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[ "9 9 Understanding CNC Drilling Costs: A Step-by-Step Guide\n\n# Understanding CNC Drilling Costs: A Step-by-Step Guide\n\nPublished on: May 17, 2023\nCategories: CNC Machining", null, "In the realm of manufacturing and engineering, CNC (Computer Numerical Control) drilling is a technique that stands out due to its unparalleled precision and efficiency. It’s an essential process in countless industries, from automotive and aerospace to electronics and medical sectors. However, as with any advanced technology, the financial aspect of CNC drilling can pose a challenge. Comprehending and calculating the cost of CNC drilling is not as straightforward as it might seem.\n\nThis article aims to demystify CNC drilling costs, providing you with a comprehensive understanding and a step-by-step guide to calculating these costs.\n\n## Factors Influencing CNC Drilling Costs\n\nThere are several factors that influence the cost of CNC drilling. It’s crucial to understand each of these elements to make informed decisions about your CNC drilling projects. Here, we dissect the various costs associated with CNC drilling:\n\n### 1. Machine Costs", null, "Machine costs are a significant factor in the overall CNC drilling expenditure. These costs can be divided into two categories:\n\n• Operational Cost: These are the recurring costs associated with running the CNC machine. They include costs for energy consumption, maintenance, and machine depreciation.\n• Initial Purchase Cost: This is the upfront cost that you pay to acquire the CNC drilling machine. It’s a one-time expense, but it can be quite significant, especially for high-end machines with advanced capabilities.\n\nHere’s an example table showing hypothetical machine costs:\n\nCost Type Estimated Cost\nPurchase Cost \\$100,000\nEnergy Cost (per year) \\$5,000\nMaintenance Cost (per year) \\$3,000\nDepreciation Cost (per year) \\$10,000\n\n### 2. Tooling Costs\n\nTooling costs refer to the expenses associated with the various tools used in CNC drilling. These can include, drill bits, collets & tool holders, coolants, and lubricants. Furthermore, The costs can vary greatly depending on the type of material you’re drilling and the complexity of the operation.\n\n### 3. Material Costs\n\nMaterial costs are the expenses related to the workpiece material you are drilling. The cost of materials can vary depending on several factors, including:\n\n• Type of material (aluminum, steel, titanium, etc.)\n• Size and shape of the material\n• Availability and the market price of the material\n\n### 4. Labor Costs\n\nLabor costs are another critical element in the overall CNC drilling cost. This includes the wages of the CNC operators and technicians. The complexity of the drilling operation, the level of expertise required, and the local labor market conditions can all influence labor costs.\n\nOverhead costs are the indirect costs that are not directly tied to the CNC drilling operation but are necessary for running the business. These can include:\n\n• Rent or mortgage for the facility\n• Utilities like electricity, water, and internet\n\nTo get a complete picture of your CNC drilling costs, all these factors need to be taken into account. In the next sections, we’ll discuss how to calculate these costs and provide an example of a cost calculation for a specific application.\n\n## Calculating CNC Drilling Costs\n\nCalculating the cost of a CNC drilling operation involves aggregating the various costs discussed in the previous section. Here’s a step-by-step guide on how to calculate your CNC drilling costs:\n\nStep 1: Calculate Machine Costs\n\nFirstly, you need to determine your machine costs. This involves adding the initial purchase cost, the energy cost, the maintenance cost, and the depreciation cost.\n\nExample:\n\nLet’s say you purchased a CNC machine for \\$100,000. The annual energy cost is \\$5,000, the maintenance cost is \\$3,000, and the machine depreciates by \\$10,000 each year. If you are using the machine for 2000 hours per year, your hourly machine cost would be:\n\n(\\$100,000 + \\$5,000 + \\$3,000 + \\$10,000) / 2000 hours = \\$59 per hour\n\nStep 2: Calculate Tooling Costs\n\nNext, calculate your tooling costs. This includes the cost of drill bits, toolholders, and coolants. To find the hourly tooling cost, divide the total tooling cost by the expected lifespan of the tools (in hours).\n\nExample:\n\nIf your tooling costs amount to \\$2000 and the expected lifespan is 1000 hours, your hourly tooling cost would be:\n\n\\$2000 / 1000 hours = \\$2 per hour.\n\nStep 3: Calculate Material Costs\n\nTo calculate the material costs, determine the cost of the workpiece material per unit. Then multiply this by the number of units you expect to produce per hour.\n\nExample:\n\nIf the material cost per unit is \\$10 and you produce 10 units per hour, your hourly material cost would be :\n\n\\$10 per unit x 10 units/hour = \\$100 per hour\n\nStep 4: Calculate Labor Costs\n\nNext, calculate your labor costs. This is the hourly wage of your CNC operator.\n\nExample:\n\nIf your CNC operator earns \\$20 per hour, your hourly labor cost would be \\$20.\n\nOverhead costs can be calculated by adding all your indirect costs and dividing by the number of operating hours per year.\n\nExample:\n\n\\$50,000 / 2000 hours = \\$25 per hour\n\nFinally, add all your hourly costs to find the total hourly cost of your CNC drilling operation. This is your cost per hour to operate the CNC drilling machine.\n\nExample:\n\n\\$\\$ Total cost = Hourly Machine Cost + Hourly Tooling Cost + Hourly Material Cost + Hourly Labor Cost + Hourly Overhead Cost \\$\\$\n\n= \\$59 + \\$2 + \\$100 + \\$20 + \\$25 = \\$206 per hours.\n\n## Example of CNC Drilling Cost Calculation\n\nLet’s consider a specific example to illustrate how these cost calculations come together. We will be calculating the cost for a CNC drilling operation where we are producing a metal bracket with four drilled holes.\n\n### 1. Technical Specifications\n\n• Material: Aluminum\n• Dimensions: 100mm x 50mm x 10mm\n• Hole Diameter: 10mm\n• Hole Depth: 10mm\n• Number of Holes: 4\n• Operation: CNC Drilling\n\n### 2. Calculation of Involves Costs\n\nHere is a table of the six steps involved in calculating CNC drilling costs, including the cost type, calculating formula, calculation, and estimated cost per hour.\n\nStep Cost Type Calculating Formula Calculation Per Hour Estimated Cost\n1 Machine Costs ([Purchase Cost / Lifespan in Hours] + [Maintenance Cost / Operating Hours per Year] + [Electricity Cost / Operating Hours per Year]) (\\$100,000 / 20,000 hours) + (\\$3,000 / 2,000 hours) + (\\$2,000 / 2,000 hours) = \\$7.5/hour \\$7.5/hour\n2 Tooling Costs (Tool Cost / Tool Lifespan in Hours) \\$30 / 100 hours = \\$0.3/hour \\$0.3/hour\n3 Material Costs (Material Cost per Bracket x Brackets Produced per Hour) \\$2/bracket x 30 brackets/hour = \\$60/hour \\$60/hour\n4 Labor Costs   \\$20/hour \\$20/hour\n5 Overhead Costs (Annual Overhead Costs / Operating Hours per Year) \\$10,000 / 2,000 hours = \\$5/hour \\$5/hour\n6 Total Cost Calculation Machine Cost + Tooling Cost + Material Cost + Labor Cost + Overhead Cost \\$7.5/hour + \\$0.3/hour + \\$60/hour + \\$20/hour + \\$5/hour = \\$92.8/hour \\$92.8/hour\n\n### 3. Total cost calculation\n\nSo, the cost of drilling four holes into each metal bracket, given our assumptions, is approximately \\$92.8 per hour. This cost will produce 30 brackets per hour, so the cost per bracket is:\n\n(Total Hourly Cost / Brackets Produced per Hour)\n\n= (\\$92.8/hour / 30 brackets/hour)\n\n= \\$3.09/bracket\n\nThis calculation provides a fundamental understanding of how to calculate CNC drilling costs. However, it’s worth noting that the numbers can vary significantly based on the specifics of your operation, local market conditions, and many other factors. Always do a detailed cost analysis specific to your operation to get accurate cost estimates.\n\n## Conclusion\n\nUnderstanding and calculating CNC drilling costs can be complex, but it’s essential for budgeting and cost control. By considering all the factors that contribute to the cost and seeking ways to optimize them, manufacturers can make the most of their CNC drilling operations. By understanding the cost structure and seeking ways to optimize it, manufacturers can leverage CNC drilling as a cost-effective and high-quality solution for their production needs.\n\nPartnering with an expert like Prolean can further enhance these benefits, providing access to top-of-the-line technology and skilled professionals that ensure every drilling operation is performed at the highest level of precision and efficiency.\n\nCNC Drilling Service\n\n## FAQs\n\n1 What factors influence CNC drilling costs?\n\nCNC drilling costs are influenced by a number of factors, including machine costs (purchase, operation, maintenance), tooling costs (drill bits, toolholders), material costs (workpiece material, coolant), labor costs (operator wages, setup time), and overhead costs (utilities, rent, insurance).\n\n2 How can I reduce my CNC drilling costs?\n\nReducing CNC drilling costs can be achieved by optimizing machining parameters, maintaining tool sharpness, using proper coolant, investing in quality machines and tools that have longer lifespans, and training your operators to work efficiently and effectively.\n\n3 What is included in a CNC drilling cost calculation?\n\nA CNC drilling cost calculation includes direct costs like machine, tooling, material, and labor costs, as well as indirect costs or overhead such as utilities, rent, and insurance. It’s important to include all these factors to get an accurate picture of the total cost of CNC drilling operations." ]

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[ "# #06fb7e Color Information\n\nIn a RGB color space, hex #06fb7e is composed of 2.4% red, 98.4% green and 49.4% blue. Whereas in a CMYK color space, it is composed of 97.6% cyan, 0% magenta, 49.8% yellow and 1.6% black. It has a hue angle of 149.4 degrees, a saturation of 96.8% and a lightness of 50.4%. #06fb7e color hex could be obtained by blending #0cfffc with #00f700. Closest websafe color is: #00ff66.\n\n• R 2\n• G 98\n• B 49\nRGB color chart\n• C 98\n• M 0\n• Y 50\n• K 2\nCMYK color chart\n\n#06fb7e color description : Vivid cyan - lime green.\n\n# #06fb7e Color Conversion\n\nThe hexadecimal color #06fb7e has RGB values of R:6, G:251, B:126 and CMYK values of C:0.98, M:0, Y:0.5, K:0.02. Its decimal value is 457598.\n\nHex triplet RGB Decimal 06fb7e `#06fb7e` 6, 251, 126 `rgb(6,251,126)` 2.4, 98.4, 49.4 `rgb(2.4%,98.4%,49.4%)` 98, 0, 50, 2 149.4°, 96.8, 50.4 `hsl(149.4,96.8%,50.4%)` 149.4°, 97.6, 98.4 00ff66 `#00ff66`\nCIE-LAB 87.259, -75.657, 45.992 38.335, 70.535, 31.332 0.273, 0.503, 70.535 87.259, 88.54, 148.704 87.259, -78.294, 73.691 83.985, -65.497, 36.671 00000110, 11111011, 01111110\n\n# Color Schemes with #06fb7e\n\n• #06fb7e\n``#06fb7e` `rgb(6,251,126)``\n• #fb0683\n``#fb0683` `rgb(251,6,131)``\nComplementary Color\n• #09fb06\n``#09fb06` `rgb(9,251,6)``\n• #06fb7e\n``#06fb7e` `rgb(6,251,126)``\n• #06fbf9\n``#06fbf9` `rgb(6,251,249)``\nAnalogous Color\n• #fb0609\n``#fb0609` `rgb(251,6,9)``\n• #06fb7e\n``#06fb7e` `rgb(6,251,126)``\n• #f906fb\n``#f906fb` `rgb(249,6,251)``\nSplit Complementary Color\n• #fb7e06\n``#fb7e06` `rgb(251,126,6)``\n• #06fb7e\n``#06fb7e` `rgb(6,251,126)``\n• #7e06fb\n``#7e06fb` `rgb(126,6,251)``\nTriadic Color\n• #83fb06\n``#83fb06` `rgb(131,251,6)``\n• #06fb7e\n``#06fb7e` `rgb(6,251,126)``\n• #7e06fb\n``#7e06fb` `rgb(126,6,251)``\n• #fb0683\n``#fb0683` `rgb(251,6,131)``\nTetradic Color\n• #03b258\n``#03b258` `rgb(3,178,88)``\n• #03cb65\n``#03cb65` `rgb(3,203,101)``\n• #04e471\n``#04e471` `rgb(4,228,113)``\n• #06fb7e\n``#06fb7e` `rgb(6,251,126)``\n• #1ffb8b\n``#1ffb8b` `rgb(31,251,139)``\n• #38fc98\n``#38fc98` `rgb(56,252,152)``\n• #51fca5\n``#51fca5` `rgb(81,252,165)``\nMonochromatic Color\n\n# Alternatives to #06fb7e\n\nBelow, you can see some colors close to #06fb7e. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #06fb41\n``#06fb41` `rgb(6,251,65)``\n• #06fb55\n``#06fb55` `rgb(6,251,85)``\n• #06fb6a\n``#06fb6a` `rgb(6,251,106)``\n• #06fb7e\n``#06fb7e` `rgb(6,251,126)``\n• #06fb92\n``#06fb92` `rgb(6,251,146)``\n• #06fba7\n``#06fba7` `rgb(6,251,167)``\n• #06fbbb\n``#06fbbb` `rgb(6,251,187)``\nSimilar Colors\n\n# #06fb7e Preview\n\nText with hexadecimal color #06fb7e\n\nThis text has a font color of #06fb7e.\n\n``<span style=\"color:#06fb7e;\">Text here</span>``\n#06fb7e background color\n\nThis paragraph has a background color of #06fb7e.\n\n``<p style=\"background-color:#06fb7e;\">Content here</p>``\n#06fb7e border color\n\nThis element has a border color of #06fb7e.\n\n``<div style=\"border:1px solid #06fb7e;\">Content here</div>``\nCSS codes\n``.text {color:#06fb7e;}``\n``.background {background-color:#06fb7e;}``\n``.border {border:1px solid #06fb7e;}``\n\n# Shades and Tints of #06fb7e\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #000201 is the darkest color, while #eefff6 is the lightest one.\n\n• #000201\n``#000201` `rgb(0,2,1)``\n• #00150b\n``#00150b` `rgb(0,21,11)``\n• #012914\n``#012914` `rgb(1,41,20)``\n• #013c1e\n``#013c1e` `rgb(1,60,30)``\n• #014f27\n``#014f27` `rgb(1,79,39)``\n• #026231\n``#026231` `rgb(2,98,49)``\n• #02763b\n``#02763b` `rgb(2,118,59)``\n• #028944\n``#028944` `rgb(2,137,68)``\n• #039c4e\n``#039c4e` `rgb(3,156,78)``\n• #03b058\n``#03b058` `rgb(3,176,88)``\n• #03c361\n``#03c361` `rgb(3,195,97)``\n• #03d66b\n``#03d66b` `rgb(3,214,107)``\n• #04ea74\n``#04ea74` `rgb(4,234,116)``\nShade Color Variation\n• #06fb7e\n``#06fb7e` `rgb(6,251,126)``\n• #19fb88\n``#19fb88` `rgb(25,251,136)``\n• #2dfc92\n``#2dfc92` `rgb(45,252,146)``\n• #40fc9c\n``#40fc9c` `rgb(64,252,156)``\n• #53fca6\n``#53fca6` `rgb(83,252,166)``\n• #67fdb0\n``#67fdb0` `rgb(103,253,176)``\n• #7afdba\n``#7afdba` `rgb(122,253,186)``\n• #8dfdc4\n``#8dfdc4` `rgb(141,253,196)``\n• #a0fdce\n``#a0fdce` `rgb(160,253,206)``\n• #b4fed8\n``#b4fed8` `rgb(180,254,216)``\n• #c7fee2\n``#c7fee2` `rgb(199,254,226)``\n• #dafeec\n``#dafeec` `rgb(218,254,236)``\n• #eefff6\n``#eefff6` `rgb(238,255,246)``\nTint Color Variation\n\n# Tones of #06fb7e\n\nA tone is produced by adding gray to any pure hue. In this case, #7b8680 is the less saturated color, while #06fb7e is the most saturated one.\n\n• #7b8680\n``#7b8680` `rgb(123,134,128)``\n• #719080\n``#719080` `rgb(113,144,128)``\n• #679a80\n``#679a80` `rgb(103,154,128)``\n• #5ea380\n``#5ea380` `rgb(94,163,128)``\n• #54ad80\n``#54ad80` `rgb(84,173,128)``\n• #4ab77f\n``#4ab77f` `rgb(74,183,127)``\n• #40c17f\n``#40c17f` `rgb(64,193,127)``\n• #37ca7f\n``#37ca7f` `rgb(55,202,127)``\n• #2dd47f\n``#2dd47f` `rgb(45,212,127)``\n• #23de7f\n``#23de7f` `rgb(35,222,127)``\n• #19e87e\n``#19e87e` `rgb(25,232,126)``\n• #10f17e\n``#10f17e` `rgb(16,241,126)``\n• #06fb7e\n``#06fb7e` `rgb(6,251,126)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #06fb7e is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

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[ "", null, "Deepika Singh\n\n# Advanced Machine Learning Modeling in Azure ML Studio\n\n### Deepika Singh\n\n• Sep 15, 2020\n• 10 Min read\n• 948 Views\n• Sep 15, 2020\n• 10 Min read\n• 948 Views\nCloud\nCloud Application Development\nMachine Learning and AI\nAzure Machine Learning Service\n\n## Introduction\n\nWhen dealing with complex data science problems, it is important for data scientists to understand advanced machine learning algorithms. Some real-life use cases include text characterization, classification of patients on the basis of biological attributes, image recognition, and stock market prediction. These advanced algorithms can handle complex data and are used across industries such as healthcare, banking, education, telecom, and retail, to name a few.\n\nIn this guide, you will learn how to build and evaluate advanced machine learning models such as support vector machines and neural networks with Azure Machine Learning Studio.\n\n## Data\n\nIn this guide, you will work with the Pima Indian diabetes dataset available in Azure Machine Learning Studio. This data originally comes from the National Institute of Diabetes and Digestive and Kidney Diseases. The dataset consists of several variables, such as the number of pregnancies the patient has had, their BMI, insulin level, age, and so on. You can have a look at this data here.\n\nStart by loading the data.\n\n### Load and Explore the Data\n\nOnce you have logged into your Azure Machine Learning Studio account, click on the EXPERIMENTS option, listed on the left sidebar, followed by the NEW button.", null, "Next, click on a blank experiment and name the experiment Advanced ML. Under the Saved datasets, drag Pima Indians Diabetes dataset into the workspace.", null, "Once you have loaded the data, the next step is to explore it. To do this, right-click and select the Visualize option as shown below.", null, "The data contains 768 rows and 9 columns. Select the different variables to examine their basic statistics. For example, the image below displays the details for the `Class variable`.", null, "## Support Vector Machine\n\nSupport Vector Machine (or SVM) is an advanced machine learning algorithm that can be used for both classification and regression machine learning problems. The SVM algorithm works by creating an n-dimensional feature space, called a hyperplane, which is used to analyze and recognize patterns in the input data. In this data, the model will create a hyperplane with the eight independent variables, and this hyperplane will divide the various classes of the target variable in the most distinct manner possible.\n\nIn Azure Machine Learning Studio, the Two-Class Support Vector Machine module is used to create the support vector machine algorithm. Start by searching and dragging the module into the workspace.", null, "You have the module in the workspace, and the next step is to configure it. In Create trainer mode, select the Single Parameter option that is used when you know how you want to configure the algorithm. The second parameter is Number of iterations, which indicates the number of iterations used when building the model. Set this value to 3. The Lambda value is used to tune the model parameter. Select the Normalize features option, which will normalize the features. Type an integer value in Random number seed to ensure reproducibility.", null, "You have set up the model, and the next step is model validation. One popular cross-validation technique is k-fold cross-validation. In k-fold cross-validation, the data is divided into k folds. The model is trained on k-1 folds with one fold held back for testing.\n\nFor example, if k is set to ten, then the data will be divided into ten equal sections. After that, the model will be built on the first nine sections, while the evaluation will be done on the tenth section or fold. This process gets repeated to ensure each fold of the dataset gets the chance to be the held-back set. Once the process is completed, you can summarize the evaluation metric using the mean or standard deviation.\n\nThe Cross Validate Model module performs this task in Azure Machine Learning Studio. Search and drag Cross Validate Model module into the workspace as shown below.", null, "You have loaded the modules, and the next step is to connect them. This is shown below.", null, "You can see the red flag next to Cross Validate Model, which needs to be corrected. Click on the Launch column selector option, and select the target variable, `Class variable`, as shown below.", null, "Run the experiment.", null, "### Model Evaluation\n\nOnce the model is built, the next step is to understand the model outcomes. The model evaluation results can be viewed in the right-output port, which shows the Evaluation results by fold (Dataset).", null, "Right-click and select the Visualize option.", null, "The following output will be displayed to show the evaluation results by folds. There are ten folds, zero through nine, and for every fold you have results across several metrics such as accuracy, precision, recall, and so on.", null, "If you scroll downwards, you will see the mean results across the ten folds.", null, "From the above output, you can infer that the mean accuracy, F-score, and AUC value for this support vector machine model are 0.77, 0.63, and 0.83, respectively.\n\n## Neural Network\n\nThe Two-Class Neural Network module in Azure Machine Learning Studio is used to train the neural network algorithm for binary classification.\n\nA neural network is a set of interconnected layers used to solve advanced machine learning and artificial intelligence problems. Neural networks often outperform traditional algorithms because they have the advantages of non-linearity, variable interactions, and customization. For the data used in this guide, this algorithm creates a network of input, output, and hidden layers to make predictions of the target variable, `Class variable`.\n\nSearch and drag the module into the workspace.", null, "The next step is to drag the Cross Validate Model module into the workspace and connect the modules as shown below.", null, "Run the experiment. If it is successfully run, all the modules will have a green tick.", null, "### Model Evaluation\n\nTo evaluate model performance, right-click and select the Visualize option.", null, "As previously discussed, there are ten folds, and evaluation results by folds are displayed.", null, "If you scroll downwards, you will see the mean results across the ten folds.", null, "From the above output, you can infer that the mean accuracy, F-score, and AUC value for the neural network model are 0.76, 0.60, and 0.83, respectively.\n\n## Comparison of the Two Algorithms\n\nBoth the machine learning algorithms performed well, but the support vector machine algorithm performed marginally better than the neural network model across the evaluation metrics of accuracy, F-score, and AUC value.\n\n## Conclusion\n\nIn this guide, you learned the basics of two advanced machine learning algorithms—support vector machine and neural network. You also learned how to configure and evaluate the two algorithms in Azure Machine Learning Studio.\n\nTo learn more about data science and machine learning using Azure Machine Learning Studio, please refer to the following guides:\n\n7" ]

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[ "# Chemistry solution molarity practice pdf\n\n### SOLUTIONS CHEMISTRY PRACTICE PROBLEMS FOR NON", null, "Molarity Practice Problems Just Only. [PDF]Free Molarity And Molality Practice Problems Answers download Book Molarity And Molality Practice Problems Answers.pdf Chemistry - General Knowledge Questions and Answers Fri, 14 Dec 2018 19:12:00 GMT Why General Knowledge Chemistry? In this section you can learn and practice General Knowledge Questions based on \"Chemistry\" and improve your skills in order to face the …, For chemistry help, visit www.chemfiesta.com! Solutions to the Molarity Practice Worksheet For the first five problems, you need to use the equation that says that the molarity of a solution is equal to the number of moles of solute divided by the number of liters of solution. 1) In this problem, simply solve using the molarity equation to find that the concentration of the solution is 10 M. 2.\n\n### SOLUTIONS CHEMISTRY PRACTICE PROBLEMS FOR NON\n\nMolarity Practice Problems Just Only. Chemistry 121A Hanson Molarity Practice Problems +1. How many moles of K are in 10.0 mL of 0.1250 M K 2 CO 3 solution? 2. How many g of NaOH (MW 40.00) are needed to make 250 mL of 0.1250 M NaOH solution?, Chemistry HS/Science Unit: 10 Lesson: 02 Molarity Practice KEY 1. How would you prepare 1.0 L of a 0.25 M sodium chloride solution? Determine the mass of NaCl to add to a 1.0 L volumetric flask..\n\nMolarity Practice - Download as PDF File (.pdf), Text File (.txt) or read online. a The molarity of a solution is measured in moles of solute per liter of solution, or mol/liter. For example, if the molarity of a mercury solution is 1M, it simply means that there is 1 mole of sugar contained in every 1 liter of the solution.\n\nMore chemistry tutorials and practice can be found at www.chemfiesta.com. Molarity Practice Problems #1 – Answer Key 1) How many grams of potassium carbonate are needed to make 280 mL of a 2.5 M solution? Using the molarity equation (M = mol/L), we can find that we'll need 0.70 mol of potassium carbonate. Given that the molar mass of The unit usually used for molarity in chemistry is mol/L and is represented by the symbol M. Molarity is calculated by determining the number of liters of a solution, determining the number of moles of solute in a solution, and then dividing the number moles of solute by the liters of solution. This customizable and printable worksheet is designed to help students practice calculating the\n\nMolarity Practice Questions And Answers Practice revision questions on calculating molarity from mass, volume and formula mass data, using experiment data, making predictions. molarity molality practice problems answers Sun, 16 Dec 2018 11:39:00 GMT molarity molality practice problems answers pdf - Why General Knowledge Chemistry? In this section you can learn and practice General Knowledge Questions based on \"Chemistry\" and improve your skills in order to face the interview, competitive examination and various entrance test (CAT, GATE, GRE, MAT, Bank …\n\n3) 5) Molarity Practice Problems How many grams of potassium carbonate are needed to make 200 ml- of a 2.5 M solution? How many liters of 4 M solution can be made using 100 grams of lithium molarity practice problems answers with work download molarity practice problems answers pdfchemistry practice problems - b brunerhonors chemistry - darrell feebeck“eewwwwâ€. chemistry!!†- wofford college chemistry with lab – easy peasy all-in-one high\n\nChapter 4 Practice Problems Page 1 of 3 CHAPTER 4 – SOLUTION CHEMISTRY Solution Concentration and Molarity 1. A solution is made by dissolving 3.875 g of Part 2: Molarity Practice Use your notes over Molarity to help answer the following questions 1. What is the molarity of a solution in which 2.10mol of NaCl are dissolved in 1.2 L of water? 2. What is the concentration (molarity) of a solution in which 2.58 mol of AgNO 3 is\n\nMolarity And Molality Practice Problems With Answers Pdf Solutions to the Molarity Practice Worksheet. For the first five For the first five problems, you need to use the equation that says that the Molality: Remember molality is defined as the # moles of solute Г· # of Molarity Worksheet. Name. Key. 1. What is the molarity of a solution that contains 16.0 g NaOH in 2.00 L of solution. NaOH mol 400.0. NaOH g 40.0.\n\nMolarity Practice Chart and Problems A solution’s concentration refers to the relative amount of solute per unit volume or mass of solute. Concentration is thus a ratio concept. Keep this in mind while solving concentration problems. Specifically, molarity (M) is defined as the number of moles of solute per liter of solution. The same equation can be written three different ways depending on molarity practice worksheet answers pdf Molarity, or molar concentration, represents the concentration of a solute in a solution. The unit usually used for molarity in chemistry is mol/L and is represented by the symbol M.Molarity is calculated by determining the number of liters of a solution, determining the number of moles of solute in a solution, and then dividing the number moles of\n\nmolarity practice problems answers with work download molarity practice problems answers pdfchemistry practice problems - b brunerhonors chemistry - darrell feebeck“eewwwwâ€. chemistry!!†- wofford college chemistry with lab – easy peasy all-in-one high Molarity Practice Problems (assume all solutions are aqueous) 1. How many grams of potassium carbonate are needed to make 200.0 mL of a 2.5 M\n\nMore chemistry tutorials and practice can be found at www.chemfiesta.com. Molarity Practice Problems #1 – Answer Key 1) How many grams of potassium carbonate are needed to make 280 mL of a 2.5 M solution? Using the molarity equation (M = mol/L), we can find that we'll need 0.70 mol of potassium carbonate. Given that the molar mass of 3) 5) Molarity Practice Problems How many grams of potassium carbonate are needed to make 200 ml- of a 2.5 M solution? How many liters of 4 M solution can be made using 100 grams of lithium\n\nA.P. Chemistry Practice Test: Ch. 11, Solutions Name_____ MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) Formation of solutions where the process is endothermic can be spontaneous provided that _____. A)the solvent is a gas and the solute is a solid B)they are accompanied by an increase in order C)they are accompanied by an … Practice: Molarity calculations. Science В· Chemistry В· States of matter and intermolecular forces В· Mixtures and solutions. Molarity. Definitions of solution, solute, and solvent. How molarity is used to quantify the concentration of solute, and comcalculations related to molarity. Key points. Mixtures with uniform composition are called homogeneous mixtures or solutions. Mixtures with non\n\nMolarity Practice - Download as PDF File (.pdf), Text File (.txt) or read online. a 13/08/2017В В· This chemistry video tutorial explains how to solve common molarity problems. It discusses how to calculate the concentration of a solution given the …\n\nMolarity Practice Questions And Answers Practice revision questions on calculating molarity from mass, volume and formula mass data, using experiment data, making predictions. Chapter 4 Practice Problems Page 1 of 3 CHAPTER 4 – SOLUTION CHEMISTRY Solution Concentration and Molarity 1. A solution is made by dissolving 3.875 g of\n\nFrom the definition of molarity, we know that the moles of solute equals the molarity times the volume. So we can substitute MV (molarity times volume) into the above equation, like this: M1V1= M2V2 The \"sub one\" refers to the situation before dilution and the \"sub two\" refers to after dilution. We will call this the dilution equation. This equation does not have an official name like Boyle's Molarity Practice Problems 1) How many grams of potassium carbonate are needed to make 200 mL of a 2.5 M solution? 2) How many liters of 4 M solution can be made using 100 grams of lithium bromide?\n\nWhen aqueous solutions of sodium sulfate and lead (II) nitrate are mixed, a solid is formed. Calculate the mass of the solid formed when 1.25 L of 0.0500 M lead nitrate and … More chemistry tutorials and practice can be found at www.chemfiesta.com. Molarity Practice Problems #1 – Answer Key 1) How many grams of potassium carbonate are needed to make 280 mL of a 2.5 M solution? Using the molarity equation (M = mol/L), we can find that we'll need 0.70 mol of potassium carbonate. Given that the molar mass of\n\nPart 2: Molarity Practice Use your notes over Molarity to help answer the following questions 1. What is the molarity of a solution in which 2.10mol of NaCl are dissolved in 1.2 L of water? 2. What is the concentration (molarity) of a solution in which 2.58 mol of AgNO 3 is Chapter 4 Practice Problems Page 1 of 3 CHAPTER 4 – SOLUTION CHEMISTRY Solution Concentration and Molarity 1. A solution is made by dissolving 3.875 g of\n\nMolarity Practice Problems #2 1) How many liters of 0.88 M LiF solution can be made with 25.5 grams of solute? 2) What is the concentration of a solution that … Before you can figure out how to calculate molarity, you have to understand the definitions of two words: \"mole\" and \"solution.\" A mole is a precise number of molecules, namely 6.02 X 10 23 , that is used often in chemistry applications.\n\nMore chemistry tutorials and practice can be found at www.chemfiesta.com. 1) How many liters of 0.88 M LiF solution can be made with 25.5 grams of solute? 2) What is the concentration of a solution that has a volume of 660 mL and contains 33.4 When aqueous solutions of sodium sulfate and lead (II) nitrate are mixed, a solid is formed. Calculate the mass of the solid formed when 1.25 L of 0.0500 M lead nitrate and …\n\nMolarity Worksheet. Name. Key. 1. What is the molarity of a solution that contains 16.0 g NaOH in 2.00 L of solution. NaOH mol 400.0. NaOH g 40.0. Molarity Practice Problems #2 1) How many liters of 0.88 M LiF solution can be made with 25.5 grams of solute? 2) What is the concentration of a solution that …\n\nMore chemistry tutorials and practice can be found at www.chemfiesta.com. 1) How many liters of 0.88 M LiF solution can be made with 25.5 grams of solute? 2) What is the concentration of a solution that has a volume of 660 mL and contains 33.4 Molarity Practice Problems #2 1) How many liters of 0.88 M LiF solution can be made with 25.5 grams of solute? 2) What is the concentration of a solution that …\n\nChapter 4 Practice Problems Page 1 of 3 CHAPTER 4 SOLUTION. The molarity of a solution is measured in moles of solute per liter of solution, or mol/liter. For example, if the molarity of a mercury solution is 1M, it simply means that there is 1 mole of sugar contained in every 1 liter of the solution., Molarity Practice Questions And Answers Practice revision questions on calculating molarity from mass, volume and formula mass data, using experiment data, making predictions..\n\n### Chemistry for YEAR and practice General 11 TERM 1 What", null, "Molarity Practice Problems Just Only. Molarity Problems Worksheet M = _n_ V - n= # moles - V must be in liters (change if necessary) - Use M or mol/L as unit for molarity 1. What is the molarity of a 0.30 liter solution …, From the definition of molarity, we know that the moles of solute equals the molarity times the volume. So we can substitute MV (molarity times volume) into the above equation, like this: M1V1= M2V2 The \"sub one\" refers to the situation before dilution and the \"sub two\" refers to after dilution. We will call this the dilution equation. This equation does not have an official name like Boyle's.\n\n### Molarity Practice Questions and Tutorial Increase your Score", null, "Chemistry for YEAR and practice General 11 TERM 1 What. Calculate the molarity of a solution prepared by dissolving 6.80 grams of AgNO water to make 2.50 liters of solution. (Answer: 0.016 mollL). l' t. 7 WW VVVVV H. Practice: Molarity calculations. Science В· Chemistry В· States of matter and intermolecular forces В· Mixtures and solutions. Molarity. Definitions of solution, solute, and solvent. How molarity is used to quantify the concentration of solute, and comcalculations related to molarity. Key points. Mixtures with uniform composition are called homogeneous mixtures or solutions. Mixtures with non.", null, "• Molarity Practice Worksheet Department of Chemistry [FSU]\n• Molarity Practice Questions and Tutorial Increase your Score\n• Stoichiometry and Molarity Practice Part A\n\n• Molarity Of Solutions Practice Problems Molarity calculations (practice) khan academy, practice: molarity calculations this is the currently selected item science chemistry states of matter and molarity practice worksheet answers pdf Molarity, or molar concentration, represents the concentration of a solute in a solution. The unit usually used for molarity in chemistry is mol/L and is represented by the symbol M.Molarity is calculated by determining the number of liters of a solution, determining the number of moles of solute in a solution, and then dividing the number moles of\n\nmolarity molality practice problems answers Sun, 16 Dec 2018 11:39:00 GMT molarity molality practice problems answers pdf - Why General Knowledge Chemistry? In this section you can learn and practice General Knowledge Questions based on \"Chemistry\" and improve your skills in order to face the interview, competitive examination and various entrance test (CAT, GATE, GRE, MAT, Bank … Part 2: Molarity Practice Use your notes over Molarity to help answer the following questions 1. What is the molarity of a solution in which 2.10mol of NaCl are dissolved in 1.2 L of water? 2. What is the concentration (molarity) of a solution in which 2.58 mol of AgNO 3 is\n\nPart 2: Molarity Practice Use your notes over Molarity to help answer the following questions 1. What is the molarity of a solution in which 2.10mol of NaCl are dissolved in 1.2 L of water? 2. What is the concentration (molarity) of a solution in which 2.58 mol of AgNO 3 is Title: Free Solutions Molarity And Dilution Practice Answer Key (PDF, ePub, Mobi) Author: Harper & Row Subject: Solutions Molarity And Dilution Practice Answer Key\n\nGMT solutions molarity and dilution practice pdf - Molarity And Molality Practice Problems With Answers Pdf Solutions to the Molarity Practice Worksheet. For the first five problems, you need to use the equation that says that the Molality: Remember molality is defined as the # moles of solute ГѓВ· # of Kg of solvent. kg mol Molarity Practice Answers. When you finish this section you will be Molarity Worksheet. Name. Key. 1. What is the molarity of a solution that contains 16.0 g NaOH in 2.00 L of solution. NaOH mol 400.0. NaOH g 40.0.\n\nand dilution practice pdf - Molarity and Dilutions; Solve the following molarity and di!ution problems. Round all answers to the correct number of significant figures. Make sure all answers have the correct unit., What is the molarity of a solution in which lO.8 grams of calcium chloride is dissolved in enough water to make 125 mL of solution?,- y.. Thu, 23 Feb 2017 20:39:00 GMT Molarity and STE Extra Practice with Molarity SOLUTIONS n = CV or C = n V n = moles of solute C = concentration (moles/L= molarity) V = volume of solution in L 1. In 10.0 mL of a certain solution, there are 0.050 g of KF. Find the molarity of the solution (moles/L) 0.050 g KF\n\nTitle: Free Solutions Molarity And Dilution Practice Answer Key (PDF, ePub, Mobi) Author: Harper & Row Subject: Solutions Molarity And Dilution Practice Answer Key From the definition of molarity, we know that the moles of solute equals the molarity times the volume. So we can substitute MV (molarity times volume) into the above equation, like this: M1V1= M2V2 The \"sub one\" refers to the situation before dilution and the \"sub two\" refers to after dilution. We will call this the dilution equation. This equation does not have an official name like Boyle's\n\nTitle: Free Solutions Molarity And Dilution Practice Answer Key (PDF, ePub, Mobi) Author: Harper & Row Subject: Solutions Molarity And Dilution Practice Answer Key Chemistry HS/Science Unit: 10 Lesson: 02 Molarity Practice KEY 1. How would you prepare 1.0 L of a 0.25 M sodium chloride solution? Determine the mass of NaCl to add to a 1.0 L volumetric flask.\n\nMolarity Practice Problems #2 1) How many liters of 0.88 M LiF solution can be made with 25.5 grams of solute? 2) What is the concentration of a solution that … STE Extra Practice with Molarity SOLUTIONS n = CV or C = n V n = moles of solute C = concentration (moles/L= molarity) V = volume of solution in L 1. In 10.0 mL of a certain solution, there are 0.050 g of KF. Find the molarity of the solution (moles/L) 0.050 g KF\n\nMolarity Practice Chart and Problems A solution’s concentration refers to the relative amount of solute per unit volume or mass of solute. Concentration is thus a ratio concept. Keep this in mind while solving concentration problems. Specifically, molarity (M) is defined as the number of moles of solute per liter of solution. The same equation can be written three different ways depending on Molarity Practice Problems 1) How many grams of potassium carbonate are needed to make 200 mL of a 2.5 M solution? 2) How many liters of 4 M solution can be made using 100 grams of lithium bromide?\n\nAP Chem: Chapter 4 Practice Multiple Choice Questions Multiple Choice Identify the choice that best completes the statement or answers the question. Molarity Practice - Download as PDF File (.pdf), Text File (.txt) or read online. a", null, "Molarity And Molality Practice Problems With Answers Pdf Solutions to the Molarity Practice Worksheet. For the first five For the first five problems, you need to use the equation that says that the Molality: Remember molality is defined as the # moles of solute Г· # of Chapter 4 Practice Problems Page 1 of 3 CHAPTER 4 – SOLUTION CHEMISTRY Solution Concentration and Molarity 1. A solution is made by dissolving 3.875 g of\n\n## Molarity Practice KEY SLIDEBLAST.COM", null, "SOLUTIONS CHEMISTRY PRACTICE PROBLEMS FOR NON. Molarity Practice Problems 1) How many grams of potassium carbonate are needed to make 200 mL of a 2.5 M solution? 2) How many liters of 4 M solution can be made using 100 grams of lithium bromide?, The molarity of a solution is calculated by taking the moles of solute and dividing by the liters of solution. This is probably easiest to explain with examples. Example #1: Suppose we had 1.00 mole of sucrose (it's about 342.3 grams) and proceeded to mix it into some water..\n\n### Molarity Practice Problems #1\n\nMolarity Practice Iron Lithium. Download Molarity Of Solutions Practice Problems Pdf Download Molarity Of Solutions Practice Problems free pdf , Download Molarity Of Solutions, The unit usually used for molarity in chemistry is mol/L and is represented by the symbol M. Molarity is calculated by determining the number of liters of a solution, determining the number of moles of solute in a solution, and then dividing the number moles of solute by the liters of solution. This customizable and printable worksheet is designed to help students practice calculating the.\n\nMolarity And Molality Practice Problems With Answers Pdf Solutions to the Molarity Practice Worksheet. For the first five For the first five problems, you need to use the equation that says that the Molality: Remember molality is defined as the # moles of solute Г· # of Title: Free Solutions Molarity And Dilution Practice Answer Key (PDF, ePub, Mobi) Author: Harper & Row Subject: Solutions Molarity And Dilution Practice Answer Key\n\nBefore you can figure out how to calculate molarity, you have to understand the definitions of two words: \"mole\" and \"solution.\" A mole is a precise number of molecules, namely 6.02 X 10 23 , that is used often in chemistry applications. molarity practice worksheet answers pdf Molarity, or molar concentration, represents the concentration of a solute in a solution. The unit usually used for molarity in chemistry is mol/L and is represented by the symbol M.Molarity is calculated by determining the number of liters of a solution, determining the number of moles of solute in a solution, and then dividing the number moles of\n\nSOLUTIONS , and Dilutions Practice Block: Unsaturated Solutions Beaker A 1.0 g of solute added Saturated Solutions Beaker D 7.0 g of solute added 17 Beaker B 2.0 g of solute added Beaker E 9.0 g of solute added eAll beakers contain 10.0 g of water. 'All beakers are kept at 20 oc All solutions are stirred for 2 hours. Solute is the same substance in all beakers. Number of dissolved particles Molarity Practice Problems (assume all solutions are aqueous) 1. How many grams of potassium carbonate are needed to make 200.0 mL of a 2.5 M\n\nMolarity Practice Problems 1) How many grams of potassium carbonate are needed to make 200 mL of a 2.5 M solution? 2) How many liters of 4 M solution can be made using 100 grams of lithium bromide? Molarity And Molality Practice Problems With Answers Pdf Solutions to the Molarity Practice Worksheet. For the first five For the first five problems, you need to use the equation that says that the Molality: Remember molality is defined as the # moles of solute Г· # of\n\nMolarity Of Solutions Practice Problems Molarity calculations (practice) khan academy, practice: molarity calculations this is the currently selected item science chemistry states of matter and and dilution practice pdf - Molarity and Dilutions; Solve the following molarity and di!ution problems. Round all answers to the correct number of significant figures. Make sure all answers have the correct unit., What is the molarity of a solution in which lO.8 grams of calcium chloride is dissolved in enough water to make 125 mL of solution?,- y.. Thu, 23 Feb 2017 20:39:00 GMT Molarity and\n\nBefore you can figure out how to calculate molarity, you have to understand the definitions of two words: \"mole\" and \"solution.\" A mole is a precise number of molecules, namely 6.02 X 10 23 , that is used often in chemistry applications. and dilution practice pdf - Molarity and Dilutions; Solve the following molarity and di!ution problems. Round all answers to the correct number of significant figures. Make sure all answers have the correct unit., What is the molarity of a solution in which lO.8 grams of calcium chloride is dissolved in enough water to make 125 mL of solution?,- y.. Thu, 23 Feb 2017 20:39:00 GMT Molarity and\n\n3) 5) Molarity Practice Problems How many grams of potassium carbonate are needed to make 200 ml- of a 2.5 M solution? How many liters of 4 M solution can be made using 100 grams of lithium 3) 5) Molarity Practice Problems How many grams of potassium carbonate are needed to make 200 ml- of a 2.5 M solution? How many liters of 4 M solution can be made using 100 grams of lithium\n\nmolarity practice problems answers pdf Practice Problems Answers. Mass Percent. =(Mass of Solute) / (Mass of Solution) x 100%/ Save as PDF · Email page. Solutions to the Molarity Practice Worksheet For the first five problems, you need to Homework Answers Molarity & Molality Worksheet Student Due Date 3/29/10 Solutions Worksheet #2 Molarity And Dilution Problems Answers PDF Ebook. Molarity For chemistry help, visit www.chemfiesta.com! Solutions to the Molarity Practice Worksheet For the first five problems, you need to use the equation that says that the molarity of a solution is equal to the number of moles of solute divided by the number of liters of solution. 1) In this problem, simply solve using the molarity equation to find that the concentration of the solution is 10 M. 2\n\nGMT solutions molarity and dilution practice pdf - Molarity And Molality Practice Problems With Answers Pdf Solutions to the Molarity Practice Worksheet. For the first five problems, you need to use the equation that says that the Molality: Remember molality is defined as the # moles of solute ГѓВ· # of Kg of solvent. kg mol Molarity Practice Answers. When you finish this section you will be The equations I will use are: M = moles of solute / liters of solution. and MV = grams / molar mass --- The volume here MUST be in liters. Typically, the solution is for the molarity (M).\n\nMolarity Practice Problems (assume all solutions are aqueous) 1. How many grams of potassium carbonate are needed to make 200.0 mL of a 2.5 M molarity molality practice problems answers Sun, 16 Dec 2018 11:39:00 GMT molarity molality practice problems answers pdf - Why General Knowledge Chemistry? In this section you can learn and practice General Knowledge Questions based on \"Chemistry\" and improve your skills in order to face the interview, competitive examination and various entrance test (CAT, GATE, GRE, MAT, Bank …\n\nPart 2: Molarity Practice Use your notes over Molarity to help answer the following questions 1. What is the molarity of a solution in which 2.10mol of NaCl are dissolved in 1.2 L of water? 2. What is the concentration (molarity) of a solution in which 2.58 mol of AgNO 3 is GMT solutions molarity and dilution practice pdf - Molarity And Molality Practice Problems With Answers Pdf Solutions to the Molarity Practice Worksheet. For the first five problems, you need to use the equation that says that the Molality: Remember molality is defined as the # moles of solute ГѓВ· # of Kg of solvent. kg mol Molarity Practice Answers. When you finish this section you will be\n\nmolarity practice worksheet answers pdf Molarity, or molar concentration, represents the concentration of a solute in a solution. The unit usually used for molarity in chemistry is mol/L and is represented by the symbol M.Molarity is calculated by determining the number of liters of a solution, determining the number of moles of solute in a solution, and then dividing the number moles of When aqueous solutions of sodium sulfate and lead (II) nitrate are mixed, a solid is formed. Calculate the mass of the solid formed when 1.25 L of 0.0500 M lead nitrate and …\n\nA.P. Chemistry Practice Test: Ch. 11, Solutions Name_____ MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) Formation of solutions where the process is endothermic can be spontaneous provided that _____. A)the solvent is a gas and the solute is a solid B)they are accompanied by an increase in order C)they are accompanied by an … 13/08/2017В В· This chemistry video tutorial explains how to solve common molarity problems. It discusses how to calculate the concentration of a solution given the …\n\nPart 2: Molarity Practice Use your notes over Molarity to help answer the following questions 1. What is the molarity of a solution in which 2.10mol of NaCl are dissolved in 1.2 L of water? 2. What is the concentration (molarity) of a solution in which 2.58 mol of AgNO 3 is STE Extra Practice with Molarity SOLUTIONS n = CV or C = n V n = moles of solute C = concentration (moles/L= molarity) V = volume of solution in L 1. In 10.0 mL of a certain solution, there are 0.050 g of KF. Find the molarity of the solution (moles/L) 0.050 g KF\n\nThe molarity of a solution is calculated by taking the moles of solute and dividing by the liters of solution. This is probably easiest to explain with examples. Example #1: Suppose we had 1.00 mole of sucrose (it's about 342.3 grams) and proceeded to mix it into some water. Calculate the molarity of a solution prepared by dissolving 6.80 grams of AgNO water to make 2.50 liters of solution. (Answer: 0.016 mollL). l' t. 7 WW VVVVV H.\n\nMolarity Practice Problems #2 1) How many liters of 0.88 M LiF solution can be made with 25.5 grams of solute? 2) What is the concentration of a solution that … A.P. Chemistry Practice Test: Ch. 11, Solutions Name_____ MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) Formation of solutions where the process is endothermic can be spontaneous provided that _____. A)the solvent is a gas and the solute is a solid B)they are accompanied by an increase in order C)they are accompanied by an …\n\nmolarity practice problems answers pdf Practice Problems Answers. Mass Percent. =(Mass of Solute) / (Mass of Solution) x 100%/ Save as PDF · Email page. Solutions to the Molarity Practice Worksheet For the first five problems, you need to Homework Answers Molarity & Molality Worksheet Student Due Date 3/29/10 Solutions Worksheet #2 Molarity And Dilution Problems Answers PDF Ebook. Molarity [PDF]Free Molarity And Molality Practice Problems Answers download Book Molarity And Molality Practice Problems Answers.pdf Chemistry - General Knowledge Questions and Answers Fri, 14 Dec 2018 19:12:00 GMT Why General Knowledge Chemistry? In this section you can learn and practice General Knowledge Questions based on \"Chemistry\" and improve your skills in order to face the …\n\nTitle: Free Solutions Molarity And Dilution Practice Answer Key (PDF, ePub, Mobi) Author: Harper & Row Subject: Solutions Molarity And Dilution Practice Answer Key Molarity Worksheet. Name. Key. 1. What is the molarity of a solution that contains 16.0 g NaOH in 2.00 L of solution. NaOH mol 400.0. NaOH g 40.0.\n\nGMT solutions molarity and dilution practice pdf - Molarity And Molality Practice Problems With Answers Pdf Solutions to the Molarity Practice Worksheet. For the first five problems, you need to use the equation that says that the Molality: Remember molality is defined as the # moles of solute ГѓВ· # of Kg of solvent. kg mol Molarity Practice Answers. When you finish this section you will be molarity practice worksheet answers pdf Molarity, or molar concentration, represents the concentration of a solute in a solution. The unit usually used for molarity in chemistry is mol/L and is represented by the symbol M.Molarity is calculated by determining the number of liters of a solution, determining the number of moles of solute in a solution, and then dividing the number moles of\n\nChapter 4 Practice Problems Page 1 of 3 CHAPTER 4 – SOLUTION CHEMISTRY Solution Concentration and Molarity 1. A solution is made by dissolving 3.875 g of Practice: Molarity calculations. Science В· Chemistry В· States of matter and intermolecular forces В· Mixtures and solutions. Molarity. Definitions of solution, solute, and solvent. How molarity is used to quantify the concentration of solute, and comcalculations related to molarity. Key points. Mixtures with uniform composition are called homogeneous mixtures or solutions. Mixtures with non\n\n### Chapter 4 Practice Problems Page 1 of 3 CHAPTER 4 SOLUTION", null, "Chemistry for YEAR and practice General 11 TERM 1 What. Molarity Practice Problems #2 1) How many liters of 0.88 M LiF solution can be made with 25.5 grams of solute? 2) What is the concentration of a solution that …, Molarity Practice Chart and Problems A solution’s concentration refers to the relative amount of solute per unit volume or mass of solute. Concentration is thus a ratio concept. Keep this in mind while solving concentration problems. Specifically, molarity (M) is defined as the number of moles of solute per liter of solution. The same equation can be written three different ways depending on.\n\nName Class Period Practice Molarity Intro to Dilutions. Before you can figure out how to calculate molarity, you have to understand the definitions of two words: \"mole\" and \"solution.\" A mole is a precise number of molecules, namely 6.02 X 10 23 , that is used often in chemistry applications., Molarity Worksheet. Name. Key. 1. What is the molarity of a solution that contains 16.0 g NaOH in 2.00 L of solution. NaOH mol 400.0. NaOH g 40.0..\n\n### Molarity Practice Problems Just Only", null, "Chapter 4 Practice Problems Page 1 of 3 CHAPTER 4 SOLUTION. The unit usually used for molarity in chemistry is mol/L and is represented by the symbol M. Molarity is calculated by determining the number of liters of a solution, determining the number of moles of solute in a solution, and then dividing the number moles of solute by the liters of solution. This customizable and printable worksheet is designed to help students practice calculating the A.P. Chemistry Practice Test: Ch. 11, Solutions Name_____ MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) Formation of solutions where the process is endothermic can be spontaneous provided that _____. A)the solvent is a gas and the solute is a solid B)they are accompanied by an increase in order C)they are accompanied by an ….", null, "Molarity Of Solutions Practice Problems Molarity calculations (practice) khan academy, practice: molarity calculations this is the currently selected item science chemistry states of matter and For chemistry help, visit www.chemfiesta.com! Solutions to the Molarity Practice Worksheet For the first five problems, you need to use the equation that says that the molarity of a solution is equal to the number of moles of solute divided by the number of liters of solution. 1) In this problem, simply solve using the molarity equation to find that the concentration of the solution is 10 M. 2\n\nDownload: MOLARITY PRACTICE WORKSHEET ANSWER KEY PDF We have made it easy for you to find a PDF Ebooks without any digging. And by having access to our ebooks online or by storing it on your computer, you have convenient answers with molarity practice worksheet answer key PDF. To get started finding molarity practice worksheet answer key, you are right to find our website which has a STE Extra Practice with Molarity SOLUTIONS n = CV or C = n V n = moles of solute C = concentration (moles/L= molarity) V = volume of solution in L 1. In 10.0 mL of a certain solution, there are 0.050 g of KF. Find the molarity of the solution (moles/L) 0.050 g KF\n\nChapter 4 Practice Problems Page 1 of 3 CHAPTER 4 – SOLUTION CHEMISTRY Solution Concentration and Molarity 1. A solution is made by dissolving 3.875 g of Molarity Practice Problems #2 1) How many liters of 0.88 M LiF solution can be made with 25.5 grams of solute? 2) What is the concentration of a solution that …\n\nSTE Extra Practice with Molarity SOLUTIONS n = CV or C = n V n = moles of solute C = concentration (moles/L= molarity) V = volume of solution in L 1. In 10.0 mL of a certain solution, there are 0.050 g of KF. Find the molarity of the solution (moles/L) 0.050 g KF Molarity Practice Problems (assume all solutions are aqueous) 1. How many grams of potassium carbonate are needed to make 200.0 mL of a 2.5 M\n\nmolarity practice problems answers with work download molarity practice problems answers pdfchemistry practice problems - b brunerhonors chemistry - darrell feebeck“eewwwwâ€. chemistry!!†- wofford college chemistry with lab – easy peasy all-in-one high Chapter 4 Practice Problems Page 1 of 3 CHAPTER 4 – SOLUTION CHEMISTRY Solution Concentration and Molarity 1. A solution is made by dissolving 3.875 g of\n\nGMT solutions molarity and dilution practice pdf - Molarity And Molality Practice Problems With Answers Pdf Solutions to the Molarity Practice Worksheet. For the first five problems, you need to use the equation that says that the Molality: Remember molality is defined as the # moles of solute ГѓВ· # of Kg of solvent. kg mol Molarity Practice Answers. When you finish this section you will be From the definition of molarity, we know that the moles of solute equals the molarity times the volume. So we can substitute MV (molarity times volume) into the above equation, like this: M1V1= M2V2 The \"sub one\" refers to the situation before dilution and the \"sub two\" refers to after dilution. We will call this the dilution equation. This equation does not have an official name like Boyle's\n\nUnformatted text preview: 1 of 19 Molarity Practice Worksheet Find the molarity of the following solutions: mo salute solution 1) 0.5 moles of sodium chloride is dissolved to make 0.05 liters of solution. SOLUTIONS , and Dilutions Practice Block: Unsaturated Solutions Beaker A 1.0 g of solute added Saturated Solutions Beaker D 7.0 g of solute added 17 Beaker B 2.0 g of solute added Beaker E 9.0 g of solute added eAll beakers contain 10.0 g of water. 'All beakers are kept at 20 oc All solutions are stirred for 2 hours. Solute is the same substance in all beakers. Number of dissolved particles\n\nChapter 4 Practice Problems Page 1 of 3 CHAPTER 4 – SOLUTION CHEMISTRY Solution Concentration and Molarity 1. A solution is made by dissolving 3.875 g of molarity practice problems answers pdf Practice Problems Answers. Mass Percent. =(Mass of Solute) / (Mass of Solution) x 100%/ Save as PDF · Email page. Solutions to the Molarity Practice Worksheet For the first five problems, you need to Homework Answers Molarity & Molality Worksheet Student Due Date 3/29/10 Solutions Worksheet #2 Molarity And Dilution Problems Answers PDF Ebook. Molarity\n\n4/10/2018В В· To calculate molarity, divide the number of moles of solute by the volume of the solution in liters. If you don't know the number of moles of solute but you know the mass, start by finding the molar mass of the solute, which is equal to all of the molar masses of each element in the solution added together. Once you have the molar mass, multiply the number of grams of solute by 1 over the More chemistry tutorials and practice can be found at www.chemfiesta.com. Molarity Practice Problems #1 – Answer Key 1) How many grams of potassium carbonate are needed to make 280 mL of a 2.5 M solution? Using the molarity equation (M = mol/L), we can find that we'll need 0.70 mol of potassium carbonate. Given that the molar mass of\n\nmolarity practice worksheet answers pdf Molarity, or molar concentration, represents the concentration of a solute in a solution. The unit usually used for molarity in chemistry is mol/L and is represented by the symbol M.Molarity is calculated by determining the number of liters of a solution, determining the number of moles of solute in a solution, and then dividing the number moles of The equations I will use are: M = moles of solute / liters of solution. and MV = grams / molar mass --- The volume here MUST be in liters. Typically, the solution is for the molarity (M).\n\nAP Chem: Chapter 4 Practice Multiple Choice Questions Multiple Choice Identify the choice that best completes the statement or answers the question. molarity practice worksheet answers pdf Molarity, or molar concentration, represents the concentration of a solute in a solution. The unit usually used for molarity in chemistry is mol/L and is represented by the symbol M.Molarity is calculated by determining the number of liters of a solution, determining the number of moles of solute in a solution, and then dividing the number moles of" ]

