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https://inches.guru/50-meters-to-inches/ | [
"# Convert 50 meters to inches\n\n50 meters = 1968.5039 inches\n\n## How many inches is 50 meters?\n\nWhat is 50 meters to inches? The answer is that 50 meters equals 1968.5039 inches.\n\nHow to convert 50 meters to inches? We know that one inch is equal to 0.0254 meters. Therefore, to convert 50 meters to inches, we divide the meter value by 0.0254. This gives us 50 meters ÷ 0.0254 = 1968.5039 inches. So, 50 meters is equal to 1968.5039 inches."
]
| [
null
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https://planfix.com/help/Calculated_data_tag_fields | [
"# Calculated data tag fields\n\nThere is a type of Calculated Field available in data tags. These calculated fields use fields from the data tags where they're added.\n\nTo use this type of calculated field, add the following type of field to the data tag:",
null,
"Sample field settings:",
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"Any standard Planfix function can be used in a calculated field formula. The value of the calculated field is automatically recalculated when you enter data in fields that are part of the formula.",
null,
"If you change the value of any of the fields that are part of the formula, the calculated field is automatically recalculated.\n\nPlease note: Calculated field values in previously entered data tags can be manually recalculated. To do this, select the necessary tasks or contacts (or all tasks or contacts) with this data tag and apply the bulk action Recalculate data tag field:",
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""
]
| [
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"https://pic.planfix.ru/pf/xC/D8MGL9.png",
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"https://pic.planfix.ru/pf/PF/7QQbYD.png",
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"https://pic.planfix.ru/pf/bq/32aPwL.png",
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"https://pic.planfix.ru/pf/Qb/W6KZ9U.png",
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https://dba.stackexchange.com/questions/279537/where-clause-for-results-appearing-after-occurence | [
"# Where Clause for Results Appearing after Occurence\n\nI'm wondering if it's possible to find records that only appear after the first occurrence of a where clause. I'm currently using PostgreSQL for this example but something not PostgreSQL specific would be preferred.\n\nIf I had a table such as this for an example\n\n``````drop table if exists dummy;\ncreate table dummy (\ncol_a varchar(10),\ncol_b varchar(10),\ncol_c varchar(10),\ntest integer\n);\n\ninsert into dummy values\n('A', 'B', 'C', 1),\n('B', 'C', 'A', 2),\n('C', 'A', 'B', 3),\n('D', 'E', 'F', 4);\n``````\n\nIf I run the bellow query I can get the data in a new order\n\n``````select * from dummy\norder by col_b;\n``````\n\nreturning\n\n``````C | A | B | 3\nA | B | C | 1\nB | C | A | 2\nD | E | F | 4\n``````\n\nIs it possible to do something so that I can only return the rows occurring after a certain where clause, something like\n\n``````select * from dummy\n/* where row appears after results 'A', 'B', 'C', 1 */\norder by col_b;\n``````\n\nTo get\n\n``````B | C | A | 2\nD | E | F | 4\n``````\n\nUPDATE\n\nThe reason I am not using `where col_b > ?` is because it doesn't cater for duplicate values\n\nIt would work until I need to match on an additional column like this\n\n``````select * from dummy\nwhere col_b > 'B'\nand col_c > 'C'\norder by col_b;\n``````\n\nIn which case it will return the last row because both values are above the cutoff but it is not cutting off after the first occurrence of the specified row.\n\nI'm trying to have a where clause that can identify the specific row being row\n\n``````A | B | C | 1\n``````\n\nIn this case\n\nit seems you are looking for something like this:\n\n``````with cte as\n(\nselect *\n,row_number() over(order by col_b) as rn\nfrom dummy\n)\nselect *\nfrom cte\nwhere rn > (select rn from cte where col_a = 'A' and col_b = 'B' and col_c = 'C' and test = 1 limit 1)\n``````\n• This is the way I was hoping to do it but it is PostgreSQL specific (or maybe it isn't but Informix doesn't have a `row_number()` equivalent). Still a great answer though Nov 11 '20 at 16:47\n• @TheLovelySausage ibm.com/support/knowledgecenter/SSGU8G_12.1.0/com.ibm.sqls.doc/… this link says you can use window functions in Informix 12.1 Nov 11 '20 at 16:56\n\nEasily.\n\n``````WITH cte AS ( SELECT col_a, col_b, col_c, test,\nROW_NUMBER() OVER (ORDER BY col_b) rn\nFROM dummy )\nSELECT col_a, col_b, col_c, test\nFROM cte\nWHERE rn > ( SELECT rn\nFROM cte\nWHERE (col_a, col_b, col_c, test) = ('A', 'B', 'C', 1 ) )\nORDER BY rn;\n``````\n\nfiddle"
]
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https://engineering.electrical-equipment.org/forum/reply/how-to-calculate-the-vodc-of-a-fullwave-bridge-rectifier-with-n-smoothing-capacitor-and-also-how-to-find-the-smoothing-capacitor-value | [
"# Re: how to calculate the Vo(dc) of a fullwave bridge rectifier with n smoothing capacitor? and also how to find the smoothing capacitor value?\n\nHome Electrical Engineering Forum General Discussion how to calculate the Vo(dc) of a fullwave bridge rectifier with n smoothing capacitor? and also how to find the smoothing capacitor value? Re: how to calculate the Vo(dc) of a fullwave bridge rectifier with n smoothing capacitor? and also how to find the smoothing capacitor value?\n\n#12299\n\n Tip: It is possible to build your power supply without adding a regulator, when you want to power up an electrical motor, a small ripple in voltage will not affect the performance, unlike the electronic circuits and devices!\n\nDesigning of +12v 5A Power Supply:\n\nTo calculate the voltage required and the transformer secondary winding we first determine the input voltage for the regulator, which is 15v, plus a 10% of this value for ripple. For a regular transformer we have to consider a bridge rectifier, as a result; we will add 1.4v. So the secondary winding should be 15+1.5+1.4=18.9 lets say 18v @ 5 Amps. Now we will calculate the capacity of the filtering capacitor. By using equation number 1 and assuming that f=60Hz, we will get C=5 x 5 / ((18-1.4) x f) =25,100µF"
]
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https://www.teachoo.com/1722/524/Ex-7.1--4---Check-whether-(5---2)--(6---4)-and-(7---2)/category/Ex-7.1/ | [
"",
null,
"",
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"",
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"Subscribe to our Youtube Channel - https://you.tube/teachoo\n\n1. Chapter 7 Class 10 Coordinate Geometry\n2. Serial order wise\n3. Ex 7.1\n\nTranscript\n\nEx 7.1 , 4 Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle. In an isosceles triangle, any 2 of the 3 sides are equal. Let the three points be P(5, −2), Q(6, 4) & R(7, −2) In order to be isosceles, Either PQ = PR or PQ = QR or PR = QR We calculate the value of PQ, QR & PR by distance formula P(5, −2), Q(6, 4) & R(7, −2) Calculating PQ x1 = 5 , y1 = −2 x2 = 6 , y2 = 4 PQ = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( 6 −5)2+(4 −(−2))2) = √(12+(6)2) = √(1+36) = √37 Now, we calculate QR P(5, −2), Q(6, 4) & R(7, −2) Calculating QR x1 = 6 , y1 = 4 x2 = 7 , y2 = −2 QR = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( 7 −6)2+(−2 −4)2) = √(12+(−6)2) = √(1+36) = √37 P(5, −2), Q(6, 4) & R(7, −2) Calculating PR x1 = 5 , y1 = −2 x2 = 7 , y2 = −2 PR = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( 7 −5)2+(−2 −(−2))2) = √(22+(−2+2)2) = √(2^2+0^2 ) = √(2^2 ) = 2 Hence PQ = √37 , QR = √37 , PR = 2 Since PQ = QR √37 = √37 It satisfies the condition of isosceles triangle Hence PQR is an isosceles triangle\n\nEx 7.1",
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https://worldtravelattractions.com/attractions/you-asked-how-do-you-calculate-travel-time-value.html | [
"You asked: How do you calculate travel time value?\n\nContents\n\nHow do you measure travel time?\n\nEstimate how fast you will go on your trip. Then, divide your total distance by your speed. This will give you an estimation of your travel time. For example, if your trip is 240 miles and you are going to be drive 40 miles an hour, your time will be 240/40 = 6 hours.\n\nHow much is travel time worth?\n\nTravel time is often worth more than monetary costs. For example, a 30 mph car trip has about 15¢ per mile operating costs compared with 25¢ per mile time costs (valued at \\$6.00 per hour with 1.2 passengers).\n\nHow do you define travel time?\n\nTravel time is a measure of the length of time necessary to move from one place to another.\n\nWhat are the components of vehicle travel time?\n\nCalculation Details – How the Data Can Be Used\n\nThe key total travel time data elements are: 1) the number of miles traveled and, 2) the speeds (both free-flow and congested) on the minor roads that are not included in the existing Urban Mobility Report dataset.\n\nIT IS SURPRISING: Do you need a green card to get a driver's permit?\n\nHow do I calculate travel distance?\n\nTo solve for distance use the formula for distance d = st, or distance equals speed times time. Rate and speed are similar since they both represent some distance per unit time like miles per hour or kilometers per hour.\n\nHow do you calculate km traveled?\n\nSpeed is distance divided by time. Simply put, if you drove 60 kilometres for one hour, it would look like this: Speed = distance (60 km) / time (1 hour) = 60km/h.\n\nWhat is the law on travel time to work?\n\nTravel time to and from work is not usually counted as working hours. However, travel as part of the employee’s duties is. … Being on standby to be called out, if the employee is at the place of work, is counted as working hours. If the employee is on call and free to pursue leisure activities, it is not.\n\nWhat is the meaning of value of time?\n\n1 INTRODUCTION: VALUE OF TIME DEFINITION AND USE 1.1 What Is the Value of Time and Why Do We Care? The value of time is a dollar amount assigned to value the benefit of a change in expected travel time or unscheduled delay resulting from transportation projects, policies, programs or events.\n\nWhat does back in time mean?\n\nto be back in time: to arrive soon enough, to return without being late.\n\nWhat is the value of saving travel time summary and conclusions?\n\nMeasuring the reduction in travel time has long been a fundamental element of the economic case for transport infrastructure investment. Reducing the amount of time spent on travel enables transport users to spend the time they have saved more productively or more enjoyably.\n\nIT IS SURPRISING: Can I apply Chile visa online?\n\nWhat is fixed delay in traffic engineering?\n\nFixed Delay- The delay to which a vehicle is subjected regardless of the amount of traffic volume and interference present. … Examples include time lost while waiting for a gap in a conflicting traffic stream, or resulting from congestion, parking maneuvers, pedestrians, and turning movement.\n\nWhat is the cost of transportation?\n\nVehicle Costs\n\nPer Household Portion of Household Total\nOther vehicle expenses \\$426 1.0%\nTotal vehicle expenses \\$7,360 17%\nPublic transport expenses \\$441 1.0%\nTotal transportation expenses \\$7,801 18.0%"
]
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https://forum.processing.org/two/discussion/comment/26380/ | [
"#### Howdy, Stranger!\n\nWe are about to switch to a new forum software. Until then we have removed the registration on this forum.\n\n# Trouble with a scatter plot - function not exists\n\nedited September 2014\n\nHello,\n\nFor a school exercise we need to make a scatterplot that displays data that contains a Y and X value, very logical:). But i'm a beginner with Open processing, so i hope you can help me to fix my code!\n\nI have read a lot in the book of Ben Fry (visualizing data), but i can't get it work.\n\nHopefully you can tell me what is wrong, and more important why.\n\nThank you for the answers, have a nice weekend!\n\n``````Table data;\nfloat dataMin, dataMax;\n\nfloat plotX1, plotY1;\nfloat plotX2, plotY2;\n\nint data1Min, data1Max;\nint[] data1;\n\nvoid setup() {\nsize(720, 405);\n\ndata = loadTable(\"test.csv\");\n\ndata1 = int(data.getRowNames());\ndata1Min = data1;\ndata1Max = data1[data1.length - 1];\n\ndataMin = 0;\ndataMax = data.getTableMax();\n\n// Corners of the plotted time series\nplotX1 = 50;\nplotX2 = width - plotX1;\nplotY1 = 60;\nplotY2 = height - plotY1;\n\nsmooth();\n}\n\nvoid draw() {\nbackground(224);\n\n// Show the plot area as a white box\nfill(255);\nrectMode(CORNERS);\nnoStroke();\nrect(plotX1, plotY1, plotX2, plotY2);\n\nstrokeWeight(7);\n// Draw the data for the first column (red)\nstroke(#c12e21);\ndrawDataPoints(0);\n}\n\n// Draw the data points\nvoid drawDataPoints(int col) {\nint rowCount = data.getRowCount();\nfor (int row = 0; row < rowCount; row++) {\nif (data.isValid(row, col)) {\nfloat value = data.getFloat(row, col);\nfloat x = map(data1[row], data1Min, data1Max, plotX1, plotX2);\nfloat y = map(value, dataMin, dataMax, plotY2, plotY1);\npoint(x, y);\n}\n}\n}\n``````\nTagged:\n\n## Answers\n\n• The class Table does not have methods (functions) called\n\n``````getRowNames(...)\ngetTableMax(...)\nisValid(...)\n``````\n\nso all of these will cause the error you mentioned.\n\nWhat makes you think they exist? Where did you see them used or documented?\n\n• Thank you for your replay, quark.\n\nI thought this where \"build in\" function in processing, but i got that (very) wrong.\n\nIs it possible to avoid these functions and let the code work? Hopefully you or an other form member can help my.\n\nI greatly appreciate your comments\n\n• I have commented out the lines affected but I can't test it because I don't have your data file.\n\nYou will have to manually set the value of dataMax because it is used later to scale the plot.\n\n``````Table data;\nfloat dataMin, dataMax;\n\nfloat plotX1, plotY1;\nfloat plotX2, plotY2;\n\nint data1Min, data1Max;\nint[] data1;\n\nvoid setup() {\nsize(720, 405);\n\ndata = loadTable(\"test.csv\");\n\n// data1 = int(data.getRowNames());\ndata1Min = data1;\ndata1Max = data1[data1.length - 1];\n\ndataMin = 0;\n\n// You would need to replace the '100' with the\n// maximum value in table\ndataMax = 100;\n\n// Corners of the plotted time series\nplotX1 = 50;\nplotX2 = width - plotX1;\nplotY1 = 60;\nplotY2 = height - plotY1;\n\nsmooth();\n}\n\nvoid draw() {\nbackground(224);\n\n// Show the plot area as a white box\nfill(255);\nrectMode(CORNERS);\nnoStroke();\nrect(plotX1, plotY1, plotX2, plotY2);\n\nstrokeWeight(7);\n// Draw the data for the first column (red)\nstroke(#c12e21);\ndrawDataPoints(0);\n}\n\n// Draw the data points\nvoid drawDataPoints(int col) {\nint rowCount = data.getRowCount();\nfor (int row = 0; row < rowCount; row++) {\n// if (data.isValid(row, col)) {\nfloat value = data.getFloat(row, col);\nfloat x = map(data1[row], data1Min, data1Max, plotX1, plotX2);\nfloat y = map(value, dataMin, dataMax, plotY2, plotY1);\npoint(x, y);\n// }\n}\n}\n``````\n• Thank you for your quick response quark!\n\ni will test it tonight and let you know\n\n• Hello quark,\n\nOn line 17: data1Min = data1; i get a NullPointerException\n\n• I have download the csv file and will look into this weekend.\n\n• Thank you quark, I appreciate it enormously!\n\n• I just remembered that this is a school exercise so I am going to help you reach the solution rather than me provide a solution. This is means you will get the satisfaction of knowing that you did it and learn some programming at the same time.\n\nWhen trying to do an exercise like this the first thing is to do is break the problem into smaller steps which can be solved in order.\n\nIn this case you have 2 initial stages\n\n1) load and confirm the validity of the data\n\n2) display the data\n\nAfter that you can consider tweaking the display to suit.\n\nSo create a sketch that loads the data into the table and confirm the number of rows and colums. I have looked at the data file and there are 2 colums in the data although I didn't count the rows.\n\nYou will be using loadTable, getRowCount and getColCount. The csv filename extension stands for 'comma separated values' and your data is sparated with semicolons so if the column count is not 2 then you may need to use a text editor to replace all ; with , Perhaps someone else is aware if it makes a difference.\n\nOnce you have loaded the data post your code here. You should also check out the reference section and see what other functions are available with Table objects.\n\n• This sketch will simply load the data and display the number of rows (100) and the number of columns (2). It will then display the data. I do not store the data in an array and I don't dynamically scale x and y axis but there is enough here to get you started :)\n\n``````Table data;\n\nvoid setup() {\nsize(720, 405);\n\ndata = loadTable(\"test.csv\");\nprintln(data.getRowCount());\nprintln(data.getColumnCount());\n\nbackground(255);\nfill(32, 32, 128);\nnoStroke();\nfor (int i = 0; i < data.getRowCount (); i++) {\nTableRow row = data.getRow(i);\n// Hard coded the scale factors just so the plot can be seen\nint x = 10 * row.getInt(0);\nint y = row.getInt(1) / 4;\nellipse(x, y, 4, 4);\nprintln(x + \" \" + y);\n}\n}\n``````\n\n**IMPORTANT NOTE: **It only worked after I used a text editor to change all the ; to , so you will have to do the same.\n\n• It only worked after I used a text editor to change all the ; to , so you will have to do the same.\n\nOr you can use loadStrings() + split() in place of loadTable(), to choose a specific separator besides ',' or '\\t'!\n\n• Thank you very much, I've been a lot closer to the solution!\n\nSign In or Register to comment."
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https://gcsephysicsninja.com/lessons/waves/ | [
"# Waves",
null,
"Welcome to the GCSE Physics Ninja Waves course.\n\n## What will I learn in this course?\n\nWhen you have completed this course, you will be able to...\n\n• Describe waves using terms such as amplitude, wavelength and frequency\n• Recall the properties of electromagnetic waves and the regions of the electromagnetic spectrum\n• Understand the uses and safety of electromagnetic waves\n• Describe how red-shift changes the wavelength and frequency of electromagnetic waves\n• Understand how waves provide evidence for the beginning of the universe\n• Give examples of mechanical waves\n• Calculate the speed of sound in a material\n• Describe the medical uses of x-rays, ultrasound and laser light\n• Understand how waves can reflect, diffract and refract\n• Explain the refractive index of a material\n• Use ray diagrams to show how converging and diverging lenses work\n• Compare the human eye and a camera\n• Understand how short sight and long sight can be corrected using lenses"
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"https://gcsephysicsninja.com/wp-content/uploads/2017/10/waves-course-image.jpg",
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https://physics.stackexchange.com/questions/397856/is-the-electric-field-that-goes-through-a-wire-generated-by-a-battery-uniform/397865 | [
"# Is the electric field that goes through a wire, generated by a battery, uniform?\n\nIf I, for example, connect a wire across the ends of a battery (short out the circuit) would the electric field going through this wire be uniform all the way through?\n\nYes.\n\nThere isn't much of an electric field within a wire, but wires do have a small amount of resistance, so you can think of the wire as a bunch of equal-length short sections, each section of which has the same (very small) resistance. The current through all the sections is the same, so the voltage across each section is the same, and dividing that by the section length gives that the magnitude of the electric field is the same through each section. (Obviously the direction of the electric field will change from place to place if the wire is curved.)\n\nAs measured along the length of the wire, yes. If there's 1A at one end, it will be the same at the other, and at any (macroscopic) point along the way.\n\nAs measured along the cross-section of the wire, maybe.\n\nIn the case of DC like a battery, I believe it will be fairly uniform, but I can't say I've really looked.\n\nFor AC, however, it is decidedly non-uniform. Every time the current reverses direction, 50 or 60 times a second in most cases, there's a time delay that causes eddy currents to form. The action of these currents creates a magnetic field that presses the current out to the surface of the wire.\n\nThis is known as the skin effect, and it has real practical outcomes. For instance, most high-voltage wires consist of an inner section of steel cable and then an outer layer of aluminum. The skin effect ensures the current stays inside the highly conductive aluminum, which is rather handy (and cheaper)."
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http://slideplayer.com/slide/6400212/ | [
"Presentation is loading. Please wait.",
null,
"# Discrete Mathematics Goals of a Discrete Mathematics Learn how to think mathematically 1. Mathematical Reasoning Foundation for discussions of methods.\n\n## Presentation on theme: \"Discrete Mathematics Goals of a Discrete Mathematics Learn how to think mathematically 1. Mathematical Reasoning Foundation for discussions of methods.\"— Presentation transcript:\n\nDiscrete Mathematics Goals of a Discrete Mathematics Learn how to think mathematically 1. Mathematical Reasoning Foundation for discussions of methods of proof 2. Combinatorial Analysis The method for counting or enumerating objects 3. Discrete Structure Abstract mathematical structures used to represent discrete objects and relationship between them What will we learn from Discrete Mathematics\n\n4. Algorithms Thinking Algorithm is the specification for solving problems. It’s design and analysis is a mathematical activity. 5. Application and Modeling Discrete Math has applications to most area of study. Modeling with it is an extremely important problem-solving skill. How to learn Discrete Mathematics? Do as many exercises as you possibly can !\n\nChapter 1 The Foundations: Logic, Sets, and Functions Rules of logic specify the precise meaning of mathematics statements. Sets are collections of objects. A function sets up a special relation between two sets. 1.1 Logic Propositions A proposition is a statement that is either true or false, but not both. Propositions 1. This class has 25 students. 2. 4+8=12 3. 5+3=7 Not propositions 1. What time is it? 2. Read this carefully. 3. x+1= 2. Examples\n\nDefinition 1. Let p be a proposition. The statement “It is not the case of p” is a proposition, called the negation of p and denoted by We let propositions be represented as p,q,r,s,…. The value of a proposition is either T(true) or F(false). p: Toronto is the capital of Canada. Examples Table 1. The Truth Table for the negation of a proposition p TFTF FTFT called connectives\n\nDefinition 2. Let p and q be proposition ｓ.The proposition “p and q”, denoted by, is the proposition that is true when both p and q are true and is false otherwise. The proposition is called the conjunction of p and q. Table 2. The Truth Table for the conjunction of two propositions p q T T F F T F TFFFTFFF Examples\n\nDefinition 3. Let p and q be proposition ｓ.The proposition “p or q”, denoted by, is the proposition that is false when both p and q are false and is true otherwise. The proposition is called the disjunction of p and q. Table 3. The Truth Table for the disjunction of two propositions p q T T F F T F TTTFTTTF Examples\n\nDefinition 4. Let p and q be proposition ｓ.The exclusive of p and q, denoted by, is the proposition that is true when exactly one of p and q is true and it is false otherwise. Table 4. The Truth Table for the exclusive or of two propositions p q T T F F T F FTTFFTTF Examples\n\nDefinition 5. Let p and q be proposition ｓ.The implication is the proposition that is false when p is true and q is false and true otherwise ， where p is called hypothesis and q is called the conclusion. Table 5. The Truth Table for the implication p q T T F F T F TFTTTFTT Examples “If p, then q” or “ p implies q”. Another example: If today is Friday, then 2+3=6.\n\nDefinition 6. Let p and q be proposition ｓ.The biconditional is the proposition that is true when p and q have the same truth values and is false otherwise. Table 6. The Truth Table for the biconditional p q T T F F T F TFFTTFFT Examples “p if and only if q”.\n\nTranslating English Sentences into Logical Expressions Example 1 You can access the Internet from campus only if you are a computer science major or you are not a freshman. a. You can access the Internet from campus. c. You are a computer science major. f. You are freshman. The sentence can be represented as Example 2 You cannot ride the roller coaster if you are under 4 feet tall unless you are older than 16 years old. q. You can ride the roller coaster. r. You are under 4 feet tall. s. You are older than 16 years old. The sentence can be represented as\n\nLogic and Bit Operations A bit has two values: 1(true) and 0(false). A variable is called a Boolean variable if its value is either true or false. Bit operations are written to be AND, OR and XOR in programming languages. Table 7. Table for the bit operations OR,AND and XOR x0011x0011 y0101y0101 01110111 00010001 01100110 Example Extend bit operations to bit strings. 01 1011 0110 11 0001 1101 11 1011 1111 bitwise OR 01 0001 0100 bitwise AND 10 1010 1011 bitwise XOR\n\n1.2 Propositional Equivalences Definition 1. A tautology is a compound proposition that is always true no matter what the values of the propositions that occur in it. A contradiction is a compound proposition that is always false ． A contingency is a proposition that is neither a tautology nor a contradiction. Table 1. Examples of a Tautology and a Contradiction. pTFpTF FTFT TTTT FFFF a contradiction a tautology Example 1.\n\nLogic Equivalences Definition 2. The proposition p and q are called logically equivalent if is a tautology. The notation denotes that p and q are logically equivalent. Using a truth table to determine whether two propositions are equivalent Example 2 Example 3 equivalent\n\nSome important equivalences.\n\nExample 4 Example 5\n\n1.3 Predicates and Quantifiers Propositional function A statement involving a variable x is P(x) is said to be a propositional function if x is a variable and P(x) becomes a proposition when a value has been assigned to x. In general, a statement involving the n variables Example 1 Let P(x) denote the statement “x>3”. What are the truth values of P(4) and P(2)? Example 2 Let Q(x,y) denote the statement “x=y+3”. What are the truth values of the propositions Q(1,2) and Q(3,0)?\n\nQuantifiers The universal quantification of P(x),denoted as is the proposition “P(x) is true for all values of x in the universe of discourse.” Solution P(x): x has studied calculus. S(x): x is in this class. The statement can be expressed as universal quantifier Example 3 Express the statement “Every student in this class has studied calculus.\n\nexistence quantifier The existential quantification of P(x),denoted as is the proposition “There exists an element x in the universe of discourse such that P(x) is true.”\n\nSolution: Every student in your school has a computer or has a friend who has a computer. Solution: There is a student none of whose friends are also friends with each other.\n\nTranslating Sentences into Logical Expressions Example 10 Express the statements “Some student in this class has visited Mexico” and “every student in this class has visited either Canada or Mexico using quantifiers. Example 11 Express the statement “Everyone has exactly one best friend” as a logical expression.\n\nExample 12 Express the statement “There is a woman who has taken a flight on every airline in the world. Negations: the negation of quantified expressions. Example 14 There is a student in the class who has taken a course in calculus. Example 13 Every student in the class has taken a course in calculus.\n\nSimilar presentations\n\nAds by Google"
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https://number.rocks/squared/1895/1509 | [
"1895/1509 squared as a fraction\n\nCalculate how much is 1895 square divided by 1509 squared\n\nThe fraction 1895/1509 square is equal to 3591025/2277081 in fraction\n\n(1895/1509)2 =\n3591025/2277081\n\nEvaluate 1895/1509 square\n\n= (1895/1509)2\n\n= 18952/15092\n\n= 3591025/2277081\n\nNext Fraction:"
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https://metric-calculator.com/convert-microgram-to-picogram.htm | [
"# Micrograms to Picograms (mcg to pg) Conversion\n\nThis converter provides conversion of micrograms to picograms (mcg to pg) and backwards.\n\nEnter micrograms or picograms for conversion:\n\nSelect conversion type:\n\nRounding options:\n\nConversion Chart\n micrograms to picograms Conversion Table: 1 mcg = 1000000 pg2 mcg = 2000000 pg3 mcg = 3000000 pg4 mcg = 4000000 pg5 mcg = 5000000 pg6 mcg = 6000000 pg7 mcg = 7000000 pg8 mcg = 8000000 pg9 mcg = 9000000 pg10 mcg = 10000000 pg11 mcg = 11000000 pg12 mcg = 12000000 pg13 mcg = 13000000 pg14 mcg = 14000000 pg15 mcg = 15000000 pg16 mcg = 16000000 pg17 mcg = 17000000 pg18 mcg = 18000000 pg19 mcg = 19000000 pg20 mcg = 20000000 pg21 mcg = 21000000 pg22 mcg = 22000000 pg23 mcg = 23000000 pg24 mcg = 24000000 pg25 mcg = 25000000 pg26 mcg = 26000000 pg27 mcg = 27000000 pg28 mcg = 28000000 pg29 mcg = 29000000 pg30 mcg = 30000000 pg31 mcg = 31000000 pg32 mcg = 32000000 pg33 mcg = 33000000 pg34 mcg = 34000000 pg35 mcg = 35000000 pg36 mcg = 36000000 pg37 mcg = 37000000 pg38 mcg = 38000000 pg39 mcg = 39000000 pg40 mcg = 40000000 pg41 mcg = 41000000 pg42 mcg = 42000000 pg43 mcg = 43000000 pg44 mcg = 44000000 pg45 mcg = 45000000 pg46 mcg = 46000000 pg47 mcg = 47000000 pg48 mcg = 48000000 pg49 mcg = 49000000 pg50 mcg = 50000000 pg 51 mcg = 51000000 pg52 mcg = 52000000 pg53 mcg = 53000000 pg54 mcg = 54000000 pg55 mcg = 55000000 pg56 mcg = 56000000 pg57 mcg = 57000000 pg58 mcg = 58000000 pg59 mcg = 59000000 pg60 mcg = 60000000 pg61 mcg = 61000000 pg62 mcg = 62000000 pg63 mcg = 63000000 pg64 mcg = 64000000 pg65 mcg = 65000000 pg66 mcg = 66000000 pg67 mcg = 67000000 pg68 mcg = 68000000 pg69 mcg = 69000000 pg70 mcg = 70000000 pg71 mcg = 71000000 pg72 mcg = 72000000 pg73 mcg = 73000000 pg74 mcg = 74000000 pg75 mcg = 75000000 pg76 mcg = 76000000 pg77 mcg = 77000000 pg78 mcg = 78000000 pg79 mcg = 79000000 pg80 mcg = 80000000 pg81 mcg = 81000000 pg82 mcg = 82000000 pg83 mcg = 83000000 pg84 mcg = 84000000 pg85 mcg = 85000000 pg86 mcg = 86000000 pg87 mcg = 87000000 pg88 mcg = 88000000 pg89 mcg = 89000000 pg90 mcg = 90000000 pg91 mcg = 91000000 pg92 mcg = 92000000 pg93 mcg = 93000000 pg94 mcg = 94000000 pg95 mcg = 95000000 pg96 mcg = 96000000 pg97 mcg = 97000000 pg98 mcg = 98000000 pg99 mcg = 99000000 pg100 mcg = 100000000 pg\n\n1 microgram (mcg) = 1000000 picogram (pg). Microgram (mcg) is a unit of Weight used in Metric system. Picogram (pg) is a unit of Weight used in Metric system. Micrograms also can be marked as Microgrammes or µg (alternative British English spelling in UK). Picograms also can be marked as Picogrammes (alternative British English spelling in UK)."
]
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https://www.colorhexa.com/def5ff | [
"# #def5ff Color Information\n\nIn a RGB color space, hex #def5ff is composed of 87.1% red, 96.1% green and 100% blue. Whereas in a CMYK color space, it is composed of 12.9% cyan, 3.9% magenta, 0% yellow and 0% black. It has a hue angle of 198.2 degrees, a saturation of 100% and a lightness of 93.5%. #def5ff color hex could be obtained by blending #ffffff with #bdebff. Closest websafe color is: #ccffff.\n\n• R 87\n• G 96\n• B 100\nRGB color chart\n• C 13\n• M 4\n• Y 0\n• K 0\nCMYK color chart\n\n#def5ff color description : Very pale blue.\n\n# #def5ff Color Conversion\n\nThe hexadecimal color #def5ff has RGB values of R:222, G:245, B:255 and CMYK values of C:0.13, M:0.04, Y:0, K:0. Its decimal value is 14611967.\n\nHex triplet RGB Decimal def5ff `#def5ff` 222, 245, 255 `rgb(222,245,255)` 87.1, 96.1, 100 `rgb(87.1%,96.1%,100%)` 13, 4, 0, 0 198.2°, 100, 93.5 `hsl(198.2,100%,93.5%)` 198.2°, 12.9, 100 ccffff `#ccffff`\nCIE-LAB 95.183, -5.54, -7.355 80.823, 88.053, 107.34 0.293, 0.319, 88.053 95.183, 9.208, 233.012 95.183, -12.716, -10.599 93.837, -10.47, -2.136 11011110, 11110101, 11111111\n\n# Color Schemes with #def5ff\n\n• #def5ff\n``#def5ff` `rgb(222,245,255)``\n• #ffe8de\n``#ffe8de` `rgb(255,232,222)``\nComplementary Color\n• #defff9\n``#defff9` `rgb(222,255,249)``\n• #def5ff\n``#def5ff` `rgb(222,245,255)``\n• #dee5ff\n``#dee5ff` `rgb(222,229,255)``\nAnalogous Color\n• #fff9de\n``#fff9de` `rgb(255,249,222)``\n• #def5ff\n``#def5ff` `rgb(222,245,255)``\n• #ffdee5\n``#ffdee5` `rgb(255,222,229)``\nSplit Complementary Color\n• #f5ffde\n``#f5ffde` `rgb(245,255,222)``\n• #def5ff\n``#def5ff` `rgb(222,245,255)``\n• #ffdef5\n``#ffdef5` `rgb(255,222,245)``\nTriadic Color\n• #deffe8\n``#deffe8` `rgb(222,255,232)``\n• #def5ff\n``#def5ff` `rgb(222,245,255)``\n• #ffdef5\n``#ffdef5` `rgb(255,222,245)``\n• #ffe8de\n``#ffe8de` `rgb(255,232,222)``\nTetradic Color\n• #92deff\n``#92deff` `rgb(146,222,255)``\n• #abe6ff\n``#abe6ff` `rgb(171,230,255)``\n• #c5edff\n``#c5edff` `rgb(197,237,255)``\n• #def5ff\n``#def5ff` `rgb(222,245,255)``\n• #f8fdff\n``#f8fdff` `rgb(248,253,255)``\n• #ffffff\n``#ffffff` `rgb(255,255,255)``\n• #ffffff\n``#ffffff` `rgb(255,255,255)``\nMonochromatic Color\n\n# Alternatives to #def5ff\n\nBelow, you can see some colors close to #def5ff. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #defdff\n``#defdff` `rgb(222,253,255)``\n• #defbff\n``#defbff` `rgb(222,251,255)``\n• #def8ff\n``#def8ff` `rgb(222,248,255)``\n• #def5ff\n``#def5ff` `rgb(222,245,255)``\n• #def2ff\n``#def2ff` `rgb(222,242,255)``\n• #def0ff\n``#def0ff` `rgb(222,240,255)``\n• #deedff\n``#deedff` `rgb(222,237,255)``\nSimilar Colors\n\n# #def5ff Preview\n\nText with hexadecimal color #def5ff\n\nThis text has a font color of #def5ff.\n\n``<span style=\"color:#def5ff;\">Text here</span>``\n#def5ff background color\n\nThis paragraph has a background color of #def5ff.\n\n``<p style=\"background-color:#def5ff;\">Content here</p>``\n#def5ff border color\n\nThis element has a border color of #def5ff.\n\n``<div style=\"border:1px solid #def5ff;\">Content here</div>``\nCSS codes\n``.text {color:#def5ff;}``\n``.background {background-color:#def5ff;}``\n``.border {border:1px solid #def5ff;}``\n\n# Shades and Tints of #def5ff\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, #000406 is the darkest color, while #f2fbff is the lightest one.\n\n• #000406\n``#000406` `rgb(0,4,6)``\n• #00121a\n``#00121a` `rgb(0,18,26)``\n• #00202d\n``#00202d` `rgb(0,32,45)``\n• #002d41\n``#002d41` `rgb(0,45,65)``\n• #003b55\n``#003b55` `rgb(0,59,85)``\n• #004968\n``#004968` `rgb(0,73,104)``\n• #00567c\n``#00567c` `rgb(0,86,124)``\n• #006490\n``#006490` `rgb(0,100,144)``\n• #0072a3\n``#0072a3` `rgb(0,114,163)``\n• #007fb7\n``#007fb7` `rgb(0,127,183)``\n• #008dca\n``#008dca` `rgb(0,141,202)``\n• #009bde\n``#009bde` `rgb(0,155,222)``\n• #00a8f2\n``#00a8f2` `rgb(0,168,242)``\nShade Color Variation\n• #06b4ff\n``#06b4ff` `rgb(6,180,255)``\n• #1abaff\n``#1abaff` `rgb(26,186,255)``\n• #2dc0ff\n``#2dc0ff` `rgb(45,192,255)``\n• #41c5ff\n``#41c5ff` `rgb(65,197,255)``\n• #55cbff\n``#55cbff` `rgb(85,203,255)``\n• #68d1ff\n``#68d1ff` `rgb(104,209,255)``\n• #7cd7ff\n``#7cd7ff` `rgb(124,215,255)``\n• #90ddff\n``#90ddff` `rgb(144,221,255)``\n• #a3e3ff\n``#a3e3ff` `rgb(163,227,255)``\n• #b7e9ff\n``#b7e9ff` `rgb(183,233,255)``\n• #caefff\n``#caefff` `rgb(202,239,255)``\n• #def5ff\n``#def5ff` `rgb(222,245,255)``\n• #f2fbff\n``#f2fbff` `rgb(242,251,255)``\nTint Color Variation\n\n# Tones of #def5ff\n\nA tone is produced by adding gray to any pure hue. In this case, #edeff0 is the less saturated color, while #def5ff is the most saturated one.\n\n• #edeff0\n``#edeff0` `rgb(237,239,240)``\n• #ecf0f1\n``#ecf0f1` `rgb(236,240,241)``\n• #ebf0f2\n``#ebf0f2` `rgb(235,240,242)``\n• #e9f1f4\n``#e9f1f4` `rgb(233,241,244)``\n• #e8f1f5\n``#e8f1f5` `rgb(232,241,245)``\n• #e7f2f6\n``#e7f2f6` `rgb(231,242,246)``\n• #e6f2f7\n``#e6f2f7` `rgb(230,242,247)``\n• #e4f3f9\n``#e4f3f9` `rgb(228,243,249)``\n• #e3f3fa\n``#e3f3fa` `rgb(227,243,250)``\n• #e2f4fb\n``#e2f4fb` `rgb(226,244,251)``\n• #e1f4fc\n``#e1f4fc` `rgb(225,244,252)``\n• #dff5fe\n``#dff5fe` `rgb(223,245,254)``\n• #def5ff\n``#def5ff` `rgb(222,245,255)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #def5ff is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population"
]
| [
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.50316,"math_prob":0.90506655,"size":3703,"snap":"2021-21-2021-25","text_gpt3_token_len":1642,"char_repetition_ratio":0.15112193,"word_repetition_ratio":0.011090573,"special_character_ratio":0.5120173,"punctuation_ratio":0.23250565,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9859733,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-25T04:31:31Z\",\"WARC-Record-ID\":\"<urn:uuid:3ffd6724-d940-44c9-bc84-e99c2e49f65e>\",\"Content-Length\":\"36337\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:10533182-b0eb-4f7a-949f-6fc431cba4ea>\",\"WARC-Concurrent-To\":\"<urn:uuid:f1d21c3e-4ebf-4f7e-a012-f4754117ce67>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/def5ff\",\"WARC-Payload-Digest\":\"sha1:CP47KEHEBHAZSE6DFSZQ7UR6YIRSX5RW\",\"WARC-Block-Digest\":\"sha1:KALRP5QCMEDPQ3TWHB3JMV2SAWLWKK23\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623488567696.99_warc_CC-MAIN-20210625023840-20210625053840-00554.warc.gz\"}"} |
https://nhomkinhnamphat.com/top-10-what-is-20-of-1200-that-will-change-your-life/ | [
"Monday, 23 Jan 2023\n\n# Top 10 what is 20% of 1200 That Will Change Your Life\n\nMục Lục\n\nBelow is information and knowledge on the topic what is 20% of 1200 gather and compiled by the nhomkinhnamphat.com team. Along with other related topics like: What is 20% of 1000, What is 15 of 1200, 20% of 1440, What is 10 of 1200, What is 30 of 1200, 15 of 1200, What is 25 percent of 1200, 25% of 1200.\n\n# What is 20% of 1200? [Solved]\n\nA percent is a ratio of a number expressed out of 100.\n\n## Answer: 20% of 1200 is 240.\n\nLet’s find 20% of 1200.\n\nExplanation:\n\n20% of 1200 can be written as 20% × 1200\n\n= 20/100 × 1200\n\n= 240\n\n## Extra Information About what is 20% of 1200 That You May Find Interested\n\nIf the information we provide above is not enough, you may find more below here.",
null,
"### What is 20% of 1200? [Solved] – Cuemath\n\n• Author: cuemath.com\n\n• Rating: 5⭐ (353905 rating)\n\n• Highest Rate: 5⭐\n\n• Lowest Rate: 3⭐\n\n• Sumary: 20% of 1200 is 240.\n\n• Matching Result: Answer: 20% of 1200 is 240. Let’s find 20% of 1200. Explanation: 20% of 1200 can be written as 20% × 1200. = 20/100 × 1200.\n\n• Intro: What is 20% of 1200? [Solved] A percent is a ratio of a number expressed out of 100. Answer: 20% of 1200 is 240. Let’s find 20% of 1200. Explanation: 20% of 1200 can be written as 20% × 1200 = 20/100 × 1200 = 240 Therefore, 20% of 1200 is 240.",
null,
"### What is 20 percent of 1200?\n\n• Rating: 5⭐ (353905 rating)\n\n• Highest Rate: 5⭐\n\n• Lowest Rate: 3⭐\n\n• Sumary: What is 20 percent of 1200? The answer is 240. Get stepwise instructions to work out “20% of 1200”.\n\n• Matching Result: Write 20% as 20/100 · Since, finding the fraction of a number is same as multiplying the fraction with the number, we have 20/100 of 1200 = 20/100 × 1200 …\n\n• Intro: What is 20 percent of 1200? 20% of 120020% of 1200 is 240Working out 20% of 1200Write 20% as 20/100Since, finding the fraction of a number is same as multiplying the fraction with the number, we have20/100 of 1200 = 20/100 × 1200Therefore, the answer is 240If you are using…\nXem Thêm: Top 10 what's it like to be a female emt That Will Change Your Life",
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"### What is 20. percent of 1200? = 240 – Percentage Calculator\n\n• Author: percentagecal.com\n\n• Rating: 5⭐ (353905 rating)\n\n• Highest Rate: 5⭐\n\n• Lowest Rate: 3⭐\n\n• Sumary: 20. percent *1200 =\n\n• Matching Result: Solution for What is 20. percent of 1200: 20. percent *1200 = (20.:100)*1200 = (20.*1200):100 = 24000:100 = 240. Now we have: 20. percent of 1200 = 240.\n\n• Intro: What is 20. percent of 1200? = 240 Solution for What is 20. percent of 1200: 20. percent *1200 = (20.:100)*1200 = (20.*1200):100 = 24000:100 = 240Now we have: 20. percent of 1200 = 240Question: What is 20. percent of 1200?Percentage solution with steps:Step 1: Our output value is 1200.Step…",
null,
"### What is 20 percent of 1200 – percentagecalculator.guru\n\n• Author: percentagecalculator.guru\n\n• Rating: 5⭐ (353905 rating)\n\n• Highest Rate: 5⭐\n\n• Lowest Rate: 3⭐\n\n• Sumary: What is 20 percent of 1200? How much is 20% of 1200?\n\n• Matching Result: 20 percent of 1200 is 240. 3. How to calculate 20 percent of 1200? Multiply 20/100 with 1200 = (20/100)*1200 = (20*1200)/ …\n\n• Intro: Percentage Calculator: What is 20 percent of 1200 – percentagecalculator.guru20 percent *1200= (20/100)*1200= (20*1200)/100= 24000/100 = 240Now we have: 20 percent of 1200 = 240Question: What is 20 percent of 1200?We need to determine 20% of 1200 now and the procedure explaining it as suchStep 1: In the given case…",
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"### 20% of 1200 Dollars – CoolConversion\n\n• Author: coolconversion.com\n\n• Rating: 5⭐ (353905 rating)\n\n• Highest Rate: 5⭐\n\n• Lowest Rate: 3⭐\n\n• Sumary: 20 is what percent of 1200? How to work out percentages? See how by using our percentage calculator online.\n\n• Matching Result: / 100 = Part / ; % = (20 x 100) / 1200 = 1.6666666666667% ; 20 is out of 1200 = 20/1200 x 100 = 1.6666666666667%.\n\n• Intro: 20% of 1200 Dollars Percentage Calculator Using this tool you can find any percentage in three ways. So, we think you reached us looking for answers like: 1) What is 20 percent (%) of 1200? 2) 20 is what percent of 1200? Or may be: 20% of 1200 Dollars? See…\nXem Thêm: Top 10 17.25 an hour is how much a year That Will Change Your Life",
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"### What is 20% off 1200 Dollars – CoolConversion\n\n• Author: coolconversion.com\n\n• Rating: 5⭐ (353905 rating)\n\n• Highest Rate: 5⭐\n\n• Lowest Rate: 3⭐\n\n• Sumary: How to calculate discount. 20% off 1200 calculator. Using this calculator you will know how to find the percent of discount of any item by just plugging in the item price and the discount in percent.\n\n• Matching Result: What is 20% off 1200 Dollars … The easiest way of calculating discount is, in this case, to multiply the normal price \\$1200 by 20 then divide it by one hundred.\n\n• Intro: What is 20% off 1200 Dollars An item that costs \\$1200, when discounted 20 percent, will cost \\$960 The easiest way of calculating discount is, in this case, to multiply the normal price \\$1200 by 20 then divide it by one hundred. So, the discount is equal to \\$240. To…",
null,
"### What is 20 percent off 1200 dollars or pounds\n\n• Author: percentage-off-calculator.com\n\n• Rating: 5⭐ (353905 rating)\n\n• Highest Rate: 5⭐\n\n• Lowest Rate: 3⭐\n\n• Sumary: Note: 1200 dollar to pound = 792 pound\n\n• Matching Result: Solution: 20% off 1200 is equal to (20 x 20) / 100 = 240. So if you buy an item at \\$1200 with 20% discounts, you will pay \\$960 and get 240 cashback rewards.\n\n• Intro: What is 20 percent off 1200 dollars| How to calculate 20% off 1200 pounds| 20% off 1200 What is 20 percent off 1200 dollars or pounds ? Note: 1200 dollar to pound = 792 pound Solution: 20% off 1200 is equal to (20 x 20) / 100 = 240. So…",
null,
"### 20 percent of 1200 – Percent Calculator\n\n• Author: percent.info\n\n• Rating: 5⭐ (353905 rating)\n\n• Highest Rate: 5⭐\n\n• Lowest Rate: 3⭐\n\n• Sumary: What is 20 percent of 1200? Step-by-step showing you how to calculate 20 percent of 1200. Detailed explanation with answer to 20% of 1200.\n\n• Matching Result: Below is a pie chart illustrating 20 percent of 1200. The pie contains 1200 parts, and the blue part of the pie is 240 parts or 20 percent of 1200. Pie chart …\n\n• Intro: 20 percent of 1200 (20% of 1200) Here we will show you how to calculate twenty percent of one thousand two hundred. Before we continue, note that 20 percent of 1200 is the same as 20% of 1200. We will write it both ways throughout this tutorial to remind you…\nXem Thêm: Top 10 how much do orthodontist get paid in hawaii That Will Change Your Life",
null,
"### What is 20 Percent of 1200? – Calculate Percentage – Quizzes.cc\n\n• Author: quizzes.cc\n\n• Rating: 5⭐ (353905 rating)\n\n• Highest Rate: 5⭐\n\n• Lowest Rate: 3⭐\n\n• Sumary: How much is 20 percent of ? Use the calculator below to calculate a percentage, either as a percentage of a number, such as 20% of or the percentage of…\n\n• Matching Result: How much is 20 percent of the following numbers? ; 20% of 1200.01 = 24000.2. 20% of 1200.02 = 24000.4. 20% of 1200.03 = 24000.6 ; 20% of 1200.26 = 24005.2. 20% of …\n\n• Intro: What is 20 Percent of ? – Calculate Percentage How much is 20 percent of ? Use the calculator below to calculate a percentage, either as a percentage of a number, such as 20% of or the percentage of 2 numbers. Change the numbers to calculate different amounts. Simply type…",
null,
"### 20 percent off 1200 Calculator\n\n• Author: percent-off.com\n\n• Rating: 5⭐ (353905 rating)\n\n• Highest Rate: 5⭐\n\n• Lowest Rate: 3⭐\n\n• Sumary: How to calculate 20 percent-off \\$1200. How to figure out percentages off a price. Using this calculator you will find that the amount after the discount is \\$960\n\n• Matching Result: Amount Saved = \\$240 (answer). In other words, a 20% discount for a item with original price of \\$1200 is equal to \\$240 (Amount Saved). Note that to …\n\n• Intro: 20 percent off 1200 CalculatorHow to calculate 20 percent-off \\$1200. How to figure out percentages off a price. Using this calculator you will find that the amount after the discount is \\$960. To find any discount, just use our Discount Calculator above. Using this calculator you can find the discount…\n\nIf you have questions that need to be answered about the topic what is 20% of 1200, then this section may help you solve it.\n\n240\n\n### How much is a 20% discount? Take…\n\nHow much does a 20% discount cost?\n\n• Take the original number and divide it by 10.\n• Subtract your doubled number from the original number.\n• You have taken 20 percent off! For \\$30, you should have \\$24.\n\n200\n\n300\n\nRate this post"
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null,
"品牌好的郴州景观亮化推荐，景观亮化哪家好\n\n品牌:利得尔,欧普照明,三峰\n\n出厂地:天峨县(六排镇)\n\n报价：面议\n\n郴州市佳境光电科技有限公司\n\n黄金会员：",
null,
"主营：路灯,室内灯具,户外灯具...\n\n•",
null,
"供应销量好的郴州led显示屏_宜章显示屏安装厂家批发\n\n品牌:利得尔,欧普照明,三峰\n\n出厂地:天峨县(六排镇)\n\n报价：面议\n\n郴州市佳境光电科技有限公司\n\n黄金会员：",
null,
"主营：路灯,室内灯具,户外灯具...\n\n•",
null,
"想买实用的郴州路灯就来郴州佳境光电_郴州显示屏\n\n品牌:利得尔,欧普照明,三峰\n\n出厂地:天峨县(六排镇)\n\n报价：面议\n\n郴州市佳境光电科技有限公司\n\n黄金会员：",
null,
"主营：路灯,室内灯具,户外灯具...\n\n•",
null,
"长沙配重块-有信誉度的配重铁厂家在湖南\n\n品牌:鼎昌铸业,鼎昌,嘉禾鼎昌\n\n出厂地:天峨县(六排镇)\n\n报价：面议\n\n嘉禾县鼎昌铸业五金厂\n\n黄金会员：",
null,
"主营：配重块,配重铁,桩机复合...\n\n•",
null,
"供应有品质的郴州交通信号灯_交通信号灯\n\n品牌:众申,众申交通设施,众申交通\n\n出厂地:天峨县(六排镇)\n\n报价：面议\n\n郴州众申交通设施开发有限公司\n\n黄金会员：",
null,
"主营：交通标线,标牌,电子警察...\n\n•",
null,
"提供优良郴州景观亮化_楼宇亮化\n\n品牌:利得尔,欧普照明,三峰\n\n出厂地:天峨县(六排镇)\n\n报价：面议\n\n郴州市佳境光电科技有限公司\n\n黄金会员：",
null,
"主营：路灯,室内灯具,户外灯具...\n\n•",
null,
"供应有品质的郴州划线-资兴安全的划线\n\n品牌:众申,众申交通设施,众申交通\n\n出厂地:天峨县(六排镇)\n\n报价：面议\n\n郴州众申交通设施开发有限公司\n\n黄金会员：",
null,
"主营：交通标线,标牌,电子警察...\n\n•",
null,
"辽宁杜鹃苗木产品哪家好-划算的杜鹃苗木出售\n\n品牌:湘南,,\n\n出厂地:天峨县(六排镇)\n\n报价：面议\n\n郴州市湘南苗木种植专业合作社\n\n黄金会员：",
null,
"主营：草皮,马尼拉草皮,台湾青...\n\n•",
null,
"屋面补漏哪家好-上哪找专业的郴州屋面补漏服务\n\n品牌:国仁,国仁防水,郴州国仁\n\n出厂地:天峨县(六排镇)\n\n报价：面议\n\n郴州国仁防水工程有限公司\n\n黄金会员：",
null,
"主营：防水工程,防水补漏,外墙...\n\n•",
null,
"标识厂家-可信赖的郴州交通标识安装就选众申交通设施\n\n品牌:众申,众申交通设施,众申交通\n\n出厂地:天峨县(六排镇)\n\n报价：面议\n\n郴州众申交通设施开发有限公司\n\n黄金会员：",
null,
"主营：交通标线,标牌,电子警察...\n\n• 没有找到合适的郴州市供应商？您可以发布采购信息\n\n没有找到满足要求的郴州市供应商？您可以搜索 批发 公司\n\n### 最新入驻厂家\n\n相关产品:\n郴州景观亮化 郴州led显示屏 郴州路灯 长沙配重块 郴州交通信号灯 郴州景观亮化 郴州划线 辽宁杜鹃苗木产品哪家好 屋面补漏哪家好 郴州交通标识安装"
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https://docs.microsoft.com/en-us/previous-versions/visualstudio/visual-studio-2013/hh749974%28v%3Dvs.120%29 | [
"# parallel_reduce Function\n\nComputes the sum of all elements in a specified range by computing successive partial sums, or computes the result of successive partial results similarly obtained from using a specified binary operation other than sum, in parallel. parallel_reduce is semantically similar to std::accumulate, except that it requires the binary operation to be associative, and requires an identity value instead of an initial value.\n\n``````template<\ntypename _Forward_iterator\n>\ninline typename std::iterator_traits<_Forward_iterator>::value_type parallel_reduce(\n_Forward_iterator_Begin,\n_Forward_iterator_End,\nconst typename std::iterator_traits<_Forward_iterator>::value_type &_Identity\n);\n\ntemplate<\ntypename _Forward_iterator,\ntypename _Sym_reduce_fun\n>\ninline typename std::iterator_traits<_Forward_iterator>::value_type parallel_reduce(\n_Forward_iterator_Begin,\n_Forward_iterator_End,\nconst typename std::iterator_traits<_Forward_iterator>::value_type &_Identity,\n_Sym_reduce_fun_Sym_fun\n);\n\ntemplate<\ntypename _Reduce_type,\ntypename _Forward_iterator,\ntypename _Range_reduce_fun,\ntypename _Sym_reduce_fun\n>\ninline _Reduce_type parallel_reduce(\n_Forward_iterator_Begin,\n_Forward_iterator_End,\nconst _Reduce_type& _Identity,\nconst _Range_reduce_fun &_Range_fun,\nconst _Sym_reduce_fun &_Sym_fun\n);\n``````\n\n## Parameters\n\n• _Forward_iterator\nThe iterator type of input range.\n\n• _Sym_reduce_fun\nThe type of the symmetric reduction function. This must be a function type with signature _Reduce_type _Sym_fun(_Reduce_type, _Reduce_type), where _Reduce_type is the same as the identity type and the result type of the reduction. For the third overload, this should be consistent with the output type of _Range_reduce_fun.\n\n• _Reduce_type\nThe type that the input will reduce to, which can be different from the input element type. The return value and identity value will has this type.\n\n• _Range_reduce_fun\nThe type of the range reduction function. This must be a function type with signature _Reduce_type _Range_fun(_Forward_iterator, _Forward_iterator, _Reduce_type), _Reduce_type is the same as the identity type and the result type of the reduction.\n\n• _Begin\nAn input iterator addressing the first element in the range to be reduced.\n\n• _End\nAn input iterator addressing the element that is one position beyond the final element in the range to be reduced.\n\n• _Identity\nThe identity value _Identity is of the same type as the result type of the reduction and also the value_type of the iterator for the first and second overloads. For the third overload, the identity value must have the same type as the result type of the reduction, but can be different from the value_type of the iterator. It must have an appropriate value such that the range reduction operator _Range_fun, when applied to a range of a single element of type value_type and the identity value, behaves like a type cast of the value from type value_type to the identity type.\n\n• _Sym_fun\nThe symmetric function that will be used in the second of the reduction. Refer to Remarks for more information.\n\n• _Range_fun\nThe function that will be used in the first phase of the reduction. Refer to Remarks for more information.\n\n## Return Value\n\nThe result of the reduction.\n\n## Remarks\n\nTo perform a parallel reduction, the function divides the range into chunks based on the number of workers available to the underlying scheduler. The reduction takes place in two phases, the first phase performs a reduction within each chunk, and the second phase performs a reduction between the partial results from each chunk.\n\nThe first overload requires that the iterator's value_type, T, be the same as the identity value type as well as the reduction result type. The element type T must provide the operator T T::operator + (T) to reduce elements in each chunk. The same operator is used in the second phase as well.\n\nThe second overload also requires that the iterator's value_type be the same as the identity value type as well as the reduction result type. The supplied binary operator _Sym_fun is used in both reduction phases, with the identity value as the initial value for the first phase.\n\nFor the third overload, the identity value type must be the same as the reduction result type, but the iterator's value_type may be different from both. The range reduction function _Range_fun is used in the first phase with the identity value as the initial value, and the binary function _Sym_reduce_fun is applied to sub results in the second phase."
]
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https://gitlab.science.ru.nl/benoit/tweetnacl/-/commit/4f1f4474e967c48f76eaec371ee920b13ec7f5cb | [
"### fix low-level\n\nparent 7faffc34\n ... ... @@ -7,12 +7,13 @@ Clight description of TweetNaCl and Coq functions producing similar behaviors. \\subsection{One proof? No, two!} In order to prove the correctness of X25519, we use VST to prove that TweetNaCl's code matches our functional specification of \\texttt{crypto\\_scalarmult} (abreviated as CSM) in Coq. Then we prove that our specification of the scalar multiplication matches the mathematical definition of elliptic curves and Theorem 2.1 by Bernstein~\\cite{Ber06}. Figure~\\ref{tk:ProofOverview} shows a graph of dependencies of the proofs considered. The mathematical proof of our specification is presented in Section~\\ref{sec3-maths}. In order to prove the correctness of X25519 in TweetNaCl code, we use VST to prove the code matches our functional specification of \\texttt{crypto\\_scalarmult} (abreviated as CSM) in Coq. Then, we prove that our specification of the scalar multiplication matches the mathematical definition of elliptic curves and Theorem 2.1 by Bernstein~\\cite{Ber06}. Figure~\\ref{tk:ProofOverview} shows a graph of dependencies of the proofs considered. The mathematical proof of our specification is presented in Section~\\ref{sec3-maths}. \\begin{figure}[h] \\centering ... ... @@ -27,7 +28,7 @@ subsequently called: \\texttt{unpack25519}; \\texttt{A}; \\texttt{Z}; \\texttt{M}; \\texttt{S}; \\texttt{car25519}; \\texttt{inv25519}; \\texttt{set25519}; \\texttt{sel25519}; \\texttt{pack25519}. We prove that the implementation of Curve25519 is \\textbf{sound} \\ie We prove the implementation of Curve25519 is \\textbf{sound}, \\ie: \\begin{itemize} \\item absence of access out-of-bounds of arrays (memory safety). \\item absence of overflows/underflow on the arithmetic. ... ... @@ -37,25 +38,26 @@ We also prove that TweetNaCl's code is \\textbf{correct}: \\item Curve25519 is correctly implemented (we get what we expect). \\item Operations on \\texttt{gf} (\\texttt{A}, \\texttt{Z}, \\texttt{M}, \\texttt{S}) are equivalent to operations ($+,-,\\times,x^2$) in \\Zfield. \\item The Montgomery ladder does compute a scalar multiplication between a natural number and a point. \\item The Montgomery ladder computes a scalar multiplication between a natural number and a point. \\end{itemize} In order to prove the soundness and correctness of \\texttt{crypto\\_scalarmult}, we first create a skeleton of the Montgomery ladder with abstract operations which can be instanciated over lists, integers, field elements... A high level specification (over a generic field $\\K$) allows use to prove the can be instantiated over lists, integers, field elements... A high level specification (over a generic field $\\K$) allows us to prove the correctness of the ladder with respect to the curves theory. This high specification does not rely on the parameters of Curve25519. By instanciating $\\K$ with $\\Zfield$, and the parameters of Curve25519 ($a = 486662, b = 1$), we define a middle level specification. Additionally we also provide a low level specification close to the \\texttt{C} code This high-level specification does not rely on the parameters of Curve25519. By instantiating $\\K$ with $\\Zfield$, and the parameters of Curve25519 ($a = 486662, b = 1$), we define a mid-level specification. Additionally we also provide a low-level specification close to the \\texttt{C} code (over lists of $\\Z$). We show this specification to be equivalent to the \\textit{semantic version} of C (\\texttt{CLight}) with VST. This low level specification gives us the soundness assurance. By showing that operations over instances ($\\K = \\Zfield$, $\\Z$, list of $\\Z$) are equivalent we bridge the gap between the low level and the high level specification equivalent, we bridge the gap between the low level and the high level specification with Curve25519 parameters. As such we prove all specifications to equivalent (Fig.\\ref{tk:ProofStructure}). As such, we prove all specifications to equivalent (Fig.\\ref{tk:ProofStructure}). This garantees us the correctness of the implementation. \\begin{figure}[h] ... ... @@ -150,7 +152,7 @@ in Coq (for the sake of simplicity we do not display the conversion in the theor For all $n \\in \\N, n < 2^{255}$ and where the bits 1, 2, 5 248, 249, 250 are cleared and bit 6 is set, for all $P \\in E(\\F{p^2})$, for all $p \\in \\F{p}$ such that $P.x = p$, there exists $Q \\in E(\\F{p^2})$ such that $Q = nP$ where $Q.x = q$ and $q$ = \\VSTe{CSM} $n$ $p$. there exists $Q \\in E(\\F{p^2})$ such that $Q = [n]P$ where $Q.x = q$ and $q$ = \\VSTe{CSM} $n$ $p$. \\end{theorem} A more complete description in Coq of Theorem \\ref{CSM-correct} with the associated conversions is as follow: ... ... @@ -280,14 +282,13 @@ Lemma Inv25519_Z_GF : forall (g:list Z), (Z16.lst (Inv25519 g)) :GF = (Inv25519_Z (Z16.lst g)) :GF. \\end{lstlisting} In TweetNaCl, \\TNaCle{inv25519} computes an inverse in $\\Zfield$. It uses the Fermat's little theorem by doing an exponentiation to $2^{255}-21$. In TweetNaCl, \\TNaCle{inv25519} computes an inverse in $\\Zfield$. It uses the Fermat's little theorem by doing an exponentiation to $2^{255}-21$. This is done by applying a square-and-multiply algorithm. The binary representation of $p-2$ implies to always do a multiplications aside for bit 2 and 4, thus the if case. of $p-2$ implies to always do multiplications aside for bits 2 and 4. To prove the correctness of the result we can use multiple strategies such as: \\begin{itemize} \\item Proving it is special case of square-and-multiply algorithm applied to a specific number and then show that this number is indeed $2^{255}-21$. \\item Proving it is a special case of square-and-multiply algorithm applied to $2^{255}-21$. \\item Unrolling the for loop step-by-step and applying the equalities $x^a \\times x^b = x^{(a+b)}$ and $(x^a)^2 = x^{(2 \\times a)}$. We prove that the resulting exponent is $2^{255}-21$. ... ... @@ -300,7 +301,7 @@ will take a long time to verify. \\subsection{Speeding up with Reflections} In order to speed up the verification, we use a technique called reflection. It provides us with flexibility such as we don't need to know the number of It provides us with flexibility, for example, we don't need to know the number of times nor the order in which the lemmas needs to be applied (chapter 15 in \\cite{CpdtJFR}). The idea is to \\textit{reflect} the goal into a decidable environment. ... ... @@ -501,7 +502,7 @@ This provide with multiple advantages: the verification by the Coq kernel can be in parallel and multiple users can work on proving different functions at the same time. For the sake of completeness we proved all intermediate functions. Memory aliasing is the next point a user should pay attention to. The way VST Memory aliasing is the next point a user should pay attention. The way VST deals with the separation logic is similar to a consumer producer problem. A simple specification of \\texttt{M(o,a,b)} will assume three distinct memory share. When called with three memory share (\\texttt{o, a, b}), the three of them will be consumed. ... ... @@ -532,11 +533,11 @@ This solution allows us to make cases analysis over possible aliasing. \\subsection{Verifiying \\texttt{for} loops} Final state of \\texttt{for} loops are usually computed by simple recursive functions. However we must define invariants which are true for each iterations. However we must define invariants which are true for each iteration. Assume we want to prove a decreasing loop where indexes go from 3 to 0. Define a function $g : \\N \\rightarrow State \\rightarrow State$ which takes as input an integer for the index and a state and return a state. It simulate the body of the \\texttt{for} loop. Define a function $g : \\N \\rightarrow State \\rightarrow State$ which takes as input an integer for the index and a state, then return a state. It simulate the body of the \\texttt{for} loop. Assume it's recursive call: $f : \\N \\rightarrow State \\rightarrow State$ which iteratively apply $g$ with decreasing index: \\begin{equation*} f ( i , s ) = ... ...\nSupports Markdown\n0% or .\nYou are about to add 0 people to the discussion. Proceed with caution.\nFinish editing this message first!"
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https://old.iasbs.ac.ir/math/name%3Dmath%26dep%3D4%26ef%3Dfa%26page%3Dpub_full-2.html | [
"",
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"\n دانشگاه تحصیلات تکمیلی علوم پایه زنجان بلوار استاد یوسف ثبوتی، پلاک 444 صندوق پستی 1159-45195 زنجان 66731-45137 ایران دورنگار: 33155142 -024 تلفن: 33151 -024 طراحی و برنامه نويسی توسط مركز كامپيوتر دانشگاه تحصيلات تكميلي علوم پايه زنجان\n دانشکده ریاضی\n 1- Ahmed, A., Anco, S. C., Asadi, E., \"Unitarily-invariant integrable systems and geometric curve flows in SU (n + 1)/U(n) and SO(2n)/U(n)\", Journal of Physics A: Mathematical and Theoretical, 51: (6), 065205-1 -065205-35, (2018).Abstract:Bi-Hamiltonian hierarchies of soliton equations are derived from geometric non-stretching (inelastic) curve flows in the Hermitian symmetric spaces \\$SU(n+1)/U(n)\\$ and \\$SO(2n)/U(n)\\$ . The derivation uses Hasimoto variables defined by a moving parallel frame along the curves. As main results, new integrable multi-component versions of the Sine–Gordon (SG) equation and the modified Korteveg–de Vries (mKdV) equation, as well as a novel nonlocal multi-component version of the nonlinear Schrödinger (NLS) equation are obtained, along with their bi-Hamiltonian structures and recursion operators. These integrable systems are unitarily invariant and correspond to geometric curve flows given by a non-stretching wave map and a mKdV analog of a non-stretching Schrödinger map in the case of the SG and mKdV systems, and a generalization of the vortex filament bi-normal equation in the case of the NLS systems."
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"https://old.iasbs.ac.ir/pictures/search_fa.jpg",
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"https://old.iasbs.ac.ir/pictures/menusplit.jpg",
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https://patrickmccallion.com/qa/quick-answer-what-type-of-number-is-z.html | [
"",
null,
"# Quick Answer: What Type Of Number Is Z?\n\n## Why does Z represent sleep?\n\nZzz is an onomatopoeic representation of snoring.\n\nIt was commonly used in media where sound effects were not an option, notably in comic books.\n\nThat’s where it got its association with sleeping, even though it wasn’t the only device used to symbolize snoring.\n\nIt’s unclear why zzz came to represent snoring..\n\n## What does a backwards Z mean?\n\nIn the Pitman Initial Teaching Alphabet (ITA), a backward ‘z’ is called ‘zess’, and is used to denote the hard ‘s’ sound used in many plural forms of nouns and third-person singular present forms of verbs (including is). The ITA is an educational aid, and is not used in normal writing to replace the standard alphabet.\n\n## Why do British people say Zed?\n\nThe primary exception, of course, is in the United States where “z” is pronounced “zee”. The British and others pronounce “z”, “zed”, owing to the origin of the letter “z”, the Greek letter “Zeta”. This gave rise to the Old French “zede”, which resulted in the English “zed” around the 15th century.\n\n## What does Z stand for in math?\n\nintegersList of Mathematical Symbols • R = real numbers, Z = integers, N=natural numbers, Q = rational numbers, P = irrational numbers. Page 1. List of Mathematical Symbols. • R = real numbers, Z = integers, N=natural numbers, Q = rational numbers, P = irrational numbers.\n\n## What is Z in number system?\n\nThe letter (Z) is the symbol used to represent integers. An integer can be 0, a positive number to infinity, or a negative number to negative infinity.\n\n## What is Z in set notation?\n\nZ denotes the set of integers; i.e. {…,−2,−1,0,1,2,…}. Q denotes the set of rational numbers (the set of all possible fractions, including the integers). R denotes the set of real numbers. C denotes the set of complex numbers.\n\n## Is Z+ the same as N?\n\nZ stands for Zahlen, which in German means numbers. When putting a + sign at the top, it means only the positive whole numbers, starting from 1, then 2 and so on. N is a little bit more complicated set. It stands for the natural numbers, and in some definitions, it starts from 0, then 1 and so on.\n\n## What does Z symbolize?\n\nAs a student of the occult (as in hidden or sacred knowledge, and not whatever dark thoughts you might associate with the word), I also checked the Hebrew alphabet, the sacred letters. Z in Hebrew is Zayin and it means ‘sword’ or ‘a weapon of the spirit. … With that, it also stands for ‘thought’ as well as ‘word. ‘\n\n## What are the 3 types of relation?\n\nTypes of RelationsEmpty Relation. An empty relation (or void relation) is one in which there is no relation between any elements of a set. … Universal Relation. … Identity Relation. … Inverse Relation. … Reflexive Relation. … Symmetric Relation. … Transitive Relation.\n\n## What is Z in relation and function?\n\nA function is a relation that has exactly one output for every possible input in the domain. … That is, if f is a function with a (or b) in its domain, then a = b implies that f(a) = f(b). For example, z – 3 = 5 implies that z = 8 because f(x) = x + 3 is a function unambiguously defined for all numbers x.\n\n## What is the example of function and relation?\n\nA function is a relation in which no two ordered pairs have the same first element. A function associates each element in its domain with one and only one element in its range. Solution: a) A = {(1, 2), (2, 3), (3, 4), (4, 5)} is a function because all the first elements are different.\n\n## What is a true number?\n\nThe real numbers include all the rational numbers, such as the integer −5 and the fraction 4/3, and all the irrational numbers, such as √2 (1.41421356…, the square root of 2, an irrational algebraic number). Included within the irrationals are the transcendental numbers, such as π (3.14159265…)."
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"https://mc.yandex.ru/watch/66673681",
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https://www.askpython.com/python-modules/numpy/numpy-conj | [
"# NumPy conj – Return the Complex Conjugate of an Input Number",
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"Hey everyone, welcome to another NumPy mathematical functions tutorial. In this tutorial, we will understand the `NumPy conj` function in detail.\n\nThe conjugate of a complex number is obtained by simply changing the sign of the imaginary part.\n\nFor example, the conjugate of a complex number 10-8j is 10+8j. We can obtain the conjugate of a complex number using the `numpy.conj()` function.\n\nSo, let’s get started.\n\nAlso read: NumPy fmax – Element-wise maximum of array elements\n\nNumPy conj is a mathematical function of the numpy library that calculates the complex conjugate of the input complex numbers.\n\n### Syntax\n\nBy definition, it sounds simple right? Now, let us look at the syntax of the function.\n\n```numpy.conj(input)\n```\n\nHere, the input can be a single complex number as well as a NumPy array of complex numbers.\n\n## Working with NumPy conj\n\nNow, let’s do some programming in python.\n\n### NumPy conj of a single complex number\n\n```import numpy as np\n\n# Complex Conjugate of a Complex number with real and imaginary parts\nprint(\"The complex conjugate of 1+6j is:\",np.conj(1+6j))\nprint(\"The complex conjugate of 1-6j is:\",np.conj(1-6j))\n\n# Complex Conjugate of a Complex number with only imaginary part\nprint(\"The complex conjugate of 0+6j is:\",np.conj(0+6j))\nprint(\"The complex conjugate of 0-6j is:\",np.conj(0-6j))\n\n# Complex Conjugate of a Complex number with only real part\nprint(\"The complex conjugate of 1 is:\",np.conj(1))\nprint(\"The complex conjugate of -1 is:\",np.conj(-1))\n```\n\n### Output\n\n```The complex conjugate of 1+6j is: (1-6j)\nThe complex conjugate of 1-6j is: (1+6j)\nThe complex conjugate of 0+6j is: -6j\nThe complex conjugate of 0-6j is: 6j\nThe complex conjugate of 1 is: 1\nThe complex conjugate of -1 is: -1\n```\n\nIn the above snippet, the NumPy library is imported using the `import` statement and the function `np.conj()` is used to calculate the complex conjugate of the input complex number.\n\nLet’s understand how the values are calculated.\n\nFor complex number `1+6j`, the conjugate is obtained by changing the sign of the imaginary part and hence the output is `1-6j`.\n\nFor complex number `1`, the conjugate will be the same as the input complex number. This is because the number 1 can be written as `1+0j`, where the imaginary part is 0, hence the output is the same as the input complex number.\n\nNow, let us pass a NumPy array of complex numbers and calculate the complex conjugate.\n\n### NumPy conj of a NumPy array of complex numbers\n\n```import numpy as np\n\na = np.array((1+3j , 0+6j , 5-4j))\n\nb = np.conj(a)\n\nprint(\"Input Array:\\n\",a)\nprint(\"Output Array:\\n\",b)\n```\n\nOutput\n\n```Input Array:\n[1.+3.j 0.+6.j 5.-4.j]\nOutput Array:\n[1.-3.j 0.-6.j 5.+4.j]\n```\n\nIn the above snippet, a NumPy array of complex numbers is created using the `np.array()` which is stored in the variable `a`. The variable `b` stores the conjugate values of the input array which is also a NumPy array.\n\nThe `np.conj()` calculates the conjugate of each element of the input array.\n\nIn the next lines, we have used print statements to print the Input Array and Output Array.\n\n### NumPy conj of a NumPy array using the NumPy eye function\n\nIn this code snippet, we will create a NumPy array using the `numpy.eye()`.\n\n```import numpy as np\n\na = np.eye(2) + 1j * np.eye(2)\n\nb = np.conj(a)\n\nprint(\"Input Array:\\n\",a)\nprint(\"Conjugated Values:\\n\",b)\n```\n\nOutput\n\n```Input Array:\n[[1.+1.j 0.+0.j]\n[0.+0.j 1.+1.j]]\nConjugated Values:\n[[1.-1.j 0.-0.j]\n[0.-0.j 1.-1.j]]\n```\n\nLet’s try to understand the above code snippet.\n\n• In the first line, we import the NumPy library using the `import` statement.\n• The function `np.eye(2)` creates a 2×2 array with the diagonal elements as 1 and other elements as 0.\n• Similarly, the expression `1j * np.eye(2)` creates a 2×2 array with the diagonal elements as 1j and other elements as 0.\n• Then, the expression `np.eye(2)` `+` `1j * np.eye(2)` adds the corresponding elements of the two arrays, which is stored in the variable a.\n• In the next line, we used the `np.conj()` function to calculate the conjugated values.\n\nThat was all about using the NumPy conj function.\n\n## Summary\n\nIn this tutorial, you learned about the NumPy conj function along with practicing different types of examples. There is one more function `numpy.conjugate()` that works exactly the same way as the `numpy.conj()` function. Happy learning and keep coding.\n\n## Reference\n\nNumPy Documentation – NumPy conj"
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https://www.physicsforums.com/threads/atmospheric-pressure-etc.921067/ | [
"# Atmospheric pressure etc.\n\nHi,\n\nThe air pressure at the surface is about 1,00,000 N/m2.\n\nIt means that on average, a virtual column of air one square meter [m2] in cross-section, measured from sea level to the top of the earth's atmosphere, has a mass of about 10,000 kilograms and weight of about 1,00,000 newtons. That weight (across one square meter) is a pressure of 1,00,000 N/m2.\n\nLet's look at the air molecules in that virtual column of air. Let's imagine that the column is closed at one end by earth's surface where we also have a pressure sensor and the other end at the top of earth's atmosphere by some other material. At any instant only a fraction of molecules will be hitting the bottom of earth's surface downward, some of them will be hitting the other end upward, some of them will be hitting the column laterally and others will be in the space between the column of box.\n\nAlthough at any instant only a fraction of molecules will be hitting the bottom of earth's surface downward, the pressure sensor will still read 1,00,000 N/m because the molecules strike the bottom of the column more frequently and harder than the top of the column. The sum total of the greater number of collisions and the greater strength of the collisions will be that the average force against the bottom will be 1,00,000 N.\n\nPlease have a look on the attachment.\n\nThe cylinder is filled with air having the mass of 1.2 kg. Please also notice that at 25 C° the density of air at sea level could also be taken to be 1.2 kg/m3. The valves 1 and 2 are closed.\n\nWhen valves 1 and 2 are opened simultaneously what would happen? I think that outside air pressure will press the mercury downward because outside air pressure is greater than that of the cylinder. What do you think? Thank you for your help.\n\n#### Attachments\n\n•",
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"pressure1.jpg\n40.7 KB · Views: 403\n\njbriggs444\nHomework Helper\nThe cylinder is filled with air having the mass of 1.2 kg. Please also notice that at 25 C° the density of air at sea level could also be taken to be 1.2 kg/m3. The valves 1 and 2 are closed.\n\nWhen valves 1 and 2 are opened simultaneously what would happen? I think that outside air pressure will press the mercury downward because outside air pressure is greater than that of the cylinder. What do you think? Thank you for your help.\nAs I understand it, you have arranged for a cylinder with a volume of 1m3 containing 1.2 kg air at a reasonable temperature. You note that the density of air in the cylinder matches the density of air at sea level. Why would you not then conclude that the pressure of air in the cylinder will match the pressure of air at sea level?\n\nHave you ever been exposed to the ideal gas law: PV=nRT?\n\nWhy would you not then conclude that the pressure of air in the cylinder will match the pressure of air at sea level?\n\nI was expecting this question but was having hard time to phrase my question in my previous post. I'll try my best what is really confusing me and why I think that the pressure should be different. I'll repeat my original post with some addition and changes. Please have a look on the both attachments. We can assume the temperature to be 25 C°.\n\nThe air pressure at the surface is about 1,00,000 N/m2.\n\nIt means that on average a virtual cylinder of air one square meter [m2] in cross-section, measured from sea level to the top of the earth's atmosphere, has a mass of about 10,000 kilograms and weight of about 1,00,000 newtons. That weight (across one square meter) constitutes pressure of 1,00,000 N/m2 at the surface of earth.\n\nPlease note that the atmosphere becomes thinner and thinner with increasing altitude, with no definite boundary between the atmosphere and outer space. The Kármán line at 100 km is often used as the border between the atmosphere and outer space. Therefore, we can say that the volume of that virtual cylinder is 1,00,000 m x 1 m2 = 1,00,000 m3.\n\nLet's look at the air molecules in that virtual cylinder containing atmospheric air. Let's imagine that the cylinder is closed at both ends - at one end by earth's surface where we also have a pressure sensor and at the other end, at the top of earth's atmosphere, by some other material. At any instant only a fraction of molecules will be hitting the bottom of earth's surface downward, some of them will be hitting the other end where the atmosphere ends, some of them will be hitting the cylinder laterally and others will be in the space between the cylinder.\n\nAlthough at any instant only a fraction of molecules will be hitting the bottom of earth's surface downward, the pressure sensor will still read 1,00,000 N/m2. This is interesting because although only a fraction hit the surface, the force still measures 1,00,000 N as if all the molecules are striking the surface. In my opinion, this is what is happening. The molecules strike the bottom of the cylinder, i.e. earth's surface, more frequently and harder than the top of the cylinder. The sum total of the greater number of collisions and the greater strength of the collisions will still be that the average force against the bottom will be 1,00,000 N.\n\nThere are greater number of collisions with the surface of earth compared to the other end for several reasons. The earth tries to pull all the air molecules in that cylinder downward toward the ground but as the force of gravity becomes weaker as the distance increases from the surface therefore its pull on the molecules far from the surface is weaker and on the other hand the air molecules also resist this compression due to gravity due to their inter-molecular repulsion and they tend to keep maximum distance from each other, and moreover the molecules have their own independent motion due to their kinetic energy and the molecules which have greater kinetic energy tend to escape the gravity pull toward the surface more easily. But overall the gravity keeps comparatively more air molecules closer to its surface and hence greater number of collisions with the surface of earth.\n\nThe striking impact of the individual air molecules with the surface of earth is more powerful for two reasons. Firstly, each molecule has its own kinetic energy and moreover it's getting pulled by gravity vertically downward which constitutes more force at the instance of impact with the surface. It's just like when you fire a bullet from your roof into a table sitting on the ground - the bullet will have forceful impact with the table compared to a bullet which hits the table along the ground. Secondly, the most important reason is cumulative effect of gained momentum. Let me explain. The bottom of the cylinder, i.e. earth's surface, is pulling all the air molecules vertically downward. A molecule which is at the top end of the cylinder, where the atmosphere ends, might strike another molecule which is below it and transfer its vertical momentum to that molecule which in turn can transfer its vertical momentum toward the ground to another molecule which is below it and it becomes a domino effect. As a result the molecules closer to the surface of earth which lie below the majority of the molecules end up with far greater individual momenta. So, the impact of each molecule would be the result of contribution of momentum from many other molecules.\n\nI assume that what I have said above is almost correct.\n\nQuestion:\nHow can a cylinder which measures only 1 m3, let's call it Cylinder #1, compared to a cylinder, let's call it Cylinder #2, which has far greater volume of 1,00,000 m3 exert an equal amount of pressure at the surface of earth?\n\nLet me try to elaborate more. If you wee to liquefy the air in both cylinders, you would end up with 1.3 kg of liquefied air in one cylinder and in the other virtual cylinder you would end up with 10,000 kilograms. How could 1.3 kg of liquefied air exert same pressure as 10,000 kg?\n\nHave you ever been exposed to the ideal gas law: PV=nRT?\nYes, I'm familiar with PV=nRT gas law but don't know how to convert 1.3 kg or 10,000 kg into moles.\n\nThank you for your help and time.\n\n#### Attachments\n\njbriggs444\nHomework Helper\nI assume that what I have said above is almost correct.\nMostly correct, but almost completely irrelevant. Pressure is what it is, regardless of whether the air is being pushed down by a 100 mile column of air above or the sealed top of a cylinder above.\n\nLet me ask you few questions to clarify it.\n\nSuppose you have a cylinder with base area 1 m2 and height 1 m which gives it a volume of 1 m3. The mass of the cylinder is 5 kg. You fill up this cylinder with 1 kg of gas X. How much the cylinder plus gas weigh? 6 kg?\n\nNow let's assume that the height of cylinder is 2 m which gives it volume of 2 m3 and mass of 8 kg. You fill the cylinder up with 2 kg of gas X. How much the cylinder plus gas weigh? 10 kg or 9 kg?\n\nThanks a lot for your help.\n\nLast edited:\njbriggs444\nHomework Helper\nSuppose you have a cylinder with base area 1 m2 and height 1 m which gives it a volume of 1 m3. The mass of the cylinder is 5 kg. You fill up this cylinder with 1 kg of gas X. How much the cylinder plus gas weigh? 6 kg?\n6 kg.\n\nBut... are you measuring this weight with a scale under the cylinder? If so, the scale will read less than 6 kg. The cylinder has buoyancy because it is in air. If the air inside and the air outside are at the same density then the weight of the air inside is exactly offset by the buoyancy from the air outside.\nNow let's assume that the height of cylinder is 2 m which gives it volume of 2 m3 and mass of 8 kg. You fill the cylinder up with 2 kg of gas X. How much the cylinder plus gas weigh? 10 kg or 9 kg?\nIts mass is 10 kg. You tell me what its down force on the scale will be.\n\n6 kg.\n\nBut... are you measuring this weight with a scale under the cylinder? If so, the scale will read less than 6 kg. The cylinder has buoyancy because it is in air. If the air inside and the air outside are at the same density then the weight of the air inside is exactly offset by the buoyancy from the air outside.\n\nI also thought that it should measure 6 kg.\n\nYes, the weight is being measured with a scale under the cylinder. The density is same inside and out.\n\nIts mass is 10 kg. You tell me what its down force on the scale will be.\n\nThe force should be 100 N assuming gravitational pull of 10 N/kg.\n\nSo, as the base area for both cases is similar, i.e. 1 m2, therefore pressure on the base for 6 kg cylinder would be 60 N/m2, and for the 10 kg cylinder it would be 100 N/m2. I hope that you agree with this.\n\nAssuming you agree with my statement above then how come 1 m high cylinder and 1,00,000 m high cylinder from one of my previous posts, i.e. post #3, would exert the same pressure.\n\nThank you.\n\njbriggs444\nHomework Helper\nAssuming you agree with my statement above then how come 1 m high cylinder and 1,00,000 m high cylinder from one of my previous posts, i.e. post #3, would exert the same pressure.\nBecause of the 99,999 m of air pressing down on the top of the cylinder.\n\n•",
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"PainterGuy\nruss_watters\nMentor\nSo, as the base area for both cases is similar, i.e. 1 m2, therefore pressure on the base for 6 kg cylinder would be 60 N/m2, and for the 10 kg cylinder it would be 100 N/m2. I hope that you agree with this.\n\nAssuming you agree with my statement above then how come 1 m high cylinder and 1,00,000 m high cylinder from one of my previous posts, i.e. post #3, would exert the same pressure.\nBecause pressure isn't weight, so you are looking at apples and asking why they aren't oranges. In the special case of the atmosphere, the weight of the column of air causes it to have the pressure we see at the bottom, but for gases in containers, the weight has essentially nothing to do with the pressure (though density does). But in neither case do you measure the pressure with a scale: you measure it with a pressure gauge.\n\n•",
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"PainterGuy\nThank you.\n\nI understand your point to some extent but I'm still confused so let me ask you few questions.\n\nBefore I ask any question I tried to calculate air pressure using PV=nRT as suggested above. Please have a look on this attachment. The molar mass of dry air is 29 g/mol so the total number of moles is (1/29 x 1300)=44.83. The temperature in Kelvin is 25 C = 298.15 K. The value of R=8.314 J/mol.K. The volume is 1 m3. The pressure would be 111125 N/m2. The value is pretty close to atmospheric pressure at sea level.\n\nQuestion 1:\nA 1 m3 cylinder containing 1.3 kg of air at 25 C will have same pressure as 'weight' of 1,00,000 m column of air with base 1 m2. The density of atmospheric air and the air in cylinder is same. How is so that both cylinder and column happen to have the same pressure? Just a coincidence?\n\nQuestion 2:\nSuppose that you have a thermally insulated 1 m3 rectangular cylinder in vacuum containing 1.3 kg of air at 25 C. The pressure would be 111125 N/m2. This pressure is result of collisions with the faces of cylinder and the molecules move around due to their 'intrinsic' motion.\n\nNow you place the cylinder on surface of earth. You weigh it and total weight would be \"weight of cylinder\" plus \"weight of air\". The weight of air is a result of collisions of air molecules with the base of cylinder although the molecules strike the top face of cylinder too but those collisions are not that forceful compared to collisions with the base because of gravity; as a matter of fact the collisions with the bottom face of cylinder are stronger by almost 13 N.\n\nThe collision of each molecule with the bottom face of cylinder consist of two components: one component due to intrinsic thermal motion and the other component by virtue of gravity. Therefore, don't you think that the bottom face should experience more pressure compared to all other faces?\n\nThank you for the guidance.\n\njbriggs444\nHomework Helper\nHow is so that both cylinder and column happen to have the same pressure? Just a coincidence?\nNo coincidence. You chose the numbers so that it would come out that way.\n\nThe second question is all over the map. It is not clear what the point is. You start with a 1 m^3 cylinder in vacuum. But you never use the fact that it is in vacuum. Instead, I suspect that you want us to consider that cylinder in an environment free from gravity. Fine. In zero g, both \"top\" and \"bottom\" inside faces are under identical pressure equal to 111125 N/m2.. The outside faces are, of course under a pressure of 0 N/m2 -- vacuum as specified.\n\nNow you put the cylinder on earth. The air inside sags a very tiny bit under gravity, arranging itself so that there is a pressure differential between top and bottom. By no coincidence, this pressure differential is just enough to support the air inside. The pressure at the inside top will be low by about 6.5 N/m2 and the pressure at the bottom will be high by about 6.5N/m2. The pressure at the top would be, perhaps, 111,119 N/m2 and the pressure at the bottom 111,131 N/m2.\n\nYou want us to \"weigh\" the cylinder. Its true weight would reflect the mass of the cylinder plus the mass of the contained air. Its measured weight will have a small discrepancy due to atmospheric buoyancy. The cylinder displaces 1 cubic meter of air massing 1.3 kg. So the measured weight will be low by 13N.\n\nYou ask whether the bottom will be under more pressure than the top. Yes, of course it will. So?\n\n•",
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"PainterGuy\nThank you for the help.\n\nI'm sorry that you might be thinking that I'm going round in circles but seriously I'm finding it hard to explain my confusion. I will give it a last try.\n\nYou already confirmed that in an environment free from gravity a 1 m3 rectangular cylinder containing 1.3 kg air will exert a pressure of 111125 N/m2 at 25 C. This pressure is due to intrinsic motion of air molecules. And when such a cylinder is placed on earth, it creates a pressure differential between top and bottom faces to account for air weight. I believe that the side faces of cylinder still experience the same average pressure of 111,125 N/m2 although the pressure will vary along the side faces with height because the density differs from top to bottom with being maximum at bottom. So, it can be concluded that two factors come into play - pressure due to intrinsic motion of air molecules and pressure differential created due to weight.\n\nWhen it comes to atmospheric pressure, it is said that the pressure is due to the weight of air as if the motion of molecules of air play no part at all. Let's suppose an ideal situation by assuming that the atmosphere is enclosed within in a sphere so that the atmosphere is enclosed between earth and that outer sphere. The distance between this sphere and earth being 100 km because the atmosphere also ends around this distance from the surface. It's just like containing the air in a container. Even if the gravity of earth disappears the pressure due to motion of molecules will still be there and it should be equal to 111,125 N/m2 but the pressure differential between top and bottom will disappear. I'm assuming ideal conditions like uniform density of air, i.e. 1.3 kg/m3, 25 C temperature, earth being perfect sphere with smooth surface etc.\n\nI understand that in reality earth is like a one sided container - the surface of earth being the only side.\n\njbriggs444\nHomework Helper\nWhen it comes to atmospheric pressure, it is said that the pressure is due to the weight of air as if the motion of molecules of air play no part at all.\nThe pressure in air is entirely due to the motion of the molecules, regardless of whether those molecules are constrained by the top of a cylinder or by the weight of all the other molecules higher up.\n\n•",
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"PainterGuy\nThank you.\n\nIf you don't mind let me ask you few questions about what I wrote in my last post.\n\nYou already confirmed that in an environment free from gravity a 1 m3 rectangular cylinder containing 1.3 kg air will exert a pressure of 111125 N/m2 at 25 C. This pressure is due to intrinsic motion of air molecules. And when such a cylinder is placed on earth, it creates a pressure differential between top and bottom faces to account for air weight. I believe that the side faces of cylinder still experience the same average pressure of 111,125 N/m2 although the pressure will vary along the side faces with height because the density differs from top to bottom with being maximum at bottom. So, it can be concluded that two factors come into play - pressure due to intrinsic motion of air molecules and pressure differential created due to weight. Question: Do you agree with this? A short answer like \"yes\" or \"no\" would suffice.\n\nWhen it comes to atmospheric pressure, it is said that the pressure is due to the weight of air as if the motion of air molecules plays no part at all. You have already directly answered this by saying \"The pressure in air is entirely due to the motion of the molecules, regardless of whether those molecules are constrained by the top of a cylinder or by the weight of all the other molecules higher up\". So, you are essentially saying that the atmosphere would still exert the same pressure, i.e. 111,125 N/m2 even if there was no gravity assuming that the atmosphere doesn't just disappear into space. Question: Do I have it correct?\n\nI assume that your answer to the question above is \"yes\". Question: Then, I would say that what is the effect of gravity on atmospheric pressure?\n\nThanks a lot for your help and patience.\n\njbriggs444\nHomework Helper\nSo, it can be concluded that two factors come into play - pressure due to intrinsic motion of air molecules and pressure differential created due to weight. Question: Do you agree with this? A short answer like \"yes\" or \"no\" would suffice.\nNo.\n\nThere is no such thing as \"intrinsic motion\" as distinct from just plain old \"motion\". There is no such thing as a pressure differential due to weight that is not also a pressure differential due to motion and density.\n\n•",
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"PainterGuy\nThank you.\n\nI have few more questions and I would request you to help me.\n\nThere is a 1000 m3 rectangular cylinder (1 m2 base x 1000 m length) in vacuum free from gravity. It contains 1300 kg air (density 1.3 kg/m3). In zero gravity, both \"top\" and \"bottom\" inside faces are under identical pressure equal to 111,125 N/m2. The pressure will be same along all other faces.\n\nNow the cylinder is put on a planet free from atmosphere with gravitational pull of 10 kg/N. The density of air is adjusted due to gravity in the sense that the air is more dense closer to bottom than at the top. Will the pressure increase only because more number of molecules are striking the bottom per unit time, or is the second reason also being that the molecules striking the bottom with more force? And are both factors equally important?\n\nIt is said that pressure acts equally in all directions. I tend to agree with this statement when it comes to gases or liquids in containers ignoring gravity. But when I think of the effect of gravity, it confuses me. Please have a look on the attachment. There are pressure sensors at locations labelled A, A', B, B' and so on. For example, \"A\" pressure sensor measures the pressure vertically downward and \" A' \" sensor measures lateral pressure; both A and A' sensors are almost at the same level. The gravity provides an extra momentum to all the molecules only vertically downward and not laterally, therefore sensors A, A' and other sensor pairs should give different readings. What do you say? Where am I going wrong?\n\nThanks a lot.\n\n#### Attachments\n\nLet's consider one air molecule in earth's atmosphere at 1 km altitude. The molecule must carry the weight of trillion air molecules above it. Here \"carry the weight\" means to transmit the weight to the next molecule below.\n\nSo our molecule punches the molecule below downwards and the molecule above upwards.\n\nNow let's assume the force of the down-punches is the same as the force of the up-punches. The net force on our molecule from other molecules is zero, so the molecule free falls downwards.\n\nNow let's assume the molecule does not fall but floats. We can see that in this case the force of the down-punches must be larger than the force of the up-punches. About one trillionth larger."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.93716604,"math_prob":0.9533441,"size":3541,"snap":"2021-31-2021-39","text_gpt3_token_len":823,"char_repetition_ratio":0.13683914,"word_repetition_ratio":0.98136646,"special_character_ratio":0.24315165,"punctuation_ratio":0.09454061,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9669961,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,1,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-23T22:00:39Z\",\"WARC-Record-ID\":\"<urn:uuid:9f6d1cb7-1ded-40bf-ad84-e1fb7e2553dd>\",\"Content-Length\":\"139706\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8403f867-2e41-434e-be11-ca3c87e60be1>\",\"WARC-Concurrent-To\":\"<urn:uuid:1a5f14af-a9db-4eb9-985b-93ac1d76f1a6>\",\"WARC-IP-Address\":\"104.26.14.132\",\"WARC-Target-URI\":\"https://www.physicsforums.com/threads/atmospheric-pressure-etc.921067/\",\"WARC-Payload-Digest\":\"sha1:7JVSAQOQR443F5SGQSUUMYRRKJA2A5QT\",\"WARC-Block-Digest\":\"sha1:VGCKYXJ33FYUTSMQ24S2BQMGW2ANWXQ6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046150067.51_warc_CC-MAIN-20210723210216-20210724000216-00100.warc.gz\"}"} |
https://download.racket-lang.org/releases/7.9/doc/teachpack/arrow.html | [
"#### 1.10Managing Control Arrows: \"arrow.rkt\"\n\n (require htdp/arrow) package: htdp-lib\n\nThe teachpack implements a controller for moving shapes across a canvass.\n\n procedure(control-left-right shape n move draw) → true shape : Shape n : number? move : (-> number? Shape Shape) draw : (-> Shape true)\nMoves shape n pixels left (negative) or right (positive).\n\n procedure(control-up-down shape n move draw) → true shape : Shape n : number? move : (-> number? Shape Shape) draw : (-> Shape true)\nMoves shape n pixels up (negative) or down (positive).\n\n procedure(control shape n move-lr move-ud draw) → true shape : Shape n : number? move-lr : (-> number? Shape Shape) move-ud : (-> number? Shape Shape) draw : (-> Shape true)\nMoves shape N pixels left or right and up or down, respectively.\n\nExample:\n ; A shape is a structure: ; (make-posn num num) ; RAD : the radius of the simple disk moving across a canvas (define RAD 10) ; move : number shape -> shape or false ; to move a shape by delta according to translate ; effect: to redraw it (define (move delta sh) (cond [(and (clear-solid-disk sh RAD) (draw-solid-disk (translate sh delta) RAD)) (translate sh delta)] [else false])) ; translate : shape number -> shape ; to translate a shape by delta in the x direction (define (translate sh delta) (make-posn (+ (posn-x sh) delta) (posn-y sh))) ; draw-it : shape -> true ; to draw a shape on the canvas: a disk with radius (define (draw-it sh) (draw-solid-disk sh RAD)) ; Run: ; this creates the canvas (start 100 50) ; this creates the controller GUI (control-left-right (make-posn 10 20) 10 move draw-it)"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.5474807,"math_prob":0.97596735,"size":1549,"snap":"2022-40-2023-06","text_gpt3_token_len":413,"char_repetition_ratio":0.1669903,"word_repetition_ratio":0.21428572,"special_character_ratio":0.2989025,"punctuation_ratio":0.16776316,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9848663,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-02T00:02:35Z\",\"WARC-Record-ID\":\"<urn:uuid:ed9c3237-fb66-4683-8da6-7cec18c23ac0>\",\"Content-Length\":\"36637\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:42a0c72a-f0bf-406d-ab59-c85bfc09b844>\",\"WARC-Concurrent-To\":\"<urn:uuid:f4342e9d-1f28-4d76-8546-98d198a0d6dd>\",\"WARC-IP-Address\":\"172.67.188.90\",\"WARC-Target-URI\":\"https://download.racket-lang.org/releases/7.9/doc/teachpack/arrow.html\",\"WARC-Payload-Digest\":\"sha1:4FULFRZAYRLCSG7VIYTYKBDLQJO3TFUP\",\"WARC-Block-Digest\":\"sha1:ATBAI5DQSV5FZJPV2NWY7LBQCLWTPSCG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030336978.73_warc_CC-MAIN-20221001230322-20221002020322-00318.warc.gz\"}"} |
https://convertoctopus.com/5631-hours-to-minutes | [
"## Conversion formula\n\nThe conversion factor from hours to minutes is 60, which means that 1 hour is equal to 60 minutes:\n\n1 hr = 60 min\n\nTo convert 5631 hours into minutes we have to multiply 5631 by the conversion factor in order to get the time amount from hours to minutes. We can also form a simple proportion to calculate the result:\n\n1 hr → 60 min\n\n5631 hr → T(min)\n\nSolve the above proportion to obtain the time T in minutes:\n\nT(min) = 5631 hr × 60 min\n\nT(min) = 337860 min\n\nThe final result is:\n\n5631 hr → 337860 min\n\nWe conclude that 5631 hours is equivalent to 337860 minutes:\n\n5631 hours = 337860 minutes\n\n## Alternative conversion\n\nWe can also convert by utilizing the inverse value of the conversion factor. In this case 1 minute is equal to 2.9598058367371E-6 × 5631 hours.\n\nAnother way is saying that 5631 hours is equal to 1 ÷ 2.9598058367371E-6 minutes.\n\n## Approximate result\n\nFor practical purposes we can round our final result to an approximate numerical value. We can say that five thousand six hundred thirty-one hours is approximately three hundred thirty-seven thousand eight hundred sixty minutes:\n\n5631 hr ≅ 337860 min\n\nAn alternative is also that one minute is approximately zero times five thousand six hundred thirty-one hours.\n\n## Conversion table\n\n### hours to minutes chart\n\nFor quick reference purposes, below is the conversion table you can use to convert from hours to minutes\n\nhours (hr) minutes (min)\n5632 hours 337920 minutes\n5633 hours 337980 minutes\n5634 hours 338040 minutes\n5635 hours 338100 minutes\n5636 hours 338160 minutes\n5637 hours 338220 minutes\n5638 hours 338280 minutes\n5639 hours 338340 minutes\n5640 hours 338400 minutes\n5641 hours 338460 minutes"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8061122,"math_prob":0.9613413,"size":1697,"snap":"2022-40-2023-06","text_gpt3_token_len":447,"char_repetition_ratio":0.21323095,"word_repetition_ratio":0.007067138,"special_character_ratio":0.3323512,"punctuation_ratio":0.052980132,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9872224,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-29T22:37:47Z\",\"WARC-Record-ID\":\"<urn:uuid:60db74c4-d3f0-41ea-8549-5c593f826c9e>\",\"Content-Length\":\"27696\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3c9c27b5-5c32-4a90-93df-f09dc84e01d5>\",\"WARC-Concurrent-To\":\"<urn:uuid:a1217a91-16e0-44f1-a839-a414c04b83cb>\",\"WARC-IP-Address\":\"172.67.171.60\",\"WARC-Target-URI\":\"https://convertoctopus.com/5631-hours-to-minutes\",\"WARC-Payload-Digest\":\"sha1:X7XT6HE5SBTL4X53QEJP4Y3FKGFNTPXW\",\"WARC-Block-Digest\":\"sha1:TZXOXENVZU2AN4MSQF67MTZWHF3YXKO2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499768.15_warc_CC-MAIN-20230129211612-20230130001612-00825.warc.gz\"}"} |
https://www.fyfllc.com/2022/12/centers-of-triangles-worksheet.html | [
"# Centers Of Triangles Worksheet\n\nCenters Of Triangles Worksheet. It is the middle most spot, equal distance from all three corners of the triangle. The centroid of a triangle is (- 1, – 2) and co-ordinates of its two vertices are and (- 8, – 12). The Download button initiates a download of the PDF math worksheet. The point the place all of the three altitudes of the triangle meet or intersect each other.\n\nIf \\(\\overrightarrow \\) is an angle bisector, find \\(\\angle ADB\\) & \\(\\angle ADC\\). Properties of the Centroid It is shaped by the intersection of the medians. Find the centroid of a triangle whose vertices are , , and . If the lengths of two sides of a triangle are added collectively, then the resultant sum is all the time greater than the size of the third side. In the next part, we have summarized a variety of the important properties of a triangle. Joining the midpoints of the three sides of the triangle results in 3 parallelograms of the same area and 4 triangles of the identical space.\n\nThe centroid is the purpose at which the medians of a triangle intersect. Using the angle sum property of a triangle, we can calculate the incenter of a triangle angle. The orthocenter, circumcenter, incenter, and centroid all lie on the similar level. Joining the midpoint of three sides divides the triangle into 4 smaller triangles of the identical space.\n\nIn the applet above, transfer the factors A, B, and C to see what changes, and what stays the identical over plenty of totally different triangles. Note that the area region illuminated by the light makes a circle that passes via the three vertices of the triangle. Therefore, the region is the circumscribed circle of the triangle.",
null,
"Consider the definitions of incenter and circumcenter, and centroid. The circumcenter is equidistant from the vertices of the triangle by the Circumcenter Theorem. Recall that by the Incenter Theorem, the incenter of a triangle is equidistant from both sides of the triangle. Since the angle bisector cuts the angle in half, the other half must additionally measure 55°. An altitude or top is every of the perpendicular strains drawn from one vertex to the opposite side . Each level in the listing is recognized by an index variety of the shape X—for instance, X is the incenter.\n\nThere are three extended sides in a triangle and each extended side ends in one exterior angle, due to this fact, a triangle has six exterior angles. In the diagram given below, angles 1, 2, three, 4, 5 and 6 are exterior angles of the triangle. Answers for the worksheet on centroid of a triangle are given beneath to examine the exact solutions of the above questions on mid-point. The centroid of a triangle is (- 1, – 2) and co-ordinates of its two vertices are and (- 8, – 12).\n\nThe data recorded about every level contains its trilinear and barycentric coordinates and its relation to strains becoming a member of other identified points. Links to The Geometer’s Sketchpad diagrams are provided for key points. The Encyclopedia additionally features a glossary of terms and definitions. Points of Concurrency Where a quantity of lines, segments rays intersect, have particular properties. Is the purpose at which the medians of a triangle intersect. A-line perpendicular to the hypotenuse from the right angle ends in three comparable triangles.",
null,
"The middle of this circle is the incenter of the triangle. Looking on the diagram, it can be seen that the roads type a triangle. Recall that the incenter of a triangle is equidistant from each side. Also, the circumcenter of a triangle is equidistant from every vertices. In a math examination, Ramsha has been given a triangle and asked to attract two circles.\n\nA right-angled triangle is a triangle during which one angle measures 90° . A right-angled triangle has one obtuse angle and two acute angles, which makes it particular among the different forms of triangles. The circumcenter of an isosceles triangle lies contained in the triangle if all the three angles of the three triangles are acute. The medians drawn from vertices of an isosceles triangle with equal angles are equal in length. We hope you loved learning about the level of concurrency with the simulations and interactive questions.\n\n• The centroid is the purpose at which the medians of a triangle intersect.\n• In the GeoGebra applet above, the factors D, E, and F characterize the centroid, circumcenter, and orthocenter of the triangle ABC.\n• If two angles in a triangle have the same measure, then the two triangles are mentioned to be congruent.\n\nIt is the purpose the place all 3 medians intersect and is usually described because the triangle’s center of gravity or as the barycent. Perpendicular bisectorof a triangle is each line drawn perpendicularly from its midpoint. In the following part, we’ll discuss the orthocenter, centroid, circumcenter, and incenter of a triangle.\n\nThe rotated triangles or the mirror picture triangles are additionally called Similar triangles because the angles and sizes are the same. The sum of the equivalent exterior angle of a triangle is at all times equal to 360 levels. Angle Sum Property of a Triangle states that the sum of all of the three angles of a triangle is the identical as 180 levels. Find the co-ordinates of the point of intersection of the medians of the triangle formed by joining the points (-1, – 2), and . Determine the coordinates of the midpoint in these pdf worksheets.\n\nFor more, and an interactive demonstration see Euler line definition. The area of a proper triangle is half the product of the base and top. If two sides of the triangle and their included angles are given. The co-ordinates of the vertices of a triangle are (4, – 3), (- 5, 2) and .\n\nYou have to have actual distance from your point to the facet, and you find that by making a perpendicular line out of your facet via middle. The perpendicular bisectors of ΔMNP meet at point O and are shown dashed. The circumscribed circle or circumcircle of a triangle can be drawn in a couple of of steps.\n\nUsing a straight edge and a compass to create the centroid or heart of gravity of a triangle. As a member, you will additionally get limitless entry to over eighty four,000 lessons in math, English, science, historical past, and more. Plus, get practice exams, quizzes, and personalised teaching that can assist you succeed. It is the center most spot, equal distance from all three corners of the triangle. The 4 common factors of concurrency are centroid, orthocenter, circumcenter, and incenter.\n\nThe planners begin by roughly finding the three clients on a sketch and finding the circumcenter of the triangle fashioned. This level is taken into account to be the center of the triangle. For every triangle, the incenter is always inside the triangle.\n\nThis theorem relies on the properties of the angles of triangle. In geometry, a triangle is a type of two-dimensional polygon, which has three sides. When the 2 sides are joined finish to finish, it known as the vertex of the triangle. Triangles possess totally different properties, and each of these properties can be studied at totally different levels of education. In this article, you will perceive what’s the incenter of a triangle, formula, properties and examples.\n\nContents\n\n## Related posts of \"Centers Of Triangles Worksheet\"\n\n#### Piecewise Functions Word Problems Worksheet\n\nPiecewise Functions Word Problems Worksheet. You might, however, outline a specific cell or differ of cells with another name. Graph piecewise word issues worksheet piecewise function. If you should proceed the search in the different worksheets in your workbook, select Workbook. You with solutions into smaller pieces is evaluated by using piecewise word problems worksheet...\n\n#### Number Bonds To 10 Worksheet\n\nNumber Bonds To 10 Worksheet. This sport consists of well designed duties to help your young mathematician practice more on the ideas of time. The kids be taught about the number of hours there are in one day, the variety of hours there are in 3 days etc. A quantity bond is a straightforward visible...\n\n#### Assets And Liabilities Worksheet\n\nAssets And Liabilities Worksheet. When you have completed these forms please return the signed documents and a banker will contact you. You might use this product as an project, quiz, or a evaluate for a check. Your answers to the six questions might be used to create a customized mannequin portfolio, starting from probably the...\n\n#### Multiplying And Dividing Integers Worksheet\n\nMultiplying And Dividing Integers Worksheet. We can even acquire the set of even integers by multiplying each integer by 2. This is a horizontal quantity line with the negatives to the left of zero and the positives to the best of zero. These worksheets teach your beginning college students the means to multiply and divide..."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9149093,"math_prob":0.9734291,"size":7265,"snap":"2023-14-2023-23","text_gpt3_token_len":1548,"char_repetition_ratio":0.21429555,"word_repetition_ratio":0.073743924,"special_character_ratio":0.20467997,"punctuation_ratio":0.10804931,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9960089,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-05-29T09:41:53Z\",\"WARC-Record-ID\":\"<urn:uuid:779c9573-6b57-43ca-affb-bfe471823477>\",\"Content-Length\":\"589614\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2a7cad7d-bc2e-4746-af3c-272a3579ddd3>\",\"WARC-Concurrent-To\":\"<urn:uuid:9ef576d5-1fea-47f8-bf35-14f74b21b4ad>\",\"WARC-IP-Address\":\"144.91.65.59\",\"WARC-Target-URI\":\"https://www.fyfllc.com/2022/12/centers-of-triangles-worksheet.html\",\"WARC-Payload-Digest\":\"sha1:OIWQVYV7OV5KU23Y372P644XAVEDR3JC\",\"WARC-Block-Digest\":\"sha1:MPQMEYOFOOKLXXVZEBBVGG5ZOJDOF2VM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224644817.32_warc_CC-MAIN-20230529074001-20230529104001-00562.warc.gz\"}"} |
https://moam.info/signal-codes-convolutional-lattice-codes_5b8b56ac097c476a018b45b3.html | [
