straight lines DE, DF, DG are which draw (12. 2.) DE, DF, DG equal to one another, and the circle perpendiculars described from the centre D, at the to AB, BC, distance of any of them, shall pass CA: And be through the extremities of the other cause the an two, and touch the straight lines AB, gle EBD is e BC, CA, because the angles at the qual to the an points E, F, G, are right angles, and gle FBD, for the straight line which is drawn froin the angle ABC the extremity of a diarneter at right is bisected by BD, and that the right angles to it, touches (16. 3.) the angle BED is equal to the right angle circle; therefore the straight lines • BÉD, the two triangles EBD, FBD AB, BC, CA do each of them touch have two angles of the one equal to the circle, and the circle EFG is intwo angles of the other, and the side scribed in the triangle ABC. Which BD, which is opposite to one of the was to be done. D PROP. V. PROB. To describe a circle about a given triangle, Let the given triangle be ABC; it Bisect (10. 1.) AB, AC in the is required to describe a circle about points D, E, and from these points ABC. draw DF, EF at right angles (il. 1.) « ForrigeFortsett »" ]

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[ "# What number is 99 in Roman numerals?\n\nYour question is: what is the number 99 in Roman numerals? Learn how to convert the normal number 99 into a correct translation of the Roman numeral.\n\nThe normal number 99 is identical to the Roman numeral XCIX\n\nXCIX = 99\n\n### How to convert 99 to Roman numerals?\n\nTo convert the number 99 into Roman numerals, the translation involves dividing the number into place values (units, tens, hundreds, thousands), like this:\n\nPlace valueNumberRoman numbers\nConversion90 + 9XC + IX\nDozens90XC\nOnes9IX\n\n## How do you write 99 in Roman numerals?\n\nTo write the number 99 as Roman numerals correctly, combine the normal converted numbers. Higher numbers should always precede the lower numbers to provide you with the correct written translation, as shown in the table above.\n\n90+9 = (XCIX) = 99\n\n100 in Roman Numerals\n\nConvert another normal number to Roman numbers." ]

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[ "# Proving a subset\n\n## Homework Statement\n\nThe problem I have been given is to prove (A - B) U C is than less or equal to (A U B U C) - (A n B)\n\n## The Attempt at a Solution\n\nI've tried starting off with just (A-B) U C. Then I would say how x ε c or x ε (A - B). Also if x ε a, then x ε c and x is not in b. If x ε c, since c is a subset of (A U B U C) , x ε (A U B U C). I don't know if this is right or where to go from here.\n\nMark44\nMentor\n\n## Homework Statement\n\nThe problem I have been given is to prove (A - B) U C is than less or equal to (A U B U C) - (A n B)\nThis is a statement about sets, so the relationship is $\\subseteq$, not ≤.\n\n## The Attempt at a Solution\n\nI've tried starting off with just (A-B) U C. Then I would say how x ε c or x ε (A - B). Also if x ε a, then x ε c and x is not in b.\nTry to be more careful with the names of the sets, which are A, B, and C, not a, b, and c.\nIf x ε c, since c is a subset of (A U B U C) , x ε (A U B U C). I don't know if this is right or where to go from here.\n\ni'd say that u can use:\n\nAUBUC= lAl + lBl + lCl - lBnCl - lAnBl - lAnCl+lAnBnCl\n\nbut i'm not 100% positive just trying to give some help :)\n\nMy professor said we can also try to prove or find a counterexample to this statement. Let A, B and C be sets. Then (A-B) U C = (A U B U C) - (A n B). I'm not really sure what she means by counterexample.\n\nHallsofIvy" ]

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[ " 植物激素对驳骨丹茎愈伤组织生长和器官再生的作用-文献传递-植物通论文库\n\n# 植物激素对驳骨丹茎愈伤组织生长和器官再生的作用\n\n1 9 8 1 年 1 2月\n\nAc t\na B i lo o g iae E xP\ne r\nim e n t a li s S i n i e a\nV o l\n.\n1 4\n,\nN o\n.\n4\nD e e e m b e\nr\n19 8 1\n\n(复旦大学生物系 )\n\n(上海园林科学研究所 )\n\nB A Z毫克 /升 + N A A 0 . 2 毫克 /升 , 分化培养\n\nN人 A 和 2 , 4一 D ) , 进行诱导发根试验 。\n\nN A A 的浓度分 0 、 0 . 5 ( L ) 、 2 . 0 ( M ) 和 6 . 0 ( H ) 4\n\n( H ) 4 种 , 相互组成 16 组 ,每组接种 10 一巧 瓶 ,\n\n3%\n, 琼脂 1% , 在 15 磅 /时 . 高压 下灭菌 20 分\n\n19 8 1年 2 月2 3日收到 .\n\n7 八 / . 口\n\n. 皿产 一\n\n4 期 植物激素对驳骨丹茎愈伤组织生长和器官再生的作用\n> 0\n.\n6时每块愈伤组织上平均出芽 15 个以\n\nB A 和 N A A 组合中 , 随 N A A 浓度上升而\n\n< 0\n.\n5 时 , 每块愈伤组织上平均发 根 l 条\n\n}七/ ll夕 l犷 .产 .\n\nl日 3 不同激素浓度组合 对愈伤组织上发根的形 晌 .\n\nN A A 浓度升高而有抑制作用 , 说明确 实\n\n.O k az a w a等还认为外源生长素对愈伤组织\n\nN A A 的处理中所得 结果 , 与 许 智 宏 等\n( 1 9 7 8) 在烟草叶上的结果不一致 , 可能还\n\n194 8)\n, 但亦有人提出 I A A 和 N A A 对 根\n\n< l 时有利根发生的结果接近 , 也和中 国\n\n0\n.\n6一 2 时有利芽 形 成 , 在 0 . 05 一 0 . 2 范\n\nH il de br ar n d t ( 1 9 7 1 ) 曾指出 , 有较高 浓\n\nB A Z 毫克 /升和 N A A 0 . 2毫克 /升的条件\n\nB A 如此 , 因此在继代培养中不需外加任\n\nN A A 的作用诱导生根 , 且随 N A A浓度的\n\n4 期 植物激素对驳骨丹茎愈伤组织生长和器官再生的作用\n\n0\n.\n2 毫克 /升 N A A诱导愈伤组织 , 附加 1\n\nIA A 和 2 , 4 一D 进行诱导发根试 验 , 除几\n4 - D 无效外 , 均能生根形成再生植株 。 根\n\nN A A 组合 , 进行愈伤组织继代培养 ,讨论\n\n0\n.\n25 一圣. 0 毫克 /升范围内 , 一都能促 进 芽\n\nN A A\n, 在 0 . 5一 6 . 0 毫克 /升的浓度 范 围\n\n【1】 中国科学院北京植物所六室激素组 . 19 7 7 . 生\n\n【2 」 许智宏等 , 19 78 . 烟草叶组织培养中器官形成\n\n【3 ] 唐定台等 , 198 0 . 植物激素对哈蜜瓜子叶形成\n\n13 2一 1 35 -\n〔4 ] 颜昌敬等 , 19 80 . 生长激素对油菜苔段分化成\n\n[ 5 ] H Ild\ne r b r a n d仁 A . C · , 19 7 1 . i n “ L es C u l t u r es\nd e T is u s d e P la n t e s\n. , ,\nPP\n.\n7 1一9 3 . C e n ·\nt er N a t io n a l d e la R e e h er e h e S e i e n t i6 q u e\n,\nP a r i s\n.\n[ 6 ] M i l l\ne r , C\n.\n0\n,\n& S k o呀 , F , 19 5 3 . C h e m ic a l\ne o n t r o l o f b u d fo\nr m a t i o n in t o b a e e o s t e m se g\n-\nm e n t s\n.\nA m e ;\n. 了. B o t · , 4 0 : 7 6 8一7 7 3 .\n[ 7 ] o k\na z a w a\n,\nY\n.\ne t a l\n. ,\n1 9 6 7\n.\nE f e e t o f a u x in\na n d k in e t in o n th e d e v e lo Pm e n t a n d d ifeT\n-\nr e n t ia t i o n o f P o t a t o t is s u e c u lt 、 l r e d i n o i t r o .\n\n[ 8 ] s k\no o g\n,\nF\n· ,\n1 9 7 1\n·\nA sp e c t s o f g r o w t h af c t o r\ni n t e r a e t i o n s i n m o r P h o g e n e s is in t o b a e e o\nt i s s u e e u l t u r e s\n.\ni n “ 玩 5 C u l t u r e s d e T is s u e\nd\ne P l a n t e s\n, , .\nP P 1 15一1 3 6 . C e n t r e N a t io n a l\nd\ne Ia R e e h e r e h e S e i e n t iif q u e\n,\nP a r 主5 .\n[ 9 ] S k o g\n,\nF\n.\n& C\n.\n0\n· 入I il le r , 1 9 5 7 . C h e m ie a l\ner g u la t i o n o f g or w t h a n d o r ga\nn fo r m a t io n\nin P la n t t退s u e s e u l t u r e d 初 。 i t r o . 今mP . oS c .\n\n[10 ] s k\no o g\n,\nF\n.\n& sT\nu i\n,\nC\n. ,\n1 9 4 8\n·\nC h e m ie a l c o n\n-\nt r o l o f g r o w th a n d b\nu d fo\nr m a t io n in\nt o b a e e o s t e m s e g m e n t s a n d e a l lu s c u l t u r e\nfn o itor\n.\nA m e r\n. 了. B o t . , 3 5 : 7 8 2一7 8 7 .\n[1 1 ] s\nz w e沙 o ws k a , A · , 1 9 7 8 : R e g u la t io n o f o r -\ng a n o g e n es is i n e e ll a n d t is s u e c u l t u r e b丫\nPh y t o h o r m o n e s\n.\ni n\n ,\nR o g u la t io n o f eD\nv e -\nlo P m e n t a l P r o e e s s in P l a n t s\n. ” E d . b y H\n.\nR\n.\nS e h u t t e a n d D\n.\nG r o s s\n.\n2 1 9一 2 3 5 .\n43 2 王凯基 倪德祥 张王方 刘家一 戴学方 包慈华 14卷\nT H E E F F E C T O F P L A N T H O R M O N E S O N C A L L U S\nG R O W T H A N D O R G A N O G E N E S I S IN B U D D L E L改\nA丁L八了了C通 L O U R . J N F7 1卫 O\nW A N G K人 x\n\nNl D\nE 一X IA N o , Z H A N o R刁 AN 。 , L I U J认心 x , D肛 X u E刁人筑。 .\n\nB人 0 lz se H U人\n\nA B S T R A C T\nhT\ne P r e s e n t ivn\ne s t ig a t i o n d e al s\nw i t h t h e P l a n t l e t of r m a t i o n i n d u e e d\nofr m t h\ne s t e m s e g m e n t s o f B u d d l e i a\na s i a t i e a L o u r\n.\nin v i t\nr o a n d t h e e fe\ne t\no f Pl a n t h o mr\no n e s o n t h e e a l l u s g r o w t h\na n d o gr a n\no g e n e s i s i n t i s s u e e u l t u r e\n.\nI t w a s of u n d ht a t P l\na n t l e t s w e r e\no b t a in e d o n M S b a s i e m e d i u m\ns u卜\nPl e m e n t e d w i t h B A (Zm g / l ) p l u s N A A\n(0\n·\n2 m g川 of r e a l l u s in d u e t i o n , BA\n( 1\n.\n0 m g / l ) fo r b u d d ife\nr e n t i t i a t i o n\n,\na n d 拓 M S b a s i e m e d i u m iw t h t h e\na d d i t i o n o f I BA\n,\nI A A\n,\nN A A o r 12\n, 4 一D\na t t h e S a m e e o n e e n t r a t i o n (0\n·\n5 m g / l )\nfo\nr r o o t fo r m a t i o n\n.\nA m o n g t h e a u x i n\nt e s t e d\n,\nN A A w a s m o r e s u P e r i o r t o a n y\no t h e r s i n or o t i n g a n d Z\n, 4\n\nD sh o w e d n o\na e t i v i t y o n r o o t fo rm\na t i o n\n.\nWh\ne n t h e e a l l i i n d u e e d fr o m s t e m\ns e gm e n t s\n,\nht cy w\ne r e t r a n s fe r e d t o ht e\nr e d ife r\ne n t i a t i o n m e d i a\n,\nM S m\ne\nd i u m\nw i t h t h e a d d i t i o n o f v a yr i n g e o n e e n\n-\nt r a t i o n o f B A a n d N A A\n,\nt h e e a l l u s\ng r o w t h a n d o gr a n o g e n e s i s w e r e r e g u l a\n·\nt e d b y q u a n t i t a t iv\ne s h i ft s i n t h e r a t i o\nb\ne tw e e n t h o s e tw o h o r m\no n e s\n.\nT h e r e\n·\ns u l t s o b t a i n e d hs o w e d th a t : 1\n.\nE x o 一\ng e n o u s a P Pl i e a t i o n o f a r e l a t iv e ly l o w\ne o n e e n t r a t i o n o f B A a n d N A A w a s\ni n d i sP e n s a b l e fo\nr e a l lu s g r o w t h o n t h e\ne u l t u r e\n.\n2\n.\nBA\ns u P P l e m e n t e d\na l o n e\ng r e a t l y s t im u l a t e d b u d fo mr\na t i o n i n\ne o n e e n t ar t i o n s r a n ig n g for m 0\n.\n2 5一 4 . 0\n\nm g / l\n,\nb u t su P P r e s s e d e o m P l e t e ly\nr o o t\ni n d u e t i o n ; w h e r e a s N A A s u P P l em e n t e d\na l o n e m a kr e d l y a e e e l e r a t e d or o t fo r -\nm a t i o n i n e o n e e n t r a t i o n s r a n g i n g fr o m\n0\n.\n5一6 . 0 m g l/\n,\nb u t s u P P r e s s e d b u d i n\n-\ndu e t i\no n w i t h i n e r e a s i n g e o n e e n t r a\n-\nt i o n s\n.\n3\n.\nI n t h e P r e s e n e e o f v a yr i n g\ne o n e e n t r a t i o n s o f B A a n d N A A i n\ne o m b i n a t i o n\n, a e y t o k i n i n\n\ng r e a t e r t h a n 0\n.\n5 w a s fo u n d t o b e fa\n-\nv o u r a b l e fo\nr b u d fo mr\na t i o n b u t s u P\n-\nP r e s s ib l e fo r or o t i n d u e t i o n : w h e r e a s\nl\na e y t o k i n i n\n\nux in\nr a t i o l o w e r t h a n 0\n.\n5\nw a s fo u n d t o b e fa v o u r a bl e r o o t fo r\n-\nm a t i o n b u t in h i b i t o yr t o t h e s h o o t\nP r o m o t i\nn g e价 e t o f e y t o ik n i n . T h e\ne o u n t e r a e t i o n s b e tw e e n a u x i n a n d\n\nb r i e if y d i s e u s s e d\n.\n\n3 4 4 王凯基 倪德祥 张王方 刘家一 戴学方 包慈华 14卷\n\nE P X LA N A T ION OP F LA T E S\nP LA T E I\ni F g\n·\n1\n.\nGl\no\nbe\n\nl ike c al 璐 P r o du e e do n th e su r f ae e o f st e me s g me n t s, 1 3 d a y s o f e u l t u , `\nF ig\n.\n2\n.\nB u d d e ve l\no\neP d fr\no m ht e e a ll su\n,\n2 we\ne ks o f c a llus e u l t u r e\n.\nF ig\n·\n3\n.\nR o t\ns d e ve l\no\neP d for m\nt h e b u ds\n,\n4 w e e ks o f ca l lus\ne u l t uer\n.\n\n4 期 植物激素对驳骨丹茎愈伤组织生长和器官再生的作用 4 5 3\n\n4 期 植物激素对驳骨丹茎愈伤组织生长和器官再生的作用 4 37\n\nL P A11 T E\nE爪 e t o fv a币 ng e n e oe n ta rt ios n oA f Ba nd NA A on t h: e iz e o fea lls u.\nNA A\n(m g /L )\nBA\n( m g /L )\nL\n( 0\n.\n5 )\nM\n( 2\n.\n0)\nH\n( 6\n.\n0)\nL ( 0\n.\n5 )\n( M 2\n.\n0)\nH( 6\n.\n0)\n:、比:,;..-I" ]

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[ "emacs-elpa-diffs\n[Top][All Lists]\n\n## [elpa] master 46e18c3: Add stream stream-delay and stream-of-directory-f\n\n From: Nicolas Petton Subject: [elpa] master 46e18c3: Add stream stream-delay and stream-of-directory-files Date: Thu, 03 Mar 2016 14:38:03 +0000\n\n```branch: master\n\n* packages/stream/stream.el (stream-delay, stream-of-directory-files):\nNew functions.\n* packages/stream/tests/stream-tests.el: Add test for stream-delay.\n---\npackages/stream/stream.el | 65 +++++++++++++++++++++++++++++++--\npackages/stream/tests/stream-tests.el | 32 ++++++++++++++++-\n2 files changed, 93 insertions(+), 4 deletions(-)\n\ndiff --git a/packages/stream/stream.el b/packages/stream/stream.el\nindex 567a9e3..4d61cf1 100644\n--- a/packages/stream/stream.el\n+++ b/packages/stream/stream.el\n@@ -152,7 +152,7 @@ range is infinite.\"\n(eq (car stream) stream--identifier)))\n\n(defun stream-empty ()\n- \"Return an empty stream.\"\n+ \"Return a new empty stream.\"\n(list stream--identifier (thunk-delay nil)))\n\n(defun stream-empty-p (stream)\n@@ -317,10 +317,69 @@ kind of nonlocal exit.\"\n(cons (stream-first stream)\n(seq-filter pred (stream-rest stream)))))))\n\n+(defmacro stream-delay (expr)\n+ \"Return a new stream to be obtained by evaluating EXPR.\n+EXPR will be evaluated once when an element of the resulting\n+stream is requested for the first time, and must return a stream.\n+EXPR will be evaluated in the lexical environment present when\n+calling this function.\"\n+ (let ((stream (make-symbol \"stream\")))\n+ `(stream-make (let ((,stream ,expr))\n+ (if (stream-empty-p ,stream)\n+ nil\n+ (cons (stream-first ,stream)\n+ (stream-rest ,stream)))))))\n+\n(cl-defmethod seq-copy ((stream stream))\n\"Return a shallow copy of STREAM.\"\n- (stream-cons (stream-first stream)\n- (stream-rest stream)))\n+ (stream-delay stream))\n+\n+(defun stream-of-directory-files-1 (directory &optional nosort recurse\n+ \"Helper for `stream-of-directory-files'.\"\n+ (stream-delay\n+ (if (file-accessible-directory-p directory)\n+ (let (files dirs (reverse-fun (if nosort #'identity #'nreverse)))\n+ (dolist (file (directory-files directory t nil nosort))\n+ (let ((is-dir (file-directory-p file)))\n+ (unless (and is-dir\n+ (member (file-name-nondirectory (directory-file-name\nfile))\n+ '(\".\" \"..\")))\n+ (push file files)\n+ (when (and is-dir\n+ (if (functionp recurse) (funcall recurse file)\nrecurse))\n+ (push file dirs)))))\n+ (apply #'stream-append\n+ (stream (funcall reverse-fun files))\n+ (mapcar\n+ (lambda (dir) (stream-of-directory-files-1 dir nosort recurse\n+ (funcall reverse-fun dirs))))\n+ (stream-empty))))\n+\n+(defun stream-of-directory-files (directory &optional full nosort recurse\n+ \"Return a stream of names of files in DIRECTORY.\n+Call `directory-files' to list file names in DIRECTORY and make\n+the result a stream. Don't include files named \\\".\\\" or \\\"..\\\".\n+The arguments FULL and NOSORT are directly passed to\n+`directory-files'.\n+\n+Third optional argument RECURSE non-nil means recurse on\n+subdirectories. If RECURSE is a function, it should be a\n+predicate accepting one argument, an absolute file name of a\n+directory, and return non-nil when the returned stream should\n+recurse into that directory. Any other non-nil value means\n+\n+Even with recurse non-nil, don't descent into directories by\n+\n+If FILTER is non-nil, it should be a predicate accepting one\n+argument, an absolute file name. It is used to limit the\n+resulting stream to the files fulfilling this predicate.\"\n+ (let* ((stream (stream-of-directory-files-1 directory nosort recurse\n+ (filtered-stream (if filter (seq-filter filter stream) stream)))\n+ (if full filtered-stream\n+ (seq-map (lambda (file) (file-relative-name file directory))\nfiltered-stream))))\n\n(provide 'stream)\n;;; stream.el ends here\ndiff --git a/packages/stream/tests/stream-tests.el\nb/packages/stream/tests/stream-tests.el\nindex 88edf91..23a54b5 100644\n--- a/packages/stream/tests/stream-tests.el\n+++ b/packages/stream/tests/stream-tests.el\n@@ -171,10 +171,40 @@\n(should (= 3 (stream-first (stream-rest (seq-filter #'cl-oddp (stream-range\n0 4))))))\n(should (stream-empty-p (stream-rest (stream-rest (seq-filter #'cl-oddp\n(stream-range 0 4)))))))\n\n+(ert-deftest stream-delay-test ()\n+ (should (streamp (stream-delay (stream-range))))\n+ (should (= 0 (stream-first (stream-delay (stream-range)))))\n+ (should (= 1 (stream-first (stream-rest (stream-delay (stream-range))))))\n+ (should (let ((stream (stream-range 3 7)))\n+ (equal (seq-into (stream-delay stream) 'list)\n+ (seq-into stream 'list))))\n+ (should (null (seq-into (stream-delay (stream-empty)) 'list)))\n+ (should (let* ((evaluated nil)\n+ (one-plus (lambda (el)\n+ (setq evaluated t)\n+ (1+ el)))\n+ (stream (seq-map one-plus (stream '(1)))))\n+ (equal '(nil 2 t)\n+ (list evaluated (stream-first stream) evaluated))))\n+ (should (let* ((a 0)\n+ (set-a (lambda (x) (setq a x)))\n+ (s (stream-delay (stream (list a))))\n+ res1 res2)\n+ (funcall set-a 5)\n+ (setq res1 (stream-first s))\n+ (funcall set-a 11)\n+ (setq res2 (stream-first s))\n+ (and (equal res1 5)\n+ (equal res2 5)))))\n+\n(ert-deftest stream-seq-copy-test ()\n(should (streamp (seq-copy (stream-range))))\n(should (= 0 (stream-first (seq-copy (stream-range)))))\n- (should (= 1 (stream-first (stream-rest (seq-copy (stream-range)))))))\n+ (should (= 1 (stream-first (stream-rest (seq-copy (stream-range))))))\n+ (should (let ((stream (stream-range 3 7)))\n+ (equal (seq-into (seq-copy stream) 'list)\n+ (seq-into stream 'list))))\n+ (should (null (seq-into (seq-copy (stream-empty)) 'list))))\n\n(ert-deftest stream-range-test ()\n(should (stream-empty-p (stream-range 0 0)))\n\n```" ]

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[ "# Finding the volume of a torus\n\nConsider a torus of average radius $R$ and cross sectional radius $r$. The volume of this shape may be evaluated analytically in cartesian coordinates as a volume of revolution: $$V = 2\\int_{R-r}^{R+r} 2\\pi x z\\;\\mathrm{d}x, \\quad \\mathrm{where}\\;z = \\sqrt{r^2 - (x-R)^2}.$$ The centre of the torus is at the origin and the $z$ axis is taken to be its symmetry axis.\n\nThe integral is tedious but yields to standard methods: $V = 2\\pi^2Rr^2$. Here we take a numerical approach with the values $R=4$, $r=1$:\n\nIn [x]: import numpy as np\nIn [x]: from scipy.integrate import quad\nIn [x]: R, r = 4, 1\nIn [x]: f = lambda x, R, r: x * np.sqrt(r**2 - (x-R)**2)\nIn [x]: V, _ = quad(f, R-r, R+r, args=(R, r))\nIn [x]: V *= 4 * np.pi\nIn [x]: Vexact = 2 * np.pi**2 * R * r**2\nIn [x]: print('V = {} (exact: {})'.format(V, Vexact))\nOut[x]: V = 78.95683520871499 (exact: 78.95683520871486)" ]

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[ "# 2020 USOJMO Problems\n\n## Day 1\n\nNote: For any geometry problem whose statement begins with an asterisk", null, "$(*)$, the first page of the solution must be a large, in-scale, clearly labeled diagram. Failure to meet this requirement will result in an automatic 1-point deduction.\n\n### Problem 1\n\nLet", null, "$n \\geq 2$ be an integer. Carl has", null, "$n$ books arranged on a bookshelf. Each book has a height and a width. No two books have the same height, and no two books have the same width. Initially, the books are arranged in increasing order of height from left to right. In a move, Carl picks any two adjacent books where the left book is wider and shorter than the right book, and swaps their locations. Carl does this repeatedly until no further moves are possible. Prove that regardless of how Carl makes his moves, he must stop after a finite number of moves, and when he does stop, the books are sorted in increasing order of width from left to right.\n\n### Problem 2\n\nLet", null, "$\\omega$ be the incircle of a fixed equilateral triangle", null, "$ABC$. Let", null, "$\\ell$ be a variable line that is tangent to", null, "$\\omega$ and meets the interior of segments", null, "$BC$ and", null, "$CA$ at points", null, "$P$ and", null, "$Q$, respectively. A point", null, "$R$ is chosen such that", null, "$PR = PA$ and", null, "$QR = QB$. Find all possible locations of the point", null, "$R$, over all choices of", null, "$\\ell$.\n\n### Problem 3\n\nAn empty", null, "$2020 \\times 2020 \\times 2020$ cube is given, and a", null, "$2020 \\times 2020$ grid of square unit cells is drawn on each of its six faces. A beam is a", null, "$1 \\times 1 \\times 2020$ rectangular prism. Several beams are placed inside the cube subject to the following conditions:\n\n• The two", null, "$1 \\times 1$ faces of each beam coincide with unit cells lying on opposite faces of the cube. (Hence, there are", null, "$3 \\cdot {2020}^2$ possible positions for a beam.)\n• No two beams have intersecting interiors.\n• The interiors of each of the four", null, "$1 \\times 2020$ faces of each beam touch either a face of the cube or the interior of the face of another beam.\n\nWhat is the smallest positive number of beams that can be placed to satisfy these conditions?\n\n## Day 2\n\n### Problem 4\n\nLet", null, "$ABCD$ be a convex quadrilateral inscribed in a circle and satisfying", null, "$DA < AB = BC < CD$. Points", null, "$E$ and", null, "$F$ are chosen on sides", null, "$CD$ and", null, "$AB$ such that", null, "$BE \\perp AC$ and", null, "$EF \\parallel BC$. Prove that", null, "$FB = FD$.\n\n### Problem 5\n\nSuppose that", null, "$(a_1,b_1),$", null, "$(a_2,b_2),$", null, "$\\dots,$", null, "$(a_{100},b_{100})$ are distinct ordered pairs of nonnegative integers. Let", null, "$N$ denote the number of pairs of integers", null, "$(i,j)$ satisfying", null, "$1\\leq i and", null, "$|a_ib_j-a_jb_i|=1$. Determine the largest possible value of", null, "$N$ over all possible choices of the", null, "$100$ ordered pairs.\n\n### Problem 6\n\nLet", null, "$n \\geq 2$ be an integer. Let", null, "$P(x_1, x_2, \\ldots, x_n)$ be a nonconstant", null, "$n$-variable polynomial with real coefficients. Assume that whenever", null, "$r_1, r_2, \\ldots , r_n$ are real numbers, at least two of which are equal, we have", null, "$P(r_1, r_2, \\ldots , r_n) = 0$. Prove that", null, "$P(x_1, x_2, \\ldots, x_n)$ cannot be written as the sum of fewer than", null, "$n!$ monomials. (A monomial is a polynomial of the form", null, "$cx^{d_1}_1 x^{d_2}_2\\ldots x^{d_n}_n$, where", null, "$c$ is a nonzero real number and", null, "$d_1$,", null, "$d_2$,", null, "$\\ldots$,", null, "$d_n$ are nonnegative integers.)\n\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.", null, "2020 USOJMO (Problems • Resources) Preceded by2019 USAJMO Followed by2021 USAJMO 1 • 2 • 3 • 4 • 5 • 6 All USAJMO Problems and Solutions" ]

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[ "# noetherianOperators(Ideal,Ideal) -- Noetherian operators of a primary component\n\n## Synopsis\n\n• Function: noetherianOperators\n• Usage:\nnoetherianOperators (I, P)\nnoetherianOperators (I, P, Strategy => \"MacaulayMatrix\")\nnoetherianOperators (I, P, Rational => false)\n• Inputs:\n• I, an ideal,\n• P, an ideal, a minimal prime of $I$\n• Outputs:\n\n## Description\n\nCompute a set of Noetherian operators for the $P$-primary component of $I$.\n\n i1 : R = QQ[x,y,t]; i2 : I1 = ideal(x^2, y^2-x*t); o2 : Ideal of R i3 : I2 = ideal((x-t)^2); o3 : Ideal of R i4 : I = intersect(I1, I2); o4 : Ideal of R i5 : noetherianOperators(I, radical I1) o5 = {| 1 |, | dy |, | tdy^2+2dx |, | tdy^3+6dxdy |} o5 : List i6 : noetherianOperators(I, radical I2) == noetherianOperators(I2) o6 = true\n\nThe optional argument Strategy can be used to choose different algorithms. Each strategy may accept additional optional arguments, see the documentation page for each strategy for details.\n\nIf the prime $P$ is known to be a rational point, the optional argument Rational can be set to true. This may offer a speed-up in the computation." ]

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[ "# Projectiles and escape velocity [closed]\n\nQ: The escape velocity for a body projected vertically upwards from the surface of earth is 11 km/s. If the body is projected at an angle of $45^\\circ$ with vertical, the escape velocity will be?\n\nMy Approach:\n\nThe new vertical velocity will be $u * \\sin(45^\\circ)$ = $u /\\sqrt2$\n\nCalculating expression for escape velocity:\n\n$$\\frac{1}{2} mv^2 = \\frac{GMm}{r}$$ $$\\frac{1}{4} mu^2 = \\frac{GMm}{r}$$ $$u = 2\\sqrt\\frac{GMm}{r}$$\n\nHence the new escape velocity = $\\sqrt2 * 11$km/s.\n\nHowever, the Correct Answer is 11km/s.\n\n## closed as off-topic by Colin McFaul, Kyle Kanos, Brandon Enright, Dan, user10851 Jun 11 '14 at 20:13\n\nThis question appears to be off-topic. The users who voted to close gave this specific reason:\n\n• \"Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better\" – Colin McFaul, Kyle Kanos, Brandon Enright, Dan, Community\nIf this question can be reworded to fit the rules in the help center, please edit the question.\n\n## 3 Answers\n\nEscape velocity is not dependent on direction. It really should be called \"escape speed\"\n\nIt's the result of the calculation \"kinetic energy of object at launch\"=\"potential energy change on leaving Earth's influence\"\n\nKinetic energy isn't a vector. It is $\\frac12 m\\vec v \\cdot \\vec v=\\frac12 m|\\vec v|^2$; a scalar quantity. Note that it is independsnt of the direction of the velocity, only the vector.\n\nRemember, if you throw a ball at 45 degrees, the Earth's gravity will attract it, but it will also speed it up. These two effects \"cancel out\"\n\nFun fact: If there was a tunnel through the Earth, and you threw a ball at escape velocity down the tunnel, it would still escape Earth's influence after passing through the tunnel.\n\nWell, this certainly is an evil trick to play on first year students!\n\nEscape velocity isn't actually a velocity at all. It's a speed, i.e., it's scalar quantity as opposed to a vector quantity. Note that when the escape \"velocity\" at r was calculated, the only assumption made was conservation of mechanical energy, and then magnitude of v is isolated from kinetic energy. No direction is presumed, and so it applies any direction.\n\nHence, v = 11km/s, projected vertically or otherwise.\n\n• Yes, or more to the point the kinetic energy isn't a vector. So long as the kinetic energy is equal to the gravitational potential energy the object will escape regardless of what direction it's moving in (unless of course it points down and isn't a neutrino :-). – John Rennie Apr 9 '13 at 9:11\n• Adding to the evilness is that in the real world Aneesh Dogra's intuition that the numbers should be different is correct because launching at an angle will result in more deceleration from air drag before clearing the atmosphere. – Dan Neely Apr 9 '13 at 15:37\n\nThe (relative) escape velocity of a mass $M$ with respect to a mass $m$ is: $$V_{rel} = \\sqrt{\\frac{2 G(M\\!+\\!m)}{r} }$$where $r$ is the relative distance of the two bodies.\n\nThe escape velocity of a mass $m$ with respect to the inertial center of mass of the two bodies is: $$V_m = \\sqrt{ \\frac{2 G M M}{ r\\, (M\\!+\\!m)} }$$ and the escape velocity of the mass $M$ with respect to the inertial center of mass is: $$V_M = \\sqrt{ \\frac{2 G m m}{ r\\, (M\\!+\\!m)} }$$\n\nThe relative velocity is the sum of the two inertial velocities: $V_{rel} = V_M + V_m$." ]

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[ "# PCB Strain Rate\n\nThis C++ Script calculates PCB strain rate according to IPC/JEDEC-9704A. After importing the script the following setup should appear:", null, "For general information about the C++ Script see -> C++ script.\n\n## Input\n\nDefine the input channel for A, B and C rosette.\n\nThis script uses the 45° rosette as seen on the picture below.", null, "## Settings\n\nThreshold type:\n\n• From PCB thickness - the threshold values are calculated automatically from PCB thickness.\n• From threshold value - use the inputted threshold value.\n\nPCB Thickness:\n\n• define the thickness of the PCB. Only used if the threshold type is “From PCB thickness”.\n\nThreshold value:\n\n• define the threshold value. Only used if threshold type is “From threshold value”.\n\nMinimal strain rate on reference curve:\n\n• x-coordinate of the minimal strain rate outputted on the reference curve. Only used for visual representation of the threshold.\n\nMaximal strain rate on reference curve:\n\n• x-coordinate of the maximal strain rate outputted on the reference curve. Only used for visual representation of the threshold.\n\nOutput epsilon A, B, C, D channels:\n\n• enable or disable output of epsilon A, B, C, D channels.\n\n## Output\n\nThe script outputs the following channels. They are organized in groups for a better overview. Epsilon outputs\n\n• Epsilon A\n• Epsilon B\n• Epsilon C\n• Epsilon D\n\nPrincipal strain outputs - strain calculated in real time\n\n• Maximum principal strain\n• Minimum principal strain\n• Diagonal strain\n\nPrincipal strain rate outputs - strain rate calculated in real time\n\n• Maximum principal strain rate\n• Minimum principle strain rate\n\nMaximum strain and strain rate outputs - the biggest value of max/min strain or strain rate compared by its absolute values\n\n• Maximum strain\n• Maximum strain rate\n\nAbsolute maximum outputs - absolute value of the maximum strain and maximum strain rate\n\n• Absolute maximum strain\n• Absolute maximum strain rate\n\nOverall strain outputs - max/min strain value over time\n\n• Overall maximum strain\n• Overall minimum strain\n\nOverall strain rate outputs - max/min strain rate value over time\n\n• Overall maximum strain rate\n• Overall minimum strain rate\n\nExceeds threshold output\n\n• Ovl\n\nReference curve outputs - used for visual representation of threshold on XY Recorder\n\n• Reference curve x\n• Reference curve y\n\nNearest point outputs - strain and the strain rate nearest in value to the threshold\n\n• Nearest strain rate to threshold\n• Nearest strain to threshold\n\nExample of the output:", null, "To get an image like this on the XY recorder choose the Graph type to be “Pairs of x-y” and in the Drawing options choose the “Log X” option. Plot the threshold by first clicking the “Reference curve x” output channel and then the “Reference curve y” output channel. Plot the maximum principal strain rate vs maximum principal strain graph first clicking the “Maximum principal strain rate” output channel and then the “Maximum principal strain” output channel." ]