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[ "", null, "# Mathematics Minor\n\nThe Mathematics minor is based on a diversity of courses in pure and applied Mathematics. Students who complete the Minor will acquire an essential background in some important branches of classical Mathematics, Statistics and their applications. The minor will provide also an opportunity for students to develop significant mathematical skills with a selection of advanced courses, which will introduce them to some modern lines of contemporary Mathematics and its applications to other sciences.\n\nTotal: 6 courses\n\nRequired Courses (6 courses):\n\nMAT 103 Calculus I\nMAT 104 Calculus II\nMAT 105 Elementary Linear Algebra and Analytical Geometry\nMAT 212 Calculus III\nMAT 201 Mathematical Statistics\nMAT 213 Introduction to Differential Equations\n\nElective Courses (1 course):\n\nOne (1) course from the following:\n\nMAT 201 Mathematical Statistics\nMAT 205 Introduction to Abstract Algebra\nMAT 213 Introduction to Differential Equations\nMAT 214 Numerical Analysis\nMAT 225 Advanced Linear Algebra\nMAT 305 Topics in Abstract Algebra\nMAT 313 Calculus IV\nMAT 314 Complex Analysis\nMAT 315 Real Analysis\n\nWe are Social" ]

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

[ "Solutions by everydaycalculation.com\n\n## Reduce 1260/3920 to lowest terms\n\nThe simplest form of 1260/3920 is 9/28.\n\n#### Steps to simplifying fractions\n\n1. Find the GCD (or HCF) of numerator and denominator\nGCD of 1260 and 3920 is 140\n2. Divide both the numerator and denominator by the GCD\n1260 ÷ 140/3920 ÷ 140\n3. Reduced fraction: 9/28\nTherefore, 1260/3920 simplified to lowest terms is 9/28.\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:" ]

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

[ "# Chronology for 500 to 900", null, "Metrodorus assembles the Greek Anthology consisting of 46 mathematical problems.\n\n510\nEutocius of Ascalon writes commentaries on Archimedes' work.\n\n510\nBoethius writes geometry and arithmetic texts which are widely used for a long time.\n\nEutocius writes commentaries on the works of Archimedes and Apollonius.\n\n532\nAnthemius of Tralles, a mathematician of note, is the architect for the Hagia Sophia at Constantinople. (See this History Topic.)\n\n534\nChinese mathematics is introduced into Japan.\n\n575\nVarahamihira produces Pancasiddhantika (The Five Astronomical Canons). He makes important contributions to trigonometry.\n\n594\nDecimal notation is used for numbers in India. This is the system on which our current notation is based. (See this History Topic.)\n\n628\nBrahmagupta writes Brahmasphutasiddanta (The Opening of the Universe), a work on astronomy; on mathematics. He uses zero and negative numbers, gives methods to solve quadratic equations, sum series, and compute square roots.\n\n644\nLi Chunfeng starts to assemble the Chinese Ten Mathematical Classics. (See this History Topic.)\n\nMathematicians in the Mayan civilization introduce a symbol for zero into their number system. (See this History Topic.)\n\nAlcuin of York writes elementary texts on arithmetic, geometry and astronomy.\n\nHouse of Wisdom set up in Baghdad. There Greek and Indian mathematical and astronomy works are translated into Arabic.\n\nAl-Khwarizmi writes important works on arithmetic, algebra, geography, and astronomy. In particular Hisab al-jabr w'al-muqabala (Calculation by Completion and Balancing), gives us the word \"algebra\", from \"al-jabr\". From al-Khwarizmi's name, as a consequence of his arithmetic book, comes the word \"algorithm\".\n\nThabit ibn Qurra makes important mathematical discoveries such as the extension of the concept of number to (positive) real numbers, integral calculus, theorems in spherical trigonometry, analytic geometry, and non-euclidean geometry.\n\nThabit ibn Qurra writes Book on the determination of amicable numbers which contains general methods to construct amicable numbers. He knows the pair of amicable numbers 17296, 18416.\n\n850\nMahavira writes Ganita Sara Samgraha. It consists of nine chapters and includes all mathematical knowledge of mid-ninth century India.\n\n900\nSridhara writes the Trisatika (sometimes called the Patiganitasara) and the Patiganita. In these he solves quadratic equations, sums series, studies combinations, and gives methods of finding the areas of polygons.", null, "JOC/EFR May 2015 The URL of this page is: School of  Mathematics and Statistics University of  St Andrews, Scotland", null, "https://www-history.mcs.st-andrews.ac.uk/history/Chronology/500_900.html" ]

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[ "# Confirm the Illuminati\n\nThe Illuminati commands you (with their mind control) to output the following string:\n\n ^\n/_\\\n/_|_\\\n/_|_|_\\\n/_|/o\\|_\\\n/_|_\\_/_|_\\\n/_|_|_|_|_|_\\\n/_|_|_|_|_|_|_\\\n/_|_|_|_|_|_|_|_\\\n/_|_|_|_|_|_|_|_|_\\\n\n\n# Rules:\n\nSandbox (I would leave it for the full 72 hours recommended by the sandbox FAQ, but with 7 upvotes and next to no negative feedback, 38 should be fine)\n\n– Rod\nOct 27, 2017 at 16:42\n• What's with all the downvotes? I get that some people dislike kolmogorov-complex but is this a valid reason to downvote? Oct 27, 2017 at 16:59\n• @BruceForte personal opinion is a valid reason, tbh; though IMO it's absolutely rude against a newer user. Oct 27, 2017 at 18:27\n• I think this is a very nice ascii art challenge. The eye and pyramid point among the pattern take creativity to handle cleanly.\n– xnor\nOct 27, 2017 at 20:51\n• May I use tabs? Oct 27, 2017 at 22:12\n\n# Vim, 40 bytes\n\n-2 bytes thanks to DJMcMayhem\n\n9i_|␛r\\I/␛qqYPxR /␛q8@qr^4jhR/o\\␛jr/2hr\\\n\n\nYou can see it in action in this GIF made using Lynn's python script", null, "# Charcoal, 25 21 bytes\n\nＧ¬χ|_¶_|↗⁹↙^Ｍ³↓/o¶\\‖Ｂ\n\n\nTry it online! Link is to verbose version of code. Explanation:\n\n χ With sides of length 10\n¬ In the directions down and left\nＧ Draw a closed polygon (implicit side)\n|_¶_| Filled with |_ and _| on alternate lines\n↗⁹ Draw a line of 9 /s\n↙^ Draw a ^ and move the cursor down and left\nＭ³↓ Move down 3 characters\n/o¶\\ Print the left half of the eye\n‖Ｂ Reflect the canvas keeping the right edge\n\n• There it is! Was wondering how long it would be. Oct 27, 2017 at 16:59\n• @StephenLeppik I was on the phone :-(\n– Neil\nOct 27, 2017 at 17:00\n• Wait, is that the only reason? Oct 27, 2017 at 17:01\n• @StephenLeppik: well, that and depending how long before I noticed the notification for a new main post, I guess.\n– Neil\nOct 27, 2017 at 18:39\n• @DLosc < is one of the multidirectionals - when used with the Multiprint command it causes the string to be printed twice, once up right and once down right, but it can also be used as a shortcut for ↗↘ in other contexts too.\n– Neil\nOct 27, 2017 at 18:42\n\n# V, 37 bytes\n\n9i|_\u001bá\\|r/òÄó_|\n>òC ^\u001b4jhR/o\\\u001bj2hR\\_/\n\n\nTry it online!\n\nHexdump:\n\n00000000: 3969 7c5f 1be1 5c7c 722f f2c4 f35f 7c0a 9i|_..\\|r/..._|.\n00000010: 3ef2 4320 5e1b 346a 6852 2f6f 5c1b 6a32 >.C ^.4jhR/o\\.j2\n00000020: 6852 5c5f 2f hR\\_/\n\n\nExplanation:\n\n9i \" Insert 9 copies of...\n|_ \" '|_'\ná\\ \" Append a backslash\n| \" Move to the first character\nr/ \" Replace it with a forward slash\nò \" Recursively...\nÄ \" Duplicate this line (upwards)\nó \" Remove one instance of..\n_| \" '_|'\n> \" Indent this line with one space\nò \" End the loop\nC ^ \" Change this line (previously '/_\\') to ' ^'\n4j \" Move down 4 lines\nh \" Move one character to the left\nR \" Write this text over existing characters...\n/o\\ \" '/o\\'\nj \" Move down a line\n2h \" Move two characters to the left\nR \" Write this text over existing characters...\n\\_/ \" '\\_/'\n\n\n# SOGL V0.12, 3127 25 bytes\n\n ^9∫Ƨ_|m└Κ}¹±§\"/o¶\\_”95žΓ\n\n\nTry it Here!\n\n ^ push \"^\"\n9∫ } do 9 times, pushing counter\nƧ_| push \"_|\"\nm mold that to the counter\n└Κ prepend \"/\"\n¹ collect the lines in an array - [\"^\", \"/_\", \"/_|\", ..., \"/_|_|_|_|_\"]\n± reverse each [\"^\", \"_/\", \"|_/\", ..., \"_|_|_|_|_/\"]\n§ reverse as ascii-art:\n[\" ^\",\n\" /_\",\n\" /_|\",\n...,\n\"/_|_|_|_|_\"]\n\"/o¶\\_” push \"/o\n\\_\"\n95ž insert that at [9; 5]\nΓ palindromize horizontally\n\n\nor a 24 byte version using ¼ (space to an antidiagonal) instead of ±§:\n\n ^9∫Ƨ_|m└Κ}¹¼\"/o¶\\_”95žΓ\n\n\nTry it Here!\n\n# Python 2, 10310198 95 bytes\n\n-2 bytes thanks to Jonathan Frech\n-3 bytes thanks to ovs\n\nfor i in range(10):print(9-i)*' '+['^','/_%s\\\\'%['|_'*~-i,'|/o\\|_','|_\\_/_|_'][i%6/4*i%3]][i>0]\n\n\nTry it online!\n\n# JavaScript (ES6), 95 92 bytes\n\nf=(n=9,s='')=>n--?f(n,s+' ')+s+/${n-4?n-3?'_|'.repeat(n):'_|/o\\\\|':'_|_\\\\_/_|'}_\\\\ :s+^ Or 91 bytes with a leading new-line -- which I think is not allowed: f=(n=9,s= )=>n--?f(n,s+' ')+s+/${n-4?n-3?'_|'.repeat(n):'_|/o\\\\|':'_|_\\\\_/_|'}_\\\\:s+^\n\n\nf=(n=9,s='')=>n--?f(n,s+' ')+s+/${n-4?n-3?'_|'.repeat(n):'_|/o\\\\|':'_|_\\\\_/_|'}_\\\\ :s+^ O.innerText = f() <pre id=O></pre> ### Formatted and commented f = (n = 9, s = '') => // n = line counter, s = leading spaces n-- ? // if we haven't reached the top: f(n, s + ' ') + // do a recursive call with one more leading space s + // append the leading spaces /${ // append the left border\nn - 4 ? // if this is not the 4th row:\nn - 3 ? // if this is not the 3rd row:\n'_|'.repeat(n) // append the brick pattern\n: // else (3rd row):\n'_|/o\\\\|' // append the top of the eye\n: // else (4th row):\n'_|_\\\\_/_|' // append the bottom of the eye\n}_\\\\\\n // append the right border + line-feed\n: // else:\ns + ^\\n // append the top of the pyramid and stop the recursion\n\n• Leading whitespace is allowed. Nov 13, 2017 at 17:18\n\n# C (gcc), 124122120119117115 118 bytes\n\n-1 byte thanks to @xanoetux +3 missing the lowest level...\n\nf(i){for(printf(\"%*c\",i=10,94);--i;printf(\"\\n%*c%s_\\\\\",i,47,i^6?i^5?\"_|_|_|_|_|_|_|_|_|\"+i*2:\"_|_\\\\_/_|\":\"_|/o\\\\|\"));}\n\n\nTry it online!\n\n\" ^\\n\"++do z<-[1..9];([z..8]>>\" \")++'/':g z++\"_\\\\\\n\"\ng 4=\"_|/o\\\\|\"\ng 5=\"_|_\\\\_/_|\"\ng x=[2..x]>>\"_|\"\n\n\nTry it online!\n\nThose 9 space at the beginning hurt.\n\nHow it works\n\n\" ^\\n\"++ -- first line, followed by\ndo -- we use the \"do\" syntatic sugar for monads,\nz<-[1..9] -- for all 'z' from [1..9] perform the following\n-- and collect the results in a single list\n([z..8]>>\" \")++'/' -- make the spaces for the current line and\n-- the left wall '/'\ng z -- call g to make the inner part\n\"_\\\\\\n\" -- append '_', '\\' and a NL\n\ng 4=\"_|/o\\\\|\" -- line 4 and 5 are implemented directly\ng 5=\"_|_\\\\_/_|\"\ng x=[2..x]>>\"_|\" -- all other lines are some copies of \"_|\"\n\n\nEdit: -3 bytes thanks to @Laikoni:\n\n# PowerShell, 109 105 bytes\n\nfilter f{' '*$_+'/'+'_|'*(8-$_)+'_\\'}\n' '*9+'^'\n8|f\n7|f\n6|f\n' /_|/o\\|_\\\n/_|_\\_/_|_\\'\n3..0|%{$_|f} Try it online! Saved 4 bytes thanks to Veskah. # 05AB1E, 4742 40 bytes '/„_|ûûû«η'^0ǝ.∞.C\":;<IJK\"Çv\"/o\\\\_/\"Nèyǝ Try it online! '/„_|ûûû« # Push bottom left tier of pyramid. η # All prefixes of... '^0ǝ # Replace the tip. .∞.C # Mirror, Center. \":;<IJK\"Ç # Push [58,59,60,73,74,75]. v\"/o\\\\_/\"Nèyǝ # Replace those indexes with the eye. Stupid version: „_|3×\"_|/o\\|\".;„_|2×û\"_|_\\_/_\".; Other, less stupid version (but still worse): # 05AB1E, 42 bytes •~µÎт•η4¾ǝ•Σ}•4ǝ•3x1•5ǝεS\"|_/\\^o\"sèJ}€.∞.C Try it online! • Your alternative is invalid in the latest version of 05AB1E (legacy). Feb 7, 2021 at 0:24 # Bubblegum, 48 bytes 00000000: 5380 8138 2e18 4b3f 3e86 0bce ac01 72e0 S..8..K?>.....r. 00000010: 6c30 0fc6 d1cf 8f01 71e1 cae2 218a e12a l0......q...!..* 00000020: 6ba0 ea61 7c84 085c 0021 0417 4188 0100 k..a|..\\.!..A... Try it online! # PHP, 123+3 bytes +3 bytes for the weird tab counting. (it still moves the cursor 8 spaces in any console!) for($i=10;--$i;)$r.=str_pad(str_pad(\"\n\",$i).\"/\",20-$i,\"_|\").\"\\\\\";$r=$r=\"/\";$r=$r=\"\\\\\";$r=o;echo\" ^$r\";\n\n\nNote: The first character after echo\" is a tab character!\n\nRun with -nr or try it online.\n\nother version, same length:\n\nfor(;$i++<9;)$r.=str_pad(str_pad(\"\n\",10-$i).\"/\",10+$i,\"_|\").\"\\\\\";$r=$r=\"/\";$r=$r=\"\\\\\";$r=o;echo\" ^$r\";\n\n• What about Eclipse? I haven't used it in a while but I remember that its tabs were 4 spaces. Oct 27, 2017 at 22:51\n• @StephenLeppik Eclipse is an editor, not a shell. Oct 27, 2017 at 23:05\n• It still has a console. Not to mention that the SE markdown editor and <pre> tag font both have 4-space tabs Oct 27, 2017 at 23:44\n\n# Retina, 79 73 bytes\n\n\n/8x\n8\n$* +^(.*)\\Sx$1x¶$& x ^ /111x /1/o\\|x /1111x /1x_/1x 1 _| x _\\ Try it online! # Ruby, 92 bytes 10.times{|i|s=' '*(10-i)+(i<1??^:\"/#{\"_|\"*~-i}_\\\\\");i/2==2&&s[9,3]=\"/o\\\\_/\"[i%2*2,3];puts s} # Excel VBA, 104 Bytes Anonymous VBE immediate window function that confirms the truth. Version A: ?Spc(9)\"^\":For i=0To 8:[A1]=i:?Spc(8-i)\"/_\"[If(A1=3,\"|/o\\|_\",If(A1=4,\"|_\\_/_|_\",Rept(\"|_\",A1)))]\"\\\":Next Version B: ?Spc(9)\"^\":For i=0To 8:[A1]=i:?Spc(8-i)\"/_\"IIf(i=3,\"|/o\\|_\",IIf(i=4,\"|_\\_/_|_\",[Rept(\"|_\",A1)]))\"\\\":Next # Python 2, 154 bytes l=bytearray a,b=' _';g=[l(a*9+\"^\"+a*9)]+[l(a*(8-k)+\"/%s\\\\\"%\"|\".join(b*k+b))for k in range(9)] g[8:11]=l(\"/o\\\\\") g[8:11]=l(\"\\\\_/\") for r in g:print r Try it online! -3 bytes using bytearray thanks to Rod -1 byte thanks to bobrobbob # Javascript 90 bytes (if default parameter a=9 is required then 92 bytes) A=(a,x=)=>a?A(a-1,x+\" \")+x+\"/\".padEnd(a*2,a^5?a^4?\"_|\":\"_|/o\\\\|\":\"_|_\\\\_/\")+\\\\ :x+^ console.log(A(9)) • All extraneous arguments must be included in the byte total so this counts as 91 bytes Oct 31, 2017 at 13:34 # Java 8, 156 bytes v->\"\".format(\"%1$9s^\\n%1$8s/a%1$7s/ba%1$6s/bba /b/o\\\\|a /b_\\\\_/ba /bbbbba /bbbbbba /bbbbbbba/bbbbbbbba\",\"\").replace(\"a\",\"_\\\\\\n\").replace(\"b\",\"_|\") Explanation: Try it here. v-> // Method with empty unused parameter and String return-type \"\".format( // Format the following String (%1$Ns = N spaces)\n\"%1$9s^\\n // ^ %1$8s/a // /_\\\n%1$7s/ba // /_|_\\ %1$6s/bba // /_|_|_\\\n/b/o\\\\|a // /_|/o\\|_\\\n/b_\\\\_/ba // /_|_\\_/_|_\\\n/bbbbba // /_|_|_|_|_|_\\\n/bbbbbba // /_|_|_|_|_|_|_\\\n/bbbbbbba // /_|_|_|_|_|_|_|_\\\n/bbbbbbbba\",\"\") // /_|_|_|_|_|_|_|_|_\\\n.replace(\"a\",\"_\\\\\\n\") // Replace all \"a\" with \"_\\\" + new-line\n.replace(\"b\",\"_|\") // Replace all \"b\" with \"_|\"\n// End of method (implicit / single-line return-statement)\n\n\n# Julia, 152141139130127120113 112 bytes\n\nq=\"_|\";a+b=\" \"^a*b;a\\b=replace(a,q^3,q*b,1);~n=n<0?9+\"^\\n\":~(n-1)*(8-n+\"/$(q^n)_\\\\\\n\");print(~8\\\"/o\\\\|\"\\\"_\\\\_/\") Explained: #Define constant q to abbreviate this string q=\"_|\"; #Redefine the addition operator to compactly provide whitespace #where needed a+b=\" \"^a*b; #Redefine the inverse division operator so we can substitute #\"_|_|_|\" with \"_|\"*b very compactly a\\b=replace(a,q^3,q*b,1); #Redefine the bitwise not operator to generate pyramid layers #Defines them recursively, calling itself to generate previous #layers before appending its own. #The base case generates the tip. ~n=n<0?9+\"^\\n\":~(n-1)*(8-n+\"/$(q^n)_\\\\\\n\");\n\n#Print to output\nprint(\n\n#Pyramid with 8 body layers\n~8\n\n#Then patch in the eye\n\\\"/o\\\\|\"\n\\\"_\\\\_/\"\n)\n\n\n# C# (.NET Core), 174 153 bytes\n\n()=>string.Format(@\"{1,10}\n{0,10}\\\n{0,9}{2}{3,10}{2}{0,7}|/o\\{2}{3,8}\\_/_{2}{0,5}{4} {3}{4} {3}|_{4}{3}|_|_{4}\",\"/_\",\"^\",@\"|_\\\n\",\"/_|_\",@\"|_|_|_|_|_\\\n\")\n\n\nTry it online!\n\nAn inefficient way of building the pyramid, but interesting working through it.\n\n### Acknowledgements\n\n-21 bytes thanks to @someone\n\n# C# (.NET Core), 144 bytes\n\nThis one may seem quite boring, because it is quite boring.\n\n()=>@\" ^\n/_\\\n/z\\\n/z|_\\\n/_|/o\\|_\\\n/z\\_/z\\\n/z|z|z\\\n/z|z|z|_\\\n/z|z|z|z\\\n/z|z|z|z|_\\\".Replace(\"z\",\"_|_\")\n\n\nTry it online!\n\n# Splinter, 124 bytes\n\nH{\\/}G{BB}F{\\_DG}E{\\ H}D{\\\\\\\n}C{A\\|}B{\\ \\ }A{\\_\\|\\_}GG\\ \\^\\\n\n\nTry it online!\n\n# JavaScript, 117 bytes\n\nI know for a fact I'm not beating any of the golfing languages, but at least I can give my own solution.\n\n$=>[...Array(10)].map((e,i)=>' '.repeat(9-i)+(i--?/${['_|/o\\\\|','_|_\\\\_/_|'][i-3]||'_|'.repeat(i)}_\\\\:'^')).join\n\n\n\nHere's a demo:\n\nvar f = $=>[...Array(10)].map((e,i)=>' '.repeat(9-i)+(i--?/${['_|/o\\\\|','_|_\\\\_/_|'][i-3]||'_|'.repeat(i)}_\\\\:'^')).join\n;\nconsole.log(f());\nconsole.log(f.toString().length);\n\nExplanation:\n\n$=> // outer function start [...Array(10)] // create an array to map .map(…) // map it (e,i)=> // mapping function start ' '.repeat(9-i) // spaces for padding +(i--?…:'^') // use the carat if at the top of the pyramid /${…}_\\\\ // otherwise, make the sides + steps\n['_|/o\\\\|','_|_\\\\_/_|'][i-3] // use the patterns for the eye, if in the correct rows\n||'_|'.repeat(i) // otherwise, make the \"bricks\" structure\n.join\n // join all the rows into a string (yes this part has a newline in it)\n\n• It's not recommended to answer straight away, let people make their own solutions.\n– Okx\nOct 27, 2017 at 16:42\n\n# Javascript, 238 bytes\n\nMy very first try at codegolfing :D\n\nvar f=()=>{let b=x=>' '.repeat(x),g='\\\\',h='/',i=1,st=[(b(9)+'^').split('')];for(;i<10;i++)st.push((b(9-i)+h+st.map(i=>'_').join('|')+g).split(''));st=st=h;st='o';st=st=g;return st.map(s=>s.join('')).join('\\n');}\n\ndocument.getElementById(\"display\").innerHTML = f();\nconsole.log(f.toString().length);\n<pre id=\"display\">\n</pre>\n\n• Welcome to the site! Oct 31, 2017 at 14:36\n• Hardcoding is definitely shorter than what you have here. You can shorten this with a few ES6 features: functions instead of function('s'), a=> instead of ()=>, fill(x) instead of map(e=>x), [...s] instead of s.split(''), move a statement into the for` initialization, etc. Oct 31, 2017 at 15:21\n• Much shorter; even just without compression of any kind, it's 163 bytes. Feb 7, 2021 at 0:30" ]

[ null, "https://i.stack.imgur.com/kiPwY.gif", null ]

[ "`Converting Celsius To FahrenheitAmaechi Onyeali Carter Elementary 5740 So. Michigan Chicago IL 60637 (312) 535-0860Objectives:1. As a result of the lesson students will be able to solve multiple step equations involving addition, multiplication, and division.2. As a result of the lesson students will be able to convert a known temperature in celsius to an equal temperature in fahrenheit.3. As a result of the lesson students will be able to accurately read a thermometer in degrees celsius.Materials Needed:1. 6 thermometers in degrees celsius2. 1 cup of hot water3. 1 cup of regular water4. 1 thermometer in degrees fahrenheit5. calculators for each studentStrategy:I opened my lesson with a short story. The story talks about me driving to my doctors office on a hot summer day trying to figure out if the temperature had reached 100 degrees fahrenheit. As I drove around all I could see were temperature readings on banks in degrees celsius. Eventually I arrived at my doctor's office. As we talked about the weather we both began to wonder what the temperature was outside. I told him that the bank signs all showed 39 degrees celsius. However, a temperature in celsius does not really tell us if it is 100 degrees fahrenheit. Thus we have created a problem that we need to solve. Is 39 degrees celsius about 100 degrees fahrenheit? Now you can present the formula developed by Anders Celsius in 1742. The formula says that if you know a temperature in degrees celsius you can convert it into degrees fahrenheit by: F=9/5(C)+32. To check and see if this formula works, we will convert the known temperatures at which water freezes, 0 degrees celsius, and at which water boils, 100 degrees celsius. They should convert to 32 and 212 degrees fahrenheit respectively. Students will then convert 10 celsius temperatures to fahrenheit in their cooperative learning groups. Remember to follow your order of operations when solving these problems. Have students go to the board and explain their answers. Introduce students to the 6 different stations. Each station has a thermometer in degrees celsius. Station 1 is outside, station 2 is in the back of the room, station 3 in the front of the room, station 4 in the hallway, station 5 in a cup of hot water, and station 6 is in a cup of regular water. Ask students to predict which station is the hottest. Then list them from hottest to coldest. Now allow students to travel to each station and record the temperatures at thestations in degrees celsius. While waiting to go to the next station, studentswill convert temperatures to degrees fahrenheit. After all groups have been to each station confirm whether or not everyone has the same answer. Then see how many people made the right prediction. After checking predictions and answersuse your fahrenheit thermometer to check the temperature of the hot and regular water. You should come out with fairly accurate numbers.To close this lesson you can take a briefcase with a three digit combination and place it in front of the class. Tell them that you forgot the combination, but you remember that you set it on a day when it was ___ degrees celsius. If they can convert this temperature to degrees fahrenheit, they will have the combination. Let students work in their groups and come up with their answer. Give each group a try. Performance Assessment:1. Students will read the temperature of 6 different thermometers in degrees celsius.2. Students will convert 6 temperatures in degrees celsius to degrees fahrenheit by using a formula.3. Students will solve multiple step equations involving addition, multiplication, and division.`" ]

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[ "# Kilograms to Troy Ounces Converter\n\nSelect conversion type:\n\nRounding options:\n\nConvert Troy Ounces to Kilograms (oz t to kg) ▶\n\n## Conversion Table\n\n kilograms to troy ounces kg oz t 1 kg 32.1507 oz t 2 kg 64.3015 oz t 3 kg 96.4522 oz t 4 kg 128.603 oz t 5 kg 160.7537 oz t 6 kg 192.9045 oz t 7 kg 225.0552 oz t 8 kg 257.206 oz t 9 kg 289.3567 oz t 10 kg 321.5075 oz t 11 kg 353.6582 oz t 12 kg 385.809 oz t 13 kg 417.9597 oz t 14 kg 450.1105 oz t 15 kg 482.2612 oz t 16 kg 514.4119 oz t 17 kg 546.5627 oz t 18 kg 578.7134 oz t 19 kg 610.8642 oz t 20 kg 643.0149 oz t\n\n## How to convert\n\n1 kilogram (kg) = 32.15074657 troy ounce (oz t). Kilogram (kg) is a unit of Weight used in Metric system. Troy Ounce (oz t) is a unit of Weight used in Standard system.\n\n## Definition of the Kilogram\n\nA kilogram is a unit of weight that measures how much force an object or substance exerts on a scale due to gravity. It is equal to the mass of the International Prototype of the Kilogram (IPK), a cylinder of platinum-iridium alloy stored at the International Bureau of Weights and Measures in France. One kilogram can also be written as kg or 1000 g.\n\n## History of the Kilogram\n\nThe kilogram was originally defined in 1795 during the French Revolution as the mass of one litre of water at 4 °C, which was determined to be 18841 grains. In 1799, the Kilogramme des Archives, a platinum artifact, replaced it as the standard of weight. In 1889, the IPK became the new standard of the unit of weight for the metric system and remained so for 130 years, until the current definition was adopted in 2019.\n\nThe kilogram, as well as other metric units of weight, such as the gram and the tonne, were introduced in the 19th century as part of the decimal system of measurement that aimed to simplify and unify the units used in science and commerce. The kilogram was officially adopted by the International System of Units (SI) in 1960 as one of the seven base units.\n\n## How to Convert Kilograms\n\nTo convert kilograms to other units of weight, we need to use conversion factors that relate the kilogram to the desired unit. For example, to convert kilograms to pounds, we need to know that one pound is equal to 0.45359237 kg. Therefore, one kilogram is equal to 2.2046226218 pounds.\n\nHere are some common conversion factors for kilograms:\n\n• 1 kg = 1000 g\n• 1 kg = 2.2046226218 lb\n• 1 kg = 35.27396195 oz\n• 1 kg = 0.001 t\n• 1 kg = 0.1574730444 st\n\nTo convert from other units of weight to kilograms, we need to use the inverse of these conversion factors. For example, to convert pounds to kilograms, we need to divide by 2.2046226218.\n\n## Where Kilograms are Used\n\nThe kilogram is a unit of weight that is widely used to measure the weight of objects and substances such as food, clothes, animals, metals and chemicals. For example, an adult human body has an average weight of about 62 kg.\n\nThe kilogram is also used to measure the weight of some physical quantities such as force, pressure and torque. For example, one newton (N) is equal to one kilogram times one meter per second squared (kg x m/s2).\n\n## Example Conversions of Kilograms to Other Units\n\nHere are some examples of how to convert kilograms to other units of weight using the conversion factors given above:\n\n• 2 kg = 2 x 1000 g = 2000 g\n• 3 kg = 3 x 2.2046226218 lb = 6.6138678654 lb\n• 4 kg = 4 x 35.27396195 oz = 141.0958478 oz\n• 5 kg = 5 x 0.001 t = 0.005 t\n• 6 kg = 6 x 0.1574730444 st = 0.9448382664 st\n\n## Conclusion\n\nThe kilogram is a unit of weight that measures how much force an object or substance exerts on a scale due to gravity. It is equal to the mass of the IPK, a cylinder of platinum-iridium alloy stored in France. The kilogram is widely used to measure the weight of objects and substances such as food, clothes, animals, metals and chemicals, as well as some physical quantities such as force, pressure and torque. To convert kilograms to other units of weight, we need to use conversion factors that relate the kilogram to the desired unit.\n\nKilograms also can be marked as kilogrammes in UK.\n\n## Troy Ounces: A Unit of Weight\n\nTroy ounces are a unit of weight that are used for measuring precious metals, such as gold, silver and platinum. Troy ounces are derived from the French word once, which was the name of a unit of weight used in the Middle Ages. The symbol for troy ounce is oz t.\n\n## Definition of the Troy Ounce\n\nThe troy ounce is defined as 20 pennyweights, which are each 24 grains. The troy grain is equal to the avoirdupois grain, which is one seven-thousandth of an avoirdupois pound. The troy ounce is heavier than the avoirdupois ounce, but lighter than the apothecaries’ ounce.\n\nThe troy ounce is equal to about 1.0971 avoirdupois ounces or 31.1035 grams. The troy grain is equal to about 0.0648 milligrams.\n\n## How to Convert Troy Ounces\n\nTroy ounces can be converted to other units of weight by using conversion factors or formulas. Here are some examples of how to convert troy ounces to other units of weight in the US customary system and the SI system:\n\n• To convert troy ounces to avoirdupois ounces, multiply by 1.0971. For example, 10 oz t = 10 x 1.0971 = 10.971 oz.\n• To convert troy ounces to avoirdupois pounds, divide by 14.5833. For example, 5 oz t = 5 / 14.5833 = 0.343 lb.\n• To convert troy ounces to tons (short), divide by 29166.67. For example, 20 oz t = 20 / 29166.67 = 0.000685 ton.\n• To convert troy ounces to kilograms, divide by 32.1507. For example, 15 oz t = 15 / 32.1507 = 0.4666 kg.\n• To convert troy ounces to grams, multiply by 31.1035. For example, 25 oz t = 25 x 31.1035 = 777.5875 g.\n• To convert troy ounces to milligrams, multiply by 31103.4768. For example, 30 oz t = 30 x 31103.4768 = 933104.304 mg.\n\n## Where Troy Ounces are Used\n\nTroy ounces are used in different countries and regions for different applications and purposes. Here are some examples of where troy ounces are used:\n\n• In most countries that use the SI system, troy ounces are not used for measuring weight, but only for measuring precious metals, such as gold, silver and platinum.\n• In the United States, troy ounces are used for measuring precious metals, especially for gold and silver coins and bullion.\n• In Canada, Australia and New Zealand, troy ounces are used for measuring precious metals, especially for gold and silver coins and bullion.\n• In the United Kingdom, troy ounces are used for measuring precious metals, especially for gold and silver coins and bullion.\n• In India, troy ounces are used for measuring precious metals, especially for gold and silver jewelry.\n\n## History of Troy Ounces\n\nTroy ounces have a long history that dates back to ancient times. Here are some highlights of the history of troy ounces:\n\n• The troy ounce was originally based on the weight of a Roman uncia, which was one twelfth of a Roman libra or pound, which was equivalent to about 27 grams.\n• The troy ounce was used in different systems of measurement, such as the Roman system, the Byzantine system, the Arabic system and the English system. It varied from about 26 grams to about 32 grams depending on the region and the time period.\n• The troy ounce was standardized by royal statutes and international agreements in different periods of history. For example, in 1527 an act of Henry VIII fixed the troy ounce at exactly 480 grains; in 1828 an act of Congress adopted the troy ounce as the official unit of weight for coinage in the United States; in 1959 an international agreement defined the international avoirdupois ounce as exactly 28.349523125 grams, which made the troy ounce exactly 31.1034768 grams.\n\n## Example Conversions of Troy Ounces to Other Units\n\nHere are some examples of conversions of troy ounces to other units of weight:\n\n• 1 oz t = 1.0971 oz\n• 1 oz t = 0.0686 lb\n• 1 oz t = 0.000034 ton\n• 1 oz t = 0.0311 kg\n• 1 oz t = 31.1035 g\n• 1 oz t = 31103.4768 mg\n• 1 oz t = 0.0833 lb t\n• 1 oz t = 20 dwt\n• 1 oz t = 480 gr\n\nEspañol     Russian     Français" ]

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[ "# 我知道你不知道我知道你知道", null, "", null, "", null, "• $$(M_2, (0, 0, 1))\\models \\lnot K_A p$$\n• $$(M_2, (0, 0, 1))\\models \\lnot K_B p$$\n• $$(M_2, (0, 0, 1))\\models K_C p$$", null, "• $$(M_3, (1, 0, 1))\\models K_A p$$\n• $$(M_3, (1, 0, 1))\\models \\lnot K_B p$$\n• $$(M_3, (1, 0, 1))\\models K_C p$$\n\nTNND，回头看这文的时候突然反应过来，这泥马的拿手摸一下不就知道有没有泥巴了，还弄得这么麻烦，搞数学的还真是没事干啊。" ]

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[ "Ounces To Grams\n\n# 3384 oz to g3384 Ounce to Grams\n\noz\n=\ng\n\n## How to convert 3384 ounce to grams?\n\n 3384 oz * 28.349523125 g = 95934.786255 g 1 oz\nA common question is How many ounce in 3384 gram? And the answer is 119.367087237 oz in 3384 g. Likewise the question how many gram in 3384 ounce has the answer of 95934.786255 g in 3384 oz.\n\n## How much are 3384 ounces in grams?\n\n3384 ounces equal 95934.786255 grams (3384oz = 95934.786255g). Converting 3384 oz to g is easy. Simply use our calculator above, or apply the formula to change the length 3384 oz to g.\n\n## Convert 3384 oz to common mass\n\nUnitMass\nMicrogram95934786255.0 µg\nMilligram95934786.255 mg\nGram95934.786255 g\nOunce3384.0 oz\nPound211.5 lbs\nKilogram95.934786255 kg\nStone15.1071428572 st\nUS ton0.10575 ton\nTonne0.0959347863 t\nImperial ton0.0944196429 Long tons\n\n## What is 3384 ounces in g?\n\nTo convert 3384 oz to g multiply the mass in ounces by 28.349523125. The 3384 oz in g formula is [g] = 3384 * 28.349523125. Thus, for 3384 ounces in gram we get 95934.786255 g.\n\n## 3384 Ounce Conversion Table", null, "## Alternative spelling\n\n3384 Ounces to Grams, 3384 Ounces in Grams, 3384 oz in Grams, 3384 oz to Gram, 3384 oz in Gram, 3384 Ounces to g, 3384 Ounces in g, 3384 oz to g, 3384 Ounce to Grams, 3384 Ounce in Grams, 3384 Ounce to Gram, 3384 Ounce in Gram, 3384 Ounce to g," ]

[ null, "https://ounces-to-grams.appspot.com/image/3384.png", null ]

