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[ "English | Español\n\n# Try our Free Online Math Solver!", null, "Online Math Solver\n\n Depdendent Variable\n\n Number of equations to solve: 23456789\n Equ. #1:\n Equ. #2:\n\n Equ. #3:\n\n Equ. #4:\n\n Equ. #5:\n\n Equ. #6:\n\n Equ. #7:\n\n Equ. #8:\n\n Equ. #9:\n\n Solve for:\n\n Dependent Variable\n\n Number of inequalities to solve: 23456789\n Ineq. #1:\n Ineq. #2:\n\n Ineq. #3:\n\n Ineq. #4:\n\n Ineq. #5:\n\n Ineq. #6:\n\n Ineq. #7:\n\n Ineq. #8:\n\n Ineq. #9:\n\n Solve for:\n\n Please use this form if you would like to have this math solver on your website, free of charge. 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[ "# Need Predictor Importance in Random Forest Expressed as a Percentage\n\n2 views (last 30 days)\nCMatlabWold on 25 Mar 2021\nCommented: CMatlabWold on 9 Apr 2021\nHi. I'm running a code to see the importance of demographics (Predictors) on my response (Complaints). I need to express the importance as percentage, as a scale of 0 to 1 (or 0% to 100%). This is the figure I am getting is attached as \"RF Importance Chart\". My predictors data is attached as \"PredictorsOnly.xlsx\" and my response data is attached as \"TotalComplaintsRF.xlsx\"\nt = templateTree('NumVariablesToSample','all',...\n'PredictorSelection','interaction-curvature','Surrogate','on');\nrng(1); % For reproducibility\nMdl = fitrensemble(X,Y,'Method','Bag','NumLearningCycles',200, ...\n'Learners',t);\nyHat = oobPredict(Mdl);\nR2 = corr(Mdl.Y,yHat)^2\nimpOOB = oobPermutedPredictorImportance(Mdl);\nfigure\nbar(impOOB)\ntitle('Unbiased Predictor Importance Estimates')\nxlabel('Predictor variable')\nylabel('Importance')\nh = gca;\nh.XTickLabel = Mdl.PredictorNames;\nh.XTickLabelRotation = 45;\nh.TickLabelInterpreter = 'none';\n\nPratyush Roy on 9 Apr 2021\nHi,\noobPermutedPredictorImportance normalizes the predictor importance by the standard error (this is common practice in the field), therefore values are not strictly scaled between 0 and 1. However one can rescale predictor importance, for example:\nimp(imp<0) = 0;\nimp = imp./sum(imp);\nHope this helps!\nCMatlabWold on 9 Apr 2021\nThank you - it works!\n\n### Categories\n\nFind more on Classification Ensembles in Help Center and File Exchange\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!" ]

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[ "# Why do charged particles attract or repel each other?\n\nContents\n\nIf a positive charge and a negative charge interact, their forces act in the same direction, from the positive to the negative charge. As a result opposite charges attract each other: The electric field and resulting forces produced by two electrical charges of opposite polarity. The two charges attract each other.\n\n## Why do charged objects attract or repel each other?\n\nOppositely charged objects will exert an attractive influence upon each other. … This repulsive force will push the two objects apart. Similarly, a negatively charged object will exert a repulsive force upon a second negatively charged object. Objects with like charge repel each other.\n\n## Why do two charges repel each other?\n\nIf we put similar charged particles nearby;they tend to repel because their electric field never reaches at the same point. Thats why they tend to repel. 1)Can an electric field terminate at positive charge?\n\n## How do charges attract and repel?\n\nAccording to Coulomb, the electric force for charges at rest has the following properties: Like charges repel each other; unlike charges attract. Thus, two negative charges repel one another, while a positive charge attracts a negative charge. The attraction or repulsion acts along the line between the two charges.\n\n## Does like charges attract each other True or false?\n\nThe like charges mean the charges of the same kind and unlike charges mean charges of different kinds. We know like charges repel each other while unlike charges attract each other. Therefore, the first statement is false and the second statement is true. So, the correct answer is “Option C”.\n\n## Can like charges attract each other?\n\nLike charges repel each other. A charged glass rod has positive charge and a charged plastic straw has a negative charge. Since opposite charges attract each other, they attract each other.\n\n## Do electrostatic charges attract or repel each other explain?\n\nAttraction and repulsion of electric charges is one of three fundamental non-contact forces in nature. … Positive and negative charged objects attract or pull each other together, while similar charged objects (2 positives or 2 negatives) repel or push each other apart.\n\n## What is attracted to negatively charged particles and repelled from positively charged particles?\n\nElectricity is the behavior of negatively and positively charged particles due to their attraction and repulsion. Like charges (two negatively charged particles or two positively charged particles) repel each other while opposite charges (a positively charged particle and a negatively charged particle) attract.\n\n## Why do electrons not repel each other?\n\nWithin Helium, for example, with 2 paired electrons they don’t repel each other because they have an opposite “spin”. The electron has a negative charge and a magnetic field at right angles to each other." ]

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[ "# help!\n\nmean absolute deviation ?\n\n1. 👍 0\n2. 👎 0\n3. 👁 85\n1. I searched Google under the key words \"mean absolute deviation\" to get these possible sources:\n\nIn the future, you can find the information you desire more quickly, if you use appropriate key words to do your own search. Also see http://hanlib.sou.edu/searchtools/.\n\n1. 👍 0\n2. 👎 0\nposted by PsyDAG\n2. thanks\n\n1. 👍 0\n2. 👎 0\n\n## Similar Questions\n\n1. ### Math\n\nPlease someone explain mean absolute deviation and give an example because I have to solve this one question that asks of me to find the mean absolute deviation of 49, 55, 52, 46, 47, 42, and 38 and I don't understand what it is.\n\nasked by Mariana on October 4, 2018\n2. ### Math\n\nNeed this answered today also explain the answers please Julia measured the high temperature in her town for one week. Using the chart below, find the mean absolute deviation for the high temperatures. Round your answer to the\n\nasked by Annie on March 13, 2018\n3. ### Chemistry\n\nCalculate the value of Z and its absolute standard deviation. All standard deviations shown are absolute. Show your work and report the standard deviation in the result with two significant digits (to confirm the correct result).\n\n4. ### Statistics\n\nThe least absolute deviation line equation for the data in the table is m = 0.1 x + 2.9 X 1,6,0.5,4,7.5,12,10,11 y 3,2.5,8,1,1.5,9,13,4 What is the sum of the absolute deviations?\n\nasked by BoB on December 12, 2018\n5. ### math\n\nDay Sunday 49 Monday 55 Tuesday 52 Wednesday 46 Thursday 47 Friday 42 Saturday 38 Temperature (degrees F) Julia measured the high temperature in her town for one week. Using the chart above, find the mean absolute deviation for\n\nasked by tori from outer space on September 10, 2019\n6. ### Math\n\nScores from the same benchmark test are collected from two algebra classes, each with 30 students enrolled. One class had a mean score of 79 with a mean absolute deviation of 5, and on the other hand a mean score of 81 with the\n\nasked by Me on March 23, 2016\n7. ### Math - Help!\n\nMrs. Higgins' Home Economics class collected data on the number of chocolate chips in cookies for two different brands, as shown in the dot plots below. The mean absolute deviation for each brand is 0.8. The difference between the\n\n8. ### Calculus (pleas help!!!)\n\nFind the absolute maximum and absolute minimum values of the function f(x)=(x−2)(x−5)^3+11 on each of the indicated intervals. Enter -1000 for any absolute extrema that does not exist. (A) Interval = [1,4] Absolute maximum =\n\n9. ### Calculus (pleas help!!!)\n\nFind the absolute maximum and absolute minimum values of the function f(x)=(x−2)(x−5)^3+11 on each of the indicated intervals. Enter -1000 for any absolute extrema that does not exist. (A) Interval = [1,4] Absolute maximum =" ]

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[ "# C + +: Topic 5 experiment (classes and functions)\n\n## C + +: Topic 5 experiment (classes and functions)\n\n### Purpose of the experiment:\n\n1. Learn the definition and call of functions.\n2. Understand the parameter transfer mechanism of function and master different parameter transfer methods.\n3. Master the method of using function to realize recursion.\n4. Master class definition, class object definition, assignment and basic access methods, and learn class member functions.\n5. Master the method of accessing classes with pointers and learn the access control of classes.\n6. Master the principle and use of class constructor and analytic function.\n\n### Experiment content:\n\n1. When passing an array as a function parameter, you must pass in an array or a pointer, not a reference. According to the above knowledge, write a program to realize bubble sorting. The basic algorithm is to put the smallest element of the last row of elements in front of each cycle, and the input parameters are array and an integer representing the number of elements to be sorted. Please review the method of passing function parameters through practice and learn the method of passing in arrays.\n2. A commonly used calculation of combinatorial number C, n and m in mathematics, please write a program to give the calculation method of combinatorial number by recursive method.\n3. Define a Book class representing a book. It is required to include the following attributes: book name, author, number of pages and price. Define an appropriate constructor. The first constructor can initialize all attributes, the second constructor can initialize the two attributes of book name and author, and the last constructor has no parameters and only outputs a statement representing a new object. Define three different objects with different constructors and output their initialized properties. Next, add a destructor whose internal implementation is to output a statement representing the recycled object. Design appropriate output statements to observe the order of recycling objects.\n4. (optional) in graphics, a mesh composed of triangles is used to represent three-dimensional objects in a hundred years. A triangular surface should contain at least the following information: the number of the surface, the color of the surface (which can be expressed by three integers), and the three-dimensional coordinates of the three vertices of the surface. Define a Node class representing vertices and a Face class representing faces according to the above format. The Face class should contain members of Node type. Define the object f of Face class, and output its number, color value and 3D coordinates of any point.\n\n#### (1) Write a program to realize bubble sorting. Please review the method of passing function parameters through practice and learn the method of passing in arrays.\n\n```#include <iostream>\n\nusing namespace std;\n\nint bubbleSort(int arr[], int len)\n{\nfor(int i = len; i > 0; i--)\nfor(int j = 0; j < len - 1; j++)\n{\nif(arr[j] > arr[j + 1])\n{\nint t = arr[j + 1];\narr[j + 1] = arr[j];\narr[j] = t;\n}\n}\n\nfor(int k = 0; k < len; k++)\n{\nif(k == len - 1)\ncout << arr[k] << endl;\nelse\ncout << arr[k] << \"<\";\n}\n\nreturn 0;\n}\n\nint main()\n{\nint len = 10;\nint a[len];\ncout << \"Please enter 10 numbers (separated by spaces):\" << endl;\n\nfor(int i = 0; i < len; i++)\ncin >> a[i];\n\nbubbleSort(a, len);\nreturn 0;\n}\n```", null, "#### (2) A commonly used calculation of combinatorial number C, n and m in mathematics, please write a program to give the calculation method of combinatorial number by recursive method.\n\n```#include <iostream>\n\nusing namespace std;\n\nint factorial(int m)\n\n{\nif(m == 1)\nreturn 1;\n\nreturn factorial(m - 1) * m;\n}\n\nint main()\n{\nint m, n;\ncout << \"Please enter Cmn Medium m and n(m>=n):\" << endl;\ncout << \"m = \";\ncin >> m;\ncout << \"n = \";\ncin >> n;\n\nif(m >= n)\ncout << \"Cmn The value of is:\" << factorial(m) / factorial(n) / factorial(m - n) << endl;\nelse\ncout << \"Wrong input!\" << endl;\n\nreturn 0;\n}\n```", null, "#### (3) Define a Book class representing a book. It is required to include the following attributes: book name, author, number of pages and price. Define an appropriate constructor. The first constructor can initialize all attributes, the second constructor can initialize the two attributes of book name and author, and the last constructor has no parameters and only outputs a statement representing a new object. Define three different objects with different constructors and output their initialized properties. Next, add a destructor whose internal implementation is to output a statement representing the recycled object. Design appropriate output statements to observe the order of recycling objects.\n\n```#include <iostream>\n\nusing namespace std;\n\nclass Book\n{\npublic:\n\nBook(string Name, string Author, int Pages, float Price);\nBook(string Name, string Author);\nBook();\n~Book();\n\nstring name;\nstring author;\nint pages;\nfloat price;\n};\n\nBook::Book(string Name, string Author, int Pages, float Price)\n{\nname = Name, author = Author, pages = Pages, price = Price;\n};\n\nBook::Book(string Name, string Author)\n{\nname = Name, author = Author;\n};\n\nBook::Book()\n{\ncout << \"A Book object has been created.\" << endl;\n};\n\nBook::~Book()\n{\ncout << \"The Book object: \" << name << \" has been deleted.\" << endl;\n};\n\nint main()\n{\nBook b1(\"Advanced mathematics\", \"Fudan University Press \", 150, 28);\nBook b2(\"linear algebra\", \"Central China Normal University Press\");\nBook b3;\ncout << \"b1:\" << b1.name << \" \" << b1.author << \" \" << b1.pages << \"page \" << b1.price << \"element \" << endl;\ncout << \"b2:\" << b2.name << \" \" << b2.author << endl;\ncout << \"b3:\" << endl;\nreturn 0;\n}\n```", null, "#### (4) (optional) in graphics, a mesh composed of triangles is used to represent three-dimensional objects in a hundred years. A triangular surface should contain at least the following information: the number of the surface, the color of the surface (which can be expressed by three integers), and the three-dimensional coordinates of the three vertices of the surface. Define a Node class representing vertices and a Face class representing faces according to the above format. The Face class should contain members of Node type. Define the object f of Face class, and output its number, color value and 3D coordinates of any point.\n\n```#include <iostream>\n\nusing namespace std;\n\nclass Node\n{\npublic:\nfloat vertex;\n};\n\nclass Face\n{\npublic:\nint number;\n//Experience the function of using pointers for these two parameters\nint* color;\nNode* node;\n\nFace(int Num, int colr, Node N);\n};\n\nFace::Face(int Num, int colour, Node N)\n{\nnumber = Num, color = colour, node = N;\n}\n\nint main()\n{\nint colour = {108, 234, 255};\nNode N = {{1, 2, 3}, {2, 1, 3}, {3, 2, 1}};\nFace f(1, colour, N);\ncout << \"f:number \" << f.number << endl;\ncout << \"color (\" << f.color << \",\" << f.color << \",\" << f.color << \")\" << endl;\ncout << \"onenode \" << f.node.vertex << \", \" << f.node.vertex << \", \" << f.node.vertex << endl;\nreturn 0;\n}\n```", null, "Tags: C++ Back-end\n\nPosted on Tue, 16 Nov 2021 23:39:25 -0500 by mjedman1" ]

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[ "# 9813 ml to grams – 9813 ml in g\n\nHere you can find 9813 ml to grams. No matter if you have been looking for 9813 ml in grams or 9813 ml in g, this is the right page for you. The result of the conversion depends on the density of the material or substance under consideration. The answer to the question 9813 ml is how many grams also varies with the pressure and the temperature. Read on below to understand everything about 9813 milliliters to grams.\n\n## Convert 9813 ml to grams\n\nTo convert 9813 ml to grams, besides the volume we have to know the substance’s density ρ in g/cm3 or in any other unit. Alternatively, when we know the material, then we can look for its density in a search engine. As explained on our home page, 9813 ml equal 9813 cm3, and as also outlined there, in common use the equation for water is 9813 milliliter = 9813 g.\n\n9813 ml to grams water = 9813 g\n9813 milliliters water to grams = 9813 g\n\nBut what about other stuff and food? In the next section you can find the mass equivalent of 9813 ml for some cooking ingredients. Below is our converter which calculates 9813 ml to g for any substance with known ρ.\n\nThe result is...\n\nBookmark us now as ml to grams.\n\nHere you can convert 9813 grams to ml.\n\n## 9813 ml in g\n\nUnless you have made use of our calculator, only for water under certain conditions do you know 9813 ml in g. And kindly note that 9813 ml to grams means the same as 9813 mL to grams.\n\n9813 ml equals how many grams for food ingredients is next:\n\nMilk = 10107.39 g, (ρ = 1.03)\nCream = 9925.85 g, (ρ = 1.0115)\nFlour = 5819.109 g, (ρ = 0.593)\nSugar = 7703.205 g, (ρ = 0.785)\nButter = 8939.643 g, (ρ = 0.911)\n\nρ is stated in g/cm3.", null, "Make sure to understand that these values for 9813 ml in grams are averages. Depending on the precise amount of fat, the quality or the kind, as well as other factors, 9813 ml to gram can be a different weight. We have explained all this in detail in our article ml to grams, where you can also find additional information on mass, volume, liter and kilogram.\n\nBesides 9813 ml to grams, similar conversions on this website include:\n\nThis concludes 9813 ml into grams. We hope you like our article and hit the sharing buttons for 9813 ml to gram. All questions and comments are really appreciated.\n\nThanks for your visit!\n\nPosted in Milliliters to Grams" ]

[ null, "https://mltograms.com/wp-content/images/9813_ml_to_grams.png", null ]

[ "# Modified Discrete Cosine Transform\n\n(Redirected from MDCT)\n\nThe Modified Discrete Cosine Transform (MDCT) is a DCT-IV transform. The MDCT tries to minimize blocking artifacts. It is common in lossy audio codecs including MP3, Vorbis, and AAC.\n\nAs a lapped transform, the MDCT is a bit unusual compared to other Fourier-related transforms in that it has half as many outputs as inputs (instead of the same number). In particular, it is a linear function F : R2n -> Rn (where R denotes the set of real numbers). The 2n real numbers x0, ..., x2n-1 are transformed into the n real numbers f0, ..., fn-1 according to the formula:", null, "The inverse MDCT is known as the IMDCT. Because there are different numbers of inputs and outputs, at first glance it might seem that the MDCT should not be invertible. However, perfect invertibility is achieved by adding the overlapped IMDCTs of subsequent overlapping blocks, causing the errors to cancel and the original data to be retrieved; this technique is known as time-domain aliasing cancellation (TDAC).\n\nThe IMDCT transforms n real numbers f0, ..., fn-1 into 2n real numbers y0, ..., y2n-1 according to the formula:", null, "The MDCT was developed by Henrique Malvar, an engineer now working for Microsoft who had considerable participation in creating the WMA format." ]

[ null, "https://wiki.hydrogenaud.io/images/3/35/MDCT.png", null, "https://wiki.hydrogenaud.io/images/7/71/IMDCT.png", null ]

[ "# An Equation in Determinants\n\nDan Sitaru has posted the following problem and its solution at the CutTheKnotMath facebook page:\n\nSolve in $M_{4}(\\mathbb{Z})$ the equation:\n\n$\\det (X^{4}+I_{4})=2013.$\n\n### Solution\n\nObserve that $X^{4}+I_{4}=(X^{2}-X\\sqrt{2}+I_{4})(X^{2}+X\\sqrt{2}+I_{4})$ from which\n\n$\\det (X^{4}+I_{4})=(m-n\\sqrt{2})(m+n\\sqrt{2})=m^{2}-2n^{2};$ $m,n\\in\\mathbb{Z}.$\n\nNow, $m^{2}-2n^{2}=2013$ is same as $m^{2}-2013=2n^{2}.$ The latter implies $m\\in 2\\mathbb{Z}+1,$ and consequently $m^{2}\\in 8\\mathbb{Z}+1.$ We'll show that this is impossible by considering two cases.\n\nIf $n\\in 2\\mathbb{Z}$ then $2n^{2}\\in 8\\mathbb{Z}$ and $m^{2}-2n^{2}\\in 8\\mathbb{Z}+1$ which means that $2013\\in 8\\mathbb{Z}+1.$ But this is not so because $2013 \\mod 8 \\equiv 5.$\n\nIf $n\\in 2\\mathbb{Z}+1$ then $n^{2}\\in 8\\mathbb{Z}$ so that $m^{2}-2n^{2}\\in 8\\mathbb{Z}+7,$ or $2013\\in 8\\mathbb{Z}+5,$ but this is as impossible as the previous case.\n\nThus the equation has no solutions.\n\n### Remark\n\nThe problem can be stated as a scalar or a polynomial equation, with only minor typographical changes.\n\nThe problem poses the question for the year 2013. The above solution will work for any year whose residue of division by $8$ is neither $1$ nor $7.$ Thus it will not work out (as it was pointed out in the comments) for $2015;$ there is enough time to investigate whether the matrix equation has or does not have a solution for the coming year. The scalar equivalent, obviously, does not have a solution in integers for most of the years.", null, "" ]

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[ "top of page\nSearch\n\n# Metals and Non-Metals Class 10 Important Questions with Answers Science Chapter 3\n\nWe have given these Important Questions for Class 10 Science Chapter 3 Metals and Non-Metals to solve different types of questions in the exam. Previous Year Questions & Important Questions of Metals and Non-Metals Class 10 Science Chapter 3 will help the students to score good marks in the board examination. Certainly! Here are 30 short-type questions based on the topics of Metals and Non-Metals for Class IX CBSE board, along with their answers:\n\n1. Q: Define a metal.\n\nA: Metals are elements that are generally solid at room temperature, have metallic luster, good conductors of heat and electricity, and are malleable and ductile.\n\n2. Q: Name a metal that is liquid at room temperature.\n\nA: Mercury (Hg)\n\n3. Q: What is the basic characteristic property of metals?\n\nA: The ability to conduct heat and electricity.\n\n4. Q: Give an example of a metal that is a poor conductor of electricity.\n\n5. Q: Define a non-metal.\n\nA: Non-metals are elements that are usually gases or solids at room temperature, have no metallic luster, and are poor conductors of heat and electricity.\n\n6. Q: Name a non-metal that is essential for respiration.\n\nA: Oxygen (O2)\n\n7. Q: Which non-metal is used in making fertilizers?\n\nA: Nitrogen (N2)\n\n8. Q: What is the product formed when a metal reacts with an acid?\n\nA: Salt and hydrogen gas.\n\n9. Q: What happens when a metal reacts with oxygen?\n\nA: The metal forms metal oxide.\n\n10. Q: Which gas is evolved when a metal reacts with dilute hydrochloric acid?\n\nA: Hydrogen gas (H2)\n\n11. Q: Which metal is used in galvanization to prevent corrosion?\n\nA: Zinc (Zn)\n\n12. Q: Name the metal that is a good conductor of electricity and is used in electrical wires.\n\nA: Copper (Cu)\n\n13. Q: What happens when a metal displaces another metal from its salt solution?\n\nA: The displaced metal forms a new salt solution, and the original metal is separated.\n\n14. Q: Why do metals not displace hydrogen from acids that are less reactive than metals?\n\nA: Metals are more reactive than hydrogen, so they displace hydrogen from acids.\n\n15. Q: What are metalloids?\n\nA: Metalloids are elements that have properties of both metals and non-metals. They are also called semi-metals.\n\n16. Q: Give an example of a metalloid.\n\nA: Silicon (Si)\n\n17. Q: Define the term 'corrosion.'\n\nA: Corrosion is the process of gradual deterioration of metals due to chemical reactions with their environment, usually involving oxygen and moisture.\n\n18. Q: How can corrosion of metals be prevented?\n\nA: By using protective coatings like painting, galvanizing, or using alloys that are less prone to corrosion.\n\n19. Q: Name a metal that does not react with water or steam.\n\nA: Gold (Au)\n\n20. Q: What happens when a non-metal reacts with oxygen?\n\nA: Non-metals form non-metallic oxides.\n\n21. Q: What is the product formed when sulfur dioxide reacts with water?\n\nA: Sulfurous acid (H2SO3)\n\n22. Q: Name a non-metal that is used to disinfect water.\n\nA: Chlorine (Cl2)\n\n23. Q: Which non-metal is used in the manufacture of fertilizers?\n\nA: Phosphorus (P)\n\n24. Q: What is the general trend in the reactivity of metals in the reactivity series?\n\nA: Metals become more reactive as we move down the reactivity series.\n\n25. Q: Which metal is stored in kerosene to prevent its reaction with air and moisture?\n\nA: Sodium (Na)\n\n26. Q: Name the process used to convert impure metals into pure metals.\n\nA: Electrolytic refining\n\n27. Q: Why are metals used to make cooking utensils?\n\nA: Metals are good conductors of heat, allowing even distribution of heat during cooking.\n\n28. Q: What are the products formed when a metal carbonate reacts with an acid?\n\nA: Salt, water, and carbon dioxide gas.\n\n29. Q: Which metal is used in the production of coins?\n\nA: Copper (Cu)\n\n30. Q: Which non-metal is used in the production of fertilizers, explosives, and medicines?\n\nA: Nitrogen (N2)\n\n31. Q: Write the chemical equation for the reaction of zinc with hydrochloric acid.\n\nA: Zn + 2HCl → ZnCl2 + H2\n\n32. Q: What is the product formed when magnesium reacts with oxygen?\n\nA: 2Mg + O2 → 2MgO\n\n33. Q: Write the chemical equation for the reaction of iron with sulfuric acid.\n\nA: Fe + H2SO4 → FeSO4 + H2\n\n34. Q: When copper reacts with oxygen in the air, what is the product formed?\n\nA: 2Cu + O2 → 2CuO\n\n35. Q: What happens when silver nitrate solution is added to sodium chloride solution?\n\nA: AgNO3 + NaCl → AgCl + NaNO3\n\n36. Q: Write the chemical equation for the reaction of lead with hydrochloric acid.\n\nA: Pb + 2HCl → PbCl2 + H2\n\n37. Q: What is the product formed when carbon dioxide is passed through lime water?\n\nA: Ca(OH)2 + CO2 → CaCO3 + H2O\n\n38. Q: Write the chemical equation for the reaction of aluminum with hydrochloric acid.\n\nA: 2Al + 6HCl → 2AlCl3 + 3H2\n\n39. Q: When dilute hydrochloric acid reacts with calcium carbonate, what are the products formed?\n\nA: CaCO3 + 2HCl → CaCl2 + CO2 + H2O\n\n40. Q: Write the chemical equation for the reaction of copper oxide with sulfuric acid.\n\nA: CuO + H2SO4 → CuSO4 + H2O\n\n41. Q: What happens when sulfur dioxide gas is passed through water?\n\nA: SO2 + H2O → H2SO3\n\n42. Q: Write the chemical equation for the reaction of lead with oxygen.\n\nA: 2Pb + O2 → 2PbO\n\n43. Q: What is the product formed when copper reacts with nitric acid?\n\nA: 3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O\n\n44. Q: Write the chemical equation for the reaction of zinc oxide with sulfuric acid.\n\nA: ZnO + H2SO4 → ZnSO4 + H2O\n\n45. Q: What happens when a metal reacts with an acid?\n\nA: Metal + Acid → Salt + Hydrogen gas\n\n46. Q: Write the chemical equation for the reaction of iron with oxygen.\n\nA: 4Fe + 3O2 → 2Fe2O3\n\n47. Q: What is the product formed when magnesium reacts with hydrochloric acid?\n\nA: Mg + 2HCl → MgCl2 + H2\n\n48. Q: Write the chemical equation for the reaction of calcium with water.\n\nA: Ca + 2H2O → Ca(OH)2 + H2\n\n49. Q: What happens when a metal carbonate reacts with an acid?\n\nA: Metal carbonate + Acid → Salt + Water + Carbon dioxide\n\n50. Q: Write the chemical equation for the reaction of aluminum with oxygen.\n\nA: 4Al + 3O2 → 2Al2O3\n\n51. Q: What is the product formed when lead reacts with hydrochloric acid?\n\nA: Pb + 2HCl → PbCl2 + H2\n\n52. Q: Write the chemical equation for the reaction of zinc with sulfuric acid.\n\nA: Zn + H2SO4 → ZnSO4 + H2\n\n53. Q: What happens when a non-metal reacts with oxygen?\n\nA: Non-metal + Oxygen → Non-metallic oxide\n\n54. Q: Write the chemical equation for the reaction of sodium with water.\n\nA: 2Na + 2H2O → 2NaOH + H2\n\n55. Q: What is the product formed when sulfur dioxide reacts with water?\n\nA: SO2 + H2O → H2SO3\n\n56. Q: Write the chemical equation for the reaction of copper with sulfuric acid.\n\nA: Cu + 2H2SO4 → CuSO4 + SO2 + 2H2O\n\n57. Q: What happens when a metal reacts with water?\n\nA: Metal + Water → Metal hydroxide + Hydrogen gas\n\n58. Q: Write the chemical equation for the reaction of lead with nitric acid.\n\nA: 3Pb + 8HNO3 → 3Pb(NO3)2 + 2NO + 4H2O\n\n59. Q: What is the product formed when carbon reacts with oxygen?\n\nA: C + O2 → CO2\n\n60. Q: Write the chemical equation for the reaction of sodium with oxygen.\n\nA: 4Na + O2 → 2Na2O\n\nRemember to revise the properties, reactions, and uses of metals and non-metals. These short questions cover various aspects of the topic, helping you to reinforce your understanding. Best of luck with your studies!" ]

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[ "# Show that the three lines with direction cosines", null, "are mutually perpendicular.\n\nWe know that\n\nIf l1, m1, n1 and l2, m2, n2 are the direction cosines of two lines; and θ is the acute angle between the two lines; then cos θ = |l1l2 + m1m2 + n1n2|\n\nIf two lines are perpendicular, then the angle between the two is θ = 90°\n\nFor perpendicular lines, | l1l2 + m1m2 + n1n2 | = cos 90° = 0, i.e.\n\n| l1l2 + m1m2 + n1n2 | = 0\n\nSo, in order to check if the three lines are mutually perpendicular, we compute | l1l2 + m1m2 + n1n2 | for all the pairs of the three lines.\n\nNow let the direction cosines of L1, L2 and L3 be l1, m1, n1; l2, m2, n2 and l3, m3, n3.\n\nFirst, consider", null, "", null, "L1 L2 ……(i)\n\nNext, consider", null, "", null, "L2 L3 …(ii)\n\nNow, consider", null, "", null, "L1 L3 …(iii)\n\nBy (i), (ii) and (iii), we have\n\nL1, L2 and L3 are mutually perpendicular.\n\nRate this question :\n\nHow useful is this solution?\nWe strive to provide quality solutions. Please rate us to serve you better." ]

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[ "### A general family of derivative free with and without memory root finding methods\n\nJPRM-Vol. 16 (2020), Issue 1, pp. 64 – 83 Open Access Full-Text PDF\nSaima Akram, Fiza Zafar, Moin-ud-Din Junja, Nusrat Yasmin\nAbstract: In this manuscript, we construct a general family of optimal derivative free iterative methods by using rational interpolation. This family is further extended to a family of with-memory methods with increased order of convergence by employing two free parameters. At each iterative step, we use a suitable variation of the free parameters. These parameters are computed by using the information from current and previous iterations so that the convergence order of the existing family is increased from $$2^{n}$$ to $$2^{n}+2^{n-1}+2^{n-2}$$ without using any additional function evaluations. To check the performance of newly developed iterative schemes with and without memory, an extensive comparison with the existing with- and without memory methods is done by taking some real world problems and standard nonlinear functions. Numerical experiments illustrate that the proposed family of methods with-memory retain better computational efficiency and fast convergence speed as compared to existing with- and without memory methods. The performance of the methods is also analyzed visually by using complex plane. Numerical and dynamical comparisons confirm that the proposed families of with and without memory methods have better efficiency, convergence regions and speed in contrast with the existing methods of the same kind.\n\n### On the generalized class of estimators for estimation of finite population mean in the presence of non-response problem\n\nJPRM-Vol. 16 (2020), Issue 1, pp. 52 – 63 Open Access Full-Text PDF\nSaba Riaz, Amna Nazeer, Javeria Abbasi, Sadia Qamar\nAbstract: This work considers a generalized class of biased estimators for the estimation of the unknown population mean of the variable of interest accompanying the issue of non-response in the study and in the auxiliary variables. The asymptotic bias and the asymptotic variance of the suggested class are acquired, up to the first degree of approximation and, compared with the linear regression estimator. The efficiency of the suggested estimators while comparing with the linear regression estimator and some other existing estimators are studied regarding percent relative efficiency (PRE). Furthermore, a simulation study also affirms the excellence of the considered class of estimators.\n\n### Characterizations of Chevalley groups using order of the finite groups\n\nJPRM-Vol. 16 (2020), Issue 1, pp. 46 – 51 Open Access Full-Text PDF\nAbid Mahboob, Taswer Hussain, Misbah Akram, Sajid Mahboob, Nasir Ali, Ali Raza\nAbstract: In this paper, we prove $$\\psi (A_{1}(4))< \\psi(G)$$, $$\\forall$$ groups which are not simple with order sixty, $$A_{1}(4)$$ is Chevalley group (Linear group) of order 60. Also we prove that $$\\psi (A_{2}(2))< \\psi(G)$$ using higher order non-simple groups of order 168.\n\n### Construction of optimal derivative free iterative methods for nonlinear equations using Lagrange interpolation\n\nJPRM-Vol. 16 (2020), Issue 1, pp. 30 – 45 Open Access Full-Text PDF\nMoin-ud-din Junjua, Saima Akram, Tariq Afzal, Ayyaz Ali\nAbstract: In this paper, we present a general family of optimal derivative free iterative methods of arbitrary high order for solving nonlinear equations by using Lagrange interpolation. The special cases of this family with optimal order of convergence two, four, eight and sixteen are obtained. These methods do not need the Newton’s or Steffensen’s iterations in the first step of their iterative schemes. The advantage of the new schemes is that they are also extendable to the iterative methods with-memory. Numerical experiments and polynomiographs are presented to confirm the theoretical results and to compare the new iterative methods with other well known methods of similar kind.\n\n### Fractional Optimal Control for a Corruption model\n\nJPRM-Vol. 16 (2020), Issue 1, pp. 11 – 29 Open Access Full-Text PDF\nEbenezer Bonyah\nAbstract: In this work, a fractional optimal control of corruption model is investigated. The variable controls are included in the model to optimize the best strategy in reducing the corruption in the society. The fraction derivative employed in the study is in Atangana–Beleanu–Caputo (ABC) sense based on generalized Mittag–Leffler. The uniqueness and existence of solution of the corruption model is established. The necessary and sufficient condition for establishing fractional optimal control in ABC sense is determined. A numerical algorithm for obtaining fractional optimal control solution is presented. The numerical solution results show that the best strategy in controlling corruption in the society is to optimize all the thee controls simultaneously.\nAbstract: For a commutative semiring $$R$$ with non-zero identity, the maximal graph of $$R$$, denoted by $$MG(R)$$, is the graph whose vertices are all elements of $$UM(R)$$ with two distinct vertices joined by an edge when there is a maximal co-ideal that contains both of them. In this paper, we study some properties of maximal graph such as planarity, radius, splitting and domination number." ]

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[ "Ch. 1 & 2  chemical formula & measurement (in forensic chem textbook)  test    [50 points]                   Chem\n\nFor problems involving calculations, show your work in an organized manner, include any relevant equation (or formula and conversion factor(s)), put the proper units in your calculations / answer, and have the proper number of significant figures in your answer.\n\nAcademic Honesty:   The answers on this test are my own and I am using only the allowed set of notes as described in the syllabus.  I have not discussed the test questions with anyone before or during the test nor have I seen the test questions prior to the exam.  If you violate any of the preceding items or do not sign, your semester grade is a F.\n\nSignature:  ___________________\n\n1.  Identify by name (not symbol) the atoms and the number of each atom in the following compounds.  [18 points]\n\na.  CH3 (CH2 )4 CH2 OH:  _______________________________________\n\nb.  Pb(NO3 )2 :  _______________________________________________\n\nc.  (NH4 )2 S:  __________________________________________________\n\n2.  Write the following numbers in scientific notation without changing the number of significant figures in the number.  [8 points]\n\na.  0.001030  =   _________________\n\nb.  3501 =  _____________________\n\nc.  371.27 =  ____________________\n\nd.  0.00018 =  ___________________\n\n3.  solve   [8 points]\n\na.  2.0  +  3.00  =  __________\n\nb.  0.125  -  0.05  =  _________\n\nc.  6  *  3.0  =  _____________\n\nd.  3.00  ÷  4.000  =  _________\n\n4.  solve  [6 points]\n\na.  25 mm2   =  __ cm2\n\nb.  35 g / second  =  __  kg / minute\n\n5.  A 45.0 gram cylinder with a density of 3.0 g / mL and a height of 1.5 meters has a radius of __ mm.  [10 points]\n\n- - - - - - - - - - - - - - - - -\n\nCh. 3, 4, & 5 atoms, molecules, & measurement  (in district adopted textbook\"retest\"           [50 points]                   Chem\n\nFor problems involving calculations, show your work in an organized manner, include any relevant equation (or formula and conversion factor(s)), put the proper units in your calculations / answer, and have the proper number of significant figures in your answer.  10 points per question._\n\n1.  A 15.0 g object with a density of 2.6 g / mL was added to a graduated cylinder containing 7.5 mL water.  What is the volume of the graduated cylinder after completely immersing the object in the graduated cylinder ?\n\n2.  fill-in the below table.\n\n Chemical name Chemical formula N2 O6 K2 CO3 Pb(OH)4 Trisulfur octachloride Iron (III) nitrate\n\n3.  fill-in the below table.\n\n symbol # protons # neutrons # electrons charge Atomic mass 80 + 2 198 32 P-3\n\n4.  solve\n\na.  25 μm / second  =  __ cm / minute\n\nb.  125 inch3   =  __  mL\n\n5.  solve\n\na.  3.0  +  4.00  =  _________\n\nb.  3.0 * 6.00  =  __________\n\nc.  3.0  ÷  7.00  =  _________\n\nd.\n\ne.  0.0125  -  0.009  =   ______" ]

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[ "", null, "Consistent Structured Prediction with Max-Min Margin Markov Networks\n\n# Consistent Structured Prediction with Max-Min Margin Markov Networks\n\nJul 12, 2020\n|\n33 views\n|\n###### Details\nMax-margin methods for binary classification such as the support vector machine (SVM) have been extended to the structured prediction setting under the name of max-margin Markov networks ($M^3N$), or more generally structural SVMs. Unfortunately, these methods are statistically inconsistent when the relationship between inputs and labels is far from deterministic. We overcome such limitations by defining the learning problem in terms of a \"max-min\" margin formulation, naming the resulting method max-min margin Markov networks ($M^4N$). We prove consistency and finite sample generalization bounds for $M^4N$ and provide an explicit algorithm to compute the estimator. The algorithm achieves a generalization error of $O(1/\\sqrt{n})$ for a total cost of $O(n)$ projection-oracle calls (which have at most the same cost as the max-oracle from $M^3N$). Experiments on multi-class classification, ordinal regression, sequence prediction and matching demonstrate the effectiveness of the proposed method. Speakers: Francis Bach, Alessandro Rudi, Alex Nowak-Vila" ]

[ null, "https://www.facebook.com/tr", null ]

[ "## Interfacing 2-Digit 7 Segment with Arduino Nano in Proteus.\n\nMarch 1, 2022\n\nThis tutorial, is to create a project with 2-digit Seven-segment display by using Arduino Nano, which include the connections with Arduino, coding and schematic with the help of Proteus software. As we have seen in previous tutorial, for interfacing a single-digit display, we need to connect the common anode pin to +5V supply or we need to connect the common cathode pin to ground, but in case of two digits display we have to drive them independently if we want to display two digits!\n\nIn this project, we use function of the 2-digits. to drive 2 digit 7-segment display we need to used two Arduino I/O pins, each driving a digit of the display individually. This setups is required to drive the common anode or cathode pin using Arduino I/O pins, so they can source enough current to light all seven segments.\n\nBefore starting Arduino coding for the project, let us know in brief about 2-digit seven segment displays. A 7-segment display is basically just a couple of regular LEDs behind a block. Each led lights up a particular segment and by lighting a specific combination of LEDs you can represent a number or some letters. Most 7-Segment display are common Cathode, which mean that each LED GND pins (Cathode) are connected together and the VCC+ pins (Anode).\n\nHere’s a diagram of the 2-digit 7 Segment display, as you can see pin 1-5 and pin 6, 9 and 10 are connected to a specific segment. Pin 7 and 8 (D1 and D2) are the common Cathode for each digit. So, by grounding D1 or D2 you select which digit you want to light up a specific segment. Of course, like any LED you need to use a resistor(10k) to limit the amount of current drawn by the LED.\n\n#### Connection required:\n\nHere, the 7-Segment display is driven directly by Arduino through the wire. Resistors need to be connected between the display and the Arduino UNO board. Depending on which number or alphabet is to be displayed, control signals are applied.\n\nFollowing schematic shows the 7-segment 2-digit display with10-pins that how to set each pin on this LED. a, b, c, d, e, f and g are not on order, so please carefully while doing a wiring. Otherwise your LED display will not show the number you want. “d1” is the power pin to support digit 1 on the right side and “d2” is the power pin to support digit 2 on the left side. “dp” is point pin next to the digit bottom.\n\n### Arduino Sketch\n\n```//Prepare binary array for all 7 segment to turn on 7 segment at position of a,b,c,d,e,f,g\nint digit = {0b0111111, 0b0000110, 0b1011011, 0b1001111, 0b1100110, 0b1101101, 0b1111101, 0b0000111, 0b1111111, 0b1101111};\n\nint digit1, digit2; // initialize individual digit to controll each segment\n\nvoid setup()\n{\nfor (int i = 2; i < 9; i++)\n{\npinMode(i, OUTPUT); // declare 0-9 th pin as output\n}\npinMode(12, OUTPUT); //declare 7 seg Digit1 pin as output\npinMode(11, OUTPUT);//declare 7 seg Digit2 pin as output\n}\nvoid loop() {\nfor (int j = 0; j <= 99; j++)// for lopp to pass value from 00-99\n{\ndigit2 = j / 10;\ndigit1 = j % 10;\nfor ( int k = 0; k < 20; k++)// For loop to control the digit control to print 00-99\n{\ndigitalWrite(11, HIGH);\ndigitalWrite(12, LOW);\ndis(digit2);\ndelay(10);\ndigitalWrite(12, HIGH);\ndigitalWrite(11, LOW);\ndis(digit1);\ndelay(10);\n}\n}\n}\nvoid dis(int num)\n{\nfor (int i = 2; i < 9; i++)\n{" ]

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[ "# OpticalFlow Class Reference\n\n`#include <OpticalFlow.h>`\n\nInheritance diagram for OpticalFlow:", null, "[legend]\n\n## Detailed Description\n\nDefinition at line 30 of file OpticalFlow.h.\n\nList of all members.\n\n## Public Member Functions\n\nOpticalFlow ()\nOpticalFlow (OpticalFlow &other)\nvirtual ~OpticalFlow ()\nvoid updateFlow ()\nUpdate the current flow vectors and increment the integrated angle.\nvoid drawFlow ()\nDraw the optical flow vectors in the camera shape space.\nvoid swapPyramids ()\nvoid initializePositions ()\nSelect the features that we are going to track.\nfloat getIntegratedFlow ()\nvoid computeRelevanceScores ()\nCompute scores for vectors to determine outliers.\n\n## Static Public Member Functions\n\nstatic bool myFlowComp (const FlowVector &v1, const FlowVector &v2)\nComparison function to sort the flow vectors by horizontal translation.\nstatic bool myFlowComp2 (const FlowVector &v1, const FlowVector &v2)\nComparison function in order to sort the flow vectors by relevance score.\nstatic bool scoreComp (const ScorePoint &p1, const ScorePoint &p2)\nstatic fmat::Column< 2 > iterativeLucasKanade (fmat::Column< 2 > center, fmat::Column< 2 > trans, RawImage &img1, RawImage &img2, int window)\n\n## Static Public Attributes\n\nstatic const unsigned int SCORE_LAYER = 0\nImage layer from which we select features to track.\nstatic const unsigned int NUM_FLOW_VECTORS = 50\nNumber of flow vectors to calculate.\nstatic const unsigned int FLOW_WINDOW = 2\nSize of width/2, height/2 of the feature when we calculate flow.\nstatic const unsigned int SCORE_WINDOW = 1\nSize of width/2, height/2 of the feature when we select them for flow calculation.\nstatic const unsigned int NUM_FRAMES = 10\nNumber of frames we remember about the history of the robot's motion. Used for removing outliers.\nstatic const unsigned int CANDIDATE_FEATURE_RESOLUTION = 30\nNumber of pixels per grid square when calculating minimum eigenvalues.\n\n## Protected Attributes\n\nfloat integratedFlow\n\n## Private Member Functions\n\nOpticalFlowoperator= (const OpticalFlow &other)\n\n## Private Attributes\n\nstd::vector< FlowVectorflowVectors\nstd::vector< DualCoding::Shape\n< DualCoding::LineData > >\nflowVectorVisuals\nstd::vector< floatpastTranslations\nRawImagePyramid pyramid1\nRawImagePyramid pyramid2\nRawImagePyramidcurrPyramid\nRawImagePyramidprevPyramid\n\n## Constructor & Destructor Documentation\n\n OpticalFlow::OpticalFlow ( )\n\nDefinition at line 6 of file OpticalFlow.cc.\n\n OpticalFlow::OpticalFlow ( OpticalFlow & other )\n virtual OpticalFlow::~OpticalFlow ( ) ` [virtual]`\n\nDefinition at line 57 of file OpticalFlow.h.\n\n## Member Function Documentation\n\n void OpticalFlow::computeRelevanceScores ( )\n\nCompute scores for vectors to determine outliers.\n\nThe lower the score, the better. Uses the idea that the robot's motion is continuous. Gives high score to vectors that are far away from the robot's history of motion.\n\nDefinition at line 19 of file OpticalFlow.cc.\n\nReferenced by updateFlow().\n\n void OpticalFlow::drawFlow ( )\n\nDraw the optical flow vectors in the camera shape space.\n\nDefinition at line 76 of file OpticalFlow.cc.\n\n float OpticalFlow::getIntegratedFlow ( )\n\nDefinition at line 83 of file OpticalFlow.h.\n\n void OpticalFlow::initializePositions ( )\n\nSelect the features that we are going to track.\n\nThe features are scored by the magnitude of their minimum eigenvalue. Then, the features wpith the largest score are selected.\n\nDefinition at line 97 of file OpticalFlow.cc.\n\nReferenced by updateFlow().\n\n fmat::Column< 2 > OpticalFlow::iterativeLucasKanade ( fmat::Column< 2 > center, fmat::Column< 2 > trans, RawImage & img1, RawImage & img2, int window ) ` [static]`\n\nDefinition at line 131 of file OpticalFlow.cc.\n\nReferenced by updateFlow().\n\n bool OpticalFlow::myFlowComp ( const FlowVector & v1, const FlowVector & v2 ) ` [static]`\n\nComparison function to sort the flow vectors by horizontal translation.\n\nDefinition at line 10 of file OpticalFlow.cc.\n\nReferenced by updateFlow().\n\n bool OpticalFlow::myFlowComp2 ( const FlowVector & v1, const FlowVector & v2 ) ` [static]`\n\nComparison function in order to sort the flow vectors by relevance score.\n\nDefinition at line 14 of file OpticalFlow.cc.\n\nReferenced by updateFlow().\n\n OpticalFlow& OpticalFlow::operator= ( const OpticalFlow & other ) ` [private]`\n static bool OpticalFlow::scoreComp ( const ScorePoint & p1, const ScorePoint & p2 ) ` [static]`\n\nDefinition at line 65 of file OpticalFlow.h.\n\nReferenced by initializePositions().\n\n void OpticalFlow::swapPyramids ( )\n\nDefinition at line 65 of file OpticalFlow.cc.\n\nReferenced by updateFlow().\n\n void OpticalFlow::updateFlow ( )\n\nUpdate the current flow vectors and increment the integrated angle.\n\nDefinition at line 29 of file OpticalFlow.cc.\n\nReferenced by DualCoding::OpticalFlowOdometry::update().\n\n## Member Data Documentation\n\n const unsigned int OpticalFlow::CANDIDATE_FEATURE_RESOLUTION = 30` [static]`\n\nNumber of pixels per grid square when calculating minimum eigenvalues.\n\nDefinition at line 52 of file OpticalFlow.h.\n\nReferenced by initializePositions().\n\n RawImagePyramid* OpticalFlow::currPyramid` [private]`\n\nDefinition at line 108 of file OpticalFlow.h.\n\nReferenced by swapPyramids(), and updateFlow().\n\n const unsigned int OpticalFlow::FLOW_WINDOW = 2` [static]`\n\nSize of width/2, height/2 of the feature when we calculate flow.\n\nThe larger this is, the better the tracking. Large values will severely decrease framerate, however.\n\nDefinition at line 41 of file OpticalFlow.h.\n\nReferenced by updateFlow().\n\n std::vector OpticalFlow::flowVectors` [private]`\n\nDefinition at line 100 of file OpticalFlow.h.\n\nReferenced by computeRelevanceScores(), drawFlow(), initializePositions(), and updateFlow().\n\n std::vector > OpticalFlow::flowVectorVisuals` [private]`\n\nDefinition at line 101 of file OpticalFlow.h.\n\nReferenced by drawFlow().\n\n float OpticalFlow::integratedFlow` [protected]`\n\nDefinition at line 97 of file OpticalFlow.h.\n\n const unsigned int OpticalFlow::NUM_FLOW_VECTORS = 50` [static]`\n\nNumber of flow vectors to calculate.\n\nDefinition at line 36 of file OpticalFlow.h.\n\nReferenced by computeRelevanceScores(), initializePositions(), and updateFlow().\n\n const unsigned int OpticalFlow::NUM_FRAMES = 10` [static]`\n\nNumber of frames we remember about the history of the robot's motion. Used for removing outliers.\n\nDefinition at line 49 of file OpticalFlow.h.\n\nReferenced by computeRelevanceScores(), and updateFlow().\n\n std::vector OpticalFlow::pastTranslations` [private]`\n\nDefinition at line 102 of file OpticalFlow.h.\n\nReferenced by computeRelevanceScores(), and updateFlow().\n\n RawImagePyramid* OpticalFlow::prevPyramid` [private]`\n\nDefinition at line 109 of file OpticalFlow.h.\n\nReferenced by swapPyramids(), and updateFlow().\n\n RawImagePyramid OpticalFlow::pyramid1` [private]`\n\nDefinition at line 106 of file OpticalFlow.h.\n\nReferenced by swapPyramids().\n\n RawImagePyramid OpticalFlow::pyramid2` [private]`\n\nDefinition at line 107 of file OpticalFlow.h.\n\nReferenced by swapPyramids().\n\n const unsigned int OpticalFlow::SCORE_LAYER = 0` [static]`\n\nImage layer from which we select features to track.\n\nDefinition at line 34 of file OpticalFlow.h.\n\nReferenced by initializePositions().\n\n const unsigned int OpticalFlow::SCORE_WINDOW = 1` [static]`\n\nSize of width/2, height/2 of the feature when we select them for flow calculation.\n\nIt is typically safe to keep this low to save computation time.\n\nDefinition at line 46 of file OpticalFlow.h.\n\nReferenced by initializePositions().\n\nThe documentation for this class was generated from the following files:\n\n Tekkotsu v5.1CVS Generated Mon May 9 04:59:13 2016 by Doxygen 1.6.3", null, "" ]

[ null, "http://www.tekkotsu.org/dox/classOpticalFlow__inherit__graph.png", null, "http://aibo2.boltz.cs.cmu.edu/tekkotsu-icon.png", null ]

[ "A point-wise separation Hahn-Banach theorem in C*-algebras\n\nLet $H$ be a Hilbert space. We denote $K(H)$ by the space of compact operators on $H$ which is a two sided ideal in $B(H)$.\n\nLet $E$ be a norm closed convex subset of positive operators in $K(H)$ and let $a$ be a non-zero positive compact operator where $a\\notin E$.\n\nQ: Is there any vector $\\zeta\\in H$ which separates $a$ and $E$, I mean there is a a positive number $\\lambda$ such that for all $x\\in E$ $$\\langle x\\zeta,\\zeta\\rangle\\leq \\lambda< \\langle a\\zeta,\\zeta\\rangle$$\n\n$2 \\times 2$ counterexample, $E = \\left\\{\\left[\\matrix{\\lambda& 0\\cr 0&\\lambda}\\right]: \\lambda \\geq 0\\right\\}$ and $A = \\left[\\matrix{1&1\\cr 1&1}\\right]$. Then for any nonzero $\\zeta$ we have $\\{\\langle B\\zeta,\\zeta\\rangle: B \\in E\\} = [0,\\infty)$, so no $\\zeta$ can separate.\nThe general idea is that you can separate $E$ and $a$ with a bounded linear functional on $K(H)$, i.e., tracing against some trace class operator, but you can't expect to do it with a single vector.\n• Dear Nik, You mean $E$ and $a$ can be separated by a positive trace class operator? – Ali Bagheri Apr 3 '16 at 16:43\n• @oeiras: no, we're talking about subsets of $K(H)$, not $B(H)$. The weak* topology isn't relevant. – Nik Weaver Apr 3 '16 at 18:25\n• @AliBagheri: no way. If $E$ is a cone, then tracing against any positive operator will yield either $\\{0\\}$ or $[0,\\infty)$. Getting $\\{0\\}$ is too much to ask for. – Nik Weaver Apr 3 '16 at 18:25\n• @NikWeaver The spaces $K(H)$ and the finite dimensional operators form a dual pair in the natural way and so we can apply the Hahn-Banach theorem to this situation. – oeiras Apr 3 '16 at 19:31" ]

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[ "# Ted is writing things\n\nOn privacy, research, and privacy research.\n\n# The magic of Gaussian noise\n\nPart of a series on differential privacy. You might want to start with the previous articles below!\n\n1. Why differential privacy is awesome presents a non-technical explanation of the definition.\n2. Differential privacy in (a bit) more detail introduces the formal definition, with very little math.\n3. Differential privacy in practice (easy version) explains how to make simple statistics differentially private.\n4. Almost differential privacy describes how to publish private histograms without knowing the categories in advance.\n5. Local vs. global differential privacy presents the two main models of differential privacy, depending on the context.\n6. The privacy loss random variable explains the real meaning of $$(\\varepsilon,\\delta)$$-differential privacy.\n7. The magic of Gaussian noise (this article) introduces Gaussian noise and its shiny properties.\n\nPreviously, we used Gaussian noise to explain the real meaning of $$\\delta$$ in $$(\\varepsilon,\\delta)$$-differential privacy. One question was left unanswered: why would anyone use Gaussian noise in the first place? The guarantees it provides aren't as strong: it gives $$(\\varepsilon,\\delta)$$-DP with $$\\delta>0$$, while Laplace noise provides pure $$\\varepsilon$$-DP. This blog post gives an answer to this question, and describes the situations in which Gaussian noise excels.\n\n# Gaussian noise is nice\n\nA first advantage of Gaussian noise is that the distribution itself behaves nicely. It's called the normal distribution for a reason: it has convenient properties, and is very widely used in natural and social sciences. People often use it to model random variables whose actual distribution is unknown. If you sum many independent random variables, you end up with a normal distribution. And these are just a few of the many properties of this fundamental distribution. Thus, most data analysts and scientists are already familiar with Gaussian noise. It's convenient when you release anonymized statistics: analysts don't need to learn too many new concepts to understand what you're doing to protect the data.\n\nA second advantage is that the Gaussian distribution has nice, thin tails. The vast majority of its probability mass is focused around its mean. Take a normal distribution with mean 0 and standard deviation $$\\sigma$$. The 68–95–99.7 rule says that a random variable sampled from this distribution will be:\n\n• in $$[-\\sigma,\\sigma]$$ with 68% probability;\n• in $$[-2\\sigma,2\\sigma]$$ with 95% probability;\n• and in $$[-3\\sigma,3\\sigma]$$ with 99.7% probability.\n\nIt even gets better as you go further away from the mean. The probability that the random variable is outside $$[-k\\sigma,k\\sigma]$$ decreases faster than $$e^{-k^2/2}$$. In practice, you're rarely surprised by the values that a Gaussian distribution takes. Even if you sample $$1,000,000$$ values, they are all probably going to be within $$[-5\\sigma,5\\sigma]$$.\n\nLaplace, by comparison, isn't quite as nice. Its tails decrease exponentially fast, but that's still much slower than Gaussian tails. Suppose you sample $$1,000,000$$ values from a Laplace distribution of standard deviation $$\\sigma$$. On average, 849 of them will be outside $$[-5\\sigma,5\\sigma]$$.\n\nOK, so Gaussian noise is nice. But that does not change a simple fact: to get a comparable level of privacy for a single statistic, Laplace is much better. Assume that we're adding noise to a simple count, of sensitivity $$1$$. This graph compares the Laplace noise needed to get $$\\varepsilon=1$$, and Gaussian noise needed to get $$\\varepsilon=1$$ and $$\\delta=10^{-5}$$.", null, "Despite its weaker privacy guarantees, the Gaussian distribution is much flatter. Its standard deviation is over 3.4, while Laplace's is less than 1.3. Thus, much more noise will need to be added, and analysts mainly care about minimizing the noise. Why, then, would Gaussian noise be a good option? The answer is simple: because it gets better when you're publishing a lot of statistics.\n\n# From one to many statistics\n\nIn most of our previous examples, we assumed that each individual appeared in a single statistic. This case is common, for example when partitioning people based on demographic information. But in many applications, this assumption does not hold. Imagine, for example, that you want to answer the question: « what types of specialized physicians did people visit in the past 10 years? »\n\nAssume we're working in the central model. We have a dataset of 〈patient ID, specialist type〉 pairs, and each record corresponds to an individual visiting a specialized physician (cardiologist, dermatologist, radiologist, etc.). We want to count the number of unique patient IDs per specialty.\n\nNote that each patient can only influence a single count once. We count distinct patient IDs: if you visit dermatologists 10 times, you will only add 1 to the \"dermatologist\" count. However, a single patient might visit many types of specialized physicians. There are many kinds of specialties, and a single patient might influence all the counts. Say there are 50 of them.\n\nHow to make these counts differentially private? A first solution is to split the privacy budget across all the counts. Here, we can split our privacy budget in 50. If we want to achieve $$\\varepsilon=1$$, we compute $$\\varepsilon'=1/50=0.02$$, and add Laplace noise of scale $$1/\\varepsilon'=50$$ to all the counts.\n\nUnfortunately, this is a lot of noise.", null, "Fortunately, this is exactly the kind of situation in which Gaussian noise shines. When a single patient can impact $$k$$ distinct statistics, we need to scale Laplace noise by $$k$$. By contrast, Gaussian noise must only be scaled by $$\\sqrt{k}$$. Comparing the two gives a much more flattering view of the power of Gaussian noise.", null, "OK, so that's the general idea. Now, why does that happen? How come composition doesn't seem to behave in the same way for Laplace and Gaussian? To understand this better, we'll first introduce the concept of sensitivity.\n\n# Different kinds of sensitivities\n\nConsider our example above. For each type of specialized physician, we count the people who consulted with one. But we won't consider this histogram as 50 different counting queries. Instead, we'll consider it as a single function, with values in $$\\mathbb{N}^{50}$$. It outputs a vector: a list of 50 coordinates, each of which corresponds to a fixed specialty. How to make such a function $$f$$ differentially private? We'll add noise, scaled by the function's sensitivity.\n\nThe sensitivity of a function measures how much its output can change when you add one record in the database. If the function returns a single number, we measure the absolute value of the difference: the sensitivity of $$f$$ is the maximum value of $$\\left|f\\left(D_1\\right)-f\\left(D_2\\right)\\right|$$. We already encountered the sensitivity before: when counting things, if each patient can change the statistic by more than $$1$$, we needed to scale the noise accordingly. The same happened for sums.\n\nHere, the function $$f$$ returns a vector. How do we measure the distance between two vectors? We have a few options. We could use the Manhattan distance, or the Euclidean distance, or even weirder stuff. As it turns out, the distance we need to use depends on which noise function we're adding. Laplace noise is scaled by the $$L^1$$ sensitivity, itself based on the Manhattan distance. Here is its definition, denoting by $$f_i(D)$$ the $$i$$-th coordinate of $$f(D)$$:\n\n$$\\Delta_1(f) = \\max \\sum_{i=1}^{50} \\left|f_i\\left(D_1\\right)-f_i\\left(D_2\\right)\\right|$$\n\nwhere the $$\\max$$ is taken over $$D_1$$ and $$D_2$$ differing in a single record. This is easy to understand: you just sum the sensitivities for each coordinate. For our function $$f$$, we have $$\\Delta_1(f)=50$$: Laplace noise needs to be scaled by 50. You might have noticed that this is equivalent to using simple composition: the scale of Laplace noise is $$\\Delta_1/\\varepsilon$$, so dividing $$\\varepsilon$$ by $$50$$ is the same as considering all coordinates together.\n\nBy contrast, Gaussian noise needs to be scaled by the $$L^2$$ sensitivity. This type of sensitivity is based on the Euclidean distance, and defined by:\n\n$$\\Delta_2(f) = \\max \\sqrt{\\sum_{i=1}^{50} {\\left|f_i\\left(D_1\\right)-f_i\\left(D_2\\right)\\right|}^2}$$\n\nstill taking the $$\\max$$ over $$D_1$$ and $$D_2$$ differing in a single record. This formula might look complicated, but the Euclidean distance is a simple concept: it's the length of a straight line between two points. If you only have two dimensions, this formula might be reminiscent of the Pythagorean theorem!\n\nThe standard deviation $$\\sigma$$ of Gaussian noise will be proportional to $$\\Delta_2(f)$$. Let's compute this value for our function. Each patient can change a single count by at most one, and each can change all counts. Thus:\n\n$$\\Delta_2(f) = \\sqrt{\\sum_{i=1}^{50} 1^2} = \\sqrt{50} \\approx 7.07.$$\n\nThe noise scales with the square root of the number of counts. This is key to Gaussian's superiority in such situations: the $$L^2$$ sensitivity grows much more slowly than the $$L^1$$ sensitivity. As a result, scaling the noise by the sensitivity hurts accuracy much less. Of course, using Gaussian noise gives you $$(\\varepsilon,\\delta)$$-DP, not pure DP, so there is still a tradeoff. As we saw in the previous article, this isn't a super scary $$\\delta$$, so it's generally worth it.\n\nThis paper (Theorem 8) gives the exact formula for calibrating Gaussian noise depending on $$\\Delta_2(f)$$, $$\\varepsilon$$ and $$\\delta$$. You need to pick $$\\sigma$$ such that the following equality holds:\n\n$$g\\left(\\frac{\\Delta_2(f)}{\\sigma},\\varepsilon\\right) = \\delta$$\n\nwhere $$g$$ is a complicated function. As you can see, increasing $$\\Delta_2(f)$$ and $$\\sigma$$ simultaneously has no effect: $$\\sigma$$ is proportional to $$\\Delta_2(f)$$. There is no analytic form for $$\\sigma$$, but since $$g$$ is monotonic, you use e.g. a binary search to approximate its value. If you want to know the exact formula, click here: .\n\nNow, why do these two noise distributions work so differently? Rather than proving this formally, here is a visual intuition. Let's look at the density function of Gaussian and Laplace noise, in two dimensions. The first is Gaussian, the second is Laplace.", null, "", null, "As you can see, Gaussian noise has a circular shape, while Laplace noise has a square shape. How indistinguishable are two points, when noise is added to both of them? With Gaussian noise, it depends on their distance from each other in a straight line. By contrast, with Laplace, it depends on how far they are in Manhattan distance.\n\nIn conclusion, whether to use Laplace or Gaussian noise depends on two things:\n\n• whether we are OK with a non-zero $$\\delta$$;\n• how many statistics a single individual can influence.\n\nThe first point is clear: if we want $$\\delta=0$$, we can't use Gaussian noise. Let's quantify the second point. If a single person can impact at most one statistic, Laplace is better. If they can impact many, Gaussian is better. Where does the boundary lie? The following graph answers this question, comparing both mechanisms by their standard deviation. We pick $$\\varepsilon=1$$ and $$\\delta=10^{-5}$$, and we assume that each person can influence each statistic at most once.", null, "For these values of $$\\varepsilon$$ and $$\\delta$$, Gaussian noise is better if each individual can influence 8 statistics or more. Of course, with different privacy parameters, the result might differ. But as the impact of a single individual grows, Gaussian noise will always end up being the least noisy choice.\n\n# Further uses\n\nWhat if each person can influence each statistic differently? Suppose, for example, that we count the number of visits to each type of physician. A single patient can add many visits to a single count. Worse, the maximum number of visits per patient can vary across physician types. Can we still use Gaussian noise? The answer (hat tip to Mark) is yes: you can scale down each statistic so the sensitivity becomes $$1$$, add noise to them, then scale them back up. This makes Gaussian noise even more powerful: if you compute many statistics on the same data, you can use the Gaussian mechanism to reduce the noise magnitude, even if the statistics are completely unrelated.\n\nFinally, Gaussian noise is heavily used in differentially private machine learning. The fundamental reason is the same. Consider methods like stochastic gradient descent, a popular algorithm in machine learning. At each iteration of this method, each data point influences many coordinates of a vector. To make it differentially private, we need to add noise to all coordinates. Thus, Gaussian noise is a good choice, for the exact same reason. Machine learning is full of more Gaussian-related goodness, but this article is long enough already.\n\nMaybe we'll come back to it in future posts! Stay tuned (on RSS or Twitter) for more updates.\n\nAll opinions here are my own, not my employers'." ]