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[ "# C++ for mathematicians: an introduction for students and professionals\n\nنویسنده:\nامتیاز دهید\n5 / 4\nبا 1 رای\nFor problems that require extensive computation, a C++ program can race through billions of examples faster than most other computing choices. C++ enables mathematicians of virtually any discipline to create programs to meet their needs quickly, and is available on most computer systems at no cost. C++ for Mathematicians: An Introduction for Students and Professionals accentuates C++ concepts that are most valuable for pure and applied mathematical research.\nThis is the first book available on C++ programming that is written specifically for a mathematical audience; it omits the language’s more obscure features in favor of the aspects of greatest utility for mathematical work. The author explains how to use C++ to formulate conjectures, create images and diagrams, verify proofs, build mathematical structures, and explore myriad examples. Emphasizing the essential role of practice as part of the learning process, the book is ideally designed for undergraduate coursework as well as self-study. Each chapter provides many problems and solutions which complement the text and enable you to learn quickly how to apply them to your own problems. An accompanying CD ROM provides all numbered programs so that readers can easily use or adapt the code as needed.\nPresenting clear explanations and examples from the world of mathematics that develop concepts from the ground up, C++ for Mathematicians can be used again and again as a resource for applying C++ to problems that range from the basic to the complex.\n۱۴۰۰/۰۱/۰۲\nاطلاعات نسخه الکترونیکی\nتعداد صفحات:\n520\nفرمت:\nPDF\n\n### کتاب‌های مرتبط\n\nدرج دیدگاه مختص اعضا است! برای ورود به حساب خود اینجا و برای عضویت اینجا کلیک کنید.\n\n## دیدگاه‌های کتاب الکترونیکی C++ for mathematicians: an introduction for students and professionals\n\nتعداد دیدگاه‌ها:\n0\nدیدگاهی درج نشده؛ شما نخستین نگارنده باشید.\nافزودن نسخه جدید", null, "کاربر گرامی!\nامکان خرید اشتراک از خارج کشور ایران، با استفاده از حساب پی‌پال فراهم شده است." ]

[ null, "https://ketabnak.com/images/covers/i_6294.webp", null ]

[ "", null, "Open access peer-reviewed chapter\n\n# Direct Numerical Simulation of Nano Channel Flows at Low Reynolds Number\n\nWritten By\n\nP. Srinivasa Rao\n\nSubmitted: 29 November 2019 Reviewed: 09 November 2020 Published: 14 December 2020\n\nDOI: 10.5772/intechopen.94949\n\nFrom the Edited Volume\n\n## Direct Numerical Simulations - An Introduction and Applications\n\nEdited by Srinivasa Rao\n\nChapter metrics overview\n\nView Full Metrics\n\n## Abstract\n\nThe governing equations of viscous fluid flow are generally represented by Navier–Stokes (NS) equations. The output of Navier Stokes equations is in essence velocity vector from which rest of the flow parameters can be calculated. It is essentially a riotous task, sometimes it becomes so unmanageable that fluid flow over simplest topologies under low Reynold’s numbers also needs the most powerful supercomputing facility to solve, if needed to model the fluid and its behavior under the turbulent conditions the best way out is to solve the averaged NS equations. However in the process of averaging Reynolds introduced certain new terms such as Reynolds Stresses. Therefore it is required to close the system of equations by relating the unknown variables with known ones. Hence we have turbulence models. Direct Numerical Simulation (DNS) is a method of solving NS equations directly that is by forfeiting the need of turbulence models as the equations are not averaged. However originally direct numerical simulation procedure does not need of additional closure equations, it is essential to have very fine grid elements and should be estimated for exceptionally small time steps to achieve precise solutions. In the present chapter an interesting flow through nano-channel problem has been discussed using the indispensable mathematical technique of computational fluid dynamics (CFD) which is DNS.\n\n### Keywords\n\n• direct numerical simulation (DNS)\n• Navier stokes (NS) equation\n• turbulent flow\n• nano-channel flow Reynolds number\n\n## 1. Introduction\n\nWhile studying any physical phenomena be it of practical or academic importance, it is not always possible to mimic the exact system. For an example, if we want to study the flow of air past a moving train engine, we might have to build a wind tunnel along with a provision for a railway track for the engine to move in and out. We should have sensitive measuring instruments fixed at different places to measure various parameters. Again, finding the right locations for fixing these instruments can only be determined by carefully performing large number of iterations because the margin for errors in real-life problems are miniscule. For example, if the paint on the car is not done correctly even in some parts, it may contribute hugely to skin-friction drag and thereby reducing the mileage of the car.\n\nOne might argue that we can use scaled models keeping the physics of the problems same. These models induce errors during extrapolation to the actual situation besides being very costly in terms of infrastructure, resources and time. If we try to write a mathematical function and solve it, we should know all the pertinent parameters and laws that govern the flow. Then we might use various techniques like dimensional analysis and arrive at an equation. The validation of this mathematical equation again, can only be done by experiments. Once we get an equation, solving it for various cases is in itself a major challenge because we almost always end up with differential, integral or Integro-Differential equations. Till date, exact solutions to some of these equations are still open problems. So, approximate solutions are desired as against exact solutions. The advent of advanced computers brought along the solution for this in the form of Computational Fluid Dynamics (CFD). We can now analyze various types of fluid flows with numerical simulations and also develop suitable simulation algorithms. In most practical problems, the state variables of the fluids can be treated as continuous functions of space for which we already have conservation laws for mass, momentum, and energy. For many fluids that are used in engineering applications, there are well-accepted constitutive relations.\n\nThe fundamental fluid mechanics is put in the form of Navier Stokes equations way back in 1800s and thereupon many attempts were made to find the general solutions for the equations but even today it is an open millennium problem [1, 2, 3]. Based on the physics of interest, we make certain approximations to Navier–Stokes equations and therefore obtain the governing partial differential equations to be solved . When tried to solve such governing equations using a numerical methods a suitable discretization procedure will be adopted and the problem will be attacked to solve using a better solver. Here we convert the mathematical model into a discrete system of algebraic equations. We even call it to be a difference equation in some gut words while the chosen method to solve the equations is Finite difference method (FDM). Many of this kind of solving methods have been evolved and to mention a few the finite element method (FEM), the finite volume method (FVM), boundary element method (BEM) are a few popular methods among so many numerical techniques. Out of many factors that influence selection, some are very crucial like problem defined on the geometry, the analyst’s preference, and the predominant trend of solution may be in a selected and particular area of the problem space. Nonetheless it will be decided later about which numerical algorithm to be adopted and solve these equations and develop a computer programs [5, 6, 7]. After the advent of enormous speed in graphics processing units the computer graphics and animations formed from large output of numerical values make the visualization of the simulation results easier to human comprehension and draw the conclusions henceforth.\n\nThe nature of fluid flow is such a complicated phenomenon no matter what governing equations we formulate and try even more numerical solutions for the same, it can well be understood with a single classic statement of legendary Nobel Laureate R. P. Feynman in his popular scientific texts, in a quote “The efforts of a child trying to dam a small stream flowing in the street and his surprise at the strange way the water works its way out has its analog in our attempts over the years to understand the flow of fluids. We have tried to dam the water up—in our understanding—by getting the laws and the equations that describe the flow” depicts how novice the formulations are. The statement given by Feynman has already been proved more than quite often times in the applications of Fluid Dynamics in modern engineering and science, and more profoundly to the computational fluid dynamists.\n\nNonetheless an additional unique feature relates to the mathematical nature of Navier-Stokes equations. Steady-state Navier-stokes equations are elliptic in nature, whereas unsteady Navier-Stokes equations are parabolic in nature. As numerous problems associated with the methods of solution of the completely defined elliptic Partial differential equations the Navier-Stokes equations generally solved as an unsteady problem even in the case if the flow is steady, using a time marching schemes [8, 9, 10]. The transient solution of Navier-Stokes problem yields the solution but not be any easy compared to the steady state flow problem in more general geometric case.\n\nTurbulent flows involve randomly fluctuating flow variables such as velocity, pressure, and temperature.\n\nWithin the defined flow decomposition in accordance with Reynolds, a flow variable at a given spatial point at a given instant can be represented as the sum of a mean value and a random fluctuation about this mean value, and the process of obtaining the average or a mean value is represented as the technique of Reynolds averaging. Therefore, for any flow variable ϕ , its spatio-temporal variation can be expressed as stated below.\n\nϕ x i t = ϕ xi ¯ + ϕ x t\n\nWhere ϕ ¯ the mean or an averaged value and ϕ is the variations in the property which is with fluctuating values or statistically steady turbulent flows.\n\nϕ x is the time average defined as\n\nϕ xi ¯ = lim T 1 T 0 T ϕ xi d t\n\nWhere T represents the averaging time interval, which must be large compared to the typical time scales of fluctuations throughout the flow length.\n\nFor unsteady flows, ϕ represents ensemble averaging defined as\n\nϕ x i , t = lim N 1 N i = 0 N ϕ x i t\n\nwhere N is the number of identical experiments.\n\nThe Reynolds averaging applied to the continuity and momentum equations of an incompressible fluid flow would be obtained in the following set of equations.\n\n( ρ v i ) ¯ xi = 0\n( ρ v i ) ¯ xi = ( ρ v i ¯ v j ) ¯ x j = p ¯ xi + ρ τ i j ¯ τijR x j\n\nWhere τ i j R = ρ v΄iv΄j ¯ is Reynolds stress tensor.\n\nThe Reynolds averaged transport equation for a scalar ϕ\n\n( ρ ϕ ) ¯ t + ( ρ v j ¯ ϕ ) ¯ x j = r ( ρ ϕ ) ¯ t + q j R x j\n\nWhere q i R = ρ v j ϕ ¯ is turbulent flux.\n\nThe higher order correlations appear in Reynold’s stress equation is due to non-linearity of Navier–Stokes equation, despite the fact that we try an attempt to derive the governing equation for higher-order terms, they would always result in equations with even higher order correlations.\n\nThe Eddy viscosity models are evolved basing on the Boussinesq proposition, which defines the Reynolds stresses which would be proportional to mean velocity gradients, as that of Deviatoric Reynolds stress. This stress is proportional to the mean rate of strain. Numerical computation of eddy viscosity possibly would be required for the solution of the set of transport equations.\n\nThus, the Reynolds averaged NS model turbulence models are going to be categorized on some of the several additional partial differential equations that are to be solved to enforce closure and find the final flow behavior in any of the defined topology. The zero-equation model is actually the one that computes the eddy viscosity solely from the Reynolds-averaged velocity field. The transport equation that is solved simultaneously with the Reynolds-averaged Navier–Stokes equations to determine the eddy viscosity is in other words known as one-equation model that refers to the family of turbulence models [11, 12, 13].\n\nThe highly established standard k -ε model which has been developed and contributed by Launder and Spalding makes use of two model equations, one for the turbulent kinetic energy k and the dissipation rate of turbulent kinetic energy per unit mass ε. The transport equations are solved using the Reynolds averaged Navier–Stokes equations. The k-€ turbulence model is well established and the most widely validated and uses the eddy Viscosity turbulence model . Diverse modifications of the standard model have been recommended to account for the near-wall flow region and low Reynolds number turbulent flows. Nevertheless the Eddy viscosity models have significant deficiencies, which are due to some eddy viscosity assumptions. The experimental measurements and numerical simulations have indicated that in the cases of three dimensional turbulent flow the eddy viscosity turns out to be a tensor quantity . Therefore, the use of a scalar eddy viscosity for computation of Reynolds stresses is might not be appropriate. It would as an alternative be recommendable to compute Reynolds stresses directly using its dynamic transport equations. This idea has been the basis of the Reynolds stress model though it the most expensive methods of the turbulence models in use in the present day CFD solutions and particularly for RANS simulations. It is relatively expensive computationally in comparison to eddy diffusivity models since it requires solving seven additional PDEs for every grid solution in this case. Hence these models are not used as widely as expected from the analysis’s the eddy diffusivity models in practical analyses.\n\nIn the later days one of the most adoptable and practically sensible methods that have surfaced is Large Eddy simulation (LES). The method in which the Large scales of motion or large eddies of fluids which possess more energy than a smaller one have been taken to be most critical in the solving for turbulent flow cases and has been proved to be more effective in most of the cases both in the experimental and industrial applications . The most effective transporters of conserved quantities viz. mass, momentum, and energy remain to be only large eddies compared to the smaller eddies. The momentum, and energy exchange is so minimal in such cases that they can be neglected. Moreover the behavior of these smaller scales of motion is universal in turbulent flows irrespective of the flow’s context and geometry.\n\nIt is moderately undemanding to capture the effect of smaller eddies through a model. The underlying principle of LES is to treat the large eddies of the flow precisely and model the more universal small scale eddies, however this method is inherently time-dependent and are highly suitable for three-dimensional simulations. Such approach is less costly than DNS but a lot more expensive than RANS model for the same flow. Large Eddy Simulation (LES) is mostly preferred method for obtaining accurate time history for high Reynolds number and complex geometry flows. It separates the larger and smaller eddies through spatial filtering operation .\n\nThe conceptual steps are somewhat like this, it begins with the spatial filter to decompose velocity field v i x t into a sum of resolved components v i x t and residual sub grid scale component v i x t v i x t represents the motion of large eddies. Such equations are very similar to the original Navier–Stokes equation except for an additional residual stress term arising from filtering. Thus it will be obtained that the filtered velocity, pressure, and temperature field using an appropriate Navier–Stokes solver. In large eddy simulation, the spatial filtering operation for any transported field is defined by a using filter function. In LES, a filter kernels has been used, this also includes a Gaussian filter, a top-hat or box filter, and also spectral cut-off filter. Filtering operation roughly implies that eddies of a size larger than cut off width are large eddies while eddies of size smaller must be modeled. The convective term in the preceding equation cannot be computed in terms of the resolved velocity field. To get an approximation for this term, let us introduce the so called sub-grid stress tensor. The sub grid scale (SGS) Reynolds stress represents large scale momentum flux caused by small or unresolved scales, and it must be modeled to ensure closure. The dissipation takes place is much smaller than the characteristic length scale considering the length scale η of the flow with its ratio L η and this being proportional to R e 3 / 4 . As we allow the Reynolds number of the flow is increased length scales range it is observed the space and time in the flow and hence it becomes wider.\n\nOn the other hand due to the rapid growth in the computational resource Direct Numerical Simulation (DNS) of turbulent flows has become one of the crucial tools in turbulence research. Though DNS has gained its importance and has been universally recognized about its technique and methodology the kind of resources it consumes to figure out turbulence in some of the fundamental geometries is in itself a complicated affair.\n\nIt is also able to provide statistical information difficult to obtain by experimental measurements. Among effects to be observed in the DNS prior to laboratory experiments about flow velocity and vorticity vectors such effects are not present in if the modeling is done gaussian fields. However the attainment of high Reynolds number flow analysis necessitates the use of subgrid scale model to represent the effects of the unresolved small-scale turbulence on the explicitly designed and simulated large-scale flow. The leading length scales are considered to be of the order of higher dimension of flow region in the fluid flows or of the size of largest Eddie leading to turbulence in the fluid flow. The nominal scale of size of eddies in the inertia forces and viscous forces are of comparable magnitude so that the energy is cascaded to the Eddie, it is directly dissipated. Consecutively to simulate whole scales in turbulent flows, the computational domain must be sufficiently larger than the largest characteristic length scale of the flow L and the grid size must be smaller than the finest turbulence scale η.\n\nThe turbulent flows are essentially three-dimensional in nature and to satisfactorily model it needs at least require L / η 3 grid points, seems to be proportional to R e 9 / 4 . The calculations must proceed for time scale of largest Eddie i.e., L/U, while the time steps must be of the order η/U which implies R e n time steps must be taken per each run of the simulation. If we calculate the computing time necessary for the range would be of years for a single run in the limits of the magnitudes of Reynolds numbers encountered in practice. Most DNS performed does not take into account the vortices with size smaller than the grid size. To be able to obtain meaningful results from DNS sufficiently fine spatial resolution must be acquired to ensure that flow phenomena taking place at scales smaller than the grid resolution are negligible. For these reasons, DNS in many cases should be regarded as “highly accurate turbulent calculations without any use of turbulent models.” The mathematical framework and numerical results therefore obtained must serve as the useful exact solutions of the physical problems only possible for limited cases and topologies for example in the region of laminar flow of fluid through a pipe and moreover that is possible if the nonlinear terms are dropped from the N-S equation which allows analytical solution in such confined and flow regions and geometries.\n\nIn the present work, turbulent flows have been selected to investigate using large eddy simulation (LES) principles and also to understand the direct numerical simulation methodology to model and study the turbulence in fluid flow in nano channels of an electronic device. The nano channel is designed to transit the fluid and therefore removes the heat from the plate. The fluid considered to be multiphase as there would be a base fluid in which the nanoparticles will be suspended and hence an Eulerian multiphase as well as VoF techniques is being implemented. The same conditions are been attempted to solve directly using the techniques of DNS code of Geris flow solver for a simple 2D model and an unified flow field has been created as an initial flow conditions to capture the turbulent flow along the length scale and adaptive mesh refinement has been adopted while selecting the flow solvers in the three dimensional flow generation in such nano channel. The problem uses a finite volume method (FVM), in which the conservation equations in their weak formulation are solved in a discrete number of cells, determining one value for the flow parameters in each cell. The visualization has been taken in paraview and the exported combined file is processed as a data file in commercial code to make a comparative study about the two well developed turbulence modeling techniques. The nano channel considered as a rectangular pipe flow shown in Figure 1. The turbulent channel flow is often referred to as a canonical flow of Poiseuille flow case, since it is one of the simplest flows one can think of to attack using the direct numerical simulation. Therefore, this test case is suitable to verify if a numerical solver is able to accurately predict the turbulent vortices near the wall of the tube particularly at the bottom of the channel as the fluid is incompressible.\n\nSimulations between all the Reynolds numbers in the range of 100 ≥ Re ≥ 320 the grids have been selected in the stream wise, the wall-normal, and the pipe length of the computational domain are L x × L y × L z = 2 × 2 × 40 as shown in the Figure 1. The grid points have been N x × N y × N z = 178 × 98 × 356 . It has been proposed as the standard condition that the computational grid is uniform in the direction of stream flow. In due course of transient time approximately 18 convective time scales it has been observed that the steady state has been reached. From this condition the time averaging has started and the flow has been simulated for more than 120 convective time steps statistically independent time levels have been considered for averaging flow velocities and vorticity in the length scale.\n\nIn the Figures 2,3 and 4 it has been shown about the convergence of the statistical data in the cases of turbulent intensity along the pipe length as the flow becomes steadily turbulent.\n\nThe gradual development from uniform turbulence to a pattern at Reynolds number180 has been accounted and presented in the Figure 4. The energy distribution plays a key role in the flow transition and henceforth is to be tracked quantitatively and with the probability distribution. The velocity gradients near the pipe walls after the flow is fully developed would exhibit the axial flow velocity as shown in Figure 2 and consecutively the root mean square velocity computed is presented in Figure 3. These plots are developed on the basis of non dimensional number along the length of the pipe. At the maximum fluid velocity along the channel it has been recorded that there is no significant raise or dip in the Reynolds number in all the runs.\n\nThe propagation velocity of the pattern is approximately that of the mean flux and is a decreasing function of Reynolds number. Examination of the time-averaged flow shows that a turbulent band is associated with two counter-rotating cells stacked in the cross-channel direction and that the turbulence is highly concentrated near the walls. Near the wall, the Reynolds stress force accelerates the fluid through a turbulent band while viscosity decelerates it; advection by the laminar profile acts in both directions. In the center, the Reynolds stress force decelerates the fluid through a turbulent band while advection by the laminar profile accelerates it. These characteristics are compared with those of turbulent-laminar banded patterns in plane Couette flow.\n\nThe average Reynolds number varies in the different cases, from 120 in the bare flow case to 512 in the fully developed convective flow conditions as a working nanofluid cooling device. Though the primary aspect of heat transfer is not considered until the turbulent flow did not notice during the flow through the channel, the swirling velocity component is completely ignored. The pressure drop is exactly getting equated to the experimentally verified case in the case of LES calculated value in almost all the cases as shown in Figure 5. Once the LES model could capture the significant turbulence properties, the problem is attempted on geris flow solver code for DNS solution for one such case to be verified with the already obtained results. Some of the results obtained are presented in Figure 6." ]

[ null, "https://cdnintech.com/web/frontend/www/assets/44.49/journals/OpenAccessLock.svg", null ]

[ "# UniAna.util.DecTac\n\nRequire Import Coq.Classes.EquivDec.\nRequire Import Decidable List.\n\nLtac decide' X :=\nlet e := fresh \"e\" in\nlet c := fresh \"c\" in\nlazymatch X with\n| ?a == ?b => destruct X as [e|c]; [destruct e|]\n| _ => lazymatch type of X with\n| ?a == ?b => destruct X as [e|c]; [destruct e|]\n| {_ === _} + {_ =/= _} => destruct X as [e|c]; [destruct e|]\nend\nend.\n\nLemma In_dec (A : Type) `{EqDec A eq} a (l : list A) : decidable (In a l).\nProof.\nunfold decidable.\ninduction l.\n- right. tauto.\n- destruct IHl.\n+ left; econstructor 2; eauto.\n+ destruct (a == a0).\n* destruct e. left; econstructor. reflexivity.\n* right. contradict H0. inversion H0; eauto. subst a0. exfalso. apply c. reflexivity.\nQed." ]

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[ "## 11144\n\n11,144 (eleven thousand one hundred forty-four) is an even five-digits composite number following 11143 and preceding 11145. In scientific notation, it is written as 1.1144 × 104. The sum of its digits is 11. It has a total of 5 prime factors and 16 positive divisors. There are 4,752 positive integers (up to 11144) that are relatively prime to 11144.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Even\n• Number length 5\n• Sum of Digits 11\n• Digital Root 2\n\n## Name\n\nShort name 11 thousand 144 eleven thousand one hundred forty-four\n\n## Notation\n\nScientific notation 1.1144 × 104 11.144 × 103\n\n## Prime Factorization of 11144\n\nPrime Factorization 23 × 7 × 199\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 5 Total number of prime factors rad(n) 2786 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 11,144 is 23 × 7 × 199. Since it has a total of 5 prime factors, 11,144 is a composite number.\n\n## Divisors of 11144\n\n1, 2, 4, 7, 8, 14, 28, 56, 199, 398, 796, 1393, 1592, 2786, 5572, 11144\n\n16 divisors\n\n Even divisors 12 4 2 2\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 16 Total number of the positive divisors of n σ(n) 24000 Sum of all the positive divisors of n s(n) 12856 Sum of the proper positive divisors of n A(n) 1500 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 105.565 Returns the nth root of the product of n divisors H(n) 7.42933 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 11,144 can be divided by 16 positive divisors (out of which 12 are even, and 4 are odd). The sum of these divisors (counting 11,144) is 24,000, the average is 1,500.\n\n## Other Arithmetic Functions (n = 11144)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 4752 Total number of positive integers not greater than n that are coprime to n λ(n) 396 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 1353 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares\n\nThere are 4,752 positive integers (less than 11,144) that are coprime with 11,144. And there are approximately 1,353 prime numbers less than or equal to 11,144.\n\n## Divisibility of 11144\n\n m n mod m 2 3 4 5 6 7 8 9 0 2 0 4 2 0 0 2\n\nThe number 11,144 is divisible by 2, 4, 7 and 8.\n\n## Classification of 11144\n\n• Arithmetic\n• Abundant\n\n### Expressible via specific sums\n\n• Polite\n• Non-hypotenuse\n\n## Base conversion (11144)\n\nBase System Value\n2 Binary 10101110001000\n3 Ternary 120021202\n4 Quaternary 2232020\n5 Quinary 324034\n6 Senary 123332\n8 Octal 25610\n10 Decimal 11144\n12 Duodecimal 6548\n20 Vigesimal 17h4\n36 Base36 8lk\n\n## Basic calculations (n = 11144)\n\n### Multiplication\n\nn×i\n n×2 22288 33432 44576 55720\n\n### Division\n\nni\n n⁄2 5572 3714.67 2786 2228.8\n\n### Exponentiation\n\nni\n n2 124188736 1383959273984 15422842149277696 171872152911550644224\n\n### Nth Root\n\ni√n\n 2√n 105.565 22.3364 10.2745 6.44775\n\n## 11144 as geometric shapes\n\n### Circle\n\n Diameter 22288 70019.8 3.9015e+08\n\n### Sphere\n\n Volume 5.79712e+12 1.5606e+09 70019.8\n\n### Square\n\nLength = n\n Perimeter 44576 1.24189e+08 15760\n\n### Cube\n\nLength = n\n Surface area 7.45132e+08 1.38396e+12 19302\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 33432 5.37753e+07 9650.99\n\n### Triangular Pyramid\n\nLength = n\n Surface area 2.15101e+08 1.63101e+11 9099.04\n\n## Cryptographic Hash Functions\n\nmd5 3103deb68465747643608bb0f506dee6 86e42aaa6ac47b0542d6d74930da4de4b66a19db d24e2d54b460fdd8d040b059d3ed9db18e7d3927b07be706a6c61c651e5acac5 2f5823d359bfae67854d89961304c08ca3f60d31faf024b29256c6b8c2eaf617b4f3e7f8c1286f2ae5b148aa90c8b213eee0bf50dcb79da67982d04b34fdd4b2 15c59577ead783e319ba6af07bb0b7eb4b0540f5" ]

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[ "# Saga 測試的範例\n\n## 測試 Saga Generator Function\n\n``````const CHOOSE_COLOR = 'CHOOSE_COLOR';\nconst CHANGE_UI = 'CHANGE_UI';\n\nconst chooseColor = (color) => ({\ntype: CHOOSE_COLOR,\npayload: {\ncolor,\n},\n});\n\nconst changeUI = (color) => ({\ntype: CHANGE_UI,\npayload: {\ncolor,\n},\n});``````\n\n``````function* changeColorSaga() {\nconst action = yield take(CHOOSE_COLOR);\nyield put(changeUI(action.payload.color));\n}``````\n\nSince Sagas always yield an Effect, and these effects have simple factory functions (e.g. put, take etc.) a test may inspect the yielded effect and compare it to an expected effect. To get the first yielded value from a saga, call its `next().value`:\n\n`````` const gen = changeColorSaga();\n\nassert.deepEqual(\ngen.next().value,\ntake(CHOOSE_COLOR),\n'it should wait for a user to choose a color'\n);``````\n\nA value must then be returned to assign to the `action` constant, which is used for the argument to the `put` effect:\n\n`````` const color = 'red';\nassert.deepEqual(\ngen.next(chooseColor(color)).value,\nput(changeUI(color)),\n'it should dispatch an action to change the ui'\n);``````\n\n`````` assert.deepEqual(\ngen.next().done,\ntrue,\n'it should be done'\n);``````\n\n### Branching Saga\n\n``````const CHOOSE_NUMBER = 'CHOOSE_NUMBER';\nconst DO_STUFF = 'DO_STUFF';\n\nconst chooseNumber = (number) => ({\ntype: CHOOSE_NUMBER,\npayload: {\nnumber,\n},\n});\n\nconst doStuff = () => ({\ntype: DO_STUFF,\n});``````\n\nNow the saga under test will put two `DO_STUFF` actions before waiting for a `CHOOSE_NUMBER` action and then putting either `changeUI('red')` or `changeUI('blue')`, depending on whether the number is even or odd.\n\n``````function* doStuffThenChangeColor() {\nyield put(doStuff());\nyield put(doStuff());\nconst action = yield take(CHOOSE_NUMBER);\nif (action.payload.number % 2 === 0) {\nyield put(changeUI('red'));\n} else {\nyield put(changeUI('blue'));\n}\n}``````\n\n``````import { put, take } from 'redux-saga/effects';\nimport { cloneableGenerator } from 'redux-saga/utils';\n\ntest('doStuffThenChangeColor', assert => {\nconst gen = cloneableGenerator(doStuffThenChangeColor)();\ngen.next(); // DO_STUFF\ngen.next(); // DO_STUFF\ngen.next(); // CHOOSE_NUMBER\n\nassert.test('user choose an even number', a => {\n// cloning the generator before sending data\nconst clone = gen.clone();\na.deepEqual(\nclone.next(chooseNumber(2)).value,\nput(changeUI('red')),\n'should change the color to red'\n);\n\na.equal(\nclone.next().done,\ntrue,\n'it should be done'\n);\n\na.end();\n});\n\nassert.test('user choose an odd number', a => {\nconst clone = gen.clone();\na.deepEqual(\nclone.next(chooseNumber(3)).value,\nput(changeUI('blue')),\n'should change the color to blue'\n);\n\na.equal(\nclone.next().done,\ntrue,\n'it should be done'\n);\n\na.end();\n});\n});``````\n\n## 測試完整的 Saga\n\nAlthough it may be useful to test each step of a saga, in practise this makes for brittle tests. Instead, it may be preferable to run the whole saga and assert that the expected effects have occurred.\n\nSuppose we have a simple saga which calls an HTTP API:\n\n``````function* callApi(url) {\nconst someValue = yield select(somethingFromState);\ntry {\nconst result = yield call(myApi, url, someValue);\nyield put(success(result.json()));\nreturn result.status;\n} catch (e) {\nyield put(error(e));\nreturn -1;\n}\n}``````\n\nWe can run the saga with mocked values:\n\n``````const dispatched = [];\n\nconst saga = runSaga({\ndispatch: (action) => dispatched.push(action);\ngetState: () => ({ value: 'test' });\n}, callApi, 'http://url');``````\n\nA test could then be written to assert the dispatched actions and mock calls:\n\n``````import sinon from 'sinon';\nimport * as api from './api';\n\ntest('callApi', async (assert) => {\nconst dispatched = [];\nsinon.stub(api, 'myApi').callsFake(() => ({\njson: () => ({\nsome: 'value'\n})\n}));\nconst url = 'http://url';\nconst result = await runSaga({\ndispatch: (action) => dispatched.push(action),\ngetState: () => ({ state: 'test' }),\n}, callApi, url).done;\n\nassert.true(myApi.calledWith(url, somethingFromState({ state: 'test' })));\nassert.deepEqual(dispatched, [success({ some: 'value' })]);\n});``````\n\nhttps://github.com/redux-saga/redux-saga/blob/master/examples/counter/test/sagas.js\n\nhttps://github.com/redux-saga/redux-saga/blob/master/examples/shopping-cart/test/sagas.js" ]

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[ "Dep. var.: $$\\Delta { \\ln }\\left( {\\frac{{L_{it + 1} }}{{L_{it} }}} \\right)$$ t = 2006 t = 2007 t = 2008", null, "" ]

[ null, "https://labourmarketresearch.springeropen.com/track/article/10.1186/s12651-019-0257-0", null ]

[ "# Extract Lines between two tags\n\nI have following sample data from text file\n\nl1\nl2\nStart :\na1\na2\na3\n-}\nl3\nl4\nl5\nStart :\na4\na5\na6\n-}\n\n\nI want to extract lines between two tages \"Start : \" and -}\n\nI tried to extract details with regexp but output was coming in one line. I want output with newline\n\nPl help.\n\nThis will get the raw content of the text file (meaning it will pull in the carriage returns and new lines). Then with that data it performs a regular expression to extract the data that you want between the \"Start :\" and the \"-}\" pulling all the matches. Then it just outputs the value for each match found. If you want to store that in a variable you would do that in the last foreach loop.\n\n$file = Get-Content -path \"C:\\text.txt\" -Raw$wantedData = ($file | select-string '(?<=Start\\s:\\n)[\\w\\s]*(?=-})' -allMatches | foreach {$_.Matches} | Foreach {$_.Value}) Probably not the best approach but this will work as long as it is always in that format: $i = \"l1 l2 Start : a1 a2 a3 -} l3 l4 l5 Start : a4 a5 a6 -}\"\n$i =$i.split(':').split('-}')\n\nforeach ($line in$i) {\nif ($line -notlike \"*start*\" ){$line.trim() }\n}\n\n\nWill there always be 3 values between start and -} ?\n\nIf so, then:\n\n$Content = Get-Content -path C:\\Text.txt$content | Select-String -Pattern \"Start :\" -Context 0,3 | foreach {\\$_.Context.DisplayPostContext}" ]

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[ "Solutions by everydaycalculation.com\n\nAdd 14/4 and 30/24\n\n1st number: 3 2/4, 2nd number: 1 6/24\n\n14/4 + 30/24 is 19/4.\n\nSteps for adding fractions\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 4 and 24 is 24\n2. For the 1st fraction, since 4 × 6 = 24,\n14/4 = 14 × 6/4 × 6 = 84/24\n3. Likewise, for the 2nd fraction, since 24 × 1 = 24,\n30/24 = 30 × 1/24 × 1 = 30/24\n4. Add the two fractions:\n84/24 + 30/24 = 84 + 30/24 = 114/24\n5. After reducing the fraction, the answer is 19/4\n6. In mixed form: 43/4\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:\nAndroid and iPhone/ iPad" ]

[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]

[ " Analytical geometry hyperbola example 4 back to math video class\n\n### Example 4 of how to find the equation of a hyperbola\n\nA hyperbola has the same focus as an ellipse. An ellipse has the equation x2/64 + y2/15 = 1. The asymptotes of the hyperbola is y = +- (3/4) x. Find the equation for the hyperbola.\n\nFrom an ellipse equation, we know its major axis, minor axis and coordinate of the focus. For a hyperbola, know the focus and asymptotes, we have the equation between the half real axis and half imaginary axis, which are two unknown variables. We need another equation to solve these two variables. The equation is that the square of the focus is equal to the sum of the square of the half major axis and the square of the half minor axis. Know the half major axis and the half minor axis, we can write the equation for the hyperbola." ]

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[ "## NCERT Solution for Class 10\n\nClass 10 Maths Chapters Chapter 1 – Real Numbers Chapter 2 – Polynomials Exercise 2.1 Exercise 2.2 Exercise 2.3 Chapter 3 – Pair Of Linear Equation in Two Variable Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5 – Multiplication Method Exercise 3.6 Chapter 4 – Quadratic Equation Exercise 4.1 Exercise 4.2 Exercise 4.3 …" ]

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[ "", null, "# Solving Multi-Step Equations\n\nRelated Topics:\nMath Worksheets\n\nExamples, videos, worksheets, stories, and solutions to help Grade 8 students learn how to Solve Multi-Step Equations.\n\nIn order to solve an equation, we need to isolate the variable on one side of the equation. In this lesson, we will learn how to solve multi-step equations. A multi-step equation is an equation that you can solve in at least three operations.\n\nIn some other lessons, we will learn how to solve one-step equations, two-step equations and equations with variables on both sides.\n\nSolving Multistep Equations\n\nSolving Multi-Step Equations\nThis video includes four examples of equations where we need to combine like terms and/or use the Distributive Property in order to solve.\n\nLearn to solve equations that involve the distributive property by first distributing through the parentheses, then combining like terms, then solving from here\nSolving Multi-Step Equations\n\nTry the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.", null, "", null, "" ]