"## Signal Codes: Convolutional Lattice Codes\n\ngers becomes larger, as explained in Sections V and VII, which increases the ...... A. R. Calderbank and N. J. A. Sloane, âNew trellis codes based on.\n\nIEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 57, NO. 8, AUGUST 2011\n\n5203\n\nSignal Codes: Convolutional Lattice Codes Ofir Shalvi, Member, IEEE, Naftali Sommer, Senior Member, IEEE, and Meir Feder, Fellow, IEEE Abstract—The coded modulation scheme proposed in this paper has a simple construction: an integer sequence, representing the information, is convolved with a fixed, continuous-valued, finite impulse response (FIR) filter to generate the codeword—a lattice point. Due to power constraints, the code construction includes a shaping mechanism inspired by precoding techniques such as the Tomlinson-Harashima filter. We naturally term these codes “convolutional lattice codes” or alternatively “signal codes” due to the signal processing interpretation of the code construction. Surprisingly, properly chosen short FIR filters can generate good codes with large minimal distance. Decoding can be done efficiently by sequential decoding or for better performance by bidirectional sequential decoding. Error analysis and simulation results indicate that for the additive white Gaussian noise (AWGN) channel, convolutional lattice codes with computationally reasonable decoders can achieve low error rate close to the channel capacity. Index Terms—Achieving AWGN capacity, coded modulation, convolutional lattice codes, lattice codes, sequential decoding, shaping.\n\nI. INTRODUCTION\n\nS\n\nEVERAL years ago we came up with a simple construction for coded modulation: pass the uncoded information sequence (represented as an integer or an odd integer sequence) through a filter to output a continuous-valued modulated codeword. To overcome the power increase at the output, we proposed to apply a shaping mechanism inspired by precoding techniques such as the Tomlinson-Harashima filter. We termed the scheme “signal codes” due to the signal processing interpretation of the code construction. Following , this paper presents and analyzes this scheme in depth. It first observes that by convolving the integer sequence with the impulse response of the filter one gets a “convolutional” lattice. The code described above, which can be naturally termed “convolutional lattice code,” is obtained by taking a finite region of this lattice using a shaping operation. It turns out that even a short length FIR filter, properly designed, can generate a lattice whose Hermite parameter (or nominal coding gain, normalized minimum distance ) is large. As demonstrated later in the paper, over the AWGN channel the scheme can work at a rate Manuscript received June 25, 2008; revised January 19, 2011; accepted March 13, 2011. Date of current version July 29, 2011. The work was supported by the Israeli Science Foundation by Grant 634/09. The material in this paper was presented at the IEEE Information Theory Workshop, Paris, France, 2003. O. Shalvi is with the Department of Electrical Engineering–Systems, TelAviv University, Tel-Aviv, Israel, and also with Anobit Technologies, Herzlia, Israel. N. Sommer is with the Department of Electrical Engineering–Systems, TelAviv University, Tel-Aviv, Israel, and also with Anobit Technologies, Herzlia, Israel. M. Feder is with the Department of Electrical Engineering–Systems, Tel-Aviv University, Tel-Aviv, Israel. Communicated by H.-A. Loeliger, Associate Editor for Coding Techniques. Color versions of one or more of the figures in this paper are available online at http://ieeexplore.ieee.org. Digital Object Identifier 10.1109/TIT.2011.2158876\n\n[or signal-to-noise ratio (SNR)] close to the channel capacity with practical decoders. This fact, together with their simple construction, makes convolutional lattice codes a viable, attractive alternative for practical coded modulation. The usage of lattice codes as a natural and elegant alternative to random Gaussian codes , in the continuous-valued space is well known. As shown in –, , , , lattice codes attain the AWGN channel capacity. Lattice codes are the Euclidean space analog of finite alphabet linear codes. Considering the rich arsenal of finite alphabet (binary) linear codes that includes algebraic codes, convolutional codes, modern capacity achieving turbo codes , and low density parity check (LDPC) codes , polar codes , and so on, one may have expected that an analog situation will exist for lattice codes. Unfortunately, this is not the case. There are some specific lattice codes based on known low dimensional classical lattices . Other constructions utilized finite alphabet algebraic (or other) codes , , to “thin out” the integer lattice by the code constraints. Yet until recently, the analogy was not utilized for designing lattice codes directly in the Euclidean space or in finding specific capacity achieving lattice codes. Convolutional lattice codes, presented here, provides the desired analogy to finite alphabet convolutional codes. The uncoded symbols are convolved with a “filter pattern” to generate a codeword. Since the codes are used over the Euclidean space (real or complex), the operations are done in the real or complex field. Noticing that the filter output has an increased power which in effect, cancels out the coding gain, the code construction includes a mechanism inspired by pre-coding techniques such as the Tomlinson-Harashima filter. This “shaping” can either guarantee that the resulting lattice point will reside in the cube corresponding to the input integer sequence (or the input PAM/QAM constellation), or even better, reside in a more power efficient shaping domain. In the latter case, the code will also have a shaping gain in addition to the lattice coding gain. Notice that this technique actually provides a general framework for constructing lattice codes directly in the Euclidean space, composed of linear (or filtering) operation and shaping operation. This construction may be an alternative to the well known techniques (constructions A-D, ) that generate lattices from finite alphabet linear codes. The role of the nonlinear shaping (pre-coding) operation should be further motivated. Shaping “whitens” the codewords. Otherwise one would expect that the codeword spectrum will be proportional to the colored filter frequency response. Good codes, and in particular capacity achieving codes, should have a white spectrum! It also clarifies the major distinction between convolutional lattice codes and other coding schemes that use linear filtering, such as Partial Response Signaling (PRS) and Faster Than Nyquist (FTN) signaling . These techniques also filter the input integer sequence, but do not employ shaping, as one of their goals is to color the output to a desired\n\n5204\n\nIEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 57, NO. 8, AUGUST 2011\n\nmay well be the best solution in terms of performance for continuous-valued channels. Convolutional lattice codes, presented here, complement the picture for cases where low delay is desired, or in cases where a somewhat lower rate can be tolerated in turn for low computational complexity, obtained by the sequential decoding. The outline of this paper is as follows. An introduction to lattices and lattice codes is presented in Section II, followed by a definition of convolutional lattice codes (Section III) and methods to design lattices for convolutional lattice codes (Section IV). Then, Section V presents several shaping algorithms that can be used for practical lattice coding for the AWGN channel, and Section VI discusses bounds on the error probability. In Section VII, a description of computationally efficient decoders is provided, followed by simulation results in Section VIII. II. LATTICES AND LATTICE CODES Fig. 1. An example of the shaping operation for a 2-D lattice.\n\nspectrum. The distinction and the advantages of convolutional lattice codes over these techniques is further discussed later. Convolutional lattice codes can be decoded efficiently by sequential decoding . Unfortunately, Viterbi decoding or backward-forward BCJR decoding cannot be used. This is since the shaping operation increases substantially the range of possible integer values for any filter tap, and hence the number of states in the Viterbi decoder. Sequential decoding may require high computational efforts to exceed the cut-off rate, yet it can guarantee low average delay, and can handle convolutional lattice codes with long filter patterns. For better performance, we also propose a more elaborate bidirectional sequential decoding, and allow larger memory for the decoding algorithm. Several algorithmic techniques are provided that further reduce the computational complexity. Error analysis and simulation results for the proposed algorithms indicate that convolutional lattice codes with computationally reasonable decoders dB from the AWGN channel can achieve low error rate at capacity. Since some of the loss is due to practical implementation compromises, the results actually indicate that the lattice itself can attain a low error probability at less than 1 dB off the optimum. Note that the proposed scheme is attractive for intersymbolinterference (ISI) channels. The code filter and the ISI filter are essentially combined, resulting in a seamless unification of equalization and decoding. The general concept of designing codes directly in the Euclidean space, originated with convolutional lattice codes, was extended a few years later with the introduction of “low density lattice codes” (LDLC) , the lattice analog of LDPC codes. In LDLC the lattice generator matrix has a sparse inverse. It was shown that by proper choice of the elements of this sparse inverse, LDLC can approach the AWGN channel capacity with iterative decoding of linear complexity in the block length. More recently LDLC’s become practical with shaping , and highly efficient decoding , . Thus, LDLC’s\n\n( ) is defined as the A real lattice of dimension in set of all linear combinations of real basis vectors, where the coefficients of the linear combination are integers\n\nis the generator matrix of the lattice, whose columns are the basis vectors which are assumed to be linearly independent . over The Voronoi region (or Voronoi cell) of a lattice point is the set of all vectors in for which this point is the closest lattice point, namely\n\nwhere\n\n. The volume of the Voronoi cell of a lattice is , or in case is square. The minimum squared distance of a lattice is defined as the minimal squared Euclidean distance between any pair of lattice points. Clearly, the minimum squared distance of a lattice equals the squared length of the shortest nonzero lattice point (1)\n\nis defined as the The kissing number of a lattice number of nearest neighbors to any lattice point. The Hermite parameter of an -dimensional lattice, also referred to as the nominal coding gain of the lattice, is defined as . This definition of the nominal coding gain properly normalizes the minimum squared distance by the density of the lattice points such that it becomes invariant to scaling. A lattice code of dimension is defined by a (possibly and a shaping region (e.g., an shifted) lattice in -dimensional sphere), where the codewords are all the lattice points that lie within the shaping region . Straightforward encoding of an integer information vector by the lattice point will not guarantee, in general, that only lattice points within the shaping region are used. Therefore, the encoding operation must be accompanied by shaping, where\n\nSHALVI et al.: SIGNAL CODES: CONVOLUTIONAL LATTICE CODES\n\n5205\n\nFig. 2. A typical lattice coding scheme for the discrete-time AWGN channel.\n\ninstead of mapping the information vector to the lattice point , it should be mapped to some other lattice point , such that the lattice points that are used as codewords belong to the shaping region. The shaping operation is the mapping of the integer vector to the integer vector . The shaping operation is illustrated in Fig. 1 for a two-dimensional lattice whose generator matrix is\n\nEach information integer is assumed to be in the range to , so a two-dimensional lattice code needs to use lattice points as codewords. If no shaping is used, the information vector is mapped directly to and the codewords will be the 81 lattice points inside the parallelogram. If a rectangular or spherical shaping region is used, each information vector should be mapped to one of the 81 lattice points inside the shown rectangle or circle, respectively, resulting in average power which is lower by 4.59 and 4.77 dB, respectively, than the no-shaping case. Note that traditionally the term “shaping gain” is defined with respect to the average energy associated with the hypercube shaping domain, and is bounded by 1.53 dB [24, Sect. IV.A]. However, the shaping operation, as defined above, can reduce the average energy by much more than 1.53 dB, since the starting point (after straightforward mapping of the integer vector to the lattice point ) may have average energy which is much higher than the average energy of a hypercube, as illustrated above for the example of Fig. 1. A typical lattice coding scheme for the discrete-time AWGN channel is summarized in Fig. 2. First, the shaping operation maps the information integer sequence to another integer sequence , as described earlier. The new sequence is encoded by multiplication with the lattice generator matrix. Then, the resulting lattice point is transmitted through the AWGN channel, which adds additive Gaussian noise with variance . , where is the avThe SNR is defined as erage energy of a single component of the transmitted lattice point . The effective coding gain of a lattice code is measured by the reduction in required SNR to achieve a certain target error probability relative to using the cubic lattice , with a hypercube shaping region, using the same data rate.\n\nAt the receiver, a maximum-likelihood (ML) decoder should find the closest lattice point within the shaping region to the noisy observation in the Euclidean space. For a general lattice, the computational complexity of finding the closest lattice point to a given vector is exponential with the lattice dimension . Therefore, lattices for practical use should have some structure that enables simple decoding. Finally, an inverse shaping operation is performed to the detected lattice point in order to recover the information integers. is defined as the set An dimensional complex lattice in of all linear combinations of a given basis of linearly indepenwith complex integer coefficients. All the dent vectors in properties of real lattices and real lattice codes that were cited above can be extended in a straightforward manner to complex lattices and complex lattice codes. III. DEFINITION OF CONVOLUTIONAL LATTICE CODES Convolutional lattice codes are defined as lattice codes which genare based on an -dimensional lattice whose erator matrix has the following Toeplitz form\n\n.. . .. . .. . .. . .. .\n\n.. . .. .\n\n.. . .. . .. . .. . .. . .. . .. . .. .\n\n.. . .. .\n\n.. . .. . .. . (2)\n\n.. . .. .\n\nare the impulse response coefficients of where a monic causal FIR filter, which will be denoted as the generating filter. As will be explained in Section V, it is worthwhile to choose a minimum-phase filter (i.e., a filter whose zeros are inside the unit circle) due to the proposed shaping methods. The -transform of the generating filter will be denoted by .\n\n5206\n\nIEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 57, NO. 8, AUGUST 2011\n\nThe components of a lattice point , where is an -dimensional vector of integers, are the convolution of the sequence of components of with the generating filter (3) for , where is assumed zero outside the range 1 to . For convenience, we shall assume at this point are real valued and is a real lattice, but the that observations below can be extended in a straightforward manner to complex lattices. We shall now show that this lattice has better (or at least than equal) nominal coding gain , which can be regarded as a latthe uncoded cubic lattice tice whose generator matrix is the identity matrix. First, we shall show that both lattices have the same density, i.e., same . For the cubic lattice , for value of every , where as discussed in Section II, for the proposed lat. As shown in Appendix A, for tice we have (This should not be surprising, since the first rows of form a lower diagonal matrix whose determinant is exactly 1). Therefore, for large , is nearly a volume preserving transformation, and the density of the proposed lattice approaches the density of the cubic lattice. is greater of equal for the It is left to show that proposed lattice than for the cubic lattice. For the cubic lat. For the proposed lattice, tice, we clearly have . denote the shortest nonzero lattice point by be the smallest index for which , where Let denotes the component of . As noted above, is the convolution of the sethe sequence of components of with the generating filter. Since quence of components of , the generating filter is monic and causal, and thus\n\nresponse signaling (PRS) and faster than Nyquist (FTN) signaling (see for an overview of these techniques). Both these techniques obtain bandwidth efficiency by introducing intentional ISI to a sequence of integer information symbols. Therefore, these schemes essentially use lattice points, where the lattice generator matrix is of the form (2). However, the basic difference is that these schemes do not employ shaping. As indicated earlier, without shaping, the coding gain of the lattice can not be utilized, since the improvement in the squared minimum distance of the lattice versus the cubic lattice is always smaller than the increase in signal power due to the filtering operation. As a result, these schemes try to achieve effective gains in other ways: in PRS, the ISI is designed to narrow the power spectrum of the transmitted signal with minimal degradation to error probability. As already noted in : “Herein lies an important fact about PRS-coded modulation: Free distance cannot be gained via linear convolutions in the complex field; the game is to lose as little as possible, while reducing the bandwidth.” In FTN, the gain is higher data rate, which is achieved by using signaling rate which is higher than the Nyquist rate of the channel, and handling the unavoidable ISI at the receiver. Convolutional lattice codes inherently differ from these two techniques, since employing the shaping operation enables to utilize the lattice coding gain, rather than changing signaling rate or signal bandwidth. Therefore, it is applicable to the simple AWGN channel, where PRS and FTN envision a different channel scenario, namely one where signals must have a certain power spectral density (PSD) shape or property. IV. LATTICE DESIGN: CHOOSING THE FILTER In order to design a convolutional lattice code, we need to choose the generating filter. We shall seek generating filters that . As will be shown later, it is beneficial to use yield high complex-valued lattices that are based on a complex-valued generating filter. For simplicity, we shall examine complex-valued , i.e., their generating filters that have a th-order zero at -transform is\n\n(4)\n\n(5)\n\nNote that it is essential to use the shaping operation, as described in Section II, in order to benefit from the improved nominal coding gain of the lattice. Otherwise, assuming that the information integers are independent, identically distributed (i.i.d.), the lattice points are filtered sequences of i.i.d. integers, than the whose power is larger by a factor of power of the i.i.d. integers. However, by considering the lattice point that corresponds to an “impulse” integer vector (with ’1’ in the first element and zero otherwise), it can be seen that the improvement in the squared minimal distance due to filtering is , so it is never upper bounded by the same term enough to justify the power increase, unless shaping is used. Several shaping methods for convolutional lattice codes will be described in Section V. In fact, incorporating the shaping operation is one of the basic differences between convolutional lattice codes and other coding schemes that employ linear filtering, such as partial\n\nwhere due to the minimum-phase restriction. This simple choice is not necessarily optimal, but the experimental results in the sequel indicate that it can lead to lattices with good coding gains. (i.e., ) it can be easily seen For . Since , the nominal coding that , it is more difficult to gain is bounded by 2 (3 dB). For analytically and a numerical search is required. find Methods that were developed for finding the minimum distance between output sequences of ISI channels can be applied here, and we have chosen to use the approach of , properly modified to our case. The resulting search algorithm, whose details are presented in Appendix B, finds all the lattice points within a given hypersphere, by developing a tree of all possible integer sequences, and truncating tree branches as soon as it can identify that all the corresponding lattice points will lie outside the hypersphere. The tree is searched in a depth first search (DFS)\n\nSHALVI et al.: SIGNAL CODES: CONVOLUTIONAL LATTICE CODES\n\nFig. 3. The squared minimum distance as a function of the filter’s zero for different value of the zero’s magnitude.\n\nFig. 4. The squared minimum distance as a function of the filter’s zero for different value of the zero’s magnitude.\n\n5207\n\nP\n\n= 2. d\n\nis shown as a function of the zero’s phase, where every plot is for a\n\nP\n\n= 3. d\n\nis shown as a function of the zero’s phase, where every plot is for a\n\nmanner, which can be easily implemented using recursion techniques. In fact, this search algorithm is equivalent to a sphere decoder , with the proper modifications due to the shift invariance of the convolution operation and the band Toeplitz structure of the generator matrix .\n\nUsing this algorithm, the squared minimum distance was . Figs. 3 and 4 show found for various values of and , respectively. The squared the results for minimum distance is shown as a function of the phase , with a different plot for each value of the magnitude .\n\n5208\n\nIEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 57, NO. 8, AUGUST 2011\n\nTABLE I HIGH CODING GAIN FILTER PATTERNS\n\nThe figures are for for relation\n\nradians, where the values can be obtained using the symmetry and the periodicity relation , where denotes the minimum . These squared distance as a function of for a fixed relations are derived in Appendix C. It can be seen that the squared minimum distance improves as the spectral null of the filter becomes deeper, either by inor by letting the zero apcreasing the number of zeros proach the unit circle more closely. However, as the spectral notch becomes deeper, the dynamic range of the shaped integers becomes larger, as explained in Sections V and VII, which increases the decoding and shaping implementation complexity. , a global The phase has a significant effect, and for a given . optimum value for exists that maximizes Several generating filters with high are summarized in Table I. For each generating filter, the table shows the squared minimal distance and the nominal coding gain of the resulting lattice (assuming the lattice dimension is large enough, as explained in Section III). The table also shows the nonzero portion of the integer vector for which is the shortest nonzero lattice point (where throughout this paper the term “nonzero portion” of a vector means the portion that starts with the first nonzero component and ends with the last nonzero component). Note that all the integer sequences in the third column of Table I maintain an interesting symmetry. Denote the , the length of the integer sequence by . Then, for ’th element and the ’th elements are equal up to comor plex conjugation followed by multiplication by (where the same factor is used for the whole sequence). It can be seen that even short filters with , , and zeros can achieve considerable nominal coding gains of more than 6, 8, and 9.5 dB, respectively, where the shortest lattice points correspond to reasonably short integer sequences. Consider the special case of . In this case, the generating , it approaches one of the filter is real-valued, and for or . well-known partial response channels As can be seen from Figs. 3 and 4, these filters have considerable (though not optimal) coding gain. However, as shown in , these filters have null error sequences, i.e., infinite-length integer sequences that yield a sequence of zeros when filtered with these filters. This singularity is not desirable, since the norm of many lattice points will be close to , making the effective coding gain much smaller than the nominal coding gain. Also,\n\nbuffers of sequential decoders will overflow with high probability due to many possible candidates with short distance from the observation. In fact, when Fig. 4 was generated, the search algorithm could not find the minimum distance for the case of and with reasonable computational complexity is not shown due to this reason, and as a result the value of for this combination. As a result, from now on we shall assume a complex , resulting in a complex lattice. Note that in case a real lattice is required (e.g., in a baseband communication system), a complex lattice can be transformed to a real lattice by using the real and imaginary parts of each lattice point component as two independent real components. demonstrates that nominal coding The behavior for gain does not necessarily guarantee effective coding gain. In fact, the nominal coding gains of dense -dimensional lattices while the effective coding gain is become infinite as bounded by channel capacity . Therefore, the generating filters will be further checked for their effective coding gain using bounds on error probability (Section VI) and numerical simulations (Section VIII). V. SHAPING As discussed in Section III and throughout, shaping is essential for convolutional lattice codes, otherwise the power increase due to the filtering operation is higher than the increase in minimal distance. As shown in Figs. 1 and 2, the shaping operation should map the information integer vector to another integer is inside a vector such that the resulting lattice point desired shaping domain, such as a hypercube or a hypersphere. We shall assume that the components of the information vector belong to either PAM or QAM constellations, which are defined as follows. An -PAM constellation is defined as the set . An -QAM constellation is defined as the set of complex integers whose real and imaginary parts belong to an -PAM constellation. Assuming equi-probable usage of the constellation values, the average energy of -PAM and -QAM constellations is and , respectively. PAM and QAM constellations use odd-valued integers in order to have a zero-mean constellation with an even number of points. Therefore, it will be more convenient to restrict also the components of the shaped integer vector to odd values. Using only odd values (instead of all integer values) is equivalent to scaling and shifting the lattice. The shaping operation has a close resemblance to the precoding operation for ISI channels, as illustrated in Fig. 5. The purpose of precoding is pre-equalizing the distortion of a linear , which is known at the transmitter, in order to channel avoid the need for equalization at the receiver. In principle, the transmitter can simply filter the data symbols with the inverse , but the inverse filtering operation can sigchannel filter nificantly increase the signal’s power and peak value. The solution is a precoder that maps the sequence of information symbols to another sequence such that the constraints at the channel . input are fulfilled after filtering the new sequence with This is exactly the required operation of the shaping algorithm\n\nSHALVI et al.: SIGNAL CODES: CONVOLUTIONAL LATTICE CODES\n\n5209\n\nFig. 5. Resemblance between shaping for lattice codes and precoding for ISI channels. (a) Preequalization for ISI channels. (b) A lattice code with shaping for the AWGN channel.\n\nfor convolutional lattice codes, where the generating filter replaces the channel inverse . Three shaping methods for convolutional lattice codes will now be proposed, where the first two are indeed based on well-known precoding schemes for ISI channels. A. Tomlinson-Harashima Shaping The first shaping method that we shall consider is based on Tomlinson-Harashima precoding , and uses a hypercube shaping domain. The components of the information vector are assumed to be i.i.d. -QAM symbols. The shaping operation is (6) , where and are the th component of the for shaped integer vector and the information vector , respectively, and is a complex integer. The inverse shaping operation (i.e., recovering the information integers from the shaped integers ) is then a simple modulo operation. The integers are chosen such that the real and imaginary parts of the comare in . Substituting (6) in the basic ponents of encoding operation of convolutional lattice codes, we get\n\nform a recursive loop, which will be stable (i.e., does not increase without bound as increases) if and only if the generis minimum phase. It is well known that ating filter except for some special cases (including for example the case ), the output of a Tomlinson-Harashima precoder is of a spectrally white sequence uniformly distributed over , . Since the for both real and imaginary parts, so its power is , the power of power of uncoded -QAM symbols is is almost the same as the uncoded signal power, albeit higher by a factor of , which is negligible for large . The power spectral density of the shaped integer sequence is proportional to , where is the Fourier transform of the generating filter. Thus, is the generating filter has a deep spectral notch (such as the filters in Table I), will be a narrow-band signal. This shaping scheme guarantees the range of the first components of the lattice point, since the filtering and shaping op. The last erations (7) and (8) are done only for components , which correspond to the “convolution tail”, can not be precisely controlled and may therefore have large magnitude. This problem is common to all the shaping methods that are proposed in this section. Three possible solutions are suggested in Appendix D which solve the problem at the price of negligible data rate degradation.\n\n(7) B. Systematic Shaping Therefore, we should choose (8) where denotes the complex integer closest to . Equations (6)–(8) are equivalent to a Tomlinson-Harashima . These equations precoder for the ISI channel\n\nThe second shaping scheme is based on flexible precoding . A lattice code that uses this shaping scheme can be regarded as “systematic,” by extending the definition of a systematic binary code, for which the information bits are part of the codeword components, in the following manner. A lattice code will be regarded as systematic if the information integers can be extracted from the corresponding noiseless lattice point by rounding the lattice point components (or part of them, in case\n\n5210\n\nIEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 57, NO. 8, AUGUST 2011\n\nof a nonsquare generator matrix). Since we use odd integers, the rounding operation will be replaced by quantizing to the closest odd integer. For the systematic shaping scheme, the shaping operation is (9) , where the complex integer sequence is for now chosen such that the real and imaginary parts of , the difference between the coded and uncoded sequences, will . By substituting (9) in the basic belong to the interval encoding operation , the required value of is (10) Equations (9)–(10) are equivalent to flexible precoding for . As for Tomlinson-Harashima the ISI channel should be minimum-phase in order to guarprecoding, antee the stability of the recursive loop. With systematic shaping, the lattice point components equal the information integers plus an additive “dither” signal, whose real and imaginary parts have magnitude less than 1. Therefore, it is indeed a systematic lattice code, as defined above. The inverse shaping operation filters to get , and then quantizes the components to get . Alternatively, can be recovered from using (10) and then can be recovered using (9). Following the same arguments that were used for TomlinsonHarashima shaping, the additive dither would generally be uniformly distributed and uncorrelated with the input sequence. Therefore, if the input to the shaping operation is -QAM symbols, the resulting shaping domain is a hypercube, where the real and imaginary parts of the lattice codewords are uniformly dis, and the same power increase tributed in factor of Tomlinson-Harashima shaping exists also here. However, systematic shaping can be combined with standard constellation shaping algorithms, such as trellis shaping or shell mapping , such that additional shaping gain of up to 1.53 dB can be potentially obtained. This can be done by applying a constellation shaping algorithm to the uncoded sequence prior to systematic shaping. The combined operation of systematic does not alter the shaping propshaping and filtering with erties of the input signal significantly, since it is equivalent to adding a small dither, so the constellation shaping gain will be retained. C. Nested Lattice Shaping The nested lattice shaping scheme tries to achieve some of the potential 1.53 dB shaping gain benefit of a hypersphere shaping domain over a hypercube shaping domain. Consider the Tomlinson-Harashima shaping operation (6). Suppose that instead of setting in a memoryless manner as in (8), we choose a sethat minimizes the energy of the resulting lattice quence point components , where . Using vector notations for (6), we have (11)\n\n. From (11), we Denote the nonshaped lattice point by . Choosing that minimizes then have is essentially finding the nearest lattice point of the scaled to the nonshaped lattice point , where the chosen lattice codeword is the difference vector between the nonshaped lat. As a result, tice point and the nearest lattice point the chosen lattice points will be uniformly distributed along the . Therefore, the resulting Voronoi cell of the coarse lattice shaping scheme is equivalent to nested lattice coding , , where the shaping domain of a lattice code is chosen as the Voronoi region of a different, “coarse” lattice, usually chosen as a scaled version of the coding lattice. Such a shaping domain has the potential to attain some of the shaping gain which is attainable by a spherical shaping domain. The complexity of finding the nearest lattice point is the same as the complexity of ML decoding in the presence of AWGN. However, unlike decoding, for shaping applications it is not critical to find the exact nearest lattice point, as the result of finding an approximate point will only be a slight penalty in signal power. Therefore, approximate algorithms may be considered. As shown in Section VIII, close-to-optimal shaping gains can be attained by nested lattice shaping using simple sub-optimal sequential decoders such as the -algorithm (see and references therein). Interestingly, it should be noted that the criterion for good shaping can be generalized to meet the needs of communications systems. For instance, the algorithm can combine power optimization with peak magnitude optimization or with short-time power optimization, or even with spectral shaping optimization that can reduce the bandwidth of the signal, similarly to PRS. D. Shaping for Nonminimum-Phase or Non-FIR Filters The proposed shaping methods incorporate nonlinear feedback loops, which are stable only if the generating filter is minimum-phase. In Appendix E, the shaping algorithms are extended to nonminimum-phase filters, and also to autoregressive, moving average (ARMA) filters, which can have both poles and zeros in the plane. Nonminimum-phase filters may have benefits for fast fading channels or for channels with impulse noise. ARMA filters are useful, for example, when the encoding operation should be combined with preequalization for an ISI channel. Joint preequalization and encoding is also described in this appendix. VI. BOUNDS ON THE PROBABILITY OF ERROR We shall consider complex lattice codes, used for the com(i.e., the plex AWGN channel with complex noise variance ). variance of the real and the imaginary parts of the noise is As explained in Section II, a ML decoder should find the closest lattice point to the noisy observation within the shaping domain. We shall assume that the decoder ignores the shaping domain boundaries, and thus performs “lattice decoding” instead of ML will be defined decoding . The probability of error as the probability that a transmitted lattice point was mistakenly detected as a different lattice point. Due to the linearity of the lattice, the probability of error for such a decoder does not depend on the transmitted lattice point, so we can calculate it under the assumption that the zero lattice point was transmitted. Then,\n\nSHALVI et al.: SIGNAL CODES: CONVOLUTIONAL LATTICE CODES\n\n5211\n\nis bounded by the union bound (, properly modified for the complex case) (12) where is the Gaussian error function. The lower and upper bounds of (12) are usually not practical since the lower bound may be too loose, where the upper bound requires an infinite can then be approximated by the union bound estisum. mate, defined as (13) is the kissing number of the lattice, i.e., the where number of nearest neighbors to any lattice point. The union bound estimate is expected to be a good approximation at high signal to noise ratios. Define the effective length of an integer vector as the length of the nonzero portion of . Due to the shift invariance of the convolution operation, for each lattice point that corresponds to an integer vector with effective length , there will be different lattice points with the same norm, corresponding to all possible shifts of the nonzero portion of the corresponding integer vector, where is the lattice dimension. Also, for each lattice point there will be 4 lattice points with the same norm that correspond to multiplications of and . Therefore, we the corresponding integer vector by , where deshall assume that notes the effective length of the integer vector that corresponds to the shortest nonzero lattice point. This assumption holds for all the filters of Table I. In case there are several lattice points with the shortest norm, except for shifts and multiplications by and , the search algorithm of Appendix B will find them, should be furand in such a case the above value of ther multiplied by the number of such points. Regarding the upper bound of (12), it is not necessary to sum over all nonzero lattice points, but only over Voronoi relevant is defined lattice points. A Voronoi relevant lattice point as a lattice point that defines a facet of the Voronoi region of the , where this facet is perpendicular to and interorigin sects it in its midpoint . The set of Voronoi-relevant vectors for which can be defined as the minimal set\n\nPractically, it is hard to find all the Voronoi relevant points of a high-dimensional lattice (A lattice of dimension can have up Voronoi relevant vectors ). Therefore, can to be approximated by truncating the infinite sum and taking into account only lattice points whose squared norm is bounded by . We then get the following approximation: (14) In order to use this approximation, the search algorithm of Appendix B can be used to generate a list of all lattice . This list is points whose squared norm is less than then used as an input to a sorting algorithm, whose details are\n\nFig. 6. A histogram of the squared norm of the lattice points for the lattice that corresponds to the third row of Table I.\n\ndescribed in Appendix F, which can determine for each lattice point whether it is Voronoi-relevant or not. As explained in Appendix B, the search algorithm finds a single representative from each group of lattice points that correspond to an integer vector with a shifted nonzero portion, and up to multiplication or . Denote the effective length (as defined above) by of this integer vector by . As noted in Appendix F, if one lattice points within such a group is of the Voronoi-relevant, then all the lattice points in the group are also Voronoi-relevant. Therefore, it is enough to sum in (14) over a single representative from each such group (which is the natural output of the search algorithm) and add a weighting to each element in the sum. coefficient of The search algorithm of Appendix B was applied to the lattice (10.2 generated by the third filter of Table I, with of the cubic lattice). This has required the examdB above ination of 500 billion tree nodes. Fig. 6 shows a histogram of the squared Euclidean distance of the 5593 lattice points that were found, where each bar corresponds to an interval of length 0.25 and shows how many lattice points had squared norm within this interval. The norms assume that all integer values are allowed for the integer vector components, without the restriction of odd integers that was imposed in Section V. When this restriction is applied, the norms should be scaled up by a factor of 4. As explained above, only a single representative is counted from each group of lattice points that corresponds to shifts and multiplicaor . tions by The leftmost bar corresponds to the (single) shortest nonzero lattice point, whose squared norm is 5.90. It can be seen that the first several shortest lattice points (whose squared norms are in the range 5.75–7.25) are discrete points, and there is no “flood” of lattice points whose norm is very close to the norm of the shortest nonzero lattice point. For higher norms, the number of lattice points grows exponentially with the Euclidean norm. The sorting algorithm of Appendix F was then applied to the output of the search algorithm in order to find which of the lattice points is Voronoi-relevant. Almost all the lattice points were\n\n5212\n\nIEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 57, NO. 8, AUGUST 2011\n\nfound to be Voronoi-relevant, and only 49 out of the 5,593 lattice points were found to be non-Voronoi-relevant. Most of those 49 points have a relatively high norm (larger than 8.75). The fact that most of the lattice points within the feasible search radius are Voronoi-relevant shows that our search did not go far from the Voronoi region of the origin. These lattice points will be used in Section VIII to compare the simulation results with the bounds of (13) and (14). Note that union bounds are not expected to be valuable, in general, beyond the channel cutoff rate . Note that convolutional lattice codes are similar in their nature to binary convolutional codes in the sense that the probability of an error event starting at symbol does not depend on (ignoring frame boundary effects), where an error event at symbol is defined as a sequence of decoder symbol errors starting at the th symbol. As in (12), this error event probaand upper bounded bility can be lower bounded by where denotes the first element of by the vector . Therefore, the probability of a frame error for a fixed generating filter approaches 1 as the frame length approaches infinity. As shown in Section VIII, for practical frame lengths of several thousands of symbols the frame error rate is still low. As attempting to approach the channel capacity bound requires very large frame length, the length of the generating filter should be increased as frame length increases (see ). VII. COMPUTATIONALLY EFFICIENT DECODERS As shown in Fig. 2, the decoder consists of two blocks. First, is detected and then an inverse shaping operation is used to calculate the corresponding . The inverse shaping operation, which is relatively simple, was defined in Section V, and in this section we propose algorithms for detecting . A. Reduced Complexity ML Decoding Consider the discrete-time AWGN channel , where is the component of the transmitted lattice point, is a sequence of zero-mean, i.i.d. complex Gaussian random and is the noisy observation (see variables with variance Fig. 2). As explained in Section II, when a lattice code is used for transmission through the AWGN channel, a ML decoder should find the closest lattice point within the shaping region to the noisy observation in the Euclidean space. Sometimes it is not simple to take the nonlinear shaping operation into account in the decoding process, and then “lattice decoding” can be used, where the decoder ignores the shaping domain (lattice decoding was assumed in Section VI when bounds on error probability were derived). With proper coding and decoding schemes, channel capacity can still be approached although lattice decoding is used . For lattice decoding, the decoder should find the values of the ’s that maximize (15) where , is the generating filter. Finding the values of the ’s is essentially an equalization problem: an integer symbol sequence was convolved with a filter, and has to be detected from the noisy convolution output.\n\nAs shown in , minimum Euclidean distance decoding can be implemented by a Viterbi Algorithm (VA) whose state is . The number of trellis branches of this VA is equal to the constellation size of , raised to the power of . Therefore, the VA is practical only if is small, and if the dynamic range of the shaped symbols is not prohibitively high. However, good codes can result in values with large range. For example, for Tomlinson-Harashima shaping (Section V-A), the sequence can be obtained by applying the filter on the transmitted sequence , which is a white sequence. As has good generating filters have deep spectral nulls, high spectral peaks and thus it significantly enhances the magnitude of . We note that since the real and imaginary parts of are in the range , the magnitude of can be bounded . In general, a straightforward VA may by be too complex, and a reduced-complexity VA decoder should be used. Reduced complexity Viterbi decoding can follow the wellknown techniques used in the context of convolutional codes and ML channel equalization. One class of such techniques is sequential decoding, e.g., the Fano and stack algorithms. Another class includes list algorithms such as the M-algorithm (see and references therein) and the T-algorithm . A third class is reduced states sequence detection (RSSD) algorithms (e.g., ). The computational complexity of sequential decoding of any tree code obeys a Pareto distribution . Such a distribution results in the computational cutoff effect, where for a given information rate, complexity increases abruptly below some cutoff SNR. Therefore, all the above reduced-complexity decoders are expected to be effective only above the cutoff SNR, which is known to be approximately 1.7 dB above the Shannon capacity for the high SNR regime of the AWGN channel . On the other hand, even when the mean or the variance of the number of computations becomes asymptotically infinite, the probability that this number will exceed a predefined threshold is still finite. Therefore, if a target finite error rate is defined, sequential decoders can achieve this error rate with finite (though probably large) complexity even beyond the cutoff rate. In Section VIII we shall show that the sequential stack decoder can be used for decoding of convolutional lattice codes close to the cutoff rate. We shall also use bidirectional sequential decoders with large complexity to demonstrate that low error rate can be achieved even more than 0.5 dB beyond the cutoff rate, with large (but still finite) computational resources. These decoding algorithms will now be further elaborated. B. The Heap-Based Stack Decoder The stack decoder is an algorithm to decode tree codes, which works as follows. A stack of previously explored paths in the tree is initialized with the root of the tree code. At each step, the path with best score in the stack is extended to all its successors, and then deleted from the stack. The successors then enter the stack. For a finite block with known termination state, the algorithm terminates when a path in the stack reaches the termination state at the end of the block. For convolutional lattice codes, each path in the tree corresponds to a sequence of integers , and its successors are sequences of the\n\nSHALVI et al.: SIGNAL CODES: CONVOLUTIONAL LATTICE CODES\n\n5213\n\nform for various values of . Any implementation of the stack decoder should use a properly chosen database that enables efficient implementation of the basic step of the decoder, such as a balanced binary tree . We propose an implementation which is based on the heap data structure . The detailed implementation is described in Appendix G. For the Tomlinson-Harashima and systematic shaping methods, the lattice point components are bounded in the . Therefore, the stack decoder can ignore interval paths for which a resulting lattice point component is outside this range. The resulting decoder is a reduced-complexity approximation of an ML decoder, since it takes into account the shaping domain boundaries. This technique is very effective for complexity reduction, and will be referred to as “x-range testing.” On the other hand, for nested lattice shaping, truncation of incorrect paths is more challenging, so the shaping gain of the encoder is traded with the decoder’s complexity. Each path in the stack is assigned a score that should reflect the likelihood of this path to be the correct path, given the noisy channel observation. Naturally, we would assign scores to the paths in the stack according to the negated squared Euclidean distance of the resulting lattice point from the noisy channel observation as in (15). However, the stack contains paths of different lengths. If we use the negated squared distance, shorter paths will get higher score, as less negative terms are accumulated. This is not desired, since we want to extend the path which coincides with the correct path, even if it is much longer than other incorrect paths in the stack. Therefore, the path scores should be defined such that the effect of path length is eliminated. This problem is addressed in Section VII-C. C. The Fano Metric For sequential decoding of binary convolutional codes, Fano suggested to subtract a bias term from each increment of the natural likelihood score, where the bias equals the code rate . Massey has shown that the score assignment problem is equivalent to decoding of a code with variable length codewords, and that the Fano metric is indeed the correct choice for stack and Fano decoding of binary convolutional codes, in the sense that the most likely path is extended in each step. Massey’s derivation can be extended to the Euclidean case, as done in for the general case of lattice decoding. Here, we follow the lines of and develop the Fano metric for convolutional lattice codes with Tomlinson-Harashima shaping. Similarly to convolutional codes, in order to extend the most likely path in each step, a bias term has to be subtracted from the score increments of (15) (16) where (17) See Appendix H for the derivation of (16) and (17). We can make an interesting observation from (17). In order for the stack algorithm (as well as the Fano algorithm) to work,\n\nthe expected value of the score of the correct path must increase along the search tree, otherwise the stack decoder may prefer shorter paths and the correct path may be thrown away . For the correct path, we have\n\n.\n\nTherefore, in order for the expected value of the path score to increase along the tree, we need to have in (16). From , resulting in . (17), we then have Now, when using a lattice code for the real-valued AWGN and noise variance , the maxchannel with power limit . imal information rate is limited by the capacity Poltyrev considered the AWGN channel without restrictions. If there is no power restriction, code rate is a meaningless measure, since it can be increased without limit. Instead, it was suggested in to use the measure of constellation density, leading to a generalized definition of the capacity as the maximal possible codeword density that can be recovered reliably. When applied to lattices, the generalized capacity implies that there exists a lattice of high enough dimension that enables transmission with arbitrary small error probability, if and only if . A lattice that achieves the generalized capacity of the AWGN channel without restrictions, also achieves the channel capacity of the power constrained AWGN channel, with a properly chosen spherical shaping region (see also ). As discussed in Section III, and taking into account that our lattice is scaled by 2 due to using only odd integers, for large lat, and the Poltyrev tice dimension we have capacity condition for complex lattices becomes . Interestingly, this is exactly the necessary condition that was developed above for the stack decoder to converge to the correct path. As this is a necessary but not sufficient condition, the stack decoder is not guaranteed to converge above capacity. Indeed, it is well known that sequential decoders can practically work only above the cutoff SNR, which is approximately 1.7 dB above capacity for the high SNR regime . (See for another example of using the Fano metric for lattice decoding.) D. Bidirectional Sequential Decoding After developing the Fano metric for the stack (or Fano) algorithms, we shall now turn to develop a bidirectional decoding scheme for convolutional lattice codes. It is well known that sequential decoding is sensitive to noise bursts . In , a bidirectional decoding algorithm was proposed for binary convolutional codes in order to reduce the complexity of decoding through a noise burst. Two stack decoders are working, where one works from the start of the block forward and the other moves from the end of the block backward. The algorithm stops when the two decoders meet at the same point. For a strong noise burst, each decoder will only have to face half the length of the burst. Assuming exponential complexity increase along the burst, the resulting complexity will be the square root of the complexity of a single decoder. For convolutional lattice codes, the two stack decoders work as follows. Each stack decoder holds a stack of previously explored paths, where each path is assigned a score according to the Fano metric, as described above. Both decoders work simultaneously. At each step, the path with best score in the stack is\n\n5214\n\nIEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 57, NO. 8, AUGUST 2011\n\nextended to all its successors and then deleted from the stack. The successors then enter the stack. Before deletion, the deleted path is compared to all the paths of the stack of the other decoder to look for a merge. A merge is declared when a path in the other decoder’s stack is found with the same state at the same time point in the data block as the current decoder, i.e., last symbols of the forward decoder match the time-reversed last symbols of the backward decoder. In order to reduce the probability of false merge indications, a match of more than symbols can be used. However, as the number of bits in each extended constellation symbol is usually large (as demonstrated in Section VIII, where 17 bits were needed to store the real or imaginary part of ), the probability of false indication is usually low enough for a match of symbols. A straightforward search for a merge will require a full pass on the whole stack every symbol. In order to avoid it, each stack entry can be assigned a hash value according to its last symbols. For each possible hash value, a linked list is maintained with all the stack entries that are assigned this value. Then, each decoder calculates the hash value that corresponds to its last symbols, and searches only the linked list of the other decoder that corresponds to this value, resulting in a much smaller search complexity. Note that in order to enable bidirectional decoding, the data must be partitioned to finite-length blocks, with known initial and final state. In principle, knowing both the initial and final states in each block requires transmitting additional overhead symbols. However, this overhead is anyway required for some of the shaping algorithms that were presented in Section V in order to properly terminate the shaping operation, as explained in Appendix D. It is desired to make the block length (or lattice dimension) as large as possible in order to make the effect of this overhead on code rate as small as possible. However, increasing the block size introduces delay to the system. In addition, the probability to have two or more distinct strong noise bursts that appear in the same block increases. In such a case, each of the two decoders will have to face a strong noise burst alone, and bidirectional decoding will no longer be effective. Therefore, the block length should be determined according to the tradeoff between these factors. Bidirectional decoding is possible for convolutional lattice codes due to the band-Toeplitz structure of the lattice generator matrix. However, decoding backward is not straightforward, as reversing the time axis causes the minimum phase generating filter to become maximum phase. Extending the paths of the stack has an effect similar to filtering with an autoregressive filter with nonstable poles, resulting in choosing extension symbols that grow without bound. This can be easily solved by filtering the codeword (in the forward direction) with the allpass , and letting the backward decoder work filter on the filtered data, which is equivalent to a code which is based on a maximum-phase generating filter. Decoding this data backward will now obey a stable recursion, where the allpass filtering does not change the power spectrum of the additive noise. VIII. SIMULATION RESULTS We shall now demonstrate the performance of convolutional lattice codes using simulations of the discrete-time\n\nAWGN channel, as shown in Fig. 2. All the simulations are for 6 information bits per (complex) symbol (equivalent to uncoded 64-QAM). Unless otherwise stated, the simulations for use the generating filter and (the third generating filter of Table I), combined with Tomlinson-Harashima shaping (Section V-A). Data is framed to finite-length blocks, where block size (lattice . The total number of blocks dimension) is that were simulated for each result is 20 000. The SNR is , where is the average power of the defined as is the variance of the complex lattice point components and noise (such that the variance of the i.i.d. real and imaginary each). parts of the noise is In Appendix D, three solutions were proposed for shaping the “convolution tail.” In this section, we shall use the second proposed scheme, where shaping and encoding are done in a values of are transmitted once continuous manner, and symbols. As shown in Appendix D for this every scheme, transmitting three ’s using 8-QAM requires 24 symbols. Therefore, the actual information rate is not 6 bits/symbol . For a given SNR, the but capacity for the discrete-time complex AWGN channel is . To achieve a capacity of 6 bits/symbol, the required SNR is 18 dB, where for 5.93 bits/symbol, the required SNR is 17.8 dB. Therefore, data framing results in a loss of 0.2 dB. This loss is essentially an implementation loss and is not related to the coding properties of the lattice. Note also that this implementation loss can be made negligible by increasing block length, or by using a more efficient coding scheme for transmitting the tail symbols. Fig. 7 shows the frame error rate (FER) versus SNR using the stack and the bidirectional stack decoders. For each deto . coder, the FER is shown for stack sizes ranging from The figure also shows the channel capacity for 5.93 bits/symbol and the error probability approximations that were presented in Section VI. The same results are also presented in Fig. 8, where for each maximal stack length, the figure shows the required . Note that this FER SNR for achieving frame error rate of value is certainly a practical value for many applications, e.g., wireless networks. It can be seen that increasing the maximal stack length improves the performance for both the stack and the bidirectional stack decoders. This can be explained as follows. When a noise burst is present, incorrect paths in the stack will temporarily have better score than the correct path. If the number of such incorrect paths exceeds the stack length, the correct path will be thrown out of the stack. Such a correct path loss (CPL) event will result with a decoding error. Figs. 7 and 8 show that for FER of and stack length which is smaller than , most of the errors result from CPL events and not from decoding to a wrong codeword that was closer to the observation in the Euclidean space, so increasing the stack length improves the FER. Fig. 7 shows only the FER and does not show the symbol error rate (SER) or bit error rate (BER), since the SER and the BER are high even when the FER is relatively low. The reason is that most frame errors are due to CPL events, as described above. In a CPL event of the proposed unidirectional algorithm, all the data symbols from the CPL start point until the end of\n\nSHALVI et al.: SIGNAL CODES: CONVOLUTIONAL LATTICE CODES\n\n5215\n\nFig. 7. Frame error rate for stack and bidirectional stack decoding, for various maximal stack lengths. Each curve is labeled with the corresponding maximal stack length.\n\nreal valued dimensions, , symbol and solving for P, we get that this rate is achievable under the above frame length and FER constraints for SNR of 18.1 dB. Therefore, the stack and bidirectional stack decoders are as close as 2.6 and 2 dB, respectively, from the minimal required SNR under these finite frame length and FER constraints. The quality of the coding scheme results from the properties of the underlying lattice, as well as from the shaping and decoding algorithms. It is beneficial to separate these factors and isolate the coding properties of the lattice itself by evaluating what would the distance to capacity be if the proposed lattice was used with ideal shaping and decoding. Since the simulations were performed with Tomlinson shaping, the input to the AWGN channel was uniformly distributed. Therefore, the actual bound on the achievable rate is not the channel capacity but the mutual information between the input and output of the AWGN channel under uniform input distribution constraint. Numerical calculation shows that 5.93 bits/symbol can be transmitted with uniform channel input distribution at SNR of 18.9 dB.1 This SNR is also shown in Figs. 7 and 8. Comparing now to 18.9 dB, the bidirectional stack decoder is 1.2 dB from the required SNR to achieve this data rate under uniform input constraint. As discussed earlier, the implementation loss due to framing is 0.2 dB. This loss relates only to the decoder implementation and not to the properties of the lattice. Also, as discussed earlier, the required SNR is further shifted by 0.3 dB due to the finite frame and the finite FER of . Taking these length of into account, the proposed lattice has a potential to work within dB from channel capacity. This is a strong indication that the\n\nSubstituting\n\nFig. 8. Required SNR to achieve frame error rate of 10 bidirectional stack decoders.\n\nfor the stack and\n\nthe block are lost, so on average an erroneous frame has half of its symbols in error. Therefore, the proposed decoders are mainly suited for data applications, where a frame with errors is ignored, regardless if it has a single error or many errors, and therefore FER is the relevant performance measurement. , and It can be seen that with a very large stack length of , the stack decoder can work as close for frame error rate of as 2.9 dB from channel capacity, where the bidirectional stack decoder can work as close as 2.3 dB from channel capacity. It is worthwhile to compare the simulation results to the maximal achievable rate under the constraints of a finite frame length and frame error probability . As shown in , this rate is , where closely approximated by is the capacity, is a characteristic of the channel referred to is the complementary Gaussian as channel dispersion, and cumulative distribution function. For the real AWGN channel and . with SNR P,\n\n1Note that the capacity loss due to the uniform distribution constraint is only 1.1 dB, as at these SNRs this capacity loss has not yet reached its asymptotic value of 1.53 dB\n\n5216\n\nIEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 57, NO. 8, AUGUST 2011\n\nFig. 9. Average and maximal number of computations for the stack decoder. Each curve is labeled with the corresponding maximal stack length.\n\nFig. 10. Average and maximal number of computations for the bidirectional stack decoder. Each curve is labeled with the corresponding maximal stack length.\n\nproposed lattice is good for AWGN coding in the sense defined in and . Fig. 7 also shows the union bound estimate (13) and the truncated upper bound (14) of Section VI, which were calculated based on the lattice points that were presented on Fig. 6. These are neither upper or lower bounds, but approximations to the probability of error, which should become more accurate as SNR increases. It can be seen that the approximations are indeed in good match with the leftmost empirical error probability curve, that corresponds to bidirectional decoding . The other curves are shifted due to with stack length of implementation-dependent errors (e.g., CPL events), where the theoretical bounds refer to an ideal ML decoder, which is approximated by the leftmost curve.\n\nTurning to complexity, we shall now examine the computational and storage requirements of the decoders. The storage is determined by the maximal stack length, where the computational complexity can be defined by the average and maximal number of computations per symbol. For this purpose, a computation is defined as the processing of a single stack entry. The number of computations per a specific symbol is calculated by dividing the total number of computations for the block that contains this symbol, by the number of symbols in the block. The maximum and average over all the 20 000 blocks of each simulation are defined as the maximal and average number of computations per symbol, respectively. Fig. 9 shows the average and maximal number of computations for the stack decoder, where Fig. 10 shows it for the\n\nSHALVI et al.: SIGNAL CODES: CONVOLUTIONAL LATTICE CODES\n\n5217\n\nFig. 11. Nested lattice shaping gain for 64-QAM and 4-QAM constellations.\n\nbidirectional stack decoder, for various maximal stack lengths. Combining the results from Figs. 8 and 10, we can see that in order for the bidirectional stack algorithm to work at FER at 2.3 dB from capacity, we need a stack of size . of The average number of computations is 80 computations per symbol, which is certainly a practical number (similar to a 64-states Viterbi decoder, or to an LDPC code with average node degree of 10 that performs 8 iterations). However, the maximal number of computations per symbol is 15 000—more than two orders of magnitude than the average. Therefore, such a decoder can be implemented with reasonable average complexity, but from time to time it will have large and unpredictable delays for the worst-case blocks. A more practical scheme might be a bidirectional stack de. FER of can be coder with maximal stack length of achieved for SNR of 20.8 dB (3 dB from capacity). The average number of computations per symbol is only 3 computations/symbol, where the maximum is 120. This is certainly a practical scheme, where the effect of nonpredictable decoding delays still exists, but it is much less severe. Note that the phenomenon of computational peaks also exists in modern iterative decoders, such as LDPC codes or Turbo codes. For these codes, it is common to have a “stopping criterion,” which stops decoding when the detected data is a valid codeword. In this case, most of the time the decoder performs a small number of iterations (e.g., 1–2), and from time to time it needs to perform more iterations (e.g., 8–16). This will result in non-uniform processing complexity. However, the “peak-to-average” of the number of computations is still significantly larger for the proposed sequential decoders. All the results so far were presented for Tomlinson-Harashima shaping. With this scheme, the codeword elements are uniformly distributed, so no shaping gain can be attained relative to uncoded QAM. However, such shaping gain can be achieved using, e.g., nested lattice shaping described in algorithm. Section V-C. Specifically, we used the following It starts from the first symbol of , and sequentially continues symbol-by-symbol. The input at stage is a list of up to candidate sequences for (where for the list\n\nis initialized with a single empty sequence). Each of the sequences is extended with all possible values for , and each , using extended sequence is assigned a score of the ’s that correspond to . The scores are sorted, sequences with smallest score are kept as input to and the the next stage. When each sequence is extended, only a finite range of values should be checked, as outside this range the will be large enough such that this path can be energy of immediately ignored. is finally chosen as the sequence with smallest score after processing of the last symbol. The storage and computational complexity of this shaping al. The storage and processing delay can be gorithm is improved if instead of waiting for the last stage, the value of is determined at stage , where is the decision delay. If is large enough, the shaping gain reduction will be minimal, where storage reduces from to . Note that for , nested lattice shaping reduces an -algorithm with , the alto Tomlinson-Harashima shaping, where for gorithm approaches a full exponential tree search that finds the exact solution for . Fig. 11 depicts the average energy when the algorithm above is used compared with the energy of uncoded (Tomlinson-Harashima shaping) the QAM symbols. For relative to uncoded -QAM, energy penalty is which is 0.07 dB for 64-QAM and 1.25 dB for 4-QAM. As increases, the shaping gain increases and reaches 1.4 dB for 64-QAM, which is close to the theoretical limit. For 4-QAM, the energy penalty of the Tomlinson-Harashima scheme is completely compensated, with additional gain of 0.2 dB. Note that most of the shaping gain can be achieved with a practical value of 100 (1.25 dB gain for 64-QAM and 0 dB for 4-QAM). Unfortunately, the computational complexity of the stack and the bidirectional stack decoders is much larger when nested lattice shaping is used, compared to the case where TomlinsonHarashima shaping is used. The reason is that for the Tomlinson-Harashima scheme, “x-range testing” can be used to dilute the stack, as described in Section VII-B. Therefore, in addition to the increased complexity at the encoder side, nested lattice shaping has also a complexity penalty at the decoder side. This is a topic for further study.\n\n5218\n\nIEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 57, NO. 8, AUGUST 2011\n\nIX. SUMMARY\n\nApplying [27, Th. 4.1] to\n\nThis paper discusses in depth convolutional lattice codes and their corresponding lattices, and provides several theoretical and practical contributions. Perhaps one unexpected observation is that these lattices, which are essentially lattices whose generator matrix is a Toeplitz band matrix, can have large coding gain with small band size (i.e., by convolving an integer sequence with a short FIR filter). By combining lattice generation with lattice shaping techniques, inspired by signal processing, the paper provides means to design lattice codes, with finite shaped power, directly in the Euclidean space. This elegant and natural concept has been followed up in designing other lattice codes, such as the low density lattice codes, without the need to go through finite alphabet error correcting codes. More specifically to convolutional lattice codes, the paper provides means to bound the error probability in lattice decoding that may be used in other contexts. The practical contributions of this paper for communication over continuous-valued channels, such as the AWGN, should also be mentioned. The paper provides a computationally efficient sequential decoder for convolutional lattice codes. Using this decoder the channel cut-off rate can be attained. This performance should not be considered lightly, as it is attained with low delay, and it outperforms trellis coded modulation techniques with similar complexity. For better performance the paper proposes a bidirectional decoder, that with large enough memory dB, even when using a hyper-cucan attain the capacity at bical shaping region rather than the optimal spherical shaping region. This part of the paper provides a chance to revisit and enhance aspects of convolutional codes, which moved away from the spotlight in the recent years. Clearly, there is room for further research and analysis of these codes. Better generating filters may be found. The decoding algorithms can be improved. Theoretically, the goal is to show that convolutional lattice codes attain capacity for the AWGN channel (probably with large filter or “constraint” length). In addition, further research should analyze and bound the error performance, at least at the level encountered in the classical analysis of convolutional codes. Finally, a challenging topic is to examine the codes over channels other than the AWGN, e.g., fading channel.\n\nTHE DETERMINANT OF\n\nAPPENDIX A FOR LARGE LATTICE DIMENSION\n\nWe would like to show that as , where is a Toeplitz matrix as in (2) whose nonzero column elements are the impulse response coefficients of a monic minwith . is a imum phase filter Hermitian Toeplitz matrix, whose elements are the autocorrelation coefficients of the filter’s impulse response\n\nwhere is assumed zero for or . for lattice dimension by Denote the eigenvalues of for . We then have (18)\n\n, we get (19)\n\nwhere\n\nis the Fourier transform of , i.e., . It is well known (e.g., [37, Ch. 12]) is of the that the right-hand side (RHS) of (19) equals 0 if form (20) and all less than unity, such that the facwith and correspond to zeros and tors poles inside the unit circle, and the factors and correspond to zeros and poles outside the unit circle. In our case, is monic, causal and minimum phase so , which is a special it is of the form case of (20). Substituting (18) in (19), we finally get\n\nwhich is the desired result. Note that this result does not hold, in is not of the form (20). For example, if general, if with , the RHS of (19) equals instead of 0. APPENDIX B FINDING THE LATTICE POINTS INSIDE A SPHERE Consider a convolutional lattice code with a given monic, , whose imcausal, and minimum-phase generating filter pulse response is . We shall now present an algorithm that finds all the lattice points whose squared norm is for a lattice dimension of . The flowchart below a given of the algorithm is shown in Fig. 12. Basically, it develops a tree , and truncates tree of all possible integer sequences branches as soon as it can identify that all the corresponding latwill have squared norms above . The tree tice points is searched in a Depth First Search (DFS) manner, which can be easily implemented using recursion techniques. Due to the shift invariance of the convolution operation, all lattice points that correspond to shifts of the nonzero portion of their corresponding integer vectors will have the same norm. Therefore, it is more efficient to find only a single lattice point from each such group. This can be done by searching only for the nonzero portion of the corresponding integer vector, and forcing it to start in the first symbol by allowing only nonzero , but whenever values for . will be allowed to be 0 for a sequence of ’s is recorded as the nonzero portion of a lattice point, it is verified that the last in the sequence is not zero. The basic step of the algorithm is a “candidate preparation” , step, where a list of candidate integers is prepared for , such that the squared norm of given the values of the resulting lattice point (and its possible extensions) can still . In order to build the list, the sequence be lower than is first convolved with the generating filter, yielding\n\nSHALVI et al.: SIGNAL CODES: CONVOLUTIONAL LATTICE CODES\n\n5219\n\nthe sequence . Since is causal, the first elewill be the first components of the resulting ments of lattice point, regardless of the values that will be chosen for . Since is monic, the ’th com. A necessary ponent of the lattice point will equal condition for the squared norm of the lattice point to be less than is that the squared sum of its first components is less than , i.e. (21) A candidate list for can then be built according to (21). The computational complexity of the tree search can be furin (21) instead ther improved by using a modified bound , where . The reason is that of the left-hand side (LHS) of (21) does not include the “convolution tail,” whose contribution to the squared norm is at least (the minimal squared value for the last symbol of the convolution, since the minimal value for a nonzero integer is 1). This will decrease the size of the candidate lists and thus reduce the total tree search complexity. As noted above, we would like to find a single representative from each group of shifted lattice points. Also, for each lattice point there will be 4 lattice points with the same norm that correspond to multiplications of the corresponding integer vector and , and we would also like to record only a single by point out of these four. In order to do that, the candidate list for the first integer is built in a different manner than for the other integer symbols. For , the candidate list contains all possible complex integers whose squared magnitude is smaller than , diluted as follows. First, as noted above, eliminating will guarantee the zero symbol from the candidate list for that shifted versions of the same integer vector will not be encountered. Also, if a complex integer is in the candidate list for , we can dilute from the list the values , and , since the resulting tree branches will correspond to the same in, , , teger vectors, up to multiplication by the constants respectively. The flow of the algorithm, as shown in Fig. 12, is as follows. The algorithm starts by building a candidate list for the first error symbol . Then, it starts passing on this list. For each possible value of , it builds a candidate list for , and so on. For a general tree node at depth , the values of are determined by the path that leads to this tree node, and the . Whenever the algorithm constructs a candidate list for is empty (immediately at its generation, candidate list for or after finished passing on it) or the sequence length exceeded the lattice dimension , the algorithm steps back one step and turns to the next candidate for . When the candidate list for is finally exhausted, the algorithm terminates. This way, the whole tree is searched in a DFS manner. In each tree node, The algorithm checks if the sequence ends with zeros, in which case it skips to the next element in the list. Such a sequence can be skipped because continuing this sequence will result in an integer vector with a gap of zeros in its nonzero portion. For such an integer vector , the corresponding lattice point is a sum of two other\n\ncorresponds to the first part of lattice points , , where the nonzero portion of (i.e., before the gap of zeros) and corresponds to the second part. The nonzero components of and are at non-overlapping indices, because the convolution “tail” of the filtered first part of the nonzero portion of does not overlap the filtered second part due to the gap of zeros. If only the shortest lattice point is required, such a point is surely and are shorter not the shortest lattice point, since both than . If all the lattice points with squared norm below are required, such a point will have squared norm larger than , so it can be skipped if . Furthermore, even if , such a point is not Voronoi-relevant, and is therefore not needed for the error bounds of Section VI, as explained in Appendix F. During the search process, it is required to record all the integer sequences for which the squared norm of the resulting . Therefore, at each tree lattice point is smaller than , the algonode, after constructing the candidate list for , zero padded to dimension rithm checks if the sequence , is a lattice point with squared norm less than . This is done by summing over all the components of the convolution of with the generating filter, including the “convolution tail,” which was calculated as part of the candidate list construction step. If the sum-of-squares of the convolution output is less , the sequence is recorded, unless its last symbol is than zero (since we want to record only the nonzero portion of the integer vector that corresponds to the lattice point, as explained above. If the last integer is 0, then either the nonzero portion of this sequence was already recorded, or it may be recorded in the future if additional nonzero symbols will be added to it). and Note that instead of calculating the convolution the partial sum at each tree node, a simple recursive update can be applied to the results of the calculations at the parent tree node, thus reducing the computational complexity. Also, instead of actually storing the candidate lists for the ’s, the appropriate candidate can be calculated at each node where only an index needs to be stored. Note also that if only the minimal distance of the code needs to be found, the computational complexity of the algorithm can : whenever a lattice be reduced by dynamically updating is recorded, point whose squared norm is smaller than is updated to the squared norm of this lattice point. We finally note that the complexity of the algorithm of Fig. 12 can be further improved by using a “backward-forward” approach. With this approach, the algorithm first builds a tailsdatabase, which stores all the possible tail sequences whose Eu. This can be done by apclidean distance is lower than plying the algorithm of Fig. 12 backwards in time. The algorithm then develops the search tree forward in time, but the condition for keeping an integer sequence in the tree is that either , or that the last its squared norm is smaller than elements of the integer sequence coincide with the first elements of a sequence from the tails database, in which case their concatenation may yield a lattice point whose squared norm is . This way, the effective search radius of the forbelow instead of , which may reward search is sult in significant complexity reduction even for relatively small . values of\n\n5220\n\nFig. 12. Algorithm for finding the lattice points inside a sphere with radius d\n\nAPPENDIX C SYMMETRY PROPERTIES OF THE SQUARED MINIMUM DISTANCE Assume that filtering an integer sequence with the filter yields the sequence . Then, it can be easily seen that filtering the integer sequence with will yield the sequence . and are Therefore, the lattices that correspond to essentially the same lattice, except for different mapping of integer vectors to lattice points and multiplication of the lattice generator matrix by a unitary matrix, which is equivalent to rotation and reflection. Using induction, the same argument is and true for the filters\n\nIEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 57, NO. 8, AUGUST 2011\n\n.\n\n. As a result, the two lattices that correspond to and have the same minimum distance, . so In the same manner, it can be easily seen that filtering the sewith will yield the quence sequence (where denotes the complex conjugate of ). and are Therefore, the lattices that correspond to essentially the same lattice, except for different mapping of integer vectors to lattice points, multiplication of the lattice generator matrix by a unitary matrix, and complex conjugation, which are equivalent to rotation and reflection. Using induction, the , as defined above, same argument is true for the filters . As a result, the two and\n\nSHALVI et al.: SIGNAL CODES: CONVOLUTIONAL LATTICE CODES\n\nlattices that correspond to imum distance, so\n\nand\n\nhave the same min.\n\n5221\n\nchosen generating filter shows that when the information symbols belong to a 64-QAM constellation, the real and imaginary parts of the ’s have a dynamic range of 17 bits (each). bits are required to store Therefore, consecutive values of . However, the narrow band nature of the sequence , as pointed out in Section V-A, can be utilized we to “compress” these values. Instead of transmitting can transmit the “compressed” values where , is the prediction error of predicting from using the pre, i.e., , and is the diction error filter prediction error of predicting from , using the prediction error filter . The dynamic range of is still 17 bits, but the dynamic range of and is now 12 and 7 bits, respectively. Therefore, the total number of bits required to store 3 conbits. In order to secutive ’s is now only protect these ’s, uncoded 8-QAM modulation is used for their transmission. This way, the ’s are protected by approximately 9 dB relative to uncoded 64-QAM. Since the gap to capacity for is approxuncoded transmission at bit error rate (BER) of imately 9 dB , the uncoded ’s will be more protected than the coded data, so the error rate due to badly detected ’s is negligible. Sending 72 bits requires 24 8-QAM symbols. If the codeword length is =2000, as used in Section VIII, then the effect of the additional 24 symbol on code rate is negligible. APPENDIX E GENERALIZATIONS FOR NONMINIMUM-PHASE AND ARMA GENERATING FILTERS 1) Nonminimum-Phase Filter Patterns: So far, we have restricted the generating filter to be a minimum-phase filter, in order for the recursive loops of the various shaping methods to be stable (Section V). We shall now show how to extend the concept to nonminimum-phase filters of the form , where is a monic is a monic minimum phase filter and and all less than unity. From maximum phase filter with Appendix A, it can be seen that we still have for large , as required (Section III), where the restriction in (2) is now removed. We can then deploy convolutional lattice coding, combined with one of the proposed shaping , methods, with the generating filter which is a minimum phase filter, and then apply an allpass filter to the encoded signal. The allpass filter does not change the signal power level or its power spectrum. Therefore, this scheme generates a lattice which is based on the , which is not minimum-phase. As the regenerating filter cursive loops of the various shaping and encoding schemes work , which is minimum-phase, stability is with the filter ensured. Note that convolving the filter pattern with an allpass filter is equivalent to multiplying a lattice generator matrix by a unitary matrix, which is equivalent to rotation and reflection of the lattice in Euclidean space, that do not change the coding-related lattice properties. Since we can transform a nonminimum-phase filter to a minimum-phase filter by allpass filtering, we should not expect non-minimum-phase filter patterns to have advantage over their minimum-phase equivalents when the AWGN\n\n5222\n\nIEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 57, NO. 8, AUGUST 2011\n\nFig. 13. Combining convolutional lattice coding with pre-equalization.\n\nchannel is considered. However, in non-AWGN channels, such as in fading channels and in impulse noise channels, mixedphase channels may be advantageous since their impulse response may be longer, thus allowing better time-diversity. 2) Autoregressive, Moving-Average (ARMA) Filter Patterns: Thus far, we have described codes which employ FIR generating filters, but the convolutional lattice code concept can be easily extended to ARMA generating filters. Suppose that we want to design a code with an ARMA generating filter , where and are monic invertible minimum phase filters. The encoding operation will then be (22) For Tomlinson-Harashima shaping, the shaping operation is\n\nIt can be easily seen that choosing\n\nresults in . The other shaping methods of Section V can be extended in a similar manner. ARMA generating filters can be particularly useful when convolutional lattice coding is combined with channel preequalization, as described in the next subsection. 3) Combining Convolutional Lattice Coding With Preequalization: Assume that coding should be used for transmission through a communications channel which introduces ISI. Convolutional lattice coding can be seamlessly combined with channel preequalization, by designing the encoder’s filter so that its convolution with the channel impulse response will be the desired convolutional lattice code generating filter, possibly up to a gain factor. However, this would work only if the channel is a minimum phase filter, since otherwise the encoder’s filter is nonminimum phase and the recursive loops of its shaping algorithms become unstable. In order to avoid this problem, an allpass filter can be applied to the transmitted signal, that converts the channel into a minimum phase system (this is a common procedure in equalization of digital communications channels , where ). Let the channel be is a gain factor, is a monic minimum phase filter, and is a monic maximum phase filter, as defined in Section A. Assume further that is stable and invertible. In order\n\nto transform the channel into its minimum phase equivalent, we to the channel input, apply the filter into a minimum transforming the combined channel is an allpass filter, i.e., , phase system. Since it does not affect the transmitted signal’s power or power spectrum. We then apply the shaping and encoding operations using , where the monic minimum phase filter is the desired convolutional lattice code generating filter. The resulting scheme is illustrated in Fig. 13. It can be easily seen that the linear system that relates to the channel output, , folds into the desired pattern , multiplied by the channel gain . Therefore, the receiver can employ a detector that is optimized for an ideal (non-ISI) channel, and the error performance will be the same as in an ideal channel with . a gain of APPENDIX F SORTING THE VORONOI-RELEVANT LATTICE POINTS Assume that for a given lattice with a generator matrix , we have found all the lattice points whose squared norm is less than , and would like to know which of them are Voronoi-relevant lattice points, as defined in Section VI. The proposed algorithm is based on the following criterion . A lattice point is a Voronoi relevant lattice point if and only if its midpoint is not a lattice point, and this midpoint has exactly two nearest lattice points, which are the origin and . An equivalent con, centered at , dition is that a hypersphere with radius should contain only two lattice points, which are the origin and . A further equivalent condition is that a hypersphere with ra, centered at , should contain only two lattice points dius whose corresponding integer vectors have even integer components, which are the origin and . Due to the linearity of the lattice, this sphere can be searched for lattice points by adding , to all the lattice points whose norm is less than or equal to and then checking if the result corresponds to an integer vector with even components. This suggests the following algorithm. – All lattice points whose\n\n# Input:\n\nsquared norm is smaller than\n\n, sorted by\n\nascending norm. – A list of corresponding integer vectors such that # Output:\n\n.\n\n– Flag bits such that\n\nVoronoi-relevant and\n\notherwise.\n\nif\n\nis\n\nSHALVI et al.: SIGNAL CODES: CONVOLUTIONAL LATTICE CODES\n\nfor\n\n5223\n\nto ; ; Fig. 14. An example of the heap data structure.\n\nwhile if\n\n;\n\nzeros) and corresponds to the second part. The nonzero components of and are at nonoverlapping indices, because the convolution “tail” of the filtered first part of the nonzero portion of does not overlap the filtered second part due to the gap of is orthogonal to , and can zeros. As a result, not be a Voronoi-relevant lattice point, since it can be easily seen has 4 nearest lattice points, which are the that its midpoint origin, , and , in contradiction to the above criterion. See for a related algorithm that uses a list of Voronoirelevant vectors for finding the nearest lattice point.\n\n;\n\nAPPENDIX G IMPLEMENTATION OF THE HEAP-BASED STACK DECODER\n\n; end ; end if\n\nelse\n\nend end This formulation is for a general lattice, assuming that the list of lattice points includes all the points in a sphere with radius . For example, if is in the list, should also be in the list. However, the list of lattice points that is generated by the algorithm of Appendix B includes only a single representative from each group of lattice points whose corresponding integer vectors have the same nonzero portion, up to a possible shift, or . It can be seen that if any and up to multiplication by single member of such a group of lattice points is Voronoi-relevant, all the members of the group are also Voronoi-relevant. Then, the following modifications to the above algorithm are is replaced by required. First, the condition or , where denotes (i.e., the portion that starts from the nonzero portion of the first nonzero component and ends with the last nonzero component). This takes into account all the lattice points which are shifted versions of the given lattice points, as well as the lattice (multiplication points that correspond to multiplication by by is already taken care of by the operation itself). In case and have different lengths, their sum is undefined and the condition is evaluated as “False.” Second, the condifor marking a lattice point as nontion , Voronoi relevant should be changed to is in the input list. since only one point out of and A special type of non-Voronoi relevant lattice points for convolutional lattice codes, that can be eliminated already during the operation of the search algorithm of Appendix B, are lattice points that correspond to integer vectors whose nonzero portion contains a subportion of consecutive zeros. For such an integer vector , the corresponding lattice point is a sum corresponds to the of two other lattice points , , where first part of the nonzero portion of (i.e., before the gap of\n\nAt each step of the stack decoder, the path with best score in the stack is extended to all its successors, and then deleted from the stack. The successors then enter the stack. In principle, an infinite stack is required, as the number of paths continuously increases. Practically, a finite stack must be used, so whenever the stack is full, the path with worst score is thrown away. Therefore, a practical stack decoder should find at each step the paths with best score and worst score in the stack. An efficient implementation of the stack algorithm can be based on the heap data structure . A heap is a data structure that stores the data in a semi-sorted manner (See an example in Fig. 14). Specifically, data is arranged in a binary complete tree (i.e., all the levels of the tree are populated, except for the lowest level, whose populated elements are located consecutively at the leftmost locations). The value of each node is larger or equal to the value of its successors. Practically, the heap is stored in a linearly addressed array, without any overhead (i.e., the root of the tree is stored in location 0 of the array, the two elements of the second level are stored in locations 1 and 2, the four elements of the third level at locations 3,4,5,6, and so on). The parent node of the element , and its two at location of the array is stored at location and , where denotes the children are at locations largest integer smaller than . In order to insert a new element to the stack, the element is initially inserted at the lowest level of the tree, adjacent to the rightmost current element. Then, the new element is moved up the path toward the root, by successively exchanging its value with the value in the node above. The operation continues until the value reaches a position where it is less than or equal to its parent, or, failing that, until it reaches the root node. Extracting the maximum element is simple, as the maximum is always at the root of the heap. However, in order to maintain a complete tree, the following procedure is used to delete the maximal element from the stack. First, the root element is deleted and replaced by the rightmost element of the bottom level of the tree. Then, its value is moved down the tree by successively exchanging it with the larger of its two children.\n\n5224\n\nIEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 57, NO. 8, AUGUST 2011\n\nstack). A symbol is deleted from the symbol memory only when no other symbol is linked to it. This way, the minimal number of symbols is stored at each point, and memory usage is optimized. Similarly to the previous approach, storage should be allocated to the symbol memory such that the additional error probability due to symbol memory overflow is negligible. APPENDIX H DERIVATION OF THE FANO METRIC Consider the following transmission model through a discrete, memoryless channel whose input and output are complex numbers in . The transmission uses a variable have lengths length code whose codewords , respectively. Let denote the th coordinate of . Let be the set of all possible , . Let complex values for the coordinate denote the cardinal number of , and let . To , having probaeach codeword bility , a random tail is appended, , producing the word where , which is sent over are independent of each the channel. It is assumed that other and of , for . Let denote . As explained in the probability distribution function of , this decoding problem is essentially the same problem of choosing the best path in each step of the stack algorithm, where the stack contains paths of different lengths. By independence, . Let denote the received word. The joint probability distribution of to a codeword and receiving is appending a tail\n\n(23) Summing over all random tails gives the marginal distribution (24) where (25) Given , the maximum a posteriori decoding rule is to choose which maximizes . Equivalently\n\ncan be maximized, as the denominator is independent of . Taking logarithms, the final statistic to be maximized by the optimum decoder is\n\n(26)\n\nSHALVI et al.: SIGNAL CODES: CONVOLUTIONAL LATTICE CODES\n\n5225\n\nInterestingly, the statistic for each codeword depends only on that portion of the received word having the same length as the codeword. We can now derive the Fano metric for the decoding of convolutional lattice codes transmitted through the AWGN channel with noise variance . For simplicity, we shall start with real valued convolutional lattice codes, and then extend the results are to the complex case. Assume that the data symbols -PAM symbols. There are possible symbols, so the a-priori is probability of a codeword of length (27)\n\n, respectively (where we have assumed that the Tomlinson-Harashima precoding causes the real and imaginary parts to be independent of each other). Substituting in (26), we get (31) again, where now we have (33)\n\nACKNOWLEDGMENT Support and discussions with E. Weinstein and D. Forney are gratefully acknowledged. REFERENCES\n\nThe numerator of the left term inside the sum of (26) is (28) In order to calculate the denominator, we shall assume that Tomlinson-Harashima shaping is used. In this case, the set , as defined above, is a finite set of values, uniformly spread in the . We shall assume that is large, such that we interval can approximate the sum of (25) by an integral\n\n(29) where . Note that the integral of (29) is a convolution between a rectangular pulse and a Gaussian. Assuming (high SNR), the Gaussian is much narrower than the rectangular pulse, so the convolution result can be approximated by a rectangular pulse with height , except for values . We of that are relatively close to the edges of the pulse at can then simply approximate (29) by the constant , assuming that the probability of being near the edges can be neglected. We then get (30) Substituting (27), (28), and (30) in (26) and organizing terms, we finally get (31) where (32) The extension of these results to complex convolutional lattice codes with -QAM input constellation and complex noise variance of is straightforward. Instead of (27), (28), and (30), we have , and\n\n E. Agrell, T. Eriksson, A. Vardy, and K. Zeger, “Closest point search in lattices,” IEEE Trans. Inf. Theory, vol. 48, pp. 2201–2214, Aug. 2002. S. A. Altekar, M. Berggren, B. E. Moision, P. H. Siegel, and J. K. Wolf, “Error-event characterization on partial-response channels,” IEEE Trans. Inf. Theory, vol. IT-45, no. 1, pp. 241–247, Jan. 1999. J. B. Anderson, “On the complexity of bounded distance decoding for the AWGN channel,” IEEE Trans. Inf. Theory, vol. IT-48, no. 5, pp. 1046–1060, May 2002. E. Arikan, “Channel polarization: A method for constructing capacity-achieving codes for symmetric binary-input memoryless channels,” IEEE Trans. Inf. Theory, vol. IT-55, no. 7, pp. 3051–3073, Jul. 2009. M. Atkinson, J. sack, N. Santoro, and T. Strothotte, “Min-Max heaps and generalized priority queues,” Commun. ACM, vol. 29, pp. 996–1000, 1986. T. Aulin, N. Rydbeck, and C. W. Sundberg, “Continuous phase modulation – Part II: Partial response signaling,” IEEE Trans. Commun., pp. 210–225, Mar. 1981. T. Aulin, “Breadth-first maximum likelihood sequence detection: Basics,” IEEE Trans. Commun., vol. 47, no. 2, pp. 208–216, Feb. 1999. L. R. Bahl, J. Cocke, F. Jelinek, and J. Raviv, “Optimal decoding of linear codes for minimizing symbol error rate,” IEEE Trans. Inf. Theory, vol. 20, pp. 284–287, 1974. C. Berrou, A. Glavieux, and P. Thitimajshima, “Near Shannon limit error-correcting coding and decoding: Turbo codes,” in Proc. IEEE Int. Conf. Commun., 1993, pp. 1064–1070. A. R. Calderbank and N. J. A. Sloane, “New trellis codes based on lattices and cosets,” IEEE Trans. Inf. Theory, vol. IT-33, pp. 177–195, Mar. 1987. A. Carlsson, “The deap: A double-ended heap to implement doubleended priority queues,” Inf. Process. Lett., vol. 26, pp. 33–36, 1987. J. H. Conway and N. J. A. Sloane, “A fast encoding method for lattice codes and quantizers,” IEEE Trans. Inf. Theory, pp. 820–824, Nov. 1983. J. H. Conway and N. J. Sloane, Sphere Packings, Lattices and Groups. New York: Springer, 1988. T. H. Cormen, C. E. Leiserson, R. L. Rivest, and C. Stein, Introduction to Algorithms. Cambridge, MA: The MIT press, 2001. R. de Buda, “The upper error bound of a new near-optimal code,” IEEE Trans. Inf. Theory, vol. IT-21, pp. 441–445, Jul. 1975. R. de Buda, “Some optimal codes have structure,” IEEE J. Sel. Areas Commun., vol. 7, pp. 893–899, Aug. 1989. U. Erez and R. Zamir, “Achieving 1/2 log(1 + SNR) on the AWGN channel with lattice encoding and decoding,” IEEE Trans. Inf. Theory, vol. 50, pp. 2293–2314, Oct. 2004. U. Erez, S. Litsyn, and R. Zamir, “Lattices which are good for (almost) everything,” IEEE Trans. Inf. Theory, vol. 51, pp. 3401–3416, Oct. 2005. M. V. Eyuboglu and S. U. H. Qureshi, “Reduced-state sequence estimation with set partitioning and decision feedback,” IEEE Trans. Commun., pp. 13–20, Jan. 1988. G. D. Forney Jr., “Maximum likelihood sequence estimation of digital sequences in the presence of intersymbol interference,” IEEE Trans. Inf. Theory, pp. 363–378, May 1972. G. D. Forney Jr., “Coset codes—Part I: Introduction and geometrical classification,” IEEE Trans. Inf. Theory, pp. 1123–1151, Sep. 1988. G. D. Forney and M. V. Eyuboglu, “Combined equalization and coding using precoding,” IEEE Commun. Mag., vol. 29, no. 12, pp. 25–34, Dec. 1991.\n\n5226\n\n G. D. Forney Jr., “Trellis shaping,” IEEE Trans. Inf. Theory, vol. IT-38, no. 2, pp. 281–300, Mar. 1992. G. D. Forney Jr. and G. Ungerboeck, “Modulation and coding for linear Gaussian channels,” IEEE Trans. Inf. Theory, pp. 2384–2415, Oct. 1998. R. G. Gallager, Low-Density Parity-Check Codes. Cambridge, MA: MIT Press, 1963. R. G. Gallager, Information Theory and Reliable Communication. New York: Wiley, 1968. R. M. Gray, “On the asymptotic eigenvalue distribution of Toeplitz matrices,” IEEE Trans. Inf. Theory, vol. 18, pp. 725–730, Nov. 1972. I. M. Jacobs and E. R. Berlekamp, “A lower bound to the distribution of computation for sequential decoding,” IEEE Trans. Inf. Theory, vol. IT-13, pp. 167–174, 1967. S. Kallel and K. Li, “Bidirectional sequential decoding,” IEEE Trans. Inf. Theory, vol. 43, pp. 1319–1326, Jul. 1997. B. Kurkoski and J. Dauwels, “Reduced-memory decoding of low-density lattice codes,” IEEE Commun. Lett., vol. 14, no. 7, Jul. 2010. R. Laroia, S. A. Tretter, and N. Farvardin, “A simple and effective precoding scheme for noise whitening on intersymbol interference channels,” IEEE Trans. Commun., vol. 41, no. 10, pp. 1460–1463, Oct. 1993. R. Laroia, N. Farvardin, and S. A. Tretter, “On optimal shaping of multidimensional constellations,” IEEE Trans. Inf. Theory, vol. 40, pp. 1044–1056, Jul. 1994. T. Linder, C. Schlegel, and K. Zeger, “Corrected proof of De Buda’s Theorem,” IEEE Trans. Inf. Theory, pp. 1735–1737, Sep. 1993. H. A. Loeliger, “Averaging bounds for lattices and linear codes,” IEEE Trans. Inf. Theory, vol. 43, pp. 1767–1773, Nov. 1997. J. Massey, “Variable-length codes and the Fano metric,” IEEE Trans. Inf. Theory, vol. IT-18, pp. 196–198, Jan. 1972. S. Mohan and J. B. Anderson, “Computationally optimal metric-first code tree search algorithms,” IEEE Trans. Commun., vol. COM-32, no. 6, pp. 710–717, Jun. 1984. A. V. Oppenheim and R. W. Schafer, Discrete-Time Signal Processing. Englewood Cliffs, NJ: Prentice-Hall, 1989. G. Poltyrev, “On coding without restrictions for the AWGN channel,” IEEE Trans. Inf. Theory, vol. 40, pp. 409–417, Mar. 1994. Y. Polyanskiy, H. Vincent Poor, and S. Verdu, “Channel coding rate in the finite blocklength regime,” IEEE Trans. Inf. Theory, vol. 56, pp. 2307–2359, May 2010. F. Rusek, “Partial Response and Faster-than-Nyquist Signaling,” Ph.D., Lund Univ., Dep. Elect. Inf. Technol., Lund, Sweden, 2007. A. Said and J. B. Anderson, “Bandwidth-efficient coded modulation with optimized linear partial-response signals,” IEEE Trans. Inf. Theory, pp. 701–713, Mar. 1998. O. Shalvi, N. Sommer, and M. Feder, “Signal codes,” in Proc. 2003 Inf. Theory Workshop, 2003, pp. 332–336. C. E. Shannon, “A mathematical theory of communication,” Bell Syst. Tech. J., vol. 27, pp. 379–423, Oct. 1948. C. E. Shannon, “Probability of error for optimal codes in a Gaussian channel,” Bell Syst. Tech. J., vol. 38, pp. 611–656, 1959. N. Shulman and M. Feder, “Improved error exponent for time-invariant and periodically time-variant convolutional codes,” IEEE Trans. Inf. Theory, vol. 46, pp. 97–103, Jan. 2000. N. Sommer, M. Feder, and O. Shalvi, “Closest point search in lattices using sequential decoding,” in Proc. Int. Symp. Inf. Theory (ISIT), 2005, pp. 1053–1057. N. Sommer, M. Feder, and O. Shalvi, “Low density lattice codes,” IEEE Trans. Inf. Theory, vol. 54, pp. 1561–1585, Apr. 2008. N. Sommer, “Capacity approaching lattice codes,” Ph.D., Tel-Aviv Univ., School of Elect. Eng., Tel-Aviv, Israel, 2008. N. Sommer, M. Feder, and O. Shalvi, “Shaping methods for low-density lattice codes,” in Proc. 2009 Inf. Theory Workshop, Taormina, Oct. 2009, pp. 238–242. N. Sommer, M. Feder, and O. Shalvi, “Finding the closest lattice point by iterative slicing,” SIAM J. Discr. Math., vol. 23, no. 2, pp. 715–731, 2009. V. Tarokh, A. Vardy, and K. Zeger, Sequential Decoding of Lattice Codes, 1996, to be published. M. Tomlinson, “New automatic equalizer employing modulo arithmetic,” Electron. Lett., pp. 138–139, Mar. 1971.\n\nIEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 57, NO. 8, AUGUST 2011\n\n R. Urbanke and B. Rimoldi, “Lattice codes can achieve capacity on the AWGN channel,” IEEE Trans. Inf. Theory, pp. 273–278, Jan. 1998. A. J. Viterbi, “Error bounds for convolutional codes and an asymptotically optimum decoding algorithm,” IEEE Trans. Inf. Theory, vol. 13, pp. 260–269, Apr. 1967. A. Viterbi and J. Omura, Principles of Digital Communication and Coding. New York: McGraw-Hill, 1979. J. M. Wozencraft and B. Reiffen, Sequential Decoding. New York: Wiley , 1961. Y. Yona and M. Feder, “Efficient parametric decoder of low density lattice codes,” in Proc. Int. Symp. Inf. Theory (ISIT), 2009.\n\nOfir Shalvi (M’04) received the B.Sc. degree in mathematics and physics from the Hebrew University’s Talpiot Program (cum laude), and the M.Sc. (summa cum laude) and the Ph.D. (with distinction) degrees in electrical engineering from Tel-Aviv University, Israel, in 1984, 1988, and 1994, respectively. He was a Postdoctoral Research Affiliate with the Department of Electrical Engineering and Computer Science, Massachusetts Institute of Technology, Cambridge, in 1994–1995, and a Visiting Professor with the School of Electrical Engineering, Tel-Aviv University, during 2005–2006. He has filed and holds more than 40 patents in the areas of signal processing and digital communications. He was a cofounder and the Chief Technology Officer (CTO) of Libit Signal Processing, a pioneer developer of cable modem technology, which was acquired in 1999 by Texas Instruments (TI), where he was elected TI Fellow. He is a cofounder and the CTO of Anobit Technologies, Herzlia, Israel, a pioneer developer of advanced signal processing technologies to the storage markets. Dr. Shalvi was the recipient of the 1990 Israeli Minister of Communications Award, the 1991 Clore Fellowship, the 1993 Wolfson Fellowship, and the 1994 Fulbright Fellowship.\n\nNaftali Sommer (M’00–SM’05) received the B.Sc., M.Sc., and Ph.D. degrees in electrical engineering from Tel-Aviv University, Israel, in 1990, 1994, and 2008, respectively. From 1990 to 1996, he was with the Israel Ministry of Defence. In 1996, he joined Libit Signal Processing, a pioneer developer of cable modem technology, as its chief engineer. After the acquisition of Libit by Texas Instruments (TI) in 1999, he was elected as TI Distinguished Member of the Technical Staff (DMTS), and was the chief scientist of TI’s cable broadband communications business unit until 2006. Since 2007, he has been the chief scientist of Anobit Technologies, Herzlia, Israel, a pioneer developer of advanced signal processing technologies for the storage markets.\n\nMeir Feder (S’81–M’87–SM’93–F’99) received the B.Sc. and M.Sc. degrees from Tel-Aviv University, Israel, and the Sc.D. degree from the Massachusetts Institute of Technology (MIT) Cambridge, and the Woods Hole Oceanographic Institution, Woods Hole, MA, all in electrical engineering in 1980, 1984, and 1987, respectively. After being a Research Associate and Lecturer with MIT, he joined the School of Electrical Engineering, Tel-Aviv University, where he is now a Professor. He had visiting appointments with the Woods Hole Oceanographic Institution, Scripps Institute, Bell Laboratories, and during 1995–1996, he has been a Visiting Professor at MIT. He is also extensively involved in the high-tech industry and cofounded several companies including Peach Networks, a developer of a unique server-based interactive TV solution which was acquired on March 2000 by Microsoft, and Amimon a leading provider of ASICs for wireless high-definition A/V connectivity at the home. Prof. Feder is a corecipient of the 1993 IEEE Information Theory Best Paper Award. He also received the 1978 “creative thinking” award of the Israeli Defense Forces, the 1994 Tel-Aviv University prize for Excellent Young Scientists, the 1995 Research Prize of the Israeli Electronic Industry, and the Research Prize in applied electronics, awarded by Ben-Gurion University. Between June 1993-June 1996, he served as an Associate Editor for Source Coding of the IEEE TRANSACTIONS ON INFORMATION THEORY."
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https://studyslide.com/doc/473156/trendy-periodic-table | [
"#### Transcript Trendy Periodic Table\n\n```Trendy Periodic Table\nPeriod vs. Row\n• The periodic table is\n__________________\nwhich are called\nPeriods.\n• The ___________\nPeriodic Table are called\nGroups or Families.\nFamilies\nMetal, Metalloid, Nonmetal\nMetals\nNonmetals\n• Become ________ ions\n(___________).\n• ______________ electricity.\n• ________________.\n• ______________.\n• Usually a __________ at room\ntemperature. ________\nmelting points.\n• Become _______ ions\n(____________).\n• _______________electricity\nor heat.\n• ______________.\n• __________________\n• Usually a _________ at room\ntemperature. _________\nmelting points.\nMetalloids:\nSome properties of metals, some properties of nonmetals.\n• Size ______________________ down a group.\n• Size generally ___________ across a period from left to right .\nIonization Energy\n• The ionization energy of an atom is the amount of\nenergy _________________________from the\ngaseous form of that atom or ion.\n• 1st ionization energy - The energy required to\nremove ___________________from a neutral\ngaseous atom.\n• For Example:\nNa(g) → Na+(g) + e-I1 = 496 kJ/mole\n• Notice that the ionization energy __________. This is\nbecause it __________ energy to remove an\nelectron.\nIonization Energy\nElectron Affinity\n• Electron Affinity is the\n_________________________________________.\n• Example:\nCl(g) + e- → Cl-(g)E.A. = -349 kJ/mole\n• Notice the sign on the energy is ______________.\nThis is because energy is usually _____________ in\nthis process, as apposed to ionization energy, which\nrequires energy.\n• A ________________corresponds to a\n____________ attraction for an electron. (An\nunbound electron has an energy of zero.)\nElectron Affinity\nElectronegativity\n• Electronegativity is an atom's ‘________' to\n___________________________. A high\nelectronegativity value implies that the\nvalence electrons are tightly held and\n___________________________________\n• Period - electronegativity _______________\nas you go from left to right across a period.\n• Group - electronegativity\n___________________ as you go down a\ngroup.\nElectronegativity\nReactivity\n• Reactivity refers to how\n_________________________________________. This is\nusually determined by how easily electrons can be removed\n(____________________) and how badly they want to take\nother atom's electrons (_____________________) because it\nis the transfer/interaction of electrons that is the basis of\nchemical reactions.\n• Metals\n– Period - reactivity ___________________ as you go from left to right\nacross a period.\n– Group - reactivity _____________________as you go down a group\n• Non-metals\n– Period - reactivity ______________ as you go from the left to the right\nacross a period.\n– Group - reactivity ___________________ as you go down the group.\n• Metals - the atomic radius of a metal is\ngenerally ______________ than the ionic"
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https://physics.stackexchange.com/questions/396375/distribution-of-energies-in-canonical-ensemble?noredirect=1 | [
"Distribution of energies in canonical ensemble\n\nFrom what I understand, in a system $S$ described by a canonical ensemble, the probability that $S$ has energy $E$ is equal to $\\frac{1}{Z}e^{-E/kT}$, where $T$ is the \"temperature\", $k$ the Boltzmann constant, and $Z$ the partition function. I have two questions:\n\n1) Is it obvious that $Z = kT$, since $\\int_0^{\\infty}e^{-E/kT}dE = kT$?\n\n2) I'm failing to understand where the size of $S$ comes into play. Is this energy distribution true whether $S$ is a system containing 1 or $10^{23}$ atoms? I understand that the heat bath should be much larger than either of these, but don't understand how the size of the system doesn't play a role.\n\nI think I'm missing something...\n\n• – Sean E. Lake Mar 29 '18 at 17:28\n\nYes, what you're missing is that the probability is defined over distinct states, not just energies. So, what $Z$ is will depend on how that is defined. For a single spin $1/2$ particle in a magnetic field there are exactly two states with energies $E_+$ and $E_-$, leading to the partition function $Z$ being $$Z = \\mathrm{e}^{-E_+/kT} + \\mathrm{e}^{-E_-/kT}.$$\nIf you're talking about a classical particle in a box with volume $V$, then distinct states are those that have distinct position $\\mathbf{x}$ and momentum $p$, producing the partition function \\begin{align} Z & = \\int \\mathrm{e}^{-p^2/(2mkT)} \\operatorname{d}^3x\\operatorname{d}^3p\\\\ & = V 4\\pi \\int_0^\\infty \\mathrm{e}^{-p^2/(2mkT)} p^2 \\operatorname{d}p. \\end{align}\nWhere the size comes into play is in both the volume, $V$, and the number of particles involved, $N$. When all of the individual particles are independent you can just raise $Z$ to the power $N$. When the particles aren't independent, you need to do more work to integrate/sum over the degrees of freedom available.\n• Even if the particle is a classical, factor $1/h^3$ must be present ( David Tong compared this situation with the vestigial effect, like the male nipple:) – Aleksey Druggist Mar 29 '18 at 7:09\n• Thanks! My understanding is that the canonical ensemble is derived by considering a small subsystem $S$ (with fixed $N$ and $V$) of a much larger one described by a microcanonical ensemble. The phase space volume in which $S$ has energy $E_s$ is proportional to the phase space volume in which the remainder of the system has energy $E-E_s$, where $E$ is the total system energy. Since phase space volume roughly increases exponentially with energy, a Taylor series expansion gives us the canonical distribution. Doesn't this allow us to consider the probability distribution of energy? – Menachem Mar 29 '18 at 14:29\n• @Menachem You still need the degeneracy of each energy level, often called $g$. – Sean E. Lake Mar 29 '18 at 17:02\n• @Menachem I'm talking about the phase space volume of individual particles, and that grows like $4\\pi p^2$. You're talking about the phase space volume of a system of particles, which grows combinatorically with $N$. Either way, the probability won't just be $Z^{-1} \\mathrm{e}^{-E/kT}$, it's $Z^{-1} g(E) \\mathrm{e}^{-E/kT}$. – Sean E. Lake Mar 29 '18 at 17:28"
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https://online-unit-converter.com/volume/convert-fluid-ounces-to-milliliters/52-oz-to-ml/ | [
"# 52 oz to ml CONVERTER. How many milliliters are in 52 fluid ounces?\n\n## 52 oz to ml\n\nThe question “What is 52 oz to ml?” is the same as “How many milliliters are in 52 oz?” or “Convert 52 fluid ounces to milliliters” or “What is 52 fluid ounces to milliliters?” or “52 fluid ounces to ml”. Read on to find free oz to milliliter converter and learn how to convert 52 oz to ml. You’ll also learn how to convert 52 oz to ml.\n\nAnswer: There are 1537.823537 ml in 52 oz.\n\nAlternatively, you can say “52 oz equals to 1537.823537 ml” or “52 oz = 1537.823537 ml” or “52 fluid ounces is 1537.823537 milliliters”.\n\n## fluid ounces to milliliter conversion formula\n\nA milliliter is equal to 0.0338140227 fluid ounces. A fluid ounce equals 29.5735295625 milliliters\nTo convert 52 fluid ounces to milliliters you can use one of the formulas:\n\nFormula 1\nMultiply 52 oz by 29.5735295625.\n52 * 29.5735295625 = 1537.823537 ml.\n\nFormula 2\nDivide 52 oz by 0.0338140227.\n52 / 0.0338140227 = 1537.823537 ml\n\nHint: no need to use a formula. Use our free oz to ml converter.\n\n## Alternative spelling of 52 oz to ml\n\nMany of our visitor spell fluid ounces and milliliters differently. Below we provide all possible spelling options.\n\n• Spelling options with “fluid ounces”: 52 fluid ounces to ml, 52 fluid ounce to ml, 52 fluid ounces to milliliters, 52 fluid ounce to milliliters, 52 fluid ounces in ml, 52 fluid ounce in ml, 52 fluid ounces in milliliters, 52 fluid ounce in milliliters.\n• Spelling options with “oz”: 52 oz to ml, 52 oz to milliliter, 52 oz to milliliters, 52 oz in ml, 52 oz in milliliter, 52 oz in milliliters.\n• Spelling options with “in”: 52 oz in ml, 52 oz in milliliter, 52 oz in milliliters, 52 oz in ml, 52 oz in milliliter, 52 oz in milliliters, 52 fluid ounces in ml, 52 fluid ounce in ml, 52 fluid ounces in milliliters, 52 fluid ounce in milliliters,\n\n## FAQ on 52 oz to ml conversion\n\nHow many milliliters are in 52 fluid ounces?\n\nThere are 1537.823537 milliliters in 52 fluid ounces.\n\n52 oz to ml?\n\n52 oz is equal to 1537.823537 ml. There are 1537.823537 milliliters in 52 fluid ounces.\n\nWhat is 52 oz to ml?\n\n52 oz is 1537.823537 ml. You can use a rounded number of 1537.823537 for convenience. In this case you can say that 52 oz is 1537.82 ml.\n\nHow to convert 52 oz to ml?\n\nUse our free fluid ounce to milliliters converter or multiply52 oz by 29.5735295625.\n52 * 29.5735295625 = 1537.823537 ml."
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https://ncatlab.org/nlab/show/factorial | [
"Contents\n\n# Contents\n\n## Definition\n\nFor $k \\in \\mathbb{N}$ a natural number, its factorial $k! \\in \\mathbb{N}$ is the number obtained by multiplying all positive natural numbers less than or equal to $k$:\n\n$k! \\;\\coloneqq\\; 1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdot \\cdots \\cdot (k-1) \\cdot k \\,.$\n\nIn combinatorics, the definition usually extends to $k = 0$ by setting $0! = 1$. This may be justified by defining $k!$ to be the number of permutations of a set with $k$ elements."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8699394,"math_prob":0.99993896,"size":590,"snap":"2019-35-2019-39","text_gpt3_token_len":128,"char_repetition_ratio":0.13481228,"word_repetition_ratio":0.0,"special_character_ratio":0.2101695,"punctuation_ratio":0.14159292,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9996793,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-17T16:36:02Z\",\"WARC-Record-ID\":\"<urn:uuid:b0baec96-c946-44d3-89c7-d0b0a237d3ba>\",\"Content-Length\":\"21879\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d0445730-92c4-4080-ba67-cbdfe1cb25ac>\",\"WARC-Concurrent-To\":\"<urn:uuid:e9184c5e-0d53-4a23-92dd-97549ad82da9>\",\"WARC-IP-Address\":\"104.27.171.19\",\"WARC-Target-URI\":\"https://ncatlab.org/nlab/show/factorial\",\"WARC-Payload-Digest\":\"sha1:4ZMULLT3ZYXIAEG5RU6PBNGKD3I3WY4S\",\"WARC-Block-Digest\":\"sha1:IKI4WSAD5U6C3U2L5VUKCY4YGB3GNLZR\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514573098.0_warc_CC-MAIN-20190917161045-20190917183045-00559.warc.gz\"}"} |
https://grad.hitbullseye.com/info-zone/speed-time-and-distance.php | [
"# Catching the Running Thief\n\nThere are various methods to approach questions from the topic Speed, Time and Distance.\nThis article demonstrates the usage of Ratios in solving various types of questions:\n###### Basics: Distance = Speed ×Time\n1. If Time is fixed, then Distance α Speed, i.e., D1/D2 = S1/S2.\n\nSo for example, if two trains start from Delhi and Chandigarh towards each other at 8 am, and meet somewhere in between at 10 am, this means both the trains have travelled for 2 hours each. So, their distances travelled will be in the ratio of their speeds.\n\n2. If Speed is fixed, then Distance α Time i.e., D1/D2 = T1/T2.\nThis basically means that if a person is moving at a fixed speed, then he will take more time in covering a greater distance as compared to the time he would take to cover a shorter distance.\n3. If Distance is fixed, then Speed α 1/Time i.e., S1/S2 = T2/T1.\nThis means that a body at a higher speed takes less time to cover the distance as compared to the time it would take to cover the same distance at a lower speed.\nSo if a body takes time T to cover a distance D at a certain speed, then it would take time T/2 to cover the same distance in case its speed is doubled.\n###### You will understand the concept better with the help of the following examples:\nExample 1:\nA thief steals a car at 9 am and drives it at a speed of 60 kmph. The police start chasing him at 10 am at the speed of 90 km/hr. How much more distance would be covered by the thief before he is caught.\nSolution:\nDistance covered by thief from 9 am to 10 am = 60 km.\nNow, after the police starts chasing him, let the distance covered by thief be Dt and that of police be Dp. So Dt/Dp = 60/90 = 2/3. So distances will be 2x and 3x.\nAs per question, 3x – 2x = 60 i.e. x = 60. So Dt = 120 km and Dp = 180 km.\n###### Extra practice:\nThe same question can also be used to find the time when the thief will be caught. Now, the thief has to run a further distance of 120 km as found above, so that means 120/60 = 2 hours. The chase started at 10am, so he will be caught at 12 Noon.\nExample 2:\nA thief steals a car at 2:30 pm and drives away at 60 kmph. The theft is discovered at 3 pm and the owner sets off in another car at 75 kmph. At what time will the thief be caught?\nSolution:\nDistance covered by thief from 2:30 pm to 3 pm = 60 × ½ = 30 km.\nAfter the chase starts, let the distance covered by thief is Dt and that of owner is Do. So Dt/Do = 60/75 = 4/5 i.e. 4x and 5x.\nNow, as per question 5x – 4x = 30 i.e. x = 30 km. So Dt = 120 km and Do = 150 km.\nThief will be caught after travelling adistance of 120kms i.e. 120/60 = 2 hours from the time the chase began i.e. 5 pm.\nExample 3:\nA train after travelling 50 km meets with an accident and then proceeds at ¾ of its former speed and arrives at its destination 35 mins late. Had the accident occurred 24 km further, it would have reached only 25 mins late. Find the speed of the train.\nSolution:\nIn the second case, the train travelled an additional 24 km with its normal speed, hence saving 10 mins. So instead of making equations, we will focus only on these 24 km.\nDistance Speed Time\ni. scenario: 24 kms 3/4 S 4/3 T\nii. Scenario: 24 kms S T\nClearly, in the first case the train takes 1/3 T extra mins i.e. 1/3 T = 10 mins.\nSo T = 30 mins. This means that the train takes 30 mins to cover 24 kmat its normal speed, hence the normal speed should be 48 kmph."