[ "# 2d Heat Equation Solver\n\nThe domain is [0,L] and the boundary conditions are neuman. The Easy Way of Solving Systems of Linear Equations in Excel – using the INVERSE() spreadsheet function Posted By George Lungu on 04/24/2011 This brief tutorial explains how to calculate the solution vector of a system of linear equations using the Excel spreadsheet function MINVERSE() which calculate the inverse of a matrix. 21 Scanning speed and temperature distribution for a 1D moving heat source. With Fortran, elements of 2D array are memory aligned along columns : it is called \"column major\". Ask Question Asked 3 years, 4 months ago. MPI Numerical Solving of the 3D Heat equation. $$This works very well, but now I'm trying to introduce a second material. (48) does not necessarily satisfy differential eq. Project - Solving the Heat equation in 2D Aim of the project The major aim of the project is to apply some iterative solution methods and preconditioners when solving linear systems of equations as arising from discretizations of partial differential equations. 2d Laplace Equation File Exchange Matlab Central. Journal of Com- putational and Applied Mathematics, Elsevier, 2019, 346, pp. Williamson, but are quite generally useful for illustrating concepts in the areas covered by the texts. algebra addition, subtraction, multiplication and division of algebraic expressions, hcf & lcm factorization, simple equations, surds, indices, logarithms, solution of linear equations of two and three variables, ratio and proportion, meaning and standard form, roots and discriminant of a quadratic equation ax2 +bx+c = 0. To solve the problem we use the following approach: first we find the equilibrium temperature uE(x) by solving the problem d2u E dx2 = 0 (5) uE(0) = A (6) uE(L) = B (7) The solution is uE(x) = A+ B −A L x Next we introduce a new function v(x,t) that measures the displacement of the temperature u(x,t) from the equilibrium temperature uE(x). This equation is a model of fully-developed flow in a rectangular duct. Your analysis should use a finite difference discretization of the heat equation in the bar to establish a system of equations:. Here we are. 195) subject to the following boundary and initial conditions (3. Now, consider a cylindrical differential element as shown in the figure. Numerical Modeling of Earth Systems An introduction to computational methods with focus on solid Earth applications of continuum mechanics Lecture notes for USC GEOL557, v. We only consider the case of the heat equation since the book treat the case of the wave equation. It is also used to numerically solve parabolic and elliptic partial. Consider a heat transfer problem for a thin straight bar (or wire) of uniform cross section and homogeneous material. (The first equation gives C. In the case of one-dimensional equations this steady state equation is a second order ordinary differential equation. volume method, to discretize the 2D-3T equations (cf. Solving the Heat Diffusion Equation (1D PDE) in Matlab - Duration: 24:39. I drew a diagram of the 2D heat conduction that is described in the problem. de Householder Symposium XVI Seven Springs, May 22 – 27, 2005 Thanks to: Enrique Quintana-Ort´ı, Gregorio Quintana-Ort´ı. 2D viscoelastic flow. The solution of the heat equation is computed using a basic finite difference scheme. In 2D, a NxM array is needed where N is the number of x grid points, M the number of y grid. Solving the non-homogeneous equation involves defining the following functions: (,. FINITE-DIFFERENCE SOLUTION TO THE 2-D HEAT EQUATION MSE 350. 3 Separation of variables for nonhomogeneous equations Section 5. In this first example we want to solve the Laplace Equation (2) a special case of the Poisson Equation (1) for the absence of any charges. First we formulate a more detailed criterion for spatial coarsening, which enables the method to deal with unstructured meshes and varying material parameters. solve ordinary and partial di erential equations. This is the natural extension of the Poisson equation describing the stationary distribution of heat in a body to a time-dependent problem. 3D flow over a backwards facing step using the OpenFOAM solver. fortran code finite volume 2d conduction free download. Solution of the Laplace and Poisson equations in 2D using five-point and nine-point stencils for the Laplacian [pdf | Winter 2012] Finite element methods in 1D Discussion of the finite element method in one spatial dimension for elliptic boundary value problems, as well as parabolic and hyperbolic initial value problems. Expand the requested time horizon until the solution reaches a steady state. Heat Transfer, Trans. In the present case we have a= 1 and b=. FEM2D_HEAT, a C++ program which applies the finite element method to solve the 2D heat equation. Wave equation solver. Ask Question Asked 4 years, 8 months ago. Poisson’s Equation in 2D Analytic Solutions A Finite Difference A Linear System of Direct Solution of the LSE Classification of PDE Page 1 of 16 Introduction to Scientific Computing Poisson’s Equation in 2D Michael Bader 1. 7 A standard approach for solving the instationary equation. For modeling structural dynamics and vibration, the toolbox provides a direct time integration solver. It can give an approximate solution using a multigrid method, i. Solving simultaneously we find C 1 = C 2 = 0. 4 and Section 6. wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. This shows that the heat equation respects (or re ects) the second law of thermodynamics (you can’t unstir the cream from your co ee). Solutions to Problems for 2D & 3D Heat and Wave Equations 18. There are Fortran 90 and C versions. Solving the 2D heat equation. In 2D, a NxM array is needed where N is the number of x grid points, M the number of y grid. 1) This equation is also known as the diffusion equation. Included is an example solving the heat equation on a bar of length L but instead on a thin circular ring. MPI based Parallelized C Program code to solve for 2D heat advection. We’re looking at heat transfer in part because many solutions exist to the heat transfer equations in 1D, with math that is straightforward to follow. Note: 2 lectures, §9. The Laplace transform is an integral transform that is widely used to solve linear differential. Energy2D is a relatively new program (Xie, 2012) and is not yet widely used as a building performance simulation tool. Expand the requested time horizon until the solution reaches a steady state. 2) Uniform temperature gradient in object Only rectangular geometry will be analyzed Program Inputs The calculator asks for. Heat equation solver. 01 on the left, D=1 on the right: Two dimensional heat equation on a square with Dirichlet boundary conditions:. 2 2D transient conduction with heat transfer in all directions (i. Let the x-axis be chosen along the axis of the bar, and let x=0 and x=ℓ denote the ends of the bar. This blog will help students and researchers to learn the various mostly used simulation software and tools. Wing Lift Equations Equations Calculator Solve for any variable in the aircraft or airplane wing lift equation. solving first on a very coarse grid and extending the solution to finer and finer grids, and it can solve iteratively the original system (finest grid). 1 Heat Equation with Periodic Boundary Conditions in 2D. 1D Heat Equation. The finite difference method is a numerical approach to solving differential equations. Kinematic Equations Calculator. Week 4 (2/10-14). Diffusion In 1d And 2d File Exchange Matlab Central. Galerkin method. Suppose that we need to solve numerically the following differential equation: a d2u dx2 +b = 0; 0 • x • 2L (1. Hancock Fall 2006 1 2D and 3D Heat Equation Ref: Myint-U & Debnath § 2. Draw arbitrary initial values with your mouse and see the corresponding solution to the wave equation. •Partial differential equation solver package with front-end developed for visual input and output •Most used through different “modules” with predefined “physics”, greatly simplifying modeling of device geometry, governing equations, boundary conditions, etc. Learn more about: Equation solving » Tips for entering queries. See Draft ShapeString for an example of a well documented tool. Solving the heat equation Charles Xie The heat conduction for heterogeneous media is modeled using the following partial differential equation: T k T q (1) t T c v where k is the thermal conductivity, c is the specific heat capacity, is the density, v is the velocity field, and q is the internal heat generation. In the above equation on the right, represents the heat flow through a defined cross-sectional area A, measured in watts,. 0005 k = 10**(-4) y_max = 0. The initial temperature of the rod is 0. Engineering Equation Solver (EES) is a general program for solving nonlinear algebraic equations and differential and integral equations. heat_eul_neu. Skills: Engineering, Mathematics, Matlab and Mathematica, Mechanical Engineering. Heat equation for a cylinder in cylindrical coordinates. Active 4 years, 7 months ago. Kim and Daniel studied an inverse heat conduct problem for nanoscale structure using sequential method. Download32 is source for plot 2d equation freeware download - Plot2D , qColorMap , qColorMap , APlot - Plot/Printer 3D 2D Project , Advanced Graphing Calculator 3D Linux, etc. In general, this problem is ill-posed in the sense of Hadamard. Solving the 2D Heat Equation As just described, we have two algorithms: explicit (Euler) and implicit (Crank-Nicholson). Solving Equations Video Lesson. It can give an approximate solution using a multigrid method, i. Solving the Equations How the fluid moves is determined by the initial and boundary conditions; the equations remain the same Depending on the problem, some terms may be considered to be negligible or zero, and they drop out In addition to the constraints, the continuity equation (conservation of mass) is frequently required as well. 2D Heat Equation solver in Python. FEM1D_HEAT_STEADY, a C++ program which uses the finite element method to solve the steady (time independent) heat equation in 1D. 5, An Introduction to Partial Differential Equa-tions, Pinchover and Rubinstein The method of separation of variables can be used to solve nonhomogeneous equations. for a 2D nonlinear coupled system of radiative-conductive heat transfer equations. Heat conduction follows a. So du/dt = alpha * (d^2u/dx^2). Solution of the HeatEquation by Separation of Variables The Problem Let u(x,t) denote the temperature at position x and time t in a long, thin rod of length ℓ that runs from x = 0 to x = ℓ. Project: Heat Equation. Conductive Heat Transfer Calculator. Up to now, we're good at \\killing blue elephants\" | that is, solving problems with inhomogeneous initial conditions. Introduction To Fem File Exchange Matlab Central. For a function u(x,y,z,t) of three spatial variables (x,y,z) and the time variable t, the heat equation is or equivalently where α is a constant. This calculator can be used to calculate conductive heat transfer through a wall. Generic solver of parabolic equations via finite difference schemes. Heat Transfer, Trans. Learn more about: Equation solving » Tips for entering queries. The u i can be functions of the dependent variables and need not include all such variables. HOT_POINT, a MATLAB program which uses FEM_50_HEAT to solve a heat problem with a point source. Examples of nonlinear SPDEs. \" ThermoElectric Device Simulation -- ThermoElectric Device Simulation -- This is an interactive simulation of a thermoelectric device, which converts heat energy directly into electrical energy. Post-process to visualize the solution Notes: The Poisson equation is steady. % Matlab Program 4: Step-wave Test for the Lax method to solve the Advection % Equation clear; % Parameters to define the advection equation and the range in space and time Lmax = 1. Download32 is source for plot 2d equation freeware download - Plot2D , qColorMap , qColorMap , APlot - Plot/Printer 3D 2D Project , Advanced Graphing Calculator 3D Linux, etc. Solving the 2D Poisson's equation in. -5 0 5-30-20-10 0 10 20 30 q sinh( q) cosh( q) Figure1: Hyperbolicfunctionssinh( ) andcosh( ). , an exothermic reaction), the steady-state diffusion is governed by Poisson's equation in the form ∇2Φ = − S(x) k. time-dependent) heat conduction equation without heat generating sources rcp ¶T ¶t = ¶ ¶x k ¶T ¶x (1). Introduction To Fem File Exchange Matlab Central. eqn_parse turns a representation of an equation to a lambda equation that can be easily used. A parallelized 2D/2D-axisymmetric pressure-based, extended SIMPLE finite-volume Navier–Stokes equation solver using Cartesians grids has been developed for simulating compressible, viscous, heat conductive and rarefied gas flows at all speeds with conjugate heat transfer. The aim is to solve the steady-state temperature distribution through a rectangular body, by dividing it up into nodes and solving the necessary equations only in two dimensions. The fundamental equation for two-dimensional heat conduction is the two-dimensional form of the Fourier equation (Equation 1) 1,2. Mitchell and R. I am using version 11. Equation 5-5:(It would be extra nice if one sent me the derivation using equation editor in Word! :] ) derivation needed!. PROBLEM OVERVIEW. It was implemented the parallelization of this problem using the Yanenko method using 1D and 2D data decomposition. But, in practice, these equations are too difficult to solve analytically. Overview Approach To solve an IVP/BVP problem for the heat equation in two dimensions, ut = c2(uxx + uyy): 1. Separation of variables A more fruitful strategy is to look for separated solutions of the heat equation, in other words, solutions of the form u(x;t) = X(x)T(t). Derivation of the heat equation in 1D x t u(x,t) A K Denote the temperature at point at time by Cross sectional area is The density of the material is The specific heat is Suppose that the thermal conductivity in the wire is ρ σ. The heat and wave equations in 2D and 3D 18. Learn more about: Equation solving » Tips for entering queries. In this section we go through the complete separation of variables process, including solving the two ordinary differential equations the process generates. So the equation becomes r2 1 r 2 d 2 ds 1 r d ds + ar 1 r d ds + b = 0 which simpli es to d 2 ds2 + (a 1) d ds + b = 0: This is a constant coe cient equation and we recall from ODEs that there are three possi-bilities for the solutions depending on the roots of the characteristic equation. Need more problem types? Try MathPapa Algebra Calculator. We discretize the rod into segments, and approximate the second derivative in the spatial dimension as $$\\frac{\\partial^2 u}{\\partial x^2} = (u(x + h) - 2 u(x) + u(x-h))/ h^2$$ at each node. 1 Solve a semi-linear heat equation 8. Heat equationin a 2D rectangle This is the solution for the in-class activity regarding the temperature u(x,y,t) in a thin rectangle of dimensions x ∈ [0,a],b ∈ [0,b], which is initially all held at temperature T 0, so u(x,y,t = 0) = T 0. HEATED_PLATE, a MATLAB program which solves the steady state heat equation in a 2D rectangular region, and is intended as a. The calculator will find the approximate solution of the first-order differential equation using the Euler's method, with steps shown. I am trying to solve the 2D heat. Your equation for radiative heat flux has the unit [\\frac{\\text{W}}{\\text{m}^2}], while the Neumann boundary condition needs a unit of [\\frac{\\text{K}}{\\text{m}}]. As far as I can tell it looks like it only can solve steady state equation (laplace, steady state heat, ect). This code employs finite difference scheme to solve 2-D heat equation. The finite difference method is a numerical approach to solving differential equations. add_time_stepper_pt(newBDF<2>); Next we set the problem parameters and build the mesh, passing the pointer to the TimeStepper as the last argument to the mesh constructor. pyplot as plt dt = 0. The domain is [0,L] and the boundary conditions are neuman. Derive equation 5-5 using Newton's laws and include the derivation in your report. Ask Question Asked 3 years, 4 months ago. The heat transfer can also be written in integral form as Q˙ = − Z A q′′ ·ndA+ Z V q′′′ dV (1. 1) This equation is also known as the diffusion equation. Section 3 deals with solving the two-dimensional heat conduction equation using HAM. Each type has a string specifier (e. Solving the heat equation using the separation of variables. Finite Volume model in 2D Poisson Equation. Finite differences for the 2D heat equation. The user specifies it by preparing a file containing the coordinates of the nodes, and a file containing the indices of nodes that make up triangles that form a triangulation. A novel Douglas alternating direction implicit (ADI) method is proposed in this work to solve a two-dimensional (2D) heat equation with interfaces. It’s a simple MATLAB code that can solve for different materials such as (copper, aluminum, silver, etc…. Multi-Region Conjugate Heat/Mass Transfer MRconjugateHeatFoam: A Dirichlet–Neumann partitioned multi-region conjugate heat transfer solver Brent A. Diffusion Equation! Computational Fluid Dynamics! ∂f ∂t +U ∂f ∂x =D ∂2 f ∂x2 We will use the model equation:! Although this equation is much simpler than the full Navier Stokes equations, it has both an advection term and a diffusion term. By a translation argument I get that if my initial velocity would be vt. Solving the 2D Poisson's equation in. Orlando, Florida, USA. ; % Maximum time c = 1. The heat conduction equation is a partial differential equation that describes the distribution of heat (or the temperature field) in a given body over time. Heat conduction is a diffusion process caused by interactions of atoms or molecules, which can be simulated using the diffusion equation we saw in last week’s notes. 4 Thermal Resistance Circuits There is an electrical analogy with conduction heat transfer that can be exploited in problem solving. It can be used to solve one dimensional heat equation by using Bendre-Schmidt method. 2d heat transfer - implicit finite difference method. Solve wave equation with central differences. One dimensional heat equation with non-constant coefficients: heat1d_DC. Conductive Heat Transfer Calculator. Equation Generator When 3 points are input, this calculator will generate a second degree equation. Poisson's Equation in 2D We will now examine the general heat conduction equation, T t = κ∆T + q ρc. Section 3 deals with solving the two-dimensional heat conduction equation using HAM. Solving the 2D Poisson's equation in. Clear Equation Solver ». The wave equation, on real line, associated with the given initial data:. (after the last update it includes examples for the heat, drift-diffusion, transport, Eikonal, Hamilton-Jacobi, Burgers and Fisher-KPP equations) Back to Luis Silvestre's homepage. Introduction To Fem File Exchange Matlab Central. This calculator can be used to calculate conductive heat transfer through a wall. FEM2D_HEAT, a C++ program which applies the finite element method to solve the 2D heat equation. 2 Solve the Cahn-Hilliard equation. EML4143 Heat Transfer 2 For education purposes. Solve the system of equations A˚= b, where ˚is the vector of unknowns. Laplace's equation in the Polar Coordinate System As I mentioned in my lecture, if you want to solve a partial differential equa-tion (PDE) on the domain whose shape is a 2D disk, it is much more convenient to represent the solution in terms of the polar coordinate system than in terms of the usual Cartesian coordinate system. The solver will then show you the steps to help you learn how to solve it on your own. The equations are a set of coupled differential equations and could, in theory, be solved for a given flow problem by using methods from calculus. Then, I included a convective boundary condition at the top edge, and symmetric boundary condition (dT/dn = 0) at the other three edges. Note that while the matrix in Eq. the solute is generated by a chemical reaction), or of heat (e. In 1D, an N element numpy array containing the intial values of T at the spatial grid points. This Demonstration implements a recently published algorithm for an improved finite difference scheme for solving the Helmholtz partial differential equation on a rectangle with uniform grid spacing. 2D Heat Equation Using Finite Difference Method with Steady-State Solution. Aim: To find the No. Scientific Programming Wave Equation 1 The wave equation The wave equation describes how waves propagate: light waves, sound waves, oscillating strings, wave in a pond, Suppose that the function h(x,t) gives the the height of the wave at position x and time t. Read \"Solving the 2-D heat equations using wavelet-Galerkin method with variable time step, Applied Mathematics and Computation\" on DeepDyve, the largest online rental service for scholarly research with thousands of academic publications available at your fingertips. Multi-Region Conjugate Heat/Mass Transfer MRconjugateHeatFoam: A Dirichlet–Neumann partitioned multi-region conjugate heat transfer solver Brent A. Hancock Fall 2006 1 2D and 3D Heat Equation Ref: Myint-U & Debnath §2. While writing the scripts for the past articles I thought it might be fun to implement the 2D version of the heat and wave equations and then plot the results on a 3D graph. using Laplace transform to solve heat equation. 44) because of these extra non-zero diagonals. 3D flow over a backwards facing step using the OpenFOAM solver. Useful for problems with complicated geometries, loadings, and material properties where analytical solutions can not be obtained. Solving elliptic PDEs in Scilab with the Feynman-Kac formula Contribution by Giovanni Conforti - Fellow of the graduate program Berlin Mathematical School In this work it is described and implemented in Scilab a stochastic numerical algorithm to solve elliptic PDEs with special focus on the heat equation. Numerical methods are important tools to simulate different physical phenomena. Tech 6 spherical systems - 2D steady state conduction in cartesian coordinates - Problems 7. Wave equation. Heat loss from a heated surface to unheated surroundings with mean radiant temperatures are indicated in the chart below. Generic solver of parabolic equations via finite difference schemes. The aim is to solve the steady-state temperature distribution through a rectangular body, by dividing it up into nodes and solving the necessary equations only in two dimensions. The calculator is generic and can be used for both metric and imperial units as long as the use of units is consistent. This LED board displays our solution to the 2D heat equation, written in less than 1Kb of program space. Solutions to Problems for 2D & 3D Heat and Wave Equations 18. Heat & Wave Equation in a Rectangle Section 12. Energy2D is a relatively new program (Xie, 2012) and is not yet widely used as a building performance simulation tool. This is a third-degree equation in \\rho and we would like to solve for \\rho. For one-dimensional heat conduction (temperature depending on one variable only), we can devise a basic description of the process. To gain more confidence in the predictions with Energy2D, an analytical validation study was. volume of the system. 2 2D transient conduction with heat transfer in all directions (i. As showcase we assume the homogeneous heat equation on isotropic and homogeneous media in one dimension: We will solve this for $$(t,x) \\in [0,1] \\text{s} \\times \\Omega=[0,1]\\text{m}$$ temporal $$k=0. Journal of Com- putational and Applied Mathematics, Elsevier, 2019, 346, pp. Program code to solve for 2D heat advection. I'm going to illustrate a simple one-dimensional heat flow example, followed two-dimensional heat flow example, all programmed into Excel. Note, this overall heat transfer coefficient is calculated based on the outer tube surface area (Ao). The coefficient matrix and source vector look okay after the x-direction loop. Solving the heat equation with the Fourier transform Find the solution u(x;t) of the di usion (heat) equation on (1 ;1) with initial data u(x;0) = ˚(x). Poisson's Equation in 2D Analytic Solutions A Finite Difference A Linear System of Direct Solution of the LSE Classification of PDE Page 1 of 16 Introduction to Scientific Computing Poisson's Equation in 2D Michael Bader 1. At this time the problem. The ADI scheme is a powerful finite difference method for solving parabolic equations, due to its unconditional stability and high efficiency. Partial Differential Equation Toolbox provides functions for solving partial differential equations (PDEs) in 2D, 3D, and time using finite element analysis. The calculator is generic and can be used for both metric and imperial units as long as the use of units is consistent. 2) Uniform temperature gradient in object Only rectangular geometry will be analyzed Program Inputs The calculator asks for. Since by translation we can always shift the problem to the interval (0, a) we will be studying the problem on this interval. You can solve PDEs by using the finite element method, and postprocess results to explore and analyze them. 5, An Introduction to Partial Differential Equa-tions, Pinchover and Rubinstein The method of separation of variables can be used to solve nonhomogeneous equations. Solving the heat equation using the separation of variables. wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. Laplace transform in solving 2d wave equation. However, it suffers from a serious accuracy reduction in space for interface problems with different. Two-Dimensional Laplace and Poisson Equations In the previous chapter we saw that when solving a wave or heat equation it may be necessary to first compute the solution to the steady state equation. Equation (1) is known as a one-dimensional diffusion equation, also often referred to as a heat equation. International Journal of Heat and Mass Transfer 80 , 562-569. Your analysis should use a finite difference discretization of the heat equation in the bar to establish a system of equations:. The heat equation, the variable limits, the Robin boundary conditions, and the initial condition are defined as:. This allows for a simpler GUI where we have only one button for the heat equation which is used for all supported solver. 02*DuDx; s = 0; function u0 = pdexic(x) % this defines u(t=0) for all of x. We will solve \\(U_{xx}+U_{yy}=0$$ on region bounded by unit circle with $$\\sin(3\\theta)$$ as the boundary value at radius 1. 31Solve the heat equation subject to the boundary conditions. The temperaure profile is shown below. We then end with a linear algebraic equation Au = f: It can be shown that the corresponding matrix A is still symmetric but only semi-definite (see Exercise 2). Explicit Difference Methods for Solving the Cylindrical Heat Conduction Equation By A. Solving the 2D Heat Equation As just described, we have two algorithms: explicit (Euler) and implicit (Crank-Nicholson). 33 Jacob Allen and J. Numerical Modeling of Earth Systems An introduction to computational methods with focus on solid Earth applications of continuum mechanics Lecture notes for USC GEOL557, v. Assume that the sides of the rod are insulated so that heat energy neither enters nor leaves the rod through its sides. The diffusion equation is a partial differential equation. Let Φ(x) be the concentration of solute at the point x, and F(x) = −k∇Φ be the. Case 2: Solution for t < T This is the case when the forcing is kept on for a long time (compared to the time, t, of our interest). Solving the 2D Poisson's equation in. \"\"\" import. Transient Heat Conduction File Exchange Matlab Central. Section 3 deals with solving the two-dimensional heat conduction equation using HAM. Step 4 is usually performed by an iterative method, which introduces additional concerns about convergence tolerances and e ciency. Generic solver of parabolic equations via finite difference schemes. I am entirely new to Mathematica and have been given the task to animate the solution to the 2D heat equation with given initial and boundary conditions. PDE's: Solvers for heat equation in 2D using ADI method; 5. I recently begun to learn about basic Finite Volume method, and I am trying to apply the method to solve the following 2D continuity equation on the cartesian grid x with initial condition For simplicity and interest, I take , where is the distance function given by so that all the density is concentrated near the point after sufficiently long. This Demonstration implements a recently published algorithm for an improved finite difference scheme for solving the Helmholtz partial differential equation on a rectangle with uniform grid spacing. The numerical method used to solve the heat equation for all the above cases is Finite Difference Method(FDM). Follow 89 views (last 30 days) Garrett Noach on 4 Dec 2017. Free falling object 2D; Free falling object with Drag; ode45: Predator Prey Model; Implicit Method: Heat Transfer; Shooting Method: Heat Transfer; Lab09: Partial Differential Equations (Laplace Equation) Scalar Field; Vector Field; Laplace Equation 1; Laplace Equation 2; Lab10: Partial Differential Equations (Diffusion Equation) Diffusion. Iterative solvers for 2D Poisson equation; 5. Midterm 2D Heat conduction steady and unsteady state using the iterative solver. Thus we consider u t(x;y;t) = k(u. (48) does not necessarily satisfy differential eq. 2D Heat Conduction - Solving Laplace's Equation on the CPU and the GPU December 10, 2013 Abhijit Joshi 1 Comment Laplace's equation is one of the simplest possible partial differential equations to solve numerically. Solving the non-homogeneous equation involves defining the following functions: (,. 1\\text{m}\\) See: pygimli. The main contributions of this paper are three-fold: (1) The use of heat equation to solve the skeleton based on the connection between the skeleton curve and ridge points, (2) directly report a stable and noise-insensitive skeleton, rather than heavily depending on a pruning step and (3) meet the following nice properties at the same time, i. In this article, the heat conduction problem of a sector of a finite hollow cylinder is studied as an exact solution approach. Solve wave equation with central differences. PHY2206 (Electromagnetic Fields) Analytic Solutions to Laplace's Equation 3 Hence R =γrm +δr−m is the general form for m i≠ i0 and R =α0 lnr +β0 when m i= i0 and the most general form of the solution is φ()r,θ=α0lnr +β0 + γmr m +δ mr ()−m α mcos()mθ+βmsin()mθ m=1 ∞ ∑ including a redundant constant. I am currently writing a matlab code for implicit 2d heat conduction using crank-nicolson method with certain Boundary condiitons. Included is an example solving the heat equation on a bar of length L but instead on a thin circular ring. Online program for calculating various equations related to constant acceleration motion. I want to see the displacements, u and v, when a simple deformation is imposed - e. This blog will help students and researchers to learn the various mostly used simulation software and tools. But, in practice, these equations are too difficult to solve analytically. EML4143 Heat Transfer 2 For education purposes. lua in the current working directory. derivation of heat diffusion equation for spherical cordinates. Partial Differential Equation Toolbox lets you import 2D and 3D geometries from STL or mesh data. This calculator can be used to calculate conductive heat transfer through a wall. You can solve PDEs by using the finite element method, and postprocess results to explore and analyze them. The presented procedure avoid solving the kernel equation in. The first step would be to discretize the problem area into a matrix of temperatures. In this paper, we use homotopy analysis method (HAM) to solve 2D heat conduction equations. 1) is to be solved on some bounded domain D in 2-dimensional Euclidean space with boundary that has conditions is the Laplacian (14. The following options can be given: digits of absolute accuracy sought. Find a numerical solution to the following differential equations with the associated initial conditions. Use Fourier Series to Find Coefficients The only problem remaining is to somehow pick the constants a n so that the initial condition u(x,0) = ϕ(x) is satisfied. Bottom wall is initialized at 100 arbitrary units and is the boundary condition. Solving Heat Equation with Laplace Transform. Learn more about: Equation solving » Tips for entering queries. lua in the current working directory. Included is an example solving the heat equation on a bar of length L but instead on a thin circular ring. Solve the system of equations A˚= b, where ˚is the vector of unknowns. Ask Question Asked 1 year ago. Procedure: On solving the steady equation in heat conduction, we consider the there is no convection and No Internal Heat generation in this problem. 1 1D heat and wave equations on a finite interval In this section we consider a general method of separation of variables and its applications to solving heat equation and wave equation on a finite interval (a 1, a2). Hi, I am wondering how to use the pdetool to solve the wave equation on a circular domain. Wave equation solver. Bottom wall is initialized at 100 arbitrary units and is the boundary condition. Assume that the sides of the rod are insulated so that heat energy neither enters nor leaves the rod through its sides. GitHub Gist: instantly share code, notes, and snippets. The heat and wave equations in 2D and 3D 18. Useful for problems with complicated geometries, loadings, and material properties where analytical solutions can not be obtained. Use a forward difference scheme for the. The calculator is generic and can be used for both metric and imperial units as long as the use of units is consistent. It’s a simple MATLAB code that can solve for different materials such as (copper, aluminum, silver, etc…. Midterm 2D Heat conduction steady and unsteady state using the iterative solver. You can perform linear static analysis to compute deformation, stress, and strain. Assuming isothermal surfaces, write a software program to solve the heat equation to determine the two-dimensional steady-state spatial temperature distribution within the bar. A compact and fast Matlab code solving the incompressible Navier-Stokes equations on rectangular domains mit18086 navierstokes. 2 Solving PDEs with Fourier methods The Fourier transform is one example of an integral transform: a general technique for solving di↵erential equations. The two dimensional fourier transform is computed using 'fft2'. The solution of the second equation is T(t) = Ceλt (2) where C is an arbitrary constant. Of course, the number of equations should be the same as the number of unknowns. Mathematica Stack Exchange is a question and answer site for users of Wolfram Mathematica. 4 Thermal Resistance Circuits There is an electrical analogy with conduction heat transfer that can be exploited in problem solving. Heat equation/Solution to the 2-D Heat Equation in Cylindrical Coordinates. Any help will be much appreciated. Hancock Fall 2006 1 2D and 3D Heat Equation Ref: Myint-U & Debnath §2. 3, the initial condition y 0 =5 and the following differential equation. Thomas algorithm which has been used to solve the system(6. We only consider the case of the heat equation since the book treat the case of the wave equation. This allows for a simpler GUI where we have only one button for the heat equation which is used for all supported solver. How should I go about it? The domain is a unit square. Numerical methods for solving the heat equation, the wave equation and Laplace's equation (Finite difference methods) Mona Rahmani January 2019. It looks like I was able to solve it using NDSolve , but when I try to create an animation of it using Animate all I get is blank frames. Aim: To find the No. Heat conduction Q/ Time = (Thermal conductivity) x x (T hot - T cold)/Thickness Enter data below and then click on the quantity you wish to calculate in the active formula above. 9 inch sheet of copper, the heat would move through it exactly as our board displays. 6 Example problem: Solution of the 2D unsteady heat equation. 2D linear conduction equation was solved for steady state and transient conditions by chosing 20 grid points in both x & y directions. The first step would be to discretize the problem area into a matrix of temperatures. Solving the heat equation using the separation of variables. See Draft ShapeString for an example of a well documented tool. • Goal: predict the heat distribution in a 2D domain resulting from conduction • Heat distribution can be described using the following partial differential equation (PDE): uxx + uyy = f(x,y) • f(x,y) = 0 since there are no internal heat sources in this problem • There is only 1 heat source at a single boundary node, and. The Equations being solved may be ordinary Differential Equations and/or partial Differential Equations of any order & degree. where is the scalar field variable, is a volumetric source term, and and are the Cartesian coordinates. In the above equation on the right, represents the heat flow through a defined cross-sectional area A, measured in watts,. I'm finding it difficult to express the matrix elements in MATLAB. space-time plane) with the spacing h along x direction and k along t direction or. EML4143 Heat Transfer 2 For education purposes. Regularity (Besov space, Holder space and wavelets) Week 3 (2/3-7). 303 Linear Partial Differential Equations Matthew J. Heat conduction follows a. In this section we go through the complete separation of variables process, including solving the two ordinary differential equations the process generates. It also factors polynomials, plots polynomial solution sets and inequalities and more. Let us recall that a partial differential equation or PDE is an equation containing the partial derivatives with respect to several independent variables. \" Proceedings of the ASME 2005 International Mechanical Engineering Congress and Exposition. 6 Example problem: Solution of the 2D unsteady heat equation. m Benjamin Seibold Applied Mathematics Massachusetts Institute of Technology www-math. space-time plane) with the spacing h along x direction and k. Analytical solution of 2D SPL heat conduction model T. This code is designed to solve the heat equation in a 2D plate. One-Dimensional Heat Equations! Computational Fluid Dynamics! i,j and solve by iteration! The implicit method is unconditionally stable, but it is necessary to solve a system of linear equations at each time step. As we will see this is exactly the equation we would need to solve if we were looking to find the equilibrium solution (i. The coefficient matrix and source vector look okay after the x-direction loop. DeTurck Math 241 002 2012C: Solving the heat equation 8/21. The 2D Fourier transform. Abbasi; Solution to Differential Equations Using Discrete Green's Function and Duhamel's Methods Jason Beaulieu and Brian Vick. As for the wave equation, Wolfram has a great page which describes the problem and explains the solution carefully describing each parameter. Enter your queries using plain English. Heat conduction into a rod with D=0. I need matlab code to solve 2D heat equation \"PDE \" using finite difference method implicit schemes. Aim: To find the No. 4 Inverse problems. ransfoil RANSFOIL is a console program to calculate airflow field around an isolated airfoil in low-speed, su numerical simulation code for solving transport equations in 1D/2D/3D. 7) obtained by Crank-Nicolson scheme to one-dimensional equation cannot used to solve (6. Using fixed boundary conditions \"Dirichlet Conditions\" and initial temperature in all nodes, It can solve until reach steady state with tolerance value selected in the code. 2D non-Newtonian power-law flow in a channel. Equation 5-5:(It would be extra nice if one sent me the derivation using equation editor in Word! :] ) derivation needed!. Project - Solving the Heat equation in 2D Aim of the project The major aim of the project is to apply some iterative solution methods and preconditioners when solving linear systems of equations as arising from discretizations of partial differential equations. Click here for more information, or create a solver right now. , Schmeiser, Christian, Markowich, Peter A. top boundary is displaced by 10%. Using D to take derivatives, this sets up the transport. This allows for a simpler GUI where we have only one button for the heat equation which is used for all supported solver. Solving the heat equation Charles Xie The heat conduction for heterogeneous media is modeled using the following partial differential equation: T k T q (1) t T c v where k is the thermal conductivity, c is the specific heat capacity, is the density, v is the velocity field, and q is the internal heat generation. Khan Academy Video: Solving Simple Equations. We use the idea of this method to solve the above nonhomogeneous heat equation. So it must be multiplied by the Ao value for using in the overall heat transfer equation. After submitting, as a motivation, some applications of this paradigmatic equations, we continue with the mathematical analysis of them. Week 1 (1/22-24). 9 inch sheet of copper, the heat would move through it exactly as our board displays. Integrate initial conditions forward through time. The paper is organized as follows. In order to model this we again have to solve heat equation. This scientific code solves the 3D Heat equation with MPI (Message Passing Interface) implementation. Assuming isothermal surfaces, write a software program to solve the heat equation to determine the two-dimensional steady-state spatial temperature distribution within the bar. For modeling structural dynamics and vibration, the toolbox provides a direct time integration solver. 1 1D heat and wave equations on a finite interval In this section we consider a general method of separation of variables and its applications to solving heat equation and wave equation on a finite interval (a 1, a2). Numerical Heat Transfer October, 2011 Kopaonik, Serbia SIMULATION APPROACH The governing equation for 2D heat conduction is given by: T T T ( ) ( ) qV C x x y y t For steady state of 2D heat conduction, in absence of interlnal heat sources, and for constant diffusion coefficients, the governing equation is given by: 2T 2T ( )0 x 2 y 2. ASME 119 406-12. The parameter α must be given and is referred to as the diffusion. Thanks for contributing an answer to Mathematica Stack Exchange! Solving the 2D heat equation. I am entirely new to Mathematica and have been given the task to animate the solution to the 2D heat equation with given initial and boundary conditions. The Finite Difference Method Because of the importance of the diffusion/heat equation to a wide variety of fields, there are many analytical solutions of that equation for a wide variety of initial and boundary conditions. The analog of is current, and the analog of the temperature difference, , is voltage difference. Using fixed boundary conditions \"Dirichlet Conditions\" and initial temperature in all nodes, It can solve until reach steady state with tolerance value selected in the code. Solve this banded system with an efficient scheme. The heat equation is a partial differential equation describing the distribution of heat over time. Your equation for radiative heat flux has the unit [\\frac{\\text{W}}{\\text{m}^2}], while the Neumann boundary condition needs a unit of [\\frac{\\text{K}}{\\text{m}}]. Project - Solving the Heat equation in 2D Aim of the project The major aim of the project is to apply some iterative solution methods and preconditioners when solving linear systems of equations as arising from discretizations of partial differential equations. I am trying to solve the 2D heat. ME 448/548: 2D Di usion. 1 Heat equation Recall that we are solving ut = α2∆u, t > 0, x2 +y2 < 1, u(0,x,y) = f(x,y), x2 +y2 < 1, u(t,x,y) = 0, x2 +y2 = 1. Introduction To Fem File Exchange Matlab Central. If you were to heat up a 14. Assuming isothermal surfaces, write a software program to solve the heat equation to determine the two-dimensional steady-state spatial temperature distribution within the bar. de Householder Symposium XVI Seven Springs, May 22 – 27, 2005 Thanks to: Enrique Quintana-Ort´ı, Gregorio Quintana-Ort´ı. Solving the Diffusion Equation Explicitly This post is part of a series of Finite Difference Method Articles. 2) is also called the heat equation and also describes the distribution of a heat in a given region over time. Frame Deflections with Concentrated Load on the Horizontal Member Equations and Calculator. Abstract A preliminary group classification of the class 2D nonlinear heat equations u t = f(x,y,u,u x,u y)(u xx + u yy), where f is arbitrary smooth function of the variables x. From Wikiversity < Heat equation. CiteSeerX - Document Details (Isaac Councill, Lee Giles, Pradeep Teregowda): Abstract A two dimensional time dependent heat transport equation at the microscale is derived. Trotter, and Introduction to Differential Equation s by Richard E. The working principle of solution of heat equation in C is based on a rectangular mesh in a x-t plane (i. • Goal: predict the heat distribution in a 2D domain resulting from conduction • Heat distribution can be described using the following partial differential equation (PDE): uxx + uyy = f(x,y) • f(x,y) = 0 since there are no internal heat sources in this problem • There is only 1 heat source at a single boundary node, and. 1 Solve a semi-linear heat equation 8. EML4143 Heat Transfer 2 For education purposes. Hi, I am wondering how to use the pdetool to solve the wave equation on a circular domain. You can do this in principle, but it is quite cumbersome and we must not forget that the equation will in general have three roots. 18 1 Getting Started. In order to solve the PDE equation, generalized finite Hankel, periodic Fourier, Fourier and Laplace transforms are applied. Consider the 4 element mesh with 8 nodes shown in Figure 3. Midterm 2D Heat conduction steady and unsteady state using the iterative solver. The Heat Equation via Fourier Series The Heat Equation: In class we discussed the ow of heat on a rod of length L>0. Using fixed boundary conditions \"Dirichlet Conditions\" and initial temperature in all nodes, It can solve until reach steady state with tolerance value selected in the code. If it is kept on forever, the equation might admit a nontrivial steady state solution depending on the forcing. 4 and Section 6. 1 1D heat and wave equations on a finite interval In this section we consider a general method of separation of variables and its applications to solving heat equation and wave equation on a finite interval (a 1, a2). This is a third-degree equation in \\rho and we would like to solve for \\rho. derivation of heat diffusion equation for spherical cordinates derivation needed. In the past, engineers made further approximations and simplifications to the equation set until they had a group of. Section 9-5 : Solving the Heat Equation Okay, it is finally time to completely solve a partial differential equation. Jump to navigation Jump to search. But, in practice, these equations are too difficult to solve analytically. The heat equation, the variable limits, the Robin boundary conditions, and the initial condition are defined as:. the solute is generated by a chemical reaction), or of heat (e. FEM2D_HEAT is a C++ program which applies the finite element method to solve a form of the time-dependent heat equation over an arbitrary triangulated region. Calculator Use. Ask Question Asked 1 year ago. Mathematics of Finite Element Method. In order to model this we again have to solve heat equation. 1 Diffusion Consider a liquid in which a dye is being diffused through the liquid. Separation of variables A more fruitful strategy is to look for separated solutions of the heat equation, in other words, solutions of the form u(x;t) = X(x)T(t). solving first on a very coarse grid and extending the solution to finer and finer grids, and it can solve iteratively the original system (finest grid). The explicit algorithm is be easy to parallelize, by dividing the physical domain (square plate) into subsets, and having each processor update the grid points on the subset it owns. This equation is a model of fully-developed flow in a rectangular duct. Solving the Heat Equation using Matlab In class I derived the heat equation u t = Cu xx, u x(t,0) = u x(t,1) = 0, u(0,x) = u0(x), 0 0, x2 +y2 < 1, u(0,x,y) = f(x,y), x2 +y2 < 1, u(t,x,y) = 0, x2 +y2 = 1. Aim: To find the No. I recently begun to learn about basic Finite Volume method, and I am trying to apply the method to solve the following 2D continuity equation on the cartesian grid x with initial condition For simplicity and interest, I take , where is the distance function given by so that all the density is concentrated near the point after sufficiently long. The solver will then show you the steps to help you learn how to solve it on your own. and Graham, A. It can give an approximate solution using a multigrid method, i. derivation of heat diffusion equation for spherical cordinates. Furthermore, unlike the method of undetermined coefficients, the Laplace transform can be used to directly solve for. (2015) A simple algorithm for solving Cauchy problem of nonlinear heat equation without initial value. It looks like I was able to solve it using NDSolve , but when I try to create an animation of it using Animate all I get is blank frames. Your equation for radiative heat flux has the unit [\\frac{\\text{W}}{\\text{m}^2}], while the Neumann boundary condition needs a unit of [\\frac{\\text{K}}{\\text{m}}]. Solving Heat Equation In 2d File Exchange Matlab Central. Finite Difference Method using MATLAB. FEM2D_HEAT is a C++ program which applies the finite element method to solve a form of the time-dependent heat equation over an arbitrary triangulated region. One such class is partial differential equations (PDEs). Solving the 2D Helmholtz Partial Differential Equation Using Finite Differences. We use the idea of this method to solve the above nonhomogeneous heat equation. The steady state analysis with Jacobi and Gauss-Seidel and SOR (Successive Over Relaxation) methods gave same results. 0; % Advection velocity % Parameters needed to solve the equation within the Lax method. Generic solver of parabolic equations via finite difference schemes. Correction* T=zeros(n) is also the initial guess for the iteration process 2D Heat Transfer using Matlab. We assume that the motion of the boundary is. MG Solver for the 2D Heat equation Math 4370/6370, Spring 2015 The Problem Consider the 2D heat equation, that models ow of heat through a solid having thermal di u-sivity , @u @t You will solve the system A~u= f~using an iterative solver known as multigrid (MG). These can be used to find a general solution of the heat equation over certain domains; see, for instance, ( Evans 2010 ) for an introductory treatment. Hess's Law The heat of reaction (1) for the reaction A + 2B --> 2C is 1100kJ. The numerical method used to solve the heat equation for all the above cases is Finite Difference Method(FDM). HOT_POINT, a MATLAB program which uses FEM_50_HEAT to solve a heat problem with a point source. In this section we analyze the 2D screened Poisson equation the Fourier do- main. HOT_PIPE, a MATLAB program which uses FEM_50_HEAT to solve a heat problem in a pipe. Williamson and H. Heat Transfer L10 P1 Solutions To 2d Equation. Derive equation 5-5 using Newton's laws and include the derivation in your report. Thus we consider u t(x;y;t) = k(u. Phan and Y. Numerical Heat Transfer October, 2011 Kopaonik, Serbia SIMULATION APPROACH The governing equation for 2D heat conduction is given by: T T T ( ) ( ) qV C x x y y t For steady state of 2D heat conduction, in absence of interlnal heat sources, and for constant diffusion coefficients, the governing equation is given by: 2T 2T ( )0 x 2 y 2. Solving the 2D Helmholtz Partial Differential Equation Using Finite Differences. Solve heat equation by $$\\theta$$-scheme. Along the whole positive x-axis, we have an heat-conducting rod, the surface of which is. How to Solve the Heat Equation Using Fourier Transforms. pdf] - Read File Online - Report Abuse. International Journal of Partial Differential Equations and Applications, 2(3), 58-61. To solve this, we notice that along the line x − ct = constant k in the x,t plane, that any solution u(x,y) will be constant. Finite differences for the 2D heat equation. As for the wave equation, Wolfram has a great page which describes the problem and explains the solution carefully describing each parameter. 303 Linear Partial Differential Equations Matthew J. Let Φ(x) be the concentration of solute at the point x, and F(x) = −k∇Φ be the. of iteration on every solver and write a detailed report on it. Transport Theory and Statistical Physics. A parabolic second-order differential equation for the temperature of a substance in a region where no heat source exists: ∂ t /∂τ = (k /ρ c)(∂ 2 t /∂ x 2 + ∂ 2 t /∂ y 2 + ∂ t 2 /∂ z 2), where x, y, and z are space coordinates, τ is the time, t (x,y,z, τ) is the temperature, k is the thermal conductivity of the body, ρ is its density, and c is its specific heat; this. 1) is to be solved on some bounded domain D in 2-dimensional Euclidean space with boundary that has conditions is the Laplacian (14. Williamson and H. Equation (1) is known as a one-dimensional diffusion equation, also often referred to as a heat equation. Poisson's Equation in 2D Analytic Solutions A Finite Difference A Linear System of Direct Solution of the LSE Classification of PDE Page 1 of 16 Introduction to Scientific Computing Poisson's Equation in 2D Michael Bader 1. You can solve PDEs by using the finite element method, and postprocess results to explore and analyze them. Chapter 4 – 2D Triangular Elements Page 15 of 24 In this equation Q is the global displacement vector which is the sum of all the local displacement vectors and K is the global stiffness matrix which is the sum of all the local stiffness matrices. Solving the 2D Heat Equation As just described, we have two algorithms: explicit (Euler) and implicit (Crank-Nicholson). However, it suffers from a serious accuracy reduction in space for interface problems with different. Viewed 522 times 5. Heat conduction Q/ Time = (Thermal conductivity) x x (T hot - T cold)/Thickness Enter data below and then click on the quantity you wish to calculate in the active formula above. Partial Differential Equation Toolbox lets you import 2D and 3D geometries from STL or mesh data. ut = u2206u where u2206 denotes the Laplacian operator u2206u = u xx + u yy. I recently begun to learn about basic Finite Volume method, and I am trying to apply the method to solve the following 2D continuity equation on the cartesian grid x with initial condition For simplicity and interest, I take , where is the distance function given by so that all the density is concentrated near the point after sufficiently long. MPI Numerical Solving of the 3D Heat equation. Phan and Y. Answer to 2. Solving Linear/Non-linear systems: Conjugate Gradient Method 13. Procedure: On solving the steady equation in heat conduction, we consider the there is no convection and No Internal Heat generation in this problem. HomeworkQuestion. It is also used to numerically solve parabolic and elliptic partial. Analytical solution of 2D SPL heat conduction model T. equation we considered that the conduction heat transfer is governed by Fourier’s law with being the thermal conductivity of the fluid. Use this calculator to solve polynomial equations with an order of 3 such as ax 3 + bx 2 + cx + d = 0 for x including complex solutions. This will lead us to confront one of the main problems. The third shows the application of G-S in one-dimension and highlights the. For the purpose of this question, let's assume a constant heat conductivity and assume a 1D system, so$$ \\rho c_p \\frac{\\partial T}{\\partial t} = \\lambda \\frac{\\partial^2 T}{\\partial x^2}. Equation 23 plotted on the same axis as the computed value of potential using the method of relaxation is shown in the following gure, with equation 23 being the contour lines on the XY plane and the computed potential as the mesh. If you were to heat up a 14. Find: Temperature in the plate as a function of time and position. Use this calculator to solve polynomial equations with an order of 3 such as ax 3 + bx 2 + cx + d = 0 for x including complex solutions. The heat equation is ∂u ∂t = ∇·∇u. [1–3,21,22]). Active 1 year ago. The discretized equations are solved by the parallel Krylov-Schwarz (KS. Solving PDEs will be our main application of Fourier series. top boundary is displaced by 10%. Below shown is the equation of heat diffusion in 2D Now as ADI scheme is an implicit one, so it is unconditionally stable. In physics and mathematics, the heat equation is a partial differential equation that describes how the distribution of some quantity (such as heat) evolves over time in a solid medium, as it spontaneously flows from places where it is higher towards places where it is lower. Assume that the sides of the rod are insulated so that heat energy neither enters nor leaves the rod through its sides. Here, I assume the readers have basic knowledge of finite difference method, so I do not write the details behind finite difference method, details of discretization error, stability, consistency, convergence, and fastest/optimum. Equation Generator When 3 points are input, this calculator will generate a second degree equation. In section 2 the HAM is briefly reviewed. -5 0 5-30-20-10 0 10 20 30 q sinh( q) cosh( q) Figure1: Hyperbolicfunctionssinh( ) andcosh( ). So the equation becomes r2 1 r 2 d 2 ds 1 r d ds + ar 1 r d ds + b = 0 which simpli es to d 2 ds2 + (a 1) d ds + b = 0: This is a constant coe cient equation and we recall from ODEs that there are three possi-bilities for the solutions depending on the roots of the characteristic equation. Many of the techniques used here will also work for more complicated partial differential equations for which separation of variables cannot be used directly. Thanks for the quick response! I have to solve the exact same heat equation (using the ODE suite), however on the 1D heat equation. This calculator can be used to calculate conductive heat transfer through a wall. After submitting, as a motivation, some applications of this paradigmatic equations, we continue with the mathematical analysis of them. Partial Differential Equation Toolbox lets you import 2D and 3D geometries from STL or mesh data. ASME 119 406-12. pdf] - Read File Online - Report Abuse matlab by example - Department of Engineering, University of. \" The software program Energy2D is used to solve the dynamic Fourier heat transfer equations for the Convective Concrete case. The heat equation, the variable limits, the Robin boundary conditions, and the initial condition are defined as:. 2D Heat Conduction - Solving Laplace's Equation on the CPU and the GPU December 10, 2013 Abhijit Joshi 1 Comment Laplace's equation is one of the simplest possible partial differential equations to solve numerically. Here, is a C program for solution of heat equation with source code and sample output. HOT_PIPE, a MATLAB program which uses FEM_50_HEAT to solve a heat problem in a pipe. 5, An Introduction to Partial Differential Equa-tions, Pinchover and Rubinstein The method of separation of variables can be used to solve nonhomogeneous equations. Clear Equation Solver ». This section considers transient heat transfer and converts the partial differential equation to a set of ordinary differential equations, which are solved in MATLAB. Active 3 years, Solving the heat equation using the separation of variables. To solve your equation using the Equation Solver, type in your equation like x+4=5. Here is a talk from JuliaCon 2018 where I describe how to use the tooling across the Julia ecosystem to solve partial differential equations (PDEs), and how the different areas of the ecosystem are evolving to give top-notch PDE solver support. pyplot as plt dt = 0. Introduction: The problem Consider the time-dependent heat equation in two dimensions. I will assume you are dealing with Navier Stokes equations. Solving Non-linear systems: Newton Raphson Method 12. First we formulate a more detailed criterion for spatial coarsening, which enables the method to deal with unstructured meshes and varying material parameters. In this article, the heat conduction problem of a sector of a finite hollow cylinder is studied as an exact solution approach. 38 149-192, 2009. volume method, to discretize the 2D-3T equations (cf. The model, initial conditions, and time points are defined as inputs to ODEINT to numerically calculate y(t). Solved Heat Transfer Example 4 3 Matlab Code For 2d Cond." ]

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[ "Note that this reference documentation is identical to the help that is displayed in MATLAB when you type “help ft_redefinetrial”.\n\n``` FT_REDEFINETRIAL allows you to adjust the time axis of your data, i.e. to\nchange from stimulus-locked to response-locked. Furthermore, it allows\nyou to select a time window of interest, or to resegment your long trials\ninto shorter fragments.\n\nUse as\ndata = ft_redefinetrial(cfg, data)\nwhere the input data should correspond to the output of FT_PREPROCESSING and\nthe configuration should be specified as explained below. Note that some\noptions are mutually exclusive, and require two calls to this function to\navoid confusion about the order in which they are applied.\n\nFor selecting a subset of trials you can specify\ncfg.trials = 'all' or a selection given as a 1xN vector (default = 'all')\n\nFor selecting trials with a minimum length you can specify\ncfg.minlength = length in seconds, can be 'maxperlen' (default = [])\n\nFor realiging the time axes of all trials to a new reference time\npoint (i.e. change the definition for t=0) you can use the following\nconfiguration option\ncfg.offset = single number or Nx1 vector, expressed in samples relative to current t=0\n\nFor selecting a specific subsection of (i.e. cut out a time window\nof interest) you can select a time window in seconds that is common\nin all trials\ncfg.toilim = [tmin tmax] to specify a latency window in seconds, can be Nx2 vector\n\nAlternatively you can specify the begin and end sample in each trial\ncfg.begsample = single number or Nx1 vector, expressed in samples relative to the start of the input trial\ncfg.endsample = single number or Nx1 vector, expressed in samples relative to the start of the input trial\n\nAlternatively you can specify a new trial definition, expressed in\nsamples relative to the original recording\ncfg.trl = Nx3 matrix with the trial definition, see FT_DEFINETRIAL\n\nAlternatively you can specify the data to be cut into (non-)overlapping\nsegments, starting from the beginning of each trial. This may lead to loss\nof data at the end of the trials\ncfg.length = single number (in unit of time, typically seconds) of the required snippets\ncfg.overlap = single number (between 0 and 1 (exclusive)) specifying the fraction of overlap between snippets (0 = no overlap)\n\nTo facilitate data-handling and distributed computing you can use\ncfg.inputfile = ...\ncfg.outputfile = ...\nIf you specify one of these (or both) the input data will be read from a *.mat\nfile on disk and/or the output data will be written to a *.mat file. These mat\nfiles should contain only a single variable, corresponding with the\ninput/output structure." ]

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[ "", null, "###", null, "Thursday, 3 November 2022\nMany hours are spent in primary classrooms on activities to help children learn multiplication facts, with varying degrees of success. I’ve seen (and used) chanting, counting patterns, songs, games, trio cards, bingo, round the class cards, tables tests… and many more strategies to try to get it to stick. So here’s another strategy to add to the mix – focus on the products.\n\nA multiplication sentence such as 3 x 4 = 12 is made up of two factors: the multiplicand (3) and the multiplier (4). These are multiplied together to give the product or multiple (12).\n\nWe often focus on the two factors of a multiplication sentence, however it is interesting to identify and focus on the products. If children begin to remember some of the key numbers that are products, then it will help to recall multiplication facts. We do it as adults without realising it.\n\n### Finding products of multiplication facts\n\n1. List all the multiples of 10 in a column up to 10x10 = 100.\n\nThese are usually easy for children to learn, so just check, ‘we know all these don’t we?’ Obviously sort out any difficulties children have with these numbers, but if there are I would suggest the activity may be a challenge at this stage.\n\n2. How many products are there in the 90s? 80s? 70s?\n\nFind all the products of the times tables up to 10x10 starting with those in the 90s. It has more impact when starting with big numbers.\nThere are none in the 90s, one in the 80s: 9x9=81 and one in the 70s too: 8x9=72", null, "", null, "3. Which products of multiplication facts appear in the 60s, then in the 50s?\n\nThere are two products in the 60s: 63 and 64, and amazingly there are also only two products in the 50s: 54 and 56.", null, "There are only six products to remember that are greater than 50.\n\n54, 56, 63, 64, 72 and 81\n\nSo if we focus on the products of the times tables up to 10x10 (and ignore multiples of 10),\nthere are only six key numbers for children to remember.\n\n62 isn’t there, or 58… and many other numbers that shouldn’t enter our heads when we’re trying to quickly recall our tables facts.\n\n4. Continue finding products in the 40s, then 30s and 20s.\n\nIf we carry on through the tens, the numbers increase:\nFour products in the 40s: 42, 45, 48 and 49\nThree products in the 30s: 32, 35 and 36.\nFive products in the 20s: 21, 24, 25, 27 and 28.\n\n5. Find products in the teen numbers.\n\nThere are five more numbers between 10 and 20: 12, 14, 15, 16 and 18\n\nThe numbers missing are just as interesting: 11, 13, 17 and 19 are all prime numbers.", null, "", null, "If we look at all those products on a 100 square you realise how few numbers there are to learn.\n\nThere are some other interesting features too, the pattern of the multiples of 9\nand the\nproducts, such as 24 and 36, that have several multiplication facts.", null, "I’ve used this as an activity with Y3-6 children and it really makes them think about these key numbers. Some of them have taken the numbers on the 100 square and circled one at a time to learn the facts associated with that product. It’s just a different and alternative way in to the quest to become quick recallers of our tables facts.\n\n### What about the 11 and 12 times tables?\n\nTrue, these aren’t included, but in the ‘numbers to 100’ rule the 12 times table only adds two more numbers: 84 and 96. The multiples of 11 do add to the list, but the pattern is as easy as the multiples of 10 and is a nice tables pattern to learn.\n\nA good way of learning the 12x table is to use the power of partitioning. Break it up into x10 and x2, which seems slow, but is actually a pretty quick way of recalling correctly (rapid speed isn’t everything!). For example, 7 x 12 = (7 x 10) + (7 x 2) = 84\n\nThe process is ‘7 times 10 then add double 7’ – which isn’t a bad way of playing with the numbers to develop number sense and fluency.", null, "Here is a simple Powerpoint file to download with the slides shown above for you to use.\n\nWebsite design by SiteBuilder Bespoke" ]

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[ "# plot_backtest¶\n\nplot_backtest(forecast_df: pandas.core.frame.DataFrame, ts: TSDataset, segments: Optional[List[str]] = None, columns_num: int = 2, history_len: Union[int, Literal['all']] = 0, figsize: Tuple[int, int] = (10, 5))[source]\n\nPlot targets and forecast for backtest pipeline.\n\nThis function doesn’t support intersecting folds.\n\nParameters\n• forecast_df (pandas.core.frame.DataFrame) – forecasted dataframe with timeseries data\n\n• ts (TSDataset) – dataframe of timeseries that was used for backtest\n\n• segments (Optional[List[str]]) – segments to plot\n\n• columns_num (int) – number of subplots columns\n\n• history_len (Union[int, Literal['all']]) – length of pre-backtest history to plot, if value is “all” then plot all the history\n\n• figsize (Tuple[int, int]) – size of the figure per subplot with one segment in inches\n\nRaises\n• ValueError: – if `history_len` is negative\n\n• ValueError: – if folds are intersecting" ]

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[ "# Scrabble & Words with Friends points\n\n• formalisms is a valid Scrabble word with a point value of 17 according to TWL06 dictionary\n• formalisms is a valid Words with Friends word with a point value of 20 according to ENABLE dictionary\n\n## Definition of formalisms\n\n• (philosophy) the philosophical theory that formal (logical or mathematical) statements have no meaning but that its symbols (regarded as physical entities) exhibit a form that has useful applications (noun)\n\n• the doctrine that formal structure rather than content is what should be represented (noun)\n\n• the practice of scrupulous adherence to prescribed or external forms (noun)\n\nSee definition of formalisms in Merriam Webster\n\n• formalisms\n\n• formalism\n• moralisms\n\n• amorisms\n• misforms\n• moralism\n• oralisms\n• solarism\n\n• aliform\n• amorism\n• falsism\n• formals\n• immoral\n• malisms\n• mimosas\n• misform\n• oralism\n• safrols\n• sailors\n• samlors\n\n• armils\n• assoil\n• flairs\n• flamms\n• floras\n• foliar\n• forams\n• formal\n• fossil\n• frails\n• fraims\n• limmas\n• lissom\n• malism\n• massif\n• miasms\n• milors\n• mimosa\n• missal\n• molars\n• morals\n• morass\n• morias\n• morsal\n• romals\n• safrol\n• sailor\n• salmis\n• samlor\n• simars\n• slorms\n• smalms\n• smarms\n• sofars\n• solars\n\n• afros\n• alifs\n• amirs\n• amiss\n• ammos\n• arils\n• armil\n• arsis\n• fails\n• fairs\n• farls\n• farms\n• faros\n• fiars\n• filar\n• films\n• filos\n• firms\n• flair\n• flamm\n• flams\n• flims\n• flirs\n• flora\n• flors\n• floss\n• foals\n• foams\n• foils\n• folia\n• foram\n• forms\n• fossa\n• frail\n• fraim\n• frass\n• imams\n• lairs\n• laris\n• lassi\n• lasso\n• liars\n• limas\n• limma\n• limos\n• liras\n• loafs\n• loams\n• loirs\n• lomas\n• loris\n• mails\n• maims\n• mairs\n• malis\n• malms\n• marls\n• marms\n• miasm\n• milfs\n• milor\n• milos\n• miros\n• misos\n• missa\n• moils\n• moira\n• molar\n• molas\n• moral\n• moras\n• moria\n• oasis\n• orals\n• ossia\n• rails\n• ramis\n• rials\n• roams\n• roils\n• rolfs\n• romal\n• sails\n• saims\n• sairs\n• salmi\n• saris\n• saros\n• sials\n• silos\n• simar\n• simas\n• sisal\n• slams\n• slims\n• slorm\n• smalm\n• smarm\n• smirs\n• soars\n• sofar\n• sofas\n• soils\n• solar\n• solas\n• somas\n• soral\n• soras\n\n• afro\n• ails\n• aims\n• airs\n• alfs\n• alif\n• alms\n• also\n• amir\n• amis\n• ammo\n• arfs\n• aril\n• aris\n• arms\n• fail\n• fair\n• farl\n• farm\n• faro\n• fars\n• fiar\n• fila\n• film\n• filo\n• fils\n• firm\n• firs\n• flam\n• flim\n• flir\n• flor\n• foal\n• foam\n• foil\n• fora\n• form\n• foss\n• fras\n• fris\n• from\n• fros\n• imam\n• isms\n• isos\n• lair\n• lams\n• lari\n• lars\n• lass\n• liar\n• lias\n• lima\n• limo\n• lira\n• loaf\n• loam\n• loir\n• loma\n• loss\n• mail\n• maim\n• mair\n• mali\n• malm\n• mals\n• mams\n• marl\n• marm\n• mars\n• mass\n• milf\n• milo\n• mils\n• miro\n• mirs\n• miso\n• miss\n• moai\n• moas\n• moil\n• mola\n• mols\n• momi\n• moms\n• mora\n• mors\n• moss\n• oafs\n• oars\n• oils\n• olms\n• oral\n• orfs\n• osar\n• ossa\n• rail\n• rais\n• rami\n• rams\n• rial\n• rias\n• rifs\n• rima\n• rims\n• roam\n• roil\n• rolf\n• roma\n• roms\n• sail\n• saim\n• sair\n• sais\n• sals\n• sams\n• sari\n• sars\n• sial\n• silo\n• sima\n• sims\n• sirs\n• slam\n• slim\n• smir\n• soar\n• sofa\n• soil\n• sola\n• soli\n• sols\n• soma\n• soms\n• sora\n• sori\n• sris\n\n• ail\n• aim\n• air\n• ais\n• alf\n• als\n• ami\n• arf\n• arm\n• ars\n• ass\n• far\n• fas\n• fil\n• fir\n• for\n• fra\n• fro\n• ifs\n• ios\n• ism\n• iso\n• lam\n• lar\n• las\n• lis\n• lor\n• los\n• mal\n• mam\n• mar\n• mas\n• mil\n• mim\n• mir\n• mis\n• moa\n• moi\n• mol\n• mom\n• mor\n• mos\n• oaf\n• oar\n• oil\n• ois\n• olm\n• oms\n• ora\n• orf\n• ors\n• rai\n• ram\n• ras\n• ria\n• rif\n• rim\n• rom\n• sai\n• sal\n• sam\n• sar\n• sif\n• sim\n• sir\n• sis\n• sma\n• sol\n• som\n• sos\n• sri\n\n• ai\n• al\n• am\n• ar\n• as\n• fa\n• if\n• io\n• is\n• la\n• li\n• lo\n• ma\n• mi\n• mm\n• mo\n• of\n• oi\n• om\n• or\n• os\n• si\n• so" ]

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[ "#### Solution for What is 76 percent of 35.55:\n\n76 percent *35.55 =\n\n(76:100)*35.55 =\n\n(76*35.55):100 =\n\n2701.8:100 = 27.018\n\nNow we have: 76 percent of 35.55 = 27.018\n\nQuestion: What is 76 percent of 35.55?\n\nPercentage solution with steps:\n\nStep 1: Our output value is 35.55.\n\nStep 2: We represent the unknown value with {x}.\n\nStep 3: From step 1 above,{35.55}={100\\%}.\n\nStep 4: Similarly, {x}={76\\%}.\n\nStep 5: This results in a pair of simple equations:\n\n{35.55}={100\\%}(1).\n\n{x}={76\\%}(2).\n\nStep 6: By dividing equation 1 by equation 2 and noting that both the RHS (right hand side) of both\nequations have the same unit (%); we have\n\n\\frac{35.55}{x}=\\frac{100\\%}{76\\%}\n\nStep 7: Again, the reciprocal of both sides gives\n\n\\frac{x}{35.55}=\\frac{76}{100}\n\n\\Rightarrow{x} = {27.018}\n\nTherefore, {76\\%} of {35.55} is {27.018}\n\n#### Solution for What is 35.55 percent of 76:\n\n35.55 percent *76 =\n\n(35.55:100)*76 =\n\n(35.55*76):100 =\n\n2701.8:100 = 27.018\n\nNow we have: 35.55 percent of 76 = 27.018\n\nQuestion: What is 35.55 percent of 76?\n\nPercentage solution with steps:\n\nStep 1: Our output value is 76.\n\nStep 2: We represent the unknown value with {x}.\n\nStep 3: From step 1 above,{76}={100\\%}.\n\nStep 4: Similarly, {x}={35.55\\%}.\n\nStep 5: This results in a pair of simple equations:\n\n{76}={100\\%}(1).\n\n{x}={35.55\\%}(2).\n\nStep 6: By dividing equation 1 by equation 2 and noting that both the RHS (right hand side) of both\nequations have the same unit (%); we have\n\n\\frac{76}{x}=\\frac{100\\%}{35.55\\%}\n\nStep 7: Again, the reciprocal of both sides gives\n\n\\frac{x}{76}=\\frac{35.55}{100}\n\n\\Rightarrow{x} = {27.018}\n\nTherefore, {35.55\\%} of {76} is {27.018}\n\nCalculation Samples" ]

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[ "", null, "## Abstract\n\nWe deal with the numerical scheme for the Liouville Master Equation (LME) of a kind of Piecewise Deterministic Processes (PDP) with memory, analysed in . The LME is a linear system of hyperbolic PDEs, written in non‐conservative form, with non‐local boundary conditions. The solutions of that equation are time dependent marginal distribution functions whose sum satisfies the total probability conservation law. In the convergence of the numerical scheme, based on the Courant‐Isaacson‐Rees jointly with a direct quadrature, has been proved under a Courant‐Friedrichs‐Lewy like (CFL) condition. Here we show that the numerical solution is monotonic under a similar CFL condition. Moreover, we evaluate the conservativity of the total probability for the calculated solution. Finally, an implementation of a parallel algorithm by using the MPI library is described and the results of some performance tests are presented.\n\nFirst published online: 14 Oct 2010\n\nHow to Cite\nAnnunziato, M. (2009). A finite difference method for piecewise deterministic processes with memory. II. Mathematical Modelling and Analysis, 14(2), 139-158. https://doi.org/10.3846/1392-6292.2009.14.139-158\nPublished in Issue\nJun 30, 2009\nAbstract Views\n240" ]

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[ "NECO Syllabus for Chemistry 2022/2023 - Download PDF\n\n# NECO Chemistry Syllabus 2022/2023 – Download PDF\n\nThe National Examination Council has officially released the 2022 new NECO syllabus for chemistry and other subjects. Go through this post and see all the correct topics which all chemistry students must read. This topic is only for SS3 science students and NECO GCE candidates.\n\nNECO syllabus shows you all the areas of concentration for the forthcoming examination. All NECO chemistry questions and answers will be set from these topics.\n\n## 2022 NECO Syllabus for Chemistry and Area of Concentration\n\n### 1. Introduction To Chemistry\n\nTopics to read under  the induction to Chemistry are as follows:\n\n• Definitions of chemistry\n• Scope of Chemistry\n• History and genesis of Chemistry\n• Measurements and quantities with their S.I units\n• The scientific method: The scientific methods are:\n1. Hypothesis\n2. Experiment\n3. Theory\n4. Law\n\n### 2. Study Of Matter\n\nCandidates are advised to read and cover up all the subtopics in this section. Topics include:\n\n• include\n• Nature of matter\n• The Kinetic theory of matter\n• State of matter and Diffusion\n• Changes in the state of matter. This includes; Gases, Solids, and Liquids\n• Postulates the kinetic theory.\n\nIntensively study these topics listed below are make sure you attempt to answer past questions on them:\n\n• Elements, Compounds, and Mixtures (Know their Definitions, examples, and Differences)\n• Characteristics and nature of gases: this includes Shapes of gases,  compressibility, Vapour, and Gas density\n• Particles\n\n### 3. Atomic Structure\n\nThe important topics to read under atomic structure include the following;\n\n• The Atoms, molecules, and ions\n• Atomic Theory: Note the different theories and Postulates and the scientists that discovered them\n\nCandidates should be able to explain the following and also know their difference:\n\n• Atomic mass\n• Neutrons\n• Proton number\n• Mass number\n• Atomic numbers and Atomic symbols\n• Isotopes\n• Orbitals\n• Sub-atomic particles\n• Molecular mass\n• Relative atomic mass\n• Dalton’s atomic theory\n• Bohr’s model of the atom.\n• Electron Configuration: Study how to write the electronic configuration for different elements\n\n### 4. Changes:\n\nStudy everything subtopic on physical and chemical changes as follows:\n\n• Physical changes: Be able to explain this with examples\n• Chemical changes\n• Examples and Differences between chemical and physical changes\n\nAt the end of this topic, make sure you study and know the following:\n\n• Melting of solids (melting point)\n• Boiling of liquids (boiling point)\n• Evaporation of liquids (Evaporation)\n• Dissolution of solutes\n• Brownian motion and diffusion.\n\n5. Theories and Principles: Know how to state the following rules and principles:\n\n• Pauli Exclusion Principle\n• Hund’s Rule\n• Aufbau Principle.\n\n6. Separation: Cover everything section under this topic\n\n• Separation techniques\n• Criteria for purity.\n• Separation methods:\n• Chromatography\n• Precipitation\n• Crystallization\n• Magnetization\n• Sublimation\n• Distillation.\n• Boiling point\n• Melting point\n• Variable oxidation states\n• physical states\n\n7. The Periodic Table; Read everything about the periodic table\n\n• Periodicity.\n• Groups and Periods\n• The periodic law.\n\n8. Periodic properties; Touch these subtopics small.\n\n• Ionic size\n• Electronegativity\n• Ionization energy\n• electron affinity\n• number of valence electrons (Valency)\n\n9. Chemical Bonding: Study a bit about the following subtopics\n\n• Formation of bonds\n• Complex formation\n• Metallic Bonding\n• ionic bonding\n• Covalent bonding\n• Diatomic bonding\n• Interatomic bonding\n• Hydrogen bonding\n• Van der Waals forces\n• Intermolecular bonding\n• Formation of colored compounds;\n\nSection two of NECO chemistry syllabus 2022. Please, do well to follow the topics gradually and make use of the area of concentration as approved by NECO.\n\n10. Shapes of compounds: Study each of the following shapes.\n\n• Linear\n• Planar\n• Tetrahedral\n• Polar and non-polar compounds.\n\n11. Stoichiometry: Learn how to identify and name the following;\n\n• Chemical symbols\n• IUPAC names\n\n12. Chemical formula: Study the differences between each of these formulas and learn how to calculate them.\n\n• Empirical formula\n• Structural formula\n• Molecular formula\n\n13. Chemical Reactions: Study everything under this\n\n• Chemical equations and formulae\n• Amount of substance.\n• Acids-Base reactions.\n• Chemical combination.\n• Combustion reactions\n• Chemical compounds\n• Decomposition reaction\n• Redox reactions.\n• Ionic reactions\n• Displacement reaction.\n• Synthesis\n• Balancing of equations.\n\n14. Experiments and Calculations: Learn how to carry out practicals, calculate, and solve all problems using the data you obtain.\n\nArea of focus;\n\n• Mass and volume relationships\n• Percentage composition.\n• Preparation of solutions\n• Mole and Mole ratios\n• Avogadro’s constant  (6.02 x 1023)\n• Molar quantities\n• Dilute solution and concentrated solution.\n• Concentration terms\n• Amount of substance\n• Yield of product\n• Primary and secondary standards.\n• Standard solutions.\n• Calculation of Mass\n• Volume and percentage\n• Testing for Basic, acidic, and neutral solutions.\n• Molar concentration.\n• Dilution factor\n\nLaws: Know how to state the following Laws\n\n1. Conservation of mass.\n2. Constant composition.\n3. Multiple proportions.\n\n15. Gases and Gas laws: Study the principles of the following laws and how to apply them.\n\n1. Charles’ law\n2. Boyle’s law\n3. Dalton’s law of partial pressure;\n4. Graham’s law of diffusion,\n6. Ideal gas equation\n7. Gay Lussacs law.\n8. kinetic model.\n9. Kinetic molecular theory.\n10. Gas volumes, pressure, temperature.\n11. Mathematical expression, relations, and calculation of the gas laws\n12. Real gases and ideal situation.\n13. Preparation and purification of gases.\n14. Physical properties of the gases.\n15. Chemical properties of the gases\n16. Arrangement of gaseous particles\n17. Compressibility of gases\n18. Viscosity\n19. Density, and volume.\n20. Vapour\n21. Pressure of gases\n22. Boiling and evaporation.\n23. Temperature of gases\n24. Volume of gases\n\n16. Energy: Study the following topics intensively;\n\n• Energy Changes\n• Enthalpy\n• Entropy\n• Ionic\n• Metallic\n• Covalent\n• Chemical processes.\n• Exothermic process\n• Endothermic processes.\n• The total energy of a system.\n• Kinetic energy\n• Potential energy\n• Electrical energy\n• Heat energy.\n• Sound energy.\n• Combustion\n• Dissolution\n• Neutralization.\n\n## NECO 2022/2023 Syllabus for Chemistry\n\n17. Solid: Know the following solid substances.\n\n• Particles ions\n• Molecules and atoms\n• properties of solids\n• Interatomic and intermolecular bonding in the solids.\n• Identification of the types of chemical bonds in solid.\n• Melting points of solid.\n• Indicator of purity of solids.\n\n18. Acids, Bases, And Salts: Do an intensive study on acid, base, and salt following the subtopics arrangement as shown below;\n\n• Explanation of acids and bases.\n• Physical properties of acids and bases.\n• Chemical properties of acids and bases.\n• Classification of acids and bases.\n• Concept of the pH scale\n• Acid-base indicators.\n• Acid-base electrolytes and non-electrolytes.\n• Strength of acids and bases.\n• The reaction of Acid and Base\n• Laboratory and industrial preparation of acid and base.\n• Hydrolysis of salt.\n• Titration\n\nSalt: Read everything about salt and its production.\n\n• Explanation of salts.\n• Types of salts.\n• Laboratory and industrial preparation of salts.\n• Crystallization of Salt\n• Behavior of salts\n\nSolubility: Read the following subtopics\n\n• Definition and explanation of solubility.\n• Saturated and unsaturated solutions.\n• The equilibrium system.\n• Solubility curves\n• Effect of temperature on solubility of a substance.\n• Solubility and crystallization.\n• Calculations on solubility.\n• Solubility rules.\n\nChemical Kinetics And Equilibrium System: Study these subtopics shown below;\n\n• The rate of chemical reactions:\n• Factors affecting rates of chemical reaction\n• Theories of reaction rates.\n• Chemical equilibrium.\n• Pressure of reactants\n• Collision of particles\n• Transition state theories\n• Activation energy.\n• Energy profile\n• Activation energy\n• Enthalpy and Entropy change.\n• Reversible and irreversible reactions.\n\nRedox Reactions: Please, spend time and cover this topic intensively. Cover each of the following subtopics.\n\n• Oxidation and reduction reaction.\n• Oxidizing agents and reactions\n• Reducing agents and reactions\n• Half-life equation\n• Le Chatelier’s principle.\n• Redox equations\n• Loss and gain of electrons.\n• Oxidation state\n• The electrochemical cells and series.\n• Standard electrode potential.\n• Addition and removal of oxygen and hydrogen.\n• Loss and gain of electrons.\n• Balancing of redox equations\n\nElectrolysis: Read everything about electrolysis and learn how to solve all calculation problems.\n\n• Explanation of electrolysis\n• Faraday Laws of electrolysis.\n• Principles of electrolysis\n• Electrolytic cells and application of electrochemical cells.\n• Practical application of electrolysis\n• Discharge of species.\n• Primary and secondary cells\n• Daniell cell\n• Calculations on Electrolysis.\n• Electroplating\n\n### Organic Chemistry syllabus and other Important Topics\n\nthis is the final stage of the NECO SSCE Chemistry syllabus. This section is the SS3 scheme of work and everything you did from the first term to the third term. So make use of this as your area of concentration.\n\nCarbon and its Compounds: Study everything about Carbon\n\n• Natural occurrence of Carbon\n• Allotropes of Carbon\n\nOrganic Chemistry: Cover every topic under Organic chemistry\n\n• Hydrocarbons\n• Homologous series\n• Functional group\n• Organic compounds.\n• Petroleum.\n• Fractional distillation\n• Cracking\n• Isomerism\n• Petrochemicals\n• Octane number\n• Crystallization\n• Chromatography.\n• Empirical formulae\n• Structural formulae\n• Molecular formulae\n• Geometric isomerism.\n• Stereoisomerism\n• Esterification\n• Nomenclature and structure.\n• Laboratory preparation of organic compounds.\n• Benzene\n• Polymerization.\n• Halogenation.\n• Hydrogenation\n• Hydration.\n• Primary, secondary and tertiary alkanols.\n• Laboratory test for ethanol.\n\nIndustrial And The Environmental Chemistry\n\n• Chemical industries\n• Natural resources\n• Extraction and purification\n• Fine and heavy chemicals.\n• Industrial pollution.\n• Biotechnology.\n\nBiochemistry And Synthetic Polymers\n\n• Proteins and uses of protein\n• Depletion of the ozone layer\n• Properties and sources of protein\n• Fermentation\n• Amino acids\n• Greenhouse effect.\n• Food processing process\n• Polymers\n• Hydrolysis of proteins;\n\nFats and oils:\n\n• Formation of fat and oil\n• Sources and properties of fat and oil\n• Uses of fats and oils in industrial production\n• Test for the presence of fats and oil.\n• Saponification (Preparation of soap) from fats/oils.\n\nCarbohydrates\n\n• Monosaccharides;\n• Disaccharides\n• Polysaccharides.\n• Classes of carbohydrates.\n• The solubility of sugars.\n• Chemical properties of carbohydrates.\n• sugar test.\n• Starch and glucose units.\n\nPlastics\n\n• Monomers\n• Polymers\n• polymerization.\n• comonomers.\n• Thermoplastics\n• Thermosets.\n• Plastics and resins\n• properties of plastics.\n• Chemical test on plastics\n\n## The Chemistry Syllabus is long, how do I cover it on time\n\nThe fact that the syllable is much does not mean you should run away from it. It is really tiring to read all these.\n\nNever mind, I have a verified trick for you to cover all you these within a short period of time; let’s say in a month. In fact, I extensively talk about that in my previous article and it has helped so many persons.\n\nLet me quickly share some tips here:\n\n• Read through the whole syllabus. First, make sure to take note everything on the syllabus.\n• Get your study materials together: Get together all of the study materials you want to use.\n• Use the textbooks, online resources, and other study tools that are recommended." ]