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[ "Project tutorial", null, "# Maze Solver Robot, using Artificial Intelligence © GPL3+\n\n\"Rex, the Robot\" will try finding how to scape from a \"labyrinth\" on the shortest and fastest way.\n\n• 47,959 views\n• 82 respects\n\n## Components and supplies", null, "Arduino Nano R3\n×1", null, "HC-06 Bluetooth Module\n×1\n TCRT5000 4CH Infrared Line Track Follower Sensor Module\n×1", null, "SparkFun RedBot Sensor - Line Follower\n×1\n ZX03 (based on TCRT5000) Reflective Infrared Sensors (Analog output)\n×2", null, "Android device\n×1", null, "RobotGeek Continuous Rotation Servo\n×2", null, "4xAA battery holder\n×2\n\n## Apps and online services\n\n### Introduction\n\nThis tutorial was developed upon my last project: Line Follower Robot - PID Control - Android Setup. Once you have a robot with line following capabilities, the next natural step is to give him some degree of intelligence. So, our dear \"Rex, the Robot\" will try now finding how to scape from a \"labyrinth\" on a shortest and fastest way (by the way, he hates the Minotaurus.\n\nFor starting, what is the difference between Maze and Labyrinth? According to http://www.labyrinthos.net, in the English-speaking world it is often considered that to be qualified as a maze, a design must have choices in the pathway. Clearly, this will include many of the modern installations in entertainment parks and tourist attractions, including our 2D maze here. Popular consensus also indicates that labyrinths have one pathway that leads inexorably from the entrance to the goal, albeit often by the most complex and winding of routes.\n\nThe majority of mazes, however complex their design may appear, were essentially formed from one continuous wall with many junctions and branches. If the wall surrounding the goal of a maze is connected to the perimeter of the maze at the entrance, the maze can always be solved by keeping one hand in contact with the wall, however many detours that may involve. Those ‘simple’ mazes are correctly known as \"`Simply-connected`\" or \"`perfect maze`\" or in other words, that contain no loops.\n\nReturning to our project, it will be split in two parts (or \"`passes`\"):\n\n• (First Pass): The robot finds its way out from a \"`non-known perfect maze` \". Does not matter where you put it inside the maze, it will find a \"`solution`\".\n• (Second Pass): Once the robot found a possible maze solution, it should optimize its solution finding the \"`shortest path from start to finish`\".\n\nThe video bellow, will show an example of Rex finding its way out. In the first time that the robot explores the maze, of course it will waste a lot of time \"`thinking`\" about what to do at any intersection. Testing the numerous possibilities, it will take several wrong paths and dead ends, what force him to run longer paths and perform unnecessary \"`U-Turns`\". During this \"`1st Pass\"`, the robot will be accumulating experiences, \"`taking notes`\" about the different intersections and eliminating the bad branches. In its \"`2nd Pass`\", the robot goes straight and quickly to the end without any mistake or doubt. Along this tutorial, we will explore in details how to do it:\n\n### Step 1: ​Bill of Materials\n\nThe list of materials are basically the same as the one one used with the Line Follower Robot, except that I included 2 extra Sensors for better accuracy on detecting the LEFT and RIGHT intersections:\n\nThe final robot is still very cheap (around \\$ 85.00):\n\n• 2 X Wood squares (80X80mm)\n• 3 X Binder clips\n• 2 X Wood wheels (diameter: 50mm)\n• 1 X Ball caster\n• 9 X Elastic bands\n• 3M Command frame strip\n• Plastic joints for sensor fix\n• 2 X sets of 4XNi-Metal Hydride Battery (5V each set)\n• 2 X SM-S4303R Continuous Rotation 360 Degree Plastic Servo\n• Arduino Nano\n• HC-06 Bluetooth module\n• 5 X Line sensors (TCRT5000 4CH Infrared Line Track Follower Sensor Module + 1 independent Track sensor)\n• 2 X ZX03 (based on TCRT5000) Reflective Infrared Sensors (Analog output)\n• 1 LED\n• 1 button\n\nNote: I used the item 7 above with analog output, because I did not have on hand sensors with digital output as those on item 6. The ideal is to have all sensors equal, if possible. Also I tested the project keeping only the original 5 sensors. It will work, but requires more sensitive adjustments on discovering intersections.\n\n### Step 2: Changes in the Body\n\nRemove the original set of 5 Line Follow Sensors and fix the new \"`Far LEFT\"` and \"`Far RIGHT`\" reflective sensors at each extreme of the support plastic bar. It is advise to have the 7 sensors as lined as possible.\n\n### Step 3: Installing and testing the new sensors\n\nThe new array of now 7 sensors, is mounted on a way that the 5 original ones are exclusively used for PID control (and detection of the \"full line\", explained later) and the 2 new ones, left to be used exclusively for LEFT and RIGHT intersection detection.\n\nAs a quick review, let's remember how the 5 original \"digital\" sensors work:\n\nIf one sensor is centered with relation to the black line, only that specific sensor will produce a HIGH. By other side, the space between sensors should be calculated to allow that 2 sensors can cover the full width of the black line simultaneously, also producing a HIGH signal on both sensors.\n\nHow the 2 new \"analog\" sensors work:\n\nIf one of the sensors is centered with relation to the black line, the output will be an analog value, usually producing an output at Arduino ADC bellow \"100\" (remember that the ADC produces an output from 0 to 1023). With lighter surfaces, the output value will be higher (I tested 500 to 600 over white paper, for example). This value must be tested on different situations of light and surface materials to define the correct THRESHOLD constant to be used in your case (see the picture here).\n\nLooking at the Arduino code, each one of the sensors will be defined with a specific name (consider that the original Line Follow Sensor more to the Left must be assigned with a label \"`0` \"):\n\n``````const int lineFollowSensor0 = 12; //Using Digital input\nconst int lineFollowSensor1 = 18; //Using Analog Pin A4 as Digital input\nconst int lineFollowSensor2 = 17; //Using Analog Pin A3 as Digital input\nconst int lineFollowSensor3 = 16; //Using Analog Pin A2 as Digital input\nconst int lineFollowSensor4 = 19; //Using Analog Pin A5 as Digital input\nconst int farRightSensorPin = 0; //Analog Pin A0\nconst int farLeftSensorPin = 1; //Analog Pin A1\n``````\n\nTo remember, the possible 5 original sensor array output when following a line are:\n\n``````1 1 1 1 1\n0 0 0 0 0\n0 0 0 0 1\n0 0 0 1 1\n0 0 0 1 0\n0 0 1 1 0\n0 0 1 0 0\n0 1 1 0 0\n0 1 0 0 0\n1 1 0 0 0\n1 0 0 0 0\n``````\n\nWith the addition of the 2 new ones, their possible outputs are:\n\n• Far LEFT Sensor: Analog Output greater or lower than a THRESHOLD\n• Far RIGHT Sensor: Analog Output greater or lower than a THRESHOLD\n\nIn order to storage the values of each sensor an array variable is created for the original 5 digital sensors:\n\n``````int LFSensor={0, 0, 0, 0, 0};\n``````\n\nAnd two integer variables for the 2 new analog sensors:\n\n``````int farRightSensor = 0;\nint farLeftSensor = 0;\n``````\n\nEach position of the array and variables will be constantly updated with the output of each one of the sensors:\n\n``````LFSensor = digitalRead(lineFollowSensor0);\n``````\n\nHaving 5 sensors, as saw in the Follower Line Robot project, permits the generation of an \"error variable\" that will help to control the robot's position over the line. Also, a variable called \"mode\" will be used for definition if the robot is following a line, over a continuous line, an intersection or no line at all.\n\nThis variable \"mode\" will be also used with the \"Far LEFT/RIGHT\" sensors. For representation, let's consider the far left and right sensors having 3 possible states:\n\n• H (higher than THRESHOLD),\n• L (smaller than THRESHOLD) and\n• X (irrelevant).\n\nFor the digital outputs , will the usual 0, 1 and we will also introduce the X:\n\n• H 0 X X X X L ==> mode = RIGHT_TURN; error = 0; (see the example at the image above)\n• L X X X X 0 H ==> mode = LEFT_TURN; error = 0;\n• X 0 0 0 0 0 X ==> mode = NO_LINE; error = 0;\n• H 0 0 0 0 1 H ==> mode = FOLLOWING_LINE; error = 4;\n• H 0 0 0 1 1 H ==> mode = FOLLOWING_LINE; error = 3;\n• H 0 0 0 1 0 H ==> mode = FOLLOWING_LINE; error = 2;\n• H 0 0 1 1 0 H ==> mode = FOLLOWING_LINE; error = 1;\n• H 0 0 1 0 0 H ==> mode = FOLLOWING_LINE; error = 0;\n• H 0 1 1 0 0 H ==> mode = FOLLOWING_LINE; error = -1;\n• H 0 1 0 0 0 H ==> mode = FOLLOWING_LINE; error = -2\n• H 1 1 0 0 0 H ==> mode = FOLLOWING_LINE; error = -3;\n• H 1 0 0 0 0 H ==> mode = FOLLOWING_LINE; error = -4;\n• X 1 1 1 1 1 X ==> mode = CONT_LINE; error = 0;\n\nSo, implementing the above logic in the function:\n\n``````void readLFSsensors()\n``````\n\nwill return the variables \"mode\" and \"error\" that will be used at the program logic. It is important to test the logic of the sensors before following with the project. The bellow function is included in the code and can be used for testing purposes:\n\n``````void testSensorLogic(void)\n{\nSerial.print (farLeftSensor);\nSerial.print (\" <== LEFT RIGH==> \");\nSerial.print (farRightSensor);\nSerial.print (\" mode: \");\nSerial.print (mode);\nSerial.print (\" error:\");\nSerial.println (error);\n}\n``````\n\n### Step 4: Solving the Maze - The Left Hand Rule\n\nAs discussed at introduction, the majority of mazes however complex their design may appear, were essentially formed from one continuous wall with many junctions and branches. If the wall surrounding the goal of a maze is connected to the perimeter of the maze at the entrance, the maze can always be solved by keeping one hand in contact with the wall, however many detours that may involve. These ‘simple’ mazes are correctly known as \"`Simply-connected` .\"\n\nSearching at Wikipedia, we learn that:\n\n\"The Wall Follower is the best-known rule for traversing mazes. It is also known as either the left-hand rule or the right-hand rule. If the maze is simply connected, that is, all its walls are connected together or to the maze's outer boundary, then by keeping one hand in contact with one wall of the maze the solver is guaranteed not to get lost and will reach a different exit if there is one; otherwise, he or she will return to the entrance having traversed every corridor next to that connected section of walls at least once.\"\n\nIn short, the Left-Hand Rule can be described like:\n\nAt every intersection, and throughout the maze, keep your left hand touching the wall on your left.\n\n• Place your left hand on the wall.\n• Begin walking forward\n• Eventually, you will reach the end of the maze. You will probably not go the shortest and most direct way, but you will get there.\n\nSo, the key here is to identify the intersections, defining what course to take based on the above rules. Specifically in our kind of 2D Maze, we can find 8 different types of intersections (see the first picture above):\n\nLooking the picture, we can realize that the possible actions at intersections are:\n\n1. At a \"Cross\":\n\n• Go to Left, or\n• Go to Right, or\n• Go Straight\n\n2. At a \"T\":\n\n• Go to Left, or\n• Go to Right\n\n3. At a \"Right Only\":\n\n• Go to Right\n\n4. At a \"Left Only\":\n\n• Go to Left\n\n5. At \"Straight or Left\":\n\n• Go to Left, or\n• Go Straight\n\n6. At \"Straight or Right\":\n\n• Go to Right, or\n• Go Straight\n\n• Go back (\"U turn\")\n\n8. At \"End of Maze\":\n\n• Stop\n\nBut, applying the \"Left-Hand Rule\", the actions will be reduced to one option each:\n\n• At a \"Cross\": Go to Left\n• At a \"T\": Go to Left\n• At a \"Right Only\": Go to Right\n• At a \"Left Only\": Go to Left\n• At a \"Straight or Left\": Go to Left\n• At a \"Straight or Right\": Go Straight\n• At a \"Dead End\": Go back (\"U turn\")\n• At the \"End of Maze\": Stop\n\nWe are almost there! \"Be calm!\"\n\nWhen the robot reaches a \"Dead End\" or the \"End of a Maze\", it is easy to identify them, because do not exist ambiguous situations (we have already implemented those actions on the Line Follower Robot, remember?). The problem is when the robot finds a \"LINE\" for example, because a line can be a \"Cross\" (1) or a \"T\" (2). Also when it reaches a \"LEFT or RIGHT TURN\", those can be the a simple turn (options 3 or 4) or options to go straight (5 or 6). To discover exactly on what type of intersection the robot is, an additional step must be taken: the robot must run an \"extra inch\" and see what is next (see the second picture above, as an example).\n\nSo, in terms of flow, the possible actions can be now describe as:\n\n• Go back (\"U turn\")\n\n2. At a \"LINE\": Run an extra inch\n\n• If there is a line: It is a \"Cross\" ==> Go to LEFT\n• If There is no line: it is a \"T\" ==> Go to LEFT\n• If there is another line: it is the \"End of Maze\" ==> STOP\n\n3. At a \"RIGHT TURN\": Run an extra inch\n\n• if there is a line: It is a Straight or Right ==> Go STRAIGHT\n• If there is no line: it is a Right Only ==> Go to RIGHT\n\n4. At a \"LEFT TURN\": Run an extra inch\n\n• if there is a line: It is a Straight or LEFT ==> Go to LEFT\n• If there is no line: it is a LEFT Only ==> Go to LEFT\n\nNote that in fact, In case of a \"LEFT TURN\", you can skip the test, because you will take LEFT anyway. I left the explanation more generic only for clarity. At the real code I will skip this test. The above 3rd picture, shows a very simple maze on my lab floor, used for test purposes.\n\n### Step 5: Implementing the \"Left Hand on the Wall\" Algorithm at Arduino Code\n\nOnce we have the `readLFSsensors()` function modified to include the extra 2 sensors, we can re-write the Loop Function to run the algorithm as described on the last Step:\n\n``````void loop()\n{\nswitch (mode)\n{\ncase NO_LINE:\nmotorStop();\ngoAndTurn (LEFT, 180);\nbreak;\ncase CONT_LINE:\nrunExtraInch();\nif (mode == CONT_LINE) mazeEnd();\nelse goAndTurn (LEFT, 90); // or it is a \"T\" or \"Cross\"). In both cases, goes to LEFT\nbreak;\ncase RIGHT_TURN:\nrunExtraInch();\nif (mode == NO_LINE) goAndTurn (RIGHT, 90);\nbreak;\ncase LEFT_TURN:\ngoAndTurn (LEFT, 90);\nbreak;\ncase FOLLOWING_LINE:\nfollowingLine();\nbreak;\n}\n}\n``````\n\nSome new functions appear here:\n\n• followingLine() is the same used with the Following Line Robot where, if it is only following a line, it must `calculatePID()``;` and control the motors depending of PID values: `motorPIDcontrol();`\n• runExtraInch(): will push the robot forward just a little bit. How much the robot will run will depend of the time that you use in the delay function, before you command the motor to stop.\n``````void runExtraInch(void)\n{\nmotorPIDcontrol();\ndelay(extraInch);\nmotorStop();\n}\n``````\n• goAndTurn (direction, angle): this special function is important because you can not turn the robot as soon you realize the type of intersection you are. Remember that we projected a Differential Robot that when making turns, it \"turns around its axe\" and so, to move 90o and continuously follow the line, the center of the wheels must be aligned with the center of intersection. Once the line of sensors is ahead of its axe, you must run the robot forward to align them. The constant of time `adjGoAndTurn` must be adjusted depending on of the distance between axe and sensor line (\"`d` \"), speed and size of the wheels (see the above picture for illustration).\n``````void goAndTurn(int direction, int degrees)\n{\nmotorPIDcontrol();\nmotorTurn(direction, degrees);\n}\n``````\n\nAt this point, the robot is in fact \"solving a maze\"! You just finish the \"First Pass\". Does not matter where you start inside a maze, you will always reach the end.\n\nBellow, a test of this phase of the project:\n\n### Step 6: Storing the Path\n\nLet's consider the example as shown in the above photo. At the chosen starting point, the robot will find 15 Intersections before reaching the end of the Maze:\n\n• Left (L)\n• Back (B)\n• Left (L)\n• Left (L)\n• Left (L)\n• Back (B)\n• Straight (S)\n• Back (B)\n• Left (L)\n• Left (L)\n• Back (B)\n• Straight (S)\n• Left (L)\n• Left (L)\n• End\n\nWhat must be done in any of those intersections, is to save the each action done exactly at same sequence that it happens. For that, let's create a new variable (array) that will store the path that the robot has taken:\n\n``````char path = \" \";\n``````\n\nWe must also create 2 indexes variables to be used together with the array:\n\n``````unsigned char pathLength = 0; // the length of the path\nint pathIndex = 0; // used to reach an specific array element.\n``````\n\nSo, if we run the example shown in the picture, we will end with:\n\n``````path = [LBLLLBSBLLBSLL]\nand pathLengh = 14\n``````\n\n### Step 7: Simplifying (optimizing) the Path\n\nLet's return to our example. Looking the first group of intersections, we realized the the first left branch is in fact a \"Dead End\" and so, if the robot instead of a \"Left-Back-Left\" only passed straight at that first intersection, a lot of energy and time would be saved! In other words, a sequence \"LBL\" in fact would be the same as \"S\". That's is exactly how the full path can be optimized. If you analyze all possibilities where a \"U turn\" is used, the set of 3 intersections where this \" U-Turn\" (\"B\") appears (\"xBx\") can be reduced to only one.\n\nThe above is only one example, bellow you can find the complete list of possibilities (try it):\n\n• LBR = B\n• LBS = R\n• RBL = B\n• SBL = R\n• SBS = B\n• LBL = S\n\nTaking the full path or our example, we can reduce it:\n\n``````path = [LBLLLBSBLLBSLL] ==> LBL = S\npath = [SLLBSBLLBSLL] ==> LBS = R\npath = [SLRBLLBSLL] ==> RBL = B\npath = [SLBLBSLL] ==> LBL = S\npath = [SSBSLL] ==> SBS = B\npath = [SBLL] ==> SBL = R\npath = [RL]\n``````\n\nAmazing! Looking at the example it is very clear that if the robot takes RIGHT at first intersection and after that, a LEFT, it will reach the End of Maze in the shortest path!\n\nThe First Path of Maze Solver total code will be consolidated in the function mazeSolve(). This function is in fact the loop() function used before, but incorporating all those steps of storing and path optimization. When the first path ended, the path[] array will have the optimized path. A new variable is introduced:\n\n``````unsigned int status = 0; // solving = 0; reach Maze End = 1\n``````\n\nBelow the First Path function:\n\n``````void mazeSolve(void)\n{\nwhile (!status)\n{\nswitch (mode)\n{\ncase NO_LINE:\nmotorStop();\ngoAndTurn (LEFT, 180);\nrecIntersection('B');\nbreak;\n\ncase CONT_LINE:\nrunExtraInch();\nif (mode != CONT_LINE) {goAndTurn (LEFT, 90); recIntersection('L');} // or it is a \"T\" or \"Cross\"). In both cases, goes to LEFT\nelse mazeEnd();\nbreak;\n\ncase RIGHT_TURN:\nrunExtraInch();\nif (mode == NO_LINE) {goAndTurn (RIGHT, 90); recIntersection('R');}\nelse recIntersection('S');\nbreak;\n\ncase LEFT_TURN:\ngoAndTurn (LEFT, 90);\nrecIntersection('L');\nbreak;\n\ncase FOLLOWING_LINE:\nfollowingLine();\nbreak;\n\n}\n}\n}\n``````\n\nHere a new function was introduced: recIntersection (direction). This function will be used for store the intersection and also to call another function simplifyPath(), that will reduce the group of 3 intersections involving an \"U-Turn\" as we saw before.\n\n``````void recIntersection(char direction)\n{\npath[pathLength] = direction; // Store the intersection in the path variable.\npathLength ++;\nsimplifyPath(); // Simplify the learned path.\n}\n``````\n\nThe CREDIT for the simplifyPath() function is to Patrick McCabe for the path Solving Code (for details, please visit https://patrickmccabemakes.com)! The strategy of Path simplification is that whenever we encounter a sequence xBx, we can simplify it by cutting out the dead end. For example, `LBL ==> S` as we saw at the example.\n\n``````void simplifyPath()\n{\n// only simplify the path if the second-to-last turn was a 'B'\nif(pathLength < 3 || path[pathLength-2] != 'B')\nreturn;\nint totalAngle = 0;\nint i;\nfor(i=1;i<=3;i++)\n{\nswitch(path[pathLength-i])\n{\ncase 'R':\ntotalAngle += 90;\nbreak;\ncase 'L':\ntotalAngle += 270;\nbreak;\ncase 'B':\ntotalAngle += 180;\nbreak;\n}\n}\n// Get the angle as a number between 0 and 360 degrees.\ntotalAngle = totalAngle % 360;\n// Replace all of those turns with a single one.\nswitch(totalAngle)\n{\ncase 0:\npath[pathLength - 3] = 'S';\nbreak;\ncase 90:\npath[pathLength - 3] = 'R';\nbreak;\ncase 180:\npath[pathLength - 3] = 'B';\nbreak;\ncase 270:\npath[pathLength - 3] = 'L';\nbreak;\n}\n// The path is now two steps shorter.\npathLength -= 2;\n\n}\n``````\n\n### Step 8: Second Pass: solving the Maze as fast as possible!\n\nThe main program: `loop ()` is simple like that:\n\n``````void loop()\n{\n\nmazeSolve(); // First pass to solve the maze\npathIndex = 0;\nstatus = 0;\n\nmazeOptimization(); // Second Pass: run the maze as fast as possible\nmode = STOPPED;\nstatus = 0; // 1st pass\npathIndex = 0;\npathLength = 0;\n}\n``````\n\nSo, when the First Pass ends, what we must to do it is only feed the robot with the optimized path array. It will start run and when a intersection is found, it will now define what to do based on what it is stored at `path[]` .\n\n``````void mazeOptimization (void)\n{\nwhile (!status)\n{\nswitch (mode)\n{\ncase FOLLOWING_LINE:\nfollowingLine();\nbreak;\ncase CONT_LINE:\nif (pathIndex >= pathLength) mazeEnd ();\nelse {mazeTurn (path[pathIndex]); pathIndex++;}\nbreak;\ncase LEFT_TURN:\nif (pathIndex >= pathLength) mazeEnd ();\nelse {mazeTurn (path[pathIndex]); pathIndex++;}\nbreak;\ncase RIGHT_TURN:\nif (pathIndex >= pathLength) mazeEnd ();\nelse {mazeTurn (path[pathIndex]); pathIndex++;}\nbreak;\n}\n}\n}\n``````\n\nTo command what to do, a new function mazeTurn(path[]) was created. The function mazeTurn (path[]) will be:\n\n``````void mazeTurn (char dir)\n{\nswitch(dir)\n{\ncase 'L': // Turn Left\ngoAndTurn (LEFT, 90);\nbreak;\n\ncase 'R': // Turn Right\ngoAndTurn (RIGHT, 90);\nbreak;\n\ncase 'B': // Turn Back\ngoAndTurn (RIGHT, 800);\nbreak;\n\ncase 'S': // Go Straight\nrunExtraInch();\nbreak;\n}\n}\n``````\n\nThe second pass is done! The below video shows the complete example worked here, first and second pass. Below the Arduino code used on this tutorial:\n\n### Step 9: Using the Android for tuning\n\nThe Android App developed for the Following Line project can also be used here (if you need, the Android App and its code are available at: Line Follower Robot - PID Control - Android Setup. The Arduino code presented at last step already includes comunication with the Android device. If you do not want to use de Android app, no problem because the code is \"`transparent` \".\n\nI used the Android a lot during the project to send test data from the robot to the device using the \"`Message Received` \" field. Several variables must be well defined in order to guarantee that the robot will turn the correct angle. The most important ones are bellow (the ones marked in Bold I changed them several times):\n\n``````const int adj = 0;\nTHRESHOLD = 150\nconst int power = 250;\nconst int iniMotorPower = 250;\nint extraInch = 200;\n``````\n\n### Step 10: Conclusion\n\nThis is the second and last part a complex project, exploring the potentiality of a line follower Robot, where Artificial Intelligence (AI) simple concepts were used to solve a maze.\n\nI am not an AI expert and based on some information that I got from web, I understood that what our little Rex, the robot did, solving the maze could be considered an application of AI. Let's take a look on the 2 sources bellow:\n\nFrom Wikipedia:\n\nThe main objective of the AI ​​systems, is to perform functions that, if a human being were to run, would be considered intelligent. It is a broad concept, and receives as many definitions as give different meanings to the word intelligence. We can think of some basic features of these systems, such as reasoning ability (apply logical rules to a set of available data to reach a conclusion ), learning (learning from the mistakes and successes so in the future act more effectively), pattern recognition (both visual and sensory patterns, as well as behavior patterns) and inference (ability to be able to apply the reasoning in situations of our daily).\n\nOr from this university paper: \"Maze Solving Robot Using Freeduino and LSRB Algorithm International Journal of Modern Engineering Research (IJMER)\"\n\nMaze-solving involves Control Engineering and Artificial Intelligence. Using a good algorithm can achieve the high efficiency of finding the shortest path. The proposed maze-solving algorithm works better and has short searching time and low spacecomplexity, and it is significant for robot’s finding path in some areas like maze-solving.\n\nThe updated files for this project can be found at GITHUB. Hope I could contribute for others to learn more about electronics, robot, Arduino, etc. For more tutorials, please visit my Blog: MJRoBot.org\n\nSaludos from the south of the world!\n\nThanks\n\nMarcelo\n\n## Code\n\n##### Code snippet #1Plain text\n```const int lineFollowSensor0 = 12; //Using Digital input\nconst int lineFollowSensor1 = 18; //Using Analog Pin A4 as Digital input\nconst int lineFollowSensor2 = 17; //Using Analog Pin A3 as Digital input\nconst int lineFollowSensor3 = 16; //Using Analog Pin A2 as Digital input\nconst int lineFollowSensor4 = 19; //Using Analog Pin A5 as Digital input\nconst int farRightSensorPin = 0; //Analog Pin A0\nconst int farLeftSensorPin = 1; //Analog Pin A1\n```\n##### Code snippet #4Plain text\n```LFSensor = digitalRead(lineFollowSensor0);\n```\n##### Code snippet #5Plain text\n```void testSensorLogic(void)\n{\nSerial.print (farLeftSensor);\nSerial.print (\" <== LEFT RIGH==> \");\nSerial.print (farRightSensor);\nSerial.print (\" mode: \");\nSerial.print (mode);\nSerial.print (\" error:\");\nSerial.println (error);\n}\n```\n##### Code snippet #6Plain text\n```void loop()\n{\nswitch (mode)\n{\ncase NO_LINE:\nmotorStop();\ngoAndTurn (LEFT, 180);\nbreak;\ncase CONT_LINE:\nrunExtraInch();\nif (mode == CONT_LINE) mazeEnd();\nelse goAndTurn (LEFT, 90); // or it is a \"T\" or \"Cross\"). In both cases, goes to LEFT\nbreak;\ncase RIGHT_TURN:\nrunExtraInch();\nif (mode == NO_LINE) goAndTurn (RIGHT, 90);\nbreak;\ncase LEFT_TURN:\ngoAndTurn (LEFT, 90);\nbreak;\ncase FOLLOWING_LINE:\nfollowingLine();\nbreak;\n}\n}\n```\n##### Code snippet #7Plain text\n```void runExtraInch(void)\n{\nmotorPIDcontrol();\ndelay(extraInch);\nmotorStop();\n}\n```\n##### Code snippet #8Plain text\n```void goAndTurn(int direction, int degrees)\n{\nmotorPIDcontrol();\nmotorTurn(direction, degrees);\n}\n```\n##### Code snippet #12Plain text\n```void mazeSolve(void)\n{\nwhile (!status)\n{\nswitch (mode)\n{\ncase NO_LINE:\nmotorStop();\ngoAndTurn (LEFT, 180);\nrecIntersection('B');\nbreak;\n\ncase CONT_LINE:\nrunExtraInch();\nif (mode != CONT_LINE) {goAndTurn (LEFT, 90); recIntersection('L');} // or it is a \"T\" or \"Cross\"). In both cases, goes to LEFT\nelse mazeEnd();\nbreak;\n\ncase RIGHT_TURN:\nrunExtraInch();\nif (mode == NO_LINE) {goAndTurn (RIGHT, 90); recIntersection('R');}\nelse recIntersection('S');\nbreak;\n\ncase LEFT_TURN:\ngoAndTurn (LEFT, 90);\nrecIntersection('L');\nbreak;\n\ncase FOLLOWING_LINE:\nfollowingLine();\nbreak;\n\n}\n}\n}\n```\n##### Code snippet #13Plain text\n```void recIntersection(char direction)\n{\npath[pathLength] = direction; // Store the intersection in the path variable.\npathLength ++;\nsimplifyPath(); // Simplify the learned path.\n}\n```\n##### Code snippet #14Plain text\n```void simplifyPath()\n{\n// only simplify the path if the second-to-last turn was a 'B'\nif(pathLength < 3 || path[pathLength-2] != 'B')\nreturn;\n\nint totalAngle = 0;\nint i;\nfor(i=1;i<=3;i++)\n{\nswitch(path[pathLength-i])\n{\ncase 'R':\ntotalAngle += 90;\nbreak;\ncase 'L':\ntotalAngle += 270;\nbreak;\ncase 'B':\ntotalAngle += 180;\nbreak;\n}\n}\n\n// Get the angle as a number between 0 and 360 degrees.\ntotalAngle = totalAngle % 360;\n\n// Replace all of those turns with a single one.\nswitch(totalAngle)\n{\ncase 0:\npath[pathLength - 3] = 'S';\nbreak;\ncase 90:\npath[pathLength - 3] = 'R';\nbreak;\ncase 180:\npath[pathLength - 3] = 'B';\nbreak;\ncase 270:\npath[pathLength - 3] = 'L';\nbreak;\n}\n\n// The path is now two steps shorter.\npathLength -= 2;\n\n}\n```\n##### Code snippet #15Plain text\n```void loop()\n{\n\nmazeSolve(); // First pass to solve the maze\npathIndex = 0;\nstatus = 0;\n\nmazeOptimization(); // Second Pass: run the maze as fast as possible\nmode = STOPPED;\nstatus = 0; // 1st pass\npathIndex = 0;\npathLength = 0;\n}\n```\n##### Code snippet #16Plain text\n```void mazeOptimization (void)\n{\nwhile (!status)\n{\nswitch (mode)\n{\ncase FOLLOWING_LINE:\nfollowingLine();\nbreak;\ncase CONT_LINE:\nif (pathIndex >= pathLength) mazeEnd ();\nelse {mazeTurn (path[pathIndex]); pathIndex++;}\nbreak;\ncase LEFT_TURN:\nif (pathIndex >= pathLength) mazeEnd ();\nelse {mazeTurn (path[pathIndex]); pathIndex++;}\nbreak;\ncase RIGHT_TURN:\nif (pathIndex >= pathLength) mazeEnd ();\nelse {mazeTurn (path[pathIndex]); pathIndex++;}\nbreak;\n}\n}\n}\n```\n##### Code snippet #17Plain text\n```void mazeTurn (char dir)\n{\nswitch(dir)\n{\ncase 'L': // Turn Left\ngoAndTurn (LEFT, 90);\nbreak;\n\ncase 'R': // Turn Right\ngoAndTurn (RIGHT, 90);\nbreak;\n\ncase 'B': // Turn Back\ngoAndTurn (RIGHT, 800);\nbreak;\n\ncase 'S': // Go Straight\nrunExtraInch();\nbreak;\n}\n}\n```\n##### Github\nhttps://github.com/Mjrovai/MJRoBot-Maze-Solver\n\n## Schematics\n\nz7IdLkxL1J66qOtphxqC.fzz\n\n#### Line Follower Robot - PID Control - Android Setup\n\nProject tutorial by MJRoBot\n\n• 36,286 views\n• 46 respects\n\n#### The \"Mars Rover\" Emulator\n\nProject tutorial by MJRoBot\n\n• 12,309 views\n• 51 respects\n\n#### Otto DIY+ Arduino Bluetooth robot easy to 3D Print\n\nProject tutorial by Team Otto builders\n\n• 77,639 views\n• 219 respects\n\n• 15,492 views\n• 57 respects" ]

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[ "Home »» Placement Papers »» All Papers »» Virtusa »» Placement Paper\n\n### VIRTUSA PAPER - AUG 2003 - HYDERABAD\n\nAUG 2003\n\n1) how many the square can divided of equal sizes.\nans 4,   u must write the reason below ,\n\nThe ans is 4squares, diagonally, 4retangles, along midpoint\n\n2)cylinder diameter 30 and height 20 , make 12 equal parts, with minimum cuts, what r the minimum cuts;\n\n3) 1*2=6 and 3*4=20 then 7*7=?\n\n4) 1,2,3,25,50,75,100 using this numbers once form 383?\n\n5) Find out the odd man out, on this 3 questions r given\n\n6) Sphere volume and surface area is given then find the radius?\n\n7) Two containers 500 and 300 r given then find 100 ml no measures r given?\n\n8)Sum of sachin test runs and product of broadman test runs\nwhich is greater ?\n\n9 ) two groups in country knights and knaves , knight will always true , knave will tell always false., then a person asks , who r u , then told \" I am knave \" , then what do u interpret?\n\n10) coding and decoding problem?\n\n11)number series?\n\nESSAY\n\n1) as india is poor country, is it nesecessary to conduct the afroasian games?\n\n2)as very talented people r migrating to IT , is other fields r getting starvation?\n\nPROGRAMMING\n\n1)written program was given and we have to find out the output ?\nthe program given is genaration of febonnacci series\n\n2)one psudo code was given of robot ,   u have to find the bugs\n\n3)considar 2 digit number , then multiply with 2, if it becomes 3 digit the take last 2 digits and repeat the process untill u got the number already generated in this sequence and  print all numbers?\n\nFor this considar array size of 50 , since maximum even number possibulities will be 50 only(2 digit numbers), this thing he is menctioning in the interview and asking reason , if u declare more size or linked lists\n\n4)considar any number if even divided by 2 and else multiply with 3 and add 1, repeat this untill u get 4, 2,1 sereis, means untill u get 1?\n\n1*3+1=4 , 4/2=2, 2/2=1 1*3+1=4 and so....on\n\n5) considat on string , write a function such that   function(string str, char omega, char alpha) ,genarate the biggest substring starting with omega and ending with alpha.    This the paper and he is giving then he will conduct the gd, in gd he asked our self to choose the topic. About hr, he is more, regural questions, why to take our company and all\n\n Comment Tweet" ]

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[ "# Draw a 3D sphere with radius with TikZ? [duplicate]\n\nDraw a sphere with radius with TikZ. According to the figure", null, "It is always hard to get answers to questions on the form \"draw this for me\" without showing any effort your self. Next time try to start a solution and add a MWE to your question, then you have much higher chance of getting someone to dig into your problem.\n\nHere I guess the problem is the equator, where you can use \\arc with different x- and y-radius.\n\n\\documentclass[border=5mm]{standalone}\n\\usepackage{tikz}\n\\begin{document}\n\\begin{tikzpicture}\n\\shade[ball color = gray!40, opacity = 0.4] (0,0) circle (2cm);\n\\draw (0,0) circle (2cm);\n\\draw (-2,0) arc (180:360:2 and 0.6);\n\\draw[dashed] (2,0) arc (0:180:2 and 0.6);\n\\fill[fill=black] (0,0) circle (1pt);\n\\draw[dashed] (0,0 ) -- node[above]{$r$} (2,0);\n\\end{tikzpicture}\n\\end{document}", null, "" ]

[ null, "https://i.stack.imgur.com/9KUc3.jpg", null, "https://i.stack.imgur.com/HfayXm.png", null ]

[ "## What is non linear regression example?\n\nOne example of how nonlinear regression can be used is to predict population growth over time. A scatterplot of changing population data over time shows that there seems to be a relationship between time and population growth, but that it is a nonlinear relationship, requiring the use of a nonlinear regression model.\n\nIs maximum likelihood estimator unbiased?\n\nMLE is a biased estimator (Equation 12).\n\n### How do you choose a nonlinear regression?\n\nGuidelines for Choosing Between Linear and Nonlinear Regression. The general guideline is to use linear regression first to determine whether it can fit the particular type of curve in your data. If you can’t obtain an adequate fit using linear regression, that’s when you might need to choose nonlinear regression.\n\nIs maximum likelihood biased?\n\nIt is well known that maximum likelihood estimators are often biased, and it is of use to estimate the expected bias so that we can reduce the mean square errors of our parameter estimates.\n\n#### How do you show Unbiasedness?\n\nAn unbiased estimator of a parameter is an estimator whose expected value is equal to the parameter. That is, if the estimator S is being used to estimate a parameter θ, then S is an unbiased estimator of θ if E(S)=θ. Remember that expectation can be thought of as a long-run average value of a random variable.\n\nWhat are some examples of non linear regression models?\n\nExamples of Non-Linear Regression Models. 1. Logistic regression model. Logistic regression is a type of non-linear regression model. It is most commonly used when the target variable or the dependent variable is categorical. For example, whether a tumor is malignant or benign, or whether an email is useful or spam.\n\n## How do you find the maximum likelihood of a regression line?\n\nThus, the principle of maximum likelihood is equivalent to the least squares criterion for ordinary linear regression. The maximum likelihood estimators ↵ and give the regression line yˆ i=ˆ↵ +ˆx i. with ˆ = cov(x,y) var(x) , and ↵ˆ determined by solving y¯ =ˆ↵ +ˆx.¯ Exercise 15.8.\n\nWhat is non linear regression in R with parameters?\n\nNon-Linear Regression in R. R Non-linear regression is a regression analysis method to predict a target variable using a non-linear function consisting of parameters and one or more independent variables. Non-linear regression is often more accurate as it learns the variations and dependencies of the data.\n\n### What is an example of maximum likelihood estimator?\n\nFor the example for the distribution of fitness effects ↵ =0.23 and =5.35 with n = 100, a simulated data set yields ↵ˆ =0.2376 and ˆ =5.690 for maximum likelihood estimator. (See Figure 15.4.) 231 Introduction to the Science of Statistics Maximum Likelihood Estimation" ]

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[ "# multiple choice help\n\nAnnouncements\n#1\nHi\n\nan help would be appreciated\n\nan electric motor of input power 100w raises a mass of 10kg vertically at a steady speed of 0.5ms. What is the efficiency of the system\n0\n4 years ago\n#2\n(Original post by GG16)\nHi\n\nan help would be appreciated\n\nan electric motor of input power 100w raises a mass of 10kg vertically at a steady speed of 0.5ms. What is the efficiency of the system\nP = WD / t = F x (D / t) = F x v\n\nCan you work out the answer from here?\n0\n4 years ago\n#3\n(Original post by GG16)\nHi\n\nan help would be appreciated\n\nan electric motor of input power 100w raises a mass of 10kg vertically at a steady speed of 0.5ms. What is the efficiency of the system\nHello there,\n\nYou should be able to calculate the output power as using the applied force (equal to the object's weight in this case) and the speed of ascent. You should use the equation", null, ". You can then find the efficiency by dividing this by the input power and if they want it as a percentage, multiply it by", null, ".\n\nI hope that this has been helpful.\n\nSmithenator5000.\n0\n4 years ago\n#4\n(Original post by GG16)\nan electric motor of input power 100w raises a mass of 10kg vertically at a steady speed of 0.5ms. What is the efficiency of the system\nThe efficiency is the useful work output divided by the input power. We have the input power of 100W. What's the useful work output?\n\n(You should get an answer of somewhere close to 50%)\n0\n4 years ago\n#5\n(Original post by GG16)\nElectric motor of input power 100 W raises mass of 10 kg vertically at steady speed 0.5 ms^-1. What is the efficiency of the system?\nPower of mass is its weight x speed (10 x 9.81 x 0.5)\nTook 100 W of electrical input power to attain that.\nEfficiency = (Output/Input) x 100% = 49.1%\n0\n7 months ago\n#6\nwhy did you times by 9.81, i dont understand\n0\n7 months ago\n#7\n(Original post by Subhawk)\nwhy did you times by 9.81, i dont understand\nThe thread is 4 years old...\n\nPower = Force * Velocity\nGet Force (weight) and you know velocity\nHence have the power actually needed to lift the object (the output)\nYou're given the input\nEfficiency is input/output * 100\n\nYou need 9.81 to calculate the weight (force)\n0\nX\n\nnew posts", null, "Back\nto top\nLatest\nMy Feed\n\n### Oops, nobody has postedin the last few hours.\n\nWhy not re-start the conversation?\n\nsee more\n\n### See more of what you like onThe Student Room\n\nYou can personalise what you see on TSR. Tell us a little about yourself to get started.\n\n### Poll\n\nJoin the discussion\n\n#### Year 12s - where are you at with making decisions about university?\n\nI’ve chosen my course and my university (26)\n27.37%\nI’ve chosen my course and shortlisted some universities (31)\n32.63%\nI’ve chosen my course, but not any universities (10)\n10.53%\nI’ve chosen my university, but not my course (3)\n3.16%\nI’ve shortlisted some universities, but not my course (5)\n5.26%\nI’m starting to consider my university options (14)\n14.74%\nI haven’t started thinking about university yet (4)\n4.21%\nI’m not planning on going to university (2)\n2.11%" ]

[ null, "https://www.thestudentroom.co.uk/latexrender/pictures/64/644d9546108c5369520fcf061b3d4933.png", null, "https://www.thestudentroom.co.uk/latexrender/pictures/34/345cfb2ea853194743256b476280d88a.png", null, "https://www.thestudentroom.co.uk/images/v2/icons/arrow_up.svg", null ]

[ "", null, "Algorithmique Numérique Distribuée Public GIT Repository\n1 #include \"xbt/sysdep.h\"\n2 #include \"smpi/smpi.h\"\n3 #include \"private.h\"\n5 #include <math.h>\n7 typedef struct s_smpi_mpi_topology {\n8   int nnodes;\n9   int ndims;\n10   int *dims;\n11   int *periodic;\n12   int *position;\n13 } s_smpi_mpi_topology_t;\n17 void smpi_topo_destroy(MPI_Topology topo) {\n18   if (topo) {\n19     if(topo->dims) {\n20       free(topo->dims);\n21     }\n22     if(topo->periodic) {\n23       free(topo->periodic);\n24     }\n25     if(topo->position) {\n26       free(topo->position);\n27     }\n28     free(topo);\n29   }\n30 }\n32 MPI_Topology smpi_topo_create(int ndims) {\n33   MPI_Topology topo = xbt_malloc(sizeof(*topo));\n34   topo->nnodes = 0;\n35   topo->ndims = ndims;\n36   topo->dims = xbt_malloc(ndims * sizeof(*topo->dims));\n37   topo->periodic = xbt_malloc(ndims * sizeof(*topo->periodic));\n38   topo->position = xbt_malloc(ndims * sizeof(*topo->position));\n39   return topo;\n40 }\n42 /* reorder is ignored, don't know what would be the consequences of a dumb\n43  * reordering but neither do I see the point of reordering*/\n44 int smpi_mpi_cart_create(MPI_Comm comm_old, int ndims, int dims[],\n45                      int periods[], int reorder, MPI_Comm *comm_cart) {\n46   int retval = MPI_SUCCESS;\n47   int i;\n48   MPI_Topology topo;\n49   MPI_Group newGroup, oldGroup;\n50   int rank, nranks,  newSize;\n54   rank = smpi_comm_rank(comm_old);\n58   newSize = 1;\n59   if(ndims != 0) {\n60     topo = smpi_topo_create(ndims);\n61     for (i = 0 ; i < ndims ; i++) {\n62       newSize *= dims[i];\n63     }\n64     if(rank >= newSize) {\n65       *comm_cart = MPI_COMM_NULL;\n66       return retval;\n67     }\n68     oldGroup = smpi_comm_group(comm_old);\n69     newGroup = smpi_group_new(newSize);\n70     for (i = 0 ; i < newSize ; i++) {\n71       smpi_group_set_mapping(newGroup, smpi_group_index(oldGroup, i), i);\n72     }\n74     topo->nnodes = newSize;\n76     memcpy(topo->dims, dims, ndims * sizeof(*topo->dims));\n77     memcpy(topo->periodic, periods, ndims * sizeof(*topo->periodic));\n79     //  code duplication... See smpi_mpi_cart_coords\n80     nranks = newSize;\n81     for (i=0; i<ndims; i++)\n82       {\n83         topo->dims[i]     = dims[i];\n84         topo->periodic[i] = periods[i];\n85         nranks = nranks / dims[i];\n86         /* FIXME: nranks could be zero (?) */\n87         topo->position[i] = rank / nranks;\n88         rank = rank % nranks;\n89       }\n91     *comm_cart = smpi_comm_new(newGroup, topo);\n92   }\n93   else {\n94     if (rank == 0) {\n95       topo = smpi_topo_create(ndims);\n96       *comm_cart = smpi_comm_new(smpi_comm_group(MPI_COMM_SELF), topo);\n97     }\n98     else {\n99       *comm_cart = MPI_COMM_NULL;\n100     }\n101   }\n102   return retval;\n103 }\n105 int smpi_mpi_cart_sub(MPI_Comm comm, const int remain_dims[], MPI_Comm *newcomm) {\n106   MPI_Topology oldTopo = smpi_comm_topo(comm);\n107   int oldNDims = oldTopo->ndims;\n108   int i, j = 0, newNDims, *newDims = NULL, *newPeriodic = NULL;\n110   if (remain_dims == NULL && oldNDims != 0) {\n111     return MPI_ERR_ARG;\n112   }\n113   newNDims = 0;\n114   for (i = 0 ; i < oldNDims ; i++) {\n115     if (remain_dims[i]) newNDims++;\n116   }\n118   if (newNDims > 0) {\n119     newDims = malloc(newNDims * sizeof(*newDims));\n120     newPeriodic = malloc(newNDims * sizeof(*newPeriodic));\n122     // that should not segfault\n123     for (i = 0 ; j < newNDims ; i++) {\n124       if(remain_dims[i]) {\n125         newDims[j] = oldTopo->dims[i];\n126         newPeriodic[j] = oldTopo->periodic[i];\n127         j++;\n128       }\n129     }\n130   }\n131   return smpi_mpi_cart_create(comm, newNDims, newDims, newPeriodic, 0, newcomm);\n132 }\n137 int smpi_mpi_cart_coords(MPI_Comm comm, int rank, int maxdims,\n138                          int coords[]) {\n139   int nnodes;\n140   int i;\n141   MPI_Topology topo = smpi_comm_topo(comm);\n143   nnodes = topo->nnodes;\n144   for ( i=0; i < topo->ndims; i++ ) {\n145     nnodes    = nnodes / topo->dims[i];\n146     coords[i] = rank / nnodes;\n147     rank      = rank % nnodes;\n148   }\n149   return MPI_SUCCESS;\n150 }\n152 int smpi_mpi_cart_get(MPI_Comm comm, int maxdims, int* dims, int* periods, int* coords) {\n153   MPI_Topology topo = smpi_comm_topo(comm);\n154   int i;\n155   for(i = 0 ; i < maxdims ; i++) {\n156     dims[i] = topo->dims[i];\n157     periods[i] = topo->periodic[i];\n158     coords[i] = topo->position[i];\n159   }\n160   return MPI_SUCCESS;\n161 }\n163 int smpi_mpi_cart_rank(MPI_Comm comm, int* coords, int* rank) {\n164   MPI_Topology topo = smpi_comm_topo(comm);\n165   int ndims = topo->ndims;\n166   int multiplier, coord,i;\n167   *rank = 0;\n168   multiplier = 1;\n172   for ( i=ndims-1; i >=0; i-- ) {\n173     coord = coords[i];\n175     /* Should we check first for args correction, then process,\n176      * or check while we work (as it is currently done) ? */\n177       if (coord >= topo->dims[i]) {\n178         if ( topo->periodic[i] ) {\n179         coord = coord % topo->dims[i];\n180         }\n181         else {\n182           // Should I do that ?\n183           *rank = -1;\n184           return MPI_ERR_ARG;\n185         }\n186       }\n187       else if (coord <  0) {\n188         if(topo->periodic[i]) {\n189           coord = coord % topo->dims[i];\n190           if (coord) coord = topo->dims[i] + coord;\n191         }\n192         else {\n193           *rank = -1;\n194           return MPI_ERR_ARG;\n195         }\n196       }\n198       *rank += multiplier * coord;\n199       multiplier *= topo->dims[i];\n200   }\n201   return MPI_SUCCESS;\n202 }\n204 int smpi_mpi_cart_shift(MPI_Comm comm, int direction, int disp,\n205                         int *rank_source, int *rank_dest) {\n206   MPI_Topology topo = smpi_comm_topo(comm);\n207   int position[topo->ndims];\n210   if(topo->ndims == 0) {\n211     return MPI_ERR_ARG;\n212   }\n213   if (topo->ndims < direction) {\n214     return MPI_ERR_DIMS;\n215   }\n217   smpi_mpi_cart_coords(comm, smpi_comm_rank(comm), topo->ndims, position);\n218   position[direction] += disp;\n220   if(position[direction] < 0 || position[direction] >= topo->dims[direction]) {\n221     if(topo->periodic[direction]) {\n222       position[direction] %= topo->dims[direction];\n223       smpi_mpi_cart_rank(comm, position, rank_dest);\n224     }\n225     else {\n226       *rank_dest = MPI_PROC_NULL;\n227     }\n228   }\n229   else {\n230     smpi_mpi_cart_rank(comm, position, rank_dest);\n231   }\n233   position[direction] =  topo->position[direction] - disp;\n234   if(position[direction] < 0 || position[direction] >= topo->dims[direction]) {\n235     if(topo->periodic[direction]) {\n236       position[direction] %= topo->dims[direction];\n237       smpi_mpi_cart_rank(comm, position, rank_source);\n238     }\n239     else {\n240       *rank_source = MPI_PROC_NULL;\n241     }\n242   }\n243   else {\n244     smpi_mpi_cart_rank(comm, position, rank_source);\n245   }\n247   return MPI_SUCCESS;\n248 }\n250 int smpi_mpi_cartdim_get(MPI_Comm comm, int *ndims) {\n251   MPI_Topology topo = smpi_comm_topo(comm);\n253   *ndims = topo->ndims;\n254   return MPI_SUCCESS;\n255 }\n259 // Everything below has been taken from ompi, but could be easily rewritten.\n261 /*\n262  * Copyright (c) 2004-2007 The Trustees of Indiana University and Indiana\n263  *                         University Research and Technology\n264  *                         Corporation.  All rights reserved.\n265  * Copyright (c) 2004-2005 The University of Tennessee and The University\n266  *                         of Tennessee Research Foundation.  All rights\n267  *                         reserved.\n268  * Copyright (c) 2004-2014 High Performance Computing Center Stuttgart,\n269  *                         University of Stuttgart.  All rights reserved.\n270  * Copyright (c) 2004-2005 The Regents of the University of California.\n271  *                         All rights reserved.\n272  * Copyright (c) 2012      Los Alamos National Security, LLC.  All rights\n273  *                         reserved.\n274  * Copyright (c) 2014      Intel, Inc. All rights reserved\n276  *\n278  *\n280  */\n283 /* static functions */\n284 static int assignnodes(int ndim, int nfactor, int *pfacts,int **pdims);\n285 static int getfactors(int num, int *nfators, int **factors);\n287 /*\n288  * This is a utility function, no need to have anything in the lower\n289  * layer for this at all\n290  */\n291 int smpi_mpi_dims_create(int nnodes, int ndims, int dims[])\n292 {\n293     int i;\n294     int freeprocs;\n295     int freedims;\n296     int nfactors;\n297     int *factors;\n298     int *procs;\n299     int *p;\n300     int err;\n302     /* Get # of free-to-be-assigned processes and # of free dimensions */\n303     freeprocs = nnodes;\n304     freedims = 0;\n305     for (i = 0, p = dims; i < ndims; ++i,++p) {\n306         if (*p == 0) {\n307             ++freedims;\n308         } else if ((*p < 0) || ((nnodes % *p) != 0)) {\n309           return  MPI_ERR_DIMS;\n311         } else {\n312             freeprocs /= *p;\n313         }\n314     }\n316     if (freedims == 0) {\n317        if (freeprocs == 1) {\n318           return MPI_SUCCESS;\n319        }\n320        return MPI_ERR_DIMS;\n321     }\n323     if (freeprocs == 1) {\n324         for (i = 0; i < ndims; ++i, ++dims) {\n325             if (*dims == 0) {\n326                *dims = 1;\n327             }\n328         }\n329         return MPI_SUCCESS;\n330     }\n332     /* Factor the number of free processes */\n333     if (MPI_SUCCESS != (err = getfactors(freeprocs, &nfactors, &factors))) {\n334       return  err;\n335     }\n337     /* Assign free processes to free dimensions */\n338     if (MPI_SUCCESS != (err = assignnodes(freedims, nfactors, factors, &procs))) {\n339       return err;\n340     }\n342     /* Return assignment results */\n343     p = procs;\n344     for (i = 0; i < ndims; ++i, ++dims) {\n345         if (*dims == 0) {\n346            *dims = *p++;\n347         }\n348     }\n350     free((char *) factors);\n351     free((char *) procs);\n353     /* all done */\n354     return MPI_SUCCESS;\n355 }\n357 /*\n358  *  assignnodes\n359  *\n360  *  Function:   - assign processes to dimensions\n361  *          - get \"best-balanced\" grid\n362  *          - greedy bin-packing algorithm used\n363  *          - sort dimensions in decreasing order\n364  *          - dimensions array dynamically allocated\n365  *  Accepts:    - # of dimensions\n366  *          - # of prime factors\n367  *          - array of prime factors\n368  *          - ptr to array of dimensions (returned value)\n369  *  Returns:    - 0 or ERROR\n370  */\n371 static int\n372 assignnodes(int ndim, int nfactor, int *pfacts, int **pdims)\n373 {\n374     int *bins;\n375     int i, j;\n376     int n;\n377     int f;\n378     int *p;\n379     int *pmin;\n381     if (0 >= ndim) {\n382        return MPI_ERR_DIMS;\n383     }\n385     /* Allocate and initialize the bins */\n386     bins = (int *) malloc((unsigned) ndim * sizeof(int));\n387     if (NULL == bins) {\n388        return MPI_ERR_NO_MEM;\n389     }\n390     *pdims = bins;\n392     for (i = 0, p = bins; i < ndim; ++i, ++p) {\n393         *p = 1;\n394      }\n396     /* Loop assigning factors from the highest to the lowest */\n397     for (j = nfactor - 1; j >= 0; --j) {\n398         f = pfacts[j];\n399         /* Assign a factor to the smallest bin */\n400         pmin = bins;\n401         for (i = 1, p = pmin + 1; i < ndim; ++i, ++p) {\n402             if (*p < *pmin) {\n403                 pmin = p;\n404             }\n405         }\n406         *pmin *= f;\n407      }\n409      /* Sort dimensions in decreasing order (O(n^2) for now) */\n410      for (i = 0, pmin = bins; i < ndim - 1; ++i, ++pmin) {\n411          for (j = i + 1, p = pmin + 1; j < ndim; ++j, ++p) {\n412              if (*p > *pmin) {\n413                 n = *p;\n414                 *p = *pmin;\n415                 *pmin = n;\n416              }\n417          }\n418      }\n420      return MPI_SUCCESS;\n421 }\n423 /*\n424  *  getfactors\n425  *\n426  *  Function:   - factorize a number\n427  *  Accepts:    - number\n428  *          - # prime factors\n429  *          - array of prime factors\n430  *  Returns:    - MPI_SUCCESS or ERROR\n431  */\n432 static int\n433 getfactors(int num, int *nfactors, int **factors) {\n434     int size;\n435     int d;\n436     int i;\n437     int sqrtnum;\n439     if(num  < 2) {\n440         (*nfactors) = 0;\n441         (*factors) = NULL;\n442         return MPI_SUCCESS;\n443     }\n444     /* Allocate the array of prime factors which cannot exceed log_2(num) entries */\n445     sqrtnum = ceil(sqrt(num));\n446     size = ceil(log(num) / log(2));\n447     *factors = (int *) malloc((unsigned) size * sizeof(int));\n449     i = 0;\n450     /* determine all occurences of factor 2 */\n451     while((num % 2) == 0) {\n452         num /= 2;\n453         (*factors)[i++] = 2;\n454     }\n455     /* determine all occurences of uneven prime numbers up to sqrt(num) */\n456     for(d = 3; (num > 1) && (d < sqrtnum); d += 2) {\n457         while((num % d) == 0) {\n458             num /= d;\n459             (*factors)[i++] = d;\n460         }\n461     }\n462     /* as we looped only up to sqrt(num) one factor > sqrt(num) may be left over */\n463     if(num != 1) {\n464         (*factors)[i++] = num;\n465     }\n466     (*nfactors) = i;\n467     return MPI_SUCCESS;\n468 }" ]

[ null, "http://info.iut-bm.univ-fcomte.fr/pub/gitweb-files/images/logo_and.png", null ]

[ "# Connotation meant by \"B of A\" and \"A's B\". What would be the differences?\n\nDo you agree with the interpretation of the following three example sentences?\n\n1. I don't want to be involved in the problems of my boss.\nThe boss probably deals with some problems. e.g. slow sales, product quality issue, customer complaint\n1. I don't want to be involved in my boss's problem.\nThis implies a similar case as above one, but with only one problem.\n1. I don't want to be involved in the problem of my boss.\nProbably the boss himself has caused trouble. e.g. a misconduct\n\nIf you agree, then\n\nQ1) Why is \"my boss's problem\" likely to mean that my boss has ONLY ONE problem?\n\nQ2) why is \"the problem of my boss\" LESS LIKELY to mean that my boss has only one problem, but MORE LIKELY to mean that the boss himself probably has caused trouble?\n\n• [how come x? Very informal. Better: Why does x mean or Why is x likely to mean etc.] Jul 25, 2020 at 13:43\n\nSounding like a native English speaker (any standard variety of English)\n\nI disagree with the interpretation provided for this simple reason:\n\nMy boss's problem [singular] and My boss's problems [plural] are the simplest and most idiomatic way of expressing what ails the boss: one problem or more problems than one.\n\nThe \"of the\" point is just awkward. In actual speech, using the \"of the\" option here sounds non-native or non-idiomatic.\n\n• I don't want to be involved with my boss's problem OR boss's problems.\n• The problem of X implies a problem caused by, or concerning X e.g. the problem of evil, whereas X's problem means a problem experienced by X. Jul 25, 2020 at 12:52\n• my boss's problem, my boss's problems, the problem or problems of my boss = all the same semantically except for one is singular and one is plural. . I don't think you can teach me anything here. Sorry. Your examples are not relevant here. Jul 25, 2020 at 12:56\n• Well, that is how I understand the problem of X. Jul 25, 2020 at 18:22\n• And you no. 3. is most naturally expressed by the problem(s) with my boss. Jul 25, 2020 at 20:25\n• @ColinFine My number 3?? 1 and 3 are the same except for plural and singular. And \"I don't want to be involved in the problem(s) with my boss.\" means something else. Jul 25, 2020 at 23:03" ]

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[ "# Download Encyclopedie des sciences mathematiques. Geometrie by Molk J. (ed.) PDF", null, "By Molk J. (ed.)\n\nBest geometry and topology books\n\nIntroduction a la Topologie\n\nCe cours de topologie a été dispensé en licence à l'Université de Rennes 1 de 1999 à 2002. Toutes les constructions permettant de parler de limite et de continuité sont d'abord dégagées, puis l'utilité de l. a. compacité pour ramener des problèmes de complexité infinie à l'étude d'un nombre fini de cas est explicitée.\n\nSpaces of Constant Curvature\n\nThis ebook is the 6th version of the vintage areas of continuing Curvature, first released in 1967, with the former (fifth) variation released in 1984. It illustrates the excessive measure of interaction among team concept and geometry. The reader will enjoy the very concise remedies of riemannian and pseudo-riemannian manifolds and their curvatures, of the illustration conception of finite teams, and of symptoms of contemporary development in discrete subgroups of Lie teams.\n\nAdditional resources for Encyclopedie des sciences mathematiques. Geometrie descriptive et elementaire\n\nExample text\n\nIf we set {x,y) = (ty)(x) (x,ye V), it is immediate t h a t ( . , . ) is a nonsingular 0-bilinear form and (j(M) = Mf for all M. If 6' and ( . , . (F,D) with «^f(F*,D°). 1 now leads to the required result. 3. (F,D) which is involutive is known as a polarity, this means i f = |(£(Jf)) for all M. A polarity is called isotropic if ilf <= | ( J f ) for all one-dimensional M. 4. Let V be a vector space of dimension n^3 over D. (V,D) admits an isotropic polarity if and only if D is commutative and dim(F) is even.\n\nAn inclusion reversing bijection £ of J^(F,D) with itself such that g(i;(M)) = i f and M n f (ilf) = 0 for all M. It is easy to see that a polarity is an orthocomplementation if and only if (D-x) n £(D-a:) = 0 for all xe F. 7 (Birkhoff-von Neumann ). (F,D). ) induce £. ) can be chosen so that (w,w) = 1 for some weV. Let a symmetric 0-bilinear form < . , . ) be called definite if (14) (i) (x,x) = 0 o x = 0, v \\ >/ (ii) (w,w) = 1 for some weV. A definite form is necessarily nonsingular. Assume now that D is one of R, C, or H.\n\nBy a frame a t 0 we mean a pair (#,{PJ ; . G t / ) such t h a t {0,{P,} yet/ } is a basis for if. There are frames at any point of i^7. From now on we fix a frame (0,{Pj}jeJ) a t 0. An element a e i f is said to lie at infinity if there is a finite set K^J such t h a t a" ]

[ null, "https://pics.kisslibrary.com/pics/199930/cover.jpg", null ]

[ "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 06 Dec 2019, 17:59", null, "### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we’ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.", null, "", null, "# 5 integers, not necessarily distinct, are chosen from the integers bet\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nDirector", null, "", null, "V\nJoined: 18 Feb 2019\nPosts: 603\nLocation: India\nGMAT 1: 460 Q42 V13", null, "GPA: 3.6\n5 integers, not necessarily distinct, are chosen from the integers bet  [#permalink]\n\n### Show Tags\n\n1\n6", null, "00:00\n\nDifficulty:", null, "", null, "", null, "95% (hard)\n\nQuestion Stats:", null, "39% (02:40) correct", null, "61% (02:29) wrong", null, "based on 56 sessions\n\n### HideShow timer Statistics\n\n5 integers, not necessarily distinct, are chosen from the integers between –(n+1) and n, inclusive, where n is a positive integer. If the probability that the product of the chosen integers is zero is 1 – (0.9375)^5, then what is the value of n?\n\nA. 7\nB. 8\nC. 9\nD. 15\nE. 16\nGMAT Tutor", null, "G\nJoined: 24 Jun 2008\nPosts: 1829\nRe: 5 integers, not necessarily distinct, are chosen from the integers bet  [#permalink]\n\n### Show Tags\n\n4\n3\nYou might first imagine how you'd solve the problem using a concrete number. Suppose we had 9 different integers, say, and one of them was zero. Then if we pick five of them with replacement, and we want a nonzero product, we need to avoid picking zero each time. We'd have an 8/9 chance to pick a nonzero number each time, and we'd have a (8/9)^5 probability of getting a nonzero product in five selections, so we'd have a 1 - (8/9)^5 probability of getting a product of zero.\n\nNow if we compare what we just did with the expression in the question, 1 - (0.9375)^5, we can see that 0.9375 should equal the probability that we pick a nonzero number. So if we have k numbers, (k-1)/k = 0.9375, or 1 - (1/k) = 0.9375, and rearranging, 1/k = 0.0625. If you know that 0.625 = 5/8, then 0.0625 = 5/80 = 1/16, so k = 16, and we must be picking from 16 different numbers here. Between -n-1 and n inclusive, we have n - (-n-1) + 1 = 2n + 2 numbers, so 2n + 2 = 16 and n = 7.\n_________________\nGMAT Tutor in Toronto\n\nIf you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com\n##### General Discussion\nManager", null, "", null, "B\nStatus: No knowledge goes waste\nJoined: 12 Jul 2019\nPosts: 63\nLocation: Norway\nConcentration: Finance, Accounting\nGPA: 3\nWE: Corporate Finance (Commercial Banking)\nRe: 5 integers, not necessarily distinct, are chosen from the integers bet  [#permalink]\n\n### Show Tags\n\nIanStewart wrote:\nYou might first imagine how you'd solve the problem using a concrete number. Suppose we had 9 different integers, say, and one of them was zero. Then if we pick five of them with replacement, and we want a nonzero product, we need to avoid picking zero each time. We'd have an 8/9 chance to pick a nonzero number each time, and we'd have a (8/9)^5 probability of getting a nonzero product in five selections, so we'd have a 1 - (8/9)^5 probability of getting a product of zero.\n\nNow if we compare what we just did with the expression in the question, 1 - (0.9375)^5, we can see that 0.9375 should equal the probability that we pick a nonzero number. So if we have k numbers, (k-1)/k = 0.9375, or 1 - (1/k) = 0.9375, and rearranging, 1/k = 0.0625. If you know that 0.625 = 5/8, then 0.0625 = 5/80 = 1/16, so k = 16, and we must be picking from 16 different numbers here. Between -n-1 and n inclusive, we have n - (-n-1) + 1 = 2n + 2 numbers, so 2n + 2 = 16 and n = 7.\n\nIanStewart can u pls tell me why did u tell that we would pick five numbers out of nine with REPLACEMENT? Why should we assume that the taken numbers will be replaced?\nGMAT Tutor", null, "G\nJoined: 24 Jun 2008\nPosts: 1829\nRe: 5 integers, not necessarily distinct, are chosen from the integers bet  [#permalink]\n\n### Show Tags\n\nHea234ven wrote:\nIanStewart can u pls tell me why did u tell that we would pick five numbers out of nine with REPLACEMENT? Why should we assume that the taken numbers will be replaced?\n\nBecause the question says: \"5 integers, not necessarily distinct, are chosen from the integers between –(n+1) and n, inclusive\". But you wouldn't see wording like this on the real test -- on the real test they'd say \"with replacement\".\n_________________\nGMAT Tutor in Toronto\n\nIf you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com\nIntern", null, "", null, "B\nJoined: 06 Jan 2015\nPosts: 30\nRe: 5 integers, not necessarily distinct, are chosen from the integers bet  [#permalink]\n\n### Show Tags\n\nnidhi\n\nstart with a simple example, What is the probability for the number 0,1,2 such that when chosen 5 times, with replacement, the product is non zero?\n\nit is \\frac{2}{3}*2/3*2/3*2/3*2/3.\n\nIf zero then 1- the before equation.\n\nso we can say if .9375^5 is zero then .0625^5 is non zero. or (625/10000)^5\nafter deducting (1/16)^5 now we know that we have to choose between 16 numbers and since n is positive and we have to go upto -(n+1) start checking with numbers\nif n is 7 we get 16 number from 7 to -8. for others the number will be higher. Thus A.\nSenior Manager", null, "", null, "P\nJoined: 07 Mar 2019\nPosts: 434\nLocation: India\nGMAT 1: 580 Q43 V27", null, "WE: Sales (Energy and Utilities)\nRe: 5 integers, not necessarily distinct, are chosen from the integers bet  [#permalink]\n\n### Show Tags\n\nIanStewart wrote:\nHea234ven wrote:\nIanStewart can u pls tell me why did u tell that we would pick five numbers out of nine with REPLACEMENT? Why should we assume that the taken numbers will be replaced?\n\nBecause the question says: \"5 integers, not necessarily distinct, are chosen from the integers between –(n+1) and n, inclusive\". But you wouldn't see wording like this on the real test -- on the real test they'd say \"with replacement\".\n\nThat's exactly where I got stuck.\n\nWhether to take probability or non zero as\n$$\\frac{(2n+1)^5}{(2n+2)^5}$$ OR $$\\frac{(2n+1)*(2n)*(2n-1)*(2n-2)*(2n-3)}{(2n+2)^5}$$\n\nAnyway it does not matter as I marked B which is wrong.\n_________________\nEphemeral Epiphany..!\n\nGMATPREP1 590(Q48,V23) March 6, 2019\nGMATPREP2 610(Q44,V29) June 10, 2019\nGMATPREPSoft1 680(Q48,V35) June 26, 2019\nCrackVerbal Quant Expert", null, "G\nJoined: 12 Apr 2019\nPosts: 313\nRe: 5 integers, not necessarily distinct, are chosen from the integers bet  [#permalink]\n\n### Show Tags\n\nThere a couple of clues that are really important here.\n\nOne, the fact that we have integers stretching from the negative end of the spectrum to the positive. Two, the fact that we have to find the probability that the product of the chosen integers is ZERO.\nThe product of any set of integers can be ZERO only when there’s at least one ZERO in the set. The fact that we have both negative and positive integers is enough to tell us that ZERO can be in the set of chosen numbers.\n\nWhen I looked at 0.9375, the first thing I did was to subtract it from 1; it gave me 0.0625, but more importantly it helped me figure out that 0.0625 is $$\\frac{1}{16}$$.\n\nFinding the probability that the product of the chosen integers is ZERO is rather difficult, this is because we do not know how many of the integers are ZERO. On the contrary, finding out the probability that the product is NOT ZERO is easy; because we know that NONE of the integers can be ZERO.\n\nTherefore,\nProbability (Product of chosen integers being ZERO) = 1 – (Product of chosen integers not ZERO).\n\nFortunately, that’s what is given to us as 1 – $$(0.9375)^5$$. If 0.0625 = $$\\frac{1}{16}$$, then 0.9375 = $$\\frac{15}{16}$$. So, we have the probability of chosen integers not being zero as $$(\\frac{15}{16})^5$$. This can only happen when there are 16 integers in total and one of them is ZERO.\n\nThis means, there need to be a total of 16 integers between –(n+1) and n. Plugging in the value of 7, we can calculate that there are a total of 16 integers between -8 and 7, inclusive. Hence, 7 has to be the answer.\nThe correct answer option is A.\n\nIn probability questions, P ( E ) = 1 – P ( E’) is a very useful concept if you know the right way of incorporating it into your solution.\n\nHope that helps!\n_________________", null, "Re: 5 integers, not necessarily distinct, are chosen from the integers bet   [#permalink] 15 Oct 2019, 07:44\nDisplay posts from previous: Sort by\n\n# 5 integers, not necessarily distinct, are chosen from the integers bet", null, "", null, "" ]