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https://intellipaat.com/community/14609/creating-a-vector-in-r-of-counts-for-number-of-times-each-element-appears-in-another-vector | [
"# Creating a vector in R of counts for number of times each element appears in another vector\n\n1 view\n\nThis is hard for me to explain, so I will just give an example instead. I have two vectors below (a & b).\n\na <- c(\"cat\",\"dog\",\"banana\",\"yogurt\",\"dog\")\n\nb <- c(\"salamander\",\"worm\",\"dog\",\"banana\",\"cat\",\"yellow\",\"blue\")\n\nWhat I would like is the following results:\n\n 0 0 2 1 1 0 0\n\nwhere each element of the result is the number of times each element of b appears in the vector a.\n\ndo.call(\"c\",lapply(b,function(x){sum(x == a)}))\n\nThis gives me what I want, but I need a vectorized/faster version of this because I am working with >20,000 records. Any help appreciated!\n\nby (25.3k points)\n\nFor a faster solution, you can use the match function to know the matching elements and table function to print the counts.i.e.,\n\na <- c(\"cat\",\"dog\",\"banana\",\"yogurt\",\"dog\")\n\nb <- c(\"salamander\",\"worm\",\"dog\",\"banana\",\"cat\",\"yellow\",\"blue\")\n\nresult <- table(factor(b, levels=b)[match(a, b, nomatch=0)])\n\nOutput:\n\nsalamander worm dog banana cat yellow blue\n\n0 0 2 1 1 0 0\n\nTo convert the output to a vector:\n\nas.vector(result)\n\n 0 0 2 1 1 0 0"
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https://se.mathworks.com/matlabcentral/answers/236993-wrong-direction-of-e-field-lines-using-quiver-function-in-a-program-to-solve-simple-laplace-proble?s_tid=prof_contriblnk | [
"# Wrong direction of E-Field lines using \"Quiver Function\" in a program to solve Simple Laplace Problem.\n\n7 views (last 30 days)\nRiyasat on 27 Aug 2015\nCommented: Star Strider on 27 Aug 2015\nThis is my program to solve a simple laplace problem where a Voltage of 100V is applied at the top and all other bounderies are 0V. The Matrix outputs the correct values (in var \"z\"). And the potential distribution seems right. The Problem is that when I use the quiver function to plot the E-Field lines then they originate from the bottom boundary (at 0V) and end at sides and the Top. But the Fields should originate from the top at (100v) and end at the sides and the bottom. I can't figure out what went wrong so any help would be appreciated. Thanks.\n%array\nu = zeros(50);\nu(1,:)=100;\nz = u;\n%loop\nfor i =1:500\nfor i = 2: size(z,1)-1;\nfor j = 2 : size(z,1)-1\nz(i,j) = (z(i-1,j)+z(i+1,j)+z(i,j-1)+z(i,j+1))/4;\nend\nend\n% getframe\nfigure(1)\nend\nimagesc(z);\nfigure(2)\naxis xy\nquiver(-ex,-ey,3);\n\nStar Strider on 27 Aug 2015\nNote the y-axis orientation between the image plot and the quiver call. If you use contourf instead of imagesc, they match. If you want zero to start at the top of the plot for both, set the 'YDir' property for each figure if you use contourf, and the quiver plot if you use imagesc:\nset(gca, 'YDir','reverse')\nIf you want to overplot the quiver arrows on top of the first plot, use the hold function.\n\nRiyasat on 27 Aug 2015\nIt worked thanks a lot. And thanks for the explanation too. Can you please explain why this discrepency between Imagesc and Contour functions?\nStar Strider on 27 Aug 2015\nMy pleasure.\nI don’t know why images start in the upper left corner, but the image convention seems to have existed before MATLAB, and follows earlier graphics conventions that I remember from the 1970s. Plots — including contour — always started in the lower left, again by convention."
]
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https://openeuphoria.org/wiki/view/updating%20oE%20%20arccosh.wc | [
"### updating oE arccosh\n\n#### arccosh\n\n```include math.e\nnamespace math\npublic function arccosh(not_below_1 a)\n```\n\ncomputes the reverse hyperbolic cosine of an object.\n\n##### Parameters:\n1. x : the object to process.\n##### Returns:\n\nAn object, the same shape as x, each atom of which was acted upon.\n\n##### Errors:\n\nSince cosh only takes values not below 1, an argument below 1 causes an error.\n\nThe hyperbolic cosine grows like the logarithm function.\n\n##### Example 1:\n```? arccosh(1) -- prints out 0\n```"
]
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https://www.jobilize.com/physics3/course/2-4-thin-lenses-geometric-optics-and-image-formation-by-openstax?qcr=www.quizover.com&page=7 | [
"# 2.4 Thin lenses (Page 8/13)\n\n Page 8 / 13\n\n## Summary\n\n• Two types of lenses are possible: converging and diverging. A lens that causes light rays to bend toward (away from) its optical axis is a converging (diverging) lens.\n• For a converging lens, the focal point is where the converging light rays cross; for a diverging lens, the focal point is the point from which the diverging light rays appear to originate.\n• The distance from the center of a thin lens to its focal point is called the focal length f .\n• Ray tracing is a geometric technique to determine the paths taken by light rays through thin lenses.\n• A real image can be projected onto a screen.\n• A virtual image cannot be projected onto a screen.\n• A converging lens forms either real or virtual images, depending on the object location; a diverging lens forms only virtual images.\n\n## Conceptual questions\n\nYou can argue that a flat piece of glass, such as in a window, is like a lens with an infinite focal length. If so, where does it form an image? That is, how are ${d}_{\\text{i}}$ and ${d}_{\\text{o}}$ related?\n\nWhen you focus a camera, you adjust the distance of the lens from the film. If the camera lens acts like a thin lens, why can it not be a fixed distance from the film for both near and distant objects?\n\nThe focal length of the lens is fixed, so the image distance changes as a function of object distance.\n\nA thin lens has two focal points, one on either side of the lens at equal distances from its center, and should behave the same for light entering from either side. Look backward and forward through a pair of eyeglasses and comment on whether they are thin lenses.\n\nWill the focal length of a lens change when it is submerged in water? Explain.\n\nYes, the focal length will change. The lens maker’s equation shows that the focal length depends on the index of refraction of the medium surrounding the lens. Because the index of refraction of water differs from that of air, the focal length of the lens will change when submerged in water.\n\n## Problems\n\nHow far from the lens must the film in a camera be, if the lens has a 35.0-mm focal length and is being used to photograph a flower 75.0 cm away? Explicitly show how you follow the steps in the [link] .\n\nA certain slide projector has a 100 mm-focal length lens. (a) How far away is the screen if a slide is placed 103 mm from the lens and produces a sharp image? (b) If the slide is 24.0 by 36.0 mm, what are the dimensions of the image? Explicitly show how you follow the steps in the [link] .\n\na. $\\frac{1}{{d}_{\\text{i}}}+\\frac{1}{{d}_{\\text{o}}}=\\frac{1}{f}⇒{d}_{\\text{i}}=3.43\\phantom{\\rule{0.2em}{0ex}}\\text{m}$ ;\nb. $m=-33.33$ , so that $\\begin{array}{c}\\left(2.40\\phantom{\\rule{0.2em}{0ex}}×\\phantom{\\rule{0.2em}{0ex}}{10}^{-2}\\phantom{\\rule{0.2em}{0ex}}\\text{m}\\right)\\left(33.33\\right)=80.0\\phantom{\\rule{0.2em}{0ex}}\\text{cm, and}\\hfill \\\\ \\left(3.60\\phantom{\\rule{0.2em}{0ex}}×\\phantom{\\rule{0.2em}{0ex}}{10}^{-2}\\phantom{\\rule{0.2em}{0ex}}\\text{m}\\right)\\left(33.33\\right)=1.20\\phantom{\\rule{0.2em}{0ex}}\\text{m}⇒0.800\\phantom{\\rule{0.2em}{0ex}}\\text{m}\\phantom{\\rule{0.2em}{0ex}}×\\phantom{\\rule{0.2em}{0ex}}1.20\\phantom{\\rule{0.2em}{0ex}}\\text{m or}\\phantom{\\rule{0.2em}{0ex}}80.0\\phantom{\\rule{0.2em}{0ex}}\\text{cm}\\phantom{\\rule{0.2em}{0ex}}\\phantom{\\rule{0.2em}{0ex}}×\\phantom{\\rule{0.2em}{0ex}}120\\phantom{\\rule{0.2em}{0ex}}\\text{cm}\\hfill \\end{array}$\n\nA doctor examines a mole with a 15.0-cm focal length magnifying glass held 13.5 cm from the mole. (a) Where is the image? (b) What is its magnification? (c) How big is the image of a 5.00 mm diameter mole?\n\nA camera with a 50.0-mm focal length lens is being used to photograph a person standing 3.00 m away. (a) How far from the lens must the film be? (b) If the film is 36.0 mm high, what fraction of a 1.75-m-tall person will fit on it? (c) Discuss how reasonable this seems, based on your experience in taking or posing for photographs.\n\na. $\\begin{array}{c}\\frac{1}{{d}_{\\text{o}}}+\\frac{1}{{d}_{\\text{i}}}=\\frac{1}{f}\\hfill \\\\ {d}_{\\text{i}}=5.08\\phantom{\\rule{0.2em}{0ex}}\\text{cm}\\hfill \\end{array}$ ;\nb. $m=-1.695\\phantom{\\rule{0.2em}{0ex}}×\\phantom{\\rule{0.2em}{0ex}}{10}^{-2}$ , so the maximum height is $\\frac{0.036\\phantom{\\rule{0.2em}{0ex}}\\text{m}}{1.695\\phantom{\\rule{0.2em}{0ex}}×\\phantom{\\rule{0.2em}{0ex}}{10}^{-2}}=2.12\\phantom{\\rule{0.2em}{0ex}}\\text{m}⇒\\text{100%}\\phantom{\\rule{0.2em}{0ex}}$ ;\nc. This seems quite reasonable, since at 3.00 m it is possible to get a full length picture of a person.\n\nA camera lens used for taking close-up photographs has a focal length of 22.0 mm. The farthest it can be placed from the film is 33.0 mm. (a) What is the closest object that can be photographed? (b) What is the magnification of this closest object?\n\nSuppose your 50.0 mm-focal length camera lens is 51.0 mm away from the film in the camera. (a) How far away is an object that is in focus? (b) What is the height of the object if its image is 2.00 cm high?\n\na. $\\frac{1}{{d}_{\\text{o}}}+\\frac{1}{{d}_{\\text{i}}}=\\frac{1}{f}⇒{d}_{\\text{o}}=2.55\\phantom{\\rule{0.2em}{0ex}}\\text{m}$ ;\nb. $\\frac{{h}_{\\text{i}}}{{h}_{\\text{o}}}=\\text{−}\\frac{{d}_{\\text{i}}}{{d}_{\\text{o}}}⇒{h}_{\\text{o}}=1.00\\phantom{\\rule{0.2em}{0ex}}\\text{m}$\n\nWhat is the focal length of a magnifying glass that produces a magnification of 3.00 when held 5.00 cm from an object, such as a rare coin?\n\nThe magnification of a book held 7.50 cm from a 10.0 cm-focal length lens is 3.00. (a) Find the magnification for the book when it is held 8.50 cm from the magnifier. (b) Repeat for the book held 9.50 cm from the magnifier. (c) Comment on how magnification changes as the object distance increases as in these two calculations.\n\na. Using $\\frac{1}{{d}_{\\text{o}}}+\\frac{1}{{d}_{\\text{i}}}=\\frac{1}{f}$ , ${d}_{\\text{i}}=-56.67\\phantom{\\rule{0.2em}{0ex}}\\text{cm}$ . Then we can determine the magnification, $m=6.67$ . b. ${d}_{\\text{i}}=-190\\phantom{\\rule{0.2em}{0ex}}\\text{cm}\\phantom{\\rule{0.2em}{0ex}}$ and $m=+20.0$ ; c. The magnification m increases rapidly as you increase the object distance toward the focal length.\n\nSuppose a 200 mm-focal length telephoto lens is being used to photograph mountains 10.0 km away. (a) Where is the image? (b) What is the height of the image of a 1000 m high cliff on one of the mountains?\n\nA camera with a 100 mm-focal length lens is used to photograph the sun. What is the height of the image of the sun on the film, given the sun is $1.40\\phantom{\\rule{0.2em}{0ex}}×\\phantom{\\rule{0.2em}{0ex}}{10}^{6}\\phantom{\\rule{0.2em}{0ex}}\\text{km}$ in diameter and is $1.50\\phantom{\\rule{0.2em}{0ex}}×\\phantom{\\rule{0.2em}{0ex}}{10}^{8}\\phantom{\\rule{0.2em}{0ex}}\\text{km}$ away?\n\n$\\begin{array}{cc}\\hfill \\frac{1}{{d}_{\\text{o}}}+\\frac{1}{{d}_{\\text{i}}}& =\\frac{1}{f}\\hfill \\\\ \\hfill {d}_{\\text{i}}& =\\frac{1}{\\left(1\\text{/}f\\right)-\\left(1\\text{/}{d}_{\\text{o}}\\right)}\\hfill \\\\ \\hfill \\frac{{d}_{\\text{i}}}{{d}_{\\text{o}}}& =6.667\\phantom{\\rule{0.2em}{0ex}}×\\phantom{\\rule{0.2em}{0ex}}{10}^{-13}=\\frac{{h}_{\\text{i}}}{{h}_{\\text{o}}}\\hfill \\\\ \\hfill {h}_{\\text{i}}& =-0.933\\phantom{\\rule{0.2em}{0ex}}\\text{mm}\\hfill \\end{array}$\n\nUse the thin-lens equation to show that the magnification for a thin lens is determined by its focal length and the object distance and is given by $m=f\\text{/}\\left(f-{d}_{\\text{o}}\\right)$ .\n\nAn object of height 3.0 cm is placed 5.0 cm in front of a converging lens of focal length 20 cm and observed from the other side. Where and how large is the image?\n\n${d}_{\\text{i}}=-6.7\\phantom{\\rule{0.2em}{0ex}}\\text{cm}$\n${h}_{\\text{i}}=4.0\\phantom{\\rule{0.2em}{0ex}}\\text{cm}$\n\nAn object of height 3.0 cm is placed at 5.0 cm in front of a diverging lens of focal length 20 cm and observed from the other side. Where and how large is the image?\n\nAn object of height 3.0 cm is placed at 25 cm in front of a diverging lens of focal length 20 cm. Behind the diverging lens, there is a converging lens of focal length 20 cm. The distance between the lenses is 5.0 cm. Find the location and size of the final image.\n\n83 cm to the right of the converging lens, $m=-2.3,{h}_{\\text{i}}=6.9\\phantom{\\rule{0.2em}{0ex}}\\text{cm}$\n\nTwo convex lenses of focal lengths 20 cm and 10 cm are placed 30 cm apart, with the lens with the longer focal length on the right. An object of height 2.0 cm is placed midway between them and observed through each lens from the left and from the right. Describe what you will see, such as where the image(s) will appear, whether they will be upright or inverted and their magnifications.\n\nعند قذف جسم إلى أعلى بسرعة إبتدائية فإنه سيصل إلى ارتفاع معين (أقصى ارتفاع) ثم يعود هابطاً نحو سطح الأرض . إذا قُذِفَ جسم إلى أعلى ووجد أن سرعته 18 م / ث عندما قطع 1/4 المسافة التي تمثل أقصى ارتفاع سيصله فالمطلوب إيجاد السرعة التي قُذِف بها بالمتر / ث . إن هذه السرعة هي واحدة من الإجابات التالية\nwhat is light\nlight is a kind of radiation That stimulates sight brightness a source of illumination.\nkenneth\nElectromagnet radiation creates space 7th, 8th, and 9th dimensions at the rate of c.\nJohn\nThat is the reason that the speed of light is constant.\nJohn\nThis creation of new space is \"Dark Energy\".\nJohn\nThe first two sets of three dimensions, 1 through 6, are \"Dark Matter\".\nJohn\nAs matter decays into luminous matter, a proton, a neutron, and an electron creat deuterium.\nJohn\nThere are three sets of three protons, 9.\nJohn\nThere are three sets of three neutrons, 9.\nJohn\nA free neutron decays into a proton, an electron, and a neutrino.\nJohn\nThere are three sets of five neutrinoes, 15.\nJohn\nNeutrinoes are two dimensional.\nJohn\nA positron is composed of the first three dimensions.\nJohn\nAn electron is composed of the second three dimensions.\nJohn\nWhat is photoelectric\nlight energy (photons) through semiconduction of N-P junction into electrical via excitation of silicon purified and cristalized into wafers with partially contaminated silicon to allow this N-P function to operate.\nMichael\ni.e. Solar pannel.\nMichael\nIf you lie on a beach looking at the water with your head tipped slightly sideways, your polarized sunglasses do not work very well.Why not?\nit has everything to do with the angle the UV sunlight strikes your sunglasses.\nJallal\nthis is known as optical physics. it describes how visible light, ultraviolet light and infrared light interact when they come into contact with physical matter. usually the photons or light upon interaction result in either reflection refraction diffraction or interference of the light.\nJallal\nI hope I'm clear if I'm not please tell me to clarify further or rephrase\nJallal\nwhat is bohrs model for hydrogen atom\nhi\nTr\nHello\nYoute\nHi\nNwangwu-ike\nhi\nSiddiquee\nhi\nOmar\nhelo\nMcjoi\nwhat is the value of speed of light\n1.79×10_¹⁹ km per hour\nSwagatika\nwhat r dwarf planet\nwhat is energy\nকাজের একক কী\nJasim\nকাজের একক কী\nJasim\nEnergy is ability so capacity to do work.\nkenneth\nfriction ka direction Kaise pata karte hai\nfriction is always in the opposite of the direction of moving object\nPunia\nA twin paradox in the special theory of relativity arises due to.....? a) asymmetric of time only b) symmetric of time only c) only time\nb) symmetric of time only\nSwagatika\nfundamental note of a vibrating string\nevery matter made up of particles and particles are also subdivided which are themselves subdivided and so on ,and the basic and smallest smallest smallest division is energy which vibrates to become particles and thats why particles have wave nature\nAlvin\nwhat are matter waves? Give some examples\naccording to de Broglie any matter particles by attaining the higher velocity as compared to light'ill show the wave nature and equation of wave will applicable on it but in practical life people see it is impossible however it is practicaly true and possible while looking at the earth matter at far\nManikant\na centeral part of theory of quantum mechanics example:just like a beam of light or a water wave\nSwagatika\nMathematical expression of principle of relativity\ngiven that the velocity v of wave depends on the tension f in the spring, it's length 'I' and it's mass 'm'. derive using dimension the equation of the wave\nWhat is the importance of de-broglie's wavelength?\nhe related wave to matter\nZahid\nat subatomic level wave and matter are associated. this refering to mass energy equivalence\nZahid\nit is key of quantum\nManikant\n\n#### Get Jobilize Job Search Mobile App in your pocket Now!",
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https://numberworld.info/50101 | [
"# Number 50101\n\n### Properties of number 50101\n\nCross Sum:\nFactorization:\nDivisors:\nCount of divisors:\nSum of divisors:\nPrime number?\nYes\nFibonacci number?\nNo\nBell Number?\nNo\nCatalan Number?\nNo\nBase 2 (Binary):\nBase 3 (Ternary):\nBase 4 (Quaternary):\nBase 5 (Quintal):\nBase 8 (Octal):\nc3b5\nBase 32:\n1gtl\nsin(50101)\n-0.89994329837888\ncos(50101)\n0.43600694914525\ntan(50101)\n-2.0640572361132\nln(50101)\n10.821796246954\nlg(50101)\n4.6998463943333\nsqrt(50101)\n223.83252668011\nSquare(50101)\n\n### Number Look Up\n\nLook Up\n\n50101 which is pronounced (fifty thousand one hundred one) is a impressive figure. The cross sum of 50101 is 7. If you factorisate the figure 50101 you will get these result . 50101 has 2 divisors ( 1, 50101 ) whith a sum of 50102. The number 50101 is a prime number. 50101 is not a fibonacci number. 50101 is not a Bell Number. The number 50101 is not a Catalan Number. The convertion of 50101 to base 2 (Binary) is 1100001110110101. The convertion of 50101 to base 3 (Ternary) is 2112201121. The convertion of 50101 to base 4 (Quaternary) is 30032311. The convertion of 50101 to base 5 (Quintal) is 3100401. The convertion of 50101 to base 8 (Octal) is 141665. The convertion of 50101 to base 16 (Hexadecimal) is c3b5. The convertion of 50101 to base 32 is 1gtl. The sine of 50101 is -0.89994329837888. The cosine of the number 50101 is 0.43600694914525. The tangent of the number 50101 is -2.0640572361132. The square root of 50101 is 223.83252668011.\nIf you square 50101 you will get the following result 2510110201. The natural logarithm of 50101 is 10.821796246954 and the decimal logarithm is 4.6998463943333. You should now know that 50101 is amazing number!"
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https://whatisconvert.com/323-square-miles-in-square-yards | [
"# What is 323 Square Miles in Square Yards?\n\n## Convert 323 Square Miles to Square Yards\n\nTo calculate 323 Square Miles to the corresponding value in Square Yards, multiply the quantity in Square Miles by 3097600.0000048 (conversion factor). In this case we should multiply 323 Square Miles by 3097600.0000048 to get the equivalent result in Square Yards:\n\n323 Square Miles x 3097600.0000048 = 1000524800.0015 Square Yards\n\n323 Square Miles is equivalent to 1000524800.0015 Square Yards.\n\n## How to convert from Square Miles to Square Yards\n\nThe conversion factor from Square Miles to Square Yards is 3097600.0000048. To find out how many Square Miles in Square Yards, multiply by the conversion factor or use the Area converter above. Three hundred twenty-three Square Miles is equivalent to one billion five hundred twenty-four thousand eight hundred point zero zero two Square Yards.\n\n## Definition of Square Mile\n\nThe square mile (abbreviated as sq mi and sometimes as mi²) is an imperial and US unit of measure for an area equal to the area of a square with a side length of one statute mile. It should not be confused with miles square, which refers to a square region with each side having the specified length. For instance, 20 miles square (20 × 20 miles) has an area equal to 400 square miles; a rectangle of 10 × 40 miles likewise has an area of 400 square miles, but it is not 20 miles square. One square mile is equal to 4,014,489,600 square inches, 27,878,400 square feet or 3,097,600 square yards.\n\n## Definition of Square Yard\n\nThe square yard is an imperial unit of area, formerly used in most of the English-speaking world but now generally eplaced by the square metre, however it i still in widespread use in the US., Canada and the U.K. It isdefined as the area of a quare with sides of one yard (three feet, thirty-six inches, 0.9144 metres) in length. There is no universally agreed symbol but the following are used: square yards, square yard, square yds, square yd, sq yards, sq yard, sq yds, sq yd, sq.yd., yards/-2, yard/-2, yds/-2, yd/-2, yards^2, yard^2, yds^2, yd^2, yards², yard², yds², yd².\n\n## Using the Square Miles to Square Yards converter you can get answers to questions like the following:\n\n• How many Square Yards are in 323 Square Miles?\n• 323 Square Miles is equal to how many Square Yards?\n• How to convert 323 Square Miles to Square Yards?\n• How many is 323 Square Miles in Square Yards?\n• What is 323 Square Miles in Square Yards?\n• How much is 323 Square Miles in Square Yards?\n• How many yd2 are in 323 mi2?\n• 323 mi2 is equal to how many yd2?\n• How to convert 323 mi2 to yd2?\n• How many is 323 mi2 in yd2?\n• What is 323 mi2 in yd2?\n• How much is 323 mi2 in yd2?"
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https://factorization.info/cube-root/9/factors-of-cube-root-of-99059.html | [
"Factors of Cube Root of 99059",
null,
"Here we will show you how to get the factors of cube root of 99059 (factors of ∛99059). We define factors of cube root of 99059 as any integer (whole number) or cube root that you can evenly divide into cube root of 99059. Furthermore, if you divide ∛99059 by a factor of ∛99059, it will result in another factor of ∛99059.\n\nFirst, we will find all the cube roots that we can evenly divide into cube root of 99059. We do this by finding all the factors of 99059 and add a radical (∛) to them like this:\n\n∛1, ∛17, ∛5827, and ∛99059\n\nNext, we will find all the integers that we can evenly divide into cube root of 99059. We do that by first identifying the perfect cube roots from the list above:\n\n∛1\n\nThen, we take the cube root of the perfect cube roots to get the integers that we can evenly divide into cube root of 99059.\n\n1\n\nFactors of cube root of 99059 are the two lists above combined. Thus, factors of cube root of 99059 (cube roots and integers) are as follows:\n\n1, ∛1, ∛17, ∛5827, and ∛99059\n\nLike we said above, cube root of 99059 divided by any of its factors, will result in another of its factors. Therefore, if you divide ∛99059 by any of factors above, you will see that it results in one of the other factors.\n\nWhat can you do with this information? For one, you can get cube root of 99059 in its simplest form. Cube root of 99059 simplified is the largest integer factor times the cube root of 99059 divided by the largest perfect cube root. Thus, here is the math to get cube root of 99059 in its simplest radical form:\n\n∛99059\n= 1 × (∛99059 ÷ ∛1)\n= ∛99059\n\nCube Root Factor Calculator\nDo you need the factors of another cube root? No problem, please enter your cube root in the box below.\n\nFactors of Cube Root of 99060\nWe hope this information was helpful. Do you want to learn more? If so, go here to get the factors of the next cube root on our list."
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"https://factorization.info/images/factors-of-cube-root.png",
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https://books.openbookpublishers.com/10.11647/obp.0075/Chapters/P29.html | [
"",
null,
"",
null,
"GO TO... Contents Copyright",
null,
"BUY THE BOOK\n\nProblem 29: Telescoping series ($✓$ $✓$) 1998 Paper II\n\n(i)\nShow that the sum ${S}_{N}$ of the first $N$ terms of the series $\\frac{1}{1.2.3}+\\frac{3}{2.3.4}+\\frac{5}{3.4.5}+\\cdots +\\frac{2n-1}{n\\left(n+1\\right)\\left(n+2\\right)}+\\cdots \\phantom{\\rule{0.3em}{0ex}}$\n\nis\n\n$\\frac{1}{2}\\left(\\frac{3}{2}+\\frac{1}{N+1}-\\frac{5}{N+2}\\right).$\n\nWhat is the limit of ${S}_{N}$ as $N\\to \\infty$?\n\n(ii)\nThe numbers ${a}_{n}$ are such that $\\frac{{a}_{n}}{{a}_{n-1}}=\\frac{\\left(n-1\\right)\\left(2n-1\\right)}{\\left(n+2\\right)\\left(2n-3\\right)}.$\n\nFind an expression for $\\frac{{a}_{n}}{{a}_{1}}$ and hence, or otherwise, evaluate $\\sum _{n=1}^{\\infty }{a}_{n}$ when ${a}_{1}=\\frac{2}{9}\\phantom{\\rule{2.77695pt}{0ex}}$.\n\nIf you haven’t the faintest idea how to do the sum, then look at the first line of the solution; but don’t look without first having had a long think about it, picking up ideas from the form of the given answer.\n\nLimits form an important part of first year university mathematics. The definition of a limit is one of the basic ideas in analysis, which is the rigorous study of calculus. At the end of part (i), no such definition is needed: you just see what happens when $N$ gets very large (some terms get very small and eventually go away).\n\nPart (ii) looks as if it might be some new idea. Since this is STEP, you will probably realise that the new series must be closely related to the series in part (i). The peculiar choice for ${a}_{1}$ ($=\\frac{2}{9}$) should make you suspect that the sum will come out to some nice round number (not in fact round in this case, but straight and thin).\n\nHaving decided how to do the first part, please don’t use the ‘cover up’ rule unless you understand why it works: mathematics at this level is not a matter of applying learned recipes. See the post-mortem for more thoughts on this matter.\n\nSolution to problem 29\n\nThe given answer to the sum suggests partial fractions. It is difficult to think of any other way of starting, so let’s convert the general term of the series to partial fractions in the hope that something good might happen. Set\n\n $\\frac{2n-1}{n\\left(n+1\\right)\\left(n+2\\right)}\\equiv \\frac{A}{n}+\\frac{B}{n+1}+\\frac{C}{n+2}\\phantom{\\rule{2.77695pt}{0ex}},$ ($\\ast$)\n\nthen use your favourite method to find $A=-\\frac{1}{2}$, $B=3$ and $C=-\\frac{5}{2}$. Note that $A+B+C+\\right)$. The series can now be written\n\n$\\begin{array}{llll}\\hfill & \\left(\\frac{A}{1}+\\frac{B}{2}+\\frac{C}{3}\\right)+\\left(\\frac{A}{2}+\\frac{B}{3}+\\frac{C}{4}\\right)+\\left(\\frac{A}{3}+\\frac{B}{4}+\\frac{C}{5}\\right)+\\phantom{\\rule{2em}{0ex}}& \\hfill & \\phantom{\\rule{2em}{0ex}}\\\\ \\hfill & +\\cdots +\\left(\\frac{A}{N-2}+\\frac{B}{N-1}+\\frac{C}{N}\\right)+\\left(\\frac{A}{N-1}+\\frac{B}{N}+\\frac{C}{N+1}\\right)+\\left(\\frac{A}{N}+\\frac{B}{N+1}+\\frac{C}{N+2}\\right).\\phantom{\\rule{2em}{0ex}}& \\hfill \\text{(}\\ast \\ast \\text{)}\\end{array}$\n\nNow we collect up terms with the same denominators and find that all the terms in the series cancel, except those with denominators 1, 2, $N+1$ and $N+2$. These exceptions sum to the required answer.\n\nThe limit as $N\\to \\infty$ is $\\frac{3}{4}$ since the other two terms obviously tend to zero.\n\nFor part (ii), we note that $\\frac{{a}_{n}}{{a}_{n-1}}=\\frac{{b}_{n}}{{b}_{n-1}},$ where ${b}_{n}$ is the general term of the series in part (i). Thus\n\nAlternatively, we can write out the $n$th term explicitly:\n\n$\\begin{array}{rcll}{a}_{n}& =& \\frac{\\left(n-1\\right)\\left(2n-1\\right)}{\\left(n+2\\right)\\left(2n-3\\right)}{a}_{n-1}=\\frac{\\left(n-1\\right)\\left(2n-1\\right)}{\\left(n+2\\right)\\left(2n-3\\right)}\\frac{\\left(n-2\\right)\\left(2n-3\\right)}{\\left(n+1\\right)\\left(2n-5\\right)}{a}_{n-2}& \\text{}\\\\ & =& \\frac{\\left(n-1\\right)\\left(2n-1\\right)}{\\left(n+2\\right)\\left(2n-3\\right)}\\phantom{\\rule{0.3em}{0ex}}\\frac{\\left(n-2\\right)\\left(2n-3\\right)}{\\left(n+1\\right)\\left(2n-5\\right)}\\phantom{\\rule{0.3em}{0ex}}\\frac{\\left(n-3\\right)\\left(2n-5\\right)}{\\left(n\\right)\\left(2n-7\\right)}\\phantom{\\rule{0.3em}{0ex}}\\cdots \\phantom{\\rule{0.3em}{0ex}}\\frac{5×11}{8×9}\\phantom{\\rule{0.3em}{0ex}}\\frac{4×9}{7×7}\\phantom{\\rule{0.3em}{0ex}}\\frac{3×7}{6×5}\\phantom{\\rule{0.3em}{0ex}}\\frac{2×5}{5×3}\\phantom{\\rule{0.3em}{0ex}}\\frac{1×3}{4×1}\\phantom{\\rule{0.3em}{0ex}}{a}_{1}& \\text{}\\\\ & =& \\frac{2n-1}{n\\left(n+1\\right)\\left(n+2\\right)}\\frac{3×2×1}{1}{a}_{1}=\\frac{12}{9}\\frac{2n-1}{n\\left(n+1\\right)\\left(n+2\\right)},& \\text{}\\end{array}$\n\nall other terms cancelling. Now using the result of the first part gives $\\sum _{1}^{\\infty }{a}_{n}=1$.\n\nPost-mortem\n\nA small point of technique: equation ($\\ast \\ast$) was made much clearer (and it saved writing) to stick with $A$, $B$ and $C$ instead of using $-\\frac{1}{2}$, $3$ and $-\\frac{5}{2}$. The method would not depend on the arithmetic values of these constants.\n\nThere are various methods for finding $A$, $B$ and $C$ in $\\left(\\ast \\right)$.\n\nOne is to set $n=1$, $n=2$ and $n=3$ consecutively and obtain three simultaneous equations.\n\nAnother is to multiply up and simplify, giving $2n-1=\\left(A+B+C\\right){n}^{2}+\\left(3A+2B+C\\right)n+2A$. You then equate coefficients of powers of $n$.\n\nA better way is to multiply up without simplifying, giving $2n-1=A\\left(n+1\\right)\\left(n+2\\right)+Bn\\left(n+2\\right)+Cn\\left(n+1\\right)$. You then choose values for $n$ that give quick results: for example, setting $n=-1$ gives $B=3$ immediately. This is of course the method behind the iniquitous ‘cover-up rule’. Note that the ‘equivalence’ sign, $\\equiv \\phantom{\\rule{0.3em}{0ex}}$, indicates an identity (something that holds for all values of $n$) rather than an equation to solve for $n\\phantom{\\rule{0.3em}{0ex}}$, so $n$ doesn’t have to be a positive integer (you could set $n=-\\frac{1}{2}$ if you fancied it)."
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null,
"https://www.openbookpublishers.com/images/open_access.png",
null,
"https://books.openbookpublishers.com/10.11647/obp.0075/Images/1.png",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9323553,"math_prob":0.9999652,"size":3024,"snap":"2022-05-2022-21","text_gpt3_token_len":700,"char_repetition_ratio":0.10993377,"word_repetition_ratio":0.010695187,"special_character_ratio":0.22784391,"punctuation_ratio":0.10509031,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999598,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-26T18:30:44Z\",\"WARC-Record-ID\":\"<urn:uuid:8df427d4-41a9-4349-bb4c-11f8e791933e>\",\"Content-Length\":\"41787\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:967812dc-4156-4176-8263-4a9ac7151b19>\",\"WARC-Concurrent-To\":\"<urn:uuid:7a4b2e29-2e5a-4ea4-8cc6-ec3501b9ed89>\",\"WARC-IP-Address\":\"13.32.204.50\",\"WARC-Target-URI\":\"https://books.openbookpublishers.com/10.11647/obp.0075/Chapters/P29.html\",\"WARC-Payload-Digest\":\"sha1:PURS6FOASINH3GAO5N2O4U77V6OE2JBE\",\"WARC-Block-Digest\":\"sha1:GTNWH5WOTYV7RQIIVCFD23OC4D4IENYA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320304959.80_warc_CC-MAIN-20220126162115-20220126192115-00578.warc.gz\"}"} |
http://sjce.journals.sharif.edu/article_1348.html | [
"# بررسی ساختار جریان آشفته اطراف آبشکن تیغهای قرار گرفته در موقعیتهای مختلف قوس 90 درجه\n\nنوع مقاله : پژوهشی\n\nنویسندگان\n\n1 دانشکده فنی و مهندسی - دانشگاه خوارزمی\n\n2 پژوهشکده مهندسی آب و دانشکده ی مهندسی عمران و محیط زیست- دانشگاه تربیت مدرس\n\n3 دانشکده مهندسی عمران و محیط زیست- دانشگاه تربیت مدرس\n\nچکیده\n\nدر نوشتار حاضر، مشخصات میدان جریان اطراف آبشکن تیغهیی قرارگرفته در قوس $^{0}$۹۰ اندازهگیری و با توجه به میزان آبشستگی ایجادشده در اطراف آبشکنهای مذکور که از مطالعات پیشین بهدست آمدهاند، ارتباط میان آبشستگی و میدان جریان اطراف آبشکنهای تیغهیی قرارگرفته در قوس $90^{0}$ بررسی شده است. نتایج بهدست آمده، وجود جریان ثانویه در قوس، گردابهی نعل اسبی در اطراف آبشکن و ناحیهی جریان برگشتی با سرعت کم را تأیید کرد. همچنین نتایج نشان داد تنشهای برشی واردبر بستر که از طریق تنشهای رینولدز و نوسانهای مؤلفههای سرعت بهدست آمدهاند، برای تعیین نقاط مستعد آبشستگی مناسب نیستند و وجود پیکهایی در طیف توانی آشفتگی در نواحی نزدیک آبشکن، نشاندهندهی توانایی جریان برای انتقال رسوبات از نواحی مذکور است. علاوه بر این، سهم پدیدهی بیرونرانی در تنشهای رینولدز در نواحی نزدیک به نوک آبشکن بیشتر بوده است که میتواند عامل اصلی انتقال رسوبات از نواحی نزدیک به نوک آبشکن در مراحل ابتدای آبشستگی بهصورت بار معلق باشد.\n\nکلیدواژهها\n\nعنوان مقاله [English]\n\n### EXPERIMENTAL INVESTIGATION OF THE FLOWFIELD AROUND STRAIGHT SPUR DIKES LOCATED IN A DIFFERENT LOCATION OF 90 BEND\n\nنویسندگان [English]\n\n• M. Mehraein 1\n• M. Ghodsian 2\n• S.M.A. Najibi 3\n1 Faculty of Engineering Kharazmi University (corresponding)\n2 Water Engineering Research Institute Faculty of Civil and Environmental Engineering, Tarbiat Modares University\n3 Dept. of Civil & Environmental Engineering\\ Tarbiat Modares University\nچکیده [English]\n\nSpur dikes deflect the approaching flow to the far bank and decreases the sediment deposition. The scour hole formation and flow field features around spur dikes are interesting subjects in hydraulic engineering. In this paper, the flow field around straight spur dikes in a 90u{1d52} sharp bend was investigated experimentally. Three different experiments were conducted in a prismatic channel with 90u{1d52} bend at Tarbiat Modares University Tehran, Iran. In each experiment, a straight spur dike was attached to the outer bank in different angles with respect to the beginning of the bend. The flow fields were measured using Vectrino apparatus. In each point, the velocity was measured in 100 HZ frequency for 3 min. Despiking process was done using previous research approach. The spectral analysis, mean flow analysis, higher correlation analysis, bed shear analysis, turbulent intensity analysis, power spectrum analysis, and quadrant analysis were conducted. Results confirm the existence of the secondary flow in the bend, horse shoe vortex, and recirculation flow region. There is no strong difference between the mean flow fields around the spur dikes. Ejection and sweep events have the main role in shear layer regions, and the interactions are the dominant processes in the downstream recirculation zone. The spectral analysis confirms the existence of the inertial subrange for some frequency range. Some peaks are observed in power spectrum graphs that may, due to the vortex shedding process, confirm the ability of the flow for sediment transport. The bed shear stress based on linear relationship between bed shear stress and turbulent kinetic energy and bed shear stress and vertical velocity fluctuation in vertical direction cannot predict the scour potential region accurately. In addition, the contribution of the ejection events in Reynolds stresses may be one of the important reasons in sediment suspension in the initial time of scour process.\n\nکلیدواژهها [English]\n\n• Experimental study\n• ADV\n• spur dike\n• 90 bend"
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https://mba.realinfo.tv/2019/11/unit-11-marginal-costing-and-profit.html | [
"# Unit 11: Marginal Costing and Profit Planning\n\nUnit 11: Marginal Costing and Profit Planning\n\nMarginal costing is one of the important tools of management not only to take decision, but also to fi x an appropriate price and to assess the level of profi tability.\n\nMarginal cost is nothing, but a change occurred in the total cost due to small change in the quantity produced.\n\nAbsorption costing technique is also known by other names as “Full costing” or “Traditional costing”.\n\nThe cost-volume-profi t analysis is a tool to show the relationship between various ingredients of profi t planning.\n\nThe ratio or percentage of contribution margin to sales is known as P/V ratio.\n\nThe crucial step in this analysis is the determination of break-even point.\n\nBEP is defi ned as the sales level at which the total revenue equals total cost.\n\nMargin of safety is the difference between the actual sales and sales at break-even point.\n\nSales beyond break-even volume brings in profi ts.\n\nBEP (Units): It is the level of units at which the fi rm neither incurs a loss nor earns profi t.\n\nBEP (Volume): It is the level of sales in Rupees at which the fi rm neither incurs a loss nor earns profi t.\n\nContribution: It is an amount of balance available after the deduction of variable cost from the sales.\n\nFixed Cost: It is a cost which is fi xed or remains the same for irrespective level of production.\n\nMarginal Cost: Change occurred in the cost of operations due to change in the level of production.\n\nPV Ratio: Profi t volume ratio which is nothing but the ratio in between the contribution and sales.\n\nVariable Cost: It varies along with the level of production."
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https://forums.wolfram.com/mathgroup/archive/2003/Feb/msg00058.html | [
"",
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"Re: To verify Cauchy-Riemann relations in complex variable graphically\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg39212] Re: To verify Cauchy-Riemann relations in complex variable graphically\n• From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>\n• Date: Tue, 4 Feb 2003 02:21:14 -0500 (EST)\n• Organization: Universitaet Leipzig\n• References: <b1idcj\\$bje\\[email protected]>\n• Sender: owner-wri-mathgroup at wolfram.com\n\n```Hi,\n\nMathematica can't draw transparent surfaces but MathGL3d can\n\nGet[\"MathGL3d`OpenGLViewer`\"]\n\nMVShow3D[z2r, MVNewScene -> True, MVAlpha -> 0.5];\nMVShow3D[z2i, MVAlpha -> 0.5];\n\nyou can get MathGL3d from\n\nhttp://phong.informatik.uni-leipzig.de/~kuska/mathgl3dv3/id3.htm\n\nRegards\nJens\n\n\"Narasimham G.L.\" wrote:\n>\n> Is it possible to have a semi transparent view of surfaces so that one\n> may verify slopes by ParametricPlot3D for Cauchy-Riemann relations?\n> The following is program for 3 functions Z^2, Z^3, Sin[Z].It was\n> expected to check slopes at the line of intersection of Re and Im parts.\n>\n> R1=x^2-y^2 ; I1= 2 x y ;\n> z2r=Plot3D[R1 , {x,-Pi,Pi},{y,-Pi,Pi} ];\n> z2i=Plot3D[I1 , {x,-Pi,Pi},{y,-Pi,Pi} ];\n> Show[z2r,z2i] ; 'Top view >> Re,Im Intxn';\n> Plot[{x ArcTan[-Sqrt+1],x ArcTan[Sqrt+1]}, {x,-Pi,Pi} ];\n>\n> R3=x^3 - 3 x y^2 ; I3= 3 x^2 y - y ^3 ;\n> z3r=Plot3D[R3 , {x,-Pi,Pi},{y,-Pi,Pi} ];\n> z3i=Plot3D[I3 , {x,-Pi,Pi},{y,-Pi,Pi} ];\n> Show[z3r,z3i] ; 'Top view >> Re,Im Intxn';\n> Plot[{x,x (-Sqrt+2) , x (-Sqrt-2) }, {x,-Pi,Pi} ];\n>\n> R2=Cosh[y] Sin[x] ; I2=Sinh[y] Cos[x] ;\n> scr=Plot3D[R2,{x,-Pi/2,Pi/2},{y,-Pi/2,Pi/2}];\n> sci=Plot3D[I2,{x,-Pi/2,Pi/2},{y,-Pi/2,Pi/2}];\n> Show[scr,sci]; 'Top view >> Re,Im Intxn';\n> Plot[{ArcTanh[Tan[x]]},{x,-Pi/2,Pi/2 }];\n> --\n> Posted via http://web2news.com\n> To contact in private, remove\n\n```\n\n• Prev by Date: Re: Fit or Interpolate\n• Next by Date: Re: Random Trouble\n• Previous by thread: RE: To verify Cauchy-Riemann relations in complex variable graphically\n• Next by thread: Re: RE: To verify Cauchy-Riemann relations in complex variable graphically"
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https://cmuc.karlin.mff.cuni.cz/cmuc1901/abs/zindul.htm | [
"## Ondřej ZindulkaStrong measure zero and meager-additive sets through the prism of fractal measures\n\nComment.Math.Univ.Carolin. 60,1 (2019) 131-155.\n\nAbstract:We develop a theory of sharp measure zero sets that parallels Borel's strong measure zero, and prove a theorem analogous to Galvin--Mycielski--Solovay theorem, namely that a set of reals has sharp measure zero if and only if it is meager-additive. Some consequences: A subset of $2^{\\omega}$ is meager-additive if and only if it is $\\mathcal E$-additive; if $f\\colon 2^{\\omega}\\to 2^{\\omega}$ is continuous and $X$ is meager-additive, then so is $f(X)$.\n\nKeywords: meager-additive; $\\mathcal E$-additive; strong measure zero; sharp measure zero; Hausdorff dimension; Hausdorff measure\n\nDOI: DOI 10.14712/1213-7243.2015.277\nAMS Subject Classification: 03E05 03E20 28A78\n\nPDF"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.80989534,"math_prob":0.9387083,"size":675,"snap":"2023-14-2023-23","text_gpt3_token_len":205,"char_repetition_ratio":0.15946348,"word_repetition_ratio":0.04494382,"special_character_ratio":0.28296295,"punctuation_ratio":0.14393939,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9979365,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-05T23:10:34Z\",\"WARC-Record-ID\":\"<urn:uuid:fd4c579a-3652-4974-8e43-6cf8408f3fde>\",\"Content-Length\":\"1500\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c4dd52c5-c462-48d9-bd3c-eb520cd787e7>\",\"WARC-Concurrent-To\":\"<urn:uuid:22f08395-171f-4f58-b145-e174b9b17262>\",\"WARC-IP-Address\":\"195.113.30.10\",\"WARC-Target-URI\":\"https://cmuc.karlin.mff.cuni.cz/cmuc1901/abs/zindul.htm\",\"WARC-Payload-Digest\":\"sha1:EZJ7SQBUUB65UTJ3YKXHDCJQHJYWHLPE\",\"WARC-Block-Digest\":\"sha1:ULAYSF4AYBVBVTPVBJWQ6ILF4YHLQZVS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224652184.68_warc_CC-MAIN-20230605221713-20230606011713-00143.warc.gz\"}"} |
https://www.engineerknow.com/2021/06/understanding-constant-and-variable-in.html | [
"# CONSTANT\n\ntf.constant is immutable value it means these value will not change in future when code is run for example you defined a bias that is not changing throughout the code execution\nNote : by default constant take dtype as int\n\nLet's see how to declare constant in tensorflow\n\na = tf.constant(value, dtype=None, shape=None)\n\ndtype is your data type you can define it by tf.int32 , tf.float32\nshape is your data /matrix shape\n\n## Defining 1D array in tensorflow\n\nimport tensorflow as tf\na = tf.constant([1,2,3,4] , dtype = tf.int32)\nprint(a)\n#print(a) give output like this\ntf.Tensor([1 2 3 4], shape=(4,), dtype=int32)\nprint(a.numpy())\n#print(a.numpy()) give output like this\n[1,2,3,4]\n\n## Defining 2D array in Tensorflow\n\nimport tensorflow as tf\na = tf.constant([[1,2,3],[4,5,6]],dtype = tf.float32)\nprint(a.numpy())\noutput =\n\n[[1. 2. 3.]\n[4. 5. 6.]]\n\nnotice there are decimal after every number in output in this case the reason is that we define the variable data type as float\n\n## Defining Shape in Tensorflow\n\nimport tensorflow as tf\na = tf.constant(0,shape=(2,3))\nprint(a.numpy())\n\noutput:\n[[0 0 0]\n[0 0 0]]\n\nif you want to know dtype or shape of your constant just type\nprint(a.shape) #note here a is my constant name\nprint(a.dtype)\n\n# VARIABLE\n\nVariable is mutable in tensorflow it means its value can be change in future while executing the code.\nExample : weights defining in neural network they will change as our code execute till they didn't get optimum value\n\n## Defining 1D array in tensorflow\n\nimport tensorflow as tf\na = tf.Variable([1,2,3,4] , dtype = tf.int32)\nprint(a)\n#print(a) give output like this\ntf.Tensor([1 2 3 4], shape=(4,), dtype=int32)\nprint(a.numpy())\n#print(a.numpy()) give output like this\n[1,2,3,4]\n\n## Defining 2D array in Tensorflow\n\nimport tensorflow as tf\na = tf.Variable([[1,2,3],[4,5,6]],dtype = tf.float32)\nprint(a.numpy())\noutput =\n\n[[1. 2. 3.]\n[4. 5. 6.]]\n\nnotice there are decimal after every number in output in this case the reason is that we define the variable data type as float\n\n## Defining Shape in Tensorflow\n\nimport tensorflow as tf\na = tf.Variable(0,shape=(2,3))\nprint(a.numpy())\n\noutput:\n[[0 0 0]\n[0 0 0]]\n\nif you want to know dtype or shape of your constant just type\nprint(a.shape) #note here a is my Variable name\nprint(a.dtype)"
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http://ma-mimo.ellintech.se/2018/04/04/30-discount-on-massive-mimo-networks-book/ | [
"# 30% Discount on “Massive MIMO Networks” Book",
null,
"The hardback version of the massive new book Massive MIMO Networks: Spectral, Energy, and Hardware Efficiency (by Björnson, Sanguinetti, Hoydis) is currently available for the special price of \\$70 (including worldwide shipping). The original price is \\$99.\n\nThis price is available until the end of April when buying the book directly from the publisher through the following link:\n\nNote: The book’s authors will give a joint tutorial on April 15 at WCNC 2018. A limited number of copies of the book will be available for sale at the conference and if you attend the tutorial, you will receive even better deal on buying the book!\n\n## 4 thoughts on “30% Discount on “Massive MIMO Networks” Book”\n\n1. Hieu Dao says:\n\nI have a question about the channel model in your new book:\n+ In my understanding, the channel response in the book is in time domain, we assume flat fading and the common model is Rayleigh fading model.\n+ But if we use OFDM, how can we model the channel? In the paper “Non cooperative cellular wireless with unlimited numbers of base station antennas”-M. Marzetta, the author talked about OFDM, they consider at one sub-carrier, and the model is still the same with Rayleigh fading model (just channel response multiply with signal) ==> Is the channel now in the frequency domain? Because I think we use convolution instead of multiply channel response with signal in frequency-selective fading.\n\n1. Emil Björnson says:\n\nMy book and also Marzetta’s works are based on the block fading channel model, which is a simplification of reality. In this model, we consider a flat-fading time-invariant channel with a coherence block, thus the channel is described by an impulse response h*delta(t) in the time domain and h in the frequency domain, where delta(t) is the Dirac delta function. So h is the coefficient that describes the channel.\n\nIn OFDM, the coherence block can be approximately viewed as spanning a certain number of subcarriers in the frequency domain and OFDM symbols in the time domain. This is what Marzetta describes in his book.\n\nIf you want to consider a more detailed model of OFDM, you will have to leave the block fading model. The following two papers show how to deal with channel variations in the frequency domain and in the time domain:\n\nThere are many other relevant papers on OFDM in Massive MIMO which you can find in the reference lists of the two papers above.\n\n2. Hieu Dao says:\n\nThere is one thing I still do not get is that the channel model. For example with a frequency selective fading L-taps, we have the channel response in time domain: h = [h_1, h_2, … h_L], each of tap h_i will follow the Rayleigh fading, h_i ~ CN(0,1). So the received signal in time domain: y[n] = x[n](*)h + w(n) with (*) is convolution. Change this model to frequency domain: Y[k] = X[k]*H[k] + W[k]. with * is conventional multiplication ==> what is the model for H[k], does it still follow ~ CN(0,1).\n==> In your book, the channel h ~ CN(0,1) which is Rayleigh fading and in my understanding, Rayleigh fading is in time domain, but when we change to frequency domain, what happens?\n\n1. Emil Björnson says:\n\nFor an L-tap channel, you can check out the first of the two papers that I referenced to. G^m_k(s) is the channel on subcarrier s between antenna m and user k. It is formed by taking the L-tap channel between antenna m and user k, and then compute an inner product with a vector that comes from a DFT matrix. Since this is a weighted sum of Gaussian channel taps, it will also be Gaussian distributed. It will be i.i.d. between antennas/users, but correlated between subcarriers.\n\nRegarding my book, let’s recall my previous answer: In my book “we consider a flat-fading time-invariant channel with a coherence block, thus the channel is described by an impulse response h*delta(t) in the time domain and h in the frequency domain, where delta(t) is the Dirac delta function. So h is the coefficient that describes the channel.” Hence, if h is iid Rayleigh fading, this occur in both the time and frequency domain."
]
| [
null,
"http://ma-mimo.ellintech.se/wp-content/uploads/2018/03/IMG_0999-e1522356401952-768x1024.jpg",
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https://socratic.org/questions/57e92857b72cff4693372cbe | [
"# How do you divide? (x^3 + y^3) div(x-y)\n\nSep 26, 2016\n\n$\\left({x}^{3} + {y}^{3}\\right) \\div \\left(x - y\\right) = {x}^{2} + x y + {y}^{2} \\text{ rem } 2 {y}^{3}$\n\n#### Explanation:\n\nWe can do it by algebraic long division:\n\n$\\text{dividend\"/\"divisor\" = \"quotient}$\n\n1. Write the dividend in the 'box' making sure that the indices are in descending powers of x. Make spaces for any missing terms.\n\n2. Divide the first term in the divisor into the term in the dividend with the highest index. Write the answer at the top,\n\n3. Multiply by BOTH terms of the divisor at the side\n\n4. Subtract\n\n5. Bring down the next term\n\nRepeat steps 2 to 5\n\n$\\textcolor{w h i t e}{\\times \\times \\times . \\times \\times \\times} \\textcolor{red}{{x}^{2}} \\text{ \" color(blue)(+xy)\" \"+color(lime)(y^2) \" rem } 2 {y}^{3}$\ncolor(white)(xxx)x-y |bar( x^3 +0x^2y+0xy^2 +y^3\" \"larrx^3divx =color(red)(x^2\n$\\textcolor{w h i t e}{\\times \\times x} - \\left(\\underline{\\textcolor{red}{{x}^{3} - {x}^{2} y}}\\right) \\text{ } \\leftarrow$ subtract\n$\\textcolor{w h i t e}{\\times \\times \\times \\times \\times} + {x}^{2} y \\text{ } \\leftarrow {x}^{2} y \\div x = \\textcolor{b l u e}{x y}$\n$\\textcolor{w h i t e}{\\times \\times \\times . x .} - \\left(\\underline{\\textcolor{b l u e}{{x}^{2} y - x {y}^{2}}}\\right) \\textcolor{w h i t e}{x} \\downarrow \\text{ } \\leftarrow$ subtract\n$\\textcolor{w h i t e}{\\times \\times \\times \\times \\times x . x . \\times x} x {y}^{2} - {y}^{3} \\text{ } \\leftarrow x {y}^{2} \\div x = \\textcolor{\\lim e}{{y}^{2}}$\ncolor(white)(xxxxxxxx..x.xxx)ul(-color(lime)((xy^2 +y^3)) \" \"larr subtract\n$\\textcolor{w h i t e}{\\times \\times \\times \\times \\times \\times \\times \\times \\times \\times} 2 {y}^{3} \\leftarrow \\text{ }$ remainder\n\n$\\left({x}^{3} + {y}^{3}\\right) \\div \\left(x - y\\right) = {x}^{2} + x y + {y}^{2} \\text{ rem } 2 {y}^{3}$"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.59384465,"math_prob":0.99997675,"size":1103,"snap":"2021-43-2021-49","text_gpt3_token_len":414,"char_repetition_ratio":0.22202002,"word_repetition_ratio":0.0,"special_character_ratio":0.34179512,"punctuation_ratio":0.05909091,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999583,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-11-28T21:59:44Z\",\"WARC-Record-ID\":\"<urn:uuid:2e32d403-d5eb-4194-bbaf-8373ac82cbb0>\",\"Content-Length\":\"36228\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1f195231-aaf3-410c-979c-4c9ca3df7393>\",\"WARC-Concurrent-To\":\"<urn:uuid:9abcfebe-46b7-45d4-a45a-53ce95a39b62>\",\"WARC-IP-Address\":\"216.239.36.21\",\"WARC-Target-URI\":\"https://socratic.org/questions/57e92857b72cff4693372cbe\",\"WARC-Payload-Digest\":\"sha1:A6EROFT4HTGKRFOF7MXQRTD7KFSR4KAD\",\"WARC-Block-Digest\":\"sha1:4CHIUO5EED5I3NP5OKT3YMP7YWBDXB3V\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964358591.95_warc_CC-MAIN-20211128194436-20211128224436-00395.warc.gz\"}"} |
https://aviation.stackexchange.com/questions/42888/is-it-possible-to-estimate-airspeed-on-top-and-bottom-sides-of-a-wing-at-zero-an | [
"# Is it possible to estimate airspeed on top and bottom sides of a wing at zero angle of attack?\n\nAssuming a flat bottom wing moving through the air at say 50km/h and at zero degree angle of attack, what will be the likely airspeed above and below the wing?\n\nI know this will be affected by camber and maybe wing chord too. But is there like a proportional estimate for specific wing camber?\n\nI haven't got much knowledge about wing maths, but a formula would be appreciated.\n\nApplying Bernoulli's equation requires the relative airflow above and below the wing. The things is, I'm trying to experiment with the fact that given atmospheric pressure, the two different airflows, and the wing area, one can actually determine the pressure difference between the top and bottom surface using Bernoulli's equation and also the total force holding the plane up by getting the product of the pressure difference and the wing area. So what I'm trying to do is estimate the lift force on my theoretical airplane model at a particular wind speed. I want to use this to estimate the minimum airspeed required for normal flight 'without flaps'.\n\n• Might be worth also asking this at physics.stackexchange.com to get an in depth answer of fluid dynamics. – Adwaenyth Aug 11 '17 at 14:06\n• We would need more information on what the rest of the wing would look like. Apart form the bottom. – Koyovis Aug 11 '17 at 14:28\n• There isn't just a single airspeed on top and another airspeed on bottom. Airspeed varies continuously over the top and bottom surfaces; for example see mh-aerotools.de/airfoils/velocitydistributions.htm – David K Aug 11 '17 at 17:26\n• That page also shows that an airfoil can be giving a lifting force even though speed of air flow over large parts of both the top and bottom is greater than the speed of the undisturbed airflow. – David K Aug 11 '17 at 17:33\n• Angle of attack has a large influence on speed of the air flow over the wing. – David K Aug 11 '17 at 17:34\n\nThe short answer is that it is possible, by way of knowing the pressure distribution and applying the Bernoulli equation to relate the pressure and the velocity.\n\nMore specifically to your question, I'm not aware of any published correlation between camber and air velocity (nor do I think it'd have much practical application), but you could come up with your own by running a variety of geometries through XFOIL for a given airfoil geometry (maybe a Clark Y of varying camber for a flat-bottomed design) and applying Bernoulli on the calculated pressure distribution for whatever freestream atmospheric conditions and chord length you want.\n\nAnother note: You asked specifically about a wing, not an airfoil. A wing will have lots of other effects contributing to variations in velocity over its surface; there is no easy way to account for all of them.\n\nOK if it is lift that you would like to compute, this is the way it is commonly done:\n\n$$L = C_L * \\frac {1}{2} \\cdot \\rho \\cdot V^2 \\cdot A$$\n\nWith\n\n• L= lift [N]\n• $C_L$ a lift coefficient depending on aerofoil shape and angle of attack $\\alpha$ [dimensionless]\n• $\\rho$ = air density [kg/$m^3$]\n• V = airspeed (of the aircraft) [m/s]\n• A = wing area [$m^2$]\n\nIf you would like to use a flat bottomed profile, you could use Clark Y. This image is a bit fuzzy but you could construct a straight line $C_L - \\alpha$ from the graph.",
null,
"Image source\n\nAt $\\alpha$ = 0 the $C_L$ is about 0.4, so at sea level, with an airspeed of 100 m/s and a wing area of 10 $m^2$ the lift would be:\n\n$$L = 0.4 \\cdot \\frac {1}{2} * 1.225 * 100^2 * 10$$\n\n= 24,500 N. So the pressure differential over the wing area is 2,450 N/m$^2$ If this would be from increased dynamic pressure over the wing only, the air speed on top of the wing would be:\n\n$$2,450 = \\frac {1}{2} \\cdot \\rho \\cdot (V_{top}^2 - V_{bottom}^2) => V_{top} = \\sqrt{\\frac{2,450 \\cdot 2 }{ 1.225} + 100^2} = 118.3 m/s$$\n\nIn that particular case, at angle of attack zero, with no skin friction or boundary layer, static pressure underneath the wing equal to atmospheric pressure etc etc, the speed under the flat wing surface would be the airspeed = 100 m/s and over the wing would be 163.2 m/s. But the creation of aerodynamic lift is way more complicated than only considering the speeds over and under a wing section and assuming that lower surface pressure = atmospheric pressure, that is why it is done with the $C_L - C_D$ graphs.\n\nApplying Bernoulli's equation requires the relative airflow above and below the wing. The things is, I'm trying to experiment with the fact that given atmospheric pressure, the two different airflows, and the wing area, one can actually determine the pressure difference between the top and bottom surface using Bernoulli's equation and also the total force holding the plane up by getting the product of the pressure difference and the wing area. So what I'm trying to do is estimate the lift force on my theoretical airplane model at a particular wind speed. I want to use this to estimate the minimum airspeed required for normal flight 'without flaps'.\n\n• Yes these are usually available in $C_L$ - $C_D$ data. You would look those up for a particular wing cross section, just take an old NACA profile. By the way, it would have been better format had you posted the contents of this answer into your question, by editing it. – Koyovis Aug 13 '17 at 4:24"
]
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null,
"https://i.stack.imgur.com/SmDrz.gif",
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]
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https://republicofsouthossetia.org/question/if-i-have-a-triangle-with-a-leg-of-4-and-a-leg-of-3-what-is-the-length-of-the-hypotenuse-15963618-97/ | [
"If I have a triangle with a leg of 4 and a leg of 3 what is the length of the hypotenuse\n\nQuestion\n\nIf I have a triangle with a leg of 4 and a leg of 3 what is the length of the hypotenuse\n\nin progress 0\n2 weeks 2022-01-03T08:58:48+00:00 1 Answer 0 views 0\n\n5\n\nStep-by-step explanation:\n\nTo solve for hypotenuse we use the Pythagorean theorem:",
null,
"In this theorem:\n\nc= hypotenuse\n\na = leg\n\nb = leg\n\nSubstitute into theorem:",
null,
"",
null,
"",
null,
"(square root both sides)",
null,
"Therefore the hypotenuse has a length of 5.\n\nPlease say thanks if correct 🙂"
]
| [
null,
"https://republicofsouthossetia.org/wp-content/ql-cache/quicklatex.com-e139c22c4390664ce45b51612a46d5f9_l3.png",
null,
"https://republicofsouthossetia.org/wp-content/ql-cache/quicklatex.com-cd90c254713bfef7ce8fa211203a5ff3_l3.png",
null,
"https://republicofsouthossetia.org/wp-content/ql-cache/quicklatex.com-d462ee6da91a35cd780746e9e78340c9_l3.png",
null,
"https://republicofsouthossetia.org/wp-content/ql-cache/quicklatex.com-16b08f5bcaf83e8cc2b23a87af750134_l3.png",
null,
"https://republicofsouthossetia.org/wp-content/ql-cache/quicklatex.com-c25911021f46360a0b6d2ac60568fcf0_l3.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.818544,"math_prob":0.8707323,"size":566,"snap":"2022-05-2022-21","text_gpt3_token_len":177,"char_repetition_ratio":0.14590748,"word_repetition_ratio":0.34951457,"special_character_ratio":0.3180212,"punctuation_ratio":0.09375,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9950134,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,4,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-18T16:01:50Z\",\"WARC-Record-ID\":\"<urn:uuid:aa9a3963-9c54-495f-9da0-5ca01263cab7>\",\"Content-Length\":\"67396\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:952ba16b-be7b-4f21-a129-b87909ef40dc>\",\"WARC-Concurrent-To\":\"<urn:uuid:6c0d2bd6-65a3-48f7-a53e-d02e4f505bcf>\",\"WARC-IP-Address\":\"198.252.99.154\",\"WARC-Target-URI\":\"https://republicofsouthossetia.org/question/if-i-have-a-triangle-with-a-leg-of-4-and-a-leg-of-3-what-is-the-length-of-the-hypotenuse-15963618-97/\",\"WARC-Payload-Digest\":\"sha1:JQQ2C3EVPAYRYGMXO4TXCVUYPJNGGSRH\",\"WARC-Block-Digest\":\"sha1:A7M5GGNRFEY2T3T27Z6MW22U6NIS4TZ6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320300934.87_warc_CC-MAIN-20220118152809-20220118182809-00155.warc.gz\"}"} |
http://www.brokencontrollers.com/faq/10105501.shtml | [
"# Apache Commons Math SpearmansCorrelation - How to use\n\nI am trying to use Apache Commons Math's SpearmansCorrelation but I am having some difficulty as I do not have enough background in mathematics/statistics.\n\nI have researched Spearman's Rank Correlation from:\n\nI have found the following examples:\n\nI would like to calculate the correlation between two lists of my class:\n\n```public class TestClass{\nint rank; // there may be two or more classes with the same rank - will need an averaging ties strategy\n}\n```\n\nFrom which I have deduced the following:\n\n```NaturalRanking naturalRanking = new NaturalRanking(NaNStrategy.FIXED,\nTiesStrategy.AVERAGE);\n\nRealMatrix dataMatrix= createRealMatrix(double[], int, int); // what do I need this matrix for and what parameters do I need to pass?\n\nSpearmansCorrelation sc = SpearmansCorrelation(dataMatrix, naturalRanking);\nsc.correlation(double[],double[]);// I have to convert my class into a list of doubles? How?\n```\n\n## Answers\n\nFor the example found in the link you provided, all you need to do is:\n\n```new SpearmansCorrelation().correlation(mathsList,englishList);\n```\n\nThe class will average ties by default.\n\n### How does an OpenGL program interface with different graphic cards\n\nFrom what I understand (correct me if I am wrong), the OpenGL api converts the function calls written by the programmer in the source code into the specific gpu driver calls of our graphic card. Th...\n\n### Query error SQL Compact Edition 3.5\n\nIn my solution I have a local database and a database connect class with a GetQuery function. This function works good but for some reason, it won't handle the query as shown below. When I run a qu..."
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.74790937,"math_prob":0.48810512,"size":1626,"snap":"2019-13-2019-22","text_gpt3_token_len":384,"char_repetition_ratio":0.15721332,"word_repetition_ratio":0.0,"special_character_ratio":0.18942189,"punctuation_ratio":0.17142858,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95498234,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-23T07:31:03Z\",\"WARC-Record-ID\":\"<urn:uuid:b9e51287-2562-43c9-811e-c9643431d2d3>\",\"Content-Length\":\"20071\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6fbfac88-94a8-4453-a2a3-e81d86fe81de>\",\"WARC-Concurrent-To\":\"<urn:uuid:c34accd1-f915-4a25-b151-3706b4672ab9>\",\"WARC-IP-Address\":\"107.180.77.63\",\"WARC-Target-URI\":\"http://www.brokencontrollers.com/faq/10105501.shtml\",\"WARC-Payload-Digest\":\"sha1:3FLJ7I2RKP7QE63FGSL5XLNG5KEY22ZY\",\"WARC-Block-Digest\":\"sha1:DOIRTWBXEOFC3DEEFM5KJHLCLQV5DQXG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232257156.50_warc_CC-MAIN-20190523063645-20190523085645-00058.warc.gz\"}"} |
https://thinkmathematics.com/products/new-syllabus-additional-mathematics-textbook-9th-edition | [
"think!Mathematics\n\n# New Syllabus Additional Mathematics Textbook (9th Edition)\n\n• Sale\n• Regular price \\$24.99\n\nNew Syllabus Additional Mathematics (NSAM) is a textbook specially designed to provide valuable learning experiences to engage the hearts and minds of students sitting for the GCE O-level examination in Additional Mathematics. Included in the textbooks are Investigation, Class Discussion, Thinking Time, and Alternative Assessments such as Journal Writing to support the teaching and learning of Mathematics.\n\nEvery chapter begins with a chapter opener that motivates students in learning the topic. Interesting stories about Mathematicians, real-life examples and applications are used to arouse students’ interest and curiosity so that they can appreciate the beauty of Mathematics in their surroundings and in the sciences.\n\nThe use of ICT helps students visualize and manipulate mathematical objects more easily, thus making the learning of mathematics more interactive.\n\nProduct Details:\n\n ISBN-13 : 9789812374998 Pages : 544 pages Consultant : Dr Yeap Ban Har Authors : Dr Joseph Yeo, Teh Keng Seng, Loh Cheng Yee, Ivy Chow Dimensions : 8.75 in x 10.75 in x 1 in Weight : 2.95 lbs\n\n Chapter 1- Simultaneous Equations, Polynomials and Partial Fractions Linear and Non-Linear Simultaneous Equations Polynomials Remainder Theorem Factor Theorem Cubic Expressions and Equations Partial Fractions Chapter 2 - Quadratic Equations and Modulus(Absolute Value) Function Sum and Product of Roots Nature of Roots of a Quadratic Equation Maximum and Minimum Values of General Quadratic Functions Quadratic Inequalities Intersection of a Line and a Curve Modulus (Absolute Value) Function Graphs of Modulus (Absolute Value) Functions Chapter 3 - Binomial Theorem Binomial Expansion of (1 + b)^n Binomial Coefficients Binomial Theorem Applications of Binomial Theorem Chapter 4 - Indices (Exponents), Surds(Radicals) and Logarithms Indicies (Exponents) Surds (Radicals) Introduction to Logarithms Laws of Logarithms and Change of Base Formula Logarithmic and Exponential Equations Graphs of Exponential and Logarithmic Functions Applications of Logarithms and Exponents Chapter 5 - Coordinate Geometry Midpoint of a Line Segment Parallel and Perpendicular Lines More Problems on Equations of Stright Lines Area of Rectilinear Figures Chapter 6 - Further Coordinate Geometry Equation of a Circle Graphs of y^2 = kx Graphs of Power Functions Chapter 7- Linear Law Why study Linear Law? Converting from Non-Linear Equation to a Linear Form Converting from a Linear Form to a Non-linear Equation Applications of Linear Law Chapter 8 - Trigonometric Functions and Equations Trigonometric Ratios and Special Angles General Angles Trigonometric Ratios and General Angles Graphs of Trigonometric Functions Further Trigonometric Graphs Graphs of y = |f(x)|, where f(x) is trigonometric Cosecant, Secant and Cotangent Ratios Trigonometric Equations Chapter 9 - Trigonometric Identities and Formulas Trigonometric Identities Proving Identities Addition Formulas Double Angle Formulas Further Proving of Identities R-Formula Chapter 10 - Proofs in Plane Geometry Basic Proofs in Plane Geometry Proofs using Congruence and Similarity Tests Midpoint Theorem Tangent-Chord Theorem (Alternate Segment Theorem) Chapter 11 - Differentiation and its Application Gradient Functions Five Rules of Differentiation Equations of Tangent and Normal to a Curve Rates of Change Chapter 12 - Further Applications of Differentiation Higher Derivatives Increasing and Decreasing Functions Stationary Points Problems on Maximum and Minimum Values Chapter 13 - Differentiation of Trigonometric, Logarithmic & Exponential Functions and their Application Derivatives of Trigonometric Functions Derivatives of Logarithmic Functions Derivatives of Exponential Functions Further Application of Differentiation Chapter 14 - Integration Integrations as a Reverse of Differentiation Two Rules of Integration Integration of a Function involving a Linear Factor Integration of Trigonometric Functions Integration of Functions of the Forms 1/x and 1/(ax + b) Integration of Exponential Functions Further Examples of Integration Chapter 15 - Application of Integration Definite Integrals Further Examples of Definite Integrals Area under a Curve Chapter 16 - Kinematics Applications of Differentiation in Kinematics Applications of Integration in Kinematics\n\nNew Syllabus Additional Mathematics (NSAM) is a textbook specially designed to provide valuable learning experiences to engage the hearts and minds of students sitting for the GCE O-level examination in Additional Mathematics. Included in the textbooks are Investigation, Class Discussion, Thinking Time, and Alternative Assessments such as Journal Writing to support the teaching and learning of Mathematics.\n\nEvery chapter begins with a chapter opener that motivates students in learning the topic. Interesting stories about Mathematicians, real-life examples and applications are used to arouse students’ interest and curiosity so that they can appreciate the beauty of Mathematics in their surroundings and in the sciences.\n\nThe use of ICT helps students visualize and manipulate mathematical objects more easily, thus making the learning of mathematics more interactive."
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https://www.jpost.com/israel/residential-building-in-safed-sustains-direct-hit | [
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https://www.deepdyve.com/lp/oxford-university-press/a-partial-differential-equation-for-the-epsilon-uniformly-quasiconvex-FD0Rd99iZq | [
"# A partial differential equation for the $\\epsilon$-uniformly quasiconvex envelope\n\nA partial differential equation for the $\\epsilon$-uniformly quasiconvex envelope Abstract Quasiconvex (QC) functions are functions whose level sets are convex. In a series of articles, Barron, Goebel and Jensen studied partial differential equations (PDEs) for QC functions. They introduced a regularization of the PDE that is more stable. In this article, we introduce a different regularization that is more amenable to numerical approximation. We prove existence and uniqueness of viscosity solutions of the PDE and demonstrate that level sets of solutions of the new PDE are uniformly convex. We build convergent finite difference approximations, comparing the QC envelope with the regularization. 1. Introduction In a series of articles from about four years ago, Barron, Goebel and Jensen introduced and studied partial differential equations (PDEs) for quasiconvexity (Barron et al., 2012a,b, 2013; Barron & Jensen, 2014). In this context, quasiconvexity means that the sublevel sets of a function are convex. The study of convexity of level sets for obstacle problems has a long history, which includes Caffarelli & Spruck (1982) and Kawohl (1985), as well as the more recent work by Colesanti & Salani (2003) and the references therein. Quasiconvex (QC) functions appear naturally in optimization, since they generalize convex functions. The property also appears in economics (Avriel et al., 2010). Earlier work by one of the authors studied a PDE for the convex envelope (Oberman, 2007), which led to a numerical method for convex envelopes (Oberman, 2008a,b). The notion of (scalar) quasiconvexity that we discuss here should not be confused with the well-known notion of quasiconvexity in vector variational problems. Our motivation for studying this PDE comes from two directions: geometric and algorithmic; it is natural to represent a set using a level set function (Sethian, 1999). The PDE allows us to find the convex hull of a set represented this way directly (in fact, every level set), rather than extracting the level set and then finding the convex hull, followed by generating a new level set function. In addition, we can use the parabolic version of the equation to progressively convexify a level set, or to make already convex level sets uniformly convex. Quasiconvexity is challenging because, unlike convexity, it is a nonlocal property (at least for functions that have flat parts). This means that, even using viscosity solutions, there is no local characterization for quasiconvexity. On the other hand, using the more stable notion of robust quasiconvexity, Barron, Goebel and Jensen showed that these functions are characterized in the viscosity sense (Crandall et al., 1992) by a PDE (Barron et al., 2013; Barron & Jensen, 2014). One motivation for this work was to build numerical solvers for the QC envelope PDE. However, we had difficulties with both the QC and the robust-QC operators: the former lacks uniqueness and the latter uses an operator defined over small slices of angles (see Fig. 3), which leads to poor accuracy when using wide-stencil finite difference schemes. An alternative, presented in Barron et al. (2013), was to use first-order nonlocal PDE solvers. In a companion article (Abbasi & Oberman, 2016), we built a nonlocal solver for the QC and robust-QC envelopes. The nonlocal PDEs can be solved explicitly and the problem implemented efficiently. By iteratively solving for the envelopes on lines in multiple directions, we approximated the solution of the problem in higher dimensions. However, we are still interested in the PDE approach, which has advantages arising from its nonlocal nature. In this article, we build on the results of Barron et al. (2012a,b) and Barron & Jensen (2014) to obtain a PDE for |$\\epsilon$|-uniformly QC functions, which we define subsequently. Note that |$\\epsilon$|-uniform quasiconvexity implies robust quasiconvexity. The choice of terminology lends itself to the fact that |$\\epsilon$|-uniformly QC functions have uniformly convex level sets. Following the argument in Barron & Jensen (2014), we establish uniqueness of viscosity solutions for the PDE. Moreover, this operator is defined for all direction vectors, which makes it amenable to discretization using wide-stencil schemes. We consider the obstacle problem for the |$\\epsilon$|-uniformly quasiconvex envelope (QCE). As is the case for robust-QC, we recover the QCE as the regularization parameter |$\\epsilon \\to 0$|. While robust-QC functions can have corners in one dimension, strictly QC functions are smoother; see Fig. 3 below, and the explicit formula for the solution in one dimension in Section 2.3. We also build and implement convergent elliptic finite difference schemes (Barles & Souganidis, 1991; Oberman, 2006) for the envelopes. These are wide-stencil finite difference schemes, which can be developed using ideas similar to Oberman (2004, 2008a,b). Solutions to these PDEs can be found using an iterative method which is equivalent to the explicit Euler discretization of the parabolic equation (Oberman, 2006). However, the method has a nonlinear Courant-Friedrichs-Lewy (CFL) condition, which restricts the step size. We find that alternating the line solver with several iterations of the parabolic PDE solver significantly improves the speed of the solution. Numerical solutions show that we obtain very similar results to the line solver for the QCE with small |$\\epsilon = h/2$|, where |$h$| is the grid resolution. We also compare large |$\\epsilon$| solutions with comparable robust QCE and find that solutions are smoother. Formally, we show that solutions are |$\\epsilon$| uniformly convex. The QC operator in two dimensions recovers the level set curvature operator. We show that our discretization of our operator agrees with the Kohn–Serfaty (Kohn & Serfaty, 2007) first-order representation of the mean curvature operator in two dimensions. See Section 4.4. 1.1 Convexity of level sets of a function We give a brief informal derivation of the operator. Given a smooth function |$u: {\\mathbb{R}^n} \\to \\mathbb{R}$|, the direction of the gradient at |$x_0$|, |$p = {\\nabla} u(x_0)$| is the normal to the sublevel set |$\\{ x \\in {\\mathbb{R}^n} \\mid u(x) \\leq u(x_0) \\}$|. The curvatures of the level set at |$x_0$| are proportional to the eigenvalues of the Hessian of |$u$|, |$M = D^2u(x_0)$|, projected onto the tangent hyperplane at |$x_0$|, |$P = P_{x_0} = \\{ v \\in {\\mathbb{R}^n} \\mid v\\cdot p = 0\\}$|. These curvatures are all positive if |$v^\\intercal M v \\geq 0$| for all |$v \\in P$|. Thus, formally, the condition for local convexity of the level set is non-negativity of the operator \\begin{equation} L_0(p,M) \\equiv \\min_{{\\vert{v}\\vert} = 1} \\left \\{ v^\\intercal M v \\mid v\\cdot p = 0 \\right. \\}. \\end{equation} (1.1) This is the operator considered in Barron et al. (2013) to study QC functions. However, for technical reasons discussed below, they chose to relax the constraint |$v \\cdot p =0$| to an inequality constraint |${\\vert{v \\cdot p}\\vert} \\leq \\epsilon$|, resulting in the operator $L_\\epsilon(p,M) \\equiv \\min_{{\\vert{v}\\vert} = 1} \\left \\{ v^\\intercal M v \\mid {\\vert{v \\cdot p}\\vert}\\le\\epsilon \\right \\}\\!.$ Our operator is obtained by instead replacing the hard constraint |${\\vert{v\\cdot p}\\vert} \\leq \\epsilon$| with a penalty in the objective function. So we define $F^\\epsilon(p,M) \\equiv \\min_{{\\vert{v}\\vert} = 1}\\left \\{ v^\\intercal M v + \\tfrac{1}{{\\epsilon}} {\\vert{v^\\intercal p}\\vert} \\right \\}\\!.$ This choice of penalty gives the operator for uniformly convex level sets, as we show subsequently. Refer to Figure 2 for a visualization of the feasible sets. Figure 3 illustrates the important differences in the notions of quasiconvexity discussed so far. Moreover, on a bounded domain, the |$\\epsilon$|-robustly QCE can have an interval of global minima, whereas our solution has a unique (global) minimum. 1.2 Basic definitions We recall some basic definitions and establish our notation. For a reference, see Boyd & Vandenberghe (2004). The set |$S$| is convex if whenever |$x$| and |$y$| are in |$S$| then so is the line segment |$[x,y]\\equiv\\{tx + (1-t)y$|: |$t \\in [0,1]\\}$|. We say that the function |$u: {\\mathbb{R}^n} \\to \\mathbb{R}$| is convex if \\begin{equation} u(tx + (1-t)y) \\le t u(x) + (1-t) u(y) \\quad \\text{ for every } x,y \\in {\\mathbb{R}^n}, \\quad 0 \\le t \\le 1. \\end{equation} (1.2) The convex envelope of a function |$g$|, hereby denoted |$g^{\\rm CE}$|, is the largest convex function majorized by |$g$|: \\begin{equation} g^{\\rm CE}(x) \\equiv \\sup \\{ v(x)\\mid v\\,\\text{is convex and}\\,v\\le g \\}. \\end{equation} (1.3) In terms of sets, if |$S$| is given by the sublevel set of a function |$u$|, \\begin{equation} S = S_\\alpha(u)\\equiv\\{x \\in {\\mathbb{R}^n} \\mid u(x)\\leq\\alpha\\}, \\end{equation} (1.4) then convexity of |$S$| is equivalent to the following condition: $u(x), u(y) \\le \\alpha \\implies u(tx + (1-t)y) \\le \\alpha \\quad \\text{ for every } 0 \\le t \\le 1.$ This condition, when applied to every level set |$\\alpha\\in\\mathbb{R}$|, characterizes quasiconvexity of the function |$u$|. That is, |$u$| is QC if every sublevel set |$S_\\alpha(u)$| is convex. Equivalently, |$u$| is QC if \\begin{equation} u(tx + (1-t)y) \\le \\max \\{ u(x), u(y) \\} \\quad \\text{ for every } x,y \\in {\\mathbb{R}^n},\\quad 0 \\le t \\le 1. \\end{equation} (1.5) Given |$g:{\\mathbb{R}^n} \\to \\mathbb{R}$|, the QCE of |$g$| is given similarly by \\begin{equation} g^{\\rm QCE}(x) \\equiv \\sup \\{ v(x)\\mid v\\,\\text{is QC and}\\,v\\le g\\}. \\end{equation} (1.6) Remark 1.1. Since the maximum of any two QC (convex) functions is QC (convex) as well, the suprema in (1.3) and (1.6) are well defined. Figure 1 provides a visual comparison between the convex envelope and the QCE in one dimension. Fig. 1. View largeDownload slide Comparison of a function (solid) and its envelopes (dashed). (a) Quasiconvex envelope. (b) Convex envelope. Fig. 1. View largeDownload slide Comparison of a function (solid) and its envelopes (dashed). (a) Quasiconvex envelope. (b) Convex envelope. Fig. 2. View largeDownload slide Constraint sets (dotted and solid) of different PDEs in two dimensions. Fig. 2. View largeDownload slide Constraint sets (dotted and solid) of different PDEs in two dimensions. Fig. 3. View largeDownload slide Comparison between solutions of different operators. Graphs arranged top to bottom: obstacle, QCE, |$\\epsilon$|-robustly QCE, and |$\\epsilon$|-uniformly QCE. Fig. 3. View largeDownload slide Comparison between solutions of different operators. Graphs arranged top to bottom: obstacle, QCE, |$\\epsilon$|-robustly QCE, and |$\\epsilon$|-uniformly QCE. 1.3 Viscosity solutions Suppose that |$\\varOmega\\subset{\\mathbb{R}^n}$| is a domain. Let |$S^n$| be the set of real symmetric |$n\\times n$| matrices and take |$N\\le M$| to denote the usual partial ordering on |$S^n$|, namely that |$N-M$| is negative semidefinite. Definition 1.2. The operator |$F(x,r,p,M):\\varOmega\\times\\mathbb{R}\\times{\\mathbb{R}^n}\\times S^{n} \\to\\mathbb{R}$| is degenerate elliptic if $F(x,r,p,M)\\le F(x,s,p,N)\\quad\\text{whenever}\\ r\\le s\\ \\text{and}\\ N\\le M.$ Remark 1.3. For brevity, we use the notation |$F[u](x)\\equiv F(x,u(x),\\nabla u(x), D^2u(x))$|. Definition 1.4 (Viscosity solutions). Suppose |$F$| is a degenerate elliptic operator, as defined above. We say that the upper semicontinuous (lower semicontinuous) function |$u:\\varOmega\\to\\mathbb{R}$| is a viscosity subsolution (supersolution) of |$F[u]=0$| in |$\\varOmega$| if for every |$\\phi\\in C^2(\\varOmega)$|, whenever |$u-\\phi$| has a strict local maximum (minimum) at |$x\\in\\varOmega$|, $F(x,u(x),\\nabla\\phi(x),D^2\\phi(x))\\le0\\ (\\ge0).$ Moreover, we say that |$u$| is a viscosity solution of |$F[u]=0$| if |$u$| is both a viscosity subsolution and a supersolution. 1.4 The QCE Use the notation |$\\lambda_{\\rm QC}[u](x) \\equiv L_0({\\nabla} u(x),D^2u(x))$|. The obstacle problem for the QCE is given by \\begin{equation} \\begin{cases} \\max\\{u-g,-\\lambda_{\\rm QC}[u](x)\\}=0, & x\\in\\varOmega,\\\\ u =g, & x\\in\\partial\\varOmega. \\end{cases} \\end{equation} (Ob) This PDE can have multiple viscosity solutions for a given |$g$| (Barron et al., 2013). However, there is a unique QC solution which is the QCE of |$g$|. Failure of uniqueness in general can be seen in the following counterexample, which is similar to Barron et al. (2013, Example 3.1). Example 1.5. Consider |$\\varOmega=\\{x^2+y^2<1\\}, u(x,y) = -y^4, v(x,y) = -(1-x^2)^2$|, so that |$u=v$| on |$\\partial\\varOmega$|. Now consider |$g\\equiv u$| on |$\\varOmega$|. We can calculate that |$\\lambda_{\\rm QC}[u]=\\lambda_{\\rm QC}[v]=0$|. In summary, |$u$| and |$v$| are both solutions of (Ob), but |$u\\neq v$| in |$\\varOmega$| (in fact |$v<u$| in |$\\varOmega$|). This contradicts the uniqueness of (Ob). It is shown in Barron et al. (2013) that continuous QC functions necessarily satisfy |$\\lambda_{\\rm QC}[u] \\geq 0$| in the viscosity sense. The converse, however, is not true: consider the function |$u(x)=-x^4$|, which is not QC yet satisfies |$\\lambda_{\\rm QC}[u]=0$| at |$x=0$|. In fact, using this as a sufficient condition is only possible when |$u$| has no local maxima (Barron et al., 2013). This operator can be used to completely characterize the set of continuous |$\\epsilon$|-robustly QC functions, defined below. Definition 1.6 (|$\\epsilon$|-robustly QC) The function |$u:\\varOmega\\to\\mathbb{R}$| is |$\\epsilon$|-robustly QC if |$u(x) +y^\\intercal x$| is QC for every |${\\vert{y}\\vert}\\leq\\epsilon$|. In particular, |$\\epsilon$|-robustly QC functions are functions whose quasiconvexity is maintained under small linear perturbations. Write |$\\lambda_{\\rm QC}^\\epsilon[u](x) \\equiv L_\\epsilon({\\nabla} u(x),D^2u(x))$|. Proposition 1.7 (Characterization of |$\\epsilon$|-robustly QC functions; Barron et al., 2013). The upper semicontinuous function, |$u:\\varOmega\\to\\mathbb{R}$|, is |$\\epsilon$|-robustly QC if and only if |$u$| is a viscosity subsolution of |$\\lambda_{\\rm QC}^\\epsilon[u]=0$|. Next, we show that viscosity subsolutions (defined below) of our operator are robustly-QC (which also implies they are QC). This allows us to avoid a technical argument from Barron et al. (2013). We believe that stronger results hold: see the formal analysis in Section 3. For simplicity, we first define the following abbreviation. Definition 1.8 (|$\\epsilon$|-uniformly QC). The function |$u\\in {\\rm USC}(\\overline\\varOmega)$| is |$\\epsilon$|-uniformly QC if it is a viscosity subsolution of |$\\epsilon-F^\\epsilon[u]=0$|. Proposition 1.9. Suppose |$u\\in {\\rm USC}(\\overline\\varOmega)$| is |$\\epsilon$|-uniformly QC. Then, |$u$| is |$\\epsilon^2$|-robustly QC. Proof. First observe that for any |$\\phi\\in C^2$| we have the following inequality: \\begin{align*} F^\\epsilon[\\phi](x) - \\frac{\\alpha}{\\epsilon} &= \\min_{{\\vert{v}\\vert}=1}\\left\\{\\tfrac{1}{{\\epsilon}}{\\vert{\\nabla \\phi(x)\\cdot v}\\vert} +v^\\intercal D^2\\phi(x)v - \\tfrac{\\alpha}{\\epsilon} \\right\\}\\\\ &\\leq \\min_{\\{{\\vert{v}\\vert}=1,{\\vert{\\nabla \\phi(x)\\cdot v}\\vert}\\le \\alpha \\}}\\left\\{v^\\intercal D^2\\phi(x)v\\right\\} = \\lambda_{QC}^{\\alpha}[\\phi](x). \\end{align*} Choosing |$\\alpha=\\epsilon^2$| we see that any viscosity subsolution of |$\\epsilon-F^\\epsilon[u]= 0$| is also a viscosity subsolution of |$-\\lambda_{\\rm QC}^{\\epsilon^2}[u]= 0$|. Thus, |$u$| is |$\\epsilon^2$|-robustly QC. □ 2. Properties of solutions In this section, we present technical arguments proving the uniqueness of solutions of our PDEs and discuss some relevant properties. 2.1 Comparison principle In this subsection, we will show that a weak comparison principle holds for the Dirichlet problem of |$h-F^\\epsilon[u]=0$| for |$h>0$| and |$\\epsilon > 0$|. Comparison also holds for the corresponding obstacle problem. The proof we present is based on the uniqueness proof of |$\\lambda_{\\rm QC}[u]=g$| for |$g>0$|, presented in Barron & Jensen (2014). The result is simpler, because our operator is continuous as a function |$(p,M)\\mapsto F^\\epsilon(p,M)$| for |$\\epsilon > 0$|. Write \\begin{equation} F^\\epsilon[u](x)=F^\\epsilon(\\nabla u(x),D^2u(x)) \\equiv \\min_{{\\vert{v}\\vert}=1}\\left\\{\\tfrac{1}{{\\epsilon}}{\\vert{\\nabla u(x)\\cdot v}\\vert} + v^\\intercal D^2u(x)v\\right\\}\\!. \\end{equation} (2.1) Note that |$-F^\\epsilon(p,M)$| is elliptic by Definition 1.2. We consider the following PDEs: (|$\\epsilon$|-QC) \\begin{gather} h(x)-F^\\epsilon[u](x) = 0, \\quad x\\in\\varOmega,\\\\ \\end{gather} (|$\\epsilon$|-QCE) \\begin{gather} \\max\\{u(x)-g(x),h(x)-F^\\epsilon[u](x)\\} = 0, \\quad x\\in\\varOmega, \\end{gather} where |$\\varOmega\\subset{\\mathbb{R}^n}$| is an open, bounded and convex domain, and |$g:\\overline\\varOmega\\to\\mathbb{R}$| is continuous. In the latter equation, |$g$| is the obstacle. We also impose the following condition on |$h$|: \\begin{equation} h:\\overline\\varOmega\\to\\mathbb{R}\\,\\text{is continuous and positive.} \\end{equation} (2.2) Enforce the following Dirichlet boundary data: \\begin{equation} u(x) = g(x),\\quad x\\in\\partial\\varOmega. \\end{equation} (Dir) Remark 2.1 (Continuity up to the boundary). In general, viscosity solutions of (ϵ-QCE) need not be continuous up to the boundary. To apply Barles & Souganidis (1991) for convergence of numerical schemes, we need a strong comparison principle, which requires that solutions be continuous up to the boundary. The following assumption ensures continuity up to the boundary There is a convex domain |$\\varOmega_L\\supset\\varOmega$| and a continuous, QC function |$g_0: \\varOmega_L \\to \\mathbb{R}$|, with $g_0(x)\\le g(x) \\quad \\text{ for } x \\in \\varOmega, \\qquad g_0(x)=g(x) \\quad \\text{ for } x\\in \\varOmega_L\\setminus\\varOmega.$ Continuity follows because we have |$g_0 \\leq u \\leq g,$| which gives |$u = g$| on |$\\partial \\varOmega$|. An alternative to this condition is to prove convergence in a neighbourhood of the boundary. The convergence proof in this setting can be found in Froese (2016). Next, we state a technical, but standard, viscosity solutions result, which gives the comparison principle in the case where we have strict subsolution and supersolution. Proposition 2.2 (Comparison principle for strict subsolutions; Crandall et al., 1992). Consider the Dirichlet problem for the degenerate elliptic operator |$F(p,M)$| on the bounded domain |$\\varOmega$|. Let |$u \\in {\\rm USC}(\\bar \\varOmega)$| be a viscosity subsolution and let |$v \\in {\\rm LSC}(\\bar \\varOmega)$| be a viscosity supersolution. Suppose further that for |$\\sigma > 0$|, \\begin{align*} F[u] + \\sigma &\\le 0 \\text{ in } \\varOmega, \\\\ F[v] &\\ge 0 \\text{ in } \\varOmega \\end{align*} holds in the viscosity sense. Then, the comparison principle holds: $\\text{ if }\\,u\\leq v\\,\\text{on}\\,\\,\\partial\\varOmega\\,\\,\\text{then}\\,u\\le v\\,\\,\\text{in}\\,\\varOmega.$ Remark 2.3. In Crandall et al. (1992, Section 5.C), it is explained how the main comparison theorem (Crandall et al., 1992, Theorem 3.3) can be applied when it is possible to perturb a subsolution to a strict subsolution. This version of the theorem is what we state in Proposition 2.2. This result was used in Bardi & Mannucci (2006, Theorem 3.1) and Bardi & Mannucci (2013) to prove a comparison principle. Using the the same perturbation technique from Barron & Jensen (2014), and consequently applying Proposition 2.2, we obtain the following comparison result. Proposition 2.4 (Comparison principle). Consider the Dirichlet problem given by (ϵ-QC) (Dir) and assume (2.2) holds. Let |$u\\in {\\rm USC}(\\overline\\varOmega)$| be a viscosity subsolution and |$v\\in {\\rm LSC}(\\overline\\varOmega)$| be a viscosity supersolution of (ϵ-QC). Then, the comparison principle holds: $\\text{ if}\\,\\,u\\leq v \\,\\, \\text{on}\\,\\partial\\varOmega \\,\\,\\text{then}\\,u\\le v \\,\\,\\text{in}\\,\\, \\varOmega.$ Proof. We will show that we can perturb |$u$| to a function |$u^\\sigma$| satisfying |$u^\\sigma\\le v$| on |$\\partial\\varOmega$| and $h-F^\\epsilon[u^\\sigma]<0 \\quad\\text{in}\\, \\varOmega$ in the viscosity sense. Applying Proposition 2.2, we will have |$u^\\sigma\\le v$| in |$\\varOmega$|. Taking |$\\sigma\\to0$| yields the desired result. Fix |$\\sigma>0$| and define the following perturbation of |$u$|, and notice that for |$x\\in\\partial\\varOmega$| we have the following relation: $u^\\sigma(x) \\equiv u(x) - \\sigma \\left(\\max_{y\\in\\partial\\varOmega}u(y)-u(x)\\right)\\le u(x) - \\sigma (u(x)-u(x)) = u(x)\\le v(x);$ that is, |$u^\\sigma\\le v$| on |$\\partial\\varOmega$|. Next, because |$u$| is a subsolution, we have $h-\\frac{1}{1+\\sigma}F^\\epsilon[u^\\sigma] \\le 0,$ leading to $h - F^\\epsilon[u^\\sigma]\\le -\\sigma h \\le -\\sigma h_{\\min} <0\\quad\\text{where}\\,h_{\\min}\\equiv \\min_{x\\in\\overline\\varOmega}h(x).$ □ We use this result to prove a weak comparison principle for the obstacle problem given by (ϵ-QCE) and (Dir). Corollary 2.5 (Comparison principle for the obstacle problem). Consider the Dirichlet problem given by (ϵ-QCE), (Dir) and assume (2.2) holds. Let |$u\\in {\\rm USC}(\\overline\\varOmega)$| be a viscosity subsolution and |$v\\in {\\rm LSC}(\\overline\\varOmega)$| be a viscosity supersolution of (ϵ-QCE). Then, the comparison principle holds: $\\text{ if}\\,\\, u\\leq v\\,\\,\\text{on}\\,\\,\\partial\\varOmega\\,\\,\\text{then}\\,\\,u\\le v\\,\\,\\text{in}\\,\\,\\varOmega.$ Proof. We begin by considering the domain |$\\varOmega_g\\equiv\\{x\\in\\varOmega:v(x)<g(x)\\}$|. Then, |$h-F^\\epsilon[v]\\ge 0$| in |$\\varOmega_g$| and |$v=g$| on |$\\partial\\varOmega_g$|, that is, |$v$| is a viscosity supersolution of |$h-F^\\epsilon[v]=0$| in |$\\varOmega_g$|. Now, by the definition of viscosity subsolutions we have |$u\\le g$| and |$h-F^\\epsilon[u]\\le 0$| in |$\\varOmega$| and thus also |$\\varOmega_g$|. This allows us to conclude that |$u\\le v$| on |$\\partial\\varOmega_g$|. Therefore, by Proposition 2.4, |$u\\le v$| in |$\\varOmega_g$|. Concluding, in |$\\varOmega\\setminus\\varOmega_g$| we necessarily have |$v\\ge g\\ge u$|. □ 2.2 The |$\\epsilon$|-uniformly QCE We formulate the |$\\epsilon$|-uniformly QCE of a function |$g$| as the unique viscosity solution of the following obstacle problem: (|$\\epsilon$|-Ob) \\begin{equation} \\begin{cases} \\max\\{u(x)-g(x),\\epsilon-F^\\epsilon[u](x)\\} = 0, & x\\in\\varOmega,\\\\ u(x) = g(x), & x\\in\\partial\\varOmega. \\end{cases} \\end{equation} Remark 2.6 (Convergence of approximate solutions). It is clear that as |$\\epsilon\\to0$|, the penalization term in |$F^\\epsilon[u]$| tends to infinity. The result is that |$F^\\epsilon[u]\\to\\lambda_{\\rm QC}[u]$| as |$\\epsilon\\to0$|. From this observation, the standard stability result of viscosity solutions and the quasiconvexity of subsolutions of |$\\epsilon-F^\\epsilon[u]=0$|, one can then apply the same argument presented in the proof of Barron et al. (2013, Theorem 5.3) to conclude that the unique viscosity solutions of (ϵ-QCE), (Dir) converge to the QCE as |$\\epsilon\\to0$|. This result allows us to compute asymptotic approximations of the QCE of a given obstacle. 2.3 Solution formula in one dimension In one dimension, |$\\epsilon-F^\\epsilon[u]$| is simply the Eikonal operator with a small diffusion term. When considering a solution |$u$| of (ϵ-Ob), whenever |$u(x) < g(x)$|, we have |$F^\\epsilon[u](x) = \\epsilon$|, which gives \\begin{equation} \\epsilon u_{xx} + {\\vert{u_x}\\vert}=\\epsilon^2. \\end{equation} (2.3) Define |$\\varOmega_g\\equiv\\{x:u(x) < g(x)\\}$|. Note that |$\\varOmega_g$| need not be connected; it can be written as the union of finitely many intervals (refer to Fig. 3). However, we can solve the equation in each interval. Lemma 2.7 (One-dimensional solution). The viscosity solution of the one-dimensional PDE for the operator (2.1) and (2.3), along with boundary conditions |$u(0)=0, u(W) = H$|, is given by the following. Set |$S\\equiv H/\\epsilon^2W$|. Then, $u^\\epsilon(x) = \\begin{cases} \\pm \\epsilon^2 x, & S=\\pm1,\\\\ \\epsilon^2x + C^+(1-\\exp(-x/\\epsilon)), & S>1,\\\\ -\\epsilon^2x + C^-(1-\\exp(x/\\epsilon)), & S<-1,\\\\ \\epsilon^2{\\vert{x-x^*}\\vert} + \\epsilon^3(\\exp(-{\\vert{x-x^*}\\vert}/\\epsilon)-1) + u_0, & {\\vert{S}\\vert}<1, \\end{cases}$ where $C^\\pm \\equiv \\frac{H\\mp\\epsilon^2 W}{\\pm\\epsilon(1-\\exp(\\mp W/\\epsilon))}$ and |$u_0\\in\\mathbb{R}$| and |$x^*\\in I = (0,W)$| are constants. Proof. If there exists |$x_0$| in the interval |$I$| such that |$u_x(x_0)=0$| then (2.3) implies |$u_{xx}(x_0)>0$|, that is, if there exists an interior critical point, then |$u$| must be strictly convex in |$I$|. This allows us to break down the analysis of (2.3), restricted to each interval, into several cases: (i) |$u_x<0$| in |$I$|, (ii) |$u_x>0$| in |$I$| and (iii) |$u_x(x_0)=0$| for some |$x_0 \\in I$|. In each case, we can solve a linear second-order ordinary differential equation. The case where |$S = \\pm 1$| is degenerate: the solution is linear. The case where |${\\vert{S}\\vert} > 1$| is not difficult. The final case, where |${\\vert{S}\\vert} < 1$| corresponds to (iii). In this case, |$u_x<0$| for |$x<x_0$| and |$u_x>0$| for |$x>x_0$|. Then, \\begin{equation} u(x) = {\\vert{x-x_0}\\vert} + (\\exp(-{\\vert{x-x_0}\\vert})-1) + u_0. \\end{equation} (2.4) For some |$u_0\\in\\mathbb{R}$|. Taylor expansion shows that |$u(x) = x^2/2 + \\mathcal{O}(x^3)$| for |$x$| near |$x_0$|. Finding |$x_0$| (or |$u_0$|) analytically is infeasible. But we can argue that the solution is correct by a continuity argument. Given the function |$u$| defined by (2.4), define the continuous function |$S_1(y)\\equiv\\frac{u(y+W_1)-u(y)}{W_1}$|. For small |$W_1$| and |$y\\ll x_0$|, we have that |$S_1(y)<-1$| (by the earlier discussion). Similarly for |$y\\gg x_0$|, we have |$S_1(y)>1$|. Therefore, by the intermediate value theorem there exists |$y^*$| such that |$S_1(y^*)=S$|. This leads to the correct choice of constants in (2.4) that achieve the boundary values. □ 3. Geometric properties of the operator To better understand the geometry of the solutions of |$\\epsilon-F^\\epsilon[u]=0$|, we present, in this section, some results regarding the convexity of the level sets of subsolutions of related PDEs. We give a proof of uniform convexity of the level sets for |$C^2$| subsolutions of the equation. The general case is beyond the scope of this article. The proof of robust quasiconvexity for viscosity solutions is technical, involving the construction of special test functions (Barron et al., 2013). Analogous results can also be found in Lions (1998) for convexity, and Carlier & Galichon (2012) for semiconcavity of solutions of parabolic equations. Next, we quantify the degree to which a function can lack quasiconvexity, given that it is uniformly QC in only certain directions. This is relevant for numerical solutions that enforce the operator in only a finite number of grid directions. A related result on semiconcavity of solutions of a parabolic equation can be found in Carlier & Galichon (2012, Section 4). We first show that we can recover a convex function by enforcing strict convexity along a prescribed set of directions. This result is of independent interest for the related numerical method (Oberman, 2008a). Next, we prove an analogous result for QC functions. In what follows, it suffices to consider subsolutions of |$\\epsilon - \\lambda_{\\rm QC}[u]=0$|. This is because of the ordering of the operators: \\begin{equation} -\\lambda_{\\rm QC}[u]\\ge-\\lambda_{\\rm QC}^{\\epsilon^2}[u]\\ge \\epsilon - F^\\epsilon[u]. \\end{equation} (3.1) The first inequality follows naturally from the restriction of the constraint set. The second inequality is evident from the proof of Proposition 1.9. 3.1 Uniform convexity of the level sets of solutions Definition 3.1 (Uniform convexity of sets). Let |$S\\subset{\\mathbb{R}^n}$| be a domain and suppose |$x,y\\in\\partial S$|. Define the midpoint |$\\bar x\\equiv(x+y)/2$|. Then we say |$S$| is uniformly convex if |${\\text{dist}}(\\bar x,\\partial S) = \\mathcal{O}(h^2)$|. Proposition 3.2 (Uniformly convex level sets of subsolutions). Suppose |$u\\in C^2(\\overline\\varOmega)$| is a subsolution of |$\\epsilon-\\lambda_{\\rm QC}[u]=0$|. Then, |$u$| has uniformly convex level sets. Proof. Suppose |$x,y\\in\\varOmega$| such that |$u(x)=u(y)=\\alpha$|. Let |$\\bar x$| be the midpoint of the line segment joining |$x$| and |$y$|, |$h = {\\vert{\\bar x - y}\\vert}$| and |$d = (\\bar x-y)/{\\vert{\\bar x -y}\\vert}$|. We see that |${\\nabla} u(\\bar x)^\\intercal d= 0$|, so that we have \\begin{align*} \\epsilon &\\le \\min_{{\\vert{v}\\vert}=1,{\\nabla} u(\\bar x)^\\intercal v=0} v^\\intercal D^2u(\\bar x) v\\\\ &\\le d^\\intercal D^2u(\\bar x) d\\\\ &= \\frac{u(\\bar x + hd) + u(\\bar x -hd) - 2u(\\bar x)}{h^2} +\\mathcal{O}(h^2)\\\\ &= \\frac{2\\alpha - 2u(\\bar x)}{h^2} + \\mathcal{O}(h^2). \\end{align*} Rearranging for |$u(\\bar x)$| yields the following inequality: $u(\\bar x) \\le \\alpha -\\frac{\\epsilon h^2}{2} + \\mathcal{O}(h^4).$ □ 3.2 Directional quasiconvexity Definition 3.3. The function |$u:\\mathbb{R}^n\\rightarrow\\mathbb{R}$| is QC along a direction |$v$| if for any |$x\\in\\mathbb{R}^n$|, the function |$\\tilde u(t)\\equiv u(x+tv)$| is QC for |$t\\in\\mathbb{R}$|. We also appeal to the following proposition, which is elementary from the definition of quasiconvexity. Proposition 3.4 The function |$u$| is QC if and only if |$u$| is QC in every direction |$v$|. Proposition 3.4 provides a convenient characterization of QC functions. In practice, however, we are confined to a grid and thus cannot enforce quasiconvexity along every direction. Therefore, we relax the notion of directional quasiconvexity to only a finite set of directions. Doing so results in the notion of approximate quasiconvexity. We can quantify the degree to which a function might lack quasiconvexity, expressed in terms of the directional resolution which we define below. Definition 3.5 (Directional resolution). Let |$\\mathcal{D}=\\{d_1,\\dots,d_n\\}$| be a set of unit vectors. Then we define the directional resolution of |$\\mathcal{D}$| as \\begin{equation} {\\rm d}\\theta \\equiv \\max_{{\\vert{w}\\vert}=1}\\min_{d\\in \\mathcal{D} }\\cos^{-1}(w^\\intercal d). \\end{equation} (3.2) In two dimensions, |${\\rm d}\\theta$| is the largest angle an arbitrary unit vector can make with any vector in |$\\mathcal{D}$|. In two dimensions, |${\\rm d}\\theta$| is simply half the maximum angle between any two direction vectors. Recalling that a necessary condition for a function |$u$| to be QC is |$-\\lambda_{\\rm QC}[u]\\le0$| in the viscosity sense (Barron et al., 2013), we have the following result. Proposition 3.6 (Approximate quasiconvexity). Suppose |$u\\in C(\\overline\\varOmega)$| and |$\\mathcal{D}$| is a set of directions with directional resolution |${\\rm d}\\theta\\leq\\frac{\\pi}{4}$|. Also, suppose |$u$| is QC along every |$d\\in \\mathcal{D}$|. Then, we have the following approximate quasiconvexity estimate: $-\\lambda_{\\rm QC}[u]\\leq\\mathcal{O}({\\rm d}\\theta).$ Proof. Without loss of generality, assume |$x=0$| and |$u(0)=0$|. Suppose |$u-\\phi$| attains a global maximum at |$x=0$|. We may assume |$\\phi$| is locally quadratic, |$\\phi(x)=x^\\intercal A x+b^\\intercal x$|, for some real, symmetric matrix |$A$|, and for |$b\\in{\\mathbb{R}^n}$|. Next, suppose |$w$| is an arbitrary unit vector satisfying |$\\nabla \\phi(0)^\\intercal w=b^\\intercal w=0$|. Decompose |$w$| as follows: $w = \\cos(\\theta)v + \\sin(\\theta) p,$ where |$v\\equiv\\text{argmin}_i(\\cos^{-1}(w^\\intercal v_i))$|, |$\\theta = \\cos^{-1}(w^\\intercal v)$| and |$p$| is some unit vector orthogonal to |$v$|. By hypothesis, |$\\theta\\le\\pi/4$|. Taking |$\\lambda_n$| to denote the largest eigenvalue of |$A$|, we observe \\begin{align*} 0=u(0) &= u\\left(\\tfrac{1}{2}(-\\cos(\\theta) v)+\\tfrac{1}{2}(\\cos(\\theta) v)\\right)\\\\ &\\leq \\max(u(-\\cos(\\theta) v),u(\\cos(\\theta) v))\\quad\\text{(by quasiconvexity)}\\\\ &\\leq \\max(\\phi(-\\cos(\\theta) v),\\phi(\\cos(\\theta) v))\\\\ &= \\max(\\phi(- w+\\sin(\\theta)p),\\phi( w-\\sin(\\theta)p))\\\\ &=w^\\intercal Aw+\\sin^2(\\theta)p^\\intercal Ap-2\\sin(\\theta)w^\\intercal Ap+\\sin(\\theta)|p^\\intercal b|\\\\ &\\leq w^\\intercal Aw+ \\sin^2(\\theta)\\lambda_n + 2\\sin(\\theta)\\lambda_n + \\sin(\\theta){\\vert{b}\\vert}\\\\ &\\leq w^\\intercal Aw+ \\sin^2({\\rm d}\\theta) \\lambda_n + 2\\sin({\\rm d}\\theta)\\lambda_n+ \\sin({\\rm d}\\theta){\\vert{b}\\vert}\\\\ &=w^\\intercal Aw + \\mathcal{O}({\\rm d}\\theta). \\end{align*} In the above calculations, we used the fact that |$y^\\intercal Ay\\leq\\lambda_n$| for every |$y$|, |$\\phi\\ge u$| in |$\\overline\\varOmega$| and the fact that |$b^\\intercal w=0$|. Taking the minimum over all unit vectors |$w$| satisfying |$b^\\intercal w=0$| yields the desired result. □ The result we present next is a follow-up to a formal result discussed in Oberman (2007) regarding the similar notion of a uniformly convex envelope. We show that by enforcing a degree of strict convexity along a prescribed set of directions, one can recover a convex function on the entire domain. The subsequent and analogous result for quasiconvexity is natural and requires only a slight modification of the proof. Before proceeding, we need to state the following equivalences, found in Cannarsa & Sinestrari (2004), which will provide a convenient framework for the proof. Lemma 3.7 (Strict convexity and semiconcavity; Cannarsa & Sinestrari, 2004). Suppose |$u\\in C(\\overline\\varOmega)$| and |$C\\ge0$|. The following three conditions are equivalent for strict convexity: (i) |$D^2u(x) \\ge CI$| in the viscosity sense; (ii) |$u - C\\frac{|\\cdot|^2}{2}$| is convex; (iii) |$u(x+h) - 2u(x) + u(x-h) \\geq C\\frac{h^2}2$| for all |$x\\in\\overline\\varOmega$| and |$h>0$|. The following three conditions are equivalent for semiconcavity: (i) |$D^2u(x) \\le CI$| in the viscosity sense; (ii) |$u - C\\frac{|\\cdot|^2}{2}$| is concave; (ii) |$u(x+h) - 2u(x) + u(x-h) \\leq C\\frac{h^2}2$| for all |$x\\in\\overline\\varOmega$| and |$h>0$|. Proposition 3.8 (Recovering convexity). Suppose |$u\\in C(\\overline\\varOmega)$| and |$\\mathcal{D}$| is a set of directions with directional resolution |${\\rm d}\\theta\\le\\pi/4$|. Suppose there exists |$C>0$| such that |$u$| satisfies the following in the viscosity sense for every |$x\\in\\overline\\varOmega$|: (i) |$D^2u(x)\\le CI$|; (ii) |$v^\\intercal D^2u(x) v \\ge C\\,{\\rm d}\\theta^2$| for every |$v\\in\\mathcal{D}$|. Then |$\\lambda_1[u]\\ge0$|. Proof. Suppose |$y$| is an arbitrary unit vector and |$x\\in\\overline\\varOmega$|. Without loss of generality we can assume that |$x=0$|. It suffices to show $\\frac{u(hy)+u(-hy)}{2}-u(0) \\ge 0\\quad\\text{for every} h>0.$ We may assume |$hy$| lies between some |$d_1,d_2$| such that |$d_i/{\\vert{d_i}\\vert}\\in\\mathcal{D}$| for each |$i$|, as shown in Fig. 4. Then, by the first hypothesis and Lemma 3.7, we can write \\begin{equation} \\frac{u(d_i)+u(-d_i)}{2} - u(0) \\ge \\frac{C\\,{\\rm d}\\theta^2{\\vert{d_i}\\vert}^2}{2},\\quad i=1,2. \\end{equation} (3.3) Next, define |$h_i\\equiv {\\vert{d_i-y}\\vert}$| and |$e_i \\equiv \\frac{d_i-y}{h}$| for |$i=1,2$|. Then we have the following inequalities by the second hypothesis and Lemma 3.7: \\begin{align} \\frac{h_1}{h_1+h_2}u(y+h_2e_2) + \\frac{h_2}{h_1+h_2}u(y-h_1e_1) - u(hy) &\\le C\\frac{h_1h_2}{2}, \\-2pt] \\end{align} (3.4) \\begin{align} \\frac{h_1}{h_1+h_2}u(-y-h_2e_2) + \\frac{h_2}{h_1+h_2}u(-y+h_1e_1) - u(-hy) &\\le C\\frac{h_1h_2}{2}. \\end{align} (3.5) Averaging (3.4) and (3.5), and substituting |d_i = y+he_i| yields \\begin{align} \\frac{h_2}{h_1+h_2}\\left(\\frac{u(d_1)+u(-d_1)}{2}\\right)+\\frac{h_1}{h_1+h_2}\\left(\\frac{u(d_2)+u(-d_2)}{2}\\right) -\\frac{u(hy)+u(-hy)}{2} \\le C\\frac{h_1h_2}{2}. \\end{align} (3.6) Now, weighting each equation in (3.3) by |h_i/(h_1+h_2)| (respectively) and averaging them yields \\begin{align} & \\frac{h_2}{h_1+h_2}\\left(\\frac{u(d_1)+u(-d_1)}{2}\\right) + \\frac{h_1}{h_1+h_2}\\left(\\frac{u(d_2)+u(-d_2)}{2}\\right) - u(0)\\nonumber\\\\[-2pt] &\\quad{} \\ge \\frac{C\\,{\\rm d}\\theta^2}{2}\\left( \\frac{h_1}{h_1+h_2}{\\vert{d_1}\\vert}^2+ \\frac{h_2}{h_1+h_2}{\\vert{d_2}\\vert}^2\\right)\\!. \\end{align} (3.7) Subtracting (3.6) from (3.7) recovers \\[ \\frac{u(hy)+u(-hy)}{2}-u(0) \\ge \\frac{C\\,{\\rm d}\\theta^2}{2}\\left( \\frac{h_1}{h_1+h_2}{\\vert{d_1}\\vert}^2+ \\frac{h_2}{h_1+h_2}{\\vert{d_2}\\vert}^2\\right)-C\\frac{h_1h_2}{2}. Defining |$\\theta_i\\equiv \\cos^{-1}(y^\\intercal d_i)$|, the worst-case scenario occurs when |$\\theta_1=\\theta_2$|, so that |${\\vert{d_1}\\vert}={\\vert{d_2}\\vert}\\equiv{\\vert{d}\\vert}$| and |$h_1=h_2\\equiv h_0$|. Then the above inequality simplifies to $\\frac{u(hy)+u(-hy)}{2}-u(0) \\ge \\frac{C}{2}\\left({\\vert{d}\\vert}^2\\,{\\rm d}\\theta^2 - h_0^2\\right)\\!.$ Now, referring to Fig. 4, we have |$h_0 = {\\vert{d}\\vert}\\sin({\\rm d}\\theta)$|, so that we get $\\frac{u(hy)+u(-hy)}{2}-u(0) \\ge \\frac{C{\\vert{d}\\vert}^2}{2}\\left({\\rm d}\\theta^2 - \\sin({\\rm d}\\theta)^2\\right)\\!,$ yielding the desired result since |${\\rm d}\\theta\\ge\\sin({\\rm d}\\theta)$|. □ Fig. 4. View largeDownload slide Illustration of the proof of recovering convexity. Fig. 4. View largeDownload slide Illustration of the proof of recovering convexity. Proposition 3.9 (Recovering quasiconvexity). Suppose |$u\\in C(\\overline\\varOmega)$|, |$\\alpha\\in\\mathbb{R}$| and |$\\mathcal{D}$| is a set of directions with directional resolution |${\\rm d}\\theta\\le\\pi/4$|. Suppose there exists |$C>0$| such that |$u$| satisfies the following in the viscosity sense for every |$x\\in\\overline\\varOmega$|: (i) |$D^2u\\le CI$|; (ii) |${\\vert{\\!{\\nabla} u}\\vert}\\le M$|; (iii) |$v^\\intercal D^2u v - \\alpha \\ge C\\,{\\rm d}\\theta^2$| whenever |$v\\in\\mathcal{D}$| satisfies |${\\vert{\\!{\\nabla} u^\\intercal v}\\vert} \\le M\\,{\\rm d}\\theta$|. Then |$\\lambda_{\\text QC}[u]\\ge\\alpha$|. Remark 3.10. Note that the third requirement is for |$u$| to be |$(M\\,{\\rm d}\\theta)$| robustly QC along directions in |$\\mathcal{D}$|. Recalling the inequalities (3.1), this can be enforced numerically by solving for subsolutions of the following: $\\alpha+C\\,{\\rm d}\\theta^2+\\sqrt{M\\,{\\rm d}\\theta} - F^{\\sqrt{M\\,{\\rm d}\\theta}}[u] = 0.$ Proof. We begin by taking an arbitrary unit vector |$y$| satisfying |${\\nabla} \\!u^\\intercal y=0$|. As before, we would like to show $\\frac{u(hy)+u(-hy)}{2}-u(0) \\ge 0\\quad\\text{for every}\\,h>0.$ The proof that follows is nearly identical to the proof of convexity. The only difference arises when trying to derive equations (3.3). In this case, we have to show |${\\nabla} \\!u^\\intercal \\left(\\frac{d_i}{{\\vert{d_i}\\vert}}\\right)\\le M{\\rm d}\\theta$| to invoke the proper inequality. This is given by the following argument for each |$i$|: \\begin{equation} {\\nabla}\\!u^\\intercal \\left(\\frac{d_i}{{\\vert{d_i}\\vert}}\\right) = {\\nabla}\\!{u}^\\intercal \\left(\\frac{y+h_ie_i}{{\\vert{d_i}\\vert}}\\right) = \\frac{h_i}{{\\vert{d_i}\\vert}}{\\nabla}\\! u ^\\intercal e_i \\le \\frac{h_i}{{\\vert{d_i}\\vert}}{\\vert{\\!{\\nabla}\\! u}\\vert}{\\vert{e_i}\\vert}\\le \\frac{h_i}{{\\vert{d_i}\\vert}}M, \\end{equation} (3.8) but |$\\frac{h_i}{{\\vert{d_i}\\vert}}=\\sin(\\theta_i)\\le\\sin({\\rm d}\\theta)\\le {\\rm d}\\theta$| for small |${\\rm d}\\theta$|. Therefore, we have ${\\nabla}\\!{u}^\\intercal \\left(\\frac{d_i}{{\\vert{d_i}\\vert}}\\right) \\le M\\,{\\rm d}\\theta.$ Then, by the third hypothesis, we are now in a position to write $\\frac{u(d_i)+u(-d_i)}{2} - u(x) \\ge \\frac{C{\\vert{d_i}\\vert}^2\\theta^2}{2}.