[ null, "https://www.onlinemathlearning.com/objects/default_image.gif", null, "https://www.onlinemathlearning.com/objects/default_image.gif", null, "https://www.onlinemathlearning.com/objects/default_image.gif", null ]

[ "### CFA Practice Question\n\nThere are 985 practice questions for this topic.\n\n### CFA Practice Question\n\nA set of four cards consists of two red cards and two black cards. The cards are shuffled thoroughly, and I am dealt two cards. I count the number of red cards X in these two cards. The random variable X has which of the following probability distributions?\n\nA. The binomial distribution with parameters n = 4 and p = 0.5\nB. The binomial distribution with parameters n = 2 and p = 0.5\nC. None of the above\n\nBecause two cards are being selected, the number of observations is two. If the observations were independent, answer B would be correct. But the observations are not independent. If the first card is red, the probability that the second card dealt is red is 1/3. If the first card is black, the probability that the second card dealt is red is 2/3. These two probabilities would have to be the same if the observations were independent.\n\nUser Comment\ndanlan Good point.\nmoma20 what a nice note\nn9845705 I don't see how it is 1/3 or 2/3 I get if 1st red then prob that 2nd red is 25/51=.4902\nAyshaAli there id only 4 cards in total 2 red and 2 black.\nfirst attempt: probability of red is 2/4 (.5)\nsecond attenpt: probability of red is dependent on the first attempt. If red was dealt then the no. of red cards will be 1 and total of cards will be 3, hence, the prob is 1/3/\nwhile if the card dealt was blak then there will be no change in total no. of red cards while the total cards are 3 in total now so prob is 2/3\naspazia do not see the difference? why are they not independent events?\nbansal imagine that there were only two cards to begin with - one red and one black...then try to see the difference....\ngaur because there are only 4 cards...its a dependant event...if there were say 7000 cards then it would be independant\no123 no...if the card was replaced it would be independant\nbobert For it to be independent it needs to be the same every time. If you roll a die, it is the same every time, 1/6, 1/6, 1/6...and so on. If you are changing the set though, as in this case taking two cards out of the deck, you will not have the same probability to receive the red cards. For all you know it is possible to get both red cards, then if you pick up the last two cards you know you have 0% chance of getting another red card. That is why this is a dependent event. If the cars were picked one at a time, and then put back, it would be independent because the picking of a card is the same each time.\ngazza77 It is dependant because the probability for each colour being drawn next will change depending on what was drawn previously\nbundy No Replacement so it can't be a Binomial distributin...ie the selction is no longer idependent after the first card is selected and not replaced\njohntan1979 Always assume dependent events (not replaced) if the probability is not stated in the question.\ndeliawmx I think the problem is that whether these two cards are dealt together one-time or one by one. Misunderstanding exists under this situation." ]

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[ "# How do you solve 4(x + 1) = 4x + 10?\n\nMar 3, 2018\n\nNo solution\n\n#### Explanation:\n\nFirst, use the distributive property on the parentheses\n$4 \\left(x + 1\\right) = 4 x + 10$\n$4 x + 4 = 4 x + 10$\n\nSince the x values on both sides of the equation has a different number being added to them, we can conclude that this equation has no solution\n\nMar 3, 2018\n\nThere is no solution.\n\n#### Explanation:\n\n$4 \\left(x + 1\\right) = 4 x + 10 \\text{ }$ 1.) Apply distributive property: $4 \\left(x + 1\\right) = 4 x + 4$\n\n$4 x + 4 = 4 x + 10 \\text{ }$ 2.) Subtract 4 from both sides\n$\\text{ \"-4\" } - 4$\n$4 x = 4 x + 6$\n\n$4 x = 4 x + 6 \\text{ }$ 3.) Subtract $4 x$ from both sides and get $0 = 6$\n$- 4 x \\text{ } - 4 x$\n$0 = 6$\n\nAfter finish solving you will get the answer $0 = 6$. However, because 0 is not equal to 6, there is no solution." ]

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We demonstrate...\n65N15 | Mathematical Methods in Physics | 65K15 | Numerical Analysis | Theoretical, Mathematical and Computational Physics | Mathematical and Computational Engineering | Mathematics, general | Mathematics | Numerical and Computational Physics, Simulation | 65N30\nJournal Article\nCalcolo, ISSN 1126-5434, 2015, Volume 53, Issue 2, pp. 189 - 199\nIn this paper, a general accelerated modulus-based matrix splitting iteration method is established, which covers the known general modulus-based matrix...\n65K15 | Modulus-based method | Numerical Analysis | Mathematics | Theory of Computation | Linear complementarity problem | Convergence | MATHEMATICS | MULTISPLITTING METHODS | Texts | Splitting | Mathematical models | Central processing units | Iterative methods\nJournal Article\nJournal Article\nNumerische Mathematik, ISSN 0945-3245, 2014, Volume 130, Issue 4, pp. 579 - 613\nJournal Article\nMathematical Programming, ISSN 0025-5610, 5/2019, Volume 175, Issue 1, pp. 241 - 262\nJournal Article\nComputational optimization and applications, ISSN 1573-2894, 2015, Volume 63, Issue 1, pp. 143 - 168\nJournal Article\nJournal of Inequalities and Applications, ISSN 1025-5834, 12/2018, Volume 2018, Issue 1, pp. 1 - 13\nThe Schwarz algorithm for a class of elliptic quasi-variational inequalities with nonlinear source terms is studied in this work. The authors prove a new error...\n65N15 | 65K15 | Mathematics | Schwarz algorithm | Quasi-variational inequalities | Error estimates | Overlapping grids | Analysis | Mathematics, general | Applications of Mathematics | Stability property | 65N30 | 05C38 | MATHEMATICS | MATHEMATICS, APPLIED | APPROXIMATION | Research\nJournal Article\nJournal Article\nDemonstratio mathematica, ISSN 2391-4661, 2018, Volume 51, Issue 1, pp. 211 - 232\nThe purpose of this paper is to study a split generalized mixed equilibrium problem and a fixed point problem for nonspreading mappings in real Hilbert...\n65K15 | 90C33 | viscosity approximation method | iterative method | 47J25 | fixed point problem | accelerated algorithm | 65J15 | split mixed equilibrium | nonspreading mapping\nJournal Article\nNumerical algorithms, ISSN 1572-9265, 2017, Volume 78, Issue 3, pp. 827 - 845\nJournal Article" ]

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[ "Solutions by everydaycalculation.com\n\n8/45 + 27/70 is 71/126.\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 45 and 70 is 630\n2. For the 1st fraction, since 45 × 14 = 630,\n8/45 = 8 × 14/45 × 14 = 112/630\n3. Likewise, for the 2nd fraction, since 70 × 9 = 630,\n27/70 = 27 × 9/70 × 9 = 243/630", 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 ]

[ "English | Español\n\n# Try our Free Online Math Solver!", null, "Online Math Solver\n\n Depdendent Variable\n\n Number of equations to solve: 23456789\n Equ. #1:\n Equ. #2:\n\n Equ. #3:\n\n Equ. #4:\n\n Equ. #5:\n\n Equ. #6:\n\n Equ. #7:\n\n Equ. #8:\n\n Equ. #9:\n\n Solve for:\n\n Dependent Variable\n\n Number of inequalities to solve: 23456789\n Ineq. #1:\n Ineq. #2:\n\n Ineq. #3:\n\n Ineq. #4:\n\n Ineq. #5:\n\n Ineq. #6:\n\n Ineq. #7:\n\n Ineq. #8:\n\n Ineq. #9:\n\n Solve for:\n\n Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg:\n\n# BROOME COMMUNITY COLLEGE Binghamton, New York\n\nCOURSE TITLE ELEMENTARY ALGEBRA AND TRIGONOMETRY MAT 096\n\nCLASS LECTURE HOURS 5\n\nLAB HOURS 0\n\nCREDIT HOURS 5\n\nDIVISION DEAN Julia Peacock\n\nDEPT. CHAIRPERSON Paul O’Heron\n\nDATE Spring 2005\n\nPREREQUISITE: MAT 092 Foundations for College Mathematics I or equivalent.\n\n## MAT 096 Objectives\n\nAfter taking MAT 096 the student should be able to:\nPerform skills in four categories: Algebra, Geometry/Trigonometry, Graphing and Problem Solving /Estimation.\nNote: Throughout the course the students are expected to solve applied problems related to the topics of the course.\n\nAlgebra:\n\nAfter taking MAT 096 the student should be able to:\n1. Solve linear equations . (a complete development)\n2. Perform operations on polynomials.\n3. Use exponent rules with rational exponents. (a complete development)\n4. Simplify rational expressions and perform operations on rational expressions including complex fractions.\n5. olve elementary rational equations.\n8. Solve and evaluate literal equations.\n9. Solve equations with absolute values.\n10. Define and evaluate functions using function notation.\n11. After a brief review, factor a monomial from a polynomial, factor trinomials, and factor special cases (difference of squares).\n12. Factor expressions in quadratic form , expressions that are sum and difference of cubes, and expressions that can be factored by\ngrouping.\n13. After a brief review, solve quadratic equations by factoring and by using the quadratic formula.\n14. Solve quadratic equations by completing the square.\n15. Solve higher degree equations by factoring.\n16. Identify parallel lines from their equations.\n17. After a brief review, solve 2 by 2 systems of linear equations by substitution and elimination.\n\nGeometry/Trigonometry:\n\nAfter taking MAT 096 the student should be able to:\n18. Classify angles and triangles using appropriate terminology .\n19. Relate the sides and angles of similar triangles.\n20. Find the area and perimeter of polygons.\n21. Convert between radians and degrees.\n22. Find reference angles for angles measured in degrees.\n23. Evaluate the six trigonometric functions of general angles measured in degrees.\n24. Understand and use right triangle trigonometry.\n\nGraphing:\n\nAfter taking MAT 096 the student should be able to:\n25. Identify and graph the following basic relations:", null, "26. Find equations of lines using point-slope form, slope- intercept form and general form.\n27. After a brief review, solve 2 by 2 systems of linear equations by graphing.\n28. Graph a parabola by finding the vertex, intercepts, and additional points.\n\nProblem Solving / Estimation:\n\nAfter taking MAT 096 the student should be able to:\n29. Solve applications problems involving 2 by 2 systems of linear equations.\n30. Solve application problems with linear and quadratic equations.\n31. Solve applications problems involving unit conversions.\n32. Solve applications using right triangle trigonometry.\n33. Solve application problems with rational equations.\n\nCalculator:\n\nAfter taking MAT 096 the student should be able to:\n34. Use TI 30X II to evaluate higher order roots , find trigonometric and inverse trigonometric values and find equivalent angle\n35. Use TI 30X II to calculate using scientific notation.\n\nComputer:\n\nAfter taking MAT 096 the student should be able to:\n36. Complete at least 6 assignments in CourseCompass.\n38. Receive email from the instructor and access their class calendar through BCCToday or WebCT.\n39. Access their own grades for this class online.\n\nFinal Exam:\n\n40. A department final will be given for MAT 096. If a faculty member chooses no to give the department final, he/she must submit\nthe final for review to the course coordinator for approval.\n41. Graphing calculators are not permitted for any portion of the final exam. It must be explicitly stated on the final and announced\nprior to the exam and at the beginning of the semester.\n42. No notecards, formula sheets, or any other aids are permitted for at least 50% of the final exam. That is, at least 50% of the final\nmust be completed without any assistance.\n\n## CATALOG COURSE DESCRIPTION:\n\nRational exponents; polynomials; factoring; functions; rational expressions; linear, quadratic and rational\nequations; graphs of basic functions; linear systems; topics in geometry; right triangle trigonometry.\n\n5 Class Hours; Prerequisite: MAT 092 Foundations for College Mathematics II or equivalent.\n\nMAT 096\nElementary Algebra and Trigonometry\nCourse outline\n\nI. Algebra and Problem Solving\nSome Basics of Algebra\nOperations and Properties of Real Numbers\nSolving Equations\nIntroduction to Problem Solving\nFormulas, Models, and Geometry\nProperties of Exponents\nScientific Notation\n\nII. Graphs, Functions, and Linear Equations\nGraphs\nFunctions\nLinear Functions: Graphs and Models\nAnother Look of Linear Graphs\nOther Equations of lines\nThe Algebra of Functions\n\nIII. Systems of Equations and Problem Solving\nSystems of Equations in Two Variables\nSolving by Substitution or Elimination\nSolving Applications: Systems of Two Equations\n\nIV. Inequalities and Problem Solving\nAbsolute-value Equations and Inequalities\n\nV. Polynomials and Polynomial Functions\nIntroduction to Polynomials and Polynomial Functions\nMultiplication of Polynomials\nCommon Factors and Factoring by Grouping\nFactoring Trinomials\nFactoring Perfect-Square Trinomials and Differences of Squares\nFactoring Sums of Differences of Cubes\nFactoring: A General Strategy\nApplications of Polynomial Equations\n\nVI. Rational Expressions, Equations, and Functions\nRational Expressions and Functions: Multiplying and Dividing\nRational Expressions and Functions: Adding and Subtracting\nComplex Rational Expressions\nRational Equations\nSolving Applications Using Rational Equations\nDivision of Polynomials\nFormulas and Applications\n\nRational Numbers as Exponents\nMultiplying, Dividing, and Simplifying Radical Expressions\nMore with Multiplication and Division\n\nIX. The Trigonometric Functions\nAngles\nAngles Relationships and Similar Triangles\nDefinitions of the Trigonometric Functions\nUsing the Definitions of the Trigonometric Functions\n\nX. Acute Angles and Right Angles\nTrigonometric Functions of Acute Angles\nTrigonometric Functions of Non-Acute Angles\nFinding Trigonometric Function Values Using a Calculator\nSolving Right Triangles\n\nXI. Radian Measure and the Circular Functions" ]

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[ "# (24a)   The Rotating Earth", null, "Index", null, "23a. The Centrifugal Force", null, "23b. Loop-the-Loop", null, "24a.The Rotating Earth", null, "24b. Rotating Frames\n\nThe Sun", null, "S-1. Sunlight & Earth", null, "S-1A. Weather", null, "S-1B. Global Climate", null, "S-2.Solar Layers", null, "S-3.The Magnetic Sun", null, "S-3A. Interplanetary\nMagnetic Fields", null, "S-4. Colors of Sunlight\n\n### The reduction of effective gravity by the Earth's rotation\n\nOne important rotating frame is the surface of the Earth, rotating with a period of about 24 hours--more accurately, 23 hrs 56.07 min or 86164 seconds. If the equatorial radius of Earth is 6378 kilometers, the circumference comes to 40074 kilometers--slightly more than the 40,000 kilometers supposed to be the pole-to-pole circumference, implied by the definition of \"one meter.\" The difference comes because the Earth bulges out at its equator. The velocity of the equator then is\n\n40074 / 86164 = 0.4651 kilometers = 465.1 meter/sec\n\nThat velocity is directed eastwards and is significantly faster than the speed of sound, which is about 335 m/s. Space rockets from Cape Canaveral need to attain about 8 km/s (in the frame of reference of the non-rotating center of the Earth ) to achieve orbit, so to give them a favorable starting velocity, such rockets are generally launched eastwards.\n\nTo derive the centrifugal acceleration on the equator (i.e. the force in Newtons on one gram mass, rotating with the Earth), we calculate in meters and seconds\n\nv2 / r = (465.1)2 / 6378000 = 216318 / 6378000 = 0,03392 m/s2\n\nComparing this to the acceleration of gravity--say 9.81 m/s2--it is only 0.00346 or 0.346%. Effective gravity on the equator is reduced by the rotation, but only by about 1/3 of a percent\n\n### The bulge of the Earth's equator", null, "Assuming the Earth is exactly spherical, we expect gravity to always point towards the center of Earth. However, the centrifugal force is perpendicular to the axis of the Earth. Except on the equator, therefore, it is not exactly opposed to gravity, but adds a small horizontal vector component, pointing towards the equator (dashed arrow in the figure). As a result, not only is effective gravity weakened, but its direction is modified--instead of pointing to the center of the Earth, is slants (ever so slightly) towards the equator.\n\nDoes this mean that if you placed a perfect ball on a very smooth horizontal surface, gravity would make it roll equatorward...? Suppose it was so. That same force would also act on the water of the ocean and make it flow equatorwards, and even the solid Earth might deform!\n\nHow long would this go on? Well, until the equatorial pile-up of material forms a \"hill\" around the Earth, rising slightly towards the equator, where its top would be. No more flow towards the equator would occur once the slope of the ground, as modified by the hill, would be exactly perpendicular to the effective direction of (modified) gravity. With such a slope, a perfect ball placed on a perfectly horizontal surface would no longer try to roll anywhere, and forces on oceans and on land would no longer try to move matter horizontally.\n\nBy now you should realize what this gives us--an explanation to the equatorial bulge of the Earth! That bulge is nothing but the \"hill\" described above, and the shape of the surface is such that the ground is always perpendicular to the effective gravity\n\nJupiter rotates in just under 10 hours and has an equatorial radius about 11 times that of Earth, creating a much stronger centrifugal force. As a result, in spite of a surface gravity 2.64 times stronger than Earth's, Jupiter is much more oblate, by about 6.5%.", null, "The oblateness of Jupiter was noted through the telescope by Giovanni Cassini (1625-1712), director of the Paris observatory, However, he and his son also claimed that Earth was different, that it bulged at the poles (more like a lemon than like a grapefruit). They based their claim on measurements on the ground, which suggested lines of latitude became closer together as one approached the pole.\n\nTheir result was later disproved by more accurate observations, but even before that, Newton by the following simple argument showed this could not be so. Assume Earth is a sphere, and imagine two deep holes extending to its center (see drawing), meeting there and filled with water. If Earth did not rotate, one may have expected the height of each water column to be the same: at the center of the Earth, by symmetry, each column would push down with the same weight, creating an equilibrium.\n\nOn a rotating Earth, however, the centrifugal force acts to reduce the weight of the water in the equatorial hole, and the water would rise there to greater height. Newton then argued that water anywhere would rise to the same level\n\n ...if our earth were not a little higher around the equator than at the poles, the seas would subside at the poles and, by ascending in the region of the equator, would flood everything there.\n\nFor more about the bulge of the Earth (including a more advanced mathematical treatment) see web site \"The Bulging Earth\".\n\n### Jet planes above the equator\n\n(Thanks to John Lowry for this neat example!)\n\nThe centrifugal force is a useful concept when figuring out equilibria in a rotating frame of reference, cases where forces balance exactly and no motion occurs.\n\nTo calculate motion in a rotating frame is much more complicated, and in general is beyond the level of this presentation. All that will be done in this section and the one that follows is point out some complications.     Let us assume the reduction of effective gravity is the same at jet-plane altitude--a small difference exists, but does not affect the argument below. Two jet airplanes fly along the equator at equal speeds--one eastwards, the other westwards. Since they fly in circles around the center of Earth, the effective gravity aboard each of them is further changed. Do both sense the same change?\n\nYou might think so, but it isn't true.\n\nIt would be the same if the Earth did not rotate (neglect here the attraction of the Sun and the rotation of the Earth around it, two effects which about balance). Without Earth rotation, in the frame of the universe, the eastward and westward motions are completely symmetric.\n\nActually, however, the speeds of the jets are measured relative to that atmosphere, which itself also rotates around the center of the Earth. Viewed from the frame of the universe, the speed of the Eastward jet is added to the rotation of the atmosphere, the one of the westward jet is subtracted.\n\nSuppose these are two supersonic Concorde airliners, flying at 465.1 meter/sec. Rather than study the motion in a rotating frame, which may or may not give correct answers, we can always go back to the non-rotating frame of the universe. In that frame, the rotational velocity of the westward jet is zero, and it senses no effective reduction of gravity. In the same frame, the eastward jet has twice the velocity, and since the centrifugal force is proportional to the velocity squared (v2), it senses four times the centrifugal force. Effective gravity aboard that jet is reduced by four times the amount felt on the ground, or 1.384%.\n\n### What it means\n\nThe above example suggests that we better watch out before applying Newton's laws in a rotating frame.\n\nNewton's first law certainly does not hold--a body not acted on by any force will not stay at rest but will fly off at a tangent. One can salvage this part of the law by adding a centrifugal force in any calculation of balanced forces. But if such forces are not balanced, if they cause motion in the rotating frame, then a modified form of Newton's laws must be used. They do exist, but are beyond the level of this course. The above example suggests some of the problems of rotating frames, and more examples are in the next section, which tells about the Coriolis effect.\n\nYou and I, and anyone else on Earth, live in a rotating frame of reference. Should we therefore refrain from using Newton's equations in everyday life? It turns out that for motions on a scale much smaller than the size of Earth, at moderate velocities, the effects are negligible.\n\nOne example concerns the famous question of whether draining water in kitchen sinks on opposite sides of the equator swirls in opposite directions, as is sometimes claimed. It turns out (see next section) that the effect in principle exists, but is much too small to affect observations. In the world wars, gunners firing big cannons corrected their aim (slightly) to account for the Earth's rotation, but you and I can usually ignore it, except for the modification of gravity by the centrifugal force.\n\nQuestions from Users:   What would happen if Earth rotated faster?\nAnd in the opposite direction:  What if the rotation of Earth could be stopped?.\nAlso: ***  Acceleration due to gravity.\n***  Does the Earth's core share the Earth's rotation?.\n***  If the Earth's Rotation would Stop...\n***  If the Earth's Rotation would Change...\n***  What makes the Earth rotate?\n\nAuthor and Curator:   Dr. David P. Stern\nMail to Dr.Stern:   stargaze(\"at\" symbol)phy6.org .\n\nLast updated: 9-22-2004\nReformatted 25 March 2006" ]

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[ " What is meant by ratio in mixture of 1:2:4 in concrete\n\n# What is meant by ratio in mixture of 1:2:4 in concrete", null, "### What is meant by ratio in mixture of 1:2:4 in concrete ...\n\nConcrete is a mixture of cement and sand and aggregate. The ratio is the volumes of each component. 1:2:4 is therefore the concrete resulting when 1 metre cubed of cement is mixed with 2 metres cubed of sand is mixed with 4 metres cubed of aggrega...\n\nget price", null, "### 1:2:4 CONCRETE RATIO MEANING - YouTube\n\nNov 11, 2017  Please watch: \"Concrete accelerators\" https://youtube/watch?v=jsxlrrw-KNw --~--In this video we will learn what is the concrete ratio!#CONSTRUCTION#B...\n\nget price", null, "### What's a 1:2:4 Mix? Concrete Construction Magazine\n\nMar 01, 1997  I have a residential-concrete specification that calls for a 1:2:4 concrete mix. What does that mean? A.: This is a little-used method for showing the amount of cement, sand, and coarse aggregate in the concrete. The specification should also say if those proportions are by weight or volume. If the proportions are based on weight, use 4 pounds ...\n\nget price", null, "### What is the W/C ratio for 1:2:4 concrete? - Quora\n\nThe 1:2:4 is nominal mix proportion of M20 grade concrete. The W/C of this M20 concrete is differ from its applications, if the concrete is Plain Cement Concrete (PCC) then the water cement ratio should be .60 and if it is RCC then the W/C ratio s...\n\nget price", null, "### Concrete Mix Ratio: What Is It? What Is 1-2-3? ...And More ...\n\nMay 17, 2020  A concrete mix ratio is usually expressed by a set of numbers separated by colons, as is the case with a 1:2:3 ratio. This tells the mixer that they need to add 1 part cement powder, 2 parts sand, and 3 parts aggregate in order to create the desired concrete consistency.\n\nget price", null, "### Concrete Mix Ratio, Types, Proportioning of Concrete Mix ...\n\nM 15 (1:2:4), M 20 (1:1.5:3) are the nominal mix ratio of concrete. Also, Read - Difference Between OPC and PPC Cement Standard Mix. The nominal concrete mix of a fixed proportion of cement, sand and aggregate provide considerable variation in strength resulting in the poor or over-rich mix.\n\nget price", null, "### Types of Concrete Mix Ratio Design and their Strengths ...\n\nNov 15, 2011  The mixes of grades M10, M15, M20 and M25 correspond approximately to the mix proportions (1:3:6), (1:2:4), (1:1.5:3) and (1:1:2) respectively. Designed Mix Ratio of Concrete In these mixes the performance of the concrete is specified by the designer but the mix proportions are determined by the producer of concrete, except that the minimum ...\n\nget price", null, "### Concrete Mix Ratios - Cement, Sand, Aggregate and Water\n\nDec 03, 2014  Concrete mix ratio: The strength of concrete mixture depends on the ratio in which these four ingredients are mixed. Concrete mix ratio of 1:3:3 - On mixing 1 part cement, 3 parts sand with 3 parts aggregate produces concrete with a compressive strength of 3000 psi.\n\nget price", null, "### Concrete Mix Ratio And Their Tpyes - 2020 Guide\n\nIn this article we are discussing about concrete mix ratio and ideal mix concrete. Making concrete, it is important to use the correct concrete mixing ratios to produce a tough, long life, durable concrete mix. To make concrete, four basic materials you need: Cement, sand, aggregate, water, and add-mixture.\n\nget price", null, "### Concrete mix ratio for various grades of concrete ...\n\nDec 08, 2017  This paste is also called as concrete. The strength of this concrete mix is determined by the proportion on which these cement, sand, stones or aggregates are mixed. There are various grades of concrete available in the market based on these ratios. Some of them are: M10, M20, M30, M35, etc. So, what really does M10 or M20 mean or represent.\n\nget price", null, "### WHAT IS THE MEANING OF RCC RATIO 1:2:4 DOES IT MEAN\n\nThe ratio of RCC 1:2:4 means one part of cement by volume, two parts of fine aggregate/sand by volume and four parts of coarse aggregate/bajri by volume. To achieve this ratio accordingly in field we make measuring box having size 1 feet x 1 feet x 1¼ feet that is the volume of box is 1¼ cft and is equal to the volume of one bag of cement.\n\nget price", null, "### How to Calculate Quantities of Cement, Sand and Aggregate ...\n\nOne bag of cement and other ingredients can produce = 400/2400 = 0.167 Cum of concrete (1:2:4) 01 bag cement yield = 0.167 cum concrete with a proportion of 1:2:4. 01 cum of concrete will require. Cement required = 1/0.167 = 5.98 Bags ~ 6 Bags. Sand required = 115/0.167 = 688 Kgs or 14.98 cft. Aggregate required = 209/0.167 = 1251 kgs or 29.96 cft\n\nget price", null, "### Types of Concrete Mix Ratio Design and their Strengths ...\n\nThe mixes of grades M10, M15, M20 and M25 correspond approximately to the mix proportions (1:3:6), (1:2:4), (1:1.5:3) and (1:1:2) respectively. Designed Mix Ratio of Concrete In these mixes the performance of the concrete is specified by the designer but the mix proportions are determined by the producer of concrete, except that the minimum ...\n\nget price", null, "### For 1:2:4 ratio the weight of cement is 50 kg then how ...\n\nIt is the ratio of cement:fine aggregate (i.e.sand) :coarse aggregate. So , for the ratio 1:2:4. If you take 50 kg of cement, Then, 2*50 = 100kg of sand. And 4*50 = 200kg of coarse aggregate.\n\nget price", null, "### What Is Grade of Concrete Concrete Mix Ratio Type of ...\n\nFor the M15 grade of concrete the mix ratio (cement: sand: aggregate) is 1 : 2: 4 and the compressive strength 15 MPa ( N/mm2 ) or 2175 psi. For the M20 grade of concrete the mix ratio (cement: sand: aggregate) is 1: 1.5 : 3 and the compressive strength 20 MPa ( N/mm2 ) or 2900 psi. The Standard Grade of Concrete\n\nget price", null, "### Concrete Mix Ratio And Their Tpyes - 2020 Guide\n\nIn this article we are discussing about concrete mix ratio and ideal mix concrete. Making concrete, it is important to use the correct concrete mixing ratios to produce a tough, long life, durable concrete mix. To make concrete, four basic materials you need: Cement, sand, aggregate, water, and add-mixture.\n\nget price", null, "### What Are Concrete Mixing Ratios? - Reference\n\nMar 25, 2020  Follow Us: Marcin Chady/CC-BY 2.0. Concrete mixing ratios are the formula for calculating the correct amount of each ingredient used, including water, cement, sand and aggregate, to produce concrete with the properties desired. A basic concrete ratio is one part cement, two parts sand, and three parts gravel, slowly mixing in water until workable.\n\nget price", null, "### Concrete Mix Ratios (Cement, Sand, Gravel)\n\nJan 26, 2021  Depending on the application concrete mix design can be complex. The below table gives a basic indication of the mix ratios used for different purposes but should be used as a guide only. Some additional things to consider when finding a suitable mix\n\nget price", null, "### Actual Concrete Mix Ratios For 3000, 3500, 4000, and 4500 ...\n\nThese are the actual concrete mix ratios for 3000, 3500, 4000, and 4500 psi concrete that I use to pour concrete floors, patios, pool decks and more. I'll show you the actual concrete batch plant ticket with the cement, sand, and aggregate break downs for the yards we used.\n\nget price", null, "### What is the mix ratio for blinding? - FindAnyAnswer\n\nApr 21, 2020  4.7/5 (2,743 Views . 20 Votes) Concrete blinding is just 2″ of regular concrete mix. It's made up of your normal ratio of cement mix, aggregates, and water. People try to use finer aggregates for this purpose because of the thickness required. It's usually done on site through hand mixtures.\n\nget price", null, "### Conventional grade C20, C25 and C30 concrete mix ratio ...\n\nThe concrete mix ratio refers to the proportion of the components in the concrete. The concrete mix ratio is usually represented by the quality of various materials per cubic meter of concrete, or amount proportion of various materials. The basic requirements for the design of concrete mix ratio are: 1, meet the strength grade of concrete design.\n\nget price", null, "### Concrete Mixing Ratios - How To Make Concrete (Cement ...\n\n3000 psi concrete mix ratio. To produce a 3000 psi cubic yard of concrete (27 cubic feet) the concrete mixture ratio is: 517 pounds of cement or (234kg) 1560 pounds of sand or (707kg) 1600 pounds of stone or (725kg) 32 - 34 gallons of water or (132L) This mixing ratio will give you a concrete mix that is strong, durable, and good for most ...\n\nget price", null, "### Concrete Calculator - Estimate Cement, Sand, Gravel ...\n\nOn this page, you can calculate material consumption viz., cement, sand, stone gravel for the following concrete mix ratios - 1:1.5:3, 1:2:4, 1:3:6, 1:4:8, 1:5:10. Once, the quantities are determined, it is easy to estimate the cost of a concrete block, driveway, patio, yard or any other structure with the price prevailing in your area ...\n\nget price", null, "### What makes concrete stronger? Mix Ratio concrete-info\n\nJul 23, 2021  The Strong Concrete Mix Ratio. The strength of the concrete depends on 3 factors:-. – The strength of aggregate. – The size of aggregate. – The water-cement ratio. The growth of strength is rapid enough to allow early stripping of moulds for rapid, economic construction on account of lesser formwork requirements.\n\nget price", null, "### Determination of appropriate mix ratios for concrete ...\n\n1:2:4 and 1:1.5:3 mix ratios respectively. To produce concrete with strength class C20/25 which is the minimum concrete strength class recommended for the construction of the load-bearing building structural members using the 1:2:4 mix ratio, Portland-limestone cement grade\n\nget price", null, "### Concrete Mix Ratio Design Types and their Strengths ...\n\nMay 10, 2019  Nominal mix ratios for these concrete mix are 1:2:4 for M15, 1:1.5:3 for M20 etc. Standard Mixes or Ratio The nominal mixes as mentioned above for cement aggregate ratio would result in wide variation of strength and also could lead to over or under-rich mixes.\n\nget price", null, "### What Is Grade of Concrete Concrete Mix Ratio Type of ...\n\nFor the M15 grade of concrete the mix ratio (cement: sand: aggregate) is 1 : 2: 4 and the compressive strength 15 MPa ( N/mm2 ) or 2175 psi. For the M20 grade of concrete the mix ratio (cement: sand: aggregate) is 1: 1.5 : 3 and the compressive strength 20 MPa ( N/mm2 ) or 2900 psi. The Standard Grade of Concrete\n\nget price", null, "### Concrete Mixing Ratios - How To Make Concrete (Cement ...\n\n3000 psi concrete mix ratio. To produce a 3000 psi cubic yard of concrete (27 cubic feet) the concrete mixture ratio is: 517 pounds of cement or (234kg) 1560 pounds of sand or (707kg) 1600 pounds of stone or (725kg) 32 - 34 gallons of water or (132L) This mixing ratio will give you a concrete mix that is strong, durable, and good for most ...\n\nget price", null, "### Concrete mix ratio for various grades of concrete ...\n\nThis paste is also called as concrete. The strength of this concrete mix is determined by the proportion on which these cement, sand, stones or aggregates are mixed. There are various grades of concrete available in the market based on these ratios. Some of them are: M10, M20, M30, M35, etc. So, what really does M10 or M20 mean or represent.\n\nget price", null, "### What Are Concrete Mixing Ratios? - Reference\n\nMar 25, 2020  Follow Us: Marcin Chady/CC-BY 2.0. Concrete mixing ratios are the formula for calculating the correct amount of each ingredient used, including water, cement, sand and aggregate, to produce concrete with the properties desired. A basic concrete ratio is one part cement, two parts sand, and three parts gravel, slowly mixing in water until workable.\n\nget price", null, "### Concrete Mix Ratios (Cement, Sand, Gravel)\n\nJan 26, 2021  Depending on the application concrete mix design can be complex. The below table gives a basic indication of the mix ratios used for different purposes but should be used as a guide only. Some additional things to consider when finding a suitable mix\n\nget price", null, "### Correct Ratios for Concrete Mixes\n\nMar 13, 2013  Large Batches of Concrete Mixes. 15 Mpa This is a low-strength concrete mix and is suitable for house foundations that are not reinforced, and for boundary walls and freestanding retaining walls.. To make 1 cubic metre of 15 Mpa concrete you will need to mix 5 1/2 bags of cement with 0,75 cubic metres of sand and 0,75 cubic metres of stone.\n\nget price", null, "### Grades of Concrete with Proportion (Mix Ratio) - Civilology\n\nConcrete grades are denoted by M10, M20, M30 according to their compressive strength. The “M” denotes Mix design of concrete followed by the compressive strength number in N/mm2. “Mix” is the respective ingredient proportions which are Cement: Sand: Aggregate Or Cement: Fine Aggregate: Coarse Aggregate. we already discussed the mix ...\n\nget price", null, "### Concrete Calculator - Estimate Cement, Sand, Gravel ...\n\nOn this page, you can calculate material consumption viz., cement, sand, stone gravel for the following concrete mix ratios - 1:1.5:3, 1:2:4, 1:3:6, 1:4:8, 1:5:10. Once, the quantities are determined, it is easy to estimate the cost of a concrete block, driveway, patio, yard or any other structure with the price prevailing in your area ...\n\nget price", null, "### Conventional grade C20, C25 and C30 concrete mix ratio ...\n\nThe concrete mix ratio refers to the proportion of the components in the concrete. The concrete mix ratio is usually represented by the quality of various materials per cubic meter of concrete, or amount proportion of various materials. The basic requirements for the design of concrete mix ratio are: 1, meet the strength grade of concrete design.\n\nget price", null, "### Methods of Proportioning Cement, Sand and Aggregates in ...\n\nSome practical values of water cement ratio for structure reinforced concrete 0.45 for 1 : 1 : 2 concrete; 0.5 for 1 : 1.5 : 3 concrete; 0.5 to 0.6 for 1 : 2 : 4 concrete. Concrete vibrated by efficient mechanical vibrators require less water cement ratio, and hence have more strength.\n\nget price", null, "### Preparing the right mix for solid concrete Farmstyle ...\n\nSelect an appropriate mix ratio from Table 2 and add the contents to the mixer in the following order: Half to two-thirds of water. Coarse aggregate. Sand. Cement. Add the remainder of the water until the desired consistency is reached. Mix the cement for at least two minutes after the last of\n\nget price", null, "### Making Concrete - Clean Water: A Layman's Guide to Water ...\n\nConcrete is made from cement, sand, gravel and water. In making concrete strong, these ingredients should usually be mixed in a ratio of 1:2:3:0.5 to achieve maximum strength. That is 1 part cement, 2 parts sand, 3 parts gravel, and 0.5 part water. (See figure 6). (A ''part'' means a unit of measure. Technically, it should be based on weight ...\n\nget price" ]