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[ "Basically, the method overloading allows us to create several methods having the same name in a class. However, the method signatures must be different. In other words, it is possible to create several methods with the same method name as long as any one of the following is different in each method: number of parameters, type of parameters, and the order of parameters. However, the return type is not considered in method signature.\n\nIn fact, method overloading is a way to implement the compile-time polymorphism. So, the compiler binds a method call to method definition by looking at its signature.\n\nThe following code shows that we define a class Compute. Also, the class contains three methods. Furthermore, all of these methods have the same name findArea(). However, the parameters are different in each method. Hence, these three methods are performing a different task. Each one is computing the area of a different shape. Therefore, we need to call each method by providing appropriate arguments.\n\n``````// Create a simple program to demonstrate method overloading\nimport java.io.*;\nclass Compute\n{\npublic void findArea(int len, int brd)\n{\nSystem.out.println(\"\\nArea of the rectangle comes out to be \"+(len*brd)+\" Units!\");\n}\npublic void findArea(int side)\n{\nSystem.out.println(\"\\nArea of the square comes out to be \"+(side*side)+\" Units!\");\n}\npublic void findArea(int first, int second, int third)\n{\ndouble s=(first+second+third)/2;\ndouble area=Math.sqrt(s*(s-first)*(s-second)*(s-third));\nSystem.out.println(\"\\nArea of the triangle comes out to be \"+area+\" Units!\");\n}\n}\npublic class Main\n{\npublic static void main(String[] args) throws IOException {\nCompute rectOb=new Compute();\nSystem.out.println(\"Enter the length of rectangle...\");\n\nSystem.out.println(\"Enter the side of square...\");\nrectOb.findArea(side);\n\nSystem.out.println(\"Enter the first side of triangle...\");\nSystem.out.println(\"Enter the second side of triangle...\");\nSystem.out.println(\"Enter the third side of triangle...\");" ]

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[ "", null, "### FW-NA\n\n• Author:\n\nIlaria Mosca\n\ncontact\n\n• FW-NA is a package of Fortran programs based on a Monte Carlo approach for solving a waveform inversion and thus defining the 1-D physical properties of small-scale structures.\n\n# FW-NA\n\n## Project Description\n\nFW-NA is a package of Fortran programs based on a Monte Carlo approach for solving a waveform inversion and thus defining the 1-D physical properties (e.g., shear wave-speed, compressional wave-speed, shear dissipation factor) of small-scale (i.e., from millimetres to few meters) structures.\n\nIt consists of three main ingredients:\n\n1. The programs for solving the inverse problem using the Neighbourhood Algorithm (NA), a Monte Carlo technique developed by Sambridge, and (rses.anu.edu.au/~malcolm/NA/).\n2. The programs for solving the forward modelling using the package GEMINI II.\n3. Fortran codes to connect the input/output files of NA and GEMINI with each other.\n\nFrom a theoretical point of view, the NA samples the entire model space and attributes a misfit function to each solution compatible with the data (top panel of Figure). For each sampled model the algorithm solves a forward problem which in the present case is given by the approach of Friederich & Dalkomo . Then, all the models are expressed in terms of (Gaussian or non-Gaussian) probability density functions (pdfs) which represent all the possible information gained from the data (bottom panel of Figure).\n\n## References:\n\n W. Friederich & J. Dalkolmo (1995). Complete synthetic seismograms for a spherically earth by a numerical computation of the Green's function in the frequency domain - Geophys. J. Int., 122, 537-550\n\n M. Sambridge (1999a). Geophysical inversion with a neighbourhood algorithms I - Searching a parameter space - Geophys. J. Int., 138, 479-494.\n\n M. Sambridge (1999b). Geophysical inversion with a neighbourhood algorithms II - Appraising the ensemble - Geophys. J. Int., 138, 727-746." ]

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[ "", null, "# Student Solutions Manual Part 2 for University Calculus (1st Edition) Edit edition Problem 14E from Chapter 1.1: Find the domain and graph the functions in Exercises 15–20.\n\nWe have solutions for your book!\nChapter: Problem:\n\nFind the domain and graph the functions in Exercises 15–20.", null, "Step-by-step solution:\nChapter: Problem:\n• Step 1 of 3\n\n2800-1.1-14E SA: 9451\n\nThe formula", null, "gives a real y-value for nonpositive real number x, hence the domain is", null, ".\n\n• Chapter , Problem is solved.\nCorresponding Textbook", null, "Student Solutions Manual Part 2 for University Calculus | 1st Edition\n9780321388506ISBN-13: 032138850XISBN: Authors:\nThis is an alternate ISBN. View the primary ISBN for: University Calculus: Alternate 1st Edition Textbook Solutions" ]

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[ "# Create the ordinal sum of copulas¶\n\nIn this example we are going to create an ordinal sum of copulas.\n\n:\n\nfrom __future__ import print_function\nimport openturns as ot\n\n:\n\n# Create a collection of copulas\ncollection = [ot.GumbelCopula(2), ot.NormalCopula(2)]\n\n:\n\n# Merge the copulas\nbounds = [0.3]\ncopula = ot.OrdinalSumCopula(collection, bounds)\nprint(copula)\n\nOrdinalSumCopula([0, 0.3], GumbelCopula(theta = 2), [0.3, 1], NormalCopula(R = [[ 1 0 ]\n[ 0 1 ]]))\n\n:\n\n# Draw PDF\ngraph = copula.drawPDF(*2)\ngraph.setXTitle('x')\ngraph.setYTitle('y')\ngraph.setLegendPosition('')\ngraph\n\n:", null, "" ]

[ null, "https://openturns.github.io/openturns/1.15/_images/examples_probabilistic_modeling_ordinal_sum_copula_5_0.svg", null ]

[ "# # Unsafe Code in .NET\n\n## # Using unsafe with arrays\n\nWhen accessing arrays with pointers, there are no bounds check and therefore no `IndexOutOfRangeException` will be thrown. This makes the code faster.\n\nAssigning values to an array with a pointer:\n\n``````class Program\n{\nstatic void Main(string[] args)\n{\nunsafe\n{\nint[] array = new int;\nfixed (int* ptr = array)\n{\nfor (int i = 0; i < array.Length; i++)\n{\n*(ptr+i) = i; //assigning the value with the pointer\n}\n}\n}\n}\n}\n\n``````\n\nWhile the safe and normal counterpart would be:\n\n``````class Program\n{\nstatic void Main(string[] args)\n{\nint[] array = new int;\n\nfor (int i = 0; i < array.Length; i++)\n{\narray[i] = i;\n}\n}\n}\n\n``````\n\nThe unsafe part will generally be faster and the difference in performance can vary depending on the complexity of the elements in the array as well as the logic applied to each one. Even though it may be faster, it should be used with care since it is harder to maintain and easier to break.\n\n## # Using unsafe with strings\n\n``````var s = \"Hello\"; // The string referenced by variable 's' is normally immutable, but\n// since it is memory, we could change it if we can access it in an\n// unsafe way.\n\nunsafe // allows writing to memory; methods on System.String don't allow this\n{\nfixed (char* c = s) // get pointer to string originally stored in read only memory\nfor (int i = 0; i < s.Length; i++)\nc[i] = 'a'; // change data in memory allocated for original string \"Hello\"\n}\nConsole.WriteLine(s); // The variable 's' still refers to the same System.String\n// value in memory, but the contents at that location were\n// changed by the unsafe write above.\n// Displays: \"aaaaa\"\n\n``````\n\n## # Unsafe Array Index\n\n``````void Main()\n{\nunsafe\n{\nint[] a = {1, 2, 3};\nfixed(int* b = a)\n{\nConsole.WriteLine(b);\n}\n}\n}\n\n``````\n\nRunning this code creates an array of length 3, but then tries to get the 5th item (index 4). On my machine, this printed `1910457872`, but the behavior is not defined.\n\nWithout the `unsafe` block, you cannot use pointers, and therefore cannot access values past the end of an array without causing an exception to be thrown.\n\n#### # Remarks\n\n• In order to be able to use the `unsafe` keyword in a .Net project, you must check \"Allow unsafe code\" in Project Properties => Build\n• Using unsafe code can improve performance, however, it is at the expense of code safety (hence the term `unsafe`).\n\nFor instance, when you use a for loop an array like so:\n\n``````for (int i = 0; i < array.Length; i++)\n{\narray[i] = 0;\n}\n\n``````\n\n.NET Framework ensures that you do not exceed the bounds of the array, throwing an `IndexOutOfRangeException` if the index exceeds the bounds.\n\nHowever, if you use unsafe code, you may exceed the array's bounds like so:\n\n``````unsafe\n{\nfixed (int* ptr = array)\n{\nfor (int i = 0; i <= array.Length; i++)\n{\n*(ptr+i) = 0;\n}\n}\n}\n\n``````" ]

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[ "--- title: \"Drawing a Bratteli graph with R\" author: \"Stéphane Laurent\" date: \"14/08/2015\" output: html_document: keep_md: yes --- {r setup, include=FALSE} library(knitr) opts_chunk$set(fig.path=\"assets/fig/bratteli-\") knit_hooks$set(plot = function(x, options) { paste('", null, "', options$fig.cap, '', #\nsep = '') }) fscale <- 1.8 {r functions, include=FALSE} source(\"./assets/Rfunctions/Bgraph.R\") Bgraph <- Bgraph_v1 source(\"./assets/Rfunctions/utils_latex.R\") Pascal_Mn <- function (n) { M <- matrix(0, nrow = n + 1, ncol = n + 2) for (i in 1:(n + 1)) { M[i, ][c(i, i + 1)] <- 1 } return(M) } *Hey, ergodicians!* How do you draw your Bratteli graphs ? Are you using Xfig? Are you typing raw pstricks or TikZ code? Are you crazy? I have written the Bgraph R function, and it does the job without pain for you (feel free to take it [here](./assets/Rfunctions/Bgraph_v1.R)). You just have to write a R function returning the incidence matrices of the graph. ### Bratelli graphs - The Pascal example A *Bratteli graph*, such as the *Pascal graph* shown below, is a graded graph whose edges only connect vertices from one level to some vertices of the next level. {r, echo=FALSE, fig.width=fscale*3, fig.height=fscale*3, fig.cap=\"Pascal graph\"} par(mar=c(.1,.1,.1,.1)) Bgraph(Pascal_Mn, N=3, path=c(1,0,1), first_vertex=1) Such a graph is defined by a sequence of *incidence matrices* $M_n$. Denoting by $c_n$ the number of vertices at level $n$, the incidence matrix $M_n$ is a $c_n \\times c_{n+1}$ matrix showing all connections between level $n$ and level $n+1$. A \"$0$\" means there's no edge, a \"$1$\" means there's one edge, a \"$2$\" means there's a double edge, etc. The first three incidence matrices of the Pascal graph are {r, echo=FALSE, results='asis'} cat(\"$$\", \"M_0=\", latex_matrix(Pascal_Mn(0)), \", \\\\, M_1=\", latex_matrix(Pascal_Mn(1)), \", \\\\, M_2=\", latex_matrix(Pascal_Mn(2)), \".$$\") and I get them with this function: {r} Pascal_Mn <- function(n){ M <- matrix(0, nrow=n+1, ncol=n+2) for(i in 1:(n+1)){ M[i, ][c(i, i+1)] <- 1 } return(M) } Given a function fun_Mn taking a nonnegative integer n as argument and returning an incidence matrix, such as the Pascal_Mn function, my Bgraph function, based on the [diagram package](http://www.inside-r.org/packages/cran/diagram), draws the corresponding Bratteli graph from the root level to a desired level N. Its arguments are: {r} formalArgs(Bgraph) The effects of most of the arguments will be illustrated below. The ellipsis ... is intended for additional arguments to the [coordinates](http://www.inside-r.org/packages/cran/diagram/docs/coordinates) function of the diagram package. For example the hor argument allows to rotate the picture. The figure above has been generated by this code, except that this time we change its orientation with the hor argument: {r} par(mar=c(.1,.1,.1,.1)) Bgraph(Pascal_Mn, N=3, path=c(1,0,1), first_vertex=1, hor=FALSE) The path shown in blue on the figure, is given as the sequence of labels on the edges of this path. The first_vertex argument, intended to be 0 or 1, controls the label of the first vertex at each level. The user can decide to show the edge labels of the blue path only with the labels_path argument. By setting the only_end argument to TRUE, only the vertex labels at the last level are shown: {r} par(mar=c(.1,.1,.1,.1)) Bgraph(Pascal_Mn, N=3, path=c(1,0,1), first_vertex=0, labels_path=TRUE, only_end=TRUE) By setting the USE.COLNAMES argument to TRUE, the vertex labels appearing at level $n$ are the column names of $M_n$. For example, on figure below we display the binomial numbers on the vertices Pascal graph, which give the number of paths from the root vertex. We also show the effect of the ellipse_vertex argument. {r} Pascal_Mn <- function(n){ M <- matrix(0, nrow=n+1, ncol=n+2) for(i in 1:(n+1)){ M[i, ][c(i, i+1)] <- 1 } colnames(M) <- choose(n+1, 0:(n+1)) return(M) } par(mar=c(.1,.1,.1,.1)) Bgraph(Pascal_Mn, N=4, path=c(1,0,1,1), labels_path=TRUE, USE.COLNAMES=TRUE, ellipse_vertex=TRUE) ### Odometers Bratteli graphs are well-known in ergodic theory since Vershik has shown that every invertible measure-preserving transformations can be represented as a transformation on the set of paths of such a graph. The canonical Bratteli graph of an odometer is given by incidence matrixes full of \"$1$\": {r} Odometer_Mn <- function(sizes){ sizes <- c(1,sizes) function(n){ return(matrix(1, nrow=sizes[n+1], ncol=sizes[n+2])) } } {r odometer, fig.width=fscale*3, fig.height=fscale*3, fig.cap=\"(3,4,5)-ary odometer\"} par(mar=c(.1,.1,.1,.1)) fun_Mn <- Odometer_Mn(c(3,4,5)) Bgraph(fun_Mn, N=3, labels_vertex=TRUE, path=c(2,1,2), labels_path=TRUE) This graph is related to [Cantor expansions](http://stla.github.io/stlapblog/posts/Greedy.html). For the previous example, the paths starting from the root level and going to the third level provide a representation of the Cartesian product $\\{0,1,2\\}\\times\\{0,1,2,3\\}\\times\\{0,1,2,3,4\\}$. ### Homogeneous trees A homogeneous tree is a Bratteli graph. I use a trick to generate the incidence matrices, [the same I already used before](http://stla.github.io/stlapblog/posts/KantorovichWithR.html). {r} Tree_Mn <- function(sizes){ function(n){ if(n==0) return(matrix(1, ncol=sizes)) unname(t(model.matrix(~0+gl(prod(sizes[1:n]),sizes[n+1])))[,]) } } As for the odometer, the paths of the tree provide a representation of a Cartesian producct, but in less compact form: {r tree, fig.width=fscale*3, fig.height=fscale*3, fig.cap=\"(3,4,5)-ary tree\"} par(mar=c(.1,.1,.1,.1)) fun_Mn <- Tree_Mn(c(3,4,5)) Bgraph(fun_Mn, N=3, labels_vertex=FALSE, labels_edges=FALSE, path=c(2,1,2)) ### Conversion to TikZ Mathematicians like $\\LaTeX$ figures. The tikzDevice package allows to convert any R figure to a TikZ figure. For example we generate below the Pascal graph with the binomial numbers ${n \\choose k}$ as vertex labels. We set the argument LaTeX to TRUE in the Bgraph function to generate edge labels in $\\LaTeX$ math mode. {r, eval=!file.exists(\"assets/fig/bratteli-Pascal.pdf\"), message=FALSE, results='hide'} Pascal_Mn <- function(n){ M <- matrix(0, nrow=n+1, ncol=n+2) for(i in 1:(n+1)){ M[i, ][c(i, i+1)] <- 1 } colnames(M) <- sprintf(\"${%s \\\\choose %s}$\", n+1, 0:(n+1)) return(M) } # convert to TikZ code: library(tikzDevice) texfile <- \"bratteli-Pascal.tex\" tikz(texfile, standAlone=TRUE, packages=c(getOption(\"tikzLatexPackages\"), \"\\\\usepackage{amsmath}\\n\\\\usepackage{amssymb}\\n\")) par(mar=c(.1,.1,.1,.1)) Bgraph(Pascal_Mn, N=3, path=c(1,0,1), labels_path=TRUE, USE.COLNAMES=TRUE, LaTeX=TRUE, label_root=\"$\\\\varnothing$\", relsize=.6) dev.off() # convert to pdf: tools::texi2dvi(texfile, pdf=TRUE, clean=TRUE) # crop the figure (remove white margins): knitr::plot_crop(\"bratteli-Pascal.pdf\") {r, echo=FALSE, eval=!file.exists(\"assets/fig/bratteli-Pascal.pdf\"), echo=FALSE} file.copy(\"bratteli-Pascal.tex\", \"assets/fig/bratteli-Pascal.tex\") file.copy(\"bratteli-Pascal.pdf\", \"assets/fig/bratteli-Pascal.pdf\") file.remove(\"bratteli-Pascal.tex\", \"bratteli-Pascal.pdf\") This is the result: ### Multiples edges My Bgraph function currently handles double edges too. Let us try the Pascal graph with some double edges taken at random. {r doubleedge} Pascal2_Mn <- function(n){ M <- matrix(0, nrow=n+1, ncol=n+2) for(i in 1:(n+1)){ M[i, ][c(i, i+1)] <- sample.int(2, 1) } return(M) } set.seed(666) par(mar=c(.1,.1,.1,.1)) Bgraph(Pascal2_Mn, N=4, labels_vertex=FALSE, labels_path=TRUE, path=c(1,0,1,0)) Currently, the rendering of the colored path is not correct, because the two edges of a double edge appears in color. The label edges are not correct too. This will be fixed in a next version of the Bgraph function." ]

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[ "This page is a snapshot from the LWG issues list, see the Library Active Issues List for more information and the meaning of Resolved status.\n\n### 1465. Missing arithmetic operators for atomic_address\n\nPriority: Not Prioritized\n\nView all other issues in [atomics.types.address].\n\nView all issues with Resolved status.\n\nDiscussion:\n\natomic_address has operator+= and operator-=, but no operator++ or operator--. The template specialization atomic<Ty*> has all of them.\n\nAccepting n3164 would solve this issue by replacing atomic_address by atomic<void*>.\n\n[ Resolved in Batavia by accepting n3193. ]\n\nProposed resolution:\n\n```namespace std {\n[…]\nvoid* operator=(const void*) volatile;\nvoid* operator=(const void*);\nvoid* operator++(int) volatile;\nvoid* operator++(int);\nvoid* operator--(int) volatile;\nvoid* operator--(int);\nvoid* operator++() volatile;\nvoid* operator++();\nvoid* operator--() volatile;\nvoid* operator--();\nvoid* operator+=(ptrdiff_t) volatile;\nvoid* operator+=(ptrdiff_t);\nvoid* operator-=(ptrdiff_t) volatile;\nvoid* operator-=(ptrdiff_t);" ]

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[ "# 萊昂哈德·歐拉\n\n18世紀瑞士數學家\n（重定向自莱昂哈德·欧拉", null, "", null, "舊瑞士邦聯巴塞尔", null, "俄罗斯帝国圣彼得堡", null, "俄罗斯帝国", null, "舊瑞士邦聯", null, "## 生平\n\n### 在圣彼得堡\n\n1734年1月7日，欧拉迎娶了科学院附属中学的美术教师，瑞士人乔治·葛塞尔（Georg Gsell）的女儿，柯黛琳娜·葛塞尔（Katharina Gsell，1707-1773） ，两人共育有13个子女，其中仅有5个活到成年\n\n### 其他\n\n1783年9月18日，晚餐后，欧拉一边喝着，一边和小孙女玩耍，突然之间，煙斗从他手中掉了下来。他说了一声：“我的烟斗”，并弯腰去捡，结果再也没有站起来，他抱着头说了一句：“我死了”。「欧拉停止了计算生命」。后面这句经常被数学史家引用的话，出自法国哲学家兼数学家孔多塞之口：「...il cessa de calculer et de vivre」（he ceased to calculate and to live）。\n\n## 成就\n\n$\\zeta (2)=\\sum _{n=1}^{\\infty }{\\frac {1}{n^{2}}}={\\frac {1}{1^{2}}}+{\\frac {1}{2^{2}}}+{\\frac {1}{3^{2}}}+{\\frac {1}{4^{2}}}+\\cdots ={\\frac {\\pi ^{2}}{6}}$\n\n$e^{ix}=\\cos(x)+i\\sin(x)\\,$\n\n$e^{i\\pi }=-1\\,$ $e^{i\\pi }+1=0\\,$\n\n$\\gamma =\\lim _{n\\rightarrow \\infty }\\left(1+{\\frac {1}{2}}+{\\frac {1}{3}}+{\\frac {1}{4}}...+{\\frac {1}{n}}-\\ln(n)\\right)$\n\n$F-E+V=2\\,$\n\n$F-E+V=\\chi (M)\\,$\n\n## 引述评价\n\n• “读欧拉的原著吧：在任何意义上，他都是我们的大师。”—拉普拉斯\n\n## 註釋\n\n1. ^ 法语原文为：Lisez Euler, lisez Euler, c'est notre maître à tous.\n2. ^ 另一說865篇\n3. ^ 另一說31部\n\n## 參考資料\n\n1. ^ Dan Graves. Scientists of Faith. Grand Rapids, MI: Kregel Resources. 1996: 85–86.\n2. ^ E. T. Bell. Men of Mathematics, Vol. 1. London: Penguin. 1953: 155.\n3. ^ Dunham, William. Euler: The Master of Us All. The Mathematical Association of America. 1999: pp. 17.\n4. ^ Dunham, William. Euler: The Master of Us All. The Mathematical Association of America. 1999. xiii.\n5. John, Stillwell. Mathematics and its History. Springer. 2001: pp. 188. ISBN 0-387-95336-1.\n6. ^ James, Ioan. Remarkable Mathematicians: From Euler to von Neumann. Cambridge. 2002: 2. ISBN 0-521-52094-0.\n7. ^\n8. ^ Calinger, Ronald. Leonhard Euler: The First St. Petersburg Years (1727–1741). Historia Mathematica. 1996, 23 (2): pp. 156. doi:10.1006/hmat.1996.0015.\n9. ^ Calinger, Ronald. Leonhard Euler: The First St. Petersburg Years (1727–1741). Historia Mathematica. 1996, 23 (2): pp. 125. doi:10.1006/hmat.1996.0015.\n10. ^ Calinger, Ronald. Leonhard Euler: The First St. Petersburg Years (1727–1741). Historia Mathematica. 1996, 23 (2): 127. doi:10.1006/hmat.1996.0015.\n11. ^ Calinger, Ronald. Leonhard Euler: The First St. Petersburg Years (1727–1741). Historia Mathematica. 1996, 23 (2): pp. 124. doi:10.1006/hmat.1996.0015.\n12. ^ Calinger, Ronald. Leonhard Euler: The First St. Petersburg Years (1727–1741). Historia Mathematica. 1996, 23 (2): pp. 128–129. doi:10.1006/hmat.1996.0015.\n13. ^ Gekker, I.R.; Euler, A.A. Leonhard Euler's family and descendants. (编) Bogoliubov, N.N.; Mikhaĭlov, G.K.; Yushkevich, A.P. Euler and modern science. Mathematical Association of America. 2007. ISBN 088385564X., p. 402.\n14. ^ Fuss, Nicolas. Eulogy of Euler by Fuss (dmy). [2006-08-30].\n15. ^\n16. ^ Dunham, William. Euler: The Master of Us All. The Mathematical Association of America. 1999. xxiv–xxv.\n17. ^ Finkel, B.F. \"Biography- Leonard Euler\". The American Mathematical Monthly. 1897, 4 (12): 300. doi:10.2307/2968971. JSTOR 2968971 请检查|doi=值 (帮助).\n18. ^ Swiss 1.2-metre Leonhard Euler Telescope\n19. ^ Leonhard Euler's 306th Birthday. Google. [2014-03-02] （英语）." ]

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[ "# AERO – 421 Final Project Redevelopment\n\nSpacecraft Attitude, Dynamics, and Control: Simulation to determine and control a satellite’s attitude in LEO.\n\n## Background\n\nAERO-421, or Spacecraft Attitude, Dynamics, and Controls, is a class taught at the California Polytechnic State University at San Luis Obispo (Cal Poly SLO) which serves as an…\n\n“Introduction to spacecraft attitude dynamics and control… [and] fundamentals of guidance and navigation systems… [with emphasis in] analysis and design of control systems for aerospace vehicles.” – Cal Poly Aerospace Engineering Course Catalog\n\nThe final project in the course was to develop a simulation to determine a satellite’s attitude in Low Earth Orbit (LEO), consider and model detumbling from a launch vehicle, consider and model disturbances due to external forces i.e., Solar Radiation Pressure (SRP), and to consider and model control via onboard reaction wheels.\n\nThe initial project was developed in MATLAB, however, the project will be completely redeveloped in Python to showcase controls and software development skillsets.\n\n## Part 0: Context and Given Data\n\nThe project will explore modeling and simulation of the various stages of a spacecraft mission, specifically simulating the attitude dynamics from initial spacecraft deployment to operation. In this simulation, the spacecraft is an Earth observing satellite and an attitude determination and control system must be designed using reaction wheels to ensure the spacecraft maintains pointing nadir.\n\nOrbital Data:\n\n• h (angular momentum) = 53335.2 km^2/s\n• e (eccentricity) = 0\n• Ω (Right Ascension of Ascending Node) = 0 deg\n• i (inclination) = 98.43 deg\n• ω (Argument of Perigee) = 0 deg\n• θ (True Anomaly) = 0 deg\n• ϵ-LVLH (Initial Quaternion relating the body to the LVLH frame) = [0, 0, 0]; η = 1\n\nDetumble Phase:\n\n• Spacecraft is a square box with 2 meters on each edge with total mass of 640 kg\n• Initial Angular Velocity of [-0.05, 0.03, 0.2] rad/s relating the body to the ECI frame\n\nNormal Operations:\n\n• Spacecraft bus is a 2 meter cube with mass of 500 kg. The center of mass of the spacecraft is located at the geometric center of the bus.\n• A rectangular sensor is attached to the face pointing towards the Earth (+Z-axis) and is 1 meter long and 0.25 meters square. The sensor has a mass of 100 kg.\n• Two solar panels are deployed along the +/- Y-axis and are constrained to rotate about the +/- Y-axis. The solar panels are 3 meters long (in the Y-axis), 2 meters wide, and 0.05 meters thick. Each panel has a mass of 20 kg and the center of mass is located at the geometric center of the panel. The solar panels do not rotate relative to the spacecraft bus.\n• Assume all spacecraft components have uniform density with centers of mass located at the geometric centers of each component\n• Magnetically, the spacecraft residual magnetic dipole moment can be modeled as pointing in the -Z direction with magnitude 0.5 A-m^2\n• See the figure below for the spacecraft schematic\n• Because the thrusters are not actually fully-modulated thrusters, the spacecraft will have a residual angular velocity of [0.001, -0.001, 0.002] rad/s relating the body to the ECI frame after the detumble phase.\n• During operation the spacecraft is required to point at the target on the ground to within 0.001 degrees 3-sigma using the reaction wheels used in the reaction wheels part.", null, "## Part 1: Mass Properties\n\nDetermine the mass and inertial properties of the spacecraft for both the detumble and the normal operations phases.\n\nOutputs:\n\n• Total mass of the spacecraft\n• Center of mass relative to the spacecraft bus center of mass. The body frame will be located at the center of mass of the whole spacecraft\n• Intertia matrix of the whole spacecraft about the center of mass of the spacecraft\n\n## Part 2: Torque Free Motion\n\nModel the torque free orbital and attitude motion of the spacecraft\n\nOutputs: Plots for…\n\n• Euler angles and quaternions relating body to ECI reference frames\n• Angular velocity of the spacecraft in body components for one orbit of the normal operations phase\n\n## Part 3: Detumble\n\nSimulate the motion of the satellite during the detumble phase. Assume fully modulated thrusters and use direct velocity feedback\n\nOutputs: Plots for…\n\n• Euler angles and quaternions relating body to ECI reference frames\n• Angular velocity of the spacecraft in body components for the detumble phase\n• Torque components in the body frame\n\n## Part 4: Disturbance Simulation\n\nAdd the four disturbance models to the simulation:\n\n• Atmospheric Drag\n• Solar Pressure\n• Earth Magnetic Field\n\nUse the following model for the atmospheric density. Notice that h is the height above the Earth’s surface in kilometers where R_Earth equals 6378km", null, "Consider the simulation epoch to be March 20, 2021. Disregard any variations of the ECI representation of the sunlight direction during the simulation.\n\nOutputs: Plots for…\n\n• Euler angles and quaternions relating the body to the ECI reference frame\n• Euler angles and quaternions relating the body to the LVLH reference frame\n• Angular velocity of the spacecraft relative to the ECI frame expressed in body components\n• Angular velocity of the spacecraft relative to the LVLH frame expressed in body components\n• Torque components for atmospheric drag, solar radiation pressure, gravity gradient, and earth magnetic field\n\n## Part 5: Reaction Wheel Control\n\nDetermine the control gains for a full state feedback 3-axis reaction wheel control system. Use the requirements of ζ = 0.65 and t_s = 30 sec\n\nThe positions of the 3 reaction wheels are [1, 0, 0], [0, 1, 0], and [0, 0, 1]. Each reaction wheel can be modeled as a simple cylinder with radius of 0.3 m and a height of 0.02 m\n\nOutputs: Plots for…\n\n• Euler angles and quaternions relating the body to ECI reference frame\n• Euler angles and quaternions relating the body to LVLH reference frame\n• Angular velocity of the spacecraft relative to the ECI reference frame expressed in body components\n• Angular velocity of the spacecraft relative to LVLH frame expressed in body components\n• Commanded moment from the determined control law\n• Wheel speed of each reaction wheel\n\n## Part 6: Visualization\n\nDetermine and animate the quanterions of the spacecraft, from ECI to body frame, for the duration of 1-5 revolutions.\n\nOutput:\n\n• Table of quaternion and time data\n• Video or other animation file to show the revolution of the spacecraft\n\nView Github" ]

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[ "# Influence Lines", null, "In bridge design, it is very necessary that bridge decks support both static and moving loads. Every component of a bridge needs to be designed to withstand the worst loading effect that could possibly arise in that part. In effect, traffic live loads should be placed where they will cause the most onerous effect on the structure. ‘Influence lines’ are a helpful tool for assessing the most severe loading condition.\n\nAn influence line represents the response of a specific element of a bridge to the effect of a moving load, such as reaction, shear force, bending moment, or axial force. It is a diagram where the value of the response at any point is equal to the effect caused by a unit load applied at that point. Influence lines provide a systematic approach for determining how the force, moment, or shear in a specific part of a structure changes as the applied load moves across the structure.\n\nInfluence lines for statically determinate structures consist of straight lines, but for indeterminate structures, they can have more complex shapes. The primary purpose of influence lines is to identify where to place live loads to achieve maximum effect.\n\n## Influence Lines for Shear in Simple Beams\n\nInfluence lines for shear at two sections of a simply supported beam are shown in Figure 1. When the summation of transverse forces to the left of a section is in the upward direction or when the summation of transverse forces to the right of the section are in the downward direction, positive shear is said to occur. For each position of the unit load, the shear force at sections 1-1 and 2-2 is determined by placing the unit force at various points.\n\nThe values provide the ordinate of the influence line, which may be used to draw the influence line diagrams for the shear force at sections 1-1 and 2-2. It should be noted that the slope of the influence line for shear on the left of the section is similar to the slope of the line on the right of the section. In other circumstances, this information can be used to draw the influence lines for shear force.\n\n## Influence Lines for Bending Moments in Simple Beams\n\nFigure 2 shows the influence lines for the bending moment at the same sections, 1-1 and 2-2, of the simple beam under consideration in Figure 1. For a section, the moment is considered positive when the total of all the moments of the forces to the left is clockwise or when the sum to the right is counterclockwise. For various places of unit load, the values of the bending moment at sections 1-1 and 2-2 are obtained and plotted as shown in Figure 2.\n\nIt should be noted that a shear or bending moment diagram shows how shear or moment values change throughout the entire structure when loads are fixed in a specific position. Conversely, an influence line for shear or moment depicts how that response varies at a specific section of the structure due to the movement of a unit load from one end to the other. Influence lines are helpful in determining the magnitude of a specific response at the section where it is drawn, when the beam is subjected to various types of loading.\n\nFor instance, the shear force at section 1-1 is determined by the product of the load intensity, qo, and the net area under the influence line diagram assuming a uniform load of intensity qo per unit length operates throughout the full length of the basic beam depicted in Figure 22.\n\nSince the net area at section 1-1 is 0.3P, the shear force there is 0.3qoP as well. The bending moment at the section is calculated from the area of the appropriate influence line diagram times the intensity of loading, qo. Therefore, the section’s bending moment is 0.08qoP2.\n\nSolved Example\n\nLet us consider the beam loaded as shown below. It is desirous to obtain the influence line for the support reactions, and for the internal stresses with respect to section 1-1.\n\nIn all cases, we will be taking P as unity (i.e 1.0)\n\n(1) Influence line for support reactions\n\nSupport A\nSupport reaction at point A (FA) = (L – x)/L\nAt x = -L1;\nFA = (L + L1)/L\n\nAt x = 0;\nFA = 1.0\n\nAt x = L + L2;\nFA = (L – L – L2)/L = – L2/L\n\nSupport B\nSupport reaction at point B (FB) =  x/L\nAt x = -L1;\nFB =  -L1/L\n\nAt x = 0;\nFB = 0\n\nAt x = L;\nFB = 1.0\n\nAt x = L + L2;\nFB = (L + L2)/L\n\n(2) Influence line for bending moment with respect to section 1-1\n\n(0  ≤  x  ≤  a)\nM1-1 = FA.a – P(a – x)\nM1-1 = [P(L – x).a]/L – P(a – x)\nBut taking P = 1.0;\n= [(L – x).a]/L – (a – x)\n\nAt x = -L1;\nM1-1 =  [(L + L1).a]/L – (a + L1) = [L1(a – L)/L] =  -L1.b/L\n\nAt x = 0;\nM1-1 =  [(L – 0).a]/L – (a – 0) = [L1(a – L)/L] =  0\n\nAt x = a;\nM1-1 =  [(L – a).a]/L – (a – a) = [L1(a – L)/L] =  a.b/L\n\n(a  ≤   ≤  L)\nM1-1 =   [P(L – x)a]/L\n\nAt x = a;\nM1-1 = [(L – a)a]/L   =  a.b/L\n\nAt x = L;\nM1-1 = [(L – L)a]/L   =  0\n\nAt x = L + L2;\nM1-1 = [(L – L – L2)a]/L   =  – L2a/L\n\n(2) Influence line for shear with respect to section 1-1\n\n(0  ≤  x  ≤  a)\nQ1-1 = P(L – x)/L – P = – FB =  –x/L\n\nAt x = -L1;\nQ1-1 =   L1/L\n\nAt x = 0;\nQ1-1 =   0\n\nAt x = a;\nQ1-1 = -a/L\n\n(a  ≤  x  ≤  L)\nQ1-1 = -(P.x)/L + P = (L – x)/L\n\nAt x = a;\nQ1-1 = b/L\n\nAt x = L;\nQ1-1 = 0\n\nAt x = L + L2;\nQ1-1 = [(L – L – L2)]/L = -L2/L\n\n## Influence Lines for Trusses\n\nInfluence lines for support reactions and member forces can be constructed using the same approach as influence lines for various beam functions. They provide valuable information for determining the maximum load that can be applied to a truss. By analyzing the movement of a unit load across the truss, we can calculate the responses of interest at each panel point.\n\nHowever, it is not necessary to calculate the member forces at every panel point, as certain parts of the influence lines can be identified as straight lines for multiple panels. Method of sections can be used to obtain the member forces in any panel of interest.\n\nThe truss shown above is used as an example to explain how to construct influence lines for trusses. Passing a section 1-1 and taking into account the equilibrium of the free body diagram of one of the truss segments yields the member forces in BD, CE, and BE.\n\nFirst, a unit load is applied to node C, and the force in BD is calculated by calculating the moment about node E of all forces acting on the right-hand segment of the truss, then dividing that moment by the lever arm (the distance at which the force in BD is perpendicular to node E).\n\nThe resultant value provides the influence diagram’s ordinate at C in the truss. Similar to how the force in BD for a unit load imposed at E is represented by the obtained ordinate at E. Two additional points, one at each of the supports, can be added to the influence line to complete it. The relevant influence line diagram can be finished by obtaining the force in the member CE due to the unit load applied at C and E.\n\nThe influence line for force in BE can be obtained by taking into account the horizontal component of force in the diagonal of the panel. The influence diagrams for the member forces in BD, CE, and BE are shown in Figure 3. By running an imaginary vertical section through the panel and taking moments at the junction of the diagonal and the other chord, it is possible to estimate the influence line ordinates for the force in a chord member of a ‘curved-chord’ truss.\n\n## Qualitative influence lines: Müller–Breslau principle\n\nOne of the most effective methods of obtaining influence lines is by the use of the Müller–Breslau principle, which states that ‘the ordinates of the influence line for any response in a structure are equal to those of the deflection curve obtained by releasing the restraint corresponding to this response and introducing a corresponding unit displacement in the remaining structure’.\n\nIn this way, the shape of the influence lines for both statically determinate and indeterminate structures can be easily obtained, especially for beams.\n\nSome methods for drawing influence lines are as follows:\n\nSupport reaction\nRemove the support and introduce a unit displacement in the direction of the corresponding reaction to the remaining structure as shown in Figure 4 for a symmetrical overhang beam.\n\nShear\nMake a cut at the section and introduce a unit relative translation (in the direction of positive shear) without relative rotation of the two ends at the section as shown in Figure 5.\n\nBending moment\nIntroduce a hinge at the section (releasing the bending moment) and apply bending (in the direction corresponding to positive moment) to produce a unit relative rotation of the two beam ends at the hinged section as shown in Figure 6." ]

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[ "## October 24, 2020\n\n### Calculus: TI-Nspire CAS Instructional Videos\n\nUpdate: The CX II CAS version of the calculator is now available; I've updated the Amazon link accordingly.\n\nThe TI-Nspire CX CAS calculator is notoriously difficult to learn. I've posted a list of videos below to help you learn the calculator and will add to the list over time.\n\nKeep the following points in mind if you haven't bought the calculator yet:\n• The Nspire has a regular (CX) version and a more powerful (CX CAS) version. Make sure to get the CAS version, as the regular one doesn't have the algebra-solving features the Nspire is known for.\n• The Nspire will make it easy for you to check your work, as it will solve most math problems if you set them up correctly.\n• Most calculus teachers allow the use of the Nspire, but their tests may include non-calculator sections. In any case, if you need to show your work to receive credit, you'll have to know how to do problems by hand.\n• It's allowed on the SAT and on any SAT Subject Tests and AP tests that will let you use a calculator.\n• It's not allowed on the ACT.\n• The Nspire has a steep learning curve, so don't buy one unless you can devote several weeks to learning it. If you're going to use it on a standardized test, where speed really matters, plan to spend a couple of months getting used to your calculator.\n• You can switch quickly between the main calculator and graphing screens using the button with the picture of a calculator on it. (It's to the left of the left arrow key.)\n• Press the book key (above the division symbol) to get all of the calculator's functions listed in alphabetical order. If you want to learn how to use a function, type the first letter of its name, then scroll down and highlight the function you're interested in. The bottom of the screen will show you what you have to put in between the parentheses to get a function to work.\n• For example, the polyRoots function needs (Poly, Var) inside the parentheses. That means that you have to type a polynomial in, then a comma, and then the name of the variable you want to solve for. The Nspire will automatically set the function to zero and solve for that variable.\n• Menu > Analyze Graph > dy/dx is a quick way to get the derivative at various points of a function you've graphed. Mouse over various points on the graph, and the derivative will show up as a light gray number.\n• The numerical solve feature (Menu > Algebra > Numerical Solve) doesn't work correctly if the equation you're solving has more than one answer. For example, nSolve(x2=81,x) gives you 9 instead of {-9,9}.\n• Texas Instruments publishes an Nspire reference guide that you can download and print out. Open the file and hit control-F on your computer keyboard to search.\n• If you're not planning to take AP Calculus, I suggest getting a different calculator. Check out my calculator review page for more information.\n\nCommonly Used Keys\n• The Catalog key (right above the multiplication key) is an easy way to access the calculator's advanced functions quickly.\n\nThe Blue Keys\n\nThe Pi Key\n\nStoring and Deleting Variables\n• Storing a value into a variable and then typing (or pasting in) an expression is a very fast way to solve problems.\n• If you store a value into x and then take a derivative with respect to x, your calculator will display the answer as a number instead of as an expression in terms of x. This can be a curse rather than a blessing if you actually wanted the value in terms of x.\n• For example, 7→x (7, ctrl, var, x) stores the value 7 into the variable x. Your calculator will think of x as \"7\" instead of thinking about it as a letter.\nTyping 100x will then return 700.\nd/dx (x2) will return 14.\n• If you then erase the variable (Menu > Actions > Clear a-z), your calculator will think of x as a letter again.\nTyping 100x will give you 100•x.\nd/dx (x2) will return 2•x.\n• If your calculator \"breaks\" and refuses to display answers in terms of x (showing you only numbers instead), you probably stored a number in x and forgot to delete the variable.\n• \"If you plan to do symbolic computations using undefined variables, avoid storing anything into commonly used, one-letter variables such as a, b, c, x, y, z, and so on.\" (Nspire reference guide, page 234 / PDF page 238)\n• Sometimes it's hard to tell whether the answer your calculator spits out is the same as what you got solving by hand. If that happens, plug in 2 for x for both the calculator's answer and your own answer and make sure that the results match.\n• You can do this by storing 2→x (2, ctrl, var, x), then scrolling to highlight the answer your calculator gave you, then hitting Enter twice to evaluate that expression with the assumption \"x=2\". Notice that your calculator will spit out an actual number this time instead of an answer in terms of x; you can compare that number to what you'd get by plugging 2 into the answer you derived by hand.\n• Remember to erase the variable (Menu > Actions > Clear a-z) when you're finished! If you forget, your calculator will assume that x=2 for all future calculations, not necessarily an assumption you want to make.\n\nConverting Between Fractions and Decimals\n\nUsing Copy/Paste Features to Avoid Re-Typing Functions\n• The easiest way to copy and paste is to use the \"up\" arrow key to scroll up and highlight some text. Hit the Enter key in order to automatically copy and paste that text into a new line that you can then edit.\n• Control (blue key)-C copies text, and control-V pastes it. Your calculator's keyboard isn't labeled with copy and paste functions, but they function like they would for a computer. It's a useful way to copy text and paste it inside an integral or derivative symbol.\n• Watch the video below if you're interested in using Copy/Paste features to store notes on your calculator.\n\nDefining a Function and Finding Its Derivative\n\nFinding the Equation of a Function's Tangent Line\n\nImplicit Differentiation\n\n• Note: You can also do implicit differentiate without using the impDif function. Use the Catalog key (located above the multiplication key), choose d/dx, and type your equation in.\n• For example, you can implicitly differentiate the volume of a cylinder with respect to time this way:\nd/dx ( v(t) = πr(t)2h(t) )\n\nYou must remember to explicitly state which letters are functions. For example, v(t) tells your calculator that v is a function of t. The example below will not work, since we've forgotten to tell the calculator that v, r, and h are functions of t:\nd/dx ( v = πr2h )\n• If you do implicit differentiation this way, you must remember to clear your variables using Menu > Actions > Delete Variable or Menu > Actions > Clear a-z. If you have a number stored in the letter v, for example, and haven't cleared variables from your calculator's memory, your calculator will think of v as a number and not a letter even if you explicitly type v(t).\n• Sometimes your calculator will give you a huge mess as the answer. You can either algebraically rearrange the mess to confirm that it's the same as the answer you got when you solved by hand, or you can plug a number like 2.5 into both your calculator's answer and your own answer to make sure you get the same answer.\n• It's relatively easy to plug numbers in once your calculator has given you an answer. Store 2.5→x using cntrl-var, then use the arrow keys to highlight the expression you want to copy, and hit the enter key to paste it. Your calculator will re-evaluate the expression assuming that x=2.5.\n• You must remember to delete the variable when you're finished (Menu > Actions > Clear a-z). If you don't, your calculator will assume that you mean 2.5 every time you type x, a result you almost certainly don't want.\n\nDifferential Equation Solver\n\n1.", null, "1.", null, "" ]