$ Proceeding identically as in the proof of Proposition 3.8 yields the desired result. □ 4. Convergent finite difference schemes In what follows, we provide numerical schemes that discretize the above PDEs. As we will see, the schemes we present here fall into a general class of degenerate elliptic finite difference schemes for which there exists a convergence framework. Before we begin, we introduce some notation, which we will carry throughout the rest of the article. We will assume we are working on the hypercube |$[-1,1]^n\\subset{\\mathbb{R}^n}$|. We write |$x=(x_1,\\dots,x_n)\\in[-1,1]^n$|. For simplicity, we discretize the domain |$[-1,1]^n$| with a uniform grid, resulting in the following spatial resolution: $h\\equiv\\frac{2}{N-1},$ where |$N$| is the number of grid points used to discretize |$[-1,1]$|. Note that we use |$h$| here to denote the spatial resolution and it is not to be confused with the |$h$| in the formulations of (ϵ-QC) and (ϵ-QCE). We use the following notation for our computational domain: $\\varOmega^h \\equiv [-1,1]^n\\cap h\\mathbb{Z}^n.$ We define a grid vector, |$v$|, as $v \\equiv \\frac{x-y}{h},\\quad x,y\\in\\varOmega^h.$ 4.1 Finite difference equations and wide stencils Consider a degenerate elliptic PDE |$F[u]=0$|, and its corresponding finite difference equation, $F_\\rho[u]=0,\\quad x\\in\\varOmega^h,$ where |$\\rho$| is the discretization parameter. In our case, we take |$\\rho=(h,\\,{\\rm d}\\theta)$| for the schemes presented hereafter. In general, |$F_\\rho[u]:C(\\varOmega^h)\\to C(\\varOmega^h)$|, where |$C(\\varOmega^h)$| is the set of grid functions |$u:\\varOmega^h\\to\\mathbb{R}$|. We assume the following form: $F_\\rho[u](x) = F_\\rho(u(x),u(x)-u(\\cdot))\\quad\\ \\text{for}\\ x\\in\\varOmega^h,$ where |$u(\\cdot)$| corresponds to the value of |$u$| at points in |$\\varOmega^h$|. For a given set of grid vectors |$V$|, we say |$F_\\rho[u]$| has stencil width|$W$| if for any |$x\\in\\varOmega^h$|, |$F_\\rho[u](x)$| depends only on the values |$u(x + hv)$|, where |$v\\in V$| and |$\\max_{v\\in V}{\\Vert{v}\\Vert}_\\infty\\le W$|. We assume stencils are symmetric about the reference point |$x$|, and have at least width 1. Example 4.1. In two dimensions, the directional resolution can be written as |${\\rm d}\\theta=\\arctan(1/W)/2$|. A stencil of width |$W=1$| would correspond to the vectors |$V=\\{(0,\\pm1), (\\pm1,0), (\\pm1,\\pm1), (\\mp1,\\pm1)\\}$|. The Dirichlet boundary conditions given by (Dir) are incorporated by setting $F_\\rho[u](x) = u(x)-g(x)\\quad\\ \\text{for}\\ x\\in\\partial\\varOmega^h.$ The PDEs we discuss involve a second-order directional derivative and a directional Eikonal operator. We use the following standard discretizations for our schemes: \\begin{align*} {\\vert{u_v^h(x)}\\vert} &\\equiv \\max\\left\\{\\frac{u(x \\pm hv)-u(x)}{h}\\right\\}\\!,\\\\ u_{vv}^h(x) &\\equiv \\frac{u(x+hv)+u(x-hv)-2u(x)}{h^2}, \\end{align*} where |$u_v^h$| and |$u_{vv}^h$| denote the finite difference approximations for the first- and second-order directional derivatives in the grid direction |$v$|. Recall that for any twice continuously differentiable function |$u$|, the standard Taylor series computation yields the following consistency estimate: \\begin{align} {\\vert{u_v^h(x)}\\vert} &= {\\vert{\\nabla u(x)^\\intercal v}\\vert} + \\mathcal{O}(h), \\\\ \\end{align} (4.1) \\begin{align} u_{vv}^h(x) &= v^\\intercal D^2u(x) v + \\mathcal{O}(h^2). \\end{align} (4.2) 4.2 A finite difference method for the PDE We begin by considering the full discretization of |$F^\\epsilon[u]$|, denoted |$F^\\epsilon_{h,{\\rm d}\\theta}[u]$|, where |$h$| is the spatial resolution and |${\\rm d}\\theta$| is the directional resolution of the set of grid vectors |$V$|. We use the spatial discretizations given by (4.1) and (4.2): \\begin{equation} F^\\epsilon_{h,{\\rm d}\\theta}[u](x)\\equiv \\min_{v\\in V}\\left\\{\\tfrac{1}{\\epsilon}{\\vert{u_v^h(x)}\\vert}+u_{vv}^h(x)\\right\\}\\quad \\text{for}\\,x\\in\\varOmega^h. \\end{equation} (4.3) We write the full discretization of (ϵ-Ob) as \\begin{equation} \\begin{cases} \\max\\{u(x)-g(x),\\epsilon - F^\\epsilon_{h,{\\rm d}\\theta}[u](x)\\} = 0, & x\\in\\varOmega^h,\\\\ u(x) = g(x), & x\\in\\partial\\varOmega^h, \\end{cases} \\end{equation} (4.4) where we make the following abbreviation: $G^\\epsilon_{h,{\\rm d}\\theta}[u]\\equiv\\max\\left\\{u-g,\\epsilon-F^\\epsilon_{h,{\\rm d}\\theta}[u]\\right.\\}.$ 4.3 Iterative solution method We implement a fixed-point solver to recover the numerical solution of (4.4). The method is equivalent to a forward Euler time discretization of the parabolic version of (ϵ-QCE), $u_{t} + G^\\epsilon[u]=0\\quad \\text{in}\\,\\varOmega,$ along with |$u = g$| on |$\\partial\\varOmega$| and |$u = u_0$| for |$t=0$|. In particular, the following iterations are performed until a steady state is reached: $\\begin{cases} u^{n+1} = u^{n} - \\delta G^\\epsilon_{h,d\\theta}[u^{n}], & n\\ge0,\\\\ u^{0} = u_0, \\end{cases}$ where satisfies the CFL condition |$\\delta\\le 1/K^{h,\\epsilon}$| (see Oberman, 2006) and |$K^{h,\\epsilon}$| is the Lipschitz constant of the scheme. For our scheme, |$K$| is given by $K^{h,\\epsilon} = \\frac{1}{\\epsilon h} + \\frac{1}{h^2},$ where the first and second terms come from the discretizations of the first- and second-order directional derivatives, respectively. 4.4 Using a first-order discretization In higher dimensions, we need the second-order term in (ϵ-QC) to guarantee quasiconvexity of the solutions. Interestingly, however, for a careful choice of |$\\epsilon$|, the discretization of only the first-order term is consistent with |$F^\\epsilon[u]$|. Indeed, observe that \\begin{align*} -\\min_{v\\in V}\\frac{{\\vert{u_v^h}\\vert}}{\\epsilon}&=-\\min_{v\\in V}\\left\\{\\frac {\\max\\{u(x\\pm hv)-u(x)\\}} {\\epsilon h}\\right\\}\\\\ &= -\\min_{v\\in V}\\left\\{ \\frac{{\\vert{\\!{\\nabla}\\! u(x)^\\intercal v}\\vert}}{\\epsilon} + \\frac{h}{2\\epsilon}v^\\intercal D^2u(x)v \\right\\}+ \\mathcal{O}\\left(\\frac{h^2}{\\epsilon}\\right). \\end{align*} Choosing |$\\epsilon = h/2$| results in $-\\min_{v\\in V}\\frac{{\\vert{u_v^h}\\vert}}{\\epsilon} = -\\min_{v\\in V}\\left\\{ \\frac{{\\vert{\\!{\\nabla}\\! u(x)^\\intercal v}\\vert}}{h/2} + v^\\intercal D^2u(x)v \\right\\}+ \\mathcal{O}(h).$ Thus, for |$\\epsilon=h/2$|, we see for fixed |$(h,{\\rm d}\\theta)$| that the discretization of the first-order term approximates |$F^\\epsilon[u]$|. Moreover, the time step in the above forward Euler discretization simplifies to |$\\delta=\\mathcal{O}(h^2)$|. 5. Convergence of numerical solutions In this section we introduce the notion of degenerate elliptic schemes and show that the solutions of the proposed numerical schemes converge to the solutions of (ϵ-Ob) as the discretization parameters tend to 0. The standard framework used to establish convergence is that of Barles & Souganidis (1991), which we state below. In particular, it guarantees that the solutions of any monotone, consistent and stable scheme converge to the unique viscosity solution of the PDE. 5.1 Degenerate elliptic schemes Consider the Dirichlet problem for the degenerate elliptic PDE |$F[u]=0$|, and recall its corresponding finite difference formulation: $\\begin{cases} F_\\rho(u(x),u(x)-u(\\cdot)) = 0, & x\\in\\varOmega^h,\\\\ u(x)-g(x) = 0, & x\\in\\partial\\varOmega^h, \\end{cases}$ where |$\\rho$| is the discretization parameter. Definition 5.1. The PDE |$F_\\rho[u]$| is a degenerate elliptic scheme if it is nondecreasing in each of its arguments. Remark 5.2. Although the convergence theory is originally stated in terms of monotone approximation schemes (schemes with non-negative coefficients), ellipticity is an equivalent formulation for finite difference operators (Oberman, 2006). In fact, degenerate ellipticity implies monotonicity and stability in the nonexpansive norm. Definition 5.3. The finite difference operator |$F_\\rho[u]$| is consistent with |$F[u]$| if for any smooth function |$\\phi$| and |$x\\in\\varOmega$| we have $F_\\rho[\\phi](x)\\to F[\\phi](x)\\quad \\text{as}\\,\\rho\\to0.$ Definition 5.4. The finite difference operator |$F_\\rho[u]$| is stable if there exists |$M>0$| independent of |$\\rho$| such that if |$F_\\rho[u]=0$| then |$\\| u \\|_\\infty\\le M$|. Remark 5.5. (Interpolating to the entire domain). The convergence theory assumes that the approximation scheme and the grid function are defined on all of |$\\varOmega$|. Although the finite difference operator acts only on functions defined on |$\\varOmega^h$|, we can extend such functions to |$\\varOmega\\setminus{\\it{\\Omega}}^h$| via piecewise interpolation. In particular, performing piecewise linear interpolation maintains the ellipticity of the scheme as well as all other relevant properties. Therefore, we can safely interchange |${\\it{\\Omega}}^h$| and |$\\varOmega$| in the discussion of convergence without any loss of generality. 5.2 Convergence of numerical approximations Next we will state the theorem for convergence of approximation schemes, tailored to elliptic finite difference schemes, and demonstrate that the proposed schemes fit in the desired framework. In particular, we will show that the schemes are elliptic, consistent and have stable solutions. Proposition 5.6 (Convergence of approximation schemes; Barles & Souganidis, 1991). Consider the degenerate elliptic PDE |$F[u]=0$|, with Dirichlet boundary conditions for which there exist a strong comparison principle. Let |$F_\\rho[u]$| be a consistent and elliptic scheme. Furthermore, assume that the solutions of |$F_\\rho[u]=0$| are bounded independently of |$\\rho$|. Then |$u^\\rho\\to u$| locally uniformly on |$\\varOmega$| as |$\\rho\\to0$|. Remark 5.7. Proposition 5.6 is an instance of the meta-theorem, which says that consistency and stability imply convergence. The proposition can also be seen as a special case of the fact that viscosity solutions are stable under perturbations of both the data and the operators, provided that the equation remains in the class of degenerate elliptic PDEs. The classical example of a perturbation is to add the Laplacian (or viscosity) to the operator: |$F^\\epsilon = F + \\epsilon {\\it{\\Delta}}$|, which is the origin of the nomenclature ‘viscosity solutions’. The finite difference approximation of |$F$| can also be regarded as a perturbation of the operator. Provided that the approximation is (i) consistent and (ii) degenerate elliptic (Oberman, 2006) or monotone and stable (Barles & Souganidis, 1991) then the perturbation converges. Lemma 5.8. (Ellipticity). The scheme given by (4.4) is elliptic. Proof. The negatives of each of the spatial discretizations above are elliptic for fixed |$v$|, so taking the minimum over all |$v$| maintains the ellipticity. Therefore, the discretization of |$F^\\epsilon_{h,{\\rm d}\\theta}[u]$| is elliptic. Moreover, the obstacle term is trivially elliptic. Hence |$G^\\epsilon_{h,{\\rm d}\\theta}[u]$|, being the maximum of two elliptic schemes, is also elliptic. □ Lemma 5.9 (Consistency). Given a smooth function |$u$| and a set of grid vectors |$V$| with directional resolution |${\\rm d}\\theta\\le\\pi/4$|, we have the following estimate: \\begin{equation} G^\\epsilon_{h,{\\rm d}\\theta}[u] - G^\\epsilon[u] = \\mathcal{O}(h+{\\rm d}\\theta). \\end{equation} (5.1) In other words, the scheme (4.4) is consistent with (ϵ-Ob). Proof. It suffices to show that |$F^\\epsilon_{h,{\\rm d}\\theta}[u] - F^\\epsilon[u] = \\mathcal{O}(h+{\\rm d}\\theta)$|. Fix |$x\\in\\overline\\varOmega$| and let |$u$| be a smooth function on |$\\overline{\\varOmega}$|. Define $w\\equiv{\\text{argmin}}_{{\\vert{v}\\vert}=1}\\left\\{ \\tfrac{1}{\\epsilon}{\\vert{\\nabla u(x)^\\intercal v}\\vert} + v^\\intercal D^2u(x) v\\right\\}.$ Let |$v\\in V$| and make the following decomposition: $v = \\cos(\\theta)w + \\sin(\\theta)p,$ for some unit vector |$p$| orthogonal to |$w$|. Performing a similar calculation to the proof in Proposition 3.6, we obtain the following consistency estimate from the directional resolution: $F^\\epsilon[u] \\le\\min_{v\\in V}\\left\\{\\tfrac{1}{\\epsilon}{\\vert{\\nabla u^\\intercal v}\\vert} + v^\\intercal D^2u v\\right\\} \\le F^\\epsilon[u] +\\mathcal{O}({\\rm d}\\theta).$ Finally, we conclude (5.1) by recalling the standard Taylor series consistency estimate from equations (4.1) and (4.2). □ Next, we establish stability of our numerical solutions by applying a discrete comparison principle for strict subsolution and supersolution. In particular, we use the following result. Proposition 5.10. (Discrete comparison principle for strict subsolutions). Let |$F_\\rho[u]$| be a degenerate elliptic scheme defined on |$\\varOmega^h$|. Then for any grid functions |$u,v$| we have $F_\\rho[u]<F_\\rho[v]\\implies u\\le v\\quad\\text{in}\\,\\varOmega^h.$ Proof. Suppose for contradiction that $\\max_{x\\in{\\it{\\Omega}}^h}\\{u(x) - v(x)\\} = u(x_0) - v(x_0) > 0.$ Then, $u(x_0) - u(x) > v(x_0) - v(x)\\quad\\text{for every}\\,x\\in\\varOmega^h,$ and so by ellipticity (Lemma 5.8), $F_\\rho[u](x_0)\\ge F_\\rho[v](x_0),$ which is the desired contradiction. □ Lemma 5.11 (Stability). The numerical solutions of (4.4) are stable. Proof. This a direct application of Proposition 5.10. Indeed, suppose |$u$| is a numerical solution such that |$G^\\epsilon_{h,{\\rm d}\\theta}[u]=0$|. Define |$g_1(x)\\equiv C+2\\epsilon^2 \\sum_{i=1}^nx_i$|, where |$C\\in\\mathbb{R}$| is chosen so that |$g_1(x)<g(x)$| for every |$x\\in\\overline\\varOmega$|. Moreover, one can check that |$(g_1)_{vv}^h(x)=0$|. Then, we observe \\begin{align*} G^\\epsilon_{h,{\\rm d}\\theta}[g_1](x) &= \\max\\left\\{g_1(x)-g(x),\\epsilon-\\min_{v\\in V}\\left\\{ \\frac{\\max\\{g_1(x\\pm hv)\\}-g_1(x)}{\\epsilon h} \\right\\}\\right\\}\\\\ &= \\max\\left\\{g_1(x)-g(x),\\epsilon-2\\epsilon\\min_{v\\in V}\\left\\vert\\sum\\nolimits_{i=1}^nv_i\\right\\vert\\right\\}\\\\ &= \\max\\left\\{g_1(x)-g(x),-\\epsilon\\right\\}, \\end{align*} where the last equality follows from the general assumption that the stencil width is at least 1 with directional resolution |${\\rm d}\\theta\\le\\pi/4$|. Next, define |$g_2(x)\\equiv\\max_{x\\in\\overline{\\it{\\Omega}}}g(x)$|. Then, in summary we have |$G^\\epsilon_{h,{\\rm d}\\theta}[g_1]<0$| and |$G^\\epsilon_{h,{\\rm d}\\theta}[g_2]\\ge0$|. Therefore, by Proposition 5.10, $\\min_{x\\in\\overline\\varOmega} g_1(x)\\le u(x)\\le g_2 \\quad\\text{for}\\ x\\in\\varOmega^h.$ Moreover, this bound is independent of |$h$| and |${\\rm d}\\theta$|. □ Proposition 5.12. The solutions of (4.4) converge locally uniformly on |$\\varOmega$| to the solutions of (ϵ-Ob) as |$h,{\\rm d}\\theta\\to0$|. Proof. By Lemmas 5.8, 5.9, 5.11 we see that (4.4) is a consistent and elliptic scheme which has stable solutions. Moreover, a weak comparison principle holds by Corollary 2.5. Therefore, by Proposition 5.6, the numerical solutions converge uniformly on compact subsets of |$\\varOmega$|. □ 6. Numerical results In this section we present the numerical results. The tolerance for the fixed-point iterations was taken to be |$10^{-6}$| and unless otherwise stated we set |$\\epsilon=h/2$|, where |$h$| is the spatial resolution of the grid. Technical conditions on |$g$| given by Remark 2.1 are needed to ensure that the boundary conditions are held in the strong sense. In practice, we violated this condition, resulting in solutions that were discontinuous at the boundary. Two natural choices for initialization of the solution are (i) the obstacle function, |$g(x)$| and (ii) a QC function below |$g$|. We found that the first choice leads to very slow convergence. In particular, the parabolic equation takes time |$\\mathcal{O}(1/\\epsilon)$| to converge. See Example 6.3 below. On the other hand, the second choice results in faster convergence: the simplest choice is simply the constant function with value the minimum of |$g$|. After one step of the iteration the boundary values are attained. Moreover, starting with this choice allows us to use the iterative method with |$\\epsilon = 0$| to find the QCE. Remark 6.1 It is an open question whether solving the |$\\epsilon=0$| time-dependent PDE with quasiconvex initial data below |$g$| will converge. To apply our convergence results, we take |$\\epsilon = h/2$|. 6.1 Results in one dimension We present examples demonstrating the convergence of approximate QCEs to the true QCE as |$\\epsilon\\to0^+$|. We also compare the iteration count when starting from below and at the obstacle. Example 6.2 (Convergence as |$\\epsilon\\to0$|). We consider the convergence of numerical solutions of (4.4) as |$\\epsilon\\to 0$| with the following obstacle function: $g(x)=\\min\\{{\\vert{x-0.5}\\vert},{\\vert{x+0.5}\\vert}-0.3\\}.$ The results are displayed in Fig. 5. Indeed, as expected we see convergence to the true QCE, from below, as |$\\epsilon\\to0$|. Fig. 5. View largeDownload slide Obstacles (solid) and the numerical solutions (dashed). Left: Example 6.2. Convergence in |$\\epsilon$|. Bottom to top: |$\\epsilon = 0.2\\,,0.1\\,,0.05\\,,0.001\\,,0.0001$|. Right: Example 6.3. Fixed-point iterations when |$u^0 \\equiv g_2$|. Fig. 5. View largeDownload slide Obstacles (solid) and the numerical solutions (dashed). Left: Example 6.2. Convergence in |$\\epsilon$|. Bottom to top: |$\\epsilon = 0.2\\,,0.1\\,,0.05\\,,0.001\\,,0.0001$|. Right: Example 6.3. Fixed-point iterations when |$u^0 \\equiv g_2$|. Example 6.3 (Visualization of iterations). Next, we consider the obstacle $g(x)= -x^2+1,$ whose QCE is simply |$g_2^{\\rm QCE}\\equiv0$|. We demonstrate the evolution of the iterations when the initial data are taken to be the obstacle. In this case, the solution corresponded closely to |$u(x,t) = \\min(g(x), c - \\epsilon t)$|, where |$c = \\max g$|. This illustrates the slow speed of convergence, and the fact that the equation degenerates to a trivial operator as |$\\epsilon \\to 0$|. Results are displayed in Fig. 5. 6.2 Numerical results in two dimensions All examples take the initial function in the iterative scheme to be the minimum value |$g_{\\rm min} = \\min g$| of the obstacle function. In all contour plots, the solid line represents the level sets of the original function and the dashed line represents the same level sets of the numerical solution. Unless otherwise stated, the two-dimensional numerical solutions shown are computed on a |$64\\times64$| grid, for illustration purposes. Computations were performed on larger-sized grids. We performed the same numerical experiments in Abbasi & Oberman (2016), and achieved similar results. This is expected, since solving our PDE with small |$\\epsilon$| is consistent with the QCE. We do not reproduce those results here to save space. The examples we present focus on the difference between the two operators for values of |$e$| close to 1. Example 6.4. (Uniform convexification of level sets). Let |$g(x)$| be the signed distance function (negative on the inside) to the square, |$S = \\left\\{x \\mid \\max_{i}{\\vert{x_i}\\vert}=1/2 \\right\\}$|. Although |$g$| is convex, its level sets are not uniformly convex. We apply the iterative procedure to |$g$| with |$\\epsilon=0.5$| so that we uniformly convexify the level sets. The results demonstrating this are displayed in Fig. 6. Fig. 6. View largeDownload slide Left: Examples 6.4, square level sets become uniformly convex with |$\\epsilon = 1/2$|. Middle: Example 6.5, the solution with |$\\epsilon=1$| is indicated by the dashed contours. Right: Example 6.6 with |$\\epsilon$| near 0. Fig. 6. View largeDownload slide Left: Examples 6.4, square level sets become uniformly convex with |$\\epsilon = 1/2$|. Middle: Example 6.5, the solution with |$\\epsilon=1$| is indicated by the dashed contours. Right: Example 6.6 with |$\\epsilon$| near 0. Example 6.5 (Nonconvex signed distance function). In this example, |$g(x)$| is the signed distance function to the curve |${\\it{\\Gamma}}$|, \\begin{align*} {\\it{\\Gamma}}(t) &= \\frac{1}{2}\\begin{cases} (\\cos(t-\\frac{\\pi}{2}),\\sin(t-\\frac{\\pi}{2})), & t\\in[0,3\\pi/2],\\9pt] \\left(0,\\frac{t-7\\pi/4}{\\pi/4}\\right)\\!, & t\\in[3\\pi/2,7\\pi/4], \\\\[9pt] \\left(\\frac{7\\pi/4-t}{\\pi/4},0\\right)\\!, & t\\in[7\\pi/4,2\\pi]. \\end{cases} \\end{align*} We compare the results of using different |\\epsilon|. In particular, we choose |\\epsilon=h/2| and |\\epsilon=1|. In the latter case, we see that the level sets are more curved. Results are displayed in Fig. 6. Example 6.6. We consider the obstacle where |g| is a cone with circular portions removed from its level sets. Take \\[ g(x) = \\begin{cases} 1, &x\\in A,\\\\ {\\vert{x}\\vert}-1/2, &x\\in {\\it{\\Omega}}\\setminus A, \\end{cases} where |$A=\\{(x_1\\pm1/2)^2+x_2^2\\le 1/16\\}\\bigcup \\{x_1^2+(x_2\\pm1/2)^2\\le1/16\\}$|. Results showing the |$g=0.7$| level set are found in Fig. 6. Example 6.7 (Comparison to the |$\\epsilon$|-robust QCE). We compare the solutions of our PDE to the solutions of the line solver presented in Abbasi & Oberman (2016), which returns the robust QCE. In particular, we use a stencil width |$W=5$|, we used the same directions in the line solver and we set |$\\epsilon=0.02$| for the |$\\epsilon$|-uniformly QCE, and we also used a regularization of |$0.02$| for the robust QC (in principle we should have used |$0.02^2$| but the matching value was better for illustration purposes). The results are found in Fig. 7. Fig. 7. View largeDownload slide Example 6.7. Top: Surface plots (inverted) of the function, robust QCE, |$\\epsilon$|-uniformly QCE. Bottom: Corresponding contour plots. Fig. 7. View largeDownload slide Example 6.7. Top: Surface plots (inverted) of the function, robust QCE, |$\\epsilon$|-uniformly QCE. Bottom: Corresponding contour plots. 6.3 Accelerating iterations using the line solver We found an effective method to reduce the computational time to find the solution. We implement the line solver for the robust QCE proposed in Abbasi & Oberman (2016), alternating with the PDE iterations. In particular, on an |$n\\times n$| grid, after every |$2n$| iterations of the PDE we apply one iteration of the line solver (with the commensurate value of |$\\epsilon$|) in each direction. Table 1 shows the number of iterations required for convergence using the same obstacle from Abbasi & Oberman (2016, Example 6.1). Note the line solver is for a different regularization of the QC operator; however, for small values of |$\\epsilon,$| the solutions are close. For larger values of |$\\epsilon,$| the operator still accelerates the solver, but it does not approach the solution to within arbitrary precision. Table 1 Number of iterations required for convergence using different solution methods |$n$| 32 64 128 256 2n PDE iterations 50 102 203 398 2n PDE It + 1 line solve 6 8 10 7 |$n$| 32 64 128 256 2n PDE iterations 50 102 203 398 2n PDE It + 1 line solve 6 8 10 7 Table 1 Number of iterations required for convergence using different solution methods |$n$| 32 64 128 256 2n PDE iterations 50 102 203 398 2n PDE It + 1 line solve 6 8 10 7 |$n$| 32 64 128 256 2n PDE iterations 50 102 203 398 2n PDE It + 1 line solve 6 8 10 7 In Table 1, we compare the number of iterations required for each method. The results were comparable across different examples. The number of iterations of the PDE operator required for convergence was typically a small constant times |$n^2$|. The computational cost of a single line solve was on the order of a small constant (say 10) times the cost of a single PDE iteration. So the combined method, requires |$\\mathcal{O}(n)$| iterations (possibly with a |$\\log n$| prefactor, which we ignore, since it is not significant at these values of |$n$|). So the combined method required (roughly) |$n$| iterations. Recall that there are |$N =n^2$| grid points. Each iteration has a cost proportional to the number of directions used and to the number of grid points, so the total cost is |$\\mathcal{O}\\left((WN^2) \\right)$| for the iterative method and |$\\mathcal{O}(WN)$| for the combined method (with constants of roughly 1 and 10, respectively). So combining the iterative solver with the line solver results in significant improvements to computation time. References Abbasi B. & Oberman A. M. ( 2016 ) Computing the quasiconvex envelope using a nonlocal line solver. Available at https://arxiv.org/abs/1612.05584. Avriel M. , Diewert W. E. , Schaible S. , & Zang I. ( 2010 ) Generalized concavity . Society for Industrial and Applied Mathematics . Bardi M. & Mannucci P. ( 2006 ) On the Dirichlet problem for non-totally degenerate fully nonlinear elliptic equations. Commun. Pure Appl. Anal. , 5 , 709 – 731 . Google Scholar Crossref Search ADS Bardi M. & Mannucci P. ( 2013 ) Comparison principles and Dirichlet problem for fully nonlinear degenerate equations of Monge–Ampère type. Forum Math , 25 , 1291 – 1330 . Google Scholar Crossref Search ADS Barles G. & Souganidis P. E. ( 1991 ) Convergence of approximation schemes for fully nonlinear second order equations. Asymptotic Anal. , 4 , 271 – 283 . Barron E. N. , Goebel R. & Jensen R. R. ( 2012a ) Functions which are quasiconvex under linear perturbations. SIAM J. Optim. , 22 , 1089 – 1108 . Google Scholar Crossref Search ADS Barron E. N. , Goebel R. & Jensen R. R. ( 2012b ) The quasiconvex envelope through first-order partial differential equations which characterize quasiconvexity of nonsmooth functions. Discrete Contin. Dyn. Syst. Ser. B , 17 , 1693 – 1706 . Google Scholar Crossref Search ADS Barron E. , Goebel R. & Jensen R. ( 2013 ) Quasiconvex functions and nonlinear pdes. Trans. Am. Math. Soc. , 365 , 4229 – 4255 . Google Scholar Crossref Search ADS Barron E. N. & Jensen R. R. ( 2014 ) A uniqueness result for the quasiconvex operator and first order PDEs for convex envelopes. In Annales de l’Institut Henri Poincare (C) Non Linear Analysis , vol. 31 . Elsevier Masson , pp. 203 – 215 . Boyd S. & Vandenberghe L. ( 2004 ) Convex Optimization . Cambridge : Cambridge University Press . Caffarelli L. A. & Spruck J. ( 1982 ) Convexity properties of solutions to some classical variational problems. Comm. Partial Differential Equations , 7 , 1337 – 1379 . Google Scholar Crossref Search ADS Cannarsa P. & Sinestrari C. ( 2004 ) Semiconcave Functions, Hamilton-Jacobi Equations, and Optimal Control , vol. 58. Springer Science & Business Media . Carlier G. & Galichon A. ( 2012 ) Exponential convergence for a convexifying equation. ESAIM Control Optim. Calc. Var. , 18 , 611 – 620 . Google Scholar Crossref Search ADS Colesanti A. & Salani P. ( 2003 ) Quasi-concave envelope of a function and convexity of level sets of solutions to elliptic equations. Math. Nachr. , 258 , 3 – 15 . Google Scholar Crossref Search ADS Crandall M. G. , Ishii H. , & Lions P. L. ( 1992 ) User’s guide to viscosity solutions of second order partial differential equations. Bull. Amer. Math. Soc. , 27 , 1 – 67 . Google Scholar Crossref Search ADS Froese B. ( 2016 ) Convergent approximation of surfaces of prescribed Gaussian curvature with weak Dirichlet conditions. Available at https://arxiv.org/abs/1601.06315. Kawohl B. ( 1985 ) Rearrangements and Convexity of Level Sets in PDE . Lecture Notes in Mathematics , vol. 1150 . Berlin, Heidelberg : Springer-Verlag , pp. 1 – 134 . Kohn R. & Serfaty S. ( 2007 ) Second-order PDE’s and deterministic games. In Proc. Internat. Congress Ind. Appl. Math. (ICIAM’07) , pp. 239 – 249 . Lions P.-L. ( 1998 ) Identification du cône dual des fonctions convexes et applications. C. R. Acad. Sci. Paris Sér. I Math. , 326 , 1385 – 1390 . Google Scholar Crossref Search ADS Oberman A. M. ( 2004 ) A convergent monotone difference scheme for motion of level sets by mean curvature. Numer. Math. , 99 , 365 – 379 . Google Scholar Crossref Search ADS Oberman A. M. ( 2006 ) Convergent difference schemes for degenerate elliptic and parabolic equations: Hamilton-Jacobi equations and free boundary problems. SIAM J. Numer. Anal. , 44 , 879 – 895 . Google Scholar Crossref Search ADS Oberman A. M. ( 2007 ) The convex envelope is the solution of a nonlinear obstacle problem. Proc. Amer. Math. Soc. , 135 , 1689 – 1694 . Google Scholar Crossref Search ADS Oberman A. M. ( 2008a ) Computing the convex envelope using a nonlinear partial differential equation. Math. Models Methods Appl. Sci. , 18 , 759 – 780 . Google Scholar Crossref Search ADS Oberman A. M. ( 2008b ) Wide stencil finite difference schemes for the elliptic Monge-Ampère equation and functions of the eigenvalues of the Hessian. Discrete Contin. Dyn. Syst. Ser. B , 10 , 221 – 238 . Google Scholar Crossref Search ADS Sethian J. A. ( 1999 ) Level Set Methods and Fast Marching Methods: Evolving Interfaces in Computational Geometry, Fluid Mechanics, Computer Vision, and Materials Science , vol. 3 . Cambridge University Press . © The authors 2017. Published by Oxford University Press on behalf of the Institute of Mathematics and its Applications. All rights reserved. This article is published and distributed under the terms of the Oxford University Press, Standard Journals Publication Model (https://academic.oup.com/journals/pages/open_access/funder_policies/chorus/standard_publication_model) http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png IMA Journal of Numerical Analysis Oxford University Press\n\n# A partial differential equation for the $\\epsilon$-uniformly quasiconvex envelope\n\nIMA Journal of Numerical Analysis, Volume 39 (1) – Jan 25, 2019\n26 pages",
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"/lp/oxford-university-press/a-partial-differential-equation-for-the-epsilon-uniformly-quasiconvex-FD0Rd99iZq\nPublisher\nOxford University Press\nISSN\n0272-4979\neISSN\n1464-3642\nDOI\n10.1093/imanum/drx068\nPublisher site\nSee Article on Publisher Site\n\n### Abstract\n\nAbstract Quasiconvex (QC) functions are functions whose level sets are convex. In a series of articles, Barron, Goebel and Jensen studied partial differential equations (PDEs) for QC functions. They introduced a regularization of the PDE that is more stable. In this article, we introduce a different regularization that is more amenable to numerical approximation. We prove existence and uniqueness of viscosity solutions of the PDE and demonstrate that level sets of solutions of the new PDE are uniformly convex. We build convergent finite difference approximations, comparing the QC envelope with the regularization. 1. Introduction In a series of articles from about four years ago, Barron, Goebel and Jensen introduced and studied partial differential equations (PDEs) for quasiconvexity (Barron et al., 2012a,b, 2013; Barron & Jensen, 2014). In this context, quasiconvexity means that the sublevel sets of a function are convex. The study of convexity of level sets for obstacle problems has a long history, which includes Caffarelli & Spruck (1982) and Kawohl (1985), as well as the more recent work by Colesanti & Salani (2003) and the references therein. Quasiconvex (QC) functions appear naturally in optimization, since they generalize convex functions. The property also appears in economics (Avriel et al., 2010). Earlier work by one of the authors studied a PDE for the convex envelope (Oberman, 2007), which led to a numerical method for convex envelopes (Oberman, 2008a,b). The notion of (scalar) quasiconvexity that we discuss here should not be confused with the well-known notion of quasiconvexity in vector variational problems. Our motivation for studying this PDE comes from two directions: geometric and algorithmic; it is natural to represent a set using a level set function (Sethian, 1999). The PDE allows us to find the convex hull of a set represented this way directly (in fact, every level set), rather than extracting the level set and then finding the convex hull, followed by generating a new level set function. In addition, we can use the parabolic version of the equation to progressively convexify a level set, or to make already convex level sets uniformly convex. Quasiconvexity is challenging because, unlike convexity, it is a nonlocal property (at least for functions that have flat parts). This means that, even using viscosity solutions, there is no local characterization for quasiconvexity. On the other hand, using the more stable notion of robust quasiconvexity, Barron, Goebel and Jensen showed that these functions are characterized in the viscosity sense (Crandall et al., 1992) by a PDE (Barron et al., 2013; Barron & Jensen, 2014). One motivation for this work was to build numerical solvers for the QC envelope PDE. However, we had difficulties with both the QC and the robust-QC operators: the former lacks uniqueness and the latter uses an operator defined over small slices of angles (see Fig. 3), which leads to poor accuracy when using wide-stencil finite difference schemes. An alternative, presented in Barron et al. (2013), was to use first-order nonlocal PDE solvers. In a companion article (Abbasi & Oberman, 2016), we built a nonlocal solver for the QC and robust-QC envelopes. The nonlocal PDEs can be solved explicitly and the problem implemented efficiently. By iteratively solving for the envelopes on lines in multiple directions, we approximated the solution of the problem in higher dimensions. However, we are still interested in the PDE approach, which has advantages arising from its nonlocal nature. In this article, we build on the results of Barron et al. (2012a,b) and Barron & Jensen (2014) to obtain a PDE for |$\\epsilon$|-uniformly QC functions, which we define subsequently. Note that |$\\epsilon$|-uniform quasiconvexity implies robust quasiconvexity. The choice of terminology lends itself to the fact that |$\\epsilon$|-uniformly QC functions have uniformly convex level sets. Following the argument in Barron & Jensen (2014), we establish uniqueness of viscosity solutions for the PDE. Moreover, this operator is defined for all direction vectors, which makes it amenable to discretization using wide-stencil schemes. We consider the obstacle problem for the |$\\epsilon$|-uniformly quasiconvex envelope (QCE). As is the case for robust-QC, we recover the QCE as the regularization parameter |$\\epsilon \\to 0$|. While robust-QC functions can have corners in one dimension, strictly QC functions are smoother; see Fig. 3 below, and the explicit formula for the solution in one dimension in Section 2.3. We also build and implement convergent elliptic finite difference schemes (Barles & Souganidis, 1991; Oberman, 2006) for the envelopes. These are wide-stencil finite difference schemes, which can be developed using ideas similar to Oberman (2004, 2008a,b). Solutions to these PDEs can be found using an iterative method which is equivalent to the explicit Euler discretization of the parabolic equation (Oberman, 2006). However, the method has a nonlinear Courant-Friedrichs-Lewy (CFL) condition, which restricts the step size. We find that alternating the line solver with several iterations of the parabolic PDE solver significantly improves the speed of the solution. Numerical solutions show that we obtain very similar results to the line solver for the QCE with small |$\\epsilon = h/2$|, where |$h$| is the grid resolution. We also compare large |$\\epsilon$| solutions with comparable robust QCE and find that solutions are smoother. Formally, we show that solutions are |$\\epsilon$| uniformly convex. The QC operator in two dimensions recovers the level set curvature operator. We show that our discretization of our operator agrees with the Kohn–Serfaty (Kohn & Serfaty, 2007) first-order representation of the mean curvature operator in two dimensions. See Section 4.4. 1.1 Convexity of level sets of a function We give a brief informal derivation of the operator. Given a smooth function |$u: {\\mathbb{R}^n} \\to \\mathbb{R}$|, the direction of the gradient at |$x_0$|, |$p = {\\nabla} u(x_0)$| is the normal to the sublevel set |$\\{ x \\in {\\mathbb{R}^n} \\mid u(x) \\leq u(x_0) \\}$|. The curvatures of the level set at |$x_0$| are proportional to the eigenvalues of the Hessian of |$u$|, |$M = D^2u(x_0)$|, projected onto the tangent hyperplane at |$x_0$|, |$P = P_{x_0} = \\{ v \\in {\\mathbb{R}^n} \\mid v\\cdot p = 0\\}$|. These curvatures are all positive if |$v^\\intercal M v \\geq 0$| for all |$v \\in P$|. Thus, formally, the condition for local convexity of the level set is non-negativity of the operator \\begin{equation} L_0(p,M) \\equiv \\min_{{\\vert{v}\\vert} = 1} \\left \\{ v^\\intercal M v \\mid v\\cdot p = 0 \\right. \\}. \\end{equation} (1.1) This is the operator considered in Barron et al. (2013) to study QC functions. However, for technical reasons discussed below, they chose to relax the constraint |$v \\cdot p =0$| to an inequality constraint |${\\vert{v \\cdot p}\\vert} \\leq \\epsilon$|, resulting in the operator $L_\\epsilon(p,M) \\equiv \\min_{{\\vert{v}\\vert} = 1} \\left \\{ v^\\intercal M v \\mid {\\vert{v \\cdot p}\\vert}\\le\\epsilon \\right \\}\\!.$ Our operator is obtained by instead replacing the hard constraint |${\\vert{v\\cdot p}\\vert} \\leq \\epsilon$| with a penalty in the objective function. So we define $F^\\epsilon(p,M) \\equiv \\min_{{\\vert{v}\\vert} = 1}\\left \\{ v^\\intercal M v + \\tfrac{1}{{\\epsilon}} {\\vert{v^\\intercal p}\\vert} \\right \\}\\!.$ This choice of penalty gives the operator for uniformly convex level sets, as we show subsequently. Refer to Figure 2 for a visualization of the feasible sets. Figure 3 illustrates the important differences in the notions of quasiconvexity discussed so far. Moreover, on a bounded domain, the |$\\epsilon$|-robustly QCE can have an interval of global minima, whereas our solution has a unique (global) minimum. 1.2 Basic definitions We recall some basic definitions and establish our notation. For a reference, see Boyd & Vandenberghe (2004). The set |$S$| is convex if whenever |$x$| and |$y$| are in |$S$| then so is the line segment |$[x,y]\\equiv\\{tx + (1-t)y$|: |$t \\in [0,1]\\}$|. We say that the function |$u: {\\mathbb{R}^n} \\to \\mathbb{R}$| is convex if \\begin{equation} u(tx + (1-t)y) \\le t u(x) + (1-t) u(y) \\quad \\text{ for every } x,y \\in {\\mathbb{R}^n}, \\quad 0 \\le t \\le 1. \\end{equation} (1.2) The convex envelope of a function |$g$|, hereby denoted |$g^{\\rm CE}$|, is the largest convex function majorized by |$g$|: \\begin{equation} g^{\\rm CE}(x) \\equiv \\sup \\{ v(x)\\mid v\\,\\text{is convex and}\\,v\\le g \\}. \\end{equation} (1.3) In terms of sets, if |$S$| is given by the sublevel set of a function |$u$|, \\begin{equation} S = S_\\alpha(u)\\equiv\\{x \\in {\\mathbb{R}^n} \\mid u(x)\\leq\\alpha\\}, \\end{equation} (1.4) then convexity of |$S$| is equivalent to the following condition: $u(x), u(y) \\le \\alpha \\implies u(tx + (1-t)y) \\le \\alpha \\quad \\text{ for every } 0 \\le t \\le 1.$ This condition, when applied to every level set |$\\alpha\\in\\mathbb{R}$|, characterizes quasiconvexity of the function |$u$|. That is, |$u$| is QC if every sublevel set |$S_\\alpha(u)$| is convex. Equivalently, |$u$| is QC if \\begin{equation} u(tx + (1-t)y) \\le \\max \\{ u(x), u(y) \\} \\quad \\text{ for every } x,y \\in {\\mathbb{R}^n},\\quad 0 \\le t \\le 1. \\end{equation} (1.5) Given |$g:{\\mathbb{R}^n} \\to \\mathbb{R}$|, the QCE of |$g$| is given similarly by \\begin{equation} g^{\\rm QCE}(x) \\equiv \\sup \\{ v(x)\\mid v\\,\\text{is QC and}\\,v\\le g\\}. \\end{equation} (1.6) Remark 1.1. Since the maximum of any two QC (convex) functions is QC (convex) as well, the suprema in (1.3) and (1.6) are well defined. Figure 1 provides a visual comparison between the convex envelope and the QCE in one dimension. Fig. 1. View largeDownload slide Comparison of a function (solid) and its envelopes (dashed). (a) Quasiconvex envelope. (b) Convex envelope. Fig. 1. View largeDownload slide Comparison of a function (solid) and its envelopes (dashed). (a) Quasiconvex envelope. (b) Convex envelope. Fig. 2. View largeDownload slide Constraint sets (dotted and solid) of different PDEs in two dimensions. Fig. 2. View largeDownload slide Constraint sets (dotted and solid) of different PDEs in two dimensions. Fig. 3. View largeDownload slide Comparison between solutions of different operators. Graphs arranged top to bottom: obstacle, QCE, |$\\epsilon$|-robustly QCE, and |$\\epsilon$|-uniformly QCE. Fig. 3. View largeDownload slide Comparison between solutions of different operators. Graphs arranged top to bottom: obstacle, QCE, |$\\epsilon$|-robustly QCE, and |$\\epsilon$|-uniformly QCE. 1.3 Viscosity solutions Suppose that |$\\varOmega\\subset{\\mathbb{R}^n}$| is a domain. Let |$S^n$| be the set of real symmetric |$n\\times n$| matrices and take |$N\\le M$| to denote the usual partial ordering on |$S^n$|, namely that |$N-M$| is negative semidefinite. Definition 1.2. The operator |$F(x,r,p,M):\\varOmega\\times\\mathbb{R}\\times{\\mathbb{R}^n}\\times S^{n} \\to\\mathbb{R}$| is degenerate elliptic if $F(x,r,p,M)\\le F(x,s,p,N)\\quad\\text{whenever}\\ r\\le s\\ \\text{and}\\ N\\le M.$ Remark 1.3. For brevity, we use the notation |$F[u](x)\\equiv F(x,u(x),\\nabla u(x), D^2u(x))$|. Definition 1.4 (Viscosity solutions). Suppose |$F$| is a degenerate elliptic operator, as defined above. We say that the upper semicontinuous (lower semicontinuous) function |$u:\\varOmega\\to\\mathbb{R}$| is a viscosity subsolution (supersolution) of |$F[u]=0$| in |$\\varOmega$| if for every |$\\phi\\in C^2(\\varOmega)$|, whenever |$u-\\phi$| has a strict local maximum (minimum) at |$x\\in\\varOmega$|, $F(x,u(x),\\nabla\\phi(x),D^2\\phi(x))\\le0\\ (\\ge0).$ Moreover, we say that |$u$| is a viscosity solution of |$F[u]=0$| if |$u$| is both a viscosity subsolution and a supersolution. 1.4 The QCE Use the notation |$\\lambda_{\\rm QC}[u](x) \\equiv L_0({\\nabla} u(x),D^2u(x))$|. The obstacle problem for the QCE is given by \\begin{equation} \\begin{cases} \\max\\{u-g,-\\lambda_{\\rm QC}[u](x)\\}=0, & x\\in\\varOmega,\\\\ u =g, & x\\in\\partial\\varOmega. \\end{cases} \\end{equation} (Ob) This PDE can have multiple viscosity solutions for a given |$g$| (Barron et al., 2013). However, there is a unique QC solution which is the QCE of |$g$|. Failure of uniqueness in general can be seen in the following counterexample, which is similar to Barron et al. (2013, Example 3.1). Example 1.5. Consider |$\\varOmega=\\{x^2+y^2<1\\}, u(x,y) = -y^4, v(x,y) = -(1-x^2)^2$|, so that |$u=v$| on |$\\partial\\varOmega$|. Now consider |$g\\equiv u$| on |$\\varOmega$|. We can calculate that |$\\lambda_{\\rm QC}[u]=\\lambda_{\\rm QC}[v]=0$|. In summary, |$u$| and |$v$| are both solutions of (Ob), but |$u\\neq v$| in |$\\varOmega$| (in fact |$v<u$| in |$\\varOmega$|). This contradicts the uniqueness of (Ob). It is shown in Barron et al. (2013) that continuous QC functions necessarily satisfy |$\\lambda_{\\rm QC}[u] \\geq 0$| in the viscosity sense. The converse, however, is not true: consider the function |$u(x)=-x^4$|, which is not QC yet satisfies |$\\lambda_{\\rm QC}[u]=0$| at |$x=0$|. In fact, using this as a sufficient condition is only possible when |$u$| has no local maxima (Barron et al., 2013). This operator can be used to completely characterize the set of continuous |$\\epsilon$|-robustly QC functions, defined below. Definition 1.6 (|$\\epsilon$|-robustly QC) The function |$u:\\varOmega\\to\\mathbb{R}$| is |$\\epsilon$|-robustly QC if |$u(x) +y^\\intercal x$| is QC for every |${\\vert{y}\\vert}\\leq\\epsilon$|. In particular, |$\\epsilon$|-robustly QC functions are functions whose quasiconvexity is maintained under small linear perturbations. Write |$\\lambda_{\\rm QC}^\\epsilon[u](x) \\equiv L_\\epsilon({\\nabla} u(x),D^2u(x))$|. Proposition 1.7 (Characterization of |$\\epsilon$|-robustly QC functions; Barron et al., 2013). The upper semicontinuous function, |$u:\\varOmega\\to\\mathbb{R}$|, is |$\\epsilon$|-robustly QC if and only if |$u$| is a viscosity subsolution of |$\\lambda_{\\rm QC}^\\epsilon[u]=0$|. Next, we show that viscosity subsolutions (defined below) of our operator are robustly-QC (which also implies they are QC). This allows us to avoid a technical argument from Barron et al. (2013). We believe that stronger results hold: see the formal analysis in Section 3. For simplicity, we first define the following abbreviation. Definition 1.8 (|$\\epsilon$|-uniformly QC). The function |$u\\in {\\rm USC}(\\overline\\varOmega)$| is |$\\epsilon$|-uniformly QC if it is a viscosity subsolution of |$\\epsilon-F^\\epsilon[u]=0$|. Proposition 1.9. Suppose |$u\\in {\\rm USC}(\\overline\\varOmega)$| is |$\\epsilon$|-uniformly QC. Then, |$u$| is |$\\epsilon^2$|-robustly QC. Proof. First observe that for any |$\\phi\\in C^2$| we have the following inequality: \\begin{align*} F^\\epsilon[\\phi](x) - \\frac{\\alpha}{\\epsilon} &= \\min_{{\\vert{v}\\vert}=1}\\left\\{\\tfrac{1}{{\\epsilon}}{\\vert{\\nabla \\phi(x)\\cdot v}\\vert} +v^\\intercal D^2\\phi(x)v - \\tfrac{\\alpha}{\\epsilon} \\right\\}\\\\ &\\leq \\min_{\\{{\\vert{v}\\vert}=1,{\\vert{\\nabla \\phi(x)\\cdot v}\\vert}\\le \\alpha \\}}\\left\\{v^\\intercal D^2\\phi(x)v\\right\\} = \\lambda_{QC}^{\\alpha}[\\phi](x). \\end{align*} Choosing |$\\alpha=\\epsilon^2$| we see that any viscosity subsolution of |$\\epsilon-F^\\epsilon[u]= 0$| is also a viscosity subsolution of |$-\\lambda_{\\rm QC}^{\\epsilon^2}[u]= 0$|. Thus, |$u$| is |$\\epsilon^2$|-robustly QC. □ 2. Properties of solutions In this section, we present technical arguments proving the uniqueness of solutions of our PDEs and discuss some relevant properties. 2.1 Comparison principle In this subsection, we will show that a weak comparison principle holds for the Dirichlet problem of |$h-F^\\epsilon[u]=0$| for |$h>0$| and |$\\epsilon > 0$|. Comparison also holds for the corresponding obstacle problem. The proof we present is based on the uniqueness proof of |$\\lambda_{\\rm QC}[u]=g$| for |$g>0$|, presented in Barron & Jensen (2014). The result is simpler, because our operator is continuous as a function |$(p,M)\\mapsto F^\\epsilon(p,M)$| for |$\\epsilon > 0$|. Write \\begin{equation} F^\\epsilon[u](x)=F^\\epsilon(\\nabla u(x),D^2u(x)) \\equiv \\min_{{\\vert{v}\\vert}=1}\\left\\{\\tfrac{1}{{\\epsilon}}{\\vert{\\nabla u(x)\\cdot v}\\vert} + v^\\intercal D^2u(x)v\\right\\}\\!. \\end{equation} (2.1) Note that |$-F^\\epsilon(p,M)$| is elliptic by Definition 1.2. We consider the following PDEs: (|$\\epsilon$|-QC) \\begin{gather} h(x)-F^\\epsilon[u](x) = 0, \\quad x\\in\\varOmega,\\\\ \\end{gather} (|$\\epsilon$|-QCE) \\begin{gather} \\max\\{u(x)-g(x),h(x)-F^\\epsilon[u](x)\\} = 0, \\quad x\\in\\varOmega, \\end{gather} where |$\\varOmega\\subset{\\mathbb{R}^n}$| is an open, bounded and convex domain, and |$g:\\overline\\varOmega\\to\\mathbb{R}$| is continuous. In the latter equation, |$g$| is the obstacle. We also impose the following condition on |$h$|: \\begin{equation} h:\\overline\\varOmega\\to\\mathbb{R}\\,\\text{is continuous and positive.} \\end{equation} (2.2) Enforce the following Dirichlet boundary data: \\begin{equation} u(x) = g(x),\\quad x\\in\\partial\\varOmega. \\end{equation} (Dir) Remark 2.1 (Continuity up to the boundary). In general, viscosity solutions of (ϵ-QCE) need not be continuous up to the boundary. To apply Barles & Souganidis (1991) for convergence of numerical schemes, we need a strong comparison principle, which requires that solutions be continuous up to the boundary. The following assumption ensures continuity up to the boundary There is a convex domain |$\\varOmega_L\\supset\\varOmega$| and a continuous, QC function |$g_0: \\varOmega_L \\to \\mathbb{R}$|, with $g_0(x)\\le g(x) \\quad \\text{ for } x \\in \\varOmega, \\qquad g_0(x)=g(x) \\quad \\text{ for } x\\in \\varOmega_L\\setminus\\varOmega.$ Continuity follows because we have |$g_0 \\leq u \\leq g,$| which gives |$u = g$| on |$\\partial \\varOmega$|. An alternative to this condition is to prove convergence in a neighbourhood of the boundary. The convergence proof in this setting can be found in Froese (2016). Next, we state a technical, but standard, viscosity solutions result, which gives the comparison principle in the case where we have strict subsolution and supersolution. Proposition 2.2 (Comparison principle for strict subsolutions; Crandall et al., 1992). Consider the Dirichlet problem for the degenerate elliptic operator |$F(p,M)$| on the bounded domain |$\\varOmega$|. Let |$u \\in {\\rm USC}(\\bar \\varOmega)$| be a viscosity subsolution and let |$v \\in {\\rm LSC}(\\bar \\varOmega)$| be a viscosity supersolution. Suppose further that for |$\\sigma > 0$|, \\begin{align*} F[u] + \\sigma &\\le 0 \\text{ in } \\varOmega, \\\\ F[v] &\\ge 0 \\text{ in } \\varOmega \\end{align*} holds in the viscosity sense. Then, the comparison principle holds: $\\text{ if }\\,u\\leq v\\,\\text{on}\\,\\,\\partial\\varOmega\\,\\,\\text{then}\\,u\\le v\\,\\,\\text{in}\\,\\varOmega.$ Remark 2.3. In Crandall et al. (1992, Section 5.C), it is explained how the main comparison theorem (Crandall et al., 1992, Theorem 3.3) can be applied when it is possible to perturb a subsolution to a strict subsolution. This version of the theorem is what we state in Proposition 2.2. This result was used in Bardi & Mannucci (2006, Theorem 3.1) and Bardi & Mannucci (2013) to prove a comparison principle. Using the the same perturbation technique from Barron & Jensen (2014), and consequently applying Proposition 2.2, we obtain the following comparison result. Proposition 2.4 (Comparison principle). Consider the Dirichlet problem given by (ϵ-QC) (Dir) and assume (2.2) holds. Let |$u\\in {\\rm USC}(\\overline\\varOmega)$| be a viscosity subsolution and |$v\\in {\\rm LSC}(\\overline\\varOmega)$| be a viscosity supersolution of (ϵ-QC). Then, the comparison principle holds: $\\text{ if}\\,\\,u\\leq v \\,\\, \\text{on}\\,\\partial\\varOmega \\,\\,\\text{then}\\,u\\le v \\,\\,\\text{in}\\,\\, \\varOmega.$ Proof. We will show that we can perturb |$u$| to a function |$u^\\sigma$| satisfying |$u^\\sigma\\le v$| on |$\\partial\\varOmega$| and $h-F^\\epsilon[u^\\sigma]<0 \\quad\\text{in}\\, \\varOmega$ in the viscosity sense. Applying Proposition 2.2, we will have |$u^\\sigma\\le v$| in |$\\varOmega$|. Taking |$\\sigma\\to0$| yields the desired result. Fix |$\\sigma>0$| and define the following perturbation of |$u$|, and notice that for |$x\\in\\partial\\varOmega$| we have the following relation: $u^\\sigma(x) \\equiv u(x) - \\sigma \\left(\\max_{y\\in\\partial\\varOmega}u(y)-u(x)\\right)\\le u(x) - \\sigma (u(x)-u(x)) = u(x)\\le v(x);$ that is, |$u^\\sigma\\le v$| on |$\\partial\\varOmega$|. Next, because |$u$| is a subsolution, we have $h-\\frac{1}{1+\\sigma}F^\\epsilon[u^\\sigma] \\le 0,$ leading to $h - F^\\epsilon[u^\\sigma]\\le -\\sigma h \\le -\\sigma h_{\\min} <0\\quad\\text{where}\\,h_{\\min}\\equiv \\min_{x\\in\\overline\\varOmega}h(x).$ □ We use this result to prove a weak comparison principle for the obstacle problem given by (ϵ-QCE) and (Dir). Corollary 2.5 (Comparison principle for the obstacle problem). Consider the Dirichlet problem given by (ϵ-QCE), (Dir) and assume (2.2) holds. Let |$u\\in {\\rm USC}(\\overline\\varOmega)$| be a viscosity subsolution and |$v\\in {\\rm LSC}(\\overline\\varOmega)$| be a viscosity supersolution of (ϵ-QCE). Then, the comparison principle holds: $\\text{ if}\\,\\, u\\leq v\\,\\,\\text{on}\\,\\,\\partial\\varOmega\\,\\,\\text{then}\\,\\,u\\le v\\,\\,\\text{in}\\,\\,\\varOmega.$ Proof. We begin by considering the domain |$\\varOmega_g\\equiv\\{x\\in\\varOmega:v(x)<g(x)\\}$|. Then, |$h-F^\\epsilon[v]\\ge 0$| in |$\\varOmega_g$| and |$v=g$| on |$\\partial\\varOmega_g$|, that is, |$v$| is a viscosity supersolution of |$h-F^\\epsilon[v]=0$| in |$\\varOmega_g$|. Now, by the definition of viscosity subsolutions we have |$u\\le g$| and |$h-F^\\epsilon[u]\\le 0$| in |$\\varOmega$| and thus also |$\\varOmega_g$|. This allows us to conclude that |$u\\le v$| on |$\\partial\\varOmega_g$|. Therefore, by Proposition 2.4, |$u\\le v$| in |$\\varOmega_g$|. Concluding, in |$\\varOmega\\setminus\\varOmega_g$| we necessarily have |$v\\ge g\\ge u$|. □ 2.2 The |$\\epsilon$|-uniformly QCE We formulate the |$\\epsilon$|-uniformly QCE of a function |$g$| as the unique viscosity solution of the following obstacle problem: (|$\\epsilon$|-Ob) \\begin{equation} \\begin{cases} \\max\\{u(x)-g(x),\\epsilon-F^\\epsilon[u](x)\\} = 0, & x\\in\\varOmega,\\\\ u(x) = g(x), & x\\in\\partial\\varOmega. \\end{cases} \\end{equation} Remark 2.6 (Convergence of approximate solutions). It is clear that as |$\\epsilon\\to0$|, the penalization term in |$F^\\epsilon[u]$| tends to infinity. The result is that |$F^\\epsilon[u]\\to\\lambda_{\\rm QC}[u]$| as |$\\epsilon\\to0$|. From this observation, the standard stability result of viscosity solutions and the quasiconvexity of subsolutions of |$\\epsilon-F^\\epsilon[u]=0$|, one can then apply the same argument presented in the proof of Barron et al. (2013, Theorem 5.3) to conclude that the unique viscosity solutions of (ϵ-QCE), (Dir) converge to the QCE as |$\\epsilon\\to0$|. This result allows us to compute asymptotic approximations of the QCE of a given obstacle. 2.3 Solution formula in one dimension In one dimension, |$\\epsilon-F^\\epsilon[u]$| is simply the Eikonal operator with a small diffusion term. When considering a solution |$u$| of (ϵ-Ob), whenever |$u(x) < g(x)$|, we have |$F^\\epsilon[u](x) = \\epsilon$|, which gives \\begin{equation} \\epsilon u_{xx} + {\\vert{u_x}\\vert}=\\epsilon^2. \\end{equation} (2.3) Define |$\\varOmega_g\\equiv\\{x:u(x) < g(x)\\}$|. Note that |$\\varOmega_g$| need not be connected; it can be written as the union of finitely many intervals (refer to Fig. 3). However, we can solve the equation in each interval. Lemma 2.7 (One-dimensional solution). The viscosity solution of the one-dimensional PDE for the operator (2.1) and (2.3), along with boundary conditions |$u(0)=0, u(W) = H$|, is given by the following. Set |$S\\equiv H/\\epsilon^2W$|. Then, $u^\\epsilon(x) = \\begin{cases} \\pm \\epsilon^2 x, & S=\\pm1,\\\\ \\epsilon^2x + C^+(1-\\exp(-x/\\epsilon)), & S>1,\\\\ -\\epsilon^2x + C^-(1-\\exp(x/\\epsilon)), & S<-1,\\\\ \\epsilon^2{\\vert{x-x^*}\\vert} + \\epsilon^3(\\exp(-{\\vert{x-x^*}\\vert}/\\epsilon)-1) + u_0, & {\\vert{S}\\vert}<1, \\end{cases}$ where $C^\\pm \\equiv \\frac{H\\mp\\epsilon^2 W}{\\pm\\epsilon(1-\\exp(\\mp W/\\epsilon))}$ and |$u_0\\in\\mathbb{R}$| and |$x^*\\in I = (0,W)$| are constants. Proof. If there exists |$x_0$| in the interval |$I$| such that |$u_x(x_0)=0$| then (2.3) implies |$u_{xx}(x_0)>0$|, that is, if there exists an interior critical point, then |$u$| must be strictly convex in |$I$|. This allows us to break down the analysis of (2.3), restricted to each interval, into several cases: (i) |$u_x<0$| in |$I$|, (ii) |$u_x>0$| in |$I$| and (iii) |$u_x(x_0)=0$| for some |$x_0 \\in I$|. In each case, we can solve a linear second-order ordinary differential equation. The case where |$S = \\pm 1$| is degenerate: the solution is linear. The case where |${\\vert{S}\\vert} > 1$| is not difficult. The final case, where |${\\vert{S}\\vert} < 1$| corresponds to (iii). In this case, |$u_x<0$| for |$x<x_0$| and |$u_x>0$| for |$x>x_0$|. Then, \\begin{equation} u(x) = {\\vert{x-x_0}\\vert} + (\\exp(-{\\vert{x-x_0}\\vert})-1) + u_0. \\end{equation} (2.4) For some |$u_0\\in\\mathbb{R}$|. Taylor expansion shows that |$u(x) = x^2/2 + \\mathcal{O}(x^3)$| for |$x$| near |$x_0$|. Finding |$x_0$| (or |$u_0$|) analytically is infeasible. But we can argue that the solution is correct by a continuity argument. Given the function |$u$| defined by (2.4), define the continuous function |$S_1(y)\\equiv\\frac{u(y+W_1)-u(y)}{W_1}$|. For small |$W_1$| and |$y\\ll x_0$|, we have that |$S_1(y)<-1$| (by the earlier discussion). Similarly for |$y\\gg x_0$|, we have |$S_1(y)>1$|. Therefore, by the intermediate value theorem there exists |$y^*$| such that |$S_1(y^*)=S$|. This leads to the correct choice of constants in (2.4) that achieve the boundary values. □ 3. Geometric properties of the operator To better understand the geometry of the solutions of |$\\epsilon-F^\\epsilon[u]=0$|, we present, in this section, some results regarding the convexity of the level sets of subsolutions of related PDEs. We give a proof of uniform convexity of the level sets for |$C^2$| subsolutions of the equation. The general case is beyond the scope of this article. The proof of robust quasiconvexity for viscosity solutions is technical, involving the construction of special test functions (Barron et al., 2013). Analogous results can also be found in Lions (1998) for convexity, and Carlier & Galichon (2012) for semiconcavity of solutions of parabolic equations. Next, we quantify the degree to which a function can lack quasiconvexity, given that it is uniformly QC in only certain directions. This is relevant for numerical solutions that enforce the operator in only a finite number of grid directions. A related result on semiconcavity of solutions of a parabolic equation can be found in Carlier & Galichon (2012, Section 4). We first show that we can recover a convex function by enforcing strict convexity along a prescribed set of directions. This result is of independent interest for the related numerical method (Oberman, 2008a). Next, we prove an analogous result for QC functions. In what follows, it suffices to consider subsolutions of |$\\epsilon - \\lambda_{\\rm QC}[u]=0$|. This is because of the ordering of the operators: \\begin{equation} -\\lambda_{\\rm QC}[u]\\ge-\\lambda_{\\rm QC}^{\\epsilon^2}[u]\\ge \\epsilon - F^\\epsilon[u]. \\end{equation} (3.1) The first inequality follows naturally from the restriction of the constraint set. The second inequality is evident from the proof of Proposition 1.9. 3.1 Uniform convexity of the level sets of solutions Definition 3.1 (Uniform convexity of sets). Let |$S\\subset{\\mathbb{R}^n}$| be a domain and suppose |$x,y\\in\\partial S$|. Define the midpoint |$\\bar x\\equiv(x+y)/2$|. Then we say |$S$| is uniformly convex if |${\\text{dist}}(\\bar x,\\partial S) = \\mathcal{O}(h^2)$|. Proposition 3.2 (Uniformly convex level sets of subsolutions). Suppose |$u\\in C^2(\\overline\\varOmega)$| is a subsolution of |$\\epsilon-\\lambda_{\\rm QC}[u]=0$|. Then, |$u$| has uniformly convex level sets. Proof. Suppose |$x,y\\in\\varOmega$| such that |$u(x)=u(y)=\\alpha$|. Let |$\\bar x$| be the midpoint of the line segment joining |$x$| and |$y$|, |$h = {\\vert{\\bar x - y}\\vert}$| and |$d = (\\bar x-y)/{\\vert{\\bar x -y}\\vert}$|. We see that |${\\nabla} u(\\bar x)^\\intercal d= 0$|, so that we have \\begin{align*} \\epsilon &\\le \\min_{{\\vert{v}\\vert}=1,{\\nabla} u(\\bar x)^\\intercal v=0} v^\\intercal D^2u(\\bar x) v\\\\ &\\le d^\\intercal D^2u(\\bar x) d\\\\ &= \\frac{u(\\bar x + hd) + u(\\bar x -hd) - 2u(\\bar x)}{h^2} +\\mathcal{O}(h^2)\\\\ &= \\frac{2\\alpha - 2u(\\bar x)}{h^2} + \\mathcal{O}(h^2). \\end{align*} Rearranging for |$u(\\bar x)$| yields the following inequality: $u(\\bar x) \\le \\alpha -\\frac{\\epsilon h^2}{2} + \\mathcal{O}(h^4).$ □ 3.2 Directional quasiconvexity Definition 3.3. The function |$u:\\mathbb{R}^n\\rightarrow\\mathbb{R}$| is QC along a direction |$v$| if for any |$x\\in\\mathbb{R}^n$|, the function |$\\tilde u(t)\\equiv u(x+tv)$| is QC for |$t\\in\\mathbb{R}$|. We also appeal to the following proposition, which is elementary from the definition of quasiconvexity. Proposition 3.4 The function |$u$| is QC if and only if |$u$| is QC in every direction |$v$|. Proposition 3.4 provides a convenient characterization of QC functions. In practice, however, we are confined to a grid and thus cannot enforce quasiconvexity along every direction. Therefore, we relax the notion of directional quasiconvexity to only a finite set of directions. Doing so results in the notion of approximate quasiconvexity. We can quantify the degree to which a function might lack quasiconvexity, expressed in terms of the directional resolution which we define below. Definition 3.5 (Directional resolution). Let |$\\mathcal{D}=\\{d_1,\\dots,d_n\\}$| be a set of unit vectors. Then we define the directional resolution of |$\\mathcal{D}$| as \\begin{equation} {\\rm d}\\theta \\equiv \\max_{{\\vert{w}\\vert}=1}\\min_{d\\in \\mathcal{D} }\\cos^{-1}(w^\\intercal d). \\end{equation} (3.2) In two dimensions, |${\\rm d}\\theta$| is the largest angle an arbitrary unit vector can make with any vector in |$\\mathcal{D}$|. In two dimensions, |${\\rm d}\\theta$| is simply half the maximum angle between any two direction vectors. Recalling that a necessary condition for a function |$u$| to be QC is |$-\\lambda_{\\rm QC}[u]\\le0$| in the viscosity sense (Barron et al., 2013), we have the following result. Proposition 3.6 (Approximate quasiconvexity). Suppose |$u\\in C(\\overline\\varOmega)$| and |$\\mathcal{D}$| is a set of directions with directional resolution |${\\rm d}\\theta\\leq\\frac{\\pi}{4}$|. Also, suppose |$u$| is QC along every |$d\\in \\mathcal{D}$|. Then, we have the following approximate quasiconvexity estimate: $-\\lambda_{\\rm QC}[u]\\leq\\mathcal{O}({\\rm d}\\theta).$ Proof. Without loss of generality, assume |$x=0$| and |$u(0)=0$|. Suppose |$u-\\phi$| attains a global maximum at |$x=0$|. We may assume |$\\phi$| is locally quadratic, |$\\phi(x)=x^\\intercal A x+b^\\intercal x$|, for some real, symmetric matrix |$A$|, and for |$b\\in{\\mathbb{R}^n}$|. Next, suppose |$w$| is an arbitrary unit vector satisfying |$\\nabla \\phi(0)^\\intercal w=b^\\intercal w=0$|. Decompose |$w$| as follows: $w = \\cos(\\theta)v + \\sin(\\theta) p,$ where |$v\\equiv\\text{argmin}_i(\\cos^{-1}(w^\\intercal v_i))$|, |$\\theta = \\cos^{-1}(w^\\intercal v)$| and |$p$| is some unit vector orthogonal to |$v$|. By hypothesis, |$\\theta\\le\\pi/4$|. Taking |$\\lambda_n$| to denote the largest eigenvalue of |$A$|, we observe \\begin{align*} 0=u(0) &= u\\left(\\tfrac{1}{2}(-\\cos(\\theta) v)+\\tfrac{1}{2}(\\cos(\\theta) v)\\right)\\\\ &\\leq \\max(u(-\\cos(\\theta) v),u(\\cos(\\theta) v))\\quad\\text{(by quasiconvexity)}\\\\ &\\leq \\max(\\phi(-\\cos(\\theta) v),\\phi(\\cos(\\theta) v))\\\\ &= \\max(\\phi(- w+\\sin(\\theta)p),\\phi( w-\\sin(\\theta)p))\\\\ &=w^\\intercal Aw+\\sin^2(\\theta)p^\\intercal Ap-2\\sin(\\theta)w^\\intercal Ap+\\sin(\\theta)|p^\\intercal b|\\\\ &\\leq w^\\intercal Aw+ \\sin^2(\\theta)\\lambda_n + 2\\sin(\\theta)\\lambda_n + \\sin(\\theta){\\vert{b}\\vert}\\\\ &\\leq w^\\intercal Aw+ \\sin^2({\\rm d}\\theta) \\lambda_n + 2\\sin({\\rm d}\\theta)\\lambda_n+ \\sin({\\rm d}\\theta){\\vert{b}\\vert}\\\\ &=w^\\intercal Aw + \\mathcal{O}({\\rm d}\\theta). \\end{align*} In the above calculations, we used the fact that |$y^\\intercal Ay\\leq\\lambda_n$| for every |$y$|, |$\\phi\\ge u$| in |$\\overline\\varOmega$| and the fact that |$b^\\intercal w=0$|. Taking the minimum over all unit vectors |$w$| satisfying |$b^\\intercal w=0$| yields the desired result. □ The result we present next is a follow-up to a formal result discussed in Oberman (2007) regarding the similar notion of a uniformly convex envelope. We show that by enforcing a degree of strict convexity along a prescribed set of directions, one can recover a convex function on the entire domain. The subsequent and analogous result for quasiconvexity is natural and requires only a slight modification of the proof. Before proceeding, we need to state the following equivalences, found in Cannarsa & Sinestrari (2004), which will provide a convenient framework for the proof. Lemma 3.7 (Strict convexity and semiconcavity; Cannarsa & Sinestrari, 2004). Suppose |$u\\in C(\\overline\\varOmega)$| and |$C\\ge0$|. The following three conditions are equivalent for strict convexity: (i) |$D^2u(x) \\ge CI$| in the viscosity sense; (ii) |$u - C\\frac{|\\cdot|^2}{2}$| is convex; (iii) |$u(x+h) - 2u(x) + u(x-h) \\geq C\\frac{h^2}2$| for all |$x\\in\\overline\\varOmega$| and |$h>0$|. The following three conditions are equivalent for semiconcavity: (i) |$D^2u(x) \\le CI$| in the viscosity sense; (ii) |$u - C\\frac{|\\cdot|^2}{2}$| is concave; (ii) |$u(x+h) - 2u(x) + u(x-h) \\leq C\\frac{h^2}2$| for all |$x\\in\\overline\\varOmega$| and |$h>0$|. Proposition 3.8 (Recovering convexity). Suppose |$u\\in C(\\overline\\varOmega)$| and |$\\mathcal{D}$| is a set of directions with directional resolution |${\\rm d}\\theta\\le\\pi/4$|. Suppose there exists |$C>0$| such that |$u$| satisfies the following in the viscosity sense for every |$x\\in\\overline\\varOmega$|: (i) |$D^2u(x)\\le CI$|; (ii) |$v^\\intercal D^2u(x) v \\ge C\\,{\\rm d}\\theta^2$| for every |$v\\in\\mathcal{D}$|. Then |$\\lambda_1[u]\\ge0$|. Proof. Suppose |$y$| is an arbitrary unit vector and |$x\\in\\overline\\varOmega$|. Without loss of generality we can assume that |$x=0$|. It suffices to show $\\frac{u(hy)+u(-hy)}{2}-u(0) \\ge 0\\quad\\text{for every} h>0.$ We may assume |$hy$| lies between some |$d_1,d_2$| such that |$d_i/{\\vert{d_i}\\vert}\\in\\mathcal{D}$| for each |$i$|, as shown in Fig. 4. Then, by the first hypothesis and Lemma 3.7, we can write \\begin{equation} \\frac{u(d_i)+u(-d_i)}{2} - u(0) \\ge \\frac{C\\,{\\rm d}\\theta^2{\\vert{d_i}\\vert}^2}{2},\\quad i=1,2. \\end{equation} (3.3) Next, define |$h_i\\equiv {\\vert{d_i-y}\\vert}$| and |$e_i \\equiv \\frac{d_i-y}{h}$| for |$i=1,2$|. Then we have the following inequalities by the second hypothesis and Lemma 3.7: \\begin{align} \\frac{h_1}{h_1+h_2}u(y+h_2e_2) + \\frac{h_2}{h_1+h_2}u(y-h_1e_1) - u(hy) &\\le C\\frac{h_1h_2}{2}, \\-2pt] \\end{align} (3.4) \\begin{align} \\frac{h_1}{h_1+h_2}u(-y-h_2e_2) + \\frac{h_2}{h_1+h_2}u(-y+h_1e_1) - u(-hy) &\\le C\\frac{h_1h_2}{2}. \\end{align} (3.5) Averaging (3.4) and (3.5), and substituting |d_i = y+he_i| yields \\begin{align} \\frac{h_2}{h_1+h_2}\\left(\\frac{u(d_1)+u(-d_1)}{2}\\right)+\\frac{h_1}{h_1+h_2}\\left(\\frac{u(d_2)+u(-d_2)}{2}\\right) -\\frac{u(hy)+u(-hy)}{2} \\le C\\frac{h_1h_2}{2}. \\end{align} (3.6) Now, weighting each equation in (3.3) by |h_i/(h_1+h_2)| (respectively) and averaging them yields \\begin{align} & \\frac{h_2}{h_1+h_2}\\left(\\frac{u(d_1)+u(-d_1)}{2}\\right) + \\frac{h_1}{h_1+h_2}\\left(\\frac{u(d_2)+u(-d_2)}{2}\\right) - u(0)\\nonumber\\\\[-2pt] &\\quad{} \\ge \\frac{C\\,{\\rm d}\\theta^2}{2}\\left( \\frac{h_1}{h_1+h_2}{\\vert{d_1}\\vert}^2+ \\frac{h_2}{h_1+h_2}{\\vert{d_2}\\vert}^2\\right)\\!. \\end{align} (3.7) Subtracting (3.6) from (3.7) recovers \\[ \\frac{u(hy)+u(-hy)}{2}-u(0) \\ge \\frac{C\\,{\\rm d}\\theta^2}{2}\\left( \\frac{h_1}{h_1+h_2}{\\vert{d_1}\\vert}^2+ \\frac{h_2}{h_1+h_2}{\\vert{d_2}\\vert}^2\\right)-C\\frac{h_1h_2}{2}. Defining |$\\theta_i\\equiv \\cos^{-1}(y^\\intercal d_i)$|, the worst-case scenario occurs when |$\\theta_1=\\theta_2$|, so that |${\\vert{d_1}\\vert}={\\vert{d_2}\\vert}\\equiv{\\vert{d}\\vert}$| and |$h_1=h_2\\equiv h_0$|. Then the above inequality simplifies to $\\frac{u(hy)+u(-hy)}{2}-u(0) \\ge \\frac{C}{2}\\left({\\vert{d}\\vert}^2\\,{\\rm d}\\theta^2 - h_0^2\\right)\\!.$ Now, referring to Fig. 4, we have |$h_0 = {\\vert{d}\\vert}\\sin({\\rm d}\\theta)$|, so that we get $\\frac{u(hy)+u(-hy)}{2}-u(0) \\ge \\frac{C{\\vert{d}\\vert}^2}{2}\\left({\\rm d}\\theta^2 - \\sin({\\rm d}\\theta)^2\\right)\\!,$ yielding the desired result since |${\\rm d}\\theta\\ge\\sin({\\rm d}\\theta)$|. □ Fig. 4. View largeDownload slide Illustration of the proof of recovering convexity. Fig. 4. View largeDownload slide Illustration of the proof of recovering convexity. Proposition 3.9 (Recovering quasiconvexity). Suppose |$u\\in C(\\overline\\varOmega)$|, |$\\alpha\\in\\mathbb{R}$| and |$\\mathcal{D}$| is a set of directions with directional resolution |${\\rm d}\\theta\\le\\pi/4$|. Suppose there exists |$C>0$| such that |$u$| satisfies the following in the viscosity sense for every |$x\\in\\overline\\varOmega$|: (i) |$D^2u\\le CI$|; (ii) |${\\vert{\\!{\\nabla} u}\\vert}\\le M$|; (iii) |$v^\\intercal D^2u v - \\alpha \\ge C\\,{\\rm d}\\theta^2$| whenever |$v\\in\\mathcal{D}$| satisfies |${\\vert{\\!{\\nabla} u^\\intercal v}\\vert} \\le M\\,{\\rm d}\\theta$|. Then |$\\lambda_{\\text QC}[u]\\ge\\alpha$|. Remark 3.10. Note that the third requirement is for |$u$| to be |$(M\\,{\\rm d}\\theta)$| robustly QC along directions in |$\\mathcal{D}$|. Recalling the inequalities (3.1), this can be enforced numerically by solving for subsolutions of the following: $\\alpha+C\\,{\\rm d}\\theta^2+\\sqrt{M\\,{\\rm d}\\theta} - F^{\\sqrt{M\\,{\\rm d}\\theta}}[u] = 0.$ Proof. We begin by taking an arbitrary unit vector |$y$| satisfying |${\\nabla} \\!u^\\intercal y=0$|. As before, we would like to show $\\frac{u(hy)+u(-hy)}{2}-u(0) \\ge 0\\quad\\text{for every}\\,h>0.$ The proof that follows is nearly identical to the proof of convexity. The only difference arises when trying to derive equations (3.3). In this case, we have to show |${\\nabla} \\!u^\\intercal \\left(\\frac{d_i}{{\\vert{d_i}\\vert}}\\right)\\le M{\\rm d}\\theta$| to invoke the proper inequality. This is given by the following argument for each |$i$|: \\begin{equation} {\\nabla}\\!u^\\intercal \\left(\\frac{d_i}{{\\vert{d_i}\\vert}}\\right) = {\\nabla}\\!{u}^\\intercal \\left(\\frac{y+h_ie_i}{{\\vert{d_i}\\vert}}\\right) = \\frac{h_i}{{\\vert{d_i}\\vert}}{\\nabla}\\! u ^\\intercal e_i \\le \\frac{h_i}{{\\vert{d_i}\\vert}}{\\vert{\\!{\\nabla}\\! u}\\vert}{\\vert{e_i}\\vert}\\le \\frac{h_i}{{\\vert{d_i}\\vert}}M, \\end{equation} (3.8) but |$\\frac{h_i}{{\\vert{d_i}\\vert}}=\\sin(\\theta_i)\\le\\sin({\\rm d}\\theta)\\le {\\rm d}\\theta$| for small |${\\rm d}\\theta$|. Therefore, we have ${\\nabla}\\!{u}^\\intercal \\left(\\frac{d_i}{{\\vert{d_i}\\vert}}\\right) \\le M\\,{\\rm d}\\theta.$ Then, by the third hypothesis, we are now in a position to write $\\frac{u(d_i)+u(-d_i)}{2} - u(x) \\ge \\frac{C{\\vert{d_i}\\vert}^2\\theta^2}{2}.$ Proceeding identically as in the proof of Proposition 3.8 yields the desired result. □ 4. Convergent finite difference schemes In what follows, we provide numerical schemes that discretize the above PDEs. As we will see, the schemes we present here fall into a general class of degenerate elliptic finite difference schemes for which there exists a convergence framework. Before we begin, we introduce some notation, which we will carry throughout the rest of the article. We will assume we are working on the hypercube |$[-1,1]^n\\subset{\\mathbb{R}^n}$|. We write |$x=(x_1,\\dots,x_n)\\in[-1,1]^n$|. For simplicity, we discretize the domain |$[-1,1]^n$| with a uniform grid, resulting in the following spatial resolution: $h\\equiv\\frac{2}{N-1},$ where |$N$| is the number of grid points used to discretize |$[-1,1]$|. Note that we use |$h$| here to denote the spatial resolution and it is not to be confused with the |$h$| in the formulations of (ϵ-QC) and (ϵ-QCE). We use the following notation for our computational domain: $\\varOmega^h \\equiv [-1,1]^n\\cap h\\mathbb{Z}^n.$ We define a grid vector, |$v$|, as $v \\equiv \\frac{x-y}{h},\\quad x,y\\in\\varOmega^h.$ 4.1 Finite difference equations and wide stencils Consider a degenerate elliptic PDE |$F[u]=0$|, and its corresponding finite difference equation, $F_\\rho[u]=0,\\quad x\\in\\varOmega^h,$ where |$\\rho$| is the discretization parameter. In our case, we take |$\\rho=(h,\\,{\\rm d}\\theta)$| for the schemes presented hereafter. In general, |$F_\\rho[u]:C(\\varOmega^h)\\to C(\\varOmega^h)$|, where |$C(\\varOmega^h)$| is the set of grid functions |$u:\\varOmega^h\\to\\mathbb{R}$|. We assume the following form: $F_\\rho[u](x) = F_\\rho(u(x),u(x)-u(\\cdot))\\quad\\ \\text{for}\\ x\\in\\varOmega^h,$ where |$u(\\cdot)$| corresponds to the value of |$u$| at points in |$\\varOmega^h$|. For a given set of grid vectors |$V$|, we say |$F_\\rho[u]$| has stencil width|$W$| if for any |$x\\in\\varOmega^h$|, |$F_\\rho[u](x)$| depends only on the values |$u(x + hv)$|, where |$v\\in V$| and |$\\max_{v\\in V}{\\Vert{v}\\Vert}_\\infty\\le W$|. We assume stencils are symmetric about the reference point |$x$|, and have at least width 1. Example 4.1. In two dimensions, the directional resolution can be written as |${\\rm d}\\theta=\\arctan(1/W)/2$|. A stencil of width |$W=1$| would correspond to the vectors |$V=\\{(0,\\pm1), (\\pm1,0), (\\pm1,\\pm1), (\\mp1,\\pm1)\\}$|. The Dirichlet boundary conditions given by (Dir) are incorporated by setting $F_\\rho[u](x) = u(x)-g(x)\\quad\\ \\text{for}\\ x\\in\\partial\\varOmega^h.$ The PDEs we discuss involve a second-order directional derivative and a directional Eikonal operator. We use the following standard discretizations for our schemes: \\begin{align*} {\\vert{u_v^h(x)}\\vert} &\\equiv \\max\\left\\{\\frac{u(x \\pm hv)-u(x)}{h}\\right\\}\\!,\\\\ u_{vv}^h(x) &\\equiv \\frac{u(x+hv)+u(x-hv)-2u(x)}{h^2}, \\end{align*} where |$u_v^h$| and |$u_{vv}^h$| denote the finite difference approximations for the first- and second-order directional derivatives in the grid direction |$v$|. Recall that for any twice continuously differentiable function |$u$|, the standard Taylor series computation yields the following consistency estimate: \\begin{align} {\\vert{u_v^h(x)}\\vert} &= {\\vert{\\nabla u(x)^\\intercal v}\\vert} + \\mathcal{O}(h), \\\\ \\end{align} (4.1) \\begin{align} u_{vv}^h(x) &= v^\\intercal D^2u(x) v + \\mathcal{O}(h^2). \\end{align} (4.2) 4.2 A finite difference method for the PDE We begin by considering the full discretization of |$F^\\epsilon[u]$|, denoted |$F^\\epsilon_{h,{\\rm d}\\theta}[u]$|, where |$h$| is the spatial resolution and |${\\rm d}\\theta$| is the directional resolution of the set of grid vectors |$V$|. We use the spatial discretizations given by (4.1) and (4.2): \\begin{equation} F^\\epsilon_{h,{\\rm d}\\theta}[u](x)\\equiv \\min_{v\\in V}\\left\\{\\tfrac{1}{\\epsilon}{\\vert{u_v^h(x)}\\vert}+u_{vv}^h(x)\\right\\}\\quad \\text{for}\\,x\\in\\varOmega^h. \\end{equation} (4.3) We write the full discretization of (ϵ-Ob) as \\begin{equation} \\begin{cases} \\max\\{u(x)-g(x),\\epsilon - F^\\epsilon_{h,{\\rm d}\\theta}[u](x)\\} = 0, & x\\in\\varOmega^h,\\\\ u(x) = g(x), & x\\in\\partial\\varOmega^h, \\end{cases} \\end{equation} (4.4) where we make the following abbreviation: $G^\\epsilon_{h,{\\rm d}\\theta}[u]\\equiv\\max\\left\\{u-g,\\epsilon-F^\\epsilon_{h,{\\rm d}\\theta}[u]\\right.\\}.$ 4.3 Iterative solution method We implement a fixed-point solver to recover the numerical solution of (4.4). The method is equivalent to a forward Euler time discretization of the parabolic version of (ϵ-QCE), $u_{t} + G^\\epsilon[u]=0\\quad \\text{in}\\,\\varOmega,$ along with |$u = g$| on |$\\partial\\varOmega$| and |$u = u_0$| for |$t=0$|. In particular, the following iterations are performed until a steady state is reached: $\\begin{cases} u^{n+1} = u^{n} - \\delta G^\\epsilon_{h,d\\theta}[u^{n}], & n\\ge0,\\\\ u^{0} = u_0, \\end{cases}$ where satisfies the CFL condition |$\\delta\\le 1/K^{h,\\epsilon}$| (see Oberman, 2006) and |$K^{h,\\epsilon}$| is the Lipschitz constant of the scheme. For our scheme, |$K$| is given by $K^{h,\\epsilon} = \\frac{1}{\\epsilon h} + \\frac{1}{h^2},$ where the first and second terms come from the discretizations of the first- and second-order directional derivatives, respectively. 4.4 Using a first-order discretization In higher dimensions, we need the second-order term in (ϵ-QC) to guarantee quasiconvexity of the solutions. Interestingly, however, for a careful choice of |$\\epsilon$|, the discretization of only the first-order term is consistent with |$F^\\epsilon[u]$|. Indeed, observe that \\begin{align*} -\\min_{v\\in V}\\frac{{\\vert{u_v^h}\\vert}}{\\epsilon}&=-\\min_{v\\in V}\\left\\{\\frac {\\max\\{u(x\\pm hv)-u(x)\\}} {\\epsilon h}\\right\\}\\\\ &= -\\min_{v\\in V}\\left\\{ \\frac{{\\vert{\\!{\\nabla}\\! u(x)^\\intercal v}\\vert}}{\\epsilon} + \\frac{h}{2\\epsilon}v^\\intercal D^2u(x)v \\right\\}+ \\mathcal{O}\\left(\\frac{h^2}{\\epsilon}\\right). \\end{align*} Choosing |$\\epsilon = h/2$| results in $-\\min_{v\\in V}\\frac{{\\vert{u_v^h}\\vert}}{\\epsilon} = -\\min_{v\\in V}\\left\\{ \\frac{{\\vert{\\!{\\nabla}\\! u(x)^\\intercal v}\\vert}}{h/2} + v^\\intercal D^2u(x)v \\right\\}+ \\mathcal{O}(h).$ Thus, for |$\\epsilon=h/2$|, we see for fixed |$(h,{\\rm d}\\theta)$| that the discretization of the first-order term approximates |$F^\\epsilon[u]$|. Moreover, the time step in the above forward Euler discretization simplifies to |$\\delta=\\mathcal{O}(h^2)$|. 5. Convergence of numerical solutions In this section we introduce the notion of degenerate elliptic schemes and show that the solutions of the proposed numerical schemes converge to the solutions of (ϵ-Ob) as the discretization parameters tend to 0. The standard framework used to establish convergence is that of Barles & Souganidis (1991), which we state below. In particular, it guarantees that the solutions of any monotone, consistent and stable scheme converge to the unique viscosity solution of the PDE. 5.1 Degenerate elliptic schemes Consider the Dirichlet problem for the degenerate elliptic PDE |$F[u]=0$|, and recall its corresponding finite difference formulation: $\\begin{cases} F_\\rho(u(x),u(x)-u(\\cdot)) = 0, & x\\in\\varOmega^h,\\\\ u(x)-g(x) = 0, & x\\in\\partial\\varOmega^h, \\end{cases}$ where |$\\rho$| is the discretization parameter. Definition 5.1. The PDE |$F_\\rho[u]$| is a degenerate elliptic scheme if it is nondecreasing in each of its arguments. Remark 5.2. Although the convergence theory is originally stated in terms of monotone approximation schemes (schemes with non-negative coefficients), ellipticity is an equivalent formulation for finite difference operators (Oberman, 2006). In fact, degenerate ellipticity implies monotonicity and stability in the nonexpansive norm. Definition 5.3. The finite difference operator |$F_\\rho[u]$| is consistent with |$F[u]$| if for any smooth function |$\\phi$| and |$x\\in\\varOmega$| we have $F_\\rho[\\phi](x)\\to F[\\phi](x)\\quad \\text{as}\\,\\rho\\to0.$ Definition 5.4. The finite difference operator |$F_\\rho[u]$| is stable if there exists |$M>0$| independent of |$\\rho$| such that if |$F_\\rho[u]=0$| then |$\\| u \\|_\\infty\\le M$|. Remark 5.5. (Interpolating to the entire domain). The convergence theory assumes that the approximation scheme and the grid function are defined on all of |$\\varOmega$|. Although the finite difference operator acts only on functions defined on |$\\varOmega^h$|, we can extend such functions to |$\\varOmega\\setminus{\\it{\\Omega}}^h$| via piecewise interpolation. In particular, performing piecewise linear interpolation maintains the ellipticity of the scheme as well as all other relevant properties. Therefore, we can safely interchange |${\\it{\\Omega}}^h$| and |$\\varOmega$| in the discussion of convergence without any loss of generality. 5.2 Convergence of numerical approximations Next we will state the theorem for convergence of approximation schemes, tailored to elliptic finite difference schemes, and demonstrate that the proposed schemes fit in the desired framework. In particular, we will show that the schemes are elliptic, consistent and have stable solutions. Proposition 5.6 (Convergence of approximation schemes; Barles & Souganidis, 1991). Consider the degenerate elliptic PDE |$F[u]=0$|, with Dirichlet boundary conditions for which there exist a strong comparison principle. Let |$F_\\rho[u]$| be a consistent and elliptic scheme. Furthermore, assume that the solutions of |$F_\\rho[u]=0$| are bounded independently of |$\\rho$|. Then |$u^\\rho\\to u$| locally uniformly on |$\\varOmega$| as |$\\rho\\to0$|. Remark 5.7. Proposition 5.6 is an instance of the meta-theorem, which says that consistency and stability imply convergence. The proposition can also be seen as a special case of the fact that viscosity solutions are stable under perturbations of both the data and the operators, provided that the equation remains in the class of degenerate elliptic PDEs. The classical example of a perturbation is to add the Laplacian (or viscosity) to the operator: |$F^\\epsilon = F + \\epsilon {\\it{\\Delta}}$|, which is the origin of the nomenclature ‘viscosity solutions’. The finite difference approximation of |$F$| can also be regarded as a perturbation of the operator. Provided that the approximation is (i) consistent and (ii) degenerate elliptic (Oberman, 2006) or monotone and stable (Barles & Souganidis, 1991) then the perturbation converges. Lemma 5.8. (Ellipticity). The scheme given by (4.4) is elliptic. Proof. The negatives of each of the spatial discretizations above are elliptic for fixed |$v$|, so taking the minimum over all |$v$| maintains the ellipticity. Therefore, the discretization of |$F^\\epsilon_{h,{\\rm d}\\theta}[u]$| is elliptic. Moreover, the obstacle term is trivially elliptic. Hence |$G^\\epsilon_{h,{\\rm d}\\theta}[u]$|, being the maximum of two elliptic schemes, is also elliptic. □ Lemma 5.9 (Consistency). Given a smooth function |$u$| and a set of grid vectors |$V$| with directional resolution |${\\rm d}\\theta\\le\\pi/4$|, we have the following estimate: \\begin{equation} G^\\epsilon_{h,{\\rm d}\\theta}[u] - G^\\epsilon[u] = \\mathcal{O}(h+{\\rm d}\\theta). \\end{equation} (5.1) In other words, the scheme (4.4) is consistent with (ϵ-Ob). Proof. It suffices to show that |$F^\\epsilon_{h,{\\rm d}\\theta}[u] - F^\\epsilon[u] = \\mathcal{O}(h+{\\rm d}\\theta)$|. Fix |$x\\in\\overline\\varOmega$| and let |$u$| be a smooth function on |$\\overline{\\varOmega}$|. Define $w\\equiv{\\text{argmin}}_{{\\vert{v}\\vert}=1}\\left\\{ \\tfrac{1}{\\epsilon}{\\vert{\\nabla u(x)^\\intercal v}\\vert} + v^\\intercal D^2u(x) v\\right\\}.$ Let |$v\\in V$| and make the following decomposition: $v = \\cos(\\theta)w + \\sin(\\theta)p,$ for some unit vector |$p$| orthogonal to |$w$|. Performing a similar calculation to the proof in Proposition 3.6, we obtain the following consistency estimate from the directional resolution: $F^\\epsilon[u] \\le\\min_{v\\in V}\\left\\{\\tfrac{1}{\\epsilon}{\\vert{\\nabla u^\\intercal v}\\vert} + v^\\intercal D^2u v\\right\\} \\le F^\\epsilon[u] +\\mathcal{O}({\\rm d}\\theta).$ Finally, we conclude (5.1) by recalling the standard Taylor series consistency estimate from equations (4.1) and (4.2). □ Next, we establish stability of our numerical solutions by applying a discrete comparison principle for strict subsolution and supersolution. In particular, we use the following result. Proposition 5.10. (Discrete comparison principle for strict subsolutions). Let |$F_\\rho[u]$| be a degenerate elliptic scheme defined on |$\\varOmega^h$|. Then for any grid functions |$u,v$| we have $F_\\rho[u]<F_\\rho[v]\\implies u\\le v\\quad\\text{in}\\,\\varOmega^h.$ Proof. Suppose for contradiction that $\\max_{x\\in{\\it{\\Omega}}^h}\\{u(x) - v(x)\\} = u(x_0) - v(x_0) > 0.$ Then, $u(x_0) - u(x) > v(x_0) - v(x)\\quad\\text{for every}\\,x\\in\\varOmega^h,$ and so by ellipticity (Lemma 5.8), $F_\\rho[u](x_0)\\ge F_\\rho[v](x_0),$ which is the desired contradiction. □ Lemma 5.11 (Stability). The numerical solutions of (4.4) are stable. Proof. This a direct application of Proposition 5.10. Indeed, suppose |$u$| is a numerical solution such that |$G^\\epsilon_{h,{\\rm d}\\theta}[u]=0$|. Define |$g_1(x)\\equiv C+2\\epsilon^2 \\sum_{i=1}^nx_i$|, where |$C\\in\\mathbb{R}$| is chosen so that |$g_1(x)<g(x)$| for every |$x\\in\\overline\\varOmega$|. Moreover, one can check that |$(g_1)_{vv}^h(x)=0$|. Then, we observe \\begin{align*} G^\\epsilon_{h,{\\rm d}\\theta}[g_1](x) &= \\max\\left\\{g_1(x)-g(x),\\epsilon-\\min_{v\\in V}\\left\\{ \\frac{\\max\\{g_1(x\\pm hv)\\}-g_1(x)}{\\epsilon h} \\right\\}\\right\\}\\\\ &= \\max\\left\\{g_1(x)-g(x),\\epsilon-2\\epsilon\\min_{v\\in V}\\left\\vert\\sum\\nolimits_{i=1}^nv_i\\right\\vert\\right\\}\\\\ &= \\max\\left\\{g_1(x)-g(x),-\\epsilon\\right\\}, \\end{align*} where the last equality follows from the general assumption that the stencil width is at least 1 with directional resolution |${\\rm d}\\theta\\le\\pi/4$|. Next, define |$g_2(x)\\equiv\\max_{x\\in\\overline{\\it{\\Omega}}}g(x)$|. Then, in summary we have |$G^\\epsilon_{h,{\\rm d}\\theta}[g_1]<0$| and |$G^\\epsilon_{h,{\\rm d}\\theta}[g_2]\\ge0$|. Therefore, by Proposition 5.10, $\\min_{x\\in\\overline\\varOmega} g_1(x)\\le u(x)\\le g_2 \\quad\\text{for}\\ x\\in\\varOmega^h.$ Moreover, this bound is independent of |$h$| and |${\\rm d}\\theta$|. □ Proposition 5.12. The solutions of (4.4) converge locally uniformly on |$\\varOmega$| to the solutions of (ϵ-Ob) as |$h,{\\rm d}\\theta\\to0$|. Proof. By Lemmas 5.8, 5.9, 5.11 we see that (4.4) is a consistent and elliptic scheme which has stable solutions. Moreover, a weak comparison principle holds by Corollary 2.5. Therefore, by Proposition 5.6, the numerical solutions converge uniformly on compact subsets of |$\\varOmega$|. □ 6. Numerical results In this section we present the numerical results. The tolerance for the fixed-point iterations was taken to be |$10^{-6}$| and unless otherwise stated we set |$\\epsilon=h/2$|, where |$h$| is the spatial resolution of the grid. Technical conditions on |$g$| given by Remark 2.1 are needed to ensure that the boundary conditions are held in the strong sense. In practice, we violated this condition, resulting in solutions that were discontinuous at the boundary. Two natural choices for initialization of the solution are (i) the obstacle function, |$g(x)$| and (ii) a QC function below |$g$|. We found that the first choice leads to very slow convergence. In particular, the parabolic equation takes time |$\\mathcal{O}(1/\\epsilon)$| to converge. See Example 6.3 below. On the other hand, the second choice results in faster convergence: the simplest choice is simply the constant function with value the minimum of |$g$|. After one step of the iteration the boundary values are attained. Moreover, starting with this choice allows us to use the iterative method with |$\\epsilon = 0$| to find the QCE. Remark 6.1 It is an open question whether solving the |$\\epsilon=0$| time-dependent PDE with quasiconvex initial data below |$g$| will converge. To apply our convergence results, we take |$\\epsilon = h/2$|. 6.1 Results in one dimension We present examples demonstrating the convergence of approximate QCEs to the true QCE as |$\\epsilon\\to0^+$|. We also compare the iteration count when starting from below and at the obstacle. Example 6.2 (Convergence as |$\\epsilon\\to0$|). We consider the convergence of numerical solutions of (4.4) as |$\\epsilon\\to 0$| with the following obstacle function: $g(x)=\\min\\{{\\vert{x-0.5}\\vert},{\\vert{x+0.5}\\vert}-0.3\\}.$ The results are displayed in Fig. 5. Indeed, as expected we see convergence to the true QCE, from below, as |$\\epsilon\\to0$|. Fig. 5. View largeDownload slide Obstacles (solid) and the numerical solutions (dashed). Left: Example 6.2. Convergence in |$\\epsilon$|. Bottom to top: |$\\epsilon = 0.2\\,,0.1\\,,0.05\\,,0.001\\,,0.0001$|. Right: Example 6.3. Fixed-point iterations when |$u^0 \\equiv g_2$|. Fig. 5. View largeDownload slide Obstacles (solid) and the numerical solutions (dashed). Left: Example 6.2. Convergence in |$\\epsilon$|. Bottom to top: |$\\epsilon = 0.2\\,,0.1\\,,0.05\\,,0.001\\,,0.0001$|. Right: Example 6.3. Fixed-point iterations when |$u^0 \\equiv g_2$|. Example 6.3 (Visualization of iterations). Next, we consider the obstacle $g(x)= -x^2+1,$ whose QCE is simply |$g_2^{\\rm QCE}\\equiv0$|. We demonstrate the evolution of the iterations when the initial data are taken to be the obstacle. In this case, the solution corresponded closely to |$u(x,t) = \\min(g(x), c - \\epsilon t)$|, where |$c = \\max g$|. This illustrates the slow speed of convergence, and the fact that the equation degenerates to a trivial operator as |$\\epsilon \\to 0$|. Results are displayed in Fig. 5. 6.2 Numerical results in two dimensions All examples take the initial function in the iterative scheme to be the minimum value |$g_{\\rm min} = \\min g$| of the obstacle function. In all contour plots, the solid line represents the level sets of the original function and the dashed line represents the same level sets of the numerical solution. Unless otherwise stated, the two-dimensional numerical solutions shown are computed on a |$64\\times64$| grid, for illustration purposes. Computations were performed on larger-sized grids. We performed the same numerical experiments in Abbasi & Oberman (2016), and achieved similar results. This is expected, since solving our PDE with small |$\\epsilon$| is consistent with the QCE. We do not reproduce those results here to save space. The examples we present focus on the difference between the two operators for values of |$e$| close to 1. Example 6.4. (Uniform convexification of level sets). Let |$g(x)$| be the signed distance function (negative on the inside) to the square, |$S = \\left\\{x \\mid \\max_{i}{\\vert{x_i}\\vert}=1/2 \\right\\}$|. Although |$g$| is convex, its level sets are not uniformly convex. We apply the iterative procedure to |$g$| with |$\\epsilon=0.5$| so that we uniformly convexify the level sets. The results demonstrating this are displayed in Fig. 6. Fig. 6. View largeDownload slide Left: Examples 6.4, square level sets become uniformly convex with |$\\epsilon = 1/2$|. Middle: Example 6.5, the solution with |$\\epsilon=1$| is indicated by the dashed contours. Right: Example 6.6 with |$\\epsilon$| near 0. Fig. 6. View largeDownload slide Left: Examples 6.4, square level sets become uniformly convex with |$\\epsilon = 1/2$|. Middle: Example 6.5, the solution with |$\\epsilon=1$| is indicated by the dashed contours. Right: Example 6.6 with |$\\epsilon$| near 0. Example 6.5 (Nonconvex signed distance function). In this example, |$g(x)$| is the signed distance function to the curve |${\\it{\\Gamma}}$|, \\begin{align*} {\\it{\\Gamma}}(t) &= \\frac{1}{2}\\begin{cases} (\\cos(t-\\frac{\\pi}{2}),\\sin(t-\\frac{\\pi}{2})), & t\\in[0,3\\pi/2],\\9pt] \\left(0,\\frac{t-7\\pi/4}{\\pi/4}\\right)\\!, & t\\in[3\\pi/2,7\\pi/4], \\\\[9pt] \\left(\\frac{7\\pi/4-t}{\\pi/4},0\\right)\\!, & t\\in[7\\pi/4,2\\pi]. \\end{cases} \\end{align*} We compare the results of using different |\\epsilon|. In particular, we choose |\\epsilon=h/2| and |\\epsilon=1|. In the latter case, we see that the level sets are more curved. Results are displayed in Fig. 6. Example 6.6. We consider the obstacle where |g| is a cone with circular portions removed from its level sets. Take \\[ g(x) = \\begin{cases} 1, &x\\in A,\\\\ {\\vert{x}\\vert}-1/2, &x\\in {\\it{\\Omega}}\\setminus A, \\end{cases} where |$A=\\{(x_1\\pm1/2)^2+x_2^2\\le 1/16\\}\\bigcup \\{x_1^2+(x_2\\pm1/2)^2\\le1/16\\}$|. Results showing the |$g=0.7$| level set are found in Fig. 6. Example 6.7 (Comparison to the |$\\epsilon$|-robust QCE). We compare the solutions of our PDE to the solutions of the line solver presented in Abbasi & Oberman (2016), which returns the robust QCE. In particular, we use a stencil width |$W=5$|, we used the same directions in the line solver and we set |$\\epsilon=0.02$| for the |$\\epsilon$|-uniformly QCE, and we also used a regularization of |$0.02$| for the robust QC (in principle we should have used |$0.02^2$| but the matching value was better for illustration purposes). The results are found in Fig. 7. Fig. 7. View largeDownload slide Example 6.7. Top: Surface plots (inverted) of the function, robust QCE, |$\\epsilon$|-uniformly QCE. Bottom: Corresponding contour plots. Fig. 7. View largeDownload slide Example 6.7. Top: Surface plots (inverted) of the function, robust QCE, |$\\epsilon$|-uniformly QCE. Bottom: Corresponding contour plots. 6.3 Accelerating iterations using the line solver We found an effective method to reduce the computational time to find the solution. We implement the line solver for the robust QCE proposed in Abbasi & Oberman (2016), alternating with the PDE iterations. In particular, on an |$n\\times n$| grid, after every |$2n$| iterations of the PDE we apply one iteration of the line solver (with the commensurate value of |$\\epsilon$|) in each direction. Table 1 shows the number of iterations required for convergence using the same obstacle from Abbasi & Oberman (2016, Example 6.1). Note the line solver is for a different regularization of the QC operator; however, for small values of |$\\epsilon,$| the solutions are close. For larger values of |$\\epsilon,$| the operator still accelerates the solver, but it does not approach the solution to within arbitrary precision. Table 1 Number of iterations required for convergence using different solution methods |$n$| 32 64 128 256 2n PDE iterations 50 102 203 398 2n PDE It + 1 line solve 6 8 10 7 |$n$| 32 64 128 256 2n PDE iterations 50 102 203 398 2n PDE It + 1 line solve 6 8 10 7 Table 1 Number of iterations required for convergence using different solution methods |$n$| 32 64 128 256 2n PDE iterations 50 102 203 398 2n PDE It + 1 line solve 6 8 10 7 |$n$| 32 64 128 256 2n PDE iterations 50 102 203 398 2n PDE It + 1 line solve 6 8 10 7 In Table 1, we compare the number of iterations required for each method. The results were comparable across different examples. The number of iterations of the PDE operator required for convergence was typically a small constant times |$n^2$|. The computational cost of a single line solve was on the order of a small constant (say 10) times the cost of a single PDE iteration. So the combined method, requires |$\\mathcal{O}(n)$| iterations (possibly with a |$\\log n$| prefactor, which we ignore, since it is not significant at these values of |$n$|). So the combined method required (roughly) |$n$| iterations. Recall that there are |$N =n^2$| grid points. Each iteration has a cost proportional to the number of directions used and to the number of grid points, so the total cost is |$\\mathcal{O}\\left((WN^2) \\right)$| for the iterative method and |$\\mathcal{O}(WN)$| for the combined method (with constants of roughly 1 and 10, respectively). So combining the iterative solver with the line solver results in significant improvements to computation time. References Abbasi B. & Oberman A. M. ( 2016 ) Computing the quasiconvex envelope using a nonlocal line solver. Available at https://arxiv.org/abs/1612.05584. Avriel M. , Diewert W. E. , Schaible S. , & Zang I. ( 2010 ) Generalized concavity . Society for Industrial and Applied Mathematics . Bardi M. & Mannucci P. ( 2006 ) On the Dirichlet problem for non-totally degenerate fully nonlinear elliptic equations. Commun. Pure Appl. Anal. , 5 , 709 – 731 . Google Scholar Crossref Search ADS Bardi M. & Mannucci P. ( 2013 ) Comparison principles and Dirichlet problem for fully nonlinear degenerate equations of Monge–Ampère type. Forum Math , 25 , 1291 – 1330 . Google Scholar Crossref Search ADS Barles G. & Souganidis P. E. ( 1991 ) Convergence of approximation schemes for fully nonlinear second order equations. Asymptotic Anal. , 4 , 271 – 283 . Barron E. N. , Goebel R. & Jensen R. R. ( 2012a ) Functions which are quasiconvex under linear perturbations. SIAM J. Optim. , 22 , 1089 – 1108 . Google Scholar Crossref Search ADS Barron E. N. , Goebel R. & Jensen R. R. ( 2012b ) The quasiconvex envelope through first-order partial differential equations which characterize quasiconvexity of nonsmooth functions. Discrete Contin. Dyn. Syst. Ser. B , 17 , 1693 – 1706 . Google Scholar Crossref Search ADS Barron E. , Goebel R. & Jensen R. ( 2013 ) Quasiconvex functions and nonlinear pdes. Trans. Am. Math. Soc. , 365 , 4229 – 4255 . Google Scholar Crossref Search ADS Barron E. N. & Jensen R. R. ( 2014 ) A uniqueness result for the quasiconvex operator and first order PDEs for convex envelopes. In Annales de l’Institut Henri Poincare (C) Non Linear Analysis , vol. 31 . Elsevier Masson , pp. 203 – 215 . Boyd S. & Vandenberghe L. ( 2004 ) Convex Optimization . Cambridge : Cambridge University Press . Caffarelli L. A. & Spruck J. ( 1982 ) Convexity properties of solutions to some classical variational problems. Comm. Partial Differential Equations , 7 , 1337 – 1379 . Google Scholar Crossref Search ADS Cannarsa P. & Sinestrari C. ( 2004 ) Semiconcave Functions, Hamilton-Jacobi Equations, and Optimal Control , vol. 58. Springer Science & Business Media . Carlier G. & Galichon A. ( 2012 ) Exponential convergence for a convexifying equation. ESAIM Control Optim. Calc. Var. , 18 , 611 – 620 . Google Scholar Crossref Search ADS Colesanti A. & Salani P. ( 2003 ) Quasi-concave envelope of a function and convexity of level sets of solutions to elliptic equations. Math. Nachr. , 258 , 3 – 15 . Google Scholar Crossref Search ADS Crandall M. G. , Ishii H. , & Lions P. L. ( 1992 ) User’s guide to viscosity solutions of second order partial differential equations. Bull. Amer. Math. Soc. , 27 , 1 – 67 . Google Scholar Crossref Search ADS Froese B. ( 2016 ) Convergent approximation of surfaces of prescribed Gaussian curvature with weak Dirichlet conditions. Available at https://arxiv.org/abs/1601.06315. Kawohl B. ( 1985 ) Rearrangements and Convexity of Level Sets in PDE . Lecture Notes in Mathematics , vol. 1150 . Berlin, Heidelberg : Springer-Verlag , pp. 1 – 134 . Kohn R. & Serfaty S. ( 2007 ) Second-order PDE’s and deterministic games. In Proc. Internat. Congress Ind. Appl. Math. (ICIAM’07) , pp. 239 – 249 . Lions P.-L. ( 1998 ) Identification du cône dual des fonctions convexes et applications. C. R. Acad. Sci. Paris Sér. I Math. , 326 , 1385 – 1390 . Google Scholar Crossref Search ADS Oberman A. M. ( 2004 ) A convergent monotone difference scheme for motion of level sets by mean curvature. Numer. Math. , 99 , 365 – 379 . Google Scholar Crossref Search ADS Oberman A. M. ( 2006 ) Convergent difference schemes for degenerate elliptic and parabolic equations: Hamilton-Jacobi equations and free boundary problems. SIAM J. Numer. Anal. , 44 , 879 – 895 . Google Scholar Crossref Search ADS Oberman A. M. ( 2007 ) The convex envelope is the solution of a nonlinear obstacle problem. Proc. Amer. Math. Soc. , 135 , 1689 – 1694 . Google Scholar Crossref Search ADS Oberman A. M. ( 2008a ) Computing the convex envelope using a nonlinear partial differential equation. Math. Models Methods Appl. Sci. , 18 , 759 – 780 . Google Scholar Crossref Search ADS Oberman A. M. ( 2008b ) Wide stencil finite difference schemes for the elliptic Monge-Ampère equation and functions of the eigenvalues of the Hessian. Discrete Contin. Dyn. Syst. Ser. B , 10 , 221 – 238 . Google Scholar Crossref Search ADS Sethian J. A. ( 1999 ) Level Set Methods and Fast Marching Methods: Evolving Interfaces in Computational Geometry, Fluid Mechanics, Computer Vision, and Materials Science , vol. 3 . Cambridge University Press . © The authors 2017. Published by Oxford University Press on behalf of the Institute of Mathematics and its Applications. All rights reserved. This article is published and distributed under the terms of the Oxford University Press, Standard Journals Publication Model (https://academic.oup.com/journals/pages/open_access/funder_policies/chorus/standard_publication_model)\n\n### Journal\n\nIMA Journal of Numerical AnalysisOxford University Press\n\nPublished: Jan 25, 2019\n\n## You’re reading a free preview. Subscribe to read the entire article.\n\n### DeepDyve is your personal research library\n\nIt’s your single place to instantly\nthat matters to you.\n\nover 18 million articles from more than\n15,000 peer-reviewed journals.\n\nAll for just $49/month ### Explore the DeepDyve Library ### Search Query the DeepDyve database, plus search all of PubMed and Google Scholar seamlessly ### Organize Save any article or search result from DeepDyve, PubMed, and Google Scholar... all in one place. ### Access Get unlimited, online access to over 18 million full-text articles from more than 15,000 scientific journals. ### Your journals are on DeepDyve Read from thousands of the leading scholarly journals from SpringerNature, Elsevier, Wiley-Blackwell, Oxford University Press and more. All the latest content is available, no embargo periods.",
null,
"DeepDyve ### Freelancer DeepDyve ### Pro Price FREE$49/month\n\\$360/year\n\nSave searches from\nPubMed\n\nCreate folders to\n\nExport folders, citations"
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https://jp.maplesoft.com/support/help/maple/view.aspx?path=LinearAlgebra/General/aboutdata&L=J | [
"",
null,
"About Data Structures - Maple Help\n\nAbout Data Structures in the LinearAlgebra Package\n\n LinearAlgebra routines operate on three principal data structures: Matrices, Vectors, and scalars. The implementation of Matrices and Vectors is based on Maple's rtable data structure. As a result, table-based arrays, matrices and vectors are not interchangeable with LinearAlgebra Matrices and Vectors.\n Note: The documentation regarding the LinearAlgebra package uses the convention that \"matrix\" (lowercase \"m\") refers to a table-based matrix used by routines in the linalg package, and \"Matrix\" (uppercase \"M\") refers to an rtable-based Matrix used by routines in the LinearAlgebra package. The convention is similar for vectors and Vectors.",
null,
"Matrices\n\n • A Matrix is implemented as an rtable with subtype option Matrix.\n • Matrices are created by using the Matrix(..) constructor command, or by using the shortcut notation ().\n • The mathematical Matrix types supported in the LinearAlgebra package are: rectangular, triangular, Hessenberg, band, and diagonal. Each of these Matrix types can be created in dense or sparse format, and the first three also come in several varieties.\n i. rectangular:\n\n symmetric, skew-symmetric, hermitian, skew-hermitian\n\n ii. triangular:\n\n upper, upper unit, lower, lower unit\n\n iii. Hessenberg (triangular plus one extra diagonal):\n\n upper, lower\n\n There is no special Matrix type for block forms of the above Matrix types. That is, while the facility exists for the construction of block Matrices of various types (e.g., block diagonal, block triangular, etc.), the blocks are flattened during the construction process, and the resulting Matrix does not preserve the block form.\n • A number of special Matrices are provided in the LinearAlgebra package (e.g., the Identity Matrix, the Zero Matrix, and the Constant Matrix). In addition, the LinearAlgebra package includes many Matrix operations. See Details of the LinearAlgebra Package for more information.\n • The type recognizer for a Matrix is type/Matrix.\n • The i, jth entry of the Matrix A is accessed, extracted, and assigned to by using the notation A[i, j]. For more information see Matrix Entry Extraction and Matrix Entry Assignment.\n • Matrix algebra expressions such as A+B, A-B, and A.B are evaluated directly. For more information on performing Matrix algebra and using the Matrix multiplication operator, see rtable_algebra and dot, respectively.\n • The map, Map, map2, and Map2 commands can be used to apply a function to each entry in a Matrix.",
null,
"Vectors\n\n • A Vector is implemented as an rtable with subtype option Vector[column] or Vector[row].\n • Vectors are created by using the Vector(..) constructor command, or by using the available shortcut notation ().\n • A number of special Vectors are provided in the LinearAlgebra package (e.g., the Unit Vector, the Zero Vector, and the Constant Vector). In addition, the LinearAlgebra package includes many Vector operations. See Details of the LinearAlgebra Package for more information.\n • The type recognizer for a Vector is type/Vector.\n • The ith entry of the Vector V is accessed, extracted, and assigned to by using the notation V[i]. For more information see Vector Entry Extraction and Vector Entry Assignment.\n • The map, Map, map2, and Map2 commands can be used to apply a function to each entry in a Vector.",
null,
"Scalars\n\n • A scalar is a Maple object of type algebraic which neither is nor includes a Matrix or a Vector.",
null,
"Compatibility with Other Maple Objects\n\n The following objects are not recognized as Matrices, Vectors or scalars.\n •\n • lists of lists\n •\n • table-based arrays\n • linalg matrices\n • linalg vectors\n • rtable-based Arrays (rtables with subtype option Array)"
]
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https://www.tutorialspoint.com/what-are-the-regular-expressions-to-finite-automata | [
"# What are the regular expressions to finite automata?\n\nData Structure AlgorithmsComputer ScienceComputers\n\nFor each of the following languages, draw the finite automata (FA) accepting it.\n\n{a,b}*{a}\n\nThe language states that the automata accept the strings containing any number of a's and b's and finally ending in a.\n\nThe finite state automaton for the language is as follows −",
null,
"{a,b}*{b,aa}{a,b*}\n\nThe language states that the automata accept the strings starting and ending with any number of a's and b's and containing any of the substrings b and aa.\n\nThe finite state automaton for the language is a follows −",
null,
"{bbb,baa}*{a}\n\nThe language states that the automata accept the strings containing any number of bbb's and baa's followed by a single a.\n\nThe final state automaton for the language is as follows −",
null,
"{a}U{b}{a}*U{a}{b}*{a}\n\nThe language states that the automata accept the strings containing a,b followed by number of a's and stings starting with a followed by number of b's and end with a.\n\nThe finite state automaton for the language is as follows −",
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https://answers.everydaycalculation.com/add-fractions/1-4-plus-75-8 | [
"# Answers\n\nSolutions by everydaycalculation.com\n\n## Add 1/4 and 75/8\n\n1st number: 1/4, 2nd number: 9 3/8\n\n1/4 + 75/8 is 77/8.\n\n#### Steps for adding fractions\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 4 and 8 is 8\n\nNext, find the equivalent fraction of both fractional numbers with denominator 8\n2. For the 1st fraction, since 4 × 2 = 8,\n1/4 = 1 × 2/4 × 2 = 2/8\n3. Likewise, for the 2nd fraction, since 8 × 1 = 8,\n75/8 = 75 × 1/8 × 1 = 75/8\n4. Add the two like fractions:\n2/8 + 75/8 = 2 + 75/8 = 77/8\n5. So, 1/4 + 75/8 = 77/8\nIn mixed form: 95/8\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#### Add Fractions Calculator\n\n+\n\n© everydaycalculation.com"
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https://www.studypug.com/ca/phys/vertical-circular-motion | [
"# Vertical circular motion\n\n### Vertical circular motion\n\n#### Lessons\n\nIn this lesson, we will learn:\n• Solving problems involving vertical circular motion\n\nNotes:\n\n• An object moving in a circular path is in circular motion. If the speed of the object is constant, it is uniform circular motion.\n• An object in uniform circular motion does experience acceleration, even though its speed is constant. Remember, acceleration is change in velocity, and a velocity is made up of speed and direction. For the object to move in a circle, the direction of its velocity must change constantly. This change in direction is the acceleration, called centripetal acceleration (\"centripetal\" means \"towards the center\"). For an object moving in a circular path, the centripetal acceleration vector is always pointed towards the center of the circle.\n• Like any other type of acceleration, centripetal acceleration is caused by a force (called centripetal force). The centripetal force vector also always points towards the center of the circle.\n• In order for an object to be moving in a circular path, the net force acting on the object must be a centripetal force (a force that always is pointed towards the center). When multiple forces act on an object in circular motion, those forces must add up to a centripetal force. It is important to understand that centripetal force is not a separate force that acts on an object. It is a net force which follows a specific rule: it always points towards the center of the circular path.\n• In a horizontal circular motion problem, any forces acting on the object in the vertical direction must balance so that $\\Sigma F_{vertical} = 0N$ (otherwise the object would accelerate vertically). Only horizontal forces will contribute to the net force causing circular motion.\nPeriod and Frequency\n\n$T = \\frac{total time}{\\# of revolutions}$\n\n$f = \\frac{\\# of revolutions}{total time}$\n\n$T = \\frac{1}{f}$\n\n$T:$ period, in seconds (s)\n\n$f:$ frequency, in hertz (Hz)\n\nCentripetal Acceleration\n\n$a_{c} = \\frac{v^{2}}{r} = \\frac{4\\pi^{2}r}{T^{2}}$\n\n$a_{c}:$ centripetal acceleration, in meters per second squared $(m/s^{2})$\n\n$v:$ velocity, in meters per second (m/s)\n\n$r:$ radius, in meters (m)\n\n$T:$ period, in seconds (s)\n\n• 1.\nUniform vertical circular motion\n\nA 0.315 kg ball attached to the end of a 0.700 m string is swung in a vertical circle so that it has a constant speed of 4.50 m/s. What is the tension in the string at the top and bottom of the circular path?\n\n• 2.\nMinimum speed to keep an object on a vertical circular path\n\nA 0.245 kg ball is attached to the end of a 0.950 m string and swung in a vertical circle. Find the minimum speed the ball needs at the top of the path to continue travelling in a circle."