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[ "Solutions by everydaycalculation.com\n\n4/45 + 3/60 is 5/36.\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 45 and 60 is 180\n2. For the 1st fraction, since 45 × 4 = 180,\n4/45 = 4 × 4/45 × 4 = 16/180\n3. Likewise, for the 2nd fraction, since 60 × 3 = 180,\n3/60 = 3 × 3/60 × 3 = 9/180\n16/180 + 9/180 = 16 + 9/180 = 25/180\n5. 25/180 simplified gives 5/36\n6. So, 4/45 + 3/60 = 5/36\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:" ]

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[ "JAMPJournal of Applied Mathematics and Physics2327-4352Scientific Research Publishing10.4236/jamp.2015.37088JAMP-57573ArticlesPhysics&Mathematics On No-Node Solutions of the Lazer-McKenna Suspension Bridge Models FangleiWang1KangbaoZhou1College of Science, Hohai University, Nanjing, China3006201503077377403 March 2015accepted 23 June 30 June 2015© Copyright 2014 by authors and Scientific Research Publishing Inc. 2014This work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/\n\nIn this paper, we are concerned with the existence and multiplicity of no-node solutions of the Lazer-McKenna suspension bridge models by using the fixed point theorem in a cone.\n\nDifferential Equations Periodic Solution Cone Fixed Point Theorem\n1. Introduction\n\nIn , the Lazer-McKenna suspension bridge models are proposed as following\n\nIf we look for no-node solutions of the form and impose a forcing term of the form, then via some computation, we can obtain the following system:\n\nIn this paper, by combining the analysis of the sign of Green's functions for the linear damped equation, together with a famous fixed point theorem, we will obtain some existence results for (1) if the nonlinearities satisfy the following semipositone condition\n\n(H) The function is bounded below, and maybe change sign, namely, there exists a sufficiently large constant M > 0 such that\n\nSuch case is called as semipositone problems, see . And one of the common techniques is the Krasnoselskii fixed point theorem on compression and expansion of cones.\n\nLemma 1.1 . Let be a Banach space，and be a cone in. Assume are open subsets of with, , Let be a completely continuous operator such that either\n\n(i); or\n\n(ii);\n\nThen, has a fixed point in\n\n2. Preliminaries\n\nIf the linear damped equation\n\nis nonresonant, namely, its unique T-periodic solution is the trivial one, then as a consequence of Fredholm’s alternative in , the nonhomogeneous equation admits a unique T-periodic solution\n\nwhich can be written as where G(t; s) is the Green’s function of (2). For convenience,\n\nwe will assume that the following standing hypothesis is satisfied throughout this paper:\n\n(H1) are T-periodic functions such that the Green’s function, associated with the linear damped equation\n\nis positive for all, and\n\n(H2) are negative T-periodic functions, and satisfy:\n\nLet E denote the Banach space with the norm for. Define K to a cone in E by where. Also, for r > 0 a positive number, let\n\nIf (H), (H1) and (H2) hold, let, (1) is transformed into\n\nwhere is chosen such that\n\nLet be a map, which defined by, where\n\nt is straightforward to verify that the solution of (1) is equivalent to the fixed point Equation\n\nLemma 2.1 Assume that (H), (H1) and (H2) hold. Then is compact and continuous.\n\nFor convenience, define, for any.\n\nLemma 2.2 Assume that (H), (H1) and (H2) hold. If, then, for i = 1, 2, the functions are continuous on, for, and\n\nLemma 2.3 Assume that (H), (H1) and (H2) hold. If, then, for i = 1, 2, the functions are continuous on, for, and\n\n3. Main Results\n\nTheorem 3.1 Assume that (H), (H1) and (H2) hold.\n\n(I) Then there exists a such that (1) has a positive periodic solution for\n\n(II) If, then for an, (1) has a positive periodic solution;\n\n(III) If, then (1) has two positive periodic solutions for all sufficiently small.\n\nProof. (I) On one hand, take R > 0 such that\n\nSet Then, for each, we have\n\nThen from the above inequalities, it follows that there exists a such that\n\nFurthermore, for any, we obtain\n\nIn the similar way, there exists a, such that and we also have\n\nSo let us choose and we can obtain\n\nOn the other hand, from the condition for all, it follows that there is a sufficient small r > 0 such that for and where is chosen such that\n\nThen, for any, we obtain\n\nSo we have\n\nTherefore, from Lemma 1.1, it follows that the operator B has at least one fixed point in, for\n\n(II) Since, then from Lemma 2.1, it follows that Define a function as By Lemma 2.5 in , it is easy to see that Thus by the definition, there is an such that where satisfying\n\nThen, for each, we have\n\nIn the similar way, for any, we also have Furthermore, from The above inequalities, we get\n\nTherefore, from Lemma 1.1, it follows that B has one fixed point in for any\n\n(III) Since, then from Lemma 2.2, it follows that By the definition, there exists such that where is chosen such that\n\nChoosing and for any, we have and\n\nThus from the above inequalities, we can get\n\nTherefore, from Lemma 1.1, it follows that the operator B has at least two fixed points in and in. Namely, system (1) has two solutions for sufficiently small\n\nCite this paper\n\nFanglei Wang,Kangbao Zhou, (2015) On No-Node Solutions of the Lazer-McKenna Suspension Bridge Models. Journal of Applied Mathematics and Physics,03,737-740. doi: 10.4236/jamp.2015.37088\n\nReferencesLazer, A.C. and McKenna, P.J. (1990) Large-Amplitude Periodic Oscillations in Suspension Bridges: Some New Connections with Nonlinear Analysis. Siam Review, 32, 537-578. http://dx.doi.org/10.1137/1032120Wang, H. (2009) Periodic Solutions to Non-Autonomous Second-Order Systems. Nonlinear Analysis: Theory, Methods & Applications, 71, 1271-1275. http://dx.doi.org/10.1016/j.na.2008.11.079Dajun, G. and Lakshmikantham, V. (1988) Nonlinear Problems in Abstract Cones.Chu, J., Fan, N. and Torres, P.J. (2012) Periodic Solutions for Second Order Singular Damped Differential Equations. Journal of Mathematical Analysis and Applications, 388, 665-675. http://dx.doi.org/10.1016/j.jmaa.2011.09.061" ]

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[ "# What is the compound interest on $98277 at 4% over 27 years? If you want to invest$98,277 over 27 years, and you expect it will earn 4.00% in annual interest, your investment will have grown to become $283,368.81. If you're on this page, you probably already know what compound interest is and how a sum of money can grow at a faster rate each year, as the interest is added to the original principal amount and recalculated for each period. The actual rate that$98,277 compounds at is dependent on the frequency of the compounding periods. In this article, to keep things simple, we are using an annual compounding period of 27 years, but it could be monthly, weekly, daily, or even continuously compounding.\n\nThe formula for calculating compound interest is:\n\n$$A = P(1 + \\dfrac{r}{n})^{nt}$$\n\n• A is the amount of money after the compounding periods\n• P is the principal amount\n• r is the annual interest rate\n• n is the number of compounding periods per year\n• t is the number of years\n\nWe can now input the variables for the formula to confirm that it does work as expected and calculates the correct amount of compound interest.\n\nFor this formula, we need to convert the rate, 4.00% into a decimal, which would be 0.04.\n\n$$A = 98277(1 + \\dfrac{ 0.04 }{1})^{ 27}$$\n\nAs you can see, we are ignoring the n when calculating this to the power of 27 because our example is for annual compounding, or one period per year, so 27 × 1 = 27.\n\n## How the compound interest on $98,277 grows over time The interest from previous periods is added to the principal amount, and this grows the sum a rate that always accelerating. The table below shows how the amount increases over the 27 years it is compounding: Start Balance Interest End Balance 1$98,277.00 $3,931.08$102,208.08\n2 $102,208.08$4,088.32 $106,296.40 3$106,296.40 $4,251.86$110,548.26\n4 $110,548.26$4,421.93 $114,970.19 5$114,970.19 $4,598.81$119,569.00\n6 $119,569.00$4,782.76 $124,351.76 7$124,351.76 $4,974.07$129,325.83\n8 $129,325.83$5,173.03 $134,498.86 9$134,498.86 $5,379.95$139,878.81\n10 $139,878.81$5,595.15 $145,473.97 11$145,473.97 $5,818.96$151,292.93\n12 $151,292.93$6,051.72 $157,344.64 13$157,344.64 $6,293.79$163,638.43\n14 $163,638.43$6,545.54 $170,183.97 15$170,183.97 $6,807.36$176,991.32\n16 $176,991.32$7,079.65 $184,070.98 17$184,070.98 $7,362.84$191,433.82\n18 $191,433.82$7,657.35 $199,091.17 19$199,091.17 $7,963.65$207,054.82\n20 $207,054.82$8,282.19 $215,337.01 21$215,337.01 $8,613.48$223,950.49\n22 $223,950.49$8,958.02 $232,908.51 23$232,908.51 $9,316.34$242,224.85\n24 $242,224.85$9,688.99 $251,913.84 25$251,913.84 $10,076.55$261,990.40\n26 $261,990.40$10,479.62 $272,470.01 27$272,470.01 $10,898.80$283,368.81\n\nWe can also display this data on a chart to show you how the compounding increases with each compounding period.\n\nAs you can see if you view the compounding chart for $98,277 at 4.00% over a long enough period of time, the rate at which it grows increases over time as the interest is added to the balance and new interest calculated from that figure. ## How long would it take to double$98,277 at 4% interest?\n\nAnother commonly asked question about compounding interest would be to calculate how long it would take to double your investment of $98,277 assuming an interest rate of 4.00%. We can calculate this very approximately using the Rule of 72. The formula for this is very simple: $$Years = \\dfrac{72}{Interest\\: Rate}$$ By dividing 72 by the interest rate given, we can calculate the rough number of years it would take to double the money. Let's add our rate to the formula and calculate this: $$Years = \\dfrac{72}{ 4 } = 18$$ Using this, we know that any amount we invest at 4.00% would double itself in approximately 18 years. So$98,277 would be worth $196,554 in ~18 years. We can also calculate the exact length of time it will take to double an amount at 4.00% using a slightly more complex formula: $$Years = \\dfrac{log(2)}{log(1 + 0.04)} = 17.67\\; years$$ Here, we use the decimal format of the interest rate, and use the logarithm math function to calculate the exact value. As you can see, the exact calculation is very close to the Rule of 72 calculation, which is much easier to remember. Hopefully, this article has helped you to understand the compound interest you might achieve from investing$98,277 at 4.00% over a 27 year investment period." ]

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[ "", null, "# 2.(i) How many atoms are present in a unit cell of a BCC lattice? (...\n\n##", null, "Question\n\nShow transcribed text\n\n##", null, "Transcribed Text\n\n2.(i) How many atoms are present in a unit cell of a BCC lattice? (ii) Find the volume density ofatoms of the BCC lattice if lattice constant is 5 Å. [2+8=10] 3. (i) Prove the lattice constant of a FCC lattice is 𝑎 =2√2𝑟 where r is the atomic radius. (ii) If the atomic radius for a FCC lattice is 0.175nm, find the volume of the unit cell. [6+4=10] 4. Find out the atomic packing factor of a BCC crystal lattice whose lattice constant is ‘a’. 5. Find out the Miller Indices of the following. [5x4=20] (i) (ii) (iii) (iv) (v) 6 6.Find out the (i) momentum and (ii) de-Broglie wavelength of an electron travelling with avelocity of 107 cm/s. Consider mass of electron as 9.1×10-31 Kg. [5+5=10] 7. Determine the probability that an energy level 3KT above Fermi level is occupied by an electron at T = 300 K. [Hint: Difference between E and EF is 3KT] 8. Calculate the (i) probability that a state in the conduction band is occupied by an electron f(EC) and calculate the (ii) thermal equilibrium electron concentration (n0) in silicon at T= 100 K. Assume the Fermi energy is 0.25 eV below the conduction band. The value of NC for silicon at T = 100 K is 2.8 x I019 cm-3. [5+5 = 10] 9. Calculate the intrinsic carrier concentration in gallium arsenide at T = 300 K. The values of NC and NV at 300 K for gallium arsenide are 4.7 x 1017 cm-3 and 7.0 x 1018 cm-3, respectively. Assume the bandgap energy of gallium arsenide is 1.42 eV.\n\n##", null, "Solution Preview\n\nThis material may consist of step-by-step explanations on how to solve a problem or examples of proper writing, including the use of citations, references, bibliographies, and formatting. This material is made available for the sole purpose of studying and learning - misuse is strictly forbidden.\n\nBy purchasing this solution you'll be able to access the following files:\nSolution.pdf.\n\n\\$10.00\nfor this solution\n\nor FREE if you\nregister a new account!\n\nPayPal, G Pay, ApplePay, Amazon Pay, and all major credit cards accepted.\n\n### Find A Tutor\n\nView available Semiconductor Physics Tutors\n\nGet College Homework Help.\n\nAre you sure you don't want to upload any files?\n\nFast tutor response requires as much info as possible." ]

[ null, "https://www.facebook.com/tr", null, "https://www.24houranswers.com/images/icons/qa-icon.png", null, "https://www.24houranswers.com/images/icons/qa-icon.png", null, "https://www.24houranswers.com/images/icons/form-icon.png", null ]

[ "000125738 001__ 125738\n000125738 005__ 20190812205231.0\n000125738 02470 $$2ISI$$a000264729000338\n000125738 037__ $$aCONF 000125738 245__$$a3D Face Recognition using Sparse Spherical Representations\n000125738 260__ $$c2008 000125738 269__$$a2008\n000125738 336__ $$aConference Papers 000125738 520__$$aThis paper addresses the problem of 3D face recognition using spherical sparse representations. We first propose a fully automated registration process that permits to align the 3D face point clouds. These point clouds are then represented as signals on the 2D sphere, in order to take benefit of the geometry information. Simultaneous sparse approximations implement a dimensionality reduction process by subspace projection. Each face is typically represented by a few spherical basis functions that are able to capture the salient facial characteristics. The dimensionality reduction step preserves the discriminant facial information and eventually permits an effective matching in the reduced space, where it can further be combined with LDA for improved recognition performance. We evaluate the 3D face recognition algorithm on the FRGC v.1.0 data set, where it outperforms classical state-of-the-art solutions based on PCA or LDA on depth face images.\n000125738 6531_ $$aLTS4 000125738 6531_$$a3D face recognition\n000125738 6531_ $$asparse spherical representations 000125738 700__$$aSala Llonch, Roser\n000125738 700__ $$0240462$$g170201$$aKokiopoulou, Effrosyni 000125738 700__$$0240452$$g163024$$aTosic, Ivana\n000125738 700__ $$0241061$$g101475$$aFrossard, Pascal 000125738 7112_$$dDecember 2008$$cTampa, Florida, USA$$aInt. Conf. on Pattern Recognition\n000125738 773__ $$tInt. Conf. on Pattern Recognition 000125738 8564_$$zURL\n000125738 8564_ $$zn/a$$uhttps://infoscience.epfl.ch/record/125738/files/3Dfaces_ICPR_final.pdf$$s211076 000125738 909C0$$xU10851$$pLTS4$$0252393\n000125738 909CO $$ooai:infoscience.tind.io:125738$$qGLOBAL_SET$$pconf$$pSTI\n000125738 937__ $$aEPFL-CONF-125738 000125738 973__$$rREVIEWED$$sPUBLISHED$$aEPFL\n000125738 980__ aCONF" ]

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[ "$y=y(x)$为轨道方程，其中$I(x,y)$是正方形车轮与“道路”的一个切点，根据道路与车轮的关系，线段AI的长度要等于弧KI的长，记该长为s，用微积分的符号表示为\n\n$$s=\\int_{x_0}^x \\sqrt{dx^2+dy^2}=\\int_{x_0}^x \\sqrt{1+\\dot{y}^2}dx$$\n（$x_0$是左端的一个零点）\n\n\\begin{aligned}AI=s,MI=1-s,RM=1 \\\\ RM \\cos\\theta+MI \\sin\\theta+y=OP\\end{aligned}\n\n$$s=1+\\dot{y}^{-1}+y\\sqrt{1+\\dot{y}^{-2}}-\\sqrt{2}\\cdot \\sqrt{1+\\dot{y}^{-2}}$$\n\n$$\\int_{x_0}^x \\sqrt{1+\\dot{y}^2}dx=1+\\dot{y}^{-1}+y\\sqrt{1+\\dot{y}^{-2}}-\\sqrt{2}\\cdot \\sqrt{1+\\dot{y}^{-2}}$$\n\n$$\\sqrt{1+\\dot{y}^2}$$\n\n$$-\\dot{y}^{-2}\\ddot{y}+(y-\\sqrt{2})\\cdot \\frac{-\\dot{y}^{-2}\\ddot{y}}{\\sqrt{1+\\dot{y}^2}}+\\sqrt{1+\\dot{y}^2}$$\n\n$$\\sqrt{2}-y=\\sqrt{1+\\dot{y}^2}$$\n\n\\begin{aligned}-x=\\int \\frac{1}{\\sqrt{(\\sqrt{2}-y)^2-1}} d(\\sqrt{2}-y) \\\\ =arccosh(\\sqrt{2}-y)+C\\end{aligned}\n\n$$y=-\\cos h(-C-x)+\\sqrt{2}=-\\cos h(C+x)+\\sqrt{2}$$\n（$cosh x$是双曲余弦函数，$cosh x=\\frac{e^x+e^{-x}}{2}$）\n\n$$y=-\\cos hx+\\sqrt{2}$$\n（取x轴上方的部分）\n\n$$y=-h \\cos h(\\frac{x}{h})+\\sqrt{a^2+h^2}$$\n\n$$y=-\\cos h x+\\sqrt{1+\\tan^2(\\frac{\\pi}{n})}$$" ]

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[ "# From Science to Engineering\n\nThe master builders, the designers, and the constructors of the Gothic cathedrals of the Middle Ages used intuition and experience to develop design rules based on simple force equilibrium and treated the material as rigid. This solution process provided the equivalent of what is now known as the lower-bound theorem of plastic limit analysis. This theorem was not proved until more than 500 years later. Modern lower-bound theorem shows that these design rules are safe. These simple design rules have existed from the earliest times for building Greek temples, Roman aqueducts and arch bridges, and domes and vaults. However, tests on real structures showed that the stresses calculated by designers with these rules could not actually be measured in practice.\n\nGalileo, in the seventeenth century, was the first to introduce recognizably modern science into the calculation of structures; he determined the breaking strength of beams but he was way ahead of his time in engineering application. In the eighteenth century, engineers moved away from his proposed “ultimate load” approach, and, until early in the nineteenth century, a formal philosophy of design had been established: a structure should remain elastic, with a safety factor on stress built into the analysis. It was an era of great advancement and a milestone in structural design but one that placed too much emphasis on the undue safety concern based on elastic response under working loads.\n\nGalileo Galilei (1564–1642) was the first to use mathematics in order to describe the law of nature, which is based on observation from his experiments. Isaac Newton (1642–1727) discovered the basic laws of physics in terms of equilibrium condition (or equation of equilibrium) and equation of motion. Robert Hooke (1635–1703) described, in mathematical form, the material response to stress, which he observed in tests. He stated the linear relationship between stress and strain (Hooke’s law or constitutive law) as a function of a material constant (elasticity modulus or Young’s modulus).\n\nMaterial continuity without discontinuities or cracks is a logical assumption in solid mechanics. This assumption leads to a mathematical description of geometric relations of a continuous medium known as continuum expressed in the form now known as compatibility conditions. For a continuum, the conditions of equilibrium (physics), constitutive (materials), and continuity (geometry) furnish the three sets of basic equations necessary for solutions in any solid mechanics problem in which structural engineering is one of its applications. In short, the mechanics analysis of a given structural problem or a proposed structural design must involve the mathematical formulation of the following three sets of equations and solutions:\n\nEquilibrium equations or motion reflecting laws of physics (e.g., Newton’s laws)\nConstitutive equations or stress–strain relations reflecting material behavior (experiments)\nCompatibility equations or kinematical relations reflecting the geometry or continuity of materials (logic)", null, "Interrelationship of the three sets of basic field equations\n\n[blogger]\n\nEngineeersdaily\n\nName\n\nEmail *\n\nMessage *" ]

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[ "scroll identifier for mobile\nmain-content\n\nÜber dieses Buch\n\nThis book is a comprehensive and accessible introduction to the cross-entropy (CE) method. The CE method started life around 1997 when the first author proposed an adaptive algorithm for rare-event simulation using a cross-entropy minimization technique. It was soon realized that the underlying ideas had a much wider range of application than just in rare-event simulation; they could be readily adapted to tackle quite general combinatorial and multi-extremal optimization problems, including many problems associated with the field of learning algorithms and neural computation. The book is based on an advanced undergraduate course on the CE method, given at the Israel Institute of Technology (Technion) for the last three years. It is aimed at a broad audience of engineers, computer scientists, mathematicians, statisticians and in general anyone, theorist or practitioner, who is interested in smart simulation, fast optimization, learning algorithms, image processing, etc. Our aim was to write a book on the CE method which was accessible to advanced undergraduate students and engineers who simply want to apply the CE method in their work, while at the same time accentu­ ating the unifying and novel mathematical ideas behind the CE method, so as to stimulate further research at a postgraduate level.\n\nInhaltsverzeichnis\n\n1. Preliminaries\n\nAbstract\nThe purpose of this preparatory chapter is to provide some initial background for this book, including a summary of the relevant definitions and concepts in probability, statistics, simulation and information theory. Readers familiar with these ideas may wish to skip through this chapter and proceed to the tutorial on the cross-entropy method in Chapter 2.\nReuven Y. Rubinstein, Dirk P. Kroese\n\n2. A Tutorial Introduction to the Cross-Entropy Method\n\nAbstract\nThe aim of this chapter is to provide a gentle and self-contained introduction to the cross-entropy (CE) method. We refer to Section 1.1 for additional background information on the CE method, including many references.\nReuven Y. Rubinstein, Dirk P. Kroese\n\n3. Efficient Simulation via Cross-Entropy\n\nAbstract\nThe performance of modern systems, such as coherent reliability systems, inventory systems, insurance risk, storage systems, computer networks and telecommunication networks is often characterized by probabilities of rare events and is frequently studied through simulation. Analytical solutions or accurate approximations for such rare-event probabilities are only available for a very restricted class of systems. Consequently, one often has to resort to simulation. However, estimation of rare-event probabilities with crude Monte Carlo techniques requires a prohibitively large numbers of trials, as illustrated in Example 1.3. Thus, new techniques are required for this type of problem. Two methods, called splitting/RESTART and importance sampling (IS) have been extensively investigated by the simulation community in the last decade.\nReuven Y. Rubinstein, Dirk P. Kroese\n\n4. Combinatorial Optimization via Cross-Entropy\n\nAbstract\nIn this chapter we show how the CE method can be easily transformed into an efficient and versatile randomized algorithm for solving optimization problems, in particular combinatorial optimization problems.\nReuven Y. Rubinstein, Dirk P. Kroese\n\n5. Continuous Optimization and Modifications\n\nAbstract\nIn this chapter we discuss how the CE method can be used for continuous multi-extremal optimization, and provide a number of modifications of the basic CE algorithm, applied to both combinatorial and continuous multiextremal optimization. Specific modifications include the use of alternative “reward/loss” functions (Section 5.2), and the formulation of a fully automated version of the CE algorithm (Section 5.3). This modified version allows automatic tuning of all the parameters of the CE algorithm. Numerical results for continuous multi-extremal optimization problems and the FACE modifications are given in Sections 5.4, and 5.5, respectively.\nReuven Y. Rubinstein, Dirk P. Kroese\n\n6. Noisy Optimization with CE\n\nAbstract\nIn this chapter we show how the CE method optimizes noisy objective functions, that is, objective functions corrupted with noise. Noisy optimization, also called stochastic optimization, can arise in various ways. For example, in the analysis of data networks , a well-known problem is to find a routing table that minimizes the congestion in the network. Typically this problem is solved under deterministic assumptions. In particular, the stream of data is represented by a constant “flow,” and the congestion at a link is measured in terms of a constant arrival rate to the link, which is assumed to depend on the routing table only. These assumptions are reasonable only when the arrival process changes very slowly relative to the average time required to empty the queues in the network. In a more realistic setting however, using random arrival and queueing processes, the optimization problem becomes noisy. Other examples of noisy optimization include stochastic scheduling and stochastic shortest/longest path problems. References on noisy optimization include [9, 26, 36, 37, 60, 78].\nReuven Y. Rubinstein, Dirk P. Kroese\n\n7. Applications of CE to COPs\n\nAbstract\nIn this chapter we show how CE can be employed to solve combinatorial optimization problems (COPs) from various fields. Our goal is to illustrate the flexibility and efficiency of the CE algorithm. We believe that the CE approach could open up a whole range of possible applications, for example in computational biology, graph theory, and scheduling Section 7.1 deals with an application of the CE method concerning the alignment of DNA sequences; this section is based partly on . In Sections 7.2–7.4, adapted from , the CE method is used to solve the single machine total weighted tardiness problem (SMTWTP), the single machine common due date problem (SMDDP), and the capacitated vehicle routing problem (CVRP). Finally, in Section 7.5 we consider various CE approaches for solving the maximum clique and related graph problems.\nReuven Y. Rubinstein, Dirk P. Kroese\n\n8. Applications of CE to Machine Learning\n\nAbstract\nIn this chapter we apply the CE method to several problems arising in machine learning, specifically with respect to optimization. In Section 8.1, adapted from , we apply CE to the well-known mastermind game. Section 8.2, based partly on , describes the application of the CE method to Markov decision processes. Finally, in Section 8.3 the CE method is applied to clustering problems. In addition to its simplicity, the advantage of using the CE method for machine learning is that it does not require direct estimation of the gradients, as many other algorithms do (for example, the stochastic approximation, steepest ascent, or conjugate gradient method). Moreover, as a global optimization procedure the CE method is quite robust with respect to starting conditions and sampling errors, in contrast to some other heuristics, such as simulated annealing or guided local search.\nReuven Y. Rubinstein, Dirk P. Kroese\n\nBackmatter\n\nWeitere Informationen" ]

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[ "The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.", null, "Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)\n A216321 phi(delta(n)), n >= 1, with phi = A000010 (Euler's totient) and delta = A055034 (degree of minimal polynomials with coefficients given in A187360). 2\n 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 4, 2, 2, 2, 2, 4, 4, 2, 6, 4, 2, 4, 10, 4, 4, 4, 6, 4, 6, 4, 8, 8, 4, 8, 4, 4, 6, 6, 4, 8, 8, 4, 12, 8, 4, 10, 22, 8, 12, 8, 8, 8, 12, 6, 8, 8, 6, 12, 28, 8, 8, 8, 6, 16, 8, 8, 20, 16, 10, 8, 24, 8, 12, 12, 8, 12, 8, 8, 24, 16, 18, 16, 40, 8, 16, 12 (list; graph; refs; listen; history; text; internal format)\n OFFSET 1,7 COMMENTS If n belongs to A206551 (cyclic multiplicative group Modd n) then there exist precisely a(n) primitive roots Modd n. For these n values the number of entries in row n of the table A216319 with value delta(n) (the row length) is a(n). Note that a(n) is also defined for the complementary n values from  A206552 (non-cyclic multiplicative group Modd n) for which no primitive root Modd n exists. See also A216322 for the number of primitive roots Modd n. LINKS FORMULA a(n) = phi(delta(n)), n >= 1, with phi = A000010 (Euler's totient) and delta = A055034 with delta(1) = 1 and delta(n) = phi(2*n)/2 if n >= 2. EXAMPLE a(8) = 2 because delta(8) = 4 and phi(4) = 2. There are 2 primitive roots Modd 8, namely 3 and 5 (see the two 4s in row n=8 of A216320). 8 = A206551(8). a(12) = 2 because delta(12) = 4 and phi(4) = 2. But there is no primitive root Modd 12, because 4 does not show up in row n=12 of A216320. 12 = A206552(1). PROG (PARI) a(n)=eulerphi(ceil(eulerphi(2*n)/2)) \\\\ Charles R Greathouse IV, Feb 21 2013 CROSSREFS Cf. A000010, A055034, A216319, A216320, A216322, A010554 (analog in modulo n case). Sequence in context: A122066 A053238 A227783 * A058263 A232398 A048669 Adjacent sequences:  A216318 A216319 A216320 * A216322 A216323 A216324 KEYWORD nonn AUTHOR Wolfdieter Lang, Sep 21 2012 STATUS approved\n\nLookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam\nContribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent\nThe OEIS Community | Maintained by The OEIS Foundation Inc.\n\nLast modified August 15 12:59 EDT 2020. Contains 336502 sequences. (Running on oeis4.)" ]

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[ "#### 200 square yards to hectares\n\nHow many hectares in 200 yd2?\n\n#### Calculate Area\n\n200 square yards to hectares. How many? What's the calculation? How much? What is 200 square yards in hectares? 200yd2 in hectares? How many hectares are there in 200 yd2? Calculate between square yards and hectares. How big is 200 square yards in hectares?" ]

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[ "# calculate $\\lim_{x \\rightarrow 0}\\left ( x^{-6}\\cdot (1-\\cos(x)^{\\sin(x)})^2 \\right )$\n\nas in topic, my task is to calculate $$\\lim_{x \\rightarrow 0}\\left ( x^{-6}\\cdot (1-\\cos x^{\\sin x})^2 \\right )$$ I do the following: (assuming that de'Hospital is pointless here, as it seems to be) I write Taylor series for $\\sin x, \\cos x$ $$\\sin x=x-\\frac{x^3}{3!}+\\frac{x^5}{5!}-\\frac{x^7}{7!}+o(x^7)$$ $$\\cos x=1-\\frac{x^2}{2!}+\\frac{x^4}{4!}-\\frac{x^6}{6!}+o(x^6)$$ I notice that denominator tells me to use sixth degree something and i get $$\\frac{\\cos x^{2 \\cdot \\sin x}- 2 \\cdot (\\cos x)^{\\sin x} + 1}{x^6}$$ and my idea here was to use Bernuolli to $\\cos x^{\\sin x}$ so that $(1+x)^n\\geq1+nx$ but it is going to be nasty and the expression $(...)\\geq(...)$ does not tell anything about that limit. Any hints? Thank you in advance.\n\n• Use repeated L'Hospital rule? – Inceptio Mar 18 '13 at 9:39\n• It will be easier if you first figure out the limit of the square root. Apply L'Hospital rule once, you get: $$\\lim_{x\\to 0}\\frac{1-\\cos(x)^{\\sin(x)}}{x^3} = \\lim_{x\\to 0} \\left( -\\cos(x)^{\\sin(x)}\\frac{\\cos(x)\\log(\\cos(x)) - \\frac{\\sin(x)^2}{\\cos(x)}}{3x^2}\\right)$$ Some of the factors like $\\cos(x)^{\\sin(x)}$ converges to 1 as $x \\to 0$, you don't need to worry around them. What's remain is manageable by Taylor expansion or L'Hospital rule. – achille hui Mar 18 '13 at 10:12\n\n## 2 Answers\n\n$$\\lim_{x \\rightarrow 0}\\left ( x^{-6}\\cdot (1-\\cos x^{\\sin x})^2 \\right )$$\n\n$$=\\lim_{x\\to 0}\\left(\\frac{\\sin x}x\\right)^6\\cdot \\lim_{x\\to 0}\\left(\\frac{1-\\cos x^{\\sin x}}{\\sin^3x}\\right)^2$$\n\nNow, Putting $y=\\sin x$ in the last limit, as $x\\to 0,y=\\sin x\\to 0$ and $\\cos x^{\\sin x}=(\\cos^2x)^{\\frac {\\sin x}2}=(1-y^2)^\\frac y2$\n\n$$\\lim_{x\\to 0}\\frac{1-\\cos x^{\\sin x}}{\\sin^3x}$$ $$=\\lim_{y\\to 0}\\frac{1-(1-y^2)^{\\frac y2}}{y^3}$$\n\n$$=\\lim_{y\\to 0}\\frac{1-\\left(1+(-y^2)\\cdot \\frac y2+(-y^2)^2\\cdot\\frac{\\frac y2\\left(\\frac y2-1\\right)}{2!}+\\cdots\\right)}{y^3}$$\n\n$$=\\lim_{y\\to 0}\\{\\frac12+O(y^2)\\}\\text { as }y\\ne0 \\text { as } y\\to 0$$\n\n$$=\\frac12$$\n\nfirst find the first terms of the series expansion of $1-\\cos(x)^{\\sin(x)}$ without using the ones of sin and cos, you will find that$$1-\\cos(x)^{\\sin(x)}=\\frac{x^3}{2}+\\text{higer trems}$$ which implies $$\\lim_{x\\to 0}\\frac{(1-\\cos(x)^{\\sin(x)})^2}{x^6}=\\lim_{x\\to 0}\\frac{\\frac{x^6}{4}+x^3(\\text{higer terms})+\\text{(higer terms)}^2}{x^6}=\\frac{1}{4}$$ since the degree of higer terms is larger than 3." ]