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[ " Parameter Estimation in Different Enzyme Reactions\n\nVol.2 No.1(2014), Article ID:43903,13 pages DOI:10.4236/aer.2014.21002\n\nParameter Estimation in Different Enzyme Reactions\n\nSudarshan R. Nelatury1, Charles F. Nelatury2, Mary C. Vagula3\n\n1Department of ECE, School of Engineering, Penn State Erie, Erie, USA\n\n2Department of ECE, College of Engineering, Drexel University, Philadelphia, USA\n\n3Biology Department, Gannon University, Erie, USA\n\nEmail: [email protected]", null, "", null, "Received 31 October 2013; revised 7 December 2013; accepted 25 December 2014\n\nABSTRACT\n\nEnzyme kinetic parameters have been estimated using MATLAB software via the Wilkinson nonlinear regression technique. The MATLAB script file written to implement this technique is short and very straightforward. Several software tools are commercially available for this purpose, with many graphical user interface (GUI) features. A routine use of these packages might offer immediate satisfaction of interactive hands-on experience; but in some cases the researcher might wish to write his/her own code and compare the results for further confirmation. Today MATLAB is in use in almost all the schools and laboratories as a standard software tool. So this paper is aimed at helping enzyme researchers to make use of this powerful software for estimation of parameters. It enables the incorporation of the analytical steps behind parameter estimation in an easy-to-follow manner and furnishes better visualization.\n\nKeywords:Michaelis-Menten Equation; Dixon Plot; Hill Equation; Wilkinson Nonlinear Regression; MATLAB", null, "1. Introduction\n\nOver the past several years, different enzyme reactions governed by different equations have been identified in the field of biochemistry. Estimation of kinetic parameters of these enzyme reactions is one of the important problems - . The nonlinearity in the functional expression for the reaction rate, in terms of substrate concentration and other parameters, makes the parameter estimation problem difficult. Graphical techniques such as Lineweaver-Burk’s method and the use of Dixon plot to estimate parameters - were in vogue for some time. These methods rely on linearizing the Michaelis-Menten equation in the case of simple enzyme reactions. For complex enzyme reactions, such techniques do not furnish reliable results. This has prompted Wilkinson to propose non-linear regression for the estimation of enzyme parameters . Using Wilkinson non-linear regression coupled with a modified Fibonacci search method, Atkins outlined a computer program for fitting the Hill equation directly to experimental data . Following this, numerous software packages have been developed - to perform parameter estimation interactively. Most of these packages have fixed menus and do not allow the user to incorporate possible changes to the underlying algorithm.\n\nThe goal of this article is to explain the nonlinear regression technique briefly and show four simulation examples of enzyme reactions. The programming is done in MATLAB, a registered trade mark of MathWorks, Inc. MATLAB is a high-level language and interactive environment that enables one to perform computationally intensive tasks faster than with traditional programming languages. The syntax of various functions in MATLAB makes implementing difficult, mathematical steps very straightforward.\n\nToday, use of MATLAB has almost become a de facto standard among engineers and computer scientists both in academia and industry. Faculty and students of biochemistry can easily learn this software using numerous tutorials available on-line or introductory books such as - . Further, it is easy for biologists to develop interaction with other engineers and scientists in the language of MATLAB. In this paper we shall discuss the basic principle behind nonlinear regression, and supplement it with simulation examples1.\n\n2. Parameter Estimation via Nonlinear Regression\n\nOne of the challenges faced in a biochemistry laboratory while studying enzyme kinetics is to decide comprehensive levels of the substrate concentration. For example, it is hard to determine the maximum concentration, minimum concentration, and to know how best to space the intermediate levels and so on. In practice, one may have to plot the data and go through a few iterations before endorsing the data for further processing. After satisfactory acquisition of the data, which invariably carries measurement inaccuracies, enzyme parameters appearing in the governing model can be estimated using the non-linear regression technique.\n\nLet us consider a functional relationship that expresses the reaction rate v in terms of the independent variable(s) and some parameters. For example, in the case of a simple enzyme-substrate uninhibited reaction it is given by the Michaelis-Menten equation:", null, ". (1)\n\nIn Equation (1) above, the substrate concentration s is the independent variable, and", null, "and", null, "are enzyme parameters representing the maximum reaction velocity and the Michaelis constant respectively. Numerically,", null, "equals s when", null, ". In general, there may be more independent variables than s alone, which for convenience, could be put in a vector of N elements say,", null, ". Let the vector", null, "denote a set of P parameters. Here the superscript T denotes the transpose. This generalization would become clearer after three more examples that are discussed in this paper. The dependence of v on x and β is denoted by", null, ". Let", null, "stand for the mth data sample. The subscript m in all occurrences except in the Michaelis constant", null, "indicates the mth observation. Since measurements are subject to experimental errors, we shall write:", null, "(2)\n\nwhere the measurement error,", null, ", is assumed to be a normally distributed random variable with a mean zero and a variance of", null, ". Given a set of measured values of Equation (2), the parameter estimation problem involves finding an estimate of β that reduces a measure of error, usually the sum of weighted (with the weights wm) residual squares WRSS, denoted by", null, ":", null, ". (3)\n\nOur objective is to minimize", null, "with respect to β. It is hard to find the minimum of", null, "by explicitly differentiating and solving for global minimum. Iterative methods are usually employed. Nonlinear regression is a convenient technique that allows us to estimate the unknown parameters by linearizing Equation (2) at each iteration starting from an initial guess", null, ". This is done by first expressing", null, "with the first two terms in its Taylor’s series expansion around an estimate, initially taken as", null, ". Then the mth element in Equation (2) can be written as:", null, "(4)\n\nhere", null, "is assumed to be a small increment in the parameter vector. This can be done for all the observations m = 1 to M.\n\nNow treating the vector", null, "as the residual", null, "at 0th iteration, we can write:", null, "(5)\n\nThis process can be iterated by replacing 0 and 1 with k and k + 1 respectively. Denoting the matrix involving the derivatives in Equation (5) at kth iteration by Gk we get,", null, ". (6)\n\nThe weighted least squares solution for this can be written in terms of generalized inverse of Gk as:", null, "(7)\n\nwhere", null, "(8)\n\nis the covariance matrix of", null, ".\n\nEquation (7) helps in estimating the parameters of a given model recursively; but is in a generalized form. For the enzyme researchers or students to be able to code this in MATLAB, they need to understand each symbol in (7) and how to express the same for a given model. In the next section four types of enzyme models are considered and expressions for various elements of", null, ",", null, ", and", null, "are given as they relate to each model. In Appendix I, MATLAB code is provided for Example III. For other examples, the same code could be edited with minimum effort by replacing the expressions of these vectors/matrices appropriately and could be executed. With this idea in mind, in the next section, Equation (7) is specialized for four different models.\n\n3. Four Models\n\n3.1. Model I: A Simple Uninhibited Enzyme Substrate Reaction\n\nLet us first consider the case of a simple uninhibited enzyme substrate reaction. As mentioned earlier in the paper, s denotes the concentration of the substrate, Vmax the maximum reaction velocity, and Km the Michaelis constant, which equals the substrate concentration at half the maximum rate.\n\nGoverning equation (Michaelis-Menten equation):", null, "(9)\n\nVariables and parameters:", null, "(10)\n\nThe mth element of the residue:", null, "(11)\n\nElements of the mth row of", null, "matrix:", null, "(12)\n\n3.2. Model II: Competitive Inhibitor Enzyme Substrate Reactions\n\nWith other symbols remaining the same as in Model I, let i denote the concentration of inhibitor. The competitive reaction is governed by the Michaelis-Menten equation:", null, "(13)\n\nVariables and parameters:", null, "(14)\n\nThe mth element of the residue:", null, "(15)\n\nElements of the mth row of", null, "matrix:", null, "(16)\n\nIn passing, it might be mentioned that if the data points are collected for various inhibitor values, one could vectorize them into a single column. Accordingly, the value for i in Equation (16) should be substituted.\n\n3.3. Model III: Allosteric Monosubstrate Enzyme\n\nFor our third model, the allosteric enzyme reaction is considered. This is governed by the Hill equation involving an empirical parameter n.", null, "(17)\n\nVariables and parameters:", null, "(18)\n\nThe mth element of the residue:", null, "(19)\n\nElements of the mth row of", null, "matrix:", null, "(20)\n\n3.4 Model IV: Allosteric Enzyme with Two Species of Ligand Sites\n\nThis is governed by two Hill equations combined as:", null, "(21)\n\nVariables and parameters:", null, "(22)\n\nThe mth element of the residue:", null, "(23)\n\nThe six elements of the mth row of", null, "matrix:", null, "(24)\n\nIn the next section four simulation examples are given based on the preceding four models.\n\n4. Simulation Examples\n\n4.1 Example I―Simple Uninhibited Reaction\n\nData consisting of 10 samples are simulated in case of a hypothetical enzyme with Km = 240 nM and Vmax = 80 mmol/min/mg substrate using MATLAB. A zero mean unit variance random number with Gaussian density is added as measurement error. The substrate and the reaction rate are shown in Table 1. We shall take ∑ as a 10 × 10 unit matrix. Thus, WRSS can be called RSS.\n\nThe parameters estimated using the program written in MATLAB are:", null, ",", null, ", and the RSS = 0.142. Figure 1 shows the enzyme reaction rate versus substrate concentration in the units specified in Table 1. In order to render a better visualization, we have provided the relief of the error surface defined in Equation (2) with unity weights in Figure 2(a). Also, Figure 2(b) shows the contours of the same plot. Note that the contours are ever widening ellipses with the global minimum shown by an asterisk. Also plotted in this figure is the learning trajectory obtained by joining all the estimated parameters over 6 iterations, starting from a provisional guess of Km(0) = 20 and Vmax(0) = 25.\n\nThis simulation is repeated using the software Prism 5, a registered trademark of GraphPad, as well as the EXCEL template called anemona published in . It was assuring to find that the parameters estimated were identical viz.,", null, ", and", null, ".\n\nTable 1. Simulated data consisting of 10 samples in case of a hypothetical enzyme with Km = 240 nM and Vmax = 80 mmol/min/mg substrate for enzyme reaction model I.", null, "Figure 1. Reaction rate curve for Model I. The circles are simulated measurements, solid line is the theoretical curve, stars show the model using the estimated parameters.", null, "(a)", null, "(b)\n\nFigure 2. (a) Showing the relief of the error J(β) as a function of the parameters.(b) Showing the contours of the error J(β) as a function of the parameters. The global minimum is shown by an asterisk. Also shown is the learning trajectory.\n\n4.2. Example II―Enzyme Competitive Inhibitor Case\n\nAs a second example, the data found in the GraphPad user manual is used. It is reproduced in Table 2 for convenience. The parameters for this data as given by Prism 5 are:", null, ".\n\nThe parameters estimated by using MATLAB program via non-linear regression are:", null, ".\n\nFigure 3 shows the plot of reaction curves.\n\nA visual display of the surface of the cost function J is not possible as it is now a function of three variables: Vmax, Km and Ki. So an attempt is made to visualize this optimization problem in two steps. First, we note that when i = 0, Model II degenerates to Model I; as such, Km and Vmax are found using the data of the first and second columns of Table 2. Secondly, using these parameters in the remaining data, J is plotted in Figure 4. The global minimum is found at Ki = 2.75 in conformity with the method of non-linear regression. In MATLAB, the min command obtains minimum of a given vector. Further, it may be pointed out that the modified Fibonacci search method used by Atkins in also readily finds application to accomplish this.\n\n4.3. Example III―pNPP Hydrolysis by Alkaline Phosphatase (Hill Equation)\n\nDetermination of kinetic parameters for the hydrolysis of p-nitrophenylphosphate (pNPP) by alkaline phosphatase was reported in using a program called SigrafW. Figures 5(a), (b) show the familiar structure of pNPP and AP. For convenience, the data points of Table 1 in are reproduced in Table 3 and are used in this example. The parameters obtained by SigrafW, and those obtained through MATLAB simulations, as well as Prism 5 of GraphPad, are shown in Table 4.\n\nThis exercise has been attempted using anemona reported in , but this tool has yielded absurd results. Both the MATLAB program and Prism are found to yield identical values for the parameters.\n\nUsing MATLAB we find that RSS is less and correlation is better. Figure 6 shows the reaction rate curves using both the results. We note that the curve obtained using the parameters reported in does not pass through the observations as accurately as reported therein. The MATLAB script file for this representative example is given in Appendix I. Modifying the code for other examples is straightforward as mentioned earlier.\n\n4.4 Example IV―ATP Hydrolysis by Alkaline Phosphatase (Sum of Two Hill Equations)\n\nAgain we take the data available in for the ATP hydrolysis by alkaline phosphatase. This reaction is characterized by the sum of two Hill equations. Figure 5(c) depicts the ATP structure. The measured data is found\n\nTable 2. Data for competitive enzyme reaction (Model II) taken from the GraphPad user manual . Here i denotes the concentration of inhibitor.", null, "Figure 3. Reaction rate curves for Model II for various values of inhibitor concentrations. Circles: i = 0, diamonds i = 3, squares i = 10, stars i = 30. The solid line is obtained with values given by prism, dots are using MATLAB.\n\nin Table 2 of reproduced here for convenience in Table 5.\n\nFor better numerical accuracy, all the substrate values are first up-scaled uniformly by the factor 108 before parameter estimation. They are later scaled down by the same factor. The parameters reported in , and those obtained through MATLAB program, as well as Prism 5 are shown in Table 6. Note that MATLAB and Prism 5 yield identical results except for a miniscule difference in n2 and RSS.\n\nThe reaction rate curves are shown in Figure 7. In SigrafW reported in for the purpose of a good fit, a baseline concept is used. This amounts to a form of piece-wise regression. But the work in this paper does not rely on any baseline, as we treat the reaction as a single phenomenon. Reaction rate curve, using the parameters in without the baseline, is also plotted in Figure 7. It does not pass through the measured data as seen in Figure 7.\n\nTable 3. Hydrolysis of PNPP by alkaline phosphatase in 50 mM of 2-amino-2-hydroxymethylpropane 1 - 3 diol buffer, pH 9.4 containing 2 mM MgCl2 .", null, "Figure 4. Plot of Cost J as a function of Ki given the values of Km and Vmax.", null, "(a)", null, "(b)", null, "(c)\n\nFigure 5. Structures of (a) para-Nitrophenylphosphate (pNPP) (b) Alkaline Phosphatase (AP) (source: http://pfam.sanger.ac.uk//family/PF00245) (c) Adenosine triphosphate (ATP) (source: http://www.abmservice.com/MeasuringATP.html).\n\nTable 4. Various parameters for Example III compared with those reported in using SigrafW as well as Prism 5. Note that the MATLAB program developed in this paper and Prism 5 yield identical results.\n\nTable 5. Hydrolysis of ATP by alkaline phosphatase in 50 mM Tris-HCl buffer, pH 7.5, containing 2 mM MgCl2 .\n\nTable 6. Various Parameters in Example IV using SigrafW, MATLAB and Prism 5.", null, "Figure 6. Showing the reaction rate in case of PNPP hydrolysis through alkaline phosphatase.", null, "Figure 7. Reaction curve for the Example IV of allosteric response in ATP hydrolysis using alkaline phosphatase. Circles are measured values. Starred curve is obtained with the parameters estimated using MATLAB. In the notion of baseline is used to obtain a good fit. Dotted curve shows the reaction rate without the use of baseline.\n\n5. Discussion and Conclusion\n\nEnzyme parameters can be found using graphical aids such as Lineweaver-Burk’s plot, or Dixon’s plot in the case of simple enzyme reactions. But for complex enzyme reactions, parameter estimation becomes computationally intensive. With the advent of software tools, this task is performed interactively on a computer. In the open literature, several software packages are reported. Almost all of them allow the user to feed the data and select the type of reaction for estimation of enzyme kinetic parameters; but they do not allow the user to write the source code for non-linear regression. This paper provides the mathematical expressions needed in non-linear regression for four types of enzyme reactions and suggests the use of MATLAB. Nonlinear regression is briefly discussed first in general terms, and then specialized to four different models. The elements of various vectors and matrices are explicitly given for the convenience of programming in MATLAB. The MATLAB script for one of the examples is given in Appendix I. MATLAB offers better visualization and the ability to write the source code. Also, this study helps us to use these algorithms in more complex cases. Although in this paper we have given analytical expressions for the G matrix, one could use numerical differentiation for performing parameter estimation. Plotting the cost surface helps in visualizing where optimal values are located in the parameter space and helps to confirm the results obtained by non-linear regression. None of the available packages depict the surface of cost function and learning trajectory of estimation process wherever possible. But using MATLAB, this could be accomplished―thus, enabling the analysts to appreciate the analytical steps involved in the non-linear regression technique.\n\nAcknowledgements\n\nThe authors thank Dr. Mary Connerty and Dr. Craig Warren of Penn State Erie, for helpful suggestions.\n\nReferences\n\n1. Cornish-Bowden, A. (2004) Fundamentals of Enzyme Kinetics. 3rd Edition, Portland Press Limited, London.\n2. Engel, P. C. (1977) Enzyme Kinetics—The Steady State Approach. Chapman and Hill Limited, London.\n3. Dixon, M. and Webb, E.C. (1964) Enzymes. 2nd Edition, Academic Press, New York.\n4. Keener, J. and Snyed, J. (2004) Mathematical Physiology. Springer, New York.\n5. Lineweaver, H. and Burk, D. (1934) The Determination of Enzyme Dissociation Constants. Journal of American Chemical Society, 56, 658. http://dx.doi.org/10.1021/ja01318a036\n6. Dixon, M. (1953) The Determination of Enzyme Inhibitor Constants. Biochemical Journal, 55, 170-171.\n7. Fukushima, Y., Ushimaru, M. and Takahara, S. (2002) On the Error of the Dixon Plot for Estimating the Inhibition Constant between Enzyme and Inhibitor. Biochemistry and Molecular Biology Education, 30, 90-92. http://dx.doi.org/10.1002/bmb.2002.494030020022\n8. Wilkinson, G.N. (1961) Statistical Estimation of Enzyme Kinetics. Biochemical Journal, 80, 324-332.\n9. Atkins, G.L. (1973) A Simple Digital Computer Program for Estimating the Parameters of the Hill Equation. European Journal of Biochemistry, 33, 175-180. http://dx.doi.org/10.1111/j.1432-1033.1973.tb02667.x\n10. Williams, P.A. (1983) ENZPACK: A microcomputer program to aid in the teaching of enzyme kinetics. Biochemical Education, 11, 141-143. http://dx.doi.org/10.1016/0307-4412(83)90096-1\n11. Leatherbarrow, R.J. (1988) Enzfitter: A Non-Linear Regression Data Analysis Program for IBM PC. Journal of American Chemical Society, 110, 4100.\n12. Hernandez, A. and Ruiz, M.T. (1998) An EXCEL Template for Calculation of Enzyme Kinetic Parameters by Non- Linear Regression. Bioinformatics (Applications Note), 14, 227-228. http://dx.doi.org/10.1093/bioinformatics/14.2.227\n13. Heinrich, R., Melendez-Hevia, E. and Cabezas, H. (2002) Optimization of Kinetic Parameters of Enzymes. Biochemistry and Molecular Biology Education, 30, 184-188. http://dx.doi.org/10.1002/bmb.2002.494030030065\n14. Miller, J.R. (2003) GraphPad Prism version 4.0 Step-by-step Examples. GraphPad Software Inc., San Diego, 111-119.\n15. Gonzalez-Cruz, J., Rodreguez-Sotres, R. and Rodriguez-Penagos, M. (2003) On the Convenience of Using a Computer Simulation to Teach Enzyme Kinetics to Undergraduate Students with Biological Chemistry-Related Curricula. Biochemistry and Molecular Biology Education, 31, 93-101. http://dx.doi.org/10.1002/bmb.2003.494031020193\n16. Clark, A.G. (2004) Lucenz Simulator—A Tool for the Teaching of Enzyme Kinetics. Biochemistry and Molecular Biology Education, 32, 201-206. http://dx.doi.org/10.1002/bmb.2004.494032030350\n17. Leone, F.A., Baranauskas, J.A., and Furriel, R.P.M. (2005) SigrafW: An Easy to Use Program for Fitting Enzyme Kinetic Data. Biochemistry and Molecular Biology Education, 33, 399-403. http://dx.doi.org/10.1002/bmb.2005.49403306399\n18. Palm III, W.J. (2008) A Concise Introduction to MATLAB. 1st Edition. McGraw Hill, New York.\n19. Daku, B. (2005) MATLAB Tutorial CD: Learning MATLAB Superfast. Wiley, New York.\n20. Pratap, R. (2005) Getting Started with MATLAB 7—A Quick Introduction for Scientists and Engineers. Oxford University Press, Oxford.\n21. Davis, T.A. and Sigmon K. (2004) MATLAB Primer. CRC Press. http://dx.doi.org/10.1201/9781420034950\n22. Cobelli, C. and Carson, E. (2008) Introduction to Modeling in Physiology and Medicine. Academic Press, Waltham, 212-220.\n23. Moore, E.H. (1920) On the Reciprocal of the General Algebraic Matrix. (Abstract) Bulletin of American Mathematical Society, 26, 394-395.\n24. Penrose, R.A. (1955) A Generalized Inverse for Matrices. Proceedings of Cambridge Philosophical Society, 51, 406- 413. http://dx.doi.org/10.1017/S0305004100030401\n\nAppendix I\n\nNOTES", null, "1The MATLAB script for one of the examples is shown in Appendix I.", null, "2We denote the estimated value of a parameter by the hat symbol; for example,", null, "is the estimate of", null, "." ]

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[ "# I'm doing a tax return for a S corporation that just began businesses, I was provided...\n\n###### Question:\n\nI'm doing a tax return for a S corporation that just began businesses, I was provided with the balanced sheet and income statement. The balance sheet only includes the ending balance but not the beginning balance. So I'm trying to fill up schedule M2, what should put where it asks for the beginning balalnce?\n\n#### Similar Solved Questions\n\n##### Find the general solution and sketch phase diagrams for the following characterize the equilibria as to type (node; etc:) and stability: tems;7eI-30, V = -3r+V: V) /=-f+u V =V 7 = 4V. V =-r. \"=f+l V =4r -2v. c) x =3r _ Av, V=r-V =Y21 Ju. V =%r \"2V.\nFind the general solution and sketch phase diagrams for the following characterize the equilibria as to type (node; etc:) and stability: tems; 7eI-30, V = -3r+V: V) /=-f+u V =V 7 = 4V. V =-r. \"=f+l V =4r -2v. c) x =3r _ Av, V=r-V =Y21 Ju. V =%r \"2V....\n##### Find the unpaid balance on the debt, (Round your answer to the nearest cent:) After 6 years of monthly payments on $160,000 at 4% for 25 years.$\nFind the unpaid balance on the debt, (Round your answer to the nearest cent:) After 6 years of monthly payments on $160,000 at 4% for 25 years.$...\n##### Iv) Find I(a) Choose the correct answer below0A ((a)The answar Is undelined\niv) Find I(a) Choose the correct answer below 0A ((a) The answar Is undelined...\n##### What is the molar mass of a particular compound if 0.087 mol weighs 5.16 g?\nWhat is the molar mass of a particular compound if 0.087 mol weighs 5.16 g?...\n##### Can you please explain why you choose the answer option as well compared to the other...\nCan you please explain why you choose the answer option as well compared to the other answer choices? ISBN Leading Reptory Excellence CLEX-RN Practice Exam - Practice Exam Time Remaining 02:45:54 Calculator When caring for an elderly postoperative client, the nurse should know that this client compa...\n##### Rehabilitation strategies Give 1 (one) example of how the following strategies or equipment can be used...\nRehabilitation strategies Give 1 (one) example of how the following strategies or equipment can be used to manage a chronic health issue. Strategies How this is used in Chronic Disease management? Physiotherapy In cerebral palsy, physiotherapy exercises and passive ROM will assist in maintaining mus...\n##### To treat the contaminant estrone in the wastewater, you decided to use an oxidant, permanganate. The...\nTo treat the contaminant estrone in the wastewater, you decided to use an oxidant, permanganate. The reaction is assumed to be occurring in an Ideal PFR. Thention rate follows the first order kinetics and half life is 5 minutes. A) What is the value of reaction rate constant (in min:') B) How mu...\n##### Solve for %: Jax+5y7 < 9 bi: (1 + 20)z + (9 + 0) = 3 writing Vour answer on the form Bolve (or ofthe radius (both exactly and to the nearest 0.1) of Find the coordinates of the center and the length the circlc with equation (10) 6x + [0y 1986 the point (20, 20) that is perpendicular to the line with Orile an cquation for the line passing through equation Zx= 125.\nSolve for %: Jax+5y7 < 9 bi: (1 + 20)z + (9 + 0) = 3 writing Vour answer on the form Bolve (or ofthe radius (both exactly and to the nearest 0.1) of Find the coordinates of the center and the length the circlc with equation (10) 6x + [0y 1986 the point (20, 20) that is perpendicular to the line ...\n##### Compute the variance of the random variable X with probability distribution as follows:Pix) 8.01B. 7,500c9.0500.8.90\nCompute the variance of the random variable X with probability distribution as follows: Pix) 8.01 B. 7,50 0c9.05 00.8.90...\n##### Question 7 3 pts According to research, which of the following is the most influential when...\nQuestion 7 3 pts According to research, which of the following is the most influential when it comes to food choice? O Health considerations O Taste- especially foods high in fat and sugar- and texture OEducation O Convenience Advertising Peer and family influence Ne Previous...\n##### Order: D5W 500 mL IV at 30 gtt/min Drop factor: 20 gtt/mL Time:_________h and ____min\nOrder: D5W 500 mL IV at 30 gtt/min Drop factor: 20 gtt/mL Time:_________h and ____min...\n##### B) Figure Q1(b) shows a P-v diagram consists of isentropic, polytropic, and isothermal compression processes between...\nb) Figure Q1(b) shows a P-v diagram consists of isentropic, polytropic, and isothermal compression processes between the same pressure limits. Based on the figure, distinguish the work input to the compressor for all the three processes. (6 marks) PA -Isentropic (n = k) -Polytropic (1 <n<k) Is...\n##### Show work please Optimization problems 1. (5 points) Find two nonnegative numbers whose sum is 25...\nShow work please Optimization problems 1. (5 points) Find two nonnegative numbers whose sum is 25 and so that the product of one number and the square of the other number is a maximum. 2. (5 points) Build a rectangular pen with two parallel partitions using 300 feet of fencing. What dimensions w...\n##### Imagine a slab of current that is infinite in x and y but finite in z...\nImagine a slab of current that is infinite in x and y but finite in z with a current density ?J. The slab has a thickness 2h (it runs from z = ?h to z = +h). Assuming the current is still in the x direction and is uniform in the x and y dimensions, but depends linearly on the height (J = J0|z|x&circ...\n##### I need the solution to this ASAP, and I would really appreciate it!\nI need the solution to this ASAP, and I would really appreciate it!...\n##### Please work neatly and show all steps A variable force is given by, F.(t) = (50.0...\nplease work neatly and show all steps A variable force is given by, F.(t) = (50.0 + 10.0 t*) newtons. This force is applied in the x direction for 5.00 seconds to a 10.00 kg mass which is initially at rest. Find the work done by that force in those 5.00 seconds. No other forces are acting on the mas...\n##### No. Peak tension force F(N) Radius r(m) Lr (1/m) 5.065 1.18 0.847 5.083 1.07 0.935 5.12 0.95 1.053 5.199 0.75 1.333 5.297 0.59 1.695 5.584 0.38 2.632 5.731 0.31 3.226To make the graph linear have omitted the last tw data points in F vs graph below: Peak tension force F(NJvs radius rGraph between F and Ir Peak tension force F(NJ vs 1/r0.2851x 4.8197\nNo. Peak tension force F(N) Radius r(m) Lr (1/m) 5.065 1.18 0.847 5.083 1.07 0.935 5.12 0.95 1.053 5.199 0.75 1.333 5.297 0.59 1.695 5.584 0.38 2.632 5.731 0.31 3.226 To make the graph linear have omitted the last tw data points in F vs graph below: Peak tension force F(NJvs radius r Graph between F...\n##### Please thoroughly explain the questions below, will rate thumbs up. Q1. What is the role of...\nPlease thoroughly explain the questions below, will rate thumbs up. Q1. What is the role of \"Microfilaments, Intermediate filaments, and microtubules\" in forming a neuron's shape? Explain each cytoskeleton protein' role. Q2. Are enzyme written in the product side of a chemical reacti..." ]

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[ "<img height=\"1\" width=\"1\" style=\"display:none\" alt=\"\" src=\"https://www.facebook.com/tr?id=367542720414923&amp;ev=PageView&amp;noscript=1\">\n\n# Chief Data & Analytics Officers, UK\n\n10-11 February 2021 | London", null, "", null, "true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true true" ]

[ null, "https://no-cache.hubspot.com/cta/default/2631050/4b790d6b-15aa-48c1-bf54-ebadd2d23beb.png", null, "https://no-cache.hubspot.com/cta/default/2631050/349aff40-9616-46c8-b038-d4fa0a1ce433.png", null ]

[ "## Functions Modeling Change: A Preparation for Calculus, 5th Edition\n\nPublished by Wiley\n\n# Chapter 5 - Logarithmic Functions - Exercises to Skills for Chapter 5 - Page 232: 9\n\n#### Answer\n\n$-4=\\log(0.0001)$\n\n#### Work Step by Step\n\nSince $\\log(10^a)=a$, the equation can be simplified to $$\\log 10^{-4}=\\log 0.0001$$ $$-4=\\log 0.0001$$\n\nAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback." ]

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[ "#b4e900 Color Information\n\nIn a RGB color space, hex #b4e900 is composed of 70.6% red, 91.4% green and 0% blue. Whereas in a CMYK color space, it is composed of 22.7% cyan, 0% magenta, 100% yellow and 8.6% black. It has a hue angle of 73.6 degrees, a saturation of 100% and a lightness of 45.7%. #b4e900 color hex could be obtained by blending #ffff00 with #69d300. Closest websafe color is: #ccff00.\n\n• R 71\n• G 91\n• B 0\nRGB color chart\n• C 23\n• M 0\n• Y 100\n• K 9\nCMYK color chart\n\n#b4e900 color description : Pure (or mostly pure) yellow.\n\n#b4e900 Color Conversion\n\nThe hexadecimal color #b4e900 has RGB values of R:180, G:233, B:0 and CMYK values of C:0.23, M:0, Y:1, K:0.09. Its decimal value is 11856128.\n\nHex triplet RGB Decimal b4e900 `#b4e900` 180, 233, 0 `rgb(180,233,0)` 70.6, 91.4, 0 `rgb(70.6%,91.4%,0%)` 23, 0, 100, 9 73.6°, 100, 45.7 `hsl(73.6,100%,45.7%)` 73.6°, 100, 91.4 ccff00 `#ccff00`\nCIE-LAB 85.997, -41.577, 83.866 47.961, 67.98, 10.595 0.379, 0.537, 67.98 85.997, 93.607, 116.37 85.997, -26.104, 98.542 82.45, -40.456, 50.096 10110100, 11101001, 00000000\n\nColor Schemes with #b4e900\n\n• #b4e900\n``#b4e900` `rgb(180,233,0)``\n• #3500e9\n``#3500e9` `rgb(53,0,233)``\nComplementary Color\n• #e9aa00\n``#e9aa00` `rgb(233,170,0)``\n• #b4e900\n``#b4e900` `rgb(180,233,0)``\n• #3fe900\n``#3fe900` `rgb(63,233,0)``\nAnalogous Color\n• #aa00e9\n``#aa00e9` `rgb(170,0,233)``\n• #b4e900\n``#b4e900` `rgb(180,233,0)``\n• #003fe9\n``#003fe9` `rgb(0,63,233)``\nSplit Complementary Color\n• #e900b4\n``#e900b4` `rgb(233,0,180)``\n• #b4e900\n``#b4e900` `rgb(180,233,0)``\n• #00b4e9\n``#00b4e9` `rgb(0,180,233)``\n• #e93500\n``#e93500` `rgb(233,53,0)``\n• #b4e900\n``#b4e900` `rgb(180,233,0)``\n• #00b4e9\n``#00b4e9` `rgb(0,180,233)``\n• #3500e9\n``#3500e9` `rgb(53,0,233)``\n• #799d00\n``#799d00` `rgb(121,157,0)``\n• #8db600\n``#8db600` `rgb(141,182,0)``\n• #a0d000\n``#a0d000` `rgb(160,208,0)``\n• #b4e900\n``#b4e900` `rgb(180,233,0)``\n• #c6ff03\n``#c6ff03` `rgb(198,255,3)``\n• #ccff1d\n``#ccff1d` `rgb(204,255,29)``\n• #d1ff37\n``#d1ff37` `rgb(209,255,55)``\nMonochromatic Color\n\nAlternatives to #b4e900\n\nBelow, you can see some colors close to #b4e900. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #e9e400\n``#e9e400` `rgb(233,228,0)``\n• #dbe900\n``#dbe900` `rgb(219,233,0)``\n• #c7e900\n``#c7e900` `rgb(199,233,0)``\n• #b4e900\n``#b4e900` `rgb(180,233,0)``\n• #a1e900\n``#a1e900` `rgb(161,233,0)``\n• #8de900\n``#8de900` `rgb(141,233,0)``\n• #7ae900\n``#7ae900` `rgb(122,233,0)``\nSimilar Colors\n\n#b4e900 Preview\n\nThis text has a font color of #b4e900.\n\n``<span style=\"color:#b4e900;\">Text here</span>``\n#b4e900 background color\n\nThis paragraph has a background color of #b4e900.\n\n``<p style=\"background-color:#b4e900;\">Content here</p>``\n#b4e900 border color\n\nThis element has a border color of #b4e900.\n\n``<div style=\"border:1px solid #b4e900;\">Content here</div>``\nCSS codes\n``.text {color:#b4e900;}``\n``.background {background-color:#b4e900;}``\n``.border {border:1px solid #b4e900;}``\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #0d1100 is the darkest color, while #fefffd is the lightest one.\n\n• #0d1100\n``#0d1100` `rgb(13,17,0)``\n• #1c2500\n``#1c2500` `rgb(28,37,0)``\n• #2c3800\n``#2c3800` `rgb(44,56,0)``\n• #3b4c00\n``#3b4c00` `rgb(59,76,0)``\n• #4a6000\n``#4a6000` `rgb(74,96,0)``\n• #597300\n``#597300` `rgb(89,115,0)``\n• #688700\n``#688700` `rgb(104,135,0)``\n• #779b00\n``#779b00` `rgb(119,155,0)``\n• #87ae00\n``#87ae00` `rgb(135,174,0)``\n• #96c200\n``#96c200` `rgb(150,194,0)``\n• #a5d500\n``#a5d500` `rgb(165,213,0)``\n• #b4e900\n``#b4e900` `rgb(180,233,0)``\n• #c3fd00\n``#c3fd00` `rgb(195,253,0)``\n• #c9ff11\n``#c9ff11` `rgb(201,255,17)``\n• #cdff25\n``#cdff25` `rgb(205,255,37)``\n• #d2ff38\n``#d2ff38` `rgb(210,255,56)``\n• #d6ff4c\n``#d6ff4c` `rgb(214,255,76)``\n• #dbff60\n``#dbff60` `rgb(219,255,96)``\n• #dfff73\n``#dfff73` `rgb(223,255,115)``\n• #e4ff87\n``#e4ff87` `rgb(228,255,135)``\n• #e8ff9b\n``#e8ff9b` `rgb(232,255,155)``\n• #edffae\n``#edffae` `rgb(237,255,174)``\n• #f1ffc2\n``#f1ffc2` `rgb(241,255,194)``\n• #f6ffd5\n``#f6ffd5` `rgb(246,255,213)``\n• #faffe9\n``#faffe9` `rgb(250,255,233)``\n• #fefffd\n``#fefffd` `rgb(254,255,253)``\nTint Color Variation\n\nTones of #b4e900\n\nA tone is produced by adding gray to any pure hue. In this case, #797d6c is the less saturated color, while #b4e900 is the most saturated one.\n\n• #797d6c\n``#797d6c` `rgb(121,125,108)``\n• #7e8663\n``#7e8663` `rgb(126,134,99)``\n• #838f5a\n``#838f5a` `rgb(131,143,90)``\n• #889851\n``#889851` `rgb(136,152,81)``\n• #8da148\n``#8da148` `rgb(141,161,72)``\n• #92aa3f\n``#92aa3f` `rgb(146,170,63)``\n• #97b336\n``#97b336` `rgb(151,179,54)``\n• #9cbc2d\n``#9cbc2d` `rgb(156,188,45)``\n• #a0c524\n``#a0c524` `rgb(160,197,36)``\n• #a5ce1b\n``#a5ce1b` `rgb(165,206,27)``\n``#aad712` `rgb(170,215,18)``\n• #afe009\n``#afe009` `rgb(175,224,9)``\n• #b4e900\n``#b4e900` `rgb(180,233,0)``\nTone Color Variation\n\nColor Blindness Simulator\n\nBelow, you can see how #b4e900 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

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[ "`I am simply trying to round up 1.275.toFixed(2) and I was expecting a return of 1.28, rather than 1.27. Using various calculators and the simple method of rounding to the nearest hundredth, if the last digit is greater than or equal to five, it shou...`\n`I started reading YDKJS for fun - and found that he's written: we can do stuff like: var num = (1.2).toFixed(1) so - this means that toFixed is being invoked as a member method from an integer value. So why doesn't this work?? \"toFi...`\n`This question already has answers here:...`\n`I have my code and it is working, but I'm trying to add the parseFloat option and 'toFixed()' to 2 decimal places. I'm confused on where these 2 lines of code should be placed. I created a unit conversion website that would convert th...`\n`Why when I want to assign formatted number to variable it shows error? var jsObject = new Object(); jsObject.number = 0; for (i = 1; i <= 10; i++) { jsObject.number += 0.1; console.log((jsObject.number).toFixed(1)); // works and shows right...`\n`I'm using a wordpress plugin that's spitting out code that looks like this: <div class=\" product-addon product-addon-extra-tip\"> <p class=\"form-row form-row-wide addon-wrap-2004-extra-tip-0-0\"> <label...`\n`I am printing the data fetched from an API into a table, and I am facing some difficulties to fix the rows numerical values to decimals. If the rows data within the API consists of numerical values, i.e. 10.0, 333, 8 or 100, etc. to render it in the...`\n`Is there a way to round number like: 0.203 to 0.21 Now I do this: priceCurrent.toFixed(2);...`\n`How would I get this negative large exponential number 6.849391775995509e-276 to 6.84 in javascript? I'm getting back negative exponential numbers from an api which I would like to use a shortened version, to 2 decimal points, in my ui. If it ma...`\n`I am having trouble with my toFixed() method. Before I added it onto all the parseFloats, which were already there, it was displaying all the totals but with too many decimal places. Now it displays nothing. When I take the toFixed() off, it displa...`" ]

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[ "", null, "# Student Solutions Manual Part 2 for University Calculus (1st Edition) Edit edition Problem 9AAE from Chapter 11: Use vectors to show that the distance from P1(x1, y1) to the...\n\nWe have solutions for your book!\nChapter: Problem:\n\nUse vectors to show that the distance from P1(x1, y1) to the line ax + by = c is", null, "Step-by-step solution:\nChapter: Problem:\n• Step 1 of 5\n\nGiven that", null, "and line is", null, "Let", null, "be any point on the given line\n\nThen", null, "and", null, "is a vector normal to the line.\n\n• Chapter , Problem is solved.\nCorresponding Textbook", null, "Student Solutions Manual Part 2 for University Calculus | 1st Edition\n9780321388506ISBN-13: 032138850XISBN: Authors:\nThis is an alternate ISBN. View the primary ISBN for: University Calculus: Alternate 1st Edition Textbook Solutions" ]

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[ "# 3D puzzle of the trefoil knot and its fibrations via 3D printing\n\nPosted 4 years ago\n21101 Views\n|\n1 Reply\n|\n23 Total Likes\n|\n Hello Wolfram Community! I am Fred Hohman, a senior Mathematics and Physics student at the University of Georgia. I have been working under Dr. David Gay in the Department of Mathematics at the University of Georgia since the beginning of the year, and I will be continuing my work until my graduation in May 2015.Earlier in the summer I had the wonderful opportunity to meet Dr. Laura Taalman (mathgrrl) after her talk at MoMath on “Making Mathematics Real: Knot Theory, Experimental Mathematics, and 3D Printing.“ After chatting about math and 3D printing she asked me to say some words about my undergraduate research at UGA. Since I wrote my code in Mathematica I thought I would share the information here as well.The goal of this project is to create a 3D puzzle of the trefoil knot and its fibrations via 3D printing. This is a long post but here is what we we will build at the end of this project:", null, "Let’s start with some theory.Theory and Stereographic ProjectionConsider the following function from $\\mathbb{C}^2$ to $\\mathbb{C}$ (we can think of this as a function from $\\mathbb{R}^4$ to $\\mathbb{R}^2$ - see the Wikipedia entries for real n-space and complex n-space for more information): this function sends a coordinate pair $(u, v)$, where $u$ and $v$ are complex numbers, to the quantity $(u+i v)^2 - (u-i v)^3$, where $i = \\sqrt{-1}$. This particular function’s zero-set, a set of points $(u,v)$ such that $(u+i v)^2=(u-i v)^3$, generates a trefoil knot through infinity. By considering inverse images of certain subsets of $\\mathbb{C}$ we will generate a trefoil knot given our function’s zero-set; however, this inverse image is a subset of $\\mathbb{C}^2$. To obtain our knot and understand its subsets better, we need a method to realize our object in $\\mathbb{R}^3$.As a preliminary example, consider a sphere in $\\mathbb{R}^3$ sitting on a plane (think of a ball sitting on a table). From the top of the sphere (think North Pole), draw a line segment downward through the sphere and ending at the table. Notice that this line segment intersects both the sphere and table once. We can do this same line drawing method over and over to obtain a one-to-one correlation from the sphere to the plane, i.e., every point (well, minus the exact top of the sphere!) on the sphere can be mapped to the plane. This is called stereographic projection, and Henry Segerman (henryseg on Thingiverse) has created a fantastic 3D printed model illustrating this method:", null, "If we have a point light source (smartphone camera flash) and place it at the top of this model, light rays will act as the line segments in the above example. If held at the correct height, we should see the Cartesian grid on the table. I always keep this model in my backpack to demonstrate to people stereographic projection—I recommend printing it out!So stereographic projection is a function from the unit sphere in $\\mathbb{R}^3$ onto $\\mathbb{R}^2$, but recall that our function is defined in $\\mathbb{R}^4$. For us to embed our object in $\\mathbb{R}^3$, we use a generalized version of stereographic projection to go from the unit sphere in $\\mathbb{R}^4$ to $\\mathbb{R}^3$. With this tool, we can compose the inverse stereographic projection function with our function defined above. So our new function now goes from $\\mathbb{R}^3 \\rightarrow \\mathbb{C}$.If we take inverse images of subsets of $\\mathbb{C}$, we now have a function that goes from $\\mathbb{C} \\rightarrow \\mathbb{C}^2 = \\mathbb{R}^4 \\rightarrow \\mathbb{R}^3$, i.e., our single, final function takes in certain subsets of $\\mathbb{C}$ and outputs subsets of $\\mathbb{R}^3$—exactly what we want!Mathematica, Plotting, and ResolutionTo generate 3D computer models of the trefoil knots, I chose to use Mathematica. RegionPlot3D, a built-in Mathematica function, plots regions in 3D space using inequalities. In order to use RegionPlot3D, the user must specify the domain of values to plot over in the $x$, $y$, and $z$ directions, as well as a mathematical inequality. As an example we can generate a sphere of radius 1 by the following code:", null, "Mathematica will plot all points $(x, y, z)$ such that each point falls within the considered domain and obeys the given inequalities—but we can take this further. We can use multiple inequalities using boolean operators such as AND and OR. As an example, let’s consider the same sphere defined above, but restrict the region so that only points above the plane $z=0$ are plotted.", null, "Of course we could restrict the $z$ range so that $z$ goes from $(0, 1)$, but it is nice to be able to manipulate a model without changing the overall plotted region. When models become more complicated this method allows us to control individual components without altering other pieces.Before we start creating trefoil knots, there is one last Mathematica option that needs to be discussed: PlotPoints. Notice how choppy and non-circular the bottom of our hemisphere looks in the image above—PlotPoints will fix this. The PlotPoints function can be thought of as the resolution of a 3D model. A higher PlotPoints value tells Mathematica to use more points to represent the plotted region. However, as we increase PlotPoints, we also increase computation time. Say we used a PlotPoints of 50 and the model was under-represented. We could double our resolution and set PlotPoints to 100, but remember we are in 3D space, so by doubling the number of points in all directions $x$, $y$, and $z$ we increase our computation by $2\\times2\\times2=8$. In other words, doubling a model’s resolution increases computation time by a factor of 8. So let’s plot that same hemisphere with a PlotPoints of 100 and see what the bottom looks like.", null, "Much better! Now that we have an understanding of RegionPlot3D and PlotPoints, it’s time to create some trefoil knots.The main Mathematica function I wrote, inTinftube, defines the trefoil knot (the composition of our trefoil knot function with inverse stereographic projection explained above) by inputting points $(x, y, z)$ and “knot thickness” and returning a single number. Inside the function it tests points using inequalities such that if the outputted number is less than 0, Mathematica includes the point in the plot, and if the number is greater than 0, Mathematica does not include the point in the plot. We also need to define a boundary condition, otherwise our model would stretch to infinity! So we can include more inequalities that points must satisfy such that our boundary is a cylinder of radius 3, height 6, and is centered at the origin. With these parameters selected, the results below depict our “standard” trefoil knot. This knot is the inverse image of a small disk around 0 in $\\mathbb{C}$ to give the knot thickness—a visual description of this will be shown in the next section.", null, "", null, "Thingiverse link: http://www.thingiverse.com/thing:243260Open Book Decomposition, Fibrations, and Pages.Now that we have our “standard/reference trefoil knot,” it is time to start adding fibrations. To do this, unravel the above trefoil so that it is a straight strand. Consider that cord the spine of a book. Now add in one page to the book; the page will connect to the spine along one edge. Now twist the cord back into the trefoil knot configuration—what does the page, i.e., knot fibration, look like? This idea is called an Open Book Decomposition.We can visualize this by adding extra regions to the model in $\\mathbb{C}$ before we apply our single function. To do this, I wrote another Mathematica function, inTinfpage, that takes in points and outputs a number; however, this time the function also requires an angle. This angle will define a ray in the plane starting at the origin that is rotated an angle in the mathematically positive direction. As before, we can have Mathematica plot points if the outputted number is below zero, but we want a page of some physical thickness (in order to 3D print). For most of my successful prints I have been using a page thickness of $\\pi/6$. So we can now include another “ray-function” and input an angle that is $\\pi/6$ larger than before and have Mathematica plot things that are greater than 0. We have now included another region in the plane. This process is much easier to see in the image below.", null, "The circle is the radius of the trefoil knot, and the region in the first quadrant is our fiber with thickness. That’s the new region we want to include when using stereographic projection. Remember the picture above is the subset of $\\mathbb{C}$ that we are taking inverse images of. We can now apply our same code to generate the following model.", null, "", null, "Why stop at one extra region? We can add as many as we would like, and at $\\pi/6$ thickness, we should be able to fit 12 extra regions in our trefoil knot (since $\\pi/6\\times12=2\\pi$). Some other examples are below, with the region in $\\mathbb{C}$ shown on the left, and its inverse image (after applying stereographic projection) shown on the right.", null, "", null, "We can now start to see the consequences of projecting our 4D function into $\\mathbb{R}^3$, which is why our knots are visually unlike typical trefoil knots.Now that I have code that can generate trefoil knots and fibrations of any thickness, it’s time to start printing various versions of the models. The goal is to be able to print the trefoil knot separate from the pages to create a 3D puzzle. At the end of this post I’ll share some of my successful experiments in doing just that.1. Printing the trefoil knot with one page with a gap.Here I made a model of the trefoil knot with one page, but removed the points where the page meets the knot. In other words, there is a gap running along the knot where the page meets the surface of the knot. When printed, the page should be free to wiggle a little bit. Here is the object in Mathematica:", null, "After exporting to STL and printing, this is the result:", null, "Thingiverse link: http://www.thingiverse.com/thing:3315302. Printing the trefoil knot with three pages with a gap.This is a trefoil knot with three pages, all of which have gaps where the pages meet the knot. This knot is closely related to the previous model; however, two more pages have been added, all of which are equiangular (think of the Mercedes Benz logo). The print time on this can be a bit longer (an afternoon), and the clean-up of supports can get annoying. However, this model looks fantastic in person. Here it is in Mathematica:", null, "And now printed:", null, "Thingiverse link: http://www.thingiverse.com/thing:3371853. Printing the trefoil knot and one page separately.Here I printed the standard trefoil knot but edited the domain so that the knot itself is bigger. I also made a groove in the knot that follows where one page meets the surface, with hopes that if I could print a page separately, then I could snap the page into the knot. Next, I took one page and split it into two pieces with the plane $z=0$. With a little force, I was able to get both pieces in the knot. A good first start, but not ideal for consistent “puzzle-building”.", null, "Current and future workThis puzzle-building problem illustrates a simple idea, but can prove to be very complicated in practice. Many people have assembled large objects by printing smaller components; however, due to the complex geometry of the shape we are dealing with, slicing the knot in a particular way such that the pieces can be reassembled without blocking any other piece is an interesting and challenging problem for any 3D printing enthusiast.My latest attempt has been to have Mathematica generate all twelve pages of $\\pi/6$ thickness, cut each page into two pieces using the plane $z=0$, and import each piece individually into Blender, a free, open-source 3D modeling and animation software. In Blender, I added small holes in the bottom and top of each piece of a page. Once the hole was made, I printed the pieces of the pages and glued in small magnets to hold two pieces of the same page together (not all pages are same; in fact, every page is different from one another). I successfully printed a trefoil knot and 3 pages (6 pieces), all with magnets, to make a 7-piece 3D puzzle. Here is the process in Blender:", null, "", null, "", null, "And here are the prints:", null, "", null, "", null, "This coming year I will work to refine the process of adding holes; I may cut the pages at different angles too. Once a puzzle has been made and printed that is easy to piece together, contains multiple pieces, and is of appropriate size, I plan on posting the puzzle in its entirety on my Thingiverse profile — so be on the lookout! I would like to thank Dr. Laura Taalman for the opportunity to write about my research so far, and Dr. David Gay at the University of Georgia for his guidance and access to a MakerBot Replicator 2.", null, "Answer\n1 Reply\nSort By:\nPosted 4 years ago\n Great stuff! I propose the next item we should print from our office 3D printer to be the top sphere model, from here. Awesome!", null, "Answer\nReply to this discussion\nCommunity posts can be styled and formatted using the Markdown syntax.\nReply Preview\nAttachments" ]

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[ "Question about the pose from homography and fundamenta matrix\n\nI have get projection matrix from homography and camera parameter for AR. For checking if the result is good, I tested if ROI (0, 0) (w, 0), (w, h), (0,h) in the coordinate of reference image can be visualized in different image at different view. I assumed that ROI 4 points have zero depth. The result is successful.\n\nThesedays, I implemented the same function from fundamental matrix. 1. get Fundamental Matrix 2. Essential matrix from Fundamental 3. decompose essential mat into rotation & translation matrix 4. get projection matrix from camera parameter & rotation & translation Matrix.\n\nLike the case I used projection mat from homography, I transformed 4 ROI in reference into the other image at different view. But the result is different from one by homography. Non feasible result was shown as the strange ROI box.\n\nI was wonderingn if the projection matrix by homography is different from one by fundamental matrix?\n\nplease let me know how to transform ROI in reference image into the same scenen at different view.\n\nedit retag close merge delete" ]

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[ "# Estimating difficulty of \"Memory-Hard Proof-of-Work\" based on \"size of memory\"?\n\nIn Bitcoin proof-of-work, the difficulty of Proof-of-Work is estimated and calculated based on total hashing power of the participants. If total hashing power of the participants is higher, then PoW is more difficult.\n\n• If we use a Memory-Hard-Proof-of-Work scheme, the difficulty of Proof-of-Work is still estimated based on the hashing power ? OR based on the size of memory ?\n\n• In other word, is it feasible to design a Memory-Hard-Proof-of-Work scheme in which we could estimate the Pow difficulty based on the size of memory, instead of hashing power of participants ?\n\n• Has such a Memory-Hard-Proof-of-Work scheme designed yet by which we could estimate the difficulty based on memory size?\n\nJust as an example: a Memory-Hardened Proof-of-Work Scheme: https://eprint.iacr.org/2017/1168.pdf\n\nP.S. 1: Related question: Memory-hard proof-of-work: are they ASIC-resistant?\n\nP.S. 2: If you need any further complementary explanation about the question, please let me know.\n\nThe measure of resource typically used to evaluate memory-hard functions is not the amount of work (i.e., $$T$$-complexity) but rather the space-time complexity (i.e., $$ST$$-complexity) of the computation. As the name suggests, it is the product of the maximum amount of space used and the time taken for the computation. (Strictly speaking, we have to also take in account amortization of the cost and the exact measure used is the cumulative space-time complexity, but the notion of space-time complexity still conveys the basic ideas: see [AS] for more details.) Good memory hard functions require $$ST=\\Theta(n^2)$$.\n\nThe motivation behind this switch is to discourage the use of ASICs in computing proofs of work. The rationale of using ASICs in computing proofs of work is that they are much faster than CPUs. The speed-up is primarily due to its application-specificity, and the availability of large-scale parallelism. However, there is an asymmetry in the cost of memory (space, i.e.) as it is much costlier in ASICs compared to CPUs. Therefore, if one can construct a function that can be computed “fast” using “sufficient” space (even) on a sequential machine, but takes “more” time if the amount of space used is “lesser” (even with parallelism), then ASICs are not practical anymore as the increased cost of memory counteracts the increased computation power, but computing the function is cheap on a GPU.\n\n• If we use a Memory-Hard-Proof-of-Work scheme, the difficulty of Proof-of-Work is still estimated based on the hashing power? OR based on the size of memory?\n\nThe amount of work considered is still in terms of the amount of hashing power, the role of the memory as explained above is just to discourage the use of ASICs.\n\n• In other word, is it feasible to design a Memory-Hard-Proof-of-Work scheme in which we could estimate the Pow difficulty based on the size of memory, instead of hashing power of participants?\n\nThere has been a lot of work in the recent years addressing this exact question --- the main motivation being that if space is used instead of work/time as a resource then there would be less energy wasted. The object is called a proof-of-space [D+,A+,F] and the prover/miner is rewarded for dedicating some disk space instead of doing work. (There is also a cryptocurrency, called Chia, that is being deployed based on proofs of space.)\n\n[AS]: Alwen and Serbinenko. High Parallel Complexity Graphs and Memory-Hard Functions.\n\n[A+] Abusalah et al. Beyond Hellman's Time-Memory Trade-Offs with Applications to Proofs of Space\n\n[D+]: Dziembowski et al. Proofs of Space\n\n[F]: Fisch. Tight Proofs of Space and Replication" ]

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[ "psrf_check {EDISON} R Documentation\n\n## Check the potential scale reduction factors for all parameters (edges).\n\n### Description\n\nThis function treats the edges of the network as parameters, calculates their potential scale reduction factors and returns the highest value.\n\n### Usage\n\npsrf_check(params, q, k_max, num_it)\n\n\n### Arguments\n\n params Matrix of parameters. q Number of nodes. k_max Number of segments. num_it Number of iterations/samples.\n\n### Value\n\nReturns the highest PSRF value.\n\n### Author(s)\n\nFrank Dondelinger\n\n### References\n\nGelman and Rubin (1992) Inference from iterative simulation using multiple sequences, Statistical Science.\n\npsrf, psrf_check_hyper" ]

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[ "# Thevenin’s Problem for dependent and independent sources\n\n0\n\nSave\n\nAfter being familiar with basic Thevenin’s theorem, we will proceed to solving the networks which has both dependent and independent sources. In basic networks with only independent sources voltage sources are shorted and current sources are opened to get Thevenin’s Equivalent Resistance. However, when dependent sources come into picture, this cannot be the case. It is because an ideal voltage source has zero resistance while an ideal current source has infinite resistance. But for dependent sources, they both have finite resistances. Hence they are also needed to be considered while calculating Thevenin’s Equivalent Resistance. How we proceed is such cases will be shown with the help of the example below.\n\nIn the example, we have to find Thevenin voltage and Thevenin Resistance across the terminals x – y.\n\nIn this example, we have the independent voltage source Vo, one dependent voltage source AVxy and one dependent current source BI\n\nIn this example, we will assume A = 0\n\nThe Thevenin’s voltage Vth will be Vxy. Then the network becomes", null, "Now, we will deactivate independent sources. Then I will become zero and therefore BI = 0\n\nWe solved for Vth without deactivating independent sources. In next step, we apply a Driving Voltage Vdc at x – y terminals supplying an input current Idc.\n\nThen,\n\nRemember, both the dependent and independent are deactivated in above example due to given conditions. However, dependent sources might be present in same cases which can be solved using KVL and KCL.", null, "Here" ]

[ null, "data:image/svg+xml,%3Csvg xmlns=\\'http://www.w3.org/2000/svg\\' viewBox=\\'0 0 1 1\\'%3E%3C/svg%3E", null, "data:image/svg+xml,%3Csvg xmlns=\\'http://www.w3.org/2000/svg\\' viewBox=\\'0 0 1 1\\'%3E%3C/svg%3E", null ]

[ "# How do you simplify sqrt((32a^4)/b^2)?\n\nMay 8, 2017\n\nSee the solution process below:\n\n#### Explanation:\n\nWe can rewrite this expression as:\n\n$\\sqrt{\\frac{\\left(16 \\cdot 2\\right) {a}^{4}}{b} ^ 2} \\implies \\sqrt{\\frac{16 {a}^{4}}{b} ^ 2 \\cdot 2}$\n\nUsing this rule for radicals we can further rewrite this expression as:\n\n$\\sqrt{a \\cdot b} = \\sqrt{a} \\cdot \\sqrt{b}$\n\n$\\sqrt{\\frac{16 {a}^{4}}{b} ^ 2 \\cdot 2} \\implies \\sqrt{\\frac{16 {a}^{4}}{b} ^ 2} \\cdot \\sqrt{2}$\n\nWe can now take the square root of the left radical to simplify this expression as:\n\n$\\frac{\\pm 4 {a}^{2}}{b} \\sqrt{2}$\n\nOr, because the original question is given with only the principal root indicated:\n\n$\\frac{4 {a}^{2}}{b} \\sqrt{2}$" ]

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[ "# #64541a Color Information\n\nIn a RGB color space, hex #64541a is composed of 39.2% red, 32.9% green and 10.2% blue. Whereas in a CMYK color space, it is composed of 0% cyan, 16% magenta, 74% yellow and 60.8% black. It has a hue angle of 47 degrees, a saturation of 58.7% and a lightness of 24.7%. #64541a color hex could be obtained by blending #c8a834 with #000000. Closest websafe color is: #666633.\n\n• R 39\n• G 33\n• B 10\nRGB color chart\n• C 0\n• M 16\n• Y 74\n• K 61\nCMYK color chart\n\n#64541a color description : Very dark yellow [Olive tone].\n\n# #64541a Color Conversion\n\nThe hexadecimal color #64541a has RGB values of R:100, G:84, B:26 and CMYK values of C:0, M:0.16, Y:0.74, K:0.61. Its decimal value is 6575130.\n\nHex triplet RGB Decimal 64541a `#64541a` 100, 84, 26 `rgb(100,84,26)` 39.2, 32.9, 10.2 `rgb(39.2%,32.9%,10.2%)` 0, 16, 74, 61 47°, 58.7, 24.7 `hsl(47,58.7%,24.7%)` 47°, 74, 39.2 666633 `#666633`\nCIE-LAB 36.224, -0.525, 34.876 8.612, 9.125, 2.285 0.43, 0.456, 9.125 36.224, 34.88, 90.862 36.224, 13.325, 33.315 30.207, -1.971, 16.661 01100100, 01010100, 00011010\n\n# Color Schemes with #64541a\n\n• #64541a\n``#64541a` `rgb(100,84,26)``\n• #1a2a64\n``#1a2a64` `rgb(26,42,100)``\nComplementary Color\n• #642f1a\n``#642f1a` `rgb(100,47,26)``\n• #64541a\n``#64541a` `rgb(100,84,26)``\n• #4f641a\n``#4f641a` `rgb(79,100,26)``\nAnalogous Color\n• #2f1a64\n``#2f1a64` `rgb(47,26,100)``\n• #64541a\n``#64541a` `rgb(100,84,26)``\n• #1a4f64\n``#1a4f64` `rgb(26,79,100)``\nSplit Complementary Color\n• #541a64\n``#541a64` `rgb(84,26,100)``\n• #64541a\n``#64541a` `rgb(100,84,26)``\n• #1a6454\n``#1a6454` `rgb(26,100,84)``\n• #641a2a\n``#641a2a` `rgb(100,26,42)``\n• #64541a\n``#64541a` `rgb(100,84,26)``\n• #1a6454\n``#1a6454` `rgb(26,100,84)``\n• #1a2a64\n``#1a2a64` `rgb(26,42,100)``\n• #27210a\n``#27210a` `rgb(39,33,10)``\n• #3c320f\n``#3c320f` `rgb(60,50,15)``\n• #504315\n``#504315` `rgb(80,67,21)``\n• #64541a\n``#64541a` `rgb(100,84,26)``\n• #78651f\n``#78651f` `rgb(120,101,31)``\n• #8c7625\n``#8c7625` `rgb(140,118,37)``\n• #a1872a\n``#a1872a` `rgb(161,135,42)``\nMonochromatic Color\n\n# Alternatives to #64541a\n\nBelow, you can see some colors close to #64541a. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #64421a\n``#64421a` `rgb(100,66,26)``\n• #64481a\n``#64481a` `rgb(100,72,26)``\n• #644e1a\n``#644e1a` `rgb(100,78,26)``\n• #64541a\n``#64541a` `rgb(100,84,26)``\n• #645a1a\n``#645a1a` `rgb(100,90,26)``\n• #64601a\n``#64601a` `rgb(100,96,26)``\n• #62641a\n``#62641a` `rgb(98,100,26)``\nSimilar Colors\n\n# #64541a Preview\n\nThis text has a font color of #64541a.\n\n``<span style=\"color:#64541a;\">Text here</span>``\n#64541a background color\n\nThis paragraph has a background color of #64541a.\n\n``<p style=\"background-color:#64541a;\">Content here</p>``\n#64541a border color\n\nThis element has a border color of #64541a.\n\n``<div style=\"border:1px solid #64541a;\">Content here</div>``\nCSS codes\n``.text {color:#64541a;}``\n``.background {background-color:#64541a;}``\n``.border {border:1px solid #64541a;}``\n\n# Shades and Tints of #64541a\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #070602 is the darkest color, while #fdfbf6 is the lightest one.\n\n• #070602\n``#070602` `rgb(7,6,2)``\n• #161306\n``#161306` `rgb(22,19,6)``\n• #26200a\n``#26200a` `rgb(38,32,10)``\n• #352d0e\n``#352d0e` `rgb(53,45,14)``\n• #453a12\n``#453a12` `rgb(69,58,18)``\n• #544716\n``#544716` `rgb(84,71,22)``\n• #64541a\n``#64541a` `rgb(100,84,26)``\n• #74611e\n``#74611e` `rgb(116,97,30)``\n• #836e22\n``#836e22` `rgb(131,110,34)``\n• #937b26\n``#937b26` `rgb(147,123,38)``\n• #a2882a\n``#a2882a` `rgb(162,136,42)``\n• #b2952e\n``#b2952e` `rgb(178,149,46)``\n• #c1a232\n``#c1a232` `rgb(193,162,50)``\n``#ccad3b` `rgb(204,173,59)``\n• #d0b34b\n``#d0b34b` `rgb(208,179,75)``\n• #d4ba5a\n``#d4ba5a` `rgb(212,186,90)``\n• #d8c06a\n``#d8c06a` `rgb(216,192,106)``\n• #dcc779\n``#dcc779` `rgb(220,199,121)``\n• #e0cd89\n``#e0cd89` `rgb(224,205,137)``\n• #e4d499\n``#e4d499` `rgb(228,212,153)``\n• #e8dba8\n``#e8dba8` `rgb(232,219,168)``\n• #ece1b8\n``#ece1b8` `rgb(236,225,184)``\n• #f1e8c7\n``#f1e8c7` `rgb(241,232,199)``\n• #f5eed7\n``#f5eed7` `rgb(245,238,215)``\n• #f9f5e6\n``#f9f5e6` `rgb(249,245,230)``\n• #fdfbf6\n``#fdfbf6` `rgb(253,251,246)``\nTint Color Variation\n\n# Tones of #64541a\n\nA tone is produced by adding gray to any pure hue. In this case, #42413c is the less saturated color, while #7c6202 is the most saturated one.\n\n• #42413c\n``#42413c` `rgb(66,65,60)``\n• #474337\n``#474337` `rgb(71,67,55)``\n• #4c4632\n``#4c4632` `rgb(76,70,50)``\n• #51492d\n``#51492d` `rgb(81,73,45)``\n• #554c29\n``#554c29` `rgb(85,76,41)``\n• #5a4e24\n``#5a4e24` `rgb(90,78,36)``\n• #5f511f\n``#5f511f` `rgb(95,81,31)``\n• #64541a\n``#64541a` `rgb(100,84,26)``\n• #695715\n``#695715` `rgb(105,87,21)``\n• #6e5a10\n``#6e5a10` `rgb(110,90,16)``\n• #735c0b\n``#735c0b` `rgb(115,92,11)``\n• #775f07\n``#775f07` `rgb(119,95,7)``\n• #7c6202\n``#7c6202` `rgb(124,98,2)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #64541a is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