]
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https://www.physicsclassroom.com/class/vectors/u3l3c.cfm | [
"Vectors - Motion and Forces in Two Dimensions - Lesson 3 - Forces in Two Dimensions\n\n# Equilibrium and Statics",
null,
"When all the forces that act upon an object are balanced, then the object is said to be in a state of equilibrium. The forces are considered to be balanced if the rightward forces are balanced by the leftward forces and the upward forces are balanced by the downward forces. This however does not necessarily mean that all the forces are equal to each other. Consider the two objects pictured in the force diagram shown below. Note that the two objects are at equilibrium because the forces that act upon them are balanced; however, the individual forces are not equal to each other. The 50 N force is not equal to the 30 N force.",
null,
"If an object is at equilibrium, then the forces are balanced. Balanced is the key word that is used to describe equilibrium situations. Thus, the net force is zero and the acceleration is 0 m/s/s. Objects at equilibrium must have an acceleration of 0 m/s/s. This extends from Newton's first law of motion. But having an acceleration of 0 m/s/s does not mean the object is at rest. An object at equilibrium is either ...\n\n• at rest and staying at rest, or\n• in motion and continuing in motion with the same speed and direction.\n\nThis too extends from Newton's first law of motion.\n\n### Analyzing a Static Equilibrium Situation\n\nIf an object is at rest and is in a state of equilibrium, then we would say that the object is at \"static equilibrium.\" \"Static\" means stationary or at rest. A common physics lab is to hang an object by two or more strings and to measure the forces that are exerted at angles upon the object to support its weight. The state of the object is analyzed in terms of the forces acting upon the object. The object is a point on a string upon which three forces were acting. See diagram at right. If the object is at equilibrium, then the net force acting upon the object should be 0 Newton. Thus, if all the forces are added together as vectors, then the resultant force (the vector sum) should be 0 Newton. (Recall that the net force is \"the vector sum of all the forces\" or the resultant of adding all the individual forces head-to-tail.) Thus, an accurately drawn vector addition diagram can be constructed to determine the resultant. Sample data for such a lab are shown below.",
null,
"Force A Force B Force C Magnitude 3.4 N 9.2 N 9.8 N Direction 161 deg. 70 deg. 270 deg\n\nFor most students, the resultant was 0 Newton (or at least very close to 0 N). This is what we expected - since the object was at equilibrium, the net force (vector sum of all the forces) should be 0 N.",
null,
"Another way of determining the net force (vector sum of all the forces) involves using the trigonometric functions to resolve each force into its horizontal and vertical components. Once the components are known, they can be compared to see if the vertical forces are balanced and if the horizontal forces are balanced. The diagram below shows vectors A, B, and C and their respective components. For vectors A and B, the vertical components can be determined using the sine of the angle and the horizontal components can be analyzed using the cosine of the angle. The magnitude and direction of each component for the sample data are shown in the table below the diagram.",
null,
"",
null,
"The data in the table above show that the forces nearly balance. An analysis of the horizontal components shows that the leftward component of A nearly balances the rightward component of B. An analysis of the vertical components show that the sum of the upward components of A + B nearly balance the downward component of C. The vector sum of all the forces is (nearly) equal to 0 Newton. But what about the 0.1 N difference between rightward and leftward forces and the 0.2 N difference between the upward and downward forces? Why do the components of force only nearly balance? The sample data used in this analysis are the result of measured data from an actual experimental setup. The difference between the actual results and the expected results is due to the error incurred when measuring force A and force B. We would have to conclude that this low margin of experimental error reflects an experiment with excellent results. We could say it's \"close enough for government work.\"\n\n###",
null,
"Analyzing a Hanging Sign\n\nThe above analysis of the forces acting upon an object in equilibrium is commonly used to analyze situations involving objects at static equilibrium. The most common application involves the analysis of the forces acting upon a sign that is at rest. For example, consider the picture at the right that hangs on a wall. The picture is in a state of equilibrium, and thus all the forces acting upon the picture must be balanced. That is, all horizontal components must add to 0 Newton and all vertical components must add to 0 Newton. The leftward pull of cable A must balance the rightward pull of cable B and the sum of the upward pull of cable A and cable B must balance the weight of the sign.\n\nSuppose the tension in both of the cables is measured to be 50 N and that the angle that each cable makes with the horizontal is known to be 30 degrees. What is the weight of the sign? This question can be answered by conducting a force analysis using trigonometric functions. The weight of the sign is equal to the sum of the upward components of the tension in the two cables. Thus, a trigonometric function can be used to determine this vertical component. A diagram and accompanying work is shown below.",
null,
"Since each cable pulls upwards with a force of 25 N, the total upward pull of the sign is 50 N. Therefore, the force of gravity (also known as weight) is 50 N, down. The sign weighs 50 N.\n\nIn the above problem, the tension in the cable and the angle that the cable makes with the horizontal are used to determine the weight of the sign. The idea is that the tension, the angle, and the weight are related. If the any two of these three are known, then the third quantity can be determined using trigonometric functions.",
null,
"As another example that illustrates this idea, consider the symmetrical hanging of a sign as shown at the right. If the sign is known to have a mass of 5 kg and if the angle between the two cables is 100 degrees, then the tension in the cable can be determined. Assuming that the sign is at equilibrium (a good assumption if it is remaining at rest), the two cables must supply enough upward force to balance the downward force of gravity. The force of gravity (also known as weight) is 49 N (Fgrav = m*g), so each of the two cables must pull upwards with 24.5 N of force. Since the angle between the cables is 100 degrees, then each cable must make a 50-degree angle with the vertical and a 40-degree angle with the horizontal. A sketch of this situation (see diagram below) reveals that the tension in the cable can be found using the sine function. The triangle below illustrates these relationships.",
null,
"### Thinking Conceptually\n\nThere is an important principle that emanates from some of the trigonometric calculations performed above. The principle is that as the angle with the horizontal increases, the amount of tensional force required to hold the sign at equilibrium decreases. To illustrate this, consider a 10-Newton picture held by three different wire orientations as shown in the diagrams below. In each case, two wires are used to support the picture; each wire must support one-half of the sign's weight (5 N). The angle that the wires make with the horizontal is varied from 60 degrees to 15 degrees. Use this information and the diagram below to determine the tension in the wire for each orientation. When finished, click the button to view the answers.",
null,
"In conclusion, equilibrium is the state of an object in which all the forces acting upon it are balanced. In such cases, the net force is 0 Newton. Knowing the forces acting upon an object, trigonometric functions can be utilized to determine the horizontal and vertical components of each force. If at equilibrium, then all the vertical components must balance and all the horizontal components must balance.\n\n### We Would Like to Suggest ...",
null,
"Sometimes it isn't enough to just read about it. You have to interact with it! And that's exactly what you do when you use one of The Physics Classroom's Interactives. We would like to suggest that you combine the reading of this page with the use of our Name That Vector Interactive, our Vector Addition Interactive, or our Vector Guessing Game Interactive. All three Interactives can be found in the Physics Interactive section of our website and provide an interactive experience with the skill of adding vectors.\n\n### Check Your Understanding\n\nThe following questions are meant to test your understanding of equilibrium situations. Click the button to view the answers to these questions.\n\n1. The following picture is hanging on a wall. Use trigonometric functions to determine the weight of the picture.",
null,
"2. The sign below hangs outside the physics classroom, advertising the most important truth to be found inside. The sign is supported by a diagonal cable and a rigid horizontal bar. If the sign has a mass of 50 kg, then determine the tension in the diagonal cable that supports its weight.",
null,
"3. The following sign can be found in Glenview. The sign has a mass of 50 kg. Determine the tension in the cables.",
null,
"4. After its most recent delivery, the infamous stork announces the good news. If the sign has a mass of 10 kg, then what is the tensional force in each cable? Use trigonometric functions and a sketch to assist in the solution.",
null,
"5. Suppose that a student pulls with two large forces (F1 and F2) in order to lift a 1-kg book by two cables. If the cables make a 1-degree angle with the horizontal, then what is the tension in the cable?",
null,
"Next Section:\n\nFollow Us"
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https://newpathworksheets.com/math/grade-2/money-1/maryland-standards | [
"## ◂Math Worksheets and Study Guides Second Grade. Money\n\n### The resources above correspond to the standards listed below:\n\n#### Maryland Standards\n\nMD.6.0. Knowledge of Number Relationships and Computation/Arithmetic: Students will describe, represent, or apply numbers or their relationships or will estimate or compute using mental strategies, paper/pencil, or technology.\n6.A.3. Knowledge of Number and Place Value: Apply knowledge of money.\n6.A.3.a. Determine the value of a given set of mixed currency up to \\$10.\n6.A.3.b. Represent money amounts up to \\$10.\n6.C.1. Number Computation: Analyze number relations and compute.\n6.C.1.f. Add and subtract money amounts up to \\$1.\nMD.7.0. Processes of Mathematics: Students demonstrate the processes of mathematics by making connections and applying reasoning to solve problems and to communicate their findings.\n7.C.1. Communications: Present mathematical ideas using words, symbols, visual displays, or technology\n7.C.1.a. Use multiple representations to express concepts or solutions\n7.C.1.d. Express solutions using concrete materials\nStandards"
]
| [
null
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https://statecancerprofiles.cancer.gov/historicaltrend/data.php?0&9928&999&7599&001&061&00&2&0&0&2&1&1&1&4 | [
"",
null,
"",
null,
"Historical Trends > Interpret\n\n## Interpretation of Historical Trends Data\n\n### Historical Trends (1975-2019)\n\nMortality, Mississippi, Ovary, All Races (incl Hisp), All Ages, Female\n\n#### Mortality, Mississippi, Ovary, All Races (incl Hisp), All Ages, Female\n\nLine graph with 45 years and 2 segments\nDuring 1975-2002, the APC1 in the rate of cancer was rising: 1.0 with a 95% confidence interval from 0.4 to 1.5.\nDuring 2002-2019, the APC1 in the rate of cancer was falling: -2.0 with a 95% confidence interval from -3.0 to -1.1.\nYearly points:\nIn 1975, the observed rate was 7.5. The estimated rate was 6.4.\nIn 1976, the observed rate was 6.3. The estimated rate was 6.4.\nIn 1977, the observed rate was 6.7. The estimated rate was 6.5.\nIn 1978, the observed rate was 7.1. The estimated rate was 6.5.\nIn 1979, the observed rate was 7.0. The estimated rate was 6.6.\nIn 1980, the observed rate was 6.6. The estimated rate was 6.7.\nIn 1981, the observed rate was 7.3. The estimated rate was 6.7.\nIn 1982, the observed rate was 6.5. The estimated rate was 6.8.\nIn 1983, the observed rate was 6.7. The estimated rate was 6.9.\nIn 1984, the observed rate was 5.4. The estimated rate was 6.9.\nIn 1985, the observed rate was 5.3. The estimated rate was 7.0.\nIn 1986, the observed rate was 7.3. The estimated rate was 7.1.\nIn 1987, the observed rate was 7.8. The estimated rate was 7.1.\nIn 1988, the observed rate was 6.8. The estimated rate was 7.2.\nIn 1989, the observed rate was 7.2. The estimated rate was 7.3.\nIn 1990, the observed rate was 6.3. The estimated rate was 7.3.\nIn 1991, the observed rate was 7.4. The estimated rate was 7.4.\nIn 1992, the observed rate was 8.8. The estimated rate was 7.5.\nIn 1993, the observed rate was 7.8. The estimated rate was 7.6.\nIn 1994, the observed rate was 7.1. The estimated rate was 7.6.\nIn 1995, the observed rate was 6.5. The estimated rate was 7.7.\nIn 1996, the observed rate was 6.3. The estimated rate was 7.8.\nIn 1997, the observed rate was 8.1. The estimated rate was 7.8.\nIn 1998, the observed rate was 8.5. The estimated rate was 7.9.\nIn 1999, the observed rate was 8.7. The estimated rate was 8.0.\nIn 2000, the observed rate was 8.7. The estimated rate was 8.1.\nIn 2001, the observed rate was 8.4. The estimated rate was 8.2.\nIn 2002, the observed rate was 8.5. The estimated rate was 8.2.\nIn 2003, the observed rate was 7.0. The estimated rate was 8.1.\nIn 2004, the observed rate was 7.9. The estimated rate was 7.9.\nIn 2005, the observed rate was 8.4. The estimated rate was 7.7.\nIn 2006, the observed rate was 8.1. The estimated rate was 7.6.\nIn 2007, the observed rate was 7.1. The estimated rate was 7.4.\nIn 2008, the observed rate was 7.1. The estimated rate was 7.3.\nIn 2009, the observed rate was 7.3. The estimated rate was 7.1.\nIn 2010, the observed rate was 6.8. The estimated rate was 7.0.\nIn 2011, the observed rate was 7.2. The estimated rate was 6.8.\nIn 2012, the observed rate was 6.6. The estimated rate was 6.7.\nIn 2013, the observed rate was 7.0. The estimated rate was 6.6.\nIn 2014, the observed rate was 5.7. The estimated rate was 6.4.\nIn 2015, the observed rate was 5.8. The estimated rate was 6.3.\nIn 2016, the observed rate was 6.3. The estimated rate was 6.2.\nIn 2017, the observed rate was 7.0. The estimated rate was 6.0.\nIn 2018, the observed rate was 5.9. The estimated rate was 5.9.\nIn 2019, the observed rate was 5.0. The estimated rate was 5.8.\n\nNotes:\n• Created by statecancerprofiles.cancer.gov on 07/02/2022 7:41 pm.\n• Regression lines calculated using the Joinpoint Regression Program (Version 4.8.0.0).\n• 1 The APC is the Annual Percent Change over the time interval. Rates used in the calculation of the APC are age-adjusted to the 2000 US standard population (19 age groups: <1, 1-4, 5-9, ... , 80-84, 85+).\n\n• Explanation of the Calculation of the Trend:\n• If the APC is less than -1.5, the trend is falling.\n• If the APC is between -1.5 and -0.5, the trend is slightly falling.\n• If the APC is between -0.5 and 0.5, the trend is statistically stable.\n• If the APC is between 0.5 and 1.5, the trend is slightly rising.\n• If the APC is greater than 1.5, the trend is rising.\n• Source: Incidence data provided by the SEER Program and the National Program of Cancer Registries SEER*Stat Database (2001-2018) - United States Department of Health and Human Services, Centers for Disease Control and Prevention (based on the 2019 submission). Rates calculated by the National Cancer Institute using SEER*Stat. Rates are age-adjusted to the 2000 US standard population (19 age groups: <1, 1-4, 5-9, ... , 80-84, 85+). Rates are for invasive cancer only (except for bladder cancer which is invasive and in situ) or unless otherwise specified. Population counts for denominators are based on Census populations as modified by NCI. The US populations included with the data release have been adjusted for the population shifts due to hurricanes Katrina and Rita for 62 counties and parishes in Alabama, Mississippi, Louisiana, and Texas. The 1969-2018 US Population Data File is used with SEER November 2020 data. Rates and trends in this graph are computed using the same standard for malignancy. For more information see malignant.html\n\nSource: Death data provided by the National Vital Statistics System public use data file. Death rates calculated by the National Cancer Institute using SEER*Stat. Death rates (deaths per 100,000 population per year) are age-adjusted to the 2000 US standard population (19 age groups: (<1, 1-4, 5-9, ... , 80-84, 85+). Population counts for denominators are based on Census populations as modified by NCI. The US populations included with the data release have been adjusted for the population shifts due to hurricanes Katrina and Rita for 62 counties and parishes in Alabama, Mississippi, Louisiana, and Texas. 1969-2018 US Population Data File is used with mortality data."
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"https://statecancerprofiles.cancer.gov/i/arrow-crumb.png",
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http://ktthaiweb-r1.pkt.cc/ | [
"",
null,
"'.\\$ice2_station.' is currently offline '; } if(\\$success!=2){ //if connection fputs(\\$fp,\"GET /status2.xsl HTTP/1.0\\r\\nUser-Agent: Icecast2 XSL Parser (Mozilla Compatible)\\r\\n\\r\\n\"); //get status2.xsl while(!feof(\\$fp)) { \\$page .= fgets(\\$fp, 1000); } fclose(\\$fp); //close connection \\$page = ereg_replace(\".*\n```\", \"\", \\$page); //extract data\n\\$page = ereg_replace(\"```\n.*\", \",\", \\$page); //extract data \\$numbers = explode(\",\",\\$page); //bomb it and extract data \\$mount = \\$numbers; \\$connections = \\$numbers; \\$stream_n = \\$numbers; \\$listeners = \\$numbers; \\$desc = \\$numbers; \\$cur_song = \\$numbers; \\$str_url = \\$numbers; \\$client_info = \\$numbers; \\$test1 = \\$numbers; //set vars that where empty and still dont know what the heck those values are ;-) \\$test2 = \\$numbers; //set vars that where empty and still dont know what the heck those values are ;-) \\$test3 = \\$numbers; //set vars that where empty and still dont know what the heck those values are ;-) \\$mount = \\$numbers; \\$connections = \\$numbers; \\$station =\\$numbers; \\$listeners = \\$numbers; \\$description = \\$numbers; \\$cur_song = \\$numbers; \\$www_url = \\$numbers; //edit html to fit your stations site, this display is for online status echo'",
null,
"",
null,
"'.\\$description.' - '.\\$cur_song.'- '.\\$listeners.'",
null,
"Listen Now\n'; } ?>"
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http://www.thefullwiki.org/Indian_mathematics | [
"# Indian mathematics: Wikis\n\nNote: Many of our articles have direct quotes from sources you can cite, within the Wikipedia article! This article doesn't yet, but we're working on it! See more info or our list of citable articles.\n\n### Did you know ...\n\nMore interesting facts on Indian mathematics\n\n# Encyclopedia\n\nBackground History of science",
null,
"Theories/sociology Historiography Pseudoscience In early cultures in Classical Antiquity In the Middle Ages In the Renaissance Scientific Revolution Natural sciences Astronomy Biology Botany Chemistry Ecology Geography Geology Paleontology Physics Mathematics Algebra Calculus Combinatorics Geometry Logic Probability Statistics Trigonometry Social sciences Anthropology Economics Linguistics Political science Psychology Sociology Technology Agricultural science Computer science Materials science Medicine Timelines Portal Categories\n\nIndian mathematics is the mathematics that emerged in South Asia, also known as the Indian subcontinent, from ancient times until the end of the 18th century. In the classical period of Indian mathematics (400 AD to 1200 AD), important contributions were made by scholars like Aryabhata, Brahmagupta, and Bhaskara II. The decimal number system in use today was first recorded in Indian mathematics. Indian mathematicians made early contributions to the study of the concept of zero as a number, negative numbers, arithmetic, and algebra. In addition, trigonometry, having evolved in the Hellenistic world and having been introduced into ancient India through the translation of Greek works, was further advanced in India, and, in particular, the modern definitions of sine and cosine were developed there. These mathematical concepts were transmitted to the Middle East, China, and Europe and led to further developments that now form the foundations of many areas of mathematics.\n\nAncient and medieval Indian mathematical works, all composed in Sanskrit, usually consisted of a section of sutras in which a set of rules or problems were stated with great economy in verse in order to aid memorization by a student. This was followed by a second section consisting of a prose commentary (sometimes multiple commentaries by different scholars) that explained the problem in more detail and provided justification for the solution. In the prose section, the form (and therefore its memorization) was not considered as important as the ideas involved. All mathematical works were orally transmitted until approximately 500 BCE; thereafter, they were transmitted both orally and in manuscript form. The oldest extant mathematical document produced on the Indian subcontinent is the birch bark Bakhshali Manuscript, discovered in 1881 in the village of Bakhshali, near Peshawar (modern day Pakistan) and is likely from the seventh century CE.\n\nA later landmark in Indian mathematics was the development of the series expansions for trigonometric functions (sine, cosine, and arc tangent) by mathematicians of the Kerala School in the fifteenth century CE. Their remarkable work, completed two centuries before the invention of calculus in Europe, provided what is now considered the first example of a power series (apart from geometric series). However, they did not formulate a systematic theory of differentiation and integration, nor is there any direct evidence of their results being transmitted outside Kerala.\n\n## Fields of Indian mathematics\n\nSome of the areas of mathematics studied in ancient and medieval India include the following:\n\n## Prehistory\n\nExcavations at Harappa, Mohenjo-daro and other sites of the Indus Valley Civilization have uncovered evidence of the use of \"practical mathematics\". The people of the IVC manufactured bricks whose dimensions were in the proportion 4:2:1, considered favorable for the stability of a brick structure. They used a standardized system of weights based on the ratios: 1/20, 1/10, 1/5, 1/2, 1, 2, 5, 10, 20, 50, 100, 200, and 500, with the unit weight equaling approximately 28 grams (and approximately equal to the English ounce or Greek uncia). They mass produced weights in regular geometrical shapes, which included hexahedra, barrels, cones, and cylinders, thereby demonstrating knowledge of basic geometry.\n\nThe inhabitants of Indus civilization also tried to standardize measurement of length to a high degree of accuracy. They designed a ruler—the Mohenjo-daro ruler—whose unit of length (approximately 1.32 inches or 3.4 centimetres) was divided into ten equal parts. Bricks manufactured in ancient Mohenjo-daro often had dimensions that were integral multiples of this unit of length.\n\n## Vedic period\n\n### Samhitas and Brahmanas\n\nThe religious texts of the Vedic Period provide evidence for the use of large numbers. By the time of the Yajurvedasaṃhitā (1200-900 BCE), numbers as high as 1012 were being included in the texts. For example, the mantra (sacrificial formula) at the end of the annahoma (\"food-oblation rite\") performed during the aśvamedha (\"horse sacrifice\"), and uttered just before-, during-, and just after sunrise, invokes powers of ten from a hundred to a trillion:\n\n\"Hail to śata (\"hundred,\" 102), hail to sahasra (\"thousand,\" 103), hail to ayuta (\"ten thousand,\" 104), hail to niyuta (\"hundred thousand,\" 105), hail to prayuta (\"million,\" 106), hail to arbuda (\"ten million,\" 107), hail to nyarbuda (\"hundred million,\" 108), hail to samudra (\"billion,\" 109, literally \"ocean\"), hail to madhya (\"ten billion,\" 1010, literally \"middle\"), hail to anta (\"hundred billion,\" 1011,lit., \"end\"), hail to parārdha (\"one trillion,\" 1012 lit., \"beyond parts\"), hail to the dawn (uśas), hail to the twilight (vyuṣṭi), hail to the one which is going to rise (udeṣyat), hail to the one which is rising (udyat), hail to the one which has just risen (udita), hail to the heaven (svarga), hail to the world (loka), hail to all.\"\n\nThe Satapatha Brahmana (ca. 7th century BCE) contains rules for ritual geometric constructions that are similar to the Sulba Sutras.\n\n### Śulba Sūtras\n\nThe Śulba Sūtras (literally, \"Aphorisms of the Chords\" in Vedic Sanskrit) (c. 700-400 BCE) list rules for the construction of sacrificial fire altars. Most mathematical problems considered in the Śulba Sūtras spring from \"a single theological requirement,\" that of constructing fire altars which have different shapes but occupy the same area. The altars were required to be constructed of five layers of burnt brick, with the further condition that each layer consist of 200 bricks and that no two adjacent layers have congruent arrangements of bricks.\n\nAccording to (Hayashi 2005, p. 363), the Śulba Sūtras contain \"the earliest extant verbal expression of the Pythagorean Theorem in the world, although it had already been known to the Old Babylonians.\"\n\nThe diagonal rope (akṣṇayā-rajju) of an oblong (rectangle) produces both which the flank (pārśvamāni) and the horizontal (tiryaṇmānī) <ropes> produce separately.\"\n\nSince the statement is a sūtra, it is necessarily compressed and what the ropes produce is not elaborated on, but the context clearly implies the square areas constructed on their lengths, and would have been explained so by the teacher to the student.\n\nThey contain lists of Pythagorean triples, which are particular cases of Diophantine equations. They also contain statements (that with hindsight we know to be approximate) about squaring the circle and \"circling the square.\"\n\nBaudhayana (c. 8th century BCE) composed the Baudhayana Sulba Sutra, the best-known Sulba Sutra, which contains examples of simple Pythagorean triples, such as: (3,4,5), (5,12,13), (8,15,17), (7,24,25), and (12,35,37) as well as a statement of the Pythagorean theorem for the sides of a square: \"The rope which is stretched across the diagonal of a square produces an area double the size of the original square.\" It also contains the general statement of the Pythagorean theorem (for the sides of a rectangle): \"The rope stretched along the length of the diagonal of a rectangle makes an area which the vertical and horizontal sides make together.\" Baudhayana gives a formula for the square root of two,",
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"$\\sqrt{2} = 1 + \\frac{1}{3} + \\frac{1}{3\\cdot4} - \\frac{1}{3\\cdot 4\\cdot 34} \\approx 1.4142156 \\cdots$\n\nThe formula is accurate up to five decimal places, the true value being",
null,
"$1.41421356 \\cdots$ This formula is similar in structure to the formula found on a Mesopotamian tablet from the Old Babylonian period (1900-1600 BCE):",
null,
"$\\sqrt{2} = 1 + \\frac{24}{60} + \\frac{51}{60^2} + \\frac{10}{60^3} = 1.41421297.$\n\nwhich expresses",
null,
"$\\sqrt{2}$ in the sexagesimal system, and which too is accurate up to 5 decimal places (after rounding).\n\nAccording to mathematician S. G. Dani, the Babylonian cuneiform tablet Plimpton 322 written ca. 1850 BCE \"contains fifteen Pythagorean triples with quite large entries, including (13500, 12709, 18541) which is a primitive triple, indicating, in particular, that there was sophisticated understanding on the topic\" in Mesopotamia in 1850 BCE. \"Since these tablets predate the Sulbasutras period by several centuries, taking into account the contextual appearance of some of the triples, it is reasonable to expect that similar understanding would have been there in India.\" Dani goes on to say:\n\n\"As the main objective of the Sulvasutras was to describe the constructions of altars and the geometric principles involved in them, the subject of Pythagorean triples, even if it had been well understood may still not have featured in the Sulvasutras. The occurrence of the triples in the Sulvasutras is comparable to mathematics that one may encounter in an introductory book on architecture or another similar applied area, and would not correspond directly to the overall knowledge on the topic at that time. Since, unfortunately, no other contemporaneous sources have been found it may never be possible to settle this issue satisfactorily.\"\n\nIn all, three Sulba Sutras were composed. The remaining two, the Manava Sulba Sutra composed by Manava (fl. 750-650 BC) and the Apastamba Sulba Sutra, composed by Apastamba (c. 600 BC), contained results similar to the Baudhayana Sulba Sutra.\n\nVyakarana\n\nAn important landmark of the Vedic period was the work of Sanskrit grammarian, Pāṇini (c. 520-460 BCE). His grammar includes early use of Boolean logic, of the null operator, and of context free grammars, and includes a precursor of the Backus–Naur form (used in the description programming languages).\n\n## Jaina Mathematics (400 BCE - 200 CE)\n\nAlthough Jainism as a religion and philosophy predates its most famous exponent, Mahavira (6th century BC), who was a contemporary of Gautama Buddha, most Jaina texts on mathematical topics were composed after the 6th century BCE. Jaina mathematicians are important historically as crucial links between the mathematics of the Vedic period and that of the \"Classical period.\"\n\nA significant historical contribution of Jaina mathematicians lay in their freeing Indian mathematics from its religious and ritualistic constraints. In particular, their fascination with the enumeration of very large numbers and infinities, led them to classify numbers into three classes: enumerable, innumerable and infinite. Not content with a simple notion of infinity, they went on to define five different types of infinity: the infinite in one direction, the infinite in two directions, the infinite in area, the infinite everywhere, and the infinite perpetually. In addition, Jaina mathematicians devised notations for simple powers (and exponents) of numbers like squares and cubes, which enabled them to define simple algebraic equations (beezganit samikaran). Jaina mathematicians were apparently also the first to use the word shunya (literally void in Sanskrit) to refer to zero. More than a millennium later, their appellation became the English word \"zero\" after a tortuous journey of translations and transliterations from India to Europe . (See Zero: Etymology.)\n\nIn addition to Surya Prajnapti, important Jaina works on mathematics included the Vaishali Ganit (c. 3rd century BCE); the Sthananga Sutra (fl. 300 BCE - 200 CE); the Anoyogdwar Sutra (fl. 200 BCE - 100 CE); and the Satkhandagama (c. 2nd century CE). Important Jaina mathematicians included Bhadrabahu (d. 298 BCE), the author of two astronomical works, the Bhadrabahavi-Samhita and a commentary on the Surya Prajinapti; Yativrisham Acharya (c. 176 BCE), who authored a mathematical text called Tiloyapannati; and Umasvati (c. 150 BCE), who, although better known for his influential writings on Jaina philosophy and metaphysics, composed a mathematical work called Tattwarthadhigama-Sutra Bhashya.\n\nPingala\n\nAmong other scholars of this period who contributed to mathematics, the most notable is Pingala (piṅgalá) (fl. 300-200 BCE), a musical theorist who authored the Chandas Shastra (chandaḥ-śāstra, also Chandas Sutra chandaḥ-sūtra), a Sanskrit treatise on prosody . There is evidence that in his work on the enumeration of syllabic combinations, Pingala stumbled upon both the Pascal triangle and Binomial coefficients, although he did not have knowledge of the Binomial theorem itself. Pingala's work also contains the basic ideas of Fibonacci numbers (called maatraameru). Although the Chandah sutra hasn't survived in its entirety, a 10th century commentary on it by Halāyudha has. Halāyudha, who refers to the Pascal triangle as Meru-prastāra (literally \"the staircase to Mount Meru\"), has this to say:\n\n\"Draw a square. Beginning at half the square, draw two other similar squares below it; below these two, three other squares, and so on. The marking should be started by putting 1 in the first square. Put 1 in each of the two squares of the second line. In the third line put 1 in the two squares at the ends and, in the middle square, the sum of the digits in the two squares lying above it. In the fourth line put 1 in the two squares at the ends. In the middle ones put the sum of the digits in the two squares above each. Proceed in this way. Of these lines, the second gives the combinations with one syllable, the third the combinations with two syllables, ...\"\n\nThe text also indicates that Pingala was aware of the combinatorial identity:",
null,
"${n \\choose 0} + {n \\choose 1} + {n \\choose 2} + \\cdots + {n \\choose n-1} + {n \\choose n} = 2^n$\nKatyayana\n\nThough not a Jaina mathematician, Katyayana (c. 3rd century BCE) is notable for being the last of the Vedic mathematicians. He wrote the Katyayana Sulba Sutra, which presented much geometry, including the general Pythagorean theorem and a computation of the square root of 2 correct to five decimal places.\n\nMathematicians of ancient and early medieval India were almost all Sanskrit pandits (paṇḍita \"learned man\"), who were trained in Sanskrit language and literature, and possessed \"a common stock of knowledge in grammar (vyākaraṇa), exegesis (mīmāṃsā) and logic (nyāya).\" Memorization of \"what is heard\" (śruti in Sanskrit) through recitation played a major role in the transmission of sacred texts in ancient India. Memorization and recitation was also used to transmit philosophical and literary works, as well as treatises on ritual and grammar. Modern scholars of ancient India have noted the \"truly remarkable achievements of the Indian pandits who have preserved enormously bulky texts orally for millennia.\"\n\n### Styles of Memorization\n\nProdigous energy was expended by ancient Indian culture in ensuring that these texts were transmitted from generation to generation with inordinate fidelity. For example, memorization of the sacred Vedas included up to eleven forms of recitation of the same text. The texts were subsequently \"proof-read\" by comparing the different recited versions. Forms of recitation included the jaṭā-pāṭha (literally \"mesh recitation\") in which every two adjacent words in the text were first recited in their original order, then repeated in the reverse order, and finally repeated again in the original order. The recitation thus proceeded as:\n\nword1word2, word2word1, word1word2; word2word3, word3word2, word2word3; ...\n\nIn another form of recitation, dhvaja-pāṭha (literally \"flag recitation\") a sequence of N words were recited (and memorized) by pairing the first two and last two words and then proceeding as:\n\nword1word2, word(N-1)wordN; word2word3, word(N-3)word(N-2); ...; word(N-1)wordN, word1word2;\n\nThe most complex form of recitation, ghana-pāṭha (literally \"dense recitation\"), according to (Filliozat 2004, p. 139), took the form:\n\nword1word2, word2word1, word1word2word3, word3word2word1, word1word2word3; word2word3, word3word2, word2word3word4, word4word3word2, word2word3word4; ...\n\nThat these methods have been effective, is testified to by the preservation of the most ancient Indian religious text, the Ṛgveda (ca. 1500 BCE), as a single text, without any variant readings. Similar methods were used for memorizing mathematical texts, whose transmission remained exclusively oral until the end of the Vedic period (ca. 500 BCE).\n\n### The Sūtra Genre\n\nMathematical activity in ancient India began as a part of a \"methodological reflexion\" on the sacred Vedas, which took the form of works called Vedāṇgas, or, \"Ancillaries of the Veda\" (7th-4th century BCE). The need to conserve the sound of sacred text by use of śikṣā (phonetics) and chandas (metrics); to conserve its meaning by use of vyākaraṇa (grammar) and nirukta (etymology); and to correctly perform the rites at the correct time by the use of kalpa (ritual) and jyotiṣa (astronomy), gave rise to the six disciplines of the Vedāṇgas. Mathematics arose as a part of the last two disciplines, ritual and astronomy (which also included astrology). Since the Vedāṇgas immediately preceded the use of writing in ancient India, they formed the last of the exclusively oral literature. They were expressed in a highly compressed mnemonic form, the sūtra (literally, \"thread\"):\n\nThe knowers of the sūtra know it as having few phonemes, being devoid of ambiguity, containing the essence, facing everything, being without pause and unobjectionable.\n\nExtreme brevity was achieved through multiple means, which included using ellipsis \"beyond the tolerance of natural language,\" using technical names instead of longer descriptive names, abridging lists by only mentioning the first and last entries, and using markers and variables. The sūtras create the impression that communication through the text was \"only a part of the whole instruction. The rest of the instruction must have been transmitted by the so-called Guru-shishya parampara, 'uninterrupted succession from teacher (guru) to the student (śisya),' and it was not open to the general public\" and perhaps even kept secret. The brevity achieved in a sūtra is demonstrated in the following example from the Baudhāyana Śulba Sūtra (700 BCE).",
null,
"",
null,
"The design of the domestic fire altar in the Śulba Sūtra\n\nThe domestic fire-altar in the Vedic period was required by ritual to have a square base and be constituted of five layers of bricks with 21 bricks in each layer. One method of constructing the altar was to divide one side of the square into three equal parts using a cord or rope, to next divide the transverse (or perpendicular) side into seven equal parts, and thereby sub-divide the square into 21 congruent rectangles. The bricks were then designed to be of the shape of the constituent rectangle and the layer was created. To form the next layer, the same formula was used, but the bricks were arranged transversely. The process was then repeated three more times (with alternating directions) in order to complete the construction. In the Baudhāyana Śulba Sūtra, this procedure is described in the following words:\n\n\"II.64. After dividing the quadri-lateral in seven, one divides the transverse [cord] in three.\nII.65. In another layer one places the [bricks] North-pointing.\"\n\nAccording to (Filliozat 2004, p. 144), the officiant constructing the altar has only a few tools and materials at his disposal: a cord (Sanskrit, rajju, f.), two pegs (Sanskrit, śanku, m.), and clay to make the bricks (Sanskrit, iṣṭakā, f.). Concision is achieved in the sūtra, by not explicitly mentioning what the adjective \"transverse\" qualifies; however, from the feminine form of the (Sanskrit) adjective used, it is easily inferred to qualify \"cord.\" Similarly, in the second stanza, \"bricks\" are not explicitly mentioned, but inferred again by the feminine plural form of \"North-pointing.\" Finally, the first stanza, never explicitly says that the first layer of bricks are oriented in the East-West direction, but that too is implied by the explicit mention of \"North-pointing\" in the second stanza; for, if the orientation was meant to be the same in the two layers, it would either not be mentioned at all or be only mentioned in the first stanza. All these inferences are made by the officiant as he recalls the formula from his memory.\n\n## The Written Tradition: Prose Commentary\n\nWith the increasing complexity of mathematics and other exact sciences, both writing and computation were required. Consequently, many mathematical works began to be written down in manuscripts that were then copied and re-copied from generation to generation.\n\n\"India today is estimated to have about thirty million manuscripts, the largest body of handwritten reading material anywhere in the world. The literate culture of Indian science goes back to at least the fifth century B.C. ... as is shown by the elements of Mesopotamian omen literature and astronomy that entered India at that time and (were) definitely not ... preserved orally.\"\n\nThe earliest mathematical prose commentary was that on the work, Āryabhaṭīya (written 499 CE), a work on astronomy and mathematics. The mathematical portion of the Āryabhaṭīya was composed of 33 sūtras (in verse form) consisting of mathematical statements or rules, but without any proofs. However, according to (Hayashi 2003, p. 123), \"this does not necessarily mean that their authors did not prove them. It was probably a matter of style of exposition.\" From the time of Bhaskara I (600 CE onwards), prose commentaries increasingly began to include some derivations (upapatti). Bhaskara I's commentary on the Āryabhaṭīya, had the following structure:\n\n• Rule ('sūtra') in verse by Āryabhaṭa\n• Commentary by Bhāskara I, consisting of:\n• Elucidation of rule (derivations were still rare then, but became more common later)\n• Example (uddeśaka) usually in verse.\n• Setting (nyāsa/sthāpanā) of the numerical data.\n• Working (karana) of the solution.\n• Verification (pratyayakaraṇa, literally \"to make conviction\") of the answer. These became rare by the 13th century, derivations or proofs being favored by then.\n\nTypically, for any mathematical topic, students in ancient India first memorized the sūtras, which, as explained earlier, were \"deliberately inadequate\" in explanatory details (in order to pithily convey the bare-bone mathematical rules). The students then worked through the topics of the prose commentary by writing (and drawing diagrams) on chalk- and dust-boards (i.e. boards covered with dust). The latter activity, a staple of mathematical work, was to later prompt mathematician-astronomer, Brahmagupta (fl. 7th century CE), to characterize astronomical computations as \"dust work\" (Sanskrit: dhulikarman).\n\n## Numerals and the Decimal Number System\n\nIt is well known that the decimal place-value system in use today was first recorded in India, then transmitted to the Islamic world, and eventually to Europe. The Syrian bishop Severus Sebokht wrote in the mid-seventh century CE about the \"nine signs\" of the Indians for expressing numbers. However, how, when, and where the first decimal place value system was invented is not so clear.\n\nThe earliest extant script used in India was the Kharoṣṭhī script used in the Gandhara culture of the north-west. It is thought to be of Aramaic origin and it was in use from the fourth century BCE to the fourth century CE. Almost contemporaneously, another script, the Brāhmī script, appeared on much of the sub-continent, and would later become the foundation of many scripts of South Asia and South-east Asia. Both scripts had numeral symbols and numeral systems, which were initially not based on a place-value system.\n\nThe earliest surviving evidence of decimal place value numerals in India and southeast Asia is from the middle of the first millennium CE. A copper plate from Gujarat, India mentions the date 595 CE, written in a decimal place value notation, although there is some doubt as to the authenticity of the plate. Decimal numerals recording the years 683 CE have also been found in stone inscriptions in Indonesia and Cambodia, where Indian cultural influence was substantial.\n\nThere are older textual sources, although the extant manuscript copies of these texts are from much later dates. Probably the earliest such source is the work of the Buddhist philosopher Vasumitra dated likely to the first century CE. Discussing the counting pits of merchants, Vasumitra remarks, \"When [the same] clay counting-piece is in the place of units, it is denoted as one, when in hundreds, one hundred.\" Although such references seem to imply that his readers had knowledge of a decimal place value representation, the \"brevity of their allusions and the ambiguity of their dates, however, do not solidly establish the chronology of the development of this concept.\"\n\nA third decimal representation was employed in a verse composition technique, later labeled Bhuta-sankhya (literally, \"object numbers\") used by early Sanskrit authors of technical books. Since many early technical works were composed in verse, numbers were often represented by objects in the natural or religious world that correspondence to them; this allowed a many-to-one correspondence for each number and made verse composition easier. According to Plofker 2009, the number 4, for example, could be represented by the word \"Veda\" (since there were four of these religious texts), the number 32 by the word \"tooth\" (since a full set consists of 32), and the number 1 by \"moon\" (since there is only one moon). So, Veda/tooth/moon would correspond to the decimal numeral 1324, as the convention for numbers was to enumerate their digits from right to left. The earliest reference employing object numbers is a ca. 269 CE Sanskrit text, Yavanajātaka (literally \"Greek horoscopy\") of Sphujidhvaja, a versification of an earlier (ca. 150 CE) Indian prose adaptation of a lost work of Hellenistic astrology. Such use seems to make the case that by the mid-third century CE, the decimal place value system was familiar, at least to readers of astronomical and astrological texts in India.\n\nIt has been hypothesized that the Indian decimal place value system was based on the symbols used on Chinese counting boards from as early as the middle of the first millennium BCE. According to Plofker 2009,\n\nThese counting boards, like the Indian counting pits, ..., had a decimal place value structure ... Indians may well have learned of these decimal place value \"rod numerals\" from Chinese Buddhist pilgrims or other travelers, or they may have developed the concept independently from their earlier non-place-value system; no documentary evidence survives to confirm either conclusion.\"\n\n## Bakhshali Manuscript\n\nThe oldest extant mathematical manuscript in South Asia is the Bakhshali Manuscript, a birch bark manuscript written in \"Buddhist hybrid Sanskrit\" in the Śāradā script, which was used in the northwestern region of the Indian subcontinent between the 8th and 12th centuries CE. The manuscript was discovered in 1881 by a farmer while digging in a stone enclosure in the village of Bakhshali, near Peshawar (then in British India and now in Pakistan). Of unknown authorship and now preserved in the Bodleian Library in Oxford University, the manuscript has been variously dated—as early as the \"early centuries of the Christian era\" and as late as between the 9th and 12th century CE. The 7th century CE is now considered a plausible date, albeit with the likelihood that the \"manuscript in its present-day form constitutes a commentary or a copy of an anterior mathematical work.\"\n\nThe surviving manuscript has seventy leaves, some of which are in fragments. Its mathematical content consists of rules and examples, written in verse, together with prose commentaries, which include solutions to the examples. The topics treated include arithmetic (fractions, square roots, profit and loss, simple interest, the rule of three, and regula falsi) and algebra (simultaneous linear equations and quadratic equations), and arithmetic progressions. In addition, there is a handful of geometric problems (including problems about volumes of irregular solids). The Bakhshali manuscript also \"employs a decimal place value system with a dot for zero.\" Many of its problems are the so-called equalization problems that lead to systems of linear equations. One example from Fragment III-5-3v is the following:\n\n\"One merchant has seven asava horses, a second has nine haya horses, and a third has ten camels. They are equally well off in the value of their animals if each gives two animals, one to each of the others. Find the price of each animal and the total value for the animals possessed by each merchant.\"\n\nThe prose commentary accompanying the example solves the problem by converting it to three (under-determined) equations in four unknowns and assuming that the prices are all integers.\n\n## Classical Period (400 - 1200)\n\nThis period is often known as the golden age of Indian Mathematics. This period saw mathematicians such as Aryabhata, Varahamihira, Brahmagupta, Bhaskara I, Mahavira, and Bhaskara II give broader and clearer shape to many branches of mathematics. Their contributions would spread to Asia, the Middle East, and eventually to Europe. Unlike Vedic mathematics, their works included both astronomical and mathematical contributions. In fact, mathematics of that period was included in the 'astral science' (jyotiḥśāstra) and consisted of three sub-disciplines: mathematical sciences (gaṇita or tantra), horoscope astrology (horā or jātaka) and divination (saṃhitā). This tripartite division is seen in Varāhamihira's sixth century compilation—Pancasiddhantika (literally panca, \"five,\" siddhānta, \"conclusion of deliberation\", dated 575 CE)—of five earlier works, Surya Siddhanta, Romaka Siddhanta, Paulisa Siddhanta, Vasishtha Siddhanta and Paitamaha Siddhanta, which were adaptations of still earlier works of Mesopotamian, Greek, Egyptian, Roman and Indian astronomy. As explained earlier, the main texts were composed in Sanskrit verse, and were followed by prose commentaries.\n\n### Fifth and Sixth Centuries\n\nSurya Siddhanta\n\nThough its authorship is unknown, the Surya Siddhanta (c. 400) contains the roots of modern trigonometry. Some authors consider that it was written under the influence of Mesopotamia and Greece.\n\nThis ancient text uses the following as trigonometric functions for the first time:\n\nIt also contains the earliest uses of:\n\n• The Hindu cosmological time cycles explained in the text, which was copied from an earlier work, gives:\n• The average length of the sidereal year as 365.2563795 days, which is only 1.4 seconds longer than the modern value of 365.2563627 days.\n• The average length of the tropical year as 365.2421756 days, which is only 2 seconds shorter than the modern value of 365.2421988 days.\n\nLater Indian mathematicians such as Aryabhata made references to this text, while later Arabic and Latin translations were very influential in Europe and the Middle East.\n\nChhedi calendar\n\nThis Chhedi calendar (594) contains an early use of the modern place-value Hindu-Arabic numeral system now used universally (see also Hindu-Arabic numerals).\n\nAryabhata I\n\nAryabhata (476-550) wrote the Aryabhatiya. He described the important fundamental principles of mathematics in 332 shlokas. The treatise contained:\n\nAryabhata also wrote the Arya Siddhanta, which is now lost. Aryabhata's contributions include:\n\nTrigonometry:\n\n• Introduced the trigonometric functions.\n• Defined the sine (jya) as the modern relationship between half an angle and half a chord.\n• Defined the cosine (kojya).\n• Defined the versine (utkrama-jya).\n• Defined the inverse sine (otkram jya).\n• Gave methods of calculating their approximate numerical values.\n• Contains the earliest tables of sine, cosine and versine values, in 3.75° intervals from 0° to 90°, to 4 decimal places of accuracy.\n• Contains the trigonometric formula sin (n + 1) x - sin nx = sin nx - sin (n - 1) x - (1/225)sin nx.\n• Spherical trigonometry.\n\nArithmetic:\n\nAlgebra:\n\n• Solutions of simultaneous quadratic equations.\n• Whole number solutions of linear equations by a method equivalent to the modern method.\n• General solution of the indeterminate linear equation .\n\nMathematical astronomy:\n\nCalculus:\n\n• Infinitesimals:\n• In the course of developing a precise mapping of the lunar eclipse, Aryabhatta was obliged to introduce the concept of infinitesimals (tatkalika gati) to designate the near instantaneous motion of the moon.\n• Differential equations:\n• He expressed the near instantaneous motion of the moon in the form of a basic differential equation.\n• Exponential function:\nVarahamihira\n\nVarahamihira (505-587) produced the Pancha Siddhanta (The Five Astronomical Canons). He made important contributions to trigonometry, including sine and cosine tables to 4 decimal places of accuracy and the following formulas relating sine and cosine functions:\n\n• sin2(x) + cos2(x) = 1\n•",
null,
"$\\sin(x)=\\cos\\left(\\frac{\\pi}{2}-x\\right)$\n•",
null,
"$\\frac{1-\\cos(2x)}{2}=\\sin^2(x)$\n\n### Seventh and Eighth Centuries",
null,
"",
null,
"Brahmagupta's theorem states that AF = FD.\n\nIn the seventh century, two separate fields, arithmetic (which included mensuration) and algebra, began to emerge in Indian mathematics. The two fields would later be called pāṭī-gaṇita (literally \"mathematics of algorithms\") and bīja-gaṇita (lit. \"mathematics of seeds,\" with \"seeds\"—like the seeds of plants—representing unknowns with the potential to generate, in this case, the solutions of equations). Brahmagupta, in his astronomical work Brāhma Sphuṭa Siddhānta (628 CE), included two chapters (12 and 18) devoted to these fields. Chapter 12, containing 66 Sanskrit verses, was divided into two sections: \"basic operations\" (including cube roots, fractions, ratio and proportion, and barter) and \"practical mathematics\" (including mixture, mathematical series, plane figures, stacking bricks, sawing of timber, and piling of grain). In the latter section, he stated his famous theorem on the diagonals of a cyclic quadrilateral:\n\nBrahmagupta's theorem: If a cyclic quadrilateral has diagonals that are perpendicular to each other, then the perpendicular line drawn from the point of intersection of the diagonals to any side of the quadrilateral always bisects the opposite side.\n\nChapter 12 also included a formula for the area of a cyclic quadrilateral (a generalization of Heron's formula), as well as a complete description of rational triangles (i.e. triangles with rational sides and rational areas).\n\nBrahmagupta's formula: The area, A, of a cyclic quadrilateral with sides of lengths a, b, c, d, respectively, is given by",
null,
"$A = \\sqrt{(s-a)(s-b)(s-c)(s-d)}$\n\nwhere s, the semiperimeter, given by:",
null,
"$s=\\frac{a+b+c+d}{2}.$\n\nBrahmagupta's Theorem on rational triangles: A triangle with rational sides a,b,c and rational area is of the form:",
null,
"$a = \\frac{u^2}{v}+v, \\ \\ b=\\frac{u^2}{w}+w, \\ \\ c=\\frac{u^2}{v}+\\frac{u^2}{w} - (v+w)$\n\nfor some rational numbers u,v, and w.\n\nChapter 18 contained 103 Sanskrit verses which began with rules for arithmetical operations involving zero and negative numbers and is considered the first systematic treatment of the subject. The rules (which included",
null,
"$a + 0 = \\ a$ and",
null,
"$a \\times 0 = 0$) were all correct, with one exception:",
null,
"$\\frac{0}{0} = 0$. Later in the chapter, he gave the first explicit (although still not completely general) solution of the quadratic equation:",
null,
"$\\ ax^2+bx=c$\n “ To the absolute number multiplied by four times the [coefficient of the] square, add the square of the [coefficient of the] middle term; the square root of the same, less the [coefficient of the] middle term, being divided by twice the [coefficient of the] square is the value. (Brahmasphutasiddhanta (Colebrook translation, 1817, page 346) ”\n\nThis is equivalent to:",
null,
"$x = \\frac{\\sqrt{4ac+b^2}-b}{2a}$\n\nAlso in chapter 18, Brahmagupta was able to make progress in finding (integral) solutions of Pell's equation,",
null,
"$\\ x^2-Ny^2=1,$\n\nwhere N is a nonsquare integer. He did this by discovering the following identity:\n\nBrahmagupta's Identity:",
null,
"$\\ (x^2-Ny^2)(x'^2-Ny'^2) = (xx'+Nyy')^2 - N(xy'+x'y)^2$ which was a generalization of an earlier identity of Diophantus: Brahmagupta used his identity to prove the following lemma:\n\nLemma (Brahmagupta): If",
null,
"$x=x_1,\\ \\ y=y_1 \\ \\$ is a solution of",
null,
"$\\ \\ x^2 - Ny^2 = k_1,$ and,",
null,
"$x=x_2, \\ \\ y=y_2 \\ \\$ is a solution of",
null,
"$\\ \\ x^2 - Ny^2 = k_2,$, then:",
null,
"$x=x_1x_2+Ny_1y_2,\\ \\ y=x_1y_2+x_2y_1 \\ \\$ is a solution of",
null,
"$\\ x^2-Ny^2=k_1k_2$\n\nHe then used this lemma to both generate infinitely many (integral) solutions of Pell's equation, given one solution, and state the following theorem:\n\nTheorem (Brahmagupta): If the equation",
null,
"$\\ x^2 - Ny^2 =k$ has an integer solution for any one of",
null,
"$\\ k=\\pm 4, \\pm 2, -1$ then Pell's equation:",
null,
"$\\ x^2 -Ny^2 = 1$\n\nalso has an integer solution.\n\nBrahmagupta did not actually prove the theorem, but rather worked out examples using his method. The first example he presented was:\n\nExample (Brahmagupta): Find integers",
null,
"$\\ x,\\ y\\$ such that:",
null,
"$\\ x^2 - 92y^2=1$\n\nIn his commentary, Brahmagupta added, \"a person solving this problem within a year is a mathematician.\" The solution he provided was:",
null,
"$\\ x=1151, \\ y=120$\n\nBhaskara I (c. 600-680) expanded the work of Aryabhata in his books titled Mahabhaskariya, Aryabhattiya Bhashya and Laghu Bhaskariya. He produced:\n\n• Solutions of indeterminate equations.\n• A rational approximation of the sine function.\n• A formula for calculating the sine of an acute angle without the use of a table, correct to 2 decimal places.\n\n### Ninth to Twelfth Centuries\n\nVirasena\n\nVirasena (9th century) was a Jain mathematician in the court of Rashtrakuta King Amoghavarsha of Manyakheta, Karnataka. He wrote the Dhavala, a commentary on Jain mathematics, which:\n\n• Deals with logarithms to base 2 (ardhaccheda) and describes its laws.\n• First uses logarithms to base 3 (trakacheda) and base 4 (caturthacheda).\n\nVirasena also gave:\n\n• The derivation of the volume of a frustum by a sort of infinite procedure.\nMahavira\n\nMahavira Acharya (c. 800-870) from Karnataka, the last of the notable Jain mathematicians, lived in the 9th century and was patronised by the Rashtrakuta king Amoghavarsha. He wrote a book titled Ganit Saar Sangraha on numerical mathematics, and also wrote treatises about a wide range of mathematical topics. These include the mathematics of:\n\nMahavira also:\n\n• Asserted that the square root of a negative number did not exist\n• Gave the sum of a series whose terms are squares of an arithmetical progression, and gave empirical rules for area and perimeter of an ellipse.\n• Solved cubic equations.\n• Solved quartic equations.\n• Solved some quintic equations and higher-order polynomials.\n• Gave the general solutions of the higher order polynomial equations:\n•",
null,
"$\\ ax^n = q$\n•",
null,
"$a \\frac{x^n - 1}{x - 1} = p$\n• Solved indeterminate cubic equations.\n• Solved indeterminate higher order equations.\nShridhara\n\nShridhara (c. 870-930), who lived in Bengal, wrote the books titled Nav Shatika, Tri Shatika and Pati Ganita. He gave:\n\nThe Pati Ganita is a work on arithmetic and mensuration. It deals with various operations, including:\n\n• Elementary operations\n• Extracting square and cube roots.\n• Fractions.\n• Eight rules given for operations involving zero.\n• Methods of summation of different arithmetic and geometric series, which were to become standard references in later works.\nManjula\n\nAryabhata's differential equations were elaborated in the 10th century by Manjula (also Munjala), who realised that the expression",
null,
"$\\ \\sin w' - \\sin w$\n\ncould be approximately expressed as",
null,
"$\\ (w' - w)\\cos w$\n\nHe understood the concept of differentiation after solving the differential equation that resulted from substituting this expression into Aryabhata's differential equation.\n\nAryabhata II\n\nAryabhata II (c. 920-1000) wrote a commentary on Shridhara, and an astronomical treatise Maha-Siddhanta. The Maha-Siddhanta has 18 chapters, and discusses:\n\n• Numerical mathematics (Ank Ganit).\n• Algebra.\n• Solutions of indeterminate equations (kuttaka).\nShripati\n\nShripati Mishra (1019–1066) wrote the books Siddhanta Shekhara, a major work on astronomy in 19 chapters, and Ganit Tilaka, an incomplete arithmetical treatise in 125 verses based on a work by Shridhara. He worked mainly on:\n\nHe was also the author of Dhikotidakarana, a work of twenty verses on:\n\nThe Dhruvamanasa is a work of 105 verses on:\n\nNemichandra Siddhanta Chakravati\n\nNemichandra Siddhanta Chakravati (c. 1100) authored a mathematical treatise titled Gome-mat Saar.\n\nBhāskara II (1114–1185) was a mathematician-astronomer who wrote a number of important treatises, namely the Siddhanta Shiromani, Lilavati, Bijaganita, Gola Addhaya, Griha Ganitam and Karan Kautoohal. A number of his contributions were later transmitted to the Middle East and Europe. His contributions include:\n\nArithmetic:\n\nAlgebra:\n\n• The recognition of a positive number having two square roots.\n• Surds.\n• Operations with products of several unknowns.\n• The solutions of:\n• Cubic equations.\n• Quartic equations.\n• Equations with more than one unknown.\n• Quadratic equations with more than one unknown.\n• The general form of Pell's equation using the chakravala method.\n• The general indeterminate quadratic equation using the chakravala method.\n• Indeterminate cubic equations.\n• Indeterminate quartic equations.\n• Indeterminate higher-order polynomial equations.\n\nGeometry:\n\nCalculus:\n\nTrigonometry:\n\n• Developments of spherical trigonometry\n• The trigonometric formulas:\n•",
null,
"$\\ \\sin(a+b)=\\sin(a) \\cos(b) + \\sin(b) \\cos(a)$\n•",
null,
"$\\ \\sin(a-b)=\\sin(a) \\cos(b) - \\sin(b) \\cos(a)$\n\n## Kerala Mathematics (1300 - 1600)\n\nThe Kerala school of astronomy and mathematics was founded by Madhava of Sangamagrama in Kerala, South India and included among its members: Parameshvara, Neelakanta Somayaji, Jyeshtadeva, Achyuta Pisharati, Melpathur Narayana Bhattathiri and Achyuta Panikkar. It flourished between the 14th and 16th centuries and the original discoveries of the school seems to have ended with Narayana Bhattathiri (1559–1632). In attempting to solve astronomical problems, the Kerala school astronomers independently created a number of important mathematics concepts. The most important results, series expansion for trigonometric functions, were given in Sanskrit verse in a book by Neelakanta called Tantrasangraha and a commentary on this work called Tantrasangraha-vakhya of unknown authorship. The theorems were stated without proof, but proofs for the series for sine, cosine, and inverse tangent were provided a century later in the work Yuktibhasa (c.1500-c.1610), written in Malayalam, by Jyesthadeva, and also in a commentary on Tantrasangraha.\n\nTheir discovery of these three important series expansions of calculus—several centuries before calculus was developed in Europe by Isaac Newton and Gottfried Leibniz—was a landmark achievement in mathematics. However, the Kerala School cannot be said to have invented calculus, because, while they were able to develop Taylor series expansions for the important trigonometric functions, they developed neither a comprehensive theory of differentiation or integration, nor the fundamental theorem of calculus. The results obtained by the Kerala school include:\n\n• The (infinite) geometric series:",
null,
"$\\frac{1}{1-x} = 1 + x + x^2 + x^3 + x^4+ \\dots$ for | x | < 1 This formula was already known, for example, in the work of the 10th century Arab mathematician Alhazen (the Latinized form of the name Ibn Al-Haytham (965-1039)).\n• A semi-rigorous proof (see \"induction\" remark below) of the result:",
null,
"$1^p+ 2^p + \\cdots + n^p \\approx \\frac{n^{p+1}}{p+1}$ for large n. This result was also known to Alhazen.\n• Intuitive use of mathematical induction, however, the inductive hypothesis was not formulated or employed in proofs.\n• Applications of ideas from (what was to become) differential and integral calculus to obtain (Taylor-Maclaurin) infinite series for sinx, cosx, and arctanx The Tantrasangraha-vakhya gives the series in verse, which when translated to mathematical notation, can be written as:",
null,
"$r\\arctan(\\frac{y}{x}) = \\frac{1}{1}\\cdot\\frac{ry}{x} -\\frac{1}{3}\\cdot\\frac{ry^3}{x^3} + \\frac{1}{5}\\cdot\\frac{ry^5}{x^5} - \\cdots ,$ where",
null,
"$y/x \\leq 1.$",
null,
"$\\sin x = x - x\\cdot\\frac{x^2}{(2^2+2)r^2} + x\\cdot \\frac{x^2}{(2^2+2)r^2}\\cdot\\frac{x^2}{(4^2+4)r^2} - \\cdot$",
null,
"$r - \\cos x = r\\cdot \\frac{x^2}{(2^2-2)r^2} - r\\cdot \\frac{x^2}{(2^2-2)r^2}\\cdot \\frac{x^2}{(4^2-4)r^2} + \\cdots ,$ where, for r = 1, the series reduce to the standard power series for these trigonometric functions, for example:\n•",
null,
"$\\sin x = x - \\frac{x^3}{3!} + \\frac{x^5}{5!} - \\frac{x^7}{7!} + \\cdots$ and\n•",
null,
"$\\cos x = 1 - \\frac{x^2}{2!} + \\frac{x^4}{4!} - \\frac{x^6}{6!} + \\cdots$\n• Use of rectification (computation of length) of the arc of a circle to give a proof of these results. (The later method of Leibniz, using quadrature (i.e. computation of area under the arc of the circle, was not used.)\n• Use of series expansion of arctanx to obtain an infinite series expression (later known as Gregory series) for π:",
null,
"$\\frac{\\pi}{4} = 1 - \\frac{1}{3} + \\frac{1}{5} - \\frac{1}{7} + \\ldots$\n• A rational approximation of error for the finite sum of their series of interest. For example, the error, fi(n + 1), (for n odd, and i = 1, 2, 3) for the series:",
null,
"$\\frac{\\pi}{4} \\approx 1 - \\frac{1}{3}+ \\frac{1}{5} - \\cdots (-1)^{(n-1)/2}\\frac{1}{n} + (-1)^{(n+1)/2}f_i(n+1)$\nwhere",
null,
"$f_1(n) = \\frac{1}{2n}, \\ f_2(n) = \\frac{n/2}{n^2+1}, \\ f_3(n) = \\frac{(n/2)^2+1}{(n^2+5)n/2}.$\n• Manipulation of error term to derive a faster converging series for π:",
null,
"$\\frac{\\pi}{4} = \\frac{3}{4} + \\frac{1}{3^3-3} - \\frac{1}{5^3-5} + \\frac{1}{7^3-7} - \\cdots$\n• Using the improved series to derive a rational expression, 104348 / 33215 for π correct up to nine decimal places, i.e. 3.141592653\n• Use of an intuitive notion of limit to compute these results.\n• A semi-rigorous (see remark on limits above) method of differentiation of some trigonometric functions. However, they did not formulate the notion of a function, or have knowledge of the exponential or logarithmic functions.\n\nThe works of the Kerala school were first written up for the Western world by Englishman C.M. Whish in 1835. According to Whish, the Kerala mathematicians had \"laid the foundation for a complete system of fluxions\" and these works abounded \"with fluxional forms and series to be found in no work of foreign countries.\"\n\nHowever, Whish's results were almost completely neglected, until over a century later, when the discoveries of the Kerala school were investigated again by C. Rajagopal and his associates. Their work includes commentaries on the proofs of the arctan series in Yuktibhasa given in two papers, a commentary on the Yuktibhasa's proof of the sine and cosine series and two papers that provide the Sanskrit verses of the Tantrasangrahavakhya for the series for arctan, sin, and cosine (with English translation and commentary).\n\nThe Kerala mathematicians included Narayana Pandit (c. 1340-1400), who composed two works, an arithmetical treatise, Ganita Kaumudi, and an algebraic treatise, Bijganita Vatamsa. Narayana is also thought to be the author of an elaborate commentary of Bhaskara II's Lilavati, titled Karmapradipika (or Karma-Paddhati). Madhava of Sangamagramma (c. 1340-1425) was the founder of the Kerala School. Although it is possible that he wrote Karana Paddhati a work written sometime between 1375 and 1475, all we really know of his work comes from works of later scholars.\n\nParameshvara (c. 1370-1460) wrote commentaries on the works of Bhaskara I, Aryabhata and Bhaskara II. His Lilavati Bhasya, a commentary on Bhaskara II's Lilavati, contains one of his important discoveries: a version of the mean value theorem. Nilakantha Somayaji (1444–1544) composed the Tantra Samgraha (which 'spawned' a later anonymous commentary Tantrasangraha-vyakhya and a further commentary by the name Yuktidipaika, written in 1501). He elaborated and extended the contributions of Madhava.\n\nCitrabhanu (c. 1530) was a 16th century mathematician from Kerala who gave integer solutions to 21 types of systems of two simultaneous algebraic equations in two unknowns. These types are all the possible pairs of equations of the following seven forms:",
null,
"$\\ x + y = a, x - y = b, xy = c, x^2 + y^2 = d, x^2 - y^2 = e, x^3 + y^3 = f, x^3 - y^3 = g$\n\nFor each case, Citrabhanu gave an explanation and justification of his rule as well as an example. Some of his explanations are algebraic, while others are geometric. Jyesthadeva (c. 1500-1575) was another member of the Kerala School. His key work was the Yukti-bhasa (written in Malayalam, a regional language of Kerala). Jyesthadeva presented proofs of most mathematical theorems and infinite series earlier discovered by Madhava and other Kerala School mathematicians.\n\n## Charges of Eurocentrism\n\nIt has been suggested that Indian contributions to mathematics have not been given due acknowledgement in modern history and that many discoveries and inventions by Indian mathematicians were known to their Western counterparts, copied by them, and presented as their own original work; and further, that this mass plagiarism has gone unrecognized due to Eurocentrism. According to G. G. Joseph:\n\n[Their work] takes on board some of the objections raised about the classical Eurocentric trajectory. The awareness [of Indian and Arabic mathematics] is all too likely to be tempered with dismissive rejections of their importance compared to Greek mathematics. The contributions from other civilizations - most notably China and India, are perceived either as borrowers from Greek sources or having made only minor contributions to mainstream mathematical development. An openness to more recent research findings, especially in the case of Indian and Chinese mathematics, is sadly missing\"\n\nThe historian of mathematics, Florian Cajori, suggested that he \"suspect[s] that Diophantus got his first glimpse of algebraic knowledge from India.\"\n\nMore recently, as discussed in the above section, the infinite series of calculus for trigonometric functions (rediscovered by Gregory, Taylor, and Maclaurin in the late 17th century) were described (with proofs) in India, by mathematicians of the Kerala School, remarkably some two centuries earlier. Some scholars have recently suggested that knowledge of these results might have been transmitted to Europe through the trade route from Kerala by traders and Jesuit missionaries. Kerala was in continuous contact with China and Arabia, and, from around 1500, with Europe. The existence of communication routes and a suitable chronology certainly make such a transmission a possibility. However, there is no direct evidence by way of relevant manuscripts that such a transmission actually took place. According to David Bressoud, \"there is no evidence that the Indian work of series was known beyond India, or even outside of Kerala, until the nineteenth century.\"\n\nBoth Arab and Indian scholars made discoveries before the 17th century that are now considered a part of calculus. However, they were not able to, as Newton and Leibniz were, to \"combine many differing ideas under the two unifying themes of the derivative and the integral, show the connection between the two, and turn calculus into the great problem-solving tool we have today.\" The intellectual careers of both Newton and Leibniz are well-documented and there is no indication of their work not being their own; however, it is not known with certainty whether the immediate predecessors of Newton and Leibniz, \"including, in particular, Fermat and Roberval, learned of some of the ideas of the Islamic and Indian mathematicians through sources we are not now aware.\" This is an active area of current research, especially in the manuscripts collections of Spain and Maghreb, research that is now being pursued, among other places, at the Centre National de Recherche Scientifique in Paris.\n\n## Notes\n\n1. ^ a b Encyclopaedia Britannica (Kim Plofker) 2007, p. 1\n2. ^ Ifrah 2000, p. 346: \"The measure of the genius of Indian civilisation, to which we owe our modern (number) system, is all the greater in that it was the only one in all history to have achieved this triumph. Some cultures succeeded, earlier than the Indian, in discovering one or at best two of the characteristics of this intellectual feat. But none of them managed to bring together into a complete and coherent system the necessary and sufficient conditions for a number-system with the same potential as our own.\"\n3. ^ Plofker 2009, pp. 44–47\n4. ^ Bourbaki 1998, p. 46: \"...our decimal system, which (by the agency of the Arabs) is derived from Hindu mathematics, where its use is attested already from the first centuries of our era. It must be noted moreover that the conception of zero as a number and not as a simple symbol of separation) and its introduction into calculations, also count amongst the original contribution of the Hindus.\"\n5. ^ Bourbaki 1998, p. 49: Modern arithmetic was known during medieval times as \"Modus Indorum\" or method of the Indians. Leonardo of Pisa wrote that compared to method of the Indians all other methods is a mistake. This method of the Indians is none other than our very simple arithmetic of addition, subtraction, multiplication and division. Rules for these four simple procedures was first written down by Brahmagupta during 7th century AD. \"On this point, the Hindus are already conscious of the interpretation that negative numbers must have in certain cases (a debt in a commercial problem, for instance). In the following centuries, as there is a diffusion into the West (by intermediary of the Arabs) of the methods and results of Greek and Hindu mathematics, one becomes more used to the handling of these numbers, and one begins to have other \"representation\" for them which are geometric or dynamic.\"\n6. ^ a b \"algebra\" 2007. Britannica Concise Encyclopedia. Encyclopædia Britannica Online. 16 May 2007. Quote: \"A full-fledged decimal, positional system certainly existed in India by the 9th century (AD), yet many of its central ideas had been transmitted well before that time to China and the Islamic world. Indian arithmetic, moreover, developed consistent and correct rules for operating with positive and negative numbers and for treating zero like any other number, even in problematic contexts such as division. Several hundred years passed before European mathematicians fully integrated such ideas into the developing discipline of algebra.\"\n7. ^ (Pingree 2003, p. 45) Quote: \"Geometry, and its branch trigonometry, was the mathematics Indian astronomers used most frequently. Greek mathematicians used the full chord and never imagined the half chord that we use today. Half chord was first used by Aryabhatta which made trigonometry much more simple. In fact, the Indian astronomers in the third or fourth century, using a pre-Ptolemaic Greek table of chords, produced tables of sines and versines, from which it was trivial to derive cosines. This new system of trigonometry, produced in India, was transmitted to the Arabs in the late eighth century and by them, in an expanded form, to the Latin West and the Byzantine East in the twelfth century.\"\n8. ^ (Bourbaki 1998, p. 126): \"As for trigonometry, it is disdained by geometers and abandoned to surveyors and astronomers; it is these latter (Aristarchus, Hipparchus, Ptolemy) who establish the fundamental relations between the sides and angles of a right angled triangle (plane or spherical) and draw up the first tables (they consist of tables giving the chord of the arc cut out by an angle θ < π on a circle of radius r, in other words the number",
null,
"$2r\\sin\\left(\\theta/2\\right)$; the introduction of the sine, more easily handled, is due to Hindu mathematicians of the Middle Ages).\"\n9. ^ Filliozat 2004, pp. 140-143\n10. ^ Hayashi 1995\n11. ^ a b Encyclopaedia Britannica (Kim Plofker) 2007, p. 6\n12. ^ Stillwell 2004, p. 173\n13. ^ Bressoud 2002, p. 12 Quote: \"There is no evidence that the Indian work on series was known beyond India, or even outside Kerala, until the nineteenth century. Gold and Pingree assert that by the time these series were rediscovered in Europe, they had, for all practical purposes, been lost to India. The expansions of the sine, cosine, and arc tangent had been passed down through several generations of disciples, but they remained sterile observations for which no one could find much use.\"\n14. ^ Plofker 2001, p. 293 Quote: \"It is not unusual to encounter in discussions of Indian mathematics such assertions as that “the concept of differentiation was understood [in India] from the time of Manjula (... in the 10th century)” [Joseph 1991, 300], or that “we may consider Madhava to have been the founder of mathematical analysis” (Joseph 1991, 293), or that Bhaskara II may claim to be “the precursor of Newton and Leibniz in the discovery of the principle of the differential calculus” (Bag 1979, 294). ... The points of resemblance, particularly between early European calculus and the Keralese work on power series, have even inspired suggestions of a possible transmission of mathematical ideas from the Malabar coast in or after the 15th century to the Latin scholarly world (e.g., in (Bag 1979, 285)). ... It should be borne in mind, however, that such an emphasis on the similarity of Sanskrit (or Malayalam) and Latin mathematics risks diminishing our ability fully to see and comprehend the former. To speak of the Indian “discovery of the principle of the differential calculus” somewhat obscures the fact that Indian techniques for expressing changes in the Sine by means of the Cosine or vice versa, as in the examples we have seen, remained within that specific trigonometric context. The differential “principle” was not generalized to arbitrary functions—in fact, the explicit notion of an arbitrary function, not to mention that of its derivative or an algorithm for taking the derivative, is irrelevant here\"\n15. ^ Pingree 1992, p. 562 Quote:\"One example I can give you relates to the Indian Mādhava's demonstration, in about 1400 A.D., of the infinite power series of trigonometrical functions using geometrical and algebraic arguments. When this was first described in English by Charles Matthew Whish, in the 1830s, it was heralded as the Indians' discovery of the calculus. This claim and Mādhava's achievements were ignored by Western historians, presumably at first because they could not admit that an Indian discovered the calculus, but later because no one read anymore the Transactions of the Royal Asiatic Society, in which Whish's article was published. The matter resurfaced in the 1950s, and now we have the Sanskrit texts properly edited, and we understand the clever way that Mādhava derived the series without the calculus; but many historians still find it impossible to conceive of the problem and its solution in terms of anything other than the calculus and proclaim that the calculus is what Mādhava found. In this case the elegance and brilliance of Mādhava's mathematics are being distorted as they are buried under the current mathematical solution to a problem to which he discovered an alternate and powerful solution.\"\n16. ^ Katz 1995, pp. 173-174 Quote:\"How close did Islamic and Indian scholars come to inventing the calculus? Islamic scholars nearly developed a general formula for finding integrals of polynomials by A.D. 1000—and evidently could find such a formula for any polynomial in which they were interested. But, it appears, they were not interested in any polynomial of degree higher than four, at least in any of the material that has come down to us. Indian scholars, on the other hand, were by 1600 able to use ibn al-Haytham's sum formula for arbitrary integral powers in calculating power series for the functions in which they were interested. By the same time, they also knew how to calculate the differentials of these functions. So some of the basic ideas of calculus were known in Egypt and India many centuries before Newton. It does not appear, however, that either Islamic or Indian mathematicians saw the necessity of connecting some of the disparate ideas that we include under the name calculus. They were apparently only interested in specific cases in which these ideas were needed. ... There is no danger, therefore, that we will have to rewrite the history texts to remove the statement that Newton and Leibniz invented calculus. Thy were certainly the ones who were able to combine many differing ideas under the two unifying themes of the derivative and the integral, show the connection between them, and turn the calculus into the great problem-solving tool we have today.\"\n17. ^ Sergent, Bernard (1997) (in French). Genèse de l'Inde. Paris: Payot. pp. 113. ISBN 2228891169.\n18. ^\n19. ^ Bisht, R. S. (1982). \"Excavations at Banawali: 1974-77\". in Possehl, Gregory L. (ed.). Harappan Civilization: A Contemporary Perspective. New Delhi: Oxford and IBH Publishing Co.. pp. 113–124.\n20. ^ a b c (Hayashi 2005, pp. 360-361)\n21. ^ A. Seidenberg, 1978. The origin of mathematics. Archive for the history of Exact Sciences, vol 18.\n22. ^ (Staal 1999)\n23. ^ a b (Hayashi 2003, p. 118)\n24. ^ a b (Hayashi 2005, p. 363)\n25. ^ Pythagorean triples are triples of integers (a,b,c) with the property: a2 + b2 = c2. Thus, 32 + 42 = 52, 82 + 152 = 172, 122 + 352 = 372 etc.\n26. ^ (Cooke 2005, p. 198): \"The arithmetic content of the Śulva Sūtras consists of rules for finding Pythagorean triples such as (3, 4, 5), (5, 12, 13), (8, 15, 17), and (12, 35, 37). It is not certain what practical use these arithmetic rules had. The best conjecture is that they were part of religious ritual. A Hindu home was required to have three fires burning at three different altars. The three altars were to be of different shapes, but all three were to have the same area. These conditions led to certain \"Diophantine\" problems, a particular case of which is the generation of Pythagorean triples, so as to make one square integer equal to the sum of two others.\"\n27. ^ (Cooke 2005, pp. 199-200): \"The requirement of three altars of equal areas but different shapes would explain the interest in transformation of areas. Among other transformation of area problems the Hindus considered in particular the problem of squaring the circle. The Bodhayana Sutra states the converse problem of constructing a circle equal to a given square. The following approximate construction is given as the solution.... this result is only approximate. The authors, however, made no distinction between the two results. In terms that we can appreciate, this construction gives a value for π of 18 (3 − 2√2), which is about 3.088.\"\n28. ^ a b c (Joseph 2000, p. 229)\n29. ^ a b (Cooke 2005, p. 200)\n30. ^ The value of this approximation, 577/408, is the seventh in a sequence of increasingly accurate approximations 3/2, 7/5, 17/12, ... to √2, the numerators and denominators of which were known as \"side and diameter numbers\" to the ancient Greeks, and in modern mathematics are called the Pell numbers. If x/y is one term in this sequence of approximations, the next is (x+2y)/(x+y). These approximations may also be derived by truncating the continued fraction representation of √2.\n31. ^ Neugebauer, O. and A. Sachs. 1945. Mathematical Cuneiform Texts, New Haven, CT, Yale University Press. p. 45.\n32. ^ Mathematics Department, University of British Columbia, The Babylonian tabled Plimpton 322.\n33. ^ Three positive integers (a,b,c) form a primitive Pythagorean triple if c2 = a2 + b2 and if the highest common factor of a,b,c is 1. In the particular Plimpton322 example, this means that 135002 + 127092 = 185412 and that the three numbers do not have any common factors. However some scholars have disputed the Pythagorean interpretation of this tablet; see Plimpton 322 for details.\n34. ^ a b (Dani 2003)\n35. ^ a b (Fowler 1996, p. 11)\n36. ^ a b (Singh 1936, pp. 623-624)\n37. ^ a b (Filliozat 2004, p. 137)\n38. ^ (Pingree 1988, p. 637)\n39. ^ (Staal 1986)\n40. ^ a b c (Filliozat 2004, p. 139)\n41. ^ a b c d e (Filliozat 2004, pp. 140-141)\n42. ^ (Yano 2006, p. 146)\n43. ^ a b c (Filliozat 2004, pp. 143-144)\n44. ^ a b (Pingree 1988, p. 638)\n45. ^ a b c (Hayashi 2003, pp. 122-123)\n46. ^ a b c (Hayashi 2003, p. 119)\n47. ^ a b Plofker 2007, p. 395\n48. ^ Plofker 2007, p. 395, Plofker 2009, pp. 47–48\n49. ^ (Hayashi 2005, p. 366)\n50. ^ a b c Plofker 2009, p. 45\n51. ^ a b c d Plofker 2009, p. 46\n52. ^ a b c d e Plofker 2009, p. 47\n53. ^ (Pingree 1978, p. 494)\n54. ^ a b Plofker 2009, p. 48\n55. ^ a b c (Hayashi 2005, p. 371)\n56. ^ (Datta 1931, p. 566)\n57. ^ (Ifrah 2000, p. 464) Quote: \"To give the second or fourth century CE as the date of this document would be an evident contradiction; it would mean that a northern derivative of Gupta writing had been developed two or three centuries before the Gupta writing itself appeared. Gupta only began to evolve into Shāradā style around the ninth century CE. In other words, the Bak(h)shali manuscript cannot have been written earlier than the ninth century CE. However, in the light of certain characteristic indications, it could not have been written any later than the twelfth century CE.\"\n58. ^ (Hayashi 2005, p. 371) Quote:\"The dates so far proposed for the Bakhshali work vary from the third to the twelfth centuries AD, but a recently made comparative study has shown many similarities, particularly in the style of exposition and terminology, between Bakhshalī work and Bhāskara I's commentary on the Āryabhatīya. This seems to indicate that both works belong to nearly the same period, although this does not deny the possibility that some of the rules and examples in the Bakhshālī work date from anterior periods.\"\n59. ^ (Ifrah 2000, p. 464)\n60. ^ a b Anton, Howard and Chris Rorres. 2005. Elementary Linear Algebra with Applications. 9th edition. New York: John Wiley and Sons. 864 pages. ISBN 0471669598.\n61. ^ (Neugebauer & Pingree (eds.) 1970)\n62. ^ Cooke, Roger (1997). \"The Mathematics of the Hindus\". The History of Mathematics: A Brief Course. Wiley-Interscience. p. 197. ISBN 0471180823. \"The word Siddhanta means that which is proved or established. The Sulva Sutras are of Hindu origin, but the Siddhantas contain so many words of foreign origin that they undoubtedly have roots in Mesopotamia and Greece.\"\n63. ^ Victor J. Katz (1995). \"Ideas of Calculus in Islam and India\", Mathematics Magazine 68 (3), p. 163-174.\n64. ^ a b c d e Joseph (2000), p. 298-300.\n65. ^ (Hayashi 2005, p. 369)\n66. ^ a b c d (Hayashi 2003, pp. 121-122)\n67. ^ (Stillwell 2004, p. 77)\n68. ^ (Stillwell 2004, p. 87)\n69. ^ a b c d e f (Stillwell 2004, pp. 72-73)\n70. ^ (Stillwell 2004, pp. 74-76)\n71. ^ a b c d e f g h i (Roy 1990)\n72. ^ a b c (Bressoud 2002)\n73. ^ a b c d e f g (Katz 1995)\n74. ^ Singh, A. N. Singh. 1936. \"On the Use of Series in Hindu Mathematics.\" Osiris 1:606-628.\n75. ^ Edwards, C. H., Jr. 1979. The Historical Development of the Calculus. New York: Springer-Verlag.\n76. ^ (Whish 1835)\n77. ^ Rajagopal, C. and M. S. Rangachari. 1949. \"A Neglected Chapter of Hindu Mathematics.\" Scripta Mathematica. 15:201-209.\n78. ^ Rajagopal, C. and M. S. Rangachari. 1951. \"On the Hindu proof of Gregory's series.\" Ibid. 17:65-74.\n79. ^ Rajagopal, C. and A. Venkataraman. 1949. \"The sine and cosine power series in Hindu mathematics.\" Journal of the Royal Asiatic Society of Bengal (Science). 15:1-13.\n80. ^ Rajagopal, C. and M. S. Rangachari. 1977. \"On an untapped source of medieval Keralese mathematics.\" Archive for the History of Exact Sciences. 18:89-102.\n81. ^ Rajagopal, C. and M. S. Rangachari. 1986. \"On Medieval Kerala Mathematics.\" Archive for the History of Exact Sciences. 35:91-99.\n82. ^ Joseph, G. G. 1997. \"Foundations of Eurocentrism in Mathematics.\" In Ethnomathematics: Challenging Eurocentrism in Mathematics Education (Eds. Powell, A. B. et al.). SUNY Press. ISBN 0791433528. p.67-68.\n83. ^ Cajori, Florian (1893). \"The Hindoos\". A History of Mathematics P 86. Macmillan & Co.. \"\"In algebra, there was probably a mutual giving and receiving [between Greece and India]. We suspect that Diophantus got his first glimpse of algebraic knowledge from India\"\"\n84. ^ a b Almeida, D. F., J. K. John, and A. Zadorozhnyy. 2001. \"Keralese Mathematics: Its Possible Transmission to Europe and the Consequential Educational Implications.\" Journal of Natural Geometry, 20:77-104.\n85. ^ Gold, D. and D. Pingree. 1991. \"A hitherto unknown Sanskrit work concerning Madhava's derivation of the power series for sine and cosine.\" Historia Scientiarum. 42:49-65.\n\n86. ^ Bourbaki, Nicolas (1998). Elements of the History of Mathematics. Berlin, Heidelberg, and New York: Springer-Verlag. 46. ISBN 3540647678.\n\n87. ^ Britannica Concise Encyclopedia (2007), entry algebra\n\n## Source Books in Sanskrit\n\n• Keller, Agathe (2006), Expounding the Mathematical Seed. Vol. 1: The Translation: A Translation of Bhaskara I on the Mathematical Chapter of the Aryabhatiya, Basel, Boston, and Berlin: Birkhäuser Verlag, 172 pages, ISBN 3764372915 .\n• Keller, Agathe (2006), Expounding the Mathematical Seed. Vol. 2: The Supplements: A Translation of Bhaskara I on the Mathematical Chapter of the Aryabhatiya, Basel, Boston, and Berlin: Birkhäuser Verlag, 206 pages, ISBN 3764372923 .\n• Neugebauer, Otto; Pingree (eds.), David (1970), The Pañcasiddhāntikā of Varāhamihira, New edition with translation and commentary, (2 Vols.), Copenhagen .\n• Pingree, David (ed) (1978), The Yavanajātaka of Sphujidhvaja, edited, translated and commented by D. Pingree, Cambridge, MA: Harvard Oriental Series 48 (2 vols.) .\n• Sarma, K. V. (ed) (1976), Āryabhaṭīya of Āryabhaṭa with the commentary of Sūryadeva Yajvan, critically edited with Introduction and Appendices, New Delhi: Indian National Science Academy .\n• Sen, S. N.; Bag (eds.), A. K. (1983), The Śulbasūtras of Baudhāyana, Āpastamba, Kātyāyana and Mānava, with Text, English Translation and Commentary, New Delhi: Indian National Science Academy .\n• Shukla, K. S. (ed) (1976), Āryabhaṭīya of Āryabhaṭa with the commentary of Bhāskara I and Someśvara, critically edited with Introduction, English Translation, Notes, Comments and Indexes, New Delhi: Indian National Science Academy .\n• Shukla, K. S. (ed) (1988), Āryabhaṭīya of Āryabhaṭa, critically edited with Introduction, English Translation, Notes, Comments and Indexes, in collaboration with K.V. Sarma, New Delhi: Indian National Science Academy ."