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[ "# 60.34 kg to lbs - 60.34 kilograms to pounds\n\nDo you want to know how much is 60.34 kg equal to lbs and how to convert 60.34 kg to lbs? Here you go. You will find in this article everything you need to make kilogram to pound conversion - theoretical and also practical. It is also needed/We also want to underline that all this article is dedicated to one amount of kilograms - exactly one kilogram. So if you want to know more about 60.34 kg to pound conversion - read on.\n\nBefore we get to the practice - this is 60.34 kg how much lbs calculation - we want to tell you some theoretical information about these two units - kilograms and pounds. So we are starting.\n\nHow to convert 60.34 kg to lbs? 60.34 kilograms it is equal 133.0269288908 pounds, so 60.34 kg is equal 133.0269288908 lbs.\n\n## 60.34 kgs in pounds\n\nWe are going to start with the kilogram. The kilogram is a unit of mass. It is a base unit in a metric system, formally known as International System of Units (in abbreviated form SI).\n\nFrom time to time the kilogram is written as kilogramme. The symbol of the kilogram is kg.\n\nThe kilogram was defined first time in 1795. The kilogram was described as the mass of one liter of water. First definition was simply but totally impractical to use.\n\nLater, in 1889 the kilogram was described using the International Prototype of the Kilogram (in short form IPK). The International Prototype of the Kilogram was made of 90% platinum and 10 % iridium. The International Prototype of the Kilogram was used until 2019, when it was replaced by another definition.\n\nThe new definition of the kilogram is based on physical constants, especially Planck constant. The official definition is: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”\n\nOne kilogram is equal 0.001 tonne. It can be also divided to 100 decagrams and 1000 grams.\n\n## 60.34 kilogram to pounds\n\nYou know a little about kilogram, so now we can move on to the pound. The pound is also a unit of mass. We want to point out that there are more than one kind of pound. What are we talking about? For example, there are also pound-force. In this article we are going to to centre only on pound-mass.\n\nThe pound is used in the Imperial and United States customary systems of measurements. Naturally, this unit is in use also in another systems. The symbol of this unit is lb or “.\n\nThe international avoirdupois pound has no descriptive definition. It is just equal 0.45359237 kilograms. One avoirdupois pound can be divided into 16 avoirdupois ounces or 7000 grains.\n\nThe avoirdupois pound was enforced in the Weights and Measures Act 1963. The definition of the pound was placed in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”\n\n### How many lbs is 60.34 kg?\n\n60.34 kilogram is equal to 133.0269288908 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.\n\n### 60.34 kg in lbs\n\nTheoretical section is already behind us. In next section we will tell you how much is 60.34 kg to lbs. Now you know that 60.34 kg = x lbs. So it is high time to get the answer. Just look:\n\n60.34 kilogram = 133.0269288908 pounds.\n\nIt is a correct result of how much 60.34 kg to pound. You may also round off this result. After it your result is exactly: 60.34 kg = 132.748 lbs.\n\nYou learned 60.34 kg is how many lbs, so let’s see how many kg 60.34 lbs: 60.34 pound = 0.45359237 kilograms.\n\nOf course, this time it is possible to also round it off. After it your outcome is as following: 60.34 lb = 0.45 kgs.\n\nWe also want to show you 60.34 kg to how many pounds and 60.34 pound how many kg results in charts. Have a look:\n\nWe will begin with a chart for how much is 60.34 kg equal to pound.\n\n### 60.34 Kilograms to Pounds conversion table\n\nKilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)\n60.34 133.0269288908 132.7480\nNow look at a chart for how many kilograms 60.34 pounds.\n\nPounds Kilograms Kilograms (rounded off to two decimal places\n60.34 0.45359237 0.45\n\nNow you learned how many 60.34 kg to lbs and how many kilograms 60.34 pound, so it is time to move on to the 60.34 kg to lbs formula.\n\n### 60.34 kg to pounds\n\nTo convert 60.34 kg to us lbs you need a formula. We will show you two formulas. Let’s start with the first one:\n\nNumber of kilograms * 2.20462262 = the 133.0269288908 result in pounds\n\nThe first formula will give you the most correct result. Sometimes even the smallest difference could be considerable. So if you need an exact result - this formula will be the best for you/option to calculate how many pounds are equivalent to 60.34 kilogram.\n\nSo move on to the second version of a formula, which also enables calculations to learn how much 60.34 kilogram in pounds.\n\nThe second version of a formula is as following, see:\n\nNumber of kilograms * 2.2 = the outcome in pounds\n\nAs you see, this formula is simpler. It can be the best choice if you want to make a conversion of 60.34 kilogram to pounds in quick way, for instance, during shopping. You only need to remember that final result will be not so accurate.\n\nNow we want to learn you how to use these two versions of a formula in practice. But before we are going to make a conversion of 60.34 kg to lbs we are going to show you another way to know 60.34 kg to how many lbs without any effort.\n\n### 60.34 kg to lbs converter\n\nAnother way to know what is 60.34 kilogram equal to in pounds is to use 60.34 kg lbs calculator. What is a kg to lb converter?\n\nConverter is an application. It is based on longer version of a formula which we showed you above. Thanks to 60.34 kg pound calculator you can easily convert 60.34 kg to lbs. You only have to enter amount of kilograms which you want to calculate and click ‘calculate’ button. The result will be shown in a second.\n\nSo try to convert 60.34 kg into lbs with use of 60.34 kg vs pound calculator. We entered 60.34 as an amount of kilograms. This is the result: 60.34 kilogram = 133.0269288908 pounds.\n\nAs you can see, our 60.34 kg vs lbs calculator is so simply to use.\n\nNow we can move on to our primary issue - how to convert 60.34 kilograms to pounds on your own.\n\n#### 60.34 kg to lbs conversion\n\nWe are going to begin 60.34 kilogram equals to how many pounds calculation with the first formula to get the most correct result. A quick reminder of a formula:\n\nNumber of kilograms * 2.20462262 = 133.0269288908 the result in pounds\n\nSo what need you do to check how many pounds equal to 60.34 kilogram? Just multiply number of kilograms, this time 60.34, by 2.20462262. It is equal 133.0269288908. So 60.34 kilogram is exactly 133.0269288908.\n\nYou can also round off this result, for example, to two decimal places. It is equal 2.20. So 60.34 kilogram = 132.7480 pounds.\n\nIt is high time for an example from everyday life. Let’s calculate 60.34 kg gold in pounds. So 60.34 kg equal to how many lbs? And again - multiply 60.34 by 2.20462262. It is 133.0269288908. So equivalent of 60.34 kilograms to pounds, if it comes to gold, is exactly 133.0269288908.\n\nIn this case you can also round off the result. This is the result after rounding off, this time to one decimal place - 60.34 kilogram 132.748 pounds.\n\nNow let’s move on to examples converted with short formula.\n\n#### How many 60.34 kg to lbs\n\nBefore we show you an example - a quick reminder of shorter formula:\n\nAmount of kilograms * 2.2 = 132.748 the outcome in pounds\n\nSo 60.34 kg equal to how much lbs? As in the previous example you need to multiply number of kilogram, this time 60.34, by 2.2. Have a look: 60.34 * 2.2 = 132.748. So 60.34 kilogram is equal 2.2 pounds.\n\nMake another calculation with use of shorer formula. Now convert something from everyday life, for example, 60.34 kg to lbs weight of strawberries.\n\nSo let’s calculate - 60.34 kilogram of strawberries * 2.2 = 132.748 pounds of strawberries. So 60.34 kg to pound mass is exactly 132.748.\n\nIf you learned how much is 60.34 kilogram weight in pounds and are able to calculate it with use of two different formulas, let’s move on. Now we are going to show you these outcomes in charts.\n\n#### Convert 60.34 kilogram to pounds\n\nWe are aware that outcomes shown in charts are so much clearer for most of you. It is totally understandable, so we gathered all these results in tables for your convenience. Due to this you can quickly compare 60.34 kg equivalent to lbs results.\n\nLet’s start with a 60.34 kg equals lbs chart for the first version of a formula:\n\nKilograms Pounds Pounds (after rounding off to two decimal places)\n60.34 133.0269288908 132.7480\n\nAnd now have a look at 60.34 kg equal pound table for the second formula:\n\nKilograms Pounds\n60.34 132.748\n\nAs you can see, after rounding off, if it comes to how much 60.34 kilogram equals pounds, the outcomes are not different. The bigger amount the more significant difference. Keep it in mind when you need to do bigger number than 60.34 kilograms pounds conversion.\n\n#### How many kilograms 60.34 pound\n\nNow you learned how to calculate 60.34 kilograms how much pounds but we will show you something more. Do you want to know what it is? What about 60.34 kilogram to pounds and ounces conversion?\n\nWe are going to show you how you can convert it step by step. Let’s begin. How much is 60.34 kg in lbs and oz?\n\nFirst things first - you need to multiply amount of kilograms, this time 60.34, by 2.20462262. So 60.34 * 2.20462262 = 133.0269288908. One kilogram is 2.20462262 pounds.\n\nThe integer part is number of pounds. So in this example there are 2 pounds.\n\nTo know how much 60.34 kilogram is equal to pounds and ounces you need to multiply fraction part by 16. So multiply 20462262 by 16. It is exactly 327396192 ounces.\n\nSo final result is 2 pounds and 327396192 ounces. It is also possible to round off ounces, for instance, to two places. Then your outcome is 2 pounds and 33 ounces.\n\nAs you can see, conversion 60.34 kilogram in pounds and ounces is not complicated.\n\nThe last calculation which we will show you is conversion of 60.34 foot pounds to kilograms meters. Both of them are units of work.\n\nTo calculate it it is needed another formula. Before we give you it, have a look:\n\n• 60.34 kilograms meters = 7.23301385 foot pounds,\n• 60.34 foot pounds = 0.13825495 kilograms meters.\n\nNow see a formula:\n\nAmount.RandomElement()) of foot pounds * 0.13825495 = the result in kilograms meters\n\nSo to calculate 60.34 foot pounds to kilograms meters you need to multiply 60.34 by 0.13825495. It is 0.13825495. So 60.34 foot pounds is exactly 0.13825495 kilogram meters.\n\nYou can also round off this result, for instance, to two decimal places. Then 60.34 foot pounds is 0.14 kilogram meters.\n\nWe hope that this calculation was as easy as 60.34 kilogram into pounds conversions.\n\nThis article was a big compendium about kilogram, pound and 60.34 kg to lbs in calculation. Thanks to this calculation you learned 60.34 kilogram is equivalent to how many pounds.\n\nWe showed you not only how to do a calculation 60.34 kilogram to metric pounds but also two another conversions - to know how many 60.34 kg in pounds and ounces and how many 60.34 foot pounds to kilograms meters.\n\nWe showed you also other solution to make 60.34 kilogram how many pounds conversions, this is with use of 60.34 kg en pound converter. It is the best solution for those of you who do not like converting on your own at all or this time do not want to make @baseAmountStr kg how lbs conversions on your own.\n\nWe hope that now all of you can do 60.34 kilogram equal to how many pounds calculation - on your own or with use of our 60.34 kgs to pounds calculator.\n\nSo what are you waiting for? Calculate 60.34 kilogram mass to pounds in the best way for you.\n\nDo you want to do other than 60.34 kilogram as pounds conversion? For example, for 5 kilograms? Check our other articles! We guarantee that calculations for other numbers of kilograms are so easy as for 60.34 kilogram equal many pounds.\n\n### How much is 60.34 kg in pounds\n\nWe want to sum up this topic, that is how much is 60.34 kg in pounds , we prepared one more section. Here you can see the most important information about how much is 60.34 kg equal to lbs and how to convert 60.34 kg to lbs . Have a look.\n\nHow does the kilogram to pound conversion look? To make the kg to lb conversion it is needed to multiply 2 numbers. Let’s see 60.34 kg to pound conversion formula . Check it down below:\n\nThe number of kilograms * 2.20462262 = the result in pounds\n\nHow does the result of the conversion of 60.34 kilogram to pounds? The exact result is 133.0269288908 lbs.\n\nThere is also another way to calculate how much 60.34 kilogram is equal to pounds with another, easier version of the formula. Have a look.\n\nThe number of kilograms * 2.2 = the result in pounds\n\nSo this time, 60.34 kg equal to how much lbs ? The result is 133.0269288908 lb.\n\nHow to convert 60.34 kg to lbs quicker and easier? It is possible to use the 60.34 kg to lbs converter , which will do whole mathematical operation for you and you will get an exact answer .\n\n#### Kilograms [kg]\n\nThe kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.\n\n#### Pounds [lbs]\n\nA pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms." ]

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[ "Statistical properties in hyperbolic dynamics, part 2\n\nPublished:\n\nThis is the second post of a series of 4 posts based on the lectures at the Houston Summer School on Dynamical Systems 2019 on Statistical properties in hyperbolic dynamics, given by Matthew Nicol, Andrew Török and William Ott. These notes are heavily edited with added comments, examples and explanations. Any mistake is of my responsibility. Some of the notation has been changed for consistency.\n\nLecture 2\n\nIn this lecture we will explore in more depth the transfer operator method, as well as Gordin’s martingale approximation method.\n\nRecall the setting from the previous lecture: let $T\\colon X\\to X$ be a measure preserving transformation of the probability space $(X,\\mu)$. For a regular enough observable $\\phi\\colon X \\to\\mathbb{R}$, we constructed the time series $X_n = \\phi\\circ T^n$. We are interested in the limit laws of this time series. For this, we will write the observable as\n\n$\\phi = \\psi + g - g\\circ T$\n\nfor functions $\\psi,g$, so that when we compute the Birkhoff sums, we obtaining\n\n$\\sum_{j=1}^n \\phi\\circ T^j = \\sum_{j=1}^n \\psi\\circ T^j + g(x) - g\\circ T^{n+1}(x).$\n\nThe idea is that the sum on the right hand side is a reverse martingale, and the two remaining terms are controllable.\n\nProbability recap\n\nLet $(\\Omega,\\mathbb{P})$ be a probility space with sigma-algebra $\\mathcal{B}$ and $\\mathcal{F}\\subset\\mathcal{B}$ a sub sigma-algebra. If $Y$ is a random variable with finite expectation, define the conditional expectation of $Y$ with respect to $\\mathcal{F}$, denoted by $\\mathbb{E}(Y | \\mathcal{F})$ to be the unique (up to zero measure sets) random variable $Z$ with the following properties:\n\n1. $Z$ is $\\mathcal{F}$-measurable\n2. for every set $A\\in\\mathcal{F}$,\n$\\int_A Z d\\mathbb{P} = \\int_A Y d\\mathbb{P}.$\n\nThe existence of $\\mathbb{E}(Y | \\mathcal{F})$ can be proved using Radon-Nikodym’s theorem We list now some easy to check properties of the conditional expectation:\n\nProperties:\n\n1. $\\mathbb{E}(\\cdot | \\mathcal{F})$ is a linear operator,\n2. $\\mathbb{E}(\\cdot | \\mathcal{F})$ is a positive operator,\n3. $\\int_{\\Omega} \\mathbb{E}(X|\\mathcal{F}) dP = \\int_{\\Omega} X dP$ for every integrable random variable $X$,\n4. if $\\mathcal{F}_1\\subset\\mathcal{F}_2\\subset\\mathcal{B}$ then $\\mathbb{E}(\\mathbb{E}(X| \\mathcal{F_2})|\\mathcal{F}_1) = \\mathbb{E}(X|\\mathcal{F}_1)$,\n5. if $X\\in\\mathcal{F}$ and $\\mathcal{B}(X),\\mathbb{E}(XY)<\\infty$, then $\\mathbb{E}(XY| \\mathcal{F}) = X\\mathbb{E}(Y|\\mathcal{F})$,\n6. if $X$ is $\\mathcal{F}$-measurable, then $\\mathbb{E}(X|\\mathcal{F}) = X$.\n\nWe can now give a definition of a martingale. A sequence of sub sigma-algebras $(\\mathcal{F}_n)$ is a filtration if $\\mathcal{F}_n\\subset\\mathcal{F}_{n+1}$ for all $n$. A stochastic process $(M_n)$ is a martingale with respect to the filtration $(\\mathcal{F}_n)$ if\n\n1. $\\mathbb{E}(|M_n|) \\lt \\infty$,\n2. $M_n$ is $\\mathcal{F}_n$-measurable,\n3. $\\mathbb{E}(M_{n+1}|\\mathcal{F}_n) = M_n$.\n\nIn this setting, we call $X_n = M_{n+1} - M_n$ a martingale difference.\n\nRemark: if $M_n = X_1 + \\dots + X_n$ and we take the filtration $\\mathcal{F}_n = \\sigma(X_1,\\dots,X_n)$, then $(M_n)$ is a martingale if and only if $\\mathbb{E}(X_{n+1}| \\mathcal{F}_n) = 0$.\n\nTransfer operator revisited\n\nRecall the definition of the transfer operator $P$ as the dual of the Koopman operator:\n\n$\\int_X \\phi\\cdot\\psi\\circ T d\\mu = \\int_X P\\phi \\cdot \\psi d\\mu$\n\nfor observables $\\phi\\in L^1$ and $\\psi\\in L^\\infty$. In this section we will discuss how we can express the action of the transfer and the Koopman operators in terms of conditional expectations. Let $\\mathcal{F}_n = T^{-n}\\mathcal{B}$, where $\\mathcal{B}$ is the sigma-algebra of the underlying phase space $X$.\n\nProposition: for every $\\phi\\in L^1$ we have:\n\n1. $\\mathbb{E}(\\phi | T^{-1}\\mathcal{B}) = UP\\phi = (P\\phi)\\circ T$ and similarly $\\mathbb{E}(\\phi | T^{-i}\\mathcal{B}) = U^iP^i\\phi$.\n2. $PU\\phi = \\phi$ and $P^iU^i \\phi = \\phi$.\n\nProof: for the first part, we have to show that $UP\\phi$ is $T^{-1}\\mathcal{B}$-measurable and that\n\n$\\int_A UP\\phi d\\mu = \\int_A \\phi d\\mu$\n\nfor every set $A\\in T^{-1}\\mathcal{B}$. Note that $(P\\phi\\circ T)^{-1}(B) = T^{-1}(P\\phi)^{-1}(B)$ for every set $B\\in\\mathcal{B}$, thus the measurability condition holds. Using invariance and duality, we have\n\n$\\int_{T^{-1}A} UP\\phi d\\mu = \\int_X (P\\phi)\\circ T 1_{T^{-1}A} d\\mu = \\int_X (P\\phi)\\circ T 1_A\\circ Td\\mu$ $\\qquad = \\int_X (P\\phi) 1_A d\\mu = \\int_X \\phi 1_A\\circ T d\\mu = \\int_{T^{-1}A}\\phi d\\mu.$\n\nThis shows the first property. The property for the iterates follows from noticing that $P^i$ is the transfer operator associated to the map $T^i$. For the second property, note that for $g\\in L^\\infty$, we have\n\n$\\int_X (PU\\phi)gd\\mu = \\int_X U\\phi g\\circ T d\\mu = \\int_X \\phi\\circ T g\\circ T d\\mu = \\int_X \\phi g d\\mu.$\n\nA standard approximation argument finishes the proof of the claim. $\\square$\n\nGordin’s method\n\nSuppose we have a Banach space $\\mathcal{B}_{\\alpha} \\subset L^2$ with norm $| \\cdot |_{\\alpha}$ such that for observables $\\phi$ with $\\int \\phi d\\mu = 0$, we have $|P^n\\phi |_{\\alpha}\\leq \\rho(n)| \\phi |_{\\alpha}$ with $r(n)$ summable. Define $g = \\sum_{n=1}^\\infty P^n \\phi$ where $\\phi\\in\\mathcal{B}_{\\alpha}$, and define $\\psi = \\phi + g - g\\circ T$. We have that $UP\\psi = \\mathbb{E}(\\psi|T^{-1}\\mathcal{B})$, hence if we show that $P\\psi = 0$, then $\\mathbb{E}(\\psi | T^{-1}\\mathcal{B}) = 0$. Note that the sum defining $g$ converges and $g\\in\\mathcal{B}_{\\alpha}$. Then\n\n$P\\psi = P\\phi + P\\sum_{n=1}^\\infty P^n\\phi - PUg = P\\phi + \\sum_{n=2}^\\infty P^n\\phi - g = 0.$\n\nLet $\\mathcal{F}_j = T^{-j}\\mathcal{B}$ and $X_j = \\psi\\circ T^j$. Then\n\n$\\mathbb{E}(X_j | \\mathcal{F}_{j+1}) = \\mathbb{E}(\\psi\\circ T^j | T^{-j-1}\\mathcal{B}) = U^{j+1}P^{j+1}U^j \\psi = U^{j+1}P\\psi = 0$\n\nso we have a reverse martingale difference. Define $G_j = T^{-(n-j)\\mathcal{B}}$ and $Y_j = \\psi\\circ T^{n-j}$ so $Y_j$ is $G_j$- measurable. The filtration $G_j$ is increasing, and if we define $M_j = \\sum_{k=1}^n Y_k$, then $(M_n)$ is a martingale with respect to the filtration $(G_j)$, which leaves us in the setting of martingales differences.\n\nLimit laws\n\nNow that we know how to write the Birkhoff sums of an observable in terms of a martingale plus a coboundary, we can transfer the limit theorems known for martingales to our dynamical setting.\n\nTheorem (Azuma-Hoeffding): Let $(M_n)$ be a martingale with respect to a filtration $(\\mathcal{F}_n)$ so then $X_n = M_{n+1}-M_n$ is a martingale difference. Suppose that $|X_n|\\leq C$ almost surely for a constant $C > 0$. Then for $\\alpha > 0$ and $n\\geq 1$\n\n$\\mathbb{P}\\left(\\dfrac{M_n}{n}> \\alpha\\right) \\leq \\exp\\left(\\dfrac{-\\alpha^2n}{2C^2}\\right).$\n\nCorollary: if $\\phi$ is a Lipschitz observable with zero mean, we have\n\n$\\mu\\left(\\dfrac{1}{n}\\sum_{k=1}^n \\phi\\circ T^k > \\alpha\\right) \\leq \\exp(-\\tau n)$\n\nfor a constant $\\tau > 0$ and every $n\\geq 1$.\n\nThe proof follows from the approximation method above taking $(M_n)$ to be the Birkhoff sums. The boundedness condition follows from the regularity of the observable. The same idea can be used to prove other limit laws, for instance Cuny and Merlevede proved here the following functional law of iterated logarithms:\n\nTheorem: Suppose $\\phi \\in L^2$ and $\\phi_n = \\phi\\circ T^n$ for $T$ ergodic. Let $\\mathcal{G}_j$ a filtration so that $\\phi_n$ is adapted to it and $\\mathbb{E}(\\phi_{n} | \\mathcal{G}_{n+1})=0$ almost surely. Then it is possible to find centered Gaussian random variables $(Z_j)$ with $\\mathbb{E}(Z_{k}^{2})=\\mathbb{E}(\\phi_{1}^{2})$ such that\n\n$\\sup_{1 \\leq k \\leq n}\\left|\\sum_{i=1}^{k} \\phi_{i}-\\sum_{i=1}^{k} Z_{i}\\right|=o(\\sqrt{n \\log \\log n})$\n\nalmost surely.\n\nNext post" ]

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[ "# Solving systems of equations by substitution\n\nPosted on by\n\n## The substitution method for solving linear systems", null, "Solving Systems of Equations... Substitution Method (NancyPi)\n\nand   online    beep beep i am a sheep   when will the new season of wentworth be on netflix   outdaughtered season 3 episode 1\n\nA way to solve a linear system algebraically is to use the substitution method. The substitution method functions by substituting the one y -value with the other. We're going to explain this by using an example. You can use the substitution method even if both equations of the linear system are in standard form. Just begin by solving one of the equations for one of its variables.\n\nThe substitution method is most useful for systems of 2 equations in 2 unknowns. The main idea here is that we solve one of the equations for one of the unknowns, and then substitute the result into the other equation. We will solve second equation for y. Note: It does not matter which equation we choose first and which second. Just choose the most convenient one first! Since the coefficient of y in equation 2 is -1, it is easiest to solve for y in equation 2.\n\nIf you're seeing this message, it means we're having trouble loading external resources on our website. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. Math Algebra I Systems of equations Solving systems of equations with substitution. Systems of equations with substitution: potato chips. Practice: Systems of equations with substitution. Substitution method review systems of equations. Next lesson.\n\nThere are three ways to solve systems of linear equations: substitution, elimination, and graphing. Take the expression you got for the variable in step 1, and plug it substitute it using parentheses into the other equation. Use the result from step 3 and plug it into the equation from step 1. I create online courses to help you rock your math class. Read more.\n\n## How Do You Solve a System of Equations Using the Substitution Method?\n\nGiven two equations of a line, we want to find if they intersect at a single point. If they do, we say that it has a unique solution which can be described as a point in the coordinate axis. - Solving a linear system in two variables by graphing works well when the solution consists of integer values, but if our solution contains decimals or fractions, it is not the most precise method.\n\n## Solving Systems of Equations Using Algebra Calculator\n\nIndex of lessons Print this page print-friendly version Find local tutors. Solving Systems by Substitution. The method of solving \"by substitution\" works by solving one of the equations you choose which one for one of the variables you choose which one , and then plugging this back into the other equation, \"substituting\" for the chosen variable and solving for the other. Then you back-solve for the first variable. Here is how it works.", null, ".\n\nwhat to do with leftover sweet and sour meatballs\n\n.\n\n.\n\n## 3 thoughts on “Solving systems of equations by substitution”\n\n1.", null, "Cerys T. on said:\n\nA Diagram Showing Two Lines Intersecting at a Point\n\n2.", null, "Wingbecenal1965 on said:\n\nIf you're seeing this message, it means we're having trouble loading external resources on our website.\n\n3.", null, "Sienna H. on said:\n\nWalk through examples of solving systems of equations with substitution." ]

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[ "# Relation between the strength and dimensionality of defect-free carbon crystals\n\n## Abstract\n\nOn the basis of ab initio simulations, the value of strength of interatomic bonds in one-, two- and three-dimensional carbon crystals is obtained. It is shown that decreasing in dimensionality of crystal gives rise to nearly linear increase in strength of atomic bonds. It is ascertained that growth of strength of the crystal with a decrease in its dimensionality is due to both a reduction in coordination number of atom and increase in the angle between the directions of atomic bonds. Based on these data, it is substantiated that the one-dimensional (1D) crystals have maximum strength, and strength of carbyne is the absolute upper limit of strength of materials.\n\n## Background\n\nAccording to the existing paradigm, there are two ways to increase strength, namely (i) by increase in the density of defects in crystal lattice (increase in a lattice distortion) or (ii) vice versa, creation of defect-free crystals. The first way to increase strength is the basis of material science of conventional “bulk” materials. Different kinds of mechanical and thermo-mechanical treatments of metals and alloys are, in most cases, aimed at increasing the distortions in the matrix, which increases the resistance to movement of dislocations, i.e., increases the strength of the metals or alloys. Removal of defects is an alternative way to increase the material strength. Nanosized crystals are an example of this . However, new factor appears in this case, which influences strength essentially. This is dimensionality . Most clearly, this effect can be demonstrated for allotropic forms of carbon. Thus, the strength of defect-free 3D crystal (diamond) is 95 GPa , of 2D-crystal (graphene)—130 GPa, and of 1D crystal (carbyne)—more than 270 GPa . These numbers are evidences for the key role of dimensionality in “governing” strength in a nanoworld. In this connection, the task of the work was to establish the regularities of influence of the crystal dimensionality on its strength and to develop ideas about the factors influencing this effect.\n\n## Methods\n\nOne-, two-, and three-dimensional carbon crystals were used as an object of investigation. To determine the strength of a one-dimensional crystal (monatomic chains of carbon atoms), ab initio calculations were employed. In this case, tension of the chain was simulated. Strength was determined for polyyne structure, which is energy-wise more favorable. In this paper, the pseudo-potential method was employed, which is successfully used for all calculations of the atomic and electronic structures of crystals and which is based on the translational symmetry. When modeling the atomic structure of infinite chains of atoms by this method, it is necessary to eliminate their interaction with each other. The model ordered structure with unit cell parameters a = b = 1 nm and с = 0.2566 nm was used to this end. Tension of chains was executed by changing parameter с. Total energies of chains were calculated by the pseudo-potential method (software package Quantum-ESPRESSO (QE)) . This method was used for modeling the mechanical properties of the chains. Pseudo-potentials for carbon were generated under the scheme Vanderbilt Ultrasoft with the package Vanderbilt code, version 7.3.4 . The exchange-correlation potential PBE was used. For the calculations, we use 51 k points in irreducible Brillouin zone. The value of the cutoff energy Ecut = 450 eV. Structural optimization of positions of the atoms in the chain in longitudinal and transverse directions was carried out within the accuracy 1 mRy/a.e. The accuracy of calculation of total energies was 0.001 eV.\n\nTo determine the strength of atomic bonds, R 2D, in two-dimensional (2D) crystal (graphene), the value of fracture stress of graphene obtained by ab initio simulation of tension of graphene sheets in two directions “zigzag” and “armchair” were used. The magnitude of atomic bond strength Fc was determined as the value of critical force acting on the bond at the time of instability of the graphene sheet at its tension:\n\n$${R}_{2\\mathrm{D}}^{\\mathrm{zig}}\\kern0.5em =\\kern0.5em {\\sigma}_{\\mathrm{c}}^{\\mathrm{zig}}{d}_0{b}_0\\kern0.5em \\times \\kern0.5em \\left(1\\kern0.5em -\\kern0.5em \\varepsilon \\right)$$\n(1)\n$${R}_{2\\mathrm{D}}^{\\mathrm{arm}}\\kern0.5em =\\kern0.5em {\\sigma}_{\\mathrm{c}}^{\\mathrm{arm}}{d}_0{b}_0\\kern0.5em \\times \\kern0.5em \\left(1\\kern0.5em -\\kern0.5em \\varepsilon \\right)\\times \\left(1\\kern0.5em +\\kern0.5em \\cos \\left({\\theta}_0\\right)\\right)\\kern0.5em \\times \\kern0.5em \\cos \\left(1\\kern0.5em -\\kern0.5em {\\theta}^{\\hbox{'}}\\right)$$\n(2)\n\nwhere $${\\sigma}_{\\mathrm{c}}^{\\mathrm{zig}}$$ and $${\\sigma}_{\\mathrm{c}}^{\\mathrm{arm}}$$ are the critical stresses of instability of graphene sheet under tension in the directions zigzag and armchair, respectively; $${b}_0\\kern0.5em =\\kern0.5em \\sqrt{3}/2{a}_0$$; a 0 = 0.1422 nm is the lattice parameter; d 0 = 0.334 nm is the effective thickness of graphene sheet; θ 0 = 120° is the angle between atomic bonds ij and jk in undeformed state; θ′ = 130° is the angle, at which instability in graphene occurs (was determined based on the geometry of the system at instability); and ε is the transverse deformation of graphene sheet.\n\nStrength of atomic bonds in three-dimensional (3D) crystal (diamond) was determined on the basis of ab initio simulations of tension of diamond crystals in the direction <111 > . The strength value was calculated as the critical magnitude of the force acting on the interatomic bond at instability of the crystal under tension:\n\n$${R}_{3\\mathrm{D}}\\kern0.5em =\\kern0.5em \\frac{3\\sqrt{3}}{2}{\\sigma}_c{a}_0^2{ \\cos}^2\\left(\\pi /2\\kern0.5em -\\kern0.5em \\theta \\right)$$\n(3)\n\nwhere σc is the critical stress of instability, a 0 is the lattice parameter, and θ is the angle between atomic bonds ij and jk in deformed state.\n\nTheoretical analysis of the orientation-dependence of strength and its value-dependence on coordination number of atom was executed within the formalism of Brenner potential , which is based on Abel pseudo-potential theory that describes well the carbon compounds. Parameterization of the potential was carried out by the set II of Brenner parameters : R = 0.139 nm, D(e) = 6.0, s = 1.22, β = 0.21 nm−1, δ = 0.5, a 0 = 0.00020813, c 0 2 = 3302, and d 0 2 = 3.52. Since the Brenner potential in initial state gives a significant overestimation of strength of atomic bonds, which reaches values of 30 nN (Fig. 1a), its modification was carried out for the crystal that consisted in correction of cutoff function fcut (rij) ≡ 1. It enabled to get the values of strength, which agree well with the results of ab initio simulations.\n\n## Results and discussion\n\nThe strength of atomic bonds in one-dimensional crystal was determined by the critical stress of instability of monatomic chain under the conditions of uniaxial tension (Fig. 2). In accordance with the data obtained, R 1D = 11.3 nN. This value is in good agreement with the evidence of 12.2 nN and with the value 12.3 nN obtained for chains containing more than ten atoms .\n\nTo determine the strength of atomic bonds in 2D-crystals, data of ab initio simulation of tension and fracture of graphene sheets in zigzag and armchair direction were used. The critical value of the force at the moment of instability of the graphene sheet was determined by Eqs. (1) and (2). According to the results obtained, the strength of atomic bonds in graphene for armchair direction was 8.9 nN and for zigzag direction—8.3 nN. It should be noted that the difference in angles between atomic bonds in the stretched sheet of graphene is the cause for different values of atomic bonds strength in graphene at transition from zigzag to armchair orientation. Thus, at the moment of instability under tension in armchair direction, the orientation of atomic bonds is characterized by angles of 130° and 130°, while for tension in zigzag direction, they are 114° and 132°, respectively (Fig. 3). Really, calculations using Brenner potential, in the first case, give the value of strength of atomic bonds equal to 8.49 nN, and in the second case, this value is 7.5 nN (Fig. 3). Accordingly, the ratio of the strengths of atomic bonds obtained by the result of ab initio calculations is 1.07 and of those obtained from the Brenner potential is 1.14. The difference does not exceed 9 %. The absolute value of these “strength anisotropy” is not much great, but it illustrates one of the specific features of interatomic interaction of carbon atoms, namely, its orientation-dependence.\n\nThe value of bond strength in diamond was obtained based on the results of ab initio simulations. The critical value of the force acting in the bonding at the time of instability of the diamond lattice was determined by Eq. (3). In compliance with the results obtained, bond strength in diamond is R 3D  = 5.2 nN.\n\nAccording to the results, change in the dimensionality of defect-free crystals essentially affects their bond strength. This effect manifests itself in a significant (more than two times) increase in the bond strength of the transition from three- to two- to one-dimensional crystals (Fig. 3). Strength values of bonds calculated using the modified Brenner potential are plotted on the same graph. Calculations were executed for both the initial crystal and taking into account changes of angles θ between the bonds when reaching a critical strain of crystal instability. The results of calculations, agree well with the ab initio data. As it is indicated in Fig. 3, the transition from three- to two- to one-dimensional crystals, change in coordination numbers is accompanied by a change in the angles θ between the directions of atomic bonds. To separate the contributions of the coordination number Z and the angle θ to this change in the strength R, Brenner formalism was used. In explicit form, the dependence of strength R on the coordination number Z may be derived only for the case of equal angles θ ijk with all neighboring atoms k:\n\n$$R\\left(Z,\\theta \\right)\\kern0.5em =\\kern0.5em {D}^{(e)}\\beta \\frac{\\sqrt{2s}}{s\\kern0.5em -\\kern0.5em 1}\\left({s}^{1/\\left(1\\kern0.5em -\\kern0.5em s\\right)}\\kern0.5em -\\kern0.5em {s}^{s/\\left(1\\kern0.5em -\\kern0.5em s\\right)}\\right)\\kern0.5em \\times \\kern0.5em {\\left(1\\kern0.5em +\\kern0.5em \\left(z\\kern0.5em -\\kern0.5em 1\\right)G\\left({\\theta}_{\\mathrm{ijk}}\\right)\\right)}^{\\frac{\\delta s}{1\\kern0.5em -\\kern0.5em s}}$$\n(4)\n\nwhere G (θ ijk ) is the angle function .\n\nUsing the relations suggested in and the magnitudes of Brenner parameters (set II from ), orientation dependencies of strength at fixed values of coordination number Z = 2, 3, 4 were built (Fig. 4).\n\nFormally, Brenner-dependence describes the change in strength with the angle within the range from 0° to 180°. But in the real objects (diamond, graphene, carbyne), variation of the angle does not exceed the range 90°–180°, so the analysis is confined to this range of values.\n\nAccording to the data in Fig. 4, the sensitivity of strength to change in the angle increases with the growth of coordination number Z. On the other hand, the sensitivity of strength to changes in coordination number increases with decreasing in angle. As it is indicated in Fig. 5, at transition from diamond to carbyne, bond strength increases 2.3 times, while 56 % of this increase in the strength is due to a decrease of the coordination number and 44 % is related to the growth of angle to the maximum possible one, which is 180°.\n\nRegularities of dependence of atomic bond strength on coordination number are demonstrated by the carbon, because it is most convenient to analyze this problem. However, these regularities are inherent also to other materials. Work gives evaluation of the strength of atomic bonds in 1D and 3D crystals for metals such as Cu, Ag, Au, Pd, and Pt. The results of calculations by different methods show that at transition from 1D- to 3D-crystals, the strength of atomic bonds should increase by two to three times.\n\nSuch a significant increase in the strength of the interatomic bond is due to the change in electronic structure of the crystal, which occurs with a decrease in the coordination number. In carbon, this manifests itself in the transition from sp3- to sp2- and sp1-hybridization. In metallic crystals, a significant change in the electronic structure is also observed. Qualitatively, this change can be interpreted as an increase in the portion of “covalent” component .\n\nThus, the material in the form of a one-dimensional crystal (monatomic chain) should have the highest level of strength. In this regard, strength of carbyne claims to be the absolute upper limit of strength of materials. According to the results of ab initio calculations, carbyne containing five atoms has a maximum strength. This strength is 13.1 nN . When the effective diameter of the chain equals 0.200 nm , its value is 417 GPa. In , for an infinite chain, the strength values 9.3–11.7 nN were obtained, and considerably smaller value of effective diameter d = 0.077 nm was utilized. This gives the strength value for infinite chain equal to 1988–2501 GPa. In , the experimental measurement of carbyne strength was executed. It was ascertained that its strength must exceed 270 GPa. This lower bound for strength of carbyne is more than two times greater than the experimental value of strength of graphene (two-dimensional crystal of carbon).\n\n## Conclusions\n\nIncrease in strength of the interatomic bonds while reducing in dimension of the crystal is a specific feature of nanosized defect-free crystals. At transition from three- to two- to one-dimensional crystal of carbon, interatomic bond strength increases almost linearly. The bond strength in carbon monatomic chain is 2.2 times higher than that in diamond. This value is of the same order as the increase in strength of metallic crystals, which may be two to three times.\n\nGrowth of crystal strength with decreasing in its dimensionality is due to two factors, namely, a decrease in coordination number and increase in the angle between the bonds. At transition from diamond to carbyne, 56 % of increase in strength is due to a decrease in coordination number and 44 %—due to increasing in angle between the bonds.\n\n## References\n\n1. 1.\n\nBei H, Shim S, Pharr GM, George EP. Effects of pre-strain on the compressive stress—strain response of Mo-alloy single-crystal micropillars. Acta Mater. 2008;56:4762–70.\n\n2. 2.\n\nShpak AP, Kotrechko S, Mazilova TI, Mikhailovskij IM. Inherent tensile strength of molybdenum nanocrystals. Sci Technol Adv Mater. 2009;10:4.\n\n3. 3.\n\nKrahnea R, Morello G, Figuerola A, Georgea C, Dekaa S, Manna L. Physical properties of elongated inorganic nanoparticles. Phys Rep. 2011;501:75–221.\n\n4. 4.\n\nKotrechko S, Timoshevskij A, Yablonovskii S, Mikhailovskij I, Mazilova T, Lidych V. The absolute upper limit of material strength and ways to reach it. Procedia Materials Science. 2014;3:391–6.\n\n5. 5.\n\nKotrechko S, Mikhailovskij I, Mazilova T, Ovsjannikov O. Strength hierarchy for nano-sized crystals. Key Eng Mater. 2014;592–593:301–6.\n\n6. 6.\n\nZhang Y, Su Y, Wang L, Kong ES, Chen X, Zhang Y. A one-dimensional extremely covalent material: monatomic carbon linear chain. Nanoscale Res Lett. 2011;6:577.\n\n7. 7.\n\nRoundy D, Cohen ML. Ideal strength of diamond, Si, and Ge. Phys Rev B. 2001;64:212103.\n\n8. 8.\n\nLee C, Wei X, Kysar JW, Hone J. Measurement of the elastic properties and intrinsic strength of monolayer graphene. Science. 2008;321:385.\n\n9. 9.\n\nMikhailovskij I, Sadanov E, Kotrechko S, Ksenofontov V, Mazilova T. Measurement of the inherent strength of carbon atomic chains. Phys Rev B. 2013;87:045410.\n\n10. 10.\n\nGiannozzi P, Baroni S, Bonini N, Calandra M, Car R, Cavazzoni C, et al. QUANTUM ESPRESSO: a modular and open-source software project for quantum simulations of materials. J Phys Condens Matter. 2009;21:5502.\n\n11. 11.\n\nGiannozzi P, S. Baroni S, Bonini N, Calandra M, Car R, Cavazzoni C, et al. http://www.quantum-espresso.org/.\n\n12. 12.\n\nPerdew JP, Burke K, Ernzerhof M. Generalized gradient approximation made simple. Phys Rev Lett. 1996;77:3865–8.\n\n13. 13.\n\nLiu F, Ming P, Li J. Ab initio calculation of ideal strength and phonon instability of graphene under tension. Phys Rev B. 2007;76:064120.\n\n14. 14.\n\nBrenner DW. Empirical potential for hydrocarbons for use in simulating the chemical vapor deposition of diamond films. Phys Rev B. 1990;42:9458.\n\n15. 15.\n\nAbell GC. Empirical chemical pseudopotential theory of molecular and metallic bonding. Phys Rev B. 1985;31:6184–96.\n\n16. 16.\n\nMing C, Meng F-X, Chen X, Zhuang J, Ning X-J. Tuning the electronic and optical properties of monatomic carbon chains. Carbon. 2014;68:487–92.\n\n17. 17.\n\nBahn SR, Jacobsen KW. Chain formation of metal atoms. Phys Rev Lett. 2001;87:266101.\n\n18. 18.\n\nSen P, Ciraci S, Buldum A, Batra IP. Structure of aluminum atomic chains. Phys Rev B. 2001;64:195420.\n\n19. 19.\n\nLiu M, Artyukhov VI, Lee H, Xu F, Yakobson BI. Carbyne from first principles: chain of c atoms, a nanorod or a nanorope. ACS Nano. 2013;7:10075–82.\n\n## Acknowledgements\n\nThe authors gratefully acknowledge the financial support from Project # 15/14-Н Program “Nanosystems, nanomaterials, and nanotechnologies” of NAS of the Ukraine.\n\n## Author information\n\nAuthors\n\n### Corresponding author\n\nCorrespondence to Andrey Timoshevskii.\n\n### Competing interests\n\nThe authors declare that they have no competing interests.\n\n### Authors’ contributions\n\nSK formulated the research problem and managed the investigations. EK executed the calculation of the critical value of interatomic interaction force for 3D crystal (diamond) and 2D crystal (graphene) and carried out a theoretical analysis of the orientation-dependence of strength and dependence of coordination number. AT and YuM carried out ab initio calculations. All authors analyzed the results and participated in the preparation of the manuscript. All authors read and approved the final manuscript.\n\n### Authors’ information\n\nSK is Doctor of Sciences in Physics and Mathematics, Professor, Head of the Department of Strength and Fracture Physics, and a highly qualified specialist in multi-scale approach to strength and reliability of materials.\n\nAT is Ph.D. in Physics and Mathematics, a senior researcher, and an eminently qualified expert in computational solid state physics.\n\nEK is BSc in Physics, a graduate of the Faculty of Physics of the National Taras Shevchenko University of Kyiv, and a researcher of the Department of Physics of Metals.\n\nYuM is Ph.D. in Physics and Mathematics, a senior researcher, and a specialist in computational solid state physics.\n\n## Rights and permissions\n\nReprints and Permissions", null, "" ]

[ null, "https://nanoscalereslett.springeropen.com/track/article/10.1186/s11671-015-0929-9", null ]

[ "# Check if a string is K-Palindrome or not\n\nA string is K-Palindrome if it becomes a palindrome on removing at most k characters from it. Write an algorithm to check if a given string is K-Palindrome or not.\n\nFor example,\n\nInput: String – ABCDBA, k = 1\nOutput: K-Palindrome\nExplanation: The string becomes a palindrome by removing either C or D from it\n\nInput: ABCDECA, k = 1\nOutput: Not a K-Palindrome\nExplanation: The string needs at-least 2-removals from it to become a palindrome\n\nBy carefully analyzing the problem, we can see that it is a variation of classic Edit Distance Problem where we need to convert the given string into its reverse by removing at most K characters from it (i.e. only delete operation is allowed).\n\nPlease note that in order to make the original string and its reverse equal, we need to perform at-most N deletions from the original string and N deletions from the reverse string. Therefore, the expression 2*N <= 2*K is satisfied if the string is K-Palindrome.\n\n## C++\n\nOutput:\n\nString is K-Palindrome\n\n## Java\n\nOutput:\n\nString is K-Palindrome\n\nThe worst case time complexity of the above solution is O(2n). The worst case happens when the string contains all different characters.\n\nThe problem has an optimal substructure and also exhibits overlapping subproblems. As both properties of dynamic programming are satisfied, we can save subproblem solutions in memory rather than computed again and again. The dynamic programming is illustrated below which runs in O(n2) time -\n\n## C++\n\nOutput:\n\nString is K-Palindrome\n\n## Java\n\nOutput:\n\nString is K-Palindrome\n\nWe can also use solve this problem by finding Longest palindromic subsequence (LPS) of a string. In order to make a string palindrome, the characters which don't contribute to LPS of the string should be removed. Therefore, a string is K-Palindrome if the difference between the length of LPS and the length of the original string is less than equal to K.\n\nFor example, LPS of the string `CABCBC` is `CBBC` and on removing A and C from it, the string becomes a palindrome.\n\n## C++\n\nOutput:\n\nString is K-Palindrome\n\n## Java\n\nOutput:\n\nString is K-Palindrome\n\nThe time complexity of above solution is O(n2) and auxiliary space used is O(n2).", null, "", null, "", null, "", null, "", null, "(1 votes, average: 5.00 out of 5)", null, "Loading...", null, "" ]

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[ "", null, "# The Nth Fibonacci kata\n\n#1\n\nDiscuss this kata. Post solutions, ask for feedback.\n\nWrite some code to generate the Fibonacci number for the nth position. Example: `int Fibonacci(int position).` The first Fibonacci numbers in the sequence are: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34." ]

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[ "Calculate Kp for each reaction.a. N2O4(g)\u001e2 NO2(g) Kc = 5.9\u001e10-3 (at 298 K)b. N2(g) + 3 H2(g)\u001e2 NH3(g) Kc = 3.7 * 108 (at 298 K)c. N2(g) + O2(g)\u001e2 NO(g) Kc = 4.10 * 10-31 (at 298 K)\n\nYou are watching: Calculate the equilibrium constant kp for the reaction at a temperature of 298 k.", null, "### Video Transcript\n\nSo this difficulty gives us number of equations and also their K C equilibrium Constance and also wants united state to discover their KP. And so first, we have the right to write the relation between these therefore k p equates to K C times. Artie, where room is the gas continuous and T is the temperature to the delta end, which is the variety of moles of product minus the variety of most reacted. And also so we'LL begin with Letter A and also the chemistry equation is end to five for gas reversible e forms to N o come gas. And also it offers us a casey the five point nine times ten come the minus 3rd and a temperature of 2 ninety eight Calvin. And so we deserve to plug numbers right into our KP equation. For this reason we have five suggest nine time ten come the minus third, and then our our worth is zero allude zero eight to one at times or temperature come ninety eight. And then, in this situation we have actually two mole of product and also one mole the reactant so 2 minus one is one. Therefore it's just this to the first. And also when we solve this, we gain that our KP is zero point one 4 four. Therefore now relocating on to letter B. This time we have an equation and also two gas plus three h to gas reversible e forms to N H. 3 gas. We have actually a Casey the three point seven times, ten come the eighth, and again, a temperature of 2 ninety eight. Calvin. And so again, if us plug into our KP equation, us take ours Casey three allude seven time ten to the eighth. Multiply that by ours gas consistent zero allude zero eight to one time or temperature come ninety eight. Calvin. And in this case, we have actually two mole of product and also formals reacted. Therefore we'LL take it this Artie to the an adverse too. And when we witnessed for this, we gain our new K ns to it is in six allude one eight times ten to the fifth. And also that's your final answer for letter B. Here, permit me Aah! Six suggest eight time ten come the fifth. So currently moving ultimately into letter C, we have finish to gas to add O come gas forms two and also oh, gas. Our casey is four suggest one time ten to the minus thirty very first and again, our temperature is come ninety eight, Calvin. Therefore we usage that very same equation We'LL start with our Casey four allude one zero time ten to the an unfavorable thirty very first are our values zero suggest zero eight come one our temperature to ninety eight Calvin. And also this time we have actually a zero below because and I'll actually compose this together ah, right. This together one one or 2 minus two We have two mole of product and also then one shopping mall of every of these reactant ce So two moles of reacting.\n\nSee more: How Do You Get To The Hinterlands, The Hinterlands\n\nAnd also so now this R T worth goes come one and also r K ns is thie exact same as ours Casey. And also so that is our last answer native Battersea." ]

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[ "", null, "# Regular continuum systems of point particles. I: systems without interaction\n\nResearch paper by V. N. Chubarikov, A. A. Lykov, V. A. Malyshev\n\nIndexed on: 08 Nov '16Published on: 08 Nov '16Published in: arXiv - Mathematical Physics\n\n#### Abstract\n\nNormally, in mathematics and physics, only point particle systems, which are either finite or countable, are studied. We introduce new formal mathematical object called regular continuum system of point particles (with continuum number of particles). Initially, each particle is characterized by the pair: (initial coordinate, initial velocity) in \\$R^{2d}\\$. Moreover, all initial coordinates are different and fill up some domain in \\$R^{d}\\$. Each particle moves via normal Newtonian dynamics under influence of some external force, but there is no interaction between particles. If the external force is bounded then trajectories of any two particles in the phase space do not intersect. More exactly, at any time moment any two particles have either different coordinates or different velocities. The system is called regular if there are no particle collisions in the coordinate space. The regularity condition is necessary for the velocity of the particle, situated at a given time at a given space point, were uniquely defined. Then the classical Euler equation for the field of velocities has rigorous meaning. Though the continuum of particles is in fact a continuum medium, the crucial notion of regularity was not studied in mathematical literature. It appeared that the seeming simplicity of the object (absence of interaction) is delusive. Even for simple external forces we could not find simple necessary and sufficient regularity conditions. However, we found a rich list of examples, one-dimensional and mufti-dimensional, where we could get regularity conditions on different time intervals. In conclusion we formulate many unsolved problems for regular systems with interaction.", null, "" ]

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[ "Hello friends, I hope you all are doing great. In today's tutorial, I am going to give you a detailed introduction to the C++ Programming language. In cross-platform programming languages, C++ is the most popular that can be used to work on low and high-level applications. Bjarne Stroustrup was the founder of C++. He modified C language to develop C++ language. Control over system resources and memory can be attained by using C++. In 2011, 2014, and 2017 it was modified to C++11, C++14, and C++17. C++ is a middle-level language. It is advantageous to both programming languages low-level (drivers, kernels) and higher-level applications (games, GUI, desktop apps etc.).\n\n## Uses of C++\n\nC++ is one of the world's most famous programming languages. It is used in today's OS, embedded systems and GUIs. It provides a clear structure to programs, permits codes to be reused and lowering development costs as its an object-oriented language. Since it is portable and can be used to create applications that can be used on multiple platforms. It is very easy to learn. As it is close to C# and Java, so switching to C++ or vice versa is very simple.\n\nIt is used in\n\n• Operating systems e.g. Linux-based OS\n• Browsers like UC browser, chrome, opera and firefox.\n• Games and graphics e.g. Photoshop\n• Clouds like Dropbox\n• Database engines like reedit\n\n## Features of C++\n\n### Rich library support:\n\n• It is a simple language as programs can be split into logical units and parts. It has rich library support and many data types.\n\n### Platform Dependent and Machine Independent:\n\n•  It is machine-independent but platform-dependent. It does not run on windows but is executable on Linux.\n\n### Middle-level language:\n\n• It is a middle-level language as we can do both low-level programmings(drivers, kernels, networking etc.) and build large-scale user applications (Media Players, Photoshop, Game Engines etc.) as a high-level language.\n\n### 3rd party libraries:\n\n• C++ has rich library support as well as 3rd party libraries (e.g. Boost libraries) for easy, smart and rapid development.\n\n### Fast execution:\n\n• For C++ speed of execution is very fast because it is compiled and highly procedural language.\n• Garbage-collection, dynamic typing etc. slow the execution of the program overall. Since there is no additional processing in C++ so it is fast than others.\n\n### Provides direct Memory-Access:\n\n• It provides pointer support to manipulate storage addresses. This helps in low-level programming ( indirect control over memory addresses).\n\n### Object-Oriented language:\n\n• It is better than C with respect to object orientation that helps it to maintain extensible programs so large-scale applications can be built easily.\n• Its friends and virtual features violate some important rules rendering it a completely object-oriented language.\n\n## Amazing facts of C++:\n\nSome interesting facts about C++ are listed below\n\n• C++ name tells us that C language modified with ++ incremental operator is C++ language\n• The most famous language C ++ is used in commercial software.\n• Four primary features of OOP are supported by C++\n1. Inheritance\n2. Encapsulation\n3. Abstraction, and\n4. Polymorphism\n• From Simula67 Programming language C++ gained the features of OOP.\n• For a C++ program to execute(at least main() function) , a function is the least requirement.\n\n## Basic concepts of C++ :\n\nBasic concepts like syntax, variables, loop type etc will be discussed here.\n\n### Syntax of C++:\n\nHere is the C++ basic program\n\n#include <iostream.h> using namespace std; int main() { cout << \"Hi this is C++\"; }\n• iostream is a header file and provides us with input & output streams.\n• namesspace std tells the compiler to use standard namespace. It can be used in 2 ways.\n• the return type of main () is int.\n• count << is used to print anything on the screen.\n\n## Data types in C++:\n\nThere are built-in as well as user-defined data types in C++.\n\n• In C++, classes are user-defined data types.\n• Built-in data types are int, float, double etc.\n• Derived data types are Array, function, pointer and reference.\n\n## C++ Program to get a sum of 3 numbers\n\n//Program to receive three integer numbers and display their sum #include <iostream> using namespace std; int main() {       int num1, num2, num3, Sum;       //variables num1, num2, num3 and sum are declared as integers cout << \"\\n Enter Number 1: \"; cin >> num1; cout << \"\\n Enter Number 2: \"; cin >> num2; cout)<< “\\n Enter Number 3: “; cin>>num3; Sum = num1 + num2 + num3; cout << \"\\n The sum of \" << num1 << num2 “ and \" << num3 << \" is \" << Sum; }\n\n## Modifiers in C++\n\nIn C++, special words(called modifiers) are used to modify built-in data types. There are four main data type modifiers in C++, they are:\n\n• Long\n• Short\n• Signed    and\n• unsigned\n\nThese modifiers are with built-in data types to make them more precise and for expanding their range.\n\n• long and short modify the maximum and minimum values that a data type can hold.\n• Signed types include both +ive and _ive numbers as is the default type.\n• Unsigned, numbers do not have any sign, so they are always positive.\n\n## Variables in C++ :\n\n• Variable is used in C++ to store any value, which can be changed in the program.\n• Variable is declared in many ways each with different memory location and functioning.\n• It is the name of the memory location allocated by the compiler to the variable.\n\nVariables are divided into two main types,\n\n• Global Variables\n• Local variables\n\n### Global variables\n\nGlobal variables are those which declared only a single time and used again and again. They are declared outside the main() function. If only declared then assigned different values at different times in program lifetime. But when they are declared and initialized at the same time then they can be assigned any value at any point in the program.\n\nFor example: Only declared, not initialized\n\ninclude <iostream> using namespace std; int x;                // Global variable declared int main() { y=10;                 // Initialized once cout <<\"first value of y = \"<< y; y=20;                 // Initialized again cout <<\"Initialized again with value = \"<< y; }\n\n### Local Variables\n\n• Local variables exist only between the curly braces, in which they are declared.\n• Outside the curly braces, they are unavailable and lead to a compile-time error.\nExample :\ninclude <iostream> using namespace std; int main() { int j=10; if(j<20)        // if condition scope starts { int m=100;   // Local variable declared and initialized }              // if condition scope ends cout << m;      // Compile time error, m not available here }\n\n## C++ Program to find the curved surface area of a cylinder (CSA) (CSA = 2 pi r * h)\n\n#include <iostream> using namespace std; int main() { float pi = 3.14, Radius, Height, CSA; cout << \"\\n Curved Surface Area of a cylinder\"; cout << \"\\n Enter radius (in cm): \"; cin >> Radius; cout << \"\\n Enter height (in cm): \"; cin >> Height; CSA = (2*pi*Radius)*Height;       system(\"cls\"); cout << \"\\n radius: \" << Radius <<\"cm\"; cout << \"\\n height: \" << Height << \"cm\"; cout << \"\\n Curved Surface Area of a Cylinder is \" << CSA <<\" sq. cm.\"; }\n\n### Output:\n\n• Curved Surface Area of a cylinder\n• Enter radius (in cm): 7\n• Enter height (in cm): 20\n• height: 20cm\n• The curved Surface Area of a Cylinder is 879.2 sq. cm.\n\n## Operators in C++ :\n\n• Operators take one or more arguments and generate a new value.\n• For example : addition (+), subtraction (-), multiplication (*) etc, are all operators.\n• These are used to perform different operations on variables and constants.\n\n## Errors in C++:\n\nThere are three types of errors that occur in C++ programming\n\n### Syntax Error:\n\n• The syntax is a set of grammatical rules to make a program. Every programming language has unique grammatical rules.\n• when grammatical rules of C++ are violated Syntax Errors occur.\n• For example: if you type as follows, C++ will throw an error.\n\ncout << “Hi welcome to C++”\n\n• As per the grammatical rules of C++, there should be a semicolon at the end of the statement. But, this statement does not end with a semicolon so Syntax Error occurs.\n\n### Logical Error:\n\nIt may be happened by the wrong use of variable or operator or order of execution etc. This means that the program is grammatically correct, but it contains some logical errors. So, “Logic Error” is also called Semantic error.\n\n### Run time error:\n\n• During the execution of the program when some illegal action takes place, run time error occurs\n• For example, if a program tries to open a file that does not exist then it will result in a run-time error.\n\n## Control Statements in C++:\n\n• The sequence of flow of instructions is changed by the use of control Statements.\n\n## Selection statement\n\n• Statements can run sequentially, selectively or iteratively in a program. Sequence, selection and iteration processes are handled by every programming language.\n• If the statements are executed sequentially then the flow is called a sequential flow. In some situations, if the statements alter the flow of execution then this flow is called a control flow.\n\n## Sequence statement\n\n• The sequential statement is executed one after the another only once from top to bottom.\n• These statements do not alter the flow of execution and are called sequential flow statements. These statements always end with a semicolon (;).\n\n## Selection statement\n\n• When a condition is provided then selection statements are used.\n• In case when the condition is true then a true block (a set of statements) is executed otherwise a false block is commanded to execute.\n• This is also called a decision statement because it helps in making decisions for the set of statements to be executed.\n• Selection statements are if, if-else and nested if statements.\n\n### if statement\n\n• The general syntax of the if statement is:\nif (expression) true-block; statement-x;\n• and flow chart for if statement is\n\n### if-else statement\n\n• The syntax of the if-else statement is given below:\nif ( expression) {       True-block; } else {       False-block; }\n• And the flow chart for if-else statement is:\n\n## C++ program to find either number is Even or Odd\n\n#include <iostream> using namespace std; int main() { int Num, rem; cout<< \"\\n enter a number: \"; cin>>Num; rem = Num % 2; if (rem==0) cout<< \"\\n The given number\" <<Num<< \" is Even\"; else cout<< \"\\n The given number \"<<Num<< \" is Odd\"; return 0; }\n\n### Output\n\n• Enter number: 12\n• The given number 12 is Even\n\n### Nested if\n\nIt has three forms\n• If nested inside if part\n• If nested inside else part\n• If nested inside both if part and else part\n\n## Iteration statement\n\n• The iteration statement is a set of statements that are executed again and again depends upon conditions.\n• If a condition evaluates to true, the set of statements (only true) is executed again and again.\n• As soon as the condition seems to be false, the repetition stops. This is also known as looping statement\n• The set of statements that are executed repeatedly is called the body of the loop.\n• The condition on which exits from the loop is called exit-condition or test-condition.\n• There are 3 kinds of loops in C++\n1. For loop\n2. While loop\n3. Do while loop\n\n### for loop\n\n• The for loop is the easiest loop which allows code to be executed again and again.\n• The general syntax is:\nfor (initialization(s); test expression; update expression(s)) {      statement 1;         statement 2;       …………. }\n\n## C++ program to sum from 1 to 5 using for loop\n\n#include <iostream> using namespace std; int main () { int i,Sum=0; for(i=1; i<=5;i++) { sum=sum+i; } cout<<\"The sum of 1 to 5 is \"<<Sum; return 0; }\n\n### Output\n\n• The sum of 1 to 5 is 15\n\n### While loop\n\n• It allows the loop statements to be executed as long as the condition is true.\n• The while loop syntax is:\nwhile ( test expression ) {       body of the loop; }\n• Flow chart for while loop is given below\n\n## C++ program to sum from 1 to 6 using while loop\n\n#include <iostream> using namespace std; int main () { int i=1,SUM=0; while(i<=6) { SUM=SUM+i; i++; } cout<<\"The sum of 1 to 6 is \"<<SUM; return 0; }\n\n### Output\n\n• The sum of 1 to 6 is 21\n\n### do-while loop\n\n• The do-while loop is used as an exit-controlled loop. In a do-while loop, after executing the body of the loop, the condition is evaluated.\n• The do-while loop syntax is:\ndo {       body of the loop; } while(condition);\n• The flow chart of the do-while loop is shown below" ]

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[ "# How to Calculate RC Car Gear Ratio: A Beginner’s Guide.\n\nRC cars are a popular hobby and pastime for enthusiasts of all ages. Whether battling friends in backyards or touring tracks, the speed and agility of these miniature cars can be a thrilling experience. However, not all RC cars are created equal. As with any motorized vehicle, there are numerous factors that can affect their performance. One of the most important of these is the gear ratio. The gearing on an RC car determines the speed and torque provided by the motor, and getting it right is essential to achieving peak performance. Knowing how to calculate your gear ratio is crucial to optimizing your RC car’s performance on any terrain or track. This skill can help you balance your speed and torque, giving you a smoother ride, faster times, and an edge over your competitors. In this article, we will take you through the basics of calculating gear ratios for RC cars. Whether you are a seasoned enthusiast or just starting, this guide will provide you with the knowledge needed to personalize your gear set up, take advantage of different terrains, and reach maximum performance.\n\n## Pinion Gear and Spur Gear\n\nCalculating gear ratio involves understanding the relationship between the pinion gear and the spur gear, each with its own job to perform. Here are some key facts to keep in mind:\n\n• The pinion gear is the small metal toothed wheel that’s directly attached to the motor.\n• The spur gear is the perpendicular larger gear that meshes with the pinion gear to control the speed and torque.\n• The gear ratio is determined by the number of teeth on each gear.\n• Increasing or decreasing the number of teeth on each gear in proportion changes the gear ratio and will affect the performance of the car.\n\nIf you’re new to RC car racing, you might want to experiment with different gears to get the best results. But before you get started, it’s important to understand the basics of how to calculate gear ratios for RC cars. There are several websites and resources dedicated to teaching you how to do it, including gear ratio calculators, online forums, and instructional videos. Some of the more popular tools that you can use include:\n\nBy using these resources, you should be able to easily determine your RC car’s gear ratio and make any necessary modifications to improve the car’s performance.\n\n### How Gear Ratio Can be Calculated?\n\nCalculating gear ratio is an important task in mechanical engineering. Here are the steps to follow:\n\n1. Count the number of teeth on both gears.\n2. Divide the number of teeth on the driven gear by the number of teeth on the driving gear.\n3. Simplify the resulting ratio by dividing the ratio by the greatest common factor.\n\nThere are various online gear ratio calculators available that can make this process easier. One such calculator is available on the csgnetwork.com website.\n\nIf you are looking for more information about gear ratios and their significance in different mechanical systems, a great resource to check out is geartechnology.com.\n\n## Calculating Gear Ratio\n\nCalculating gear ratio for an RC car requires a bit of basic math. Here are the steps to follow:\n\n1. Count the number of teeth on both the pinion gear and spur gear.\n2. Divide the total number of teeth on the spur gear by the total number of teeth on the pinion gear.\n3. The result is the gear ratio.\n\nFor example, if the spur gear has 60 teeth and the pinion gear has 20 teeth, the gear ratio is 3:1 (60/20 = 3).\n\nIn addition to the gear ratio calculation formula, there are a few other things to keep in mind:\n\n• RC cars that are geared for high speed will have a smaller pinion gear and a larger spur gear.\n• RC cars that are geared for more torque will have a larger pinion gear and a smaller spur gear.\n• Differentials, which distribute power to the wheels, can also impact gear ratio.\n\nTo simplify gear ratio calculation, you can also refer to gear ratio charts that are available from many RC car manufacturers, including Traxxas and HPI. These charts provide detailed information about the right gear ratios for different RC car models.\n\nHere’s an example of a gear ratio chart:\n\nVehicle Model Pinion Gear Spur Gear Gear Ratio\nTraxxas Slash 4×4 23 86 3.74:1\nHPI Trophy Truggy 13 47 3.62:1\nTeam Associated RC10B6 31 76 2.45:1\n\nWith a bit of research and practice, calculating gear ratios for RC cars can be a fun and rewarding experience. As you learn more about gear ratios, you’ll begin to appreciate the many factors that go into optimizing an RC car’s performance.\n\n### What is the transmission gear ratio of a RC car?\n\nThe transmission gear ratio of a RC car refers to the ratio of the number of teeth on the pinion gear (connected to the motor) compared to the number of teeth on the spur gear (connected to the wheels). This ratio determines the speed and torque of the car. For example, a lower gear ratio means higher top speed but less torque, while a higher gear ratio means lower top speed but more torque. It is important to choose the right gear ratio based on your RC car and the type of racing or driving you plan to do.\n\nThere are various websites and products that offer information and tools to help you choose the right gear ratio for your RC car, such as RC Universe and RC Gear Ratio Calculator.\n\nRC car enthusiasts can adjust their car’s gear ratio to suit different terrains, track layouts, or racing styles. Here are some ways to adjust gear ratios:\n\n• Change the size of the pinion and spur gears.\n• Adjust the tire size by changing the diameter or thickness.\n• Modify the motor to increase or decrease its power output.\n• Change the differential oil weight to change the power distribution to the wheels.\n\nHowever, it’s important to note that adjusting gear ratio can impact the car’s overall performance. Increasing the gear ratio will make the car faster but sacrifice torque, while decreasing the gear ratio will make the car slower and provide more torque.\n\nOne way to determine the right gear ratio for an RC car is to experiment with different setups and test them on the track. Alternatively, you can refer to online resources such as gear ratio charts and YouTube videos where RC car enthusiasts share their experience and insights on the best gear ratios for a specific car model or racing style.\n\nWhile some RC car manufacturers offer gear ratio charts and calculators on their websites, there are many independent online tools available that provide accurate gear ratio calculations. Examples include rcscoringpro.com, gearratiocalculator.net, and gearrc.com. These calculators only require inputting the number of teeth on the pinion and spur gears to generate the gear ratio, saving time and avoiding errors.\n\n### What is Gear Ratio in RC Car Gears?\n\nGear ratio is the relationship between the number of teeth on two gears that are meshed or engaged with each other. In an RC car, the gear ratio determines the speed and torque of the vehicle. A higher gear ratio will provide better acceleration but lower top speed, while a lower gear ratio will provide higher top speed but lower acceleration.\n\nIt is important to choose the right gear ratio for your RC car to achieve the desired performance. Some popular online RC hobby stores like Tower Hobbies and AMain Hobbies offer a wide range of gears and gear ratios for RC cars.\n\nIn addition, there are various tutorials and guides available online to help you understand and calculate gear ratios for your RC car. Some helpful resources include the RC Car Action magazine and the RC Universe forum.\n\n## Using Gear Ratio Calculators\n\nGear ratio calculators are useful tools that simplify the process of calculating gear ratios for RC cars. They are available in different formats, including mobile apps and online calculators. Here are some benefits of using gear ratio calculators:\n\n• They save time and eliminate errors associated with manual gear ratio calculation.\n• They can be used by both novice and experienced RC car enthusiasts.\n• They provide accurate gear ratio calculations to optimize car performance.\n• Some gear ratio calculators also suggest ideal gear ratios based on the car model and use case.\n\nTo use a gear ratio calculator online, you need to input the number of teeth on the pinion gear and the spur gear. Once the values are entered, the calculator will generate the gear ratio result. RC car enthusiasts can use this result to adjust the gear ratio to suit their track or terrain. Some popular gear ratio calculators include:\n\n• RCScoringPro Gear Ratio Calculator: A free online calculator that generates gear ratio results and suggests ideal pinion and spur gear sizes based on the input.\n• MyRCBox Gear Ratio Calculator: An online calculator that uses the spur gear and pinion gear sizes on the car to determine the gear ratio and RPM of the motor and wheels.\n• Robinson Racing Gear Ratio Chart: A gear ratio chart that provides gear ratio values for different pinion gear and spur gear tooth configurations.\n\nOverall, gear ratio calculators are important tools for calculating ratios accurately and delivering optimal car performance.\n\n### How do you calculate ideal gear ratio?\n\nTo calculate the ideal gear ratio, you will need the following information:\n\n• The desired output speed in revolutions per minute (RPM)\n• The input speed of the motor in RPM\n\nOnce you have this information, the formula to calculate the gear ratio is as follows:\n\nGear ratio = Output speed (RPM) / Input speed (RPM)\n\nFor example, if the desired output speed is 600 RPM and the motor’s input speed is 1200 RPM, the gear ratio would be:\n\nGear ratio = 600/1200 = 0.5\n\nThere are also online gear ratio calculators available, such as the one provided by Omni Gear, that can make this calculation easier.\n\n## Conclusion\n\nCalculating the gear ratio is an essential skill for RC car enthusiasts seeking optimal performance from their cars. As discussed in this article, RC car gear ratio calculation is a relatively simple process involving the number of teeth on the pinion and spur gears. Adjusting the gear ratio can impact speed and torque, and gear ratio calculators are valuable tools for simplifying this process. By utilizing these valuable tools, enthusiasts can be precise and accurate when it comes to gear ratio selection.\n\nIn conclusion, the importance of gear ratio cannot be overstated. It’s a vital factor that determines the balance between speed and torque, which impacts an RC car’s overall performance. Proper gear ratio calculation and selection can enhance car performance and give an edge of competitors on the track. Therefore, understanding the fundamentals of gear ratio calculation, adjusting as necessary, and using gear ratio calculators can make a significant difference in RC car performance. With this knowledge, RC car enthusiasts can identify the right gear ratio that meets their specific setup and preferences, and ultimately have a more enjoyable and fulfilling experience with their RC car." ]