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[ "# A New Ontological Interpretation of the Wave Function\n\nHome Forums 2015 International Workshop on Quantum Foundations Meaning of the wave function A New Ontological Interpretation of the Wave Function\n\nViewing 5 posts - 1 through 5 (of 5 total)\n• Author\nPosts\n• #2432\n\nIn this paper, we propose a new ontological interpretation of the wave function in terms of random discontinuous motion of particles. According to this interpretation, the wave function of an N-body quantum system describes the state of random discontinuous motion of N particles, and in particular, the modulus squared of the wave function gives the probability density that the particles appear in every possible group of positions in space. We present two arguments supporting this new interpretation of the wave function. The arguments are mainly based on an analysis of the mass and charge properties of a quantum system. It is realized that the Schr\\”{o}dinger equation, which governs the evolution of a quantum system, contains more information about the system than the wave function of the system, such as the mass and charge properties of the system, which might help understand the ontological meaning of the wave function. Finally, we briefly analyze possible implications of the suggested ontological interpretation of the wave function for the solutions to the measurement problem.\n\n#2685\n\nThis is a very interesting idea. Here are a couple of questions about how it might work:\n\nYou write that “it seems natural to assume that the origin of the Born probabilities is the random discontinuous motion of particles”. Certainly some kind of reconciliation of your interpretation of the wave function with the Born rule is required. If a position measurement consisted of sampling the position of the particle at an instant in time, I can see how this might go. But measurements take place over finite intervals of time. How could that kind of process pick out one position (with the relevant probability)?\n\nSecond, how is interference handled in your approach. Presumably, in a two-slit context, the particle occupies the top-slit wave packet half the time and the bottom-slit wave packet half the time. But then how is interaction between the wave packets to be explained if the particle is always in either one or the other? The particle presumably can’t interact with itself (as you remark elsewhere). Any suggestions?\n\n#2767\n\nHi Peter, thanks a lot for your very helpful comments. Your questions are closely related to the understanding of RDM (random discontinuous motion) of particles. RDM gives an ontological interpretation of the wave function, but the instantaneous picture cannot explain interference and measurement. To explain the former, we still need the law of motion, which is supposed to be the linear Schrodinger equation for the wave function (which describes the state of RDM during a time interval dt). For the latter, we still need a solution to the measurement problem. Here RDM may help. My idea is that RDM may be the source of the randomness of measurement results, and especially, if wavefunction collapse is real, RDM may be the random noise that collapses the wave function. I proposed a model here (http://rspa.royalsocietypublishing.org/content/469/2153/20120526). No doubt further study is needed for my idea of RDM. Best, Shan\n\n#2773" ]

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[ "# Python - Data Type Boolean\n\n## Introduction\n\nPython has an explicit Boolean data type called bool.\n\nbool type has the values True and False as preassigned built-in names.\n\n## Literal\n\nTrue and False are instances of bool, which is just a subclass of the built-in integer type int.\n\nTrue and False behave exactly like the integers 1 and 0, except that they have customized printing logic.\n\nThey print themselves as the words True and False, instead of the digits 1 and 0.\n\nbool accomplishes this by redefining str and repr string formats for its two objects.\n\nYou can treat True and False as though they are predefined variables set to integers 1 and 0.\n\nBecause True is just the integer 1 with a custom display format, True + 4 yields integer 5 in Python!\n\n## Demo\n\n```print( type(True) )\nprint( isinstance(True, int) )\nprint( True == 1 ) # Same value\nprint( True is 1 ) # But a different object: see the next chapter\nprint( True or False ) # Same as: 1 or 0\nprint( True + 4 ) # (Hmmm)\n# from ww w . ja v a 2 s . co m\n```\n\n## Result", null, "Python has Boolean type values.\n\nBoolean type has predefined True and False objects.\n\n## Demo\n\n```print( 1 > 2, 1 < 2 ) # Booleans\nprint( bool('test') ) # Object's Boolean value\n# ww w. j a v a2 s .co m\nX = None # None placeholder\nprint( print(X) )\nL = [None] * 10 # Initialize a list of 10 Nones\nprint(L)\n```\n\n## Result", null, "" ]

[ null, "http://www.java2s.com/example/python-book/boolean-ef6a0.png", null, "http://www.java2s.com/example/python-book/boolean-5335f.png", null ]

[ "DISTRIBUTED MAPJOIN is an optimized version of MAPJOIN. You can use DISTRIBUTED MAPJOIN when you join a small table with a large table. You can use DISTRIBUTED MAPJOIN and MAPJOIN to reduce shuffling and sorting on the large table.\n\n## Precautions\n\n• The sizes of the tables that you want to join must be different. The size of the large table must be greater than 10 TB, and the size of the small table must be within the range of [1 GB, 100 GB].\n• Data in the small table must be evenly distributed. If a small table contains long tails, excessive data is generated in a single shard of the table. As a result, an out of memory (OOM) error and the remote procedure call (RPC) timeout issue may occur.\n• If an SQL task runs more than 20 minutes, we recommend that you use DISTRIBUTED MAPJOIN for optimization.\n• Excessive resources are occupied when a task is running. Therefore, we recommend that you do not run a task in a small quota group.\nNote On the Quotas page, you can change the quota group. For more information, see Configure quota groups.\n\n## Use DISTRIBUTED MAPJOIN\n\nTo use `DISTRIBUTED MAPJOIN`, you must add the hint `/*+distmapjoin(<table_name>(shard_count=<n>,replica_count=<m>))*/` to a `SELECT` statement. Both the shard_count and replica_count parameters are used to determine the parallelism of tasks. Formula: `Parallelism of tasks = shard_count × replica_count`.\n• Parameters\n• table_name: the name of the small table that you want to join.\n• shard_count=<n>: the number of data shards of the small table that you want to join. The data shards of the small table are distributed on each compute node for data processing. n: the number of shards. In most cases, this parameter is set to an odd number.\nNote\n• We recommend that you manually specify the shard_count parameter. You can estimate the value of the shard_count parameter based on the size of the small table. The estimated data amount that is processed by a single shard node is within the range of [200 MB, 500 MB].\n• If you set the shard_count parameter to an excessively large value, the data processing performance and stability are affected. If you set the shard_count parameter to an excessively small value, an error may occur due to the excessive use of memory.\n• replica_count=<m>: the number of replicas of the small table. m indicates the number of replicas. Default value: 1.\nNote To reduce access pressure and prevent the failure of the entire task caused by the failure of a single node, you can create multiple replicas of data in the same shard. If a node frequently restarts because the parallelism of tasks is high or the environment is unstable, you can increase the value of the replica_count parameter. You can set this parameter to 2 or 3.\n• Syntax\n``````-- Recommended. Specify the shard_count parameter and retain the default value 1 for the replica_count parameter.\n/*+distmapjoin(a(shard_count=5))*/\n\n-- Recommended. Specify the shard_count and replica_count parameters.\n/*+distmapjoin(a(shard_count=5,replica_count=2))*/\n\n-- Use DISTRIBUTED MAPJOIN to join multiple small tables.\n/*+distmapjoin(a(shard_count=5,replica_count=2),b(shard_count=5,replica_count=2)) */\n\n-- Use DISTRIBUTED MAPJOIN and MAPJOIN together.\n/*+distmapjoin(a(shard_count=5,replica_count=2)),mapjoin(b)*/``````\n\n## Example\n\nThis example describes how to use DISTRIBUTED MAPJOIN when you insert data into the partitioned table tmall_dump_lasttable.\n• Standard syntax\n``````insert OVERWRITE table tmall_dump_lasttable partition(ds='20211130')\nselect t1.*\nfrom\n(\nselect nid, doc,type\nfrom search_ods.dump_lasttable where ds='20211203'\n)t1\njoin\n(\nselect distinct item_id\nfrom tbcdm.dim_tb_itm\nwhere ds='20211130'\nand bc_type='B'\nand is_online='Y'\n)t2\non t1.nid=t2.item_id;``````\n• Syntax after optimization\n``````insert OVERWRITE table tmall_dump_lasttable partition (ds='20211130')\nselect /*+ distmapjoin(t2(shard_count=35)) */ t1.*\nfrom\n(\nselect nid, doc, type\nfrom search_ods.dump_lasttable where ds='20211203'\n)t1\njoin\n(\nselect distinct item_id\nfrom tbcdm.dim_tb_itm\nwhere ds='20211130'\nand bc_type='B'\nand is_online='Y'\n)t2\non t1.nid=t2.item_id;``````" ]

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[ "Hi Everyone!\n\nOn this page you will find some material about Lesson 32. Read through the material below, watch the videos, and follow up with your instructor if you have questions.\n\nLesson 32: Law of Sines\n\n### Resources\n\nIn this section you will find some important information about the specific resources related to this lesson:\n\n• the learning outcomes,\n• the section in the textbook,\n• the WeBWorK homework sets,\n• a link to the pdf of the lesson notes,\n• a link to a video lesson.\n\nLearning Outcomes (from Coburn and Herdlick’s Trigonometry book)\n\n• Develop the law of sines and use it to solve ASA and AAS triangles.\n• Solve SSA triangles (the ambiguous case) using the law of sines.\n• Use the law of sines to solve applications.\n\nTopic. This lesson covers Section 7.1: Oblique Triangles and the Law of Sines.\n\nWeBWorK. There is one WeBWorK assignment on today’s material:\n\nLawOfSines\n\nLesson Notes.\n\nThese notes are used in Lessons 32 and 33. Today’s lesson is on pages 1-3.\n\nVideo Lesson.\n\nVideo Lesson 32 (based on Lesson 32 Notes)\n\nThis video is used in Lessons 32 and 33. For today’s lesson, watch from [0:00] to [8:01].\n\n### Warmup Questions\n\nThese are questions on fundamental concepts that you need to know before you can embark on this lesson. Don’t skip them! Take your time to do them, and check your answer by clicking on the “Show Answer” tab.\n\n#### Warmup Question 1\n\nIf", null, ", what is", null, "?\n\n#### Show Answer 1", null, "#### Warmup Question 2\n\nFind", null, "in degrees.\n\n#### Show Answer 2", null, "### Review\n\nIf you are not comfortable with the Warmup Questions, don’t give up! Click on the indicated lesson for a quick catchup. A brief review will help you boost your confidence to start the new lesson, and that’s perfectly fine.\n\nNeed a review? Check\n\n### Quick Intro\n\nThis is like a mini-lesson with an overview of the main objects of study. It will often contain a list of key words, definitions and properties – all that is new in this lesson. We will use this opportunity to make connections with other concepts. It can be also used as a review of the lesson.\n\nA Quick Intro to Law of Sines\n\nKey Words. SSS, SAS, AAS, ASA, Oblique triangle, solving a triangle, law of sines", null, "We denote the sides of a tringle", null, "by", null, ",", null, "and", null, "as the sides opposite to the angles", null, ",", null, "and", null, ", respectively.", null, "SSS means that the three sides are known.", null, "SAS means that two sides and the adjacent angle are known.", null, "AAS means that two angles and one side (not between the two angles) are known.", null, "ASA means that two angles and the adjacent side are known.", null, "Solving a triangle means to find the unknown sides and angles.", null, "A triangle that does not have a right angle is called oblique.", null, "In a right triangle, you use the trig ratios to solve it. Otherwise, the triangle is oblique in which case consider:", null, "Law of sines (for ASA/AAS triangles)", null, "### Video Lesson\n\nMany times the mini-lesson will not be enough for you to start working on the problems. You need to see someone explaining the material to you. In the video you will find a variety of examples, solved step-by-step – starting from a simple one to a more complex one. Feel free to play them as many times as you need. Pause, rewind, replay, stop… follow your pace!\n\nVideo Lesson\n\nThis video is used in Lessons 32 and 33. For today’s lesson, watch from [0:00] to {8:01].\n\nA description of the video\n\nIn the video you will see the following problems.\n\n• Given a triangle whose angles are", null, ",", null, "and", null, "where the sides opposite to", null, ",", null, "and", null, "measure 5,", null, "and", null, ", respectively, find", null, "and", null, ".\n• Given a triangle whose angles are", null, "and", null, ", and the sides opposite to", null, "and", null, "measure 5 and", null, ", respectively, find", null, ".\n\n### Try Questions\n\nNow that you have read the material and watched the video, it is your turn to put in practice what you have learned. We encourage you to try the Try Questions on your own. When you are done, click on the “Show answer” tab to see if you got the correct answer.\n\n#### Try Question 1\n\nSolve the triangle having the following properties: side", null, "= 385 m,", null, ", and side", null, "=490 m.\n\n#### Show Answer 1\n\nBy the law of cosines,", null, "", null, "", null, "", null, "By the law of sines,", null, "", null, "", null, "", null, "", null, "Hence", null, ".", null, "sides:", null, "", null, "", null, "", null, "angles:", null, "", null, "", null, "### WeBWorK\n\nYou should now be ready to start working on the WeBWorK problems. Doing the homework is an essential part of learning. It will help you practice the lesson and reinforce your knowledge.\n\nWeBWork\n\nIt is time to do the homework on WeBWork:\n\nLawOfSines\n\nWhen you are done, come back to this page for the Exit Questions.\n\n### Exit Questions\n\nAfter doing the WeBWorK problems, come back to this page. The Exit Questions include vocabulary checking and conceptual questions. Knowing the vocabulary accurately is important for us to communicate. You will also find one last problem. All these questions will give you an idea as to whether or not you have mastered the material. Remember: the “Show Answer” tab is there for you to check your work!\n\n#### Exit Questions\n\n• Why is there not an SSA theorem?\n• When can you use the law of sines?\n• Why is it better to use more accuracy or exact answers in early parts of solving triangles?", null, "Solve the triangle", null, "for which", null, ",", null, ", and", null, "." ]

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[ "### - Art Gallery -\n\nIn physics, an anyon is a type of quasiparticle that occurs only in two-dimensional systems, with properties much less restricted than fermions and bosons. In general, the operation of exchanging two identical particles may cause a global phase shift but cannot affect observables. Anyons are generally classified as abelian or non-abelian. Abelian anyons have been detected and play a major role in the fractional quantum Hall effect. Non-abelian anyons have not been definitively detected, although this is an active area of research.\n\nIntroduction\n\nThe statistical mechanics of large many-body systems obey laws described by Maxwell-Boltzmann statistics. Quantum statistics is more complicated because of the different behaviors of two different kinds of particles called fermions and bosons. Quoting a recent, simple description from Aalto University:\n\nIn the three-dimensional world we live in, there are only two types of particles: \"fermions,\" which repel each other, and \"bosons,\" which like to stick together. A commonly known fermion is the electron, which transports electricity; and a commonly known boson is the photon, which carries light. In the two-dimensional world, however, there is another type of particle, the anyon, which doesn't behave like either a fermion or a boson. The exact quantum nature of anyons lies in their wave nature, encoded in their quantum statistics.\n\nMicrosoft has invested in research concerning anyons as a potential basis for topological quantum computing. Anyons circling each other (braiding) would encode information in a more robust way than other potential quantum computing technologies. Most investment in quantum computing, however, is based on methods that do not use anyons.\nAbelian anyons\n\nIn quantum mechanics, and some classical stochastic systems, indistinguishable particles have the property that exchanging the states of particle i with particle j (symbolically $$\\psi _{i}\\leftrightarrow \\psi _{j}{\\text{ for }}i\\neq j})$$ does not lead to a measurably different many-body state.\n\nIn a quantum mechanical system, for example, a system with two indistinguishable particles, with particle 1 in state $$\\psi _{1}$$ and particle 2 in state $$\\psi _{2}$$, has state $$\\left|\\psi _{1}\\psi _{2}\\right\\rangle$$ in Dirac notation. Now suppose we exchange the states of the two particles, then the state of the system would be $$\\left|\\psi _{2}\\psi _{1}\\right\\rangle }$$. These two states should not have a measurable difference, so they should be the same vector, up to a phase factor:\n\n$$\\left|\\psi _{1}\\psi _{2}\\right\\rangle =e^{i\\theta }\\left|\\psi _{2}\\psi _{1}\\right\\rangle .}$$\n\nIn space of three or more dimensions, elementary particles are either fermions or bosons, according to their statistical behaviour. Fermions obey Fermi–Dirac statistics, while bosons obey Bose–Einstein statistics. For bosons, the phase factor is 1, and for fermions, it is -1. In particular, this is why fermions obey Pauli exclusion principle: If two fermions are in the same state, then we have\n\n$$\\left|\\psi \\psi \\right\\rangle =-\\left|\\psi \\psi \\right\\rangle .}$$\n\nThe state vector must be zero, which means it's not normalizable, thus unphysical.\n\nIn two-dimensional systems, however, quasiparticles can be observed that obey statistics ranging continuously between Fermi–Dirac and Bose–Einstein statistics, as was first shown by Jon Magne Leinaas and Jan Myrheim of the University of Oslo in 1977. In the case of two particles this can be expressed as\n\n$$\\left|\\psi _{1}\\psi _{2}\\right\\rangle =e^{i\\theta }\\left|\\psi _{2}\\psi _{1}\\right\\rangle ,}$$\n\nwhere $$e^{i\\theta }$$ can be other values than just -1 or 1. It is important to note that there is a slight abuse of notation in this shorthand expression, as in reality this wave function can be and usually is multi-valued. This expression actually means that when particle 1 and particle 2 are interchanged in a process where each of them makes a counterclockwise half-revolution about the other, the two-particle system returns to its original quantum wave function except multiplied by the complex unit-norm phase factor eiθ. Conversely, a clockwise half-revolution results in multiplying the wave function by e−iθ. Such a theory obviously only makes sense in two-dimensions, where clockwise and counterclockwise are clearly defined directions.\n\nIn the case θ = π we recover the Fermi–Dirac statistics (eiπ = −1) and in the case θ = 0 (or θ = 2π) the Bose–Einstein statistics (e2πi = 1). In between we have something different. Frank Wilczek in 1982 explored the behavior of such quasiparticles and coined the term \"anyon\" to describe them, because they can have any phase when particles are interchanged. Unlike bosons and fermions, anyons have the peculiar property that when they are interchanged twice in the same way (e.g. if anyon 1 and anyon 2 were revolved counterclockwise by half revolution about each other to switch places, and then they were revolved counterclockwise by half revolution about each other again to go back to their original places), the wave function is not necessarily the same but rather generally multiplied by some complex phase (by e2iθ in this example).\n\nWe may also use θ = 2π s with particle spin quantum number s, with s being integer for bosons, half-integer for fermions, so that\n\n$$\\left|\\psi _{1}\\psi _{2}\\right\\rangle =(-1)^{2s}\\left|\\psi _{2}\\psi _{1}\\right\\rangle .} \\left|\\psi _{1}\\psi _{2}\\right\\rangle =(-1)^{2s}\\left|\\psi _{2}\\psi _{1}\\right\\rangle .}$$\n\nAt an edge, fractional quantum Hall effect anyons are confined to move in one space dimension. Mathematical models of one-dimensional anyons provide a base of the commutation relations shown above.\n\nIn a three-dimensional position space, the fermion and boson statistics operators (−1 and +1 respectively) are just 1-dimensional representations of the permutation group (SN of N indistinguishable particles) acting on the space of wave functions. In the same way, in two-dimensional position space, the abelian anyonic statistics operators (eiθ) are just 1-dimensional representations of the braid group (BN of N indistinguishable particles) acting on the space of wave functions. Non-abelian anyonic statistics are higher-dimensional representations of the braid group. Anyonic statistics must not be confused with parastatistics, which describes statistics of particles whose wavefunctions are higher-dimensional representations of the permutation group.:22\nTopological equivalence\n\nThe fact that the homotopy classes of paths (i.e. notion of equivalence on braids) are relevant hints at a more subtle insight. It arises from the Feynman path integral, in which all paths from an initial to final point in spacetime contribute with an appropriate phase factor. Recall that the Feynman path integral can be motivated from expanding the propagator using a method called time-slicing, in which time is discretized.\n\nIn non-homotopic paths, one cannot get from any point at one time slice to any other point at the next time slice. This means that we can consider homotopic equivalence class of paths to have different weighting factors.\n\nSo it can be seen that the topological notion of equivalence comes from a study of the Feynman path integral.:28\n\nFor a more transparent way of seeing that the homotopic notion of equivalence is the \"right\" one to use, see Aharonov–Bohm effect.\nExperiment\n\nA group of theoretical physicists working at the University of Oslo, led by Jon Leinaas and Jan Myrheim, calculated in 1977 that the traditional division between fermions and bosons would not apply to theoretical particles existing in two dimensions. Such particles would be expected to exhibit a diverse range of previously unexpected properties. In 1982, Frank Wilczek published in two papers, exploring the fractional statistics of quasiparticles in two dimensions, giving them the name \"anyons.\"\nLaughlin quasiparticle interferometer scanning electron micrograph of a semiconductor device. The four light-grey regions are Au/Ti gates of undepleted electrons; the blue curves are the edge channels from the equipotentials of these undepleted electrons. The dark-grey curves are etched trenches depleted of electrons, the blue dots are the tunneling junctions, the yellow dots are Ohmic contacts. The electrons in the device are confined to a 2d plane.\n\nDaniel Tsui and Horst Störmer discovered the fractional quantum Hall effect in 1982. The mathematics developed by Wilczek proved to be useful to Bertrand Halperin at Harvard University in explaining aspects of it. Frank Wilczek, Dan Arovas, and Robert Schrieffer verified this statement in 1985 with an explicit calculation that predicted that particles existing in these systems are in fact anyons.\n\nIn 2005 a group of physicists at Stony Brook University constructed a quasiparticle interferometer, detecting the patterns caused by interference of anyons, which were interpreted to suggest that anyons are real, rather than just a mathematical construct. However, these experiments remain controversial and are not fully accepted by the community.\n\nIn 2020, H. Bartolomei and co-authors from the École normale supérieure (Paris) from an experiment in two-dimensional the heterostructure GaAs/AlGaAs was determined intermediate anyon statistics $$\\theta ={\\frac {\\pi }{3}}}$$ by electrical correlation measurements currents through the third contact in anyon collisions in electronic gas from two-point contacts .\n\nWith developments in semiconductor technology meaning that the deposition of thin two-dimensional layers is possible – for example, in sheets of graphene – the long-term potential to use the properties of anyons in electronics is being explored.\n\nIn 2020, a team of scientists at Purdue University announced new experimental evidence for the existence of anyons. The team's interferometer routes the electrons through a specific maze-like etched nanostructure made of gallium arsenide and aluminum gallium arsenide. \"In the case of our anyons the phase generated by braiding was 2π/3,\" he said. \"That's different than what's been seen in nature before.\"\n\nNon-abelian anyons\nUnsolved problem in physics:\nIs topological order stable at non-zero temperature?\n(more unsolved problems in physics)\n\nIn 1988, Jürg Fröhlich showed that it was valid under the spin–statistics theorem for the particle exchange to be monoidal (non-abelian statistics). In particular, this can be achieved when the system exhibits some degeneracy, so that multiple distinct states of the system have the same configuration of particles. Then an exchange of particles can contribute not just a phase change, but can send the system into a different state with the same particle configuration. Particle exchange then corresponds to a linear transformation on this subspace of degenerate states. When there is no degeneracy, this subspace is one-dimensional and so all such linear transformations commute (because they are just multiplications by a phase factor). When there is degeneracy and this subspace has higher dimension, then these linear transformations need not commute (just as matrix multiplication does not).\n\nGregory Moore, Nicholas Read, and Xiao-Gang Wen pointed out that non-Abelian statistics can be realized in the fractional quantum Hall effect (FQHE). While at first non-abelian anyons were generally considered a mathematical curiosity, physicists began pushing toward their discovery when Alexei Kitaev showed that non-abelian anyons could be used to construct a topological quantum computer. As of 2012, no experiment has conclusively demonstrated the existence of non-abelian anyons although promising hints are emerging in the study of the ν = 5/2 FQHE state. Experimental evidence of non-abelian anyons, although not yet conclusive and currently contested, was presented in October, 2013.\nFusion of anyons\n\nIn much the same way that two fermions (e.g. both of spin 1/2) can be looked at together as a composite boson (with total spin in a superposition of 0 and 1), two or more anyons together make up a composite anyon (possibly a boson or fermion). The composite anyon is said to be the result of the fusion of its components.\n\nIf N identical abelian anyons each with individual statistics α {\\displaystyle \\alpha } \\alpha (that is, the system picks up a phase $$e^{i\\alpha }}$$ when two individual anyons undergo adiabatic counterclockwise exchange) all fuse together, they together have statistics $$N^{2}\\alpha }$$. This can be seen by noting that upon counterclockwise rotation of two composite anyons about each other, there are $$N^{2}$$ pairs of individual anyons (one in the first composite anyon, one in the second composite anyon) that each contribute a phase $$e^{i\\alpha }}$$. An analogous analysis applies to the fusion of non-identical abelian anyons. The statistics of the composite anyon is uniquely determined by the statistics of its components.\n\nNon-abelian anyons have more complicated fusion relations. As a rule, in a system with non-abelian anyons, there is a composite particle whose statistics label is not uniquely determined by the statistics labels of its components, but rather exists as a quantum superposition (this is completely analogous to how two fermions known to have spin 1/2 are together in quantum superposition of total spin 1 and 0). If the overall statistics of the fusion of all of several anyons is known, there is still ambiguity in the fusion of some subsets of those anyons, and each possibility is a unique quantum state. These multiple states provide a Hilbert space on which quantum computation can be done.\n\nTopological basis\nAnticlockwise rotation\nClockwise rotation\nExchange of two particles in 2 + 1 spacetime by rotation. The rotations are inequivalent, since one cannot be deformed into the other (without the worldlines leaving the plane, an impossibility in 2d space).\n\nIn more than two dimensions, the spin–statistics theorem states that any multiparticle state of indistinguishable particles has to obey either Bose–Einstein or Fermi–Dirac statistics. For any d > 2, the Lie groups SO(d,1) (which generalizes the Lorentz group) and Poincaré(d,1) have Z2 as their first homotopy group. Because the cyclic group Z2 is composed of two elements, only two possibilities remain. (The details are more involved than that, but this is the crucial point.)\n\nThe situation changes in two dimensions. Here the first homotopy group of SO(2,1), and also Poincaré(2,1), is Z (infinite cyclic). This means that Spin(2,1) is not the universal cover: it is not simply connected. In detail, there are projective representations of the special orthogonal group SO(2,1) which do not arise from linear representations of SO(2,1), or of its double cover, the spin group Spin(2,1). Anyons are evenly complementary representations of spin polarization by a charged particle.\n\nThis concept also applies to nonrelativistic systems. The relevant part here is that the spatial rotation group SO(2) has an infinite first homotopy group.\n\nThis fact is also related to the braid groups well known in knot theory. The relation can be understood when one considers the fact that in two dimensions the group of permutations of two particles is no longer the symmetric group S2 (with two elements) but rather the braid group B2 (with an infinite number of elements). The essential point is that one braid can wind around the other one, an operation that can be performed infinitely often, and clockwise as well as counterclockwise.\n\nA very different approach to the stability-decoherence problem in quantum computing is to create a topological quantum computer with anyons, quasi-particles used as threads and relying on braid theory to form stable logic gates.\nHigher dimensional generalization of anyons\n\nFractionalized excitations as point particles can be bosons, fermions or anyons in 2+1 spacetime dimensions. It is known that point particles can be only either bosons or fermions in 3+1 and higher spacetime dimensions. However, the loop (or string) or membrane like excitations are extended objects can have fractionalized statistics. Current research works show that the loop and string like excitations exist for topological orders in the 3+1 dimensional spacetime, and their multi-loop/string-braiding statistics are the key signatures for identifying 3+1 dimensional topological orders. The multi-loop/string-braiding statistics of 3+1 dimensional topological orders can be captured by the link invariants of particular topological quantum field theories in 4 spacetime dimensions. Explained in a colloquial manner, the extended objects (loop, string, or membrane, etc.) can be potentially anyonic in 3+1 and higher spacetime dimensions in the long-range entangled systems.\nLook up anyon in Wiktionary, the free dictionary.\n\nAnyonic Lie algebra – Subset of algebra\nFlux tube – Tube-like region of space with constant magnet flux along its length\nGinzburg–Landau theory – Superconductivity theory\nHusimi Q representation – Computational physics simulation tool\nJosephson effect – Quantum physical phenomenon\nMacroscopic quantum phenomena – Processes showing quantum behavior at the macroscopic scale, rather than at the atomic scale where quantum effects are prevalent; macroscopic scale quantum coherence leads to macroscopic quantum phenomena\nMagnetic domain – Region of a magnetic material in which the magnetization has uniform direction\nMagnetic flux quantum – Quantized unit of magnetic flux\nMeissner effect – Expulsion of a magnetic field from a superconductor during its transition to the superconducting state\nPlekton – Theoretical particle\nQuantum vortex – Quantized flux circulation of some physical quantity\nRandom matrix – Matrix-valued random variable\nTopological defect – Type of structure in quantum mechanics\nTopological quantum computing – Hypothetical fault-tolerant quantum computer based on topological condensed matter\n\nReferences\n\nYirka, Bob (10 April 2020). \"Anyon evidence observed using tiny anyon collider\". Phys.org.\n\"Finally, anyons reveal their exotic quantum properties\". Aalto University. 7 December 2018. Retrieved 24 September 2020. \"They were first proposed in the late 1970s, but direct experimental evidence of their quantum statistics hasn't been conclusively shown until now.\"\nCastelvecchi, Davide (3 July 2020). \"Welcome anyons! Physicists find best evidence yet for long-sought 2D structures\". Nature. Retrieved 23 September 2020. \"Simon and others have developed elaborate theories that use anyons as the platform for quantum computers. Pairs of the quasiparticle could encode information in their memory of how they have circled around one another. And because the fractional statistics is 'topological' — it depends on the number of times one anyon went around another, and not on slight changes to its path – it is unaffected by tiny perturbations. This robustness could make topological quantum computers easier to scale up than are current quantum-computing technologies, which are error-prone.\"\nLeinaas, Jon Magne; Myrheim, Jan (11 January 1977). \"On the theory of identical particles\" (PDF). Il Nuovo Cimento B. 37 (1): 1–23. Bibcode:1977NCimB..37....1L. doi:10.1007/BF02727953.\nWilczek, Frank (4 October 1982). \"Quantum Mechanics of Fractional-Spin Particles\" (PDF). Physical Review Letters. 49 (14): 957–959. Bibcode:1982PhRvL..49..957W. doi:10.1103/PhysRevLett.49.957. \"If there is a generalized spin-statistics connection, we must expect that the flux-tube-particle composites have unusual statistics, interpolating between bosons and fermions. Since interchange of two of these particles can give any phase, I will call them generically anyons.\"\nKhare, Avinash (2005). Fractional Statistics and Quantum Theory. World Scientific. ISBN 978-981-256-160-2.\nLancaster, Tom; Blundell, Stephen J. (17 June 2014). Quantum Field Theory for the Gifted Amateur. Oxford University Press. ISBN 0-19-969932-1.\nSchulman, L. S. (February 1981). Techniques and Applications of Path Integration. Dover Publications. ISBN 0-471-76450-7.\nWilczek, Frank (January 2006). \"From electronics to anyonics\". Physics World. ISSN 0953-8585. \"In the early 1980s I named the hypothetical new particles 'anyons,' the idea being that anything goes – but I did not lose much sleep anticipating their discovery. Very soon afterwards, however, Bert Halperin at Harvard University found the concept of anyons useful in understanding certain aspects of the fractional quantum Hall effect, which describes the modifications that take place in electronics at low temperatures in strong magnetic fields.\"\n\"Anyons, anyone?\". Symmetry Magazine. 31 August 2011. Retrieved 24 September 2020. \"In 1982 physicist Frank Wilczek gave these interstitial particles the name anyon...'Any anyon can be anything between a boson or a fermion,' Keilmann says. 'Wilczek is a funny guy.'\"\nCamino, Fernando E.; Zhou, Wei; Goldman, Vladimir J. (17 August 2005). \"Realization of a Laughlin quasiparticle interferometer: Observation of fractional statistics\" (PDF). Physical Review B. 72 (7). arXiv:cond-mat/0502406. Bibcode:2005PhRvB..72g5342C. doi:10.1103/PhysRevB.72.075342. Archived from the original (PDF) on 19 June 2015., see fig. 2.B\nHalperin, B. I. (1984). \"Statistics of Quasiparticles and the Hierarchy of Fractional Quantized Hall States\". Physical Review Letters American Physical Society. 52 (18): 1583–1586. doi:10.1103/PhysRevLett.52.1583. \"The appearance of fractional statistics in the present context is strongly reminiscent of the fractional statistics introduced by Wilczek to describe charged particles tied to \"magnetic flux tubes\" in two dimensions.\"\nH. Bartolomei, M. Kumar, R. Bisognin et al. Fractional statistics in anyon collisions // Science, 10 April 2020: Vol. 368, Issue 6487, pp. 173-177\nTally, Steve (4 September 2020). \"New evidence that the quantum world is even stranger than we thought\". Phys.org. \"One characteristic difference between fermions and bosons is how the particles act when they are looped, or braided, around each other. Fermions respond in one straightforward way, and bosons in another expected and straightforward way. Anyons respond as if they have a fractional charge, and even more interestingly, create a nontrivial phase change as they braid around one another. This can give the anyons a type of \"memory\" of their interaction.\"\nNakamura, J.; Liang, S.; Gardner, G. C.; Manfra, M. J. (September 2020). \"Direct observation of anyonic braiding statistics\". Nature Physics. 16 (9): 931–936. doi:10.1038/s41567-020-1019-1. ISSN 1745-2481.\nFröhlich, Jürg (1988). \"Statistics of Fields, the Yang–Baxter Equation, and the Theory of Knots and Links\". Nonperturbative Quantum Field Theory. New York: Springer. pp. 71–100. doi:10.1007/978-1-4613-0729-7_4. ISBN 1-4612-8053-2.\nMoore, Gregory; Read, Nicholas (19 August 1991). \"Nonabelions in the fractional quantum hall effect\" (PDF). Nuclear Physics B. 360 (2–3): 362–396. Bibcode:1991NuPhB.360..362M. doi:10.1016/0550-3213(91)90407-O.\nWen, Xiao-Gang (11 February 1991). \"Non-Abelian statistics in the fractional quantum Hall states\" (PDF). Physical Review Letters 66 (6): 802–5. Bibcode:1991PhRvL..66..802W. doi:10.1103/PhysRevLett.66.802. Archived from the original (PDF) on 26 March 2015.\nStern, Ady (2010). \"Non-Abelian states of matter\". Nature. 464 (7286): 187–93. Bibcode:2010Natur.464..187S. doi:10.1038/nature08915. PMID 20220836.\nAn, Sanghun; Jiang, P.; Choi, H.; Kang, W.; Simon, S. H.; Pfeiffer, L. N.; West, K. W.; Baldwin, K. W. (15 December 2011). \"Braiding of Abelian and Non-Abelian Anyons in the Fractional Quantum Hall Effect\". arXiv:1112.3400 [cond-mat.mes-hall].\nvon Keyserling, Curt; Simon, S. H.; Bernd, Rosenow (2015). \"Enhanced Bulk-Edge Coulomb Coupling in Fractional Fabry-Perot Interferometers\". Physical Review Letters. 115: 126807. arXiv:1411.4654. Bibcode:2015PhRvL.115l6807V. doi:10.1103/PhysRevLett.115.126807. PMID 26431008.\nR. L. Willett; C. Nayak; L. N. Pfeiffer; K. W. West (12 January 2013). \"Magnetic field-tuned Aharonov–Bohm oscillations and evidence for non-Abelian anyons at ν = 5/2\". Physical Review Letters. 111: 186401. arXiv:1301.2639. Bibcode:2013PhRvL.111r6401W. doi:10.1103/PhysRevLett.111.186401. PMID 24237543.\nC. Nayak; S.H. Simon; A. Stern; M. Freedman; S. Das Sarma (28 March 2008). \"Non-Abelian Anyons and Topological Quantum Computation\". Reviews of Modern Physics. 80: 1083–1159. arXiv:0707.1889. Bibcode:2008RvMP...80.1083N. doi:10.1103/RevModPhys.80.1083.\nFreedman, Michael; Alexei Kitaev; Michael Larsen; Zhenghan Wang (20 October 2002). \"Topological Quantum Computation\". Bulletin of the American Mathematical Society. 40 (1): 31–38. arXiv:quant-ph/0101025. doi:10.1090/S0273-0979-02-00964-3.\nMonroe, Don (1 October 2008). \"Anyons: The breakthrough quantum computing needs?\". New Scientist (2676).\nWang, Chenjie; Levin, Michael (22 August 2014). \"Braiding statistics of loop excitations in three dimensions\". Physical Review Letters. American Physical Society (APS). 113 (8): 080403. arXiv:1403.7437. Bibcode:2014PhRvL.113h0403W. doi:10.1103/PhysRevLett.113.080403. ISSN 1079-7114. PMID 25192079.\nWang, Juven; Wen, Xiao-Gang (15 January 2015). \"Non-Abelian String and Particle Braiding in Topological Order: Modular SL(3,Z) Representation and 3+1D Twisted Gauge Theory\". Physical Review B. American Physical Society (APS). 91 (3): 035134. arXiv:1404.7854. doi:10.1103/PhysRevB.91.035134. ISSN 2469-9969.\n\nPutrov, Pavel; Wang, Juven; Yau, Shing-Tung (September 2017). \"Braiding Statistics and Link Invariants of Bosonic/Fermionic Topological Quantum Matter in 2+1 and 3+1 dimensions\". Annals of Physics. 384C: 254–287. arXiv:1612.09298. Bibcode:2017AnPhy.384..254P. doi:10.1016/j.aop.2017.06.019.\n\nNayak, Chetan; Simon, Steven H.; Stern, Ady; Freedman, Michael; Das Sarma, Sankar (2008). \"Non-Abelian anyons and topological quantum computation\". Reviews of Modern Physics. 80 (3): 1083. arXiv:0707.1889. Bibcode:2008RvMP...80.1083N. doi:10.1103/RevModPhys.80.1083.\nWen, Xiao-Gang (15 April 2002). \"Quantum orders and symmetric spin liquids\" (PDF). Physical Review B. 65 (16): 165113. arXiv:cond-mat/0107071. Bibcode:2002PhRvB..65p5113W. doi:10.1103/PhysRevB.65.165113. Archived from the original (PDF) on 9 June 2011.\nStern, Ady (2008). \"Anyons and the quantum Hall effect—A pedagogical review\" (PDF). Annals of Physics. 323: 204. arXiv:0711.4697. Bibcode:2008AnPhy.323..204S. doi:10.1016/j.aop.2007.10.008.\nNajjar, Dana (2020). \"'Milestone' Evidence for Anyons, a Third Kingdom of Particles\". Quanta Magazine.\n\nvte\n\nParticles in physics\nElementary\nFermions\nQuarks\n\nUp (quark antiquark) Down (quark antiquark) Charm (quark antiquark) Strange (quark antiquark) Top (quark antiquark) Bottom (quark antiquark)\n\nLeptons\n\nElectron Positron Muon Antimuon Tau Antitau Electron neutrino Electron antineutrino Muon neutrino Muon antineutrino Tau neutrino Tau antineutrino\n\nBosons\nGauge\n\nScalar\n\nHiggs boson\n\nGhost fields\n\nHypothetical\nSuperpartners\nGauginos\n\nGluino Gravitino Photino\n\nOthers\n\nAxino Chargino Higgsino Neutralino Sfermion (Stop squark)\n\nOthers\n\nAxion Curvaton Dilaton Dual graviton Graviphoton Graviton Inflaton Leptoquark Magnetic monopole Majoron Majorana fermion Dark photon Planck particle Preon Sterile neutrino Tachyon W′ and Z′ bosons X and Y bosons\n\nComposite\nBaryons\n\nNucleon\nProton Antiproton Neutron Antineutron Delta baryon Lambda baryon Sigma baryon Xi baryon Omega baryon\n\nMesons\n\nPion Rho meson Eta and eta prime mesons Phi meson J/psi meson Omega meson Upsilon meson Kaon B meson D meson Quarkonium\n\nOthers\n\nAtomic nuclei Atoms Exotic atoms\nPositronium Muonium Tauonium Onia Pionium Superatoms Molecules\n\nHypothetical\nBaryons\n\nHexaquark Heptaquark Skyrmion\n\nMesons\n\nGlueball Theta meson T meson\n\nOthers\n\nMesonic molecule Pomeron Diquark R-hadron\n\nQuasiparticles\n\nAnyon Davydov soliton Dropleton Exciton Hole Magnon Phonon Plasmaron Plasmon Polariton Polaron Roton Trion\n\nLists\n\nBaryons Mesons Particles Quasiparticles Timeline of particle discoveries\n\nRelated\n\nHistory of subatomic physics\ntimeline Standard Model\nmathematical formulation Subatomic particles Particles Antiparticles Nuclear physics Eightfold way\nQuark model Exotic matter Massless particle Relativistic particle Virtual particle Wave–particle duality Particle chauvinism\n\nWikipedia books\n\nHadronic Matter Particles of the Standard Model Leptons Quarks\n\nPhysics Encyclopedia\n\nWorld\n\nIndex" ]

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[ "## Phaser audio effect\n\nJuly 4, 20112 comments Coded in C\n\nThis phaser implementation was created using cascaded 2nd order variable\nnotch filters to take advantage of the strong phase shift produced\nnear the center notch frequency. The Q parameter determines the\nfrequency band for the notch filter, and for lower Q values the\nfrequency band will be wider and the phase shift effect around the\nnotch frequency will be more pronounced.\n\nTo generate the output the input and the phase shifted signal are\nscaled and added, this will produce phase cancellations or enhancements\nin some frequencies present on the input signal as the center frequency notch\nis varied in a frequency range.\n\nIn this code is used the code in my previous entrey at http://www.dsprelated.com/showcode/173.php\n\nas a building block to implement the variable notch filter stages.\n\nHere is a sample of how it sounds like:\n\nNotice that this code is not optimized to any particular DSP architecture but you can use it as a reference code to further optimize it.\n\nI hope you find this useful.\n\n``````/*\nPhaser audio effect:\n\nXin Yout\n-------------------------------[dir_mix]--------------------->[+]--------->\n| ^\n| |\n|-->[VNS1]-->[VNS2]-->[VNS3]...-->[VNSN]-->[pha_mix]------\n\n^ ^ ^ ^\n| | | |\n|--------|--------|---...------\n|\n[LFO]\n\nVNS = Variable notch stage\n\n*/\n\n#include \"br_iir.h\"\n#include \"Phaser.h\"\n\n/*This defines the phaser stages\nthat is the number of variable notch blocks\n*/\n#define PH_STAGES 20\n\nstatic short center_freq; /*Center frequency counter*/\nstatic short samp_freq; /*Sampling frequency*/\nstatic short counter; /*Smaple counter*/\nstatic short counter_limit; /*Smaple counter limit*/\nstatic short control; /*LFO Control*/\nstatic short max_freq; /*Maximum notch center frequency*/\nstatic short min_freq; /*Minimum notch center frequency*/\nstatic double pha_mix; /*Filtered signal mix*/\nstatic short f_step; /*Sweep frequency step*/\nstatic double dir_mix; /*Direct signal mix*/\nstatic struct br_filter H[PH_STAGES]; /*Array of notch filters stages*/\n\n/*\nThis funtion initializes the phaser control variables\nand the variable notch filter coefficients array\n*/\nvoid Phaser_init(short effect_rate,short sampling,short maxf,short minf,short Q,double gainfactor,double pha_mixume,short freq_step, double dmix) {\n/*Initialize notch filter coefficients set array*/\nbr_iir_init(sampling,gainfactor,Q,freq_step, minf);\n\n/*Initializes the phaser control variables*/\ncenter_freq = 0;\nsamp_freq = sampling;\ncounter = effect_rate;\ncontrol = 0;\ncounter_limit = effect_rate;\n\n/*Convert frequencies to integer indexes*/\nmin_freq = 0;\nmax_freq = (maxf - minf)/freq_step;\n\npha_mix = pha_mixume;\nf_step = freq_step;\ndir_mix = dmix;\n}\n\n/*\nThis function does the actual phasing processing\n1. It takes the input sample and pass it trough the\n2. It takes tha output of the cascaded notch filters\nand scales it, scales the input sample and generate\nthe output effect sample.\n*/\ndouble Phaser_process(double xin) {\ndouble yout;\nint i;\n\nyout = br_iir_filter(xin,&H);\n\nfor(i = 1; i < PH_STAGES; i++) {\nyout = br_iir_filter(yout,&H[i]);\n}\n\nyout = dir_mix*xin + pha_mix*yout;\n\nreturn yout;\n}\n\n/*\nThis function makes vary the center notch frequency\nin all the cascaded notch filter stages by a simulated\ntriangle wave LFO that goes up and down\n*/\nvoid Phaser_sweep(void) {\nint i;\n\nif (!--counter) {\nif (!control) {\ncenter_freq+=f_step;\n\nif (center_freq > max_freq) {\ncontrol = 1;\n}\n}\nelse if (control) {\ncenter_freq-=f_step;\n\nif (center_freq == min_freq) {\ncontrol = 0;\n}\n}\nfor(i = 0; i < PH_STAGES; i++) {\nbr_iir_setup(&H[i],center_freq);\n}\ncounter = counter_limit;\n}\n}\n\n/************\n\nPhaser.h\n\n***********/\n\n#ifndef __PHASER_H__\n#define __PHASER_H__\n\nextern void Phaser_init(short effect_rate,short sampling,short maxf,short minf,short Q,double gainfactor,double mix_volume,short freq_step, double dmix);\nextern double Phaser_process(double xin);\nextern void Phaser_sweep(void);\n\n#endif``````" ]

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[ "# Angular Momentum(Q&M)\n\n## Homework Statement\n\nWhat are the possible measurements for Lz\n\n## Homework Equations\n\n$$\\psi(\\theta,\\phi) = \\sqrt{\\frac{3}{4 \\pi}} sin(\\phi) sin(\\theta)$$\nprobability Lz quantum\n\n## The Attempt at a Solution\n\nWell I'm sure I can expand $$sin(\\phi)= \\frac{e^{i \\phi}-e^{-i \\phi}}{2 i}$$\nGetting m=1,-1.\n$$\\psi(\\theta,\\phi) = \\sqrt{\\frac{3}{4 \\pi}} sin(\\theta) \\frac{e^{i \\phi}-e^{-i \\phi}}{2 i}$$\nShould the the probability be the coefficents mod squared?\n\nYes, but you have to keep track of normalization; the coefficients should be those of normalized eigenfunctions of Lz. I suggest writing psi in terms of spherical harmonics.\n\nyou are working with two quantum numbers, If I remember correctly you have add up the normalized coefficients squared by increasing their orbital quantum number l up to m.\n\nok. So:\n$$\\psi=\\frac{\\sqrt{2}i}{2}\\left(Y_{1,1}+Y_{1,-1}\\right)$$ with everything sorted\nI have equal probability of measuring $$\\pm \\hbar$$ for $$L_z$$, right?\n\nWait, what I typed can't be right. m has to be -1,0,1 since l=1. so in terms of spherical harmonics I should have one more term $$Y_{1,0}$$ and then normalize that. That makes more sense.\n\nmalawi_glenn\nHomework Helper\nYou don't need a Y(1,0) A priori.\n\nIf your psi in post 4 equals the psi in Relevant equations in your first post, then it is ok.\n\nYou also check the normalizability of your result in post #4 by simply integrating it over $r^2 d\\Omega$ and see if you indeed get 1.\n\nY(1,0) is proportional to cos(theta), and you only have sin(theta), so you should not expect a Y(1,0) term.\n\nYou don't need a Y(1,0) A priori.\n\nIf your psi in post 4 equals the psi in Relevant equations in your first post, then it is ok.\n\nYou also check the normalizability of your result in post #4 by simply integrating it over $r^2 d\\Omega$ and see if you indeed get 1.\n\nY(1,0) is proportional to cos(theta), and you only have sin(theta), so you should not expect a Y(1,0) term.\nok.\nIn post four the coefficents work out since$$\\Sigma |c_k|^2=1$$\n\ndiazona\nHomework Helper\nWait, what I typed can't be right. m has to be -1,0,1 since l=1. so in terms of spherical harmonics I should have one more term $$Y_{1,0}$$ and then normalize that. That makes more sense.\nYou could consider the coefficient of $$Y^1_0$$ to be 0 ;-)" ]

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[ "# Abstract:\n\nThe analysis is concerned with the evaluation of the upper bound to the capacity of a linear time-varying channel obtained by finding the capacity of the same channel when its behavior at all times is known a priori. Numerical evaluation turns out to depend on the ditribution of the eigenvalues of a random infinite-dimensional matrix, which is apparently an unsolved mathematical problem. Evaluation is possible in two special ases first, where the channel is not time-varying in which case the capacity formula reduces to that formerly known second, when the channel has no memory, that is y ax u, where a is random and n is white and gaussian. 9author" ]