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https://www.wyzant.com/resources/answers/835723/c-programming-first-fully-parenthesize-the-following-expressions-according- | [
"Kat H.\n\n# C Programming. First, fully parenthesize the following expressions according to the precedence and associativity rules.\n\nFirst, fully parenthesize the following expressions according to the precedence and associativity rules. Then, replacing the variables and constants with the appropriate type names, show how the type of the expression is derived by replacing the highest precedence expressions with its resulting type.\n\nThe variables are:\n\nchar c;\n\nint i;\n\nunsigned u;\n\nfloat f;\n\nFor example: i = u+1; parenthesizes as (i = (u+1));\n\nThe types are\n\n(int = (unsigned + int));\n\nthen\n\n(int = (unsigned)); /* usual arithmetic conversions */\n\nthen\n\n(int); /* assignment */\n\na. c = u * f + 2.6L;\n\nb. u += --f / u % 3;\n\nc. i <<= u * - ++f;\n\nd. u = i + 3 + 4 + 3.1;\n\ne. u = 3.1 + i + 3 + 4;\n\nf. c == (i << - --f) & 0xf;\n\nBy:\n\nTutor\n4.9 (30)\n\nMath and computer tutor/teacher\n\n## Still looking for help? Get the right answer, fast.\n\nGet a free answer to a quick problem.\nMost questions answered within 4 hours.\n\n#### OR\n\nChoose an expert and meet online. No packages or subscriptions, pay only for the time you need."
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https://logic.umwblogs.org/syntax-in-propositional-logic-exercises/ | [
"# Syntax in propositional logic — exercises\n\nSyntax in propositional logic Exercises\n\nSyntax-in-propositional-logic-Exercises-solutions\n\nFill in the truth tables for the operators\n\np q p . q p v q p ﬤ q p ≡ q ~p\n\nT T\n\nT F\n\nF T\n\nF F\n\nCalculate the truth value of these statements\n\n1. ~[(A v B) (D > E)]\n\nf t f f\n\n1. (A > C) v [(H > G) > (N v M)]\n\nf t t f t t\n\n1. [(F > B) v (K v G)] ~ P\n\nt t f f t\n\n[(A v B) > (~B > A)] > [(A • B) • (~B• ~A)]\n\nt f f t t f f t\n\nName the kind of statement each is, or circle errors that make it unreadable.\n\nAssuming that A, C and H, and R are true, and that B, D, and J are false, calculate the truth value when possible.\n\n1. (A v B) > (C ~D) v H\n\n1. A v [B > (C . ~D)] v H\n\n1. [(A v B) > (C v ~D)] v H\n\n1. (A v B) > (C ~D) ≡ H ~ v (R . J)\n\n1. (A v B) > (C ~D) ≡ H v (R . J)\n\n1. [(A v B) > (C ~D)] ≡ [H v (R . J)]\n\n1. {[(A v B) > (C > D)] ≡ H} v (R . J)\n\n1. (A v B) > {[(~C v ~D) ≡ H] v (R . J)}\n\n1. [(A v B)] > {(C > D) ≡ [H v (R > J)]}\n\n1. {(A v B) > [(C ~D) ≡ H]} > (R ≡ J)"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.58369756,"math_prob":0.9787517,"size":922,"snap":"2020-34-2020-40","text_gpt3_token_len":414,"char_repetition_ratio":0.1764706,"word_repetition_ratio":0.104477614,"special_character_ratio":0.49457702,"punctuation_ratio":0.067226894,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.999974,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-07T14:54:39Z\",\"WARC-Record-ID\":\"<urn:uuid:1e07be0f-648e-4613-a247-74f0e327e533>\",\"Content-Length\":\"30405\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4a379f2e-e50e-4477-ab9d-4256ae88a6ce>\",\"WARC-Concurrent-To\":\"<urn:uuid:35cf8d06-b166-40a9-b4e8-598b4324f329>\",\"WARC-IP-Address\":\"157.230.215.55\",\"WARC-Target-URI\":\"https://logic.umwblogs.org/syntax-in-propositional-logic-exercises/\",\"WARC-Payload-Digest\":\"sha1:2QFO2ZPWMCA5IPLXW24XLOWXOC7MRLTT\",\"WARC-Block-Digest\":\"sha1:BXRZKHPE6VJBRALPBTHLV3QSJCBPGLZ2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439737204.32_warc_CC-MAIN-20200807143225-20200807173225-00164.warc.gz\"}"} |
https://www.daytraderland.com/daytrading-indicators/moving-averages/ | [
"# Moving Averages",
null,
"One of the first things that a daytrader meets in his search after indicators for his technical analysis is the moving averages. But unfortunately, many people get it wrong on how to use the moving averages, how they are calculated correctly and also the disadvantages of this method.\n\nIn this article we will go through those three things and end it with almost an hour long video, which goes into even more depth on this important indicator.\n\n## The basics\n\nA moving average is simply the sum of the data divided by the number of data and this calculation is then done “moving” every day. Does it sound confusing? Then take a look here.\n\nHere is the price for a given stock 5 days in a row:\n\nDay 1: 100\nDay 2: 110\nDay 3: 120\nDay 4: 130\nDay 5: 140\n\nTo get the average price for those 5 days we do the following:\n\n(100 + 110 + 120 + 130 + 140) / 5 =\n\n600 / 5 = 120\n\nThis means that the average price for the last 5 days is 120.\n\n## So what is a moving average?\n\nLet us say that we on the 6th day get a stock price at:\n\nDay 6: 150\n\nWe still want to calculate 5 days of average price, and we do this by not including data from day 1 and only use data from day 2 up to day 6 like this:\n\nDag 2: 110\nDag 3: 120\nDag 4: 130\nDag 5: 140\nDag 6: 150\n\nWe do the same calculation as before:\n\n(110 + 120 + 130 + 140 + 150) / 5 =\n\n650 / 5 = 130\n\n## What can we use the moving average for?\n\nNormally, I wouldn’t sit and do these calculations by hand every day as we have computers to do this for us.\nAnd we can use these moving averages to get an idea of the general trend of the stock. Of course, we can also see this by just looking at the chart, but the moving average removes some of the noise from the daily swings and gives a more stable picture about whether the trend is up or down.\n\nThere are many ways to interpret these average prices. Some traders buy if the price of a stock rises above a certain moving average. Other buy or sell when to moving averages cross each other etc.\n\n## What is the disadvantage of this method?\n\nNo matter what moving averages you use, then you have to remember that they only describe the past movement. That means, that it is a so-called lagging indicator. Which means, that it will always be a little behind.\n\nThat is why you should always be careful to not build your whole strategy on one certain moving average, but you can use it as a form of support/resistance (Link to article).\n\n## The most commonly used moving averages\n\nFor short-term daytraders and the likes, 10 and 20 period moving averages are widely used. In this case it is not the last 10 or 20 days of data you are looking at, but rather the last 10 price bars, for example 10 x 1 minute or 20 x 5 minute.\n\nSo the timescale is different – but the method the same.\n\nOn the higher timeframes many use the 20, 50 and 200 daily moving average.\n\nIn the videos about this subject you can see much more clearly how the different types of moving averages are calculated and the big variety of use of moving averages in both investing, swingtrading and daytrading.\n\nIf you would like to know more about the moving average indicator, then take a look at these two videos describing what a moving average is and how they can be used advantageously.",
null,
"#### Hans Henrik Nielsen\n\nHans Henrik Nielsen is a very experienced day and swing trader, teacher and publisher, who in the last couple of years has produced 80-134% profit on his day trading."
]
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"https://www.daytraderland.com/wp-content/uploads/2017/08/base-12.png",
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"https://secure.gravatar.com/avatar/71bef1083487f7204c3545f669d3b011",
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https://onlinemonks.com/web-stories/modden-brief-price-prediction-analysis-forecast-28-Dec-2022/ | [
"LIVE CURRENT PRICE (28 DEC 2022)\n\nALL TIME HIGH\n\nTOTAL MARKET CAP\n\n## MODDEN (MDDN)\n\n### Current Rank #8713\n\nPRICE PREDICTION 2022\n\nThe price of 1 Modden is expected to reach at a minimum level of \\$0.005 in 2022. The MDDN price can reach a maximum level of \\$0.005 with the average price of \\$0.005 throughout 2022.\n\n## MODDEN\n\nPRICE FORECAST 2023\n\nIn 2023 the price of Modden is predicted to reach at a minimum level of \\$0.007. The MDDN price can reach a maximum level of \\$0.009 with the average trading price of \\$0.008.\n\n## MODDEN\n\nPRICE ANALYSIS 2024\n\nThe price of Modden is predicted to reach at a minimum level of \\$0.011 in 2024. The Modden price can reach a maximum level of \\$0.013 with the average price of \\$0.011 throughout 2024.\n\n## MODDEN\n\nPRICE TARGET 2025\n\nThe price of Modden is predicted to reach at a minimum value of \\$0.015 in 2025. The Modden price could reach a maximum value of \\$0.019 with the average trading price of \\$0.016 throughout 2025.\n\n## MODDEN\n\nPRICE PREDICTION 2026\n\nIn 2026 the price of Modden is predicted to reach at a minimum level of \\$0.023. The MDDN price can reach a maximum level of \\$0.027 with the average trading price of \\$0.023.\n\n## MODDEN\n\nPRICE FORECAST 2027\n\nthe MDDN price could reach a maximum possible level of \\$0.038 with the average forecast price of \\$0.032.\n\n## MODDEN\n\nPRICE ANALYSIS 2028\n\nIn 2028 the price of Modden is expected to reach at a minimum price value of \\$0.043. The MDDN price can reach a maximum price value of \\$0.053 with the average value of \\$0.045.\n\n## MODDEN\n\nPRICE TARGET 2029\n\nThe price of Modden is predicted to reach at a minimum value of \\$0.062 in 2029. The Modden price could reach a maximum value of \\$0.074 with the average trading price of \\$0.064 throughout 2029.\n\n## MODDEN\n\nPRICE PREDICTION 2030\n\nthe MDDN price could reach a maximum possible level of \\$0.11 with the average forecast price of \\$0.093.\n\n## MODDEN\n\nPRICE FORECAST 2031\n\nIn 2031 the price of Modden is forecasted to be at around a minimum value of \\$0.13. The Modden price value can reach a maximum of \\$0.16 with the average trading value of \\$0.13 in USD.\n\n## MODDEN\n\nProject202 Brief, Price Prediction, Analysis & Forecast 2022-2031"
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https://www.global-sci.org/intro/article_detail/getBib?article_id=419 | [
"@Article{IJNAM-14-243, author = {}, title = {An \\$L^∞\\$ Bound for the Cahn-Hilliard Equation with Relaxed Non-Smooth Free Energy}, journal = {International Journal of Numerical Analysis and Modeling}, year = {2016}, volume = {14}, number = {2}, pages = {243--254}, abstract = {\n\nPhase field models are widely used to describe multiphase systems. Here a smooth indicator function, called phase field, is used to describe the spatial distribution of the phases under investigation. Material properties like density or viscosity are introduced as given functions of the phase field. These parameters typically have physical bounds to fulfil, e.g. positivity of the density. To guarantee these properties, uniform bounds on the phase field are of interest. In this work we derive a uniform bound on the solution of the Cahn-Hilliard system, where we use the double-obstacle free energy, that is relaxed by Moreau-Yosida relaxation.\n\n}, issn = {2617-8710}, doi = {https://doi.org/}, url = {http://global-sci.org/intro/article_detail/ijnam/419.html} }"
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https://proofwiki.org/wiki/Definition:Uniformly_Continuous_Real_Function | [
"# Definition:Uniform Continuity/Real Function\n\n## Definition\n\nLet $I \\subseteq \\R$ be a real interval.\n\nA real function $f: I \\to \\R$ is said to be uniformly continuous on $I$ if and only if:\n\nfor every $\\epsilon > 0$ there exists $\\delta > 0$ such that the following property holds:\nfor every $x, y \\in I$ such that $\\size {x - y} < \\delta$ it happens that $\\size {\\map f x - \\map f y} < \\epsilon$.\n\nFormally: $f: I \\to \\R$ is uniformly continuous if and only if the following property holds:\n\n$\\forall \\epsilon > 0: \\exists \\delta > 0: \\paren {x, y \\in I, \\size {x - y} < \\delta \\implies \\size {\\map f x - \\map f y} < \\epsilon}$\n\nIt can be seen that this says exactly the same thing as the definition for metric spaces if $\\R$ is considered a metric space under the Euclidean metric."
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https://mathematica.stackexchange.com/questions/138842/creating-a-multiline-function-with-localized-variables | [
"# Creating a multiline function with localized variables\n\nI would like to create a function with several expressions in the code body.\n\nFor example, I would like to write something like this, but which will actually work.\n\nmyFunction[x]:=\ny=x\ny=y+3;\n(y+3)^3\n\n\nI know that I can do this :\n\nmyFunction[x]:=(y=x;y=y+3;(y+3)^3)\n\n\nBut it will be hardly writable if I have a lot of expression to evaluate.\n\nHow can I do it?\n\n(I just installed Mathematica and read some tutorials on the internet so my knowledge is veeeeeery basic).\n\n• Use Module[{y},...] to make y a local variable. You can use newlines, they have no syntactical meaning. – Felix Feb 28 '17 at 15:00\n• You mean I should write : myFunction[x]:=Module[{y},y=x;y=y+3;(y+3)^3] And I can do newline between my \";\" ? – StarBucK Feb 28 '17 at 15:04\n• Take a look reference.wolfram.com/language/howto/… and mathematica.stackexchange.com/a/39464/5478 for more basic tutorials. – Kuba Feb 28 '17 at 15:04\n• @Felix in cells they have, unless they are inside expressions. – Kuba Feb 28 '17 at 15:08\n• Yes, within Module (or simply parentheses for that matter) you can use a newline for spacing, but you still need ; (CompoundExpression) or you end up multiplying expressions unintentionally. Please see (41091) for more. – Mr.Wizard Feb 28 '17 at 15:11\n\nYou should write your function like so:\n\nmyFunction[x_] :=\nModule[{y},\ny = x;\ny = y + 3;\n(y + 3)^3]\n\n\nNote the underscore in x_. This makes x into a formal argument that will not be confused with any definition of x you might have made.\n\nThen, even when x and y have global values they will not interfere with either the proper definition of myFunction nor with calls to it.\n\nx = 42; y = 43;\nmyFunction\n\n\n343"
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https://www.physicsforums.com/threads/calculating-contact-forces-three-objects.636116/ | [
"# Calculating Contact Forces- Three objects\n\n## Homework Statement\n\nIf you have a system with three boxes, two side by side and one on top of the first box and there is a horizontal force acting on the boxes, do you have to account for the third box in calculating the contact force between two of them? Say if box 1, box 2, and box 3 (three is on top of box 1) had masses of 10 kg, 5 kg, and 6 kg, respectively. And the horizontal force was equal to 20 N and you were asked to calculate the contact force between boxes 1 and 2...would you have to divide 20N by the sum of all THREE masses or just the masses of ONE and TWO in order to get the acceleration to calculate contact force?\n\n## Homework Equations\n\nYou would need to use F = ma (a = F/m) to get the acceleration in the horizontal direction and then to get the contact force.\n\n## The Attempt at a Solution\n\nI thought that you would have to take the box on top into account so a = 20 N / (10 kg + 5 kg + 6 kg) = 0.95 m/s^2\n\nAnd then to get the contact force between boxes 1 and 2, you would multiply 0.95 m/s^2 by 5 kg to get 4.76 N.\n\nLast edited:\n\nPhanthomJay\nHomework Helper\nGold Member\nYes you are correct\n\nCWatters\nHomework Helper\nGold Member\nYes correct but...\n\nIf the force is applied to the lower boxes then you may need to look at the friction force between upper and lower boxes. If you try to accelerate the lower boxes too fast they will \"leave the top box behind\". The top box can only accelerate as fast as the friction force/it's mass. That might be less than the applied force/total mass.\n\nPhanthomJay"
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https://www.51cto.com/article/747588.html | [
"# 深度学习神经网络之图像分类应用实战\n\nKeras、PyTorch和TensorFlow等框架有助于定制深度神经网络的艰难构建、训练、验证和部署过程。在现实生活中创建深度学习应用程序时，这几款框架显然成为首选。",
null,
"",
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"",
null,
"#### 1. 加载数据\n\n`import numpy as npimport pandas as pdimport osfrom os.path import joinfrom tensorflow.keras.preprocessing.image import load_img, img_to_arrayfrom sklearn.model_selection import train_test_split`\n\n`cats_dir = \"data\\\\cats\"all_cats_path = [join(cats_dir,filename) for filename in os.listdir(cats_dir)]images_dir = \"data\\\\random_images\"images_path = [join(images_dir,filename) for filename in os.listdir(images_dir)]all_paths = all_cats_path + images_pathdf = pd.DataFrame({ 'path': all_paths, 'is_cat': [1 if path in all_cats_path else 0 for path in all_paths] })`\n\n`X = df.pathY = df.is_catX_train, X_test, y_train, y_test = train_test_split(X,Y, test_size=.2 , shuffle= True)X_train = [load_img(img_path,target_size=(128,128)) for img_path in X_train]X_train = np.array([img_to_array(img) for img in X_train])X_test = [load_img(img_path,target_size=(128,128)) for img_path in X_test]X_test = np.array([img_to_array(img) for img in X_test])(X_train.shape,X_test.shape)`",
null,
"4X_train和X_test向量的形状\n\n#### 2. 初始化参数\n\n`def initialize(layers_dimensions): parameters = {} L = len(layers_dimensions) for l in range (1,L): parameters['w' + str(l)] = np.random.randn(layers_dimensions[l],layers_dimensions[l-1]) / np.sqrt(layers_dimensions[l-1]) parameters['b' + str(l)] = np.zeros((layers_dimensions[l],1)) return parameters`",
null,
"#### 3. 正向传播\n\n`def linear_forward(activation, weight, bias): Z = np.dot(weight,activation) + bias cache = (activation, weight, bias) return Z, cache`",
null,
"`def sigmoid(Z): activation = 1/ (1+ np.exp(-Z)) cache = Z return activation, cache def relu(Z): activation = np.maximum(0,Z) cache = Z return activation, cache`\n\n`def sigmoid_activation(previous_activation, weight, bias): Z, linear_cache = linear_forward(previous_activation,weight, bias) activation, activation_cache = sigmoid(Z) cache = (linear_cache,activation_cache) return activation, cache def relu_activation(previous_activation, weight, bias): Z, linear_cache = linear_forward(previous_activation,weight, bias) activation, activation_cache = relu(Z) cache = (linear_cache,activation_cache) return activation, cache`\n\n`def l_layer_model_forward(data, parameters): caches = [] activation = data n_layers = len(parameters)//2 for layer in range (1,n_layers): previous_activation = activation activation, cache = relu_activation(previous_activation, weight = parameters['w' + str(layer)], bias = parameters['b' + str(layer)]) caches.append(cache) last_activation, cache = sigmoid_activation(activation, weight = parameters['w' + str(layer+1)], bias = parameters['b' + str(layer+1)]) caches.append(cache) return last_activation, caches`\n\n#### 4. 交叉熵损失函数",
null,
"`def cross_entropy_cost(last_activation,true_label): m = true_label.shape cost = -1/m * np.sum(np.dot(true_label,np.log(last_activation).T) + np.dot(1-true_label, np.log(1-last_activation).T)) cost = np.squeeze(cost) return cost`\n\n#### 5. 反向传播",
null,
"`def linear_backward(dZ, cache): previous_activation, weight, bias = cache m = previous_activation.shape dw = 1/m * np.dot(dZ, previous_activation.T) db = 1/m * np.sum(dZ, keepdims = True, axis = 1) dpreviousactivation = np.dot(weight.T,dZ) return dpreviousactivation, dw, db`",
null,
"`def relu_backward(dactivation, cache): Z = cache dZ = np.array(dactivation, copy=True) dZ[Z <= 0] = 0 return dZ def sigmoid_backward(dactivation, cache): Z = cache s = 1/(1+np.exp(-Z)) dZ = dactivation * s * (1-s) return dZ`\n\n`def linear_activation_backward(dactivation, cache, activation): linear_cache, activation_cache = cache if activation == 'relu': dZ = relu_backward(dactivation, activation_cache) dprevious_activation, dw, db = linear_backward(dZ,linear_cache) elif activation == 'sigmoid': dZ = sigmoid_backward(dactivation, activation_cache) dprevious_activation, dw, db = linear_backward(dZ,linear_cache) return dprevious_activation, dw, db`",
null,
"`def l_layer_model_backward(last_activation, true_labels, caches): gradients = {} n_layers = len(caches) true_labels = true_labels.reshape(last_activation.shape) dlast_activation = -(np.divide(true_labels, last_activation) - np.divide(1 - true_labels, 1 - last_activation)) current_cache = caches[n_layers-1] dprevious_activation, dw_temp, db_temp = linear_activation_backward(dlast_activation,current_cache,'sigmoid') gradients[\"da\" + str(n_layers-1)] = dprevious_activation gradients[\"dw\" + str(n_layers)] = dw_temp gradients[\"db\" + str(n_layers)] = db_temp for layer in reversed(range(n_layers-1)): current_cache = caches[layer] dprevious_activation, dw_temp, db_temp = linear_activation_backward(gradients[\"da\" + str(layer + 1)],current_cache,'relu') gradients[\"da\" + str(layer)] = dprevious_activation gradients[\"dw\" + str(layer+1)] = dw_temp gradients[\"db\" + str(layer+1)] = db_temp return gradients`\n\n#### 6. 参数更新\n\n`def update_parameters(parameters, gradients, learning_rate): parameters = parameters.copy() n_layers = len(parameters) // 2 for layer in range (n_layers): parameters[\"w\" + str(layer+1)] = parameters[\"w\" + str(layer+1)] - learning_rate * gradients[\"dw\" + str(layer+1)] parameters[\"b\" + str(layer+1)] = parameters[\"b\" + str(layer+1)] - learning_rate * gradients[\"db\" + str(layer+1)] return parameters`\n\n#### 7. 预处理矢量\n\n`layers_dimensions = [49152, 20, 7, 5, 1]X_train_flatten = X_train.reshape(X_train.shape, -1).TX_test_flatten = X_test.reshape(X_test.shape, -1).TX_train = X_train_flatten/255.X_test = X_test_flatten/255.y_train = np.array(y_train)y_test = np.array(y_test)Y_train = y_train.reshape(-1,1).TY_test = y_test.reshape(-1,1).T print(f'X train shape: {X_train.shape}')print(f'Y train shape: {Y_train.shape}')print(f'X test shape: {X_test.shape}')print(f'Y test shape: {Y_test.shape}')`",
null,
"#### 8. 训练\n\n`def l_layer_model(X, Y, layers_dimensions, learning_rate = 0.0075, iterations = 3000, print_cost=False): costs = [] parameters = initialize(layers_dimensions) for i in range(0, iterations): last_activation, caches = l_layer_model_forward(X, parameters) cost = cross_entropy_cost(last_activation, Y) gradients = l_layer_model_backward(last_activation, Y, caches) parameters = update_parameters(parameters, gradients, learning_rate) if print_cost and i % 50 == 0 or i == iterations - 1: print(f\"Cost after iteration {i}: {np.squeeze(cost)}\") if i % 100 == 0 or i == iterations: costs.append(cost) return parameters, costs`\n\n`parameters, costs = l_layer_model(X_train, Y_train, layers_dimensions, iterations = 2500, print_cost = True)`",
null,
"12成本输出值越来越小\n\n#### 9. 预测\n\n`def predict(X, y, parameters): m = X.shape p = np.zeros((1,m)) probs, _ = l_layer_model_forward(X, parameters) for i in range(0, probs.shape): if probs[0,i] > 0.5: p[0,i] = 1 else: p[0,i] = 0 print(\"Accuracy: \" + str(np.sum((p == y)/m))) return p pred_test = predict(X_test, Y_test , parameters)`",
null,
"13调用预测函数\n\n#### 译者介绍",
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"",
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"",
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"",
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"",
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"",
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"51CTO技术栈公众号",
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""
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| {"ft_lang_label":"__label__zh","ft_lang_prob":0.9779814,"math_prob":0.9944143,"size":7615,"snap":"2023-14-2023-23","text_gpt3_token_len":6832,"char_repetition_ratio":0.06135856,"word_repetition_ratio":0.0,"special_character_ratio":0.154826,"punctuation_ratio":0.025974026,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99920744,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-22T02:48:23Z\",\"WARC-Record-ID\":\"<urn:uuid:41ba98ee-df9d-4aa0-8900-c2550ca054e0>\",\"Content-Length\":\"298303\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a3c111ec-8a38-4815-88a6-90680c3833da>\",\"WARC-Concurrent-To\":\"<urn:uuid:64f4e7bb-8d8a-4ff0-b55c-a1de27c7d7f0>\",\"WARC-IP-Address\":\"203.107.44.140\",\"WARC-Target-URI\":\"https://www.51cto.com/article/747588.html\",\"WARC-Payload-Digest\":\"sha1:O4Q3JURT3LV66REJWBZKGCML64X7JW7N\",\"WARC-Block-Digest\":\"sha1:BF5HLBBIGLTQSYKG7H7UBESBUCUKAIT4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296943749.68_warc_CC-MAIN-20230322020215-20230322050215-00100.warc.gz\"}"} |
https://dml.cz/handle/10338.dmlcz/119341 | [
"# Article\n\nKeywords:\nultracompleteness; Čech-completeness; countable type; pointwise countable type\nSummary:\nWe prove that if $X^n$ is a union of $n$ subspaces of pointwise countable type then the space $X$ is of pointwise countable type. If $X^\\omega$ is a countable union of ultracomplete spaces, the space $X^\\omega$ is ultracomplete. We give, under CH, an example of a Čech-complete, countably compact and non-ultracomplete space, giving thus a partial answer to a question asked in [BY2].\nReferences:\n[Ar] Arhangel'skiĭ A.V.: Bicompact sets and the topology of spaces. Dokl. Akad. Nauk SSSR 150 (1963), 9-12. MR 0150733\n[BY1] Buhagiar D., Yoshioka I.: Ultracomplete topological spaces. preprint. MR 1924245 | Zbl 1019.54015\n[BY2] Buhagiar D., Yoshioka I.: Sums and products of ultracomplete topological spaces. Topology Appl., to appear. MR 1919293 | Zbl 1019.54015\n[BGT] Balogh Z, Gruenhage G., Tkachuk V.: Additivity of metrizability and related properties. Topology Appl. (1998), 84 91-103. MR 1611277 | Zbl 0991.54032\n[Lo] López de Luna M.: Some new results on Čech-complete spaces. Topology Proceedings, vol.24, 1999. MR 1876382\n[Pa] Pasynkov B.A.: Almost metrizable topological groups (in Russian). Dokl. Akad. Nauk SSSR (1965), 161.2 281-284. MR 0204565\n[PT] Ponomarev V.I., Tkachuk V.V.: The countable character of $X$ in $\\beta X$ compared with the countable character of the diagonal in $X\\times X$. Vestnik Moskovskogo Universiteta, Matematika, 42 (1987), 5 16-19. MR 0913263 | Zbl 0652.54003\n[Tk] Tkachuk V.V.: Finite and countable additivity topological properties in nice spaces. Trans. Amer. Math. Soc. (1994), 341 585-601. MR 1129438\n[Tk1] Tkachenko M.G.: On a property of bicompacta. Seminar on General Topology, ed. by P.S. Alexandroff, Mosc. Univ. P.H., Moscow, 1981, pp.149-156. MR 0656955 | Zbl 0491.54002"
]
| [
null
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.6363753,"math_prob":0.8718255,"size":1726,"snap":"2020-45-2020-50","text_gpt3_token_len":594,"char_repetition_ratio":0.13066202,"word_repetition_ratio":0.0,"special_character_ratio":0.3441483,"punctuation_ratio":0.25789472,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9501447,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-23T16:24:03Z\",\"WARC-Record-ID\":\"<urn:uuid:4ff6e602-287e-4bc9-ab76-6e41567ad7a1>\",\"Content-Length\":\"15711\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:90017ab5-febe-4e85-ba45-336096699420>\",\"WARC-Concurrent-To\":\"<urn:uuid:d4ade938-6c3b-44bf-a67a-5928d048875c>\",\"WARC-IP-Address\":\"147.251.6.150\",\"WARC-Target-URI\":\"https://dml.cz/handle/10338.dmlcz/119341\",\"WARC-Payload-Digest\":\"sha1:4LV53YKFFWGKH3PRHQLPCV34GYCDZ7ED\",\"WARC-Block-Digest\":\"sha1:3GPANIJAJV3JVGC6W6RWURMF3LT6F3PA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141163411.0_warc_CC-MAIN-20201123153826-20201123183826-00283.warc.gz\"}"} |
https://stackoverflow.com/questions/13912282/subset-xts-object-by-time-of-day | [
"Subset xts object by time of day\n\nA simple question: I know how to subset time series in `xts` for years, months and days from the help: `x['2000-05/2001']` and so on.\n\nBut how can I subset my data by hours of the day? I would like to get all data between 07:00 am and 06:00 pm. I.e., I want to extract the data during business time - irrelevant of the day (I take care for weekends later on). Help has an example of the form:\n\n``````.parseISO8601('T08:30/T15:00')\n``````\n\nBut this does not work in my case. Does anybody have a clue?\n\n• Can you please give a reproducible example? – agstudy Dec 17 '12 at 11:29\n• If your `xts` object is called `x` then something like `y <- x[\"T09:30/T11:00\"]` works for me to get a slice of the morning session, for example. – SlowLearner Dec 17 '12 at 11:31\n• @agstudy sample.time = timeDate('2012-01-01 00:00:00')+15*60*(1:500) data = 1:500 data.ts = xts(data,order.by=sample.time) data.ts[\"T09:30/T11:00\"] – Richard Dec 17 '12 at 12:13\n• @SlowLearner You are right .. it simply works ... I was confused because I just used it on the time index and not on the xts object. Aplied to the object it simply works. data.ts[\"T09:30/T11:00\"] works, but sample.time[\"T09:30/T11:00\"] does not. – Richard Dec 17 '12 at 12:15\n• @SlowLearner ... I would accept, if your comment were an answer. ... – Richard Dec 17 '12 at 12:16\n\nIf your `xts` object is called `x` then something like `y <- x[\"T09:30/T11:00\"]` works for me to get a slice of the morning session, for example.\n\nFor some reason to cut xts time of day using `x[\"T09:30/T11:00\"]` is pretty slow, I use the method from R: Efficiently subsetting dataframe based on time of day and data.table time subset vs xts time subset to make a faster function with similar syntax:\n\n``````cut_time_of_day <- function(x, t_str_begin, t_str_end){\n\ntstr_to_sec <- function(t_str){\n#\"09:00:00\" to sec of day\nas.numeric(as.POSIXct(paste(\"1970-01-01\", t_str), \"UTC\")) %% (24*60*60)\n}\n\n#POSIX ignores leap second\n#sec_of_day = as.numeric(index(x)) %% (24*60*60) #GMT only\nsec_of_day = {lt = as.POSIXlt(index(x)); lt\\$hour *60*60 + lt\\$min*60 + lt\\$sec} #handle tzone\nsec_begin = tstr_to_sec(t_str_begin)\nsec_end = tstr_to_sec(t_str_end)\n\nreturn(x[ sec_of_day >= sec_begin & sec_of_day <= sec_end,])\n}\n``````\n\nTest:\n\n``````n = 100000\ndtime <- seq(ISOdate(2001,1,1), by = 60*60, length.out = n)\nattributes(dtime)\\$tzone <- \"CET\"\nx = xts((1:n), order.by = dtime)\n\ny2 <- cut_time_of_day(x,\"07:00:00\", \"09:00:00\")\ny1 <- x[\"T07:00:00/T09:00:00\"]\n\nidentical(y1,y2)\n``````\n• Thanks for the demonstration of how much faster xts' time-of-day subsetting can be! I've created an issue to work on improving it. – Joshua Ulrich Jun 10 '17 at 16:26\n• for those who are wondering ... the custom function above is still about 1000x faster than the internal xts function. – ricardo Mar 30 '18 at 6:50\n• @ricardo: I get results that are hundreds of times faster, but not close to 1000x. Can you share your benchmark and `sessionInfo()` output? Feel free to email it to me. – Joshua Ulrich May 29 '18 at 18:23"
]
| [
null
]
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http://cnsjhf.com/qspevdu_d001016003 | [
"• 安徽\n• 安庆市\n• 蚌埠市\n• 亳州市\n• 巢湖市\n• 池州市\n• 滁州市\n• 阜阳市\n• 合肥市\n• 淮北市\n• 淮南市\n• 黄山市\n• 六安市\n• 马鞍山市\n• 宿州市\n• 铜陵市\n• 芜湖市\n• 宣城市\n• 广西\n• 百色市\n• 北海市\n• 崇左市\n• 防城港市\n• 贵港市\n• 桂林市\n• 河池市\n• 贺州市\n• 来宾市\n• 柳州市\n• 南宁市\n• 钦州市\n• 梧州市\n• 玉林市\n• 河南\n• 安阳市\n• 鹤壁市\n• 焦作市\n• 开封市\n• 洛阳市\n• 漯河市\n• 南阳市\n• 平顶山市\n• 濮阳市\n• 三门峡市\n• 商丘市\n• 新乡市\n• 信阳市\n• 许昌市\n• 郑州市\n• 周口市\n• 驻马店市\n• 吉林\n• 白城市\n• 白山市\n• 长春市\n• 吉林市\n• 辽源市\n• 四平市\n• 松原市\n• 通化市\n• 延边朝鲜族自治州\n• 广东\n• 潮州市\n• 东莞市\n• 佛山市\n• 广州市\n• 河源市\n• 惠州市\n• 江门市\n• 揭阳市\n• 茂名市\n• 梅州市\n• 清远市\n• 汕头市\n• 汕尾市\n• 韶关市\n• 深圳市\n• 阳江市\n• 云浮市\n• 湛江市\n• 肇庆市\n• 中山市\n• 珠海市\n• 辽宁\n• 鞍山市\n• 本溪市\n• 朝阳市\n• 大连市\n• 丹东市\n• 抚顺市\n• 阜新市\n• 葫芦岛市\n• 锦州市\n• 辽阳市\n• 盘锦市\n• 沈阳市\n• 铁岭市\n• 营口市\n• 湖北\n• 鄂州市\n• 恩施土家族苗族自治州\n• 黄冈市\n• 黄石市\n• 荆门市\n• 荆州市\n• 直辖行政单位\n• 十堰市\n• 随州市\n• 武汉市\n• 咸宁市\n• 襄阳市\n• 孝感市\n• 宜昌市\n• 江西\n• 抚州市\n• 赣州市\n• 吉安市\n• 景德镇市\n• 九江市\n• 南昌市\n• 萍乡市\n• 上饶市\n• 新余市\n• 宜春市\n• 鹰潭市\n• 浙江\n• 杭州市\n• 湖州市\n• 嘉兴市\n• 金华市\n• 丽水市\n• 宁波市\n• 衢州市\n• 绍兴市\n• 台州市\n• 温州市\n• 舟山市\n• 青海\n• 果洛藏族自治州\n• 海北藏族自治州\n• 海东地区\n• 海南藏族自治州\n• 海西蒙古族藏族自治州\n• 黄南藏族自治州\n• 西宁市\n• 玉树藏族自治州\n• 甘肃\n• 白银市\n• 定西市\n• 甘南藏族自治州\n• 嘉峪关市\n• 金昌市\n• 酒泉市\n• 兰州市\n• 临夏回族自治州\n• 陇南市\n• 平凉市\n• 庆阳市\n• 天水市\n• 武威市\n• 张掖市\n• 贵州\n• 安顺市\n• 毕节市\n• 贵阳市\n• 六盘水市\n• 黔东南苗族侗族自治州\n• 黔南布依族苗族自治州\n• 黔西南布依族苗族自治州\n• 铜仁地区\n• 遵义市\n• 陕西\n• 安康市\n• 宝鸡市\n• 汉中市\n• 商洛市\n• 铜川市\n• 渭南市\n• 西安市\n• 咸阳市\n• 延安市\n• 榆林市\n• 西藏\n• 阿里地区\n• 昌都地区\n• 拉萨市\n• 林芝地区\n• 那曲地区\n• 日喀则地区\n• 山南地区\n• 宁夏\n• 固原市\n• 石嘴山市\n• 吴忠市\n• 银川市\n• 中卫市\n• 福建\n• 福州市\n• 龙岩市\n• 南平市\n• 宁德市\n• 莆田市\n• 泉州市\n• 三明市\n• 厦门市\n• 漳州市\n• 内蒙古\n• 阿拉善盟\n• 巴彦淖尔市\n• 包头市\n• 赤峰市\n• 鄂尔多斯市\n• 呼和浩特市\n• 呼伦贝尔市\n• 通辽市\n• 乌海市\n• 乌兰察布市\n• 锡林郭勒盟\n• 兴安盟\n• 云南\n• 保山市\n• 楚雄彝族自治州\n• 大理白族自治州\n• 德宏傣族景颇族自治州\n• 迪庆藏族自治州\n• 红河哈尼族彝族自治州\n• 昆明市\n• 丽江市\n• 临沧市\n• 怒江傈僳族自治州\n• 曲靖市\n• 思茅市\n• 文山壮族苗族自治州\n• 西双版纳傣族自治州\n• 玉溪市\n• 昭通市\n• 新疆\n• 阿克苏地区\n• 阿勒泰地区\n• 巴音郭楞蒙古自治州\n• 博尔塔拉蒙古自治州\n• 昌吉回族自治州\n• 哈密地区\n• 和田地区\n• 喀什地区\n• 克拉玛依市\n• 克孜勒苏柯尔克孜自治州\n• 直辖行政单位\n• 塔城地区\n• 吐鲁番地区\n• 乌鲁木齐市\n• 伊犁哈萨克自治州\n• 黑龙江\n• 大庆市\n• 大兴安岭地区\n• 哈尔滨市\n• 鹤岗市\n• 黑河市\n• 鸡西市\n• 佳木斯市\n• 牡丹江市\n• 七台河市\n• 齐齐哈尔市\n• 双鸭山市\n• 绥化市\n• 伊春市\n• 香港\n• 香港\n• 九龙\n• 新界\n• 澳门\n• 澳门\n• 其它地区\n• 台湾\n• 台中市\n• 台南市\n• 高雄市\n• 台北市\n• 基隆市\n• 嘉义市\n•",
null,
"三星装饰优质施工工艺-广东保利天玺室内装修装潢设计费用\n\n品牌:三星装饰,三星家装,三星软装\n\n出厂地:广东省 佛山市\n\n报价：面议\n\n佛山市三星装饰设计工程有限公司\n\n黄金会员：",
null,
"主营：室内装修设计,室内家装设计,室内软装设计...\n\n•",
null,
"玻璃隔热涂料厂家_瑞佩姆智能涂料高质量的隔热涂料供应\n\n品牌:瑞佩姆,至能,滋润\n\n出厂地:广东省 佛山市\n\n报价：面议\n\n佛山市瑞佩姆智能涂料有限公司\n\n黄金会员：",
null,
"主营：防水涂料,隔热涂料,内外墙底漆,智能外墙...\n\n•",
null,
"对开门哪家好-买不锈钢对开门就来振宏门业\n\n品牌:振宏,,\n\n出厂地:广东省 佛山市\n\n报价：面议\n\n佛山市禅城区振宏门业加工厂\n\n黄金会员：",
null,
"主营：对讲门,推拉门,对开门,不锈钢门,不锈钢...\n\n•",
null,
"佛山皮床品牌-品质有保障的皮床米兰家居供应\n\n品牌:米兰,米兰家居,\n\n出厂地:广东省 佛山市\n\n报价：面议\n\n广东米兰家居有限公司\n\n黄金会员：",
null,
"主营：皮沙发,布艺沙发,皮床,布艺床,床垫\n\n•",
null,
"废塑胶回收公司|提供佛山可靠的回收各种有色金属\n\n品牌:森润能物资,,\n\n出厂地:广东省 佛山市\n\n报价：面议\n\n佛山市南海区森润能物资回收部\n\n黄金会员：",
null,
"主营：废铁打包,锌铁价格,回收金银铜铝铁,回收...\n\n•",
null,
"佛山塑料瓶厂家_佛山优良的大口方形透明瓶供应\n\n品牌:思盾,,\n\n出厂地:广东省 佛山市\n\n报价：面议\n\n佛山市思盾包装制品有限公司\n\n黄金会员：",
null,
"主营：塑料桶,塑料瓶,化工桶,塑料水塔,洗洁精...\n\n•",
null,
"批发配送食材_广东哪里有口碑好的食材配送\n\n品牌:客兴餐饮,客兴团膳,客兴工作餐\n\n出厂地:广东省 佛山市\n\n报价：面议\n\n佛山市三水客兴餐饮管理服务有限公司\n\n黄金会员：",
null,
"主营：食材配送,食堂承包,农副产品配送,饭堂承...\n\n•",
null,
"加热普通碳钢系列电磁加热器_电磁加热器推荐\n\n品牌:全桥,,\n\n出厂地:广东省 佛山市\n\n报价：面议\n\n广东佛山市全桥电器有限公司\n\n黄金会员：",
null,
"主营：电磁加热器价格,电磁采暖炉厂家,电磁炉机...\n\n•",
null,
"浙江香蕉伞厂家推广_哪里有卖出色的香蕉伞\n\n品牌:金凤港,,\n\n出厂地:广东省 佛山市\n\n报价：面议\n\n佛山市金凤港帐篷有限公司\n\n黄金会员：",
null,
"主营：扳手伞,侧力伞,帝国伞,吊篮,空中舞星\n\n•",
null,
"新鲜土猪|口碑好的土猪肉哪里有卖\n\n品牌:皇永顺,,\n\n出厂地:广东省 佛山市\n\n报价：面议\n\n佛山市皇永顺食品有限公司\n\n黄金会员：",
null,
"主营：土猪肉,腊肠,腊肉,猪肉丸,土猪餐厅\n\n• 没有找到合适的佛山市供应商？您可以发布采购信息\n\n没有找到满足要求的佛山市供应商？您可以搜索 批发 公司\n\n### 最新入驻厂家\n\n相关产品:\n三星装饰优质施工工艺 玻璃隔热涂料厂家 对开门哪家好 佛山皮床品牌 废塑胶回收公司 佛山塑料瓶厂家 批发配送食材 加热普通碳钢系列电磁加热器 浙江香蕉伞厂家推广 新鲜土猪"
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null,
"http://image-ali.bianjiyi.com/1/2018/1129/15/15434766478579.jpg",
null,
"http://www.shangwuwang.com/Public/Images/ForeApps/grade2.png",
null,
"http://image-ali.bianjiyi.com/1/2018/0719/11/15319698531903.jpg",
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null,
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null,
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https://www.20sim.com/webhelp/toolboxes_frequency_domain_toolbox_linear_system_editor_editor_bode_plots.php | [
" 20-sim webhelp > Toolboxes > Control Toolbox > Controller Design Editor > Bode Plots\n\n# Bode Plots\n\nBode Plots show the amplitude and phase of a linear system as function of the frequency. Bode plots can be shown for every 20-sim model through Linearization. During Linearization you are asked to enter the input variable and output variable for which linearization should be performed. After that the linear system is calculated and shown in the Linear System Editor. From the Linear System Editor you can generate a bode plot. These actions can also be predefined using the Frequency Response command of the Properties menu.",
null,
"20-sim automatically generates a range of logarithmically displayed frequencies, based on the system dynamics. With the Plot Properties command (right mouse menu), you can change this horizon and recalculate the bode response (click the Bode command again).\n\nThe magnitude part of the plot can be displayed in dB or in absolute values. The phase part can be displayed in radians or degrees. The frequency can be displayed in radians per second or in Hz.\n\n## Plot Options\n\nUsing the toolbar or the right mouse menu, you can use various options:\n\n • Plot Properties: Set the plot properties.\n • Numerical Values: Inspect numerical values.\n • Magnitude (dB): Display magnitude in decibels\n • Magnitude (-): Display magnitude in absolute values."
]
| [
null,
"https://www.20sim.com/webhelp/bodeplot_zoom80.jpg",
null
]
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https://origin.geeksforgeeks.org/tilt-binary-tree/?ref=lbp | [
"# Tilt of Binary Tree\n\n• Difficulty Level : Medium\n• Last Updated : 30 Jun, 2021\n\nGiven a binary tree, return the tilt of the whole tree. The tilt of a tree node is defined as the absolute difference between the sum of all left subtree node values and the sum of all right subtree node values. Null nodes are assigned tilt to be zero. Therefore, tilt of the whole tree is defined as the sum of all nodes’ tilt.\nExamples:\n\n```Input :\n1\n/ \\\n2 3\nOutput : 1\nExplanation:\nTilt of node 2 : 0\nTilt of node 3 : 0\nTilt of node 1 : |2-3| = 1\nTilt of binary tree : 0 + 0 + 1 = 1\n\nInput :\n4\n/ \\\n2 9\n/ \\ \\\n3 5 7\nOutput : 15\nExplanation:\nTilt of node 3 : 0\nTilt of node 5 : 0\nTilt of node 7 : 0\nTilt of node 2 : |3-5| = 2\nTilt of node 9 : |0-7| = 7\nTilt of node 4 : |(3+5+2)-(9+7)| = 6\nTilt of binary tree : 0 + 0 + 0 + 2 + 7 + 6 = 15```\n\nThe idea is to recursively traverse tree. While traversing, we keep track of two things, sum of subtree rooted under current node, tilt of current node. Sum is needed to compute tilt of parent.\n\n## C++\n\n `// CPP Program to find Tilt of Binary Tree` `#include ` `using` `namespace` `std;` `/* A binary tree node has data, pointer to` `left child and a pointer to right child */` `struct` `Node {` ` ``int` `val;` ` ``struct` `Node *left, *right;` `};` `/* Recursive function to calculate Tilt of` ` ``whole tree */` `int` `traverse(Node* root, ``int``* tilt)` `{` ` ``if` `(!root)` ` ``return` `0;` ` ``// Compute tilts of left and right subtrees` ` ``// and find sums of left and right subtrees` ` ``int` `left = traverse(root->left, tilt);` ` ``int` `right = traverse(root->right, tilt); ` ` ``// Add current tilt to overall` ` ``*tilt += ``abs``(left - right);` ` ``// Returns sum of nodes under current tree` ` ``return` `left + right + root->val;` `}` `/* Driver function to print Tilt of whole tree */` `int` `Tilt(Node* root)` `{` ` ``int` `tilt = 0;` ` ``traverse(root, &tilt);` ` ``return` `tilt;` `}` `/* Helper function that allocates a` `new node with the given data and` `NULL left and right pointers. */` `Node* newNode(``int` `data)` `{` ` ``Node* temp = ``new` `Node;` ` ``temp->val = data;` ` ``temp->left = temp->right = NULL;` ` ``return` `temp;` `}` `// Driver code` `int` `main()` `{` ` ``/* Let us construct a Binary Tree` ` ``4` ` ``/ \\` ` ``2 9` ` ``/ \\ \\` ` ``3 5 7 */` ` ``Node* root = NULL;` ` ``root = newNode(4);` ` ``root->left = newNode(2);` ` ``root->right = newNode(9);` ` ``root->left->left = newNode(3);` ` ``root->left->right = newNode(8);` ` ``root->right->right = newNode(7);` ` ``cout << ``\"The Tilt of whole tree is \"` `<< Tilt(root);` ` ``return` `0;` `}`\n\n## Java\n\n `// Java Program to find Tilt of Binary Tree ` `import` `java.util.*;` `class` `GfG {` `/* A binary tree node has data, pointer to ` `left child and a pointer to right child */` `static` `class` `Node { ` ` ``int` `val; ` ` ``Node left, right; ` `}` `/* Recursive function to calculate Tilt of ` `whole tree */` `static` `class` `T{` ` ``int` `tilt = ``0``;` `}` `static` `int` `traverse(Node root, T t ) ` `{ ` ` ``if` `(root == ``null``) ` ` ``return` `0``; ` ` ``// Compute tilts of left and right subtrees ` ` ``// and find sums of left and right subtrees ` ` ``int` `left = traverse(root.left, t); ` ` ``int` `right = traverse(root.right, t); ` ` ``// Add current tilt to overall ` ` ``t.tilt += Math.abs(left - right); ` ` ``// Returns sum of nodes under current tree ` ` ``return` `left + right + root.val; ` `} ` `/* Driver function to print Tilt of whole tree */` `static` `int` `Tilt(Node root) ` `{ ` ` ``T t = ``new` `T(); ` ` ``traverse(root, t); ` ` ``return` `t.tilt; ` `} ` `/* Helper function that allocates a ` `new node with the given data and ` `NULL left and right pointers. */` `static` `Node newNode(``int` `data) ` `{ ` ` ``Node temp = ``new` `Node(); ` ` ``temp.val = data; ` ` ``temp.left = ``null``;` ` ``temp.right = ``null``; ` ` ``return` `temp; ` `} ` `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` ` ``/* Let us construct a Binary Tree ` ` ``4 ` ` ``/ \\ ` ` ``2 9 ` ` ``/ \\ \\ ` ` ``3 5 7 */` ` ``Node root = ``null``; ` ` ``root = newNode(``4``); ` ` ``root.left = newNode(``2``); ` ` ``root.right = newNode(``9``); ` ` ``root.left.left = newNode(``3``); ` ` ``root.left.right = newNode(``8``); ` ` ``root.right.right = newNode(``7``); ` ` ``System.out.println(``\"The Tilt of whole tree is \"` `+ Tilt(root)); ` `}` `} `\n\n## Python3\n\n `# Python3 Program to find Tilt of ` `# Binary Tree ` `# class that allocates a new node ` `# with the given data and ` `# None left and right pointers. ` `class` `newNode:` ` ``def` `__init__(``self``, data):` ` ``self``.val ``=` `data ` ` ``self``.left ``=` `self``.right ``=` `None` `# Recursive function to calculate` `# Tilt of whole tree ` `def` `traverse(root, tilt):` ` ``if` `(``not` `root): ` ` ``return` `0` ` ``# Compute tilts of left and right subtrees ` ` ``# and find sums of left and right subtrees ` ` ``left ``=` `traverse(root.left, tilt) ` ` ``right ``=` `traverse(root.right, tilt) ` ` ``# Add current tilt to overall ` ` ``tilt[``0``] ``+``=` `abs``(left ``-` `right) ` ` ``# Returns sum of nodes under ` ` ``# current tree ` ` ``return` `left ``+` `right ``+` `root.val` `# Driver function to print Tilt` `# of whole tree ` `def` `Tilt(root):` ` ``tilt ``=` `[``0``]` ` ``traverse(root, tilt) ` ` ``return` `tilt[``0``]` `# Driver code ` `if` `__name__ ``=``=` `'__main__'``:` ` ` ` ``# Let us construct a Binary Tree ` ` ``# 4 ` ` ``# / \\ ` ` ``# 2 9 ` ` ``# / \\ \\ ` ` ``# 3 5 7 ` ` ``root ``=` `None` ` ``root ``=` `newNode(``4``) ` ` ``root.left ``=` `newNode(``2``) ` ` ``root.right ``=` `newNode(``9``) ` ` ``root.left.left ``=` `newNode(``3``) ` ` ``root.left.right ``=` `newNode(``8``) ` ` ``root.right.right ``=` `newNode(``7``) ` ` ``print``(``\"The Tilt of whole tree is\"``, ` ` ``Tilt(root))` `# This code is contributed by PranchalK`\n\n## C#\n\n `// C# Program to find Tilt of Binary Tree ` `using` `System;` `class` `GfG ` `{` `/* A binary tree node has data, pointer to ` `left child and a pointer to right child */` `public` `class` `Node ` `{ ` ` ``public` `int` `val; ` ` ``public` `Node left, right; ` `}` `/* Recursive function to calculate Tilt of ` `whole tree */` `public` `class` `T` `{` ` ``public` `int` `tilt = 0;` `}` `static` `int` `traverse(Node root, T t ) ` `{ ` ` ``if` `(root == ``null``) ` ` ``return` `0; ` ` ``// Compute tilts of left and right subtrees ` ` ``// and find sums of left and right subtrees ` ` ``int` `left = traverse(root.left, t); ` ` ``int` `right = traverse(root.right, t); ` ` ``// Add current tilt to overall ` ` ``t.tilt += Math.Abs(left - right); ` ` ``// Returns sum of nodes under current tree ` ` ``return` `left + right + root.val; ` `} ` `/* Driver function to print Tilt of whole tree */` `static` `int` `Tilt(Node root) ` `{ ` ` ``T t = ``new` `T(); ` ` ``traverse(root, t); ` ` ``return` `t.tilt; ` `} ` `/* Helper function that allocates a ` `new node with the given data and ` `NULL left and right pointers. */` `static` `Node newNode(``int` `data) ` `{ ` ` ``Node temp = ``new` `Node(); ` ` ``temp.val = data; ` ` ``temp.left = ``null``;` ` ``temp.right = ``null``; ` ` ``return` `temp; ` `} ` `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` ` ``/* Let us construct a Binary Tree ` ` ``4 ` ` ``/ \\ ` ` ``2 9 ` ` ``/ \\ \\ ` ` ``3 5 7 */` ` ``Node root = ``null``; ` ` ``root = newNode(4); ` ` ``root.left = newNode(2); ` ` ``root.right = newNode(9); ` ` ``root.left.left = newNode(3); ` ` ``root.left.right = newNode(8); ` ` ``root.right.right = newNode(7); ` ` ``Console.WriteLine(``\"The Tilt of whole tree is \"` `+ Tilt(root)); ` `}` `}` `// This code contributed by Rajput-Ji`\n\n## Javascript\n\n ``\n\nOutput:\n\n`The Tilt of whole tree is 15`\n\nComplexity Analysis:\n\n• Time complexity: O(n), where n is the number of nodes in binary tree.\n• Auxiliary Space: O(n) as in worst case, depth of binary tree will be n.\n\nMy Personal Notes arrow_drop_up\nRecommended Articles\nPage :"
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https://cbselibrary.com/tag/9780618595419/ | [
"## Larson Algebra 2 Solutions Chapter 14 Trigonometric Graphs, Identities and Equations Exercise 14.7\n\nLarson Algebra 2 Solutions Chapter 14 Trigonometric Graphs, Identities and Equations Exercise 14.7 Larson Algebra 2 Solutions Chapter 14 Trigonometric Graphs, Identities and Equations Exercise 14.7 1E Chapter 14 Trigonometric Graphs, Identities and Equations Exercise 14.7 1GP Chapter 14 Trigonometric Graphs, Identities and Equations Exercise 14.7 1Q Chapter 14 Trigonometric Graphs, Identities and Equations Exercise … Read more\n\n## Larson Algebra 2 Solutions Chapter 14 Trigonometric Graphs, Identities and Equations Exercise 14.6\n\nLarson Algebra 2 Solutions Chapter 14 Trigonometric Graphs, Identities and Equations Exercise 14.6 Larson Algebra 2 Solutions Chapter 14 Trigonometric Graphs, Identities and Equations Exercise 14.6 1E Chapter 14 Trigonometric Graphs, Identities and Equations Exercise 14.6 1GP Chapter 14 Trigonometric Graphs, Identities and Equations Exercise 14.6 2E Chapter 14 Trigonometric Graphs, Identities and Equations Exercise … Read more\n\n## Larson Algebra 2 Solutions Chapter 14 Trigonometric Graphs, Identities and Equations Exercise 14.5\n\nLarson Algebra 2 Solutions Chapter 14 Trigonometric Graphs, Identities and Equations Exercise 14.5 Larson Algebra 2 Solutions Chapter 14 Trigonometric Graphs, Identities and Equations Exercise 14.5 1E Chapter 14 Trigonometric Graphs, Identities and Equations Exercise 14.5 1GP Chapter 14 Trigonometric Graphs, Identities and Equations Exercise 14.5 1Q Chapter 14 Trigonometric Graphs, Identities and Equations Exercise … Read more\n\n## Larson Algebra 2 Solutions Chapter 14 Trigonometric Graphs, Identities and Equations Exercise 14.4\n\nLarson Algebra 2 Solutions Chapter 14 Trigonometric Graphs, Identities and Equations Exercise 14.4 Larson Algebra 2 Solutions Chapter 14 Trigonometric Graphs, Identities and Equations Exercise 14.4 1E Chapter 14 Trigonometric Graphs, Identities and Equations Exercise 14.4 1GP Chapter 14 Trigonometric Graphs, Identities and Equations Exercise 14.4 1MR Chapter 14 Trigonometric Graphs, Identities and Equations Exercise … Read more\n\n## Larson Algebra 2 Solutions Chapter 14 Trigonometric Graphs, Identities and Equations Exercise 14.3\n\nLarson Algebra 2 Solutions Chapter 14 Trigonometric Graphs, Identities and Equations Exercise 14.3 Larson Algebra 2 Solutions Chapter 14 Trigonometric Graphs, Identities and Equations Exercise 14.3 1E Chapter 14 Trigonometric Graphs, Identities and Equations Exercise 14.3 1GP Chapter 14 Trigonometric Graphs, Identities and Equations Exercise 14.3 2E Chapter 14 Trigonometric Graphs, Identities and Equations Exercise … Read more"
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https://aihubprojects.com/python-programming-examples/ | [
"# PYTHON PROGRAMMING EXAMPLES",
null,
"##### WRITE A PROGRAM TO CHECK PRIME NUMBERS\n`````` >>> a = int (input(\"enter a number : \"))\n>>> c = 0\n>>> if (a<1):\nprint(\"invalid input ! Try next number \")\n>>> if(a==1):\nprint (\"1 is not prime\")\n>>> else:\nfor i in range(2,a,1):\nif (a%i==0):\nc=1\nif (c==0):\nprint(a,\" is prime\")\nelse:\nprint(a,\" is not prime\")\n\nOutput:\nenter a number : 9\n9 is not prime``````\n##### WRITE A PROGRAM TO FIND FIBONACCI SERIES\n``````# print fibonacci series\n>>> a = 1\n>>> b = 0\n>>> n = int(input(\"how many fibonacci series ?? \"))\n>>> while (n>=1):\nc = a+b\na = b\nb = c\nprint (c, end=\",\")\nn=n-1\n\nOutput:\nhow many fibonacci series ?? 5\n1,1,2,3,5``````\n\n``````# print nth fibonacci series value\n>>> a = 1\n>>> b = 0\n>>> n = int(input(\"how many fibonacci series ?? \"))\n>>> while (n>=1):\nc = a+b\na = b\nb = c\nn=n-1\n>>> print (c,\"is the required output\")\n\nOutput:\nhow many fibonacci series ?? 6\n8 is the required output``````\n##### WRITE A PROGRAM TO CHECK ARMSTRONG NUMBER\n`````` >>> #153 is an Armstrong number.\n>>> #1*1*1 + 5*5*5 + 3*3*3 = 153\n>>> sum = 0\n>>> c = 0\n>>> num = int(input(\"enter a number : \"))\n>>> temp = num\n>>> while(num>0): # to find length of number i.e. 153= 3\na = num%10\nnum= num//10\nc=c+1\n>>> num = temp\n>>> while(num>0): # 1*1*1+5*5*5+3*3*3 = 153\na = num%10\nnum= (num-a)/10\nsum = int(sum + a**c)\n>>> if (sum == temp):\nprint(\"Armstrong Number\")\n>>> else:\nprint (\" Not an armstrong number\")\n\nOutput:\nenter a number : 153\nArmstrong Number``````\n\n##### WRITE A PROGRAM TO FIND HCF OF TWO NUMBER\n`````` >>> num1 = int (input (\"enter first number : \"))\n>>> num2 = int (input (\"enter second number : \"))\n>>> if num1 < num2:\nsmall = num1\n>>> else:\nsmall = num2\n>>> for i in range(1, small+1):\nif((num1 % i == 0) and (num2 % i == 0)):\nhcf = i\n>>> print(\" HCF or GCD of \",num1,\"&\", num2,\"is\",hcf)\n\nOutput:\nenter first number : 3\nenter second number : 5\nHCF or GCD of 3 & 5 is 1``````\n##### WRITE A PROGRAM TO FIND LCM OF TWO NUMBER\n``````>>> num1 = int (input (\"enter first number : \"))\n>>> num2 = int (input (\"enter second number : \"))\n>>> if num1 < num2:\nsmall = num1\n>>> else:\nsmall = num2\n>>> for i in range(1, small+1):\nif((num1 % i == 0) and (num2 % i == 0)):\nhcf = i\n>>> lcm = int ((num1*num2)/hcf)\n>>> print(\" LCM of \",num1,\"&\", num2,\"is\",lcm)\n\nOutput:\nenter first number : 4\nenter second number : 8\nLCM of 4 & 8 is 8``````\n\n##### WRITE A PROGRAM TO FIND FACTOR OF A NUMBER\n``````# 25= 5*5\n>>> a = int(input(\"enter a number : \"))\n>>> for i in range(2,a):\nfor j in range (2,a+1):\nif (a%j==0):\nprint (j,end=\",\")\nbreak\na = int(a/j)\n\nOutput:\nenter a number : 4\n2,2``````\n##### WRITE A NUMBER TO FIND FACTORIAL OF A NUMBER\n``````>>> # 5! = 5*4*3*2*1\n>>> a = int(input(\"enter a number : \"))\n>>> c = 1\n>>> if (a==0):\nprint (\" factorial of 0 is 1 \")\n>>> else:\nfor i in range(1,a+1):\nc=c*i\n>>> print (\"\\n factorial of \",a,\"is\",c)\n\nOutput:\nenter a number : 4\nfactorial of 4 is 24``````\n##### WRITE A PROGRAM TO PRINT ALL VOWELS OF STRING\n``````>>> a = str(input(\"enter a string \")).lower()\n>>> b = len(a)\n>>> c = 0\n>>> while (c<=b-1):\nif a[c]=='a' or a[c]=='e' or a[c]=='i' or a[c]=='o' or a[c]=='u' :\nprint(a[c])\nc=c+1\n\nOutput:\nenter a string Diwas\ni\na``````\n\n##### WRITE A PROGRAM TO CHECK PALINDROME STRING\n``````>>> a = str(input(\"enter a string : \"))\n>>> b = (a[::-1])\n>>> if (a==b):\nprint(\"Palindrome number\")\n>>> else:\nprint(\" Not Palindrome number\")\n\nOutput:\nenter a string : sawid\nNot Palindrome number``````\n##### WRITE A PROGRAM TO SORT WORDS IN ALPHABETICAL ORDER\n``````>>> a = \"apple cow dog ant\"\n>>> a = a.split()\n>>> a.sort()\n>>> for word in a:\nprint(word)\n\nOutput:\nant\napple\ncow\ndog``````\n##### WRITE A PROGRAM TO FIND SUM OF ITEMS OF LIST\n``````>>> a = [1,2,3,4,5]\n>>> sum = 0\n>>> for i in a:\nsum = sum + i\n>>> print (\" sum of items in list is : \",sum)\n\nOutput:\nsum of items in list is : 15``````\n\nThis page is contributed by Diwas & Sunil . If you like AIHUB and would like to contribute, you can also write an article & mail your article to [email protected] . See your articles appearing on AI HUB platform and help other AI Enthusiast.\n\n### Recommended Posts",
null,
"Highly motivated, strong drive with excellent interpersonal, communication, and team-building skills. Motivated to learn, grow and excel in Data Science, Artificial Intelligence, SEO & Digital Marketing\n\nView all posts by Diwas Pandey →\n\n### 4 Comments on “PYTHON PROGRAMMING EXAMPLES”\n\n1. Aw, this was an exceptionally nice post. Spending some time and actual effort to generate a good article… but\nwhat can I say… I put things off a whole lot and don’t seem\nto get nearly anything done. adreamoftrains best web hosting company\n\n2.",
null,
"website hosting services says:\n\nHello to all, how is the whole thing, I think every one is getting more from this website, and your views are pleasant in support of\nnew people."
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"https://aihubprojects.com/wp-content/uploads/2020/04/h-735x400.jpg",
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"https://secure.gravatar.com/avatar/5b441e4a39e4a1f93acf1ba0495dee19",
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"https://secure.gravatar.com/avatar/64cabb227909dca52cd1eec37340587c",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.6415739,"math_prob":0.98467785,"size":4238,"snap":"2021-04-2021-17","text_gpt3_token_len":1435,"char_repetition_ratio":0.15682569,"word_repetition_ratio":0.252071,"special_character_ratio":0.41222274,"punctuation_ratio":0.14055794,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99736285,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-23T20:36:20Z\",\"WARC-Record-ID\":\"<urn:uuid:7aff4d1b-92de-48c9-acec-5d8231600729>\",\"Content-Length\":\"104909\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e0d80347-c6bc-4572-861e-0c0c32b6d923>\",\"WARC-Concurrent-To\":\"<urn:uuid:3c62c89f-22bc-46a8-a0cb-55b41f829976>\",\"WARC-IP-Address\":\"172.67.221.139\",\"WARC-Target-URI\":\"https://aihubprojects.com/python-programming-examples/\",\"WARC-Payload-Digest\":\"sha1:65LTKQOREA7JP3QQVNQG6WNAAAPDICXE\",\"WARC-Block-Digest\":\"sha1:7RZNDPHCNH22DQMEDXXZPLRRXG5GRHMJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703538431.77_warc_CC-MAIN-20210123191721-20210123221721-00646.warc.gz\"}"} |
https://toolshed.g2.bx.psu.edu/repos/shellac/sam_consensus_v3/file/4f3585e2f14b/env/lib/python3.9/site-packages/networkx/algorithms/approximation/tests/test_approx_clust_coeff.py | [
"### view env/lib/python3.9/site-packages/networkx/algorithms/approximation/tests/test_approx_clust_coeff.py @ 0:4f3585e2f14bdraftdefaulttip\n\nauthor shellac Mon, 22 Mar 2021 18:12:50 +0000\nline wrap: on\nline source\n```\nimport networkx as nx\nfrom networkx.algorithms.approximation import average_clustering\n\n# This approximation has to be be exact in regular graphs\n# with no triangles or with all possible triangles.\n\ndef test_petersen():\n# Actual coefficient is 0\nG = nx.petersen_graph()\nassert average_clustering(G, trials=int(len(G) / 2)) == nx.average_clustering(G)\n\ndef test_petersen_seed():\n# Actual coefficient is 0\nG = nx.petersen_graph()\nassert average_clustering(\nG, trials=int(len(G) / 2), seed=1\n) == nx.average_clustering(G)\n\ndef test_tetrahedral():\n# Actual coefficient is 1\nG = nx.tetrahedral_graph()\nassert average_clustering(G, trials=int(len(G) / 2)) == nx.average_clustering(G)\n\ndef test_dodecahedral():\n# Actual coefficient is 0\nG = nx.dodecahedral_graph()\nassert average_clustering(G, trials=int(len(G) / 2)) == nx.average_clustering(G)\n\ndef test_empty():\nG = nx.empty_graph(5)\nassert average_clustering(G, trials=int(len(G) / 2)) == 0\n\ndef test_complete():\nG = nx.complete_graph(5)\nassert average_clustering(G, trials=int(len(G) / 2)) == 1\nG = nx.complete_graph(7)\nassert average_clustering(G, trials=int(len(G) / 2)) == 1\n```"
]
| [
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https://artofproblemsolving.com/wiki/index.php/1973_USAMO_Problems/Problem_1 | [
"# 1973 USAMO Problems/Problem 1\n\n## Problem\n\nTwo points",
null,
"$P$ and",
null,
"$Q$ lie in the interior of a regular tetrahedron",
null,
"$ABCD$. Prove that angle",
null,
"$PAQ < 60^\\circ$.\n\n## Solutions\n\n### Solution 1\n\nLet the side length of the regular tetrahedron be",
null,
"$a$. Link and extend",
null,
"$AP$ to meet the plane containing triangle",
null,
"$BCD$ at",
null,
"$E$; link",
null,
"$AQ$ and extend it to meet the same plane at",
null,
"$F$. We know that",
null,
"$E$ and",
null,
"$F$ are inside triangle",
null,
"$BCD$ and that",
null,
"$\\angle PAQ = \\angle EAF$\n\nNow let’s look at the plane containing triangle",
null,
"$BCD$ with points",
null,
"$E$ and",
null,
"$F$ inside the triangle. Link and extend",
null,
"$EF$ on both sides to meet the sides of the triangle",
null,
"$BCD$ at",
null,
"$I$ and",
null,
"$J$,",
null,
"$I$ on",
null,
"$BC$ and",
null,
"$J$ on",
null,
"$DC$. We have",
null,
"$\\angle EAF < \\angle IAJ$\n\nBut since",
null,
"$E$ and",
null,
"$F$ are interior of the tetrahedron, points",
null,
"$I$ and",
null,
"$J$ cannot be both at the vertices and",
null,
"$IJ < a$,",
null,
"$\\angle IAJ < \\angle BAD = 60$. Therefore,",
null,
"$\\angle PAQ < 60$.\n\nSolution with graphs posted at\n\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.",
null,
""
]
| [
null,
"https://latex.artofproblemsolving.com/4/b/4/4b4cade9ca8a2c8311fafcf040bc5b15ca507f52.png ",
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"https://latex.artofproblemsolving.com/9/8/6/9866e3a998d628ba0941eb4fea0666ac391d149a.png ",
null,
"https://latex.artofproblemsolving.com/f/9/e/f9efaf9474c5658c4089523e2aff4e11488f8603.png ",
null,
"https://latex.artofproblemsolving.com/d/8/2/d82cb802aef179231600293427a9b39af036b05b.png ",
null,
"https://latex.artofproblemsolving.com/c/7/d/c7d457e388298246adb06c587bccd419ea67f7e8.png ",
null,
"https://latex.artofproblemsolving.com/e/3/b/e3bb4eeb347e74e025fd175c22f1f7275af0555e.png ",
null,
"https://latex.artofproblemsolving.com/d/7/a/d7ada763e99852a6e44a15d46cce3c3709a944d5.png ",
null,
"https://latex.artofproblemsolving.com/f/a/2/fa2fa899f0afb05d6837885523503a2d4df434f9.png ",
null,
"https://latex.artofproblemsolving.com/7/5/6/7563031eab26ad18658cf2f6a8e4948861bb9013.png ",
null,
"https://latex.artofproblemsolving.com/a/0/5/a055f405829e64a3b70253ab67cb45ed6ed5bb29.png ",
null,
"https://latex.artofproblemsolving.com/f/a/2/fa2fa899f0afb05d6837885523503a2d4df434f9.png ",
null,
"https://latex.artofproblemsolving.com/a/0/5/a055f405829e64a3b70253ab67cb45ed6ed5bb29.png ",
null,
"https://latex.artofproblemsolving.com/d/7/a/d7ada763e99852a6e44a15d46cce3c3709a944d5.png ",
null,
"https://latex.artofproblemsolving.com/0/7/2/072b031c4ff67af3d7dcc1d2270ad56c899eccfb.png ",
null,
"https://latex.artofproblemsolving.com/d/7/a/d7ada763e99852a6e44a15d46cce3c3709a944d5.png ",
null,
"https://latex.artofproblemsolving.com/f/a/2/fa2fa899f0afb05d6837885523503a2d4df434f9.png ",
null,
"https://latex.artofproblemsolving.com/a/0/5/a055f405829e64a3b70253ab67cb45ed6ed5bb29.png ",
null,
"https://latex.artofproblemsolving.com/c/1/2/c128243ea8125d0bb007b8132dcfa44461df73a7.png ",
null,
"https://latex.artofproblemsolving.com/d/7/a/d7ada763e99852a6e44a15d46cce3c3709a944d5.png ",
null,
"https://latex.artofproblemsolving.com/0/2/7/027f4a11d6090f9eac0ce2488df6384dad1263ea.png ",
null,
"https://latex.artofproblemsolving.com/a/b/b/abb5588023cfa1ac14643e9778699f03eecc57a3.png ",
null,
"https://latex.artofproblemsolving.com/0/2/7/027f4a11d6090f9eac0ce2488df6384dad1263ea.png ",
null,
"https://latex.artofproblemsolving.com/6/c/5/6c52a41dcbd739f1d026c5d4f181438b75b76976.png ",
null,
"https://latex.artofproblemsolving.com/a/b/b/abb5588023cfa1ac14643e9778699f03eecc57a3.png ",
null,
"https://latex.artofproblemsolving.com/2/7/e/27e525b776c49bbe7d0ae66cfc7ab8f58f40624e.png ",
null,
"https://latex.artofproblemsolving.com/5/8/a/58af38e8286edcd5a57363d7c65661421eb0ec58.png ",
null,
"https://latex.artofproblemsolving.com/f/a/2/fa2fa899f0afb05d6837885523503a2d4df434f9.png ",
null,
"https://latex.artofproblemsolving.com/a/0/5/a055f405829e64a3b70253ab67cb45ed6ed5bb29.png ",
null,
"https://latex.artofproblemsolving.com/0/2/7/027f4a11d6090f9eac0ce2488df6384dad1263ea.png ",
null,
"https://latex.artofproblemsolving.com/a/b/b/abb5588023cfa1ac14643e9778699f03eecc57a3.png ",
null,
"https://latex.artofproblemsolving.com/e/0/8/e088b2e35b6fd2238a980d3fa1073963ee582399.png ",
null,
"https://latex.artofproblemsolving.com/7/0/9/7099eeab6cc3d3f355737dd3716b8a64d59dd18f.png ",
null,
"https://latex.artofproblemsolving.com/c/7/f/c7fba0d5b781cffd07ff692d3adcd0064b065a41.png ",
null,
"https://wiki-images.artofproblemsolving.com//8/8b/AMC_logo.png",
null
]
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https://excelxor.com/tag/three-dimensional/ | [
"# Incrementing Indirect Column References Within SUMIF(S)/COUNTIF(S) 13\n\nMost Excel users are aware that, when a formula containing relative column references is copied to further columns, those references are updated accordingly. So, for example, the formula:\n\n`=SUMIFS(C:C,\\$A:\\$A,\"X\",\\$B:\\$B,\"X\")`\n\nwhen dragged to the right, will become, successively:\n\n`=SUMIFS(D:D,\\$A:\\$A,\"X\",\\$B:\\$B,\"X\")`\n`=SUMIFS(E:E,\\$A:\\$A,\"X\",\\$B:\\$B,\"X\")`\n\netc., etc.\n\nAnd so we have a relatively (no pun intended) simple means by which we can obtain a conditional sum from successive columns.\n\nBut what if the range we wish to increment is being referenced indirectly? For example, what if we are using a version of the above, but in which the sheet being referenced is dynamic, viz:\n\n`=SUMIFS(INDIRECT(\"'\"&\\$A\\$1&\"'!C:C\"),INDIRECT(\"'\"&\\$A\\$1&\"'!A:A\"),\"X\",INDIRECT(\"'\"&\\$A\\$1&\"'!B:B\"),\"Y\")`\n\nwhere A1 contains the sheet name (e.g. “Sheet1”) which is to be referenced at any given time?\n\n# VLOOKUP Across Several Worksheets (2) – Multiple Search Criteria 9\n\nI recently made the post here, in which I presented a solution to the problem of returning a value based upon matching a single criterion in a given column across multiple worksheets.\n\nIn this follow-up post I will look at the analogous case in which we are not matching a single criterion, but several. As mentioned in the first instalment, I will look at two solutions to this problem, one in which we make use of an extra “helper” column in each of the relevant worksheets, and one in which we do without such aids.\n\n# VLOOKUP Across Several Worksheets (1) – One Search Criterion 4\n\nMost people with an average level of ability in Excel are perfectly capable of using VLOOKUP when this operation is performed over a range within a single worksheet.\n\nBut what happens when we wish to extend our search to multiple worksheets, and so return the first match from whichever sheet happens to be the first which contains our search value(s)?\n\nIn this post I will present a solution for such cases in which we have a single criterion to be matched in a given column across multiple worksheets.\n\nIn the next instalment (to follow shortly) I will also look at cases in which we are not matching a single criterion, but several. In this situation by far the simplest method is to use an extra “helper” column in each of the relevant worksheets in which we first perform a concatenation of the fields of interest. By doing this we ensure that it is a relatively straightforward case of extending the solution designed for one criterion to work also with multiple criteria.\n\n# Collating from multiple sheets based on conditions 30\n\nSome of us may be familiar with the standard technique using INDEX, SMALL, etc. which, given a single-column or single-row array, we can use to return a list of only those values which satisfy one or more criteria of our choosing.\n\nIn a previous post (see here) I outlined a method which, given a range consisting of more than one column, returned a single column consisting of all non-blank entries from that range. It can easily be verified that the single condition within this formula (i.e. that the entry be non-blank) can be extended to multiple criteria and so, effectively, we now have at our disposable the means with which to generate single-column lists from both one- and two-dimensional arrays.\n\nBut can we go one further yet again? “Three-dimensional” is the collective term often applied to those formulas in Excel which are capable of operating over not just single columns or rows, nor yet ranges consisting of multiple columns or rows (two-dimensional), but which also function effectively over multiple worksheets.\n\n# COUNTIFS: Multiple “OR” criteria for one or two criteria_Ranges 111\n\nIn this post I would like to clear up what appears to me to be a rather widespread misunderstanding of how COUNTIFS/SUMIFS operate, in particular when we pass arrays consisting of more than one element as the Criteria to one or even two of the Criteria_Ranges.\n\nThis latter technique is used when the criteria in question are to be considered as “OR” criteria, which is not to be confused with cases where we wish the criteria passed to be calculated rather as “AND” criteria.\n\nFor example, given the following data:\n\n# Advanced Formula Challenge #5: Results and Discussion 5\n\nLast week I set readers the challenge which can be found here.\n\nThis is a reasonably complex problem, and certainly so if we want to present a solution which is relatively concise. However, despite its complexity (and arguably lack of practical use), the solution demonstrates some important techniques for working with strings, and so is not without merit.\n\nThe required set-up is as follows:"
]
| [
null
]
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https://the-algorithms.com/algorithm/rail-fence-cipher?lang=php | [
"#### Rail Fence Cipher\n\nA\np\n```\"\"\" https://en.wikipedia.org/wiki/Rail_fence_cipher \"\"\"\n\ndef encrypt(input_string: str, key: int) -> str:\n\"\"\"\nShuffles the character of a string by placing each of them\nin a grid (the height is dependent on the key) in a zigzag\nformation and reading it left to right.\n\n>>> encrypt(\"Hello World\", 4)\n'HWe olordll'\n\n>>> encrypt(\"This is a message\", 0)\nTraceback (most recent call last):\n...\nValueError: Height of grid can't be 0 or negative\n\n>>> encrypt(b\"This is a byte string\", 5)\nTraceback (most recent call last):\n...\nTypeError: sequence item 0: expected str instance, int found\n\"\"\"\ntemp_grid: list[list[str]] = [[] for _ in range(key)]\nlowest = key - 1\n\nif key <= 0:\nraise ValueError(\"Height of grid can't be 0 or negative\")\nif key == 1 or len(input_string) <= key:\nreturn input_string\n\nfor position, character in enumerate(input_string):\nnum = position % (lowest * 2) # puts it in bounds\nnum = min(num, lowest * 2 - num) # creates zigzag pattern\ntemp_grid[num].append(character)\ngrid = [\"\".join(row) for row in temp_grid]\noutput_string = \"\".join(grid)\n\nreturn output_string\n\ndef decrypt(input_string: str, key: int) -> str:\n\"\"\"\nGenerates a template based on the key and fills it in with\nthe characters of the input string and then reading it in\na zigzag formation.\n\n>>> decrypt(\"HWe olordll\", 4)\n'Hello World'\n\n>>> decrypt(\"This is a message\", -10)\nTraceback (most recent call last):\n...\nValueError: Height of grid can't be 0 or negative\n\n>>> decrypt(\"My key is very big\", 100)\n'My key is very big'\n\"\"\"\ngrid = []\nlowest = key - 1\n\nif key <= 0:\nraise ValueError(\"Height of grid can't be 0 or negative\")\nif key == 1:\nreturn input_string\n\ntemp_grid: list[list[str]] = [[] for _ in range(key)] # generates template\nfor position in range(len(input_string)):\nnum = position % (lowest * 2) # puts it in bounds\nnum = min(num, lowest * 2 - num) # creates zigzag pattern\ntemp_grid[num].append(\"*\")\n\ncounter = 0\nfor row in temp_grid: # fills in the characters\nsplice = input_string[counter : counter + len(row)]\ngrid.append(list(splice))\ncounter += len(row)\n\noutput_string = \"\" # reads as zigzag\nfor position in range(len(input_string)):\nnum = position % (lowest * 2) # puts it in bounds\nnum = min(num, lowest * 2 - num) # creates zigzag pattern\noutput_string += grid[num]\ngrid[num].pop(0)\nreturn output_string\n\ndef bruteforce(input_string: str) -> dict[int, str]:\n\"\"\"Uses decrypt function by guessing every key\n\n>>> bruteforce(\"HWe olordll\")\n'Hello World'\n\"\"\"\nresults = {}\nfor key_guess in range(1, len(input_string)): # tries every key\nresults[key_guess] = decrypt(input_string, key_guess)\nreturn results\n\nif __name__ == \"__main__\":\nimport doctest\n\ndoctest.testmod()\n```",
null,
"",
null,
""
]
| [
null,
"https://the-algorithms.com/logo_t.svg",
null,
"https://the-algorithms.com/powered-by-vercel-t.svg",
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http://mfs.uimech.org/uim2006.1.009,en | [
"",
null,
"",
null,
"ISSN 2542–0380 Труды Института механики им. Р.Р. Мавлютова Электронный научный журнал | Electronic Scientific Journal Proceedings of the Mavlyutov Institute of Mechanics",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"им. Р.Р. Мавлютова\nУФИЦ РАН",
null,
"",
null,
"Mikhaylenko С.I., Khizbullina S.F. About one efficient pipeline parallel algorithm for solving problems of continuum mechanics Proceedings of the Institute of Mechanics of Ufa Branch of RAS. 4 (2006). 90–102.\n2006. Vol. 4. Issue 1, Pp. 90–102\nURL: http://proc.uimech.org/uim2006.1.009,en\nDOI: 10.21662/uim2006.1.009\nAbout one efficient pipeline parallel algorithm for solving problems of continuum mechanics\nMikhaylenko С.I., Khizbullina S.F.\nInstitute of Mechanics, Ufa\n\nAbstract\n\nThe paper proposes a technique for pipelining a computational process in the spatial decomposition of the computational domain for constructing efficient parallel algorithms for numerical solution of hydrodynamic problems oriented to cluster computing systems. Methods for achieving high efficiency of a parallel application based on a finite-difference explicit or semi-explicit numerical scheme are shown. An expression is written to determine the minimum size of the calculated area, at which the efficiency of the parallel program approaches unity.\n\nKeywords\n\nparallel algorithm,\ncluster computing system,\nspatial decomposition,\nmathematical modeling,\nhydrodynamics"
]
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https://medialive.com.pl/4689/22-04-2016/equation-for-ball-mill.html | [
"# equation for ball mill\n\n## equation for ball mill",
null,
"### Bond Work Index - an overview | ScienceDirect Topics\n\nThe ball mill work index laboratory test is conducted by grinding an ore sample prepared to passing 336 mm (6 mesh) to product size in the range of 45-150 µm (325-100 mesh), thus determining the ball mill work index (Wi B or BWi) The work index calculations across a narrow size range are conducted using the appropriate laboratory work ,\n\nGet Price",
null,
"### Making a Ball Mill - YouTube\n\nMar 31, 2018· The milling of the materials is a very useful procedure in various domains in life as well as in the constructive domain Click HERE to subscribe to Make it,\n\nGet Price",
null,
"### equation formula critical speed of the ball mill\n\nball mill critical speed equation dent all eu The formula to calculate critical speed is given below n c sqtdd n c critical speed of the mill d mill diameter specified in meters d diameter of the ball in practice ball mills are driven at a speed of of the critical speed, the ,\n\nGet Price",
null,
"### formula for critical speed of ball mill\n\nAxial transport in dry ball mills - ScienceDirect Ball mills are used for grinding of rocks, cement clinker and lizenithne from 10 to , one-dimensional diffusion equation with the diffusion coefficient decreasing with , the mill speed was 75% of the critical speed required to centrifuge particl\n\nGet Price",
null,
"### formula of calculating rpm of ball mill\n\nformula of calculating rpm of ball mill See sample quotcalculating drive rpmsquot formula below the left side of the data chart shows a cut away of a standard rework sheet after the tooling is reworked on a typical standard mill configuration, write down the reworked throat diameter of the first breakdown pass, first fin pass, and first sizing pass\n\nGet Price",
null,
"### Ball Mill Critical Speed\n\nA Ball Mill Critical Speed (actually ball, rod, AG or SAG) is the speed at which the centrifugal forces equal gravitational forces at the mill shell’s inside surface and no balls will fall from its position onto the shell The imagery below helps explain what goes on inside a mill as speed vari Use our online formula The mill speed is typically defined as the percent of the Theoretical ,\n\nGet Price",
null,
"### Calculate and Select Ball Mill Ball Size for Optimum Grinding\n\nIn Grinding, selecting (calculate) the correct or optimum ball size that allows for the best and optimum/ideal or target grind size to be achieved by your ball mill is an important thing for a Mineral Processing Engineer AKA Metallurgist to do Often, the ball used in ball mills is oversize “just in case” Well, this safety factor can cost you much in recovery and/or mill liner wear and ,\n\nGet Price",
null,
"### Calculation of energy required for grinding in a ball mill ,\n\nThe grinding-product size, P, in a Bond ball mill, which is given by the aperture size which passes 80% of the grinding product as a function of the aperture size of the test screen P k, can be expressed by the formula P= P k K 2\n\nGet Price",
null,
"### Grinding Mills - 911Metallurgist\n\nFig 4—Graphical representation of ball charge within a grinding mill and the factors influencing horsepower calculations, as given in the horsepower equation The equation shown in the graph below will be helpful in determining the percent charge in specific mills, knowing Q (see Fig 4)\n\nGet Price",
null,
"### Basic Math For Ballnose Tools | Modern Machine Shop\n\nNov 01, 2003· The larger the stepover, the larger the height of the scallops between passes will be To hold this height below a certain limit, find the right stepover distance using the formula below Here h is the height of these peaks—in inches, assuming the diameter is in inches—and D is the full diameter of the ball:\n\nGet Price",
null,
"### Ball Mill - SlideShare\n\nNov 18, 2008· We discuss the types of ball mill, the basic principles of the ball mill, how it works, the details of design including equations for optimum dimensions in all cases, some manufacturers for the ball mill, and estimation of the cost the ball mill 6 1 Introduction Ball mill is an efficient tool for grinding many materials into fine powder\n\nGet Price",
null,
"### Milling Speed and Feed Calculator - CustomPartNet\n\nMilling operations remove material by feeding a workpiece into a rotating cutting tool with sharp teeth, such as an end mill or face mill Calculations use the desired tool diameter, number of teeth, cutting speed, and cutting feed, which should be chosen based on the specific cutting conditions, including the workpiece material and tool material\n\nGet Price",
null,
"### (PDF) Effect of circulating load and classification ,\n\nThe ball mill is the most common ore grinding technology today, and probably more than 50% of the total world energy consumption for ore grinding is consumed in ball mills\n\nGet Price",
null,
"### Make Black Powder Quick and Easy — Skylighter, Inc\n\nThe one, critical machine which makes this method of manufacturing black powder possible is the ball mill The Quick & Easy Black Powder Ball Mill project presents an overview of a good, small, economical ball mill and provides important directions and safety information for its correct use The directions below are tailored to the use of the ,\n\nGet Price",
null,
"### Ball Nose Finishing Mills Speed & Feed Calculator - DAPRA\n\nBall Nose Finishing Mills Speed & Feed Calculator Instructions: Fill in the blocks shaded in blue with your application information The calculator will automatically provide the necessary speed and feed in the green fields For assistance setting up your milling program, ,\n\nGet Price",
null,
"### Ball Mills - Mine EngineerCom\n\nThis formula calculates the critical speed of any ball mill Most ball mills operate most efficiently between 65% and 75% of their critical speed Photo of a 10 Ft diameter by 32 Ft long ball mill in a Cement Plant Photo of a series of ball mills in a Copper Plant, grinding the ore for flotation Image of cut away ball mill, showing material ,\n\nGet Price",
null,
"### Bond formula for the grinding balls size calculation\n\nOct 19, 2017· The grinding balls diameter determined by the Bond formula has a recommendatory character and serves as a starting point for calculating the necessary proportion grinding media feeding a new mill More precisely adjust the ball load in the mill can only by industrial test performing\n\nGet Price",
null,
"### Modelling SAG milling power and specific energy ,\n\nJan 01, 2015· The lever arm is a function of the rotational speed (N/Nc) and the power is a function of the rotational speed (N/Nc) as indicated before in the basic equation Of course, ball charge plus ore retained inside the mill plus water retained inside the mill ,\n\nGet Price",
null,
"### TECHNICAL NOTES 8 GRINDING R P King\n\nLG A mill power equation for SAG mills Minerals and Metallurgical Processing Feb 1990 pp57-62 Gross power No load power Net power drawn by the charge (813) The net power is calculated from Net power KD25L e! c/ Watts (814) In equation 814, D is the diameter inside the mill liners and Le is the effective length of the mill including the ,\n\nGet Price",
null,
"### THE OPTIMAL BALL DIAMETER IN A MILL\n\nThe ball impact energy on grain is proportional to the ball diameter to the third power: 3 E K 1 d b (3) The coefficient of proportionality K 1 directly depends on the mill diameter, ball mill loading, milling rate and the type of grinding (wet/dry) None of the characteristics of the material being ground have any influence on K 1\n\nGet Price",
null,
"### Ball mill - Wikipedia\n\nA ball mill, a type of grinder, is a cylindrical device used in grinding (or mixing) materials like ores, chemicals, ceramic raw materials and paintsBall mills rotate around a horizontal axis, partially filled with the material to be ground plus the grinding medium Different materials are used as media, including ceramic balls, flint pebbles, and stainless steel balls\n\nGet Price",
null,
"### (PDF) A quick method for bond work index approximate value ,\n\nBond ball mill and grind for an arbitrary number of mill revolutions (N 1 = 50, 100 or 150 revolutions) After grinding, screen the sample on the comparative sieve and determine\n\nGet Price",
null,
"### COMMON EQUATIONS FOR OPTIMAL PERFORMANCE\n\nBall Nose Effective , Feed rate is the speed of the end mill’s movement correspondent to the workpiece The feed rate is measured in inches per minute (IPM) , COMMON EQUATIONS FOR OPTIMAL PERFORMANCE Too high of a speed or too light of a feed leads to reduction in tool life\n\nGet Price",
null,
"### Ball mill - SlideShare\n\nApr 24, 2015· Ball mill consist of a hollow cylindrical shell rotating about its axis Axis of the shell horizontal or at small angle to the horizontal It is partially filled with balls made up of Steel,Stainless steel or rubber Inner surface of the shell is lined with abrasion resistant materials such as Manganese,Steel or rubber Length of the mill is ,\n\nGet Price",
null,
"### Ball Mills - an overview | ScienceDirect Topics\n\nIt is possible to make an approximate calculation of the capacity of a ball mill by means of the equation: N = ( 0104 D 3 L ρ b ap ϕ 0 88 + 01 L n ) 1 η 1 η 2 where ρ bap is the apparent density of the balls; l is the degree of filling of the mill by balls; n is revolutions per minute; η ,\n\nGet Price",
null,
"Nov 18, 2014· This must be taken into account when energy consumption is calculated according to the Bond formula unit energy cost extra increased as circulating load decreased from 250% However, unit energy cost decreased as circulating most suitable circulating load for amorphous silica used experiments is 250% because mill capacity is decreased\n\nGet Price",
null,
"### Optimization of mill performance by using\n\nmill absorbed power and ball filling degree As indicated on the graph, a small variation in power could be the result of a significant variation of balls filling degree As the ball wear rate depends directly on the surface of the media charge, a small variation in power will lead to an\n\nGet Price",
null,
"### Making a Ball Mill - YouTube\n\nThe milling of the materials is a very useful procedure in various domains in life as well as in the constructive domain Click HERE to subscribe to Make it,\n\nGet Price",
null,
"### Calculate Top Ball Size of Grinding Media - Equation & Method\n\nAn online calculator lets you calculate Top Ball Size of Grinding Media for your mill Use this Equation & Method to properly grind your ore\n\nGet Price"
]
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https://chemistry.stackexchange.com/questions/16748/converting-from-grams-to-molecules | [
"# Converting from Grams to Molecules\n\nI was doing some problems from the textbook on converting between units using dimensional analysis and I came across this problem.\n\nA vat of Hydrogen Peroxide ($\\ce{H2O2}$) contains 455 grams of oxygen atoms. How many molecules of ($\\ce{H2O2}$) are in the vat?\n\nSorry if I didn`t format the question correctly, I am a still a newbie. But how would I set this up? I would not like the answer to this question but I would like some input on how would I get how many molecules of hydrogen peroxide from grams of oxygen atoms? Any help would be much appreciated. Thanks in advance!\n\nA vat of Hydrogen Peroxide ($\\ce{H2O2}$) contains 455 grams of oxygen atoms. How many molecules of ($\\ce{H2O2}$) are in the vat?\n1. How many moles of oxygen atoms are in 455 gms of oxygen atoms? Let's call the answer $x$.\n2. How many moles of $\\ce{O2}$ are in $x$? Let's call the answer $y$.\n3. How many moles of $\\ce{H2O2}$ could you make from $y$ moles of $\\ce{O2}$? Let's call the answer $z$.\n4. 1 mole of any substance contains Avogadro's number of molecules. How many molecules are in $z$ moles?"
]
| [
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https://blog.csdn.net/weixin_32364911/article/details/117156387 | [
"# c语言中将结构体写入文件,C语言中将结构体写入文件\n\nfwrite(&some_struct,sizeof somestruct,1,fp);",
null,
"",
null,
"#include\n\n#include\n\ntypedef\n\nstruct {\n\nchar c;\n\nint h;\n\nshort n;\n\nlong m;\n\nfloat f;\n\ndouble d1;\n\nchar *s;\n\ndouble d2;\n\n}st;\n\nint main(\n\nvoid)\n\n{\n\nFILE *fp;\n\nst sa,sb;\n\nchar *str=\n\n\"\n\nabcdefg\n\n\";\n\nsa.c=\n\n'\n\nK\n\n';\n\nsa.h=-\n\n3;\n\nsa.n=\n\n20;\n\nsa.m=\n\n100000000;\n\nsa.f=\n\n33.32f;\n\nsa.d1=\n\n78.572;\n\nsa.d2=\n\n33.637;\n\nsa.s=str;\n\nfp=fopen(\n\n\"\n\nst.txt\n\n\",\n\n\"\n\nw+\n\n\");\n\nif(!fp)\n\n{\n\nprintf(\n\n\"\n\nerrror!\\n\n\n\");\n\nexit(-\n\n1);\n\n}\n\nprintf(\n\n\"\n\nsa:c=%c,h=%d,n=%d,m=%d,f=%f,d1=%f,s=%s,d2=%f\\n\n\n\",sa.c,sa.h,sa.n,sa.m,sa.f,sa.d1,sa.s,sa.d2);\n\nprintf(\n\n\"\n\nsizeof(sa)=%d:&c=%x,&h=%x,&n=%x,&m=%x,&f=%x,&d1=%x,&s=%x,&d2=%x\\n\n\n\",\n\nsizeof(sa),&sa.c,&sa.h,&sa.n,&sa.m,&sa.f,&sa.d1,&sa.s,&sa.d2);\n\nfwrite(&sa,\n\nsizeof(sa),\n\n1,fp);\n\nrewind(fp);\n\nsizeof(sb),\n\n1,fp);\n\nprintf(\n\n\"\n\nsb:c=%c,h=%d,n=%d,m=%d,f=%f,d1=%f,s=%s,d2=%f\\n\n\n\",sb.c,sb.h,sb.n,sb.m,sb.f,sb.d1,sb.s,sb.d2);\n\nfclose(fp);\n\nreturn\n\n0;\n\n}",
null,
"",
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"",
null,
"#include\n\n#include\n\ntypedef\n\nstruct {\n\nchar c;\n\nint h;\n\nshort n;\n\nlong m;\n\nfloat f;\n\ndouble d1;\n\nchar *s;\n\ndouble d2;\n\n}st;\n\nint main(\n\nvoid)\n\n{\n\nFILE *fp;\n\nst sb;\n\nfp=fopen(\n\n\"\n\nst.txt\n\n\",\n\n\"\n\nr\n\n\");\n\nif(!fp)\n\n{\n\nprintf(\n\n\"\n\nerrror!\\n\n\n\");\n\nexit(-\n\n1);\n\n}\n\nsizeof(sb),\n\n1,fp);\n\nprintf(\n\n\"\n\nsb:c=%c,h=%d,n=%d,m=%d,f=%f,d1=%f,s=%s,d2=%f\\n\n\n\",sb.c,sb.h,sb.n,sb.m,sb.f,sb.d1,sb.s,sb.d2);\n\nprintf(\n\n\"\n\nsizeof(sb)=%d:&c=%x,&h=%x,&n=%x,&m=%x,&f=%x,&d1=%x,&s=%x,&d2=%x\\n\n\n\",\n\nsizeof(sb),&sb.c,&sb.h,&sb.n,&sb.m,&sb.f,&sb.d1,&sb.s,&sb.d2);\n\nfclose(fp);\n\nreturn\n\n0;\n\n}",
null,
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null,
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null,
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null,
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null,
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https://nrich.maths.org/7670/clue | [
"#### You may also like",
null,
"### Real(ly) Numbers\n\nIf x, y and z are real numbers such that: x + y + z = 5 and xy + yz + zx = 3. What is the largest value that any of the numbers can have?",
null,
"In y = ax +b when are a, -b/a, b in arithmetic progression. The polynomial y = ax^2 + bx + c has roots r1 and r2. Can a, r1, b, r2 and c be in arithmetic progression?",
null,
"### Polynomial Relations\n\nGiven any two polynomials in a single variable it is always possible to eliminate the variable and obtain a formula showing the relationship between the two polynomials. Try this one.\n\n# Interpolating Polynomials\n\n##### Age 16 to 18 Challenge Level:\n\nTo find four points that a quadratic couldn't possibly fit, remember that quadratics only have one turning point.\n\nTo find the quadratic that fits three points, make sure you understand how you can add and subtract graphs, and what happens to the result. Don't try to fit all three points at once – fit two and then “fix” your line to fit the third.\n\nUniqueness: The Factor Theorem states that if $p$ is a polynomial and $p(a) = 0$, then there is a polynomial $q$ such that $p(x) = (x-a)q(x)$. What does this mean about the degrees of $p$ and $q$?\n\nFinally, to prove two polynomials are equal, try proving their difference is zero."