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[ "#### Need Help?\n\nGet in touch with us", null, "", null, "# Compare Fractions: Same Numerator and Denominator Method\n\n### Key Concepts\n\n• Use models to compare fractions: Same Denominator\n• Use models to compare fractions: Same Numerator\n\n## Introduction:\n\nThe same denominator method\n\n• The denominator is the bottom number in a fraction.\n• It shows how many equal parts the item is divided.\n\nWhen two fractions have the same denominator, they are easy to compare:\n\nFor instance, 4/9 is less than, 5/9 (because 4 < 5)\n\n4/9 is 4 times the unit fraction 1/9\n\n5/9 is 5 times the unit fraction 1/9\n\nDraw 𝟏/𝟗 strips,\n\nHere, we can observe that denominators are the same in both fractions.\n\nSo, we must compare numerators.\n\nNumerator 4 is less than 5 (4 < 5)\n\n∴4/9 is less than 5/9\n\nComparison using symbols: 4/9 < 5/9\n\nExample\n\nWhich is greater, 3/6 or 2/6 ?\n\n3/6 is 3 of the unit fraction 1/6\n\n2/6 is 2 of the unit fraction 1/6\n\nSo, 3/6 is greater than 2/6\n\nComparison using symbols: 3/6 > 2/6\n\nReal-life Example\n\nA Pizza was divided into three equal parts (slices).\n\nIn the diagram, the whole is represented with the fraction 3/3\n\nIf you take out one part, the remaining portion represents 2/3\n\nIn this example, both denominators are the same, so we must compare numerators.\n\nThe numerator 3 is greater than 2.\n\n∴ 3/3 is greater than 2/3.\n\nUsing 𝟏/𝟑 strips:\n\nComparison using symbols: 3/3 > 2/3\n\n### 13.4 Use models to compare fractions: Same Numerator\n\nLook at the models\n\nThey each have 1 piece shaded in. So, all we have to compare is the size of each piece\n\nWhich pie has the largest shaded part?\n\n• The pie which is partitioned into halves\n• The denominator with the smallest number gives you the biggest size pieces\n\n### Comparison using symbols:\n\n1/2 > 1/4 > 1/6\n\nWhat happens to the denominator as your pieces get smaller?\n\n• The number in the denominator gets bigger\n• The same numerator is equal to the same number of pieces being referred to\n• So, if the fractions refer to the same whole, and the numerators are the same, then the number of pieces being compared is the same.\n\nExample\n\nTom and Jerry each made a pie. The pies were the same size. Tom cut his pie into 8 slices; Jerry cut his into 6 slices. They each ate 2 slices of their own pie. Who ate more?\n\nDraw a model. Write the fraction that each ate\n\nDid they eat the same number of pieces?\n\nAre the numerators the same?\n\nWe use fraction strips to compare the size of the pieces or compare the denominators.\n\nSixths are bigger than eighths, so\n\n2/6 is bigger than 2/8.\nSo, Jerry ate more pie.\n\n## Exercise:\n\n1. Rani reads 1/6 of a book in the morning; she reads 4/6 of the book in the afternoon. What fraction of the book does she read?\n2. What is the equivalent fraction of 3/4 with denominator 20?\n3. Raj has 26 toffees. He gave one-half to his friend. How many toffees did he give to his friend?\n4. Write  two comparison statements about the fractions shown below.\n\n5. Which is  greater ¼  or 1/6 ?  Draw fraction strips to complete the diagram and answer the question.\n\n6. Maria and Nina each ordered a small pizza. Maria ate 3/8 of her pizza. Nina ate 3/6 of her pizza. Who ate more pizza?\n\n7. Explain Why is 1/6 greater than 1/8 but less than 1/3?\n\n8. Two pizzas were each cut into sixths. Ashraf, Drew, and Katie shared the pizzas equally. How many sixths did each friend get?\n\n9. Eric and Frank want to share 4/3 feet of rope equally. What length of rope should each friend get? Explain how to use a drawing to help solve the problem.\n\n10. Ronald spent the day making a painting for his friend. At the end of the day, Ronald finished ¼ of the painting. If he is able to finish as much of a painting each day he works, how long will it take Ronald to make 2 whole paintings?\n\n### What we have learned:\n\n• How to compare fractions that refer to the same-sized whole and have the same denominator by comparing their numerators\n• How to compare fractions that refer to the same-sized whole and have the same numerator by comparing their denominators\n• How to use symbols (>, <, =) to compare fractions with different numerators and denominators\n• Recognize that to compare two fractions both must refer to the same whole\n• How to draw area models to compare two fractions\n• How to compare fractions that refer to the same-sized whole and have the same denominator by comparing their numerators\n• How to compare fractions that refer to the same-sized whole and have the same numerator by comparing their denominators\n• How to use symbols (>, <, =) to compare fractions with different numerators and denominators\n• Recognize that to compare two fractions both must refer to the same whole\n• How to draw area models to compare two fractions\n\n### Concept Map", null, "", null, "" ]

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[ "# Basics#\n\nDeerLab applications to dipolar EPR spectroscopy relies on a few central quantities. They include distance distributions and model functions, background and their model functions, time-domain signals, and kernels. They are described in this section.\n\nAll functions in DeerLab use the same units: all distances are in units of nanometers, and all times in units of microseconds.\n\n## Distance distributions#\n\nA distance distribution", null, "between two spins is represented by a pair of vectors: a distance vector r (in nanometers) and a vector of densities P (in inverse nanometers). The distance vector r can have linearly or non-linearly increasing values, but must have positive non-zero values. The elements P[i] are the distance distribution values at r[i] and are posiive or zero. Outside of the range defined by r, the distribution P is assumed to be zero, i.e. the distribution is truncated to the range r. The distance distribution P is normalized such that the integral over the range of the provided r equals one:", null, "DeerLab distinguishes between non-parametric and parametric distance distributions.\n\nNon-parametric distance distributions\n\nThese provide the more general definition of distance distributions. They have no particular shape and are represented by the vectors P and r. For example, you can generate P and r by an external program for spin label rotamer modeling or from molecular dynamics simulations.\n\nParametric distance distributions\n\nThese have specific shapes that are determined by a few parameters. DeerLab provides many parametric distance distribution built-in models. All these model’s names start with the prefix dd_ (dd stands for “distance distribution”). They take a distance vector r and a parameter vector param as inputs and return the distance distribution as a vector P. Here is an example:\n\nr = np.linspace(1.5,6,200) # distance vector, in nm\nrmean = 3 # nm, mean of Gaussian\nsigma = 0.2 # nm, standard deviation of Gaussian\nP = dl.dd_gauss(r, rmean, sigma) # Gaussian distribution\nplt.plot(r,P)\n\n\nTo programmatically get information on a particular distance distribution model and its parameters, print the model, which will return information on the model itself, its call signature, and a parameter table containing all the model parameters, its boundaries, start values and other useful information\n\n>>>print(dl.dd_gauss)\nDescription: Gaussian distribution model\nSignature: (r, mean, std)\nConstants: [r]\nParameter Table:\n======= ======= ======= ======== ======== ======= ====================\nName Lower Upper Type Frozen Units Description\n======= ======= ======= ======== ======== ======= ====================\nmean 1 20 nonlin No nm Mean\nstd 0.05 2.5 nonlin No nm Standard deviation\n======= ======= ======= ======== ======== ======= ====================\n\n\nIn least-squares fitting, non-parametric distance distributions make fewer assumptions about the distribution than parametric distance distributions. They are more flexible and introduce less bias.\n\n## Dipolar background#\n\nIn DeerLab, all inter-molecular contributions to the dipolar modulation (i.e. the echo modulation function due to randomly distributed spins in the sample that are not part of the spin-labeled protein or object) are referred to as the dipolar background. DeerLab has a range of parametric models for the background. All these background model’s names start with the prefix bg_. They take the time axis vector t (in microseconds) and a parameter vector param as inputs. The output is a background vector B defined over t. To get information on the model and its parameters, print the model\n\n>>>print(dl.bg_hom3d)\nDescription: Background from homogeneous distribution of spins in a 3D medium\nSignature: (t, conc, lam)\nConstants: [t]\nParameter Table:\n====== ======= ======= ======== ======== ======= ====================\nName Lower Upper Type Frozen Units Description\n====== ======= ======= ======== ======== ======= ====================\nconc 0.01 5e+03 nonlin No μM Spin concentration\nlam 0 1 nonlin No Pathway amplitude\n====== ======= ======= ======== ======== ======= ====================\n\n\nDeerLab’s background models fall into two categories, physical and phenomenological:\n\nPhysical background models\n\nDescribe particular distributions of spin labels in space and depend on physical parameters such as spin concentration, exclusion distances, and fractal dimensionality. The most common background model is deerlab.bg_hom3d, which describes the signal due to a homogeneous three-dimensional distribution of spins of a given concentration. A background due to a homogeneous distribution of spins in fractal dimensions is available with deerlab.bg_homfractal, and excluded-volume effects can be accounted for using deerlab.bg_hom3dex to model the background.\n\nt = np.linspace(-0.1,4,200) # time, in microseconds\nlam = 0.4 # modulation depth\nconc = 70 # spin concentration, in µM\nB = dl.bg_hom3d(t,conc,lam) # homogeneous 3D background\nplt.plot(t,B)\n\nPhenomenological background models\n\nRepresent various mathematical functions that are intended to mimic the background decay, without reference to a particular spatial distribution of spins. The parameters of these models do no have a direct physical meaning. Some examples include deerlab.bg_exp, which models the background decay as a simple exponential function, or deerlab.bg_strexp which model the background decay as a stretched exponential function.\n\nt = np.linspace(-0.1,4,200) # time, in microseconds\nkappa = 0.35 # decay rate, in inverse microseconds\nB = dl.bg_exp(t,kappa) # exponential background\nplt.plot(t,B)\n\n\nIn general, it is preferable to use physical instead of phenomenological models.\n\n## Experiments#\n\nDeerLab supports a wide range of dipolar EPR experiments (4-pulse DEER, 4-pulse DEER, RIDME, etc). Experiments differ in the number and nature of their modulated dipolar pathways. Each of these pathways leads to a dipolar modulation contribution to the total dipolar signal, with specific amplitude and refocusing times. The overall dipolar signal is the sum of an unmodulated contribution and a contribution from all modulated pathways, each of which with its own amplitude, refocusing time, and harmonic. For each supported experiment, there is a dedicated experiment :ref: constructor <modelsref_ex> starting with ex_, which generate experimental information on the pathway refocusing times and amplitudes based on the type of experiment and the experimental pulse sequence delays. This information can later be used to refine and constrain the dipolar models.\n\n## Dipolar kernels#\n\nOne of the core functions of DeerLab’s dipolar EPR applications is dipolarkernel. It constructs the kernel that provides the connection between the distance distribution and the time-domain dipolar signal via", null, "The simplest dipolar kernel just requires the time-vector t and distance-vector r\n\nt = np.linspace(0,6,300) # time axis, in µs\nr = np.linspace(2,7,300) # distance axis, in nm\nK0 = dl.dipolarkernel(t,r) # dipolar kernel matrix\n\n\nK0 is the kernel matrix. It assumes no orientation selection and absence of exchange couplings. To calculate the dipolar signal corresponding to a distance distribution P according to the equation above, use\n\nV = K0@P # calculate signal from distribution\n\n\nThe above K0 is the most elementary kernel, giving a single dipolar evolution function centered at time zero, with modulation depth 1, and without any background decay. The kernel can also account for the background and the dipolar pathways. Then, operation V=K@P will return the complete time-domain dipolar signal. Here is an example for a 4-pulse DEER signal\n\nlam = 0.4 # modulation depth\nB = dl.bg_hom3d(t,200,lam) # background (inter-molecular modulation function)\nK = dl.dipolarkernel(t,r,mod=lam,bg=B) # kernel matrix, including lam and B\nV = K@P # calculate signal from distribution\nplt.plot(t,V) # plotting\n\n\nFor experiments with more than one modulated dipolar pathway (such as 5-pulse DEER), modulation amplitudes and refocusing times for each pathway must be provided to dipolarkernel. Additionally, the background must be provided as a callable function that takes only time and modulation amplitude and encapsulates all other parameters. For example, for a 5-pulse DEER signal\n\nLam0 = 0.5 # amplitude of the unmodulated component\nlam1 = 0.4 # amplitude of the primary modulated pathway\nlam2 = 0.1 # amplitude of the secondary modulated pathway\ntref1 = 0.0 # refocusing time of the primary pathway, in µs\ntref2 = 3.1 # refocusing time of the secondary pathway, in µs\n\n# Dipolar pathways of the 5-pulse DEER experiment\npathways = [[Lam0],\n[lam1,tref1],\n[lam2,tref2]]\nBfcn = lambda t,lam: dl.bg_hom3d(t,200,lam) # Function for background\nK = dl.dipolarkernel(t,r,pathways=pathways,bg=Bfcn) # 5-pulse DEER dipolar kernel\n\n\nThe function dipolarkernel also has options to set the excitation bandwidth, to select the internal calculation method, and more.\n\n## Dipolar signals#\n\nDipolar signals are the results of the many different dipolar EPR spectroscopy experiments. They represent the data from which distance distributions can be infered. DeerLab provides the tools for simulating dipolar signals originating from different experiments.\n\nTo generate complete time-domain signals from a distance distribution and a background decay, use the function dipolarkernel (described above) and apply it to the distance distribution:\n\nK = dl.dipolarkernel(t,r,mod=lam,bg=B) # generate dipolar kernel\nV = K@P # generate dipolar signal\nplt.plot(t,V)\n\n\nIt is possible to add noise to simulated data by using the whitegaussnoise function:\n\nsigma = 0.05 # noise level\nV = K@P + dl.whitegaussnoise(t,sigma) # add some noise\n\n\nWith this, uncorrelated Gaussian noise with standard deviation sigma is added to the noise-free signal.\n\nAdding a phase rotation is also possible, yielding a complex-valued signal with non-zero imaginary component. The phase shift on the noise has to be taken into account too:\n\nphase = np.pi/4 # phase shift, radians\nV = K@P*exp(-1j*phase) # add a phase shift\nrnoise = dl.whitegaussnoise(t,sigma) # noise of real component noise\ninoise = dl.whitegaussnoise(t,sigma) # noise of imaginary component\nV = V + rnoise + inoise # complex-valued noisy signal" ]

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[ "# Cool Cell\n\nCool Cell(CoCe) is cell-based language by User:ChuckEsoteric08. It was created because author wanted to create version of Turing-machine which compiles to Quests, but later it become BF with limited memory which can implement BF easily\n\n## Descrtiption\n\nNumber of cells in implementation should be any number between 29999 and 30001(meaning there should be always only 30000 cells). Every cell is 8-bit and it will set to 0 if after incrementing value will be more than 255 and to 255 if number will be less than zero. Commands always seperated with `;`. `[X]` and `[Y]` should be any number. In the start of the program you need to input number which will be used in `p[X]` command. Comment start with `#` and it needs to be after `a[X]` command. Counting of cell starts from 1( first cell number is 1, second cell numember is 2 and so on).\n\nCommand Meaning\n`s[X]` point to cell `[X]`\n`r[X]` Dereference cell `[X]` and point to N-th cell, where N is value of that cell\n`i[X]` increment pointed cell `[X]` times\n`d[X]` same as previous command, but decrement\n`p[X]` put every digit in inputed number in tape starting from cell `[X]`\n`a[X]` declare label `[X]`\n`g[X]` goto label `[X]`\n`z[X]` if pointed cell is zero goto label `[X]`\n`o[X]` same as previous command, but if not zero\n`f[X]t[Y]` if pointed cell is `[X]`(It can be any digit from 0 to 9), then goto `[Y]`\n`t[X]` output number `[X]`\n`h[X]` print `[X]` and halt\n\n## Examples\n\n### Hello World\n\n```tHello, World!;\n```\n\n### Truth Machine\n\n```p1;s1;oloop;h0;aloop;#a;t1;gloop;\n```\n\nIf first digit in input is not 1 or 0 then it is 1\n\n```p1;s1;f1tloop;z0;hError!;a0;#a;h0;aloop;#a;t1;gloop;\n```\n\nIt will print \"Error!\" if first digit in input is not 1 or 0\n\n```i288;\n```\n\nSet cell to zero\n\n```d123;\n```\n\nSet cell to 255\n\n## Computational class\n\nIt is Bounded-storage machine, because it has limited memory" ]

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[ "", null, "", null, "www.caigou.com.cn", null, "<\n•", null, ">", null, "", null, "• 通用分析仪器\n• 分光光度计\n• 表面张力仪\n• 液氮/超低温控制系统\n• 数字滴定仪\n• 数字式分注器\n• 密度测试系统\n• 高温控制系统\n• 荧光测试系统\n• 实验室通用设备\n• 水浴系统\n• 超声波仪器\n• 二氧化碳培养箱\n• 恒温恒湿箱\n• 光照培养/人工气候箱\n• 高温炉\n• 真空系统/真空泵\n• 生化（BOD）培养箱\n• 蒸馏水器\n• 孵化器\n• 电热恒温培养/干燥箱\n• 加热/磁力搅拌器\n• 高低温浴槽\n• 天平\n• 震荡/混合器\n• 置顶式搅拌器\n• 实验室仪器\n• 超声波清洗器\n• 昆虫动物类仪器\n• 滴定仪\n• 动物行为分析记录仪\n• 动物昆虫电生理测量仪\n• 虫害监测站\n• 动物无线跟踪系统\n• 昆虫采样器\n• 动物呼吸作用测量系统\n• 虫害声音探测器\n• 昆虫计数器\n• 土壤动物分离漏斗\n• 喷雾塔\n• 野外动物侦察照相机\n• 昆虫工具\n• 昆虫工具\n• 昆虫工具\n• 工业应用类仪器\n• 野外便携式掌上电脑\n• 加速度记录仪\n• 炉温记录仪\n• 热特性分析仪\n• 热电偶记录仪\n• 电流，电压传感器\n• 体育运动类仪器\n• 人体运动生理监测系统\n• 汽车运行状态记录仪\n• 食品检测类仪器\n• 谷物水分\n• 折光糖度计\n• 色彩色差计\n• 果实硬度计\n• 酒品分析系统\n• 酒品蒸馏系统\n• 环境水质分析仪器\n• 现场测试试纸\n• 多参数水质测试盒/箱\n• 水质/溶液取样器\n• 离子计\n• 微型电极\n• 多参数水质分析系统\n• 连续流动分析仪\n• PH/ORP计\n• 电导率/TDS/盐度计\n• 溶解氧（DO）计\n• 浊度计\n• 系统集成类仪器\n• 热通量传感器\n• 雪厚/水位传感器\n• 温度传感器\n• 土壤水分传感器\n• 数据采集器\n• 光照辐射传感器\n• 气体传感器\n• 土壤水分温度电导率传感...\n• 叶面湿度传感器\n• 水文水利类仪器\n• 藻类检测仪\n• 水下探测仪\n• 水体采样器\n• 水质检测仪\n• 自动水文站\n• 声纳深度计\n• 水位记录仪\n• 流速流量测量仪\n• 水体温度记录仪\n• 底泥采样器\n• 气象环境类仪器\n• 温度测量仪\n• 雪特性分析仪\n• 温湿度计\n• 风速风量测量仪\n• 环境气体测量仪\n• 蒸发量(ET)测量仪\n• 海拔气压计\n• 光照辐射测量仪\n• 自动气象站\n• 温度记录仪\n• 温湿度记录仪\n• 自动记录雨量计\n• 风蚀监测系统\n• 涡动协方差系统\n• 雨滴测量仪\n• 波文比测量系统\n• 热像仪\n• 手持式气象仪\n• 负氧离子测量仪\n• 干湿沉降收集器\n• 植物类仪器\n• 植物水势仪\n• 植物茎流测量系统\n• 光合作用测量系统\n• 光谱分析系统\n• 树木年轮分析仪\n• 木质检测仪\n• 便携式叶绿素仪\n• 植物冠层分析仪\n• 植物根系分析仪\n• 测高测距仪\n• 木材密度分析\n• 凋落物水分仪\n• 植物气孔计\n• 植物生理生态监测系统\n• 树木测径仪\n• 植物叶面积仪\n• 植物氮分析仪\n• 植物氧电极\n• 植物导水率测量仪\n• 植物生长箱\n• 叶面根系图像分析\n• 植物采样设备\n• 植物荧光测量仪\n• 植物气穴测量仪\n• 数字式显微镜\n• 树木生长锥\n• 树木断层检测\n• 植物倒伏测试仪\n• 植物组织比色卡\n• 植物荧光成像系统\n• 木材水分\n• 植物多酚测定仪\n• 土壤类仪器\n• 土壤紧实度仪\n• 泥炭探测仪\n• 便携式土壤水分仪\n• 土壤水势仪\n• 土壤温度测量仪\n• 土壤污染调查系统\n• 土壤三相测量仪\n• 土壤通气性测定仪\n• 土壤水分温度电导率速测...\n• 土壤颗粒分析仪\n• 人工降雨模拟器\n• 激光微地貌扫描仪\n• 土壤硬度仪\n• 土壤原位pH计\n• 土壤原位电导率仪\n• 非饱和导水率测量仪\n• 土壤取样器\n• 土壤呼吸测量系统\n• 土壤养分仪\n• 土壤入渗仪\n• 土壤重金属分析仪\n• 地表径流测量系统\n• 土壤溶液取样器\n• 土壤比色卡\n• 土壤水分监测系统\n• 土壤剪切仪\n• 土壤水分特征曲线测定仪\n• 石材水分测量仪\n• 热门产品\n同类产品推荐\n• 产品名称价格地区公司名称更新时间" ]

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[ "Cody\n\n# Problem 895. Generate N equally spaced intervals between -L and L\n\nGiven N and L, return a list of numbers (in ascending order) that divides the interval [-L L] into N equal-length segments.\n\nFor example, if L = 3 and N = 2\n\n` y = [ -3 0 3 ]`\n\n### Solution Stats\n\n47.01% Correct | 52.99% Incorrect\nLast Solution submitted on Mar 29, 2020" ]

[ null ]

[ "# Creating Trajectories with Splines\n\n## Introduction\n\nThis example shows hows to create a trajectory for waypoints in MATLAB with Splines. Trajectories are used to navigate between waypoints smoothly with respect to time.\n\nThis smoothness is important because it means that the acceleration or velocity has no discontinuities. Since acceleration is generally tied to the torque input, this means that motors are ramped up and down in a continuous smooth manner that reduces stress on the motors.\n\nFirst, we'll show how to generate a spline between two points, then move onto an n-point example. There are three methods that we can use:\n\n• Cubic spline polynomial and solving for the coefficients\n• \"spline\" function\n• \"interp1\" function\n\nWe will start off by showing the cubic spline polynomial method, followed by spline and briefly show interp1. Spline and interp1 implementations are almost identical. Spline has the added advantage of being able to set the slope at the endpoints.\n\n```x = [0 1];\ny = [0 1];\nxq = x(1):0.1:x(end);\nyq = spline(x,[0 y 0],xq);\nplot(xq,yq,':.b',x,y,'or');\naxis([-0.5 1.5 -0.5 1.5]);\n```", null, "## Two Point Example\n\nStarting with the cubic spline polynomial:\n\nWe have two points.\n\n```plot(x,y,'or');\naxis([-0.5 1.5 -0.5 1.5]);\n```", null, "We would like to connect the two points with a line.\n\n```plot(x,y,x,y,'or');\naxis([-0.5 1.5 -0.5 1.5]);\n```", null, "We would also like to connect the two points with a spline so that we can specify the slope at the endpoints.\n\n```plot(xq,yq,':.b',x,y,'or');\naxis([-0.5 1.5 -0.5 1.5]);\n```", null, "We are going to start with x(t). Let's say we have four boundary conditions:\n\n```x(t0) = x0; % position of initial coordinate\nx(tf) = xf; % position of final coordinate\nxdot(t0) = xdot0; % slope of initial coordinate\nxdot(tf) = xdotf; % slope of final coordinate```\n```where:\nt0 = 0; tf = 10;\nx0 = 1; xf = 11;\nxdot0 = 1; xdotf = 1;```\n\nThe general equation for a spline is;\n\n```syms x xdot a0 a1 a2 a3 t\nx = a0+a1*t+a2*t^2+a3*t^3;\nxdot = a1+2*a2*t+3*a3*t^2;\n```\n\nInserting the values of the initial conditions for t, we get x and xdot as a function of a1,a2,a3,a4.\n\n```subs(x,t,0) % = 1\nsubs(x,t,10) % = 11\nsubs(xdot,t,0) % = 1\nsubs(xdot,t,10) % = 1\n```\n```\nans =\n\na0\n\nans =\n\na0 + 10*a1 + 100*a2 + 1000*a3\n\nans =\n\na1\n\nans =\n\na1 + 20*a2 + 300*a3\n\n```\n\nIn matrix form, this becomes C*a = bc, where:\n\n```xbc = [1 11 1 1]'; % boundary conditions for y\nC = [1 0 0 0; 1 10 100 1000; 0 1 0 0; 0 1 20 300]; % coefficients from spline equations\n```\n\nSolving for the spline coefficients a, we obtain the following:\n\n```a = C\\xbc; % a = [a1 a2 a3 a4]';\ndisplay(a);\n```\n```a =\n\n1\n1\n0\n0\n\n```\n\nThe cubic spline equation then becomes:\n\n```syms x;\nf = a(1)+a(2)*x+a(3)*x.^2+a(4)*x.^3;\ndisplay(f);\n```\n```\nf =\n\nx + 1\n\n```\n\ncalculate spline interpolated data using calculated coefficients\n\n```tbc = [0 10]; % [x0 xf]\nxbc = [1 11];\nt = 0:0.1:10;\nx = a(1)+a(2)*t+a(3)*t.^2+a(4)*t.^3;\n\n% plot spline\nplot(t,x,tbc,xbc,'or');\naxis([-1 11 0 12]);\ntitle('x vs. t for Cubic Spline Equation')\n```", null, "This obtains the spline for the x vs. t motion. We would then have to do this for y vs. t as well.\n\n```ybc = [1 5 0 0]';\nC = [1 0 0 0; 1 10 100 1000; 0 1 0 0; 0 1 20 300]; % coefficients from spline equations\na = C\\ybc; % a = [a1 a2 a3 a4]';\n\ntbc = [0 10];\nybc = [1 5];\nt = 0:0.1:10;\ny = a(1)+a(2)*t+a(3)*t.^2+a(4)*t.^3;\n\n% plot spline\nplot(t,y,tbc,ybc,'or');\naxis([-1 12 0 6]);\ntitle('y vs. t for Cubic Spline Equation')\n```", null, "After obtaining these two trajectories with respect to time, we can then show the y vs. x trajectory\n\n```plot(x,y,xbc,ybc,'or');\naxis([-1 12 0 6]);\ntitle('y vs. x for Cubic Spline Equation')\n```", null, "MATLAB offers a function called \"spline\" which does the polynomial coefficient solving for you.\n\nIf we did not want to specify the slope boundary conditions, the equation would be:\n\ny = spline(xbc,ybc,x)\n\nThe result would be a straight line between both points.\n\nTo specify the slopes we use the following equation:\n\n```tbc = [0 10];\nxbc = [1 3];\nxdotbc = [0 0];\nt = 0:0.1:10;\nx = spline(tbc,[xdotbc(1) xbc xdotbc(2)],t);\n\nplot(t,x,tbc,xbc,'or');\ntitle('Spline Equation')\nxlabel('t')\nylabel('x')\naxis([-1 11 0 4]);\n```", null, "## N-Point Example\n\nThis is a \"Cubic polynomials for a path with via points\" problem\n\n```points = [0 0; 1 0; 1 1; 0 1; 0 2; 1 2];\nx = points(:,1);\ny = points(:,2);\n\nplot(x,y,x,y,'o');\naxis([-0.5 1.5 -0.5 2.5]);\ntitle('Waypoints for N-Point example');\n```", null, "We can not fit a spline directly to the x-y data since the x data values are not monotonically increasing.\n\n```% Instead, we need to introduce an independent variable that is always\n% monotonically increasing, like time!\n```\n\nFirst, let's define the waypoints\n\n```t = [0 1 2 3 4 5];\npoints = [0 0; 1 0; 1 1; 0 1; 0 2; 1 2];\nx = points(:,1);\ny = points(:,2);\n```\n\nThen, we'll plot the original waypoints in t-x, t-y and x-y space\n\n```figure;\nplot(t,x,t,x,'o');\ntitle('x vs. t');\naxis([-0.5 5.5 -0.5 1.5]);\n\nfigure;\nplot(t,y,t,y,'o');\ntitle('y vs. t');\naxis([-0.5 5.5 -0.5 2.5]);\n\nfigure;\nplot(x,y,x,y,'o');\ntitle('y vs. x');\naxis([-0.5 1.5 -0.5 2.5]);\n```", null, "", null, "", null, "To calculate the spline for waypoints, we will use the spline function\n\n```tq = 0:0.1:5;\nslope0 = 0;\nslopeF = 0;\nxq = spline(t,[slope0; x; slopeF],tq);\nyq = spline(t,[slope0; y; slopeF],tq);\n\n% plot spline in t-x, t-y and x-y space\nfigure;\nplot(t,x,'o',tq,xq,':.');\naxis([-0.5 5.5 -0.5 1.5]);\ntitle('x vs. t');\n\nfigure;\nplot(t,y,'o',tq,yq,':.');\naxis([-0.5 5.5 -0.5 2.5]);\ntitle('y vs. t');\n\nfigure;\nplot(x,y,'o',xq,yq,':.');\naxis([-0.5 1.5 -0.5 2.5]);\ntitle('y vs. x');\n```", null, "", null, "", null, "## Interp1\n\nInterp1 can be used to create a trajectory as long as the user does not need to define the slope at the beginning and end.\n\nIf this were the desired trajectory for a mobile robot, the robot would be requested to go from v = 0 to v = \"some value\" immediately causing the required torque to spike.\n\n```xq = interp1(t,x,tq,'spline');\nyq = interp1(t,y,tq,'spline');\n\nfigure;\nplot(t,x,'o',tq,xq,':.');\naxis([-0.5 5.5 -0.5 1.5]);\ntitle('x vs. t');\n\nfigure;\nplot(t,y,'o',tq,yq,':.');\naxis([-0.5 5.5 -0.5 2.5]);\ntitle('y vs. t');\n\nfigure;\nplot(x,y,'o',xq,yq,':.');\naxis([-0.5 1.5 -0.5 2.5]);\ntitle('y vs. x');\n\nclose all\n```\n\nOpen example for Trajectory Planning\n\nOpen Example" ]

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[ "# 10th Maths Book Back Algebra Ex 3.15\n\n## Samacheer Kalvi 10th Maths Book Back Solution:\n\nTamil Nadu 10th Maths Book Back Answers Unit 3 – Algebra Ex 3.15 are provided on this page. Samacheer Kalvi Maths Book Back Solutions/ Guide available for all Units. TN Samacheer Kalvi 10th Maths Book consists of 8 Units and each unit book back solutions given below topics wise with Questions and Answers. The complete Samacheer Kalvi Books Back Answers/Solutions are available on our site.\n\nThe Samacheer kalvi 10th Maths solutions are useful to enhance your skills. Candidates who prepared for the Competitive and board exams 10th Maths Book Back Answers in English and Tamil Medium. The 10th Maths Unit 3 Algebra consists of 19 units. Each Unit Book Back Answers provide topic-wise on this page. 10th Maths Book Back Answers are prepared according to the latest syllabus. The 10th Maths Book Back Algebra Ex 3.15 Answers in English.\n\n### 10th Maths Book Back Answers/Solutions:\n\nTN Samacheer Kalvi 10th Maths Unit 3 Chapter 15 Book Back Exercise has given below. The 10th Maths Book Back Solutions Guide is uploaded below,\n\n### Exercise 3.15 Algebra\n\n1.Graph the following quadratic equations and state their nature of solutions,\n(i) x2 – 9x + 20 = 0\nSolution:", null, "Step 1:\nPoints to be plotted : (-4, 72), (-3, 56), (-2, 42), (-1, 30), (0, 20), (1, 12), (2, 6), (3, 2), (4, 0)\nStep 2:\nThe point of intersection of the curve with x axis is (4, 0)\nStep 3:", null, "The roots are real & unequal\n∴ Solution {4, 5}\n\nStep 1: Points to be plotted : (-4, 36), (-3, 25), (-2, 16), (-1, 9), (0, 4), (1, 1), (2, 0), (3, 1), (4, 4)\nStep 2: The point of intersection of the curve with x axis is (2, 0)\nStep 3:", null, "Since there is only one point of intersection with x axis, the quadratic equation x2 – 4x + 4 = 0 has real and equal roots.\n∴ Solution{2, 2}\n\n(iii) x2 + x + 7 = 0\nLet y = x2 + x + 7\nStep 1:", null, "Step 2:\nPoints to be plotted: (-4, 19), (-3, 13), (-2, 9), (-1, 7), (0, 7), (1, 9), (2, 13), (3, 19), (4, 27)\nStep 3:\nDraw the parabola and mark the co-ordinates of the parabola which intersect with the x-axis.", null, "Step 4:\nThe roots of the equation are the points of intersection of the parabola with the x axis. Here the parabola does not intersect the x axis at any point.\nSo, we conclude that there is no real roots for the given quadratic equation,\n\n(iv) x2 – 9 = 0\nLet y = x2 – 9\nStep 1:", null, "Step 2:\nThe points to be plotted: (-4, 7), (-3, 0), (-2, -5), (-1, -8), (0, -9), (1,-8), (2, -5), (3, 0), (4, 7)\nStep 3:\nDraw the parabola and mark the co-ordinates of the parabola which intersect the x-axis.", null, "Step 4:\nThe roots of the equation are the co-ordinates of the intersecting points (-3, 0) and (3, 0) of the parabola with the x-axis which are -3 and 3 respectively.\nStep 5:\nSince there are two points of intersection with the x axis, the quadratic equation has real and unequal roots.\n∴ Solution{-3, 3}\n\n(v) x2 – 6x + 9 = 0\nLet y = x2 – 6x + 9\nStep 1:", null, "Step 2:\nPoints to be plotted: (-4, 49), (-3, 36), (-2, 25), (-1, 16), (0, 9), (1, 4), (2, 1), (3, 0), (4, 1)\nStep 3:\nDraw the parabola and mark the co-ordinates of the intersecting points.", null, "Step 4:\nThe point of intersection of the parabola with x axis is (3, 0)\nSince there is only one point of intersection with the x-axis, the quadratic equation has real and equal roots. .\n∴ Solution (3, 3)\n\n(vi) (2x – 3)(x + 2) = 0\n2x2 – 3x + 4x – 6 = 0\n2x2 + 1x – 6 = 0\nLet y = 2x2 + x – 6 = 0\nStep 1:", null, "Step 2:\nThe points to be plotted: (-4, 22), (-3, 9), (-2, 0), (-1, -5), (0, -6), (1, -3), (2, 4), (3, 15), (4, 30)\nStep 3:\nDraw the parabola and mark the co-ordinates of the intersecting point of the parabola with the x-axis.", null, "Step 4:\nThe points of intersection of the parabola with the x-axis are (-2, 0) and (1.5, 0).\nSince the parabola intersects the x-axis at two points, the, equation has real and unequal roots.\n∴ Solution {-2, 1.5}\n\n2.Draw the graph of y = x2 – 4 and hence solve x2 – x – 12 = 0\nSolution:", null, "", null, "", null, "Point of intersection (-3, 5), (4, 12) solution of x2 – x – 12 = 0 is -3, 4\n\n3.Draw the graph of y = x2 + x and hence solve x2 + 1 = 0.\nSolution:", null, "Draw the parabola by the plotting the points (-4, 12), (-3, 6), (-2, 2), (-1, 0), (0, 0), (1, 2), (2, 6), (3, 12), (4, 20), (5, 30)", null, "To solve: x2 + 1 = 0, subtract x2 + 1 = 0 from y = x2 + x.\nx2 + 1 = 0 from y = x2 + x", null, "Plotting the points (-2, -3), (0, -1), (2, 1) we get a straight line. This line does not intersect the parabola. Therefore there is no real roots for the equation x2 + 1 = 0.\n\n4.Draw the graph of y = x2 + 3x + 2 and use it to solve x2 + 2x + 1 = 0.\nSolution:", null, "Draw the parabola by plotting the point (-4, 6), (-3, 2), (-2, 0), (-1, 0), (0, 2), (1, 6), (2, 12), (3, 20), (4, 30).", null, "To solve x2 + 2x + 1 = 0, subtract x2 + 2x + 1 = 0 from y = x2 + 3x + 2", null, "Draw the straight line by plotting the points (-2, -1), (0, 1), (2, 3)\nThe straight line touches the parabola at the point (-1,0)\nTherefore the x coordinate -1 is the only solution of the given equation\n\n5.Draw the graph of y = x2 + 3x – 4 and hence use it to solve x2 + 3x – 4 = 0. y = x2 + 3x – 4\nSolution:", null, "Draw the parabola using the points (-4, 0), (-3, -4), (-2, -6), (-1, -6), (0, -4), (1, 0), (2, 6), (3, 14), (4, 24).", null, "To solve: x2 + 3x – 4 = 0 subtract x2 + 3x – 4 = 0 from y = x2 + 3x – 4 ,", null, "The points of intersection of the parabola with the x axis are the points (-4, 0) and (1, 0), whose x – co-ordinates (-4, 1) is the solution, set for the equation x2 + 3x – 4 = 0.\n\n6.Draw the graph of y = x2 – 5x – 6 and hence solve x2 – 5x – 14 = 0.\nSolution:", null, "Draw the parabola using the points (-5, 44), (-4, 30), (-3, 18), (-2, 8), (-1, 0), (0, -6), (1, -10), (2, -12), (3, -12), (4, -10)", null, "To solve the equation x2 – 5x – 14 = 0, subtract x2 – 5x – 14 = 0 from y = x2 – 5x – 6.", null, "The co-ordinates of the points of intersection of the line and the parabola forms the solution set for the equation x2 – 5x – 14 = 0.\n∴ Solution {-2, 7}\n\n7.Draw the graph of y = 2x2 – 3x – 5 and hence solve 2x2 – 4x – 6 = 0. y = 2x2 – 3x – 5\nSolution:", null, "Draw the parabola using the points (-4, 39), (-3, 22), (-2, 9), (-1, 0), (0, -5), (1, -6), (2, -3), (3, 4), (4, 15).", null, "To solve 2x2 – 4x – 6 = 0, subtract it from y = 2x2 – 3x – 5", null, "Draw a straight line using the points (-2, -1), (0, 1), (2, 3). The points of intersection of the parabola and the straight line forms the roots of the equation.\nThe x-coordinates of the points of intersection forms the solution set.\n∴ Solution {-1, 3}\n\n8.Draw the graph of y = (x – 1)(x + 3) and hence solve x2 – x – 6 = 0.\nSolution:\ny = (x – 1)(x + 3) = x2 – x + 3x – 3 = 0\ny = x2 + 2x – 3", null, "Draw the parabola using the points (-4, 5), (-3, 0), (-2, -3), (-1,-4), (0, -3), (1, 0), (2, 5), (3, 12), (4, 21)", null, "To solve the equation x2 – x – 6 = 0, subtract x2 – x – 6 = 0 from y = x2 – 2x – 3.", null, "Plotting the points (-2, -3), (-1, 0), (0, 3), (2, 9), we get a straight line.\nThe points of intersection of the parabola with the straight line gives the roots of the equation. The co¬ordinates of the points of intersection forms the solution set.\n∴ Solution {-2, 3}" ]

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[ "# Why is $A^0 = \\{ \\emptyset \\}$?\n\nLet $A$ be a set and denote $\\underbrace{A \\times A \\times \\cdots \\times A}_\\text{n times}$ by $A^n$. Why is it that $A^0 = \\{ \\emptyset \\}$ ?\n\nBy your definition, $A^0$ should be the set of all tuples of length zero of elements of $A$. There is exactly one zero-tuple. One might want to call it $()$, apparently the source that you use calls it $\\emptyset$, which is justified for the reasons that Asaf has given.\n\nBecause $A^B$ is the set of all functions from $B$ into $A$. Given that $0$ is $\\varnothing$, how many functions are there from the empty set into $A$?\n\n• It might pay to expand this a drop to indicate that the usual interpretation of a natural number in this context is a von Neumann ordinal, and perhaps to note that the $\\varnothing$ in the result is a function represented by its graph. – dfeuer Dec 12 '13 at 4:32\n\nAsaf Karagila has given the technical answer, so I will give a more hand-wavy answer.\n\nIn the real numbers, for example, we want $x^{m+n} = x^{m}\\cdot x^{n}$ to hold, which means that we are forced into saying that $x^{0} = 1$. For Cartesian products, we also want $A^{m+n} = A^{m} \\times A^{n}$ to hold, and so we are forced to say that $A^{0} = \\left\\{\\varnothing\\right\\}$.\n\nIn order to see that this convention makes sense, just notice that, as sets, $A$ and $A\\times\\left\\{\\varnothing\\right\\}$ are essentially the same, since there is an obvious bijection sending $a\\mapsto\\left(a,\\varnothing\\right)$. So we can say, loosely, that they are equal. Technically, they are different, but the same is true that, technically, $A^{m+n}$ is different from $A^{m}\\times A^{n}$.\n\nWe can't have, say, $A^{0} = \\varnothing$ since $A\\times\\varnothing$ is always $\\varnothing$, but that's also why we can't say that $x^{0} = 0$ in $\\mathbb{R}$." ]

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[ "# Million to Thousand Converter\n\nI think all will be knowing what is Million? and What is Thousand?, In this article you can learn how to convert Million into Thousand by using our Free Conversion tool.\n\n## Convert Million to Thousand\n\n### What is Million?\n\n1 million in numerical is represented as 1000000. The Indian equivalent of a million is ten lakh rupees. It is not a standard quantity because it is not in common usage. Different words are used all across the world to express the same number. ‘How much does a million mean?’ or ‘How many rupees does a million mean?’ are a few common queries among Indians who encounter foreign number terminology for the first time.\n\n### What is Thousand?\n\nA large but unspecified number, amount, or quantity. Pilgrims gather there by the hundreds. There will be hundreds of people there. Hundreds of tree species face extinction.\n\nSummary:-\n\nWe all know that 1 Crore = 100 Lakhs, Therefore 1 Million = 1000K [Thousand].\n\n## Steps to Convert Million into Thousand\n\nJust the Read the below step-by-step guideliness to use our converter.\n\n1. Enter the Number You want to Convert into Thousand\n2. Result will be Displayed in Thousand Now\n\n## Examples of Million to Thousand\n\n Million Thousand 1 1000 2 2000 3 3000 4 4000 5 5000 10 10000 20 20000 30 30000 40 40000 50 50000 100 100000 200 200000\n\n## Questions Asked on Million to Thousand\n\nQuestion:- How much is 1 million to Thousand?\n\nAnswer:- We know that 10 million is called 100 lakhs and also 10 Lakhs is called 1 Million, So that 1 Million is Equal to 1000K [Thousand].\n\nQuestion:-Fill in the blank: 1 Million = ________ Thousand\n\nAnswer:- 1 Million = 1000K Thousand\n\nQuestion:- How many Thousand make three Millions?\n\nAnswer:- Everyone will know that 100 lakhs is known as 10 Million and also 10 Million is called as 1 Lakhs, Therefore that Three Million is Equal to 3000K Thousand.\n\nQuestion:-Fill in the blank: 100 Million = ________ Thousand\n\nAnswer:- 100 Million = 10000K Thousand\n\nQuestion:-What is the formula to convert Million into Thousand\n\nAnswer:- Million into Thousand can be converted by multiplying the value with 1000\n\nOther Million Converters" ]

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[ "# Need help on these proof questions\n\nWatch\nAnnouncements\n#1\nQuestions: http://prntscr.com/rmuy7e\n\nFor Q11 c) I think there is a typo in the question as I can only prove it if it's - a + b - c + d:\nn = 1000a + 100b + 10c + d\nn = 1001a - a + 99b + b + 11c - c + d\nn = 11(91a + 9b + c) - a + b - c + d\n\nFor Q11 a) and Q12, I'm not sure it's even possible to 'prove'. In the textbook it just says 'position value' for Q11a and 'consider the rationality of sqrt(2)^sqrt(2)' for Q12.\n0\n1 week ago\n#2\n(Original post by TSR360)\nQuestions: http://prntscr.com/rmuy7e\n\nFor Q11 c) I think there is a typo in the question as I can only prove it if it's - a + b - c + d:\nn = 1000a + 100b + 10c + d\nn = 1001a - a + 99b + b + 11c - c + d\nn = 11(91a + 9b + c) - a + b - c + d\n\nFor Q11 a) and Q12, I'm not sure it's even possible to 'prove'. In the textbook it just says 'position value' for Q11a and 'consider the rationality of sqrt(2)^sqrt(2)' for Q12.\nQ11c) Both are correct as -121 is divisible by 11, just as is 121. You could get the book answer by flipping the 1001 or 999 in the original subtraction.\nQ12) the example shows that irrational^irrational can be rational. It can also be irrational. Obviously the example given is rational. If sqrt(2)^sqrt(2) is rational, result is proved. If it's irrational, you've just shown when you raise it to sqrt(2) it's rational.\nNot really sure what your problem is with Q11a)?\nLast edited by mqb2766; 1 week ago\n0\n#3\n(Original post by mqb2766)\nQ11c) Both are correct as -121 is divisible by 11, just as is 121. You could get the book answer by flipping the 1001 or 999 in the original subtraction.\nQ12) the example shows that irrational^irrational can be rational. It can also be irrational. Obviously the example given is rational. If sqrt(2)^sqrt(2) is rational, result is proved. If it's irrational, you've just shown when you raise it to sqrt(2) it's rational.\nNot really sure what your problem is with Q11a)?\nI don't think Q11a is possible to prove. The best 'proof' I can come up with is proof by example, but in the textbook it says Position value. How do I prove a position value?\n0\n1 week ago\n#4\n(Original post by TSR360)\nI don't think Q11a is possible to prove. The best 'proof' I can come up with is proof by example, but in the textbook it says Position value. How do I prove a position value?\nThey said explain, so you just write down the position value representation?\n23 = 20+3 = 2*10 + 3\nI think that's all that's required.\nLast edited by mqb2766; 1 week ago\n0\nX\n\nnew posts", null, "Back\nto top\nLatest\nMy Feed\n\n### Oops, nobody has postedin the last few hours.\n\nWhy not re-start the conversation?\n\nsee more\n\n### See more of what you like onThe Student Room\n\nYou can personalise what you see on TSR. Tell us a little about yourself to get started.\n\n### Poll\n\nJoin the discussion\n\nYes (306)\n34.27%\nNo (587)\n65.73%" ]

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[ "# What is the Least Common Multiple of 9 and 25?\n\nLeast common multiple or lowest common denominator (lcd) can be calculated in two way; with the LCM formula calculation of greatest common factor (GCF), or multiplying the prime factors with the highest exponent factor.\n\nLeast common multiple (LCM) of 9 and 25 is 225.\n\nLCM(9,25) = 225\n\nLCM Calculator and\nand\n\n## Least Common Multiple of 9 and 25 with GCF Formula\n\nThe formula of LCM is LCM(a,b) = ( a × b) / GCF(a,b).\nWe need to calculate greatest common factor 9 and 25, than apply into the LCM equation.\n\nGCF(9,25) = 1\nLCM(9,25) = ( 9 × 25) / 1\nLCM(9,25) = 225 / 1\nLCM(9,25) = 225\n\n## Least Common Multiple (LCM) of 9 and 25 with Primes\n\nLeast common multiple can be found by multiplying the highest exponent prime factors of 9 and 25. First we will calculate the prime factors of 9 and 25.\n\n### Prime Factorization of 9\n\nPrime factors of 9 are 3. Prime factorization of 9 in exponential form is:\n\n9 = 32\n\n### Prime Factorization of 25\n\nPrime factors of 25 are 5. Prime factorization of 25 in exponential form is:\n\n25 = 52\n\nNow multiplying the highest exponent prime factors to calculate the LCM of 9 and 25.\n\nLCM(9,25) = 32 × 52\nLCM(9,25) = 225" ]

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[ "1. Math\n2. Statistics And Probability\n3. matlab notes b find likelihood for each sequence 19 for...\n\n# Question: matlab notes b find likelihood for each sequence 19 for...\n\n###### Question details\nMATLAB", null, "", null, "Notes: B)\nFind likelihood for each sequence 1-9 for each hypothesis.\nFirst hypothesis with the fair coin (theta value = 0.5): what is the probability that we can get each sequence, 1 through 9.\nSecond hypothesis with weighted coin(don’t know theta value): what is likelihood of getting sequence 1 through 9. Using equation to find all possible hundred theta values. For each theta value we need to compute likelihood.\nThen we sum them up across all possible theta values. Then we get likelihood under the second hypothesis\n###### Solution by an expert tutor", null, "" ]

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[ "My Research and Inventions", null, "###### click here to see all content or button bellow for specific topic", null, "### ee-interaction", null, "### ee-interaction", null, "### ee-interaction", null, "### ee-interaction", null, "### ee-interaction", null, "### ee-interaction\n\nDrift Charge Current and Diffusive Spin Current.\n\n### Spin and Charge Transport. Classical model of the spin-up/spin-down band.\n\n#####", null, "For the modified model, which includes all above-mentioned facts, click here or here", null, "The same content can be foundin", null, "V. Zayets Phys. Rev. B 86, 174415 (2012) (clich here", null, "to download pdf);or http://arxiv.org/abs/1205.1278", null, "### Abstract:\n\n#### Drift current is current drifted by an external applied electrical field. The drift current flows from a drain towards a source. Diffusion spin current is a diffusion of the spins from a region of the larger spin accumulation towards a region of the smaller spin accumulation. The properties of diffusion and drift currents are drastically different.\n\n##### about difference between drift and diffussion currents you can read here\n\nThere is a significant difference between flows of the charge and the spin in a solid. In general the spins flow as a diffusion current and charge flows as a drift current\n\n## Drift and Diffusion Currents\n\nIn a material there are two possible types of electron transport: diffusion and drift.\n\nThe drift current is a flow of charge particles along an applied electrical field.\n\nThe diffusion is a flow of uncharged particles from the region of their high concentration into regions of their lower concentration. In the case when the same amounts of electrons of opposite spins are flowing in opposite directions, in total there is no transport of charge, but there is only a transport of spin. Such transport can be considered as a flow of uncharged particles, which carry only spin. Therefore, spin may diffuse from region of higher spin concentrations into regions of smaller spin concentrations.\n\nIt should be noticed the fact that charged particles can not diffuse. The diffusion of charge particles creates a charge accumulation, which induces an electrical field, which stops further diffusion. However, the charged particles of opposite polarities can diffuse together in the same direction or the charged particles of same polarity can diffuse together in opposite directions. In both cases the diffusion does not cause a charge accumulation.\n\n## Drift and Diffusion Currents in Different materials\n\nThe properties of diffusion and drift currents may be substantially different for different materials. Generally, in metals the diffusion and drift currents do not interact with each other and they flow rather independently. In contrast, in the semiconductors there is a significant interaction between the diffusion and drift currents. It leads to several interesting effects. For example, in the semiconductors the spin diffusion length may be elongated or shorten, when drift charge current flows along spin diffusion. (Theory is here. Comparison with experiment is here).\n\nThe table below compares the basic properties of diffusion spin current and drift charge current in non-magnetic metals (near constant density of states near Fermi energy), ferromagnetic metals (different density of states for electrons with spin along and opposite to direction of the metal magnetization), non-degenerated semiconductor (any material, in which density of states is not constant near Fermi energy ) and the metals in which electron-electron interaction is strong.\n\n# ferromagnetic metals", null, "", null, "# metals with strong e-e interaction", null, "", null, "## Spin chemical potential and charge chemical potential", null, "Fig.1 The charge current flows between the charge source and the charge drain", null, "Fig.2 The spin current is generated at the spin source and freely flows in all directions\n\nThe spin chemical potential", null, "describes the flow of the spin, the charge chemical potential", null, "describes flow of the charge. The", null, "", null, "are the chemical potential for spin-up and spin-down electrons.\n\nThe spin chemical potential", null, "is linearly proportional to the spin accumulation in the material.\n\nThe charge chemical potential", null, "describes distribution of electrical field in the material and charge accumulation is proportional to the second derivation of the charge chemical potential.", null, "## External Electrical Field\n\nWhen the external electrical field is applied, the chemical potentials for both spin-up and spin-down electrons obey", null, "Therefore", null, "Form (3), there is a gradient of", null, "along electrical field. That means that the charge is drifted along the applied electric field.", null, "describes a diffusion spin current. The diffusion spin current is a flow of spins from a region of the larger spin accumulation towards a region of smaller spin accumulation.", null, "and the spin current are not sensitive to the applied electrical field.\n\n## Source and Drain", null, "Fig.2 The flows of spin and charge currents\n\nEven though, spin and charge are undivided features of an electron, there is a significant difference between flows of charge and spin in a solid. In contrast to the charge current, which always flows from a source to a drain, the flow of the spin current does not require a drain. After having been emitted from a source, the spin current propagates in material until it decays.\n\n## Spin and charge accumulations\n\nThere is a significant difference between a charge accumulation and a spin accumulation. The charge accumulation induces an electrical field, which affects in the long range the charge distribution, electrical potential and charge current. In contrast, the magnetic field induced by the spin accumulation is weak. In the long range it does not affect the spin or charge distributions.\n\n## Intrinsic and effective spin diffusion lengths\n\nThe intrinsic spin diffusion length", null, "is the spin diffusion length, which is used in spin/charge transport equations. The intrinsic spin diffusion length is proportional to the spin relaxation rate as", null, "where", null, "are the spin relaxation rate and the conductivity, respectively.\n\nThe effective spin diffusion length", null, "is proportional to the decay rate of spin current, the spin accumulation and the spin chemical potential along the flow direction of the spin current.", null, "In the case of non-magnetic metals", null, "In the case of ferromagnetic metals", null, "In the case of semiconductors the effective spin diffusion length depends on the charge current flowing parallel to the spin current. Therefore, the effective spin diffusion length can be controlled by the charge current .", null, "It should be noted that the spin drain affects the effective spin diffusion length.\n\nWhen the spin flow in one direction (There is no flow in opposite direction)", null, "## Spin and charge currents in different materials\n\nNon-magnetic metal\n\nThere is no interaction between the spin and charge currents.\n\nDrift current.", null, "Drift current does not induce a spin torque", null, "Diffusion current.", null, "Diffussion current does induce a spin torque", null, "ferromagnetic metal\n\nThere is no interaction between the dc spin and charge currents, but there is an interaction between the ac spin and charge currents, because of a charge accumulation along the flow direction of diffusion spin current.\n\nDrift current.\n\nDrift current does not induce a spin torque. It has a spin-current component.", null, "Diffusion current.\n\nDiffusion current induces a spin torque. There is a charge accumulation along flow of diffusive current.", null, "semiconductors\n\nThe conductivity of a semiconductor is spin-dependent only in the presence of a spin accumulation. There is a substantial interaction between the diffusion and drift currents in semiconductors.\n\nDrift current.\n\nThe properties of the drift current differ depending on whether the drift current is flowing in the presence of a diffusive spin current or not.\n\nWhen there is no diffusive current, the properties of the drift current is described by Eqn. (10). In the presence of a diffusive current, the properties of the drift current is described by Eqn. (12).\n\nDiffusion current.\n\nIt induces a spin torque. There is a charge accumulation along flow of diffusive current. The properties of the diffusion current are different depending whether the diffusion current flowing in presence of drift spin current or not. For example, the effective diffusion length depends on the magnitude of the drift current.\n\nThe diffusion current in semiconductor satisfies", null, "## Ohm's law for spin current\n\nIn the simplest case of spin/charge transport in non-magnetic metals, the charge and spin currents satisfy", null, "where", null, "are electrical potential, conductivity and voltage, respectively. Eqn. (9a) is the Ohm's law, which describes the relation between the electrical voltage and the charge current. Similarly, Eqn. (9b) could be considered as the Ohm's law for spin current. When the spin current flows only in one direction (There is no flow in opposite direction) , Eqn. (9b) is simplified to", null, "", null, "can be treated as spin voltage. Then, if we define spin resistance sR for non-magnetic metals as", null, "and the Ohm's law for spin current will be", null, "in the case of ferromagnetic metals the spin resistance sR will be", null, "In materials with low spin resistance (low resistance +short spin diffusion length) (example: ferromagnetic metal Fe, Co or metals:Pt), a small", null, "induces a large spin current. These materials have a strong spin-drain strength, but a small spin-detection efficiency. In contrast, in materials with high spin resistance (high resistance +long spin diffusion length), even when spin current is weak the", null, "is large. These materials have a weak spin-drain strength, but a significant spin-detection efficiency.\n\n## Spin-transfer torque\n\nSpin-transfer torque is proportional to the spin relaxation rate. It only occurs when there is a spin accumulation. Therefore, only a spin diffusive current induces the spin transfer torque. The drift current does not induce any spin-transfer torque. Even in case of ferromagnetic metals, when a drift current has a spin-current component, the drift current does not induce the spin-transfer torque.\n\nThe signs of the spin-transfer torque are different for material magnetizations aligned either parallel or anti-parallel to the spin polarization of diffusive spin current. In the case of ferromagnetic metals, which have a spontaneous magnetization, the spin-transfer torque may reverse the magnetization direction or may cause magnetization precession. In the case of semiconductors or non-magnetic metals, the spin-transfer torque may cause precession of the induced magnetization.\n\nThe spin transfer torque occurs because along the flow of spin diffusive current the number of electrons, which flip their spin in the direction parallel to the magnetization of material and in opposite direction, are different.\n\nLet us consider a case of a spin accumulation when the number of spin up electrons is above equilibrium and the number of spin-down electrons is below equilibrium. Along the flow of the diffusion spin current the number of flipped spin-up electrons is larger than the number of flipped spin-down electrons. Therefore, the spin accumulation decays along the flow direction. The spin-transfer torque is proportional to the decay rate of the spin accumulation. Therefore, it is proportional to the difference of the flipped spin-down and spin-up electrons:", null, "where DS is the density of states and", null, "is reduction of spin chemical potential per flow distance", null, ".\n\nSubstituting (5) into (15) gives", null, "Therefore, the spin transfer torque is linearly proportional to", null, ". The shorter the effective spin diffusive length is, the stronger the spin transfer torque becomes. It should be noticed that the strength of the spin-transfer torque in semiconductors can be modulated by a charge current (See Eqn. (8) )." ]

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[ "1. Home\n2. Mathematics & Statistics\n3. Developmental Math\n4. Intermediate Algebra\n5. Intermediate Algebra For College Students with Integrated Review", null, "", null, "# Intermediate Algebra For College Students with Integrated Review, 10th edition", null, "• Allen R. Angel\n• Dennis Runde\n\n10th edition\n\n## Overview\n\nNOTE: Before purchasing, check with your instructor to confirm the correct ISBN. Several versions of the MyLab™ and Mastering™ platforms exist for each title, and registrations are not transferable. To register for and use MyLab or Mastering, you may also need a Course ID, which your instructor will provide.\n\nUsed books, rentals, and purchases made outside of Pearson\nIf purchasing or renting from companies other than Pearson, the access codes for the MyLab platform may not be included, may be incorrect, or may be previously redeemed. Check with the seller before completing your purchase.\n\nFor courses in Intermediate Algebra.\n\nThis package includes MyLab Math.\n\nMaking math clear\n\nThe Angel/Runde author team has developed a text and MyLab™ program that uses short, concise sentences and clearly laid-out examples to ensure that all students develop a true understanding of algebra. In this revision, the authors retain this hallmark clarity of language to build the foundation of understanding algebra, while revamping exercise sets and creating new MyLab Math resources to give students a comprehensive learning and practice environment.\n\nAngel/Runde offers market-leading content written by author-educators, tightly integrated with the #1 choice in digital learning – MyLab Math. Bringing the authors’ voice and approach into the MyLab course gives students the clarity, motivation, and understanding they need to master algebra.\n\nReach every student by pairing this text with MyLab Math\n\nMyLab™ Math is the teaching and learning platform that empowers instructors to reach every student. By combining trusted authors’ content with digital tools and a flexible platform, MyLab personalizes the learning experience and improves results for each student.\n\n0135215978 / 9780135215975   Intermediate Algebra for College Students with Integrated Review and Worksheets Plus MyLab Math - Access Card Package, 10/e\n\nPackage consists of:\n\n• 0134758994 / 9780134758992  Intermediate Algebra for College Students\n• 0135224497 / 9780135224496 Student Workbook for Intermediate Algebra for College Students with Integrated Review\n• 0135231086 / 9780135231081  MyLab Math with Pearson eText - Standalone Access Card - for Intermediate Algebra for College Students with Integrated Review\n\nChapter 1: Basic Concepts\n\nIntegrated Review Topics:\n\n• Identify which numbers belong to a set\n• Subtraction of Real Numbers\n• Multiplication and Division of Real Numbers\n• Exponents, Parenthesis, and the Order of Operation\n\n1.1 Study Skills for Success in Mathematics, and Using a Calculator\n\n1.2 Sets and Other Basic Concepts\n\n1.3 Properties of and Operations with Real Numbers\n\n1.4 Order of Operations\n\nMid-Chapter Test: Sections 1.1–1.4\n\n1.5 Exponents\n\n1.6 Scientific Notation\n\nChapter 2. Equations and Inequalities\n\nIntegrated Review Topics:\n\n• Combining Like Terms\n• The Addition Property of Equality\n• The Multiplication Property of Equality\n• Solving Linear Equations with a Variable on Only Side of the Equation\n• Solving Linear Equations with a Variables on Both Sides of the Equation\n• Inequalities in One Variable\n\n2.1 Solving Linear Equations\n\n2.2 Problem Solving and Using Formulas\n\n2.3 Applications of Algebra\n\nMid-Chapter Test: Sections 2.1–2.3\n\n2.5 Solving Linear Inequalities\n\n2.6 Solving Equations and Inequalities Containing Absolute Values\n\nChapter 3. Graphs and Functions\n\nIntegrated Review Topics:\n\n• The Cartesian Coordinate System and Linear Equations in Two Variables\n• The Slope of a Line\n• Slope-Intercept Form and Point-Slope Forms of a Linear Equation\n• Graphing Linear Equations\n• Functions\n\n3.1 Graphs\n\n3.2 Functions\n\n3.3 Linear Functions: Graphs and Applications\n\n3.4 The Slope-Intercept Form of a Linear Equation\n\nMid-Chapter Test: Sections 3.1–3.4\n\n3.5 The Point-Slope Form of a Linear Equation\n\n3.6 The Algebra of Functions\n\n3.7 Graphing Linear Inequalities\n\nChapter 4. Systems of Equations and Inequalities\n\nIntegrated Review Topics:\n\n• Solving Systems of Equations Graphically\n• Solving Systems of Equations by Addition\n• Solving Systems of Equations by Substitution\n\n4.1 Solving Systems of Linear Equations in Two Variables\n\n4.2 Solving Systems of Linear Equations in Three Variables\n\n4.3 Systems of Linear Equations: Applications and Problem Solving\n\nMid-Chapter Test: Sections 4.1–4.3\n\n4.4 Solving Systems of Equations Using Matrices\n\n4.5 Solving Systems of Equations Using Determinants and Cramer’s Rule\n\n4.6 Solving Systems of Linear Inequalities\n\nChapter 5. Polynomials and Polynomial Functions\n\nIntegrated Review Topics:\n\n• Factoring Trinomials of the form ax2+bx+c, a=1\n• Factoring Trinomials of the form ax2+bx+c, a is not equal to 1\n• Special Factoring Formulas and a General Review of Factoring\n• Solving Quadratic Equations Using Factoring\n• Addition and Subtraction of Polynomials\n• Multiplication of Polynomials\n\n5.1 Addition and Subtraction of Polynomials\n\n5.2 Multiplication of Polynomials\n\n5.3 Division of Polynomials and Synthetic Division\n\n5.4 Factoring a Monomial from a Polynomial and Factoring by Grouping\n\nMid-Chapter Test: Sections 5.1–5.4\n\n5.5 Factoring Trinomials\n\n5.6 Special Factoring Formulas\n\n5.7 A General Review of Factoring\n\n5.8 Polynomial Equations\n\nChapter 6. Rational Expressions and Equations\n\nIntegrated Review Topics:\n\n• Fractions\n• Factoring Trinomials of the form ax2+bx+c, a=1\n• Factoring Trinomials of the form ax2+bx+c, a=1\n\n6.1 The Domains of Rational Functions and Multiplication and Division of Rational Expressions\n\n6.2 Addition and Subtraction of Rational Expressions\n\n6.3 Complex Fractions\n\n6.4 Solving Rational Equations\n\nMid-Chapter Test: Sections 6.1–6.4\n\n6.5 Rational Equations: Applications and Problem Solving\n\n6.6 Variation\n\nChapter 7. Roots, Radicals, and Complex Numbers\n\nIntegrated Review Topics:\n\n• Evaluating Square Roots\n• Simplifying Square Roots\n• Adding, Subtracting and Multiplying Square Roots\n• Higher Roots and Rational Exponents\n• Complex Numbers\n\n7.2 Rational Exponents\n\nMid-Chapter Test: Sections 7.1–7.4\n\n7.7 Complex Numbers\n\nIntegrated Review Topics:\n\n• Solve Quadratic Equations Using Factoring\n\n8.1 Solving Quadratic Equations by Completing the Square\n\n8.3 Quadratic Equations: Applications and Problem Solving\n\nMid-Chapter Test: Sections 8.1–8.3\n\n8.4 Writing Equations in Quadratic Form\n\n8.6 Quadratic and Other Inequalities in One Variable\n\nChapter 9. Exponential and Logarithmic Functions\n\nIntegrated Review Topics:\n\n• Exponents\n• Negative Exponents\n\n9.1 Composite and Inverse Functions\n\n9.2 Exponential Functions\n\n9.3 Logarithmic Functions\n\n9.4 Properties of Logarithms\n\nMid-Chapter Test: Sections 9.1–9.4\n\n9.5 Common Logarithms\n\n9.6 Exponential and Logarithmic Equations\n\n9.7 Natural Exponential and Natural Logarithmic Functions\n\nChapter 10. Conic Sections\n\nIntegrated Review Topics:\n\n• Solve Systems of equations by the Addition Method\n\n10.1 The Parabola and the Circle\n\n10.2 The Ellipse\n\nMid-Chapter Test: Sections 10.1–10.2\n\n10.3 The Hyperbola\n\n10.4 Nonlinear Systems of Equations and Their Applications\n\nChapter 11. Sequences, Series, and the Binomial Theorem\n\nIntegrated Review Topics:\n\n• Exponents, Parentheses, and Order of Operations\n• Multiplication of Polynomials\n\n11.1 Sequences and Series\n\n11.2 Arithmetic Sequences and Series\n\n11.3 Geometric Sequences and Series\n\nMid-Chapter Test: Sections 11.1–11.3\n\n11.4 The Binomial Theorem\n\nAppendix: Geometric Formulas\n\n## For teachers\n\nAll the material you need to teach your courses.\n\nDiscover teaching material" ]