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[ "", null, "", null, "", null, "`# adl-mapstack.pkg`\n`#`\n`# Another implementation of pushdown-stack of key-val maps.`\n`# (These are used to track syntactic scopes, pushing a new`\n`# map when we enter a scope and popping it when we leave.)`\n`#`\n`# This gets used in:`\n`#     `src/lib/compiler/back/low/tools/arch/adl-symboltable.pkg\n\n`# Compiled by:`\n`#     `src/lib/compiler/back/low/tools/arch/make-sourcecode-for-backend-packages.lib\n\n`###                   \"It is better wither to be silent, or to`\n`###                    say things of more value than silence.`\n`###                    Sooner throw a pearl at hazard than`\n`###                    an idle or useless word; and do not say`\n`###                    a little in many words, but a great deal in a few.\"`\n`###`\n`###                                          -- Pythagoras `\n\n`api Adl_Mapstack {`\n`    #`\n`    Mapstack(X);`\n\n`    exception SYMBOLTABLE; `\n\n`    symboltable:  String ->     Mapstack(X);`\n`    envir:        String -> Ref(Mapstack(X));`\n`    #`\n`    get:              Mapstack(X) -> String -> X;`\n`    lookup:       Ref(Mapstack(X))    -> String -> X;`\n`    #`\n`    get' :        Mapstack(X) -> X -> String -> X;`\n`    put:          Mapstack(X) -> (String, X) -> Mapstack(X);`\n`    #`\n`    set:        Ref(Mapstack(X)) -> (String, X) -> Void;`\n`    #`\n`    apply:      ((String, X) -> Void)      -> Mapstack(X) -> Void;`\n`    map:        ((String, X) -> Y)         -> Mapstack(X) -> List(Y);`\n`    fold:       ((String, X, Y) -> Y) -> Y -> Mapstack(X) -> Y;`\n`    #`\n`    union:      (Mapstack(X), Mapstack(X)) -> Mapstack(X);`\n`    unions:     List( Mapstack(X) )        -> Mapstack(X);`\n`    #`\n`    empty:      Mapstack(X);`\n`    #`\n`    bind:         (String, X) -> Mapstack(X);`\n`    consolidate:  Mapstack(X) -> Mapstack(X);`\n`};`\n\n`stipulate`\n`    package h = hashtable;`\n`herein`\n\n`    package adl_mapstack`\n`    :       Adl_Mapstack`\n`    {`\n`        #`\n`        Mapstack(X)`\n`          #`\n`          = EMPTY `\n`          ``| TABLE    (h::Hashtable (String,X))`\n`          ``| OVERRIDE (Mapstack(X), Mapstack(X))`\n`          ``| NAMING   (String, X)`\n`          ;`\n\n`        exception SYMBOLTABLE; `\n\n`        fun symboltable name = EMPTY;`\n`        fun envir name = REF EMPTY;`\n`        empty = EMPTY;`\n\n`        fun get (NAMING   (k, v)) x =>  if (x == k)  v;  else  raise exception SYMBOLTABLE;  fi;`\n`            get (OVERRIDE (a, b)) x =>  get b x except _ = get a x;                                     # Should this be   get a x except _ = get b x;   ?  If not, why not?   2011-05-05 CrT`\n`            get (TABLE         t) x =>  h::look_up t x;`\n`            #`\n`            get EMPTY _  => raise exception SYMBOLTABLE;`\n`        end;`\n\n`        fun get' symboltable default x`\n`            =`\n`            get symboltable x`\n`            except`\n`                _ = default;`\n\n`        fun lookup (REF symboltable) x`\n`            =`\n`            get symboltable x;`\n\n`        fun union (a, EMPTY) =>  a;`\n`            union (EMPTY, b) =>  b;`\n`            union (a, b)     =>  OVERRIDE (a, b);`\n`        end;`\n\n`        fun put symboltable x =  union (symboltable, NAMING x);`\n`        fun set symboltable x =  symboltable := put *symboltable x;`\n\n`        fun flatten symboltable`\n`            = `\n`            {   t = h::make_hashtable (hash_string::hash_string, (==)) { size_hint => 13, not_found_exception => SYMBOLTABLE };`\n`                #`\n`                put = h::set t;`\n\n`                f symboltable`\n`                where`\n`                    fun f EMPTY             =>  ();`\n`                        f (NAMING x)        =>  put x;`\n`                        f (OVERRIDE (a, b)) =>  {  f a;  f b;  };`\n`                        f (TABLE t)         =>  h::keyed_apply put t;`\n`                    end;`\n`                end;`\n\n`                t;`\n`            };`\n\n`        fun apply f symboltable`\n`            =`\n`            h::keyed_apply f (flatten symboltable);`\n\n`        fun map f symboltable`\n`            =`\n`            list::map f (h::keyvals_list (flatten symboltable));`\n\n`        fun fold f x symboltable`\n`            =`\n`            h::foldi f x (flatten symboltable);`\n\n`        fun unions dicts`\n`            =`\n`            fold_backward union EMPTY dicts;`\n\n`        fun consolidate symboltable`\n`            =`\n`            TABLE (flatten symboltable);`\n\n`        bind = NAMING;`\n`    };`\n`end;`", null, "", null, "", null, "" ]

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[ "Solutions by everydaycalculation.com\n\nSubtract 9/7 from 18/5\n\n1st number: 3 3/5, 2nd number: 1 2/7\n\n18/5 - 9/7 is 81/35.\n\nSteps for subtracting fractions\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 5 and 7 is 35\n2. For the 1st fraction, since 5 × 7 = 35,\n18/5 = 18 × 7/5 × 7 = 126/35\n3. Likewise, for the 2nd fraction, since 7 × 5 = 35,\n9/7 = 9 × 5/7 × 5 = 45/35\n4. Subtract the two fractions:\n126/35 - 45/35 = 126 - 45/35 = 81/35\n5. In mixed form: 211/35\n\n-" ]

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[ "# Work on Sliding Object#\n\n## Part 1#\n\nThe work required to accelerate an object on a frictionless surface from a speed of 2$$v$$ to a speed of 3$$v$$ is:\n\n• 6 times the work required to accelerate the object from $$v$$ = 0 to $$v$$\n\n• 4 times the work required to accelerate the object from $$v$$ = 0 to $$v$$\n\n• 5 times the work required to accelerate the object from $$v$$ = 0 to $$v$$\n\n• Equal to the work required to accelerate the object from 3$$v$$ to 4$$v$$\n\n• Not known without knowledge of the acceleration", null, "" ]

[ null, "https://raw.githubusercontent.com/firasm/bits/master/by-nc-sa.png", null ]

[ "# mil; thou to ångström conversion\n\nConversion number between mil; thou [mil] and ångström [Å] is 254000. This means, that mil; thou is bigger unit than ångström.\n\n### Contents [show][hide]", null, "Switch to reverse conversion:\nfrom ångström to mil; thou conversion\n\n### Enter the number in mil; thou:\n\nDecimal Fraction Exponential Expression\n [mil]\neg.: 10.12345 or 1.123e5\n\nResult in ångström\n\n?\n precision 0 1 2 3 4 5 6 7 8 9 [info] Decimal: Exponential:\n\n### Calculation process of conversion value\n\n• 1 mil; thou = (exactly) (0.0000254) / (1*10^-10) = 254000 ångström\n• 1 ångström = (exactly) (1*10^-10) / (0.0000254) = 3.9370078740157 × 10-6 mil; thou\n• ? mil; thou × (0.0000254  (\"m\"/\"mil; thou\")) / (1*10^-10  (\"m\"/\"ångström\")) = ? ångström\n\n### High precision conversion\n\nIf conversion between mil; thou to metre and metre to ångström is exactly definied, high precision conversion from mil; thou to ångström is enabled.\n\nDecimal places: (0-800)\n\nmil; thou\nResult in ångström:\n?\n\n### mil; thou to ångström conversion chart\n\n Start value: [mil; thou] Step size [mil; thou] How many lines? (max 100)\n\nvisual:\nmil; thouångström\n00\n102540000\n205080000\n307620000\n4010160000\n5012700000\n6015240000\n7017780000\n8020320000\n9022860000\n10025400000\n11027940000\nCopy to Excel\n\n## Multiple conversion\n\nEnter numbers in mil; thou and click convert button.\nOne number per line.\n\nConverted numbers in ångström:\nClick to select all\n\n## Details about mil; thou and ångström units:\n\nConvert Mil; thou to other unit:\n\n### mil; thou\n\nDefinition of mil; thou unit: ≡ 1×10−3 in .\n\nConvert ångström to other unit:\n\n### ångström\n\nDefinition of ångström unit: ≡ 1×10−10 m . ≡ 0.1 nm", null, "" ]

[ null, "http://conversion.org/images/switch.png", null, "http://conversion.org/menufiles/top.png", null ]

[ "### Sample Problem\n\nGiven 1 in = 2.54 cm, how many centimeters are in a foot?\n\ncm\n\n#### Solution\n\nOne foot = 12 inches", null, "Notice how the relation 1 in = 2.54 cm is expressed as 2.54 cm/in, or in words, 2.54 cm per 1 in. Metric conversion units are not usually considered for significant figures because the factors are exact (1m = 100 cm for instance). However because inches are in the English system, the conversion factor from metric is not exact, so significant figures are considered." ]

[ null, "https://s3.amazonaws.com/ck12bg.ck12.org/curriculum/105136/1.png", null ]

[ "# On balance sheet derivatives of trigonometric functions\n\nBalance sheet\n\n## On balance sheet derivatives of trigonometric functions\n\nOn balance sheet derivatives of trigonometric functions. Derivatives of Trig Functions We’ ll give the derivatives of trigonometric the trig functions in this section. balance Be trigonometric able to balance compute the derivatives of the inverse trigonometric functions speci cally, tan x , cos x, sin x sec x. Concept of exponential and logarithmic functions. derivatives of some standard functions and then adjust those formulas to make them antidifferentiation formulas. Derivatives of Exponential Logarithm Functions In this section we will get the derivatives of the exponential logarithm functions.\nMath Cheat Sheet for Derivatives. The default behavior of your system allows trigonometric functions without sheet parentheses. x ( ) cosx = \u0001 sinx Version 2 of the Limit Definition of the Derivative Function in Section 3. How are derivatives accounted in balance sheet? Pre- Calculus Homework Section 99. Know how to apply logarithmic di erentiation to compute the derivatives of functions g( x) of the form ( f( x) ) where f g are non- constant functions of x. Logarithmic differentiation, derivative of functions expressed in parametic balance forms. The basic trigonometric functions include the following 6 functions: sine ( sinx) tangent ( tanx), cotangent ( cotx), secant ( secx) , cosine ( cosx) cosecant ( cscx).\n2 – sheet Derivatives of Trigonometric Functions Take the derivative of Pre- Calculus Homework Section 99. 4 7 Inverse Trigonometric Functions. Steps for solving trig equations using inverses ex 2 solve a trig equation contain inverse functions composite inverse trig functions. There are rules we can follow trigonometric to find many derivatives. Find the derivatives derivatives of trigonometric functions: = 4sin + 5cos = sin cos = 2sec + sheet tan = ˘ ˇ ˆ˙ ˝ ˇ = sheet sheet sin ˛ 3 − cos ˛ 3 balance = ˆ˙ ˝ ˛ ˇ. The trigonometric ratios balance of the angles 30º 45º , 60º are often used in mechanics other balance branches of mathematics. a derivative is a financial instrument which derives its value of some other financial instrument or variable the value from which a derivative derives its values is called Financial derivatives -. financial sheet derivatives.\nFor each of these functions, there is an inverse trigonometric function. Note: the little mark ’ means \" Derivative of\" , f g are. This worksheet is balance a great resource for the 5th 7th Grade, , 6th Grade 8th Grade. f( x) xn 1 x ex cos x sin x 1 1 + x2 F( x) = ∫ f( x) dx xn + 1 balance n trigonometric + 1 ln x ex sin balance x- cos trigonometric x tan- 1 x ( There is a more extensive list of balance anti- differentiation formulas on page 406 of the text. 2 – Derivatives of Trigonometric Functions Take the derivative of the given equations with respect to x. To create your new password, just click the link in the email we sent you. Trigonometric functions of arcs from 0 to ± 2p balance Calculation of values of trigonometric functions, Trigonometric reduction formulas - the reference angles Basic relationships between trigonometric functions of the same angle\" See more. 90 Chapter 4 Trigonometry Complex Numbers Note Ordinary functions require parentheses around the function argument while trigonometric functions commonly do trigonometric sheet not. So it is useful to calculate derivatives them and know their values by heart.\n\nDerivatives of the Basic Sine and Cosine Functions 1) D. Derivatives of Trigonometric Functions. 84% Asset Turnover = Revenue. § Proof of 1) Let fx( ) = sheet sinx. x ( ) sinx = cosx 2) D. In fact, they do not even use Limit Statement # 2 in Part C.\n\nHere balance are useful rules to help you work out the derivatives of many functions ( with examples below). In this case, the triangle is isosceles. Derivatives And balance Integrals Involving Inverse Trig Functions. Briefly discuss the impact of sheet the changes in asset turnover and financial leverage on the change BALANCE SHEET DATAFixed Assets \\$ 41 \\$ 70 Total AssetsWorking CapitalTotal debt 16 0 Total Shareholder equitySolution: a) For Operating margin = EBIT/ sheet Revenue Operating margin 6. a d b y A p t t u s. Inverse Trigonometric Functions Wikipedia. 2 Part sheet A provides us with more elegant proofs. ASC 606 for dummies whitepaper.\nEvaluating Expressions Involving Inverse Sine Cosine And. On balance sheet derivatives of trigonometric functions. Trigonometric Ratios Worksheets This Trigonometry Worksheet will balance produce trigonometric ratio problems. sheet Previously in India derivatives was accounted derivatives in balance sheet as per Accounting Standard- 30 derivatives respectively, to IAS 39, 31, IAS 32 , 32 that correspond IFRS 7. If you want parentheses to be required for. Continuity derivative of composite functions, chain rule, differentiability, derivatives of inverse trigonometric functions derivative of implicit functions. For example: The slope of a constant value ( like 3) is always 0; The slope of balance a line like 2x is 2 , 3x is 3 etc; so on.\n\nHence the opposite side adjacent sides are equal say 1 unit. Derivatives of logarthmic and exponential functions.\n\n## Derivatives trigonometric\n\nCheck the formula sheet of integration. Topics include Basic Integration Formulas Integral of special functions Integral by Partial Fractions Integration by Parts Other Special Integrals Area as a sum Properties of definite integration Integration of Trigonometric Functions, Properties of Definite Integration are all mentioned here. You are currently. Chapter 4: Trigonometric Functions is not available.\n\n``on balance sheet derivatives of trigonometric functions``\n\nChapter 5: Analytic Trigonometry is not available. Limits, Derivatives. Trigonometric Functions." ]

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[ "### re: Advent of Code 2019 Solution Megathread - Day 5: Sunny with a Chance of Asteroids VIEW POST\n\nInitial kotlin solution (I'm going to clean this up a LOT, but in the interest of sharing what I learn, I promise not to pave this post over.)\n\n``````import arrow.core.None\nimport arrow.core.Option\nimport arrow.core.Some\nimport arrow.core.some\nimport java.nio.file.Files\nimport java.nio.file.Paths\n\nobject Day05 {\nsealed class Mode {\nobject Immediate : Mode()\nobject Position : Mode()\n}\n\nfun Char.toMode(): Mode {\nreturn when (this) {\n'0' -> Mode.Position\n'1' -> Mode.Immediate\nelse -> throw Error(\"Bad mode \\$this\")\n}\n}\n\nfun parseInstruction(instruction: String): Instruction {\nreturn Day05.Instruction(\nopcode = instruction.takeLast(2).toInt(),\nparamModeInput = instruction.take(\nkotlin.math.max(\n0,\ninstruction.count() - 2\n)\n).map { it.toMode() }.reversed()\n)\n}\n\nclass Instruction(val opcode: Int, paramModeInput: List<Mode>) {\nval paramModes: Array<Mode> = arrayOf(Mode.Position, Mode.Position, Mode.Position, Mode.Position)\n\ninit {\nparamModeInput.forEachIndexed { i, it ->\nparamModes[i] = it\n}\n}\n\noverride fun toString(): String {\nreturn \"Instruction[\\$opcode, \\${paramModes.contentToString()}]\"\n}\n}\n\nfun handle(pointer: Int, input: Int, code: Array<String>): Option<Int> {\nval instr = parseInstruction(code[pointer])\nval inputs = code.drop(pointer + 1).take(3)\nval params = inputs\n.zip(instr.paramModes).map { (value, mode) ->\nwhen (mode) {\nMode.Immediate -> value.toInt()\nMode.Position -> {\ncode.getOrNull(value.toInt())?.toInt() ?: -1000000\n}\n}\n}\nreturn when (instr.opcode) {\n1 -> {\ncode[inputs.toInt()] = (params + params).toString()\n(pointer + 4).some()\n}\n2 -> {\ncode[inputs.toInt()] = (params * params).toString()\n(pointer + 4).some()\n}\n3 -> {\ncode[inputs.toInt()] = input.toString()\n(pointer + 2).some()\n}\n4 -> {\nprintln(\"output:\\${params}\")\n(pointer + 2).some()\n}\n5 -> {\nwhen (params) {\n0 -> pointer + 3\nelse -> params\n}.some()\n}\n6 -> {\nwhen (params) {\n0 -> params\nelse -> pointer + 3\n}.some()\n}\n7 -> {\ncode[inputs.toInt()] = when {\nparams < params -> \"1\"\nelse -> \"0\"\n}\n(pointer + 4).some()\n}\n8 -> {\ncode[inputs.toInt()] = when {\nparams == params -> \"1\"\nelse -> \"0\"\n}\n(pointer + 4).some()\n}\n99 -> Option.empty<Int>()\nelse -> throw Error(\"Unknown opcode: \\${instr.opcode}\")\n}\n}\n\ntailrec fun step(\ncode: Array<String>,\ninput: Int,\ninstructionPointer: Option<Int> = 0.some()\n) {\nreturn when (instructionPointer) {\nis None -> Unit\nis Some<Int> -> {\nval nextInstruction = handle(instructionPointer.t, input, code)\nstep(code, input, nextInstruction)\n}\n}\n}\n\nconst val FILENAME = \"src/main/resources/day05.txt\"\n}\n\nfun main() {\n// Part 01\nDay05.step(\ninput = 1\n)\n\n// Part 02\nDay05.step(\ninput = 5\n)\n}\n\n``````\n\nOk, as promised, here's the code I wanted to write, but I was halfway done before I saw it, and I wanted the answer so bad I didn't give myself the time to rewrite.\n\n``````import arrow.core.Either\nimport arrow.core.Option\nimport arrow.core.flatMap\nimport arrow.core.getOrElse\nimport arrow.core.left\nimport arrow.core.right\nimport arrow.core.some\nimport java.nio.file.Files\nimport java.nio.file.Paths\n\nsealed class Mode {\nobject Immediate : Mode()\nobject Position : Mode()\n}\n\nsealed class Instruction {\nabstract val opcodeFormat: String\nabstract fun execute(pointer: Int, code: Array<Int>, params: List<Int>): Either<Option<String>, Int>\n\nopen fun findInputs(code: Array<Int>, pointer: Int) =\ncode.drop(pointer + 1)\n.take(3)\n.zip(opcodeFormat.format(code[pointer] / 100).map {\nwhen (it) {\n'0' -> Mode.Position\n'1' -> Mode.Immediate\nelse -> throw Error(\"Bad mode \\$it\")\n}\n}.reversed())\n.map { (it, mode) ->\nwhen (mode) {\nMode.Position -> code[it]\nMode.Immediate -> it\n}\n}\n\nsealed class ThreeParameterInstruction : Instruction() {\noverride val opcodeFormat = \"1%02d\"\n\noverride fun execute(pointer: Int, code: Array<Int>, params: List<Int>): Either<Option<String>, Int> {\ncode[params] = params + params\nreturn (pointer + 4).right()\n}\n}\n\nclass Multiply : ThreeParameterInstruction() {\noverride fun execute(pointer: Int, code: Array<Int>, params: List<Int>): Either<Option<String>, Int> {\ncode[params] = params * params\nreturn (pointer + 4).right()\n}\n}\n\nclass LessThan : ThreeParameterInstruction() {\noverride fun execute(pointer: Int, code: Array<Int>, params: List<Int>): Either<Option<String>, Int> {\ncode[params] = when {\nparams < params -> 1\nelse -> 0\n}\nreturn (pointer + 4).right()\n}\n}\n\nclass Equal : ThreeParameterInstruction() {\noverride fun execute(pointer: Int, code: Array<Int>, params: List<Int>): Either<Option<String>, Int> {\ncode[params] = when {\nparams == params -> 1\nelse -> 0\n}\nreturn (pointer + 4).right()\n}\n}\n}\n\nsealed class TwoParameterInstruction : Instruction() {\noverride val opcodeFormat = \"%02d\"\n\nclass JumpIfTrue : TwoParameterInstruction() {\noverride fun execute(pointer: Int, code: Array<Int>, params: List<Int>) =\nwhen (params) {\n0 -> pointer + 3\nelse -> params\n}.right()\n}\n\nclass JumpIfFalse : TwoParameterInstruction() {\noverride fun execute(pointer: Int, code: Array<Int>, params: List<Int>) =\nwhen (params) {\n0 -> params\nelse -> pointer + 3\n}.right()\n}\n}\n\nclass SetFromInput(private val input: Int) : Instruction() {\noverride val opcodeFormat = \"1\"\noverride fun execute(pointer: Int, code: Array<Int>, params: List<Int>): Either<Option<String>, Int> {\ncode[params] = input\nreturn (pointer + 2).right()\n}\n}\n\nclass Output : Instruction() {\noverride val opcodeFormat = \"0\"\noverride fun execute(pointer: Int, code: Array<Int>, params: List<Int>): Either<Option<String>, Int> {\nprintln(\"output: \\${params}\")\nreturn (pointer + 2).right()\n}\n}\n\nobject End : Instruction() {\noverride val opcodeFormat: String\nget() = throw Error(\"No opcode format for End instructions.\")\n\noverride fun findInputs(code: Array<Int>, pointer: Int) = emptyList<Int>()\noverride fun execute(pointer: Int, code: Array<Int>, params: List<Int>): Either<Option<String>, Int> =\nOption.empty<String>().left()\n}\n}\n\nobject Day05 {\n\nprivate fun parseInstruction(instruction: String, input: Int) =\nwhen (instruction.takeLast(2).toInt()) {\n2 -> Instruction.ThreeParameterInstruction.Multiply().right()\n3 -> Instruction.SetFromInput(input).right()\n4 -> Instruction.Output().right()\n5 -> Instruction.TwoParameterInstruction.JumpIfTrue().right()\n6 -> Instruction.TwoParameterInstruction.JumpIfFalse().right()\n7 -> Instruction.ThreeParameterInstruction.LessThan().right()\n8 -> Instruction.ThreeParameterInstruction.Equal().right()\n99 -> Instruction.End.right()\nelse -> \"Problem parsing instruction \\$instruction\".some().left()\n}\n\nprivate fun handleCodePoint(pointer: Int, input: Int, code: Array<Int>) =\nparseInstruction(code[pointer].toString(), input).flatMap { instr ->\ninstr.execute(pointer, code, instr.findInputs(code, pointer))\n}\n\ntailrec fun step(\ncode: Array<Int>,\ninput: Int,\ninstructionPointer: Either<Option<String>, Int> = Either.right(0)\n): String = when (instructionPointer) {\nis Either.Left<Option<String>> -> instructionPointer.a.getOrElse { \"Program terminated successfully.\" }\nis Either.Right<Int> -> {\nval nextInstruction = handleCodePoint(instructionPointer.b, input, code)\nstep(code, input, nextInstruction)\n}\n}\n\nconst val FILENAME = \"src/main/resources/day05.txt\"\n}\n\nfun main() {\n.first()\n.split(\",\")\n.map { it.toInt() }\n\n// Part 01\nprintln(\"Part 01\")\nDay05.step(code = problemInput.toTypedArray(), input = 1)\n\n// Part 02\nprintln(\"\\nPart 02\")\nDay05.step(code = problemInput.toTypedArray(), input = 5)\n\nprintln(\"\\nDay 02\")\n.split(\",\")\n.map { it.toInt() }.toTypedArray()\nDay05.step(code = day02Code, input = 5)\nprintln(day02Code.take(10))\n}\n``````\n\nDefinitely can say this implementation has a lot of `class`. 🤣\n\nYeah, I wish there was a more elegant way of doing discriminated unions in Kotlin. I'm used to the lightweight kinds/types of Haskell. Arrow-KT lets you do basic stuff pretty nicely, but it gets ugly the more you try to be full functional style only.\n\nI'm also disappointed in the type erasure.\n\ncode of conduct - report abuse", null, "", null, "" ]

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[ "# Factors affecting the resistance", null, "#### Length of the material\n\nThe resistance of a substantial is directly proportional to the length. The resistance of lengthier wire is a lot of. It is denoted by “l”.\n\n#### Cross sectional area\n\nThe resistance of a substantial is reciprocally proportional to the cross-sectional space of the fabric a lot of cross-sectional area permitted the path of additional number of free electrons, giving less resistance. It is denoted by 'a'.\n\n#### The nature of the material and its type\n\nIt contains a more number of free electrons or not, distresses the value of the resistance.Therefore material that is conductor has less resistance whereas associate nonconductor has terribly high resistance.\n\n#### Temperature\n\nThe temperature of the material affects the value of the resistance. Typically the resistance of the material will increase as its temperature will increase. Typically impact of tiny changes in temperature on the resistance is not thought of as it's negligibly tiny.\n\nSo for a bound material at a bound temperature we have a tendency to will write a mathematical expression as,\nR ∝ (l/a)\nand impact of nature of material is thought of through the constant of proportion denoted by p (rho) known as resistance or specific resistance of the material. So finally,\nR=(ρl)/a\nWhere,\nL = length in meters\nA = cross sectional area in sq.meter\nρ = resistivity in ohm–meters\nR = Resistance in ohms", null, "1 Comments", null, "Comments\n\n#### Read Comment Policy We have Zero Tolerance to Spam. Chessy Comments and Comments with Links will be deleted immediately upon our review.\n\n1.", null, "" ]

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[ "Archive for 2015\n\n## Programming an estimation command in Stata: A better OLS command\n\nI use the syntax command to improve the command that implements the ordinary least-squares (OLS) estimator that I discussed in Programming an estimation command in Stata: A first command for OLS. I show how to require that all variables be numeric variables and how to make the command accept time-series operated variables.\n\nThis is the seventh post in the series Programming an estimation command in Stata. I recommend that you start at the beginning. See Programming an estimation command in Stata: A map to posted entries for a map to all the posts in this series. Read more…\n\nCategories: Programming Tags:\n\n## Programming an estimation command in Stata: A first command for OLS\n\n$$\\newcommand{\\betab}{\\boldsymbol{\\beta}} \\newcommand{\\xb}{{\\bf x}} \\newcommand{\\yb}{{\\bf y}} \\newcommand{\\Xb}{{\\bf X}}$$I show how to write a Stata estimation command that implements the ordinary least-squares (OLS) estimator by explaining the code. I use concepts that I introduced in previous #StataProgramming posts. In particular, I build on Programming an estimation command in Stata: Using Stata matrix commands and functions to compute OLS objects, in which I recalled the OLS formulas and showed how to compute them using Stata matrix commands and functions and on\nProgramming an estimation command in Stata: A first ado command, in which I introduced some ado-programming concepts. Although I introduce some local macro tricks that I use all the time, I also build on Programing an estimation command in Stata: Where to store your stuff.\n\nThis is the sixth post in the series Programming an estimation command in Stata. I recommend that you start at the beginning. See Programming an estimation command in Stata: A map to posted entries for a map to all the posts in this series. Read more…\n\nCategories: Programming Tags:\n\n## Programming an estimation command in Stata: Using Stata matrix commands and functions to compute OLS objects\n\n$$\\newcommand{\\epsilonb}{\\boldsymbol{\\epsilon}} \\newcommand{\\ebi}{\\boldsymbol{\\epsilon}_i} \\newcommand{\\Sigmab}{\\boldsymbol{\\Sigma}} \\newcommand{\\betab}{\\boldsymbol{\\beta}} \\newcommand{\\eb}{{\\bf e}} \\newcommand{\\xb}{{\\bf x}} \\newcommand{\\zb}{{\\bf z}} \\newcommand{\\yb}{{\\bf y}} \\newcommand{\\Xb}{{\\bf X}} \\newcommand{\\Mb}{{\\bf M}} \\newcommand{\\Eb}{{\\bf E}} \\newcommand{\\Xtb}{\\tilde{\\bf X}} \\newcommand{\\Vb}{{\\bf V}}$$I present the formulas for computing the ordinary least-squares (OLS) estimator, and I discuss some do-file implementations of them. I discuss the formulas and the computation of independence-based standard errors, robust standard errors, and cluster-robust standard errors. I introduce the Stata matrix commands and matrix functions that I use in ado-commands that I discuss in upcoming posts.\n\nThis is the fifth post in the series Programming an estimation command in Stata. I recommend that you start at the beginning. See Programming an estimation command in Stata: A map to posted entries for a map to all the posts in this series. Read more…\n\nCategories: Programming Tags:\n\n## xtabond cheat sheet\n\nRandom-effects and fixed-effects panel-data models do not allow me to use observable information of previous periods in my model. They are static. Dynamic panel-data models use current and past information. For instance, I may model current health outcomes as a function of health outcomes in the past— a sensible modeling assumption— and of past observable and unobservable characteristics.\n\nToday I will provide information that will help you interpret the estimation and postestimation results from Stata’s Arellano–Bond estimator xtabond, the most common linear dynamic panel-data estimator. Read more…\n\nCategories: Statistics Tags:\n\n## Programming an estimation command in Stata: A first ado-command\n\nI discuss the code for a simple estimation command to focus on the details of how to implement an estimation command. The command that I discuss estimates the mean by the sample average. I begin by reviewing the formulas and a do-file that implements them. I subsequently introduce ado-file programming and discuss two versions of the command. Along the way, I illustrate some of the postestimation features that work after the command.\n\nThis is the fourth post in the series Programming an estimation command in Stata. I recommend that you start at the beginning. See Programming an estimation command in Stata: A map to posted entries for a map to all the posts in this series. Read more…\n\nCategories: Programming Tags:\n\n## Using mlexp to estimate endogenous treatment effects in a probit model\n\nI use features new to Stata 14.1 to estimate an average treatment effect (ATE) for a probit model with an endogenous treatment. In 14.1, we added new prediction statistics after mlexp that margins can use to estimate an ATE.\n\nI am building on a previous post in which I demonstrated how to use mlexp to estimate the parameters of a probit model with sample selection. Our results match those obtained with biprobit; see [R] biprobit for more details. In a future post, I use these techniques to estimate treatment-effect parameters not yet available from another Stata command. Read more…\n\nCategories: Statistics Tags:\n\n## Programming an estimation command in Stata: Global macros versus local macros\n\nI discuss a pair of examples that illustrate the differences between global macros and local macros. You can view this post as a technical appendix to the previous post in the #StataProgramming series, which introduced global macros and local macros.\n\nIn every command I write, I use local macros to store stuff in a workspace that will not alter a user’s data and to make my code easier to read. A good understanding of the differences between global macros and local macros helps me to write better code. The essential differences between global macros and local macros can be summarized in two points. Read more…\n\nCategories: Programming Tags:\n\n## Fixed effects or random effects: The Mundlak approach\n\nToday I will discuss Mundlak’s (1978) alternative to the Hausman test. Unlike the latter, the Mundlak approach may be used when the errors are heteroskedastic or have intragroup correlation. Read more…\n\nCategories: Statistics Tags:\n\n## Programming an estimation command in Stata: Where to store your stuff\n\nIf you tell me “I program in Stata”, it makes me happy, but I do not know what you mean. Do you write scripts to make your research reproducible, or do you write Stata commands that anyone can use and reuse? In the series #StataProgramming, I will show you how to write your own commands, but I start at the beginning. Discussing the difference between scripts and commands here introduces some essential programming concepts and constructions that I use to write scripts and commands.\n\nThis is the second post in the series Programming an estimation command in Stata. I recommend that you start at the beginning. See Programming an estimation command in Stata: A map to posted entries for a map to all the posts in this series. Read more…\n\nCategories: Programming Tags:\n\n## Probit model with sample selection by mlexp\n\nOverview\n\nIn a previous post, David Drukker demonstrated how to use mlexp to estimate the degree of freedom parameter in a chi-squared distribution by maximum likelihood (ML). In this post, I am going to use mlexp to estimate the parameters of a probit model with sample selection. I will illustrate how to specify a more complex likelihood in mlexp and provide intuition for the probit model with sample selection. Our results match the heckprobit command; see [R] heckprobit for more details. Read more…\n\nCategories: Statistics Tags:" ]