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null,
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https://discourse.julialang.org/t/too-many-allocations/55478 | [
"# Too many allocations?\n\nI’m thinking about optimizing performance and my mind initially goes to reducing the number of allocations. Below is a function with a loop that executes ~ 1,000,000 times.\nWhen I “time” the routine, I find that it executes ~9 million reallocations. I just have no idea whether this is a lot and I should try to reduce it or whether this is about right.\n\n``````function RK2(pend::pendulum;poincare::Bool= false)\nr = [pend.thetaInitial,pend.omegaInitial]\nt = 0.0\nN = round(Integer,pend.tMax/pend.dt)\ntheta = zeros(Float64,N+1,1)\nomega = zeros(Float64,N+1,1)\nsaveTheta = zeros(Float64,M,1)\nsaveOmega = zeros(Float64,M,1)\nomega = zeros(Float64,N+1,1)\n\ntime = zeros(Float64,N+1,1)\ntheta = pend.thetaInitial\nomega = pend.omegaInitial\ntime = 0.0\ncounter = 2\nindex = 1\nfor i=1:N\nk1 = pend.dt .* getDerivs(pend,r,t)\n# test = getDerivs(pend,r,t + 1/2 * pend.dt)\nk2 = pend.dt .* getDerivs(pend,r .+ 0.5 .* k1,t + 1/2 * pend.dt)\n\nr += k2\nif r > π && pend.resets\nr -= 2π\nelseif r < -π && pend.resets\nr += 2π\nend\n\nif inPhase(pend,t)\nsaveTheta[index] = r\nsaveOmega[index] = r\nindex += 1\nend\nt+= pend.dt\ntheta[counter] = r\nomega[counter] = r\ntime[counter] = t\ncounter += 1\nend\nif poincare\nsaveTheta,saveOmega\nelse\ntime,theta,omega\nend\nend\n``````\n\nYou need to give runnable code, otherwise it is impossible to say.\n\n1 Like\n``````mutable struct pendulum\nthetaInitial::Float64\nomegaInitial::Float64\ndt::Float64\ntMax::Float64\nl::Float64\nq::Float64\nFd::Float64\nresets::Bool\noffset::Float64\ng::Float64\n\nfunction pendulum(thetaInitial::Float64, omegaInitial::Float64, dt::Float64,tMax::Float64,l::Float64,q::Float64,Fd::Float64,omegaD::Float64;resets::Bool=true,offset::Float64 = 0.0,g::Float64=9.8)\n#Only here to handle the defaults\n\nend\nend\n\nfunction getDerivs(pend::pendulum, vec::Array{Float64,1},t::Float64)\n\ndOmega = -pend.g/pend.l * sin(vec) - pend.q * vec + pend.Fd * sin(pend.omegaD * t)\n\nreturn [vec,dOmega]\n\nend\n\nfunction inPhase(pend::pendulum,t::Float64)\nif abs(pend.omegaD * (t - pend.offset)/( 2π ) - round(Integer,pend.omegaD * (t - pend.offset)/(2π) ) ) < pend.dt * pend.omegaD/(4π)\nreturn true\nelse\nreturn false\nend\n\nend\nfunction RK4(pend::pendulum;poincare::Bool= false)\nr = [pend.thetaInitial,pend.omegaInitial]\nt = 0.0\nN = round(Integer,pend.tMax/pend.dt)\ntheta = zeros(Float64,N+1,1)\nomega = zeros(Float64,N+1,1)\nsaveTheta = zeros(Float64,M,1)\nsaveOmega = zeros(Float64,M,1)\nomega = zeros(Float64,N+1,1)\n\ntime = zeros(Float64,N+1,1)\ntheta = pend.thetaInitial\nomega = pend.omegaInitial\ntime = 0.0\ncounter = 2\nindex = 1\nfor i=1:N\nk1 = pend.dt .* getDerivs(pend,r,t)\n# test = getDerivs(pend,r,t + 1/2 * pend.dt)\nk2 = pend.dt .* getDerivs(pend,r .+ 0.5 .* k1,t + 1/2 * pend.dt)\n\nr += k2\nif r > π && pend.resets\nr -= 2π\nelseif r < -π && pend.resets\nr += 2π\nend\n\nif inPhase(pend,t)\nsaveTheta[index] = r\nsaveOmega[index] = r\nindex += 1\nend\nt+= pend.dt\ntheta[counter] = r\nomega[counter] = r\ntime[counter] = t\ncounter += 1\nend\nif poincare\nsaveTheta,saveOmega\nelse\ntime,theta,omega\nend\nend\n\nusing Profile\nusing Plots\nplotly()\n\nthetaInitial = 0.2\nomegaInitial = 0.0\ndt = 0.001\ntMax = 16000.0\nl = 9.8\nq = 0.5\nFd = 1.2\nresets = false\noffset = 0.0\n@time saveTheta,saveOmega = RK4(myPendulum,poincare=true)\n\nscatter(saveTheta,saveOmega,ms=1)\n#timeTwo,thetaTwo = RK4(mySecondPendulum)\n#plot!(timeOne,thetaTwo)\n\n#plot!(timeOne,thetaTwo,yaxis=:log,ylim = (1e-6,10))\n#Profile.print()\n``````\n\nThis is my first attempt at a Julia code. If you see anything that is bad practice or have any suggestions, I welcome them.\n\nThanks\n\nThis stands out to me as an easy thing to improve. You’re returning two values (which is fine), but you’re returning them by constructing an `Array` to hold those two elements. Constructing a new `Array` is rather expensive (if you do it a million times, at least), and that’s one of your sources of frequent allocations.\n\nInstead, you can return a `Tuple`, which is basically free to construct. That would look like:\n\n``````return vec, dOmega\n``````\n\nor equivalently, `(vec, dOmega)` because the parentheses are optional here\n\n5 Likes\n\nThank you. Great Idea. Can I use a tuple to do vector math upon returning from the function or will I have to do it component by component.\n\nI should mention that this code is already much faster than it’s python counterpart. I’m just looking to build intuition for optimizing Julia code. Thanks in advance.\n\nYou can probably preallocate and re-use `k1` and `k2`. `.=` assigns in place, `=` make a new array which would allocate. Also `omegaD = Float64(2/3)` the `Float64` is uneccesary as `2/3` already returns a `Float64`.\n\n3 Likes\n\nGood Idea. I had that thought but I neglected the `.=` so I didn’t get an improvement.\n\nYou can broadcast over a tuple (e.g. you can do `x .+ 1` if `x` is a tuple), but if you want to do anything more vector-like, then check out GitHub - JuliaArrays/StaticArrays.jl: Statically sized arrays for Julia . An SVector from StaticArrays is as cheap as a tuple (it actually is a tuple under the hood) but acts like a proper vector.\n\n3 Likes\n\nYou guys are all awesome. I preallocated `k1` and `k2` and returned a tuple instead of an array. This reduced the number of allocations from 9 M to about 1 M and sped the code up by a factor of ~5. The code was already crazy fast and now it’s ridiculous fast. Thanks a ton for your quick and helpful replies.\n\n4 Likes\n\nYou could use Parameters.jl to initialize the structure instead of the inner constructor (which, in that case, can be also an outer constructor, that is more common). And the name of the struct is recommended to be in upper case (`struct Pendulum`):\n\n``````julia> using Parameters\n\njulia> @with_kw mutable struct A\nx :: Float64\nb :: Bool = false\nend\nA\n\njulia> a = A(x=1.0)\nA\nx: Float64 1.0\nb: Bool false\n\n``````\n\nFrom your code I could not see why your `Pendulum` has to be mutable. Immutable structus will be faster, in general.\n\nAlso: `Array{Float64,1}` is the same as `Vector{Float64}` (but you probably want just `AbstractVector` in the function signature. That will allow your function to receive `views` and vectors of other types of numbers (like `Float32`).\n\nThis line probably allocates:\n\nshould better be `r .+= k2` (with the dot).\n\nAlso, are you sure you want `theta = zeros(Float64,N+1,1)` instead of `zeros(N+1)`? which is a vector, not a two-dimensional array in which the second dimension has only one element?\n\n2 Likes\n\nGreat suggestions. Thank you!. I’m not familiar with Parameters.jl so I’ll read up on it. Inner constructors are not typical? Is it just a question of style or are there efficiency reasons?\n\nGenerally they are used when you need to impose constraints for the parameters, and/or if the constructor is self-referential (in which case an outer constructor would loop until segfault). This is very well written here:\n\nhttps://docs.julialang.org/en/v1/manual/constructors/\n\n2 Likes\n\nWhat mental routine do you go through for determining how many allocations is appropriate and expected. (i.e. When to stop optimizing and just let it go.) In this case, I’m executing a loop 1 M times and I’m getting ~ 1 M reallocations. Seems reasonable???\n\nIf you do these changes, you reduce even further the allocations:\n\n`````` rtmp = zeros(2) # preallocate a temporary vector here\nfor i=1:N\nk1 = pend.dt .* getDerivs(pend,r,t)\nrtmp .= r .+ 0.5 .* k1 # compute it here\nk2 = pend.dt .* getDerivs(pend,rtmp,t + 1/2 * pend.dt)\nr .+= k2 # add the dot\n\n``````\n``````julia> @time saveTheta,saveOmega = RK4(myPendulum,poincare=true)\n0.977168 seconds (19 allocations: 488.308 MiB, 4.39% gc time)\n\n``````\n\n(now the loop does not allocate anything).\n\nSome tips here on how to find the allocations: Disabling allocations\n\nMy impression here is: loops that perform critical tasks should only allocate if one is clearly aware of the reason of the allocation. Otherwise it is a good idea to search where the allocation is and try to solve it. Of course, if the execution time is still a problem at all.\n\n3 Likes\n\nIt can be hard to have such a mental model for allocations if a function is complicated and does many things.\n\nMy strategy is to split code into functions (where it makes sense) and profile the program to see where time is spent. Then, I can focus on one function at a time and manually look for what operations might be allocating or generally inefficient. Most of the time, those allocations can be minimized or eliminated, e.g. by passing in preallocated arrays and using in-place operations.\n\n1 Like\n\nThis seems like a classic case where you should be using StaticArrays: a small array whose size (2) is a compile-time constant. This would also eliminate heap allocations, and would probably be faster.\n\nIn general, my feeling is that if your innermost computational loop has any allocations proportional to the number of loop iterations, you are doing something “wrong”.\n\n8 Likes\n\nThe above probably doesn’t hurt performance much, but the pattern is quite strange:\n\n``````function compare(a, b)\nif a > b\nreturn true\nelse\nreturn false\nend\nend\n``````\n\ncan be replaced with\n\n``````function compare(a, b)\nreturn a > b\nend\n``````\n\nbecause `a > b` already is a boolean value. You’re basically writing “if x is true, return true, else return false”, instead of just returning “x” in the first place.\n\nAnother style thing: `round(Integer, x)` should probably be `round(Int, x)`. It doesn’t seem to hurt, it’s just something I’ve never seen before, as `Integer` is an abstract type.\n\n3 Likes\n\nShouldn’t this return an error (as confusing as it could be for a new user)?\n\nWhy? I think it makes sense as it is:\n\n``````julia> f(x) = typeof(round(Integer, x))\nf (generic function with 1 method)\n\njulia> f(1.1)\nInt64\n\njulia> f(big(1.1))\nBigInt\n\n``````\n2 Likes"
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null
]
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https://excel.officetuts.net/en/training/introduction-to-number-formatting | [
"# Introduction to number formatting\n\nIf a cell is not formatted to a particular type, Excel will read that value and try to guess what format should it be. For example, if you enter the value 12 or 2.4, Excel will treat this value as a decimal number. If you enter 4 1/5 it will be treated as a simple fraction.\n\n### CAUTION\n\nExcel treats entered value as a number, only if it doesn’t contain any additional characters. Otherwise, Excel will treat this value as text. For example, 5A will be treated as text, not as number 5.\n\n## Number formats\n\nIn the Format Cells Ctrl + Shift + F window, in the Number tab, you can change data formatting by selecting one of the available predefined formats.\n\n### General\n\nWhen you create a new worksheet, each cell has a default cell format- General. All cells from a newly created worksheet will use this data type.\n\nThese are a few rules that characterize this type:\n\n1. Excel removes leading zeroes. Number 03.00 will be stored as 3. One exception to this rule is when the value is between –1 and 1 (e.g. .45 will be converted to 0.45).\n2. If the number is a decimal fraction, then Excel displays it in a way to fit it in the cell. If the fractional part does not fit in the cell, then it is rounded. In a similar way, Excel deals with integers. When the value is long, Excel displays it in the scientific notation. When the value is even longer it uses ### instead of displaying the number.\n\n### Example 1:\n\nThe numbers in cells B2, C2, D2 are the same as numbers in cells B3, C3, D3.",
null,
"### Number\n\nAs the name suggests, this format is used for formatting numbers. Here, you can specify the number of decimal places so that the fractional part always stays in the same place.\n\nIn this type, you can also use the 1000 separator, which is useful for very large numbers.\n\n### Example 2:\n\nLook at the following example. Cell B2 is formatted without, and cell B3 with the thousands separator.",
null,
"### Currency\n\nThe currency format includes the currency symbol, and the value is displayed with commas.\n\n### Accounting\n\nThe accounting format is very similar to the currency format. The difference is that in the accounting format the currency sign is placed just at the left edge of the cell and not, as in the currency format near the number. Besides, in the accounting format, the space between the value and the right edge is greater than in the currency format.\n\n### Percentage\n\nThe percentage format converts numbers to percentages, for example:\n\n1 to 100%\n\n0.321 to 32.1%\n\n2.12 is 212%\n\nIn the percentage format, you can set the number of decimal places.\n\n### Fraction\n\nWith the fraction format, you can convert a number to a fraction. When you format cells to fractional type you can enter a fraction, such as 1/3. Otherwise, when the cell is of the general type it will be treated not as a fraction, but as a date (January 3, 2014 (current year)). Therefore, in such cases, you should precede this type by using 0 at the beginning (0 1/3).\n\n### Scientific\n\nScientific notation displays the value in an abbreviated form. For example, the number 1234567890 will be displayed as 1.23E+09. You can also set the number of decimal places to increase precision.\n\n### Special\n\nThis is the formatting for data, such as Zip CodePhone Number or Social Security Number."
]
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null,
"data:image/gif;base64,R0lGODlhAQABAAD/ACwAAAAAAQABAAACADs=",
null,
"data:image/gif;base64,R0lGODlhAQABAAD/ACwAAAAAAQABAAACADs=",
null
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8732923,"math_prob":0.9625504,"size":3180,"snap":"2019-35-2019-39","text_gpt3_token_len":742,"char_repetition_ratio":0.15522671,"word_repetition_ratio":0.014010508,"special_character_ratio":0.2408805,"punctuation_ratio":0.13244048,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96701735,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-17T13:02:32Z\",\"WARC-Record-ID\":\"<urn:uuid:17c069c2-dbee-445b-bd06-ad7e9b989140>\",\"Content-Length\":\"42957\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6d590b3a-c250-4e0f-acae-5f7cc455a22b>\",\"WARC-Concurrent-To\":\"<urn:uuid:a6cdff65-a9dd-4e5f-935e-bc65eac179d8>\",\"WARC-IP-Address\":\"104.28.0.88\",\"WARC-Target-URI\":\"https://excel.officetuts.net/en/training/introduction-to-number-formatting\",\"WARC-Payload-Digest\":\"sha1:RDCVKIAIWYJTHPJPCYZMZGNXODPIFKNK\",\"WARC-Block-Digest\":\"sha1:7PEBYBZ4X4VKVQB5O3ZU7QQZQHO4ODWK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514573071.65_warc_CC-MAIN-20190917121048-20190917143048-00238.warc.gz\"}"} |
https://prettycheapjewelry.savingadvice.com/2013/08/19/rich-man-in-my-dreams-was-warren-buffet_104285/ | [
"User Real IP - 34.207.247.69\n```Array\n(\n => Array\n(\n => 182.68.68.92\n)\n\n => Array\n(\n => 101.0.41.201\n)\n\n => Array\n(\n => 43.225.98.123\n)\n\n => Array\n(\n => 2.58.194.139\n)\n\n => Array\n(\n => 46.119.197.104\n)\n\n => Array\n(\n => 45.249.8.93\n)\n\n => Array\n(\n => 103.12.135.72\n)\n\n => Array\n(\n => 157.35.243.216\n)\n\n => Array\n(\n => 209.107.214.176\n)\n\n => Array\n(\n => 5.181.233.166\n)\n\n => Array\n(\n => 106.201.10.100\n)\n\n => Array\n(\n => 36.90.55.39\n)\n\n => Array\n(\n => 119.154.138.47\n)\n\n => Array\n(\n => 51.91.31.157\n)\n\n => Array\n(\n => 182.182.65.216\n)\n\n => Array\n(\n => 157.35.252.63\n)\n\n => Array\n(\n => 14.142.34.163\n)\n\n => Array\n(\n => 178.62.43.135\n)\n\n => Array\n(\n => 43.248.152.148\n)\n\n => Array\n(\n => 222.252.104.114\n)\n\n => Array\n(\n => 209.107.214.168\n)\n\n => Array\n(\n => 103.99.199.250\n)\n\n => Array\n(\n => 178.62.72.160\n)\n\n => Array\n(\n => 27.6.1.170\n)\n\n => Array\n(\n => 182.69.249.219\n)\n\n => Array\n(\n => 110.93.228.86\n)\n\n => Array\n(\n => 72.255.1.98\n)\n\n => Array\n(\n => 182.73.111.98\n)\n\n => Array\n(\n => 45.116.117.11\n)\n\n => Array\n(\n => 122.15.78.189\n)\n\n => Array\n(\n => 14.167.188.234\n)\n\n => Array\n(\n => 223.190.4.202\n)\n\n => Array\n(\n => 202.173.125.19\n)\n\n => Array\n(\n => 103.255.5.32\n)\n\n => Array\n(\n => 39.37.145.103\n)\n\n => Array\n(\n => 140.213.26.249\n)\n\n => Array\n(\n => 45.118.166.85\n)\n\n => Array\n(\n => 102.166.138.255\n)\n\n => Array\n(\n => 77.111.246.234\n)\n\n => Array\n(\n => 45.63.6.196\n)\n\n => Array\n(\n => 103.250.147.115\n)\n\n => Array\n(\n => 223.185.30.99\n)\n\n => Array\n(\n => 103.122.168.108\n)\n\n => Array\n(\n => 123.136.203.21\n)\n\n => Array\n(\n => 171.229.243.63\n)\n\n => Array\n(\n => 153.149.98.149\n)\n\n => Array\n(\n => 223.238.93.15\n)\n\n => Array\n(\n => 178.62.113.166\n)\n\n => Array\n(\n => 101.162.0.153\n)\n\n => Array\n(\n => 121.200.62.114\n)\n\n => Array\n(\n => 14.248.77.252\n)\n\n => Array\n(\n => 95.142.117.29\n)\n\n => Array\n(\n => 150.129.60.107\n)\n\n => Array\n(\n => 94.205.243.22\n)\n\n => Array\n(\n => 115.42.71.143\n)\n\n => Array\n(\n => 117.217.195.59\n)\n\n => Array\n(\n => 182.77.112.56\n)\n\n => Array\n(\n => 182.77.112.108\n)\n\n => Array\n(\n => 41.80.69.10\n)\n\n => Array\n(\n => 117.5.222.121\n)\n\n => Array\n(\n => 103.11.0.38\n)\n\n => Array\n(\n => 202.173.127.140\n)\n\n => Array\n(\n => 49.249.249.50\n)\n\n => Array\n(\n => 116.72.198.211\n)\n\n => Array\n(\n => 223.230.54.53\n)\n\n => Array\n(\n => 102.69.228.74\n)\n\n => Array\n(\n => 39.37.251.89\n)\n\n => Array\n(\n => 39.53.246.141\n)\n\n => Array\n(\n => 39.57.182.72\n)\n\n => Array\n(\n => 209.58.130.210\n)\n\n => Array\n(\n => 104.131.75.86\n)\n\n => Array\n(\n => 106.212.131.255\n)\n\n => Array\n(\n => 106.212.132.127\n)\n\n => Array\n(\n => 223.190.4.60\n)\n\n => Array\n(\n => 103.252.116.252\n)\n\n => Array\n(\n => 103.76.55.182\n)\n\n => Array\n(\n => 45.118.166.70\n)\n\n => Array\n(\n => 103.93.174.215\n)\n\n => Array\n(\n => 5.62.62.142\n)\n\n => Array\n(\n => 182.179.158.156\n)\n\n => Array\n(\n => 39.57.255.12\n)\n\n => Array\n(\n => 39.37.178.37\n)\n\n => Array\n(\n => 182.180.165.211\n)\n\n => Array\n(\n => 119.153.135.17\n)\n\n => Array\n(\n => 72.255.15.244\n)\n\n => Array\n(\n => 139.180.166.181\n)\n\n => Array\n(\n => 70.119.147.111\n)\n\n => Array\n(\n => 106.210.40.83\n)\n\n => Array\n(\n => 14.190.70.91\n)\n\n => Array\n(\n => 202.125.156.82\n)\n\n => Array\n(\n => 115.42.68.38\n)\n\n => Array\n(\n => 102.167.13.108\n)\n\n => Array\n(\n => 117.217.192.130\n)\n\n => Array\n(\n => 205.185.223.156\n)\n\n => Array\n(\n => 171.224.180.29\n)\n\n => Array\n(\n => 45.127.45.68\n)\n\n => Array\n(\n => 195.206.183.232\n)\n\n => Array\n(\n => 49.32.52.115\n)\n\n => Array\n(\n => 49.207.49.223\n)\n\n => Array\n(\n => 45.63.29.61\n)\n\n => Array\n(\n => 103.245.193.214\n)\n\n => Array\n(\n => 39.40.236.69\n)\n\n => Array\n(\n => 62.80.162.111\n)\n\n => Array\n(\n => 45.116.232.56\n)\n\n => Array\n(\n => 45.118.166.91\n)\n\n => Array\n(\n => 180.92.230.234\n)\n\n => Array\n(\n => 157.40.57.160\n)\n\n => Array\n(\n => 110.38.38.130\n)\n\n => Array\n(\n => 72.255.57.183\n)\n\n => Array\n(\n => 182.68.81.85\n)\n\n => Array\n(\n => 39.57.202.122\n)\n\n => Array\n(\n => 119.152.154.36\n)\n\n => Array\n(\n => 5.62.62.141\n)\n\n => Array\n(\n => 119.155.54.232\n)\n\n => Array\n(\n => 39.37.141.22\n)\n\n => Array\n(\n => 183.87.12.225\n)\n\n => Array\n(\n => 107.170.127.117\n)\n\n => Array\n(\n => 125.63.124.49\n)\n\n => Array\n(\n => 39.42.191.3\n)\n\n => Array\n(\n => 116.74.24.72\n)\n\n => Array\n(\n => 46.101.89.227\n)\n\n => Array\n(\n => 202.173.125.247\n)\n\n => Array\n(\n => 39.42.184.254\n)\n\n => Array\n(\n => 115.186.165.132\n)\n\n => Array\n(\n => 39.57.206.126\n)\n\n => Array\n(\n => 103.245.13.145\n)\n\n => Array\n(\n => 202.175.246.43\n)\n\n => Array\n(\n => 192.140.152.150\n)\n\n => Array\n(\n => 202.88.250.103\n)\n\n => Array\n(\n => 103.248.94.207\n)\n\n => Array\n(\n => 77.73.66.101\n)\n\n => Array\n(\n => 104.131.66.8\n)\n\n => Array\n(\n => 113.186.161.97\n)\n\n => Array\n(\n => 222.254.5.7\n)\n\n => Array\n(\n => 223.233.67.247\n)\n\n => Array\n(\n => 171.249.116.146\n)\n\n => Array\n(\n => 47.30.209.71\n)\n\n => Array\n(\n => 202.134.13.130\n)\n\n => Array\n(\n => 27.6.135.7\n)\n\n => Array\n(\n => 107.170.186.79\n)\n\n => Array\n(\n => 103.212.89.171\n)\n\n => Array\n(\n => 117.197.9.77\n)\n\n => Array\n(\n => 122.176.206.233\n)\n\n => Array\n(\n => 192.227.253.222\n)\n\n => Array\n(\n => 182.188.224.119\n)\n\n => Array\n(\n => 14.248.70.74\n)\n\n => Array\n(\n => 42.118.219.169\n)\n\n => Array\n(\n => 110.39.146.170\n)\n\n => Array\n(\n => 119.160.66.143\n)\n\n => Array\n(\n => 103.248.95.130\n)\n\n => Array\n(\n => 27.63.152.208\n)\n\n => Array\n(\n => 49.207.114.96\n)\n\n => Array\n(\n => 102.166.23.214\n)\n\n => Array\n(\n => 175.107.254.73\n)\n\n => Array\n(\n => 103.10.227.214\n)\n\n => Array\n(\n => 202.143.115.89\n)\n\n => Array\n(\n => 110.93.227.187\n)\n\n => Array\n(\n => 103.140.31.60\n)\n\n => Array\n(\n => 110.37.231.46\n)\n\n => Array\n(\n => 39.36.99.238\n)\n\n => Array\n(\n => 157.37.140.26\n)\n\n => Array\n(\n => 43.246.202.226\n)\n\n => Array\n(\n => 137.97.8.143\n)\n\n => Array\n(\n => 182.65.52.242\n)\n\n => Array\n(\n => 115.42.69.62\n)\n\n => Array\n(\n => 14.143.254.58\n)\n\n => Array\n(\n => 223.179.143.236\n)\n\n => Array\n(\n => 223.179.143.249\n)\n\n => Array\n(\n => 103.143.7.54\n)\n\n => Array\n(\n => 223.179.139.106\n)\n\n => Array\n(\n => 39.40.219.90\n)\n\n => Array\n(\n => 45.115.141.231\n)\n\n => Array\n(\n => 120.29.100.33\n)\n\n => Array\n(\n => 112.196.132.5\n)\n\n => Array\n(\n => 202.163.123.153\n)\n\n => Array\n(\n => 5.62.58.146\n)\n\n => Array\n(\n => 39.53.216.113\n)\n\n => Array\n(\n => 42.111.160.73\n)\n\n => Array\n(\n => 107.182.231.213\n)\n\n => Array\n(\n => 119.82.94.120\n)\n\n => Array\n(\n => 178.62.34.82\n)\n\n => Array\n(\n => 203.122.6.18\n)\n\n => Array\n(\n => 157.42.38.251\n)\n\n => Array\n(\n => 45.112.68.222\n)\n\n => Array\n(\n => 49.206.212.122\n)\n\n => Array\n(\n => 104.236.70.228\n)\n\n => Array\n(\n => 42.111.34.243\n)\n\n => Array\n(\n => 84.241.19.186\n)\n\n => Array\n(\n => 89.187.180.207\n)\n\n => Array\n(\n => 104.243.212.118\n)\n\n => Array\n(\n => 104.236.55.136\n)\n\n => Array\n(\n => 106.201.16.163\n)\n\n => Array\n(\n => 46.101.40.25\n)\n\n => Array\n(\n => 45.118.166.94\n)\n\n => Array\n(\n => 49.36.128.102\n)\n\n => Array\n(\n => 14.142.193.58\n)\n\n => Array\n(\n => 212.79.124.176\n)\n\n => Array\n(\n => 45.32.191.194\n)\n\n => Array\n(\n => 105.112.107.46\n)\n\n => Array\n(\n => 106.201.14.8\n)\n\n => Array\n(\n => 110.93.240.65\n)\n\n => Array\n(\n => 27.96.95.177\n)\n\n => Array\n(\n => 45.41.134.35\n)\n\n => Array\n(\n => 180.151.13.110\n)\n\n => Array\n(\n => 101.53.242.89\n)\n\n => Array\n(\n => 115.186.3.110\n)\n\n => Array\n(\n => 171.49.185.242\n)\n\n => Array\n(\n => 115.42.70.24\n)\n\n => Array\n(\n => 45.128.188.43\n)\n\n => Array\n(\n => 103.140.129.63\n)\n\n => Array\n(\n => 101.50.113.147\n)\n\n => Array\n(\n => 103.66.73.30\n)\n\n => Array\n(\n => 117.247.193.169\n)\n\n => Array\n(\n => 120.29.100.94\n)\n\n => Array\n(\n => 42.109.154.39\n)\n\n => Array\n(\n => 122.173.155.150\n)\n\n => Array\n(\n => 45.115.104.53\n)\n\n => Array\n(\n => 116.74.29.84\n)\n\n => Array\n(\n => 101.50.125.34\n)\n\n => Array\n(\n => 45.118.166.80\n)\n\n => Array\n(\n => 91.236.184.27\n)\n\n => Array\n(\n => 113.167.185.120\n)\n\n => Array\n(\n => 27.97.66.222\n)\n\n => Array\n(\n => 43.247.41.117\n)\n\n => Array\n(\n => 23.229.16.227\n)\n\n => Array\n(\n => 14.248.79.209\n)\n\n => Array\n(\n => 117.5.194.26\n)\n\n => Array\n(\n => 117.217.205.41\n)\n\n => Array\n(\n => 114.79.169.99\n)\n\n => Array\n(\n => 103.55.60.97\n)\n\n => Array\n(\n => 182.75.89.210\n)\n\n => Array\n(\n => 77.73.66.109\n)\n\n => Array\n(\n => 182.77.126.139\n)\n\n => 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124.253.195.172\n)\n\n => Array\n(\n => 203.142.218.149\n)\n\n => Array\n(\n => 157.43.165.180\n)\n\n => Array\n(\n => 39.40.242.57\n)\n\n => Array\n(\n => 103.92.43.150\n)\n\n => Array\n(\n => 39.42.133.202\n)\n\n => Array\n(\n => 119.160.66.11\n)\n\n => Array\n(\n => 138.68.3.7\n)\n\n => Array\n(\n => 210.56.125.226\n)\n\n => Array\n(\n => 157.50.4.249\n)\n\n => Array\n(\n => 124.253.81.162\n)\n\n => Array\n(\n => 103.240.235.141\n)\n\n => Array\n(\n => 132.154.128.20\n)\n\n => Array\n(\n => 49.156.115.37\n)\n\n => Array\n(\n => 45.133.7.48\n)\n\n => Array\n(\n => 122.161.49.137\n)\n\n => Array\n(\n => 202.47.46.31\n)\n\n => Array\n(\n => 192.140.145.148\n)\n\n => Array\n(\n => 202.14.123.10\n)\n\n => Array\n(\n => 122.161.53.98\n)\n\n => Array\n(\n => 124.253.114.113\n)\n\n => Array\n(\n => 103.227.70.34\n)\n\n => Array\n(\n => 223.228.175.227\n)\n\n => Array\n(\n => 157.39.119.110\n)\n\n => Array\n(\n => 180.188.224.231\n)\n\n => Array\n(\n => 132.154.188.85\n)\n\n => Array\n(\n => 197.210.227.207\n)\n\n => Array\n(\n => 103.217.123.177\n)\n\n => Array\n(\n => 124.253.85.31\n)\n\n => Array\n(\n => 123.201.105.97\n)\n\n => Array\n(\n => 39.57.190.37\n)\n\n => Array\n(\n => 202.63.205.248\n)\n\n => Array\n(\n => 122.161.51.100\n)\n\n => Array\n(\n => 39.37.163.97\n)\n\n => Array\n(\n => 43.231.57.173\n)\n\n => Array\n(\n => 223.225.135.169\n)\n\n => Array\n(\n => 119.160.71.136\n)\n\n => Array\n(\n => 122.165.114.93\n)\n\n => Array\n(\n => 47.11.77.102\n)\n\n => Array\n(\n => 49.149.107.198\n)\n\n => Array\n(\n => 192.111.134.206\n)\n\n => Array\n(\n => 182.64.102.43\n)\n\n => Array\n(\n => 124.253.184.111\n)\n\n => Array\n(\n => 171.237.97.228\n)\n\n => Array\n(\n => 117.237.237.101\n)\n\n => Array\n(\n => 49.36.33.19\n)\n\n => Array\n(\n => 103.31.101.241\n)\n\n => Array\n(\n => 129.0.207.203\n)\n\n => Array\n(\n => 157.39.122.155\n)\n\n => Array\n(\n => 197.210.85.120\n)\n\n => Array\n(\n => 124.253.219.201\n)\n\n => Array\n(\n => 152.57.75.92\n)\n\n => Array\n(\n => 169.149.195.121\n)\n\n => Array\n(\n => 198.16.76.27\n)\n\n => Array\n(\n => 157.43.192.188\n)\n\n => Array\n(\n => 119.155.244.221\n)\n\n => Array\n(\n => 39.51.242.216\n)\n\n => Array\n(\n => 39.57.180.158\n)\n\n => Array\n(\n => 134.202.32.5\n)\n\n => Array\n(\n => 122.176.139.205\n)\n\n => Array\n(\n => 151.243.50.9\n)\n\n => Array\n(\n => 39.52.99.161\n)\n\n => Array\n(\n => 136.144.33.95\n)\n\n => Array\n(\n => 157.37.205.216\n)\n\n => Array\n(\n => 217.138.220.134\n)\n\n => Array\n(\n => 41.140.106.65\n)\n\n => Array\n(\n => 39.37.253.126\n)\n\n => Array\n(\n => 103.243.44.240\n)\n\n => Array\n(\n => 157.46.169.29\n)\n\n => Array\n(\n => 92.119.177.122\n)\n\n => Array\n(\n => 196.240.60.21\n)\n\n => Array\n(\n => 122.161.6.246\n)\n\n => Array\n(\n => 117.202.162.46\n)\n\n => Array\n(\n => 205.164.137.120\n)\n\n => Array\n(\n => 171.237.79.241\n)\n\n => Array\n(\n => 198.16.76.28\n)\n\n => Array\n(\n => 103.100.4.151\n)\n\n => Array\n(\n => 178.239.162.236\n)\n\n => Array\n(\n => 106.197.31.240\n)\n\n => Array\n(\n => 122.168.179.251\n)\n\n => Array\n(\n => 39.37.167.126\n)\n\n => Array\n(\n => 171.48.8.115\n)\n\n => Array\n(\n => 157.44.152.14\n)\n\n => Array\n(\n => 103.77.43.219\n)\n\n => Array\n(\n => 122.161.49.38\n)\n\n => Array\n(\n => 122.161.52.83\n)\n\n => Array\n(\n => 122.173.108.210\n)\n\n => Array\n(\n => 60.254.109.92\n)\n\n => Array\n(\n => 103.57.85.75\n)\n\n => Array\n(\n => 106.0.58.36\n)\n\n => Array\n(\n => 122.161.49.212\n)\n\n => Array\n(\n => 27.255.182.159\n)\n\n => Array\n(\n => 116.75.230.159\n)\n\n => Array\n(\n => 122.173.152.133\n)\n\n => Array\n(\n => 129.0.79.247\n)\n\n => Array\n(\n => 223.228.163.44\n)\n\n => Array\n(\n => 103.168.78.82\n)\n\n => Array\n(\n => 39.59.67.124\n)\n\n => Array\n(\n => 182.69.19.120\n)\n\n => Array\n(\n => 196.202.236.195\n)\n\n => Array\n(\n => 137.59.225.206\n)\n\n => Array\n(\n => 143.110.209.194\n)\n\n => Array\n(\n => 117.201.233.91\n)\n\n => Array\n(\n => 37.120.150.107\n)\n\n => Array\n(\n => 58.65.222.10\n)\n\n => Array\n(\n => 202.47.43.86\n)\n\n => Array\n(\n => 106.206.223.234\n)\n\n => Array\n(\n => 5.195.153.158\n)\n\n => Array\n(\n => 223.227.127.243\n)\n\n => Array\n(\n => 103.165.12.222\n)\n\n => Array\n(\n => 49.36.185.189\n)\n\n => Array\n(\n => 59.96.92.57\n)\n\n => Array\n(\n => 203.194.104.235\n)\n\n => Array\n(\n => 122.177.72.33\n)\n\n => Array\n(\n => 106.213.126.40\n)\n\n => Array\n(\n => 45.127.232.69\n)\n\n => Array\n(\n => 156.146.59.39\n)\n\n => Array\n(\n => 103.21.184.11\n)\n\n => Array\n(\n => 106.212.47.59\n)\n\n => Array\n(\n => 182.179.137.235\n)\n\n => Array\n(\n => 49.36.178.154\n)\n\n => Array\n(\n => 171.48.7.128\n)\n\n => Array\n(\n => 119.160.57.96\n)\n\n => Array\n(\n => 197.210.79.92\n)\n\n => Array\n(\n => 36.255.45.87\n)\n\n => Array\n(\n => 47.31.219.47\n)\n\n => Array\n(\n => 122.161.51.160\n)\n\n => Array\n(\n => 103.217.123.129\n)\n\n => Array\n(\n => 59.153.16.12\n)\n\n => Array\n(\n => 103.92.43.226\n)\n\n => Array\n(\n => 47.31.139.139\n)\n\n => Array\n(\n => 210.2.140.18\n)\n\n => Array\n(\n => 106.210.33.219\n)\n\n => Array\n(\n => 175.107.203.34\n)\n\n => Array\n(\n => 146.196.32.144\n)\n\n => Array\n(\n => 103.12.133.121\n)\n\n => Array\n(\n => 103.59.208.182\n)\n\n => Array\n(\n => 157.37.190.232\n)\n\n => Array\n(\n => 106.195.35.201\n)\n\n => Array\n(\n => 27.122.14.83\n)\n\n => Array\n(\n => 194.193.44.5\n)\n\n => Array\n(\n => 5.62.43.245\n)\n\n => Array\n(\n => 103.53.80.50\n)\n\n => Array\n(\n => 47.29.142.233\n)\n\n => Array\n(\n => 154.6.20.63\n)\n\n => Array\n(\n => 173.245.203.128\n)\n\n => Array\n(\n => 103.77.43.231\n)\n\n => Array\n(\n => 5.107.166.235\n)\n\n => Array\n(\n => 106.212.44.123\n)\n\n => Array\n(\n => 157.41.60.93\n)\n\n => Array\n(\n => 27.58.179.79\n)\n\n => Array\n(\n => 157.37.167.144\n)\n\n => Array\n(\n => 119.160.57.115\n)\n\n => Array\n(\n => 122.161.53.224\n)\n\n => Array\n(\n => 49.36.233.51\n)\n\n => Array\n(\n => 101.0.32.8\n)\n\n => Array\n(\n => 119.160.103.158\n)\n\n => Array\n(\n => 122.177.79.115\n)\n\n => Array\n(\n => 107.181.166.27\n)\n\n => Array\n(\n => 183.6.0.125\n)\n\n => Array\n(\n => 49.36.186.0\n)\n\n => Array\n(\n => 202.181.5.4\n)\n\n => Array\n(\n => 45.118.165.144\n)\n\n => Array\n(\n => 171.96.157.133\n)\n\n => Array\n(\n => 222.252.51.163\n)\n\n => Array\n(\n => 103.81.215.162\n)\n\n => Array\n(\n => 110.225.93.208\n)\n\n => Array\n(\n => 122.161.48.200\n)\n\n => Array\n(\n => 119.63.138.173\n)\n\n => Array\n(\n => 202.83.58.208\n)\n\n => Array\n(\n => 122.161.53.101\n)\n\n => Array\n(\n => 137.97.95.21\n)\n\n => Array\n(\n => 112.204.167.123\n)\n\n => Array\n(\n => 122.180.21.151\n)\n\n => Array\n(\n => 103.120.44.108\n)\n\n => Array\n(\n => 49.37.220.174\n)\n\n => Array\n(\n => 1.55.255.124\n)\n\n => Array\n(\n => 23.227.140.173\n)\n\n => Array\n(\n => 43.248.153.110\n)\n\n => Array\n(\n => 106.214.93.101\n)\n\n => Array\n(\n => 103.83.149.36\n)\n\n => Array\n(\n => 103.217.123.57\n)\n\n => Array\n(\n => 193.9.113.119\n)\n\n => Array\n(\n => 14.182.57.204\n)\n\n => Array\n(\n => 117.201.231.0\n)\n\n => Array\n(\n => 14.99.198.186\n)\n\n => Array\n(\n => 36.255.44.204\n)\n\n => Array\n(\n => 103.160.236.42\n)\n\n => Array\n(\n => 31.202.16.116\n)\n\n => Array\n(\n => 223.239.49.201\n)\n\n => Array\n(\n => 122.161.102.149\n)\n\n => Array\n(\n => 117.196.123.184\n)\n\n => Array\n(\n => 49.205.112.105\n)\n\n => Array\n(\n => 103.244.176.201\n)\n\n => Array\n(\n => 95.216.15.219\n)\n\n => Array\n(\n => 103.107.196.174\n)\n\n => Array\n(\n => 203.190.34.65\n)\n\n => Array\n(\n => 23.227.140.182\n)\n\n => Array\n(\n => 171.79.74.74\n)\n\n => Array\n(\n => 106.206.223.244\n)\n\n => Array\n(\n => 180.151.28.140\n)\n\n => Array\n(\n => 165.225.124.114\n)\n\n => Array\n(\n => 106.206.223.252\n)\n\n => Array\n(\n => 39.62.23.38\n)\n\n => Array\n(\n => 112.211.252.33\n)\n\n => Array\n(\n => 146.70.66.242\n)\n\n => Array\n(\n => 222.252.51.38\n)\n\n => Array\n(\n => 122.162.151.223\n)\n\n => Array\n(\n => 180.178.154.100\n)\n\n => Array\n(\n => 180.94.33.94\n)\n\n => Array\n(\n => 205.164.130.82\n)\n\n => Array\n(\n => 117.196.114.167\n)\n\n => Array\n(\n => 43.224.0.189\n)\n\n => Array\n(\n => 154.6.20.59\n)\n\n => Array\n(\n => 122.161.131.67\n)\n\n => Array\n(\n => 70.68.68.159\n)\n\n => Array\n(\n => 103.125.130.200\n)\n\n => Array\n(\n => 43.242.176.147\n)\n\n => Array\n(\n => 129.0.102.29\n)\n\n => Array\n(\n => 182.64.180.32\n)\n\n => Array\n(\n => 110.93.250.196\n)\n\n => Array\n(\n => 139.135.57.197\n)\n\n => Array\n(\n => 157.33.219.2\n)\n\n => Array\n(\n => 205.253.123.239\n)\n\n => Array\n(\n => 122.177.66.119\n)\n\n => Array\n(\n => 182.64.105.252\n)\n\n => Array\n(\n => 14.97.111.154\n)\n\n => Array\n(\n => 146.196.35.35\n)\n\n => Array\n(\n => 103.167.162.205\n)\n\n => Array\n(\n => 37.111.130.245\n)\n\n => Array\n(\n => 49.228.51.196\n)\n\n => Array\n(\n => 157.39.148.205\n)\n\n => Array\n(\n => 129.0.102.28\n)\n\n => Array\n(\n => 103.82.191.229\n)\n\n => Array\n(\n => 194.104.23.140\n)\n\n => Array\n(\n => 49.205.193.252\n)\n\n => Array\n(\n => 222.252.33.119\n)\n\n => Array\n(\n => 173.255.132.114\n)\n\n => Array\n(\n => 182.64.148.162\n)\n\n => Array\n(\n => 175.176.87.8\n)\n\n => Array\n(\n => 5.62.57.6\n)\n\n => Array\n(\n => 119.160.96.229\n)\n\n => Array\n(\n => 49.205.180.226\n)\n\n => Array\n(\n => 95.142.120.59\n)\n\n => Array\n(\n => 183.82.116.204\n)\n\n => Array\n(\n => 202.89.69.186\n)\n\n => Array\n(\n => 39.48.165.36\n)\n\n => Array\n(\n => 192.140.149.81\n)\n\n => Array\n(\n => 198.16.70.28\n)\n\n => Array\n(\n => 103.25.250.236\n)\n\n => Array\n(\n => 106.76.202.244\n)\n\n => Array\n(\n => 47.8.8.165\n)\n\n => Array\n(\n => 202.5.145.213\n)\n\n => Array\n(\n => 106.212.188.243\n)\n\n => Array\n(\n => 106.215.89.2\n)\n\n => Array\n(\n => 119.82.83.148\n)\n\n => Array\n(\n => 123.24.164.245\n)\n\n => Array\n(\n => 187.67.51.106\n)\n\n => Array\n(\n => 117.196.119.95\n)\n\n => Array\n(\n => 95.142.120.66\n)\n\n => Array\n(\n => 156.146.59.35\n)\n\n => Array\n(\n => 49.205.213.148\n)\n\n => Array\n(\n => 111.223.27.206\n)\n\n => Array\n(\n => 49.205.212.86\n)\n\n => Array\n(\n => 103.77.42.103\n)\n\n => Array\n(\n => 110.227.62.25\n)\n\n => Array\n(\n => 122.179.54.140\n)\n\n => Array\n(\n => 157.39.239.81\n)\n\n => Array\n(\n => 138.128.27.234\n)\n\n => Array\n(\n => 103.244.176.194\n)\n\n => Array\n(\n => 130.105.10.127\n)\n\n => Array\n(\n => 103.116.250.191\n)\n\n => Array\n(\n => 122.180.186.6\n)\n\n => Array\n(\n => 101.53.228.52\n)\n\n => Array\n(\n => 39.57.138.90\n)\n\n => Array\n(\n => 197.156.137.165\n)\n\n => Array\n(\n => 49.37.155.78\n)\n\n => Array\n(\n => 39.59.81.32\n)\n\n => Array\n(\n => 45.127.44.78\n)\n\n => Array\n(\n => 103.58.155.83\n)\n\n => Array\n(\n => 175.107.220.20\n)\n\n => Array\n(\n => 14.255.9.197\n)\n\n => Array\n(\n => 103.55.63.146\n)\n\n => Array\n(\n => 49.205.138.81\n)\n\n => Array\n(\n => 45.35.222.243\n)\n\n => Array\n(\n => 203.190.34.57\n)\n\n => Array\n(\n => 205.253.121.11\n)\n\n => Array\n(\n => 154.72.171.177\n)\n\n => Array\n(\n => 39.52.203.37\n)\n\n => Array\n(\n => 122.161.52.2\n)\n\n => Array\n(\n => 82.145.41.170\n)\n\n => Array\n(\n => 103.217.123.33\n)\n\n => Array\n(\n => 103.150.238.100\n)\n\n => Array\n(\n => 125.99.11.182\n)\n\n => Array\n(\n => 103.217.178.70\n)\n\n => Array\n(\n => 197.210.227.95\n)\n\n => Array\n(\n => 116.75.212.153\n)\n\n => Array\n(\n => 212.102.42.202\n)\n\n => Array\n(\n => 49.34.177.147\n)\n\n => Array\n(\n => 173.242.123.110\n)\n\n => Array\n(\n => 49.36.35.254\n)\n\n => Array\n(\n => 202.47.59.82\n)\n\n => Array\n(\n => 157.42.197.119\n)\n\n => Array\n(\n => 103.99.196.250\n)\n\n => Array\n(\n => 119.155.228.244\n)\n\n => Array\n(\n => 130.105.160.170\n)\n\n => Array\n(\n => 78.132.235.189\n)\n\n => Array\n(\n => 202.142.186.114\n)\n\n => Array\n(\n => 115.99.156.136\n)\n\n => Array\n(\n => 14.162.166.254\n)\n\n => Array\n(\n => 157.39.133.205\n)\n\n => Array\n(\n => 103.196.139.157\n)\n\n => Array\n(\n => 139.99.159.20\n)\n\n => Array\n(\n => 175.176.87.42\n)\n\n => Array\n(\n => 103.46.202.244\n)\n\n => Array\n(\n => 175.176.87.16\n)\n\n => Array\n(\n => 49.156.85.55\n)\n\n => Array\n(\n => 157.39.101.65\n)\n\n => Array\n(\n => 124.253.195.93\n)\n\n => Array\n(\n => 110.227.59.8\n)\n\n => Array\n(\n => 157.50.50.6\n)\n\n => Array\n(\n => 95.142.120.25\n)\n\n => Array\n(\n => 49.36.186.141\n)\n\n => Array\n(\n => 110.227.54.161\n)\n\n => Array\n(\n => 88.117.62.180\n)\n\n => Array\n(\n => 110.227.57.8\n)\n\n => Array\n(\n => 106.200.36.21\n)\n\n => Array\n(\n => 202.131.143.247\n)\n\n => Array\n(\n => 103.46.202.4\n)\n\n => Array\n(\n => 122.177.78.217\n)\n\n => Array\n(\n => 124.253.195.201\n)\n\n => Array\n(\n => 27.58.17.91\n)\n\n => Array\n(\n => 223.228.143.162\n)\n\n => Array\n(\n => 119.160.96.233\n)\n\n => Array\n(\n => 49.156.69.213\n)\n\n => Array\n(\n => 41.80.97.54\n)\n\n => Array\n(\n => 122.176.207.193\n)\n\n => Array\n(\n => 45.118.156.6\n)\n\n => Array\n(\n => 157.39.154.210\n)\n\n => Array\n(\n => 103.48.197.173\n)\n\n => Array\n(\n => 103.46.202.98\n)\n\n => Array\n(\n => 157.43.214.102\n)\n\n => Array\n(\n => 180.191.125.73\n)\n\n => Array\n(\n => 157.39.57.227\n)\n\n => Array\n(\n => 119.152.107.157\n)\n\n => Array\n(\n => 103.166.58.254\n)\n\n => Array\n(\n => 139.135.54.30\n)\n\n => Array\n(\n => 185.199.102.101\n)\n\n => Array\n(\n => 59.103.204.4\n)\n\n => Array\n(\n => 202.14.123.34\n)\n\n => Array\n(\n => 103.48.197.243\n)\n\n => Array\n(\n => 39.34.158.129\n)\n\n => Array\n(\n => 156.146.59.38\n)\n\n => Array\n(\n => 110.227.51.220\n)\n\n => Array\n(\n => 157.45.123.212\n)\n\n => Array\n(\n => 110.227.52.20\n)\n\n => Array\n(\n => 43.242.176.136\n)\n\n => Array\n(\n => 223.233.71.146\n)\n\n => Array\n(\n => 216.73.160.38\n)\n\n)\n```\nRich Man In my Dreams was Warren Buffet: The Frugal Artisan\n Layout: Blue and Brown (Default) Author's Creation\n Home > Rich Man In my Dreams was Warren Buffet\n\n# Rich Man In my Dreams was Warren Buffet\n\nAugust 19th, 2013 at 03:03 pm\n\nA few nights ago there was a rich man in my dreams! Sit back and have a laugh at the story...\n\nI was at a conference for small business owners. It was a typical fancy hotel banquet room. Only about 100 people were there, we were all invited because we were supposed to be bright entrepreneurial stars.\n\nEach person was allowed to ask one question. My turn and I was asking some dumb boring question like: What are the best ways to increase your market. And he was starting to answer the same boring principles we already knew.\n\nWhen I interrupted and said, \"What I REALLY want to do is give my potential customers some incentive like... WIN A LUNCH DATE WITH WARREN BUFFET! Would you be willing to do that?\"\n\nWhereupon he brightened up and answered, \"EXACTLY! See this person has used this opportunity of being at this conference to get the next opportunity. This is the essence of entrepreneurialship!\"\n\nI think he said yes, but next in my dream I was with him one-on-one and asking him to sign a paper (for doing the lunch I think), but he was OLD and was kind of falling asleep with the pen in his hand. I felt sorry for him! And walked down the hall toward his hotel room with him for a minute.\n\nThen I woke up!\n\nHA! LOVE IT!\n\nSo now you know the secret to a successful small business",
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"Say goodbye to summer vacation, school starts tomorrow.\n\nThe good: I ride the bus to work, and save gobs of money on gas.\n\nThe bad: Less free time to take care of my jewelry business, housework, and exercise.\n\nHere's where we went the last weekend of summer vacation 2013:\n\nMorning Light at Trail Pass - Eastern Sierras",
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"Backpacked in about 1.5 miles, much shorter than the planned 5 miles because I was so out of shape! I also felt altitude sickness at about 10,000 feet. Still the trip was salvaged by camping at this pass, and it was my 13 yo daughters first back pack anyway which she enjoyed.\n\n### 1 Responses to “Rich Man In my Dreams was Warren Buffet”\n\n1. Joan.of.the.Arch Says:\n\nDreams are great! Yours sounds like one to stick with you for a long time!\n\n(Note: If you were logged in, we could automatically fill in these fields for you.)\n Name: * Email: Will not be published. Subscribe: Notify me of additional comments to this entry. URL: Verification: * Please spell out the number 4. [ Why? ]\n\nvB Code: You can use these tags: [b] [i] [u] [url] [email]"
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https://www.physicsforums.com/threads/thermistors-and-potential-divider-circuits.72601/ | [
"# Thermistors and potential divider circuits\n\nhey ill be quick about this, i really need help with a potential divider circuit. basically i have a resistor and a thermistor in series. 5 v input, and the values given for the thermistor are in volts, i need them in ohms. so i need to use the voltage across the thermistor, the total input voltage and the resistance of the resistor to find the resistance of the thermistor ata temperature. can anyone help?\n\nRelated Introductory Physics Homework Help News on Phys.org\nHi\n\nYou can use the formula to find the resistance of the thermistor.\n\nX / R+X * 5(volts) = V\n\nWhere R is the resistance of the resistor and V is the voltage across the thermistor.\n\nThen solve for X which is the resistance of the thermistor."
]
| [
null
]
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https://www.python-course.eu/matplotlib_gridspec.php | [
"## Matplotlib Tutorial: Gridspec",
null,
"We have seen in the last chapter of our Python tutorial on Matplotlib how to create a figure with multiple axis or subplot. To create such figures we used the subplots function. We will demonstrate in this chapter how the submodule of matplotlib gridspec can be used to specify the location of the subplots in a figure. In other words, it specifies the location of the subplots in a given GridSpec. GridSpec provides us with additional control over the placements of subplots, also the the margins and the spacings between the individual subplots. It also allows us the creation of axes which can spread over multiple grid areas.\n\nWe create a figure and four containing axes in the following code. We have covered this in our previous chapter.\n\nimport matplotlib\nimport matplotlib.pyplot as plt\nimport matplotlib.gridspec as gridspec\n\nfig, axes = plt.subplots(ncols=2, nrows=2, constrained_layout=True)",
null,
"We will create the previous example now by using GridSpec. At first, we have to create a figure object and after this a GridSpec object. We pass the figure object to the parameter figure of GridSpec. The axes are created by using add_subplot. The elements of the gridspec are accessed the same way as numpy arrays.\n\nimport matplotlib\nimport matplotlib.pyplot as plt\nimport matplotlib.gridspec as gridspec\n\nfig = plt.figure(constrained_layout=True)\nspec = gridspec.GridSpec(ncols=2, nrows=2, figure=fig)",
null,
"The above example is not a good usecase of GridSpec. It does not give us any benefit over the use of subplots. In principle, it is only more complicated in this case.\n\nThe importance and power of Gridspec unfolds, if we create subplots that span rows and columns.\n\nWe show this in the following example:\n\nimport matplotlib\nimport matplotlib.pyplot as plt\nimport matplotlib.gridspec as gridspec\n\nfig = plt.figure(constrained_layout=True)\nax1.set_title('gs[0, :]')\nax2.set_title('gs[1, :-1]')\nax3.set_title('gs[1:, -1]')\nax4.set_title('gs[-1, 0]')\nax5.set_title('gs[-1, -2]')\n\nOutput: :\nText(0.5, 1.0, 'gs[-1, -2]')",
null,
":mod:~matplotlib.gridspec is also indispensable for creating subplots of different widths via a couple of methods.\n\nThe method shown here is similar to the one above and initializes a uniform grid specification, and then uses numpy indexing and slices to allocate multiple \"cells\" for a given subplot.\n\nfig = plt.figure(constrained_layout=True)\noptional_params = dict(xy=(0.5, 0.5),\nxycoords='axes fraction',\nva='center',\nha='center')\n\nax.annotate('GridSpec[0, 0]', **optional_params)\n\nOutput: :\nText(0.5, 0.5, 'GridSpec[1:, 1:]')",
null,
"Another option is to use the width_ratios and height_ratios parameters. These keyword arguments are lists of numbers. Note that absolute values are meaningless, only their relative ratios matter. That means that width_ratios=[2, 4, 8] is equivalent to width_ratios=[1, 2, 4] within equally wide figures. For the sake of demonstration, we'll blindly create the axes within for loops since we won't need them later.\n\nfig5 = plt.figure(constrained_layout=True)\nwidths = [2, 3, 1.5]\nheights = [1, 3, 2]\nheight_ratios=heights)\nfor row in range(3):\nfor col in range(3):\nlabel = 'Width: {}\\nHeight: {}'.format(widths[col], heights[row])\nax.annotate(label, (0.1, 0.5), xycoords='axes fraction', va='center')",
null,
"# GridSpec using SubplotSpec\n\nYou can create GridSpec from the :class:~matplotlib.gridspec.SubplotSpec, in which case its layout parameters are set to that of the location of the given SubplotSpec.\n\nNote this is also available from the more verbose .gridspec.GridSpecFromSubplotSpec.\n\nfig10 = plt.figure(constrained_layout=True)\n\ngs00 = gs0.subgridspec(2, 3)\ngs01 = gs0.subgridspec(3, 2)\n\nfor a in range(2):\nfor b in range(3):",
null,
"### Psd Demo\n\nPlotting Power Spectral Density (PSD) in Matplotlib.\n\nThe PSD is a common plot in the field of signal processing. NumPy has many useful libraries for computing a PSD. Below we demo a few examples of how this can be accomplished and visualized with Matplotlib.\n\nimport matplotlib.pyplot as plt\nimport numpy as np\nimport matplotlib.mlab as mlab\nimport matplotlib.gridspec as gridspec\n\n# Fixing random state for reproducibility\nnp.random.seed(42)\n\ndt = 0.01\nt = np.arange(0, 10, dt)\nnse = np.random.randn(len(t))\nr = np.exp(-t / 0.05)\n\ncnse = np.convolve(nse, r) * dt\ncnse = cnse[:len(t)]\ns = 0.1 * np.sin(2 * np.pi * t) + cnse\n\nplt.subplot(211)\nplt.plot(t, s)\nplt.subplot(212)\nplt.psd(s, 512, 1 / dt)\n\nplt.show()",
null,
"This type of flexible grid alignment has a wide range of uses. I most often use it when creating multi-axes histogram plots like the ones shown here:\n\n# Create some normally distributed data\nmean = [0, 0]\ncov = [[1, 1], [1, 2]]\nx, y = np.random.multivariate_normal(mean, cov, 3000).T\n\n# Set up the axes with gridspec\nfig = plt.figure(figsize=(6, 6))\ngrid = plt.GridSpec(4, 4, hspace=0.2, wspace=0.2)\ny_hist = fig.add_subplot(grid[:-1, 0], xticklabels=[], sharey=main_ax)\nx_hist = fig.add_subplot(grid[-1, 1:], yticklabels=[], sharex=main_ax)\n\n# scatter points on the main axes\nmain_ax.plot(x, y, 'ok', markersize=3, alpha=0.2)\n\n# histogram on the attached axes\nx_hist.hist(x, 40, histtype='stepfilled',\norientation='vertical', color='gray')\nx_hist.invert_yaxis()\n\ny_hist.hist(y, 40, histtype='stepfilled',\norientation='horizontal', color='gray')\ny_hist.invert_xaxis()",
null,
""
]
| [
null,
"https://www.python-course.eu/images/gridspec_with_images.png",
null,
"data:image/png;base64,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 ",
null,
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 ",
null,
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 ",
null,
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 ",
null,
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 ",
null,
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 ",
null,
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 ",
null,
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https://kupdf.net/download/mastering-physics-answers_59cab88c08bbc50952686f67_pdf | [
"September 27, 2017 | Author: curiousbookworm | Category: Acceleration, Force, Euclidean Vector, Velocity, Friction\n\n#### Description\n\nMP02: Motion Diagrams Velocity and Acceleration of a Power Ball Learning Goal: To understand the distinction between velocity and acceleration with the use of motion diagrams. In common usage, velocity and acceleration both can imply having considerable speed. In physics, they are sharply defined concepts that are not at all synonymous. Distinguishing clearly between them is a prerequisite to understanding motion. Moreover, an easy way to study motion is to draw a motion diagram, in which the position of the object in motion is sketched at several equally spaced instants of time, and these sketches (or snapshots) are combined into one single picture. In this problem, we make use of these concepts to study the motion of a power ball. This discussion assumes that we have already agreed on a coordinate system from which to measure the position of objects as a function of time. Let\n\nand\n\n(also called the position vector)\n\nbe velocity and acceleration, respectively.\n\nHarvaran Ghai Consider the motion of a power ball that is dropped on the floor and bounces back. In the following questions, you will describe its motion at various points in its fall in terms of its velocity and acceleration. Part A You drop a power ball on the floor. The motion diagram of the ball is sketched in the figure the magnitude of the velocity of the ball is increasing, decreasing, or not changing. increasing\n\n. Indicate whether\n\ndecreasing not changing Correct\n\nWhile the ball is in free fall, the magnitude of its velocity is increasing, so the ball is accelerating. Part B\n\nSince the length of is directly proportional to the length of , the vector connecting each dot to the next could represent velocity vectors as well as position vectors, as shown in the figure here . Indicate whether the velocity and acceleration of the ball are, respectively, positive (upward), negative, or zero. Use P, N, and Z for positive (upward), negative, and zero, respectively. Separate the letters for velocity and acceleration with a comma. N,N Correct\n\nPart C Now, consider the motion of the power ball once it bounces upward. Its motion diagram is shown in the figure here . Indicate whether the magnitude of the velocity of the ball is increasing, decreasing, or not changing. increasing decreasing not changing Correct\n\nSince the magnitude of the velocity of the ball is decreasing, the ball must be accelerating (specifically, slowing down). Part D The next figure shows the velocity vectors corresponding to the upward motion of the power ball. Indicate whether its velocity and acceleration, respectively, are positive (upward), negative, or zero. Use P, N, and Z for positive (upward), negative, and zero, respectively. Separate the letters for velocity and acceleration with a comma. P,N Correct\n\nPart E The power ball has now reached its highest point above the ground and starts to descend again. The motion diagram representing the velocity vectors is the same as that after the initial release, as shown in the figure of Part B. Indicate whether the velocity and acceleration of the ball at its highest point are positive (upward), negative, or zero. Use P, N, and Z for positive (upward), negative, and zero, respectively. Separate the letters for velocity and acceleration with a comma. Z,N Correct\n\nThese examples should show you that the velocity and acceleration can have opposite or similar signs or that one of them can be zero while the other has either sign. Try hard to think carefully about them as distinct physical quantities when working with kinematics.\n\nMotion of Two Rockets Learning Goal: To learn to use images of an object in motion to determine velocity and acceleration. Two toy rockets are traveling in the same direction (taken to be the x axis). A diagram is shown of a timeexposure image where a stroboscope has illuminated the rockets at the uniform time intervals indicated.\n\nHarvaran Ghai Part A At what time(s) do the rockets have the same velocity? at time\n\nonly\n\nat time\n\nonly\n\nat times\n\nand\n\nat some instant in time between\n\nand\n\nat no time shown in the figure Correct\n\nPart B At what time(s) do the rockets have the same x position? at time\n\nonly\n\nat time\n\nonly\n\nat times\n\nand\n\nat some instant in time between\n\nand\n\nat no time shown in the figure Correct\n\nPart C At what time(s) do the two rockets have the same acceleration? at time\n\nonly\n\nat time\n\nonly\n\nat times\n\nand\n\nat some instant in time between\n\nand\n\nat no time shown in the figure Correct\n\nPart D The motion of the rocket labeled A is an example of motion with uniform (i.e., constant) __________. and nonzero acceleration velocity displacement time Correct\n\nPart E The motion of the rocket labeled B is an example of motion with uniform (i.e., constant) __________. and nonzero acceleration velocity displacement time Correct\n\nPart F At what time(s) is rocket A ahead of rocket B? before after before\n\nonly only and after\n\nbetween and at no time(s) shown in the figure Correct\n\nPSS 1.1: (Almost) a Dozen Diagrams Learning Goal: To practice ProblemSolving Strategy 1.1 for constructing motion diagrams. A car is traveling with constant velocity along a highway. The driver notices he is late for work so he stomps down on the gas pedal and the car begins to accelerate. The car has just achieved double its initial velocity when the driver spots a policeman behind him and applies the brakes. The car then decelerates, coming to rest at a stoplight ahead. In this problem, you will be asked several questions related to construction of a motion diagram for this situation and a few others.\n\nHarvaran Ghai Represent the moving object as a particle. Make simplifying assumptions when interpreting the problem statement. MODEL:\n\nVISUALIZE:\n\nA complete motion diagram consists of:\n\nThe position of the object in each frame of the film, shown as a dot. Use five or six dots to make the motion clear but without overcrowding the picture. More complex motions may need more dots.\n\nThe average velocity vectors, found by connecting each dot in the motion diagram to the next with a vector arrow. There is one velocity vector linking each set of two position dots. Label the row of velocity vectors .\n\nThe average acceleration vectors, found using Tactics Box 1.3. There is one acceleration vector linking each set of two velocity vectors. Each acceleration vector is drawn at the dot between the two velocity vectors it links. Use to indicate a point at which the acceleration is zero. Label the row of acceleration vectors .\n\nModel It is appropriate to use the particle model for the car. You should also make some simplifying assumptions. Part A Which of the following simplifying assumptions is it reasonable to make in this problem? A. During each of the three different stages of its motion, the car is moving with constant (possibly zero) acceleration. B. During each of the three different stages of its motion, the car is moving with constant (possibly zero) velocity. C. The highway is straight (i.e., there are no curves). D. The highway is level (i.e., there are no hills or valleys). Enter the letters of all the correct answers in alphabetical order. Do not use commas. For example, if you think that assumptions C and D are reasonable, enter CD. ACD Correct\n\nVisualize Now draw a motion diagram, including all the elements listed in the problemsolving strategy. Use your diagram to answer the following questions. In interpreting the diagrams that follow, assume that the car is moving in a straight line to the right. Refer to this set of motion diagrams\n\nPart B Which of the diagrams best describes the position and the velocity of the car before the driver notices he is late? A B C Correct\n\nPart C Which of the diagrams best describes the position and the velocity of the car after the driver hits the gas, but before he notices the policeman? A B C Correct\n\nPart D Which of the diagrams best describes the position and the velocity of the car after the driver notices the policeman? A B C Correct\n\nPart E Which of these diagrams problem introduction?\n\nmost accurately depicts the acceleration of the car during the events described in the\n\nAssume that the car is initially moving to the right. A B C Correct\n\nNow let's use our results for the car and apply them to some other problems. Consider these three situations: \n\nA train has its brakes released and pulls out of the station, slowly picking up speed. A sled is given a quick push along a horizontal surface; the sled comes to a stop after covering some distance.\n\nA motorcycle is moving along a straight highway at 105 km/h (the legal speed limit in many states).\n\nPart F Of the three situations described, which object corresponds to the position and velocity diagram shown here? the train the sled\n\nthe motorcycle Correct\n\nNote that the diagram shown is not a complete motion diagram; it lacks the vector representing the acceleration of the object. Part G Of the three situations described, which object corresponds to the motion diagram shown here? the train\n\nthe sled the motorcycle Correct\n\nLet us now consider another scenario. A car and a truck are moving at the same velocity along a straight highway. Both drivers apply the brakes at the same moment. The car and truck both come to a stop. The car takes less time to stop than the truck. Refer to the motion diagrams shown here\n\nAssume that both cars are moving to the right. Part H Which of the three diagrams shown best describes the motion of the car and truck after the brakes have been applied? A B C Correct\n\nPart I Diagram (B) is incorrect because, according to it: The car and truck move in different directions. The car is moving at constant velocity. The truck is speeding up. The car and the truck have the same acceleration. Correct\n\nPart J Diagram (C) is incorrect because, according to it: The car and truck move in different directions. The car is moving at constant velocity. The truck is speeding up. The car and the truck have the same acceleration. Correct\n\nMP04: Using Motion Diagrams Curved Motion Diagram The motion diagram shown here represents a pendulum released from rest at an angle of 45 from the vertical. The dots in the motion diagram represent the positions of the pendulum bob at eleven moments separated by equal time intervals. The green arrows represent the average velocity between adjacent dots. Also given is a \"compass rose\" of directions in which the different directions are labeled with the letters of the alphabet.\n\nHarvaran Ghai Part A What is the direction of the acceleration of the pendulum bob at moment 5? Enter the letter of the arrow with this direction from the compass rose in the figure. Type Z if the acceleration vector has zero length. A Correct\n\nPart B What is the direction of the acceleration of the pendulum bob at moments 0 and 10? Enter the letters of the arrows with these directions from the compass rose in the figure, separated by commas. Type Z if the acceleration vector has zero length. directions at moment 0, moment 10 =D,F Correct\n\nPart C In which of the following other scenarios could the motion reasonably be represented by the motion diagram in the introduction? A. A weight placed on the rim of a bicycle wheel that is being held off the ground so it can rotate freely B. An airplane pulling out of a dive C. A race car rounding a turn D. A marble released part way up the inside surface of a smoothly rounded bowl Enter the letters of all possible correct scenarios in alphabetical order. Do not use commas. AD Correct\n\nPart D Assume that the diagram in the problem introduction represents the motion of a ball tied on the end of a stringthat is, a pendulum. Also assume that the interval between each time step in the diagram is 0.10 s. The total time it takes for this pendulum to swing back and forth, also called the period of the pendulum, is then approximately 2 s. An interesting fact about the pendulum is that its period is essentially independent of the weight of the ball (or whatever other object is used). It depends only on the length of the string (with longer period for longer strings) and the strength of the force of gravity (which is essentially constant over the surface of the earth). Based on observation or comparison with other reallife pendula, estimate the length of the string needed for the pendulum to have a period of 2 s. Express in meters. A factor of 3 error is allowed in either direction. =1.0 Correct\n\nPhysics can often seem to be a science of very precise answers. However, having a solid grasp of the fundamental concepts allows a physicist to make reasonably accurate estimates like this one very quickly. Even if you were ultimately looking for a more precise answer, an initial estimate gives you a way of checking whether the result of a long calculation is reasonable.\n\nAverage Velocity from a Position vs. Time Graph Learning Goal: To learn to read and interpolate on a graph of position versus time and to change units.\n\nIn this problem you must find the average velocity from a graph of indicate the average velocity over the time interval from to . Thus interval from 1 to 3 s.\n\n. We will use the notation\n\nto\n\nis the average velocity over the time\n\nHarvaran Ghai Part A Find the average velocity over the time interval from 0 to 1 second. Answer to the nearest integer. =0 Correct\n\nPart B Find the average velocity over the time interval from 1 to 3 seconds. Answer to the nearest integer. =20 Correct\n\nPart C Now that you have\n\nand\n\n, find\n\n.\n\nGive your answer to three significant figures. =13.3 Correct\n\nNote that\n\nis not equal to the simple arithmetic average of\n\nand\n\nfor time intervals of different length. You would have to double the weight given to interval twice as long.\n\n, because they are averages because it is for an\n\nPart D Find the average velocity over the time interval from 1 to 5 seconds. You will need to interpolate to find the position at time . Do not simply eyeball the position or you will likely not be able to obtain the solution to the desired accuracy. Round your answer to two significant figures. =6.7 Correct\n\nPart E Obtaining this answer required some interpolation on the graph. Now see if you can express this result in terms of kilometers per hour. Express your answer to the nearest integer. =24 Correct\n\nPart F Find the average velocity over the time interval from 2.5 to 6.0 seconds. Express your answer to two significant figures. Correct\n\n=8.6\n\nRunning and Walking Tim and Rick both can run at speed and walk at speed distance half.\n\n, with\n\n. They set off together on a journey of\n\n. Rick walks half of the distance and runs the other half. Tim walks half of the time and runs the other\n\nHarvaran Ghai Part A\n\nHow long does it take Rick to cover the distance\n\n?\n\nExpress the time taken by Rick in terms of , = Correct\n\n, and\n\n.\n\nPart B Find Rick's average speed for covering the distance Express Rick's average speed in terms of , = Correct\n\n.\n\n, and\n\n.\n\nPart C How long does it take Tim to cover the distance? Express the time taken by Tim in terms of , = Correct\n\n, and\n\n.\n\nPart D Who covers the distance more quickly? Think logically, but without using the detailed answers in the previous parts. Rick Tim Neither. They cover the distance in the same amount of time. Correct\n\nPart E In terms of given quantities, by what amount of time, , does Tim beat Rick? It will help you check your answer if you simplify it algebraically and check the special case Express the difference in time, = Correct\n\nPart F\n\nin terms of ,\n\n, and\n\n.\n\n.\n\nIn the special case that Correct\n\n, what would be Tim's margin of victory\n\n?\n\nGraph of v(t) for a Sports Car The graph questions.\n\nshows the velocity of a sports car as a function of time . Use the graph to answer the following\n\nHarvaran Ghai Part A Find the maximum velocity of the car. Express your answer in meters per second to the nearest integer. =55 Correct\n\nPart B During which time interval is the acceleration positive? Indicate the most complete answer. to to to to Correct\n\nto\n\nPart C Find the maximum acceleration of the car. Express your answer in meters per second squared to the nearest integer. =30 Correct\n\nPart D Find the minimum magnitude of the acceleration of the car. Express your answer in meters per second squared to the nearest integer. =0 Correct\n\nPart E Find the distance traveled by the car between 0 and 2 s. Express your answer in meters to the nearest integer. =55 Correct\n\nRearending Drag Racer To demonstrate the tremendous acceleration of a top fuel drag racer, you attempt to run your car into the back of a dragster that is \"burning out\" at the red light before the start of a race. (Burning out means spinning the tires at high speed to heat the tread and make the rubber sticky.) You drive at a constant speed of toward the stopped dragster, not slowing down in the face of the imminent collision. The dragster driver sees you coming but waits until the last instant to put down the hammer, accelerating from the starting line at constant acceleration, . Let the time at which the dragster starts to accelerate be\n\n.\n\nHarvaran Ghai\n\nPart A\n\nWhat is , the longest time after the dragster begins to accelerate that you can possibly run into the back of the dragster if you continue at your initial velocity? = Correct\n\nPart B Assuming that the dragster has started at the last instant possible (so your front bumper almost hits the rear of the dragster at\n\n), find your distance from the dragster when he started. If you calculate positions on the way to\n\nthis solution, choose coordinates so that the position of the drag car is 0 at . Remember that you are solving for a distance (which is a magnitude, and can never be negative), not a position (which can be negative). = Correct\n\nPart C Find numerical values for\n\nand\n\nin seconds and meters for the (reasonable) values\n\nand . Separate your two numerical answers by commas, and give your answer to two significant figures. , Correct\n\n=0.54,7.2 s, m\n\n(26.8 m/s)\n\nThe blue curve shows how the car, initially at accelerating drag car (red) at\n\n, continues at constant velocity (blue) and just barely touches the\n\n.\n\nA small source of light is located at a distance from a vertical wall. An opaque object with a height of\n\nmoves toward the wall with constant velocity of magnitude . At time\n\n, the object is located at the source .\n\nPart A\n\nHarvaran Ghai\n\nFind an expression for , the magnitude of the velocity of the top of the object's shadow, at time . Express the speed of the top of the object's shadow in terms of , , , and .\n\n= Correct\n\nRocket Height A rocket, initially at rest on the ground, accelerates straight upward from rest with constant net acceleration , until time , when the fuel is exhausted.\n\nHarvaran Ghai\n\nPart A Find the maximum height\n\nthat the rocket reaches (neglecting air resistance).\n\nExpress the maximum height in terms of , , and/or . Note that in this problem, is a positive number equal to the magnitude of the acceleration due to gravity. = a*t1*((a*t1)/g)(.5g((a*t1)/g)2)+(.5(t1)2*a) Correct\n\nPart B If the rocket's net acceleration is\n\nfor\n\nExpress your answer numerically in meters, using\n\n, what is the maximum height the rocket will reach? .\n\n=1470 m Correct\n\nA Flower Pot Falling Past a Window As you look out of your dorm window, a flower pot suddenly falls past. The pot is visible for a time , and the vertical length of your window is . Take down to be the positive direction, so that downward velocities are positive and the acceleration due to gravity is the positive quantity . Assume that the flower pot was dropped by someone on the floor above you (rather than thrown downward).\n\nPart A From what height above the bottom of your window was the flower pot dropped?\n\nHarvaran Ghai\n\nExpress your answer in terms of\n\n, , and .\n\n= Correct\n\nPart B If the bottom of your window is a height above the ground, what is the velocity of the pot as it hits the ground? You may introduce the new variable , the speed at the bottom of the window, defined by\n\n. Express your answer in terms of some or all of the variables\n\n= Correct\n\n, ,\n\n, and .\n\nResolving Vector Components with Trigonometry\n\nOften a vector is specified by a magnitude and a direction; for example, a rope with tension\n\nexerts a force of\n\nmagnitude in a direction 35 degrees north of east. This is a good way to think of vectors; however, to calculate results with vectors, it is best to select a coordinate system and manipulate the components of the vectors in that coordinate system.\n\nHarvaran Ghai\n\nPart A\n\nFind the components of the vector with length and angle with respect to the x axis as shown, named . Don't forget that when multiplying two factors, you must include a multiplication symbol; also, the cos and sin functions must have parentheses around their arguments. For example, a vector might take the form p*sin(Q),m*cos(N). Write the components in the form x,y. = Correct\n\nPart B Find the components of the vector with length and angle with respect to the x axis as shown, named . Write the components in the form x,y. = Correct\n\nNotice that vectors\n\nand have the same form despite their placement with respect to the y axis on the drawing.\n\nPart C Find the components of the vector with length and angle as shown, named . Express your answer in terms of and . Write the components in the form x,y. = Correct\n\nTracking a Plane A radar station, located at the origin of xz plane, as shown in the figure\n\n, detects an airplane coming straight at\n\nthe station from the east. At first observation (point A), the position of the airplane relative to the origin is position vector\n\nhas a magnitude of 360\n\ntracked for another 123\n\nand is located at exactly 40\n\n. The\n\nabove the horizon. The airplane is\n\nin the vertical eastwest plane for 5.0 , until it has passed directly over the station\n\nand reached point B. The position of point B relative to the origin is (the magnitude of is 880 ). The contact points are shown in the diagram, where the x axis represents the ground and the positive z direction is upward.\n\nHarvaran Ghai Part A\n\nDefine the displacement of the airplane while the radar was tracking it: components of\n\n. What are the\n\n?\n\nExpress in meters as an ordered pair, separating the x and z components with a comma, to two significant figures. =1100,26 Correct\n\nTwo Forces Acting at a Point Two forces,\n\nand\n\n, act at a point.\n\nnegative x axis in the second quadrant. negative x axis in the third quadrant.\n\nhas a magnitude of 9.80 and is directed at an angle of 60.0 above the has a magnitude of 6.40 and is directed at an angle of 53.2 below the\n\nHarvaran Ghai Part A What is the x component of the resultant force? Express your answer in newtons. 8.73 Correct\n\nPart B What is the y component of the resultant force? Express your answer in newtons. 3.36 Correct\n\nPart C What is the magnitude of the resultant force? Express your answer in newtons. 9.36 Correct\n\nA Push or a Pull? Learning Goal: To understand the concept of force as a push or a pull and to become familiar with everyday forces. A force can be simply defined as a push or a pull exerted by one object upon another. Although such a definition may not sound too scientific, it does capture three essential properties of forces: \n\nEach force is created by some object. Each force acts upon some other object.\n\nThe action of a force can be visualized as a push or a pull.\n\nSince each force is created by one object and acts upon another, forces must be described as interactions. The proper words describing the force interaction between objects A and B may be any of the following:\n\n\"Object A acts upon object B with force .\"\n\n\"Object A exerts force upon object B.\"\n\n\"Force is applied to object B by object A.\"\n\n\"Force due to object A is acting upon object B.\"\n\nOne of the biggest mistakes you may make is to think of a force as \"something an object has.\" In fact, at least two objects are always required for a force to exist. Each force has a direction: Forces are vectors. The main result of such interactions is that the objects involved change their velocities: Forces cause acceleration. However, in this problem, we will not concern ourselves with accelerationnot yet.\n\nHarvaran Ghai Some common types of forces that you will be dealing with include the gravitational force (weight), the force of tension, the force of friction, and the normal force. It is sometimes convenient to classify forces as either contact forces between two objects that are touching or as longrange forces between two objects that are some distance apart. Contact forces include tension, friction, and the normal force. Longrange forces include gravity and electromagnetic forces. Note that such a distinction is useful but not really fundamental: For instance, on a microscopic scale the force of friction is really an electromagnetic force. In this problem, you will identify the types of forces acting on objects in various situations. First, consider a book resting on a horizontal table. Part A Which object exerts a downward force on the book? the book itself the earth the surface of the table Correct\n\nPart B The downward force acting on the book is __________. a contact force a longrange force Correct\n\nPart C What is the downward force acting on the book called? tension normal force weight friction Correct\n\nPart D Which object exerts an upward force on the book? the book itself the earth the surface of the table Correct\n\nPart E The upward force acting on the book is __________. a contact force a longrange force Correct\n\nPart F What is the upward force acting on the book called? tension normal force weight friction\n\nCorrect\n\nNow consider a different situation. A string is attached to a heavy block. The string is used to pull the block to the right along a rough horizontal table. Part G Which object exerts a force on the block that is directed toward the right? the block itself the earth the surface of the table the string Correct\n\nPart H The force acting on the block and directed to the right is __________. a contact force a longrange force Correct\n\nTo exert a tension force, the string must be connected to (i.e., touching) the block. Part I What is the force acting on the block and directed to the right called? tension normal force weight friction Correct\n\nPart J Which object exerts a force on the block that is directed toward the left? the block itself the earth the surface of the table the string Correct\n\nPart K The force acting on the block and directed to the left is __________. a contact force a longrange force Correct\n\nPart L What is the force acting on the block and directed to the left called? tension normal force weight friction Correct\n\nNow consider a slightly different situation. The same block is placed on the same rough table. However, this time, the string is disconnected and the block is given a quick push to the right. The block slides to the right and eventually stops. The following questions refer to the motion of the block after it is pushed but before it stops. Part M How many forces are acting on the block in the horizontal direction? 0 1 2 3 Correct\n\nOnce the push has commenced, there is no force acting to the right: The block is moving to the right because it was given a velocity in this direction by some force that is no longer applied to the block (probably, the normal force exerted by a student's hand or some spring launcher). Once the contact with the launching object has been lost, the only horizontal force acting on the block is directed to the leftwhich is why the block eventually stops. Part N What is the force acting on the block that is directed to the left called? tension normal force\n\nweight friction Correct\n\nThe force of friction does not disappear as long as the block is moving. Once the block stops, fricion becomes zero (assuming the table is perfectly horizontal).\n\nFree-Body Diagrams: Introduction Learning Goal: To learn to draw freebody diagrams for various reallife situations. Imagine that you are given a description of a reallife situation and are asked to analyze the motion of the objects involved. Frequently, that analysis would involve finding the acceleration of the objects. That, in turn, requires that you find the net force. To find the net force, you must first identify all of the forces involved and then add them as vectors. Such a procedure is not always trivial. It is helpful to replace the sketch of the situation by the drawing of the object (represented as a particle) and all the forces applied to it. Such a drawing is called a freebody diagram. This problem will walk you through several examples of freebody diagrams and will demonstrate some of the possible pitfalls. Here is the general strategy for drawing freebody diagrams: Identify the object of interest. This may not always be easy: A sketch of the situation may contain many objects, each of which has a different set of forces acting on it. Including forces acting on different objects in the same diagram will lead to confusion and a wrong solution. Draw the object as a dot. Draw and clearly label all the forces acting on the object of interest. The forces should be shown as vectors originating from the dot representing the object of interest. There are two possible difficulties here: omitting some forces and drawing the forces that either don't exist at all or are applied to other objects. To avoid these two pitfalls, remember that every force must be applied to the object of interest by some other objector, as some like to say, \"every force must have a source.\" \n\nOnce all of the forces are drawn, draw the coordinate system. The origin should coincide with the dot representing the object of interest and the axes should be chosen so that the subsequent calculations of vector components of the forces will be relatively simple. That is, as many forces as possible must be either parallel or perpendicular to one of the axes.\n\nHarvaran Ghai It should come as good news that, even though real life can present us with a wide variety of situations, we will be mostly dealing with a very small number of forces. Here are the principal ones of interest: \n\nWeight, or the force due to gravity. Weight acts on every object and is directed straight down unless we are considering a problem involving the nonflat earth (e.g., satellites). Normal force. The normal force exists between two surfaces that are pressed against each other; it is always perpendicular to the surfaces.\n\nForce of tension. Tension exists in strings, springs, and other objects of finite length. It is directed along the string or a spring. Keep in mind that a spring can be either compressed or stretched whereas a string can only be stretched.\n\nForce of friction. A friction force exists between two surfaces that either move or have a tendency to move relative to each other. Sometimes, the force of air drag, similar in some ways to the force of friction, may come into play. These forces are directed so that they resist the relative motion of the surfaces. Keep in mind that to simplify problems you often assume friction is negligible on smooth surfaces. In addition, the word friction commonly refers to resistive forces other than air drag that are caused by contact between surfaces so you can ignore air drag in problems unless you are told to consider its effects.\n\nThe following examples should help you learn to draw freebody diagrams. We will start with relatively simple situations in which the object of interest is either explicitly suggested or fairly obvious. Part A A hockey puck slides along a horizontal, smooth icy surface at a constant velocity as shown. diagram for the puck. Which of the following forces are acting on the puck? A. weight B. friction C. force of velocity D. force of push E. normal force\n\nDraw a freebody\n\nF. air drag G. acceleration Type the letters corresponding to all the correct answers in alphabetical order. Do not use commas. For instance, if you think that only answers C and D are correct, type CD. AE Correct\n\nThere is no such thing as \"the force of velocity.\" If the puck is not being pushed, there are no horizontal forces acting on it. Of course, some horizontal force must have acted on it before, to impart the velocityhowever, in the situation described, no such \"force of push\" exists. Also, the air drag in such cases is assumed to be negligible. Finally, the word \"smooth\" usually implies negligible surface friction. Your freebody diagram should look like the\n\none shown here. Part B Consider a block pulled by a horizontal rope along a horizontal surface at a constant velocity as shown. The tension in the rope is nonzero. Draw a freebody diagram for the block. Which of the following forces are acting on the block? A. weight B. friction C. force of velocity D. force of tension E. normal force F. air drag G. acceleration\n\nType the letters corresponding to all the correct answers in alphabetical order. Do not use commas. For instance, if you think that only answers C and D are correct, type CD.\n\nABDE Correct\n\nBecause the velocity is constant, there must be a force of friction opposing the force of tension. Since the block is moving, it is kinetic friction. Your freebody diagram should look like that shown here.\n\nConsider the following situation in parts C F. A block is resting on a slope as shown.\n\nPart C Which of the following forces are acting on the block? A. weight B. kinetic friction\n\nC. static friction D. force of push E. normal force Type the letters corresponding to all the correct answers in alphabetical order. Do not use commas. For instance, if you think that only answers C and D are correct, type CD. ACE Correct\n\nPart D What is the direction of the force due to gravity acting on the block? vertically upward vertically downward perpendicular to the slope\n\nupward along the slope downward along the slope Correct\n\nPart E What is the direction of the normal force acting on the block? vertically upward\n\nvertically downward perpendicular to the slope upward along the slope downward along the slope Correct\n\nPart F Draw the freebody diagram for the block. What is the direction of the force of friction acting on the block? vertically upward\n\nvertically downward perpendicular to the slope upward along the slope downward along the slope Correct\n\nWithout friction, the block would slide down the slope; so the force of static friction must oppose such a motion and be directed upward along the slope. Your freebody diagram should look like that shown here.\n\nNow consider a block sliding up a rough slope after having been given a quick push as shown. Part G Which of the following forces are acting on the block? A. weight B. kinetic friction\n\nC. static friction D. force of push E. normal force F. the force of velocity Type the letters corresponding to all the correct answers in alphabetical order. Do not use commas. For instance, if you think that only answers C and D are correct, type CD. ABE Correct\n\nThe word \"rough\" implies the presence of friction. Since the block is in motion, it is kinetic friction. Once again, there is no such thing as \"the force of velocity.\" However, it seems a tempting choice to some students since the block is going up. Part H Draw the freebody diagram for the block. What is the direction of the force of friction acting on the block? vertically upward vertically downward perpendicular to the slope upward along the slope downward along the slope Correct\n\nThe force of kinetic friction opposes the upward motion of the block. Your freebody diagram should look like the\n\none shown here. Part I Now consider a block being pushed up a smooth slope. The force pushing the block is parallel to the slope. Which of the following forces are acting on the block? A. weight B. kinetic friction C. static friction D. force of push E. normal force Type the letters corresponding to all the correct answers in alphabetical order. Do not use commas. For instance, if you think that only answers C and D are correct, type CD. ADE Correct\n\nYour freebody diagram should look like the one shown here.\n\nby the palm of the hand of the person pushing the block.\n\nThe force of push is the normal force exerted, possibly,\n\nIn all the previous situations just described, the object of interest was explicitly given. Let us consider a situation where choosing the objects for which to draw the freebody diagrams is up to you. Two blocks of masses and are connected by a light string that goes over a light frictionless pulley. The block of mass is sliding to the right on a rough horizontal surface of a lab table. Part J To solve for the acceleration of the blocks, you will have to draw the freebody diagrams for which objects? A. the block of mass B. the block of mass C. the connecting string D. the pulley E. the table F. the earth Type the letters corresponding to all the correct answers in alphabetical order. Do not use commas. For instance, if you think that only answers C and D are correct, type CD. AB Correct\n\nPart K Draw the freebody diagram for the block of mass none one two\n\n. How many forces are exerted on this block?\n\nthree four Correct\n\nYour freebody diagram should look like that shown here.\n\nPart L Draw the freebody diagram for the block of mass none one two three four Correct\n\n. How many forces are exerted on this block?\n\nYour freebody diagram should look like that shown here.\n\nUnderstanding Newton's Laws\n\nHarvaran Ghai\n\nPart A An object cannot remain at rest unless which of the following holds? The net force acting on it is zero. The net force acting on it is constant and nonzero. There are no forces at all acting on it. There is only one force acting on it. Correct\n\nIf there is a net force acting on a body, regardless of whether it is a constant force, the body accelerates. If the body is at rest and the net force acting on it is zero, then it will remain at rest. The net force could be zero either because there are no forces acting on the body at all or because several forces are acting on the body but they all cancel out. Part B If a block is moving to the left at a constant velocity, what can one conclude? There is exactly one force applied to the block. The net force applied to the block is directed to the left. The net force applied to the block is zero. There must be no forces at all applied to the block. Correct\n\nIf there is a net force acting on a body, regardless of whether the body is already moving, the body accelerates. If a body is moving with constant velocity, then it is not accelerating and the net force acting on it is zero. The net force could be zero either because there are no forces acting on the body at all or because several forces are acting on the body but they all cancel out. Part C A block of mass is acted upon by two forces: you say about the block's motion? It must be moving to the left.\n\n(directed to the left) and\n\n(directed to the right). What can\n\nIt must be moving to the right. It must be at rest. It could be moving to the left, moving to the right, or be instantaneously at rest. Correct\n\nThe acceleration of an object tells you nothing about its velocitythe direction and speed at which it is moving. In this case, the net force on (and therefore the acceleration of) the block is to the right, but the block could be moving left, right, or in any other direction. Part D A massive block is being pulled along a horizontal frictionless surface by a constant horizontal force. The block must be __________. continuously changing direction moving at constant velocity moving with a constant nonzero acceleration moving with continuously increasing acceleration Correct\n\nSince there is a net force acting, the body does not move at a constant velocity, but it accelerates instead. However, the force acting on the body is constant. Hence, according to Newton's 2nd law of motion, the acceleration of the body is also constant. Part E Two forces, of magnitude and , are applied to an object. The relative direction of the forces is unknown. The net force acting on the object __________. A. cannot be equal to B. cannot be equal to C. cannot be directed the same way as the force of\n\nD. must be greater than Enter the letters of all the correct answers in alphabetical order. Do not use commas. For example, if you think only the last option is correct, enter D. A Correct\n\nConceptual Questions on Newton's 1st and 2nd Laws Learning Goal: To understand the meaning and the basic applications of Newton's 1st and 2nd laws. In this problem, you are given a diagram representing the motion of an objecta motion diagram. The dots represent the object's position at moments separated by equal intervals of time. The dots are connected by arrows representing the object's average velocity during the corresponding time interval. Your goal is to use this motion diagram to determine the direction of the net force acting on the object. You will then determine which force diagrams and which situations may correspond to such a motion.\n\nHarvaran Ghai Part A What is the direction of the net force acting on the object at position A? upward downward to the left to the right The net force is zero.\n\nCorrect\n\nThe velocity vectors connecting position A to the adjacent positions appear to have the same magnitude and direction. Therefore, the acceleration is zeroand so is the net force. Part B What is the direction of the net force acting on the object at position B? upward downward to the left to the right The net force is zero. Correct\n\nThe velocity is directed to the right; however, it is decreasing. Therefore, the acceleration is directed to the leftand so is the net force. Part C What is the direction of the net force acting on the object at position C? upward downward to the left to the right The net force is zero. Correct\n\nThe horizontal component of the velocity does not change. The vertical component of the velocity increases. Therefore, the accelerationand the net forceare directed straight downward. The next four questions are related to the force diagrams numbered 1 to 6. These diagrams represent the forces acting on a moving object. The number next to each arrow represents the magnitude of the force in newtons. Part D Which of these diagrams may possibly correspond to the situation at point A on the motion diagram? Type, in increasing order, the numbers corresponding to the correct diagrams. Do not use commas. For instance, if you think that only diagrams 3 and 4 are correct, type 34. 6 Correct\n\nPart E\n\nWhich of these diagrams may possibly correspond to the situation at point B on the motion diagram? Type, in increasing order, the numbers corresponding to the correct diagrams. Do not use commas. For instance, if you think that only diagrams 3 and 4 are correct, type 34. 35 Correct\n\nPart F Which of these diagrams may possibly correspond to the situation at point C on the motion diagram? Type, in increasing order, the numbers corresponding to the correct diagrams. Do not use commas. For instance, if you think that only diagrams 3 and 4 are correct, type 34. 24 Correct\n\nPart G Which of these diagrams correspond to a situation where the moving object (not necessarily the one shown in the motion diagram) is changing its velocity? Type, in increasing order, the numbers corresponding to the correct diagrams. Do not use commas. For instance, if you think that only diagrams 3 and 4 are correct, type 34. 12345 Correct\n\nConsider the following situations: A. A car is moving along a straight road at a constant speed. B. A car is moving along a straight road while slowing down. C. A car is moving along a straight road while speeding up. D. A hockey puck slides along a smooth (i.e., frictionless) icy surface. E. A hockey puck slides along a rough concrete surface. F. A cockroach is speeding up from rest. G. A rock is thrown horizontally; air resistance is negligible. H. A rock is thrown horizontally; air resistance is substantial. I.\n\nA rock is dropped vertically; air resistance is negligible.\n\nJ.\n\nA rock is dropped vertically; air resistance is substantial.\n\nPart H Which of these situations describe the motion shown in the motion diagram at point A? Type the letters corresponding to all the right answers in alphabetical order. Do not use commas. For instance, if you think that only situations C and D are correct, type CD. AD Correct\n\nPart I Which of these situations describe the motion shown in the motion diagram at point B? Type the letters corresponding to all the right answersin alphabetical order. Do not use commas. For instance, if you think that only situations C and D are correct, type CD. BE Correct\n\nPart J Which of these situations describe the motion shown in the motion diagram at point C? Type the letters corresponding to all the right answers in alphabetical order. Do not use commas. For instance, if you think that only situations C and D are correct, type CD. G Correct\n\nA World-Class Sprinter Worldclass sprinters can accelerate out of the starting blocks with an acceleration that is nearly horizontal and has magnitude\n\n.\n\nHarvaran Ghai Part A How much horizontal force must a sprinter of mass 54.0 acceleration? Express your answer in newtons.\n\nexert on the starting blocks to produce this\n\n=810 Correct\n\nPart B Which body exerts the force that propels the sprinter, the blocks or the sprinter? the blocks the sprinter Correct\n\nTo start moving forward, sprinters push backward on the starting blocks with their feet. As a reaction, the blocks push forward on their feet with a force of the same magnitude. This external force accelerates the sprinter forward.\n\nBlock on an Incline A block lies on a plane raised an angle from the horizontal. Three forces act upon the block: gravity; , the normal force; and block from sliding .\n\n, the force of friction. The coefficient of friction is large enough to prevent the\n\nPart A Consider coordinate system a, with the x axis along the plane. Which forces lie along the axes? only only only and and and and Correct\n\n, the force of\n\nand\n\nPart B Which forces lie along the axes of the coordinate system b, in which the y axis is vertical?\n\nHarvaran Ghai\n\nonly only only and and and and Correct\n\nand\n\nNow you are going to ignore the general rule (actually, a strong suggestion) that you should pick the coordinate system with the most vectors, especially unknown ones, along the coordinate axes. You will find the normal force, , using vertical coordinate system b. In these coordinates you will find the magnitude and y equations, each multiplied by a trigonometric function.\n\nappearing in both the x\n\nPart C Because the block is not moving, the sum of the y components of the forces acting on the block must be zero. Find an expression for the sum of the y components of the forces acting on the block, using coordinate system b. Express your answer in terms of some or all of the variables\n\n, and .\n\nCorrect\n\nPart D Because the block is not moving, the sum of the x components of the forces acting on the block must be zero. Find an expression for the sum of the x components of the forces acting on the block, using coordinate system b. Express your answer in terms of some or all of the variables\n\n, and .\n\nCorrect\n\nPart E To find the magnitude of the normal force, you must express\n\nin terms of\n\nequations you found in the two previous parts, find an expression for = Correct\n\nsince\n\ninvolving\n\nis an unknown. Using the and but not\n\n.\n\nCongratulations on working this through. Now realize that in coordinate system a, which is aligned with the plane, the ycoordinate equation is for\n\n, which leads immediately to the result obtained here\n\nCONCLUSION: A thoughtful examination of which coordinate system to choose can save a lot of algebra.\n\nHanging Chandelier A chandelier with mass is attached to the ceiling of a large concert hall by two cables. Because the ceiling is covered with intricate architectural decorations (not indicated in the figure, which uses a humbler depiction), the workers who hung the chandelier couldn't attach the cables to the ceiling directly above the chandelier. Instead, they attached the cables to the ceiling near the walls. Cable 1 has tension Cable 2 has tension\n\nand makes an angle of with the ceiling.\n\nand makes an angle of with the ceiling.\n\nHarvaran Ghai Part A Find an expression for\n\n, the tension in cable 1, that does not depend on\n\nExpress your answer in terms of some or all of the variables acceleration due to gravity . = Correct\n\n.\n\n, , and , as well as the magnitude of the\n\nProblem 5.11\n\nHarvaran Ghai\n\nPart A An astronaut's weight on earth is 805 . What is his weight on Mars, where 309 N Correct\n\nProblem 5.12 A woman has a mass of\n\n.\n\nHarvaran Ghai\n\nPart A What is her weight on earth? 539 N Correct\n\nPart B What is her mass on the moon, where 55.0 kg Correct\n\nPart C What is her weight on the moon? 89.1 N Correct\n\nProblem 5.14 The figure shows the velocity graph of a\n\npassenger in an elevator.\n\nHarvaran Ghai Part A What is the passenger's apparent weight at 1040 N Correct\n\nPart B At t = ? 735 N Correct\n\nPart C At t = ? 585 N Correct\n\nPushing a Chair along the Floor A chair of weight 120 lies atop a horizontal floor; the floor is not frictionless. You push on the chair with a force of = 35.0 directed at an angle of 41.0 below the horizontal and the chair slides along the floor.\n\nHarvaran Ghai Part A Using Newton's laws, calculate , the magnitude of the normal force that the floor exerts on the chair. Express your answer in newtons. =143 Correct\n\nBoard Pulled Out from under a Box A small box of mass is sitting on a board of mass and length . The board rests on a frictionless horizontal surface. The coefficient of static friction between the board and the box is . The coefficient of kinetic friction between the board and the box is, as usual, less than . Throughout the problem, use for the magnitude of the acceleration due to gravity. In the hints, use magnitude of the friction force between the board and the box.\n\nfor the\n\nHarvaran Ghai\n\nPart A\n\nFind , the constant force with the least magnitude that must be applied to the board in order to pull the board out from under the the box (which will then fall off of the opposite end of the board). Express your answer in terms of some or all of the variables answer. = Correct\n\n, , and . Do not include\n\nin your\n\nFriction Force on a Dancer on a Drawbridge A dancer is standing on one leg on a drawbridge that is about to open. The coefficients of static and kinetic friction between the drawbridge and the dancer's foot are\n\nand\n\n, respectively. represents the normal force exerted on\n\nthe dancer by the bridge, and represents the gravitational force exerted on the dancer, as shown in the drawing . For all the questions, we can assume that the bridge is a perfectly flat surface and lacks the curvature characteristic of most bridges.\n\nHarvaran Ghai Part A Before the drawbridge starts to open, it is perfectly level with the ground. The dancer is standing still on one leg. What is the x component of the friction force,\n\n?\n\nExpress your answer in terms of some or all of the variables ,\n\n, and/or\n\n=0 Correct\n\nThis shows a very important point. When you are not told that an object is slipping or on the verge of slipping, then the friction force is determined using Newton's laws of motion in conjunction with the observed motion and the other forces on the object. Under these circumstances the friction force is limited by or but is otherwise not necessarily related to or . Part B\n\nThe drawbridge then starts to rise and the dancer continues to stand on one leg. The drawbridge stops just at the\n\npoint where the dancer is on the verge of slipping. What is the magnitude Express your answer in terms of some or all of the variables , in your answer. = Correct\n\nof the frictional force now?\n\n, and/or\n\n. The angle should not appear\n\nPart C Then, because the bridge is old and poorly designed, it falls a little bit and then jerks. This causes the person to start to slide down the bridge at a constant speed. What is the magnitude\n\nof the frictional force now?\n\nExpress your answer in terms of some or all of the variables , in your answer.\n\n, and/or\n\n. The angle should not appear\n\n= Correct\n\nPart D The bridge starts to come back down again. The dancer stops sliding. However, again because of the age and design of the bridge it never makes it all the way down; rather it stops half a meter short. This half a meter corresponds to an angle\n\ndegree (see the diagram, which has the angle exaggerated). What is the force of friction\n\nExpress your answer in terms of some or all of the variables , , = Correct\n\n, and/or\n\nnow?\n\nSkydiving A sky diver of mass 80.0\n\n(including parachute) jumps off a plane and begins her descent.\n\nThroughout this problem use 9.80\n\nfor the magnitude of the acceleration due to gravity.\n\nHarvaran Ghai Part A At the beginning of her fall, does the sky diver have an acceleration? No; the sky diver falls at constant speed. Yes and her acceleration is directed upward. Yes and her acceleration is directed downward. Correct\n\nThis Error! Hyperlink reference not valid. shows the sky diver (not to scale) with her position, speed, and acceleration graphed as functions of time. You can see how her acceleration drops to zero over time, giving constant speed after a long time.\n\nPart B At some point during her free fall, the sky diver reaches her terminal speed. What is the magnitude of the drag force due to air resistance that acts on the sky diver when she has reached terminal speed? Express your answer in newtons. =784 Correct\n\nPart C For an object falling through air at a high speed , the drag force acting on it due to air resistance can be expressed as\n\n, where the coefficient\n\ndepends on the shape and size of the falling object and on the density of air. For a human\n\nbody, the numerical value for Using this value for\n\n, what is the terminal speed\n\n. of the sky diver?\n\nExpress you answer in meters per second. =56.0 Correct\n\nRecreational sky divers can control their terminal speed to some extent by changing their body posture. When oriented in a headfirst dive, a sky diver can reach speeds of about 54 meters per second (120 miles per hour). For maximum drag and stability, sky divers often will orient themselves \"bellyfirst.\" In this position, their terminal speed is typically around 45 meters per second (100 miles per hour). Part D When the sky diver descends to a certain height from the ground, she deploys her parachute to ensure a safe landing. (Usually the parachute is deployed when the sky diver reaches an altitude of about 900 after deploying the parachute, does the skydiver have a nonzero acceleration? No; the sky diver keeps falling at constant speed. Yes and her acceleration is directed downward. Yes and her acceleration is directed upward. Correct\n\nPart E\n\n3000 .) Immediately\n\nWhen the parachute is fully open, the effective drag coefficient of the sky diver plus parachute increases to 60.0 . What is the drag force Express your answer in newtons.\n\nacting on the sky diver immediately after she has opened the parachute?\n\n=1.88×105 Correct\n\nPart F What is the terminal speed of the sky diver when the parachute is opened? Express your answer in meters per second. =3.61 Correct\n\nA typical \"student\" parachute for recreational skydiving has a drag coefficient that gives a terminal speed for landing of about 2 meters per second (5 miles per hour). If this seems slow based on video or reallife sky divers you have seen, that may be because the sky divers you saw were using highperformance parachutes; these offer the sky divers more maneuverability in the air but increase the terminal speed up to 4 meters per second (10 miles per hour).\n\nAn Object Accelerating on a Ramp Learning Goal: Understand that the acceleration vector is in the direction of the change of the velocity vector. In one dimensional (straight line) motion, acceleration is accompanied by a change in speed, and the acceleration is always parallel (or antiparallel) to the velocity. When motion can occur in two dimensions (e.g. is confined to a tabletop but can lie anywhere in the xy plane), the definition of acceleration is\n\nin the limit\n\n.\n\nIn picturing this vector derivative you can think of the derivative of a vector as an instantaneous quantity by thinking of the velocity of the tip of the arrow as the vector changes in time. Alternatively, you can (for small approximate the acceleration as\n\n. Obviously the difference between in the same direction,\n\nwill be parallel to\n\nand\n\nis another vector that can lie in any direction. If it is longer but . On the other hand, if\n\nhas the same magnitude as\n\nbut is in a slightly different direction, then in both magnitude and direction, hence\n\nwill be perpendicular to . In general, can have any direction relative to\n\ncan differ from .\n\nThis problem contains several examples of this.Consider an object sliding on a frictionless ramp as depicted here. The object is already moving along the ramp toward position 2 when it is at position 1. The following questions concern the direction of the object's acceleration vector, . In this problem, you should find the direction of the acceleration vector by drawing the velocity vector at two points near to the position you are asked about. Note that since the object moves along the track, its velocity vector at a point will be tangent to the track at that point. The acceleration vector will point in the same direction as the vector difference of the two velocities. (This is a result of the equation\n\ngiven above.)\n\nHarvaran Ghai Part A Which direction best approximates the direction of when the object is at position 1? straight up downward to the left downward to the right straight down Correct\n\nPart B Which direction best approximates the direction of when the object is at position 2? straight up upward to the right\n\nstraight down downward to the left Correct\n\nEven though the acceleration is directed straight up, this does not mean that the object is moving straight up. Part C Which direction best approximates the direction of when the object is at position 3? upward to the right to the right straight down downward to the right Correct\n\nProblem 6.3 A particle's trajectory is described by\n\nand\n\n, where is in s.\n\nHarvaran Ghai Part A What is the particle's speed at t 2.00 m/s Correct\n\nPart B What is the particle's speed at = 4.50 ? 12.6 m/s Correct\n\nPart C What is the particle's direction of motion, measured from the xaxis, at 270 counterclockwise from the +x axis Correct\n\n0 ?\n\nPart D What is the particle's direction of motion, measured from the xaxis, at = 4.50 ? 11.4 counterclockwise from the +x axis Correct\n\nProjectile Motion Tutorial Learning Goal: Understand how to apply the equations for 1dimensional motion to the y and x directions separately in order to derive standard formulae for the range and height of a projectile.\n\nA projectile is fired from ground level at time , at an angle with respect to the horizontal. It has an initial speed . In this problem we are assuming that the ground is level.\n\nHarvaran Ghai Part A Find the time Express = Correct\n\nit takes the projectile to reach its maximum height.\n\nin terms of\n\n, , and (the magnitude of the acceleration due to gravity).\n\nPart B Find\n\n, the time at which the projectile hits the ground.\n\nExpress the time in terms of = Correct\n\n, , and .\n\nPart C Find\n\n, the maximum height attained by the projectile.\n\nExpress the maximum height in terms of = Correct\n\n, , and .\n\nPart D Find the total distance (often called the range) traveled in the x direction; in other words, find where the projectile lands. Express the range in terms of = Correct\n\n, , and .\n\nThe actual formula for\n\nis less important than how it is obtained:\n\n1. 2.\n\nConsider the x and y motion separately. Find the time of flight from the ymotion\n\n3.\n\nFind the xposition at the end of the flight this is the range.\n\nIf you remember these steps, you can deal with many variants of the basic problem, such as: a cannon on a hill that fires horizontally (i.e. the second half of the trajectory), a projectile that lands on a hill, or a projectile that must hit a moving target.\n\nHorizontal Cannon on a Cliff A cannonball is fired horizontally from the top of a cliff. The cannon is at height ball is fired with initial horizontal speed .\n\nabove ground level, and the\n\nHarvaran Ghai Part A Assume that the cannon is fired at\n\nand that the cannonball hits the ground at time\n\nposition of the cannonball at the time\n\n?\n\nExpress the y position of the cannonball in terms of answer. Correct\n\n=\n\n. What is the y\n\n. The quantities\n\nand should not appear in your\n\nPart B Given that the projectile lands a distance Express the initial speed in terms of = Correct\n\nfrom the cliff, as shown, find the initial speed of the projectile, , , and\n\n.\n\nPart C What is the y position of the cannonball when it is a distance Express the position of the cannonball in terms of = Correct\n\nonly.\n\nfrom the hill?\n\n.\n\nProblem 6.13 A boat takes 3.70\n\nto travel 15.0\n\ndown a river, then 5.30\n\nto return.\n\nHarvaran Ghai\n\nPart A How fast is the river flowing? 0.612 km/h Correct\n\nCrossing a River\n\nA swimmer wants to cross a river, from point A to point B. The distance\n\ndistance\n\n(from C to B) is 150\n\nswimmer makes an angle of the figure.\n\n(from A to C) is 200\n\n, and the speed of the current in the river is 5 (0.785\n\n, the\n\n. Suppose that the\n\n) with respect to the line from A to C, as indicated in\n\nHarvaran Ghai Part A To swim directly from A to B, what speed , relative to the water, should the swimmer have? Express the swimmer's speed numerically, to three significant figures, in units of kilometers per hour. =4.04 Correct\n\nAnother way to do this problem, without using any kinematics, would be to add the swimmer's and river's velocities vectorially, and set the angle that this vector makes with AC or the river bank equal to that which AB makes with the same.\n\nSpeed of a Bullet A bullet is shot through two cardboard disks attached a distance , as shown.\n\napart to a shaft turning with a rotational period\n\nHarvaran Ghai Part A Derive a formula for the bullet speed in terms of , , and a measured angle between the position of the hole in the first disk and that of the hole in the second. If required, use , not its numeric equivalent. Both of the holes lie at the same radial distance from the shaft. measures the angular displacement between the two holes; for instance, means that the holes are in a line and means that when one hole is up, the other is down. Assume that the bullet must travel through the set of disks within a single revolution. = Correct\n\nDirection of Acceleration of Pendulum Learning Goal: To understand that the direction of acceleration is in the direction of the change of the velocity, which is unrelated to the direction of the velocity. The pendulum shown makes a full swing from to . Ignore friction and assume that the string is massless. The eight labeled arrows represent directions to be referred to when answering the following questions.\n\nHarvaran Ghai Part A Which of the following is a true statement about the acceleration of the pendulum bob, . is equal to the acceleration due to gravity. is equal to the instantaneous rate of change in velocity. is perpendicular to the bob's trajectory. is tangent to the bob's trajectory. Correct\n\nPart B What is the direction of when the pendulum is at position 1? Enter the letter of the arrow parallel to . H Correct\n\nPart C What is the direction of at the moment the pendulum passes position 2? Enter the letter of the arrow that best approximates the direction of . C Correct\n\nWe know that for the object to be traveling in a circle, some component of its acceleration must be pointing radially inward. Part D What is the direction of when the pendulum reaches position 3? Give the letter of the arrow that best approximates the direction of . F Correct\n\nPart E As the pendulum approaches or recedes from which position(s) is the acceleration vector almost parallel to the velocity vector . position 2 only positions 1 and 2 positions 2 and 3 positions 1 and 3 Correct\n\nBanked Frictionless Curve, and Flat Curve with Friction A car of mass traveling at speed enters a banked turn covered with ice. The road is banked at an angle , and there is no friction between the road and the car's tires. A cross section of the curve is shown in the diagram.\n\nHarvaran Ghai Part A What is the radius of the turn (assuming the car continues in uniform circular motion around the turn)? Express your answer in terms of some or all of the variables acceleration due to gravity . = Correct\n\n, , , as well as the magnitude of the\n\nPart B Now, suppose that the curve is level ( ) and that the ice has melted, so that there is a coefficient of static friction between the road and the car's tires. What is , the minimum value of the coefficient of static friction between the tires and the road required to prevent the car from slipping? Assume that the car's speed is still and that the radius of the curve is given by . Express your answer in terms of some or all of the variables , , acceleration due to gravity . = Correct\n\n, as well as the magnitude of the\n\nAt the Test Track You want to test the grip of the tires on your new race car. You decide to take the race car to a small test track to experimentally determine the coefficient of friction. The racetrack consists of a flat, circular road with a radius of 45\n\n. Error! Hyperlink reference not valid. shows the result of driving the car around the track at various speeds.\n\nPart A What is , the coefficient of static friction between the tires and the track? Express your answer to two significant figures. =0.95 Correct\n\nHarvaran Ghai\n\nConical Pendulum I\n\nA bob of mass\n\nis suspended from a fixed point with a massless string of length (i.e., it is a pendulum). You\n\nare to investigate the motion in which the string moves in a cone with halfangle .\n\nHarvaran Ghai Part A What tangential speed, , must the bob have so that it moves in a horizontal circle with the string always making an angle from the vertical? Express your answer in terms of some or all of the variables gravity . = Correct\n\n, , and , as well as the acceleration due to\n\nPart B How long does it take the bob to make one full revolution (one complete trip around the circle)? Express your answer in terms of some or all of the variables gravity .\n\n, , and , as well as the acceleration due to\n\nCorrect\n\nProblem 7.18 Part A What is the acceleration due to gravity of the sun at the distance of the earth's orbit?\n\nHarvaran Ghai\n\n6.00×10−3 Correct\n\nProblem 7.19 The passengers in a roller coaster car feel 50% heavier than their true weight as the car goes through a dip with a 40.0 radius of curvature.\n\nHarvaran Ghai Part A What is the car's speed at the bottom of the dip? 14.0 Correct\n\nProblem 7.23 A car speeds up as it turns from traveling due south to heading due east. When exactly halfway around the curve, the car's acceleration is 3.40\n\n, 40.0 north of east.\n\nHarvaran Ghai Part A\n\nWhat is the radial component of the acceleration at that point? 3.39 Correct\n\nPart B What is the tangential component of the acceleration at that point? 0.296 Correct\n\nA Satellite in Orbit A satellite used in a cellular telephone network has a mass of 1850 700\n\nand is in a circular orbit at a height of\n\nabove the surface of the earth.\n\nHarvaran Ghai Part A What is the gravitational force\n\non the satellite?\n\nTake the gravitational constant to be = 6.67×10−11 and the radius of the Earth to be = 6.38×106 .\n\n, the mass of the earth to be\n\nExpress your answer in newtons. =1.47×104 Correct\n\nPart B What fraction is this of the satellite's weight at the surface of the earth? Take the freefall acceleration at the surface of the earth to be = 9.80 0.811 Correct\n\n.\n\n= 5.97×1024\n\nAlthough it is easy to find the weight of the satellite using the constant acceleration due to gravity, it is instructional to consider the weight calculated using the law of gravitation:\n\n. Dividing the gravitational force on\n\nthe satellite by , we find that the ratio of the forces due to the earth's gravity is simply the square of the ratio of the earth's radius to the sum of the earth's radius and the height of the orbit of the satellite above the earth, . This will also be the fraction of the weight of, say, an astronaut in an orbit at the same altitude. Notice that an astronaut's weight is never zero. When people speak of \"weightlessness\" in space, what they really mean is \"free fall.\"\n\nGravitational Acceleration inside a Planet Consider a spherical planet of uniform density . The distance from the planet's center to its surface (i.e., the planet's radius) is . An object is located a distance from the center of the planet, where located inside of the planet.)\n\n. (The object is\n\nHarvaran Ghai\n\nPart A Find an expression for the magnitude of the acceleration due to gravity,\n\n, inside the planet.\n\nExpress the acceleration due to gravity in terms of , , , and , the universal gravitational constant. = Correct\n\nPart B Rewrite your result for function of R.\n\nin terms of\n\nExpress your answer in terms of = Correct\n\n, the gravitational acceleration at the surface of the planet, times a\n\n, , and\n\n.\n\nNotice that increases linearly with , rather than being proportional to center of the planet, as required by symmetry.\n\n. This assures that it is zero at the\n\nPart C Find a numerical value for\n\n, the average density of the earth in kilograms per cubic meter. Use\n\nthe radius of the earth, Calculate your answer to four significant digits.\n\n=5497\n\n, and a value of at the surface of\n\n.