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[ "Number of samples\n\nIt is important to note that the number of samples cannot be changed after calibration, otherwise the calibration is invalidated. The following steps are used to calculate the number of samples necessary:\n\n1. Calculate the wavelength. For a 5 MHz probe in shear mode in steel, the wavelength is approximately 0.65 mm\n2. Calculate the UT path. If the start is at 10 mm and the range is set at 30 mm then the UT path is 46.7 mm at 50°\n3. Calculate the number of samples. The UT path is divided by 15% of the wavelength. Hence, the approximate number of samples needed is 479\n4. If the calculated number of samples is bigger than the maximum number of samples permitted by the instrument, the UT path must be shrunk\n5. Some instruments offer an auto-set for the number of samples. It is, in general, a good idea to use that setting\n\nIt should be noted that the start and range change for every angle when in True Depth mode." ]

[ null ]

[ "# 5.4: A Molecular View of Elements and Compounds\n\nLearning Objectives\n• Classify substances as atomic elements, molecular elements, molecular compounds, or ionic compounds\n\n## Atomic Elements\n\nMost elements exist with individual atoms as their basic unit. It is assumed that there is only one atom in a formula if there is no numerical subscript on the right side of an element’s symbol.\n\n## Molecular Elements\n\nThere are many substances that exist as two or more atoms connected together so strongly that they behave as a single particle. These multi-atom combinations are called molecules. The smallest part of a substance that has the physical and chemical properties of that substance. A molecule is the smallest part of a substance that has the physical and chemical properties of that substance. In some respects, a molecule is similar to an atom. A molecule, however, is composed of more than one atom.\n\n Hydrogen, H Oxygen Nitrogen Fluorine Chlorine Bromine Iodine\n\nSome elements exist naturally as molecules. For example, hydrogen and oxygen exist as two-atom molecules. Other elements also exist naturally as diatomic molecules. A molecule with only two atoms (Table $$\\PageIndex{1}$$). As with any molecule, these elements are labeled with a molecular formula, a formal listing of what and how many atoms are in a molecule. (Sometimes only the word formula is used, and its meaning is inferred from the context.) For example, the molecular formula for elemental hydrogen is H2, with H being the symbol for hydrogen and the subscript 2 implying that there are two atoms of this element in the molecule. Other diatomic elements have similar formulas: O2, N2, and so forth. Other elements exist as molecules—for example, sulfur normally exists as an eight-atom molecule, S8, while phosphorus exists as a four-atom molecule, P4 (Figure $$\\PageIndex{1}$$).", null, "Figure $$\\PageIndex{1}$$: Molecular Art of S8 and P4 Molecules. If each green ball represents a sulfur atom, then the diagram on the left represents an S8 molecule. The molecule on the right shows that one form of elemental phosphorus exists, as a four-atom molecule.\n\nFigure $$\\PageIndex{1}$$ shows two examples of how we will be representing molecules in this text. An atom is represented by a small ball or sphere, which generally indicates where the nucleus is in the molecule. A cylindrical line connecting the balls represents the connection between the atoms that make this collection of atoms a molecule. This connection is called a chemical bond.\n\n## Ionic Compounds\n\nThe elements in the periodic table are divided into specific groupings; the metals, the non-metals, the semi-metals, and so on. These groupings are largely based on physical properties and on the tendency of the various elements to bond with other elements by forming either an ionic or a covalent bond. As a general rule of thumb, compounds that involve a metal binding with either a non-metal or a semi-metal will display ionic bonding. Thus, the compound formed from sodium and chlorine will be ionic (a metal and a non-metal). The basic unit of ionic compounds are called formula units.\n\n## Molecular Compounds\n\nCompounds that are composed of only non-metals or semi-metals with non-metals will display covalent bonding and will be classified as molecular compounds. Nitrogen monoxide (NO) will be a covalently bound molecule (two non-metals) and silicon dioxide (SiO2) will also be a covalently bound molecule (a semi-metal and a non-metal). The basic unit of molecular compounds are called molecules.\n\nExample $$\\PageIndex{1}$$\n\nProvide the classification (i.e. atomic element, molecular element, molecular compound, or ionic compound) of each substance.\n\n1. Fe\n2. PCl3\n3. LiBr\n4. P4\n5. oxygen gas\n\nSOLUTION\n\n1. Fe, iron is an element that is represented with no subscript, so it is an atomic element.\n2. PCl3 is made up of two nonmetals, so it is a molecular compound.\n3. LiBr is made up of lithium, a metal and bromine a nonmetal, so it is an ionic compound.\n4. P4 is a substance that is made up of four atoms of the same element, so it is a molecular element.\n5. The formula for oxygen gas is O2 so it is a molecular element.\nExercise $$\\PageIndex{1}$$\n\nProvide the classification (i.e. atomic element, molecular element, molecular compound, or ionic compound) of each substance.\n\n1. I2\n2. He\n3. H2O\n4. Al\n5. CuCl\nmolecular element\natomic element" ]

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[ "", null, "# The graph of a function f is shown. Which graph is an antiderivative of f? F", null, "philosphy111of 2021-11-18 Answered\nThe graph of a function f is shown. Which graph is an antiderivative of f?", null, "You can still ask an expert for help\n\n• Questions are typically answered in as fast as 30 minutes\n\nSolve your problem for the price of one coffee\n\n• Math expert for every subject\n• Pay only if we can solve it", null, "Coldst\nStep 1\nLet $g\\left(x\\right)$ be the antiderivative of $f\\left(x\\right)$. Then we must have\n$g\\left(x\\right)=\\int f\\left(x\\right)dx$\n$⇒{g}^{\\prime }\\left(x\\right)=f\\left(x\\right)$\nNow, since $f\\left(x\\right)$ is negative that means the ${g}^{\\prime }\\left(x\\right)$ is negative in this interval. So, g(x) (the antiderivative of $f\\left(x\\right)$ must be decreasing in this interval. This scenario is shown in two graphs a and b.\nAgain we notice that $f\\left(x\\right)$ is positive that means, ${g}^{\\prime }\\left(x\\right)$ is positive in . So, the function $g\\left(x\\right)=$ antiderivative of $f\\left(x\\right)$ must be increasing in this interval and from the graph we can see that the graph b is increasing after the point $x=p$.\nSo, the correct graph for the antiderivative of $f\\left(x\\right)$ is b.", null, "" ]

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[ "Contents\n\n# Contents\n\n## Idea\n\nA basic fact in number theory is that the natural numbers may be decomposed into the rational numbers and the p-adic integers for all prime numbers $p$. Dually in arithmetic geometry this says that Spec(Z) has a cover by all its formal disks and the complements of finitely many points, a fact that is crucial in the geometric interpretation of the function field analogy and which motivates for instance the geometric Langlands correspondence. (See below.)\n\nLifting this statement to stable homotopy theory and “higher arithmetic geometry” the arithmetic fracture theorem says that stable homotopy types (and suitably tame plain homotopy types) canonically decompose into their rationalization and their p-completion for all primes $p$, hence into their images in rational homotopy theory and p-adic homotopy theory. Since these images are typically simpler than the original homotopy type itself, this decomposition is a fundamental computational tool in stable homotopy theory, often known under the slogan of “working one prime at a time”. (See below.)\n\nOne finds that this arithmetic fracturing in stable homotopy theory is really a statement about the Bousfield localization of spectra with respect to the Moore spectrum for $\\mathbb{Q}$ and that of $\\mathbb{Q}/\\mathbb{Z}$. Viewed this way there is a more general fracture theorem which says that for any suitable pair $E,F$ of spectra/homology theories the Bousfield localization at their coproduct decomposes into the separate Bousfield localizations. This generalized fracture theorem appears for instance in chromatic homotopy theory for localization at Morava K-theory and Morava E-theory. (See below.)\n\nIn cohesive homotopy theory every stable homotopy type canonically sits in a fracture square formed from the localizations exhibited by the shape modality and the flat modality. For differential cohesion over infinitesimal cohesion this is a higher geometric analog of the classical arithmetic fracture. (See below.)\n\n## Statement\n\n### In number theory and arithmetic geometry\n\nThe statement in number theory/arithmetic geometry is the following:\n\n###### Proposition\n\nThe integers $\\mathbb{Z}$ are the fiber product of all the p-adic integers $\\underset{p\\;prime}{\\prod} \\mathbb{Z}_p$ with the rational numbers $\\mathbb{Q}$ over the rationalization of the former, hence there is a pullback diagram in CRing of the form\n\n$\\array{ && \\mathbb{Q} \\\\ & \\swarrow && \\nwarrow \\\\ \\mathbb{Q}\\otimes_{\\mathbb{Z}}\\left(\\underset{p\\;prime}{\\prod} \\mathbb{Z}_p \\right) && && \\mathbb{Z} \\\\ & \\nwarrow && \\swarrow \\\\ && \\underset{p\\;prime}{\\prod} \\mathbb{Z}_p } \\,.$\n\nEquivalently this is the fiber product of the rationals with the integral adeles $\\mathbb{A}_{\\mathbb{Z}}$ over the ring of adeles $\\mathbb{A}_{\\mathbb{Q}}$\n\n$\\array{ && \\mathbb{Q} \\\\ & \\swarrow && \\nwarrow \\\\ \\mathbb{A}_{\\mathbb{Q}} && && \\mathbb{Z} \\\\ & \\nwarrow && \\swarrow \\\\ && \\mathbb{A}_{\\mathbb{Z}} } \\,,$\n\nSince the ring of adeles is the rationalization of the integral adeles $\\mathbb{A}_{\\mathbb{Q}} = \\mathbb{Q} \\otimes_{\\mathbb{Z}} \\mathbb{A}_{\\mathbb{Z}}$, this is also (by the discussion here) a pushout diagram in CRing, and in fact in topological commutative rings (for $\\mathbb{Q}$ with the discrete topology and $\\mathbb{A}_{\\mathbb{Z}}$ with its profinite/completion topology).\n\nAn original discussion is (Sullivan 05, prop. 1.18). Review includes (Riehl 14, lemma 14.4.2).\n\n###### Remark\n\nUnder the function field analogy we may think of\n\n• $Spec(\\mathbb{Z})$ as an arithmetic curve over F1;\n\n• $\\mathbb{A}_{\\mathbb{Z}}$ as the ring of functions on the formal disks around all the points in this curve;\n\n• $\\mathbb{Q}$ as the ring of functions on the complement of a finite number of points in the curve;\n\n• $\\mathbb{A}_{\\mathbb{Q}}$ is the ring of functions on punctured formal disks around all points, at most finitely many of which do not extend to the unpunctured disk.\n\nUnder this analogy the arithmetic fracture square of prop. says that the curve $Spec(\\mathbb{Z})$ has a cover whose patches are the complement of the curve by some points, and the formal disks around these points.\n\nThis kind of cover plays a central role in number theory, see for instance the following discussions:\n\n### In homotopy theory\n\nIn homotopy theory the corresponding statement is that homotopy types may be decomposed into that of rational homotopy types and p-complete homotopy types of p-local homotopy types.\n\n###### Proposition\n\nLet $p$ be a prime number. Let $X$ be a homotopy type/∞-groupoid satisfying at least one of the following sufficient conditions\n\nThen $X$ is the homotopy fiber product\n\n$X \\simeq X_{\\mathbb{Q}} \\underset{(X_p^\\wedge)_{\\mathbb{Q}}}{\\times} X_p^\\wedge$\n\nof its rationalization $X_{\\mathbb{Q}}$ with its p-completion $X_p^\\wedge$ over the rationalization $(X_p^\\wedge)_{\\mathbb{Q}}$of the $p$-completions.\n\nThis originates around (Bousfield-Kan 72, VI.8.1). A detailed more modern account is in (May-Ponto, theorem 13.1.4). A quick survey is in (Riehl 14, theorem 14.4.14).\n\n### In stable homotopy theory\n\nSimilar statements hold in stable homotopy theory for spectra. There is a stable version of\n\nbut more generally there are fracture squares for the coproduct homology theory $E \\vee F$ whenever $F$-localization is $E$-acyclic:\n\nwarning: a condition missing in the following, see the comment section of this MO reply. Somebody should add the relevant clause here…\n\n#### The arithmetic fracture square for spectra\n\nFor $p$ a prime number write\n\n###### Proposition\n\n(Sullivan arithmetic square)\n\nFor every spectrum $X$ the canonical square\n\n$\\array{ && L_{\\mathbb{Q}}X \\\\ & \\swarrow && \\nwarrow \\\\ L_{\\mathbb{Q}} \\left( \\prod_p L_p X \\right) && && X \\\\ & \\nwarrow && \\swarrow \\\\ && \\prod_p L_p X }$\n\nis a homotopy pushout (hence also a homotopy pullback).\n\nOriginal statements of this include (Bousfield 79, Sullivan 05, prop. 3.20). Review includes (van Koughnett 13, prop. 4.5, Bauer 11, lemma 2.1).\n\n###### Remark\n\n(“one prime at a time”)\n\nThe impact of prop. is that it decomposes the study of (stable) homotopy theory into that of\n\n1. p-adic homotopy theory for each prime $p$.\n\nBoth the rationalization $X_{\\mathbb{Q}}$ and the p-completion $X_{p}^\\wedge$ are typically much easier to analyze than $p$ itself and so the fracture theorem gives a way to decompose the remaining hard part of study of homotopy types into that of $p$-local/$p$-complete spaces. This procedure is known in homotopy theory as working “one prime at a time”.\n\nMore generally:\n\n###### Proposition\n\nThe product of all p-completions is equivalently the Bousfield localization of spectra at the wedge sum $\\vee_p S \\mathbb{F}_p$ of all Moore spectra\n\n$\\prod_p L_p X \\simeq L_{\\vee_p S \\mathbb{F}_p} X \\,.$\n\nMoreover there is a Bousfield equivalence\n\n$S (\\mathbb{Q}/\\mathbb{Z}) \\simeq_{Bousf} \\vee_p S \\mathbb{F}_p \\,,$\n\nand therefore also an equivalence\n\n$\\prod_p L_p X \\simeq L_{S (\\mathbb{Q}/\\mathbb{Z})} X \\,.$\n\nThe first statement originates around (Bousfield 79, prop. 2.6), review includes (van Koughnett 13, prop. 4.4, Bauer 11, below prop. 2.2); the second is highlighted in (Strickland 12, MO comment).\n\n###### Remark\n\nBy the discussion at Bousfield localization of spectra and at localization of a space, the rationalization and the p-completion maps on spectra are homotopy cofibers of $E$-acyclifications $G_E(X) \\to X$, for $E = S \\mathbb{Q} \\simeq H \\mathbb{Q}$ and $E = S \\mathbb{F}_p$ the Moore spectra of $\\mathbb{Q}$ and of the cyclic group/finite field $\\mathbb{F}_p = \\mathbb{Z}/(p)$, respectively (e.g. Lurie 10, lecture 20).\n\nIncluding this into the statement of prop. says that for spectra $X$ satisfying sufficient conditions as above, then the canonical diagram\n\n$\\array{ && X_{\\mathbb{Q}} && \\longleftarrow && G_{S (\\mathbb{Q}/\\mathbb{Z})}(X) \\\\ & \\swarrow && \\nwarrow && \\swarrow \\\\ (\\prod_p X_p^\\wedge)_{\\mathbb{Q}} && && X \\\\ & \\nwarrow && \\swarrow && \\nwarrow \\\\ && \\prod_p X_p^\\wedge && \\longleftarrow && G_{H\\mathbb{Q}}(X) }$\n\nis homotopy exact, in that\n\n1. the square is a homotopy pullback and hence also a homotopy pushout (this is prop. );\n\n2. the diagonals are homotopy cofiber sequences and hence also homotopy fiber sequences (by this proposition at Bousfield localization of spectra);\n\n3. the top and bottom outer composite sequences are homotopy fiber sequences (and hence homotopy cofiber sequences) (by applying the pasting law to the previous two items).\n\nTogether this is like two-thirds of a differential cohomology hexagon – to the extent that $p$-adic completion is adjoint to $p$-torsion approximation. This is indeed the case, as the next proposition asserts\n\nNotice that in view of remark then $X_p^\\wedge$ is like the restriction of $X$ from Spec(Z) to all formal disks around the points $(p)$, and hence $G_{S\\mathbb{F}_p}$ is like the restriction to the “complement of all formal disks”. Finally $X_{\\mathbb{Q}}$ may be understood as the restriction to the Ran space of $Spec(\\mathbb{Z})$ (Gaitsgory 11), roughly the colimit of the restriction of $X$ to the complement of finitely many points, as this set of points ranges through all points.\n\nIn view of remark we may regard the following fact as an refinement of the traditional arithmetic fracture theorem.\n\n###### Proposition\n\nLet $A$ be an E-∞ ring and let $\\mathfrak{a} \\subset \\pi_0 A$ be a finitely generated ideal in its underlying commutative ring.\n\n$\\array{ \\underoverset{ A Mod_{\\mathfrak{a}comp}^{op}} {A Mod_{\\mathfrak{a}tors}^{op}} {\\simeq} &\\stackrel{\\overset{\\Pi_{\\mathfrak{a}}}{\\longleftarrow}}{\\stackrel{\\hookrightarrow}{\\underset{\\flat_{\\mathfrak{a}}}{\\longleftarrow}}}& A Mod^{op} }$\n\nwhere\n\n• $A Mod$ is the stable (∞,1)-category of modules, i.e. of ∞-modules over $A$;\n\n• $A Mod_{\\mathfrak{a}tors}$ and $A Mod_{\\mathfrak{a} comp}$ are the full sub-(∞,1)-categories of $\\mathfrak{a}$-torsion and of $\\mathfrak{a}$-complete $A$-∞-modules, respectively;\n\n• $(-)^{op}$ denotes the opposite (∞,1)-category;\n\n• the equivalence of (∞,1)-categories on the left is induced by the restriction of $\\Pi_{\\mathfrak{a}}$ and $\\flat_{\\mathfrak{a}}$.\n\n###### Proof\n\nThis is effectively the content of (Lurie “Proper morphisms”, section 4):\n\n• the existence of $\\Pi_{\\mathfrak{a}}$ is corollary 4.1.16 and remark 4.1.17\n\n• the existence of $\\flat_{\\mathfrak{a}}$ is lemma 4.2.2 there;\n\n• the equivalence of sub-$\\infty$-categories is proposition 4.2.5 there.\n\n###### Corollary\n\nThe traditional arithmetic fracture square of prop. , regarded as in remark , is the left part of the “differential cohomology diagram” induced by the adjoint modality $(\\Pi_{\\mathfrak{a}} \\dashv \\flat_{\\mathfrak{a}} )$ of prop. , for the special case that $X = \\mathbb{S}$ is the sphere spectrum and $\\mathfrak{a} = (p)$ a prime ideal\n\n$\\array{ && \\Pi_{\\mathfrak{a}dR} X && \\stackrel{\\mathbf{d}}{\\longrightarrow} && \\flat_{\\mathfrak{a}dR} X \\\\ & \\nearrow & & \\searrow & & \\nearrow && \\searrow \\\\ \\Pi_{\\mathfrak{a}dR} \\flat X && \\Downarrow && X && \\Downarrow && \\Pi_{\\mathfrak{a}} \\flat_{\\mathfrak{a}dR} X \\\\ & \\searrow & & \\nearrow & & \\searrow && \\nearrow \\\\ && \\flat_{\\mathfrak{a}} X && \\longrightarrow && \\Pi_{\\mathfrak{a}} X } \\,,$\ncohesion modalitysymbolinterpretation\nflat modality$\\flat$formal completion at\nshape modality$ʃ$torsion approximation\ndR-shape modality$ʃ_{dR}$localization away\ndR-flat modality$\\flat_{dR}$adic residual\n$\\array{ && localization\\;away\\;from\\;\\mathfrak{a} && \\stackrel{}{\\longrightarrow} && \\mathfrak{a}\\;adic\\;residual \\\\ & \\nearrow & & \\searrow & & \\nearrow && \\searrow \\\\ \\Pi_{\\mathfrak{a}dR} \\flat_{\\mathfrak{a}} X && && X && && \\Pi_{\\mathfrak{a}} \\flat_{\\mathfrak{a}dR} X \\\\ & \\searrow & & \\nearrow & & \\searrow && \\nearrow \\\\ && formal\\;completion\\;at\\;\\mathfrak{a}\\; && \\longrightarrow && \\mathfrak{a}\\;torsion\\;approximation } \\,,$\n###### Remark\n\nThe special case of prop. where $A$ is the Eilenberg-MacLane spectrum of a plain commutative ring, and hence – by the stable Dold-Kan correspondence – the case where $A$-∞-modules are equivalently chain complexes, has a longer tradition in the existing literature. This we highlight separately below.\n\n###### Remark\n\nBy the discussion at completion of modules in the section Monoidalness, the adjoint modality of prop. is a monoidal (∞,1)-functor without, possibly, respect the tensor unit in $A Mod$. This means that $(\\Pi_{\\mathfrak{a}}\\dashv \\flat_{\\mathfrak{a}})$ passes to “commutative ∞-monoids-without unit” in $A Mod$, hence to (formal duals of) nonunital E-∞ algebras. By this proposition (Lurie “Algebra”, prop. 5.2.3.15) nonunital E-∞ rings are equivalent to augmented E-∞ rings over the sphere spectrum, hence this is E-∞ arithmetic geometry under $Spec(\\mathbb{S})$.\n\nNotice that in addition $\\Pi_{\\mathfrak{a}}$ here should preserve finite products (because by the discussion at completion of a module – monoidalness the underlying $\\Pi_{\\mathfrak{a}} \\colon A Mod \\to A Mod$ preserves all small (∞,1)-colimits and because by this proposition finite coproducts in $CRng(A Mod)$ are computed in the underlying $A Mod$.\n\nTherefore we may think of $\\Pi_{\\mathfrak{a}}$ as a shape modality and of $\\flat_{\\mathfrak{a}}$ as a sharp modality on affine E-∞-arithmetic geometry under $Spec(\\mathbb{S})$ – namely on formal duals of nonunital E-∞ rings .\n\n(It may be entertaining to note that on the level of ∞-groups of units then E-∞ arithmetic geometry under $Spec(\\mathbb{S})$ translates to abelian ∞-groups of twists over the sphere spectrum – which has been argued to be the homotopy-theoretic incarnation of superalgebra, see at superalgebra – abstract idea for more on this.)\n\nFor more discussion of this see also differential cohesion and idelic structure.\n\n#### The arithmetic fracture square for chain complexes\n\nWe discuss here arithmetic fracturing on chain complexes of modules. Under the stable Dold-Kan correspondence this is a special case of prop. , but this special case has a longer tradition in the literature – going back to Grothendieck local duality and sometimes called Greenlees-May duality due to (Greenlees-May 92) – and we point to these original proofs.\n\n###### Definition\n\nLet $A$ be a commutative ring, let $\\mathfrak{a} \\subset A$ be be an ideal generated by a single regular element (i.e. not a zero divisor). Write $A Mod_{\\infty}^{op}$ for the opposite (∞,1)-category of the (∞,1)-category of modules over $A$.\n\nWrite\n\n• $\\flat_{\\mathfrak{a}}\\colon A Mod_\\infty^{op} \\to A Mod_{\\infty}^{op}$ for the derived functor of formal completion (adic completion) of modules at $\\mathfrak{a}$;\n\nwith canonical natural transformation\n\n$\\epsilon_{\\mathfrak{a}} \\colon \\flat_{\\mathfrak{a}} \\longrightarrow id$\n\n• $\\Pi_{\\mathfrak{a}} \\colon A Mod_\\infty^{op} \\to A Mod_\\infty^{op}$ for the total derived functor of the $\\mathfrak{a}$-torsion approximation-functor;\n\nwith canonical natural transformation\n\n$\\eta_{\\mathfrak{a}}\\colon id \\longrightarrow \\Pi_{\\mathfrak{a}}$\n\nFinally write\n\n$(A Mod_\\infty^{op})^{\\mathfrak{a}com}, (A Mod_\\infty^{op})^{\\mathfrak{a}tor} \\hookrightarrow A Mod_\\infty$\n\nfor the full (∞,1)-subcategories of objects $X$ for which, $\\epsilon_{\\mathfrak{a}}(X)$ or $\\eta_{\\mathfrak{a}}(X)$ is an equivalence in an (∞,1)-category, respectively.\n\n###### Proposition\n1. The transformation $\\epsilon_{\\mathfrak{a}}$ exhibits $(A Mod_\\infty^{op})^{\\mathfrak{a}com}\\hookrightarrow A Mod_\\infty$ as a reflective (∞,1)-subcategory, hence $\\flat_{\\mathfrak{a}}$ as an idempotent (∞,1)-monad.\n\n2. The transformation $\\eta_{\\mathfrak{a}}$ exhibits $(A Mod_\\infty^{op})^{\\mathfrak{a}tor}\\hookrightarrow A Mod_\\infty$ as a co-reflective $(\\infty,1)$-category, hence $\\Pi_{\\mathfrak{a}}$ as an idempotent $(\\infty,1)$-comonad.\n\n3. Restricted to these sub-$(\\infty,1)$-categories both $\\flat_{\\mathfrak{a}}$ as well as $\\Pi_{\\mathfrak{a}}$ become equivalences of (∞,1)-categories, hence exhibiting $(\\Pi_{\\mathfrak{a}} \\dashv \\flat_{\\mathfrak{a}})$ as a higher adjoint modality.\n\n###### Proof\n\nThis is a paraphrase of the results in (Dwyer-Greenlees 99) and (Porta-Shaul-Yekutieli 10) from the language of derived categories to (∞,1)-category theory.\n\nFirst of all, by our simplifying assumption that $\\mathfrak{a}$ is generated by a single regular element, the running assumption of “weak proregularity” in (Porta-Shaul-Yekutieli 10, def.3.21) is satisfied.\n\nThen in view of (Porta-Shaul-Yekutieli 10, corollary 3.31) the statement of (Porta-Shaul-Yekutieli 10, theorem 6.12) is the characterization of reflectors-category#CharacterizationOfReflectors) as discussed at reflective sub-(∞,1)-category, and formally dually so for the coreflection. With the fully faithfulness that goes with this the equivalence of the two inclusions on the level of homotopy categories given by (Hovey-PalieriS-trickland 97, 3.3.5, Dwyer-Greenlees 99, theorem 2.1 Porta-Shaul-Yekutieli 10, theorem 6.11) implies the canonical equivalence of the two sub-(∞,1)-categories and this means that $\\Pi_{\\mathfrak{a}}$ and $\\flat_{\\mathfrak{a}}$ are the adjoint pair induced from the reflection/coreflection adjoint triple.\n\nThis adjoint triple is stated more explicitly in (Dwyer-Greenlees 99, section 4), see also (Porta-Shaul-Yekutieli 10, end of remark 6.14).\n\nTherefore arithmetic fracture squares in the homotopy theory of chain complexes are induced by this as in corollary above.\n\n#### General fracture squares of spectra\n\nBy prop. the arithmetic fracture square of prop. is equivalently of the form\n\n$\\array{ && L_{H\\mathbb{Q}}X \\\\ & \\swarrow && \\nwarrow \\\\ L_{H\\mathbb{Q}} L_{S \\mathbb{Q}/\\mathbb{Z}} X && && X \\\\ & \\nwarrow && \\swarrow \\\\ && L_{S \\mathbb{Q}/\\mathbb{Z}} X } \\,.$\n\nIn this form the statement holds much more generally:\n\n###### Proposition\n\nLet $E, F, X$ be spectra such that the $F$-localization of $X$ is $E$-acyclic, i.e. $E_\\bullet(L_F X) \\simeq 0$, then the canonical square diagram\n\n$\\array{ && L_F X \\\\ & \\swarrow && \\nwarrow \\\\ L_F L_E X && && L_{E\\vee F} X \\\\ & \\nwarrow && \\swarrow \\\\ && L_E X }$\n\nis a homotopy pullback (and hence by stability also a homotopy pushout).\n\ne.g. (Bauer 11, prop. 2.2)\n\n###### Remark\n\nThe general version of the fracture statement in prop. is used frequently in chromatic homotopy theory for decomposition in Morava K-theory and Morava E-theory-localizations. For example there is a chromatic fracture square:\n\n$\\array{ && L_{E(n-1)} X \\\\ & \\swarrow && \\nwarrow \\\\ L_{E(n-1)} L_{K(n)} X && && L_{E(n)} X \\\\ & \\nwarrow && \\swarrow \\\\ && L_{K(n)} X }$\n\nIn particular it is used for instance in the construction of tmf, see example below.\n\nFrom another perspective:\n\n###### Claim\n\nSuppose that $L \\colon Spectra \\to Spectra$ is a smashing localization given by smash product with some spectrum $T$. Write $F$ for the homotopy fiber\n\n$F \\longrightarrow \\mathbb{S} \\longrightarrow T \\,.$\n\nThen there is a fracture diagram of operations\n\n$\\array{ T \\wedge (-) && \\longleftarrow && [T,-] \\\\ & \\nwarrow && \\swarrow \\\\ && \\mathbb{S} \\\\ & \\swarrow & & \\nwarrow \\\\ [F,-] && \\longleftarrow && F \\wedge (-) }$\n\nwhere $[F,-]$ and $T \\wedge (-) \\colon Spectra \\to Spectra$ are idempotent (∞,1)-monads and $[T,-]$, $F \\wedge (-)$ are idempotent $\\infty$-comonads, the diagonals are homotopy fiber sequences.\n\n###### Example\n\nFor $T = S \\mathbb{Z}[p^{-1}]$ the Moore spectrum of the integers localized away from $p$, then\n\n$F = \\Sigma^{-1} S (\\mathbb{Z}[p^{-1}]/\\mathbb{Z}) \\to \\mathbb{S} \\to S \\mathbb{Z}[p^{-1}]$\n\nand hence\n\n• $\\Sigma^{-1} S (\\mathbb{Z}[p^{-1}]/\\mathbb{Z}) \\wedge (-)$ is $p$-torsion approximation;\n\n• $[\\Sigma^{-1} S (\\mathbb{Z}[p^{-1}]/\\mathbb{Z}),-]$ is $p$-completion;\n\n• $S \\mathbb{Z}[p^{-1}] \\wedge (-)$ is $p$-rationalization;\n\n• $[T,-]$ is forming $p$-adic residual.\n\n$\\array{ && localization\\;away\\;from\\;\\mathfrak{a} && \\stackrel{}{\\longrightarrow} && \\mathfrak{a}\\;adic\\;residual \\\\ & \\nearrow & & \\searrow & & \\nearrow && \\searrow \\\\ && && X && && \\\\ & \\searrow & & \\nearrow & & \\searrow && \\nearrow \\\\ && formal\\;completion\\;at\\;\\mathfrak{a}\\; && \\longrightarrow && \\mathfrak{a}\\;torsion\\;approximation } \\,,$\n\nWith (Bousfield 79, prop.2.5)\n\n#### For $E_\\infty$-modules\n\n###### Proposition\n\nLet $A$ be an E-∞ ring and let $\\mathfrak{a} \\subset \\pi_0 A$ be a finitely generated ideal in its underlying commutative ring.\n\n$\\array{ \\underoverset{ A Mod_{\\mathfrak{a}comp}^{op}} {A Mod_{\\mathfrak{a}tors}^{op}} {\\simeq} &\\stackrel{\\overset{\\Pi_{\\mathfrak{a}}}{\\longleftarrow}}{\\stackrel{\\hookrightarrow}{\\underset{\\flat_{\\mathfrak{a}}}{\\longleftarrow}}}& A Mod^{op} }$\n\nwhere\n\n• $A Mod$ is the stable (∞,1)-category of modules, i.e. of ∞-modules over $A$;\n\n• $A Mod_{\\mathfrak{a}tor}$ and $A Mod_{\\mathfrak{a} comp}$ are the full sub-(∞,1)-categories of $\\mathfrak{a}$-torsion and of $\\mathfrak{a}$-complete $A$-∞-modules, respectively;\n\n• $(-)^{op}$ denotes the opposite (∞,1)-category;\n\n• the equivalence of (∞,1)-categories on the left is induced by the restriction of $\\Pi_{\\mathfrak{a}}$ and $\\flat_{\\mathfrak{a}}$.\n\nThis is effectively the content of (Lurie “Completions”, section 4):\n\n### In cohesive (stable) homotopy theory\n\nIn cohesive homotopy theory every stable homotopy type $X$ sits in a fracture square of the form\n\n$\\array{ && \\Pi_{dR} X && \\longrightarrow && \\flat_{dR} X \\\\ & \\nearrow & & \\searrow && \\nearrow \\\\ \\Pi_{dR} \\flat X && && X \\\\ & \\searrow & & \\nearrow && \\searrow \\\\ && \\flat X && \\longrightarrow && \\Pi X }$\n\nwhere $\\flat$ is the flat modality and $\\Pi_{dR}$ the homotopy fiber of the unit $X\\to \\Pi X$ of the shape modality. This is the left part of the differential cohomology hexagon for $X$, see there for details.\n\nHere $\\Pi_{dR} X$ is such that for any other stable cohesive homotopy type $\\hat E$ then functions $\\Pi_{dR} X \\to \\hat E$ are equivalent to functions $X \\to \\flat_{dR} \\hat E$, where $\\hat E \\to \\flat_{dR} \\hat E$ is a generalized form of rationalization in the sense discussed at differential cohomology hexagon. In particular if $\\hat E$ is a Hopkins-Singer-type differential cohomology refinement of a plain spectrum $E$, then $E\\to \\flat_{dR} E$ is its ordinary rationalization given by the Chern character and $\\hat E \\to \\flat_{dR} \\hat E$ is the corresponding map on Chern curvature forms.\n\nMoreover, if the ambient cohesion is differential cohesion over a base of infinitesimal cohesion, then the flat modality $\\flat$ takes any space $X$ to the union of all its formal disks. (See at differential cohesion and idelic structure.) Accordingly the collection of functions $\\flat X \\to \\hat E$ in this case behave like the product of all formal power series of $\\hat E$-valued functions around all global points of $X$, analogous to remark .\n\nAn example of this are synthetic differential ∞-groupoids regarded as cohesive over their formal moduli problems, as its its complex analytic incarnation by synthetic differential complex analytic ∞-groupoids. In this context if $X = \\Sigma$ is a complex curve then $\\flat \\Sigma$ is precisely the analog of the integral adeles as it is predicted by the function field analogy.\n\n## Examples\n\n###### Example\n\nThe construction of the tmf-spectrum – the spectrum of global sections of the derived Deligne-Mumford stack of derived elliptic curves – as described in (Behrens 13) proceeds by first applying the arithmetic fracture square of prop. , prop. to decompose the moduli stack of elliptic curves into rational and $p$-adic curves, and then in a second step in applying in turn the general fracture square of prop. for Morava K-theory to the remaining $p$-adic pieces.\n\nSee at tmf – Decomposition via arithmetic fracture squares for more on this.\n\nRelated MO-discussion:\n\nDiscussion of rational functions as functions on the Ran space is in\n\nDiscussion of $\\mathfrak{a}$-adic completion and $\\mathfrak{a}$-torsion approximation as derived idempotent (co-)monads on a derived category of chain complexes of modules – Greenlees-May duality – is in\n\nbuilding on\n\nDiscussion of this in stable homotopy theory and the full generality of higher algebra is in\n\nAnd in the context of commutative DG-rings in\n\nThis and further generalization is in\n\nDiscussion in homotopy type theory:\n\n• Luis Scoccola, Nilpotent Types and Fracture Squares in Homotopy Type Theory (arXiv:1903.03245)\n\nExposition from the point of view of modal homotopy theory:\n\nFormalization in cohesive modal homotopy type theory homotopy type theory:" ]

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Your NRIC/Passport number may not be eligible because you have not fulfilled the 60 days cooling-off period OR your account was not in ACTIVE status when the report was generated.\n• ACTIVE status means that your Tune Talk credit validity has not expired.\n• Eligibility to the PA Insurance coverage is subject to accuracy of subscriber's personal information as per his/her Mykad/Passport/Police ID/Army ID entered at point of registration.\n• Tune Talk reserves the right to amend the terms and conditions without prior notice.\n\n## POLICY WORDING, NOMINATION & CLAIM FORM\n\nPA Claim Form\n(for accidents on/\nafter 1 August 2018)\n\nResponse time: 0.62092208862305\n1. select * from `rainlab_translate_attributes` where `locale` = ? and `model_id` = ? and `model_type` = ? limit 1\n2. select * from `rainlab_translate_attributes` where `locale` = ? and `model_id` = ? and `model_type` = ? limit 1\n3. select * from `rainlab_translate_attributes` where `locale` = ? and `model_id` = ? and `model_type` = ? 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[ null, "https://www.facebook.com/tr", null ]

[ "# JavaScript Date Object\n\nDescribes the JavaScript Date Object including properties, constructors, and methods.\n\n### Properties\n\n• prototype - For creating more properties.\n\n### Constructors\n\n• Date() - Use the current date and time to create an instance of the object date.\n• Date(dateString) - Use the date specified by the string to create the instance of the date object. String format is \"month day, year hours:minutes:seconds\".\n• Date(year, month, day) - Create an instance of date with the specified values. Year is 0 to 99.\n• Date(year, month, day, hours, minutes, seconds) - Create an instance of date with the specified values.\n\n### Methods\n\n• getDate() - Get the day of the month. It is returned as a value between 1 and 31.\n```var curdate = new Date()\nvar mday = curdate.getDate()\ndocument.write(mday + \"<BR>\")\n```\n\nThe above code prints the day of the month.\n\n• getDay() - Get the day of the week as a value from 0 to 6\n```var curdate = new Date()\nvar wday = curdate.getDate()\ndocument.write(wday + \"<BR>\")\n```\n\nThe above code prints the day of the week.\n\n• getHours() - The value returned is 0 through 23.\n```var curdate = new Date()\nvar hours = curdate.getHours()\ndocument.write(hours + \"<BR>\")\n```\n\nThe above code prints the hours since midnight.\n\n• getMinutes() - The value returned is 0 through 59.\n```var curdate = new Date()\nvar minutes = curdate.getMinutes()\ndocument.write(minutes + \"<BR>\")\n```\n\nThe above code prints the minutes past the hour.\n\n• getMonth() - Returns the month from the date object as a value from 0 through 11.\n```var curdate = new Date()\nvar month = curdate.getMonth()\ndocument.write(month + \"<BR>\")\n```\n\nThe above code prints the numeric value of the month.\n\n• getSeconds() - The value returned is 0 through 59.\n```var curdate = new Date()\nvar seconds = curdate.getSeconds()\ndocument.write(seconds + \"<BR>\")\n```\n\nThe above code prints the seconds since the last minute.\n\n• getTime() - The number of milliseconds since January 1, 1970. this function allows you to manipulate the date object based on a millisecond value then convert it back to the form you want. In the example below, it is used to set a future expiration time of a cookie.\n```var futdate = new Date()\nvar expdate = futdate.getTime()\nexpdate += 3600*1000 //expires in 1 hour(milliseconds)\nfutdate.setTime(expdate)\n```\n• getTimeZoneOffset() - Time zone offset in hours which is the difference between GMT and local time.\n```var curdate = new Date()\nvar offset = curdate.getTimeZoneOffset()\ndocument.write(offset + \"<BR>\")\n```\n\nThe above code prints the number of hours different between your timezone and GMT. This value may change with daylight savings time..\n\n• getYear() - Returns the numeric four digit value of the year.\n```var curdate = new Date()\nvar year = curdate.getYear()\ndocument.write(year + \"<BR>\")\n```\n\nThe above code prints the numeric value of the year which is currently 2000.\n\n• parse() - The number of milliseconds after midnight January 1, 1970 till the given date espressed as a string in the example which is IETF format.\n```var curdate = \"Wed, 18 Oct 2000 13:00:00 EST\"\nvar dt = Date.parse(curdate)\ndocument.write(dt + \"<BR>\")\n```\n• setDate(value) - Set the day of the month in the date object as a value from 1 to 31.\n• setHours(value) - Set the hours in the date object with a value of 0 through 59.\n• setMinutes(value) - Set the minutes in the date object with a value of 0 through 59.\n• setMonth(value) - Set the month in the date object as a value of 0 through 11.\n• setSeconds(value) - Set the seconds in the date object with a value of 0 through 59.\n• setTime(value) - Sets time on the basis of number of milliseconds since January 1, 1970. The below example sets the date object to one hour in the future.\n```var futdate = new Date()\nvar expdate = futdate.getTime()\nexpdate += 3600*1000 //expires in 1 hour(milliseconds)\nfutdate.setTime(expdate)\n```\n• setYear(value) - Set the year in the date instance as a 4 digit numeric value.\n• toGMTString() - Convert date to GMT format in a form similar to \"Fri, 29 Sep 2000 06:23:54 GMT\".\n```var curdate = new Date()\ndstring = curdate.toGMTString()\ndocument.write(dstring + \"<BR>\" + curdate.toLocaleString() + \"<BR>\")\n```\n\nThe above example produces:\n\n```Wed, 18 Oct 2000 18:08:11 UTC\n10/18/2000 14:08:11\n```\n• toLocaleString() - Convert date to local time zone format. See the example, above.\n• UTC() - Based on a comma delimited string, the number of milliseconds after midnight January 1, 1970 GMT is returned. The syntax of the string is \"year, month, day [, hrs] [, min] [, sec]\". An example is \"2000, 9, 29, 5, 43, 0\" for Sept 29, 2000 at 5:43:0. The string is considered to be GMT. The hours, minutes, and seconds are optional.\n```document.write(Date.UTC(2000, 9, 29, 5, 43, 0) + \"\")\n```\n\nThe above example produces:\n\n```972798180000\n```" ]

[ null ]

[ "# 6.3: Finding Probabilities for the Normal Distribution\n\n$$\\newcommand{\\vecs}{\\overset { \\rightharpoonup} {\\mathbf{#1}} }$$ $$\\newcommand{\\vecd}{\\overset{-\\!-\\!\\rightharpoonup}{\\vphantom{a}\\smash {#1}}}$$$$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$$$\\newcommand{\\AA}{\\unicode[.8,0]{x212B}}$$\n\nThe Empirical Rule is just an approximation and only works for certain values. What if you want to find the probability for x values that are not integer multiples of the standard deviation? The probability is the area under the curve. To find areas under the curve, you need calculus. Before technology, you needed to convert every x value to a standardized number, called the z-score or z-value or simply just z. The z-score is a measure of how many standard deviations an x value is from the mean. To convert from a normally distributed x value to a z-score, you use the following formula.\n\nDefinition $$\\PageIndex{1}$$: z-score\n\n$z=\\dfrac{x-\\mu}{\\sigma} \\label{z-score}$\n\nwhere $$\\mu$$= mean of the population of the x value and $$\\sigma$$= standard deviation for the population of the x value\n\nThe z-score is normally distributed, with a mean of 0 and a standard deviation of 1. It is known as the standard normal curve. Once you have the z-score, you can look up the z-score in the standard normal distribution table.\n\nDefinition $$\\PageIndex{2}$$: standard normal distribution\n\nThe standard normal distribution, z, has a mean of $$\\mu =0$$ and a standard deviation of $$\\sigma =1$$.", null, "Figure $$\\PageIndex{1}$$: Standard Normal Curve\n\nLuckily, these days technology can find probabilities for you without converting to the zscore and looking the probabilities up in a table. There are many programs available that will calculate the probability for a normal curve including Excel and the TI-83/84. There are also online sites available. The following examples show how to do the calculation on the TI-83/84 and with R. The command on the TI-83/84 is in the DISTR menu and is normalcdf(. You then type in the lower limit, upper limit, mean, standard deviation in that order and including the commas. The command on R to find the area to the left is pnorm(z-value or x-value, mean, standard deviation).\n\nExample $$\\PageIndex{1}$$ general normal distribution\n\nThe length of a human pregnancy is normally distributed with a mean of 272 days with a standard deviation of 9 days (Bhat & Kushtagi, 2006).\n\n1. State the random variable.\n2. Find the probability of a pregnancy lasting more than 280 days.\n3. Find the probability of a pregnancy lasting less than 250 days.\n4. Find the probability that a pregnancy lasts between 265 and 280 days.\n5. Find the length of pregnancy that 10% of all pregnancies last less than.\n6. Suppose you meet a woman who says that she was pregnant for less than 250 days. Would this be unusual and what might you think?\n\nSolution\n\na. x = length of a human pregnancy\n\nb. First translate the statement into a mathematical statement.\n\nP (x>280)\n\nNow, draw a picture. Remember the center of this normal curve is 272.", null, "Figure for Example $$\\PageIndex{1}$$b\n\nTo find the probability on the TI-83/84, looking at the picture you realize the lower limit is 280. The upper limit is infinity. The calculator doesn’t have infinity on it, so you need to put in a really big number. Some people like to put in 1000, but if you are working with numbers that are bigger than 1000, then you would have to remember to change the upper limit. The safest number to use is $$1 \\times 10^{99}$$, which you put in the calculator as 1E99 (where E is the EE button on the calculator). The command looks like:\n\n$$\\text{normalcdf}(280,1 E 99,272,9)$$", null, "Figure $$\\PageIndex{3}$$: TI-83/84 Output for Example $$\\PageIndex{1}$$b\n\nTo find the probability on R, R always gives the probability to the left of the value. The total area under the curve is 1, so if you want the area to the right, then you find the area to the left and subtract from 1. The command looks like:\n\n$$1-\\text { pnom }(280,272,9)$$\n\nThus, $$P(x>280) \\approx 0.187$$\n\nThus 18.7% of all pregnancies last more than 280 days. This is not unusual since the probability is greater than 5%.\n\nc. First translate the statement into a mathematical statement.\n\nP (x<250)\n\nNow, draw a picture. Remember the center of this normal curve is 272.", null, "Figure for Example $$\\PageIndex{1}$$c\n\nTo find the probability on the TI-83/84, looking at the picture, though it is hard to see in this case, the lower limit is negative infinity. Again, the calculator doesn’t have this on it, put in a really small number, such as $$-1 \\times 10^{99}=-1 E 99$$ on the calculator.", null, "Figure $$\\PageIndex{5}$$: TI-83/84 Output for Example $$\\PageIndex{1}$$c\n\n$$P(x<250)=\\text { normalcdf }(-1 E 99,250,272,9)=0.0073$$\n\nTo find the probability on R, R always gives the probability to the left of the value. Looking at the figure, you can see the area you want is to the left. The command looks like:\n\n$$P(x<250)=\\text { pnorm }(250,272,9)=0.0073$$\n\nThus 0.73% of all pregnancies last less than 250 days. This is unusual since the probability is less than 5%.\n\nd. First translate the statement into a mathematical statement.\n\nP (265<x<280)\n\nNow, draw a picture. Remember the center of this normal curve is 272.", null, "Figure for Example $$\\PageIndex{1}$$d\n\nIn this case, the lower limit is 265 and the upper limit is 280.\n\nUsing the calculator", null, "Figure $$\\PageIndex{7}$$: TI-83/84 Output for Example $$\\PageIndex{1}$$d\n\n$$P(265<x<280)=\\text { normalcdf }(265,280,272,9)=0.595$$\n\nTo use R, you have to remember that R gives you the area to the left. So $$P(x<280)=\\text { pnom }(280,272,9)$$ is the area to the left of 280 and $$P(x<265)=\\text { pnom }(265,272,9)$$ is the area to the left of 265. So the area is between the two would be the bigger one minus the smaller one. So, $$P(265<x<280)=\\text { pnorm }(280,272,9)-\\text { pnorm }(265,272,9)=0.595$$. Thus 59.5% of all pregnancies last between 265 and 280 days.\n\ne. This problem is asking you to find an x value from a probability. You want to find the x value that has 10% of the length of pregnancies to the left of it. On the TI-83/84, the command is in the DISTR menu and is called invNorm(. The invNorm( command needs the area to the left. In this case, that is the area you are given. For the command on the calculator, once you have invNorm( on the main screen you type in the probability to the left, mean, standard deviation, in that order with the commas.", null, "Figure $$\\PageIndex{8}$$: TI-83/84 Output for Example $$\\PageIndex{1}$$e\n\nOn R, the command is qnorm(area to the left, mean, standard deviation). For this example that would be qnorm(0.1, 272, 9)\n\nThus 10% of all pregnancies last less than approximately 260 days.\n\nf. From part (c) you found the probability that a pregnancy lasts less than 250 days is 0.73%. Since this is less than 5%, it is very unusual. You would think that either the woman had a premature baby, or that she may be wrong about when she actually became pregnant.\n\nExample $$\\PageIndex{2}$$ general normal distribution\n\nThe mean mathematics SAT score in 2012 was 514 with a standard deviation of 117 (\"Total group profile,\" 2012). Assume the mathematics SAT score is normally distributed.\n\n1. State the random variable.\n2. Find the probability that a person has a mathematics SAT score over 700.\n3. Find the probability that a person has a mathematics SAT score of less than 400.\n4. Find the probability that a person has a mathematics SAT score between a 500 and a 650.\n5. Find the mathematics SAT score that represents the top 1% of all scores.\n\nSolution\n\na. x = mathematics SAT score\n\nb. First translate the statement into a mathematical statement.\n\nP (x>700)\n\nNow, draw a picture. Remember the center of this normal curve is 514.", null, "Figure for Example $$\\PageIndex{2}$$b\n\nOn TI-83/84: $$P(x>700)=\\text { normalcdf }(700,1 E 99,514,117) \\approx 0.056$$\n\nOn R: $$P(x>700)=1-\\text { pnorm }(700,514,117) \\approx 0.056$$\n\nThere is a 5.6% chance that a person scored above a 700 on the mathematics SAT test. This is not unusual.\n\nc. First translate the statement into a mathematical statement.\n\nP (x<400)\n\nNow, draw a picture. Remember the center of this normal curve is 514.", null, "Figure for Example $$\\PageIndex{2}$$c\n\nOn TI-83/84: $$P(x<400)=\\text { normalcdf }(-1 E 99,400,514,117) \\approx 0.165$$\n\nOn R: $$P(x<400)=\\operatorname{pnorm}(400,514,117) \\approx 0.165$$\n\nSo, there is a 16.5% chance that a person scores less than a 400 on the mathematics part of the SAT.\n\nd. First translate the statement into a mathematical statement.\n\nP (500<x<650)\n\nNow, draw a picture. Remember the center of this normal curve is 514.", null, "Figure for Example $$\\PageIndex{2}$$d\n\nOn TI-83/84: $$P(500<x<650)=\\text { normalcdf }(500,650,514,117) \\approx 0.425$$\n\nOn R: $$P(500<x<650)=\\text { pnorm }(650,514,117)-\\text { pnorm }(500,514,117) \\approx 0.425$$\n\nSo, there is a 42.5% chance that a person has a mathematical SAT score between 500 and 650.\n\ne. This problem is asking you to find an x value from a probability. You want to find the x value that has 1% of the mathematics SAT scores to the right of it. Remember, the calculator and R always need the area to the left, you need to find the area to the left by 1 - 0.01 = 0.99.\n\nOn TI-83/84: $$\\text{invNorm}(.99,514,117) \\approx 786$$\n\nOn R: $$\\text{qnorm}(.99,514,117) \\approx 786$$\n\nSo, 1% of all people who took the SAT scored over about 786 points on the mathematics SAT.\n\n## Homework\n\nExercise $$\\PageIndex{1}$$\n\n1. Find each of the probabilities, where z is a z-score from the standard normal distribution with mean of $$\\mu =0$$ and standard deviation $$\\sigma =1$$. Make sure you draw a picture for each problem.\n1. P (z<2.36)\n2. P (z>0.67)\n3. P (0<z<2.11)\n4. P (-2.78<z<1.97)\n2. Find the z-score corresponding to the given area. Remember, z is distributed as the standard normal distribution with mean of $$\\mu =0$$ and standard deviation $$\\sigma =1$$.\n1. The area to the left of z is 15%.\n2. The area to the right of z is 65%.\n3. The area to the left of z is 10%.\n4. The area to the right of z is 5%.\n5. The area between -z and z is 95%. (Hint draw a picture and figure out the area to the left of the -z.)\n6. The area between -z and z is 99%.\n3. If a random variable that is normally distributed has a mean of 25 and a standard deviation of 3, convert the given value to a z-score.\n1. x = 23\n2. x = 33\n3. x = 19\n4. x = 45\n4. According to the WHO MONICA Project the mean blood pressure for people in China is 128 mmHg with a standard deviation of 23 mmHg (Kuulasmaa, Hense & Tolonen, 1998). Assume that blood pressure is normally distributed.\n1. State the random variable.\n2. Find the probability that a person in China has blood pressure of 135 mmHg or more.\n3. Find the probability that a person in China has blood pressure of 141 mmHg or less.\n4. Find the probability that a person in China has blood pressure between 120 and 125 mmHg.\n5. Is it unusual for a person in China to have a blood pressure of 135 mmHg? Why or why not?\n6. What blood pressure do 90% of all people in China have less than?\n5. The size of fish is very important to commercial fishing. A study conducted in 2012 found the length of Atlantic cod caught in nets in Karlskrona to have a mean of 49.9 cm and a standard deviation of 3.74 cm (Ovegard, Berndt & Lunneryd, 2012). Assume the length of fish is normally distributed.\n1. State the random variable.\n2. Find the probability that an Atlantic cod has a length less than 52 cm.\n3. Find the probability that an Atlantic cod has a length of more than 74 cm.\n4. Find the probability that an Atlantic cod has a length between 40.5 and 57.5 cm.\n5. If you found an Atlantic cod to have a length of more than 74 cm, what could you conclude?\n6. What length are 15% of all Atlantic cod longer than?\n6. The mean cholesterol levels of women age 45-59 in Ghana, Nigeria, and Seychelles is 5.1 mmol/l and the standard deviation is 1.0 mmol/l (Lawes, Hoorn, Law & Rodgers, 2004). Assume that cholesterol levels are normally distributed.\n1. State the random variable.\n2. Find the probability that a woman age 45-59 in Ghana, Nigeria, or Seychelles has a cholesterol level above 6.2 mmol/l (considered a high level).\n3. Find the probability that a woman age 45-59 in Ghana, Nigeria, or Seychelles has a cholesterol level below 5.2 mmol/l (considered a normal level).\n4. Find the probability that a woman age 45-59 in Ghana, Nigeria, or Seychelles has a cholesterol level between 5.2 and 6.2 mmol/l (considered borderline high).\n5. If you found a woman age 45-59 in Ghana, Nigeria, or Seychelles having a cholesterol level above 6.2 mmol/l, what could you conclude?\n6. What value do 5% of all woman ages 45-59 in Ghana, Nigeria, or Seychelles have a cholesterol level less than?\n7. In the United States, males between the ages of 40 and 49 eat on average 103.1 g of fat every day with a standard deviation of 4.32 g (\"What we eat,\" 2012). Assume that the amount of fat a person eats is normally distributed.\n1. State the random variable.\n2. Find the probability that a man age 40-49 in the U.S. eats more than 110 g of fat every day.\n3. Find the probability that a man age 40-49 in the U.S. eats less than 93 g of fat every day.\n4. Find the probability that a man age 40-49 in the U.S. eats less than 65 g of fat every day.\n5. If you found a man age 40-49 in the U.S. who says he eats less than 65 g of fat every day, would you believe him? Why or why not?\n6. What daily fat level do 5% of all men age 40-49 in the U.S. eat more than?\n8. A dishwasher has a mean life of 12 years with an estimated standard deviation of 1.25 years (\"Appliance life expectancy,\" 2013). Assume the life of a dishwasher is normally distributed.\n1. State the random variable.\n2. Find the probability that a dishwasher will last more than 15 years.\n3. Find the probability that a dishwasher will last less than 6 years.\n4. Find the probability that a dishwasher will last between 8 and 10 years.\n5. If you found a dishwasher that lasted less than 6 years, would you think that you have a problem with the manufacturing process? Why or why not?\n6. A manufacturer of dishwashers only wants to replace free of charge 5% of all dishwashers. How long should the manufacturer make the warranty period?\n9. The mean starting salary for nurses is $67,694 nationally (\"Staff nurse -,\" 2013). The standard deviation is approximately$10,333. Assume that the starting salary is normally distributed.\n1. State the random variable.\n2. Find the probability that a starting nurse will make more than $80,000. 3. Find the probability that a starting nurse will make less than$60,000.\n4. Find the probability that a starting nurse will make between $55,000 and$72,000.\n5. If a nurse made less than $50,000, would you think the nurse was under paid? Why or why not? 6. What salary do 30% of all nurses make more than? 10. The mean yearly rainfall in Sydney, Australia, is about 137 mm and the standard deviation is about 69 mm (\"Annual maximums of,\" 2013). Assume rainfall is normally distributed. 1. State the random variable. 2. Find the probability that the yearly rainfall is less than 100 mm. 3. Find the probability that the yearly rainfall is more than 240 mm. 4. Find the probability that the yearly rainfall is between 140 and 250 mm. 5. If a year has a rainfall less than 100mm, does that mean it is an unusually dry year? Why or why not? 6. What rainfall amount are 90% of all yearly rainfalls more than? Answer 1. a. $$P(z<2.36)=0.9909$$, b. $$P(z>0.67)=0.2514$$, c. $$P(0<z<2.11)=0.4826$$, d. $$P(-2.78<z<1.97)=0.9729$$ 3. a. -0.6667, b. -2.6667, c. -2, d. 6.6667 5. a. See solutions, b. $$P(x<52 \\mathrm{cm})=0.7128$$, c. $$P(x>74 \\mathrm{cm})=5.852 \\times 10^{-11}$$, d. $$P(40.5 \\mathrm{cm}<x<57.5 \\mathrm{cm})=0.9729$$, e. See solutions, f. 53.8 cm 7. a. See solutions, b. $$P(x>110 \\mathrm{g})=0.0551$$ c. $$P(x<93 \\mathrm{g})=0.0097$$, d. $$P(x<65 \\mathrm{g}) \\approx 0$$ or $$5.57 \\times 10^{-19}$$, e. See solutions, f. 110.2 g 9. a. See solutions, b. $$P(x>\\ 80,000)=0.1168$$, c. $$P(x>\\ 80,000)=0.2283$$, d. $$P(\\ 55,000<x<\\ 72,000)=0.5519$$, e. See solutions, f.$73,112\n\nThis page titled 6.3: Finding Probabilities for the Normal Distribution is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Kathryn Kozak via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request." ]

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[ "Skip to main content\n\n# What rule does Empower use for rounding significant digits? - WKB2758\n\nArticle number: 2758\n\n• Empower\n\n## ANSWER\n\nWhen rounding is required, Empower looks at the digit in the decimal place to the right of the specified precision in the unrounded number.\n\n• If this digit is less than than five, it and all digits to its right are eliminated and the preceding digit remains unchanged.\n• If this digit is equal to or greater than five, it and all digits to its right are eliminated and the preceding digit is increased by 1.\n\nFor example, with a specified precision of 4, the number 7.2525515 would be rounded up to 7.2526, while the number 7.2525415 would be rounded down to 7.2525.\n\nEmpower uses the printf functionality in C++ for rounding.\n\nFor reference, the USP Rounding Rules are as follows:\n\nWhen rounding is required, consider only one digit in the decimal place to the right of the last place in the limit expression. If this digit is less than five, it is eliminated and the preceding digit is unchanged. If this digit is equal to or greater than five, it is eliminated and the preceding digit is increased by 1.\n\nNot able to find a solution? Click here to request help.\n\n• Was this article helpful?" ]

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[ "# hector_slam/gmapping with T265 (odometry) and RPLidar\n\nHello everybody\n\nI am trying to make SLAM using a 2D laser scanner (RPLidar A1) supporting it with a Realsense camera T265, which provides an accurate odometry.\n\nThe arrangement of both devices is as follows:", null, "The realsense ROS package provides two main frames: camera_odom_frame and camera_pose_frame. This last one is the one that moves around when the camera moves through the space. This is the tf tree created by the camera package:", null, "I have tried to use these two devices with two SLAM packages: hector_slam and gmapping.\n\nhector_slam\n\nThis package works well when carrying out SLAM without odometry. My problem comes when I try to include the odometry information. After launching the LIDAR and the camera packages, I launch hector_slam using the following launch file:\n\n<?xml version=\"1.0\"?>\n<launch>\n<arg name=\"tf_map_scanmatch_transform_frame_name\" default=\"scanmatcher_frame\"/>\n\n<!-- <arg name=\"base_frame\" default=\"base_footprint\"/> -->\n\n<!-- <arg name=\"odom_frame\" default=\"nav\"/> -->\n<arg name=\"odom_frame\" default=\"camera_odom_frame\"/>\n\n<arg name=\"pub_map_odom_transform\" default=\"true\"/>\n<arg name=\"scan_subscriber_queue_size\" default=\"5\"/>\n<arg name=\"scan_topic\" default=\"scan\"/>\n<arg name=\"map_size\" default=\"2048\"/>\n\n<node pkg=\"hector_mapping\" type=\"hector_mapping\" name=\"hector_mapping\" output=\"screen\">\n\n<!-- Frame names -->\n<param name=\"map_frame\" value=\"map\" />\n<param name=\"base_frame\" value=\"$(arg base_frame)\" /> <param name=\"odom_frame\" value=\"$(arg odom_frame)\" />\n\n<!-- Tf use -->\n<param name=\"use_tf_scan_transformation\" value=\"true\"/>\n<param name=\"use_tf_pose_start_estimate\" value=\"false\"/>\n<param name=\"pub_map_odom_transform\" value=\"$(arg pub_map_odom_transform)\"/> <!-- Map size / start point --> <param name=\"map_resolution\" value=\"0.050\"/> <param name=\"map_size\" value=\"$(arg map_size)\"/>\n<param name=\"map_start_x\" value=\"0.5\"/>\n<param name=\"map_start_y\" value=\"0.5\" />\n<param name=\"map_multi_res_levels\" value=\"2\" />\n\n<!-- Map update parameters -->\n<param name=\"update_factor_free\" value=\"0.4\"/>\n<param name=\"update_factor_occupied\" value=\"0.9\" />\n<param name=\"map_update_distance_thresh\" value=\"0.4\"/>\n<param name=\"map_update_angle_thresh\" value=\"0.06\" />\n<param name=\"laser_z_min_value\" value = \"-1.0\" />\n<param name=\"laser_z_max_value\" value = \"1.0\" />\n\n<param name=\"scan_subscriber_queue_size\" value=\"$(arg scan_subscriber_queue_size)\"/> <param name=\"scan_topic\" value=\"$(arg scan_topic)\"/>\n\n<!-- Debug parameters -->\n<!--\n<param name=\"output_timing\" value=\"false\"/>\n<param name=\"pub_drawings\" value=\"true\"/>\n<param name=\"pub_debug_output\" value=\"true\"/>\n-->\n<param name=\"tf_map_scanmatch_transform_frame_name\" value=\"\\$(arg tf_map_scanmatch_transform_frame_name)\" />\n</node>\n\n<node pkg=\"tf\" type=\"static_transform_publisher\" name=\"map_to_camera_odom_broadcaster\" args=\"0 0 0 0 0 0 map camera_odom_frame 100\"/>\n<node pkg=\"tf\" type=\"static_transform_publisher\" name=\"base_to_camera_broadcaster\" args=\"0 0.05 -0.10 3.1415/2 0 0 base_link camera_pose_frame 100\"/>\n<node pkg=\"tf\" type=\"static_transform_publisher\" name=\"base_to_laser_broadcaster\" args=\"0 0 0 0 0 0 base_link laser 100\"/>\n</launch>\n\n\nNote the tf nodes launched at the end of the file. This launcher throws this error:\n\n[ERROR] [1591184110.747014370]: Transform failed during publishing of map_odom transform: Could not find a connection between 'camera_odom_frame' and 'base_link' because they are not part of the same tree.Tf has two or more ...\nedit retag close merge delete\n\nno access to your google drive to see TF trees - that may be an issue on my side but I've never run into before using these forums.\n\nI upvoted the post so you may now have the karma needed to post TF trees directly.\n\nSort by » oldest newest most voted\n\nHector: You pass, thus create, base_link in the hector launchfile, but it should likely be the camera_base_frame.\n\n<arg name=\"base_frame\" default=\"base_link\"/> ---try----> <arg name=\"base_frame\" default=\"camera_base_frame\"/>\n\nI would also suggest, looking into changing the camera specific odom and base frame-IDs to more conventional (for the time beeing maybe even dummy ones with Zero static transform publishers) odom and base_link. Because if you develop further you might actually introduce a \"real\" robot base_link (thats not equal to camera_base) and sensorfusion-odometry. hector gives odometry as well, might make sense to fuse that with the camera odom in future upgrades.\n\nmore\n\nThank you very much for the answer.\n\nAs you suggest, replacing base_link by camera_pose_frame makes it work.\n\nAdditionally, I have realized that actually the disconnection problem is that in the definition of the tf tree, it is important the definition parent->child, that is, the direction of the arrows (which I didn't know before). The tf tree cannot be defined with a frame with two inbound arrows. Changing the 2nd last tf in the launch file to this:\n\n <node pkg=\"tf\" type=\"static_transform_publisher\" name=\"base_to_camera_broadcaster\" args=\"0 -0.05 0.10 -3.1415/2 0 0 camera_pose_frame base_link 100\"/>\n\n\nNow the base_link is the child of camera_pose_frame (not the other way around), and the tf between camera_odom_frame and camera_pose_frame does not break:", null, "Everything is well connected.\n\nRegards\n\nmore\n\n1\n\nargs=\"0 -0.05 0.10 -3.1415/2\n\ndoes that /2 really work?" ]

[ null, "https://answers.ros.org/upfiles/15912017893930124.png", null, "https://answers.ros.org/upfiles/15912018469689131.png", null, "https://answers.ros.org/upfiles/15912569815018301.png", null ]