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[ "# Soal Uas / Ukk Matematika Kelas 2 Sd Semester 2 (Www.Bimbelbrilian.Com)\n\nApproved & Edited by ProProfs Editorial Team\nThe editorial team at ProProfs Quizzes consists of a select group of subject experts, trivia writers, and quiz masters who have authored over 10,000 quizzes taken by more than 100 million users. This team includes our in-house seasoned quiz moderators and subject matter experts. Our editorial experts, spread across the world, are rigorously trained using our comprehensive guidelines to ensure that you receive the highest quality quizzes.\n| By Catherine Halcomb\nC\nCatherine Halcomb\nCommunity Contributor\nQuizzes Created: 1518 | Total Attempts: 5,459,314\nQuestions: 25 | Attempts: 3,102", null, "", null, "Settings", null, "", null, ".\n\n• 1.\n\n### 7 x 5 = ...\n\n• A.\n\n25\n\n• B.\n\n35\n\n• C.\n\n45\n\nB. 35\nExplanation\nThe correct answer is 35 because when you multiply 7 by 5, you get the product of 35.\n\nRate this question:\n\n• 2.\n\n### 9 x 4 = . . .\n\n• A.\n\n32\n\n• B.\n\n34\n\n• C.\n\n36\n\nC. 36\nExplanation\nMultiplying 9 by 4 gives us a product of 36.\n\nRate this question:\n\n• 3.\n\n### 56 : 8 = . . .\n\n• A.\n\n7\n\n• B.\n\n8\n\n• C.\n\n9\n\nA. 7\nExplanation\nThe given question is asking for the missing number in the equation 56 divided by 8 equals what. When we divide 56 by 8, we get the answer 7. Therefore, the missing number in the equation is 7.\n\nRate this question:\n\n• 4.\n\n### 72 : 8 = . . .\n\n• A.\n\n8\n\n• B.\n\n9\n\n• C.\n\n10\n\nB. 9\nExplanation\nThe given equation is 72 divided by 8. When we divide 72 by 8, we get the quotient as 9. Therefore, the correct answer is 9.\n\nRate this question:\n\n• 5.\n\n### 12 x 13 = ...\n\n• A.\n\n120\n\n• B.\n\n156\n\n• C.\n\n146\n\nB. 156\nExplanation\nThe correct answer is 156 because when you multiply 12 by 13, you get the product of 156.\n\nRate this question:\n\n• 6.\n\n### 15 x 10 = ...\n\n• A.\n\n155\n\n• B.\n\n154\n\n• C.\n\n150\n\nC. 150\nExplanation\nThe correct answer is 150 because when we multiply 15 by 10, we are essentially adding 15 to itself 10 times. So, 15 + 15 + 15 + 15 + 15 + 15 + 15 + 15 + 15 + 15 equals 150.\n\nRate this question:\n\n• 7.\n\n### ( 9 x 5 ) + ( 8 x 2 ) = ...\n\n• A.\n\n61\n\n• B.\n\n62\n\n• C.\n\n63\n\nA. 61\nExplanation\nThe given equation involves multiplication and addition. We multiply 9 by 5 to get 45, and we multiply 8 by 2 to get 16. Then, we add these two results together, which gives us 61. Therefore, the correct answer is 61.\n\nRate this question:\n\n• 8.\n\n### ( 7 x 10 ) - ( 8 x 5 ) = ...\n\n• A.\n\n50\n\n• B.\n\n40\n\n• C.\n\n30\n\nC. 30\nExplanation\nThe given expression is a simple subtraction problem. We first calculate 7 multiplied by 10, which equals 70. Then, we calculate 8 multiplied by 5, which equals 40. Finally, we subtract 40 from 70, giving us an answer of 30.\n\nRate this question:\n\n• 9.\n\n### ( 49 : 7 ) x 9 = ...\n\n• A.\n\n72\n\n• B.\n\n63\n\n• C.\n\n81\n\nB. 63\nExplanation\nThe given equation can be solved by first performing the division operation inside the parentheses. 49 divided by 7 equals 7. Then, multiplying the result by 9 gives us 63. Therefore, the correct answer is 63.\n\nRate this question:\n\n• 10.\n\n### 150 : ( 6 x 5) = ....\n\n• A.\n\n4\n\n• B.\n\n5\n\n• C.\n\n6\n\nB. 5\nExplanation\nThe given equation states that 150 is equal to 6 multiplied by 5. To find the missing number, we need to divide 150 by 6, resulting in 25. Therefore, the missing number is 5.\n\nRate this question:\n\n• 11.\n\n### 30 x .... = 210\n\n• A.\n\n8\n\n• B.\n\n9\n\n• C.\n\n7\n\nC. 7\nExplanation\nTo find the missing number, we need to divide 210 by one of the given numbers (8, 9, or 7). When we divide 210 by 7, the result is 30. Therefore, the missing number is 7.\n\nRate this question:\n\n• 12.\n\n### 90 : .... = 9\n\n• A.\n\n5\n\n• B.\n\n11\n\n• C.\n\n10\n\nC. 10\nExplanation\nThe pattern in the given question is that the first number is multiplied by 10 and then divided by the second number to get the third number. In this case, 90 is multiplied by 10 and divided by 9 to get 10. Therefore, the missing number that completes the pattern is 10.\n\nRate this question:\n\n• 13.\n\n• A.\n\n4 buah\n\n• B.\n\n5 buah\n\n• C.\n\n8 buah\n\nA. 4 buah\nExplanation\nThe correct answer is 4 buah because a segi empat, which translates to a quadrilateral in English, is a polygon with four sides. Therefore, the number of sides in a segi empat is 4.\n\nRate this question:\n\n• 14.\n\n### Bangun yang memiliki 3 buah sudut dan 3 buah sisi adalah ....\n\n• A.\n\nSegi empat\n\n• B.\n\nPersegi panjang\n\n• C.\n\nSegitiga\n\nC. Segitiga\nExplanation\nA triangle is a polygon with three sides and three angles. It is the only option given that fits this description. A square has four sides and four angles, while a rectangle has four sides but only two pairs of congruent angles. Therefore, the correct answer is a triangle.\n\nRate this question:\n\n• 15.\n\n### Nama bangun diatas adalah ....\n\n• A.\n\nSegitiga\n\n• B.\n\nJajar genjang\n\n• C.\n\nTrapesium\n\nC. Trapesium\nExplanation\nThe given shape is a trapezium because it has only one pair of parallel sides. A trapezium is a quadrilateral with one pair of parallel sides. In this case, the top and bottom sides of the shape are parallel to each other, while the other two sides are not. Therefore, the correct answer is trapesium.\n\nRate this question:\n\n• 16.\n\n### Bangun jajar genjang memiliki ....\n\n• A.\n\n3 sudut\n\n• B.\n\n4 sudut\n\n• C.\n\n5 sudut\n\nB. 4 sudut\nExplanation\nA jajar genjang is a quadrilateral with opposite sides that are parallel. Since a quadrilateral has four sides, a jajar genjang also has four angles. Therefore, the correct answer is \"4 sudut.\"\n\nRate this question:\n\n• 17.\n\n### Bangun lingkaran memiliki sisi sebanyak ....\n\n• A.\n\n3 buah\n\n• B.\n\n1 buah\n\n• C.\n\n4 buah\n\nB. 1 buah\nExplanation\nThe correct answer is \"1 buah\" because a circle does not have sides. A circle is a two-dimensional shape that is defined by a curved line called the circumference. It does not have any straight sides like other polygons such as triangles, squares, or rectangles. Therefore, it is incorrect to say that a circle has any number of sides.\n\nRate this question:\n\n• 18.\n\n### Bangun diatas dibentuk dari rangkaian bangun....  .\n\n• A.\n\nSegiempat\n\n• B.\n\nSegitiga\n\n• C.\n\nPersegi panjang\n\nB. Segitiga\nExplanation\nThe given answer, \"Segitiga\" (triangle), is correct because the question states that the structure is formed from a series of shapes. Among the options provided, a triangle is the only shape that can be used to form a structure by connecting its sides. A square (segiempat) and a rectangle (persegi panjang) have straight sides that can be connected to form a structure, but they cannot create the angles and slopes that a triangle can. Therefore, a triangle is the most suitable shape to form the given structure.\n\nRate this question:\n\n• 19.\n\n### Nama bangun di atas adalah ....\n\n• A.\n\nTrapesium\n\n• B.\n\nLayang-layang\n\n• C.\n\nJajar genjang\n\nC. Jajar genjang\nExplanation\nThe given shape is a parallelogram with opposite sides parallel and equal in length. It has four sides and opposite angles are equal. Therefore, the correct answer is \"Jajar genjang\" which translates to \"parallelogram\" in English.\n\nRate this question:\n\n• 20.\n\n### Benda yang berbentuk lingkaran adalah ....\n\n• A.\n\nUang logam\n\n• B.\n\nPapan tulis\n\n• C.\n\nBuku tulis\n\nA. Uang logam\nExplanation\nUang logam adalah benda yang memiliki bentuk lingkaran. Hal ini karena uang logam biasanya dibuat dengan menggunakan mesin pencetak khusus yang memberikan bentuk bulat atau lingkaran pada uang tersebut. Bentuk lingkaran pada uang logam juga memudahkan dalam penggunaan dan penyimpanan uang tersebut.\n\nRate this question:\n\n• 21.\n\n### Yang tidak termasuk bangun datar adalah ....\n\n• A.\n\nSegitiga\n\n• B.\n\nLingkaran\n\n• C.\n\nKubus\n\nC. Kubus\nExplanation\nA cube is not a two-dimensional shape, unlike a triangle and a circle. A cube is a three-dimensional shape with six square faces. Therefore, it does not belong to the category of flat shapes or plane figures.\n\nRate this question:\n\n• 22.\n\n### Buku tulis dan meja berbentuk ....\n\n• A.\n\nTrapesium\n\n• B.\n\nPersegi panjang\n\n• C.\n\nLingkaran\n\nB. Persegi panjang\nExplanation\nThe correct answer is \"Persegi panjang\" because a book and a table are both rectangular in shape, which is characteristic of a rectangle. A trapezium has only one pair of parallel sides, which does not match the given objects. A circle is a completely different shape and does not resemble a book or a table. Therefore, the most suitable option is \"Persegi panjang\" or rectangle.\n\nRate this question:\n\n• 23.\n\n### Segi empat memiliki sisi-sisi yang ....\n\n• A.\n\nSama panjang\n\n• B.\n\nBerbeda\n\n• C.\n\nTidak sama\n\nA. Sama panjang\nExplanation\nA square has all sides of equal length, which means that all the sides of a square are the same length. Therefore, the correct answer is \"Sama panjang\" which translates to \"Equal length\" in English.\n\nRate this question:\n\n• 24.\n\n### Bangun datar yang memiliki 3 buah sudut adalah ....\n\n• A.\n\nPersegi\n\n• B.\n\nLingkaran\n\n• C.\n\nSegitiga\n\nC. Segitiga\nExplanation\nA segitiga is a two-dimensional shape that has three sides and three angles. It is the only option given that fits the criteria of having three angles. A persegi has four sides and four angles, while a lingkaran is a circle and does not have any angles. Therefore, the correct answer is segitiga.\n\nRate this question:\n\n• 25.\n\n### Bangun datar yang sisi-sisinya sama panjang adalah ....\n\n• A.\n\nJajar genjang\n\n• B.\n\nTrapesium\n\n• C.\n\nSegiempat", null, "Back to top" ]

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[ "# SolverSave Function\n\nSaves the Solver problem specifications on the worksheet.\n\n## Syntax\n\n`SolverSave( SaveArea)`\n\nSaveArea Required Variant. The range of cells where the Solver model is to be saved. If this is a single-cell range, Solver uses as many cells as it needs to save the model, in a column starting with the specified cell. If this is a multi-cell range, Solver uses only cells within that range, even if the model cannot be entirely saved, The range represented by the SaveArea argument can be on any worksheet, but you must specify the worksheet if it is not the active sheet. For example, `SolverSave(\"Sheet2!A1:A3\")` saves the model on Sheet2 even if Sheet2 is not the active sheet.\n\n## Example\n\nThe SolverSave function saves the current problem to a range on the active worksheet.\n\n``````Sub SolverDemo()\nWorksheets(\"Sheet1\").Activate\nRange(\"F4\").Formula = \"=SUM(C4:E4)\"\nRange(\"F5\").Formula = \"=SUM(C5:E5)\"\nRange(\"F6\").Formula = \"=SUM(C6:E6)\"\nRange(\"F7\").Formula = \"=SUM(F4:F6)\"\nSolverReset\nSolverOptions Precision:=0.001\nSolverOK SetCell:=Range(\"F7\"), MaxMinVal:=3, ValueOf:=50, ByChange:=Range(\"C4:E6\")" ]

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[ "ox_ec.h | Developers\n\n# ox_ec.h\n\nElliptic curve cryptography syscalls. More...\n\n## Data Structures\n\nstruct  cx_curve_weierstrass_s\nWeierstrass curve defined by y^3 = x^2 + a*x + b over GF(p). More...\nstruct  cx_curve_twisted_edwards_s\nTwisted Edwards curve defined by a*x^2 + y^2 = 1 + d*x2*y2 over GF(q). More...\nstruct  cx_curve_montgomery_s\nMontgomery curve defined by B*y^2= x^3 + A*x^2 + x over GF(q). More...\nstruct  cx_curve_domain_s\nAbstract type for elliptic curve domain. More...\nstruct  cx_ec_point_s\nElliptic curve point. More...\n\n## Macros\n\n#define CX_MAX_DOMAIN_LENGTH   66\nLargest domain parameters length. More...\n#define HAVE_SECP256K1_CURVE\nEnables the Koblitz curve Secp256k1. More...\n#define HAVE_SECP256R1_CURVE\nEnables the verifiably random curve Secp256r1. More...\n#define HAVE_SECP384R1_CURVE\nEnables the verifiably random curve Secp384r1. More...\n#define HAVE_SECP521R1_CURVE\nEnables the verifiably random curve Secp521r1. More...\n#define HAVE_BRAINPOOL_P256R1_CURVE\nEnables the curve BrainpoolP256r1. More...\n#define HAVE_BRAINPOOL_P256T1_CURVE\nEnables the twisted curve BrainpoolP256t1. More...\n#define HAVE_BRAINPOOL_P320R1_CURVE\nEnables the curve BrainpoolP320r1. More...\n#define HAVE_BRAINPOOL_P320T1_CURVE\nEnables the twisted curve BrainpoolP320t1. More...\n#define HAVE_BRAINPOOL_P384R1_CURVE\nEnables the curve BrainpoolP384r1. More...\n#define HAVE_BRAINPOOL_P384T1_CURVE\nEnables the twisted curve BrainpoolP384t1. More...\n#define HAVE_BRAINPOOL_P512R1_CURVE\nEnables the curve BrainpoolP512r1. More...\n#define HAVE_BRAINPOOL_P512T1_CURVE\nEnables the twisted curve BrainpoolP512t1. More...\n#define HAVE_ED25519_CURVE\nEnables the twisted Edwards curve Ed25519. More...\n#define HAVE_ED448_CURVE\nEnables the twisted Edwards curve Ed448. More...\n#define HAVE_CV25519_CURVE\nEnables the Montgomery curve Curve25519. More...\n#define HAVE_CV448_CURVE\nEnables the Montgomery curve Curve448. More...\n#define HAVE_STARK256_CURVE\nEnables the Stark curve. More...\n#define CX_ECCINFO_PARITY_ODD   1\nIndicates the parity of a point coordinate. More...\n#define CX_ECCINFO_xGTn   2\n#define CX_CURVE_256K1   CX_CURVE_SECP256K1\nAllowed identifier for Secp256k1. More...\n#define CX_CURVE_256R1   CX_CURVE_SECP256R1\nLegacy identifier for Secp256r1. More...\n#define CX_CURVE_NISTP256   CX_CURVE_SECP256R1\nLegacy identifier for Secp256r1. More...\n#define CX_CURVE_NISTP384   CX_CURVE_SECP384R1\nAllowed identifier for Secp384r1. More...\n#define CX_CURVE_NISTP521   CX_CURVE_SECP521R1\nAllowed identifier for Secp521r1. More...\n#define CX_CURVE_RANGE(i, dom)\nReturns true if the curve identifier is in the specified range. More...\n#define CX_CURVE_IS_WEIERSTRASS(c)\nReturns true if the curve is a short Weierstrass curve. More...\n#define CX_CURVE_IS_TWISTED_EDWARDS(c)\nReturns true if the curve is a twisted Edwards curve. More...\n#define CX_CURVE_IS_MONTGOMERY(c)\nReturns true if the curve is a Montgomery curve. More...\nCurve domain parameters. More...\n\n## Typedefs\n\ntypedef enum cx_curve_e cx_curve_t\nConvenience type. More...\ntypedef struct cx_curve_weierstrass_s cx_curve_weierstrass_t\nConvenience type. More...\ntypedef struct cx_curve_twisted_edwards_s cx_curve_twisted_edwards_t\nConvenience type. More...\ntypedef struct cx_curve_montgomery_s cx_curve_montgomery_t\nConvenience type. More...\ntypedef struct cx_curve_domain_s cx_curve_domain_t\nConvenience type. More...\ntypedef struct cx_ec_point_s cx_ecpoint_t\nConvenience type. More...\ntypedef enum cx_curve_dom_param_s cx_curve_dom_param_t\n\n## Enumerations\n\nenum  cx_curve_e {\nCX_CURVE_NONE, CX_CURVE_WEIERSTRASS_START = 0x20, CX_CURVE_SECP256K1 = 0x21, CX_CURVE_SECP256R1 = 0x22,\nCX_CURVE_SECP384R1 = 0x23, CX_CURVE_SECP521R1 = 0x24, CX_CURVE_BrainPoolP256T1 = 0x31, CX_CURVE_BrainPoolP256R1 = 0x32,\nCX_CURVE_BrainPoolP320T1 = 0x33, CX_CURVE_BrainPoolP320R1 = 0x34, CX_CURVE_BrainPoolP384T1 = 0x35, CX_CURVE_BrainPoolP384R1 = 0x36,\nCX_CURVE_BrainPoolP512T1 = 0x37, CX_CURVE_BrainPoolP512R1 = 0x38, CX_CURVE_BLS12_381_G1 = 0x39, CX_CURVE_FRP256V1 = 0x41,\nCX_CURVE_Stark256 = 0x51, CX_CURVE_WEIERSTRASS_END = 0x6F, CX_CURVE_TWISTED_EDWARDS_START = 0x70, CX_CURVE_Ed25519 = 0x71,\nCX_CURVE_Ed448 = 0x72, CX_CURVE_TWISTED_EDWARDS_END = 0x7F, CX_CURVE_MONTGOMERY_START = 0x80, CX_CURVE_Curve25519 = 0x81,\nCX_CURVE_Curve448 = 0x82, CX_CURVE_MONTGOMERY_END = 0x8F\n}\nList of supported elliptic curves. More...\nenum  cx_curve_dom_param_s {\nCX_CURVE_PARAM_NONE = 0, CX_CURVE_PARAM_A = 1, CX_CURVE_PARAM_B = 2, CX_CURVE_PARAM_Field = 3,\nCX_CURVE_PARAM_Gx = 4, CX_CURVE_PARAM_Gy = 5, CX_CURVE_PARAM_Order = 6, CX_CURVE_PARAM_Cofactor = 7\n}\nIdentifiers of the domain parameters. More...\n\n## Functions\n\nSYSCALL cx_err_t cx_ecdomain_size (cx_curve_t curve, size_t *length)\nGets the bit length of each parameter of the curve. More...\nSYSCALL cx_err_t cx_ecdomain_parameters_length (cx_curve_t cv, size_t *length)\nGets the byte length of each parameter of the curve. More...\nSYSCALL cx_err_t cx_ecdomain_parameter (cx_curve_t cv, cx_curve_dom_param_t id, uint8_t *p, uint32_t p_len)\nGets a specific parameter of the curve. More...\nSYSCALL cx_err_t cx_ecdomain_parameter_bn (cx_curve_t cv, cx_curve_dom_param_t id, cx_bn_t p)\nStores a specific parameter of the curve as a BN. More...\nSYSCALL cx_err_t cx_ecdomain_generator (cx_curve_t cv, uint8_t *Gx, uint8_t *Gy, size_t len)\nGets the generator of the curve. More...\nSYSCALL cx_err_t cx_ecdomain_generator_bn (cx_curve_t cv, cx_ecpoint_t *P)\nGets the generator of the curve and stores it in the point structure. More...\nSYSCALL cx_err_t cx_ecpoint_alloc (cx_ecpoint_t *P, cx_curve_t cv)\nAllocates memory for a point on the curve. More...\nSYSCALL cx_err_t cx_ecpoint_destroy (cx_ecpoint_t *P)\nDestroys a point on the curve. More...\nSYSCALL cx_err_t cx_ecpoint_init (cx_ecpoint_t *P, const uint8_t *x, size_t x_len, const uint8_t *y, size_t y_len)\nInitializes a point on the curve. More...\nSYSCALL cx_err_t cx_ecpoint_init_bn (cx_ecpoint_t *P, const cx_bn_t x, const cx_bn_t y)\nInitializes a point on the curve with the BN indexes of the coordinates. More...\nSYSCALL cx_err_t cx_ecpoint_export (const cx_ecpoint_t *P, uint8_t *x, size_t x_len, uint8_t *y, size_t y_len)\nExports a point. More...\nSYSCALL cx_err_t cx_ecpoint_export_bn (const cx_ecpoint_t *P, cx_bn_t *x, cx_bn_t *y)\nExports a point using BN indexes of the coordinates. More...\nSYSCALL cx_err_t cx_ecpoint_compress (const cx_ecpoint_t *P, uint8_t *xy_compressed, size_t xy_compressed_len, uint32_t *sign)\nComputes the compressed form of a point. More...\nSYSCALL cx_err_t cx_ecpoint_decompress (cx_ecpoint_t *P, const uint8_t *xy_compressed, size_t xy_compressed_len, uint32_t sign)\nComputes the affine coordinates of a point given its compressed form. More...\nSYSCALL cx_err_t cx_ecpoint_add (cx_ecpoint_t *R, const cx_ecpoint_t *P, const cx_ecpoint_t *Q)\nAdds two points on a curve. More...\nSYSCALL cx_err_t cx_ecpoint_neg (cx_ecpoint_t *P)\nComputes the opposite of a point. More...\nSYSCALL cx_err_t cx_ecpoint_rnd_scalarmul (cx_ecpoint_t *P, const uint8_t *k, size_t k_len)\nPerforms a secure scalar multiplication. More...\nSYSCALL cx_err_t cx_ecpoint_rnd_scalarmul_bn (cx_ecpoint_t *P, const cx_bn_t bn_k)\nPerforms a secure scalar multiplication given the BN index of the scalar. More...\nSYSCALL cx_err_t cx_ecpoint_rnd_fixed_scalarmul (cx_ecpoint_t *P, const uint8_t *k, size_t k_len)\nPerforms a secure scalar multiplication with a fixed scalar length. More...\nSYSCALL cx_err_t cx_ecpoint_scalarmul (cx_ecpoint_t *P, const uint8_t *k, size_t k_len)\nPerforms a scalar multiplication. More...\nSYSCALL cx_err_t cx_ecpoint_scalarmul_bn (cx_ecpoint_t *P, const cx_bn_t bn_k)\nPerforms a scalar multiplication given the BN index of the scalar. More...\nSYSCALL cx_err_t cx_ecpoint_double_scalarmul (cx_ecpoint_t *R, cx_ecpoint_t *P, cx_ecpoint_t *Q, const uint8_t *k, size_t k_len, const uint8_t *r, size_t r_len)\nPerforms a double scalar multiplication. More...\nSYSCALL cx_err_t cx_ecpoint_double_scalarmul_bn (cx_ecpoint_t *R, cx_ecpoint_t *P, cx_ecpoint_t *Q, const cx_bn_t bn_k, const cx_bn_t bn_r)\nPerforms a double scalar multiplication given the BN indexes of the scalars. More...\nSYSCALL cx_err_t cx_ecpoint_cmp (const cx_ecpoint_t *P, const cx_ecpoint_t *Q, bool *is_equal)\nCompares two points on the same curve. More...\nSYSCALL cx_err_t cx_ecpoint_is_on_curve (const cx_ecpoint_t *R, bool *is_on_curve)\nChecks whether a given point is on the curve. More...\nSYSCALL cx_err_t cx_ecpoint_is_at_infinity (const cx_ecpoint_t *R, bool *is_at_infinity)\nChecks whether a given point is the point at infinity. More...\n\n## Detailed Description\n\nElliptic curve cryptography syscalls.\n\nThis file contains elliptic curves definitions and functions.\n\n## CX_CURVE_256K1\n\n #define CX_CURVE_256K1   CX_CURVE_SECP256K1\n\nAllowed identifier for Secp256k1.\n\n## CX_CURVE_256R1\n\n #define CX_CURVE_256R1   CX_CURVE_SECP256R1\n\nLegacy identifier for Secp256r1.\n\nValue:\ncx_curve_t curve; \\\nunsigned int bit_size; \\\nunsigned int length; \\\nconst uint8_t *a; \\\nconst uint8_t *b; \\\nconst uint8_t *p; \\\nconst uint8_t *Gx; \\\nconst uint8_t *Gy; \\\nconst uint8_t *n; \\\nconst uint8_t *h; \\\nconst uint8_t *Hn; \\\nconst uint8_t *Hp; \\\nenum cx_curve_e cx_curve_t\nConvenience type.\nDefinition: ox_ec.h:241\n\nCurve domain parameters.\n\nThe parameters are common to cx_curve_weierstrass_s, cx_curve_twisted_edwards_s, and cx_curve_montgomery_s.\n\n• `curve:` Curve identifier. See cx_curve_e\n• `bit_size:` Curve size in bits\n• `length:` Component lenth in bytes\n• `a:` a coefficient of the curve equation\n• `b:` b (Weierstrass or Montgomery) or d (twisted Edwards) coefficient of the curve equation\n• `p:` Prime specifying the base field\n• `Gx:` x-coordinate of the base point\n• `Gy:` y-coordinate of the base point\n• `n:` Curve order: order of the group generated by G\n• `h:` Cofactor i.e. h = |E(GF(p))|/n\n• `Hn:` Second Montgomery constant for the curve order\n• `Hp:` Second Montgomery constant for the field characteristic p\n\n## CX_CURVE_IS_MONTGOMERY\n\n #define CX_CURVE_IS_MONTGOMERY ( c )\n\nReturns true if the curve is a Montgomery curve.\n\n## CX_CURVE_IS_TWISTED_EDWARDS\n\n #define CX_CURVE_IS_TWISTED_EDWARDS ( c )\n\nReturns true if the curve is a twisted Edwards curve.\n\n## CX_CURVE_IS_WEIERSTRASS\n\n #define CX_CURVE_IS_WEIERSTRASS ( c )\n\nReturns true if the curve is a short Weierstrass curve.\n\n## CX_CURVE_NISTP256\n\n #define CX_CURVE_NISTP256   CX_CURVE_SECP256R1\n\nLegacy identifier for Secp256r1.\n\n## CX_CURVE_NISTP384\n\n #define CX_CURVE_NISTP384   CX_CURVE_SECP384R1\n\nAllowed identifier for Secp384r1.\n\n## CX_CURVE_NISTP521\n\n #define CX_CURVE_NISTP521   CX_CURVE_SECP521R1\n\nAllowed identifier for Secp521r1.\n\n## CX_CURVE_RANGE\n\n #define CX_CURVE_RANGE ( i, dom )\n\nReturns true if the curve identifier is in the specified range.\n\n## CX_ECCINFO_PARITY_ODD\n\n #define CX_ECCINFO_PARITY_ODD   1\n\nIndicates the parity of a point coordinate.\n\n## CX_ECCINFO_xGTn\n\n #define CX_ECCINFO_xGTn   2\n\n## CX_MAX_DOMAIN_LENGTH\n\n #define CX_MAX_DOMAIN_LENGTH   66\n\nLargest domain parameters length.\n\n## HAVE_BRAINPOOL_P256R1_CURVE\n\n #define HAVE_BRAINPOOL_P256R1_CURVE\n\nEnables the curve BrainpoolP256r1.\n\n## HAVE_BRAINPOOL_P256T1_CURVE\n\n #define HAVE_BRAINPOOL_P256T1_CURVE\n\nEnables the twisted curve BrainpoolP256t1.\n\n## HAVE_BRAINPOOL_P320R1_CURVE\n\n #define HAVE_BRAINPOOL_P320R1_CURVE\n\nEnables the curve BrainpoolP320r1.\n\n## HAVE_BRAINPOOL_P320T1_CURVE\n\n #define HAVE_BRAINPOOL_P320T1_CURVE\n\nEnables the twisted curve BrainpoolP320t1.\n\n## HAVE_BRAINPOOL_P384R1_CURVE\n\n #define HAVE_BRAINPOOL_P384R1_CURVE\n\nEnables the curve BrainpoolP384r1.\n\n## HAVE_BRAINPOOL_P384T1_CURVE\n\n #define HAVE_BRAINPOOL_P384T1_CURVE\n\nEnables the twisted curve BrainpoolP384t1.\n\n## HAVE_BRAINPOOL_P512R1_CURVE\n\n #define HAVE_BRAINPOOL_P512R1_CURVE\n\nEnables the curve BrainpoolP512r1.\n\n## HAVE_BRAINPOOL_P512T1_CURVE\n\n #define HAVE_BRAINPOOL_P512T1_CURVE\n\nEnables the twisted curve BrainpoolP512t1.\n\n## HAVE_CV25519_CURVE\n\n #define HAVE_CV25519_CURVE\n\nEnables the Montgomery curve Curve25519.\n\n## HAVE_CV448_CURVE\n\n #define HAVE_CV448_CURVE\n\nEnables the Montgomery curve Curve448.\n\n## HAVE_ED25519_CURVE\n\n #define HAVE_ED25519_CURVE\n\nEnables the twisted Edwards curve Ed25519.\n\n## HAVE_ED448_CURVE\n\n #define HAVE_ED448_CURVE\n\nEnables the twisted Edwards curve Ed448.\n\n## HAVE_SECP256K1_CURVE\n\n #define HAVE_SECP256K1_CURVE\n\nEnables the Koblitz curve Secp256k1.\n\n## HAVE_SECP256R1_CURVE\n\n #define HAVE_SECP256R1_CURVE\n\nEnables the verifiably random curve Secp256r1.\n\n## HAVE_SECP384R1_CURVE\n\n #define HAVE_SECP384R1_CURVE\n\nEnables the verifiably random curve Secp384r1.\n\n## HAVE_SECP521R1_CURVE\n\n #define HAVE_SECP521R1_CURVE\n\nEnables the verifiably random curve Secp521r1.\n\n## HAVE_STARK256_CURVE\n\n #define HAVE_STARK256_CURVE\n\nEnables the Stark curve.\n\n## cx_curve_dom_param_t\n\n typedef enum cx_curve_dom_param_s cx_curve_dom_param_t\n\n## cx_curve_domain_t\n\n typedef struct cx_curve_domain_s cx_curve_domain_t\n\nConvenience type.\n\n## cx_curve_montgomery_t\n\n typedef struct cx_curve_montgomery_s cx_curve_montgomery_t\n\nConvenience type.\n\n## cx_curve_t\n\n typedef enum cx_curve_e cx_curve_t\n\nConvenience type.\n\nSee cx_curve_e.\n\n## cx_curve_twisted_edwards_t\n\n typedef struct cx_curve_twisted_edwards_s cx_curve_twisted_edwards_t\n\nConvenience type.\n\n## cx_curve_weierstrass_t\n\n typedef struct cx_curve_weierstrass_s cx_curve_weierstrass_t\n\nConvenience type.\n\n## cx_ecpoint_t\n\n typedef struct cx_ec_point_s cx_ecpoint_t\n\nConvenience type.\n\nSee cx_ec_point_s.\n\n## cx_curve_dom_param_s\n\n enum cx_curve_dom_param_s\n\nIdentifiers of the domain parameters.\n\nEnumerator\nCX_CURVE_PARAM_NONE\n\nNo parameter.\n\nCX_CURVE_PARAM_A\n\nFirst coefficient of the curve.\n\nCX_CURVE_PARAM_B\n\nSecond coefficient of the curve.\n\nCX_CURVE_PARAM_Field\n\nCurve field.\n\nCX_CURVE_PARAM_Gx\n\nx-coordinate of the curve's generator\n\nCX_CURVE_PARAM_Gy\n\ny-coordinate of the curve's generator\n\nCX_CURVE_PARAM_Order\n\nOrder of the generator.\n\nCX_CURVE_PARAM_Cofactor\n\nCofactor.\n\n## cx_curve_e\n\n enum cx_curve_e\n\nList of supported elliptic curves.\n\nEnumerator\nCX_CURVE_NONE\n\nUndefined curve.\n\nCX_CURVE_WEIERSTRASS_START\n\nLow limit (not included) of Weierstrass curve ID.\n\nCX_CURVE_SECP256K1\n\nSecp256k1.\n\nCX_CURVE_SECP256R1\n\nSecp256r1.\n\nCX_CURVE_SECP384R1\n\nSecp384r1.\n\nCX_CURVE_SECP521R1\n\nSecp521r1.\n\nCX_CURVE_BrainPoolP256T1\n\nBrainpoolP256t1.\n\nCX_CURVE_BrainPoolP256R1\n\nBrainpoolP256r1.\n\nCX_CURVE_BrainPoolP320T1\n\nBrainpoolP320t1.\n\nCX_CURVE_BrainPoolP320R1\n\nBrainpoolP320r1.\n\nCX_CURVE_BrainPoolP384T1\n\nBrainpoolP384t1.\n\nCX_CURVE_BrainPoolP384R1\n\nBrainpool384r1.\n\nCX_CURVE_BrainPoolP512T1\n\nBrainpoolP512t1.\n\nCX_CURVE_BrainPoolP512R1\n\nBrainpoolP512r1.\n\nCX_CURVE_BLS12_381_G1\n\nBLS12-381 G1.\n\nCX_CURVE_FRP256V1\n\nANSSI FRP256.\n\nCX_CURVE_Stark256\n\nStark.\n\nCX_CURVE_WEIERSTRASS_END\n\nHigh limit (not included) of Weierstrass curve ID.\n\nCX_CURVE_TWISTED_EDWARDS_START\n\nLow limit (not included) of Twisted Edwards curve ID.\n\nCX_CURVE_Ed25519\n\nEd25519.\n\nCX_CURVE_Ed448\n\nEd448.\n\nCX_CURVE_TWISTED_EDWARDS_END\n\nHigh limit (not included) of Twisted Edwards curve ID.\n\nCX_CURVE_MONTGOMERY_START\n\nLow limit (not included) of Montgomery curve ID.\n\nCX_CURVE_Curve25519\n\nCurve25519.\n\nCX_CURVE_Curve448\n\nCurve448.\n\nCX_CURVE_MONTGOMERY_END\n\nHigh limit (not included) of Montgomery curve ID.\n\n## cx_ecdomain_generator()\n\n SYSCALL cx_err_t cx_ecdomain_generator ( cx_curve_t cv, uint8_t * Gx, uint8_t * Gy, size_t len )\n\nGets the generator of the curve.\n\nParameters\n [in] cv Curve identifier. [out] Gx Buffer to store the x-coordinate of the generator. [out] Gy Buffer to store the y-coordinate of the generator. [in] len Byte length of each coordinate.\nReturns\nError code:\n• CX_OK on success\n• CX_EC_INVALID_CURVE\n• CX_INVALID_PARAMETER\n\n## cx_ecdomain_generator_bn()\n\n SYSCALL cx_err_t cx_ecdomain_generator_bn ( cx_curve_t cv, cx_ecpoint_t * P )\n\nGets the generator of the curve and stores it in the point structure.\n\nParameters\n [in] cv Curve identifier. [out] P Pointer to the structure where to store the generator.\nReturns\nError code:\n• CX_OK on success\n• CX_EC_INVALID_CURVE\n• CX_NOT_LOCKED\n• CX_INVALID_PARAMETER\n• CX_INVALID_PARAMETER_SIZE\n• CX_EC_INVALID_POINT\n\n## cx_ecdomain_parameter()\n\n SYSCALL cx_err_t cx_ecdomain_parameter ( cx_curve_t cv, cx_curve_dom_param_t id, uint8_t * p, uint32_t p_len )\n\nGets a specific parameter of the curve.\n\nParameters\n [in] cv Curve identifier. [in] id Parameter identifier. [out] p Buffer where to store the parameter. [in] p_len Length of the buffer.\nReturns\nError code:\n• CX_OK on success\n• CX_EC_INVALID_CURVE\n• CX_INVALID_PARAMETER\n\n## cx_ecdomain_parameter_bn()\n\n SYSCALL cx_err_t cx_ecdomain_parameter_bn ( cx_curve_t cv, cx_curve_dom_param_t id, cx_bn_t p )\n\nStores a specific parameter of the curve as a BN.\n\nParameters\n [in] cv Curve identifier. [in] id Parameter identifier. [out] p BN where to store the parameter.\nReturns\nError code:\n• CX_OK on success\n• CX_EC_INVALID_CURVE\n• CX_NOT_LOCKED\n• CX_INVALID_PARAMETER\n• CX_INVALID_PARAMETER_SIZE\n\n## cx_ecdomain_parameters_length()\n\n SYSCALL cx_err_t cx_ecdomain_parameters_length ( cx_curve_t cv, size_t * length )\n\nGets the byte length of each parameter of the curve.\n\nParameters\n [in] cv Curve identifier. [out] length Byte length of each parameter.\nReturns\nError code:\n• CX_OK on success\n• CX_EC_INVALID_CURVE\n\n## cx_ecdomain_size()\n\n SYSCALL cx_err_t cx_ecdomain_size ( cx_curve_t curve, size_t * length )\n\nGets the bit length of each parameter of the curve.\n\nParameters\n [in] curve Curve identifier. [out] length Bit length of each parameter.\nReturns\nError code:\n• CX_OK on success\n• CX_EC_INVALID_CURVE\n\n SYSCALL cx_err_t cx_ecpoint_add ( cx_ecpoint_t * R, const cx_ecpoint_t * P, const cx_ecpoint_t * Q )\n\nAdds two points on a curve.\n\nEach point should not be the point at infinity. If one of the point is the point at infinity then the function returns a CX_EC_INFINITE_POINT error.\n\nParameters\n [out] R Pointer to the result point. [in] P Pointer to the first point to add. The point must be on the curve. [in] Q Pointer to the second point to add. The point must be on the curve.\nReturns\nError code:\n• CX_OK on success\n• CX_NOT_LOCKED\n• CX_INVALID_PARAMETER\n• CX_EC_INVALID_CURVE\n• CX_EC_INVALID_POINT\n• CX_EC_INFINITE_POINT\n• CX_MEMORY_FULL\n\n## cx_ecpoint_alloc()\n\n SYSCALL cx_err_t cx_ecpoint_alloc ( cx_ecpoint_t * P, cx_curve_t cv )\n\nAllocates memory for a point on the curve.\n\nParameters\n [in] P Pointer to a point. [in] cv Curve on which the point is defined.\nReturns\nError code:\n• CX_OK on success\n• CX_EC_INVALID_CURVE\n• CX_NOT_LOCKED\n• CX_INVALID_PARAMETER\n• CX_MEMORY_FULL\n\n## cx_ecpoint_cmp()\n\n SYSCALL cx_err_t cx_ecpoint_cmp ( const cx_ecpoint_t * P, const cx_ecpoint_t * Q, bool * is_equal )\n\nCompares two points on the same curve.\n\nParameters\n [in] P First point to compare. [in] Q Second point to compare. [out] is_equal Boolean which indicates whether the two points are equal or not: 1 if the points are equal 0 otherwise\nReturns\nError code:\n• CX_OK on success\n• CX_NOT_LOCKED\n• CX_INVALID_PARAMETER\n• CX_EC_INVALID_CURVE\n• CX_EC_INFINITE_POINT\n• CX_MEMORY_FULL\n\n## cx_ecpoint_compress()\n\n SYSCALL cx_err_t cx_ecpoint_compress ( const cx_ecpoint_t * P, uint8_t * xy_compressed, size_t xy_compressed_len, uint32_t * sign )\n\nComputes the compressed form of a point.\n\nThe compressed form depends on the curve type. For a Weierstrass or a Montgomery curve, the compressed form consists of the x-coordinate and a prefix. For a Twisted Edwards curve the compressed form consists of a y-coordinate and a prefix.\n\nParameters\n [in] P Pointer to the point to be compressed. [out] xy_compressed Buffer to hold the compressed coordinate. [in] xy_compressed_len Length of the compressed coordinate in bytes. This should be equal to the length of one coordinate. [out] sign Pointer to the sign of the hidden coordinate: correspond to the least significant bit of the y-coordinate for a Weierstrass or Montgomery curve and of the x-coordinate for a Twisted Edwards curve.\nReturns\nError code:\n• CX_OK on success\n• CX_NOT_LOCKED\n• CX_INVALID_PARAMETER\n• CX_EC_INVALID_CURVE\n• CX_EC_INFINITE_POINT\n• CX_MEMORY_FULL\n\n## cx_ecpoint_decompress()\n\n SYSCALL cx_err_t cx_ecpoint_decompress ( cx_ecpoint_t * P, const uint8_t * xy_compressed, size_t xy_compressed_len, uint32_t sign )\n\nComputes the affine coordinates of a point given its compressed form.\n\nParameters\n [out] P Pointer to the point. [in] xy_compressed Pointer to the buffer holding the compressed coordinate. [in] xy_compressed_len Length of the compressed coordinate in bytes. This should be equal to the length of one coordinate. [in] sign Sign of the coordinate to recover.\nReturns\nError code:\n• CX_OK on success\n• CX_NOT_LOCKED\n• CX_INVALID_PARAMETER\n• CX_EC_INVALID_CURVE\n• CX_MEMORY_FULL\n• CX_NO_RESIDUE\n\n## cx_ecpoint_destroy()\n\n SYSCALL cx_err_t cx_ecpoint_destroy ( cx_ecpoint_t * P )\n\nDestroys a point on the curve.\n\nParameters\n [in] P Pointer to the point to destroy. If the pointer is NULL, nothing is done.\nReturns\nError code:\n• CX_OK on success\n• CX_NOT_LOCKED\n• CX_INVALID_PARAMETER\n• CX_INTERNAL_ERROR\n\n## cx_ecpoint_double_scalarmul()\n\n SYSCALL cx_err_t cx_ecpoint_double_scalarmul ( cx_ecpoint_t * R, cx_ecpoint_t * P, cx_ecpoint_t * Q, const uint8_t * k, size_t k_len, const uint8_t * r, size_t r_len )\n\nPerforms a double scalar multiplication.\n\nThis implements the Straus-Shamir algorithm for computing R = [k]P + [r]Q. This should be used only for non-secret computations.\n\nParameters\n [out] R Pointer to the result. [in] P Pointer to the first point. [in] Q Pointer to the second point. [in] k Pointer to the first scalar. [in] k_len Length of the first scalar. [in] r Pointer to the second scalar. [in] r_len Length of the second scalar.\nReturns\nError code:\n• CX_OK on success\n• CX_NOT_LOCKED\n• CX_INVALID_PARAMETER\n• CX_EC_INVALID_POINT\n• CX_EC_INVALID_CURVE\n• CX_MEMORY_FULL\n• CX_EC_INFINITE_POINT\n\n## cx_ecpoint_double_scalarmul_bn()\n\n SYSCALL cx_err_t cx_ecpoint_double_scalarmul_bn ( cx_ecpoint_t * R, cx_ecpoint_t * P, cx_ecpoint_t * Q, const cx_bn_t bn_k, const cx_bn_t bn_r )\n\nPerforms a double scalar multiplication given the BN indexes of the scalars.\n\nThis implements the Straus-Shamir algorithm for computing R = [k]P + [r]Q. This should be used only for non-secret computations.\n\nParameters\n [out] R Pointer to the result. [in] P Pointer to the first point. [in] Q Pointer to the second point. [in] bn_k BN index of the first scalar. [in] bn_r BN index of the second scalar.\nReturns\nError code:\n• CX_OK on success\n• CX_NOT_LOCKED\n• CX_INVALID_PARAMETER\n• CX_EC_INVALID_POINT\n• CX_EC_INVALID_CURVE\n• CX_MEMORY_FULL\n• CX_EC_INFINITE_POINT\n\n## cx_ecpoint_export()\n\n SYSCALL cx_err_t cx_ecpoint_export ( const cx_ecpoint_t * P, uint8_t * x, size_t x_len, uint8_t * y, size_t y_len )\n\nExports a point.\n\nFills two distinct buffers with the x-coordinate and the y-coordinate of the point. If the point is not in affine representation, it will be normalized first.\n\nParameters\n [in] P Pointer to the point to export. [out] x Buffer for the x-coordinate. [in] x_len Length of the x buffer. [out] y Buffer for the y-coordinate. [in] y_len Length of the y buffer.\nReturns\nError code:\n• CX_OK on success\n• CX_NOT_LOCKED\n• CX_INVALID_PARAMETER\n• CX_EC_INVALID_CURVE\n• CX_EC_INFINITE_POINT\n• CX_MEMORY_FULL\n\n## cx_ecpoint_export_bn()\n\n SYSCALL cx_err_t cx_ecpoint_export_bn ( const cx_ecpoint_t * P, cx_bn_t * x, cx_bn_t * y )\n\nExports a point using BN indexes of the coordinates.\n\nParameters\n [in] P Pointer to the point to export. [out] x Pointer to the BN index of the x-coordinate. [out] y Pointer to the BN index of the y-coordinate.\nReturns\nError code:\n• CX_OK on success\n• CX_NOT_LOCKED\n• CX_INVALID_PARAMETER\n• CX_EC_INVALID_CURVE\n• CX_EC_INFINITE_POINT\n• CX_MEMORY_FULL\n\n## cx_ecpoint_init()\n\n SYSCALL cx_err_t cx_ecpoint_init ( cx_ecpoint_t * P, const uint8_t * x, size_t x_len, const uint8_t * y, size_t y_len )\n\nInitializes a point on the curve.\n\nParameters\n [in] P Pointer to the point to initialize. [in] x x-coordinate of the point. This must belong to the curve field. [in] x_len Length of the x-coordinate. This must be at most equal to the curve's domain number of bytes. [in] y y-coordinate of the point. This must belong to the curve field. [in] y_len Length of the y-coordinate. This must be at most equal to the curve's domain number of bytes.\nReturns\nError code:\n• CX_OK on success\n• CX_NOT_LOCKED\n• CX_INVALID_PARAMETER\n• CX_EC_INVALID_CURVE\n\n## cx_ecpoint_init_bn()\n\n SYSCALL cx_err_t cx_ecpoint_init_bn ( cx_ecpoint_t * P, const cx_bn_t x, const cx_bn_t y )\n\nInitializes a point on the curve with the BN indexes of the coordinates.\n\nParameters\n [in] P Pointer to the point to initialize. [in] x BN index of the x-coordinate. The coordinate must belong to the base field. [in] y BN index of the y-coordinate. The coordinate must belong to the base field.\nReturns\nError code:\n• CX_OK on success\n• CX_NOT_LOCKED\n• CX_INVALID_PARAMETER\n• CX_EC_INVALID_CURVE\n\n## cx_ecpoint_is_at_infinity()\n\n SYSCALL cx_err_t cx_ecpoint_is_at_infinity ( const cx_ecpoint_t * R, bool * is_at_infinity )\n\nChecks whether a given point is the point at infinity.\n\nThe point at infinity has a z-coordinate equal to 0.\n\nParameters\n [in] R Pointer to the point to check. [out] is_at_infinity Boolean which indicates whether the point is at infinity or not: 1 if the point is at infinity 0 otherwise\nReturns\nError code:\n• CX_OK on success\n• CX_NOT_LOCKED\n• CX_INVALID_PARAMETER\n• CX_EC_INVALID_CURVE\n\n## cx_ecpoint_is_on_curve()\n\n SYSCALL cx_err_t cx_ecpoint_is_on_curve ( const cx_ecpoint_t * R, bool * is_on_curve )\n\nChecks whether a given point is on the curve.\n\nParameters\n [in] R Pointer to the point to check. [out] is_on_curve Boolean which indicates whether the point is on the curve or not: 1 if the point is on the curve 0 otherwise\nReturns\nError code:\n• CX_OK on success\n• CX_NOT_LOCKED\n• CX_INVALID_PARAMETER\n• CX_EC_INVALID_CURVE\n• CX_EC_INFINITE_POINT\n• CX_MEMORY_FULL\n\n## cx_ecpoint_neg()\n\n SYSCALL cx_err_t cx_ecpoint_neg ( cx_ecpoint_t * P )\n\nComputes the opposite of a point.\n\nThe point should not be the point at infinity, otherwise the function returns a CX_EC_INFINITE_POINT error.\n\nParameters\n [in,out] P Pointer to a point of the curve. This will hold the result.\nReturns\nError code:\n• CX_OK on success\n• CX_NOT_LOCKED\n• CX_INVALID_PARAMETER\n• CX_EC_INVALID_CURVE\n• CX_EC_INVALID_POINT\n• CX_MEMORY_FULL\n• CX_EC_INFINITE_POINT\n\n## cx_ecpoint_rnd_fixed_scalarmul()\n\n SYSCALL cx_err_t cx_ecpoint_rnd_fixed_scalarmul ( cx_ecpoint_t * P, const uint8_t * k, size_t k_len )\n\nPerforms a secure scalar multiplication with a fixed scalar length.\n\nParameters\n [in,out] P Pointer to a point on a curve. This will hold the result. [in] k Pointer to the scalar. The scalar is an integer at least equal to 0 and at most equal to the order of the curve minus 1. [in] k_len Length of the scalar. This should be equal to the domain length.\nReturns\nError code:\n• CX_OK on success\n• CX_NOT_LOCKED\n• CX_INVALID_PARAMETER\n• CX_EC_INVALID_POINT\n• CX_EC_INVALID_CURVE\n• CX_EC_INFINITE_POINT\n• CX_MEMORY_FULL\n\n## cx_ecpoint_rnd_scalarmul()\n\n SYSCALL cx_err_t cx_ecpoint_rnd_scalarmul ( cx_ecpoint_t * P, const uint8_t * k, size_t k_len )\n\nPerforms a secure scalar multiplication.\n\nParameters\n [in,out] P Pointer to a point on a curve. This will hold the result. [in] k Pointer to the scalar. The scalar is an integer at least equal to 0 and at most equal to the order of the curve minus 1. [in] k_len Length of the scalar. This should be equal to the domain length.\nReturns\nError code:\n• CX_OK on success\n• CX_NOT_LOCKED\n• CX_INVALID_PARAMETER\n• CX_EC_INVALID_POINT\n• CX_EC_INVALID_CURVE\n• CX_EC_INFINITE_POINT\n• CX_MEMORY_FULL\n\n## cx_ecpoint_rnd_scalarmul_bn()\n\n SYSCALL cx_err_t cx_ecpoint_rnd_scalarmul_bn ( cx_ecpoint_t * P, const cx_bn_t bn_k )\n\nPerforms a secure scalar multiplication given the BN index of the scalar.\n\nParameters\n [in,out] P Pointer to a point on a curve. This will hold the result. [in] bn_k BN index of the scalar. The scalar is an integer at least equal to 0 and at most equal to the order of the curve minus 1.\nReturns\nError code:\n• CX_OK on success\n• CX_NOT_LOCKED\n• CX_INVALID_PARAMETER\n• CX_EC_INVALID_POINT\n• CX_EC_INVALID_CURVE\n• CX_EC_INFINITE_POINT\n• CX_MEMORY_FULL\n\n## cx_ecpoint_scalarmul()\n\n SYSCALL cx_err_t cx_ecpoint_scalarmul ( cx_ecpoint_t * P, const uint8_t * k, size_t k_len )\n\nPerforms a scalar multiplication.\n\nWarning\nThis should be called only for non critical purposes. It is recommended to use cx_ecpoint_rnd_scalarmul or cx_ecpoint_rnd_fixed_scalarmul rather than this function.\nParameters\n [in,out] P Pointer to a point on a curve. This will hold the result. [in] k Pointer to the scalar. The scalar is an integer at least equal to 0 and at most equal to the order of the curve minus 1. [in] k_len Length of the scalar.\nReturns\nError code:\n• CX_OK on success\n• CX_NOT_LOCKED\n• CX_INVALID_PARAMETER\n• CX_EC_INVALID_POINT\n• CX_EC_INVALID_CURVE\n• CX_EC_INFINITE_POINT\n• CX_MEMORY_FULL\n\n## cx_ecpoint_scalarmul_bn()\n\n SYSCALL cx_err_t cx_ecpoint_scalarmul_bn ( cx_ecpoint_t * P, const cx_bn_t bn_k )\n\nPerforms a scalar multiplication given the BN index of the scalar.\n\nWarning\nThis should be called only for non critical purposes. It is recommended to use cx_ecpoint_rnd_scalarmul_bn rather than this function.\nParameters\n [in,out] P Pointer to a point on a curve. This will hold the result. [in] bn_k BN index of the scalar. The scalar is an integer at least equal to 0 and at most equal to the order of the curve minus 1.\nReturns\nError code:\n• CX_OK on success\n• CX_NOT_LOCKED\n• CX_INVALID_PARAMETER\n• CX_EC_INVALID_POINT\n• CX_EC_INVALID_CURVE\n• CX_EC_INFINITE_POINT\n• CX_MEMORY_FULL" ]

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[ "", null, "# NCERT Solutions Class 10 Mathematics Solutions for Arithmetic Progressions - Exercise 5.2 in Chapter 5 - Arithmetic Progressions\n\n30th term of the A.P: 10, 7, 4, …, is\n\nGiven here,\n\nA.P. = 10, 7, 4, …\n\nTherefore, we can find,\n\nFirst term, a = 10\n\nCommon difference, d = a2 − a= 7−10 = −3\n\nAs we know, for an A.P.,\n\nan = a +(n−1)d\n\nPutting the values;\n\na30 = 10+(30−1)(−3)\n\na30 = 10+(29)(−3)\n\na30 = 10−87 = −77\n\nHence, the correct answer is option C.\n\nRelated Questions\n\nLido\n\nCourses\n\nTeachers\n\nBook a Demo with us\n\nSyllabus\n\nMaths\nCBSE\nMaths\nICSE\nScience\nCBSE\n\nScience\nICSE\nEnglish\nCBSE\nEnglish\nICSE\nCoding\n\nTerms & Policies\n\nSelina Question Bank\n\nMaths\nPhysics\nBiology\n\nAllied Question Bank\n\nChemistry\nConnect with us on social media!" ]

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[ "# Problem: The mole is defined as the amount of a substance containing the same number of particles as exactly 12 grams of C-12. The amu is defined as 1/12 of the mass of an atom of C-12.Why is it important that both of these definitions reference the same isotope?\n\n###### FREE Expert Solution\n\nmole is defined as the amount of a substance containing the same number of particles as exactly 12 grams of C-12\n\namu is defined as 1/12 of the mass of an atom of C-12\n\n99% (77 ratings)", null, "###### Problem Details\n\nThe mole is defined as the amount of a substance containing the same number of particles as exactly 12 grams of C-12. The amu is defined as 1/12 of the mass of an atom of C-12.\n\nWhy is it important that both of these definitions reference the same isotope?\n\nFrequently Asked Questions\n\nWhat scientific concept do you need to know in order to solve this problem?\n\nOur tutors have indicated that to solve this problem you will need to apply the Isotopes concept. You can view video lessons to learn Isotopes. Or if you need more Isotopes practice, you can also practice Isotopes practice problems." ]

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[ "# Statistically prove the effectiveness of a treatment using GLM repeated measurements\n\nI have two lots of samples: one is the control lot and the other undergoes some treatment. I did three measurements for the samples: one at the initial time (T1) and the other two later.\n\n Descriptive Statistics\nLot Mean Std. Deviation N\nT1 1.00 124.3043 3.21127 23\n2.00 124.1333 1.94286 30\nTotal 124.2075 2.54467 53\nT2 1.00 112.8261 5.81262 23\n2.00 107.7000 7.42387 30\nTotal 109.9245 7.18398 53\nT3 1.00 90.3478 8.47783 23\n2.00 114.7000 4.43458 30\nTotal 104.1321 13.77852 53\n\n\nThe plotted data look like this:", null, "What I did:\n\nUse IBM SPSS (ver 20) and run General Linear Model/Repeated Measures. Established the factor to MeasureTime with three levels and the measurement. Between subjects factor was set to the lot number (1 or 2), Model full factorial, Contrasts to MeasureTime, Simple, Reference category set to first.\n\nSPSS output a lot of statistics. Almost all are relevant (sig < 0.000) except Mauchly's Test of Sphericity.\n\nWhat I want to know:\n\n1. How can I see if the mean differences in T1/T2/T3 are significant? (I expect that T1 is not and T3 is.)\n2. What statistic (from SPSS output) tells me that the treatment works?\n\nP.S. I read other related questions suggested by Stack Exchange, but no luck with my questions.\n\nI think that ANOVA would work for a factor with multiple levels, but as I read on other questions, in case of repeated measurements, ANOVA assumptions fail. The data are not independent.\n\nFor post hoc comparison in repeated measures: Actually, it seems that SPSS does not offer a specific post hoc test for repeated measures ANOVA and the reason probably is that the repeated measures are dependent thus not appropriate for those methods. A workaround is to apply 3 paired samples t - test between all combinations of $T_i$ and $T_j$ making a correction for p, that is considering a statistically significant result when $p < \\alpha / 3$, instead of $p < \\alpha$." ]

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[ "", null, "", null, "### Home > CCAA8 > Chapter 5 Unit 6 > Lesson CCA: 5.1.1 > Problem5-17\n\n5-17.\n\nThe dartboard shown at right is in the shape of an equilateral triangle. It has a smaller equilateral triangle in the center, which was made by joining the midpoints of the three edges. If a dart hits the board at random, what is the probability that:\n\n1. The dart hits the center triangle?\n\nHow many smaller triangles make up the bigger triangle?\n\nThe probability is $\\frac{1}{4}$.\n\n2. The dart misses the center triangle but hits the board?\n\nHow many smaller triangles can be hit if the center one is missed?", null, "" ]

[ null, "https://homework.cpm.org/dist/7d633b3a30200de4995665c02bdda1b8.png", null, 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", 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[ "## linear programming , Operation Research\n\nAssignment Help:\nA paper mill produces two grades of paper viz., X and Y. Because of raw material restrictions, it cannot produce more than 400 tons of grade X paper and 300 tons of grade Y paper in a week. There are 160 production hours in a week. It requires 0.20 and 0.40 hours to produce a ton of grade X and Y papers. The mill earns a profit of Rs. 200 and Rs. 500 per ton of grade X and Y paper respectively. Formulate this as a Linear Programming Problem.\n\n#### Why two regression lines - correlation regression analysis, Why Two regress...\n\nWhy Two regression Lines Once regression line  cannot  minimize the sum  of square of deviations  for both  the x and Y series  unless the relationship  between  them  indicat\n\n#### Uses of mode - measure of central tendency, Uses of Mode The  use of t...\n\nUses of Mode The  use of the mode is recommended in the following  situations: a.When a quick  and approximate measures of  central tendency desired. b.When the  measure\n\n#### Simple random sampling - sampling decisions, Simple Random Sampling  ...\n\nSimple Random Sampling  This  is  simplest  and most  popular  technique  of sampling. In it each  unit  of the  population has equal  chance  of being  included in the samp\n\n#### Linear programming, B A paper mill produces two grades of paper viz., X and...\n\nB A paper mill produces two grades of paper viz., X and Y. Because of raw material restrictions, it cannot produce more than 400 tons of grade X paper and 300 tons of grade Y paper\n\n#### Introduction to probability distribution , Introduction to Probability Dist...\n\nIntroduction to Probability Distribution By  theoretical distribution we mean  a frequency distribution  which  is obtained  in relation to a random  variable by some  mathem\n\n#### Simple method, Maximize Z= 3x1 + 2X2 Subject to the constraints: X1+ X2 4 X...\n\nMaximize Z= 3x1 + 2X2 Subject to the constraints: X1+ X2 4 X1+ X2 2 X1, X2 0on..\n\n#### Histogram and historigram, #qudistinguish between histogram and historigram...\n\n#qudistinguish between histogram and historigram estion..\n\n#### Explain what do you understand by dynamic programming, Question: (a) (i...\n\nQuestion: (a) (i) Explain what do you understand by ‘Dynamic Programming'. (ii) Describe the dynamic programming approach to solve the shortest route problem. (iii) Outli\n\n#### Linear Programming, b. A paper mill produces two grades of paper viz., X an...\n\nb. A paper mill produces two grades of paper viz., X and Y. Because of raw material restrictions, it cannot produce more than 400 tons of grade X paper and 300 tons of grade Y pape\n\n#### Techniques of operations research, There is no unique set of problems which...\n\nThere is no unique set of problems which can be solved by using operations Research Models r techniques. Several operations Research Models or techniques can be grouped into some b", null, "", null, "" ]

[ null, "http://www.expertsmind.com/questions/CaptchaImage.axd", null, "http://www.expertsmind.com/prostyles/images/3.png", null ]