\n\nfor\n\nCorrect\n\nNewton's 3rd Law Discussed Learning Goal: To understand Newton's 3rd law, which states that a physical interaction always generates a pair of forces on the two interacting bodies. In Principia, Newton wrote: To every action there is always opposed an equal reaction: or, the mutual actions of two bodies upon each other are always equal, and directed to contrary parts. (translation by Cajori) The phrase after the colon (often omitted from textbooks) makes it clear that this is a statement about the nature of force. The central idea is that physical interactions (e.g., due to gravity, bodies touching, or electric forces) cause forces to arise between pairs of bodies. Each pairwise interaction produces a pair of opposite forces, one acting on each body. In summary, each physical interaction between two bodies generates a pair of forces. Whatever the physical cause of the interaction, the force on body A from body B is equal in magnitude and opposite in direction to the force on body B from body A. Incidentally, Newton states that the word \"action\" denotes both (a) the force due to an interaction and (b) the changes in momentum that it imparts to the two interacting bodies. If you haven't learned about momentum, don't worry; for now this is just a statement about the origin of forces. Mark each of the following statements as true or false. If a statement refers to \"two bodies\" interacting via some force, you are not to assume that these two bodies have the same mass.\n\nPart A Every force has one and only one 3rd law pair force. true false Correct\n\nPart B The two forces in each pair act in opposite directions. true false Correct\n\nPart C\n\nHarvaran Ghai\n\nThe two forces in each pair can either both act on the same body or they can act on different bodies. true false Correct\n\nPart D The two forces in each pair may have different physical origins (for instance, one of the forces could be due to gravity, and its pair force could be due to friction or electric charge). true false Correct\n\nPart E The two forces of a 3rd law pair always act on different bodies. true false Correct\n\nPart F Given that two bodies interact via some force, the accelerations of these two bodies have the same magnitude but opposite directions. (Assume no other forces act on either body.) true false Correct\n\nNewton's 3rd law can be summarixed as follows: A physical interaction (e.g., gravity) operates between two interacting bodies and generates a pair of opposite forces, one on each body. It offers you a way to test for real forces (i.e., those that belong on the force side of )there should be a 3rd law pair force operating on some other body for each real force that acts on the body whose acceleration is under consideration. Part G According to Newton's 3rd law, the force on the (smaller) moon due to the (larger) earth is greater in magnitude and antiparallel to the force on the earth due to the moon. greater in magnitude and parallel to the force on the earth due to the moon. equal in magnitude but antiparallel to the force on the earth due to the moon. equal in magnitude and parallel to the force on the earth due to the moon. smaller in magnitude and antiparallel to the force on the earth due to the moon.\n\nsmaller in magnitude and parallel to the force on the earth due to the moon. Correct\n\nA Book on a Table A book weighing 5 N rests on top of a table.\n\nHarvaran Ghai Part A A downward force of magnitude 5 N is exerted on the book by the force of the table gravity . inertia Correct\n\nPart B An upward force of magnitude _____ is exerted on the _____ by the table. 5 N / book Correct\n\nPart C Do the downward force in Part A and the upward force in Part B constitute a 3rd law pair? yes no\n\nCorrect\n\nPart D The reaction to the force in Part A is a force of magnitude _____, exerted on the _____ by the _____. Its direction is _____ . 5 N / earth / book / upward Correct\n\nPart E The reaction to the force in Part B is a force of magnitude _____, exerted on the _____ by the _____. Its direction is _____. 5 N / table / book / downward Correct\n\nPart F Which of Newton's laws dictates that the forces in Parts A and B are equal and opposite? Newton's 1st or 2nd law Newton's 3rd law Correct\n\nSince the book is at rest, the net force on it must be zero (1st or 2nd law). This means that the force exerted on it by the earth must be equal and opposite to the force exerted on it by the table. Part G Which of Newton's laws dictates that the forces in Parts B and E are equal and opposite? Newton's 1st or 2nd law Newton's 3rd law Correct\n\nBlock on an Incline Adjacent to a Wall A wedge with an inclination of angle rests next to a wall. A block of mass is sliding down the plane, as shown. There is no friction between the wedge and the block or between the wedge and the horizontal surface.\n\nHarvaran Ghai\n\nPart A Find the magnitude, Express = Correct\n\n, of the sum of all forces acting on the block.\n\nin terms of and\n\n, along with any necessary constants.\n\nPart B Find the magnitude, Express = Correct\n\n, of the force that the wall exerts on the wedge.\n\nin terms of and\n\n, along with any necessary constants.\n\nYour answer to Part B could be expressed as either as gets very small or as approaches 90 degrees (\n\nor\n\n. In either form, we see that\n\nradians), the contact force between the wall and the wedge\n\ngoes to zero. This is what we should expect; in the first limit ( small), the block is accelerating very slowly, and all horizontal forces are small. In the second limit ( about 90 degrees), the block simply falls vertically and exerts no horizontal force on the wedge.\n\nPSS 8.1: Dashing up the Slope\n\nLearning Goal: To practice ProblemSolving Strategy 8.1 for problems involving the dynamics of an interacting systems of objects. A girl of mass is walking up a slippery slope while pulling a sled of unknown mass; the slope makes an angle with the horizontal. The coefficient of static friction between the girl's boots and the slope is ; the friction between the sled and the slope is negligible. It turns out that the girl can pull the sled up the slope with acceleration up to without slipping down the slope. Find the mass of the sled . Assume that the rope connecting the girl and the sled is kept parallel to the slope at all times.\n\nHarvaran Ghai MODEL:\n\nIdentify which objects are systems and which are part of the environment. Make simplifying assumptions.\n\nVISUALIZE:\n\nPictorial representation: Show important points in the motion with a sketch. You may want to give each system a separate coordinate system. Define symbols and identify what you are trying to find. Include acceleration constraints as part of the pictorial model. Physical representation: Identify all forces acting on each system and all actionreaction pairs. Draw a separate freebody diagram for each system. Connect the force vectors of actionreaction pairs with dotted\n\nlines. Use subscript labels to distinguish forces, such as and , that act independently on more than one system. SOLVE:\n\nUse Newton's 2nd and 3rd laws:\n\n \n\nWrite the equations of Newton's 2nd law for each system using the force information from the freebody diagrams. Equate the magnitudes of actionreaction pairs.\n\nInclude the acceleration constraints, the friction model, and other quantitative information relevant to the problem.\n\nSolve for the acceleration, then use kinematics to find velocites and positions.\n\nASSESS:\n\nCheck that your result has the correct units, is reasonable, and answers the question.\n\nModel Start by making simplifying assumptions appropriate for the situation. Part A Which of the following objects qualify as systems in this problem? A. the slope B. the girl\n\nC. the earth D. the sled E. the air List alphabetically all the letters corresponding to the systems. Do not use commas. For instance, if you think the slope and sled qualify as systems, type AD. BD Correct\n\nThe slope, the earth, and the air all qualify as part of the environment. They each exert external forces on the two systems (the sled and the girl). Unlike for the two systems, however, it is not important to keep track of all the forces acting on elements of the environment. Part B Which of the following simplifying assumptions are reasonable? A. The air resistance acting on the girl is negligible. B. The air resistance acting on the girl equals the force of friction acting on her. C. The air resistance acting on the sled is negligible. D. The normal force acting on the sled is negligible. E. The weight of the sled is a constant. F. The weight of the sled increases as the sled accelerates. List alphabetically all the letters corresponding to reasonable assumptions. Do not use commas. For instance, if you think A and B are reasonable, type AB. ACE Correct\n\nPart C Which of the following simplifying assumptions are reasonable? A. The rope connecting the sled and the girl is massless. B. The rope connecting the sled and the girl is unstretchable. C. The tension in the rope connecting the sled and the girl is zero. D. The sled has the same acceleration as the girl. E. The sled has greater acceleration than the girl. F. The sled has smaller acceleration than the girl. List alphabetically all the letters corresponding to reasonable assumptions. Do not use commas. For instance, if you think A and B are reasonable, type AB. ABD Correct\n\nVisualize Now draw a sketch that includes the freebody diagrams for each system and the appropriate coordinate system. Use your sketch to answer the following questions. For all questions, assume that the slope angles downhill to the left:\n\n.\n\nPart D Which freebody diagram for the girl is correct? Note that the forces are not labeled; however, they should be labeled on your diagram. You are looking for the correct number of forces in the correct directions. Don't worry about relative magnitudes at this point.\n\na b c d e Correct\n\nPart E Which freebody diagram for the sled is correct? Note that the forces are not labeled; however, they should be labeled on your diagram. You are looking for the correct number of forces in the correct directions. Don't worry about relative magnitudes at this point.\n\na b c d e Correct\n\nPart F Which of these coordinate systems is most convenient for solving this problem? (The same coordinate system is appropriate for both the sled and the girl.)\n\na b c d\n\nCorrect\n\nPart G You should have identified the pairs of actionreaction forces on your freebody diagrams. Which of the following pairs of forces form actionreaction pairs, according to Newton's 3rd law? A. The weight of the girl and the normal force on the girl B. The weight of the sled and the normal force on the sled C. The weight of the girl and the weight of the sled D. The force of friction on the girl and the tension of the string E. The weight of the sled and the tension of the string F. The weight of the sled and the gravitational force applied by the sled to the earth List alphabetically all the letters corresponding to the actionreaction pairs in this problem. Do not use commas. For instance, if you think A and B are both valid actionreaction pairs, type AB. F Correct\n\nAn actionreaction pair between objects A and B is always a pair of forces and . In our situation, if we assume that the girl and the sled act directly on each other (a reasonable assumption since the mass of the string is negligible), then the forces \"girl on the sled\" and \"sled on the girl\" would form an actionreaction pair. Each of the other forces mentioned in this question does have a reaction force, of course. However, the objects on which such reaction forces are acting are part of the environment: For instance, the reaction force to the weight of the girl is the gravitational force applied by the girl to the earth, which is part of the environment.\n\nThere are many variations on how you might draw good pictorial and physical representations for this problem.\n\nHere is one example.\n\nSolve Now use the information and the insights that you have accumulated to construct the necessary mathematical expressions and to derive the solution. Part H Find the mass of the sled . Express the sled's mass in terms of the given quantities and , the magnitude of the acceleration due to gravity. = Correct\n\nAssess When you work on a problem on your own, without the computerprovided feedback, only you can assess whether your answer seems right. The following questions will help you practice the skills necessary for such an assessment. Part I Intuitively, what would happen if there were very little (or no) static friction between the girl and the slope?\n\nVery little force would be required to pull a heavy sled up the slope. The girl would slip down the slope and never be able to pull the sled up. The girl would be able to pull the sled up the slope with a very large acceleration. The girl would be able to pull the sled up only at constant velocity. Correct\n\nIf is very close to or equal to zero, the formula you derived in the Solve section would give a negative value for the mass of the sled. Since the mass of the sled must be positive, such an answer simply means that the formula is not applicable: The girl would not be able to pull the sled up the slope, no matter how small the mass of the sled. Part J Intuitively, what would happen if the \"slope\" were horizontal and the mass of the sled were equal to the mass of the girl? The girl would be able to pull the sled with acceleration greater than . The girl would slip along the surface and not be able to pull the sled. The girl would be able to pull the sled with up to some maximum acceleration that depends on the friction between her and the slope. The girl would be able to pull the sled only at constant velocity. Correct\n\nIf the slope is horizontal,\n\nand\n\n. Your formula then becomes\n\n,\n\nand, if\n\n, it follows that the maximum acceleration is\n\nPart K Which of the following expressions have the dimensions of mass? A. B.\n\nC.\n\nD.\n\n.\n\nE. List alphabetically all the letters corresponding to expressions with the correct dimensions. Do not use commas. For instance, if you think A and B have the correct dimensions, type AB. ACE Correct\n\nNote that trigonometric functions and the coefficient of static friction are dimensionless: They do not affect the dimension or units of the final answer.\n\nPulling Three Blocks Three identical blocks connected by ideal strings are being pulled along a horizontal frictionless surface by a horizontal force . block has mass .\n\nThe magnitude of the tension in the string between blocks B and C is . Assume that each\n\nHarvaran Ghai\n\nPart A What is the magnitude of the force? Express the magnitude of the force in terms of . = Correct\n\nKinetic Friction in a Block-and-Pulley System Consider the system shown in the figure . Block A has weight and block B has weight . Once block B is set into downward motion, it descends at a constant speed. Assume that the mass and friction of the pulley are negligible.\n\nHarvaran Ghai Part A Calculate the coefficient of kinetic friction between block A and the table top. = Correct\n\nPart B A cat, also of weight , falls asleep on top of block A. If block B is now set into downward motion, what is the magnitude of its acceleration? = Correct\n\nTwo Masses, a Pulley, and an Inclined Plane Block 1, of mass , is connected over an ideal (massless and frictionless) pulley to block 2, of mass , as shown. Assume that the blocks accelerate as shown with an acceleration of magnitude and that the coefficient of kinetic friction between block 2 and the plane is .\n\nHarvaran Ghai\n\nPart A Find the ratio of the masses\n\n.\n\nExpress your answer in terms of some or all of the variables , , and , as well as the magnitude of the acceleration due to gravity . = Correct\n\nThe Impulse-Momentum Theorem Learning Goal: To learn about the impulsemomentum theorem and its applications in some common cases. Using the concept of momentum, Newton's second law can be rewritten as\n\n, (1)\n\nwhere\n\nis the net force\n\nacting on the object, and\n\nis the rate at which the object's momentum is changing.\n\nIf the object is observed during an interval of time between times and , then integration of both sides of equation (1) gives\n\n. (2)\n\nThe right side of equation (2) is simply the change in the object's momentum\n\n. The left side is called the\n\nimpulse of the net force and is denoted by . Then equation (2) can be rewritten as\n\n. This equation is known as the impulsemomentum theorem. It states that the change in an object's momentum is equal to the impulse of the net force acting on the object. In the case of a constant net force direction of motion, the impulsemomentum theorem can be written as\n\nacting along the\n\n. (3) Here , , and are the components of the corresponding vector quantities along the chosen coordinate axis. If the motion in question is twodimensional, it is often useful to apply equation (3) to the x and y components of motion separately.\n\nHarvaran Ghai The following questions will help you learn to apply the impulsemomentum theorem to the cases of constant and varying force acting along the direction of motion. First, let us consider a particle of mass moving along the x axis. The net force is acting on the particle along the x axis. is a constant force. Part A The particle starts from rest at that\n\n. What is the magnitude of the momentum of the particle at time ? Assume\n\n.\n\nExpress your answer in terms of any or all of = Correct\n\nPart B\n\n, , and .\n\nThe particle starts from rest at\n\n. What is the magnitude of the velocity of the particle at time ? Assume that\n\n. Express your answer in terms of any or all of = Correct\n\n, , and .\n\nPart C The particle has momentum of magnitude seconds later?\n\nat a certain instant. What is\n\nExpress your answer in terms of any or all of = Correct\n\n, , and\n\n, the magnitude of its momentum\n\n.\n\nPart D The particle has momentum of magnitude seconds later?\n\nat a certain instant. What is\n\nExpress your answer in terms of any or all of = Correct\n\n, , and\n\n, the magnitude of its velocity\n\n.\n\nLet us now consider several twodimensional situations. A particle of mass\n\nis moving in the positive x direction at speed . After a certain constant force is applied to the\n\nparticle, it moves in the positive y direction at speed\n\n.\n\nPart E Find the magnitude of the impulse delivered to the particle. Express your answer in terms of and . Use three significant figures in the numerical coefficient. = Correct\n\nPart F Which of the vectors below best represents the direction of the impulse vector ?\n\n1 2 3 4 5 6 7 8 Correct\n\nPart G\n\nWhat is the angle between the positive y axis and the vector as shown in the figure?\n\n26.6 degrees 30 degrees 60 degrees 63.4 degrees\n\nCorrect\n\nPart H If the magnitude of the net force acting on the particle is , how long does it take the particle to acquire its final velocity,\n\nin the positive y direction?\n\nExpress your answer in terms of = Correct\n\n, , and . If you use a numerical coefficient, use three significant figures.\n\nSo far, we have considered only the situation in which the magnitude of the net force acting on the particle was either irrelevant to the solution or was considered constant. Let us now consider an example of a varying force acting on a particle. Part I A particle of mass\n\nkilograms is at rest at\n\nseconds. A varying force\n\nis acting on the particle between speed of the particle at\n\nseconds.\n\nseconds and\n\nseconds. Find the\n\nExpress your answer in meters per second to three significant figures. =43 Correct\n\nProblem 9.11 A 500 airtrack glider collides with a spring at one end of the track. The figure and the force exerted on the glider by the spring.\n\nshows the glider's velocity\n\nHarvaran Ghai Part A How long is the glider in contact with the spring? 0.167 s Correct\n\nFilling the Boat A boat of mass 250\n\nis coasting, with its engine in neutral, through the water at speed 3.00\n\nrain. The rain is falling vertically, and it accumulates in the boat at the rate of 10.0\n\nwhen it starts to\n\n.\n\nHarvaran Ghai Part A What is the speed of the boat after time 2.00 has passed? Assume that the water resistance is negligible. Express your answer in meters per second. 2.78 Correct\n\nPart B Now assume that the boat is subject to a drag force due to water resistance. Is the component of the total momentum of the system parallel to the direction of motion still conserved? yes no Correct\n\nThe boat is subject to an external force, the drag force due to water resistance, and therefore its momentum is not conserved. Part C The drag is proportional to the square of the speed of the boat, in the form . What is the acceleration of the boat just after the rain starts? Take the positive axis along the direction of motion. Express your answer in meters per second per second. −1.80×10−2 Correct\n\nPSS 9.1: Tools of the Trade Learning Goal: To practice ProblemSolving Strategy 9.1 for problems involving conservation of momentum. An astronaut performs maintenance work outside her spaceship when the tether connecting her to the spaceship breaks. The astronaut finds herself at rest relative to the spaceship, at a distance from it. To get back to the ship, she decides to sacrifice her favorite wrench and hurls it directly away from the spaceship at a speed relative to the spaceship. What is the distance between the spaceship and the wrench by the time the astronaut reaches the spaceship? The mass of the astronaut is\n\n; the mass of the wrench is\n\n.\n\nHarvaran Ghai MODEL:\n\nClearly define the system.\n\nIf possible, choose a system that is isolated ( ) or within which the interactions are sufficiently short and intense that you can ignore external forces for the duration of the interaction (the impulse approximation). Momentum is conserved.\n\nIf it is not possible to choose an isolated system, try to divide the problem into parts such that momentum is conserved during one segment of the motion. Other segments of the motion can be analyzed using Newton's laws or, as you'll learn in Chapters 10 and 11, conservation of energy.\n\nDraw a beforeandafter pictorial representation. Define symbols that will be used in the problem, list known values, and identify what you are trying to find. VISUALIZE:\n\nThe mathematical representation is based on the law of conservation of momentum: form, this is SOLVE:\n\n. In component\n\n,\n\nASSESS:\n\nCheck if your result has the correct units, is reasonable, and answers the question.\n\nModel We start by choosing the objects that would make up the system. In this case, it is possible to identify the system that is isolated. Part A In addition to the astronaut, which of the following are components of the system that should be defined to solve the problem? A. the spaceship B. the wrench C. the earth Enter the letter(s) of the correct answer(s) in alphabetical order. Do not use commas. For example, if you think the system consists of all the objects listed, enter ABC. B Correct\n\nPart B Which of the following reasons best explains why the astronaut + wrench can be considered an isolated system? The mass of the wrench is much smaller than that of the astronaut. The force that the astronaut exerts on the wrench is very small. The force that the astronaut exerts on the wrench is very large. The force that the spaceship exerts on the wrench is very small. The force that the spaceship exerts on the wrench is very large. Correct\n\nVisualize Now draw a beforeandafter pictorial representation including all the elements listed in the problemsolving strategy. Be sure that your sketch is clear and includes all necessary symbols, both known and unknown. By the time the astronaut reaches the spaceship, the wrench will have covered a certain distance; on your pictorial representation, label this distance\n\n.\n\nPart C After the wrench is thrown, the astronaut and the wrench move in opposite directions. in the same direction. in perpendicular directions. Correct\n\n, and\n\nis correct?\n\nCorrect\n\nHere is an example of what a good beforeandafter pictorial representation might look like for this problem.\n\nSolve Now use the information and the insights that you have accumulated to construct the necessary mathematical expressions and to derive the solution. Part E Find the final distance between the spaceship and the wrench. Express the distance in terms of the given variables. You may or may not use all of them.\n\n= Correct\n\nAssess When you work on a problem on your own, without the computerprovided feedback, only you can assess whether your answer seems right. The following questions will help you practice the skills necessary for such an assessment. Part F Intuitively, which of the following statements are correct? A. For realistic values of the quantities involved, it is possible that . B. If the astronaut threw a space pen instead of a wrench, the pen would travel further than the wrench would in the time it takes the astronaut to reach the ship. (Assume the space pen weighs less than the wrench). C. If the astronaut were more massive, the wrench would travel further in the time it takes the astronaut to reach the ship. Type the letters corresponding to the correct answers. Do not use commas. For instance, if you think that only expressions C and D have the units of distance, type CD. BC Correct\n\ncould only be zero if . As you can see from your answer, this would only happen if the mass of the astronaut were zero, which is obviously unrealistic. Part G Which of the following mathematical expressions have the units of distance, where A. B.\n\nC.\n\nD.\n\nE.\n\nF.\n\nand\n\nare distances?\n\nType the letters corresponding to the correct answers. Do not use commas. For instance, if you think that only expressions C and D have the units of distance, type CD. ABE Correct\n\nColliding Cars In this problem we will consider the collision of two cars initially moving at right angles. We assume that after the collision the cars stick together and travel off as a single unit. The collision is therefore completely inelastic. Two cars of masses and collide at an intersection. Before the collision, car 1 was traveling eastward at a speed of , and car 2 was traveling northward at a speed of . After the collision, the two cars stick together and travel off in the direction shown.\n\nHarvaran Ghai Part A First, find the magnitude of , that is, the speed of the twocar unit after the collision. Express in terms of , , and the cars' initial speeds and . = Correct\n\nPart B Find the tangent of the angle .\n\nExpress your answer in terms of the momenta of the two cars, = Correct\n\nand\n\n.\n\nPart C Suppose that after the collision, ; in other words, is The magnitudes of the momenta of the cars were equal. The masses of the cars were equal.\n\n. This means that before the collision:\n\nThe velocities of the cars were equal. Correct\n\nCollision at an Angle Two cars, both of mass , collide and stick together. Prior to the collision, one car had been traveling north at speed\n\n, while the second was traveling at speed at an angle south of east (as indicated in the figure). After the\n\ncollision, the twocar system travels at speed\n\nat an angle east of north.\n\nHarvaran Ghai Part A Find the speed\n\nof the joined cars after the collision.\n\nExpress your answer in terms of and .\n\n=\n\nCorrect\n\nPart B What is the angle with respect to north made by the velocity vector of the two cars after the collision? Express your answer in terms of . Your answer should contain an inverse trigonometric function. = Correct\n\nA Girl on a Trampoline A girl of mass second. At height For this problem, use\n\nkilograms springs from a trampoline with an initial upward velocity of meters above the trampoline, the girl grabs a box of mass\n\nkilograms.\n\nmeters per second per second for the magnitude of the acceleration due to gravity.\n\nPart A What is the speed of the girl immediately before she grabs the box? Express your answer numerically in meters per second. =4.98 Correct\n\nmeters per\n\nHarvaran Ghai\n\nPart B What is the speed of the girl immediately after she grabs the box? Express your answer numerically in meters per second. =3.98 Correct\n\nPart C Is this \"collision\" elastic or inelastic? elastic inelastic Correct\n\nIn inelastic collisions, some of the system's kinetic energy is lost. In this case the kinetic energy lost is converted to heat energy in the girl's muscles as she grabs the box, and sound energy. Part D What is the maximum height that the girl (with box) reaches? Measure trampoline. Express your answer numerically in meters.\n\nwith respect to the top of the\n\n=2.81 Correct\n\nCircling Ball A ball of mass is attached to a string of length . It is being swung in a vertical circle with enough speed so that the string remains taut throughout the ball's motion. Assume that the ball travels freely in this vertical circle with negligible loss of total mechanical energy. To avoid confusion, take the upward direction to be positive throughout the problem. At the top and bottom of the vertical circle, label the ball's speeds and , and label the corresponding tensions in the string\n\nand\n\nand\n\nhave magnitudes\n\nand\n\n.\n\nHarvaran Ghai\n\nPart A\n\nFind , the difference between the magnitude of the tension in the string at the bottom relative to that at the top of the circle. Express the difference in tension in terms of and . The quantities and should not appear in your final answer. = Correct\n\nThe method outlined in the hints is really the only practical way to do this problem. If done properly, finding the difference between the tensions,\n\n, can be accomplished fairly simply and elegantly.\n\nBungee Jumping Kate, a bungee jumper, wants to jump off the edge of a bridge that spans a river below. Kate has a mass\n\n, and the\n\nsurface of the bridge is a height above the water. The bungee cord, which has length when unstretched, will first straighten and then stretch as Kate falls. Assume the following: \n\nThe bungee cord behaves as an ideal spring once it begins to stretch, with spring constant . Kate doesn't actually jump but simply steps off the edge of the bridge and falls straight downward.\n\nKate's height is negligible compared to the length of the bungee cord. Hence, she can be treated as a point particle.\n\nUse for the magnitude of the acceleration due to gravity.\n\nHarvaran Ghai Part A How far below the bridge will Kate eventually be hanging, once she stops oscillating and comes finally to rest? Assume that she doesn't touch the water. Express the distance in terms of quantities given in the problem introduction. = Correct\n\nPart B If Kate just touches the surface of the river on her first downward trip (i.e., before the first bounce), what is the spring constant ? Ignore all dissipative forces. Express in terms of , , = Correct\n\n, and .\n\nDancing Balls Four balls, each of mass , are connected by four identical relaxed springs with spring constant . The balls are simultaneously given equal initial speeds directed away from the center of symmetry of the system.\n\nHarvaran Ghai\n\nPart A As the balls reach their maximum displacement, their kinetic energy reaches __________. a maximum zero neither a maximum nor zero Correct\n\nPart B Use geometry to find , the distance each of the springs has stretched from its equilibrium position. (It may help to draw the initial and the final states of the system.) Express your answer in terms of , the maximum displacement of each ball from its initial position. = Correct\n\nPart C Find the maximum displacement of any one of the balls from its initial position. Express in terms of some or all of the given quantities , , and = Correct\n\n.\n\nElastic Collision in One Dimension Block 1, of mass , moves across a frictionless surface with speed . It collides elastically with block 2, of mass , which is at rest ( ). After the collision, block 1 moves with speed , while block 2 moves with speed . Assume that , so that after the collision, the two objects move off in the direction of the first object before the collision.\n\nPart A This collision is elastic. What quantities, if any, are conserved in this collision? kinetic energy only momentum only kinetic energy and momentum Correct\n\nPart B What is the final speed of block 1? Express in terms of , , and . = Correct\n\nPart C What is the final speed of block 2? Express in terms of , , and . = Correct\n\nSinking the 9-Ball\n\nHarvaran Ghai\n\nJeanette is playing in a 9ball pool tournament. She will win if she sinks the 9ball from the final rack, so she needs to line up her shot precisely. Both the cue ball and the 9ball have mass , and the cue ball is hit at an initial speed of . Jeanette carefully hits the cue ball into the 9ball off center, so that when the balls collide, they move away from each other at the same angle from the direction in which the cue ball was originally traveling (see figure). Furthermore, after the collision, the cue ball moves away at speed , while the 9ball moves at speed . For the purposes of this problem, assume that the collision is perfectly elastic, neglect friction, and ignore the spinning of the balls.\n\nHarvaran Ghai Part A Find the angle that the 9ball travels away from the horizontal, as shown in the figure. Express your answer in degrees to three significant figures. =45.0 Correct\n\nNote that the angle between the final velocities of the two balls is . It turns out that in any elastic collision between two objects of equal mass, one of which is initially at rest, the angle between the final velocities of the two objects will be ninety degrees.\n\nEnergy in a Spring Graphing Question A toy car is held at rest against a compressed spring, as shown in the figure. across the room. Let\n\nWhen released, the car slides\n\nbe the initial position of the car. Assume that friction is negligible.\n\nHarvaran Ghai Part A Sketch a graph of the total energy of the spring and car system. There is no scale given, so your graph should simply reflect the qualitative shape of the energy vs. time plot. HORIZONTAL LINE\n\nCorrect\n\nPart B Sketch a plot of the elastic potential energy of the spring from the point at which the car is released to the equilibrium position of the spring. Make your graph consistent with the given plot of total energy (the gray line given in the graphing window). 4 POINTS GRADUALLY SLOPING DOWN\n\nCorrect\n\nPart C Sketch a graph of the car's kinetic energy from the moment it is released until it passes the equilibrium position of the spring. Your graph should be consistent with the given plots of total energy (gray line in graphing window) and potential energy (gray parabola in graphing window). 5 POINTS GRADUALLY SLOPING UP OPPOSITE OF GIVEN\n\nCorrect\n\nGraphing Gravitational Potential Energy A 1.00\n\nball is thrown directly upward with an initial speed of 16.0\n\nA graph of the ball's gravitational potential energy vs. height, , for an arbitrary initial velocity is given in Part A. The zero point of gravitational potential energy is located at the height at which the ball leaves the thrower's hand. For this problem, take\n\nas the acceleration due to gravity.\n\nHarvaran Ghai Part A Draw a line on the graph representing the total energy of the ball. HORIZONTAL LINE AT 126\n\nCorrect\n\nPart B Using the graph, determine the maximum height reached by the ball. Express your answer to one decimal place. 12.8 Correct\n\nThe ball reaches its maximum height when its velocity (and therefore kinetic energy) is zero, so all of its energy is potential. This occurs at the height at which the total energy and potential energy graphs intersect. Part C Draw a new gravitational potential energy vs. height graph to represent the gravitational potential energy if the ball had a mass of 2.00 reference. Take\n\n. The graph for a 1.00\n\nball with an arbitrary initial velocity is provided again as a\n\nas the acceleration due to gravity. (100,5) (150, 7.5)\n\nCorrect\n\nFor a ball with twice the mass, you should expect the plot of potential energy vs. height to have twice the slope.\n\nUnderstanding Work and Kinetic Energy Learning Goal: To learn about the WorkEnergy Theorem and its basic applications. In this problem, you will learn about the relationship between the work done on an object and the kinetic energy of that object. The kinetic energy of an object of mass moving at a speed is defined as . It seems reasonable to say that the speed of an objectand, therefore, its kinetic energycan be changed by performing work on the object. In this problem, we will explore the mathematical relationship between the work done on an object and the change in the kinetic energy of that object.\n\nHarvaran Ghai\n\nFirst, let us consider a sled of mass being pulled by a constant, horizontal force of magnitude along a rough, horizontal surface. The sled is speeding up. Part A How many forces are acting on the sled? one two three four Correct\n\nPart B The work done on the sled by the force of gravity is __________. zero negative positive Correct\n\nPart C The work done on the sled by the normal force is __________. zero negative positive Correct\n\nPart D The work done on the sled by the pulling force is __________. zero negative positive Correct\n\nPart E The work done on the sled by the force of friction is __________. zero negative positive\n\nCorrect\n\nPart F The net work done on the sled is __________. zero negative positive Correct\n\nPart G In the situation described, the kinetic energy of the sled __________. remains constant decreases increases Correct\n\nLet us now consider the situation quantitatively. Let the mass of the sled be acting on the sled be\n\nand the magnitude of the net force\n\n. The sled starts from rest.\n\nConsider an interval of time during which the sled covers a distance and the speed of the sled increases from to . We will use this information to find the relationship between the work done by the net force (otherwise known as the net work) and the change in the kinetic energy of the sled. Part H Find the net force acting on the sled. Express your answer in terms of some or all of the variables = Correct\n\n, , , and\n\nPart I Find the net work\n\ndone on the sled.\n\nExpress your answer in terms of some or all of the variables = Correct\n\nand .\n\n.\n\nPart J Use to find the net work done on the sled. Express your answer in terms of some or all of the variables = Correct\n\n, , and\n\n.\n\nYour answer can also be rewritten as\n\nor , where\n\nand\n\nare, respectively, the initial and the final kinetic energies of the sled. Finally, one can write .\n\nThis formula is known as the WorkEnergy Theorem. The calculations done in this problem illustrate the applicability of this theorem in a particlar case; however, they should not be interpreted as a proof of this theorem. Nevertheless, it can be shown that the WorkEnergy Theorem is applicable in all situations, including those involving nonconstant forces or forces acting at an angle to the displacement of the object. This theorem is quite useful in solving problems, as illustrated by the following example. Here is a simple application of the WorkEnergy Theorem. Part K A car of mass\n\naccelerates from speed to speed\n\nwhile going up a slope that makes an angle with the\n\nhorizontal. The coefficient of static friction is , and the acceleration due to gravity is . Find the total work done on the car by the external forces. Express your answer in terms of the given quantities. You may or may not use all of them. = Correct\n\nWork Done by a Spring Consider a spring, with spring constant , one end of which is attached to a wall. unstretched, with the unconstrained end of the spring at position\n\nThe spring is initially\n\n.\n\nHarvaran Ghai\n\nPart A The spring is now compressed so that the unconstrained end moves from\n\nto\n\n. Using the work integral\n\n, find the work done by the spring as it is compressed. Express the work done by the spring in terms of and . = Correct\n\nWork from a Constant Force Learning Goal: To understand how to compute the work done by a constant force acting on a particle that moves in a straight line.\n\nIn this problem, you will calculate the work done by a constant force. A force is considered constant if independent of . This is the most frequently encountered situation in elementary Newtonian mechanics.\n\nis\n\nHarvaran Ghai\n\nPart A\n\nConsider a particle moving in a straight line from initial point B to final point A, acted upon by a constant force\n\n.\n\nThe force (think of it as a field, having a magnitude and direction at every position ) is indicated by a series of identical vectors pointing to the left, parallel to the horizontal axis. The vectors are all identical only because the force is constant along the path. The magnitude of the force is , and the displacement vector from point B to point A is (of magnitude , making and angle (radians) with the positive x axis). Find\n\n, the work that the force\n\nperforms on the particle as it moves from point B to point A. Express the work in terms of , , and . Remember to use radians, not degrees, for any angles that appear in your answer. = Correct\n\nThis result is worth remembering! The work done by a constant force of magnitude , which acts at an angle of with respect to the direction of motion along a straight path of length , is\n\n. This equation\n\ncorrectly gives the sign in this problem. Since is the angle with respect to the positive x axis (in radians), ; hence\n\nPart B Now consider the same force acting on a particle that travels from point A to point B. vector now points in the opposite direction as it did in Part A. Find the work Express your answer in terms of , , and .\n\nThe displacement\n\ndone by in this case.\n\n= Correct\n\nThe Work Done in Pulling a Supertanker Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.60×106 , one at an angle 14.0 west of north, and the other at an angle 14.0 east of north, as they pull the tanker a distance 0.890 north.\n\ntoward the\n\nHarvaran Ghai Part A What is the total work done by the two tugboats on the supertanker? Express your answer in joules. 2.76×109 Correct\n\nWork on a Sliding Block A block of weight sits on a frictionless inclined plane, which makes an angle with respect to the horizontal, as shown.\n\nPart A\n\nA force of magnitude , applied parallel to the incline, pulls the block up the plane at constant speed.\n\nHarvaran Ghai\n\nThe block moves a distance up the incline. The block does not stop after moving this distance but continues to move with constant speed. What is the total work done on the block by all forces? (Include only the work done after the block has started moving, not the work needed to start the block moving from rest.) Express your answer in terms of given quantities. =0 Correct\n\nPart B What is , the work done on the block by the force of gravity as the block moves a distance up the incline? Express the work done by gravity in terms of the weight and any other quantities given in the problem introduction. = Correct\n\nPart C What is\n\n, the work done on the block by the applied force as the block moves a distance up the incline?\n\nExpress your answer in terms of and other given quantities. = Correct\n\nPart D What is , the work done on the block by the normal force as the block moves a distance up the inclined plane? Express your answer in terms of given quantities. =0 Correct\n\nPotential Energy Graphs and Motion Learning Goal: To be able to interpret potential energy diagrams and predict the corresponding motion of a particle. Potential energy diagrams for a particle are useful in predicting the motion of that particle. These diagrams allow one to determine the direction of the force acting on the particle at any point, the points of stable and unstable equilibrium, the particle's kinetic energy, etc.\n\nConsider the potential energy diagram shown. The curve represents the value of potential energy as a function of the particle's coordinate . The horizontal line above the curve represents the constant value of the total energy of the particle . The total energy is the sum of kinetic (\n\n) and potential ( ) energies of the particle.\n\nThe key idea in interpreting the graph can be expressed in the equation\n\nwhere is the x component of the net force as function of the particle's coordinate . Note the negative sign: It means that the x component of the net force is negative when the derivative is positive and vice versa. For instance, if the particle is moving to the right, and its potential energy is increasing, the net force would be pulling the particle to the left. If you are still having trouble visualizing this, consider the following: If a massive particle is increasing its gravitational potential energy (that is, moving upward), the force of gravity is pulling in the opposite direction (that is, downward). If the x component of the net force is zero, the particle is said to be in equilibrium. There are two kinds of equilibrium: \n\nStable equilibrium means that small deviations from the equilibrium point create a net force that accelerates the particle back toward the equilibrium point (think of a ball rolling between two hills). Unstable equilibrium means that small deviations from the equilibrium point create a net force that accelerates the particle further away from the equilibrium point (think of a ball on top of a hill).\n\nIn answering the following questions, we will assume that there is a single varying force acting on the particle along the x axis. Therefore, we will use the term force instead of the cumbersome x component of the net force.\n\nHarvaran Ghai\n\nPart A The force acting on the particle at point A is __________. directed to the right directed to the left equal to zero Correct\n\nConsider the graph in the region of point A. If the particle is moving to the right, it would be \"climbing the hill,\" and the force would \"pull it down,\" that is, pull the particle back to the left. Another, more abstract way of thinking about this is to say that the slope of the graph at point A is positive; therefore, the direction of is negative. Part B The force acting on the particle at point C is __________. directed to the right directed to the left equal to zero Correct\n\nPart C The force acting on the particle at point B is __________. directed to the right directed to the left equal to zero Correct\n\nThe slope of the graph is zero; therefore, the derivative Part D The acceleration of the particle at point B is __________. directed to the right directed to the left equal to zero Correct\n\n, and\n\n.\n\nIf the net force is zero, so is the acceleration. The particle is said to be in a state of equilibrium. Part E If the particle is located slightly to the left of point B, its acceleration is __________. directed to the right directed to the left equal to zero Correct\n\nPart F If the particle is located slightly to the right of point B, its acceleration is __________. directed to the right directed to the left equal to zero Correct\n\nAs you can see, small deviations from equilibrium at point B cause a force that accelerates the particle further away; hence the particle is in unstable equilibrium. Part G Name all labeled points on the graph corresponding to unstable equilibrium. List your choices alphabetically, with no commas or spaces; for instance, if you choose points B, D, and E, type your answer as BDE. BF Correct\n\nPart H Name all labeled points on the graph corresponding to stable equilibrium. List your choices alphabetically, with no commas or spaces; for instance, if you choose points B, D, and E, type your answer as BDE. DH Correct\n\nPart I Name all labeled points on the graph where the acceleration of the particle is zero. List your choices alphabetically, with no commas or spaces; for instance, if you choose points B, D, and E, type your answer as BDE. BDFH Correct\n\nYour answer, of course, includes the locations of both stable and unstable equilibrium. Part J Name all labeled points such that when a particle is released from rest there, it would accelerate to the left. List your choices alphabetically, with no commas or spaces; for instance, if you choose points B, D, and E, type your answer as BDE. AE Correct\n\nPart K Consider points A, E, and G. Of these three points, which one corresponds to the greatest magnitude of acceleration of the particle? A E G Correct\n\nKinetic energy If the total energy of the particle is known, one can also use the graph of kinetic energy of the particle since\n\n. As a reminder, on this graph, the total energy is shown by the horizontal line. Part L What point on the graph corresponds to the maximum kinetic energy of the moving particle? D Correct\n\nIt makes sense that the kinetic energy of the particle is maximum at one of the (force) equilibrium points. For example, think of a pendulum (which has only one force equilibrium pointat the very bottom). Part M At what point on the graph does the particle have the lowest speed? B Correct\n\nAs you can see, many different conclusions can be made about the particle's motion merely by looking at the graph. It is helpful to understand the character of motion qualitatively before you attempt quantitative problems. This problem should prove useful in improving such an understanding.\n\nPotential Energy Calculations Learning Goal: To understand the relationship between the force and the potential energy changes associated with that force and to be able to calculate the changes in potential energy as definite integrals. Imagine that a conservative force field is defined in a certain region of space. Does this sound too abstract? Well, think of a gravitational field (the one that makes apples fall down and keeps the planets orbiting) or an electrostatic field existing around any electrically charged object. If a particle is moving in such a field, its change in potential energy does not depend on the particle's path and is determined only by the particle's initial and final positions. Recall that, in general, the component of the net force acting on a particle equals the negative derivative of the potential energy function along the corresponding axis:\n\n. Therefore, the change in potential energy can be found as the integral\n\n,\n\nwhere\n\nis the change in potential energy for a particle moving from point 1 to point 2, is the net force acting\n\non the particle at a given point of its path, and\n\nis a small displacement of the particle along its path from 1 to 2.\n\nEvaluating such an integral in a general case can be a tedious and lengthy task. However, two circumstances make it easier: 1. Because the result is pathindependent, it is always possible to consider the most straightforward way to reach point 2 from point 1. 2. The most common realworld fields are rather simply defined.\n\nIn this problem, you will practice calculating the change in potential energy for a particle moving in three common force fields. Note that, in the equations for the forces, is the unit vector in the x direction, is the unit vector in the y direction, and is the unit vector in the radial direction in case of a spherically symmetrical force field.\n\nHarvaran Ghai\n\nPart A Consider a uniform gravitational field (a fair approximation near the surface of a planet). Find\n\n, where\n\nand Express your answer in terms of Correct\n\n=\n\n, ,\n\n.\n\n, and .\n\nPart B Consider the force exerted by a spring that obeys Hooke's law. Find\n\n, where\n\n, and the spring constant is positive. Express your answer in terms of , Correct\n\n=\n\n, and\n\n.\n\nPart C Finally, consider the gravitational force generated by a spherically symmetrical massive object. The magnitude and direction of such a force are given by Newton's law of gravity:\n\n, where\n\n; ,\n\n, and\n\nare constants; and\n\n. Find\n\n.\n\nExpress your answer in terms of ,\n\nCorrect\n\n=\n\n, , and .\n\nAs you can see, the change in potential energy of the particle can be found by integrating the force along the particle's path. However, this method, as we mentioned before, does have an important restriction: It can only be applied to a conservative force field. For conservative forces such as gravity or tension the work done on the particle does not depend on the particle's path, and the potential energy is the function of the particle's position. In case of a nonconservative forcesuch as a frictional or magnetic forcethe potential energy can no longer be defined as a function of the particle's position, and the method that you used in this problem would not be applicable.\n\nA Mass-Spring System with Recoil and Friction An object of mass is traveling on a horizontal surface. There is a coefficient of kinetic friction between the object and the surface. The object has speed when it reaches\n\nand encounters a spring. The object compresses\n\nthe spring, stops, and then recoils and travels in the opposite direction. When the object reaches trip, it stops.\n\non its return\n\nHarvaran Ghai Part A Find , the spring constant. Express in terms of , = Correct\n\n, , and .\n\nDragging a Board A uniform board of length and mass lies near a boundary that separates two regions. In region 1, the coefficient of kinetic friction between the board and the surface is , and in region 2, the coefficient is . The positive direction is shown in the figure.\n\nHarvaran Ghai\n\nPart A\n\nFind the net work done by friction in pulling the board directly from region 1 to region 2. Assume that the board moves at constant velocity. Express the net work in terms of = Correct\n\n, , ,\n\n, and\n\n.\n\nThis answer makes sense because it is as if the board spent half its time in region 1, and half in region 2, which on average, it in fact did. Part B What is the total work done by the external force in pulling the board from region 1 to region 2? (Again, assume that the board moves at constant velocity.) Express your answer in terms of = Correct\n\n, , ,\n\n, and\n\n.\n\nDrag on a Skydiver A skydiver of mass jumps from a hot air balloon and falls a distance before reaching a terminal velocity of magnitude . Assume that the magnitude of the acceleration due to gravity is .\n\nPart A\n\nHarvaran Ghai\n\nWhat is the work\n\ndone on the skydiver, over the distance , by the drag force of the air?\n\nExpress the work in terms of , , = Correct\n\n, and the magnitude of the acceleration due to gravity .\n\nPart B Find the power supplied by the drag force after the skydiver has reached terminal velocity . Express the power in terms of quantities given in the problem introduction. = Correct\n\nPower Dissipation Puts a Drag on Racing The dominant form of drag experienced by vehicles (bikes, cars, planes, etc.) at operating speeds is called form drag. It increases quadratically with velocity (essentially because the amount of air you run into increases with and so does the amount of force you must exert on each small volume of air). Thus , where is the crosssectional area of the vehicle and\n\nis called the coefficient of drag.\n\nHarvaran Ghai Part A Consider a vehicle moving with constant velocity . Find the power dissipated by form drag. Express your answer in terms of = Correct\n\n, , and speed .\n\nPart B A certain car has an engine that provides a maximum power . Suppose that the maximum speed of the car, , is limited by a drag force proportional to the square of the speed (as in the previous part). The car engine is now modified, so that the new power Assume the following:\n\nis 10 percent greater than the original power (\n\n \n\nThe top speed is limited by air drag. The magnitude of the force of air drag at these speeds is proportional to the square of the speed.\n\nBy what percentage, , is the top speed of the car increased? Express the percent increase in top speed numerically to two significant figures. Correct\n\n=3.2 %\n\nYou'll note that your answer is very close to onethird of the percentage by which the power was increased. This dependence of small changes on each other, when the quantities are related by proportionalities of exponents, is common in physics and often makes a useful shortcut for estimations.\n\nEnergy of a Spacecraft Very far from earth (at ), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force of the earth were to act on it (i.e., neglect the forces from the sun and other solar system objects), the spacecraft would eventually crash into the earth. The mass of the earth is and its radius is . Neglect air resistance throughout this problem, since the spacecraft is primarily moving through the near vacuum of space.\n\nHarvaran Ghai Part A Find the speed of the spacecraft when it crashes into the earth. Express the speed in terms of = Correct\n\n, and the universal gravitational constant .\n\nPart B Now find the spacecraft's speed when its distance from the center of the earth is Express the speed in terms of and . = Correct\n\nOrbiting Satellite\n\n, where\n\n.\n\nA satellite of mass density . ( constant.\n\nis in a circular orbit of radius\n\naround a spherical planet of radius\n\nis measured from the center of the planet, not its surface.) Use\n\nfor the universal gravitational\n\nHarvaran Ghai\n\nPart A Find the kinetic energy of this satellite,\n\n.\n\nExpress the satellite's kinetic energy in terms of , = Correct\n\n, ,\n\n, and .\n\nPart B Find , the gravitational potential energy of the satellite. Take the gravitational potential energy to be zero for an object infinitely far away from the planet. Express the satellite's gravitational potential energy in terms of = Correct\n\n, ,\n\nPart C What is the ratio of the kinetic energy of this satellite to its potential energy? Express\n\nin terms of parameters given in the introduction.\n\n, and .\n\n=0.500 Correct\n\nThe result of this problem may be expressed as\n\nwhere\n\nis the exponent of the force law (i.e.\n\n). This is a specical case of a general and powerful theroem of advanced classical mechanics known as the Virial Theorem. The theorem applies to the average of the kinetic and potential energies of of any one or multiple objects moving over any closed (or almost closed) path that returns very close to itself provided that all objects interact via potentials with the same power law dependence on their separation. Thus it applies to stars in a galaxy, or masses tied together with springs (where\n\nsince the force law is\n\n).\n\nProblem 12.28 The space shuttle is in a 250\n\nhigh circular orbit. It needs to reach a 700\n\nHubble Space Telescope for repairs. The shuttle's mass is 8.00×104\n\nhigh circular orbit to catch the\n\n.\n\nHarvaran Ghai\n\nPart A How much energy is required to boost it to the new orbit? 1.53×1011 J Correct\n\nAn Exhausted Bicyclist An exhausted bicyclist pedals somewhat erratically, so that the angular velocity of his tires follows the equation\n\n, where represents time (measured in seconds).\n\nHarvaran Ghai Part A\n\nThere is a spot of paint on the front tire of the bicycle. Take the position of the spot at time\n\nto be at angle\n\nradians with respect to an axis parallel to the ground (and perpendicular to the axis of rotation of the tire) and measure positive angles in the direction of the tire's rotation. What angular displacement has the spot of paint undergone between time 0 and 2 seconds? Express your answer in radians. =0.793 Correct\n\nPart B Express the angular displacement undergone by the spot of paint at\n\nseconds in degrees.\n\n=45.5 Correct\n\nPart C What distance has the spot of paint moved in 2 seconds if the radius of the tire is 50 centimeters? Express your answer in centimeters. =39.7 Correct\n\nPart D Which one of the following statements describes the motion of the spot of paint at seconds? The angular acceleration of the spot of paint is constant and the magnitude of the angular speed is decreasing. The angular acceleration of the spot of paint is constant and the magnitude of the angular speed is increasing. The angular acceleration of the spot of paint is positive and the magnitude of the angular speed is decreasing. The angular acceleration of the spot of paint is positive and the magnitude of the angular speed is increasing. The angular acceleration of the spot of paint is negative and the magnitude of the angular speed is decreasing. The angular acceleration of the spot of paint is negative and the magnitude of the angular speed is increasing. Correct\n\nA Spinning Grinding Wheel\n\nAt time\n\na grinding wheel has an angular velocity of 23.0\n\n32.0 433\n\n. It has a constant angular acceleration of\n\nuntil a circuit breaker trips at time = 1.90 . From then on, the wheel turns through an angle of as it coasts to a stop at constant angular deceleration.\n\nHarvaran Ghai Part A Through what total angle did the wheel turn between Express your answer in radians.\n\nand the time it stopped?\n\n534 Correct\n\nPart B At what time does the wheel stop? Express your answer in seconds. 12.2 Correct\n\nPart C What was the wheel's angular acceleration as it slowed down? Express your answer in radians per second per second. 8.11 Correct\n\nFinding Torque A force of magnitude , making an angle with the x axis, is applied to a particle located at point A, at Cartesian coordinates (0, 0) in the figure. The vector and the four reference points (i.e., A, B, C, and D) all lie in the xy plane. Rotation axes A D lie parallel to the z axis and pass through each respective reference point.\n\nThe torque of a force acting on a particle having a position vector with respect to a reference point (thus points from the reference point to the point at which the force acts) is equal to the cross product of and , . The magnitude of the torque is\n\n, where is the angle between and ; the direction of\n\nis perpendicular to both and . For this problem\n\n; negative torque about a reference point corresponds\n\nto clockwise rotation. You must express in terms of , , and/or when entering your answers.\n\nHarvaran Ghai\n\nPart A What is the torque due to force about the point A? Express the torque about point A at Cartesian coordinates (0, 0). =0 Correct\n\nPart B What is the torque\n\ndue to force about the point B? (B is the point at Cartesian coordinates (0, ), located a\n\ndistance from the origin along the y axis.) Express the torque about point B in terms of , , , , and/or other given coordinate data. = Correct\n\nPart C What is the torque along the x axis?\n\nabout the point C, located at a position given by Cartesian coordinates ( , 0), a distance\n\nExpress the torque about point C in terms of , , , , and/or other given coordinate data. = Correct\n\nPart D What is the torque axis?\n\nabout the point D, located at a distance from the origin and making an angle with the x\n\nExpress the torque about point D in terms of , , , , and/or other given coordinate data. = Correct\n\nNote that the cross product\n\nwhich simplifies to\n\ncan also be expressed as a thirdorder determinant\n\nwhen and lie in the xy plane.\n\nPivoted Rod with Unequal Masses A thin rod of mass and length is allowed to pivot freely about its center, as shown in the diagram. A small sphere of mass is attached to the left end of the rod, and a small sphere of mass is attached to the right end. The spheres are small enough that they can be considered point particles. The gravitational force acts downward, with the magnitude of the gravitational acceleration equal to .\n\nPart A\n\nHarvaran Ghai\n\nWhat is the moment of inertia of this assembly about the axis through which it is pivoted? Express the moment of inertia in terms of = Correct\n\n, and . Remember, the length of the rod is\n\n, not .\n\nPart B Suppose the rod is held at rest horizontally and then released. (Throughout the remainder of this problem, your answer may include the symbol , the moment of inertia of the assembly, whether or not you have answered the first part correctly.) What is the angular acceleration of the rod immediately after it is released? Take the counterclockwise direction to be positive. Express in terms of some or all of the variables\n\n,\n\n, , , and . = Correct\n\nPulling a String to Accelerate a Wheel A bicycle wheel is mounted on a fixed, frictionless axle, as shown\n\n. A massless string is wound around the\n\nwheel's rim, and a constant horizontal force of magnitude starts pulling the string from the top of the wheel starting at time\n\nwhen the wheel is not rotating. Suppose that at some later time the string has been pulled\n\nthrough a distance . The wheel has moment of inertia , where is a dimensionless number less than 1, is the wheel's mass, and is its radius. Assume that the string does not slip on the wheel.\n\nHarvaran Ghai\n\nPart A\n\nFind , the angular acceleration of the wheel, which results from pulling the string to the left. Use the standard convention that counterclockwise angular accelerations are positive. Express the angular acceleration, , in terms of , , = Correct\n\n, and (but not\n\n).\n\nPart B\n\nThe force pulling the string is constant; therefore the magnitude of the angular acceleration of the wheel is constant for this configuration. Find the magnitude of the angular velocity of the wheel when the string has been pulled a distance . Note that there are two ways to find an expression for ; these expressions look very different but are equivalent. Express the angular velocity of the wheel in terms of the displacement , the magnitude of the applied force, and the moment of inertia of the wheel , if you've found such a solution. Otherwise, following the hints for this part should lead you to express the angular velocity of the wheel in terms of the displacement , the wheel's radius , and . = Correct\n\nThis solution can be obtained from the equations of rotational motion and the equations of motion with constant acceleration. An alternate approach is to calculate the work done over the displacement by the force and equate this work to the increase in rotational kinetic energy of rotation of the wheel Part C Find , the speed of the string after it has been pulled by over a distance . Express the speed of the string in terms of , , , and = Correct\n\n; do not include , , or\n\nNote that this is the speed that an object of mass\n\n(which is less than\n\n) would attain if pulled a distance by a\n\nforce with constant magnitude .\n\nA Bar Suspended by Two Vertical Strings A rigid, uniform, horizontal bar of mass\n\nand length is supported by two identical massless strings.\n\nBoth\n\nstrings are vertical. String A is attached at a distance from the left end of the bar and is connected to the ceiling; string B is attached to the left end of the bar and is connected to the floor. A small block of mass is supported against gravity by the bar at a distance from the left end of the bar, as shown in the figure. Throughout this problem positive torque is that which spins an object counterclockwise. Use for the magnitude of the acceleration due to gravity.\n\nHarvaran Ghai\n\nPart A Find\n\n, the tension in string A.\n\nExpress the tension in string A in terms of , = Correct\n\nPart B Find\n\n, the magnitude of the tension in string B.\n\n, , ,\n\n, and .\n\nExpress the magnitude of the tension in string B in terms of = Correct\n\n, and .\n\nPart C If the bar and block are too heavy the strings may break. Which of the two identical strings will break first? string A string B Correct\n\nPart D If the mass of the block is too large and the block is too close to the left end of the bar (near string B) then the horizontal bar may become unstable (i.e., the bar may no longer remain horizontal). What is the smallest possible value of such that the bar remains stable (call it Express your answer for\n\nin terms of\n\n= Correct\n\n)?\n\n, , and .\n\nPart E Note that since , as computed in the previous part, is not necessarily positive. If stable no matter where the block of mass is placed on it. Assuming that\n\n, , and are held fixed, what is the maximum block mass\n\nstable? In other words, what is the maximum block mass such that Answer in terms of = Correct\n\n, , and .\n\nA Person Standing on a Leaning Ladder\n\n, the bar will be\n\nfor which the bar will always be ?\n\nand length rests against a smooth wall.\n\nA doityourself enthusiast of mass\n\nstands on the ladder a distance from the bottom (measured along the ladder). The ladder makes an angle with the ground. There is no friction between the wall and the ladder, but there is a frictional force of magnitude between the floor and the ladder. is the magnitude of the normal force exerted by the wall on the ladder, and is the magnitude of the normal force exerted by the ground on the ladder. Throughout the problem, consider counterclockwise torques to be positive. None of your answers should involve (i.e., simplify your trig functions).\n\nHarvaran Ghai Part A What is the minimum coeffecient of static friction ladder does not slip? Express = Correct\n\nin terms of\n\nrequired between the ladder and the ground so that the\n\n, , , and .\n\nPart B Suppose that the actual coefficent of friction is one and a half times as large as the value of the ladder?\n\n. Under these circumstances, what is the magnitude of the force of friction that the floor applies to\n\nExpress your answer in terms of and . = Correct\n\n. That is,\n\n, , , , and . Remember to pay attention to the relation of force\n\nA Rolling Hollow Sphere A hollow spherical shell with mass 1.80 the horizontal.\n\nrolls without slipping down a slope that makes an angle of 39.0 with\n\nHarvaran Ghai Part A Find the magnitude of the acceleration\n\nof the center of mass of the spherical shell.\n\nTake the freefall acceleration to be = 9.80\n\n.\n\n=3.70 Correct\n\nPart B Find the magnitude of the frictional force acting on the spherical shell. Take the freefall acceleration to be = 9.80\n\n.\n\n=4.44 Correct\n\nThe frictional force keeps the spherical shell stuck to the surface of the slope, so that there is no slipping as it rolls down. If there were no friction, the shell would simply slide down the slope, as a rectangular box might do on an inclined (frictionless) surface. Part C Find the minimum coefficient of friction needed to prevent the spherical shell from slipping as it rolls down the slope. =0.324 Correct\n\nWeight and Wheel Consider a bicycle wheel that initially is not rotating. A block of mass\n\nis attached to the wheel and is allowed to\n\nfall a distance . Assume that the wheel has a moment of inertia about its rotation axis.\n\nHarvaran Ghai Part A Consider the case that the string tied to the block is attached to the outside of the wheel, at a radius . Find\n\n, the angular speed of the wheel after the block has fallen a distance , for this case.\n\nExpress\n\nin terms of\n\n, , ,\n\n= Correct\n\n, and .\n\nPart B Now consider the case that the string tied to the block is wrapped around a smaller inside axle of the wheel of radius\n\n. Find\n\nExpress = Correct\n\n, the angular speed of the wheel after the block has fallen a distance , for this case.\n\nin terms of\n\n, , ,\n\n, and .\n\nPart C Which of the following describes the relationship between\n\nCorrect\n\nand\n\n?\n\nThis is related to why gears are found on the inside rather than the outside of a wheel.\n\nRecord and Turntable Learning Goal: To understand how to use conservation of angular momentum to solve problems involving collisions of rotating bodies. Consider a turntable to be a circular disk of moment of inertia rotating at a constant angular velocity around an axis through the center and perpendicular to the plane of the disk (the disk's \"primary axis of symmetry\"). The axis of the disk is vertical and the disk is supported by frictionless bearings. The motor of the turntable is off, so there is no external torque being applied to the axis. Another disk (a record) is dropped onto the first such that it lands coaxially (the axes coincide). The moment of inertia of the record is . The initial angular velocity of the second disk is zero. There is friction between the two disks. After this \"rotational collision,\" the disks will eventually rotate with the same angular velocity.\n\nPart A What is the final angular velocity, Express = Correct\n\nPart B\n\nin terms of , , and\n\nHarvaran Ghai , of the two disks? .\n\nBecause of friction, rotational kinetic energy is not conserved while the disks' surfaces slip over each other. What is the final rotational kinetic energy,\n\n, of the two spinning disks?\n\nExpress the final kinetic energy in terms of , , and the initial kinetic energy angular velocities should appear in your answer. = Correct\n\nof the twodisk system. No\n\nSome of the energy was converted into heat and sound as the frictional force, torque acted, stopping relative motion. Part C Assume that the turntable deccelerated during time before reaching the final angular velocity ( is the time interval between the moment when the top disk is dropped and the time that the disks begin to spin at the same angular velocity). What was the average torque,\n\n, acting on the bottom disk due to friction with the record?\n\nExpress the torque in terms of ,\n\n.\n\n= Correct\n\n, and\n\nProblem 13.87 During most of its lifetime, a star maintains an equilibrium size in which the inward force of gravity on each atom is balanced by an outward pressure force due to the heat of the nuclear reactions in the core. But after all the hydrogen \"fuel\" is consumed by nuclear fusion, the pressure force drops and the star undergoes a gravitational collapse until it becomes a neutron star. In a neutron star, the electrons and protons of the atoms are squeezed together by gravity until they fuse into neutrons. Neutron stars spin very rapidly and emit intense pulses of radio and light waves, one pulse per rotation. These \"pulsing stars\" were discovered in the 1960s and are called pulsars.\n\nHarvaran Ghai Part A A star with the mass and size of our sun rotates once every 34.0 days. After undergoing gravitational collapse, the star forms a pulsar that is observed by astronomers to emit radio pulses every 0.100 . By treating the neutron star as a solid sphere, deduce its radius. 6.46×104 m Correct\n\nPart B What is the speed of a point on the equator of the neutron star? Your answer will be somewhat too large because a star cannot be accurately modeled as a solid sphere. 4.06×106 m/s Correct\n\nAnalyzing Simple Harmonic Motion This Error! Hyperlink reference not valid. shows two masses on springs, each accompanied by a graph of its position versus time.\n\nHarvaran Ghai Part A What is an expression for , the position of mass I as a function of time? Assume that position is measured in meters and time is measured in seconds. Express your answer as a function of . Express numerical constants to three significant figures. = Correct\n\nPart B What is , the position of mass II as a function of time? Assume that position is measured in meters and time is measured in seconds. Express your answer as a function of . Express numerical constants to three significant figures. = Correct\n\nHarmonic Oscillator Acceleration Learning Goal: To understand the application of the general harmonic equation to finding the acceleration of a spring oscillator as a function of time. One end of a spring with spring constant is attached to the wall. The other end is attached to a block of mass . The block rests on a frictionless horizontal surface. The equilibrium position of the left side of the block is defined to be\n\n. The length of the relaxed spring is .\n\nThe block is slowly pulled from its equilibrium position to some position the block is released with zero initial velocity. The goal of this problem is to determine the acceleration of the block and .\n\nalong the x axis. At time\n\nas a function of time in terms of ,\n\nIt is known that a general solution for the position of a harmonic oscillator is , where , , and are constants.\n\nYour task, therefore, is to determine the values of , , and in terms of , connection between\n\nand\n\n,and\n\nand then use the\n\nto find the acceleration.\n\nHarvaran Ghai Part A Combine Newton's 2nd law and Hooke's law for a spring to find the acceleration of the block time. Express your answer in terms of , = Correct\n\n, and the coordinate of the block\n\nas a function of\n\n.\n\nThe negative sign in the answer is important: It indicates that the restoring force (the tension of the spring) is always directed opposite to the block's displacement. When the block is pulled to the right from the equilibrium position, the restoring force is pulling back, that is, to the leftand vice versa. Part B Using the fact that acceleration is the second derivative of position, find the acceleration of the block function of time. Express your answer in terms of , , and\n\n.\n\nas a\n\n= Correct\n\nPart C Find the angular frequency . Express your answer in terms of and = Correct\n\n.\n\nNote that the angular frequency and, therefore, the period of oscillations depend only on the intrinsic physical characteristics of the system ( and amplitude of the motion.\n\n). Frequency and period do not depend on the initial conditions or the\n\nEnergy of Harmonic Oscillators Learning Goal: To learn to apply the law of conservation of energy to the analysis of harmonic oscillators. Systems in simple harmonic motion, or harmonic oscillators, obey the law of conservation of energy just like all other systems do. Using energy considerations, one can analyze many aspects of motion of the oscillator. Such an analysis can be simplified if one assumes that mechanical energy is not dissipated. In other words, , where is the total mechanical energy of the system,\n\nis the kinetic energy, and is the potential energy.\n\nHarvaran Ghai As you know, a common example of a harmonic oscillator is a mass attached to a spring. In this problem, we will consider a horizontally moving block attached to a spring. Note that, since the gravitational potential energy is not changing in this case, it can be excluded from the calculations. For such a system, the potential energy is stored in the spring and is given by\n\n, where is the force constant of the spring and is the distance from the equilibrium position. The kinetic energy of the system is, as always,\n\n, where\n\nis the mass of the block and is the speed of the block.\n\nWe will also assume that there are no resistive forces; that is,\n\n.\n\nConsider a harmonic oscillator at four different moments, labeled A, B, C, and D, as shown in the figure Assume that the force constant , the mass of the block, the following questions.\n\n, and the amplitude of vibrations, , are given. Answer\n\nPart A Which moment corresponds to the maximum potential energy of the system? A B\n\n.\n\nC D Correct\n\nPart B Which moment corresponds to the minimum kinetic energy of the system? A B C D Correct\n\nWhen the block is displaced a distance from equilibrium, the spring is stretched (or compressed) the most, and the block is momentarily at rest. Therefore, the maximum potential energy is course,\n\n. Recall that\n\n. At that moment, of\n\n. Therefore,\n\n. In general, the mechanical energy of a harmonic oscillator equals its potential energy at the maximum or minimum displacement. Part C Consider the block in the process of oscillating. at the equilibrium position. at the amplitude displacement. If the kinetic energy of the block is increasing, the block must be\n\nmoving to the right. moving to the left. moving away from equilibrium. moving toward equilibrium.\n\nCorrect\n\nPart D Which moment corresponds to the maximum kinetic energy of the system? A B\n\nC D Correct\n\nPart E Which moment corresponds to the minimum potential energy of the system? A B C D Correct\n\nWhen the block is at the equilibrium position, the spring is not stretched (or compressed) at all. At that moment, of course,\n\n. Meanwhile, the block is at its maximum speed (\n\nthen be written as\n\n. Recall that\n\nand that\n\n. Recalling what we found out before,\n\n, we can now conclude that\n\n, or\n\n. Part F At which moment is A B\n\n?\n\n). The maximum kinetic energy can at the equilibrium position. Therefore,\n\nC D Correct\n\nPart G Find the kinetic energy\n\nof the block at the moment labeled B.\n\nExpress your answer in terms of and . = Correct\n\nEnergy of a Spring An object of mass\n\nattached to a spring of force constant oscillates with simple harmonic motion. The maximum\n\ndisplacement from equilibrium is and the total mechanical energy of the system is .\n\nHarvaran Ghai Part A What is the system's potential energy when its kinetic energy is equal to\n\nCorrect\n\nPart B What is the object's velocity when its potential energy is\n\n?\n\n?\n\nCorrect\n\nGravity on Another Planet After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 55.0 explorer finds that the pendulum completes 108 full swing cycles in a time of 136 .\n\n. The\n\nHarvaran Ghai Part A What is the value of the acceleration of gravity on this planet? Express your answer in meters per second per second. =13.7 Correct\n\n136/108…2pi/ans…ans*0.55\n\nThe Fish Scale The scale of a spring balance reading from 0 to 205 has a length of 13.5 the spring oscillates vertically at a frequency of 2.95\n\nPart A Ignoring the mass of the spring, what is the mass Express your answer in kilograms.\n\n. A fish hanging from the bottom of\n\n.\n\nHarvaran Ghai of the fish?\n\n=4.42 Correct\n\nVertical Mass-and-Spring Oscillator A block of mass is attached to the end of an ideal spring. Due to the weight of the block, the block remains at rest when the spring is stretched a distance from its equilibrium length. constant .\n\nThe spring has an unknown spring\n\nHarvaran Ghai Part A What is the spring constant ? Express the spring constant in terms of given quantities and , the magnitude of the acceleration due to gravity. = Correct\n\nPart B Suppose that the block gets bumped and undergoes a small vertical displacement. Find the resulting angular frequency of the block's oscillation about its equilibrium position. Express the frequency in terms of given quantities and , the magnitude of the acceleration due to gravity. = Correct\n\nIt may seem that this result for the frequency does not depend on either the mass of the block or the spring constant, which might make little sense. However, these parameters are what would determine the extension of the spring when the block is hanging:\n\nOne way of thinking about this problem is to consider both and as unknowns. By measuring and (both fairly simple measurements), and knowing the mass, you can determine the value of the spring constant and the acceleration due to gravity experimentally."
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"## Modeling Nonideal Reactors\n\nThe use of RTDs for the prediction of reactor performance is a complex subject and the reader is encouraged to consult other sources for more complete coverage. \"v\" The techniques commonly used in environmental engineering are relatively straightforward, however, and thus we will look at them briefly.\n\n4.4.1 Continuous Stirred Tank Reactors in Series Model\n\nThe simplest way to model a reactor with a nonideal flow pattern is as a series of CSTRs and this technique will be used extensively in this book. The basis for doing this may be seen by considering the response of such a system to a step input of tracer. Consider a chain of N CSTRs, each with volume V, receiving a flow F, giving each CSTR a mean hydraulic residence time t (Figure 4.5). At time zero the feed to the first tank is switched to one with a tracer concentration SI(1. The response",
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"Figure 4.5 CSTRs in series.\n\nfrom the first tank is the feed to the second tank, the response of the second is the feed for the third, etc. If we write and solve the mass balance equations for each (with no reaction term since the tracer is assumed to be inert), we obtain the following expression:\n\nTracer concentration profiles, i.e., F(t) curves, for various numbers of tanks in series are shown in Figure 4.6 while the corresponding £(t) curves are shown in Figure 4.7. The curve for N = 1 is the classical response of a single CSTR. More significantly, however, the curve for N = ^ is the classical response for a PFR, i.e., a step change in effluent concentration after one HRT. This suggests that the step response for N CSTRs in series will lie somewhere between that of a single CSTR and a PFR, with the pattern depending on the number of tanks in the chain. Furthermore, this implies that a real reactor which has the response of neither a CSTR nor a PFR can be simulated as N CSTRs in series. The easiest way of determining the appropriate value for N, which is sufficiently accurate in many cases, is to plot either the F(t) or the £(t) curve for the reactor in question and compare it to the curves in Figures 4.6 and 4.7, thereby selecting the value of N which corresponds most closely.\n\nThe tanks in series model has been used frequently in environmental engineering practice. For example, Murphy and Boyko\" found that many conventional activated sludge systems had RTDs that were equivalent to three to five CSTRs in series.",
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"Figure 4.6 Responses of N CSTRs in series to a step tracer input, t in the abscissa is the total HRT for N CSTRs in series.",
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"Figure 4.7 Responses of N CSTRs in series to an impulse tracer input, t in the abscissa is the total HRT for N CSTRs in series.\n\nTime, t\n\nFigure 4.7 Responses of N CSTRs in series to an impulse tracer input, t in the abscissa is the total HRT for N CSTRs in series.\n\n### 4.4.2 Axial Dispersion Model\n\nAn alternative approach is to superimpose some degree of backmixing upon a plug flow of fluid. The magnitude of the backmixing is assumed to be independent of the position in the reactor and is expressed by the axial dispersion coefficient, D|, which is analogous to the coefficient of molecular diffusion in Fick's law of diffusion. Modeling the superimposed backmixing by axial dispersion requires adding another transport term to the mass balance equation for the differential element of the PFR in Figure 4.2. In addition to the advective transport term used in Eq. 4.7, a term for transport by axial dispersion must be included, thereby increasing the number of terms in the resulting partial differential equation:\n\ndSA d2SA F 3Sa\n\nThe equation is usually rewritten to include the term D,/vL (the dispersion number):\n\nin which v is the longitudinal velocity through the basin (F/AJ, L is the basin length, z is dimensionless distance along the basin (x/L) and 9 is dimensionless lime (t/-r).\n\nWhen the dispersion number is zero, there is no axial dispersion and therefore plug flow, whereas when it is infinitely large, complete backmixing exists and the reactor behaves as a CSTR.\n\nThe effect of the value of the dispersion number on the RTD may be seen by solving Eq. 4.23 with the appropriate initial and boundary conditions for a step input of an inert tracer.8 The solution, expressed in the form of an F(t) curve, is shown in Figure 4.8. To characterize the mixing pattern in a reactor by this technique, we need a way to select the appropriate value for the dispersion number. One way is to evaluate the derivative of the F(t) function at one mean residence time:\n\n0 0"
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https://se.mathworks.com/matlabcentral/profile/authors/10122683 | [
"Community Profile",
null,
"# Asliddin Komilov\n\nLast seen: 2 månader ago Active since 2019\n\n#### Statistics\n\n•",
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"#### Content Feed\n\nView by\n\ncontour level appointing problem\nI have changed the part of the code, but still don't see my minima: for pp=1:rowNum*colmNum subplot (rowNum,colmNum,...\n\n| accepted\n\nQuestion\n\ncontour level appointing problem\nHello everyone! I appointed the contour level to the minimum value and 0.1 (code below) but in some I get only 0.1 without the ...\n\n### 2\n\nQuestion\n\nextrapolation of contour data set\nHello everyone! I have made a contour from my data (both is attached), but I need have data in contour that is from 0 to 0,1. ...\n\n### 1\n\nR-square and the F statistic... error\nThanks, it seems I got rid of the warning but have another one: \"Warning: X is rank deficient to within machine precision. \" An...\n\nQuestion\n\nR-square and the F statistic... error\nHi everyone, I am getting this error for the each line (Warning: R-square and the F statistic are not well-defined unless X has ...\n\n### 2\n\nQuestion\n\n2 unknowns in matching datasets\nHello everyone. I have 3 sets of data and need to find \"a\" and \"b\": Y1=a*Y2+b*Y3; there are 2 unknowns so I would need 2 equa...\n\n### 1\n\nextrapolate data to cross x axis\nsorry for the mess, I thought I have saved one variable. This is exactly what I needed. thank you so much\n\nQuestion\n\nextrapolate data to cross x axis\nI have this curve and need to extrapolate it so both ends cross x axis, I have used Vq = INTERP1(X,V,Xq,'linear\",'extrap') and t...\n\n### 3\n\nCreation mixed level orthogonal arrays for orthogonal array testing in matlab\nif you create multidimentional output you can make the ortogonal arrays using ind2sub\n\nungefär ett år ago | 0\n\nQuestion\n\nranges of dimensions when using nan\ndear community, I have a 6D result Arr(crb, cefb, crr, crp, ctb, ccb) where Arr(Arr>1)=nan. How can I find the values of crb...\n\nungefär ett år ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nprocessing experimental data with polyfit\nI have this code, it works, but with some warnigs and it doesn't look like other codes I see here, because I am not a programmer...\n\nnästan 2 år ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nhow to use taguchi array\nI used this code [Ta] = TaguchiArray(3,101) (from https://www.mathworks.com/matlabcentral/fileexchange/71628-taguchiarray) t...\n\nnästan 2 år ago | 0 answers | 0\n\n### 0\n\nQuestion\n\nextract matrix from matrix with the values of the axis\nx=linspace(0,8,91); y=linspace(1,3,56); I have a matrix DL1 (91x46) and need to extract DL2 =DL1(DL1<=2) and the values of the...\n\nungefär 2 år ago | 1 answer | 0\n\n### 1\n\nQuestion\n\n2d data set distribution\nI have this data set. Is it possible to find the distribution of the data to see the most and the least common values? thanks\n\nungefär 2 år ago | 0 answers | 0\n\n### 0\n\nthe same range color bar looks different\nThe upper limits of the color bars were slightly different and the gradient changed when I set them to the same value (100) (see...\n\nungefär 2 år ago | 0\n\nQuestion\n\nthe same range color bar looks different\nWhy these two similar range colorbars look differently (see attachment). How can I make them look the same? thanks\n\nungefär 2 år ago | 2 answers | 0\n\n### 2\n\nQuestion\n\nsensitivity analysis of data with multiple variables\nI have a set of data that depends on 3 variables: Sys_LCOE(jj,i,ii) ii=1:1:length(YF); i=1:1:length(PVprice); jj=1:1:length(...\n\nungefär 2 år ago | 0 answers | 0\n\n### 0\n\nQuestion\n\nhow to fit 3 lines with one equation\nI have these 3 lines with x and y, that correspond to C=[25 40 60], is it possible generate an equation that will be valid for a...\n\nungefär 2 år ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nmaximum points on the surface\nOn this beautiful plot I need a line that shows the maximum points https://www.mathworks.com/matlabcentral/answers/uploaded_f...\n\nmer än 2 år ago | 1 answer | 0\n\n### 1\n\nfitting 2d data set fit function is not working\nthanks, it is perfect but the fit gives this warning: Warning: Equation is badly conditioned. Remove repeated data points or tr...\n\nmer än 2 år ago | 0\n\nQuestion\n\nfitting 2d data set fit function is not working\nHi, this is my data set and I wanted to try to get a fit for it. well, did not work with fit([x,y],z,'poly23'); first it sa...\n\nmer än 2 år ago | 2 answers | 0\n\n### 2\n\ninterp1 find values at axis zero problem The grid vectors must contain unique points.\nI have deleted data that was >0.024, so its fixed thank you everyone\n\nmer än 2 år ago | 0\n\nQuestion\n\ninterp1 find values at axis zero problem The grid vectors must contain unique points.\nVoc=interp1(I,V,0); it is a simple code, it runs with the first 2 sets of data but it doesn't run with the third. someone he...\n\nmer än 2 år ago | 2 answers | 0\n\n### 2\n\nhow to have the same plot area size when some axis names become 2 lines\nOk having two lines in the axis name does the job, thanks\n\nmer än 2 år ago | 0\n\nQuestion\n\nhow to have the same plot area size when some axis names become 2 lines\nI am generating plots with different axis names, and when long names become 2 lines they make the plot area smaller, but I need ...\n\nmer än 2 år ago | 3 answers | 0\n\n### 3\n\nQuestion\n\noptimizing for loop that is been called often\ncan this loop be optimized? It is called 1m times and is the only thing I couldn't optimize yet. Oper_year=1:1:Lifetime; [co...\n\nmer än 2 år ago | 1 answer | 0\n\n### 1\n\nQuestion\n\n5 equations with 5 unknowns\nthis is the code, and it says: Undefined function or variable 'Rp'. Vmp=0.57 Voc=0.66 Isc=0.0067 Imp=0.0062 k=1.38e-23 q=1...\n\nmer än 2 år ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nsfit data array output\nHi, I have this: SS_fit_Xs=fit(Wave_length,real_Solar_spectrum,'poly9'); then I can plot it so easily like this: plot(SS_fit_...\n\nnästan 3 år ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nhow to change axis of a matix?\nmy matrix is M(x,y,z) where: x=1:90 y=1:3:120 z=1:40 and c=y./z and I need to convert M into N(x,c). Any ideas?\n\nnästan 3 år ago | 2 answers | 0"
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https://visualday.aulablog.com/quickstart.php?cat=buy-written-persuasive-speeches&postID=554&calculus-homework-help-app | [
"Seleccionar página\n\n## Wolfram Course Assistant Apps",
null,
"• Pearson introduces Aida, an AI-powered calculus tutor\n• Top 5 Apps to Help With Homework\n• Calculus homework help app",
null,
"## Calculus Course Assistant\n\nuniversity of iowa creative homework help predicate logic taureg warriors ancient africa calculus homework help app history cv writing service macclesfield homework help help writing acceptance homework help rate homework calculation goemetry homework primary help homework help co uk victorians industrial revolutionhtml homework help app resume writing services tips homework help the woods tx goldsmiths creative writing and calculus homework help app education? Take calculus? Next, you need calculus homework help app a s assistant for homework help on the Wolfram Calculus course. Obtained from a review of the world's leading homework help site for math software, this definitive app for calculus addresses socratic math solutions, solves and tests homework problems calculus homework help app with the help of homework. To help you learn the concept of calculus. Forget the canned example! Wolfram Calculus Course Assistant solves specific calculus problems on the fly. The Aida A Calculus app from Pearson is the first AI mobile teacher to provide instant feedback on your steps, help with transformed homework relationships, and hundreds of videos and examples of home finance assistance. like Photomath, for help at work. Another problem is the lack of liquidity in the learning pages calculus homework help app by entering a specific topic page and returning to the calculus homework help app previous page. Top cheap professional resume services calculus homework help app Homework Help Apps Homework Help Ocean Adam Rowe September, : AM Everyone goes back to school, and this time, they have other multistep equations with fractions and decimals to help with homework. in."
]
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"http://sodajerkcountry.com/img/5e5ce6bc9b73dcb53ee7bdf16974c0a4.png",
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"http://sozopolistyleclub.com/img/e762a03e4c24509dbcadd13bd3c4810a.png",
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https://allexamreview.com/transmission-and-distribution-part-10/ | [
"# Transmission And Distribution Part 10\n\n## Transmission And Distribution Part 10",
null,
"Category –EE Online Test\n\nAttempt Free Transmission And Distribution Part 10 Here. Read The Important Electrical MCQ From Below.\n\nQ1. There is a greater possibility of occurrence of corona during\n\nA. dry weather\nB. winter\nC. summer heat\nD. humid weather\n\nAns. D\n\nQ2. In A.C.S.R. conductors, the insulation between aluminum and steel conductors is\n\nA. insulin\nB. bitumen\nC. varnish\nD. no insulation is required\n\nAns. D\n\nQ3. Corona usually occurs when the electrostatic stress in air around the conductor exceeds\n\nA. 6.6 kV (r.m.s. value)/cm\nB. 11 kV (r.m.s. value)/cm\nC. 22 kV (maximum value)/cm\nD. 30 kV (maximum value)/cm\n\nAns. D\n\nQ4. In a tap changing transformer, the tappings are provided on\n\nA. primary winding\nB. secondary winding\nC. high voltage winding\nD. any of the above\n\nAns. B\n\nTransmission And Distribution Part 10\n\nQ5. The angular displacement between two interconnected stations is mainly due to\n\nA. armature reactance of both alternators\nB. reactance of the interconnector\nC. synchronous reactance of both the alternators\nD. all of the above\n\nAns. B\n\nQ6. The skin effect of a conductor will reduce as the\n\nA. resistivity of conductor material increases\nB. permeability of conductor material increases\nC. diameter increases\nD. frequency increases\n\nAns. A\n\nQ7. Which of the following distribution systems is the most economical ?\n\nA. A.C. 1-phase system\nB. A.C. 3-phase 3 wire system\nC. A.C. 3-phase 4 wire system\nD. Direct current system\n\nAns. D\n\nQ8. Power loss due to corona is not directly proportional to\n\nA. spacing between conductors\nB. supply voltage frequency\nC. phase-neutral voltage\nD. all of the above\n\nAns. A\n\nTransmission And Distribution Part 10\n\nQ9. In the design of a distributor which of the following is the major consideration ?\n\nA. Voltage drop\nB. Current carrying capacity\nC. Frequency\nD. kVA of system\n\nAns. A\n\nQ10. Which of the following cause transient disturbances ?\n\nA. Faults\nC. Switching operations\nD. Any of the above\n\nAns. D"
]
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https://www.programminginpython.com/category/math-programs/ | [
"## Python Program to find the LCM of two numbers\n\nHello everyone, welcome back to programminginpython.com. Here I am going to tell a simple logic by which we can find the LCM of two numbers in python. LCM means Least Common Multiple, for a given …\n\n## Python program to print Fibonacci sequence\n\nHello everyone, welcome to programminginpython.com. Here I am going to discuss a python program that prints the Fibonacci sequence of a given number. Fibonacci sequence is calculated by summing up all the numbers below that …\n\n## Find square root of a number using exponential operation\n\nHello everyone, welcome back! Here we learn about a python program which finds the square root of a given number. We find the square-root of a number using exponential operations. Here I calculate the square-root …"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.88094914,"math_prob":0.96996623,"size":1653,"snap":"2021-43-2021-49","text_gpt3_token_len":357,"char_repetition_ratio":0.13887204,"word_repetition_ratio":0.3457627,"special_character_ratio":0.20871143,"punctuation_ratio":0.09726444,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9983505,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-05T04:03:28Z\",\"WARC-Record-ID\":\"<urn:uuid:891d2d9d-a76f-451a-904e-e47c7edcb4c1>\",\"Content-Length\":\"77130\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dfa377de-bb7a-4ba5-96cd-cc5a15bb866e>\",\"WARC-Concurrent-To\":\"<urn:uuid:7b83b7de-764d-484c-a8a6-f17287e93c52>\",\"WARC-IP-Address\":\"128.199.31.198\",\"WARC-Target-URI\":\"https://www.programminginpython.com/category/math-programs/\",\"WARC-Payload-Digest\":\"sha1:4D2KSHGGZ7ML5J2QETETLO43JEES6VQX\",\"WARC-Block-Digest\":\"sha1:D425ACHALY22EU3N7VU3ZPOSDILFXLNF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363135.71_warc_CC-MAIN-20211205035505-20211205065505-00067.warc.gz\"}"} |
https://win-vector.com/2016/05/30/on-ranger-respect-unordered-factors/ | [
"# On ranger respect.unordered.factors\n\nIt is often said that “R is its packages.”\n\nOne package of interest is ranger a fast parallel C++ implementation of random forest machine learning. Ranger is great package and at first glance appears to remove the “only 63 levels allowed for string/categorical variables” limit found in the Fortran randomForest package. Actually this appearance is due to the strange choice of default value `respect.unordered.factors=FALSE` in `ranger::ranger()` which we strongly advise overriding to `respect.unordered.factors=TRUE` in applications.\n\nTo illustrate the issue we build a simple data set (split into training and evaluation) where the dependent (or outcome) variable `y` is given as the sum of how many input level codes end in an odd digit minus how many input level codes end in an even digit.\n\nSome example data is given below\n\n``` print(head(dTrain)) ## x1 x2 x3 x4 y ## 77 lev_008 lev_004 lev_007 lev_011 0 ## 41 lev_016 lev_015 lev_019 lev_012 0 ## 158 lev_007 lev_019 lev_001 lev_015 4 ## 69 lev_010 lev_017 lev_018 lev_009 0 ## 6 lev_003 lev_014 lev_016 lev_017 0 ## 18 lev_004 lev_015 lev_014 lev_007 0 ```\n\nGiven enough data this relation is easily learnable. In our example we have only 100 training rows and 20 possible levels for each input variable- so we at best get a noisy impression of how each independent (or input) variable affects `y`.\n\nWhat the default ranger default training setting `respect.unordered.factors=FALSE` does is decide that string-valued variables (such as we have here) are to be treated as “ordered”. This allows ranger to skip any of the expensive re-encoding of such variables as contrasts, dummies or indicators. This is achieved in ranger by only using ordered cuts in its underlying trees and is equivalent to re-encoding the categorical variable as the numeric order codes. These variables are thus essentially treated as numeric, and ranger appears to run faster over fairly complicated variables.\n\nThe above is good if all of your categorical variables were in fact known to have ordered relations with the outcome. We must emphasize that this is very rarely the case in practice as one of the main reasons for using categorical variables is that we may not a-priori know the relation between the variable levels and outcome and would like the downstream machine learning to estimate the relation. The default `respect.unordered.factors=FALSE` in fact weakens the expressiveness of the ranger model (which is why it is faster).\n\nThis is simpler to see with an example. Consider fitting a ranger model on our example data (all code/data shared including classification and use of parallel here).\n\nIf we try to build a ranger model on the data using the default settings we get the following:\n\n``` # default ranger model, treat categoricals as ordered (a very limiting treatment) m1 <- ranger(y~x1+x2+x3+x4, data=dTrain, write.forest=TRUE) ```",
null,
"Keep in mind the 0.24 R-squared on test.\n\nIf we set `respect.unordered.factors=TRUE` ranger takes a lot longer to run (as it is doing more work in actually respecting the individual levels of our categorical variables) but gets a much better result (test R-squared 0.54).\n\n``` m2 <- ranger(y~x1+x2+x3+x4, data=dTrain, write.forest=TRUE, respect.unordered.factors=TRUE) ```",
null,
"The loss of modeling power seen with the default `respect.unordered.factors=FALSE` is similar to the undesirable loss of modeling power seen if one hash-encodes categorical levels. The default behavior of `ranger` is essentially equivalent to calling `as.numeric(as.factor())` on the categorical columns. Everyone claims they would never do such a thing (hash or call `as.numeric()`), but we strongly suggest inspecting your team’s work for these bad but tempting shortcuts.\n\nIf even one of the variables had 64 or more levels ranger would throw an exception and not complete training (as the randomForest library also does).\n\nThe correct way to feed large categoricals to a random forest model remains to explicitly introduce the dummy/indicators yourself or re-encode them as impact/effect sub models. Both of these are services supplied by the vtreat package so we demonstrate the technique here.\n\n``` # vtreat re-encoded model ct <- vtreat::mkCrossFrameNExperiment(dTrain, c('x1','x2','x3','x4'), 'y') newvars <- ct\\$treatments\\$scoreFrame\\$varName[(ct\\$treatments\\$scoreFrame\\$code=='catN') & (ct\\$treatments\\$scoreFrame\\$sig<1)] m3 <- ranger(paste('y',paste(newvars,collapse=' + '),sep=' ~ '), data=ct\\$crossFrame, write.forest=TRUE) dTestTreated <- vtreat::prepare(ct\\$treatments,dTest, pruneSig=c(),varRestriction=newvars) dTest\\$rangerNestedPred <- predict(m3,data=dTestTreated)\\$predictions WVPlots::ScatterHist(dTest,'rangerNestedPred','y', 'ranger vtreat nested prediction on test', smoothmethod='identity',annot_size=3) ```",
null,
"The point is a test R-squared of 0.6 or 0.54 is a lot better than an R-squared of 0.24. You do not want to achieve 0.24 if 0.6 is within easy reach. So at the very least when using ranger set `respect.unordered.factors=TRUE`; for unordered factors (the most common kind) the default is making things easy for ranger at the expense of model quality.\n\nInstructions explaining the use of `vtreat` can be found here.\n\nTagged as:",
null,
"### jmount\n\nData Scientist and trainer at Win Vector LLC. One of the authors of Practical Data Science with R.\n\n### 9 replies ›\n\n1.",
null,
"Carlos Ortega says:\n\nThanks for this illustrative example that demonstrate how powerful can be “vtreat” an clarifies up a lot the meaning of “respect.unordered.factors” in “ranger” package.\n\nIn this example you outcome “y” can be considered as a categorical variable (multilevel), but you consider it as “numeric”.\n\nI am trying to fully understand if “designTreatmentsC” is just for *binary* categorical outcomes. For multilevel class of problems, the outcome should it be converted to numeric?.\nAs far as I see, it can be coded without any kind of limitation (I mention this because xgboost for instante require values higher than 0).\n\nThanks again,\nCarlos Ortega.\n\n1.",
null,
"John Mount says:\n\nCarlos, thanks for the interesting discussion!\n\nAs background we have the following.\n\n`vtreat::designTreatmentsN` and `vtreat::mkCrossFrameNExperiment` are for numeric or regression problems. `vtreat::designTreatmentsC` and `vtreat::mkCrossFrameCExperiment` are for binary categorization problems (in this case `y` can one of many types: character, factor, numeric and we treat it as a binary outcome using the user-supplied `outcometarget` which says what value of `y` is considered “TRUE” considering all other values to be “FALSE”). Currently `vtreat` does not directly support multi-class problems (though one could try to emulate such as a series of binary classification problems).\n\nFor the problem at hand (the `ranger` example) we are treating `y` as a numeric outcome to be regressed against. As you noticed this is not exploiting the domain fact that in this example `y` can only take on the values `-4`, `-3`, `-2`, `-1`, `0`, `1`, `2`, `3`, `4`. With this many values regression isn’t a bad approximation (and it does get the domain advantage of being able to immediately exploit the order relations in `y`). The most powerful way to encode this problem would be some hybrid that exploits both the moderate number of possible values (multi-class classification) and the order relations.\n\nSo really I see at least four types of predictive problems: regression, binary classification, unordered multinomial classification, and ordered multinomial classification. `vtreat` directly supports the first two. It would be nice to also directly support unordered multinomial classification (the math is easy, just would require some code changes). And for this example problem a system that supported ordered multinomial outcomes would likely be the most powerful (though it is uncommon to see this implemented in general packages). Ordered multinomial could also be done, but it would take a bit more engineering.\n\n1.",
null,
"Carlos Ortega says:\n\nOK. Thanks.\nI wanted to apply “vtreat” to a three class problem, by considering it as a regression problem.\nThe dataset has some columns with high cardinality and some other with NAs so *vtreat* was a good choice to handle all these thing together.\n\nSo far, I could handle the high cardinality with the hash-trick feature approach, but seeing your package, I saw a possibillity to treat everything in one shot. The algorithm I am using (after trying and assessing their performance, is a randomForest through “ranger”, that manages multiclass classification without any problem.\n\nAlthough it is not a very recommendable way to proceed I am going to try to model it as a regression to take advantage of “vtreat” as it is now.\n\nThanks again,\nCarlos Ortega.\n\n2.",
null,
"John Mount says:\n\nSince it is a three class problem (not too many outcome classes) I would suggest also trying building 3 “one versus rest” binary classifiers ( https://en.wikipedia.org/wiki/Multiclass_classification#One-vs.-rest ). It is slow (you end up running ranger 3 times) to get three probabilities ( pA versus pNotA, pB versus pNotB, and pC versus pNotC) and then re-normalizing them to sum to 1 as your “multiclass” classifier (pA/(pA+pB+pC), pB/(pA+pB+pC), pC/(pA+pB+pC); it is an abuse as if pA,pB,pC were really disjoint and complete probabilities on the same event they would already add up to 1). It is a bit slower (and theoretically a bit less powerful than an all in one multiclass classifier) but a good method all the same.\n\n3.",
null,
"Carlos Ortega says:\n\nThanks!\nI used “vtreat” with the dataset I referred to you yesterday.\n\nOne of the doubts I got, following your example, was the reason to use just “catN” as a filter for “newvars” variable.\n\nIn my case, I tried first just “catN” that got a good scoring, but also including the “clean” variables the scoring got better. Besides high cardinality, my dataset also has many “NAs” and “vtreat” run smoothly all over it.\n\nI want to try also with no filter and use all the new variables created by “vtreat”. In my case, that will provide many more new variables, although very sparse.\n\nAnd also, I used “ranger” with “classification = TRUE” when I saw all of this.\n\nI will let you know how this behaves.\n\nI am a little confuse about is behind “catN”, “catD”, etc..\nAnd in the different vignettes I could not find many details. If you please could point me out to the adequate place…\n\nThanks again,\nCarlos Ortega.\n\n2.",
null,
"Carlos Ortega says:\n\nSorry, I just found that you have a particular vignette with a description of the meaning of “catN”, “catD”, etc.. (different “VariableTypes”).\n\nThanks,\nCarlos.\n\n1.",
null,
"John Mount says:\n\nNot a problem, hope `vtreat` is working well for you. For anyone else interested in what the variable types are here is a link to an explanation: http://winvector.github.io/vtreathtml/vtreatVariableTypes.html . Normally you just take all the new variables that turn out to be significant (especially all the “`clean`” pass-throughs). Dealing with `NA` is one of `vtreat`‘s core services Nina Zumel was written a bit on this here http://winvector.github.io/DataPrep/EN-CNTNT-Whitepaper-Data-Prep-Using-R.pdf .\n\n3.",
null,
"Marvin N. Wright says:\n\nThank you for this post! Inspired by this, I’ve implemented the approach described by Hastie et al. in their book “The Elements of Statistical Learning”, chapter 9.2.4 (see also https://github.com/imbs-hl/ranger/issues/36#issuecomment-203967512).\n\nSince ranger v0.4.5 (available at https://github.com/imbs-hl/ranger) this method is used by default. I tried your example and it was as fast as the model with respect.unordered.factors=FALSE but as good as the model with respect.unordered.factors=TRUE.\n\nPlease note that there are now 3 options for respect.unordered.factors, see the ranger R help for details.\n\n1.",
null,
"John Mount says:\n\nHi Marvin, thanks for you note! Really neat to hear from a ranger developer! The “split by sorting and scanning” idea shown in 9.2.4 is pretty common in combinatorial optimization (I was a bit surprised to see the book says it is hard to prove theorems about it, but I guess the sorting step makes thing hard to reason about).\n\nIt would be great if more tree based methods didn’t require pre-encoding categorical variables to work well. As you have found there are some great ideas out there.\n\n(edit: Wow! Version 0.4.5 came out 2016-05-31, so you are not kidding about taking some inspiration! Really neat, great stuff.)"
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https://answers.everydaycalculation.com/add-fractions/60-2-plus-21-12 | [
"Solutions by everydaycalculation.com\n\n1st number: 30 0/2, 2nd number: 1 9/12\n\n60/2 + 21/12 is 127/4.\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 2 and 12 is 12\n\nNext, find the equivalent fraction of both fractional numbers with denominator 12\n2. For the 1st fraction, since 2 × 6 = 12,\n60/2 = 60 × 6/2 × 6 = 360/12\n3. Likewise, for the 2nd fraction, since 12 × 1 = 12,\n21/12 = 21 × 1/12 × 1 = 21/12\n4. Add the two like fractions:\n360/12 + 21/12 = 360 + 21/12 = 381/12\n5. 381/12 simplified gives 127/4\n6. So, 60/2 + 21/12 = 127/4\nIn mixed form: 313/4\n\nMathStep (Works offline)",
null,
"Download our mobile app and learn to work with fractions in your own time:"
]
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null,
"https://answers.everydaycalculation.com/mathstep-app-icon.png",
null
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https://www.clutchprep.com/chemistry/practice-problems/73590/a-flexible-container-at-an-initial-volume-of-7-14-l-contains-2-51-mol-of-gas-mor | [
"# Problem: A flexible container at an initial volume of 7.14 L contains 2.51 mol of gas. More gas is then added to the container until it reaches a final volume of 17.3 L. Assuming the pressure and temperature of the gas remain constant, calculate the number of moles of gas added to the container.\n\n###### FREE Expert Solution\n97% (90 ratings)",
null,
"###### Problem Details\n\nA flexible container at an initial volume of 7.14 L contains 2.51 mol of gas. More gas is then added to the container until it reaches a final volume of 17.3 L. Assuming the pressure and temperature of the gas remain constant, calculate the number of moles of gas added to the container.",
null,
""
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null,
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null,
"https://lightcat-files.s3.amazonaws.com/problem_images/6c4238043470c4ab-1527252489614.jpg",
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https://www.zora.uzh.ch/id/eprint/108314/ | [
"",
null,
"# Measurement of resonant and $CP$ components in $\\overline{B}_s^0\\rightarrow J/\\psi \\pi^+\\pi^-$ decays\n\nLHCb Collaboration; Bernet, R; Müller, K; Steinkamp, O; Straumann, U; Vollhardt, A; et al (2014). Measurement of resonant and $CP$ components in $\\overline{B}_s^0\\rightarrow J/\\psi \\pi^+\\pi^-$ decays. Physical Review D (Particles, Fields, Gravitation and Cosmology), 89:092006.\n\n## Abstract\n\nThe resonant structure of the decay $\\overline{B}_s^0\\to J/\\psi\\pi^+\\pi^-$ is studied using data corresponding to 3 fb$^{-1}$ of integrated luminosity from $pp$ collisions by the LHC and collected by the LHCb detector. Five interfering $\\pi^+\\pi^-$ states are required to describe the decay: $f_0(980),f_0(1500),f_0(1790),f_2(1270)$, and $f_2^{\\prime}(1525)$. An alternative model including these states and a non-resonant $J/\\psi \\pi^+\\pi^-$ component also provides a good description of the data. Based on the different transversity components measured for the spin-2 intermediate states, the final state is found to be compatible with being entirely $CP$-odd. The $CP$-even part is found to be $<2.3$% at 95% confidence level. The $f_0(500)$ state is not observed, allowing a limit to be set on the absolute value of the mixing angle with the $f_0(980)$ of $<7.7^{\\circ}$ at 90% confidence level, consistent with a tetraquark interpretation of the $f_0(980)$ substructure.\n\n## Abstract\n\nThe resonant structure of the decay $\\overline{B}_s^0\\to J/\\psi\\pi^+\\pi^-$ is studied using data corresponding to 3 fb$^{-1}$ of integrated luminosity from $pp$ collisions by the LHC and collected by the LHCb detector. Five interfering $\\pi^+\\pi^-$ states are required to describe the decay: $f_0(980),f_0(1500),f_0(1790),f_2(1270)$, and $f_2^{\\prime}(1525)$. An alternative model including these states and a non-resonant $J/\\psi \\pi^+\\pi^-$ component also provides a good description of the data. Based on the different transversity components measured for the spin-2 intermediate states, the final state is found to be compatible with being entirely $CP$-odd. The $CP$-even part is found to be $<2.3$% at 95% confidence level. The $f_0(500)$ state is not observed, allowing a limit to be set on the absolute value of the mixing angle with the $f_0(980)$ of $<7.7^{\\circ}$ at 90% confidence level, consistent with a tetraquark interpretation of the $f_0(980)$ substructure.\n\n## Statistics\n\n### Citations\n\nDimensions.ai Metrics\n75 citations in Web of Science®\n61 citations in Scopus®\n\n### Altmetrics\n\nDetailed statistics\n\nItem Type: Journal Article, refereed, original work 07 Faculty of Science > Physics Institute 530 Physics Physical Sciences > Nuclear and High Energy Physics Physical Sciences > Physics and Astronomy (miscellaneous) English 2014 24 Feb 2015 14:34 30 Jul 2020 17:24 American Physical Society 1550-2368 Hybrid https://doi.org/10.1103/PhysRevD.89.092006\n\n##",
null,
"",
null,
"",
null,
""
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null,
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null,
"https://www.zora.uzh.ch/108314/1.hassmallThumbnailVersion/1402.6248v1.pdf",
null,
"https://www.zora.uzh.ch/108314/1.haspreviewThumbnailVersion/1402.6248v1.pdf",
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