[ "# scanpy.tl.dpt\n\nscanpy.tl.dpt(adata, n_dcs=10, n_branchings=0, min_group_size=0.01, allow_kendall_tau_shift=True, neighbors_key=None, copy=False)\n\nInfer progression of cells through geodesic distance along the graph [Haghverdi16] [Wolf19].\n\nReconstruct the progression of a biological process from snapshot data. `Diffusion Pseudotime` has been introduced by [Haghverdi16] and implemented within Scanpy [Wolf18]. Here, we use a further developed version, which is able to deal with disconnected graphs [Wolf19] and can be run in a `hierarchical` mode by setting the parameter `n_branchings>1`. We recommend, however, to only use `dpt()` for computing pseudotime (`n_branchings=0`) and to detect branchings via `paga()`. For pseudotime, you need to annotate your data with a root cell. For instance:\n\n```adata.uns['iroot'] = np.flatnonzero(adata.obs['cell_types'] == 'Stem')\n```\n\nThis requires to run `neighbors()`, first. In order to reproduce the original implementation of DPT, use `method=='gauss'` in this. Using the default `method=='umap'` only leads to minor quantitative differences, though.\n\nNew in version 1.1.\n\n`dpt()` also requires to run `diffmap()` first. As previously, `dpt()` came with a default parameter of `n_dcs=10` but `diffmap()` has a default parameter of `n_comps=15`, you need to pass `n_comps=10` in `diffmap()` in order to exactly reproduce previous `dpt()` results.\n\nParameters\nadata : `AnnData`\n\nAnnotated data matrix.\n\nn_dcs : `int` (default: `10`)\n\nThe number of diffusion components to use.\n\nn_branchings : `int` (default: `0`)\n\nNumber of branchings to detect.\n\nmin_group_size : `float` (default: `0.01`)\n\nDuring recursive splitting of branches (‘dpt groups’) for `n_branchings` > 1, do not consider groups that contain less than `min_group_size` data points. If a float, `min_group_size` refers to a fraction of the total number of data points.\n\nallow_kendall_tau_shift : `bool` (default: `True`)\n\nIf a very small branch is detected upon splitting, shift away from maximum correlation in Kendall tau criterion of [Haghverdi16] to stabilize the splitting.\n\nneighbors_key : `Optional`[`str`] (default: `None`)\n\nIf not specified, dpt looks .uns[‘neighbors’] for neighbors settings and .obsp[‘connectivities’], .obsp[‘distances’] for connectivities and distances respectively (default storage places for pp.neighbors). If specified, dpt looks .uns[neighbors_key] for neighbors settings and .obsp[.uns[neighbors_key][‘connectivities_key’]], .obsp[.uns[neighbors_key][‘distances_key’]] for connectivities and distances respectively.\n\ncopy : `bool` (default: `False`)\n\nCopy instance before computation and return a copy. Otherwise, perform computation inplace and return `None`.\n\nReturn type\nReturns\n\nDepending on `copy`, returns or updates `adata` with the following fields.\n\nIf `n_branchings==0`, no field `dpt_groups` will be written.\n\n`dpt_pseudotime``pandas.Series` (`adata.obs`, dtype `float`)\n\nArray of dim (number of samples) that stores the pseudotime of each cell, that is, the DPT distance with respect to the root cell.\n\n`dpt_groups``pandas.Series` (`adata.obs`, dtype `category`)\n\nArray of dim (number of samples) that stores the subgroup id (‘0’, ‘1’, …) for each cell. The groups typically correspond to ‘progenitor cells’, ‘undecided cells’ or ‘branches’ of a process.\n\nNotes\n\nThe tool is similar to the R package `destiny` of [Angerer16]." ]

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[ "Guru\n\n# PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents P and Q intersect at a point T. Find the length TP.\n\n• 1\n\nThis is a very Important NCERT Maths question Circles chapter 10 example 3, well I have solved the complete exercises but I am not able to solve this example, can someone please solve this question for me, here PQ a chord of length 8 cm of a circle of radius 5 cm is given to us. The tangents P and Q intersect each other at a point T. You have to find the length TP.\n\nShare" ]

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[ "Søk Bilder Maps Play YouTube Nyheter Gmail Disk Mer »\nLogg på\n Bøker Bok", null, "If from the ends of a side of a triangle, there be drawn two straight lines to a point within the triangle, these shall be less than the other two sides of the triangle, but shall contain a greater angle.", null, "The Elements of Euclid: Viz. the First Six Books, Together with the Eleventh ... - Side 26\nav Euclid - 1810 - 518 sider\nUten tilgangsbegrensning - Om denne boken", null, "The Practical Teacher, Volum 2\n\n1883\n...3- If from the ends of a side of the triangle FGH there be drawn t wo straight lines to the point К within the triangle, these shall be less than the...of the triangle, but shall contain a greater angle. Prop. 21, Bk. I. Needlework. FEMALES. One hour allowed for this Exerçât. THIRD YEAR. Pupil Teachers...\nUten tilgangsbegrensning - Om denne boken", null, "Geometry for schools, comprising books i. and ii. of Euclid, with some ...\n\nEuclides - 1883 - 96 sider\n...the middle point of the base. This line is called a median. PROP. 21. THEOR. If from the ends of one side of a triangle there be drawn two straight lines to a point within the triangle, these two lines shall be less than the two sides of the triangle, but shall contain a greater angle. Given...\nUten tilgangsbegrensning - Om denne boken", null, "The text of Euclid's geometry, book 1, uniformly and systematically arranged ...\n\nEuclides - 1884\n...opposite to it. Axiom 2. If equals be added to equals the wholes are equal. PROPOSITION XXI. THEOREM. If from the ends of the side of a triangle there be...of the triangle, but shall contain a greater angle. In the triangle ABC ; GIVEN that D is a point within it, and BD and DC are joined ; IT is REQUIRED...\nUten tilgangsbegrensning - Om denne boken", null, "Stewart's guide-book to the queen's scholarship and certificate examinations ...\n\nStewart W. and co - 1884\n...line MO, and therefore the angle NOM shall coincide with angle KMO, and is therefore equal to it. QED If from the ends of the side of a triangle there be...of the triangle, but shall contain a greater angle. Write out the enunciations of any previous propositions employed in this proof. See Euclid, Prop. 21,...\nUten tilgangsbegrensning - Om denne boken", null, "Stewart's specific subjects. Euclid. [1st] (-3rd stage). [With 2 issues of ...\n\nStewart W. and co - 1884\n...than CA ; and BC, CA, greater than AB. XXI. — If from the ends of the side of a triangle, there lie drawn two straight lines to a point within the triangle,...angle. Let the two straight lines BD, CD, be drawn from B,C, the ends of the side BC of the triangle ABC, to the point D within it; BD and DC are less than...\nUten tilgangsbegrensning - Om denne boken", null, "Relfe brothers' Euclid sheets, propositions 1-26, book 1\n\nEuclides - 1884\n...1—26, BOOK I, ARE NOW PUBLISHED IN A SIMILAR FORM TO THIS. PROPOSITION XXL If from the ends of a side of a triangle, there be drawn two straight lines...the triangle, but shall contain a greater angle. Let be a triangle, and from the points , , the ends of the side , let the two straight lines , be drawn...\nUten tilgangsbegrensning - Om denne boken", null, "An introduction to geometry, consisting of Euclid's Elements, book ..., Volum 1\n\nEuclides - 1884\n...greater than AB. Wherefore any two sides, &c. QED 59. PROPOSITION XXI. — THEOREM. If from the ends of a side of a triangle there be drawn two straight lines...within the triangle, these shall be less than the two sides of the triangle, but shall contain a greater angle. Let ABC be a triangle, and D a point...\nUten tilgangsbegrensning - Om denne boken", null, "The Elements of Euclid, books i. to vi., with deductions, appendices and ...\n\nEuclides - 1884\n...semi-perimeter is less, than the sum of the three medians. PROPOSITION 21. THEOREM. If from the ends of any side of a triangle there be drawn two straight lines to a point within the triangle, these straight lines shall be together less than the other two sides of the triangle, but shall contain a...\nUten tilgangsbegrensning - Om denne boken", null, "The Practical Teacher, Volum 4\n\n1885\n...that there are several ways in which the line may be drawn. Prop. 2, Bk. I. 2. If from the ends of a side of a triangle there be drawn two straight lines...other two sides of the triangle, but shall contain я greater angle. Prop. 21, Bk. I. 3. Show that the vertex of an isosceles triangle and the intersections...\nUten tilgangsbegrensning - Om denne boken", null, "Senate Documents, Otherwise Publ. as Public Documents and Executive ..., Volum 3\n\n...Into how many parts is it possible, by mean« of Euclid's constructions, to divide a right angle t 'i- If from the ends of the side of a triangle there be...straight lines to a point within the triangle, these straight lines shall be less than the other two sides of thr triangle. A triangle and a quadrilateral,...\nUten tilgangsbegrensning - Om denne boken" ]

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[ "# Question Video: Subtracting Two Numbers up to 9 Mathematics • Kindergarten\n\nA boy had 7 pencils, but he gave 3 pencils to his sister. How many pencils does he have now?\n\n01:40\n\n### Video Transcript\n\nA boy had seven pencils. But he gave three to his sister. How many pencils does he have now?\n\nWe know the boy had seven pencils to begin with. And we’re also told that he gave three to his sister. And we’re asked to work out how many pencils he has now. Which operation do we need to use to help us solve the problem? Do we need to add, subtract, multiply, or divide?\n\nWe could use a model to help us think about the problem. We know there are seven pencils. And we know the boy gives three to his sister. So he will have four pencils left. It’s a subtraction. Seven subtract three equals four. If a boy has seven pencils and he gives three to his sister, he will have four pencils left. Seven take away three equals four.\n\nTo help us answer this question, first we selected the numbers from the question, seven and three. Then, we used the model to help us decide which operation to use. Then, we subtracted three from seven to give us our answer of four." ]

[ null ]

[ "", null, "OpenCV  3.4.1 Open Source Computer Vision\ncv::DataType< _Tp > Class Template Reference\n\nTemplate \"trait\" class for OpenCV primitive data types. More...\n\n#include \"traits.hpp\"\n\n## Detailed Description\n\n### template<typename _Tp> class cv::DataType< _Tp >\n\nTemplate \"trait\" class for OpenCV primitive data types.\n\nNote\nDeprecated. This is replaced by \"single purpose\" traits: traits::Type and traits::Depth\n\nA primitive OpenCV data type is one of unsigned char, bool, signed char, unsigned short, signed short, int, float, double, or a tuple of values of one of these types, where all the values in the tuple have the same type. Any primitive type from the list can be defined by an identifier in the form CV_<bit-depth>{U|S|F}C(<number_of_channels>), for example: uchar CV_8UC1, 3-element floating-point tuple CV_32FC3, and so on. A universal OpenCV structure that is able to store a single instance of such a primitive data type is Vec. Multiple instances of such a type can be stored in a std::vector, Mat, Mat_, SparseMat, SparseMat_, or any other container that is able to store Vec instances.\n\nThe DataType class is basically used to provide a description of such primitive data types without adding any fields or methods to the corresponding classes (and it is actually impossible to add anything to primitive C/C++ data types). This technique is known in C++ as class traits. It is not DataType itself that is used but its specialized versions, such as:\n\ntemplate<> class DataType<uchar>\n{\ntypedef uchar value_type;\ntypedef int work_type;\ntypedef uchar channel_type;\nenum { channel_type = CV_8U, channels = 1, fmt='u', type = CV_8U };\n};\n...\ntemplate<typename _Tp> DataType<std::complex<_Tp> >\n{\ntypedef std::complex<_Tp> value_type;\ntypedef std::complex<_Tp> work_type;\ntypedef _Tp channel_type;\n// DataDepth is another helper trait class\nenum { depth = DataDepth<_Tp>::value, channels=2,\ntype=CV_MAKETYPE(depth, channels) };\n};\n...\n\nThe main purpose of this class is to convert compilation-time type information to an OpenCV-compatible data type identifier, for example:\n\n// allocates a 30x40 floating-point matrix\nMat A(30, 40, DataType<float>::type);\nMat B = Mat_<std::complex<double> >(3, 3);\n// the statement below will print 6, 2 , that is depth == CV_64F, channels == 2\ncout << B.depth() << \", \" << B.channels() << endl;\n\nSo, such traits are used to tell OpenCV which data type you are working with, even if such a type is not native to OpenCV. For example, the matrix B initialization above is compiled because OpenCV defines the proper specialized template class DataType<complex<_Tp> > . This mechanism is also useful (and used in OpenCV this way) for generic algorithms implementations.\n\nNote\nDefault values were dropped to stop confusing developers about using of unsupported types (see #7599)\n\nThe documentation for this class was generated from the following file:" ]

[ null, "https://docs.opencv.org/3.4.1/opencv-logo-small.png", null ]

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[ "# [Python] Prime number generator\n\nGod may not play dice with the universe, but something strange is going on with the prime numbers.\n\nPaul Erdos", null, "Photo by Markus Spiske from Pexels\n\nA prime number is a positive number greater than 1, which has no positive integer divisors except 1 and itself.  Therefore, numbers like 2, 3, 5, 7, 11 are all prime numbers.\n\nAs a result of the definition of a prime number, one might probably decide to check if a given number, x, is prime by trying to divide it by all the integers starting from 2 and smaller than x. If no such number divides x evenly, then it can be concluded that x is a prime number. Although such a method can be used for very small numbers, but it will be very time consuming and also subject to errors for large numbers. For instance, if x=100,000,000, then the program must check each and every number between 2 and 99,999,999 to see if it divides x evenly or not.\n\nBetter methods can nevertheless be used to calculate prime numbers. One of the most efficient algorithms for calculating prime numbers is called Sieve of Eratosthenes. Put simply, this algorithm starts from the very first prime number, 2, and marks every multiple of it smaller than x as composite. Then the algorithm continues with the next prime number. The remaining numbers will be primes. The following animation, adapted from Wikipedia, clearly illustrates this algorithm.", null, "Let’s have a quick look at the code which does this for us:\n\n``````\n\n\"\"\"Econowmics.com\"\"\"\n\ndef Sieve(n):\n\"\"\"This function calculates all prime numbers up to a given limit(n)\"\"\"\n\n#Defining the inintial list, all equal to True\nprimes = [True for i in range(n+1)]\nprimes = primes = False\n\n#The starting number\np = 2\n\n#Running a loop to check all numbers below the square root of the given number\nwhile p ** 2 < n:\n#Check to see if it is equal to True (if the number is prime or not)\nif primes[p]:\n#If it was a prime number, kill every other multiple of it\nfor number in range(p*2, n+1, p):\nprimes[number] = False\n\n#Increment the starting number by one to keep the loop working\np += 1\n\nprime_numbers = [i for i in range (n+1) if primes[i]==True]\n#If you want to see the list of prime numbers in the output\n#return prime_numbers\n\n#If you want to see how many prime numbers are there in a given range\nreturn len(prime_numbers)``````\n\nFirst, we define a list equal to the size of n, with all elements equal to True. As the algorithm runs, those indexes of this list that are marked as composite will be changed to False. Before the loop, we set the first two instances, namely 0 and 1, which are not primes, into False.\n\nThen, we define the starting number, p=2.\n\nIn the next step, we start a loop to check all the numbers below the square root of the number we are looking for. The reason why we only check the numbers below the square root can be seen from this photo:", null, "The factors repeat themselves after the square root of the given number.\n\nAfter the square root, the numbers just repeat themselves and it would be of no use to check the other numbers.\n\nNow, if the number in the index that we are checking is actually a prime, we set all of its multiples as composite, i.e. set them to False (not primes).\n\nThen we increment p by 1 to check the next number. This goes on until the condition of the while-loop turns False.\n\nIn the last step, we can return the result that we want. For example, here two options are given. Depending on each one suits you best, you have to uncomment it. Here, we check for the total amount of primes less than a given number.\n\nLet’s test the function:\n\n``````import time\n\n#marking the starting time\nstart = time.time()\n\n#For-loop to evaluate the function\nfor i in [10,100,1000,10000,100000,1000000,10000000,100000000]:\nprint (\"There are \" + str(Sieve(i)) + \" prime numbers below \" + str(i) + \".\")\n\n#marking the end of time and calculating the total duration\nend = time.time()\nduration = end - start\nprint (duration)``````\n\nWhich results in:", null, "### Further reading on prime numbers\n\nThe Largest Known prime by Year: A Brief History\n\nPrime Numbers–Why are They So Exciting?\n\nWhy Prime Numbers Still Surprise and Mystify Mathematicians\n\nunusual and physical methods for finding prime numbers" ]

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[ "# How to increase the speed of exhaust gas?\n\nI have a tube, at the bottom of the tube is burning wood and an air inlet. How can i increase the speed of the exhaust gas coming out of the tube? I thought about \"bottle-necking\" the exhaust pipe to make it go faster (Bernoulli's principle) but is there any other way?\n\n• What do you need the higher exhaust speed for? Apr 19, 2016 at 22:36\n\nThe speed of exhaust gases from a stack is governed (approximately) by:\n\n$V=C\\sqrt{2gH\\frac{T_i-T_o}{T_i}}$, where:\n\n$C$ is the discharge coefficient (typically between 0.65-0.7, but can vary based on the inlet and outlet geometry)\n\n$g$ is the acceleration due to gravity, approximately $9.81 m/s^2$ or $32.2 ft/s^2$\n\n$H$ is the height of the chimney\n\n$T_i$ is the average temperature of the gas in the stack (in $K$ or $^{\\circ}R$)\n\n$T_o$ is the temperature of the air outside\n\nYou have several ways to increase the velocity of the gases, the easiest of which should be increasing the length of your chimney.\n\nSources:\n\n• Thank you so much, this is exactly what i have been looking for! Do you know the name of this equation, or what to search for in Google so i can find it? Apr 19, 2016 at 22:52\n• @Mohdak I took the equations from the wiki pages I mentioned above, but used the relation $V=\\frac{Q}{A}$, assuming incompressible flow in the stack Apr 19, 2016 at 23:10\n• You just saved the life of a final year electrical engineering student. thank you so much! Apr 19, 2016 at 23:16\n• Keep in mind increasing the length may result in additional heat loss, reducing the average temperature, offsetting some of your gains. Appropriate insulation should mitigate the effect. Apr 19, 2016 at 23:36\n\nIf the tube is acting as a chimney, then i would first try a longer tube, and add lots of insulation to the outside of it too.\n\nThe hotter gases inside it would be less dense and want to rise faster, the longer tube gives the gases more time to accelerate.\n\nBasically stuff to increase the draft of the chimney\n\ni am not sure about bottelnecking what does it means but the only natural way to increase the exhaust velocities is to place a NOZZLE at the outlet as it converts the pressure energy into kinetic energy(From Steady Flow Energy Equation) and to have more kinetic energy you should have high rate of combustion that will produce high pressure resulting into high velocities through a nozzle. for the non natural processes you can use a centrifugal/reciprocating pump to achieve higher rate of discharge" ]

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[ "", null, "# Teradata – Merging two change history tables\n\nBy Roland Wenzlofsky\n\nMarch 24, 2014\n\nHave you ever been in the situation of facing a bad physical data model where the columns of an object are scattered randomly across tables, and you would like to bring them together where they belong to?\n\nWhile merging is a straightforward operation for not-historical tables, it demands more than just the execution of a merge SQL statement once historization is introduced to at least one of the tables.\n\nChange history tables are at the high end. But dear reader, rejoice! You are relieved from designing this demanding task from scratch – here is the solution taking a small dummy example to walk through all steps:\n\nCREATE MULTISET VOLATILE TABLE <JOIN_TYPE_LOOKUP> (\nJOIN_TYPE CHAR(4)\n) ON COMMIT PRESERVE ROWS;\n\nINSERT INTO <JOIN_TYPE_LOOKUP> VALUES (‘before’);\nINSERT INTO <JOIN_TYPE_LOOKUP> VALUES (‘between’);\nINSERT INTO <JOIN_TYPE_LOOKUP> VALUES (‘after’);\n\n— Two historized test tables\n\nCREATE MULTISET VOLATILE TABLE <TABLE_A> (\nID_A INTEGER,\nRSD DATE,\nRED DATE\n) ON COMMIT PRESERVE ROWS;\n\nCREATE MULTISET VOLATILE TABLE <TABLE_B> (\nID_B INTEGER,\nRSD DATE,\nRED DATE\n) ON COMMIT PRESERVE ROWS;\n\nINSERT INTO <TABLE_A> VALUES (1, ‘2014-01-01’, ‘2014-01-31’);\nINSERT INTO <TABLE_B> VALUES (1, ‘2014-01-31’, ‘9999-12-31’);\n\nSELECT\n——— Block for columns from table A ———\nCASE\nWHEN JOIN_TYPE = ‘before’ AND a.RSD < b.RSD THEN ID_A\nWHEN JOIN_TYPE = ‘before’ AND b.RSD < a.RSD THEN NULL\nWHEN JOIN_TYPE = ‘between’ THEN ID_A\nWHEN JOIN_TYPE = ‘after’ AND a.RED > b.RED THEN ID_A\nWHEN JOIN_TYPE = ‘after’ AND b.RED > a.RED THEN NULL\nEND AS ID_A,\n\n——— Block for columns from table B ———\nCASE\nWHEN JOIN_TYPE = ‘before’ AND a.RSD < b.RSD THEN NULL\nWHEN JOIN_TYPE = ‘before’ AND b.RSD < a.RSD THEN ID_B\nWHEN JOIN_TYPE = ‘between’ THEN ID_B\nWHEN JOIN_TYPE = ‘after’ AND a.RED > b.RED THEN NULL\nWHEN JOIN_TYPE = ‘after’ AND b.RED > a.RED THEN ID_B\nEND AS ID_B,\n\n——– Identify new RSD/RED values ——–\nCASE\nWHEN JOIN_TYPE = ‘before’ AND a.RSD < b.RSD THEN a.RSD\nWHEN JOIN_TYPE = ‘before’ AND b.RSD < a.RSD THEN b.RSD\nWHEN JOIN_TYPE = ‘between’ AND a.RSD < b.RSD THEN b.RSD\nWHEN JOIN_TYPE = ‘between’ AND b.RSD <= a.RSD THEN a.RSD\nWHEN JOIN_TYPE = ‘after’ AND a.RED < b.RED THEN a.RED + 1\nWHEN JOIN_TYPE = ‘after’ AND b.RED < a.RED THEN b.RED + 1\nWHEN JOIN_TYPE = ‘between’ THEN (coalesce(a.RSD, b.RSD))\nEND AS new_RSD,\nCASE\nWHEN JOIN_TYPE = ‘before’ AND a.RSD < b.RSD THEN b.RSD – 1\nWHEN JOIN_TYPE = ‘before’ AND b.RSD < a.RSD THEN a.RSD – 1\nWHEN JOIN_TYPE = ‘between’ AND a.RED < b.RED THEN a.RED\nWHEN JOIN_TYPE = ‘between’ AND b.RED <= a.RED THEN b.RED\nWHEN JOIN_TYPE = ‘after’ AND a.RED < b.RED THEN b.RED\nWHEN JOIN_TYPE = ‘after’ AND b.RED < a.RED THEN a.RED\nWHEN JOIN_TYPE = ‘between’ THEN (coalesce(a.RED, b.RED))\nEND AS new_RED\n—————————————————\nfrom <TABLE_A> a\nFULL OUTER JOIN\n<TABLE_B> b\nON (a.RSD, a.RED+1) OVERLAPS (b.RSD, b.RED+1)\n— further join conditions\nCROSS JOIN <JOIN_TYPE_LOOKUP> h\nWHERE\nnew_RSD IS NOT NULL\nAND new_RED IS NOT NULL;\n\nA note on the terminology used:\n\nI assume that you are familiar with the concept of volatile tables and the basic concepts in historisation. RSD and RED stand for the record start date and the record end date." ]

[ null, "https://www.facebook.com/tr", null ]

[ "Learn\n\nRow-major order for 2D arrays refers to a traversal path which moves horizontally through each row starting at the first row and ending with the last.\n\nAlthough we have already looked at how 2D array objects are stored in Java, this ordering system conceptualizes the 2D array into a rectangular matrix and starts the traversal at the top left element and ends at the bottom right element.\n\nHere is a diagram which shows the path through the 2D array:", null, "This path is created by the way we set up our nested loops. In the previous exercise, we looked at how we can traverse the 2D array by having nested loops in a variety of formats, but if we want to control the indices, we typically use standard for loops.\n\nLet’s take a closer look at the structure of the nested for loops when traversing a 2D array:\n\nGiven this 2D array of strings describing the element positions:\n\n``````String[][] matrix = {{\"\", \"\", \"\"},\n{\"\", \"\", \"\"},\n{\"\", \"\", \"\"},\n{\"\", \"\", \"\"}};``````\n\nLets keep track of the total number of iterations as we traverse the 2D array:\n\n``````int stepCount = 0;\n\nfor(int a = 0; a < matrix.length; a++) {\nfor(int b = 0; b < matrix[a].length; b++) {\nSystem.out.print(\"Step: \" + stepCount);\nSystem.out.print(\", Element: \" + matrix[a][b]);\nSystem.out.println();\nstepCount++;\n}\n}``````\n\nHere is the output of the above code:\n\n``````Step: 0, Element: \nStep: 1, Element: \nStep: 2, Element: \nStep: 3, Element: \nStep: 4, Element: \nStep: 5, Element: \nStep: 6, Element: \nStep: 7, Element: \nStep: 8, Element: \nStep: 9, Element: \nStep: 10, Element: \nStep: 11, Element: \n``````\n\nThe step value increases with every iteration within the inner for loop. Because of this, we can see the order in which each element is accessed. If we follow the step value in the output shows us that the elements are accessed in the same order as the row-major diagram above. Now why is that?\n\nThis is because in our for loop, we are using the number of rows as the termination condition within the outer for loop header `a < matrix.length;` Additionally, we are using the number of columns `b < matrix[a].length` as the termination condition for our inner loop. Logically we are saying: “For every row in our matrix, iterate through every single column before moving to the next row”. This is why our above example is traversing the 2D array using row-major order.\n\nHere is a diagram showing which loop accesses which part of the 2D array for row-major order:", null, "Why Use Row-Major Order?\n\nRow-major order is important when we need to process data in our 2D array by row. You can be provided data in a variety of formats and you may need to perform calculations of rows of data at a time instead of individual elements. Let’s take one of our previous checkpoint exercises as an example. You were asked to calculate the sum of the entire 2D array of integers by traversing and accessing each element. Now, if we wanted to calculate the sum of each row, or take the average of each row, we can use row-major order to access the data in the order that we need. Let’s look at an example!\n\nGiven a 6X3 2D array of doubles:\n\n``````double[][] data = {{0.51,0.99,0.12},\n{0.28,0.99,0.89},\n{0.05,0.94,0.05},\n{0.32,0.22,0.61},\n{1.00,0.95,0.09},\n{0.67,0.22,0.17}};``````\n\nCalculate the sum of each row using row-major order:\n\n``````double rowSum = 0.0;\nfor(int o = 0; o < data.length; o++) {\nrowSum = 0.0;\nfor(int i = 0; i < data[o].length; i++) {\nrowSum += data[o][i];\n}\nSystem.out.println(\"Row: \" + o +\", Sum: \" + rowSum);\n}``````\n\nThe output of the above code is:\n\n``````Row: 0, Sum: 1.62\nRow: 1, Sum: 2.16\nRow: 2, Sum: 1.04\nRow: 3, Sum: 1.15\nRow: 4, Sum: 2.04\nRow: 5, Sum: 1.06``````\n\nAn interesting thing to note is that, due to the way 2D arrays are structured in Java, enhanced for loops are always in row-major order. This is because an enhanced for loop iterates through the elements of the outer array which causes the terminating condition to be the length of the 2D array which is the number of rows.\n\n### Instructions\n\n1.\n\nYou are provided with some runner lap data. Take a look at the loops we’re using to iterate through this 2D array. Replace the incorrect `for` loop headers to perform row-major traversal. Use the iterators `outer` and `inner` for the outer and inner loops.\n\n2.\n\nEnter the missing line of code within the nested `for` loop to sum up the values for each row in the runner data.\n\nWe’ve already created a variable named `runnerTime` that you can use to sum these values.\n\n3.\n\nWe’ve given you a variable named `averageVal` that currently stores `0`. Edit that line of code to find the average time of each runner." ]

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[ "### Why do we need a valuation?\n\nA share valuation may be needed for any of the following:\n\n• For any sale or purchase of shares\n• As a price for a merger or acquisition\n• For tax purposes - e.g. capital gains, inheritance tax\n• For disclosure purposes - e.g. if directors are given shares as part of their remuneration\n• For other legal purposes, such as divorce settlements\n\n### How to value a business\n\nThere are three basic ways of valuing a business\n\n• cash based methods - the theoretical premise here is that the value of the company should be equal to the discounted value of future cash flows.\n• market based methods - where we assume that the market is efficient, so use market information (such as share prices and P/E ratios) for the target company and other companies. The assumption is that the market values businesses consistently so, if necessary, the value of one company can be used to find the value of another.\n• asset based methods - the firm's assets form the basis for the company's valuation. Asset based methods are difficult to apply to companies with high levels of intangible assets, but we shall look at methods of trying to value intangible as well as tangible assets.\n\n## Cash based methods\n\n### The free cash flow method\n\nFree cash flows can used to find the value of a firm. This value can be used:\n\n• to determine the price in a merger or acquisition\n• to identify a share price for the sale of a block of shares\n• to calculate the \"shareholder value added\" (SVA) by management from one period to another.\n\nCalculating the value\n\nTechnically, in order for the value of the business to be accurately determined, free cash flow for all future years should be estimated. However rather than attempting to predict the free cash flows for every year, in practice a short cut method is applied.\n\nFuture cash flows are divided into two time periods:\n\n• Those that occur during the \"planning horizon\".\n• Those that occur after the planning horizon.", null, "The planning horizon\n\nIn competitive industries, a business may have a period of competitive advantage where it can earn excess returns on capital by maintaining a commercial advantage over the competition. However this period is unlikely to last indefinitely. Returns are likely to reach a steady state where the business earns on average its cost of capital but no more.\n\nThe planning horizon (which may last up to ten years or more) is the period during which the returns are expected to be higher than the cost of finance.\n\nIn period beyond the planning horizon it is usually assumed that the returns earned will continue at their current rate for the remainder of the investors' time horizon. This may be a given number of years or in perpetuity. Alternatively the value of the cash flows may be expressed as a lump sum using a P/E ratio.\n\nCalculating free cash flows from accounting information\n\nIdentifying free cash flows for an entire company or business unit is a complex task and is usually determined from the already prepared accounting information and is found by working back from profits as follows:", null, "This method gives the level of free cash flow to the firm as a whole.\n\nFree cash flow to equity\n\nThe above approach calculates free cash flows before deducting either interest or dividend payments.\n\nThe free cash flow to equity only can be calculated by taking the free cash flow calculated above and:\n\n• deducting debt interest paid\n• deducting any debt repayments\n• adding any cash raised from debt issues.\n\nIn practical terms, the free cash flow to equity determines the dividend capacity of a firm i.e. the amount the firm can afford to payout as a dividend.\n\nForecasting growth in free cash flows\n\nThe methods above have identified a figure for the free cash flow of the business based on its current financial statements. In order to value the business, the future free cash flows need to be forecast and then discounted.\n\nTo forecast the likely growth rate for the free cash flows, the following three methods can be used:\n\n• Historical estimates - For example, if the business has achieved growth of 5% per annum each year for the last five years, 5% may be a sensible growth rate to apply to future free cash flows.\n• Analyst forecasts - Particularly for listed companies, market analysts regularly produce forecasts of growth. These independent estimates could be a useful indicator of the likely future growth rate.\n• Fundamental analysis - The formula for Gordon's growth approximation (g = r × b) can be used to calculate the likely future growth rate, where r is the company's return on equity (cost of equity) and b is the earnings retention rate. The formula is based on the assumption that growth will be driven by the reinvestment of earnings.\n\nUse of free cash flow to equity (FCFE) in valuation\n\nThe previous calculations have found equity value by:\n\n• discounting free cash flow to present value using the WACC, and then deducting debt value.\n\nAlternatively, the value of equity can be found directly by:\n\n• discounting free cash flow TO EQUITY at the cost of equity.\n\nIn the simplest case (if FCFE is assumed to be growing at a constant rate into perpetuity), the following formula can be applied:\n\nFCFE0(1 + g) / (ke - g)\n\nThe formula is based on the dividend valuation model theory\n\nThe economic value added (EVATM) valuation model was developed by a firm called Stern Stewart. It combines:\n\n• the logic of the NPV model\n• the use of accounting data produced by a firm.\n\nEVATM is an estimate of \"economic\" profit - the amount by which earnings exceed the required minimum rate of return that investors could get from other securities of comparable risk.\n\nIt follows the same principles as residual income by deducting from profits a charge for the opportunity cost of the capital invested.\n\nEVATM is defined as:\n\nNOPAT - rC\n\nwhere:\n\nNOPAT = Net operating profit after tax\n\nr = WACC\n\nC = firm's invested capital\n\nThe EVATM is then \"capitalised\". Assuming the earnings will continue in perpetuity, the capitalised value is EVATM / WACC\n\nThen, the value of the firm = C + capitalised EVATMand the value of equity = value of the firm less value of the total debt outstanding.\n\nThe NOPAT and C figures are not simply taken from the company's financial statements. There are over 160 different adjustments that maybe required to bring the figures into closer alignment with the true underlying profitability and level of capital invested - i.e. to remove the distortions caused by conforming to the governing GAAP - although in practice only between 5 and 15 of them are usually needed to achieve a meaningful result.\n\n### The dividend valuation model (DVM)\n\nTheory: The value of the company/share is the present value of the expected future dividends discounted at the shareholders' required rate of return.\n\nAssuming a constant growth rate in dividends, g:\n\nP0 = D0(1 + g) / (Ke - g)\n\nIf the growth pattern of dividends is not expected to be stable, but will vary over time, the formula can be adapted.\n\nKe = cost of equity.\n\ng = constant rate of growth in dividends, expressed as a decimal.\n\nD0(1 + g) = dividend just paid adjusted for one year's growth.\n\nThe model is highly sensitive to changes in assumptions:\n\n• Where growth is high relative to the shareholders' required return, the share price is very volatile.\n• Even a minor change in investors' expectations of growth rates can cause a major change in share price contributing to the share price crashes seen in recent years.\n\nDVM is more suitable for valuing minority stakes, since it only considers dividends. In practice the model does tend to accurately match actual stock market share prices.\n\n## Market based methods\n\n### Stock market value (market capitalisation)\n\nFor a listed company, the stock market value of the shares (or \"market capitalisation\") is the starting point for the valuation process.\n\nIn a perfectly efficient market, the market price of the shares would be fair at all times, and would accurately reflect all information about a company. In reality, share prices tend to reflect publicly available information.\n\nThe market share price is suitable when purchasing a minority stake. However, a premium usually has to be paid above the current market price in order to acquire a controlling interest.\n\n### The price-earnings ratio (P/E) method\n\nThe P/E method is a very simple method of valuation. It is the most commonly used method in practice.\n\nValue of equity = PAT × Suitable industry PE ratio.\n\nValue of a share = EPS × Suitable industry PE ratio.\n\nPE ratio\n\nThe PE ratio applied should be that of a company (or average of several companies), that are similar with respect to:\n\n• business risk - i.e. in the same industry\n• finance risk - i.e. have approximately the same level of gearing\n• growth - are growing at about the same rate.\n\nIn practice this may be difficult to find, and the parties involved in the acquisition will then negotiate the applied PE ratio up or down depending on the specific company circumstances.\n\nIn particular, if using a quoted company's PE to value an unquoted business, a substantial discount is often applied to reflect the lower marketability of unquoted shares (around 25% in practice).\n\nProfit figure\n\nThe PAT must be:\n\n• maintainable earnings going forward - some adjustment will be needed for changes anticipated as a result of the acquisition or future trading conditions (e.g. forecast synergies)\n\nWeaknesses of the model\n\n• It is applied to accounting earnings which are more subjective than cash flows.\n• It assumes that the market is actually valuing earnings rather than some other aspect of the company's output - dividends, earnings growth, risk etc.\n• It assumes that the market does accurately value shares\n\n### Earnings yield approach\n\nThe earnings yield is simply the inverse of the PE ratio:\n\nEarnings yield = EPS/Price per share\n\nIt can therefore be used to value the shares or market capitalisation of a company in exactly the same way as the PE ratio:\n\nValue of company = Total earnings  / Earnings yield\n\nValue per share = EPS  / Earnings yield\n\n### Market to book ratio (based on Tobin's Q)\n\nMarket value of target company = Market to book ratio × book value of target company's assets\n\nwhere market to book ratio = (Market capitalisation/Book value of assets) for a comparator company (or take industry average)\n\nThis method assumes a constant relationship between market value of the equity and the book value of the firm.\n\nProblems with the model\n\n• Choosing an appropriate comparator - should we use industry average, or an average of similar firms only?\n• The ratio the market applies is not constant throughout its business cycle, so strictly the comparator should be taken only from other companies at the same stage.\n\n## Asset based methods\n\n### The basic model\n\nThe traditional asset based valuation method is to take as a starting point the value of all the firm's balance sheet assets less any liabilities.\n\nAsset values used can be:\n\n• book value - the book value of assets can easily be found from the financial statements. However, it is unlikely that book values (which are based on historic cost accounting principles) will be a reliable indicator of current market values.\n• replacement cost - the buyer of a business will be interested in the replacement cost, since this represents the alternative cost of setting up a similar business from scratch (organic growth versus acquisition).\n• net realisable value - the seller of a business will usually see the realisable value of assets as the minimum acceptable price in negotiations.\n\nHowever:\n\n• replacement cost is not easy to identify in practice, and\n• the business is more than just the sum of its constituent parts. In fact the value of the tangible assets in many businesses is minimal since much of the value comes from the intangible assets and goodwill (e.g. compare a firm of accountants with a mining company).\n\n### Intangible asset valuation methods\n\nDefinition of intangible assets\n\nIntangible assets are those assets that cannot be touched, weighed or physically measured. They include:\n\n• assets such as patents with legal rights attached\n• intangibles such as goodwill, purchased and valued as part of a previous acquisition\n• relationships, networks and skills built up by the business over time.\n\nA major flaw with the basic asset valuation model is that it does not take account of the true value of intangibles.\n\nBasic intangible valuation method\n\nThe simplest way of incorporating intangible value into the process is by the following basic formula:\n\nFirm value = [book or replacement cost of the real assets] + [multiplier × annual profit or turnover]\n\nThe multiplier is negotiated between the parties to compensate for goodwill.\n\nEffectively, some attempt is being made to estimate the extra value generated by the intangible assets, above the value of the firm's tangible assets.\n\nThis simple formula provides the basis for the two main intangible valuation methods: CIV (Calculated Intangible Value) and Lev's method.\n\n### Calculated intangible value (CIV)\n\nThis method is based on comparing (benchmarking) the return on assets earned by the company with:\n\n• a similar company in the same industry or\n• the industry average.\n\nThe method is similar to the residual income technique. It calculates the company's value spread - the profit it earns over the return on assets that would be expected for a firm in that business.\n\nMethod\n\n(1) A suitable competitor (similar in size, structure etc.) is identified and their return on assets calculated:\n\nOperating profit/Assets employed\n\n(2) If no suitable similar competitor can be identified, the industry average return may be used.\n\n(3) The company's value spread is then calculated.", null, "(4) Assuming that the value spread would be earned in perpetuity, the Calculated Intangible Value (CIV) is found as follows:\n\n• Find the post-tax value spread.\n• Divide the post-tax value spread by the cost of capital to find the present value of the post-tax value spread as a perpetuity (the CIV).\n\n(5) The CIV is added to the net asset value to give an overall value of the firm.\n\nProblems with the CIV model:\n\n• Finding a similar company in terms of industry, similar asset  portfolio, similar cost gearing etc.\n• Since the competitor firm presumably also has intangibles, CIV actually measures the surplus intangible value our company has over that of the competitor rather than over its own asset value.\n\nLev's knowledge earnings method\n\nAn alternative method of valuing intangible assets involves isolating the earnings deemed to be related to intangible assets, and capitalising them. However it is more complex than the CIV model in how it determines the return to intangibles and the future growth assumptions made.\n\nIn practice, this model does produce results that are close to the actual traded share price, suggesting that is a good valuation technique.\n\nHowever, it is often criticised as over complex given that valuations are in the end dependent on negotiation between the parties.", null, "", null, "", null, "Created at 9/25/2012 11:20 AM  by System Account  (GMT) Greenwich Mean Time : Dublin, Edinburgh, Lisbon, London Last modified at 11/13/2012 4:04 PM  by System Account  (GMT) Greenwich Mean Time : Dublin, Edinburgh, Lisbon, London", null, "", null, "### Rating :", null, "Ratings & Comments  (Click the stars to rate the page)", null, "### Tags:\n\n valuation;asset valuation;discounted cash flows;free cash flow;FCFE;free cash flow to equity;p/e;PE ratio;EVA;DVM;dividend valuation model;intangible assets;calculated intangible value;CIV;Lev's knowledge earnings method;Market to book ratio;Tobin's Q;market capitalisation;replacement cost;net realisable value", null, "### Recent Discussions\n\n There are no items to show in this view." ]

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[ "What is the distance between (8, 2) and (4, 7) ?\n\nDec 29, 2015\n\nIts $\\sqrt{41}$\n\nExplanation:\n\nWe first need to calculate the difference in x axis and in y axis:\n$\\triangle x = 8 - 4 = 4$\n$\\triangle y = 7 - 2 = 5$\nHaving this values, we now have a triangle rectangle with angles in both points. Now, just apply Pythagoras' formula:\n${\\text{line}}^{2} = \\triangle {x}^{2} + \\triangle {y}^{2}$\n${\\text{line}}^{2} = {4}^{2} + {5}^{2}$\n$\\text{line} = \\sqrt{41}$" ]

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[ "Hierarchical Re-estimation of Topic Models for Measuring Topical Diversity\n\nOur paper \"Hierarchical Re-estimation of Topic Models for Measuring Topical Diversity\", with Hosein Azarbonyad, Tom Kenter, Maarten Marx, Jaap Kamps, and Maarten de Rijke, has been accepted as a long paper at The 39th European Conference on Information Retrieval (ECIR'17). \\o/", null, "Quantitative notions of topical diversity in text documents are useful in several contexts, e.g., to assess the interdisciplinarity of a research proposal or to determine the interestingness of a document. An influential formalization of diversity has been introduced in biology. It decomposes diversity in terms of elements that belong to categories within a population and formalizes the diversity of a population", null, "as the expected distance between two randomly selected elements of the population:", null, "where", null, "and", null, "are the proportions of categories", null, "and", null, "in the population and", null, "is the distance between", null, "and", null, ".\n\nThis notion of diversity had been adapted to quantify the topical diversity of a text document. Words are considered elements, topics are categories, and a document is a population. When using topic modeling for measuring topical diversity of text document", null, ", We can model elements based on the probability of a word", null, "given", null, ",", null, ", categories based on the probability of", null, "given topic", null, ",", null, ", and populations based on the probability of", null, "given", null, ",", null, ". In probabilistic topic modeling, at estimation time, these distributions are usually assumed to be sparse.\n\n1. First, the content of a document is assumed to be generated by a small subset of words from the vocabulary (i.e.,", null, "is sparse).\n2. Second, each topic is assumed to contain only some topic-specific related words (i.e.,", null, "is sparse).\n3. Finally, each document is assumed to deal with a few topics only (i.e.,", null, "is sparse).\n\nWhen approximated using currently available methods,", null, "and", null, "are often dense rather than sparse. Dense distributions cause two problems for the quality of topic models when used for measuring topical diversity: generality and impurity. General topics mostly contain general words and are typically assigned to most documents in a corpus. Impure topics contain words that are not related to the topic. Generality and impurity of topics both result in low quality", null, "distributions.", null, "Different topic re-estimation approaches. TM is a topic modeling approach like, e.g., LDA. DR is document re-estimation, TR is topic re-estimation, and TAR is topic assignment re-estimation.\n\nIn this research, we propose HiTR, a hierarchical re-estimation process for making the distributions", null, ",", null, "and", null, "more sparse. We re-estimate the parameters of these distributions so that general, collection-wide items are removed and only salient items are kept. We re-estimate the parameters of these distributions so that general, collection-wide items are removed and only salient items are kept. For the re-estimation, we use the concept of parsimonization to extract only essential parameters of each distribution.\n\nOur main contributions are:\n\n• We propose a hierarchical re-estimation process for topic models to address two main problems in estimating topical diversity of text documents, using a biologically inspired definition of diversity.\n• We study the efficacy of each level of re-estimation, and improve the accuracy of estimating topical diversity, outperforming the current state-of-the-art on a publicly available dataset commonly used for evaluating document diversity.\n\nFor more details on the results of experiments, please take a look at our paper:" ]

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[ "### BEEE 1st Year (Unit 1)\n\n## Unit.1: Electrical and Magnetic Circuits\n\nChapter:1.DC & AC Circuits\n\nIntroduction to DC and AC circuits, Active and passive two terminal elements, Ideal and Practical sources, Ohms law, Voltage-Current relations for resistor, inductor, capacitor, Kirchhoff’s laws, Analysis of R-L, R-C and R-L-C series circuits.\n\nChapter:2.Magnetic Circuits and Transformers\n\nMagnetic effects of electric current, Law of Electromagnetic Induction, Self-Inductance, Mutual Inductance, Single Phase Transformer: Construction, Working principle, Efficiency.\n\n## Lecture Topic 1.1.1\n\nIntroduction to DC & AC CIRCUITS\n\n1. Introduction: As we know there are two types of supply which are alternating current and direct current. Both the currents are used in their suitable applications. AC supply mostly is preferred for our daily home appliances like washing machines, refrigerator, drilling machines, transformers while DC supply is used for laptops, cell phones, battery charging applications. Both the supplies have their advantages and disadvantages.\n\nAC supply: It is basically defined as a supply in which magnitude and directions both vary with respect to time.\n\nDC supply: It is defined as a supply in which magnitude and directions both are constant at any instant of time.\n\n2. Basic electrical elements:\n\nResistance: It is the property of resistor which opposes the flow of charge or free electrons. It is represented by R and its unit is in ohm.\n\nInductance: It is property in which property of inductor by virtue of which it opposes any change of magnitude and direction of electric current passing through conductor. It stores energy in the form of magnetic field. It is represented by L and its unit is henry.\n\nCapacitance: It is the property of two plate capacitor which stores energy in the form of electric field. It is represented by C and its unit is in farad.\n\n3. Types of elements: On the basis of different parameters there are different types of elements such as:\n\n1. Active and passive\n2. Linear and non-linear\n3. Unilateral and bi-lateral\n\nActive and passive elements: Active elements are those elements in which they have their own source of energy. They do not require any external source of energy.\n\nexp. Voltage source, current source\n\nPassive elements are those which need external source of energy.\n\nexp. resistance, inductance and capacitance\n\nLinear and non-linear elements: Linear elements are those elements in which they show linear characteristics.\n\nexp. Resistance, inductance, capacitance\n\nNon-linear elements are those which do not follow ohm’s law hence these are called non-linear elements.\n\nexp. Diode, transistor\n\nUnilateral and bilateral elements: Unilateral elements are those in which conduction of current is in one direction.\n\nexp. Diode, rectifier\n\nBilateral element are those in which conduction of current is in both direction.\n\nexp. Voltage source, current source, resistance, inductance, capacitance.\n\n4.Classification of sources:", null, "Fig.1. Classification of sources\n\nIdeal and practical voltage source: An ideal voltage source is that source in which any variation in current does not impact on voltage source otherwise it is called practical voltage source.\n\nFig.2. Ideal and practical voltage source\n\nIdeal and practical current source: An ideal current source is defined as a source in which any variation in voltage does not impact on current otherwise it is called practical current source.\n\nFig.3. Ideal and practical current source\n\nAbove figures show the characteristics of ideal and practical voltage and current source. For practical due to presence of internal resistance there will be some slight drop in response. In practical voltage source internal resistance is connected in series while for practical current source it is connected in parallel with current source.\n\nIndependent and dependent source:", null, "Fig.4. Types of different sources\n\nIndependent source: These are those sources in which any electrical quantity does not depend upon any other electrical quantity like voltage, current.\n\nFig.5. Independent voltage source\n\nDependent source: It is defined as a source in which any electrical quantity depends upon other electrical quantity like voltage, current.", null, "Fig.6. Independent current source\n\n5. Ohm’s law: It is defined as in any closed circuit if the physical properties like pressure, temperature etc. are not changed then voltage drop in the circuit is directly proportional to the current flow in the circuit.", null, "Applications of ohm’s law:\n\n1. Applicable to linear circuits only.\n2. Applicable to bilateral elements only.\n3. Use for determining current in the circuit.\n4. Determine voltage current relationship in the circuit.\n\nSummary\n\nWe have discussed following topics as in chapter-1.\n\n• Introduction to dc and ac circuits.\n• Various types of elements.\n• Differentiation between ac and dc supply.\n• Different types of sources used in electric circuits.\n• Dependent and independent sources.\n• Ideal and practical voltage source.\n• Ideal and practical current source.\n• Introduction to ohm’s laws.\n• Limitations of ohm’s law.\n\nPractice Questions\n\nQ.1 Why we need electricity and what are the types of supply?\n\nAns. As we need electricity in our daily life we require to turn on our home appliances like television, drilling machine, refrigerator etc. so we require electricity. Without it there is no life. There are two types of supply such ac and dc.\n\nQ.2 What are the types of elements?\n\nAns. There are different types of elements such as\n\n1. Active and passive\n2. Linear and non-linear\n3. Unilateral and bilateral\n\nQ.3 What is the difference between ideal and practical voltage source?\n\nAns. An ideal voltage source is that source in which any variation in current does not impact on voltage source otherwise it is called practical voltage source.\n\nQ.4 What do you mean by dependent source?\n\nAns. It is defined as a source in which any electrical quantity depends upon other electrical quantity like voltage, current.\n\n## Lecture Topic 1.1.2\n\nIntroduction to Kirchhoff’s Laws\n\nIntroduction: Kirchhoff’s laws were introduced by Gustav Kirchhoff in 1845. They introduced Kirchhoff’s laws which were based on conservation of energy and conservation of charge. There are basically two types of laws such as Kirchhoff’s voltage law and Kirchhoff’s current law.\n\nKirchhoff’s voltage law: It states that in any closed loop circuit the algebraic sum of all voltage drops is zero.\n\nΣV = 0\n\nOr\n\nIn other words, in any closed loop path the algebraic sum of all voltage drops and total emf sources is zero.\n\nΣV + Σemf = 0\n\nIntroduction to Kirchhoff’s Laws\n\nIntroduction: Kirchhoff’s laws were introduced by Gustav Kirchhoff in 1845. They introduced kirchhoff’s laws which were based on conservation of energy and conservation of charge. There are basically two types of laws such as kirchhoff’s voltage law and kirchhoff’s current law.\n\nKirchhoff’s voltage law: It states that in any closed loop circuit the algebraic sum of all voltage drops is zero.\n\nΣV = 0\n\nOr\n\nIn other words, in any closed loop path the algebraic sum of all voltage drops and total emf sources is zero.\n\nΣV + Σemf = 0\n\nFrom above it is shown that\n\nΣV = Σ(V1 + V2 + V3)\n\nExp. Find out the current in the circuit by using kvl.\n\nSol. We will consider that when we go from (-) polarity to (+) polarity we will take value as positive.\n\n45-5xI-10xI-7.5xI = 0\n\n(22.5×10^3)x I = 45\n\nI = 45/( 22.5×10^3)\n\nI = 2mA  Ans\n\nSummary\n\n1. Introduction to kirchhoff’s law.\n2. Studied about different terminologies used in kirchhoff’s law.\n3. Various examples based on kirchhoff’s voltage law.\n4. Difference between KVL & KCL.\n5. We have learnt applications of kirchhoff’s law.\n\nPractice Question\n\nQ.1 What do you mean by kvl?\n\nAns. In any closed loop circuit the algebraic sum of potential drop is zero. It is based on conservation of energy.\n\nQ.2 What are the applications of kirchhoff’s law?\n\nSol. We can\n\n• Evaluate current in any loop.\n• Measure voltage across any branch\n• Reduce complexity of the circuit.\n\n## Lecture Topic [1.1.3]\n\nIntroduction to Kirchhoff’s Laws\n\nKirchhoff’s current law: It states that at any node the algebraic sum of currents is zero.\n\nΣI = 0\n\nOr\n\nAt any node or junction the algebraic sum of incoming voltage and outgoing voltage is zero.\n\nΣIinc = ΣIout", null, "Fig.1 KCL circuit\n\nFrom figure we will consider incoming current is positive and outgoing current is negative.\n\ni1+i2+i6 = i3 +i4 +i5\n\nExp. Find current I3 in given figure if I1 = 3A, I2 =5A, I4 = 1A\n\nSol. We will write equation:\n\nI1 +I2 = I3 +I4\n\n3 + 5 = I3 + 1\n\nI3 = 7Amp  Ans\n\nSummary\n\n• Introduction to Kirchhoff’s law.\n• Studied about different terminologies used in Kirchhoff’s law.\n• Various examples based on Kirchhoff’s current law.\n• Difference between KVL & KCL.\n• We have learnt applications of Kirchhoff’s law.\n\nPractice Question\n\nQ.1 Explain kcl in your own statement.\n\nAns. It states that at any node the algebraic sum of currents is zero. It is based on conservation of charge.\n\nQ.2 What are the applications of Kirchhoff’s law?\n\nSol. We can\n\n• Evaluate current in any loop.\n• Measure voltage across any branch\n• Reduce complexity of the circuit.\n\n## Lecture Topic [1.1.4]\n\nAnalysis of Series RL Circuit:\n\nA circuit that contains a pure resistance R ohms connected in series with a coil having a pure inductance of L (Henry) is known as RL Series Circuit. When an AC supply voltage V is applied, the current, I flows in the circuit.\n\nSo, IR and IL will be the current flowing in the resistor and inductor respectively, but the amount of current flowing through both the elements will be same as they are connected in series with each other. The circuit diagram of RL Series Circuit is shown below:", null, "Fig.1 Series RL circuit\n\nWhere,\n\n• VR – voltage across the resistor R\n• VL – voltage across the inductor L\n• V – Total voltage of the circuit\n###### Phasor Diagram of the RL Series Circuit\n\nThe phasor diagram of the RL Series circuit is shown below:", null, "Fig.2 Phasor dig of RL circuit\n\nThe following steps are given below which are followed to draw the phasor diagram step by step:\n\n• Current I is taken as a reference.\n• The Voltage drop across the resistance VR = IR is drawn in phase with the current I.\n• The voltage drop across the inductive reactance VL =IXL is drawn ahead of the current I. As the current lags voltage by an angle of 90 degrees in the pure Inductive circuit.\n• The vector sum of the two voltages drops VR and VL is equal to the applied voltage V.\n\nWhere,\n\nZ is the total opposition offered to the flow of alternating current by an RL Series circuit and is called impedance of the circuit. It is measured in ohms (Ω).\n\n###### Phase Angle\n\nIn RL Series circuit the current lags the voltage by 90 degrees angle known as phase angle. It is given by the equation:", null, "Series RC Circuit Analysis:\n\nA circuit that contains pure resistance R ohms connected in series with a pure capacitor of capacitance C farads is known as RC Series Circuit. A sinusoidal voltage is applied and current I flows through the resistance (R) and the capacitance (C) of the circuit.", null, "Fig.3 Series RC circuit\n\nThe following steps are used to draw the phasor diagram of RC Series circuit\n\n• Take the current I (r.m.s value) as a reference vector\n• Voltage drop in resistance VR = IR is taken in phase with the current vector\n• Voltage drop in capacitive reactance VC = IXC is drawn 90 degrees behind the current vector, as current leads voltage by 90 degrees (in the pure capacitive circuit)\n• The vector sum of the two voltage drops is equal to the applied voltage V (r.m.s value).\n###### Phase angle\n\nFrom the phasor diagram shown above, it is clear that the current in the circuit leads the applied voltage by an angle ϕ and this angle is called the phase angle.", null, "Summary\n\nWe have discussed following topics such as:\n\n1.    Analysis of RL Circuit(Phasor Diagram , Impedance, and Current in RL Circuit)\n\n2.    Analysis of RC Circuit (Phasor Diagram , Impedance,\n\nand Current in RC Circuit)\n\n## Lecture Topic [1.1.5]\n\nSeries RLC Circuit Analysis\n\nWhen a pure resistance of R ohms, a pure inductance of L Henry and a pure capacitance of C farads are connected together in series combination with each other then RLC Series Circuit is formed. As all the three elements are connected in series so, the current flowing through each element of the circuit will be the same as the total current I flowing in the circuit.\n\nFig.1. Series RLC Circuit\n\nIn the RLC Series circuit\n\nXL = 2πfL and XC = 1/2πfC\n\nWhen the AC voltage is applied through the RLC Series circuit the resulting current I flows through the circuit, and thus the voltage across each element will be:\n\n• VR = IR that is the voltage across the resistance R and is in phase with the current I.\n• VL = IXL that is the voltage across the inductance L and it leads the current I by an angle of 90 degrees.\n• VC = IXC that is the voltage across capacitor C and it lags the current I by an angle of 90 degrees.\n###### Phasor Diagram of RLC Series Circuit\n\nThe phasor diagram of the RLC series circuit when the circuit is acting as an inductive circuit that means (VL>VC) is shown below and if (VL< VC) the circuit will behave as a capacitive circuit.\n\nSteps to draw phasor diagram:\n\n• Take current I as the reference as shown in the figure above\n• The voltage across the inductor L that is VL is drawn leads the current I by a 90-degree angle.\n• The voltage across the capacitor c that is Vc is drawn lagging the current I by a 90-degree angle because in capacitive load the current leads the voltage by an angle of 90 degrees.\n• The two vector VL and VC are opposite to each other.", null, "", null, "It is the total opposition offered to the flow of current by an RLC Circuit and is known as Impedance of the circuit.\n\nPhase Angle\n\nFrom the phasor diagram, the value of phase angle will be", null, "The three cases of RLC Series Circuit\n\n• When XL > XC, the phase angle ϕ is positive. The circuit behaves as RL series circuit in which the current lags behind the applied voltage and the power factor is lagging.\n• When XL < XC, the phase angle ϕ is negative, and the circuit acts as a series RC circuit in which the current leads the voltage by 90 degrees.\n• When XL = XC, the phase angle ϕ is zero, as a result, the circuit behaves like a purely resistive circuit. In this type of circuit, the current and voltage are in phase with each other. The value of the power factor is unity.\n###### Impedance Triangle of RLC Series Circuit\n\nWhen the quantities of the phasor diagram are divided by the common factor I then the right angle triangle is obtained known as impedance triangle. The impedance triangle of the RL series circuit, when (XL > XC) is shown below:\n\nIf the inductive reactance is greater than the capacitive reactance than the circuit reactance is inductive giving a lagging phase angle otherwise it will be called as leading phase angle.", null, "Fig.3. Impedance Triangle\n\n###### Applications of RLC Series Circuit\n\nThe following are the application of the RLC circuit:\n\n• It acts as a variable tuned circuit\n• It acts as a low pass, high pass, bandpass, bandstop filters depending upon the type of frequency.\n• The circuit also works as an oscillator\n• Voltage multiplier and pulse discharge circuit\n\n## Lecture Topic 1.2.1\n\nIntroduction to Magnetic Circuits\n\n• A magnetic circuit is made up of one or more closed loop paths containing a magnetic flux. The flux is usually generated by permanent magnets or electromagnets and confined to the path by magnetic cores consisting of ferromagnetic materials like iron, although there may be air gaps or other materials in the path.\n• Magnetic circuits are employed to efficiently channel magnetic fields in many devices such aselectricmotors,generators,transformers,relays,liftingelectromagnets,SQUIDs,galvanometers, and magnetic recording heads.\n• The concept of a “magnetic circuit” exploits a one-to-one correspondence between the equations of the magnetic field in a non saturated ferromagnetic material to that of an electrical circuit.\n• Using this concept the magnetic fields of complex devices such as transformers can be quickly solved using the methods and techniques developed for electrical circuits.\n\nSome examples of magnetic circuits are:\n\n·        horseshoe magnet with iron keeper (low-reluctance circuit)\n\n·        horseshoe magnet with no keeper (high-reluctance circuit)\n\n·        electric motor (variable-reluctance circuit)\n\n·        some types of pickup cartridge (variable-reluctance circuits)\n\nmagnetic circuit is made up of magnetic materials having high permeability such as iron, soft steel, etc. Magnetic circuits are used in various devices like electric motor, transformers, relays, generators galvanometer, etc.\n\nFig.1 Magnetic Circuit\n\nConsider a solenoid having N turns wound on an iron core. The magnetic flux of ø Weber sets up in the core when the current of I ampere is passed through a solenoid as shown above.\n\nLet, l = mean length of the magnetic circuit\n\nA = cross-sectional area of the core\n\nµr = relative permeability of the core\n\nNow the flux density in the core material\n\nMagnetising force in the core According to work law, the work done in moving a unit pole once round the magnetic circuit is equal to the ampere-turns enclosed by the magnetic circuit.\n\nAccording to work law, the work done in moving a unit pole once round the magnetic circuit is equal to the ampere-turns enclosed by the magnetic circuit.", null, "The above equation explains the following points:\n\n1. Directly proportional to the number of turns (N) and current (I).It shows that the flux increase if the number of turns or current increases and decreases when either of the two quantitydecreases.NI is the magnetomotive force (MMF).\n2. Inversely proportional tol/aµ0µr,where(l/aµ0µr)is known as reluctance. The lower the reluctance, the higher will be the flux and vice- verse.\n\nDifferent terms used in Magnetic Circuits\n\nMagnetic Field or Magnetic Induction (B)\n\nMagnet or Electromagnet produces a Magnetic field. The field where the magnet attracts or repels magnetic materials such as iron, steel, etc. it may be defined as a force on a moving charge,\n\nF = q x v x B\n\nWhere\n\n·        F = Force,\n\n·        V = Speed of Particles,\n\n·        B = magnitude of the field.\n\nMagnetic Field Strength (H)\n\nThe amount of magnetizing force (how much force it has to magnetize, magnetic materials such as iron, steel, etc) is called Magnetic field strength which is denoted by (H). It is inversely proportional to the length of wire and directly proportional to the current passing through it. The SI unit of Magnetic Field Strength is Ampere/meter (A/m) and it is a vector quantity and the SI formula for Magnetic Field strength is\n\nH = NI / 1c\n\nWhere 1c = magnetic path in meter.\n\nMagnetic Flux (Φ)\n\nIn simple words, Magnetic field x area perpendicular to the magnetic field (B) is called Magnetic Flux which is denoted by Φ or Φm or ΦB. Or it is the amount of magnetic field or magnetic lines of force passing through a surface like conducting area, space, air, etc. The SI Unit of magnetic flux is Wb (Weber). The Formula for finding magnetic flux in the SI system is;\n\nΦ = BAc\n\nWhere\n\nAc = area in m2\n\nAnd CGS unit and formula for Magnetic Flux is Maxwell (M) and Φ = BAc Ac = area in cm2 respectively.\n\nMagnetization (M)\n\nThe state of a material being magnetized or the process in which magnetic materials are magnetized. It is the density of permanent magnet or electromagnet dipole moments in magnetic materials. Or the magnetic moment (m) per unit volume (v) by a magnetic field is called Magnetization. The SI Unit of Magnetization is Ampere/meter (A/m) and it is also a vector quantity. The SI formula for Magnetization is\n\nM = m/V\n\nWhere,\n\nm = Total magnetic moment\n\nAnd V= volume in m3.\n\nThe CGS unit and formula of Magnetization is Emu/cm3 and M = m/V respectively, where, m = Total magnetic moment, V = volume in cm3 and EMU = Electromagnetic units. It may also be defined in term of M = (N/V) x m → M = nm ……. (N/V) = n. Where, “m” is the magnetic moment and “n” is the number density of magnetic moments.\n\nMagnetic Permeability of vacuum\n\n§ It is the amount of resistance encountered to the magnetic field when forming in a vacuum.\n\n§ The SI unit of Permeability is (H·m−1), or Newton per ampere squared (N·A−2). The SI unit and formulas of Magnetic Permeability of vacuum is Newton/Ampere2 and µ○ = 4πx10-7 ≈ 1.2566370614 H·m−1 respectively. The CGS unit of magnetic permeability of vacuum is 1.\n\nInductance (L)\n\nInductance is the property of conductor, coil or wire which opposes the change of current flowing through it. The change of current flowing through a conductor produces a voltage called Back EMF or Electro motive force.\n\nEven The change of current flowing through a conductor or coil produces voltage through it which is called Self induced EMF and in any nearby coils or conductors which is called Mutual inductance. The SI unit of Inductance (L) is Henry “H” and formula is\n\nL = µ○ µ N2 Ac/1c\n\nWhere\n\n·        N = Turns\n\n·        Ac = Area in m2\n\n·        1c = magnetic path in meter\n\nCGS unit and formula of Inductance is Henry “H” (Joseph Henry) and L = 0.4π µN2Ac/1c x10-8 respectively\n\nwhere;\n\n·        L = Inductance\n\n·        N = Turns\n\n·        Ac = Area in cm2\n\n·        1c = magnetic path in cm.\n\nSelf Inductance formula\n\nL = µ○ (N2xA)/l\n\nWhere:\n\n·        L = in Henries\n\n·        μο = the Permeability of Free Space (4.π.10-7)\n\n·        N = the Number of turns\n\n·        A = the Inner Core Area (π.r 2) in m2\n\n·        l = the length of the Coil in meters\n\nMutual Inductance formula\n\nM = μο μrN1N2A/l\n\nWhere:\n\n·        µo = the permeability of free space (4.π.10-7)\n\n·        µr = the relative permeability of the soft iron core\n\n·        N = in the number of coil turns\n\n·        A = in the cross-sectional area in m2\n\n·        l = the coils length in meters\n\nVoltage or E.M.F (V)\n\nThe Electric Potential Difference between two points is called Voltage. Or the work done per unit charge in a static electric field to move the charge between two points, so the equation becomes as\n\nWhere;\n\n·        V = Voltage\n\n·        E = Energy in joules\n\n·        q = Charge in Coulombs\n\nOr the electric potential energy per unit charge is called Voltage.\n\nIn Ohm’s Law, Voltage = V = I x R, Where I = Current in amperes and R = Resistance in Ohms (Ω)\n\nThe SI unit of Voltage is the Volt (V) or Joules per Coulomb. Where 1V = 1Joule/1Coulomb\n\nThe SI formula of Voltage is\n\nV = -N dΦ/dt\n\nWhere;\n\n·        N = number of coil Turns\n\n·        dΦ = rate of the Change in flux\n\n·        t = time\n\nLaws of Electromagnetic Induction\n\nFaraday’s law of electromagnetic induction (referred to as Faraday’s law) is a basic law of electromagnetism predicting how a magnetic field will interact with an electric circuit to produce an electromotive force (EMF). This phenomenon is known as electromagnetic induction.\n\nFaraday’s law states that a current will be induced in a conductor which is exposed to a changing magnetic field. Lenz’s law of electromagnetic induction states that the direction of this induced current will be such that the magnetic field created by the induced current opposes the initial changing magnetic field which produced it. The direction of this current flow can be determined using Fleming’s right-hand rule.\n\nAny change in the magnetic field of a coil of wire will cause an emf to be induced in the coil. This emf induced is called induced emf and if the conductor circuit is closed, the current will also circulate through the circuit and this current is called induced current.\n\nMethod to change the magnetic field:\n\n1.    By moving a magnet towards or away from the coil\n\n2.    By moving the coil into or out of the magnetic field\n\n3.    By changing the area of a coil placed in the magnetic field\n\n4.    By rotating the coil relative to the magnet\n\nIt states that the magnitude of emf induced in the coil is equal to the rate of change of flux that linkages with the coil. The flux linkage of the coil is the product of the number of turns in the coil and flux associated with the coil.\n\n• Power transformers function based on Faraday’s law\n• The basic working principle of the electrical generator is Faraday’s law of mutual induction.\n• The Induction cooker is the fastest way of cooking. It also works on the principle of mutual induction. When current flows through the coil of copper wire placed below a cooking container, it produces a changing magnetic field. This alternating or changing magnetic field induces an emf and hence the current in the conductive container, and we know that the flow of current always produces heat in it.\n• Electromagnetic Flow Meter is used to measure the velocity of certain fluids. When a magnetic field is applied to an electrically insulated pipe in which conducting fluids are flowing, then according to Faraday’s law, an electromotive force is induced in it. This induced emf is proportional to the velocity of fluid flowing.\n• Form bases of Electromagnetic theory, Faraday’s idea of lines of force is used in well known Maxwell’s equations. According to Faraday’s law, change in magnetic field gives rise to change in electric field and the converse of this is used in Maxwell’s equations.\n\nApplications\n\nMagnetic circuits are employed to efficiently channel magnetic fields in many devices such as\n\n1. electric motors\n2. generators\n3. transformers and relays\n4. lifting electromagnets\n5. SQUIDs\n6. galvanometers and magnetic recording heads.\n\nSummary\n\nIn this section we have focused on following topics such as:\n\n• Introduction to magnetic circuit.\n• Different terms used in magnetic circuit.\n• Concept of mmf and magnetic flux.\n• Concept of magnetic permeance and magnetic permeability.\n• Types of magnetic circuit.\n• Applications of magnetic circuit.\n\n## Lecture Topic 1.2.2\n\nSelf inductance and Mutual inductance\n\nSelf-inductance or in other words inductance of the coil is defined as the property of the coil due to which it opposes the change of current flowing through it. Inductance is attained by a coil due to the self-induced emf produced in the coil itself by changing the current flowing through it.\n\nIf the current in the coil is increasing, the self-induced emf produced in the coil will oppose the rise of current, that means the direction of the induced emf is opposite to the applied voltage.Self inductance and Mutual inductance\n\nSelf-inductance or in other words inductance of the coil is defined as the property of the coil due to which it opposes the change of current flowing through it. Inductance is attained by a coil due to the self-induced emf produced in the coil itself by changing the current flowing through it.\n\nIf the current in the coil is increasing, the self-induced emf produced in the coil will oppose the rise of current, that means the direction of the induced emf is opposite to the applied voltage.", null, "If the current in the coil is decreasing, the emf induced in the coil is in such a direction as to oppose the fall of current; this means that the direction of the self-induced emf is same as that of the applied voltage. Self-inductance does not prevent the change of current, but it delays the change of current flowing through it.\n\nThis property of the coil only opposes the changing current (alternating current) and does not affect the steady current that is (direct current) when flows through it. The unit of inductance is Henry (H).\n\nExpression For Self Inductance\n\nYou can determine the self-inductance of a coil by the following expression.\n\nThe above expression is used when the magnitude of self-induced emf (e) in the coil and the rate of change of current (dI/dt) is known.\n\nPutting the following values in the above equations as e = 1 V, and dI/dt = 1 A/s then the value of Inductance will be L = 1 H.\n\nHence, from the above derivation, a statement can be given that a coil is said to have an inductance of 1 Henry if an emf of 1 volt is induced in it when the current flowing through it changes at the rate of 1 Ampere/second.\n\nThe expression for Self Inductance can also be given as:\n\nFrom the above discussion, the following points can be drawn about Self Inductance\n\n·        The value of the inductance will be high if the magnetic flux is stronger for the given value of current.\n\n·        The value of the Inductance also depends upon the material of the core and the number of turns in the coil or solenoid.\n\n·        The higher will be the value of the inductance in Henry, the rate of change of current will be lower.\n\n·        1 Henry is also equal to 1 Weber/ampere\n\nThe solenoid has large self-inductance.\n\nMutual Inductance :\n\nIt is defined as the property of the coil due to which it opposes the change of current in the other coil, or you can say in the neighbouring coil.\n\nWhen the current in the neighbouring coil changes, the flux sets up in the coil and because of this, changing flux emf is induced in the coil called Mutually Induced emf and the phenomenon is known as Mutual Inductance.\n\nLet us understand the phenomenon of Mutual Inductance by considering an example as shown in the above figure.\n\nTwo coils namely coil A and coil B are placed nearer to each other. When the switch S is closed, and the current flows in the coil, it sets up the flux φ in the coil A and emf is induced in the coil and if the value of the current is changed by varying the value of the resistance (R), the flux linking with the coil B also changes because of this changing current.", null, "Fig.2. Mutual Inductance\n\nThus this phenomenon of the linking flux of the coil A with the other coil, B is called Mutual Inductance.\n\nFor determining the Mutual Inductance between the two coils, the following expression is used", null, "This expression is used when the magnitude of mutually induced emf in the coil and the rate of change of current in the neighbouring coil is known.\n\nIf emf = 1 volt and dI1/dt = 1 ampere then putting this value in the equation (1) we get the value of mutual inductance as M=1 Henry\n\nHence, from the above statement, you can define Mutual Inductance as “the two coils are said to have a mutual inductance of one Henry if an emf of 1 volt is induced in one coil or say primary coil when the current flowing through the other neighbouring coil or secondary coil is changing at the rate of 1 ampere/second”.", null, "", null, "The value of Mutual Inductance (M) depends upon the following factors\n\n·        Number of turns in the secondary or neighboring coil\n\n·        Cross-sectional area\n\n·        Closeness of the two coils\n\nMutual Coupling In the Magnetic Circuit\n\nWhen on a magnetic core, two or more than two coils are wound, the coils are said to be mutually coupled. The current, when passed in any of the coils wound around the magnetic core, produces flux which links all the coils together and also the one in which current is passed. Hence, there will be both self-induced emf and mutual induced emf in each of the coils.\n\nThe best example of the mutual inductance is the transformer, which works on the principle of Faraday’s Law of\n\nElectromagnetic Induction.\n\nFaraday’s law of electromagnetic induction states that “the magnitude of voltage is directly proportional to the rate of change of flux.” which is explained in the topic Faraday’s Law of Electromagnetic Induction.\n\nSummary\n\nIn this section we have focused on following topics such as:\n\n• Introduction to faraday’s law of electromagnetic induction.\n• Concept of Lenz law.\n• Self inductance phenomenon.\n• Mutual inductance phenomenon.\n• Coefficient of coupling.\n\nNumerical Practice\n\n1.           A long solenoid has 500 turns. When a current of 2 A is passed through it, the resulting magnetic flux linked with each turn of the solenoid is 4×10−3 Wb. Find the self-inductance of the solenoid.\n\n2.           Two coils connected in series–aiding fashion have a total inductance of 600mH. When connected in a series-opposing configuration, the coils have a total inductance of 400mH. If the inductance of one coil is three times the other , Find L1,L2 ,M and the value of coupling coefficient.\n\nPractice Questions\n\nQ.1 Explain the farday’s law of electromagnetic induction .\n\nQ.2 Explain the difference between self inductance and mutual inductance.\n\n## Lecture Topic 1.2.3\n\nTransformer\n\nA transformer is defined as a passive electrical device that transfers electrical energy from one circuit to another through the process of electromagnetic induction. It is most commonly used to increase (‘step up’) or decrease (‘step down’) voltage levels between circuits.\n\nWorking principle of electrical power transformer is very easy, it is similar to that of mutual induction. A transformer is a static (or stationary) piece of apparatus by means of which electric power in one circuit is transformed into electric power of the same frequency in another circuit.\n\n• It can raise or lower the voltage in a circuit but with a corresponding decrease or increase in current.\n• The physical basis of a power transformer is mutual induction between two circuits linked by a common magnetic flux. In its simplest form, it consists of two inductive coils which are electrically separated but magnetically linked through a path of low reluctance as shown in the figure below.\n• The two coils possess high mutual inductance. If one coil is connected to a source of alternating voltage, an alternating flux is set up in the laminated core, most of which is linked with the other coil in which it produces mutually-induced e.m.f. (according to Faraday’s Laws of Electromagnetic Induction = M.dI/dt).\n• If the second coil circuit is closed, a current flow in it and so electric energy is transferred (entirely magnetically)from the first coil to the second coil.\n• The first coil, in which electric energy is fed from the a.c.supply mains is called primary winding and the other from which energy is drawn out, is called secondary winding.", null, "Fig.1 Transformer\n\n• The power transformer consists of two coils having mutual inductance and a laminated steel core. The two coils are insulated from each other and the steel core. Other necessary parts are some suitable container for assembled core and windings; a suitable medium for insulating the core and its windings from its container; suitable bushings(either of porcelain, oil-filled or capacitor-type) for insulating and bringing out the terminals of windings from the tank.\n• In all types of power transformers, the core is constructed of transformer sheet steel laminations assembled to provide a continuous magnetic path with a minimum of air-gap included.\n• The steel used is of high silicon content, sometimes heat treated to produce a high permeability and a low hysteresis loss at the usual operating flux densities. The eddy current loss is minimised by laminating the core, the laminations being insulated from each other by a light coat of core-plate varnish or by an oxide layer on the surface.\n• The thickness of laminations varies from 0.35 mm for a frequency of 50 Hz to 0.5 mm for a frequency of 25 Hz. The core laminations (in the form of strips) are joined as shown in Figure.\n\nIt is seen that the joints in the alternate layers are staggered in order to avoid the presence of narrow gaps right through the cross-section of the core. Such staggered joints are said to be ‘imbricate.\n\nThe two types are known as\n\n(i) core-type transformer (ii) shell-type transformers\n\nCore Types Transformer: Another recent development is spiral-core or wound-core type, the trade name being Spiral Core transformer. In the so called core type transformers, the windings surround a considerable part of the core whereas, in shell type transformers, the core surrounds a considerable portion of the windings as shown schematically in the figures(a) and (b) respectively.\n\nIn the simplified diagram for the core type transformers, the primary and secondary winding are shown located on the opposite legs (or limbs) of the core, but in actual construction, these are always interleaved to reduce leakage flux.\n\nAs shown in Figure, half the primary and half the secondary winding have been placed side by side or concentrically on each limb, not primary on one limb (or leg) and the secondary on the other.\n\nShell Types Transformer:\n\nHV and LV windings are wound longitudinally along the core alternately. The HV coils are sandwiched between two LV coils as shown in the figure below\n\nince, both HV and LV coils are wound on the central limb surface, the quantity of conductor required for woundings of shell type transformer is less than that of a similar core type transformer.\n\nFig.3 Shell type Transformer\n\nThe design of shell-type is a bit complex compared to core-type due to its winding structure. Any defect in the inner windings can only be attended after removing all outer windings hence the winding maintenance jobs are very hard in shell type transformer.\n\nTransformation Ratio\n\nThe transformer transformation ratio or transformer turns ratio (K) is the quotient value obtained by dividing the number of turns of the primary winding (N1) and the number of turns of the secondary winding (N2).\n\nThen                                      K = N1/N2\n\nK = V1/V2 = I2/I1 =N1/N2\n\nVoltage Transformation Ratio\n\nA transformer with an equal number of turns on its primary and secondary windings will have a secondary voltage only slightly less than the primary applied voltage, and its voltage ratio is said to be 1:1. If, however, the secondary winding have only one-half as many turns as the primary winding, the secondary voltage will be only one-half as great as the primary voltage. The voltage ratio then will be 2:1. The primary and secondary generated electromotive forces are proportional to the primary and secondary turns respectively.\n\nSummary\n\nAs we have discussed following topics:\n\n• Faraday’s Law of Electromagnetic Induction\n• Types of induced emf\n• Self induced emf\n• Mutual induced emf\n• Introduction of single phase transformer\n• Working Principle of Single Phase Transformer\n• Different types of transformer\n• Step up transformer\n• Step down transformer\n• Construction of Transformer\n•  core type\n• shell type\n• Transformation ratio\n\n1.) What does a single-phase mean?\n\nAnswer:- A single-phase system or circuit which generates or uses single alternating voltage\n\n2.) Do houses use single-phase supply?\n\nAnswer:- Generally, homes are supplied with single-phase supply\n\n3.) On which principles does the single-phase transformer operate?\n\n4.) What is the transformer “Turns Ratio”?\n\nAnswer:- NP/NS = VP/VS = n = Turns Ratio\n\n5.) Give two uses of a single-phase transformer\n\nAnswer:- In television sets for voltage regulation\n\nTo step-up power in home inverters\n\n## Lecture Topic 1.2.4\n\nLosses in transformer\n\nThere are various losses exist in a transformer which occur due to the heat present in the losses. These losses reduce the efficiency of the working of transformer.\n\nBelow chart shows the description of transformer:", null, "Core losses or Iron losses\n\n•     Iron losses are caused by the alternating flux in the core of the transformer as this loss occurs in the core it is also known as Core loss. Iron loss is further divided into hysteresis and eddy current loss.\n\n•     Eddy current loss and hysteresis loss depend upon the magnetic properties of the material used for the construction of core.\n\n•     Hence these losses are also known as core losses or iron losses.\n\nHysteresis loss in transformer: Hysteresis loss is due to reversal of magnetization in the transformer core. This loss depends upon the volume and grade of the iron, frequency of magnetic reversals and value of flux density.\n\nWhere\n\n·        KȠ is a proportionality constant which depends upon the volume and quality of the material of the core used in the transformer,\n\n·        f is the supply frequency,\n\n·        Bmax is the maximum or peak value of the flux density.\n\nThe iron or core losses can be minimized by using silicon steel material for the construction of the core of the transformer.\n\nEddy Current Loss\n\nWhen the flux links with a closed circuit, an emf is induced in the circuit and the current flows, the value of the current depends upon the amount of emf around the circuit and the resistance of the circuit.\n\nSince the core is made of conducting material, these EMFs circulate currents within the body of the material.\n\nThese circulating currents are called Eddy Currents. They will occur when the conductor experiences a changing magnetic field.\n\nAs these currents are not responsible for doing any useful work, and it produces a loss (I2R loss) in the magnetic material known as an Eddy Current Loss.\n\nThe eddy current loss is minimized by making the core with thin laminations.\n\nThe equation of the eddy current loss is given as:", null, "Where,\n\n·        Ke – coefficient of eddy current. Its value depends upon the nature of magnetic material like volume and resistivity of core material, the thickness of laminations\n\n·        Bm – maximum value of flux density in wb/m2\n\n·        T – thickness of lamination in meters\n\n·        F – frequency of reversal of the magnetic field in Hz\n\n·        V – the volume of magnetic material in m3\n\nCopper Loss Or Ohmic Loss\n\nThese losses occur due to ohmic resistance of the transformer windings. If I1 and I2 are the primary and the secondary current. R1 and R2 are the resistance of primary and secondary winding then the copper losses occurring in the primary and secondary winding will be I12R1 and I22R2 respectively.\n\nTherefore, the total copper losses will be", null, "These losses varied according to the load and known hence it is also known as variable losses. Copper losses vary as the square of the load current.\n\nStray Loss\n\nThe occurrence of these stray losses is due to the presence of leakage field.\n\nThe percentage of these losses are very small as compared to the iron and copper losses so they can be neglected.\n\nDielectric Loss\n\nDielectric loss occurs in the insulating material of the transformer that is in the oil of the transformer, or in the solid insulations.\n\nWhen the oil gets deteriorated or the solid insulation gets damaged, or its quality decreases, and because of this, the efficiency of the transformer gets affected.\n\nSummary\n\nAs we have discussed following topics:\n\n• Construction of transformer\n• Ideal transformer\n• Losses in transformer\n• Core losses or Iron losses\n• Copper loss\n• Stray losses\n• Dielectric loss\n• How to reduced core losses\n• How to reduced hysteresis losses\n\n## Lecture Topic 1.2.5\n\nTransformer Efficiency:\n\nThe Efficiency of the transformer is defined as the ratio of useful output power to the input power. The input and output power are measured in the same unit. Its unit is either in Watts (W) or KW. Transformer efficiency is denoted by Ƞ.\n\nWhere,\n\n·        V2 – Secondary terminal voltage\n\n·        I2 – Full load secondary current\n\n·        Cosϕ2 – power factor of the load\n\n·        Pi – Iron losses = hysteresis losses + eddy current losses\n\n·        Pc – Full load copper losses = I22Res\n\nConsider, the x is the fraction of the full load. The efficiency of the transformer regarding x is expressed as", null, "given power factor the Transformer efficiency depends upon the load current I2. In equation (1), the numerator is constant and the transformer efficiency will be maximum if the denominator with respect to the variable I2 is equated to zero.", null, "i.e Copper losses = Iron losses\n\nThus, the transformer will give the maximum efficiency when their copper loss is equal to the iron loss.", null, "From equation (2) the value of output current I2 at which the transformer efficiency will be maximum is given as", null, "If x is the fraction of full load KVA at which the efficiency of the transformer is maximum then,\n\nCopper losses = x2Pc (where Pc is the full load copper losses)\n\nIron losses = Pi\n\nFor maximum efficiency\n\nx2 Pc = Pi\n\nTherefore", null, "Thus, output KVA corresponding to maximum efficiency", null, "Putting the value of x from the above equation (3) in equation (4) we will get,", null, "The above equation (5) is the maximum efficiency condition of the transformer.\n\n#### Transformer Voltage Regulation\n\nWe have seen in this series of tutorials about the transformer, that when the primary winding of a transformer is energized, it produces a secondary voltage and current at an amount determined by the transformers turns ratio, (TR). So if a single-phase transformer has a step down turns ratio of 2:1 and 240V is applied to the high voltage primary winding, we would expect to see an output terminal voltage on the secondary winding of 120 VAC because we have assumed it to be an ideal transformer.\n\nHowever in the real world this is not always true as being a wound magnetic circuit, all transformers suffer from losses consisting of I2R copper losses and magnetic core losses which would reduce this ideal secondary value by a few percent to say 117 VAC, and this is normal. But there is also another value related to transformers (and electrical machines) which also has an affect on this secondary voltage value when the transformer is supplying full power, and this is called “regulation”.\n\nVoltage Regulation of single-phase transformers is the percentage (or per unit value) change in its secondary terminal voltage compared to its original no-load voltage under varying secondary load conditions. In other words, regulation determines the variation in secondary terminal voltage which occurs inside the transformer as a result of variations in the transformers connected load thereby affecting its performance and efficiency if these losses are high and the secondary voltage becomes too low.\n\nWhen there is no-load connected to the transformers secondary winding, that is its output terminals are open-circuited, there is no closed-loop condition, so there is no output load current (IL = 0) and the transformer acts as one single winding of high self-inductance. Note that the no-load secondary voltage is a result of the fixed primary voltage and the turns ratio of the transformer.\n\nLoading the secondary winding with a simple load impedance causes a secondary current to flow, at any power factor, through the internal winding of the transformer. Thus voltage drops due to the windings internal resistance and its leakage reactance causes the output terminal voltage to change.\n\nA transformers voltage regulation change between its secondary terminal voltage from a no-load condition when IL = 0, (open circuit) to a fully-loaded condition when IL = IMAX (maximum current) for a constant primary voltage is given as:\n\nNumerical Practice\n\n1.     A single-phase transformer has 400 primary and 1000 secondary turns. The net cross-sectional area of the core is 60 cm2. If the primary winding be connected to a 50-Hz supply at 520 V, calculate (i) the peak value of flux density in the core (ii) the voltage induced in the secondary winding.\n\n2.    A single phase transformer has 500 turns in the primary and 1200 turns in the secondary. The cross-sectional area of the core is 80 sq. cm. If the primary winding is connected to a 50 Hz supply at 500 V, calculate (i) Peak flux-density, and (ii) Voltage induced in the secondary.\n\nSummary\n\nAs we have discussed following topics:\n\n• Various types of losses\n• Why is a transformer rated in KVA but not in KW?\n• Transformer Efficiency\n• Condition for maximum efficiency of transformer\n• All Day Efficiency of a Transformer\n• Voltage Regulation\n• Expression of Voltage Regulation of Transformer\n• Applications of transformer\n\nCheck out other subject study materials\n\n* The material and content uploaded on this website are for general information and reference purposes only. Please do it by your own first. COPYING MATERIALS IS STRICTLY PROHIBITED.\n\n## More from PROGIEZ\n\n### Programming Data Structure And Algorithms Using Python | Week 2" ]