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[ "The impact of quantum effects on relativistic electron motion in a chaotic regime\n\n# The impact of quantum effects on relativistic electron motion in a chaotic regime\n\nA.V. Bashinov Institute of Applied Physics, Russian Academy of Sciences, 603950 Nizhny Novgorod, Russia University of Nizhny Novgorod, 603950 Nizhny Novgorod, Russia    A.V. Kim Institute of Applied Physics, Russian Academy of Sciences, 603950 Nizhny Novgorod, Russia University of Nizhny Novgorod, 603950 Nizhny Novgorod, Russia    A.M. Sergeev Institute of Applied Physics, Russian Academy of Sciences, 603950 Nizhny Novgorod, Russia University of Nizhny Novgorod, 603950 Nizhny Novgorod, Russia\nJuly 16, 2019\n###### Abstract\n\nWe consider the impact of quantum effects on electron dynamics in a plane linearly polarized standing wave with relativistic amplitudes. For this purpose analysis the Lyapunov characteristic exponent spectrum with and without allowance for the classic radiation reaction force has been analyzed. Based on this analysis it is concluded that the contraction effect of phase space in the stochastic regime due to the radiation reaction force in the classical form doesn’t occur when the quantum nature of hard photon emission is taken into account. It is shown that electron bunch kinetics has a diffusion solution rather than the d’Alambert type solution as in the classic description.It is also revealed that the electron motion can be described using the Markov chain formalism. This method gives exact characteristics of electron bunch evolution, such as motion of the center of mass and electron bunch dimensions.\n\n42.60.-m\n\n## I Introduction\n\nNowadays, much attention is paid to the interaction of relativistically strong laser fields with matter, when radiation losses play an important role. The interest in this subject is connected not only with the fundamental problem of charged particle dynamics in electromagnetic fields, but also with possible applications, such as laser-based electron accelerators and hard photon sources Hooker (2013); Phuoc et al. (2012). It was recently found that the dynamics of charged particles in ultrarelativistic electromagnetic fields changes drastically when radiative reaction force is taken into account. As was shown in Refs.Ji et al. (2014); Gonoskov et al. (2014), at super strong laser fields charged particles are drawn into a high-field region, thus leading to efficient generation of gamma rays and particle beams. The need to consider the back impact of the emitted photon !fields! is dictated by the facts that the recoil force becomes comparable with the Lorentz force and a particle can lose a substantial part of its energy in one act of emission. Thus, the effect of radiation losses can dramatically change the trajectory of the particle over the laser field period. In some cases, small changes of particle trajectories over the laser period caused by weak radiative losses can be accumulated and significantly change particle motion for a time much longer than the optical cycle. This follows from the theorem of phase space contraction in the presence of dissipative forces, that has been proved in Ref.Tamburini et al. (2011) for the case of charged particle motion in an electromagnetic field, with the classical radiation reaction force taken into consideration. Particular examples of this effect were considered in the field of traveling Bashinov et al. (2013) and standing waves Lehmann and Spatschek (2012). In the latter case, it is connected with stochastic heating Sheng et al. (2002); Mendonça and Doveil (1982); Sentoku et al. (2002). However, due to the quantum nature, emission occurs randomly in the form of discrete photons that may randomize motion. As a consequence, the phase volume may be not compressed and even increases in spite of dissipation. Comparative analysis of an electron beam interaction with a counterpropagting laser pulse, taking into consideration radiative losses in classical and quantum form showed that at certain electron energies and laser field intensity, the phase space volume can be expanded during the interaction Neitz and Di Piazza (2013). In this paper, we analyze the characteristics of electron motion in the field of a standing linearly polarized plane wave taking into account radiative losses in conformity with the quantum and classical descriptions. We consider the wave amplitudes at which an electron loses a small part of its energy in one act of photon emission but its motion can qualitatively change due to discreteness of the photon emission. A comparative analysis of electron dynamics with and without radiation reaction force in classical and quantum forms is presented.\n\n## Ii Electron motion in the classical case\n\nFor taking into account radiative losses in the classical case, we add to the equation of motion the radiation reaction force in the Landau-Lifshitz form, keeping the most important term proportional to the squared Lorentz factor Tamburini et al. (2010):\n\n dpdt=−eE−ecv×B− δe2γ2vωmc2[(E+1cv×B)2−1c2(E⋅v)2], (1) drdt=pmγ, (2)\n\nwhere are the momentum, velocity, Lorentz factor, charge and mass of the electron, correspondingly; is the velocity of light; is the time in the laboratory frame of reference; ; and is the laser frequency. Without loss of generality, we assume that the electric field is directed along the Z axis and the magnetic field is directed along the X axis:\n\n E=E0cos(ωt+φ0)cos(ky)z, (3) B=E0sin(ωt+φ0)sin(ky)x, (4)\n\nwhere the vacuum wave number and is the initial wave phase. The electron density is considered to be low enough, so that the effect of the particles influence on each other can be neglected in comparison with the impact of the external field. Dimensionless variables are used for simplicity; we also assume that at the initial moment of time, the particle moves in the ZY plane. In this case, its motion always occurs in this plane:\n\n γdρydτ=−ρza0sin(τ+φ0)sin(η)−δρya20Θ(τ,η), (5) γdρzdτ=ρya0sin(τ+φ0)sin(η)− γa0cos(τ+φ0)cos(η)−δρza20Θ(τ,η), (6) dηdτ=ρy/γ, (7) dζdτ=ρz/γ, (8)\n\nwhere , , . To analyze the electron motion without radiation reaction force it suffices to assume . Note that, although further analysis is performed for the wavelength , when , the system depends to a good accuracy only on one similarity parameter Bulanov et al. (2010).\n\nIn a general case, analytical solution of the equation of motion (5-8) in the field of a plane linearly polarized standing wave hasn’t been found, even without radiation reaction, so the spectrum of characteristic Lyapunov exponents is used to analyze electron motion with the help of a modified generalized Benettin algorithm Benettin et al. (1980a),Benettin et al. (1980b). In order to find all possible modes of motion, it is necessary to consider all possible initial conditions. Because of the periodicity of the equations along the Y axis it is sufficient to set the initial position of the electrons along the axis within one wavelength. Taking into account the radiative reaction force, it is reasonable to assume that the absolute value of the transverse momentum (along ) during steady motion doesn’t exceed , and the absolute value of the longitudinal momentum doesn’t exceed . However, significant computing resources are needed for good resolution of phase space by initial conditions. We confine ourselves to a single particle, which has zero momentum at the initial moment of time and is located between the electric and magnetic field nodes (in the calculations, the particle was set midway between the nodes). Thus, we haven’t obtained all possible modes of motion (for example, stable trajectory in the region of the electric field antinode at a relativistic initial momentum along the electric field Kaplan and Pokrovsky (2005)), but such initail position and momentum allow the particle to sit on attractors, which are discussed in the work Lehmann and Spatschek (2012). To reduce the influence of the initial conditions, the calculation of the Lyapunov characteristic exponents starts only after 30000 periods of laser field. As shown by calculations, this time is sufficient for the particle trajectory to sit on the attractor, if it exists, even with . Figure 1 shows the obtained spectra of Lyapunov characteristic exponents as a function of .", null, "Figure 1: Lyapunov characteristic exponents of system (5-7) with (a) and without (Θ=0) (b) radiation reaction force. According to value/По старшинству λ1 (red), λ2 (black), λ3 (blue), λ4 (magenta). The sum of the exponents is λg (green).\n\nFirst, note that, as the dimension of system (5-8) is 4, there exist chaotic attractors. Second, the results of numerical modeling are consistent with the theorem of phase space contraction Tamburini et al. (2011), taking into account the classic radiation reaction: the sum of the Lyapunov exponents is negative, despite the fact that the maximum Lyapunov exponent can be positive, the Lyapunov exponents equal zero. Consequently, system (5-8) has a chaotic or a regular attractor. Note that there is a range of amplitudes where the attractor is regular, which is conformed by the conclusion in Lehmann and Spatschek (2012). The stochastic heating is suppressed in this amplitude range. Since the sum of Lyapunov exponents is negative, one of them is assumed to be zero, which improves accuracy of estimations of the other Lyapunov exponents. Average squared error of the exponents in this case did not exceed 1.\n\nBut without the radiation reaction, system (5-7) may have attractors under certain initial conditions Sheng et al. (2002), Kaplan and Pokrovsky (2005). However, the calculations havn’t given us definite information at what amplitudes attractors may arise. The exact value of the sum of the exponents could not be established because , i.e. the error is much greater than the value itself. This occurs as the largest and the smallest absolute values of the Lyapunov exponents are opposite in sign , and the second and third parameters are close to zero . However, the value of the sum of the Lyapunov exponents isn’t of great significance, most important is that the senior indicator is positive and there is chaotic motion without radiation friction Lehmann and Spatschek (2012).\n\nNote that the spectrum of the characteristic Lyapunov exponents is calculated for a certain trajectory that originates under certain initial conditions. However, according to the multiplicative ergodic theorem Oseledets (1968), this spectrum can be attributed to the attractor as a whole. We have considered the attarctor which electron initially at rest reaches. Moreover if the initial momentum is , the particle goes to the same attractor Lehmann and Spatschek (2012). We also note that access to the attractor, as follows from numerical calculations, does not depend on the initial phase of the wave. The initial phase affects the steady component of the pulse along the electric field (), but due to radiation losses the steady component disappears in some time that will be estimated later. As a result, the average value of is equal to 0, and that’s why the initial phase has no influence on a possibility of reaching the attractor. However, there are points in phase space starting at which an electron cann’t reach the attractor. The magnetic field antinodes (where n is an integer) are stable at small perturbations. In the nonrelativistic case, the electron motion near this point can be described as damped oscillations in the ponderomotive potential due to radiation reaction force. It can be assessed in dimensionless units that along with and are sufficient for stability. The momentum can be assessed as , since a trajectory is rotated in a magnetic field with consecutive increasing and decreasing of energy due to the electric field work. Therefore, a particle with zero initial momentum is captured by the antinode of the magnetic field, if . Due to the small size of this area around the antinode of the magnetic field, from which the particle will not go away, only a small fraction of the uniformly distributed particles is captured by this area. There is also an unstable point of equilibrium , however, the arbitrarily small perturbations allow the particle to reach the attractor. Thus, only a small amount of particles does not reach the found attractor. In the case without radiation reaction, the initial phase of the standing wave comes into play. For comparison of the results of the calculation of Lyapunov exponents with and without radiation reaction, the initial phase is chosen to be 0, which corresponds to a zero steady component of the transversal momentum. Without radiation reaction, the antinodes of the electric field are also unstable. The magnetic field antinodes are in indifferent equilibrium. Starting with an arbitrary position outside a small region around the magnetic field antinode, the trajectories are qualitatively similar (for a long period of time, the trajectories fill the same space of phase projection ). So, the Lyapunov exponent spectrum of a trajectory can be generalized to the others.\n\nNote that each characteristic parameter corresponds to the basis vector and along these vectors the trajectory perturbation increases or decreases as , depending on the sign of . According to numerical simulation, the same basis vector (9) corresponds to the highest nonzero Lyapunov exponent with and without radiation reaction. The vector is determined to an accuracy of 5 in the range of standing plane wave amplitudes . Disturbances along this vector decrease most slowly (the Lyapunov exponent is negative, the radiation reaction is taken into account) or increase most rapidly (the Lyapunov exponent is positive, the radiation reaction is neglected):\n\n →rb=(1ipy,0ipz,0it,0iy) (9)\n\nCompare the dynamics of electrons in the field of a standing plane wave with and without radiation reaction. We consider the wave amplitudes at which a regular attractor is possible assuming the radiation reaction to be taken into account. If the attractor is chaotic, no significant difference is observed in the motion of an electron with and without radiation reaction force. Moreover, the maximum Lyapunov exponents are approximately the same in these two cases.", null, "Figure 2: Trajectory of an electron in the field of a linearly polarized standing wave with amplitude a=136 initially placed between electric and magnetic field nodes with zero momentum after 3000T and γ and χ as a function of time without radiation losses.", null, "Figure 3: Trajectory of an electron like in Fig.2 but with classical radiation reaction force.\n\nFor numerical analysis it is convenient to consider the wave amplitude corresponding to the shortest time needed for the particle to approach the attractor; it occurs when . In this case, the largest non-zero Lyapunov exponent . Figure 2,3 shows the trajectories of the electron in the field of a plane wave with amplitude , with and without radiation reaction taken into account, respectively, after 3000 field periods. This time is sufficient for the particle motion to converge to the attractor. The particle motion along the attractor is directed, the averaged transversal momentum is zero (there is no drift in the transversal direction), and the longitudinal momentum oscillates around with an amplitude of and an averaged longitudinal velocity of . Note that the averaged value of energy is . Without radiation reaction, the electron motion is partially similar to the electron motion with radiation reaction taken into account, but there is a possibility for the electron to be trapped by a magnetic field antinode. When trapped, it stays in this state for random time, drifting along the electric field, thus making the motion chaotic. This is the reason of directed motion loss. So, if we consider the electron bunch motion in the field of a standing wave with allowance for the radiation reaction, then we will have a \"wave-like solution\"’ (the electrons are halved and move in opposite directions along the Y axis). In this case, after a while there are no electrons in the initial region. However, without radiation reaction, electron bunch spreading is observed (a \"diffusion like solution\") due to the random change of electron motion direction. Let us analyze the electron bunch motion. 1000 electrons are initially uniformly distributed within the region and . The initial momentum is zero. Figure 4-6 shows the results of calculations. The quantum parameter Ritus (1985),Nikishov (1985), i.e. the classical approach is applicable for description of the electron motion:\n\n χ=eℏm3c4√(mcγE+p×H)2−(p⋅E)2 (10)", null, "Figure 4: Elelctron energy distribution in the field of a plane wave with amplitude a=136 (a), and phase space projection on the pypz plane (b) with classical description of radiation.", null, "Figure 5: Electron energy distribution in the field of a plane wave with amplitude a=136 (a), and phase space projection on the pypz plane (b) without radiation reaction.\n\nThe energy spectrum of the electrons, with the radiation reaction taken into account, is discrete, and the spectrum line thickness is less than 0.2MeV. The number of spectrum lines is determined by the time needed for the electron to approach the attractor, as well as by the fact that the electron motion is quasiperiodic and its period equals several periods of the standing wave ( in the case considered). The average energy of the electrons along the attractor is 80MeV. The projection of the phase space portrait on the plane is a line that is the result of contraction of the phase space in Fig.4. In particular, this figure confirms that the major part of particles initially uniformly distributed within the range converge to the same attractor. In other words, for the majority of particles their initial position is of no significance, which justifies our choice of the initial conditions for estimating Lyapunov exponents. The phase portrait without radiation reaction is \"blurred\", the spectrum becomes continuous and shifted to lower energies due to a random change of electron motion direction, and the average energy decreases to 60 MeV (Fig. 5). In the absence of radiation reaction force, several particles can reach higher energies. Part of the particles have , which is impossible if an electron moves along the attractor.", null, "Figure 6: Time dependence of electron distribution along Y and Z axes for a=136 with (a) and without (b) radiation reaction force.\n\nThe electron energy spectrum reflects the specific features of particle motion. There are divergent electron beams along the characteristic curves with radiation reaction. The electron distribution along the Z axis () is determined only by the initial stage of approaching to the attractor in Fig. 6a), 7 and doesn’t change after that. Thus, the electron temperature tends to zero. Note that in , the particles escape from the area where they were initially located. Figure 7a) confirms that the position of the leading edge of the electrons corresponds to .", null, "Figure 7: Electron distribution along Y and Z axes after 2000 (red dashed line) and 5000 (black dashed line) wave periods.\n\nIf the radiative losses are neglected, the diffusive spreading of the electron distribution along the Y and Z axes is observed due to particle trapping by the antinode of the magnetic field and changing direction of motion (Fig.6b), 8). In this case, the particle motion can be described by the diffusion equation for the linear density of the particles:\n\n ∂Ne∂t=D∂2Ne∂r2, (11)\n\nwhere is implied to be or . As can be seen from Fig. 6, the diffusion along the Y axis is much faster, so it is possible to consider the diffusion in each direction independently. Since the electrons are initially within the half wavelength range and their following motion is considered during 5000 periods of the standing wave, during this time a typical scale of electron distribution becomes . Then, it is possible to use the fundamental solution, assuming the diffusion coefficient to be constant and the initial distribution to be a delta function multiplied by the number of particles (12):\n\n Ner=N0√4πD∥,⊥texp(−r24D∥,⊥t), (12)\n\nwhere is the linear electron density along the Y,Z axes. The evolution of the electron density profile along the Y and Z axes is shown in Fig. 8.", null, "Figure 8: Diffusion of electrons along Y (a) and Z (b) axes. Black and red lines correspond to time t=2000T and t=5000T. Chain line corresponds to numerical calculation of 1000 trajectories of electrons initially distributed in the λ/2 range along Y and Z axes. Solid line corresponds to solution (12)\n\nCalculations show that, if the wave amplitude , then the longitudinal diffusion coefficient , and the transverse one .\n\nComparison of the electron motion with and without radiation reaction points that the qualitative differences in the motion occur only after the particles approach the attractor, and it takes hundreds of wave periods and continuous action of dissipative force in time. However, an electron emits a portion of its energy, quanta, and each act of emission \"‘knocks\"’(or pushes) an electron, thus leading to lengthening of the time of convergence to the attractor or disappearance of the attractor at all. To understand the influence of the discreteness of photon emission on electron motion, we can estimate how often the electron emits and what portion of its energy turns into radiation. For this purpose it is necessary to estimate the parameter . If the electromagnetic field is specified by (3,4), can be written as\n\n χ=ℏωmc2a√(ρ2y+1)cos2(τ+φ0)cos2(η)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯+ρ2sin2(τ+φ0)sin2(η)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯−0.5ρyγsin(2η)sin(2(τ+φ0)) (13)\n\nRough estimation is , which demonstrates a good fit with the value of along the trajectory with and without classical radiation reaction (Fig.2,3). In the ultrarelativistic case, an average photon energy can be estimated as Landau and Lifshitz (1962), and the classic radiation power holds valid in the case . Hence, the average number of photons during the period can be assessed to be , and part of electron energy lost in one act of photon emission is . Note that, as follows from calculations, each of 1000 electrons in the wave field with amplitude during 5000 field periods emits on the average. Rough estimations are in a good agreement with numerical results. Let us determine an average number of photons emitted by an electron over the wave period according to the probability of photon emission per unit time in the quantum case Bayer et al. (1973):\n\n W=13√3πe2ℏcmc2ℏγ∞∫05u2+7u+5(1+u)3K2/3(2u/3χ)du. (14)\n\nWhen , the expression (14) reduces to . This gives an estimate of a photon number , and each act of photon emission takes away about 0.02 part of electron’s energy. Thus, wave periods are sufficient for an electron to lose a steady component of the momentum along the electric field. It may seem that the classical description can correctly describe electron motion due to a small fraction of electron energy converted to photon energy, but discreteness of emission plays an important role. On the one hand, the energy loss must lead to formation of an attractor. On the other hand, randomness leads to divergence of the particles in phase space. So, existence of an attractor is determined by the competition of these two factors. Estimations show that the particles move about 0.17 wave periods between the acts of photon emission without radiation reaction, and in this case the maximal Lyapunov exponent is . This means that the particles which are close in phase space diverge in . The time of particle convergence due to the radiation reaction may be estimated to be . The difference of these times destroys the attractor, so it is necessary to consider radiation as a quantum mechanism.\n\nThe importance of a quantum mechanism may also have a different explanation. Suppose that there is a regular attractor with allowance for quantum description of photon emission. Then, the attractor will be like in the classical case neglecting discreteness of radiation, since if , then radiation losses are the same in the classical and quantum cases. Hence, the value of the characteristic Lyapunov exponents must be the same too. In this case, as was shown earlier, the largest non-zero Lyapunov exponent is , which corresponds to the basis unit vector (9). Perturbation along this vector during the time period between the radiation events decreases as . However, taking into account the discrete nature of radiation, it follows that the particle during is affected only by the Lorentz force, and in this case the largest non-zero Lyapunov exponent , and the perturbations grow along the vector (9) as . The result is that the perturbations grow as , since , and a regular attractor cannot exist. Consequently, the discreteness of radiation plays a qualitatively important role. Note that the system of equations (5-8) when is determined by one similarity parameter . Thus, in contrast to the classical case, there is no regular attractor and suppression of stochastic heating, if and or .\n\n## Iii Electron motion taking into account discreteness of radiation\n\nLet us use the tried-and-true approach for the quantum description of radiation and classical description of electron motion Duclous et al. (2011). Electron motion is described by equations 1,2, but without radiation reaction. Each particle should propagate a randomly assigned optical length in the electromagnetic field before photon emission. If the random number , then the optical path . Once , then the procedure of photon emission is started. The choice of the photon energy is carried out with the help of the Monte Carlo Rejection sampling method Neal (2003). To do so, the auxiliary function is selected, such that where is the fraction of the electron energy converted to the photon energy:\n\n M(η)=⎧⎪⎨⎪⎩0.325η2/3,η≤0.870.18(1−η)1/3,0.87<η≤1 (15)\n C(χ)=1.6e2ℏcmc2ℏωγχ2/3. (16)\n\nHere, is calculated according to the momentum and strength of magnetic and electric fields in the constant-field approximation; is defined as follows Bayer et al. (1973):\n\n dW(χ,η)dη=1π√3e2ℏcmc2ℏωγ⎡⎢ ⎢ ⎢ ⎢⎣∞∫2η3(1−η)χK5/3(y)dy+ η21−ηK2/3(2η3(1−η)χ)]. (17)\n\nThe second random number allows determining from the equation . Using the third random number it is possible to determine whether the photon is emitted ; otherwise, it is necessary to choose a random number again. A properly chosen function is as close as possible to the function , which reduces the number of attempts to emit a photon. Also, it is easy to determine , i.e. to find the inverse function.\n\nLet’s consider a typical trajectory of an electron using the quantum description of radiation (Fig.9).", null, "Figure 9: Electron trajectory as in Fig.2 but with quantum description of photon emission.\n\nTrajectories in cases of quantum description and without radiation losses are similar, as is evident from comparison of Figs. 9, 2, but there is one qualitative difference. Without radiation reaction, there is an integral of motion , because the equation of motion does not depend on the coordinate . Therefore, the motion along the z coordinate is different for different initial phases of a standing wave, the electrons can have a steady component of momentum along the Z axis, but this component vanishes both with allowance for classical and quantum radiation losses.", null, "Figure 10: Electron energy distribution in the field of a plane wave with amplitude a=136 (a), and phase space projection on the pypz plane (b) with quantum description of photon emission.\n\nLet us now examine electron bunch motion under the initial conditions same as in the classical case. Qualitatively, the motion of the electron bunch coincides with the case without radiation friction, which confirms the conclusion of the importance of discreteness and that the motion is generally determined by the Lorentz force rather than the radiation reaction (Fig.10,11). Figure 11 shows that part of particles may have energy greater than . The spectrum has an analogous feature, when the radiation reaction is not taken into consideration. This is due to the discrete nature of radiation, because particle does not undergo radiation damping between the acts of emission. The electron bunch motion can also be described using the fundamental solution 12 (Fig.12). Calculations demonstrate that the diffusion coefficients are smaller than in the case without radiation friction, as the electron changes its direction more often. Thus, the diffusion decreases and the electron bunch \"spreads\" slower.", null, "Figure 11: Electron distribution along Y and Z axes as a function of time for a=136 using quantum description of photon emission.\n\nThe longitudinal diffusion coefficient is , and the transversal diffusion coefficient . In order to determine why the diffusion is like this, note that the motion of an electron in a plane wave with quantum description of radiative losses is chaotic. It is found that the electron retains the character of the motion each half period, and the character of motion in the next half period is determined by the present one. Such motion can be regarded to be a Markov chain.", null, "Figure 12: Diffusion of electrons along Y (a) and Z (b) axes at quantum description of photon emission. Black and red lines correspond to time t=2000T and t=5000T. Chain line corresponds to numerical calculation of 1000 trajectories of electrons distributed initially in λ range along Y and Z axes. Solid line corresponds to solution (12)\n\n## Iv Markov chain\n\nThe process described by the Markov chain is defined by a transition matrix which sets the probabilities of the states of the n step from the known state of the n-1 step. We use the formalism of Markov chains to determine the width of the electron distribution along the Y and Z axes. There are 3 states of motion along the Y axis (all the states are shown in Fig. 13). 1) A particle shifts by in the direction opposite to the Y axis, while shifting by along the Z axis. The probability of changing the direction with respect to Z without changing the motion along Y is 0.35. The probability of the transition to the second state is 0.023. 2) On the average, the particle propagating along Z does not move along Y. The probability of changing the direction of motion along the Z axis is 0.4. The probability of transition to state 3) or 1) is 0.15. State 3) is the same as state 1), but the electron moves along the Y axis. Numerical simulations allow determining the probability of the transition between the states (indicated above the arrows in Fig. 13), as well as determining the states. Assume that is the state vector of electron motion along the Y axis corresponding to the n step. and denote the probability of the motion along the Y axis and in the opposite direction with respect to the Y axis correspondingly, denotes the probability of rest along the Y axis. Then the transition matrix from state i to j is written in the form\n\n Pij=⎛⎜⎝0.9770.02300.150.70.1500.0230.977⎞⎟⎠ (18)\n\nwhich means that =. Let’s calculate the mathematical expectation and the variance of electron deviation along the Y axis in n steps . The mean value of the sum is the sum of the mean values, i.e. . Initially, the particles are distributed uniformly in states 1), 2), and 3) according to numerical simulations of electron bunch dynamics, i.e. . because of the equivalent motion in both directions along the Y axis and because of the symmetry of matrix and the vector of initial states. Thus, as evidenced by Fig. 12, on the average, motion occurs without drift of the center of mass. Let’s calculate the variance . Note that the events become weakly correlated after a sufficiently large number of steps and the state vector at the -th step . In order to find , it is necessary to solve the equation in conjunction with the condition that the sum of the values in the rows of equilibrium transition matrix (a matrix with three identical columns) is equal to 1. Because of the symmetry of motion in both directions along the Y axis, the equilibrium transition matrix has the form , , and . Hence, , , and . It took steps for the equilibrium state to be formed. The electron bunch dynamics is observed during 5000 wave periods, i.e. steps. The first term in the expression of variance can be calculated as , where is the displacement during half the wave period along the Y axis. Let’s calculate the second term. Assume that is the state vector at the i step. For finding the correlation of states i and j, it is necessary to know the state vector using the conditional probability, if we know the value of random variable . These probabilities are expressed in the form of elements of the matrix , where the column indicates which states will be at the j step. If the state k is at the i step, then . The approximation is used because , hence . Note that tends to zero for large i, since components 1 and 3 of the vector A are similar in magnitude and opposite in sign, in other words the dependence on initial conditions disappears and the motions along the Y axis are equal in both directions. Thus, for good accuracy it is sufficient to take into account a certain quantity of terms . It suffices to take into account only 500 terms in view of the fact that the equilibrium state is achieved in 500 steps. Finally, to a good accuracy . Generally, the variance is due to correlations and . Based on the root function of the number of steps (twice the number of periods) it becomes obvious that \"spreading\" of an electron bunch occurs at large n with constant dimensionless diffusion coefficient . The value of the variance is in a good agreement with the characteristic width of the electron distribution along the Y axis in Fig. 12 and with the diffusion coefficient. Figure 14 shows the distribution of electrons along the Y and Z axes derived from the Markov process according to Fig. 13 in comparison with the numerical simulation of electron motion taking into account the quantum description of radiation.", null, "Figure 14: Electron distribution over 2000 wave periods (red line) and over 5000 wave periods (black line) along Y a) and Z b) axes as a result of Markov chain process(solid line) and as a result of numerical solution of motion equation (dashed line) taking into account quantum description of radiation.\n\nSimilar analysis can be made with the electron motion along the Z axis. In this case, the state vector consists of four elements. The result is the following: and , which agrees well with the results presented in Fig. 12.\n\nIt should be noted that the consideration of the electron motion as a process described by a Markov chain allows us to understand the transition to the normal radiative trapping Lehmann and Spatschek (2012),Gonoskov et al. (2014). The state of becomes absorbing, i.e. in terms of the transition matrix we can say that , and tend to zero with increasing .\n\n## V Conclusion\n\nThe electron motion in a plane standing wave is considered within the framework the quantum and classical description of the radiation reaction. It is shown that in the case of small , the classical description of photon emission can lead to qualitatively wrong electron dynamics despite the small radiative losses. The Lyapunov characteristic exponents obtained in this case show that there is a regular attractor in the phase space, whereas without radiative losses the motion is chaotic. In the quantum case, the character of motion changes drastically due to the discrete nature of photon emission, which randomizes particle motion. This leads to the additional divergence in the phase space, which completely destroys formation of a regular attractor. A new method is proposed for describing electron kinetics in this case using Markov chains. The electron distribution obtained by this method is in a good agreement with the results of direct calculations.\n\nThe research is partly supported by the grant (the agreement of August 27, 2013 №02.В.49.21.0003 between The Ministry of Education and Science of the Russian Federation and Lobachevsky State University of NizhniNovgorod) and RFBR (grant №14-02-31495 mola).\n\n## References\n\nYou are adding the first comment!\nHow to quickly get a good reply:\n• Give credit where it’s due by listing out the positive aspects of a paper before getting into which changes should be made.\n• Be specific in your critique, and provide supporting evidence with appropriate references to substantiate general statements.\n• Your comment should inspire ideas to flow and help the author improves the paper.\n\nThe better we are at sharing our knowledge with each other, the faster we move forward.\nThe feedback must be of minimum 40 characters and the title a minimum of 5 characters", null, "", null, "", null, "" ]

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[ "# How do you solve 3x-1<2 or -5x<-20 ?\n\nMay 19, 2015\n\nWith the first inequality, first add $1$ to both sides to get:\n\n$3 x < 3$\n\nThen divide both sides by $3$ to get:\n\n$x < 1$\n\nWith the second inequality, first add $5 x$ to both sides to get:\n\n$0 < 5 x - 20$\n\nThen add $20$ to both sides to get:\n\n$20 < 5 x$\n\nFinally divide both sides by $5$ to get:\n\n$4 < x$\n\nPutting these together we have:\n\n$x < 1$ or $x > 4$\n\nNotice what I did not do. I did not divide both sides of $- 5 x < - 20$ by $- 5$. If you divide both sides by a negative number then you must also reverse the inequality, which would have given us $x > 4$ in shorter order, but might have required more explanation.\n\nIn general you can perform any of the following operations and preserve the truth of an inequality:\n\n(1) Add or subtract the same value on both sides.\n(2) Multiply or divide both sides by the same positive value.\n(3) Multiply or divide both sides by the same negative value and reverse the inequality." ]

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[ "Cody\n\n# Problem 1895. Count ones\n\nSolution 1954007\n\nSubmitted on 30 Sep 2019 by SuNe\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1   Pass\nx = 1123421; y_correct = 3; assert(isequal(your_fcn_name(x),y_correct))\n\n2   Pass\nx = 1111; y_correct = 4; assert(isequal(your_fcn_name(x),y_correct))\n\n3   Pass\nx = 345450; y_correct = 0; assert(isequal(your_fcn_name(x),y_correct))" ]

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[ "Books Books", null, "The Circumference of every circle is supposed to be divided into 360 equal parts, called Degrees ; and each degree into 60 Minutes, each minute into 60 Seconds, and so on. Hence a semicircle contains 180 degrees, and a quadrant 90 degrees. 58. The Measure...", null, "A Course of Mathematics: For the Use of Academies as Well as Private Tuition - Page 349\nby Charles Hutton - 1812", null, "## A Treatise on Mensuration, Both in Theory and Practice\n\nCharles Hutton - Measurement - 1788 - 703 pages\n...femicircle contains 1 80 degrees, and a quadrant 90 degrees. 57. The meafureofa right-lined angle, is. an arc of any circle contained between the two...which form that angle, the angular point being the center; and it is eftimated by the number of degrees contained '' ' in that arc. Hence a right-angle...", null, "## Mathematics: Compiled from the Best Authors and Intended to be the ..., Volume 1\n\nMathematics - 1801 - 426 pages\n...semicircle contains 1 80 degrees, and a quadrant 90 degrees. / 57. The measure of a rightlined angle is an arc of any circle, contained between the two....which form that angle, the angular point being the <?entrc ; and it is estimated by the nuaii bcr of degrees, contained in that arc. Hence a, right angle...", null, "## Mathematics: Compiled from the Best Authors, and Intended to be ..., Volume 1\n\nMathematics - 1808 - 470 pages\n...semicircle contains 1 80 degrees, and a quadrant 90 degrees, 57. The measure of a rightlined angle is an arc of any circle, contained between the two...degrees, contained in that arc. Hence a right angle is an angle of 90 degrees. 58. Identical figures are such, as have all the sides and all the angles...", null, "## The principles of architecture, Volume 1\n\nPeter Nicholson - 1809\n...ab, and bc, will be the two sides containing that angle. 53. The measure of any right lined angle, is an arc of any circle contained between the two lines which form the angle, the angular point being in the centre, as K. Thus . if the arc bed be double of the arc...", null, "## A Course of Mathematics ...: Composed for the Use of the Royal Military ...\n\nCharles Hutton - Mathematics - 1811\n...seinicircle contains 1 80 degrees, and a quadrant 90 degrees. 3. Th'e Measure of an angle (Def. 57, Geom.) is an arc of any circle contained between the...of the circle, is an angle of 90 degrees ; and the stim of the three angles of every triangle, or two right angles, is equal to 180 degrees. Therefore,...", null, "## A Course of Mathematics ...: Composed for the Use of the Royal Military ...\n\nCharles Hutton - Mathematics - 1811\n...on. Hence a semicircle contains 1 80 degrees, and a quadrant 90 degrees. 58. The Measure of an angle, is an arc of any circle contained between the two lines which form that angle, the an gi ilar point being the centre; and it is estimated by the number of degrees contained in that arc....", null, "## Encyclopaedia Perthensis; or, Universal dictionary of Knowledge ..., Volume 10\n\nEncyclopaedia Perthensis - 1816\n...a femicircle contains 180 degres, and a quadrant 90 degrees. 41. The MEASURE of a RECTILINEAL ANGLE is an arc of any circle contained between the two...angle, the angular point being the centre, and it is eftimated by the number of degrees in that arc. Fig. 15. 4t. IDENTICAL FIGURES are fuch as liavc all...", null, "## Encyclopaedia Perthensis; Or Universal Dictionary of the Arts ..., Volume 10\n\n...a femicircle contains 180 degres, and a quadrant 90 degrees. 41. fhf MEASURE of a RECTILINEAL ANGLS is an arc of any circle contained between the two...angle, the angular point being the centre, and it is eftimated by the number of degrees in that arc. Fig. 25. 42. IDENTICAL FIGURES are fuch as have all...", null, "## A Course of Mathematics: For the Use of Academies, as Well as Private ...\n\nCharles Hutton - Mathematics - 1822 - 618 pages\n...each degree into 60 Minutes, each minute into 60 Seconds, and so on. Hence a semicircle contains 100 degrees, and a quadrant 90 degrees. 3. The measure...centre ; and it is estimated by the number of degrees continued in that arc. Hence, a right angle, being measured by a quadrant, or quarter of the circle,...", null, "" ]

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[ "## Solutions To Text Problems\n\nA market is a group of buyers (who determine demand) and a group of sellers (who determine supply) of a particular good or service. A perfectly competitive market is one in which there are many buyers and many sellers of an identical product so that each has a negligible impact on the market price. 2. Here is an example of a monthly demand schedule for pizza The demand curve is graphed in Figure 1. The demand curve is graphed in Figure 1. Number of Pizza Slices Demanded Figure 1 Number of Pizza...\n\n## Figure 3\n\nWhen the Mexican orange market is opened to trade, the new equilibrium price is PW, the quantity consumed is Q the quantity produced domestically is Q, and the quantity imported is Qd - Q. Consumer surplus increases from A to A B D E. Producer surplus decreases from B C to C. Total surplus changes from A B C to A B C D E, an increase of D E. 9. a. Figure 10 shows the market for grain in an exporting country. The world price is Pw. b. An export tax will reduce the effective world price...\n\n## Figure 5\n\nCosts are shown in the following table 7. a. Costs are shown in the following table b. If the price is 50, the firm will minimize its loss by producing 4 units. This would give the firm a loss of 40. If the firm shuts down, it will earn a loss equal to its fixed cost 100 . c. If the firm produces 1 unit, its loss will still be 100. However, because the marginal costs of the second and third unit are lower than the price, the firm could reduce its loss by producing more units. 8. a. The...\n\n## Problems and Applications\n\nFigure 9 shows the effects of the minimum wage. In the absence of the minimum wage, the market wage would be wi and Q1 workers would be employed. With the minimum wage wm imposed above wu the market wage is wm, the number of employed workers is Q, and the number of workers who are unemployed is Q3 - Q. Total wage payments to workers are shown as the area of rectangle ABCD, which equals wm times Q. b. An increase in the minimum wage would decrease employment. The size of the effect on...\n\n## At different prices to different customers\n\nExample Readalot Publishing Company 2. The firm pays an author 2 million for the right to publish a book. Assume that the cost of printing the book is zero. 3. The firm knows that there are two types of readers. a. There are 100,000 die-hard fans living in Australia of the author willing to pay up to 30 for the book. b. There are 400,000 other readers living in the United States who are willing to pay up to 5 for the book. 4. How should the firm set its price a. If the firm sets its price..." ]