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[ "# AoPS Wiki:AoPS-Mathlinks Rules and Tips\n\n## Mathematical Discussion\n\nThe Art of Problem Solving Forum & MathLinks are a shared community for discussing mathematical problems on various levels. You can share problems with others, solve others' problems, create your own problems and post them, and more.\n\n## Posting problems on AoPS/MathLinks\n\n### Posting problems\n\nIf you want to post a new problem, post it in a new topic. If you have several problems to post, post every problem in a new topic.\n\nDiscussing different problems in one topic leads to confusion.\n\nPlease double check your problem for mistakes. A simple typo in a formula can make a problem incomprehensible.\n\nPlease also give your topic a useful name. Names should describe something. It is better to write names like \"prove concurrency in a right-angled triangle\", \"a^4+b^4+c^4\\geq a^3b+b^3c+c^3a\", \"a^2+b^2=20c^2\", \"f(xf(y))f(y)=yf(xy)\" than the following ones: \"hard geometry\", \"please help me!!\", \"what to do with this one?\", \"don't know how to solve it\".\n\n### Where to post\n\nThere are different forums to post problems in.\n\nIf you have a problem, you have to choose the right forum according to its level:\n\n• High School Basics is for problems that would be considered part of the standard U.S. high school curriculum. This generally means algebra and geometry. Very easy combinatorics or number theory problems would also be appropriate for this forum. Problems in this forum should be at the SAT or easy AMC 10/12 level. Problems in this forum should usually not require complicated proofs.\n• Intermediate Topics is for problems that are harder than \"basic\", but not yet approaching the olympiad level. Some examples are harder AMC 10/12 problems or easier AIME problems. Generally these will be problems that do not require complicated proofs.\n• Pre-Olympiad is for problems slightly under the olympiad level and for very easy olympiad problems. Many problems at this level will require proofs.\n• Olympiad Section is for olympiad-level problems.\n• College Playground is a special forum for problems that require knowledge that is usually not given in schools - for instance, linear algebra, calculus, topology.\n\nThe Olympiad Section is subdivided into subforums:\n\nDecide which topic your problem belongs to, and post it into the respective subforum.\n\nEach of these subforums is subdivided into subsubforums:\n\n• Unsolved Problems: Post your problem here if you don't know its solution and you are searching for it, but you know there is a solution (e.g. because the problem was given at an olympiad).\n• Proposed & Own Problems: Post your problem here if you know the solution and you want to share the problem with others.\n• Open Questions: Post your problem here if it is a conjecture, i.e. you don't have its solution and you don't even know whether it has ever been solved.\n• Solved Problems: This is the archive for former \"Unsolved\" problems which were then solved. Don't post new problems here, it's an archive!\n• Theorems and Formulas: This is for important and useful theorems.\n\n### How to post\n\nAll posts should:\n\n• be correct: strive for no typos and errors, and double-check your message before you post it. If you find a typo after posting, please edit your message to correct it.\n• be readable: use LaTeX (see below) on your mathematical statements. Write in complete sentences.\n• be interesting and/or helpful for someone: a reply that just contains \"it's easy\" or \"I solved it\" interests no one. Every post should have some content.\n• don't quote whole pages just to make a one-line answer. You don't need to quote previous response(s) unless you are responding to a specific portion of someone's earlier post.\n• be appropriate and on-topic: be nice, don't shout or insult, and post only things that fit the topic.\n• always practice good netiquette.\n\n### Posting solutions\n\nWherever you see a problem on the forum, you can post a solution to it if you want.\n\nThe solution needs not be 100% detailed, but it must be understandable. Posts consisting only of words like \"The problem is easy\" or \"Cauchy-Schwarz solves it\" have no value and may be deleted.\n\nPlease post solutions at your level! If you are an Olympiad-level student, do not post solutions in the \"High School Basics\" forum -- leave those problems for students at that level. However, you may comment (politely, of course!) on other students' solutions, and/or provide hints if the students seem stuck.\n\n### Other cases of posting\n\nYou are not limited to posting and solving problems. You can write whatever you want, assumed that it is useful and relevant to the forum. For instance, you can generalize problems, point out mistakes in others' proofs, ask questions about others' solutions, simplify others' solutions, and so on. But please note that the mathematical sections of AoPS/MathLinks are for mathematical discussion only. For non-mathematical discussions, there are the Round Table, the Games & Fun Factory and some more. Non-mathematical discussions in mathematical sections of AoPS/MathLinks can be considered offtopic and removed.\n\n### Running competitions and homework\n\nThis may be obvious, but: Don't post problems from running homework competitions! You are supposed to solve them on your own, so making others solve them for you is cheating!\n\nIf you post school or university homework problems, please indicate that your problems are homework. In most cases, you will get no complete solutions, but hints and other help.\n\n### Using LaTeX\n\nLaTeX makes it possible to include formulas in your posts. For instance, by writing\n\n$a^2 + a^3b$\n\nyou get", null, "$a^{2}+a^{3}b$, and by writing\n\n$\\frac{a}{b} + 2! = 3^{10}$\n\nyou get", null, "$\\frac{a}{b}+2! = 3^{10}$. Learning LaTeX is very easy by this short tutorial and by testing it out on the forum. Testing can be done in the Test Forum.\n\n### Searching on the forum\n\nBefore you post a new problem, you should ask yourself whether this problem has already been discussed on AoPS/MathLinks. You can find this out using two functions of the forum:\n\nThe Search function helps you find topics in the forum by some keywords. For instance, if your problem is from the USAMO 2002, you can type \"USAMO 2002\" into the search field. If your problem is about a triangle, its incircle and some altitudes, you can try typing \"triangle incircle altitude*\" into the search field (the * in \"altitude*\" is there to find both \"altitude\" and \"altitudes\"). You can also restrict the search to some subforums if necessary, so if you look for a Number Theory problem, try restricting the search to the Number Theory subforum if at first there were too many results.\n\nThe Resources section contains lists of problems of several olympiads. If you know what competition your problem is from, you can look up this competition in the Resources section. If you find your problem there, click on the problem number on the left of the page, and you get a thread with this problem.\n\nYou are not supposed to spend half an hour searching for your problem on AoPS/MathLinks, but it is helpful if you try it at least once, better twice.\n\nInvalid username\nLogin to AoPS" ]

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[ "# Cold bubble\n\n## The density current test case\n\nGeometry:\nComputational domain extends in horizontal direction from -25.6 to 25.6 km and in vertical direction from 0 to 6.4 km.\nIntegration time: t=1800s\nProfile: Temperature\n\n$$\\theta = 3000\\,K$$\n$$\\Delta T = \\begin{cases} 0.0 \\text{if}\\quad L \\gt 1.0, \\\\ -15.0 \\cos(L \\pi/2) \\text{if}\\quad L \\le 1.0 \\end{cases}$$\nwhere\n$$L = ((\\frac{x-x_{c}}{x_{r}})^{2} + (\\frac{z-z_{c}}{z_{r}}^{2}))^{1/2}$$\nand\n$$x_{c}=0.0.\\,km ; x_{r}=4.0\\,km ; z_{c}=3.0 \\,km$$ and $$z_{r}=2.0 \\,km$$\n\n## results\n\n top plane: density upper middle plane: horizontal veocity lower middle plane: vertical velocity bottom plane: temperature perturbation" ]

[ null ]

[ "", null, "# plotting lines in the \"background\" of other lines\n\nhi all,\n\ni am plotting several lines together in a plot. i’d like some of the lines to be “behind” the other ones – i.e., when the two lines intersect, i want the most recently plotted line to be on top of the previously plotted lines.\n\nin general, this is true but it seems to be violated by plt.axhline. for example if i have\n\nplt.axhline(…plot horizontal line…, color=‘red’)\nplt.plot(…plot 1st line…)\nplt.plot(…plot 2nd line…)\n\nthen the horizontal line appears ON TOP of the other two lines – i want it to be the opposite, i want the horizontal line to be in the background. how can i do this?\n\nthank you" ]

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[ "", null, "diffalg(deprecated)/print_ranking - Maple Help\n\ndiffalg\n\n print_ranking\n print a message describing the ranking of a differential polynomial ring.", null, "Calling Sequence print_ranking (R)", null, "Parameters\n\n R - differential polynomial ring", null, "Description\n\n • Important: The diffalg package has been deprecated. Use the superseding package DifferentialAlgebra instead.\n • The print_ranking command prints a message describing the ranking defined on a differential polynomial ring R set up with the differential_ring command.\n • The ranking of a differential polynomial ring R is a total ordering over the set of all the derivatives of the differential indeterminates of R that is compatible with derivation (see ranking)\n • The command with(diffalg,print_ranking) allows the use of the abbreviated form of this command.", null, "Examples\n\nImportant: The diffalg package has been deprecated. Use the superseding package DifferentialAlgebra instead.\n\n > $\\mathrm{with}\\left(\\mathrm{diffalg}\\right):$\n > $p≔u\\left[x,y\\right]+v\\left[x,x\\right];$$q≔v\\left[x\\right]+v\\left[y,y\\right]$\n ${p}{≔}{{u}}_{{x}{,}{y}}{+}{{v}}_{{x}{,}{x}}$\n ${q}{≔}{{v}}_{{x}}{+}{{v}}_{{y}{,}{y}}$ (1)\n > $Q≔\\mathrm{differential_ring}\\left(\\mathrm{derivations}=\\left[x,y\\right],\\mathrm{ranking}=\\left[\\mathrm{grlexA}\\left[u,v\\right]\\right]\\right):$\n > $\\mathrm{print_ranking}\\left(Q\\right)$\n In lists, leftmost elements are greater than rightmost ones. The derivatives of [u, v] are ordered by grlexA: _U [tau] > _V [phi] when     |tau| > |phi| or     |tau| = |phi| and _U > _V w.r.t. the list of indeterminates or     |tau| = |phi| and _U = _V and tau > phi w.r.t. [x, y]\n > $\\mathrm{leader}\\left(p,Q\\right),\\mathrm{leader}\\left(q,Q\\right)$\n ${{u}}_{{x}{,}{y}}{,}{{v}}_{{y}{,}{y}}$ (2)\n > $R≔\\mathrm{differential_ring}\\left(\\mathrm{derivations}=\\left[x,y\\right],\\mathrm{ranking}=\\left[\\mathrm{grlexB}\\left[u,v\\right]\\right]\\right)$\n ${R}{≔}{\\mathrm{PDE_ring}}$ (3)\n > $\\mathrm{print_ranking}\\left(R\\right)$\n In lists, leftmost elements are greater than rightmost ones. The derivatives of [u, v] are ordered by grlexB: _U [tau] > _V [phi] when     |tau| > |phi| or     |tau| = |phi| and tau > phi w.r.t. [x, y] or     tau = phi and _U > _V w.r.t. the list of indeterminates\n > $\\mathrm{leader}\\left(p,R\\right),\\mathrm{leader}\\left(q,R\\right)$\n ${{v}}_{{x}{,}{x}}{,}{{v}}_{{y}{,}{y}}$ (4)\n > $S≔\\mathrm{differential_ring}\\left(\\mathrm{derivations}=\\left[x,y\\right],\\mathrm{ranking}=\\left[\\mathrm{lex}\\left[u,v\\right]\\right]\\right)$\n ${S}{≔}{\\mathrm{PDE_ring}}$ (5)\n > $\\mathrm{print_ranking}\\left(S\\right)$\n In lists, leftmost elements are greater than rightmost ones. The derivatives of [u, v] are ordered by lex: _U [tau] > _V [phi] when     tau > phi for the lex. order [x, y] or     tau = phi and _U > _V w.r.t. the list of indeterminates\n > $\\mathrm{leader}\\left(p,S\\right),\\mathrm{leader}\\left(q,S\\right)$\n ${{v}}_{{x}{,}{x}}{,}{{v}}_{{x}}$ (6)\n > $T≔\\mathrm{differential_ring}\\left(\\mathrm{derivations}=\\left[x,y\\right],\\mathrm{indeterminates}=\\left\\{u,v\\right\\},\\mathrm{leaders_of}\\left(\\left[p,q\\right]\\right)=\\left[u\\left[x,y\\right],v\\left[x\\right]\\right]\\right)$\n ${T}{≔}{\\mathrm{PDE_ring}}$ (7)\n > $\\mathrm{print_ranking}\\left(T\\right)$\n In lists, leftmost elements are greater than rightmost ones. The derivatives of [u, v] are ordered by weights: Weights are [u = 3, v = 0, x = 3, y = 1] _U [tau] > _V [phi] when     weight (_U [tau]) > weight (_V [phi]) or     weights are equal and _U > _V w.r.t. the list of indeterminates or     weights and indeterminates are equal and         tau > phi for the lex. order [x, y]" ]

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[ "# Higher category theory\n\nIn mathematics, higher category theory is the part of category theory at a higher order, which means that some equalities are replaced by explicit arrows in order to be able to explicitly study the structure behind those equalities. Higher category theory is often applied in algebraic topology (especially in homotopy theory), where one studies algebraic invariants of spaces, such as their fundamental weak ∞-groupoid.\n\n## Strict higher categories\n\nAn ordinary category has objects and morphisms. A 2-category generalizes this by also including 2-morphisms between the 1-morphisms. Continuing this up to n-morphisms between (n-1)-morphisms gives an n-category.\n\nJust as the category known as Cat, which is the category of small categories and functors is actually a 2-category with natural transformations as its 2-morphisms, the category n-Cat of (small) n-categories is actually an (n+1)-category.\n\nAn n-category is defined by induction on n by:\n\n• A 0-category is a set,\n• An (n+1)-category is a category enriched over the category n-Cat.\n\nSo a 1-category is just a (locally small) category.\n\nThe monoidal structure of Set is the one given by the cartesian product as tensor and a singleton as unit. In fact any category with finite products can be given a monoidal structure. The recursive construction of n-Cat works fine because if a category C has finite products, the category of C-enriched categories has finite products too.\n\nWhile this concept is too strict for some purposes in for example, homotopy theory, where \"weak\" structures arise in the form of higher categories, strict cubical higher homotopy groupoids have also arisen as giving a new foundation for algebraic topology on the border between homology and homotopy theory; see the book \"Nonabelian algebraic topology\" referenced below.\n\n## Weak higher categories\n\nIn weak n-categories, the associativity and identity conditions are no longer strict (that is, they are not given by equalities), but rather are satisfied up to an isomorphism of the next level. An example in topology is the composition of paths, where the identity and association conditions hold only up to reparameterization, and hence up to homotopy, which is the 2-isomorphism for this 2-category. These n-isomorphisms must well behave between hom-sets and expressing this is the difficulty in the definition of weak n-categories. Weak 2-categories, also called bicategories, were the first to be defined explicitly. A particularity of these is that a bicategory with one object is exactly a monoidal category, so that bicategories can be said to be \"monoidal categories with many objects.\" Weak 3-categories, also called tricategories, and higher-level generalizations are increasingly harder to define explicitly. Several definitions have been given, and telling when they are equivalent, and in what sense, has become a new object of study in category theory.\n\n## Quasi-categories\n\nWeak Kan complexes, or quasi-categories, are simplicial sets satisfying a weak version of the Kan condition. André Joyal showed that they are a good foundation for higher category theory. Recently, in 2009, the theory has been systematized further by Jacob Lurie who simply calls them infinity categories, though the latter term is also a generic term for all models of (infinity,k) categories for any k.\n\n## Simplicially enriched categories\n\nSimplicially enriched categories, or simplicial categories, are categories enriched over simplicial sets. However, when we look at them as a model for (infinity,1)-categories, then many categorical notions (e.g., limits) do not agree with the corresponding notions in the sense of enriched categories. The same for other enriched models like topologically enriched categories.\n\n## Topologically enriched categories\n\nTopologically enriched categories (sometimes simply topological categories) are categories enriched over some convenient category of topological spaces, e.g. the category of compactly generated Hausdorff topological spaces.\n\n## Segal categories\n\nThese are models of higher categories introduced by Hirschowitz and Simpson in 1998, partly inspired by results of Graeme Segal in 1974." ]

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[ "Former Member\nSep 14, 2011 at 06:07 AM\n\n# Testing values in logic\n\n23 Views\n\nI have used logic to post values to accounts based on the account number. Now I need to post values based on account values. Logically what I want to do is this;\n\nif account-1 value = account-2 value post account-1 value to account-3\n\nIf Excel this would be a simple formula. In script logic I am struggling to find examples that reference the account values rather than the account ids.\n\nThanks\n\nCliff" ]

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Hydraulic engineering (fitting data with power to fit a curve of the form engineering applications in differential and integral calculus 79. https://en.wikipedia.org/wiki/List_of_numerical-analysis_software Cgn 3421 - computer methods gurley numerical methods lecture 5 - curve fitting techniques page 96 of 102 re-write these equations, and put into matrix form.\n\n23/08/2016в в· application, curve fitting, engineering, matlab, polynomial, tutorial. polynomials and curve fitting in matlab from shameer ahmed koya the research and application of curve and surface fitting parallel least squares algorithm . engineering can be calculated and make scientific\n\n8/04/2011в в· a quick application of linear systems, and matlab's tools for solving them, involving finding the line of best fit through four points in the plane. of electrical engineering jabalpur it has many applications including energy i used my curve fitting algorithm to study effect of\n\nThe research and application of curve and surface fitting parallel least squares algorithm . engineering can be calculated and make scientific laplace transforms are an important tool with many applications in engineering such as control system and automation. regression and curve fitting.\n\nCurve fitting in matlab. matlab has two functions, polyfit and polyval, which can quickly and easily fit a set of data points with a polynomial. curve fitting is the process of constructing a curve, or for graphical and image applications geometric fitting seeks to provide the best visual fit;\n\nApplications of frequency domain curve-fitting in the efdd technique engineering and civil applications of frequency domain curve-fitting in the efdd a curve-fitting technique is chebyshev polynomial curve-fitting technique for strain-gage applications. department of aerospace and mechanical engineering the\n\nLabtalk scripting guide > analysis and applications > curve fitting \\$ = system. path. program \\$ + \"samples\\curve fitting\\polynomial the new fitted curve set curve fitting with maple maple includes a graph showing both the points in your data as well as the curve returned by the choosen curve fitting engineering\n\nIn the context of curve fitting, a linear curve is a curve that has a linear dependence on the curve parameters. examples of linear curves are: lines curve fitting app. the curve fitting generate matlab code to fit and plot curves and surfaces, accelerating the pace of engineering and science.\n\nApplications and use of laplace transform scribd is the world's documents similar to applications and use of laplace transform in the field of engineering. 6/04/2014в в· (for curve fitting more that 10 points.please the buy paid version which curve fit ★ application -researchers -engineers -engineering", null, "Parametric and nonparametric curve fitting both in mechanical engineering at the time series prediction and automotive and environmental applications. labtalk scripting guide > analysis and applications > curve fitting \\$ = system. path. program \\$ + \"samples\\curve fitting\\polynomial the new fitted curve set" ]

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[ "# Parameter estimations of geometric extreme exponential distribution based on dual generalized order statistics\n\nChansoo Kim, Young Han Bae, Woosuk Kim\n\nResearch output: Contribution to journalArticle\n\n### Abstract\n\nIn this study, we consider the maximum likelihood and Bayes esti-mation of the parameters of geometric extreme exponential distribution based on dual generalized order statistics. However, the Bayes esti-mator does not exist in an explicit form for the parameters. We usedan approximation based on Lindley method for obtaining Bayes esti-mates under squared error loss function. We also discuss the asymptotic variance-covariance matrix of maximum likelihood estimators of two pa-rameters. Through Monte Carlo simulation, we compare the maximum likelihood and Bayes estimates of the parameters. And we include one real data analysis.\n\nOriginal language English (US) 3173-3185 13 Applied Mathematical Sciences 10 61-64 https://doi.org/10.12988/ams.2016.69240 Published - Jan 1 2016\n\n### Fingerprint\n\nGeneralized Order Statistics\nBayes\nExponential distribution\nParameter estimation\nMaximum likelihood\nParameter Estimation\nExtremes\nStatistics\nSquared Error Loss Function\nBayes Estimate\nVariance-covariance Matrix\nAsymptotic Variance\nCovariance matrix\nMaximum Likelihood Estimate\nMaximum Likelihood Estimator\nMaximum Likelihood\nData analysis\nMonte Carlo Simulation\nApproximation\n\n### All Science Journal Classification (ASJC) codes\n\n• Applied Mathematics\n\n### Cite this\n\n@article{5664974ae9e548998a360d44d4f789e0,\ntitle = \"Parameter estimations of geometric extreme exponential distribution based on dual generalized order statistics\",\nabstract = \"In this study, we consider the maximum likelihood and Bayes esti-mation of the parameters of geometric extreme exponential distribution based on dual generalized order statistics. However, the Bayes esti-mator does not exist in an explicit form for the parameters. We usedan approximation based on Lindley method for obtaining Bayes esti-mates under squared error loss function. We also discuss the asymptotic variance-covariance matrix of maximum likelihood estimators of two pa-rameters. Through Monte Carlo simulation, we compare the maximum likelihood and Bayes estimates of the parameters. And we include one real data analysis.\",\nauthor = \"Chansoo Kim and Bae, {Young Han} and Woosuk Kim\",\nyear = \"2016\",\nmonth = \"1\",\nday = \"1\",\ndoi = \"10.12988/ams.2016.69240\",\nlanguage = \"English (US)\",\nvolume = \"10\",\npages = \"3173--3185\",\njournal = \"Applied Mathematical Sciences\",\nissn = \"1312-885X\",\npublisher = \"Hikari Ltd.\",\nnumber = \"61-64\",\n\n}\n\nIn: Applied Mathematical Sciences, Vol. 10, No. 61-64, 01.01.2016, p. 3173-3185.\n\nResearch output: Contribution to journalArticle\n\nTY - JOUR\n\nT1 - Parameter estimations of geometric extreme exponential distribution based on dual generalized order statistics\n\nAU - Kim, Chansoo\n\nAU - Bae, Young Han\n\nAU - Kim, Woosuk\n\nPY - 2016/1/1\n\nY1 - 2016/1/1\n\nN2 - In this study, we consider the maximum likelihood and Bayes esti-mation of the parameters of geometric extreme exponential distribution based on dual generalized order statistics. However, the Bayes esti-mator does not exist in an explicit form for the parameters. We usedan approximation based on Lindley method for obtaining Bayes esti-mates under squared error loss function. We also discuss the asymptotic variance-covariance matrix of maximum likelihood estimators of two pa-rameters. Through Monte Carlo simulation, we compare the maximum likelihood and Bayes estimates of the parameters. And we include one real data analysis.\n\nAB - In this study, we consider the maximum likelihood and Bayes esti-mation of the parameters of geometric extreme exponential distribution based on dual generalized order statistics. However, the Bayes esti-mator does not exist in an explicit form for the parameters. We usedan approximation based on Lindley method for obtaining Bayes esti-mates under squared error loss function. We also discuss the asymptotic variance-covariance matrix of maximum likelihood estimators of two pa-rameters. Through Monte Carlo simulation, we compare the maximum likelihood and Bayes estimates of the parameters. And we include one real data analysis.\n\nUR - http://www.scopus.com/inward/record.url?scp=85009259827&partnerID=8YFLogxK\n\nUR - http://www.scopus.com/inward/citedby.url?scp=85009259827&partnerID=8YFLogxK\n\nU2 - 10.12988/ams.2016.69240\n\nDO - 10.12988/ams.2016.69240\n\nM3 - Article\n\nAN - SCOPUS:85009259827\n\nVL - 10\n\nSP - 3173\n\nEP - 3185\n\nJO - Applied Mathematical Sciences\n\nJF - Applied Mathematical Sciences\n\nSN - 1312-885X\n\nIS - 61-64\n\nER -" ]

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[ "# How do you add \\frac { - 1} { 8d } + \\frac { 7} { 6d }?\n\nMay 2, 2017\n\nSee the solution process below:\n\n#### Explanation:\n\nTo add two fractions they must be over a common denominator. For this problem the Lowest Common Denominator is $24 d$. Therefore, before adding we must multiply each fraction by the appropriate form of $1$ to put them each over this common denominator:\n\n$\\left(\\frac{3}{3} \\times - \\frac{1}{8 d}\\right) + \\left(\\frac{4}{4} \\times \\frac{7}{6 d}\\right) \\implies \\left(\\frac{3 \\times - 1}{3 \\times 8 d}\\right) + \\left(\\frac{4 \\times 7}{4 \\times 6 d}\\right)$\n\n$\\implies - \\frac{3}{24 d} + \\frac{28}{24 d}$\n\nWe can now add the numerators over the common denominators:\n\n$- \\frac{3}{24 d} + \\frac{28}{24 d} \\implies \\frac{- 3 + 28}{24 d} \\implies \\frac{25}{24 d}$" ]

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[ "# preface\n\nInfix expression is the expression we often use in addition, subtraction, multiplication and division. Suffix expression is a computer-friendly expression. The computer can use two stacks to calculate the input calculation string\n\n# 1, Suffix expression\n\nSuffix expression is also called inverse Polish. It is not difficult to realize the inverse Polish algorithm, but why convert the seemingly simple infix expression into complex inverse Polish? The reason is that this simple is relative to the human thinking structure. For computers, the middle order expression is a very complex structure. In contrast, the inverse Polish structure is relatively simple and easy to understand in the view of the computer. Because the memory structure commonly used by computers is a stack structure, it executes the order of first in and last out.\nThe suffix expression of (a+b)c-(a+b)/e is ab+cab+e/-\n\n# 2, Infix to suffix mode\n\nFirst, two stacks need to be allocated, one is stack S1 (including an end symbol) as a temporary storage operator, and the other is stack S2 (empty stack) as a storage result (inverse Polish). Stack S1 can be put into the operator with the lowest priority #. Note that infix should end with the operator with the lowest priority. Other characters can be specified, not necessarily # not required. Take characters from the left end of infix, and proceed to the following steps one by one:\n\n1. If the extracted character is an operand, the complete operand is analyzed, and the operand is directly sent to S2 stack.\n2. If the extracted character is an operator, compare the operator with the top element of S1 stack. If the priority of the operator (excluding the bracket operator) is higher than that of the top operator of S1 stack (including the left bracket), put the operator into S1 stack. Otherwise, pop up the top operator of S1 stack and put it into S2 stack until the top operator of S1 stack (including the left bracket) is lower than that of S1 stack (not including) stop popping the operator when it is equal to the priority of the operator, and finally send the operator to S1 stack.\n3. If the extracted character is \"(\", it will be directly sent to the top of S1 stack.\n4. If the extracted character is \")\", the operators between \"(\" closest to the top of S1 stack will be out of the stack one by one and sent to S2 stack in turn. At this time, \"(\" will be discarded.\n5. Repeat steps 1 to 4 above until all input characters are processed.\n6. If the extracted character is \"#\", all operators in S1 stack (excluding \"#\") will be out of the stack one by one and sent to S2 stack in turn.\n\nCalculation example", null, "# 3, Infix to suffix, calculation and Implementation\n\n```public class testPolan {\n\npublic static void main(String[] args) {\nString exp = \"1+((12+0)*4)-5\";\nList<String> infixExpressionList = toInfixExpressionList(exp);\nList<String> suffixExpressionList = parseSuffixExpressionList(infixExpressionList);\nSystem.out.println(\"suffix List\" + suffixExpressionList);\nSystem.out.printf(exp + \" = \" + calculate(suffixExpressionList));\n}\n\n//Method: convert infix to suffix\npublic static List<String> parseSuffixExpressionList(List<String> ls) {\n//Define two stacks\nStack<String> s1 = new Stack<String>();//Symbol stack\n//Because there is no pop in s2 stack, it will be output in reverse order\n//Therefore, it is troublesome. We use List2 directly\nList<String> s2 = new ArrayList<String>();\n\n//Traversal ls\nfor (String item : ls) {\n//If it's a number, add s2\nif (item.matches(\"\\\\d+\")) {\n} else if (item.equals(\"(\")) {\ns1.push(item);\n} else if (item.equals(\")\")) {\nwhile (!s1.peek().equals(\"(\")) {\n}\ns1.pop();//Pop up parentheses\n} else {\n//The operator priority of item is less than or equal to the top of the stack\n//The priority method is missing, see below\nwhile (s1.size() != 0 && Operation.getValue(s1.peek()) >= Operation.getValue(item)) {\n}\n//You also need to push the item onto the stack\ns1.push(item);\n}\n}\n//Add the rest of s1 to s2\nwhile (s1.size() != 0) {\n}\nreturn s2;\n}\n\n//Method: convert infix expression to corresponding List\npublic static List<String> toInfixExpressionList(String s) {\n//First, define a list to store the data corresponding to the infix expression\nList<String> ls = new ArrayList<String>();\nint i = 0;//This is a pointer to traverse the infix expression string\nString str;//Multi bit splicing\nchar c;//Each character traversed is put into c\n\ndo {\n//If c is not a number, we add it to ls\nif ((c = s.charAt(i)) < 48 || (c = s.charAt(i)) > 57) {\ni++;\n} else {\n//Consider multi bit number\nstr = \"\";\nwhile (i < s.length() && (c = s.charAt(i)) >= 48 && (c = s.charAt(i)) <= 57) {\nstr += c;\ni++;\n}\n}\n} while (i < s.length());\nreturn ls;\n}\n\n//Scan the expression from left to right. When encountering a number, push the number onto the stack\n//When an operator is encountered, two numbers at the top of the stack are popped up and calculated\n//Note the order: secondary top element - processing - > stack top element\n//And put the result on the stack, and repeat the above process until the rightmost end of the expression\n//The final result is the calculation result\n\npublic static int calculate(List<String> ls) {\nStack<String> stack = new Stack<String>();\n//Traversal ls\nfor (String item : ls) {\nif (item.matches(\"\\\\d+\")) {\n//Push\nstack.push(item);\n} else {\nint num2 = Integer.parseInt(stack.pop());\nint num1 = Integer.parseInt(stack.pop());\nint res = 0;\nif (item.equals(\"+\")) {\nres = num1 + num2;\n} else if (item.equals(\"-\")) {\nres = num1 - num2;\n} else if (item.equals(\"*\")) {\nres = num1 * num2;\n} else if (item.equals(\"/\")) {\nres = num1 / num2;\n} else {\nthrow new RuntimeException(\"error\");\n}\nstack.push(res + \"\");\n}\n}\nreturn Integer.parseInt(stack.pop());\n}\n\n//Class: return priority number\nstatic class Operation {\nprivate static int ADD = 1;\nprivate static int SUB = 1;\nprivate static int MUL = 2;\nprivate static int DIV = 2;\n\n//Method to return the corresponding priority number\npublic static int getValue(String oper) {\nint result = 0;\nswitch (oper) {\ncase \"+\":\nbreak;\ncase \"-\":\nresult = SUB;\nbreak;\ncase \"*\":\nresult = MUL;\nbreak;\ncase \"/\":\nresult = DIV;\ndefault:\n// System.out.println(\"unable to parse\");\nbreak;\n}\nreturn result;\n}\n}\n\n}\n```\n\n# 4, Summary\n\nCome on!\n\nKeywords: Algorithm data structure\n\nAdded by jrolands on Fri, 10 Sep 2021 03:25:11 +0300" ]

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[ "# How to extract numbers from within an array?\n\n1 view\n\nUsing Python to try and pull forecast dew point data for a single location, with the goal being to come up with code that can pull the forecasted dew points and write it into a csv file. My question is to see if there's a way to print only the data within the array without showing the coordinates and other items.\n\nComputer is a Windows running Python 3.6 via Anaconda. I've been able to focus in on just a few forecast data points I'm looking for (shown as [12:15] in the code) but no matter what it produces the coordinates and other data that I don't exactly need. I tried messing around with xarray.DataArray.drop a bit but I couldn't figure out exactly what I would need to drop.\n\nimport csv\n\nimport matplotlib.pyplot as plt\n\nimport pandas as pd\n\nimport xarray as xr\n\nimport datetime as dt\n\nfrom datetime import datetime, timedelta\n\ndayFile = datetime.now() - timedelta(days=1)\n\ndayFile  = dayFile.strftime(\"%Y%m%d\")\n\nds = xr.open_dataset(url)\n\nlati = 41.4; loni = 100.8\n\ndsloc = ds.sel(lon=loni, lat=lati, method='nearest')\n\ndata_to_write_day_0 = dsloc['dpt2m'][12:15]\n\nprint(data_to_write_day_0)\n\nEdit: Expected output would be producing the forecast numbers only (I.e. 283.729, 284.827, 283.282). I guess in other words the desire is to break up the array so that only the forecasted values are printed out.\n\nby (41.4k points)\n\nHere, for accessing the data we will use the values attribute.\n\nValues returns the data as a numpy.ndarray:\n\nfor x in data_to_write_day_0.values: print(x)\n\nOutput:\n\n283.1779\n\n282.57788\n\n282.7585" ]

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[ "# Interesting question about finding order with orbit stabilizer theorem?\n\nI found this question, about finding the order of the rotational symmetry group in two ways.\n\nIt's by using the orbit-stabilizer theorem on a triangle, and by using it on a square.", null, "I know that the orbit stabilizer theorem is the one below, but I don't get how we get a different order even though it's all the same group in the end.\n\n$$\\left|{\\operatorname{Orb} \\left({x}\\right)}\\right| = \\left[{G : \\operatorname{Stab} \\left({x}\\right)}\\right] = \\dfrac {\\left|{G}\\right|} {\\left|{\\operatorname{Stab} \\left({x}\\right)}\\right|}$$\n\nWhy are the results that we get different for picking a triangle and picking a square?" ]

[ null, "https://i.stack.imgur.com/PlNug.png", null ]

[ "# What are the possibilities?\n\n## Reproduce the diagram on a piece of paper and answer the following questions. Provide the reasoning that justifies your answer in each case.\n\n1. Determine the measure of .\n2. Find the lengths of BC and AC.\n3. Slide points P and Q a bit. Assume that Q is able to move infinitely to the right. Then determine the range of possible lengths for CP and CQ.\n4. Sketch two possible triangles which have a side 7 units in length adjacent to a angle and a side 8 units in length opposite the same angle. If this is not possible, explain why." ]

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[ "# TIC TAC TOE javascript\n\nCreating a tic tac toe game. I am trying to read columns and rows of the tic tac toe to check who won. I am using `\\$scope.checkResults()` to grab the columns and rows and pass them into the `allTheSame` function to check if the values are equal which is not working. Here is the codepen link. http://codepen.io/theMugician/pen/ojJrRp\n\n``````var app = angular.module(\"ticTacToe\", []);\napp.controller(\"MainCtrl\", function(\\$scope){\nvar cell = \\$(\".square\");\n\\$scope.player = \"\";\n\\$scope.AI = \"\";\n// changed special chars to X and O as the if statement failed.\nvar cross = \"✖\";\nvar circle = \"◯\";\n\n/*** Choose a shape ***/\n\\$scope.choosePlayer = function(e) {\n\\$scope.player = \\$(e.currentTarget).text();\n\\$('.choose').css('top', '-2000px');\n\\$('#wrapper').css('top', '-600px');\n\\$('#wrapper').css('opacity', '1');\n//these if statements failed before (AI was always empty)\nif(\\$scope.player === cross){\n\\$scope.AI = circle;\n}else if(\\$scope.player === circle){\n\\$scope.AI = cross;\n}\n}\n\n/*** Shape Cells ***/\n\\$scope.cells = [ { value: '' }, { value: '' }, { value: '' },\n{ value: '' }, { value: '' }, { value: '' } ,\n{ value: '' }, { value: '' }, { value: '' }\n];\n// made a ref to scope cells\n\\$scope.emptyCells = \\$scope.cells;\n\n/*** Make a move ***/\n\\$scope.move = function(cell){\ncell.value = \\$scope.player;\nvar round = 0;\n/*** AI makes a move ***/\nwhile(round < 1){\n// filtered to get only available cells (for performance)\n\\$scope.emptyCells = \\$scope.cells.filter(function(cell){\nreturn cell.value === '';\n});\n// got random cell according to empty cells\nvar randomCell = \\$scope.emptyCells[Math.floor((Math.random()*(\\$scope.emptyCells.length-1))+1)];\nif(randomCell.value === \"\" ){\nrandomCell.value = \\$scope.AI;\nround = 1;\n}else{\nround = 0;\n}\n}\n\\$scope.checkResults();\n};\n\n//checks if values are the same\nfunction allthesame(arr){\nvar L= arr.length-1;\nwhile(L){\nif(arr[L--]!==arr[L]) return false;\n}\n}\n\n//checks Columns and rows\n\\$scope.checkResults = function(){\nvar allCells = \\$scope.cells;\n// check rows\nvar cellRows = [];\nwhile(allCells > 0){\ncellRows.push(allCells.splice(0,3));\n}\nfor(var i = 0; i < cellRows.length; i++){\nallTheSame(cellRows[i]);\n}\n// check columns\nvar cellCols = [];\nwhile(allCells > 0){\ncellCols.push(allCells.splice(0));\ncellCols.push(allCells.splice(3));\ncellCols.push(allCells.splice(6));\n}\nwhile(cellCols > 0){\ncellCols.push(cellCols.splice(0,3));\n}\nfor(var i = 0; i < cellCols.length; i++){\nallTheSame(cellCols[i]);\n}\n}\n\n\\$scope.reset = function(){\n\\$scope.cells = [ { value: '' }, { value: '' }, { value: '' },\n{ value: '' }, { value: '' }, { value: '' } ,\n{ value: '' }, { value: '' }, { value: '' }\n];\n}\n});\n``````\n\nSo you have an array of 9 values and need to compare to 8 possible winning arrangements, 3 vertical, 3 horizontal and 2 diagonal.. an array iteration against the follow list of \"winning\" combinations might be what you want to do.\n\n``````{0,1,2}\n{3,4,5}\n{6,7,8}\n{0,3,6}\n{1,4,7}\n{2,5,8}\n{0,4,8}\n{2,4,6}\n``````\n\nI was just on my way out the door I can post some code later but I think saving this list as a comparison item for your cells might be the easier method.\n\nHere is a reference article I grabbed quickly: this should be easy to convert to javascript: http://www.codeproject.com/Articles/2400/Tic-Tac-Toe-in-C\n\n• I'll try your solution and see what I can come up with. Nov 28, 2015 at 21:31\n\nUsing Frodo's advice I came up with a solution which is a lot simpler than what I was doing before.\n\nI stored all winning combinations into an array. Then I checked each winning combo to see if each cell's value in that combo matched. Simple as that.\n\n``````var winningNums = [\n[0,1,2],\n[3,4,5],\n[6,7,8],\n[0,3,6],\n[1,4,7],\n[2,5,8],\n[0,4,8],\n[2,4,6]\n];\n\n//checks if values are the same\n\\$scope.checkResults = function(){\nvar allCells = \\$scope.cells;\nfor(var i = 0; i < winningNums.length; i++){\nvar a = winningNums[i],b=winningNums[i],c=winningNums[i];\nvar cell1 = allCells[a].value, cell2 = allCells[b].value, cell3 = allCells[c].value;\nif(cell1 == \"\" || cell2 == \"\" || cell3 == \"\" ){\nbreak;\n}\nif(cell1 === cell2 && cell2 === cell3 ){\nvar winnerDiv = \"<div><h1>\" + cell1 + \" is the winner</h1></div>\";\n\\$(\n\"#wrapper\").append(winnerDiv);\n}\n\n}\n\n}\n``````\n\n# This Answer Does not Use `jQuery`\n\nUsing an array similar to what's shown in the above 👆🏾answer, we can just use a few different array methods:\n\n``````const WINNING_INDICES = [\n[0, 1, 2],\n[0, 3, 6],\n[0, 4, 8],\n[1, 4, 7],\n[2, 5, 8],\n[2, 4, 6],\n[3, 4, 5],\n[6, 7, 8],\n];\n\nfunction check4Winner(board, letter) {\nreturn WINNING_INDICES.some(winningIndices =>\nwinningIndices.every(winningIndex => board[winningIndex] === letter),\n);\n}\n\nconsole.log(check4Winner(['X', 'O', 'O', 'X', 'O', 'X', 'X', '', '']), \"X\");\n``````\n\nHere, I have passed in a hard-coded board that's a flat array. It represents the nine squares on a Tic-Tac-Toe board.\n\nSummarily, we want to see if there is at least one instance (`some`) where all the indices of a given element (`winningIndices`) of `WINNING_INDICES`, when used with `board`, match the given letter (`\"X\"`).\n\nThis is just the function of my code to determine who won in the game. I also use the winning combination numbers in my code.\n\n`````` var i = -1;\nvar WinnerChecker = setInterval(() =>{\nvar p1 = 0, p2 = 0;\ni = i<TileCombos.length ? ++i : 0\nfor(let j = 0; j<3; j++){\nif(player[TileCombos[i][j]].innerHTML == chk.innerHTML){\np1 += 1;\n}\nif(player[TileCombos[i][j]].innerHTML == chk.innerHTML){\np2 += 1;\n}\n}\nif(p1 == 3){\nconsole.log('PLAYER 1 WON');\n}\nif(p2 == 3){\nconsole.log('PLAYER 2 WON');\n}\n\n})\n``````" ]