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[ "# #042708 Color Information\n\nIn a RGB color space, hex #042708 is composed of 1.6% red, 15.3% green and 3.1% blue. Whereas in a CMYK color space, it is composed of 89.7% cyan, 0% magenta, 79.5% yellow and 84.7% black. It has a hue angle of 126.9 degrees, a saturation of 81.4% and a lightness of 8.4%. #042708 color hex could be obtained by blending #084e10 with #000000. Closest websafe color is: #003300.\n\n• R 2\n• G 15\n• B 3\nRGB color chart\n• C 90\n• M 0\n• Y 79\n• K 85\nCMYK color chart\n\n#042708 color description : Very dark (mostly black) lime green.\n\n# #042708 Color Conversion\n\nThe hexadecimal color #042708 has RGB values of R:4, G:39, B:8 and CMYK values of C:0.9, M:0, Y:0.79, K:0.85. Its decimal value is 272136.\n\nHex triplet RGB Decimal 042708 `#042708` 4, 39, 8 `rgb(4,39,8)` 1.6, 15.3, 3.1 `rgb(1.6%,15.3%,3.1%)` 90, 0, 79, 85 126.9°, 81.4, 8.4 `hsl(126.9,81.4%,8.4%)` 126.9°, 89.7, 15.3 003300 `#003300`\nCIE-LAB 12.572, -20.624, 14.882 0.819, 1.494, 0.475 0.294, 0.536, 1.494 12.572, 25.433, 144.187 12.572, -10.611, 12.594 12.224, -9.428, 6.253 00000100, 00100111, 00001000\n\n# Color Schemes with #042708\n\n• #042708\n``#042708` `rgb(4,39,8)``\n• #270423\n``#270423` `rgb(39,4,35)``\nComplementary Color\n• #122704\n``#122704` `rgb(18,39,4)``\n• #042708\n``#042708` `rgb(4,39,8)``\n• #04271a\n``#04271a` `rgb(4,39,26)``\nAnalogous Color\n• #270412\n``#270412` `rgb(39,4,18)``\n• #042708\n``#042708` `rgb(4,39,8)``\n• #1a0427\n``#1a0427` `rgb(26,4,39)``\nSplit Complementary Color\n• #270804\n``#270804` `rgb(39,8,4)``\n• #042708\n``#042708` `rgb(4,39,8)``\n• #080427\n``#080427` `rgb(8,4,39)``\n• #232704\n``#232704` `rgb(35,39,4)``\n• #042708\n``#042708` `rgb(4,39,8)``\n• #080427\n``#080427` `rgb(8,4,39)``\n• #270423\n``#270423` `rgb(39,4,35)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #021003\n``#021003` `rgb(2,16,3)``\n• #042708\n``#042708` `rgb(4,39,8)``\n• #063e0d\n``#063e0d` `rgb(6,62,13)``\n• #095511\n``#095511` `rgb(9,85,17)``\n• #0b6c16\n``#0b6c16` `rgb(11,108,22)``\nMonochromatic Color\n\n# Alternatives to #042708\n\nBelow, you can see some colors close to #042708. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #092704\n``#092704` `rgb(9,39,4)``\n• #062704\n``#062704` `rgb(6,39,4)``\n• #042705\n``#042705` `rgb(4,39,5)``\n• #042708\n``#042708` `rgb(4,39,8)``\n• #04270b\n``#04270b` `rgb(4,39,11)``\n• #04270e\n``#04270e` `rgb(4,39,14)``\n• #042711\n``#042711` `rgb(4,39,17)``\nSimilar Colors\n\n# #042708 Preview\n\nThis text has a font color of #042708.\n\n``<span style=\"color:#042708;\">Text here</span>``\n#042708 background color\n\nThis paragraph has a background color of #042708.\n\n``<p style=\"background-color:#042708;\">Content here</p>``\n#042708 border color\n\nThis element has a border color of #042708.\n\n``<div style=\"border:1px solid #042708;\">Content here</div>``\nCSS codes\n``.text {color:#042708;}``\n``.background {background-color:#042708;}``\n``.border {border:1px solid #042708;}``\n\n# Shades and Tints of #042708\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #000301 is the darkest color, while #f1fef2 is the lightest one.\n\n• #000301\n``#000301` `rgb(0,3,1)``\n• #021504\n``#021504` `rgb(2,21,4)``\n• #042708\n``#042708` `rgb(4,39,8)``\n• #06390c\n``#06390c` `rgb(6,57,12)``\n• #084b0f\n``#084b0f` `rgb(8,75,15)``\n• #095c13\n``#095c13` `rgb(9,92,19)``\n• #0b6e17\n``#0b6e17` `rgb(11,110,23)``\n• #0d801a\n``#0d801a` `rgb(13,128,26)``\n• #0f921e\n``#0f921e` `rgb(15,146,30)``\n• #11a422\n``#11a422` `rgb(17,164,34)``\n• #13b525\n``#13b525` `rgb(19,181,37)``\n• #14c729\n``#14c729` `rgb(20,199,41)``\n• #16d92c\n``#16d92c` `rgb(22,217,44)``\n• #1be833\n``#1be833` `rgb(27,232,51)``\n• #2de942\n``#2de942` `rgb(45,233,66)``\n• #3feb52\n``#3feb52` `rgb(63,235,82)``\n• #51ed62\n``#51ed62` `rgb(81,237,98)``\n• #62ef72\n``#62ef72` `rgb(98,239,114)``\n• #74f182\n``#74f182` `rgb(116,241,130)``\n• #86f392\n``#86f392` `rgb(134,243,146)``\n• #98f4a2\n``#98f4a2` `rgb(152,244,162)``\n• #a9f6b2\n``#a9f6b2` `rgb(169,246,178)``\n• #bbf8c2\n``#bbf8c2` `rgb(187,248,194)``\n``#cdfad2` `rgb(205,250,210)``\n• #dffce2\n``#dffce2` `rgb(223,252,226)``\n• #f1fef2\n``#f1fef2` `rgb(241,254,242)``\nTint Color Variation\n\n# Tones of #042708\n\nA tone is produced by adding gray to any pure hue. In this case, #151615 is the less saturated color, while #012a05 is the most saturated one.\n\n• #151615\n``#151615` `rgb(21,22,21)``\n• #131813\n``#131813` `rgb(19,24,19)``\n• #111a12\n``#111a12` `rgb(17,26,18)``\n• #101b11\n``#101b11` `rgb(16,27,17)``\n• #0e1d10\n``#0e1d10` `rgb(14,29,16)``\n• #0c1f0e\n``#0c1f0e` `rgb(12,31,14)``\n• #0b200d\n``#0b200d` `rgb(11,32,13)``\n• #09220c\n``#09220c` `rgb(9,34,12)``\n• #07240b\n``#07240b` `rgb(7,36,11)``\n• #062509\n``#062509` `rgb(6,37,9)``\n• #042708\n``#042708` `rgb(4,39,8)``\n• #022907\n``#022907` `rgb(2,41,7)``\n• #012a05\n``#012a05` `rgb(1,42,5)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #042708 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

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[ "Publication\nESORICS 2021\nConference paper\n\n# More Efficient Amortization of Exact Zero-Knowledge Proofs for LWE\n\n## Abstract\n\nWe propose a practical zero-knowledge proof system for proving knowledge of short solutions s, e to linear relations A s + e= u mod q which gives the most efficient solution for two naturally-occurring classes of problems. The first is when A is very \"tall\", which corresponds to a large number of LWE instances that use the same secret s. In this case, we show that the proof size is independent of the height of the matrix (and thus the length of the error vector e) and rather only linearly depends on the length of s. The second case is when A is of the form A' tensor I, which corresponds to proving many LWE instances (with different secrets) that use the same samples A'. The length of this second proof is square root in the length of s, which corresponds to a square root of the length of all the secrets. Our constructions combine recent advances in \"purely\" lattice-based zero-knowledge proofs with the Reed-Solomon proximity testing ideas present in some generic zero-knowledge proof systems -- with the main difference that the latter are applied directly to lattice instances without going through intermediate problems.\n\n04 Oct 2021\n\nESORICS 2021" ]

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[ "# On the Nambu fermion-boson relations for superfluid He 3\n\nJ. A. Sauls, Takeshi Mizushima\n\nResearch output: Contribution to journalArticle\n\n10 Scopus citations\n\n## Abstract\n\nSuperfluid He3 is a spin-triplet (S=1), p-wave (L=1) BCS condensate of Cooper pairs with total angular momentum J=0 in the ground state. In addition to the breaking of U(1) gauge symmetry, separate spin or orbital rotation symmetry is broken to the maximal subgroup SO(3)S×SO(3)L→SO(3)J. The fermions acquire mass mF≡Δ, where Δ is the BCS gap. There are also 18 bosonic excitations: 4 Nambu-Goldstone modes and 14 massive amplitude Higgs modes. The bosonic modes are labeled by the total angular momentum J{0,1,2}, and parity under particle-hole symmetry c=±1. For each pair of angular momentum quantum numbers J,Jz, there are two bosonic partners with c=±1. Based on this spectrum, Nambu proposed a sum rule connecting the fermion and boson masses for BCS-type theories, which for He3-B is MJ,+2+MJ,-2=4mF2 for each family of bosonic modes labeled by J, where MJ,c is the mass of the bosonic mode with quantum numbers (J,c). The Nambu sum rule (NSR) has recently been discussed in the context of Nambu-Jona-Lasinio models for physics beyond the standard model to speculate on possible partners to the recently discovered Higgs boson at higher energies. Here, we point out that the Nambu fermion-boson mass relations are not exact. Corrections to the bosonic masses from (i) leading-order strong-coupling corrections to BCS theory, and (ii) polarization of the parent fermionic vacuum lead to violations of the sum rule. Results for these mass corrections are given in both the T→0 and T→Tc limits. We also discuss experimental results, and theoretical analysis, for the masses of the Jc=2± Higgs modes and the magnitude of the violation of the NSR.\n\nOriginal language English (US) 094515 Physical Review B 95 9 https://doi.org/10.1103/PhysRevB.95.094515 Published - Mar 22 2017\n\n## ASJC Scopus subject areas\n\n• Electronic, Optical and Magnetic Materials\n• Condensed Matter Physics" ]

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[ "# How do you read the symbol \"$\\in$\"?\n\nA variable in an equation may be replaced by any of the numbers in its domain. The resulting equation may be either true or false.\n\nHere is another way to show that the domain of a variable $y$ is $\\lbrace$$0, 1, 2, 3$$\\rbrace$:\n\n$$y\\in\\lbrace 0, 1, 2, 3\\rbrace$$\n\n(Read $\"y$ $\\color\\red{\\text{belongs to}}$ the set whose members are $0, 1, 2, 3\"$.)\n\nReplacing each variable in an an open sentence by each of the values in its domain is a way to find solutions of the open sentence.\n\nQuestion: To me, the red typed words: $\"\\color\\red{\\text{belongs to}}\"$ says the variable belongs to the set; while the the other two statements, which use the possessive $\\text{\"its domain\"}$, says the set belongs to the variable. Is there a non-contradictory interpretation that I am missing?\n\nWould \"is a member of\" be a such a non-contradictory interpretation and reading of the symbol? Then the statement would read: \"$y$ is a member of the set whose members are $0, 1, 2, 3$\". Which seems kind of wordy.\n\n• \"Its domain\" relates to the \"open sentence\", not to the set. A set does not have a domain, a domain is a set.\n– user65203\nSep 18, 2014 at 7:48\n• no \"its domain\" relates to the variable, the domain is the set of values that the variable can take. Sep 18, 2014 at 9:36\n• its about as contradictory as the statement \"I am a $member$ of $my$ family.\" Sep 18, 2014 at 13:39\n\nThe relation $\\in$ is typed \\in (in TeX) and often pronounced \"is in\", or \"is a member of\", or \"is an element of\". But many variations occur occuring to the particular set, so you might pronounce $z\\in\\mathbf C$ as \"$z$ is a complex number\" rather than \"$z$ is an element of the set of complex numbers\".\n\nYour confusion is due to the two different roles of a variable. A variable is a name, a lexical object introduced in a mathematical text, usually specifying what kind of value it designates by specifying a set of allowed values, its domain. The domain is an attribute of the name of the variable. However, when reasoning about the variable, one assumes that it designates one specific element of its domain, the (current) value of the variable. When talking about $y$ in this way, one always means the value of (the name) $y$, not the name $y$ itself. It is the value of $y$ that is an element of (is a member of, is in) the domain of (the name) $y$.\n\n\"You belong to me\", \"I belong to you\"... Possession always causes confusion.\n\nA set, by definition, is a collection of elements. A given element $x$ can either be in a given collection, or not in the collection. I read \"$y\\in\\{0,1,2,3\\}$\" as \"$y$ is one of the elements $0,1,2,3$\". No belonging involved.\n\nAs for \"$\\in$\" see the other answers. Concerning the idea of a variable:\n\nEach letter, say $y$, denoting a variable comes a priori with its domain $D_y$, a certain set. We are allowed to replace this $y$ in the formula by any element $a\\in D_y$ and obtain a proposition about constants which is either true or false.\n\nIs the 'variable' in 'let $y=f(x)$' free, bound, or neither?\nThe most general ways to \"pronounce\" $\\in$ certainly are \"is an element of\" or \"is a member of\". However, in a case like this one where the set is not only finite but also very small it might make sense to read \"$y \\in \\{1,2,3\\}$\" as \"$y$ is either $1$ or $2$ or $3$\" or \"$y$ is one of the values $1$, $2$, and $3$\".\nThis is in accordance with, say, reading \"$y \\in \\mathbb{N}$\" as \"$y$ is a natural number\"." ]

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[ "# Numerical optimization and analysis on vibration characteristics of bicycles based on the novel CA-PSO algorithm\n\n## Jin Zhang1, Jiang Hua Gao2, Jian Guo Wu3\n\n1Department of Physical Education, Changchun Finance College, Changchun 130022, China\n\n2Student Affairs Office, Rizhao Polytechnic, Rizhao 276826, China\n\n3School of Mathematics and Statistics, Hunan University of Finance and Economics, Changsha 410205, China\n\n3Corresponding author\n\nJournal of Vibroengineering, Vol. 19, Issue 6, 2017, p. 4018-4032. https://doi.org/10.21595/jve.2017.17932\nReceived 3 November 2016; received in revised form 28 July 2017; accepted 31 July 2017; published 30 September 2017\n\nCopyright © 2017 JVE International Ltd. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.\nViews 149\nCrossRef Citations 0\nAbstract.\n\nRegarding the published researches on bicycles, they fail to design fatigue characteristics of the bicycle. In addition, dynamics and fatigue characteristics are not further improved by using advanced optimization algorithms. Aiming at these questions, this paper tries to optimize dynamics and fatigue characteristics of the bicycle through combining finite element model with advanced algorithms. The advanced algorithm applies ideas of cellular automation (CA) to Particle Swarm Optimization (PSO), and then a hybrid CA-PSO algorithm is proposed. Moreover, the finite element model is also validated by experimental test. Computational results show that: the maximum stress of bicycles is mainly distributed on the frame, especially on joints of different round pipes at different moments mainly because a dead corner is at the joint, and the dead corner can easily cause stress concentration. Under alternating forces, the stress concentration at joints will cause fatigue damage. Therefore, the service life of this position will be the shortest. As a result, the dynamics and fatigue characteristics of the joint position are taken as the optimized objective. In order to verify the optimized effectiveness of the proposed CA-PSO algorithm in the paper, the widely used PSO algorithm and PSO-GA algorithm are also used to optimize the bicycle. When the traditional PSO algorithm is used to optimize the bicycle, the root-mean-square value and maximum difference of vibration accelerations are decreased by 11.9 % and 14.3 %. When the PSO-GA algorithm is used to optimize the bicycle, the root-mean-square value and maximum difference of vibration accelerations are decreased by 20.3 % and 12.9 %. When the proposed CA-PSO algorithm is used to optimize the bicycle, the root-mean-square value and maximum difference of vibration accelerations are decreased by 27.1 % and 18.6 %. Compared with other two kinds of PSO algorithms, optimized effects of vibration accelerations are very obvious. In addition, the fatigue life of the original structure is 5 years, while the fatigue life of the optimized bicycle is 7 years. Therefore, the fatigue life is improved obviously.\n\nKeywords: dynamics, fatigue life, bicycles, vibration accelerations, root-mean-square value, maximum difference, PSO algorithm, PSO-GA algorithm, CA-PSO algorithm.\n\n#### 1. Introduction\n\nWith the improvement of people’s material culture life level, bicycles are not only a sports activity instrument, but also a traffic tool [1-6]. A fitness bicycle can simulate force and work of riding in the outdoor, so as to achieve fitness effects similar to highway bicycles. An interactive fitness bicycle integrates multiple technologies such as computer technology, sensing technology and ergonomics, managing to provide riders with omni-sensory stimulation and simulate bicycle riding more vividly. How to provide immersive experience of real scenes during the riding is not only embodied in vision, but more importantly, it aims to achieve consistency between riding loads between the fitness bicycle and field riding. Highway bicycle is an endurance project with physical energy orientation. The riding is characterized by single-cycle continuity, long duration, large riding strength and high strength in sprinting. Because of these characteristics, corresponding requirements are proposed for power output of a rider at different stages. On the other hand, close combination of physical energy consumption and technology is the key to win a match. Riding of the highway bicycle has complicated dynamic principles. While trampling the pedals to drive a bicycle, a rider shall overcome many types of resistance caused by the bicycle body, ground and air. Real-time changes in factors such as acceleration, path and wind direction during riding will comprehensively affect the force and work of the rider. During riding, a bicycle bears large alternating force for a long time. Therefore, it is necessary to study its force bearing process [7-12].\n\nAt present, some scholars have made a lot of researches to analyze forces of bicycles. Based on the finite element method, Du conducted modal analysis on a bicycle frame and conducted experimental verification through experimental test, obtaining vibration information of the frame on the top 8 orders. Based on dynamic analysis, Cai established a dynamic mathematic model of highway bicycles, and analyzed characteristics and rules under different riding states through model solution. Chen established a bicycle virtual sample machine, where the frame was processed as a flexible body to make the simulation approach actual situations, and safety of the bicycle under dangerous impact loads was analyzed. Feng designed and analyzed frame materials and pipe thickness using ergonomics and mechanics of materials, so as to make the bicycle achieve function transformation and satisfy dynamic requirements. During bicycle riding, jolt and vibration would be caused easily; frequent braking will generate instant impact effects. When the instant impact force is accumulated to certain times, weak links of the frame will suffer from fatigue damage if the frame does not have sufficient strength. Therefore, above researches failed to design fatigue characteristics of the bicycle. In addition, dynamics and fatigue characteristics were not further improved by optimization algorithms.\n\nAiming at these problems, the paper tries to optimize dynamics and fatigue characteristics of the bicycle through combining finite element model and advanced algorithms. The paper applies ideas of cellular automation (CA) [17-20] to Particle Swarm Optimization (PSO) [21-24], and proposes a hybrid CA-PSO algorithm to study the communication structure, information transmission and inheritance mechanism of populations in the PSO algorithm. Particle swarm is deemed as a CA model, where each particle only conducts information communication with neighbors determined by a neighbor function. In this way, information can be propagated slowly in the population; population diversity can be kept; search space can be explored; and local information of each particle can be fully excavated. Very obvious effects can be obtained when the algorithm is used to optimize dynamics and fatigue characteristics of the bicycle.\n\n#### 2. Finite element model of bicycles\n\nSOLIDWORKS software was used to establish a CAD model according to dimensions provided by a bicycle design drawing. In order to ensure mesh division effects in subsequent work, some chamfers and unimportant structures were neglected during modeling. Finally, a geometric model of the bicycle was obtained, as shown in Fig. 1. The bicycle is made of a frame system, a transmission system and a wheel system. The bicycle frame system is divided into a frame part and a front fork part. The frame is divided into a front triangle part and a rear triangle part. The front triangle is made of an upper pipe, a lower pipe and a seat pipe. The rear triangle is made of a seat stay and a chain stay. The transmission system is made of pedals, a chain wheel, a chain, a flywheel, a transmission, etc. The wheel system is mainly composed of tires and tire beads.\n\nSolid elements were used to conduct mesh division of the bicycle. During mesh division, relations between division accuracy and computational scale should be weighed, and the two factors must be considered comprehensively during determining the mesh quantity. During selecting basic elements of the finite element model, the addressed aspects must be taken into account: (1) The bicycle frame is basically composed of steel pipes with various cross section shapes, while the steel pipes have small cross section size, wall thickness and length. Therefore, these structures could be dispersed by shell elements. Hexahedral solid elements have very high interpolation accuracy, but mesh division will be difficult and computational time will increase if they are used to disperse frame pipes. The computational model based on a shell structure will have a larger pre-processing workload and longer computation time than a rod system structure, but it can accurately simulate actual situations of local parts such as joints and holes. Therefore, high-accuracy stress distribution could be obtained, and stress concentration at local positions could be computed as well. This could not be achieved by the rod system structure. The bicycle frame is made of Q235 with elasticity modulus of 210 GPa, density of 7800 kg/m3 and Poisson’s ratio of 0.3. It is an isotropous linear elastic material. As for the tire rubber material, density is 1.5 kg/m3, elasticity modulus is 7.8 GPa and Poisson’s ratio of 0.47. The bicycle finite element model contains 56902 elements and 69026 nodes. Fig. 2(a) shows a complete finite element model of the bicycle. Fig. 2(b) shows a finite element model of some parts and components of the bicycle.\n\nFig. 1. Simplified geometric model of bicycles", null, "Fig. 2. Finite element model of complete bicycles, parts and components", null, "a) Finite element model of the complete bicycle", null, "b) Finite element model of parts and components\n\n#### 3.1. Numerical computation of vibration characteristics of bicycles\n\nA lot of practices show that damage forms of a material or structure under alternating forces are very different from those under static forces. Under the alternating force, even if the maximum stress of the material or structure is lower than the yield limit, the structure will suffer from sudden fractures at a moment under long-term repeated effects. Even a material with high plasticity does not have obvious macroscopic plastic deformation before the fracture. Therefore, it is necessary to apply real alternating forces as the excitation during studying dynamics and fatigue characteristics of bicycles. With the published references concerning excitation loads as the reference, excitation loads at the bicycle saddle were obtained, as shown in Fig. 3. It is shown in the figure that the excitation loads presented weak periodic features, wherein the maximum absolute value was 31.2 m/s2 and the corresponding times was 8 s. The minimum value was 2.1 m/s2 and corresponding time was 1.7 s.\n\nFig. 3. Excitation loads of bicycles", null, "Excitation forces are uniformly applied to the bicycle saddle, and then dynamic responses are computed. Strain distribution contours at different moments are extracted, as shown in Fig. 4. It is shown in the figure that the bicycle saddle suffers from the most serious deformation at different moments as the excitation force is applied at the position. Secondly, the bicycle frame suffers from obvious deformation as the frame is mainly composed of round steel pipes with low stiffness. The bicycle handlebar suffers from the minimum deformation at all the moments because this position is far from the excitation force and has a high stiffness.\n\nStress contours at different moments are also extracted, as shown in Fig. 5. It is shown in the figure that the maximum stress of bicycles is mainly distributed on the frame, especially on joints of different round pipes of the frame at different moments. The bicycle frame is composed through welding of hollow round steel pipes with relatively low stiffness. Maximum stress is generated at joints of different round pipes mainly because that a dead corner is at the joint, could easily cause stress concentration and belongs to a dangerous region. Forces are uniform at the bicycle handlebar and tires and do not have obvious stress concentration phenomena. Under alternating forces, the stress concentration at joints will generate fatigue damage.\n\nDue to material non-uniformity or geometrical shape of the structure, the structure suffers from very high stress in some regions while bearing external loads. Long-term alternating stress effects amplified the local stress value and gradually formed microscopic cracks in the local regions with high stress. Tips of microscopic cracks suffer from serious stress concentration, so the cracks begin expanding gradually and finally form macroscopic cracks. Joints of different round pipes are stress risky regions as the different round pipes are generally connected by welding. During welding, continuous welding joints are adopted along the rims, so that strength of the cross section is weakened. Fig. 6 and Fig. 7 represent fatigue damage and life of the bicycle respectively. It is shown in Fig. 6 that the round pipe joints suffer from the most serious fatigue damage. Therefore, the service life of this position is the shortest according to the fatigue life distribution in Fig. 7.\n\nFig. 4. Strain contours of bicycles at different moments", null, "a) Time = 1 s", null, "b) Time = 3 s", null, "c) Time = 5 s", null, "d) Time = 7 s\n\nFig. 5. Stress contours of bicycles at different moments", null, "a) Time = 1 s", null, "b) Time = 3 s", null, "c) Time = 5 s", null, "d) Time = 7 s\n\n#### 3.2. Experimental verification of the computational model of bicycles\n\nDynamics and fatigue characteristics of the bicycle are analyzed by the computational model. However, the computational model has very complicated boundary conditions and mesh division. Therefore, the correctness of the computational model must be verified. Before experimental verification, vibration acceleration curves of corresponding positions of the bicycle must be extracted for comparison with experimental data. Fig. 8 shows vibration acceleration values at different positions. It is shown in Fig. 8 that the vibration response curves of the bicycle fluctuated very obviously; the maximum vibration acceleration responses are at the round pipe joints; the minimum vibration acceleration is at the bicycle handle; these distribution results are consistent with rules of stress and strain. In addition, it could be found that the vibration acceleration on the seat stay round pipe presents certain periodicity. At the joints, the maximum vibration acceleration is 8.5 m/s2, and the minimum vibration acceleration is –65 m/s2. Obviously, the maximum difference of vibration accelerations is 73.5 m/s2. Such obvious difference will seriously affect safety performance of the bicycle.\n\nFig. 6. Fatigue damage distribution of bicycles", null, "a)", null, "b)\n\nFig. 7. Fatigue life distribution of bicycles", null, "a)", null, "b)\n\nFig. 9 shows experimental test on the bicycle vibration accelerations. The experiment aims to measure acceleration responses of the handlebar, seat stay and joint position during bicycle riding on the pavement. Experimental results are compared with those of numerical simulation, so the correctness and validity of the established simulation model could be verified. In the experiment, an LMS signal collection system and an ICP acceleration sensor are adopted. During the experiment, the tested sample bicycle is ridden on the pavement, as shown in Fig. 10(a). Sensors are set on the handlebar, seat stay and joint position, as shown in Fig. 10(b). Meanwhile, a small automobile equipped with testing instruments and equipment traveled on the flat pavement side by side with the same speed. In this way, the riding state of the tested bicycle could be ensured, and measurement errors caused by pavement jolt to instruments and equipment used in experimental testing could be avoided. In this way, the key problem in the testing experiment of bicycle riding could be solved.\n\nFig. 8. Vibration acceleration of different positions of bicycles", null, "a) Handlebar", null, "b) Seat stay", null, "c) Joint position\n\nFig. 9. Experimental test of vibration accelerations of bikes", null, "Signals collected by sensors are input into software in the computer. The tested data is processed using the method [26, 27], so vibration accelerations of the bicycle handlebar, seat stay and joint position are obtained. The vibration accelerations are compared with numerical simulation results, as shown in Fig. 11. It is shown in Fig. 11 that vibration accelerations basically had the consistent changing trends between the experimental test and the numerical simulation while the accelerations did not differ a lot. Therefore, the numerical computation model established in the paper is feasible.\n\nFig. 10. Equipment and environment of experimental test", null, "a) Test environment", null, "b) Test position\n\nFig. 11. Comparison of vibration accelerations between experiment and simulation", null, "a) Handlebar", null, "b) Seat stay", null, "c) Joint position\n\n#### 4. Numerical optimization of vibration characteristics of bicycles\n\nIt is shown that large stress concentration and fatigue damage will easily appear at round pipe joints of the bicycle. The maximum vibration acceleration response also appears at this position. Therefore, the position is taken as the optimization object. Maximum difference and root-mean-square value of the vibration acceleration response are taken as optimization object. Mass is taken as a constraint. Thickness of the round steel pipe and the welding region area of the joint position are taken as design variables. Among many intelligent optimization algorithms, PSO not only has memorability, but also has the capabilities of communication, response, cooperation and self-learning. Therefore, it has excellent local search ability. Compared with other algorithms, it has fewer parameter settings, could be achieved easily and could converge more quickly under most cases. However, the PSO algorithm could easily fall into local extreme values during optimization. CA and PSO algorithms come from different fields, so they are essentially different. CA is a discrete computation model used for computation. PSO algorithm is an optimization tool applied to the optimization field. However, through comparison of CA model and PSO algorithm, many similar characteristics of them could be found. At first, both of them are composed of multi-individual sets. Individuals in CA are called as cells. Individuals in the PSO algorithm are called as particles. Each individual has inherent characteristics which could distinguish it from other individuals. Each cell in CA has a cell state respectively. In PSO algorithm, information of each particle, such as speed, position, fitness, individual optimal position and global optimal position, is taken as the particle characteristic. Current state of each cell is changed through information communications with its neighbors. Similarly, each particle communicates with other particles in the population to update its inherent characteristics. In CA, transfer rules are generally used to guide evolution and update of cells. In the PSO algorithm, particles are updated according to a speed formula and position formula. In addition, both of them are running in discrete time dimensions. Therefore, the paper applies CA ideas to the PSO algorithm to study the communication structure, and information transmission and inheritance mechanism of populations in the PSO algorithm. Particle swarm is deemed as a CA model. Each particle only conducts information communications with neighbors which are determined by a neighbor function. In this way, information is propagated slowly in the population; population diversity could be kept; search space could be explored; and local information of each particle could be fully excavated.\n\nDuring defining CA, cells are fixed at each discrete time point. The cell state at next moment depends on its own state and impacts of neighbor cell states. In one-dimensional cellular automation, radius is generally used to determine neighbors, namely all the cells within the radius are deemed as neighbors. Neighbor definition of two-dimensional cellular automation is more complicated than one-dimensional cellular automation. Four forms are shown in Fig. 12. In general, Moore is selected as a cell neighbor definition form in the CA model.\n\nThe hybrid cellular automation-particle swarm optimization algorithm (CA-PSO) is based on characteristics of a particle swarm. At first, particles are generated randomly. Then, each particle is placed in a two-dimensional mesh topology structure disorderly, as shown in Fig. 13(a). Each particle is deemed as a cell. The same neighbor amount is assigned to each cell, and the cell amount is equal to the population size. As shown in Fig. 13(b), a Moore neighbor structure is used and each cell has 8 neighbor individuals, where the grey circle represents a cell individual, the blue circle represents a center cell individual, and imaginary line frames represent its neighbor individuals. Orange individuals are overlaying neighbors between two neighbors. In fact, overlaying of the neighbor individuals provides a recessive migration mechanism for the algorithm. The mechanism can make the optimal individual get diffused gently in the whole population, so that selection pressure of the algorithm could be reduced, and diversity of the algorithm could be improved.\n\nFlow diagram of Hybrid CA-PSO algorithm is shown in Fig. 14, and it is mainly realized by the following steps: (1) random initialization is conducted on a particle swarm; particles are distributed in a two-dimensional mesh structure to generate corresponding neighbors, target vectors of each particle are computed, and non-inferior solutions are stored in an external document. (2) Fitness values of particles are computed according to their positions. (3) Extreme values of particles are initialized, and the global population extreme values are solved. Fitness values computed at the previous step is used to initialize individual extreme values of particles. Individual extreme values of each particle are compared, and the global population extreme values are solved. (4) Positions and speeds of the particles are updated. (5) All the particles are evaluated; through the previous step, all the particle states are updated, so fitness values under new states shall be computed. (6) Each particle neighbor is evaluated respectively, and the fitness value of each particle neighbor is computed. (7) Fitness values of particles and their neighbors are compared; if the particle has neighbors with better fitness, the neighbor state shall be used to replace the state of next moment; otherwise, the current particle state will be kept unchanged. (8) Extreme values of particle individuals are updated. Current fitness values and individual extremes values of the particle are compared. If the current fitness values are better than individual extreme values, then the individual extreme values will be updated. (9) The global population extreme values are updated. The current global extreme values are compared with individual extreme values of each particle. The global extreme values will be updated. (10) Whether the termination condition is satisfied. If the condition is not satisfied, steps (4)-(9) will be executed repeatedly. Otherwise, the circulation will be ended, and the optimal solution will be output.\n\nFig. 12. Neighbor definition form of two-dimensional cellular automation", null, "a) Von Neumann", null, "b) Moore", null, "c) Extended Moore", null, "d) Margolus\n\nFig. 13. Schematic diagram of population and neighbor structure", null, "a) Network topology structure", null, "b) Cellular neighbor definition\n\nFig. 14. Flow diagram of the proposed CA-PSO algorithm", null, "In order to verify the effectiveness of the CA-PSO algorithm proposed by the paper, the widely used PSO algorithm and PSO-GA algorithm are also used to optimize the bicycle. Errors and population iteration processes of three kinds of algorithms are shown in Fig. 15 and Fig. 16. An optimization process will converge when the set critical error of 0.01 or a termination generation is reached. It is shown in Fig. 15 that the traditional PSO algorithm cannot converge during the whole iteration process and falls into local extreme values. The PSO-GA optimization algorithm converges to the set critical error when the iteration is conducted to 480th generation. The CA-PSO algorithm proposed by the paper already converges to the set critical error when the iteration is conducted to the 320th generation. Therefore, the optimization algorithm proposed by the paper has high optimization efficiency and will not always fall into local extreme values. Root-mean-square value and maximum difference of original vibration accelerations of the bicycle are 29.5 m/s2 and 73.5 m/s2. When the traditional PSO algorithm is used to optimize the bicycle, the obtained root-mean-square value and maximum difference of optimal vibration accelerations are 26 m/s2 and 63 m/s2. The root-mean-square value and maximum difference of accelerations are decreased by 11.9 % and 14.3 %, respectively. The optimization process falls into local extreme values, but optimization effects are still obvious. When the PSO-GA algorithm is used to optimize the bicycle, the obtained root-mean-square value and maximum difference of the optimal vibration acceleration are 23.5 m/s2 and 64 m/s2. The root-mean-square value and maximum difference of accelerations are decreased by 20.3 % and 12.9 %. Obviously, compared with the PSO algorithm, the optimization effects of root-mean-square value of vibration accelerations are obvious, but the optimization effects of maximum difference are reduced. When the CA-PSO algorithm proposed by the paper is used to optimize the bicycle, the iteration has a smooth stage when the number of iterations is within 160–250 because it has fallen into the local extreme value, but it soon jumped out from this local extreme value. As a result, the obtained root-mean-square value and maximum difference of optimal vibration accelerations are 21.5 m/s2 and 59.8 m/s2. The root-mean-square value and maximum difference of vibration accelerations are decreased by 27.1 % and 18.6 %. Compared with other two kinds of PSO algorithms, optimization effects of the root-mean-square value and maximum difference of vibration accelerations are very obvious.\n\nFig. 15. Error iteration processes of three kinds of PSO algorithms", null, "Fig. 16. Population iteration processes of three PSO algorithms", null, "a) PSO algorithm", null, "b) PSO-GA algorithm", null, "c) CA-PSO algorithm\n\nAccording to the optimized parameters, modeling is conducted on the bicycle again, and time-domain vibration accelerations are computed. Computational results are compared with original results, as shown in Fig. 17. It is shown in the figure that the optimized acceleration values are obviously better than those of the original structure at most time points. However, at some time points, the optimized accelerations are smaller than the original result. This is normal as the optimization objective in the paper is root-mean-square value and maximum difference of the vibration acceleration rather than vibration acceleration at each time point. In addition, the fatigue life of the optimized bicycle is also computed. The fatigue life of the original structure is 5 years while the fatigue life of the optimized bicycle is 7 years. Therefore, the fatigue life is improved obviously.\n\nFig. 17. Comparison of vibration accelerations before and after optimization", null, "#### 5. Conclusions\n\nThe paper tries to optimize dynamics and fatigue characteristics of the bicycle through combining finite element model with advanced algorithms. The advanced algorithm applies ideas of cellular automation (CA) to Particle Swarm Optimization (PSO), and then a hybrid CA-PSO algorithm is proposed. The conclusion has been detailed as follows:\n\na) Damage forms of a material or structure under alternating forces are very different from those under static forces. Under the alternating force, even if the maximum stress of the material or structures is lower than the yield limit, the structure will suffer from sudden fractures at a moment under long-term repeated effects. Even a material with high plasticity does not have obvious macroscopic plastic deformation before the fracture. Therefore, it is necessary to apply real alternating forces as the excitation during studying dynamics and fatigue characteristics of bicycles.\n\nb) The maximum stress of bicycles is mainly distributed on the frame, especially on joints of different round pipes of the frame at different moments mainly because that a dead corner is at the joint, and the dead corner can easily cause stress concentration and belongs to a dangerous region. Forces are uniform at the bicycle handlebar and tires and do not have obvious stress concentration phenomena. Under alternating forces, the stress concentration at joints will generate fatigue damage. Therefore, the service life of this position is the shortest.\n\nc) The vibration response curves of the bicycle fluctuated very obviously; the maximum vibration acceleration responses are at the round pipe joints; the minimum vibration acceleration is at the bicycle handle; these distribution results are consistent with rules of stress and strain. In addition, it could be found that the vibration acceleration on the seat stay round pipe presents certain periodicity. Moreover, vibration accelerations basically have the consistent changing trends between the experimental test and the numerical simulation. Therefore, the numerical computation model established in the paper is feasible.\n\nd) In order to verify the effectiveness of the CA-PSO algorithm proposed by the paper, the widely used PSO algorithm and PSO-GA algorithm are also used to optimize the bicycle. When the traditional PSO algorithm is used to optimize the bicycle, the root-mean-square value and maximum difference of accelerations are decreased by 11.9 % and 14.3 %, respectively. When the PSO-GA algorithm is used to optimize the bicycle, the root-mean-square value and maximum difference of accelerations are decreased by 20.3 % and 12.9 %. When the proposed CA-PSO algorithm in the paper is used to optimize the bicycle, the root-mean-square value and maximum difference of accelerations are decreased by 27.1 % and 18.6 %. Compared with other two kinds of PSO algorithms, optimization effects of the root-mean-square value and maximum difference of vibration accelerations are very obvious.\n\ne) The optimized acceleration values are obviously better than those of the original structure at most time points. However, at some time points, the optimized accelerations are smaller than the original result. This is normal as the optimization objective in the paper is root-mean-square value and maximum difference of the vibration acceleration rather than vibration acceleration at each time point. In addition, the fatigue life of the original structure is 5 years, while the fatigue life of the optimized bicycle is 7 years. Therefore, the fatigue life is improved obviously.\n\n1. Erdoğan G., Battarra M., Calvo R. W. An exact algorithm for the static rebalancing problem arising in bicycle sharing systems. European Journal of Operational Research, Vol. 245, Issue 3, 2015, p. 667-679. [Publisher]\n2. Heinen E., Maat K., Van Wee B. The role of attitudes toward characteristics of bicycle commuting on the choice to cycle to work over various distances. Transportation Research Part D: Transport and Environment, Vol. 16, Issue 2, 2011, p. 102-109. [Publisher]\n3. Gylling M., Heikkilä J., Jussila K., et al. Making decisions on offshore outsourcing and backshoring: A case study in the bicycle industry. International Journal of Production Economics, Vol. 162, 2015, p. 92-100. [Publisher]\n4. Lathia N., Ahmed S., Capra L. Measuring the impact of opening the London shared bicycle scheme to casual users. Transportation Research Part C: Emerging Technologies, Vol. 22, 2012, p. 88-102. [Publisher]\n5. Bröde P., De Bruyne G., Aerts J. M., et al. Head sweat rate prediction for thermal comfort assessment of bicycle helmets. Extreme Physiology and Medicine, Vol. 4, Issue 1, 2015, p. A85. [Publisher]\n6. Bachand Marleau J., Lee B., El Geneidy A. Better understanding of factors influencing likelihood of using shared bicycle systems and frequency of use. Transportation Research Record: Journal of the Transportation Research Board, Vol. 2314, 2012, p. 66-71. [Publisher]\n7. Ferrer Roca V., Roig A., Galilea P., et al. Influence of saddle height on lower limb kinematics in well-trained cyclists: static vs. dynamic evaluation in bicycle fitting. The Journal of Strength and Conditioning Research, Vol. 26, Issue 11, 2012, p. 3025-3029. [Publisher]\n8. Han F. X., Zhao S. J., Zhang L., et al. Survey of strategies for switching off base stations in heterogeneous networks for greener 5G system. IEEE Access, Vol. 4, 2016, p. 4959-4973. [Publisher]\n9. Chen C. H., Wu Y. K., Chan M. S., et al. The force output of handle and pedal in different bicycle-riding postures. Research in Sports Medicine, Vol. 24, Issue 1, 2016, p. 54-66. [Publisher]\n10. Caya A., Champoux Y., Drouet J. M. Dynamic behaviour and measurement accuracy of a bicycle brake hood force transducer. Procedia Engineering, Vol. 34, 2012, p. 526-531. [Publisher]\n11. Yang K., Yang N., Xing C. W., et al. Space-time network coding with transmit antenna selection and maximal-ratio combining. IEEE Transactions on Wireless Communications, Vol. 14, Issue 4, 2015, p. 2106-2117. [Publisher]\n12. Yan X. Z. X. U. R. B. U., Xiaofan W. U. Optimal design of bicycle frame parameters considering biomechanics. Chinese Journal of Mechanical Engineering, Vol. 24, Issue 1, 2011, p. 1-5. [Search CrossRef]\n13. Du W. H., Zhang L., Zhang D. W. Dynamic characteristics study of electric bicycle vibration reduction frame. Proceedings of the 3rd International Conference on Computational Intelligence and Industrial Application, Vol. 9, 2010, p. 363-366. [Search CrossRef]\n14. Cai R., Chen L., He S. J., Jiang C. M., Li R. Analysis on the dynamic and simulation of road cycling. China Sport Science and Technology, Vol. 50, Issue 1, 2014, p. 125-128. [Search CrossRef]\n15. Chen R. X., Wang H. F. Bicycle frame analysis based on Adams and Ansys. Light Industry Machinery, Vol. 29, Issue 6, 2011, p. 18-24. [Search CrossRef]\n16. Feng Z. X., Wu X. L., Wang J. J. Ergonomic design of new exercise bicycle frame based on stress analysis. Machinery Design and Manufacture, Vol. 2, 2016, p. 127-130. [Search CrossRef]\n17. Alizadeh R. A dynamic cellular automaton model for evacuation process with obstacles. Safety Science, Vol. 49, Issue 2, 2011, p. 315-323. [Publisher]\n18. Guan D. J., Li H. F., Inohae T., et al. Modeling urban land use change by the integration of cellular automaton and Markov model. Ecological Modelling, Vol. 222, Issue 20, 2011, p. 3761-3772. [Publisher]\n19. Hoekstra A., Kroc J., Sloot P. Introduction to modeling of complex systems using cellular automata. Simulating Complex Systems by Cellular Automata, 2010, https://doi.org/10.1007/978-3-642-12203-3_1. [Search CrossRef]\n20. Zheng Y., Jia B., Li X. G., et al. Evacuation dynamics with fire spreading based on cellular automaton. Physica A: Statistical Mechanics and its Applications, Vol. 390, Issue 18, 2011, p. 3147-3156. [Publisher]\n21. Wang Y., Lv J., Zhu L., et al. Crystal structure prediction via particle-swarm optimization. Physical Review B, Vol. 82, 2010, p. 9-94116. [Publisher]\n22. Kulkarni R. V., Venayagamoorthy G. K. Particle swarm optimization in wireless-sensor networks: a brief survey. IEEE Transactions on Systems, Man, and Cybernetics, Part C, Applications and Reviews, Vol. 41, Issue 2, 2011, p. 262-267. [Publisher]\n23. Ding G., Tan Z., Wu J., et al. Indoor fingerprinting localization and tracking system using particle swarm optimization and Kalman filter. IEICE Transactions on Communications, Vol. 98, Issue 3, 2015, p. 502-514. [Publisher]\n24. Rini D. P., Shamsuddin S. M., Yuhaniz S. S. Particle swarm optimization: technique, system and challenges. International Journal of Computer Applications, Vol. 14, Issue 1, 2011, p. 19-26. [Publisher]\n25. Xiao P., Wu J. S., Cowan C. F. N. MIMO detection schemes with interference and noise estimation enhancement. IEEE Transactions on Communications, Vol. 59, Issue 1, 2011, p. 26-32. [Publisher]\n26. Hu H. G., Wu J. S. New constructions of codebooks nearly meeting the Welch bound with equality. IEEE Transactions on Information Theory, Vol. 60, Issue 2, 2014, p. 1348-1355. [Publisher]" ]