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[ "", null, "Query Report – Inventory Aging\n\nInventory Aging Report\n\nDeclare @a char(60)\n\nSet @a  = (select distinct t20.docdate from oinm t20 where t20.docdate = ‘[%2]’)\n\nSelect  distinct c.code, c.name\n\n,SUM(c.bal) [Bal],sum(c.Val) [Val],\n\n(case when avg(c.[FullStock]) > 0 then avg(c.[Pric]) / avg(c.[FullStock]) end ) [AvgPrice],\n\n(case when SUM(c.[0-30 Days]) >= SUM(c.bal)  then SUM(c.bal)  else SUM(c.[0-30 Days]) end) [0-30 Days],\n\n(case when SUM(c.[0-30 Days]) >= SUM(c.bal)  then 0 else\n\n(case when (sum(c.[30-60 Days]) + SUM(c.[0-30 Days])) >=  SUM(c.bal) then (SUM(c.bal) – sum(c.[0-30 Days]))\n\nelse sum(c.[30-60 Days]) end)end)   [30-60 Days],\n\n(case when SUM(c.[0-30 Days]) >= SUM(c.bal)  then 0 else\n\n(case when (sum(c.[30-60 Days]) + SUM(c.[0-30 Days])) >=  SUM(c.bal) then 0 else\n\n(case when (sum(c.[30-60 Days]) + SUM(c.[0-30 Days]) + sum(c.[60-90 Days])) >= SUM(c.bal) then\n\n(SUM(c.bal) – (sum(c.[0-30 Days]) + sum(c.[30-60 Days]))) else\n\nsum(c.[60-90 Days])end)end) end)   [60-90 Days],\n\n(case when SUM(c.[0-30 Days]) >= SUM(c.bal)  then 0 else\n\n(case when (sum(c.[30-60 Days]) + SUM(c.[0-30 Days])) >=  SUM(c.bal) then 0 else\n\n(case when (sum(c.[30-60 Days]) + SUM(c.[0-30 Days]) + sum(c.[60-90 Days])) >= SUM(c.bal) then 0 else\n\n(case when (sum(c.[30-60 Days]) + SUM(c.[0-30 Days]) + sum(c.[60-90 Days]) + sum(c.[90-180 Days])) >= SUM(c.bal) then\n\n(SUM(c.bal) – (sum(c.[0-30 Days]) + sum(c.[30-60 Days]) + sum(c.[60-90 Days]))) else sum(c.[90-180 Days])\n\nend)end) end) end)  [90-180 Days],\n\n(case when SUM(c.[0-30 Days]) >= SUM(c.bal)  then 0 else\n\n(case when (sum(c.[30-60 Days]) + SUM(c.[0-30 Days])) >=  SUM(c.bal) then 0 else\n\n(case when (sum(c.[30-60 Days]) + SUM(c.[0-30 Days]) + sum(c.[60-90 Days])) >= SUM(c.bal) then 0 else\n\n(case when (sum(c.[30-60 Days]) + SUM(c.[0-30 Days]) + sum(c.[60-90 Days]) + sum(c.[90-180 Days])) >= SUM(c.bal) then 0 else\n\n(case when (sum(c.[30-60 Days]) + SUM(c.[0-30 Days]) + sum(c.[60-90 Days]) +\n\nsum(c.[90-180 Days]) +  sum(c.[180-360 Days]) ) >= SUM(c.bal) then\n\n(SUM(c.bal) – (sum(c.[0-30 Days]) + sum(c.[30-60 Days]) + sum(c.[60-90 Days]) + sum(c.[90-180 Days])))\n\nelse sum(c.[180-360 Days])end)end)end)end)end) [180-360 Days],\n\n(case when SUM(c.[0-30 Days]) >= SUM(c.bal)  then 0 else\n\n(case when (sum(c.[30-60 Days]) + SUM(c.[0-30 Days])) >=  SUM(c.bal) then 0 else\n\n(case when (sum(c.[30-60 Days]) + SUM(c.[0-30 Days]) + sum(c.[60-90 Days])) >= SUM(c.bal) then 0 else sum(c.[Above 360 Days])\n\nend)end)end) [Above 360 Days]\n\nfrom (\n\nselect b.code, b.name,\n\nb.Bal, b.Val,b.[Pric],b.[FullStock],\n\nisnull(case when b.days <=30 then b.Bal end,0)[0-30 Days] ,\n\nisnull(case when b.days <= 30 then b.val end,0)[0-30 Days-Value] ,\n\nisnull(case when b.days between 31 and 60 then b.Bal end,0) [30-60 Days],\n\nisnull(case when b.days between 61 and 90 then b.Bal end,0) [60-90 Days],\n\nisnull(case when b.days between 91 and 180 then b.Bal end,0) [90-180 Days],\n\nisnull(case when b.days between 181 and 360 then b.Bal end,0) [180-360 Days],\n\nisnull(case when b.days >= 361 then b.Bal end,0) [Above 360 Days]\n\nfrom\n\n(\n\nselect a.code,a.name,\n\na.bal,a.val,datediff(dd,dt,@a)’days’,a.[Receipt],a.[Pric],a.[FullStock]\n\nfrom\n\n(\n\nselect max(t0.itemcode)’Code’,max(t0.Dscription)’Name’,\n\n(select sum(isnull(t11.transvalue,0))  from oinm t11 where (t11.itemcode = (t0.itemcode))\n\nand (t11.docdate <= ‘[%2]’ )) [Pric],\n\n(select (sum(isnull(t11.inqty,0)) – sum(isnull(t11.outqty,0)) ) from oinm t11 where (t11.itemcode = (t0.itemcode))\n\nand (t11.docdate <= ‘[%2]’ )) [FullStock],\n\n–(t0.Warehouse)’Wh’,\n\n(sum(isnull(t0.inqty,0))) [Receipt],\n\n(sum(isnull(t0.inqty,0)) – sum(isnull(t0.outqty,0)) )’Bal’,sum(isnull(t0.transvalue,0))’Val’,(t0.docdate)’dt’\n\nfrom oinm t0 inner join oitm t1 on t0.itemcode=t1.itemcode inner join oitb ob on ob.itmsgrpcod = t1.itmsgrpcod\n\nwhere  (ob.itmsgrpnam = ‘[%1]’)\n\nand (t0.docdate <= ‘[%2]’)\n\ngroup by t0.itemcode,t0.docdate\n\n)a\n\n)b\n\n)c where c.bal>0\n\ngroup by c.code, c.name\n\nAssigned Tags\n\nYou must be Logged on to comment or reply to a post.", null, "Dear sir pales help me chart of accounts queries\n\nfor ex:=\n\napril may jun july  in all mounts balancess in  mount wise pivot table querys\n\n5004 purchase of milk a/c\n\nplase tell me\n\n<removed by moderator>\n\nvenubabu", null, "I think this Query will full fill your requirement.\n\nDeclare @test VARCHAR(100)\n\nSelect @test= AcctCode from dbo.OACT S0 where S0.AcctCode= '[%0]'\n\nselect *\n\nfrom\n\n(select  datename(month, T0.RefDate) As Mth,T1.[Account]\n\n,SUM(T1.[Debit] - T1.[Credit]) as [Opening Balance]\n\nfrom OJDT T0  INNER JOIN\n\nJDT1 T1 ON T0.TransId = T1.TransId AND T1.[Account] = @test AND T0.RefDate between '20130401' and '20140331'\n\ngroup by datename(month, T0.RefDate),T1.[Account]\n\n) PP\n\npivot( sum([Opening Balance]) for mth in ([April],[May],[June],[July],[August],[September],[October],[November],[December],[January],[February],[March])) as X", null, "Swapnil Vichare\n\nTHANKING YOU\n\nG VENU BABU\n\nAnd another problem\n\nDear sir pales help me chart of accounts queries\n\nfor ex:=\n\napril may jun july  in all mounts balancess in  mount wise pivot table querys\n\n5004 purchase of milk a/c BANALNCESS IN B.P CODE WISE\n\nplase tell me\n\nVENUBABU", null, "Hi,\n\npurchase of milk a/c BANALNCESS IN B.P CODE WISE  ? ? ?", null, "chart of account 5004 balancess in year (mounth wise) report is ok\n\nbut another Query is\n\n1. chart of account  balance's in year (mounts wise balance's) project code wise query\n\n2.chart of account balance's in year (mounts wise balances) vendor code wise query (based on journal eatery)\n\nvenubabu", null, "plase help me\n\nvenubabu", null, "This query gives you project wise & Account wise report.\n\nIf you required a report for all accounts and required selection of Project Code then just remove  \"AND T1.[Account] = @AccCode\" From Join Between OJDT & JDT1\n\nDeclare @AccCode VARCHAR(100)\n\nSelect @AccCode= AcctCode from dbo.OACT S0 where S0.AcctCode= '[%0]'\n\nDeclare @PrjCode VARCHAR(100)\n\nSelect @PrjCode= PrjCode from dbo.OPRJ S1 where S1.PrjCode= '[%1]'\n\nselect *\n\nfrom\n\n(select  datename(month, T0.RefDate) As Mth,T1.[Account]\n\n,SUM(T1.[Debit] - T1.[Credit]) as [Opening Balance]\n\nfrom OJDT T0  INNER JOIN\n\nJDT1 T1 ON T0.TransId = T1.TransId AND T1.[Account] = @AccCode AND T0.RefDate between '20130401' and '20140331' Inner Join\n\nOACT T2 ON T1.Account=T2.AcctCode Left Outer Join\n\nOPRJ T3 ON T2.Project=T3.PrjCode\n\nWhere T3.PrjCode =@PrjCode\n\ngroup by datename(month, T0.RefDate),T1.[Account]\n\n) PP\n\npivot( sum([Opening Balance]) for mth in ([April],[May],[June],[July],[August],[September],[October],[November],[December],[January],[February],[March])) as X", null, "DEAR SIR REPORT IS EXECUTE BUT OUT PUT NO DATA\n\nHELP ME CHANGE QUERY\n\nMY PROBLEM\n\n1.ONE GL A/C CODE IS  SELECTED DISPLAY THE PROJECT CODE\n\nWISE,ALL MOUNT'S BALANCES  IN SAP BUSINESS ON QUERY\n\n(OR)\n\n2.GL A/C EX:- 5004 IS FIXED AND PROJECT CODE EX:-(CHI,MAP,GKD, NVP) IS FIXED\n\nAND DISPLAY REPORT TITLES  (PROJECT CODE ,ALL MONTHS)\n\n3.ONE GL A/C CODE IS  SELECTED ,PROJECT CODE  SELECTED\n\nVENDOR CODE WISE MOUTH (Business Partners)WISE  BALANCES  BASE IN (Journal Entry)   QUERY\n\nPEASE\n\nHELP ME CHANGE QUERY\n\nVENUBABU", null, "Declare @AccCode VARCHAR(100)\n\nSelect @AccCode= AcctCode from dbo.OACT S0 where S0.AcctCode= '[%0]'\n\nselect *\n\nfrom\n\n(select  datename(month, T0.RefDate) As Mth,T1.[Account]\n\n,SUM(T1.[Debit] - T1.[Credit]) as [Opening Balance]\n\nfrom OJDT T0  INNER JOIN\n\nJDT1 T1 ON T0.TransId = T1.TransId AND T1.[Account] = @AccCode AND T0.RefDate between '20130401' and '20140331' Inner Join\n\nOACT T2 ON T1.Account=T2.AcctCode Left Outer Join\n\nOPRJ T3 ON T2.Project=T3.PrjCode\n\nWhere (T3.PrjCode ='[%2]'  OR '[%2]' =' ')\n\ngroup by datename(month, T0.RefDate),T1.[Account]\n\n) PP\n\npivot( sum([Opening Balance]) for mth in ([April],[May],[June],[July],[August],[September],[October],[November],[December],[January],[February],[March])) as X\n\nThis Query will gives you proper report But it is important to select Proper Project Code for that selected Account.\n\nThere is other provision in this query, If you does not know project code for that account then leave project code field blank .", null, "Is you use segmentation in Cart of account in your company.", null, "QUERY IS NOT WORKING SIR\n\n5004 – IS  PURCHASE OF MILK A/CIN CHART OF ACCOUNT CODE\n\nMY PURCHASE OF MILK IN BRANCHS IS THER\n\nEX:  APRIAL MOUNTH\n\n5004- CHART OF ACCOUNT BALANCESS IN 100000/- MAIN HEAD TOTAL\n\nBRANCHS PROJECTCODE OR GROUPMAME WISE\n\nGROUPNAME   PROJECT CODE  AMOUNT\n\nCHITTOOR-                         CHI         50000\n\nGURRAMKONDA             GKD       25000\n\nTOTAL  100000\n\nTHIS QUERY IS EXECUT IN CURRANT ANSWER IN BETWEEN DAYS\n\nSELECT T0.[Project],(sum(T0.Debit)-sum(T0.Credit))as 'value' FROM JDT1 T0 WHERE T0.Account='5004' and T0.RefDate>=[%0] and T0.RefDate<=[%1] GROUP BY T0.Project\n\n # Project Code value 1 CHI 2,349,671.12 2 GAV 926,025.83 3 GKD 1,146,867.70 4 GPP 434,241.00 5 HYD 979,550.00 6 KLK 651,926.00 7 KOD 4,661,521.33 8 KSP 3,208,826.18 9 MAP 2,202,668.56 10 NVP 1,368,536.86 11 SHA 3,129,242.22 12 VAR 3,672,568.59\n\nBUT MY PROBLEM IS MOUNTH WISE BALANCESS IN PIOVT TABLE QUERY IN ALL MOUNTHES IN BASE ON UPQUERY\n\nVENUBABU", null, "This Query definitely solve your issue for Project wise balance of chart of account 🙂\n\nDeclare @AccCode VARCHAR(100) Select @AccCode= AcctCode from dbo.OACT S0 where S0.AcctCode= '[%0]'     select * from (select  datename(month, T0.RefDate) As Mth,T1.[Account],T1.[Project]  ,SUM(T1.[Debit] - T1.[Credit]) as [Opening Balance]     from OJDT T0  INNER JOIN      JDT1 T1 ON T0.TransId = T1.TransId AND T1.[Account] = @AccCode AND T0.RefDate between '20130401' and '20140331' Inner Join      OACT T2 ON T1.Account=T2.AcctCode Left Outer Join      OPRJ T3 ON T2.Project=T3.PrjCode Where (T3.PrjCode ='[%2]'  OR '[%2]' =' ')    group by datename(month, T0.RefDate),T1.[Account],T1.[Project]      ) PP pivot( sum([Opening Balance]) for mth in ([April],[May],[June],[July],[August],[September],[October],[November],[December],[January],[February],[March])) as X/body>", null, "This Query  solve my issue for Project wise balance of chart of account\n\ni am use fill query sir\n\nTHANKING YOU\n\nvenubabu", null, "Till now i doesn't understand your 2 Query requirement.\n\nWhat u exactly want? 😕\n\nPlease explain 2nd query issue in brief.", null, "Select  P.CardCode,P.CardName,\n\n as [Jan],\n\n as [Feb],\n\n as [Mar],\n\n as [Apr],\n\n as [May],\n\n as [Jun],\n\n as [Jul],\n\n as [Aug],\n\n as [Sep],\n\n as [Oct],\n\n as [Nov],\n\n as [Dec]\n\nfrom((select T1.CardCode ,T1.CardName,(sum(T0.Debit)-sum(T0.Credit))as 'value' ,month(t0.docdate) as 'Month'\n\nfrom JDT1 T0 Inner join OCRD T1 on T1.CardCode=T0.ShortName\n\ninner join OCRG T2 on T1.GroupCode=T2.GroupCode\n\nwhere T0.ContraAct='2657' and  T2.GroupName=[%0]\n\nand T1.CardType='S' and T2.GroupType='S'\n\ngroup by T1.CardCode,T1.CardName) as A\n\nPivot((sum(T0.Debit)-sum(T0.Credit))as 'value' FOR [Month] IN (,,,,,,,,,,,)) P\n\norder by T1.CardCode\n\nQUERY IS PROBLEM SIR\n\nTHIS Query is execute me\n\nsolow the vendor wise balancess report problem\n\nplase help me", null, "Select * from(select T1.CardCode ,T1.CardName,(sum(T0.Debit)-sum(T0.Credit))as 'value'\n\nfrom JDT1 T0 Inner join OCRD T1 on T1.CardCode=T0.ShortName\n\ninner join OCRG T2 on T1.GroupCode=T2.GroupCode\n\nwhere T0.ContraAct='2657' and  T2.GroupName=[%0]\n\nand T1.CardType='S' and T2.GroupType='S'\n\ngroup by T1.CardCode,T1.CardName) as A\n\nwhere A.value<>0\n\norder by A.CardCode\n\nthis query total balances in all mount's but  i am expect query in mounts wise balancess query\n\npales help me\n\nvenubabu", null, "This query will solve your issue for monthwise balance of BP, Just select Group name for it.\n\nselect *\n\nfrom\n\n(SELECT datename(month, T0.RefDate) As Mth,T2.CardCode,T2.CardName,SUM(T1.Credit-T1.Debit) AS [Value]  FROM\n\nOJDT T0  INNER JOIN\n\nJDT1 T1 ON T0.TransId = T1.TransId AND T1.ContraAct='2657' Inner Join\n\nOCRD T2 ON T1.ShortName=T2.CardCode AND T2.CardType='S' Inner Join\n\nOCRG T3 ON T2.GroupCode=T3.GroupCode AND T3.GroupType='S'\n\nWhere T3.GroupName ='[%0]'\n\ngroup by datename(month, T0.RefDate),T2.CardCode,T2.CardName\n\n) PP\n\npivot( sum([Value]) for mth in ([April],[May],[June],[July],[August],[September],[October],[November],[December],[January],[February],[March])) as X", null, "If you want total report for all groups then use this query and keep Group name field blank .\n\nselect *\n\nfrom\n\n(SELECT datename(month, T0.RefDate) As Mth,T2.CardCode,T2.CardName,T3.GroupName,SUM(T1.Credit-T1.Debit) AS [Value]  FROM\n\nOJDT T0  INNER JOIN\n\nJDT1 T1 ON T0.TransId = T1.TransId AND T1.ContraAct='2657' Inner Join\n\nOCRD T2 ON T1.ShortName=T2.CardCode AND T2.CardType='S' Inner Join\n\nOCRG T3 ON T2.GroupCode=T3.GroupCode AND T3.GroupType='S'\n\nWhere (T3.GroupName ='[%0]' OR '[%0]'=' ')\n\ngroup by datename(month, T0.RefDate),T2.CardCode,T2.CardName,T3.GroupName\n\n) PP\n\npivot( sum([Value]) for mth in ([April],[May],[June],[July],[August],[September],[October],[November],[December],[January],[February],[March])) as X", null, "i think your both problems r solved, if u have any issues then tell me about that.\n\n•    1)  This Query definitely solve your issue for Project wise balance of chart of account 🙂\n\nDeclare @AccCode VARCHAR(100)\n\nSelect @AccCode= AcctCode from dbo.OACT S0 where S0.AcctCode= '[%0]'\n\nselect *\n\nfrom\n\n(select  datename(month, T0.RefDate) As Mth,T1.[Account],T1.[Project]\n\n,SUM(T1.[Debit] - T1.[Credit]) as [Opening Balance]\n\nfrom OJDT T0  INNER JOIN\n\nJDT1 T1 ON T0.TransId = T1.TransId AND T1.[Account] = @AccCode AND T0.RefDate between '20130401' and '20140331' Inner Join\n\nOACT T2 ON T1.Account=T2.AcctCode Left Outer Join\n\nOPRJ T3 ON T2.Project=T3.PrjCode\n\nWhere (T3.PrjCode ='[%2]'  OR '[%2]' =' ')\n\ngroup by datename(month, T0.RefDate),T1.[Account],T1.[Project]\n\n) PP\n\npivot( sum([Opening Balance]) for mth in ([April],[May],[June],[July],[August],[September],[October],[November],[December],[January],[February],[March])) as X\n\n• 2)          This query will solve your issue for monthwise balance of BP, Just select Group name for     it.\n\nIf you want total report for all groups then use this query and keep Group name field blank .\n\nselect *\n\nfrom\n\n(SELECT datename(month, T0.RefDate) As Mth,T2.CardCode,T2.CardName,T3.GroupName,SUM(T1.Credit-T1.Debit) AS [Value]  FROM\n\nOJDT T0  INNER JOIN\n\nJDT1 T1 ON T0.TransId = T1.TransId AND T1.ContraAct='2657' Inner Join\n\nOCRD T2 ON T1.ShortName=T2.CardCode AND T2.CardType='S' Inner Join\n\nOCRG T3 ON T2.GroupCode=T3.GroupCode AND T3.GroupType='S'\n\nWhere (T3.GroupName ='[%0]' OR '[%0]'=' ')\n\ngroup by datename(month, T0.RefDate),T2.CardCode,T2.CardName,T3.GroupName\n\n) PP\n\npivot( sum([Value]) for mth in ([April],[May],[June],[July],[August],[September],[October],[November],[December],[January],[February],[March])) as X", null, "This Query  solve my issue for vendor  balance\n\ni am use fill query sir\n\nTHANKING YOU\n\nvenubabu", null, "dear\n\nsir\n\nmy name is venubabu\n\nmy qualification is m,com\n\ni am accountant in jyothi dairy pvt ltd Hyderabad\n\n4 years  in working in sap business one software in accounting purpose my job\n\ni am change this job software side (implantation side sap business one)\n\nbasic qualification tell me in implantation side sap business one\n\nwhat are you course training tell me sir\n\nvenubabu", null, "Hi Naga Srinu,\n\nThanks & Regards,\n\nNagarajan", null, "dear sir\n\nhelp me\n\nquery manager in queries is execute report  BUT EXECUTE direct in FIXED REPORT (PLD,EXCEL)  PLASE TELL ME\n\nVENUBABU", null, "Hi,\n\nTo get answers for problem, you should post as discussion.\n\nRegards,\n\nNagarajan", null, "SIR\n\nMY problem is queries is setting in execute to excel format or any stranded report is pose-bull\n\ntell me in process\n\nvenubabu", null, "Hi,\n\nDo you know write query in SAP B1?", null, "yas", null, "I don't exactly understand your problem. 😕\n\nI think you required to Export the output of your report in Excel format.\n\nRun your Query in Query manager and click on Excel icon in tool bar to Export your report in Excel Format.", null, "DEAR\n\nsir\n\nBP CODE ,BP NAME ,OFFSET ACCOUNT  5657, 4006 ,1607 ,1662, 1686 BALANCES\n\nEX:- SELECTED BETWEEN DAYS 01.04.13 TO 15.04.13\n\nCS001-VENUBABU-                                   5000-  600-  8000-  4000- 1500\n\nIN BETWEEN DAY\n\nHELP ME\n\nVENUBABU", null, "Kennedy T\nBlog Post Author\n\nRaise in questing not here read rules of engagement first", null, "dear sir plase help me\n\nSelect * from(select T1.CardCode ,T1.CardName,(sum(T0.Debit)-sum(T0.Credit))as 'purchase',\n\nFrom  JDT1 T0 Inner join OCRD T1 on T1.CardCode=T0.ShortName\n\ninner join OCRG T2 on T1.GroupCode=T2.GroupCode\n\n(where T0.Account='2511' as 'purchase' and where T0.Account='1857' as 'advance'),\n\nwhere T2.GroupName=[%0] and ((T0.RefDate>=[%1] and T0.RefDate<=[%2]) or (T0.RefDate<[%1]))\n\nand T1.CardType='S' and T2.GroupType='S'\n\ngroup by T1.CardCode,T1.CardName\n\norder by A.CardCode\n\nthis report not excute sir\n\nmy problem vendor chart of accountent balaancess in bussiness partener master data  parchase ac 2511 and advance  ac 1857 two account balancess is there query is excute  2511 balancess and 1857 balancess\n\nex:-total bp balance 15000/- 2511 ac -10000/-and 1857 ac  +25000/-\n\nplase help me this report", null, "What a complete Inventory Aging Report. Thumbs up.", null, "Hi,\n\nNice post.\n\nThanks & Regards\n\nAnkit Chauhan", null, "Hi Sap Gurus,\n\nCould you please provide info cube & Multi provider cube on inventory ageing report\n\nregards,\n\nChand", null, "Hi Kennedy\n\nThis report gives stock aging details of old stock also. For eg. My current stock qty is 500 is in 0-30 days but my overall aging is 30-60 days is 2000 Qty. This report shows data for both 0-30 days as 500 qty and 30-60 days as 2000 Qty.\n\nPlease let me know how to make visible only the 0-30 days Stock Qty which is 500 when I choose Particular date.\n\nThis is the Query Which I tested\n\nDeclare @a char(60)\n\nSet @a = (select distinct t20.docdate from oinm t20 where t20.docdate = '20181231')\n\nSelect distinct c.code, c.name\n\n,SUM(c.bal) [Bal],sum(c.Val) [Val],\n\n(case when avg(c.[FullStock]) > 0 then avg(c.[Pric]) / avg(c.[FullStock]) end ) [AvgPrice],\n\n(case when SUM(c.[0-30 Days]) >= SUM(c.bal) then SUM(c.bal) else SUM(c.[0-30 Days]) end) [0-30 Days],\n\n(case when SUM(c.[0-30 Days]) >= SUM(c.bal) then 0 else\n\n(case when (sum(c.[30-60 Days]) + SUM(c.[0-30 Days])) >= SUM(c.bal) then (SUM(c.bal) - sum(c.[0-30 Days]))\n\nelse sum(c.[30-60 Days]) end)end) [30-60 Days],\n\n(case when SUM(c.[0-30 Days]) >= SUM(c.bal) then 0 else\n\n(case when (sum(c.[30-60 Days]) + SUM(c.[0-30 Days])) >= SUM(c.bal) then 0 else\n\n(case when (sum(c.[30-60 Days]) + SUM(c.[0-30 Days]) + sum(c.[60-90 Days])) >= SUM(c.bal) then\n\n(SUM(c.bal) - (sum(c.[0-30 Days]) + sum(c.[30-60 Days]))) else\n\nsum(c.[60-90 Days])end)end) end) [60-90 Days],\n\n(case when SUM(c.[0-30 Days]) >= SUM(c.bal) then 0 else\n\n(case when (sum(c.[30-60 Days]) + SUM(c.[0-30 Days])) >= SUM(c.bal) then 0 else\n\n(case when (sum(c.[30-60 Days]) + SUM(c.[0-30 Days]) + sum(c.[60-90 Days])) >= SUM(c.bal) then 0 else\n\n(case when (sum(c.[30-60 Days]) + SUM(c.[0-30 Days]) + sum(c.[60-90 Days]) + sum(c.[90-180 Days])) >= SUM(c.bal) then\n\n(SUM(c.bal) - (sum(c.[0-30 Days]) + sum(c.[30-60 Days]) + sum(c.[60-90 Days]))) else sum(c.[90-180 Days])\n\nend)end) end) end) [90-180 Days],\n\n(case when SUM(c.[0-30 Days]) >= SUM(c.bal) then 0 else\n\n(case when (sum(c.[30-60 Days]) + SUM(c.[0-30 Days])) >= SUM(c.bal) then 0 else\n\n(case when (sum(c.[30-60 Days]) + SUM(c.[0-30 Days]) + sum(c.[60-90 Days])) >= SUM(c.bal) then 0 else\n\n(case when (sum(c.[30-60 Days]) + SUM(c.[0-30 Days]) + sum(c.[60-90 Days]) + sum(c.[90-180 Days])) >= SUM(c.bal) then 0 else\n\n(case when (sum(c.[30-60 Days]) + SUM(c.[0-30 Days]) + sum(c.[60-90 Days]) +\n\nsum(c.[90-180 Days]) + sum(c.[180-360 Days]) ) >= SUM(c.bal) then\n\n(SUM(c.bal) - (sum(c.[0-30 Days]) + sum(c.[30-60 Days]) + sum(c.[60-90 Days]) + sum(c.[90-180 Days])))\n\nelse sum(c.[180-360 Days])end)end)end)end)end) [180-360 Days],\n\n(case when SUM(c.[0-30 Days]) >= SUM(c.bal) then 0 else\n\n(case when (sum(c.[30-60 Days]) + SUM(c.[0-30 Days])) >= SUM(c.bal) then 0 else\n\n(case when (sum(c.[30-60 Days]) + SUM(c.[0-30 Days]) + sum(c.[60-90 Days])) >= SUM(c.bal) then 0 else sum(c.[Above 360 Days])\n\nend)end)end) [Above 360 Days]\n\nfrom (\n\nselect b.code, b.name,\n\nb.Bal, b.Val,b.[Pric],b.[FullStock],\n\nisnull(case when b.days <=30 then b.Bal end,0)[0-30 Days] ,\n\nisnull(case when b.days <= 30 then b.val end,0)[0-30 Days-Value] ,\n\nisnull(case when b.days between 31 and 60 then b.Bal end,0) [30-60 Days],\n\nisnull(case when b.days between 61 and 90 then b.Bal end,0) [60-90 Days],\n\nisnull(case when b.days between 91 and 180 then b.Bal end,0) [90-180 Days],\n\nisnull(case when b.days between 181 and 360 then b.Bal end,0) [180-360 Days],\n\nisnull(case when b.days >= 361 then b.Bal end,0) [Above 360 Days]\n\nfrom\n\n(\n\nselect a.code,a.name,\n\na.bal,a.val,datediff(dd,dt,@a)'days',a.[Receipt],a.[Pric],a.[FullStock]\n\nfrom\n\n(\n\nselect max(t0.itemcode)'Code',max(t0.Dscription)'Name',\n\n(select sum(isnull(t11.transvalue,0)) from oinm t11 where (t11.itemcode = (t0.itemcode))\n\n) [Pric],\n\n(select (sum(isnull(t11.inqty,0)) - sum(isnull(t11.outqty,0)) ) from oinm t11 where (t11.itemcode = (t0.itemcode))\n\nand (t11.docdate <= '20181231' )) [FullStock],\n\n(t0.Warehouse)'Wh',\n\n(sum(isnull(t0.inqty,0))) [Receipt],\n\n(sum(isnull(t0.inqty,0)) - sum(isnull(t0.outqty,0)) )'Bal',sum(isnull(t0.transvalue,0))'Val',(t0.docdate)'dt'\n\nfrom oinm t0 inner join oitm t1 on t0.itemcode=t1.itemcode inner join oitb ob on ob.itmsgrpcod = t1.itmsgrpcod\n\ngroup by t0.itemcode,t0.docdate,t0.Warehouse\n\n)a\n\n)b\n\n)c where c.bal>0 and c.Code='6PAB-84x130PNA'\n\ngroup by c.code, c.name" ]

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[ "# Getting started with matplotlib\n\n## Imperative vs. Object-oriented Syntax #\n\nMatplotlib supports both object-oriented and imperative syntax for plotting. The imperative syntax is intentionally designed to be very close to Matlab syntax.\n\nThe imperative syntax (sometimes called ‘state-machine’ syntax) issues a string of commands all of which act on the most recent figure or axis (like Matlab). The object-oriented syntax, on the other hand, explicitly acts on the objects (figure, axis, etc.) of interest. A key point in the zen of Python states that explicit is better than implicit so the object-oriented syntax is more pythonic. However, the imperative syntax is convenient for new converts from Matlab and for writing small, “throwaway” plot scripts. Below is an example of the two different styles.\n\n``````import matplotlib.pyplot as plt\nimport numpy as np\n\nt = np.arange(0, 2, 0.01)\ny = np.sin(4 * np.pi * t)\n\n# Imperative syntax\nplt.figure(1)\nplt.clf()\nplt.plot(t, y)\nplt.xlabel('Time (s)')\nplt.ylabel('Amplitude (V)')\nplt.title('Sine Wave')\nplt.grid(True)\n\n# Object oriented syntax\nfig = plt.figure(2)\nfig.clf()\nax.plot(t, y)\nax.set_xlabel('Time (s)')\nax.set_ylabel('Amplitude (V)')\nax.set_title('Sine Wave')\nax.grid(True)\n``````\n\nBoth examples produce the same plot which is shown below.", null, "## Installation and Setup #\n\nThere are several ways to go about installing matplotlib, some of which will depend on the system you are using. If you are lucky, you will be able to use a package manager to easily install the matplotlib module and its dependencies.\n\n# Windows #\n\nOn Windows machines you can try to use the pip package manager to install matplotlib. See here for information on setting up pip in a Windows environment.\n\n# OS X #\n\nIt is recommended that you use the pip package manager to install matplotlib. If you need to install some of the non-Python libraries on your system (e.g. `libfreetype`) then consider using homebrew.\n\nIf you cannot use pip for whatever reason, then try to install from source.\n\n# Linux #\n\nIdeally, the system package manager or pip should be used to install matplotlib, either by installing the `python-matplotlib` package or by running `pip install matplotlib`.\n\nIf this is not possible (e.g. you do not have sudo privileges on the machine you are using), then you can install from source using the `--user` option: `python setup.py install --user`. Typically, this will install matplotlib into `~/.local`.\n\n## Debian/Ubuntu #\n\n`sudo apt-get install python-matplotlib`\n\n## Fedora/Red Hat #\n\n`sudo yum install python-matplotlib`\n\n# Troubleshooting #\n\nSee the matplotlib website for advice on how to fix a broken matplotlib.\n\n## Two dimensional (2D) arrays #\n\nDisplay a two dimensional (2D) array on the axes.\n\n``````import numpy as np\nfrom matplotlib.pyplot import imshow, show, colorbar\n\nimage = np.random.rand(4,4)\nimshow(image)\ncolorbar()\nshow()\n``````", null, "## Customizing a matplotlib plot #\n\n``````import pylab as plt\nimport numpy as np\n\nplt.style.use('ggplot')\n\nfig = plt.figure(1)\nax = plt.gca()\n\n# make some testing data\nx = np.linspace( 0, np.pi, 1000 )\ntest_f = lambda x: np.sin(x)*3 + np.cos(2*x)\n\n# plot the test data\nax.plot( x, test_f(x) , lw = 2)\n\n# set the axis labels\nax.set_ylabel(r'\\$f(x)\\$', fontsize=14, labelpad=25, rotation=0)\n\n# set axis limits\nax.set_xlim(0,np.pi)\n\nplt.draw()\n``````", null, "``````# Customize the plot\nax.grid(1, ls='--', color='#777777', alpha=0.5, lw=1)\nax.tick_params(labelsize=12, length=0)\nax.set_axis_bgcolor('w')\n# add a legend\nleg = plt.legend( ['text'], loc=1 )\nfr = leg.get_frame()\nfr.set_facecolor('w')\nfr.set_alpha(.7)\nplt.draw()\n``````", null, "" ]

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[ "# Field Update in Workflow Rule\n\nI need to set a field on the `Order` object using a `Workflow Field Update`, and the updated value should be formatted as follows:\n\n``````<Account JDE Ship to number> - YYMMDDHHmmSS\n``````\n\nI have tried to the following formula:\n\n``````Account.JDE_ship_to_account_no__c - DATETIMEVALUE()\n``````\n\nBut the formula editor gives me this error:\n\nError: Incorrect number of parameters for function 'DATETIMEVALUE()'.` Expected 1, received 0\n\nHow can I make this field update work?\n\n• Where does the date you want to put in there come from? – Alexander Aeons Torn Feb 9 '17 at 13:53\n• What is `Account JDE Shipto number`? Do you want your field to look like `XX999 - YYMMDDHHmmSS`, assuming XX999 is your `Account JDE Shipto number` and YYMMDDHHmmSS a date/time stamp of last value update on some field? – o-lexi Feb 9 '17 at 18:26\n\nYou should use something like the one here:\n\nFormula\n\n``````Account.JDE_ship_to_account_no__c + '-' +\nRight( Text( Year( Today() ) ), 2) +\nLpad( Text( Month( Today() )), 2, '0' ) +\nLpad( Text( Day( Today() )), 2, '0' ) +\nSUBSTITUTE(Left( Right( Text( Now() ), 9), 8), ':', '')\n``````\n\nResult", null, "Warning!\n\nDon't be scared, just note that the time part will return the value in GMT not in your local time. Details and some very useful tips on working with date-time formulas can be found here.\n\nDATETIMEVALUE\n\nUse: DATETIMEVALUE(expression) and replace expression with a date/time or text value, merge field, or expression.Description: Returns a year, month, day and GMT time\n\nField Name & '-' & YYYYMMDDHHSS\n\nY know how to print the YYYYMMDD but i'm not sure how to print the hours and seconds on an easy way since there on build in function yet :\n\nYou can check more discussion about that here: https://developer.salesforce.com/forums/?id=906F00000008v7gIAA\n\nFor YYYYMMDD assuming you always want 2 digits on MM & DD I would use :\n\n``````TEXT(YEAR(DATEVALUE(NOW()))) & IF(LEN(TEXT(MONTH(DATEVALUE(NOW()))))=1,'0'&TEXT(MONTH(DATEVALUE(NOW()))),TEXT(MONTH(DATEVALUE(NOW())))) & IF(LEN(TEXT(DAY(DATEVALUE(NOW()))))=1,'0'&TEXT(DAY(DATEVALUE(NOW()))),TEXT(DAY(DATEVALUE(NOW()))))\n``````\n\nBut probably there are more elegant ways of doing it.\n\n• How to add the field as before , that field is in Number data type – SFDC Feb 9 '17 at 14:16\n• but you guided thank you but i want this format YYMMDDHHmmSS – SFDC Feb 9 '17 at 14:36\n\nDATETIMEVALUE() expect a parameter Date as as String to convert to a date\n\nExample :\n\n``````DATETIMEVALUE(\"2005-11-15 17:00:00\")\n``````\n\nTo get the current timestamp use the function NOW()" ]

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[ "Coming up with properties is hard.\n\nFor example, say you want to implement set coverage. Write a function that takes a list of sets, and returns the smallest list of these sets, where every element in the large set is now in the small set.\n\n``````large_set_of_sets = [{1}, {1, 2}, {2}]\nreduce_sets(large_set_of_sets) # [{1, 2}]\n``````\n\nWe’re going to walk through how to implement properties that test set coverage. Once we’ve written properties that give boundaried so it’s behavior, hypothesis will discover bugs in how we’ve written our code and give us concrete examples of bugs to fix.\n\n## What are our properties?\n\nWhat are the properties of `reduce_sets` ? If you want to steal from Discrete Math or CS, here are a couple:\n\n1. The number of sets in the output is the same or less than the number of sets from the input.\n2. Every number in the list of input sets is in the list of output sets.\n3. Every set of numbers in the output list was also in the input list.\n4. There are no duplicate sets in the output.\n5. No output set is a subset of any other output set\n6. No output set is a subset of the other sets combined.\n7. The output set is the smallest possible number of sets with the previous properties still satisfied.\n\n(I got most of these from Jessica Kerr)\n\nThe last property is particularly hard, for example the following two solutions are valid on all the properties but the last property:\n\n``````large_set_of_sets = [{1}, {1, 2}, {2}]\nreduce_sets(large_set_of_sets) # [{1}, {2}] -- covers all the elements\nreduce_sets(large_set_of_sets) # [{1, 2}] -- but this is smaller\n``````\n\nBut we’ll get to why it’s hard at the end.\n\nLet’s work through how we’d implement these properties.\n\n## A Unit Test\n\nWe’ll be using hypothesis with pytest.\n\nLet’s turn our first example of how set coverage works into a unit test on its own.\n\n``````from setcoverage import reduce_sets\n\ndef test_example():\ns = [{1}, {1, 2}, {2}]\nassert reduce_sets(s) == [{1, 2}]\n``````\n\nAnd here’s the minimum amount of code it takes to make the test pass:\n\n``````def reduce_sets(sets):\nreturn [{1, 2}]\n``````\n\nLook! Set Coverage in one line! Running the tests proves our success and genius:\n\n``````\\$ tox -e py27\npy27 installed: enum34==1.1.6,hypothesis==3.4.1,py==1.4.31,pytest==2.9.2,setcoverage==0.0.1,wheel==0.24.0\npy27 runtests: PYTHONHASHSEED='4047888479'\npy27 runtests: commands | py.test\n======================= test session starts =======================\nplatform darwin -- Python 2.7.10, pytest-2.9.2, py-1.4.31, pluggy-0.3.1\nplugins: hypothesis-3.4.1\ncollected 1 items\n\ntests/test_thing.py .\n\n==================== 1 passed in 0.01 seconds =====================\n_____________________________ summary _____________________________\npy27: commands succeeded\ncongratulations :)\n``````\n\n## Our First Property: The Cardinality Constraint\n\nThe first property we’re going to implement on our set coverage code is this: The output list of sets should have the same number of sets or fewer a number of sets than the input list of sets.\n\nWe know this has to be a property of a function that solves set coverage because if it didn’t have this property, that would mean it sometimes returns more sets than it was given. That goes against the premise of what it does – finding the smallest number of sets.\n\nHere’s what that property looks like in python:\n\n``````import hypothesis as h\nfrom hypothesis import strategies as st\n\n# ...\n\n@h.given(st.lists(st.sets(st.integers())))\ndef test_that_output_has_fewer_sets_than_input(sets):\noutput = reduce_sets(sets)\nassert len(output) <= len(sets)\n``````\n\nNow we’ll let hypothesis find us a failing test case by just re-running our test suite.\n\n``````\n======================= test session starts =======================\nplatform darwin -- Python 2.7.10, pytest-2.9.2, py-1.4.31, pluggy-0.3.1\nplugins: hypothesis-3.4.1\ncollected 2 items\n\ntests/test_thing.py .F\n\n============================ FAILURES =============================\n___________ test_that_output_has_fewer_sets_than_input ____________\n\n@h.given(st.lists(st.sets(st.integers())))\n> def test_that_output_has_fewer_sets_than_input(sets):\n\ntests/test_thing.py:13:\n_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _\n.tox/py27/lib/python2.7/site-packages/hypothesis/core.py:520: in wrapped_test\nprint_example=True, is_final=True\n.tox/py27/lib/python2.7/site-packages/hypothesis/executors.py:58: in default_new_style_executor\nreturn function(data)\n.tox/py27/lib/python2.7/site-packages/hypothesis/core.py:110: in run\nreturn test(*args, **kwargs)\n_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _\n\nsets = []\n\n@h.given(st.lists(st.sets(st.integers())))\ndef test_that_output_has_fewer_sets_than_input(sets):\noutput = reduce_sets(sets)\n> assert len(output) <= len(sets)\nE assert 1 <= 0\nE + where 1 = len([set([1, 2])])\nE + and 0 = len([])\n\ntests/test_thing.py:15: AssertionError\n--------------------------- Hypothesis ----------------------------\nFalsifying example: test_that_output_has_fewer_sets_than_input(sets=[])\n=============== 1 failed, 1 passed in 0.10 seconds ================\n_____________________________ summary _____________________________\nERROR: py27: commands failed\n``````\n\nHypothesis found data which will show our implementation doesn’t satisfy that property. Let’s zoom in on the error itself to see what Hypothesis tried.\n\n``````sets = []\n\n@h.given(st.lists(st.sets(st.integers())))\ndef test_that_output_has_fewer_sets_than_input(sets):\noutput = reduce_sets(sets)\n> assert len(output) <= len(sets)\nE assert 1 <= 0\nE + where 1 = len([set([1, 2])])\nE + and 0 = len([])\n\ntests/test_thing.py:15: AssertionError\n--------------------------- Hypothesis ----------------------------\nFalsifying example: test_that_output_has_fewer_sets_than_input(sets=[])\n``````\n\nAt the beginning of the failure test case, we can see the input data clearly labeled: `sets = []`. This means that Hypothesis tried to run the test with an empty list. If we look a bit closer at the assertion that failed we can see how the property we define interacts with the input data:\n\n``````> assert len(output) <= len(sets)\nE assert 1 <= 0\nE + where 1 = len([set([1, 2])])\nE + and 0 = len([])\n``````\n\nThis is a feature of pytest that extends Python’s built-in assertion statement to provide useful debug information. It’s showing us how the empty list has zero elements, where our function is still returning data to solve its own use-case.\n\nLet’s return to our implementation and write the minimum amount of code to pass both the test case and this property:\n\n``````def reduce_sets(sets):\n# remove first and last item\nreturn sets[1:][:-1]\n``````\n\nIt just so happens that slice syntax will return empty lists if you over-slice and on our test case the input only needs the first and last items pruned to pass.\n\n``````======================= test session starts =======================\nplatform darwin -- Python 2.7.10, pytest-2.9.2, py-1.4.31, pluggy-0.3.1\nplugins: hypothesis-3.4.1\ncollected 2 items\n\ntests/test_thing.py ..\n\n==================== 2 passed in 0.31 seconds =====================\n_____________________________ summary _____________________________\npy27: commands succeeded\ncongratulations :)\n``````\n\n# Property Two: All elements in the Sets are in the Output from the Input\n\nThe next property of a function that does set coverage is that no items are lost in the output list. To implement this property, we’ll just say that the set of all items coming in are the same as the set of all items coming out.\n\n``````from functools import reduce\n\n# ...\n\n@h.given(st.lists(st.sets(st.integers())))\ndef test_no_items_lost(sets):\nunion = lambda a, b: a.union(b)\nall_input = reduce(union, sets, set())\nall_output = reduce(union, reduce_sets(sets), set())\nassert all_input == all_output\n``````\n\nEasy enough, let’s see what Hypothesis finds:\n\n``````======================= test session starts =======================\nplatform darwin -- Python 2.7.10, pytest-2.9.2, py-1.4.31, pluggy-0.3.1\nplugins: hypothesis-3.4.1\ncollected 3 items\n\ntests/test_thing.py ..F\n\n============================ FAILURES =============================\n_______________________ test_no_items_lost ________________________\n\n@h.given(st.lists(st.sets(st.integers())))\n> def test_no_items_lost(sets):\n\ntests/test_thing.py:20:\n_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _\n.tox/py27/lib/python2.7/site-packages/hypothesis/core.py:520: in wrapped_test\nprint_example=True, is_final=True\n.tox/py27/lib/python2.7/site-packages/hypothesis/executors.py:58: in default_new_style_executor\nreturn function(data)\n.tox/py27/lib/python2.7/site-packages/hypothesis/core.py:110: in run\nreturn test(*args, **kwargs)\n_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _\n\nsets = [set()]\n\n@h.given(st.lists(st.sets(st.integers())))\ndef test_no_items_lost(sets):\nunion = lambda a, b: a.union(b)\nall_input = reduce(union, sets, set())\nall_output = reduce(union, reduce_sets(sets), set())\n> assert all_input == all_output\nE assert set() == set([])\nE Extra items in the left set:\nE 0\nE Use -v to get the full diff\n\ntests/test_thing.py:24: AssertionError\n--------------------------- Hypothesis ----------------------------\nFalsifying example: test_no_items_lost(sets=[{0}])\n``````\n\nLooks like our pruning strategy from before is removing valuable sets. Let’s come up with a way to satisfy all the properties we’ve written so far:\n\n``````def reduce_sets(sets):\nif sets == []:\nreturn sets\n# create a set of the items found\no = set()\nfor s in sets:\no = o.union(s)\nreturn [o]\n``````\n\n## Property Three: Every output set can be found from the list of input sets\n\nThis property basically is making sure no novel sets are created from our function. We can implement this property like this:\n\n``````@h.given(st.lists(st.sets(st.integers())))\ndef test_all_sets_come_from_input(sets):\noutput = reduce_sets(sets)\nforbidden_sets = [s for s in output if s not in sets]\nassert len(forbidden_sets) == 0\n``````\n\nBy running our tests, we can find a case that this fails on:\n\n``````sets = [set(), set()]\n\n@h.given(st.lists(st.sets(st.integers())))\ndef test_all_sets_come_from_input(sets):\noutput = reduce_sets(sets)\nforbidden_sets = [s for s in output if s not in sets]\n> assert len(forbidden_sets) == 0\nE assert 1 == 0\nE + where 1 = len([set([0, 1])])\n\ntests/test_thing.py:31: AssertionError\n--------------------------- Hypothesis ----------------------------\nFalsifying example: test_all_sets_come_from_input(sets=[{0}, {1}])\n``````\n\nIt’s clear why this is happening since our implementation is strictly creating a new superset from the sets its given.\n\nLet’s come up with a new way to satisfy all the tests we’ve seen so far:\n\n``````def _without(objects, i):\n\"\"\"\nget a list of objects without\nthe element at i\n\"\"\"\nreturn objects[:i] + objects[i+1:]\n\ndef _prune_one(sets):\n\"\"\" attempt to remove a single item \"\"\"\nfor i, e in enumerate(sets):\nfor s in _without(sets, i):\nif s == e:\nreturn _without(sets, i)\nreturn sets\n\ndef reduce_sets(sets):\none_removed = _prune_one(sets)\ntwo_removed = _prune_one(one_removed)\nreturn two_removed\n``````\n\nOur implementation just looks for items to safely remove, and removes them. I’m only doing this twice because I want only want to solve for the properties we’ve seen so far.\n\nNote: This implementation will break that first unit test we wrote. To prevent this unit test from distracting us, let’s mark it for skipping so that pytest won’t run it.\n\n``````import pytest\n\n# ...\n\n@pytest.mark.skip(reason='focusing on prop tests for now')\ndef test_example():\ns = [{1}, {1, 2}, {2}]\nassert reduce_sets(s) == [{1, 2}]\n``````\n\nWhen you’re skipping a test, it shows up in the suite progress as the letter ‘s’:\n\n``````======================= test session starts =======================\nplatform darwin -- Python 2.7.10, pytest-2.9.2, py-1.4.31, pluggy-0.3.1\nplugins: hypothesis-3.4.1\ncollected 5 items\n\ntests/test_thing.py s....\n\n=============== 4 passed, 1 skipped in 1.12 seconds ===============\n_____________________________ summary _____________________________\npy27: commands succeeded\ncongratulations :)\n``````\n\n## Property Four: There are no duplicate sets in the output.\n\nThe previous solution removes duplicates, but now let’s write a property to make sure all duplicates are removed.\n\n``````@h.given(st.lists(st.sets(st.integers())))\ndef test_no_duplicate_sets(sets):\noutput = reduce_sets(sets)\nfor e_ix, e in enumerate(output):\nfor s_ix, s in enumerate(output):\nif e_ix == s_ix:\ncontinue\nassert e != s, \"duplicate in {}\".format(output)\n``````\n\nGives us:\n\n``````sets = [set([]), set([]), set([]), set([])]\n\n@h.given(st.lists(st.sets(st.integers())))\ndef test_no_duplicate_sets(sets):\noutput = reduce_sets(sets)\nfor e_ix, e in enumerate(output):\nfor s_ix, s in enumerate(output):\nif e_ix == s_ix:\ncontinue\n> assert e != s, \"duplicate in {}\".format(output)\nE AssertionError: duplicate in [set([]), set([])]\nE assert set([]) != set([])\n\ntests/test_thing.py:41: AssertionError\n--------------------------- Hypothesis ----------------------------\nFalsifying example: test_no_duplicate_sets(sets=[set(), set(), set(), set()])\n``````\n\nLooks like our code fails for inputs with many duplicate sets, which we can cover by just running the prune helper we wrote for as many times as there are sets passed in.\n\n``````def reduce_sets(sets):\no = sets\nfor n in range(len(sets)):\no = _prune_one(o)\nreturn o\n``````\n\n## Property Five: No set in the output list is a subset of any other set in that list.\n\nWe ultimately want to remove anything we can from the list of sets passed in. Removing duplicates is a good, but we can do better than that because anything that is a subset of another set can also be tossed out. By removing a subset, we will always have the same number of elements covered in our output.\n\nLet’s write a property to look for sets that are subsets of others in the output.\n\n``````@h.given(st.lists(st.sets(st.integers())))\ndef test_no_subsets(sets):\noutput = reduce_sets(sets)\nfor e_ix, e in enumerate(output):\nfor s_ix, s in enumerate(output):\nif e_ix == s_ix:\ncontinue\nassert not e.issubset(s), \"subset found in {}\".format(output)\n``````\n\nAnd Hypothesis tells us:\n\n``````sets = [set(), set([])]\n\n@h.given(st.lists(st.sets(st.integers())))\ndef test_no_subsets(sets):\noutput = reduce_sets(sets)\nfor e_ix, e in enumerate(output):\nfor s_ix, s in enumerate(output):\nif e_ix == s_ix:\ncontinue\n> assert not e.issubset(s), \"subset found in {}\".format(output)\nE AssertionError: subset found in [set(), set([])]\nE assert not True\nE + where True = <built-in method issubset of set object at 0x10cbab3f0>(set())\nE + where <built-in method issubset of set object at 0x10cbab3f0> = set([]).issubset\n\ntests/test_thing.py:53: AssertionError\n--------------------------- Hypothesis ----------------------------\nFalsifying example: test_no_subsets(sets=[{0}, set()])\n``````\n\nWe could solve this immediately with a one-line change, but I want to play with this failure case a bit first. The empty set are subsets of all other sets, so let’s write a property especially for that. Our function should never return anything with an empty set in it.\n\n``````@h.given(st.lists(st.sets(st.integers())))\ndef test_no_empty_sets(sets):\noutput = reduce_sets(sets)\nassert set([]) not in output\n``````\n\nAnd to solve for it:\n\n``````def reduce_sets(sets):\no = [s for s in sets if s != set([])]\nfor n in range(len(sets)):\no = _prune_one(o)\nreturn o\n``````\n\nNow we have a more interesting failure case on our original property:\n\n``````sets = [set(), set([0, 1])]\n\n@h.given(st.lists(st.sets(st.integers())))\ndef test_no_subsets(sets):\noutput = reduce_sets(sets)\nfor e_ix, e in enumerate(output):\nfor s_ix, s in enumerate(output):\nif e_ix == s_ix:\ncontinue\n> assert not e.issubset(s), \"subset found in {}\".format(output)\nE AssertionError: subset found in [set(), set([0, 1])]\nE assert not True\nE + where True = <built-in method issubset of set object at 0x1077de4d8>(set([0, 1]))\nE + where <built-in method issubset of set object at 0x1077de4d8> = set().issubset\n\ntests/test_thing.py:59: AssertionError\n--------------------------- Hypothesis ----------------------------\nFalsifying example: test_no_subsets(sets=[{0}, {0, 1}])\n``````\n\nSo we can see that `{0}` is sneaking into our output even though element `0` is covered by the set `{0, 1}`. Let’s tweak our pruning function to prune for elements that can find their superset. This still satisfies the previous property because duplicate sets are both supersets and subsets of each other.\n\n``````def _prune_one(sets):\n\"\"\" attempt to remove a single item \"\"\"\nfor i, e in enumerate(sets):\nfor s in _without(sets, i):\nif s.issuperset(e): # changed from 's == e'\nreturn _without(sets, i)\nreturn sets\n\ndef reduce_sets(sets):\no = [s for s in sets if s != set([])]\nfor n in range(len(sets)):\no = _prune_one(o)\nreturn o\n``````\n\nAnd now running our test suite again:\n\n``````======================= test session starts =======================\nplatform darwin -- Python 2.7.10, pytest-2.9.2, py-1.4.31, pluggy-0.3.1\nplugins: hypothesis-3.4.1\ncollected 7 items\n\ntests/test_thing.py s......\n\n=============== 6 passed, 1 skipped in 1.75 seconds ===============\n_____________________________ summary _____________________________\n``````\n\nSuccess! Everything passes! We can even lift that `skip` mark we added before.\n\n## Property Six: No output set is a subset of the other sets combined.\n\nThis property is actually just a small extension of the previous one. Checking to see if a set has a superset isn’t going far enough. If our function had some sort of input data and it returned `[{1, 2}, {3, 4}, {2, 3}]` you can tell that it’s incomplete because the set `{2, 3}` can be covered from elements found in the other two sets. Even though it isn’t a strict subset of either, it is redundant and safe-to-prune because all of its contents are found in other sets.\n\nThe implementation of this property looks like this:\n\n``````@h.given(st.lists(st.sets(st.integers(min_value=-1, max_value=5), max_size=3), max_size=8))\ndef test_no_sibling_superset_cover_a_set_in_output(sets):\nunion = lambda a, b: a.union(b)\nwithout = lambda o, i: o[:i] + o[i+1:]\noutput = reduce_sets(sets)\nfor i, e in enumerate(output):\nsiblings_superset = reduce(union, without(output, i), set())\nassert not siblings_superset.issuperset(e)\n``````\n\nAnd after multiple trials I finally got it to crash:\n\n``````sets = [set([0, 2]), set([0, 1]), set([1, 2])]\n\n@h.given(st.lists(st.sets(st.integers(min_value=-1, max_value=5), max_size=3), max_size=8))\ndef test_no_sibling_superset_cover_a_set_in_output(sets):\nunion = lambda a, b: a.union(b)\nwithout = lambda o, i: o[:i] + o[i+1:]\noutput = reduce_sets(sets)\nfor i, e in enumerate(output):\nsiblings_superset = reduce(union, without(output, i), set())\n> assert not siblings_superset.issuperset(e)\nE assert not True\nE + where True = <built-in method issuperset of set object at 0x1042cd878>(set([0, 2]))\nE + where <built-in method issuperset of set object at 0x1042cd878> = set([0, 1, 2]).issuperset\n\ntests/test_thing.py:68: AssertionError\n--------------------------- Hypothesis ----------------------------\nFalsifying example: test_no_sibling_superset_cover_a_set_in_output(sets=[{0, 2}, {0, 1}, {1, 2}])\n``````\n\nThe fix to this error is very similar to the last property we implemented. We actually just have to check to make sure a given set is not a subset of all the other sets combined and it will have the same effect as if we were just checking the sibling sets individually. Because math.\n\n``````def _union_all(sets):\nreturn reduce(lambda a, b: a.union(b), sets, set())\n\ndef _prune_one(sets):\n\"\"\" attempt to remove a single item \"\"\"\nfor i, e in enumerate(sets):\ns = _union_all(_without(sets, i))\nif s.issuperset(e):\nreturn _without(sets, i)\nreturn sets\n\ndef reduce_sets(sets):\no = [s for s in sets if s != set([])]\nfor n in range(len(sets)):\no = _prune_one(o)\nreturn o\n``````\n\nAnd there we have it, one step closer to our set coverage.\n\n## Property Seven: The output list has the smallest possible number of sets\n\nThis property is tricky, or at least tedious. How do we prove that a list of sets is the smallest possible number of sets? To show how our solution is incomplete, we have to come up with a specific failing test case:\n\n``````def test_hard_coverage():\ns = [\n{1, 2}, {3, 4},\n{1}, {2, 3}, {4}\n]\nassert reduce_sets(s) == [{1, 2}, {3, 4}]\n``````\n\nAnother possible solution could be `[{1}, {2, 3}, {4}]`, among others, but no solution is as small in length as the one in the test.\n\nRunning pytest with the `-v` flag, we can see a completely different solution was found.\n\n``````\\$ tox -e py27 -- -v\n...\n============================ FAILURES =============================\n_______________________ test_hard_coverage ________________________\n\ndef test_hard_coverage():\ns = [\n{1, 2}, {3, 4},\n{1}, {2, 3}, {4}\n]\n> assert reduce_sets(s) == [{1, 2}, {3, 4}]\nE assert [set(), se...3]), set()] == [set([1, 2]), set([3, 4])]\nE At index 0 diff: set() != set([1, 2])\nE Left contains more items, first extra item: set()\nE Full diff:\nE - [set(), set([2, 3]), set()]\nE + [set([1, 2]), set([3, 4])]\n\ntests/test_thing.py:75: AssertionError\n=============== 1 failed, 8 passed in 2.35 seconds ================\n``````\n\nOur code greedily converges on the wrong solution. It finds something that satisfies all the other properties, but it isn’t the smallest solution possible. The reason it fails is that it first considers the two sets `{1, 2}` and `{3, 4}` for removal. Because the last 3 elements have all the elements of the first two, the two ideal sets get removed.\n\nThe only way to find an ideal solution would be to generate all possible solutions that satisfy the above properties, and then grab the smallest one. In other words, to implement the checker of this property, you have to implement set coverage yourself, which is the heart of what defines an NP-Complete problem. If you could implement the solution to this in polynomial time, you win Computer Science. Every problem as hard as this one could be solved by your solution too.\n\nSo this is the one property I will not implement with Hypothesis. It would have to be a strict reimplementation of the solution, maybe if there were more than one way to implement it we could validate that the two solutions are equivalent with enough data, but for now a few pointed unit tests will have to do.\n\nHere’s a pretty slow implementation of set coverage:\n\n``````from itertools import combinations, chain\n\n# ...\n\ndef brute_force_set_coverage(sets):\n\"\"\"\nmost naiive approach to set coverage.\n\nTry all solutions\nFilter non-covering solutions\nChoose smallest solution\n\"\"\"\no = [s for s in sets if s != set([])]\nif len(o) == 0:\nreturn o\ngoal_set = _union_all(o)\nsolution_generators = [combinations(o, i + 1) for i in range(len(o))]\nall_possible_solutions = chain.from_iterable(solution_generators)\nvalid_solutions = [i for i in all_possible_solutions if _union_all(i) == goal_set]\nreturn list(sorted((i for i in valid_solutions), key=len))\n``````\n\nBut before we can swap it out for tests, we have to adjust how our generators work. Right now, this version will take a much longer time to find a solution as the list of sets grows in size, so lets put a hard limit on how big sets get.\n\nFirst let’s refactor our hypothesis generators to source from the same function:\n\n``````def list_of_sets():\nreturn st.lists(st.sets(st.integers()))\n\n@h.given(list_of_sets())\ndef test_that_output_has_fewer_sets_than_input(sets):\noutput = brute_force_set_coverage(sets)\nassert len(output) <= len(sets)\n\n@h.given(list_of_sets())\ndef test_no_items_lost(sets):\nunion = lambda a, b: a.union(b)\nall_input = reduce(union, sets, set())\nall_output = reduce(union, brute_force_set_coverage(sets), set())\nassert all_input == all_output\n\n``````\n\nAnd then put some limitations on the size of the lists and sets that get generated by this generator.\n\n``````def list_of_sets():\nreturn st.lists(\nst.sets(\nst.integers(max_value=8),\nmax_size=8\n),\nmax_size=16\n)\n``````\n\nEven by limiting it to these small numbers, the entire test suite can take several seconds longer than it was before:\n\n``````======================= test session starts =======================\nplatform darwin -- Python 2.7.10, pytest-2.9.2, py-1.4.31, pluggy-0.3.1" ]

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[ "# Crystal calculations\n\nA project log for Pulse to Tone Converter\n\nConverts pulse dialing of your POTS telephone to DTMF dialing allowing you to use it with your VOIP's FXS-port.\n\nWhat crystal package to use : small 3.2x2.5mm or big HC49/US?  One parameter determining the choice is the drive level of the crystal.  If the drive level is too high, i.e. >200µW typically, the small crystal can't be used.\n\nThe drive level is calculated as follows:\n\nWhere:\n\n• Vpp = peak-to-peak voltage, measured differentiallly over the crystal\n• f = crystal frequency\n• CL = load capacitance, see the Pierce oscillator on Wikipedia how you calculate it.\n• ESR = equivalent series resistor, from the crystal's datasheet\n\n## Example\n\nCalculation of CL\n\n• C1 = C2 = 18pF\n• Cs = 9pF\n• Ci = 6pF, Co=16pF (datasheet Attiny85)\n\nCalculation of drive level\n\n• f = 8MHz\n• ESR=80ohm, from the crystal's datasheet\n• Vpp=1.7Vpp, measured with differential probe over the crystal\n\nConclusion\n\nThe smaller crystal package can be used safely.\n\nAnother parameter that determines the choice is availability of the crystals.  For 8MHz, HC49US crystals have better specs, are more widely used and are cheaper than the 3.2x2.5mm package crystals.\n\n## Discussions", null, "" ]

[ null, "https://analytics.supplyframe.com/trackingservlet/impression", null ]

[ "# Scatterplot\n\nA Scatterplot displays the relationship between 2 numeric variables. Each dot represents an observation. Their position on the X (horizontal) and Y (vertical) axis represents the values of the 2 variables.\n\nUsing the `ggplot2` package\n\nScatterplots are built with ggplot2 thanks to the `geom_point()` function. Discover a basic use case in graph #272, and learn how to custom it with next examples below.\n\nUsing base R\n\nBase R is also a good option to build a scatterplot, using the `plot()` function. The chart #13 below will guide you through its basic usage. Following examples allow a greater level of customization.\n\nSpecial use case: Manhattan plots\n\nA Manhattan plot is a particular type of scatterplot used in genomics. The X axis displays the position of a genetic variant on the genome. Each chromosome is usually represented using a different color. The Y axis shows p-value of the association test with a phenotypic trait.\n\nRelated chart types" ]

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[ "$\\newcommand{\\Cat}{{\\rm Cat}}$ $\\newcommand{\\A}{\\mathcal A}$ $\\newcommand{\\freestar}{ \\framebox[7pt]{\\star} }$\n\n## 3. Geometry\n\n1. #### Problem 3.1.\n\n[Qing Han] (Problem originally proposed by Fang-Hua Lin) Let $(T^2,g_0) \\subseteq (\\mathbb R^3, g_{\\mathrm{eucl}})$ be an isometric embedding and $g$ be a metric on $T^2$ which is $C^0$ close to $g_0$. Is there an isometric embedding of $(T^2,g)$ in $\\mathbb R^3$?\n\nCite this as: AimPL: Nonlinear PDEs in real and complex geometry, available at http://aimpl.org/nonlinpdegeom." ]

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[ "Question\n(a) A defibrillator sends a 6.00-A current through the chest of a patient by applying a 10,000-V potential as in the figure below. What is the resistance of the path? (b) The defibrillator paddles make contact with the patient through a conducting gel that greatly reduces the path resistance. Discuss the difficulties that would ensue if a larger voltage were used to produce the same current through the patient, but with the path having perhaps 50 times the resistance. (Hint: The current must be about the same, so a higher voltage would imply greater power. Use this equation for power: $P=I^2R$)\nQuestion Image", null, "Figure 20.41 The capacitor in a defibrillation unit drives a current through the heart of a patient.\n\na) $1.67 \\textrm{ k}\\Omega$\n\nb) The patient would get burned by the increased power.\n\nSolution Video\n\n# OpenStax College Physics Solution, Chapter 20, Problem 7 (Problems & Exercises) (1:57)", null, "", null, "" ]

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[ "# What Is The Square Root Of 625\n\nWhat Is The Square Root Of 625. Divide 625 by 5, 625 ÷ 5 = 125. Calculate the square root square root of 625.", null, "find the square root of 625/1296 Brainly.in from brainly.in\n\nThe square root of 256 is expressed as √256 in the radical form and as (256) ½ or (256) 0.5 in the exponent form. Find the product of the factors obtained in step 3. For example, 25 is the square root of.\n\n### What Is The Square Root Of 625 In Simplest Radical Form?\n\nWhat type of number is 625? Pull terms out from under the radical, assuming positive real numbers. It is the positive solution of.\n\n### Hence, The Correct Answer Is The Option (B).\n\nDivide 25 by 5, 25 ÷ 5 = 5. The square root of 62 cannot be simplified. Find the product of the factors obtained in step 3.\n\n### Method Of Prime Factorization Makes It More Clear, I.e.\n\nA square root of a number 'x' is a number y such that y 2 = x, in other words, a number y whose square is y. Click here👆to get an answer to your question ️ the square root of 625 is: The solution above and other related solutions were provided by the find the square root of a.\n\n### 625 Is A Perfect Square.\n\nAll square root calculations can be converted to a number (called the base) with a fractional exponent. Divide 125 by 5, 125 ÷ 5 = 25. Take one factor from each group.\n\n### The Square Root Of 256 Is Expressed As √256 In The Radical Form And As (256) ½ Or (256) 0.5 In The Exponent Form.\n\nCheck out the work below for reducing 625 into simplest radical form. What is a 8 table? The square root of 256 is 16." ]

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[ "Related Articles\nClass 12 NCERT Solutions – Mathematics Part I – Chapter 5 Continuity And Differentiability – Exercise 5.5 | Set 2\n• Last Updated : 05 Apr, 2021\n\n### (x cos x)x + (x sin x)1/x\n\nSolution:\n\nGiven: (x cos x)x + (x sin x)1/x\n\nLet us considered y = u + v\n\nWhere, u = (x cos x)x and v = (x sin x)1/x\n\nSo, dy/dx = du/dx + dv/dx ………(1)\n\nSo first we take u = (x cos x)\n\nOn taking log on both sides, we get\n\nlog u = log(x cos x)\n\nlog u = xlog(x cos x)\n\nNow, on differentiating w.r.t x, we get", null, "", null, "", null, "", null, "………(2)\n\nNow we take u =(x sin x)1/x\n\nOn taking log on both sides, we get\n\nlog v = log (x sin x)1/x\n\nlog v = 1/x log (x sin x)\n\nlog v = 1/x(log x + log sin x)\n\nNow, on differentiating w.r.t x, we get", null, "", null, "", null, "", null, "………(3)\n\nNow put all the values from eq(2) and (3) into eq(1)", null, "### Question 12. xy + yx = 1\n\nSolution:\n\nGiven: xy + yx = 1\n\nLet us considered\n\nu = xy and v = yx\n\nSo,", null, "………(1)\n\nSo first we take u = xy\n\nOn taking log on both sides, we get\n\nlog u = log(xy)\n\nlog u = y log x\n\nNow, on differentiating w.r.t x, we get", null, "", null, "", null, "………(2)\n\nNow we take v = yx\n\nOn taking log on both sides, we get\n\nlog v = log(y)x\n\nlog v = x log y\n\nNow, on differentiating w.r.t x, we get", null, "", null, "", null, "………(3)\n\nNow put all the values from eq(2) and (3) into eq(1)", null, "", null, "", null, "### Question 13. yx = xy\n\nSolution:\n\nGiven: yx = xy\n\nOn taking log on both sides, we get\n\nlog(yx) = log(xy)\n\nxlog y = y log x\n\nNow, on differentiating w.r.t x, we get", null, "", null, "", null, "", null, "", null, "", null, "### Question 14. (cos x)y = (cos y)x\n\nSolution:\n\nGiven: (cos x)y = (cos y)x\n\nOn taking log on both sides, we get\n\ny log(cos x) = x log (cos y)\n\nNow, on differentiating w.r.t x, we get", null, "", null, "", null, "", null, "", null, "### Question 15. xy = e(x – y)\n\nSolution:\n\nGiven: xy = e(x – y)\n\nOn taking log on both sides, we get\n\nlog(xy) = log ex – y\n\nlog x + log y = x – y\n\nNow, on differentiating w.r.t x, we get", null, "", null, "", null, "", null, "", null, "### Question 16. Find the derivative of the function given by f(x) = (x + 1)(x + x2)(1 + x4)(1 + x8) and hence find f'(1).\n\nSolution:\n\nGiven: f(x) = (x + 1)(x + x2)(1 + x4)(1 + x8)\n\nFind: f'(1)\n\nOn taking log on both sides, we get\n\nlog(f(x)) = log(1 + x) + log(1 + x2) + log(1 + x4) + log(1 + x8)\n\nNow, on differentiating w.r.t x, we get", null, "", null, "∴ f'(1) = 2.2.2.2.", null, "", null, "f'(1) = 120\n\n### Do they all give the same answer?\n\nSolution:\n\n(i) By using product rule", null, "", null, "dy/dx = (3x4 – 15x3 + 24x2 + 7x2 – 35x + 56) + (2x4 + 14x2 + 18x – 5x3 – 35x – 45)\n\ndy/dx = 5x4 – 20x3 + 45x2 – 52x + 11\n\n(ii) By expansion\n\ny = (x2 – 5x + 8)(x3 + 7x + 9)\n\ny = x5 + 7x3 + 9x2 – 5x4 – 35x2 – 45x + 8x3 + 56x + 72\n\ny = x5 – 5x4 + 15x3 – 26x2 + 11x + 72\n\ndy/dx = 5x4 – 20x3 + 45x2 – 52x + 11\n\n(iii) By logarithmic expansion\n\nTaking log on both sides\n\nlog y = log(x2 – 5x + 8) + log(x3 + 7x + 9)\n\nNow on differentiating w.r.t. x, we get", null, "", null, "", null, "dy/dx = 2x4 + 14x2 + 18x – 5x3 – 35x – 45 + 3x4 – 15x3 + 24x2 + 7x2 – 35x + 56\n\ndy/dx = 5x4 – 20x3 + 45x2 – 52x + 11\n\nAnswer is always same what-so-ever method we use.\n\n###", null, "Solution:\n\nLet y = u.v.w.\n\nMethod 1: Using product Rule", null, "", null, "", null, "Method 2: Using logarithmic differentiation\n\nTaking log on both sides\n\nlog y = log u + log v + log w\n\nNow, Differentiating w.r.t. x", null, "", null, "", null, "My Personal Notes arrow_drop_up" ]

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[ "# Adaptive discretization-based algorithms for semi-infinite programs with unbounded variables\n\nThe proof of convergence of adaptive discretization-based algorithms for semi-infinite programs (SIPs) usually relies on compact host sets for the upper- and lower-level variables. This assumption is violated in some applications, and we show that indeed convergence problems can arise when discretization-based algorithms are applied to SIPs with unbounded variables. To mitigate these convergence problems, we first examine the underlying assumptions of adaptive discretization-based algorithms. We do this paradigmatically using the lower-bounding procedure of [Mitsos. Optimization, 2011], which uses the algorithm proposed by [Blankenship and Falk. JOTA, 1976]. It is noteworthy that the considered procedure and assumptions are essentially the same in the broad class of adaptive discretization-based algorithms. We give sharper, slightly relaxed, assumptions with which we achieve the same convergence guarantees. We show that the convergence guarantees also hold for certain SIPs with unbounded variables based on these sharpened assumptions. However, these sharpened assumptions may be difficult to prove a priori. For these cases, we propose additional, stricter, assumptions which might be easier to prove and which imply the sharpened assumptions. Using these additional assumptions, we present numerical case studies with unbounded variables. Finally, we review which applications are tractable with the proposed additional assumptions.\n\n## Citation\n\nhttps://doi.org/10.1007/s00186-022-00792-y" ]

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[ "# Thread: Orienting View based on Transpose Action Line\n\n1. ##", null, "Very cool, thanks for including the code to learn from also!", null, "2. ##", null, "Teriyaki, I think you were close. Rotations can be tricky...\nI am not going to talk much about trigonometry, I am not a mathematician, but I think this problem can be understood without using very advanced maths.\n\nIt helped me a lot to think of a sphere of R=1.\nWith Pixolpick you get the x,y,z components of a vector that you can imagine as the radius of the unit sphere in that particular orientation. That vector can also be defined by two angles, T and P.\nLet's say T is the vertical angle, this angle is defined by the value of the y component of your unit vector so T=ACOS(Uy).\nThen the P angle depends of the values in the X and Z axis. P=ATAN2(Ux,Uz) ... atan2 takes care of all the different cases like when Uz is equal to 0, so we don't have to worry about it...\nThese two angles can also be called spherical coordinates and rotating the camera by those angles should do the trick.\nNow we have to convert the angles to the equivalent distance (in pixels) to simulate a drag click that moves our camera the exact amount.\n\nFor the vertical rotation I found 1 degree is equal to 2.777777 pixels while the horizontal rotation 1 degree is equal to 1.111111 . This means we need to move our pen more pixels to rotate vertically, that's actually a very cool idea from the pixo guys because very often we just want to rotate our models horizontally to get a good 3D feeling (at least this is why I think they didn't use the same number)\nThis method should obviously have a rounding error because pixels are integers but I think it's not very noticeable", null, "3. ##", null, "Thanks for the explanations - I will work to understand this", null, "", null, "4. ##", null, "I am looking for the same function. Thank you for the CamControl plugin. It works almost perfectly. But as you mentioned, it is not exactly normal, causing irregular surfaces when a surface is repositioned and using the Trim-Front brush for example.\n\nDo you know if an exact AlignCameraTosurfaceNormal feature has been added to zbrush2018? Or have another suggestion?\n\nThank you", null, "####", null, "Posting Permissions\n\n• You may not post new threads\n• You may not post replies\n• You may not post attachments\n• You may not edit your posts\n•" ]

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