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[ "QGIS API Documentation  2.14.0-Essen\nQgsStatisticalSummary Class Reference\n\nCalculator for summary statistics for a list of doubles. More...\n\n`#include <qgsstatisticalsummary.h>`\n\n## Public Types\n\nenum  Statistic {\nCount = 1, Sum = 2, Mean = 4, Median = 8,\nStDev = 16, StDevSample = 32, Min = 64, Max = 128,\nRange = 256, Minority = 512, Majority = 1024, Variety = 2048,\nFirstQuartile = 4096, ThirdQuartile = 8192, InterQuartileRange = 16384, All = Count | Sum | Mean | Median | StDev | Max | Min | Range | Minority | Majority | Variety | FirstQuartile | ThirdQuartile | InterQuartileRange\n}\nEnumeration of flags that specify statistics to be calculated. More...\n\n## Public Member Functions\n\nQgsStatisticalSummary (const QgsStatisticalSummary::Statistics &stats=All)\nConstructor for QgsStatisticalSummary. More...\n\nvirtual ~QgsStatisticalSummary ()\n\nvoid calculate (const QList< double > &values)\nCalculates summary statistics for a list of values. More...\n\nint count () const\nReturns calculated count of values. More...\n\ndouble firstQuartile () const\nReturns the first quartile of the values. More...\n\ndouble interQuartileRange () const\nReturns the inter quartile range of the values. More...\n\ndouble majority () const\nReturns majority of values. More...\n\ndouble max () const\nReturns calculated maximum from values. More...\n\ndouble mean () const\nReturns calculated mean of values. More...\n\ndouble median () const\nReturns calculated median of values. More...\n\ndouble min () const\nReturns calculated minimum from values. More...\n\ndouble minority () const\nReturns minority of values. More...\n\ndouble range () const\nReturns calculated range (difference between maximum and minimum values). More...\n\nvoid reset ()\nResets the calculated values. More...\n\ndouble sampleStDev () const\nReturns sample standard deviation. More...\n\nvoid setStatistics (const Statistics &stats)\nSets flags which specify which statistics will be calculated. More...\n\ndouble statistic (Statistic stat) const\nReturns the value of a specified statistic. More...\n\nStatistics statistics () const\nReturns flags which specify which statistics will be calculated. More...\n\ndouble stDev () const\nReturns population standard deviation. More...\n\ndouble sum () const\nReturns calculated sum of values. More...\n\ndouble thirdQuartile () const\nReturns the third quartile of the values. More...\n\nint variety () const\nReturns variety of values. More...\n\n## Static Public Member Functions\n\nstatic QString displayName (Statistic statistic)\nReturns the friendly display name for a statistic. More...\n\n## Detailed Description\n\nCalculator for summary statistics for a list of doubles.\n\nStatistics are calculated by calling calculate and passing a list of doubles. The individual statistics can then be retrieved using the associated methods. Note that not all statistics are calculated by default. Statistics which require slower computations are only calculated by specifying the statistic in the constructor or via setStatistics.\n\nNote\n\nDefinition at line 39 of file qgsstatisticalsummary.h.\n\n## Member Enumeration Documentation\n\nEnumeration of flags that specify statistics to be calculated.\n\nEnumerator\nCount\n\nCount.\n\nSum\n\nSum of values.\n\nMean\n\nMean of values.\n\nMedian\n\nMedian of values.\n\nStDev\n\nStandard deviation of values.\n\nStDevSample\n\nSample standard deviation of values.\n\nMin\n\nMin of values.\n\nMax\n\nMax of values.\n\nRange\n\nRange of values (max - min)\n\nMinority\n\nMinority of values.\n\nMajority\n\nMajority of values.\n\nVariety\n\nVariety (count of distinct) values.\n\nFirstQuartile\n\nFirst quartile.\n\nThirdQuartile\n\nThird quartile.\n\nInterQuartileRange\n\nInter quartile range (IQR)\n\nAll\n\nDefinition at line 44 of file qgsstatisticalsummary.h.\n\n## Constructor & Destructor Documentation\n\n QgsStatisticalSummary::QgsStatisticalSummary ( const QgsStatisticalSummary::Statistics & stats = `All` )\n\nConstructor for QgsStatisticalSummary.\n\nParameters\n stats flags for statistics to calculate\n\nDefinition at line 28 of file qgsstatisticalsummary.cpp.\n\n QgsStatisticalSummary::~QgsStatisticalSummary ( )\nvirtual\n\nDefinition at line 34 of file qgsstatisticalsummary.cpp.\n\n## Member Function Documentation\n\n void QgsStatisticalSummary::calculate ( const QList< double > & values )\n\nCalculates summary statistics for a list of values.\n\nParameters\n values list of doubles\n\nDefinition at line 62 of file qgsstatisticalsummary.cpp.\n\n int QgsStatisticalSummary::count ( ) const\ninline\n\nReturns calculated count of values.\n\nDefinition at line 102 of file qgsstatisticalsummary.h.\n\n QString QgsStatisticalSummary::displayName ( QgsStatisticalSummary::Statistic statistic )\nstatic\n\nReturns the friendly display name for a statistic.\n\nParameters\n statistic statistic to return name for\n\nDefinition at line 237 of file qgsstatisticalsummary.cpp.\n\n double QgsStatisticalSummary::firstQuartile ( ) const\ninline\n\nReturns the first quartile of the values.\n\nThe quartile is calculated using the \"Tukey's hinges\" method.\n\nthirdQuartile\ninterQuartileRange\n\nDefinition at line 166 of file qgsstatisticalsummary.h.\n\n double QgsStatisticalSummary::interQuartileRange ( ) const\ninline\n\nReturns the inter quartile range of the values.\n\nThe quartiles are calculated using the \"Tukey's hinges\" method.\n\nfirstQuartile\nthirdQuartile\n\nDefinition at line 180 of file qgsstatisticalsummary.h.\n\n double QgsStatisticalSummary::majority ( ) const\ninline\n\nReturns majority of values.\n\nThe majority is the value with most occurances in the list This is only calculated if Statistic::Majority has been specified in the constructor or via setStatistics.\n\nminority\n\nDefinition at line 159 of file qgsstatisticalsummary.h.\n\n double QgsStatisticalSummary::max ( ) const\ninline\n\nReturns calculated maximum from values.\n\nDefinition at line 123 of file qgsstatisticalsummary.h.\n\n double QgsStatisticalSummary::mean ( ) const\ninline\n\nReturns calculated mean of values.\n\nDefinition at line 110 of file qgsstatisticalsummary.h.\n\n double QgsStatisticalSummary::median ( ) const\ninline\n\nReturns calculated median of values.\n\nThis is only calculated if Statistic::Median has been specified in the constructor or via setStatistics.\n\nDefinition at line 115 of file qgsstatisticalsummary.h.\n\n double QgsStatisticalSummary::min ( ) const\ninline\n\nReturns calculated minimum from values.\n\nDefinition at line 119 of file qgsstatisticalsummary.h.\n\n double QgsStatisticalSummary::minority ( ) const\ninline\n\nReturns minority of values.\n\nThe minority is the value with least occurances in the list This is only calculated if Statistic::Minority has been specified in the constructor or via setStatistics.\n\nmajority\n\nDefinition at line 152 of file qgsstatisticalsummary.h.\n\n double QgsStatisticalSummary::range ( ) const\ninline\n\nReturns calculated range (difference between maximum and minimum values).\n\nDefinition at line 127 of file qgsstatisticalsummary.h.\n\n void QgsStatisticalSummary::reset ( )\n\nResets the calculated values.\n\nDefinition at line 39 of file qgsstatisticalsummary.cpp.\n\n double QgsStatisticalSummary::sampleStDev ( ) const\ninline\n\nReturns sample standard deviation.\n\nThis is only calculated if Statistic::StDev has been specified in the constructor or via setStatistics.\n\nstDev\n\nDefinition at line 139 of file qgsstatisticalsummary.h.\n\n void QgsStatisticalSummary::setStatistics ( const Statistics & stats )\ninline\n\nSets flags which specify which statistics will be calculated.\n\nSome statistics are always calculated (eg sum, min and max).\n\nParameters\n stats flags for statistics to calculate\nstatistics\n\nDefinition at line 83 of file qgsstatisticalsummary.h.\n\n double QgsStatisticalSummary::statistic ( QgsStatisticalSummary::Statistic stat ) const\n\nReturns the value of a specified statistic.\n\nParameters\n stat statistic to return\nReturns\ncalculated value of statistic\n\nDefinition at line 197 of file qgsstatisticalsummary.cpp.\n\n Statistics QgsStatisticalSummary::statistics ( ) const\ninline\n\nReturns flags which specify which statistics will be calculated.\n\nSome statistics are always calculated (eg sum, min and max).\n\nsetStatistics\n\nDefinition at line 76 of file qgsstatisticalsummary.h.\n\n double QgsStatisticalSummary::stDev ( ) const\ninline\n\nReturns population standard deviation.\n\nThis is only calculated if Statistic::StDev has been specified in the constructor or via setStatistics.\n\nsampleStDev\n\nDefinition at line 133 of file qgsstatisticalsummary.h.\n\n double QgsStatisticalSummary::sum ( ) const\ninline\n\nReturns calculated sum of values.\n\nDefinition at line 106 of file qgsstatisticalsummary.h.\n\n double QgsStatisticalSummary::thirdQuartile ( ) const\ninline\n\nReturns the third quartile of the values.\n\nThe quartile is calculated using the \"Tukey's hinges\" method.\n\nfirstQuartile\ninterQuartileRange\n\nDefinition at line 173 of file qgsstatisticalsummary.h.\n\n int QgsStatisticalSummary::variety ( ) const\ninline\n\nReturns variety of values.\n\nThe variety is the count of unique values from the list. This is only calculated if Statistic::Variety has been specified in the constructor or via setStatistics.\n\nDefinition at line 145 of file qgsstatisticalsummary.h.\n\nThe documentation for this class was generated from the following files:" ]

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[ "# Points\n\nPoints are similar to XP, however they are completely separate, and are per server. You can control how many points are given out, as well as the cooldown between when points are given out.\n\nYou can reward you users for getting points with point roles.\n\n## Changing how many points are given out\n\nTo set how many points are given out each time, execute the following command:\n\n``````setup set points [amount]\n``````\n\nSo for example, to give 25 points each time you would do:\n\n``````setup set points 25\n``````\n\n## Changing the cooldown\n\nTo change the cooldown, do the following command:\n\n``````setup set pointscooldown [time]\n``````\n\nThe time is in seconds.\n\nSo if you wanted a 10 seconds cooldown, you would do:\n\n``````setup set pointscooldown 10\n``````" ]

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[ "", null, "", null, "### Home > CC1MN > Chapter 4 > Lesson 4.1.5 > Problem4-55\n\n4-55.\n\nThe first four multiples of $5$ are $5$, $10$, $15$, and $20$.\n\n1. What are the first six multiples of $10$?\n\nTo find the first six multiples of $10$, you simply multiply $10$ by the numbers $1$ through $6$ separately and create a list of the products you find similar to the list written above.\n\nThe first six multiples of $10$ are $10,20,30,40,50$, and $60$.\n\n2. What are the first six multiples of $8$?\n\nThe first six multiples of $8$ are $8,16,24,32,40$, and $48$.\n\n3. What is the least common multiple of $10$ and $8$?\n\nThe Least Common Multiple of two or more integers is the lowest positive integer that is divisible by both (or all ) of the integers. For example, the multiples of $3$ and $5$ are shown below in the table.  $15$ is the Least Common Multiple because it is the smallest positive integer divisible by both $3$ and $5$. Use this idea to find the least common multiple of $10$ and $8$.\n\n$\\begin{array}{c|c|c|c|c|c} \\quad 3 \\; & \\; 6 \\; & \\; 9 \\; & \\; 12 \\; & \\; \\mathbf{15} \\; & \\; 18 \\quad\\\\ \\hline \\quad 5 \\; & \\; 10 \\; & \\; \\mathbf{15} \\; & \\; 20 \\; & \\; 25 \\; & \\; 30 \\quad\\\\ \\end{array}$\n\n4. What is the greatest common factor of $10$ and $8$\n\nThe Greatest Common Factor of two or more integers is the greatest positive integer that is a factor of both (or all) of the integers. For example, the factors of $18$ are $1$, $2$, $3$, $6$, and $18$ and the factors of $12$ are $1$, $2$, $3$, $4$, $6$, and $12$, so the Greatest Common Factor of $12$ and $18$ is $6$.\n\nThe factors of $10$ are: $1,\\ 2,\\ 5$, and $10$. The factors of $8$ are: $1,2,4,$ and $8$. Can you find the greatest common factor now?" ]

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SMAK/8hRgLL8S6SwvMcWDQzvascJkuopwm+szYqyA2SH3kRum89v6EE33NrjKLdwLy0Ffh2G4qUg32uVon3YtWxXrWXUEd8FCqftTH765n3cuqEC7zXUczvGyW8W5TzFrwvFmda1k/5wn0wEqelQJ7qWX/XlHC9Jr6z9hLrr0LRKws9tPhJS4FKutaTFjbUcSQcIhO48vcP7F9sZHWJhA58zshvpW/D9SoNNFAIMkRXQ27yHInWkL+ADa2LqTyGCXv+6ciz9GLs7aWfxLT3s4GIAxq8x5n2oALpQCB38X7PeXlw5bNM/2mmfdY59jz/38HjPr7BfFwVk4ejeXxG4NhHeN2XJJr/AOWJlfWOK/IO7D0v8fbv4z0Xnvlv3vNAfsf07+exh6ic+cR5Ae9jPVbYvijwbhDvMZv32jMmz0fy/FsK1P+TmZ9rCjz7VF7nm72ou7vElAfK6RGWq0/4tzL9PwJ1Au/04zH3QnDrLyRaCvkVvtvZRd7tRL7/13gOzv2l9OwGRPndXCBfuO8nipSFfbffKpBmBtNMLXKtk5gOsUTDlKYU/WmhZ2MIvbNCefqQ00BmaG3tE9Nozab2HCLoNY5G7Fp3owNp0T0wpgzFoFLYjB6Mnfn/VeYRDc6lEi0aM9GxEDZhwybcZxeoBfHbYMVT2ABZLX8bCqam/WlMPr4i+eF7Q4rkGaMbtuS76QqUWcJpxOud/HY69cfm91iS6IWedY38xgUsDuXxVd7+/VlvhrNsXmR5oSG+nedMi7EyJ/P4ZCoSqx2PyFjHE5Ry6ppb31c639P2tIirPCX4VxKtBgjMo/W1PZ/9Uzy2wrnODvRWYA6HCQEr3JbDigIWHIJGtyWxX0GPgA+U89Ysq3JRRyXGWrJZx1BA3vYyciiVsLWO8rgd03YG6vBRVODvcu6D7+MevosMFTYowntQcPw7Xt6+4xDnElrmyOsJLG8onU85dXIrJ1+2TXHzdQzzNTNG0Z1MRWwyvYAhq34sy+Ub/BbfiCnT8/jemjYy40PxHrTQQ+iqoFtoNK2PI9kQ7BtDtLDkf+6QiA806D8q4X7PsdFMDED5X83GaIFEa7uPpxxPUsAwv9O9cgZ+xgZ/R/4iNuA2ktN0yc++57pZz2BjEfIQuKMFisUjWCI7xcmDK+PZ+LrXQgO8k5Nmd8fC/j6f3ffQxE3qkw4QKkj8Jv7+kff6MJXDHzLNZVSQfNgpi4VKneuheJjPY8t5MvfPoQJkn/dwrx52eN/Dt0jYq1incc4H+X6XkbAv9JTmDsfrcEGJ5eBiJz4b0OwoE6FvN84zVgz2/UKp2I1ltAOf78tU9A/y6rDN77leHd6dym09CXGYo1TdSDKczfLYieV3GdOc79WhfRwyv5RpbZ14gG3M9Z4HzObrvJh81Xn58pXJcY6XZq8i3w6I+rSYNJ93PAgdou52xQAQ+kBgKt1icV6GIbRKFhS5DhqDtwcg/2igPsftMyVa/jXDjxgW5ZU8dnbAbbmazzWPv3B7TqIS00wLxMeOtH58wHrbtBf5X+TkwZW5bMh90niNx+fTMsJ8BLMc5aAv+CS9Bkv4PHNYlktIpo+wrp8ZOHcij83l/0nOsTbut+X8hkN+9nlej7G0xCGkE7l9Cb0IHSyTu0ggQqKPc69+m5ZoOTiGHoV5zO+kfqzLackHvM7n9g2S78I4WnpOKLXUq8OoEyfxnYEcd2G63aiItbKePM93i/7w7xm5m+lOdK5tn/XPVBiX8ZyX6alq4/UPCTwL7v8vL1+TuB+KcqhLwN77Nf6eUEKZTQ54C1EPz1JaUgw0oW/oRUlg2V5cJE2t89HH4T5q300DUPZoHBpp3TweOD6dpPftwHtKxlhLL3M7zl39TU8Bgqvwq45VWA7K6a6B5VoT2P9bx5rsSx3awfG2LA0cn0Kiv9Xb30yLKMuyWUhLb8uY+6Sc56ktMW9Qlmx/+gOB4w+R3DeR9fvdq0g8C3jfH5dxT6Q71lEGXqVC8MF+qstx5fG04wWqLaH+LCVxAkMdi1eoWL0WOOde/m7r7NveO+biLXrAzohRxEL5Wu7UK1/p2oyKwTpes4WK+ogSPJH+PBoHSnwMgULRL4Qeck03SnhseiXRzgbxMDZSxQjIRr+jEX8wcBxW0jkFnqm/Yee1XynhaG7sn0Fr3Y+E7o7xSNh+8IXesQdo2XzMs0pgOW1HC/8fZea/EjETbzl5b+jDdWwjG+dpQUAUgsf+GmhA4SlBlwC6CeBih2v1iAq+5yaSWafk+9r9et1CIqnzvrMsLbZVtCi/U+I94fL9AOsBvAD3U2Hqr9EdWQlH2u/rELVfx0PR+weQjLO08oHhzjUk5juxdci2aU1F6sPdVJifCRwL5etAyceCvOwd+yy/ZVjyCGJDtwCi8A8t0Hb+kt/w1x3FxSrcwEyJjw1SKCpiZbkNUKjRapJ8UE9fAGviSoeQYXku4wf+ai8UljQVgNmelfgTiSJJB7rsu6T8/stNaNW6VuC32OgsCxAXgv4w8c+1THc3G3jr3kMU9GllNN7AFWwwk16D9b2YhlJilCrrceiLhZ4sUDcLwbpGf+80pCdy/3SpzOp5SckPLQzFBXQ7+xMBJe0JiVzXeEfnUvF4usg9j3eIK81fBGIhIvxyqVwAq1uXMT/FWueZP8P8WgLzyxJW7OZMm6FX5EQqP4gHedF7t+uKKJZJpwxD9WFXfjdZJ13I6j/Cy9dYenf8fPllfadThw5mHZoRk2d8n2OoKEyi9wWWOUZ9wN3/fxLFZWj/uaLfCT2k9Q7nR+AT+v5s4NNO5QSp3sCPI4TFrNCVBAgGQTBnOhbs1AEue7dhKddDcDLFByL7vyw9o5mHsnFBfy2Gtu1GBeyjtDhmUukpB3EL8/y0DEJ3yyJbobIsFWioD2KjbUdVII5hCZ9tl148R2/ec7H3D+/Xj0jGu7Px372AEjhC8gFwv+bvoxL1Ce9A6/3+CtdlfP+PxRybwW/Px3HSc8hZG7/9s5xyK/ZuE166uHNQhhO8c690lA6LYwKeDHjIEIB7tqeYjGd5tku+L38W0+9PBXtujBJyNQkdVvr/UuGCAYKA1/kyMF5DxSAk9BcC+6C9fs2z8rDvssBHBFxVwPqp7qdnRV6OYkOOhV2WD3DZ9+WDfZtKSZKNACwjuPxulsi1HipTuG2voyJzjuOt+G82pMky84358Z+UvFswUaB+FPKgDFRZHk6yhJvddjesIrmfxkb9mQrlLdGH57CW4mkkzY+TBBbFXOMztEThfXrEsW7RdQOX/cR+IPRuWq7dfKcZEtmdjlLhA11hiB9AVx2i4D9EMjy1l+82UeQcxGu8QuPCkm1XgXwlWc7IF0ZOTAmktYGHs0jCwJtMj2NHSj641QW6l+5gvUM3GQJz0RXWQkLfSqlJsaEI/a8kR/+jQXAV+o7gEkRf4BdjyBxE9KCEg6T6E8v4cR0vPYOjBgJtzsddI4XXhk94FsgvJN//Xw5gZaCf7mj+XyDR+OjeAIQxu49lYPu+OyTvUrWKRZzClw4oA+scS7FURcK6SuGh2JPfQkbyoyKg/F1c5L2Ugg5aZPUSjhOwM9+JxA/Vs+WNbo6LJBri9ouYdLYb4SXvuawCcBjLaWUF6/JKWqpryzgHwai3OSQICxf90RjG+ZyTrt3xMoUwxClnW286vPplFVeLmwsQ+h+db+JNtmeH0ZvldtHVOJb8K3z+JOuntcqhPP1Qes7SZ2daRJ5ukXyA73S2Ux9QalL0Br2xkBBA9ZeYY0fzY/lpDJkDP6FLKjUAz3ujQ2YDjVX8qEfHNFZoQOACnik9I2t7a9kulfUnl7mOjXBvrldXgTKw0elLnEbYTuoyJuacTZ3ycz0WwLiYc6ZQibya/3eSfDQxJtV5lMdhrf+A+xE1vW8FnnEFSQllHJo2eRRJqU16Dvfzgbw9zXNs95Gr6CHP+3H7C95zXeeU38H94G0q1zho8Ej0CSo2/ph7G/W+eUybMc6rD1lHWdk65t7betcOKQhW6XhM8rP8uXBHDZxHb8iD/D2f+6Gc7FqgDOyshlYpvVYpSbGhCd0O8elNANzj1EIH0ipevJGU/Rx6K+okP3TMfS/Q2g8gma8ONKC9xfW0gEAMN/XhOi1lpE1Lz0AsDEeyE7Xc5+x/mL8TAoQKIjuJ2+5qfU84SpAfXTyWFu2+TkNvXaVv0Br7jSP4/6pDin3FUsfiDAUens73PUcKj2e3jf43aFmGukg+T6JEEOTtged6vsBztffxOftSJ9P0PgBwU3/CMyDWkZxPCNSHL3h1QBzP0XHSc6w3vAC7sx17rEi+YO3b2QWP8IwU6+GZS0+DW9b4P9/zBMV5by6nV+g6Cfe3KxQlo7f91a+wgt9awCoKWfbHSt9dmO8VrGUjdj01fFikGGJUS9I6hA3Kd6Uy0dYWi9lgurOR9QYns4FLBOoUvAovelb1+ZJ3PW5FTwkaW7g1f+aR80zWL/R7wmWJvkaMrf86FYGF9LZYPMWG9Bg2pldTYRlH5RPW3WtsNF1X6eUSng4XZT+Lv2OkbxMPZfme9yPBQIGzUd/HOXkBcZQy2uFJWuoXBAh1IrevlfA0txNIdgfwHSxwjkHhCc15kKLy9Eg/fw/38N1/gs/2WYcwf05FBvVkRyp9GP+Ncd8Y5vaW5GeNBG6gVwZu9XtZHkizN89JUZl9roR8WSt9Ar/FQ6lkH+5Y578LnIeI/RlUsnBea8z1URf+UKaCrFBUlNCFHzg+kMvYKMW5YGHJ3yzR0JvVXgPUHEhf7rKmdpUjH0PLuEbcilH93c8PMkFUMmaz+hLFAtbk2bJ+P7V1B5Y6ZrsupkxDQ4CaS3hmt6xPLZBuCQndXmszkqePZ+ideMuziibz3EMCxPQyFZ63A+ckaeH5i6y8SOsObtmjqBRkJD9TnY+H+Qyb0AK8xiub5hiLtNqpey4xoovqFF7ncIcMrKcDBHaHsy/pvOOQJY5vDv26OzvvAwqDndp2ZsxzQcnBzHbbsq5d6NxnP8m7631MjyF06wIfVoa3z9az2oCVPo1K7aFU6OxznMO6jzI8V9aPTH+ZyqXr3XiLRHozy+hG716/ooLgoqlIvv7A+ngg68WmrE9xAYb30usxjnVyRoF7rIkp16GiY9EVG4jQhZYSgt8QbIbpRnciQWXo9kODfZ/0nOjEupum8eNIO/mZ1wt33Q9oSaWdRnCJlD4U6kESjjseGNd4dgO8g8tpBdg5vrtpOaCBn+OlvZ3l83AZStc0elSKWZFX0QouZLV08nqjC3gNkpJ3f2Jq3qmyflBQgiSGYw9IeEz0clpoIL6DmS8ohugT/rX07IKwjeJRJDpEem9BpegR75x2PkMhFze8J6eTIBd75DGNhNEZ4/24hPfw83gTlbOJJJkEy+D2wPtZRpJHw7405tuBBXi8971cwW8t7n2jfqPvfU/nPFiIr0p+oZQQad8Xc715VC7WluF5g7W8jazvIreAgnUWyTLlKaCnsqxQJ7Zk+T7EfS0xyuIEltFeJMc3SMx/jsnXdgXydSYV03rWtWl8f3HBhVA4v0KPwhpHMYIy9XiRMprH72ZlActeoehpcWWz5Q3/3WrX0wZ7kUmiKjjC62w25NdrtVIoFJXG/KemayEo+tVCH3x0noiN/XlaCg87UigUCoVi47HQFQqFQqFQbHzQgAuFQqFQKJTQFQqFQqFQKKErFAqFQqGoCP4jwADQNvw20jA5ogAAAABJRU5ErkJggg==", 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[ "# 5.2. Equality¶\n\n## 5.2.1. Loose Equality With `==`¶\n\nIn the section Booleans, we learned about the comparison operators `==` and `!=`, which test whether two values are equal or not equal, respectively. However, there are some quirks with using the `==` operator, which occur when we use `==` to compare different data types.\n\nExample\n\n ```1 2 3``` ```console.log(7 == \"7\"); console.log(0 == false); console.log(0 == ''); ```\n\nConsole Output\n\n```true\ntrue\ntrue\n```\n\nIn order to properly make a comparison, the two operands must be the same type. If the two operands to `==` are of different data types, JavaScript will implicitly convert the operands so that the values are of the same data type before comparing the two. For this reason, the `==` operator is often said to measure loose equality.\n\nType conversions with `==` are carried out according to a complex set of rules, and while many of these conversions make some sense, others do not.\n\nFor example, `Number(\"7\")` returns `7`, so it makes some sense that `7 == \"7\"` returns `true`. However, the following example leaves us scratching our heads.\n\nExample\n\n ```1 2 3``` ```console.log('0' == 0); console.log(0 == ''); console.log('0' == ''); ```\n\nConsole Output\n\n```true\ntrue\nfalse\n```\n\nThe `==` operator is non-transitive. We think of equality as being transitive; for example, if A and B are equal and B and C are equal, then A and C are also equal. However, the example above demonstrates that that is not the case for the `==` operator.\n\nSince `==` does not follow rules that we typically associate with equality, unexpected results may occur if `==` is used in a program. Thankfully, JavaScript provides another operator that returns more predictable results.\n\n## 5.2.2. Strict Equality With `===`¶\n\nThe operator `===` compares two operands without converting their data types. In other words, if `a` and `b` are of different data types (say, `a` is a string and `b` is a number) then `a === b` will always be false.\n\nExample\n\n ```1 2 3``` ```console.log(7 === \"7\"); console.log(0 === false); console.log(0 === ''); ```\n\nConsole Output\n\n```false\nfalse\nfalse\n```\n\nFor this reason, the `===` operator is often said to measure strict equality.\n\nJust as equality operator `==` has the inequality operator `!=`, there is also a strict inquality operator, `!==`. The boolean expression `a !== b` returns `true` when the two operands are of different types, or if they are of the same type and have different values.\n\nTip\n\nUSE `===` AND `!==` WHENEVER POSSIBLE. In this book we will use these strict operators over the loose operators from now on.\n\nQuestion\n\nWhat is the result of the following boolean expression?\n\n```4 == \"4\"\n```\n1. `true`\n2. `false`\n3. `\"true\"`\n4. `\"false\"`\n\nQuestion\n\nWhat is the difference between `==` and `===`?\n\n1. There is no difference. They work exactly the same.\n2. Only `===` throws an error if its arguments are of different types.\n3. `==` converts values of different types to be the same type, while `===` does not.\n4. `==` works with all data types, while `===` does not." ]

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