[ "Sophie, I think that the inequalities from this Lecture 27 are the \"correct\" ones. A nice way to remember which way the unit condition should go is to note that there the greater or equal side is of the form \\$$f(-)\\$$, both in \\$$f(x) \\otimes f(x') \\leq f(x\\otimes x')\\$$ and in \\$$I \\leq f(I)\\$$. These two inequalities are like the multiplication and the unit of a monoid: in the case of the multiplication, you have *two* things going in and *one* thing coming out, which corresponds to two \\$$f\\$$'s on the left and one on the right. The unit of a monoid is just a constant, which means that it has *no* thing going in and again *one* thing coming out; just like the \\$$f\\$$'s in \\$$I\\leq f(I)\\$$.\n\nIn case that all this sounds a bit mysterious or confusing to you now, let me just say that it will become clear further along the course! (Assuming that lax monoidal functors will be covered.)" ]

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[ "Classic comparisons\n\nMeans (t-tests)\n\nProportions\n\nVariances\n\nCorrelations\n\nMcNemar’s test\n\nClustered data\n\nANOVA\n\nOne-way models\n\nTwo-way models\n\nRepeated-measures models\n\nLinear regression\n\nSlope test\n\nCoefficient-of-determination test\n\nPartial-correlation test\n\nContingency tables\n\nCochran—Mantel—Haenszel test\n\nMatched case—control studies\n\nCochran—Armitage trend test\n\nCluster randomized designs\n\nOne-sample mean\n\nTwo-sample means\n\nOne-sample proportion\n\nTwo-sample proportions\n\nLog-rank test\n\nSurvival analysis\n\nLog-rank test of survival functions\n\nCox proportional hazards model\n\nExponential regression\n\nClustered Data\n\nSolve for\n\nPower\n\nSample size\n\nNumber of clusters\n\nCluster size\n\nEffect size (minimum detectable effect)\n\nDesigns\n\nMatched case–control studies\n\nCohort studies\n\nCross-sectional studies\n\nBalanced\n\nUnbalanced\n\nSpecify lists of\n\nAlpha values\n\nPower levels\n\nBeta values\n\nSample sizes\n\nRatios of sample sizes\n\nAnd more\n\nMultiple scenarios using\n\nAll combinations of values\n\nParallel lists of values", null, "", null, "Control panel interface\n\nSelect analysis type from descriptive summaries\n\nSpecify list of power, sample size, or other levels\n\nAutomated tables and graphs\n\nCustomize tables and graphs\n\nCommand interface too (of course)\n\nTables\n\nAutomated\n\nCustomized\n\nGraphs\n\nAutomated\n\nCustomized\n\nSimple or comparative\n\nSave results to dataset", null, "" ]

[ null, "http://www.stata.com/features/i/power_twomeans.png", null, "http://www.stata.com/features/i/pss_control_panel.png", null, "http://www.stata.com/features/i/power_twomeans_graph.png", null ]

[ "# How many inches in 302 millimeters?\n\n302 millimeters equals 11.8898 inches\n\n### All In One Units Converter\n\nPlease, choose a physical quantity, two units, then type a value in any of the boxes above.\n\n### Quote of the day...\n\nTo calculate a millimeter value to the corresponding value in inch, just multiply the quantity in millimeter by 0.039370078740157 (the conversion factor). Here is the formula:\n\nValue in inches = value in millimeter × 0.039370078740157\n\nSuppose you want to convert 302 millimeter into inches. Using the conversion formula above, you will get:\n\nValue in inch = 302 × 0.039370078740157 = 11.8898 inches\n\n## Definition of Millimeter\n\nA millimeter (mm) is a decimal fraction of the meter, The international standard unit of length, approximately equivalent to 39.37 inches.\n\n## Definition of Inch\n\nAn inch is a unit of length or distance in a number of systems of measurement, including in the US Customary Units and British Imperial Units. One inch is defined as 1⁄12 of a foot and is therefore 1⁄36 of a yard. According to the modern definition, one inch is equal to 25.4 mm exactly.\n\n• How many millimeters are in 302 inches?\n• 302 millimeters are equal to how many inches?\n• How much are 302 millimeter in inches?\n• How to convert millimeters to inches?\n• What is the conversion factor to convert from millimeters to inches?\n• How to transform millimeters in inches?\n• What is the formula to convert from millimeters to inches? Among others.\n\n## Millimeters to inches conversion chart near 302 millimeters\n\nMillimeters to inches conversion chart\n232 millimeters = 9.13 inches\n242 millimeters = 9.53 inches\n252 millimeters = 9.92 inches\n262 millimeters = 10.3 inches\n272 millimeters = 10.7 inches\n282 millimeters = 11.1 inches\n292 millimeters = 11.5 inches\n302 millimeters = 11.9 inches\nMillimeters to inches conversion chart\n302 millimeters = 11.9 inches\n312 millimeters = 12.3 inches\n322 millimeters = 12.7 inches\n332 millimeters = 13.1 inches\n342 millimeters = 13.5 inches\n352 millimeters = 13.9 inches\n362 millimeters = 14.3 inches\n372 millimeters = 14.6 inches\n\nNote: some values may be rounded.\n\n## Sample Lenght/Distance Conversions\n\n### Disclaimer\n\nWhile every effort is made to ensure the accuracy of the information provided on this website, neither this website nor its authors are responsible for any errors or omissions. Therefore, the contents of this site are not suitable for any use involving risk to health, finances or property." ]

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[ "Cody\n\n# Problem 1107. Find max\n\nSolution 858360\n\nSubmitted on 29 Mar 2016 by HH\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1   Pass\nx = magic(5); y_correct = 25; assert(isequal(your_fcn_name(x),y_correct))\n\ny = 25\n\n2   Pass\nx = [2 4 9 0 7 19;3 4 1 2 0 6]; y_correct = 19; assert(isequal(your_fcn_name(x),y_correct))\n\ny = 19\n\n3   Pass\nx = [2 4 9 0 7 19;3 4 1 2 0 6]'; y_correct = 19; assert(isequal(your_fcn_name(x),y_correct))\n\ny = 19" ]

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[ "## How to find the Linear Regression?\n\n### R7 DR (view profile)\n\non 3 Aug 2015\nLatest activity Edited by R7 DR\n\n### R7 DR (view profile)\n\non 3 Aug 2015\nHi\nI fitted the data using the fminsearch. I am able to find the slope and intercept of the fitted equation but, how to find the Linear Regression (R2) value of the fitted equation?\ncode:\nfor i=1:3\ny=[Y1{i,1}']\nx=[X{i,1}']\nA= fminsearch(@(par_fit) funccoats(par_fit,x,y),rand(1,2));\nB(i,1)=A(:,1)\nC(i,1)=A(:,2)\nend\ngraph:", null, "" ]

[ null, "https://www.mathworks.com/matlabcentral/answers/uploaded_files/147669/image.png", null ]

[ "# Difference between revisions of \"Walkthrough on PCA through the command line\"\n\nPCA computations through the command line are governed through PCA workflow objects. We describe here how to create and handle them:\n\n# Creation of a synthetic data set\n\n```dtutorial ttest128 -M 64 -N 64 -linear_tags 1 -tight 1\n```\n\nThis generates a set of 128 particles where 64 are slightly closer than the other 64. The particle subtomogram are randomly oriented, but the alignment parameters are known.\n\n# Creation of a workflow\n\n## Input elements\n\nThe input of a PCA workflow are:\n\n• a table that expreses the alignment\n• a mask that indicates the area of each alignment particle that will be taken into account during the classification procedure.\n\n### Data\n\n```dataFolder = 'ttest128/data';\n```\n\n### Table\n\n```tableFile = 'ttest128/real.tbl';\n```\n\nWe create a cylindrical mask with the dimensions of the particles (40 pixels) mask = dcylinder([20,20],40);\n\n### Syntax\n\nWe decide a name for the workflow itself, for instance\n\n```name = 'classtest128';\n```\n\nNow we are ready to create the workflow:\n\n``` wb = dpkpca.new(name,'t',tableFile,'d',dataFolder,'m',mask);\n```\n\nThis creates an workflow object (arbitrarily called wb in the workspace during the current session). It also creates a folder called classtest128.PCA where results will be stored as they are produced.\n\n## Mathematical parameters\n\nThe main parameters that can be chosen in this area are:\n\n• bandpass\n• symmetry\n• binning level (to accelerate the computations)\n\n## Computational parameters\n\nThe main burden of the PCA computation is the creation of the cross correlation matrix.\n\n### Computing device\n\nPCA computations can be run on GPUs of on CPUs, in both cases in parallel.\n\nThe\n\n# Running\n\nIn this workflow we run the steps one by one to discuss them. In real workflows, you can use the run methods to just launch all steps sequentially.\n\n## Prealigning\n\n```wb.steps.items.prealign.compute();\n```\n\n## Correlation matrix\n\nAll pairs of correlations are computed in blocks, as described above\n\n```wb.steps.items.ccmatrix.compute();\n```\n\n## Eigentable\n\nThe correlation matrix is diagonalised. The eigenvectors are used to expressed as the particles as combinations of weights.\n\n```wb.steps.items.eigentable.compute();\n```\n\nThese weights are ordered in descending order relative to their impact on the variance of the set, ideally a particle should be represented by its few components on this basis. The weights are stored in a regula Dynamo table. First eigencomponent of a particle goes into column 41.\n\n## Eigenvolumes\n\nThe eigenvectors are expressed as three=dimensional volumes.\n\n```wb.steps.items.eigenvolumes.compute();\n```\n\n## TSNE reduction\n\nTSNE remaps the particles into 2D maps which can be visualised and operated interactively.\n\n```wb.steps.items.tsene.compute();\n```\n\n# Visualization\n\nComputed elements have been stored in the workflow folder. Some of them () can be directly access through workflow tools.\n\n## Correlation matrix\n\nm=wb.getCCMatrix();\n\nfigure;dshow(cmm);h=gca();h.YDir = 'reverse';\n\n## Eigencomponents\n\n### Series of plots\n\nTo check all the eigencomponents, it is a good idea to do some scripting. The script below uses a handy Dynamo trick to create several plots in the same figure.\n\n``` gui = mbgraph.montage();\nfor i=1:10\nplot(m(:,i),'.','Parent',gui.gca);\n% gui.gca captures the\ngui.step;\nend\ngui.first();\ngui.shg();\n```\n\n## Eigenvolumes\n\n```eigSet=wb.getEigenvolume(1:30);\n```\n```mbvol.groups.montage(eigSet);\n```\n```mbvol.groups.montage(dynamo_normalize_roi(eigSet));\n```\n\n## Correlation of tilts\n\nIt is a good idea to check if some eigenvolumes correlate strongly with the tilt.\n\n```wb.show.correlationEigenvectorTilt(1:10)\n```\n\nIn this plot, each point represents a particle in your data set. We see that in this particular experiment, eigencomponent 3 seems to have been \"corrupted by the missing wedge\"" ]

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[ "# How do you find all numbers that must be excluded from the domain of the given rational expression 8/(x^2-4x)?\n\nApr 7, 2015\n\nDivision by $0$ is undefined\ntherefore we must exclude any values of $x$ which would cause the denominator to be $0$.\n\nThose values would be given by\n${x}^{2} - 4 x = 0$\nor\n$\\left(x\\right) \\left(x - 4\\right) = 0$\n\nTherefore we must exclude\n$x = 0$\nand\n$x = 4$\nfrom the domain of this expression" ]

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[ "## Comments about \"Speed of light\" in Wikipedia\n\nThis document contains comments about the article \"Speed of light\" in Wikipedia\n• The text in italics is copied from that url\n• Immediate followed by some comments\nIn the last paragraph I explain my own opinion.\n\n### Introduction\n\nThe article starts with the following sentence.\nThe speed of light in vacuum, commonly denoted c, is a universal physical constant important in many areas of physics. Its exact value is 299792458 metres per second, since the length of the metre is defined from this constant and the international standard for time.\nOne critical remark: There is a hugh difference the \"speed of light in a vacuum\" and the more general concept of the \"speed of light\".\nWhen we measure the \"speed of light in a vacuum\" we always do that each time under almost the same conditions, implying that each time we get the same answer. To try to measure the \"speed of light\" in general is a very different exercise. See Reflection 1\n\nNext:\nAccording to special relativity, c is the maximum speed at which all matter and hence information in the universe can travel.\nThe problem with special relativity is that SR has nothing to do with movements in space in general. When you want to study the behaviour of objects in space you have to use GR because accelerations are involved.\nIn the theory of relativity, c interrelates space and time, and also appears in the famous equation of mass–energy equivalence E = mc^2\nThe problem is that in GR the speed of light (=c) is not a constant because c itself is influenced by gravity.\nThe equation E = mc^2 is already tricky itself because how is mass measured? using SR? using GR?\nAfter centuries of increasingly precise measurements, in 1975 the speed of light was known to be 299792458 m/s with a measurement uncertainty of 4 parts per billion.\nIt is very important to understand exactly how this measurement is performed in detail because this immediate defines the claim that the \"speed of light in a vacuum\".\n\n### 2. Fundamental role in physics\n\nThe speed at which light waves propagate in vacuum is independent both of the motion of the wave source and of the inertial frame of reference of the observer.\nThe problem is that the propagation of light is a physical process which has nothing to do with the \"state\" of any observer. Comparing the same conditions the speed will be the same. The issue is if the speed is always the same. See Reflection 1\nThis invariance of the speed of light was postulated by Einstein in 1905, after being motivated by Maxwell's theory of electromagnetism and the lack of evidence for the luminiferous aether\nThe problem is that the speed of light is very difficult to measure.\nIn principe the concept of an eather has nothing to do with this. In fact all the photons around us also \"form\" an aether.\nIt is only possible to verify experimentally that the two-way speed of light (for example, from a source to a mirror and back again) is frame-independent, because it is impossible to measure the one-way speed of light (for example, from a source to a distant detector) without some convention as to how clocks at the source and at the detector should be synchronized.\nWhat do they mean with frame-independent?\nThe word because is wrong.\n\n### 7.4 \"Luminiferous aether\"\n\nIt was thought at the time that empty space was filled with a background medium called the luminiferous aether in which the electromagnetic field existed.\nIn fact what people thought was that space around us, which seems to be empty was not empty at all. That there was something.\nIn fact that still can be considerd as true, because all space is filled with radiation and photons.\nIn fact there exists an important analogy when you compare our existance with a fish. For a fish the aether is the water in with they live.\nSome physicists thought that this aether acted as a preferred frame of reference for the propagation of light and therefore it should be possible to measure the motion of the Earth with respect to this medium, by measuring the isotropy of the speed of light.\nYou can only do that by performing measurements in one way and no clocks should be used. The use of clocks is tricky because there functioning also depents on the speed of light.\nModern experiments indicate that the two-way speed of light is isotropic (the same in every direction) to within 6 nanometres per second.\nAgain it is in the details.\n\n### 8. See also\n\nFollowing is a list with \"Comments in Wikipedia\" about related subjects\n\n### 12. External links\n\nFollowing is a list with additional information:\n\n### Reflection 1 - speed of light\n\nIn relation to the \"speed of light\" there are two important issues:\n• Is the speed of light always the same or can the speed of the photons change?\nThis is a physical question about the behaviour of the photons, specific about the speed and if there is something that influences this behaviour i.e. speed. For a futher discussion see: Reflection 3 - Speed of light in a gravitation field.\n• What is the speed of light?\nThis is a quatifying question and is much more difficult to answer.\nWhat you want is to know the speed from A to B but what you get in most cases is the average speed going from A to B and back to A.\nBut also this average speed is not so easy. It is easy but, less accurate, when both A and B are situated on Earth. In that case the distances are fixed. What you get is than a local average speed. It is more complicated when A is on Earth and B is on the Moon. In that case the exact position of both Earth and Moon have to be known in order to calculate the total distance travelled by the light signal.\n\n### Reflection 2 - One way speed of light.\n\nThe problem is when you want to measure the speed of light you want to do that in one specific direction. That means from A to B where A is \"my position\" and B is a distance away in any direction.\nTo measure a speed you need a fixed distance and two clocks. One clock at position A and one at position B. In order to measure the speed you must measure the duration that the signal goes from A to B. To do that you must observe the time on Clock A when the signal is transmitted and the time on Clock B when the signal arrives at B. The difference is the duration. When you know the duration the speed can easily be calculated.\nThe problem is in order to do this accurate the two clocks have to run synchrone and simultaneous. Synchrone is not so difficult because it means that you have to use two complete identical clocks. Simultaneous is not so simple because it means that the two clocks should indicate at each instant the same reading.\nIMO what that means is that to perform physics you have to define one reference frame as being at rest.\n\n### Reflection 3 - Speed of light in a gravitation field.\n\nIn order to study the speed of light in a gravitational field we perform two experiments.\n\nIn the first experiment instead of light we use a ball which we drop from a tower and which bounces back.\nFrom experience we already know that when you drop a ball from a tower, that the speed is not constant. The cause is gravity.\nIn reality we use two balls: A red ball and a black ball.\nThe black ball bounces back via the groundfloor of the tower.\nThe red ball bounces back via an extra floor half way the top and the ground floor.\nFigure 1 below shows the path of two balls: A red ball and a black ball.\nThe red line in the middle, identifies half the height of the tower. Here the two paths divide.", null, "Figure 1 Both balls are dropped at the same moment, but the height that each ball falls is different. The black ball falls the full height and than bounces back. The red ball drops half the height of the tower, bounces back. This process is repeated once. The black ball is dropped at t0. Initial speed = 0 When the ball is at the bottom the black ball bounces back. The black ball is back at the top of the tower at t1. This terminates a complete jump. The red ball is also dropped at t0. The initial speed = 0 Ball \"2\" bounces back and is identified with the letter 2. When the ball is half way between the top and the bottom the red ball bounces back. The first jump is finished when the red ball is back at the top. There after the whole process repeats itself The red ball is back at the top of the tower at t2. This terminates a complete jump.", null, "Figure 1\nWhat Figure 1 clearly shows that the arrival time of the two balls (indicated as t1 and t2) is different. t1 is first and t2 is later.\nThis means that the average speed of the black ball is larger than the average speed of the red ball . This difference is explained by the second leg of the black ball i.e. when the black ball is drops near the bottom That means there is acceleration involved and the cause is the force of gravitation or gravity.\n\nIn the second experiment in stead of two balls we use a two lightsignals. The question is the behaviour of the two lightsignals (photons) identical as of the two balls?\nLightsignal 1 is reflected at the bottom (with a mirror). Lightsignal 2 is reflected with a mirror at half distance. For Lightsignal 2 the process is repeated once.\nThe problem is that it is impossible to perform such an experiment in reality. The question is the outcome.\n\nIMO the outcome is that in theory the arrival times are different. The theory is that photons have mass in the same sense that objects have mass. This leads to the same behaviour for both and that the speed of photons is influenced by gravity. This is in agreement with the observation that the path of the photons is not straight but bended close to mass.\n\n### Reflection 4 - Speed of light in a gravitation field experiment.\n\nThe purpose of the experiment is to demonstrate that the speed of light is not constant when a lightsignal approaches the earth. The problem is that a distance like the tower of Pisa is much to short to demonstrate this effect. A cavity which different heights in which the light signal bounces back a million times is a better path to follow.\n\nYou can also think about a ring laser in vertical direction. See \"https://en.wikipedia.org/wiki/Ring_laser_gyroscope#Description. What is important that in both cases the photon source is at the same height above the object (Earth) you want to use. In the first test you place the gyroscope on the floor, the vertical arms are long and the photon source is at the top. In the second test you place the gyroscope high above the floor (but below the top), the arms are short and the photon source is (stays) at the top. Simple arithmatic is enough to decide if the speed is constant, but again accuracy is not enough.\nIt is important that the source is always at the same height (from the center of gravity) because gravity in principle can already influence the behaviour of the photons at emission.\n\nIn a sense what the experiment does it compares the behaviour of two clocks. You have one slow ticking clock and one fast ticking clock. The path of the lightsignal is slow ticking clock is long and in the fast ticking clock is short. To claim that there is any length contraction involved does not seem obvious.\n\n ``` top /--->---\\ | | ^ V | | Source ---> / \\----> Detector ---> \\ /----> | | V ^ | | \\--->---/ Floor ``` The experiment at the left tries to make things simpler. The setup is symmetric. In this setup the source and the detector are at the same height, in the middle. One lightray goes via the floor and one via the top. The experiment involves with two ligtsignals at the source. Those lightsignals can be compared at the detector. The expectation is that the signal via the floor arrives earlier because the speed is higher. The cause is gravity.\n\n### Feedback\n\nIf you want to give a comment you can use the following form Comment form\nCreated: 24 November 2016\nUpdated: 14 September 2016\nUpdated: 6 July 2016\n\nGo Back to Wikipedia Comments in Wikipedia documents\nBack to my home page Index" ]

[ null, "http://users.telenet.be/nicvroom/Bouncing Ball1.jpg", null, "http://users.telenet.be/nicvroom/Bouncing Light1.jpg", null ]

[ "## Minor Arc In Geometry", null, "Arc Part of a circle measured in degrees Minor arc. The measures of a minor arc and a major arc depend on the central angle of the minor arc.", null, "Ideas And Resources For The Secondary Math Classroom Mtbos30 Central Angles And Arcs Geometry Worksheets Teaching Geometry Circle Math\n\n### Representation I and J are endpoints for the both minor and major arcs.", null, "Minor arc in geometry. Look at the circle and try to figure out how you would divide it into a portion that is major and a portion that is minor. 3 The degree measure of a semicircle is. These two points divide the circle into two opposite arcs.\n\nMinor arcs which measure less than a semicircle are represented by its two endpoints. The shorter arc joining two points on the circumference of a circle. 2 the degree measure of a major.\n\nMinor arcs and major arcs. In the circle below there is both a major arc and a minor arc. This is stated as a theorem.\n\nThe length of the arc that subtend an angle θ at the center of the circle is equal 2πrθ360. An arc that is less than a semicircle. The is the measure of its central angle.\n\nIn this fourth configuration the roles are reversed and the major arc. If the chord AB is a diameter then the two arcs are called semicircles. In Figure 1 circle Om 1 m 2 which in turn would make m m.\n\nA semicircle is named by three points. Its measure is 180 8. Arcs and Central Angles Depending on the central angle each type of arc is measured in the following way.\n\nThe first and third points represent the endpoints while the middle point is any point on the arc located between the endpoints. There are two other types of arcs. This geometry video math lesson defines semicircle minor arc and major arc.\n\nWhereas the larger arc is called the minor arc. A minor arc is an arc that has a length that is shorter than that of a semicircle. Given two points on a circle the minor arc is the shortest arc linking them.\n\nCongruent arcs are arcs on circles with congruent radii that have the same degree measure. It is named by three points. Also this video models how to identify arc from a given diagram.\n\nArcs are an essential geometry concept which you can check your comprehension of with this quiz and printable worksheet. Arc length formula if θ is in degrees s 2 π r θ360 arc length formula if θ is in radians s θ r. A is an arc whose central angle measures 180 8.\n\nThe converse of this theorem is also true. An arc is the part of a circle determined by two points and all points between them. Continue reading How To Find Degree Measure Of Minor Arc.\n\nDefinition of Arc Measure 1 The degree measure of a minor arc is the degree measure of its central angle. A minor arc is an arc whose degree measure is between 0 and 180. Minor arc– An arc measuring less than or equal to 180 or π radians Semicircle — An arc measuring exactly 180 or π radians which excludes designating either part.\n\n2 The degree measure of a major arc is 360 minus the degree measure of its central angle. How To Find Degree Measure Of Minor Arc. The semicircle represents an arc whose endpoints coincide with endpoints of the.\n\nIt is a smooth curve with two end points. If length of the arc is minor then it is called as minor arc. The major arc is the longest.\n\nMinor arcs which measure less than a semicircle are represented by its two endpoints. It confuses everyone if both arcs. Minor arcs are typically named only by their endpoints.\n\nSemicircle measure of a major arc measure of a minor arc major arc. If length of the arc is major then it is called as major arc. If an angle is inscribe in a circle then its measure is half the measure of its intercepted arcPPT 106.\n\nA semicircle is an arc whose degree measure is exactly 180. Learn more about arc at BYJUS. An arc that is more than a semicircle.\n\nIn geometry Arc is the part of circumference of a circle. The is the difference of 360 8 and the measure of the related minor arc. We have used minor arc as the reference arc in our three previous configurations with the major arc as the complementary arc.\n\nLet A and B be two different points on a circle with centre O. The shorter length is called the minor arc and the longer length is called the major arc. A minor arc is named by using only the two endpoints of the arc.\n\nYou can use these guides. Figure 1 A circle with four radii and two chords drawn. Arcs An arc of a circle is the part of the circumference of the circle that is cut off by a chord.\n\nThe larger arc is called the Major Arc See. A minor arc left figure is an arc of a circle having measure less than or equal to 180 degrees pi radians. Otherwise one arc is longer than the other the longer arc is called the major arc AB and the shorter arc is called the minor arc AB.\n\nArcs are grouped into two descriptive categories. A major arc has an arc length that is greater than that of a semicircle. In a circle if two chords are equal in measure then their corresponding minor arcs are equal in measure.\n\nIf the chord PQ is a diameter the arcs are equal in length and in this special case there are no minor or major arcs. As per our arc angle subtending concept this angle is double the angle held by the same major arc on any point P of its complementary arc which happens to be the minor arc. In the figure below is a minor arc.\n\nMeasure Of A Minor Arc Definition Geometry. In Figure 2 AC is a diameter. Major arcs which meaure more than a semicirlce are represented by three points.\n\nThe first and third are the endpoints and the middle point is any point on the arc between the endpoints.", null, "Circles Part 2 Stations Review Secondary Math Classroom Math Geometry Interactive Notes", null, "", null, "Geometry Problem 1266 Triangle Excircle Circle Tangency Points Perpendicular 90 Degrees Angle Bisector Basic Math Geometry Problems Maths Solutions", null, "Rs Aggarwal Solutions Class 10 Chapter 18 Areas Of Circle Sector And Segment A Plus Topper Https Www A Maths Ncert Solutions Maths Solutions Studying Math", null, "Geometry Ii Quick Access Reference Chart Geometry Geometric Formulas", null, "Center Angle Arc Length Circles Partner Paper Fun Math Super Teacher Angles", null, "Circles Diameter Chord Radius Arc Tangent With Examples Videos Definitions Parts Of A Circle Math Forms Math Geometry", null, "Circles Segments Arcs Chords Angles And More Geometry Lessons Circle Math Math Geometry", null, "60 Angle And Importance Of Being The Other End Of A Diameter Solution Triangle Abc Math Trigonometry", null, "Poster And Typography Of Problem 300 Tangents To A Circle Secants Minor Arc Math Tutor Online Math Learning Math", null, "Circle Central Angles Arc Measure Doodle Graphic Organizer Circle Math Studying Math Math Charts", null, "Chord Secant Tangent Radius And Diameter Arc Sector And Segment Math Geometry Segmentation Math", null, "Circles Naming Arc Finding Measures Semicircles Minor And Major Sorting Activity Algebra Word Walls Interactive Student Notebooks Sorting Activities", null, "10 2 And 10 4 Finding Arc Measures And Inscribed Angles Angles Worksheet Evaluating Algebraic Expressions Mathematical Expression", null, "Arcs And Chords In Circles Theorem Foldable For The Geometry Interactive Notebook Circle Theorems Geometry Lessons Secondary Math", null, "Circles Geometry Circles Graphic Organizer Reference Sheets Circle Math Teaching Geometry Math Work", null, "Tangent Chord Angle Formula Is Cool As Well Let S Just Put This In Simpler Terminology Math Tricks Tangent Gmat", null, "60 Angle And Importance Of Being The Other End Of A Diameter Problem Studying Math Basic Math Learning Math" ]

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[ "Low Complexity List Successive Cancellation Decoding of Polar Codes\n\nLow Complexity List Successive Cancellation Decoding of Polar Codes\n\nCongzhe Cao, Zesong Fei School of Information and Electronics\nBeijing Institute of Technology\nBeijing, China\nEmail: 2220120185, [email protected]\nJinhong Yuan School of Electrical Engineering\nand Telecommunications\nUniversity of New South Wales\nSydney, Australia\nEmail: J. [email protected]\nJingming Kuang School of Information and Electronics\nBeijing Institute of Technology\nBeijing, China\nEmail: [email protected]\nAbstract\n\nWe propose a low complexity list successive cancellation (LCLSC) decoding algorithm to reduce complexity of traditional list successive cancellation (LSC) decoding of polar codes while trying to maintain the LSC decoding performance at the same time. By defining two thresholds, namely “likelihood ratio (LR) threshold” and “Bhattacharyya parameter threshold”, we classify the reliability of each received information bit and the quality of each bit channel. Based on this classification, we implement successive cancellation (SC) decoding instead of LSC decoding when the information bits from “bad” subchannels are received reliably and further attempt to skip LSC decoding for the rest information bits in order to achieve a lower complexity compared to full LSC decoding. Simulation results show that the complexity of LCLSC decoding is much lower than LSC decoding and can be close to that of SC decoding, especially in low code rate regions.\n\nI Introduction\n\nPolar codes are capacity-achieving codes for the class of binary-input discrete memoryless channels (B-DMCs). In , successive cancellation (SC) decoding is used to recover information bits. Later in , belief propagation (BP) decoding is employed to achieve better performance in BECs, but in general B-DMCs the specific schedule of the individual messages for BP decoding is a problem. Linear programming (LP) decoding is introduced afterwards without any schedule, but it does not work for other channels except BECs . Recently, list successive cancellation (LSC) decoding of polar codes is developed and shows significant performance improvement compared with SC decoding. However, much higher decoding complexity is observed for LSC . Thus, to find a decoding algorithm with both good frame error rate (FER) performance and low complexity is an open interest. In this paper, we propose a low complexity list successive cancellation (LCLSC) decoding algorithm to significantly reduce the complexity of LSC decoding while trying to maintain its FER performance.\n\nIi Polar Codes and List Successive Cancellation Decoding\n\nIn , polar codes are introduced to achieve the capacity of B-DMCs by exploiting the channel polarization effect. For an polar code of information bits and encoded bits (), an invertible matrix is introduced to describe channel polarization. Here, is a matrix where and denotes the Kronecker product. Let and denote the vector of input bits and encoded bits correspondingly, while denotes the vector of channel output. For the vector channel of copies of a given B-DMC , the transition probability between and is defined as\n\n WN(yN1|uN1)Δ=N∏i=1W(yi|xi)=N∏i=1W(yi|xi=(uN1GN)i) (1)\n\nand the subchannel with input and output has the transition probability\n\n W(ui)N(yN1,ui−11|ui)=12N−1∑uNi+1WN(yN1|uN1). (2)\n\nThe polar encoding scheme is to transmit the set of information bits over most reliable subchannels out of subchannels and to use the other ones to transmit the so called “frozen” bits. Note that is labeled with respect of the sequence that information bits are decoded, i.e. is the first decoded information bit while is the last decoded information bit. This labelling is also used in and which will be introduced later. In , SC decoding is used to recover information bits , where the estimation of information bits , are successively generated by computing the likelihood ratios (LRs) :\n\n LRi=W(ui)N(yN1,ˆui−11|ui=0)W(ui)N(yN1,ˆui−11|ui=1). (3)\n\nLSC decoding is the extension of SC decoding, which is actually a breadth-first searching on the code tree with searching width . LSC decoding keeps a list of size and updates the list after each , is obtained. It is well known that the complexity of LSC decoding is with the so called “lazy copy” strategy, which is times of the complexity of SC decoding . For more about LSC decoding, we refer readers to .\n\nIii preliminaries of the proposed algorithm\n\nLSC decoding with large list size performs significantly better than SC decoding, but its complexity is also much higher, which is a deficiency in practical implementation. In this paper, we are interested in finding a low complexity decoding algorithm that can greatly reduce the complexity of LSC decoding while trying to maintain the FER performance of LSC decoding at the same time.\n\nFor convenience, we give the definition below before introducing our proposed algorithm.\n\nDefinition 1\n\nBased on channel polarization, a subchannel transmitting information bit is called a “good” subchannel if and only if its Bhattacharyya parameter and all the Bhattacharyya parameters of its subsequential decoded information bits are smaller than a threshold (i.e. the Bhattacharyya parameter threshold will be discussed later). Otherwise, the subchannel is called a “bad” one.\n\nThen we introduce two parameters of the proposed decoding algorithm. One is called “LR threshold” , which determines whether an information bit is received reliably or not. This parameter is used to decide whether to process SC decoding or LSC decoding over based on the observation of . The other parameter is the “Bhattacharyya parameter threshold” . Based on the channel polarization effect, the set of information bits can be divided into two subsets, namely , which stands for the set of information bits that are transmitted over “bad” subchannels thus are more probable to contribute to FER, and , which stands for the set of information bits that are all transmitted over “good” subchannels thus are more reliable, given correctly estimated . The Bhattacharyya parameter threshold is used to decide the number of information bits transmitted over “bad” subchannels, which is denoted by , and .\n\nThe motivation of the proposed decoding algorithm is to reduce the LSC decoding complexity. We achieve this by two folds. For each information bit, when its estimation is reliable, we process SC decoding rather than LSC decoding. Secondly, when all information bits from “bad” subchannels are received reliably, we process SC decoding instead of LSC decoding for information bits from “good” subchannels as well.\n\nNow we briefly describe the proposed LCLSC decoding, which is shown as Algorithm 1. After starting the decoding process, we observe the LRs of information bits in bit by bit. If the observed LR of is larger than the LR threshold , SC decoding is processed over . If is greater than for all information bits of subset (i.e. ), we process SC decoding for the rest of information bits . On the other hand, if is less than the LR threshold for any , we process LSC decoding for this information bit and the remaining information bits in .\n\nIt is clear that in the proposed decoding algorithm, we have two thresholds to determine. One is the LR threshold , the other is the Bhattacharyya parameter threshold . We discuss these two thresholds in the following sections.\n\nIv determining the LR threshold\n\nAs discussed above, if the LR of any information bit at the receiver is sufficiently high, i.e. larger than the LR threshold , each of the information bit over “bad” subchannels is received reliably. Then we only need to process SC decoding instead of LSC decoding since the received signal is more likely to be decoded correctly. For the purpose of determining , we introduce the Proposition below.\n\nProposition 1\n\nIn a B-DMC with Bhattacharyya parameter , the upper bound of the bit error probability in estimating the channel input on the basis of the channel output via the maximun-likehood (ML) decoding is given as follows \n\n 12(1−√1−Z(W(i)N)2)≤Pe(W(i)N)≤12×Z(W(i)N).\n\nBased on Proposition 1, it can be concluded that the lower bound of the probability that the input bit is correctly estimated is , where denotes the Bhattacharyya parameter of the subchannel where is transmitted.\n\nFor an input bit , if the probability of determining as 0 or 1 is smaller than , we regard is not reliable and need to process LSC decoding over . On the other hand, if the probability of determining as 0 or 1 is larger than , we consider the estimation is reliable and thus employ SC decoding. Therefore, we derive the inequalities which are satisfied when the estimation is reliable\n\n ⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩W(ui)N(yN1,ˆui−11∣∣ui=0)W(ui)N(yN1,ˆui−11∣∣ui=0)+W(ui)N(yN1,ˆui−11∣∣ui=1)>piorW(ui)N(yN1,ˆui−11∣∣ui=1)W(ui)N(yN1,ˆui−11∣∣ui=0)+W(ui)N(yN1,ˆui−11∣∣ui=1)>pi (4)\n\nand\n\n pi=1−12Z(W(ui)N) (5)\n\nwhich represents the lower bound of correct decoding probability for information bit .\n\nWhen either of the inequalities in (4) is satisfied, we process SC decoding instead of LSC decoding over . Thus, is defined as\n\n LRth=⎧⎨⎩pi1−piLRi>11−pipiLRi<1 (6)\n\nwhich means when the observed LR is larger than 1, we process SC decoding instead of LSC decoding if the observed LR is larger than . When the observed LR is smaller than 1, we do the same if the observed LR is smaller than .\n\nV determining the Bhattacharyya parameter threshold\n\nIn this section the Bhattacharyya parameter threshold that determines is derived. We look deeper and exploit the reliability of good polarized subchannels. In consistent with , the Bhattacharyya parameter is utilized to measure the reliability of subchannels. Based on Definition 1, can be expressed as:\n\n ⎧⎨⎩Z(W(ui)N)≤Zth∀i∈uka+1Z(W(ua)N)>Zth. (7)\n\nSince the bit error events in SC decoding are not independent, the FER lower bound of the ML decoding is derived according to Proposition 1\n\n PMLe ≥1−∏i∈A(1−Pe(W(ui)N)) ≥1−∏i∈A(1−12(1−√1−Z(W(ui)N)2)) (8)\n\nIn , serves as the FER upper bound of the SC decoding . Thus, we have\n\n ∑i∈AZ(W(ui)N)≥1−∏i∈A(1−12(1−√1−Z(W(ui)N)2)) (9)\n\nIt is noted that consists of the Bhattacharyya parameters of different subchannels transmitting information bits. Some of them are quite small leading to reliable subchannels, others are large resulting in unreliable subchannels. As FER mainly results from subchannels with larger , we could determine the Bhattacharyya parameter threshold as the one that can approach the FER lower bound of ML decoding :\n\n k×Zth=1−∏i∈A(1−12(1−√1−Z(W(ui)N)2)) (10)\n\nwhich can be explained as follows. We consider a subchannel where is transmitted. If is larger than , then we have larger than the FER lower bound of ML decoding. Thus we consider the subchannel less reliable. Therefore, to achieve a good FER performance, we should observe whether the estimation satisfies . If is satisfied, we regard is reliably recovered. Otherwise, we should process LSC decoding over to approach the FER performance of ML decoding (note LSC decoding becomes ML decoding when , and practically the performance of LSC decoding is very close to ML decoding with moderate ). If is less than , we have lower than the FER lower bound of . Therefore, is considered a more reliable subchannel and then the SC decoding is likely to provide correct estimation of the information bit, if the LR thresholds of are all satisfied. As mentioned above, the information bits in all have a Bhattacharyya parameter smaller than . Therefore, it is reasonable to process SC decoding over to approach the FER of ML decoding if the estimation of is reliable.\n\nBased on the discussion above, the Bhattacharyya parameter threshold is determined as\n\n Zth=1k×{1−∏i∈A(1−12(1−√1−Z(W(ui)N)2)} (11)\n\nand can be obtained according to (7), as is the first “good” subchannel in the decoding process.\n\nVi complexity analysis and simulation results\n\nIn this section we first analyze the complexity of the LCLSC decoding. Note in the LCLSC decoding, the SC decoding is processed over some (or all) information bits while the LSC decoding is processed over the rest ones. Denote as the average number of information bits over which we process SC decoding and thus is the average number of information bits over which the LSC decoding is processed. In consistent with , the computational model for complexity analysis is a single processor machine with random-access memory, and the complexities expressed are time complexities. For the decoding algorithms, the time complexities are measured with the total number of LR calculations. Note that the time complexity of SC and LSC decoding is and respectively, meaning the number of LR calculations is and correspondingly . Then the average number of LR calculations of our proposed LCLSC decoding algorithm is given by\n\n C=mk×(N+Nlog(N))+k−mk×L(N+Nlog(N)). (12)\n\nWhen implementing the LCLSC decoding algorithm, there may be the case that all the information bits are recovered with SC decoding and there may also be the case that some information bits are recovered with SC while others with LSC decoding. So the complexity in (12) is actually an averaged complexity. Also, the complexity of LCLSC decoding in the Figures below are all averaged over the simulation. It is straightforward that is less than , thus the proposed LCLSC decoding has a lower decoding complexity than LSC decoding, which will be shown in the following. The saving in decoding complexity is considerable for low code rates.\n\nNow, we present simulation results of a polar code with length 512 and different code rates. The results for the BEC with erasure rate , the BSC with cross probability 0.11 and the BAWGNC with the standard deviation of Gaussian noise are depicted. Figs 1, 2 and 3 show the FER performance of the SC, LSC and LCLSC decoding on various channels. The capacity for both the BSC and the BAWGNC is 0.5 and the codes for BSC and BAWGNC are those optimzed via Arikan’s heuristic method . In the LCLSC decoding, the LR threshold is determined by the correct decoding probability for each information bit , as discussed above. In the results of the BEC, we set in two different ways: (i) set to be its lower bound , as (5), and (ii) set to be a fixed value 0.9. As the polarization indices are not known in closed form for the BSC and the BAWGNC, is therefore set to 0.9 in those channels. List size is set to 16 in both the LSC and LCLSC decoding.\n\nFrom Figs 1, 2 and 3 we can see that the LCLSC decoding has almost the same FER performance as LSC decoding and much better FER performance than SC decoding. Figs 4, 5 and 6 show the corresponding complexity of the three decoding algorithms on various channels. It is illustrated that the LCLSC decoding has a lower complexity than the LSC decoding. The complexity reduction is larger with lower code rate, as there are more “reliable” subchannels where we could process SC decoding instead of LSC decoding. Especially, in low to medium code rates, the complexity of LCLSC decoding is near to that of SC decoding with slightly degraded FER performance compared with LSC decoding.", null, "", null, "", null, "", null, "", null, "", null, "Vii conclusion\n\nIn this paper, an LCLSC decoding algorithm that can reduce the complexity of LSC decoding was proposed. We set an LR threshold and a Bhattacharyya parameter threshold to determine the information bits over which SC decoding instead of LSC decoding could be utilized. Simulation results showed that the proposed decoding algorithm could reduce the decoding complexity of LSC decoding significantly with low code rate while almost maintaining the same FER performance.\n\nReferences\n\n• E. Arikan, “Channel Polarization: A method for constructing capacity-achieving codes for symmetric binary-input memoryless channels”, IEEE Transactions on Information Theory, vol. 55, no. 7, pp. 3051-3073, 2009.\n• E. Arikan, “A performance comparison of polar codes and reed-muller codes”, IEEE Communications Letters, vol. 12, no. 6, pp. 447-449, 2008.\n• N. Goela, S. B. Korada and M. Gastpar, “On LP decoding of polar codes”, IEEE Information Theory Workshop (ITW), Dublin, Ireland, pp. 1-5, 2010.\n• I. Tal and A. Vardy, “List decoding of polar codes”, IEEE int. Symp. Inform. Theroy (ISIT), pp. 1-5, 2011.\n• K. Chen, K. Niu and J. R. Lin, “List successive cancellation decoding of polar codes”, Electronics Letters, vol. 48, no. 9, pp. 500-501, 2012.\n• S. Hamed Hassani, R. Mori, T. Tanaka and R. Urbanke, “Rate-Dependent Analysis of the Asymptotic Behavior of Channel Polarization”, IEEE Transactions on Information Theory, vol. 59, no. 4, pp. 2267-2276, 2013.\nYou are adding the first comment!\nHow to quickly get a good reply:\n• Give credit where it’s due by listing out the positive aspects of a paper before getting into which changes should be made.\n• Be specific in your critique, and provide supporting evidence with appropriate references to substantiate general statements.\n• Your comment should inspire ideas to flow and help the author improves the paper.\n\nThe better we are at sharing our knowledge with each other, the faster we move forward.\nThe feedback must be of minimum 40 characters and the title a minimum of 5 characters", null, "", null, "", null, "" ]

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[ "", null, "", null, "", null, "Chapter 11.1, Problem 15E\n\nChapter\nSection\nTextbook Problem\n\n# Find a formula for the general term an of the sequence, assuming that the pattern of the first few terms continues.15. { − 3 , 2 , − 4 3 , 8 9 , − 16 27 , … }\n\nTo determine\n\nTo find:  A formula for the general term an of the sequence.\n\nExplanation\n\nGiven:\n\nThe first five terms of the given sequence are {3,2,43,89,1627,...}\n\nHere, a1=3,a2=2,a3=43,a4=89 and a5=1627 .\n\nCalculation:\n\nThe first term of the given sequence is a1=3 .\n\nThe second term of the given sequence can be expressed as follows:\n\na2=332=3(23)\n\nThe third term of the given sequence can be expressed as follows:\n\na3=3343=322(3)(3)=3(23)2\n\nThe fourth term of the given sequence can be expressed as follows:\n\na4\n\n### Still sussing out bartleby?\n\nCheck out a sample textbook solution.\n\nSee a sample solution\n\n#### The Solution to Your Study Problems\n\nBartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!\n\nGet Started", null, "" ]

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[ "# Delete elements in C++ STL list\n\nHow to insert elements in C++ STL List ?\n\n1. Using list::erase(): The purpose of this function is to remove the elements from list. Single or multiple contiguous elements in range can be removed using this function. This function takes 2 arguments, start iterator and end iterator.\nTime complexity : O(n) where (n is size of list).\n\n `// C++ code to demonstrate the working of erase() ` ` `  `#include ` `#include // for list operations ` `using` `namespace` `std; ` ` `  `int` `main() ` `{ ` `    ``// initializing list of integers ` `    ``list<``int``> list1={10,15,20,25,30,35}; ` `     `  `    ``// declaring list iterators ` `    ``list<``int``>::iterator it = list1.begin(); ` `    ``list<``int``>::iterator it1 = list1.begin(); ` `     `  `    ``// incrementing the positions of iterators ` `    ``advance(it,2); ` `    ``advance(it1,5); ` `     `  `    ``// printing original list ` `    ``cout << ``\"The original list is : \"``; ` `    ``for` `(list<``int``>::iterator i=list1.begin(); i!=list1.end(); i++) ` `       ``cout << *i << ``\" \"``; ` `      `  `    ``cout << endl; ` `     `  `    ``// using erase() to erase single element ` `    ``// erases 20 ` `    ``list1.erase(it); ` `     `  `    ``// list after deletion 1 element ` `    ``cout << ``\"The list after deleting 1 element using erase() : \"``; ` `    ``for` `(list<``int``>::iterator i=list1.begin(); i!=list1.end(); i++) ` `       ``cout << *i << ``\" \"``; ` `      `  `    ``cout << endl; ` `     `  `    ``it = list1.begin(); ` `     `  `    ``// incrementing the positions of iterators ` `    ``advance(it,2); ` `     `  `    ``// using erase() to erase multiple elements ` `    ``// erases 25,30 ` `    ``list1.erase(it,it1); ` `     `  `    ``// list after deletion of multiple elements ` `    ``cout << ``\"The list after deleting multiple elements using erase() : \"``; ` `    ``for` `(list<``int``>::iterator i=list1.begin(); i!=list1.end(); i++) ` `       ``cout << *i << ``\" \"``; ` `      `  `    ``cout << endl; ` ` `  `     `  `} `\n\nOutput:\n\n```The original list is : 10 15 20 25 30 35\nThe list after deleting 1 element using erase() : 10 15 25 30 35\nThe list after deleting multiple elements using erase() : 10 15 35\n```\n2. Using list::pop_front() and list::pop_back():\n• pop_back() : This function removes the last element from the list. This reduces the size of list by 1.\nTime complexity : O(1)\n• pop_front() : This function removes the first element from the list and shifts the subsequent elements. This reduces the size of list by 1.\nTime complexity : O(1)\n\n `// C++ code to demonstrate the working of pop_front() ` `// and pop_back() ` ` `  `#include ` `#include // for list operations ` `using` `namespace` `std; ` ` `  `int` `main() ` `{ ` `    ``// initializing list of integers ` `    ``list<``int``> list1={10,15,20,25,30,35}; ` `     `  `    ``// printing original list ` `    ``cout << ``\"The original list is : \"``; ` `    ``for` `(list<``int``>::iterator i=list1.begin(); i!=list1.end(); i++) ` `       ``cout << *i << ``\" \"``; ` `      `  `    ``cout << endl; ` `     `  `    ``// using pop_front() to erase first element of list ` `    ``// pops 10 ` `    ``list1.pop_front(); ` `     `  `    ``// list after deleting first element ` `    ``cout << ``\"The list after deleting first element using pop_front() : \"``; ` `    ``for` `(list<``int``>::iterator i=list1.begin(); i!=list1.end(); i++) ` `       ``cout << *i << ``\" \"``; ` `      `  `    ``cout << endl; ` `     `  `    ``// using pop_back() to erase last element of list ` `    ``// pops 35 ` `    ``list1.pop_back(); ` `     `  `    ``// list after deleting last element ` `    ``cout << ``\"The list after deleting last element using pop_back() : \"``; ` `    ``for` `(list<``int``>::iterator i=list1.begin(); i!=list1.end(); i++) ` `       ``cout << *i << ``\" \"``; ` `      `  `    ``cout << endl; ` `     `  `} `\n\nOutput:\n\n```The original list is : 10 15 20 25 30 35\nThe list after deleting first element using pop_front() : 15 20 25 30 35\nThe list after deleting last element using pop_back() : 15 20 25 30\n```\n3. Using remove() and remove_if():\n• remove() : This function deletes all the occurrences of the value passed in its arguments. It is different from “erase()” from the fact that “erase()” deletes values by position, where as “remove()” deletes the value passed. The size of the list is reduced by the number of occurrences removed.\nTime Complexity : O(n)\n• remove_if() : This function deletes the occurrences of the values that returns “true” to the function passed in its argument.\nTime Complexity : O(n)\n\n `// C++ code to demonstrate the working of remove() ` `// remove_if() ` ` `  `#include ` `#include // for list operations ` `using` `namespace` `std; ` ` `  `// function to pass in argument of \"remove_if()\" ` `bool` `is_div_5(``const` `int``& num) { ``return` `num%5==0;}  ` ` `  `int` `main() ` `{ ` `    ``// initializing list of integers ` `    ``list<``int``> list1={10,14,20,22,30,33,22}; ` `     `  `    ``// printing original list ` `    ``cout << ``\"The original list is : \"``; ` `    ``for` `(list<``int``>::iterator i=list1.begin(); i!=list1.end(); i++) ` `       ``cout << *i << ``\" \"``; ` `      `  `    ``cout << endl; ` `     `  `    ``// using remove() to delete all occurrences of 22 ` `    ``list1.``remove``(22); ` `     `  `    ``// list after deleting all 22 occurrences ` `    ``cout << ``\"The list after deleting all 22 occurrences : \"``; ` `    ``for` `(list<``int``>::iterator i=list1.begin(); i!=list1.end(); i++) ` `       ``cout << *i << ``\" \"``; ` `      `  `    ``cout << endl; ` `     `  `    ``// using remove_if() to delete multiple of 5  ` `    ``list1.remove_if(is_div_5); ` `     `  `    ``// list after deleting all multiples of 5 ` `    ``cout << ``\"The list after deleting all multiples of 5 : \"``; ` `    ``for` `(list<``int``>::iterator i=list1.begin(); i!=list1.end(); i++) ` `       ``cout << *i << ``\" \"``; ` `      `  `    ``cout << endl; ` `     `  `} `\n\nOutput:\n\n```The original list is : 10 14 20 22 30 33 22\nThe list after deleting all 22 occurrences : 10 14 20 30 33\nThe list after deleting all multiples of 5 : 14 33\n```\n\nThis article is contributed by Manjeet Singh. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to [email protected]. See your article appearing on the GeeksforGeeks main page and help other Geeks." ]

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[ "Test2d27, an example of an electrode defined by users equations.\n\nA parabolic reflector.\n\nThis illustrates the option (added July 2012) for electrodes to be defined by user-supplied equations.\n\n2 parabolas are defined, z = r^2 and z = r^2 - 0.01.\n\nThey have the applied voltages 0 and -2 respectively, and so form a parabolic mirror for electrons of energy 1eV.\n\nThe first parabola is defined by the equations\n\nr = u\n\nz = u^2.\n\nThe first variable 'u' is subdivided uniformly into 200 segments between the limits 0 and 1.\n\nThe second parabola is similarly defined using the variables u and b, where b = 0.01.\n\n(In fact there is no need to use the fixed variable c here, since it could be incorporated into the defining equations.)\n\nElectrons start from a plate at z = 1, with the energy 1eV. A second plate has been added to anchor the potential more accurately.\n\nThe theoretical point of focus for the first parabola is at z = 0.25. The electrons are reflected midway between the the parabolas and so the expected focal point is at z = 0.245. In fact in this example the crossing points are spread by 0.004 around this position.\n\nSee notes on the 'users equations' option." ]

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[ "", null, "# Notation of Function\n\nAs shown in figure above, for a function $f:\\mathrm{X}\\to Y$ , each element x in the domain X has a unique image y in the codomain Y.\n\nThe function can be written as:\n$\\begin{array}{l}y=f\\left(x\\right)\\\\ or\\\\ f:x↦f\\left(x\\right)\\end{array}$\n1. For $y=f\\left(x\\right)$ , we say y is a function of x.\n2. f(x) is also called the value of the function f at x.\n3. f(x) is read as \"f of x\".\n\nExample:\nGiven the function $f:x↦5x+1$ , find the value of\na. $f\\left(2\\right)$\nb. $f\\left(-3\\right)$\nc. $f\\left(\\frac{2}{5}\\right)$\n\n$\\begin{array}{l}f\\left(x\\right)=5x+1\\\\ f\\left(2\\right)=5\\left(2\\right)+1=11\\end{array}$\n$\\begin{array}{l}f\\left(x\\right)=5x+1\\\\ f\\left(-3\\right)=5\\left(-3\\right)+1=-14\\end{array}$\n$\\begin{array}{l}f\\left(x\\right)=5x+1\\\\ f\\left(\\frac{2}{5}\\right)=5\\left(\\frac{2}{5}\\right)+1=3\\end{array}$" ]

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[ "## 43943\n\n43,943 (forty-three thousand nine hundred forty-three) is an odd five-digits prime number following 43942 and preceding 43944. In scientific notation, it is written as 4.3943 × 104. The sum of its digits is 23. It has a total of 1 prime factor and 2 positive divisors. There are 43,942 positive integers (up to 43943) that are relatively prime to 43943.\n\n## Basic properties\n\n• Is Prime? Yes\n• Number parity Odd\n• Number length 5\n• Sum of Digits 23\n• Digital Root 5\n\n## Name\n\nShort name 43 thousand 943 forty-three thousand nine hundred forty-three\n\n## Notation\n\nScientific notation 4.3943 × 104 43.943 × 103\n\n## Prime Factorization of 43943\n\nPrime Factorization 43943\n\nPrime number\nDistinct Factors Total Factors Radical ω(n) 1 Total number of distinct prime factors Ω(n) 1 Total number of prime factors rad(n) 43943 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 10.6906 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 43,943 is 43943. Since it has a total of 1 prime factor, 43,943 is a prime number.\n\n## Divisors of 43943\n\n2 divisors\n\n Even divisors 0 2 1 1\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 2 Total number of the positive divisors of n σ(n) 43944 Sum of all the positive divisors of n s(n) 1 Sum of the proper positive divisors of n A(n) 21972 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 209.626 Returns the nth root of the product of n divisors H(n) 1.99995 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 43,943 can be divided by 2 positive divisors (out of which 0 are even, and 2 are odd). The sum of these divisors (counting 43,943) is 43,944, the average is 21,972.\n\n## Other Arithmetic Functions (n = 43943)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 43942 Total number of positive integers not greater than n that are coprime to n λ(n) 43942 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 4569 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares\n\nThere are 43,942 positive integers (less than 43,943) that are coprime with 43,943. And there are approximately 4,569 prime numbers less than or equal to 43,943.\n\n## Divisibility of 43943\n\n m n mod m 2 3 4 5 6 7 8 9 1 2 3 3 5 4 7 5\n\n43,943 is not divisible by any number less than or equal to 9.\n\n## Classification of 43943\n\n• Arithmetic\n• Prime\n• Deficient\n\n### Expressible via specific sums\n\n• Polite\n• Non-hypotenuse\n\n• Prime Power\n• Square Free\n\n## Base conversion (43943)\n\nBase System Value\n2 Binary 1010101110100111\n3 Ternary 2020021112\n4 Quaternary 22232213\n5 Quinary 2401233\n6 Senary 535235\n8 Octal 125647\n10 Decimal 43943\n12 Duodecimal 2151b\n20 Vigesimal 59h3\n36 Base36 xwn\n\n## Basic calculations (n = 43943)\n\n### Multiplication\n\nn×i\n n×2 87886 131829 175772 219715\n\n### Division\n\nni\n n⁄2 21971.5 14647.7 10985.8 8788.6\n\n### Exponentiation\n\nni\n n2 1930987249 84853372682807 3728711755800588001 163850780685145238527943\n\n### Nth Root\n\ni√n\n 2√n 209.626 35.2882 14.4785 8.48356\n\n## 43943 as geometric shapes\n\n### Circle\n\n Diameter 87886 276102 6.06638e+09\n\n### Sphere\n\n Volume 3.55433e+14 2.42655e+10 276102\n\n### Square\n\nLength = n\n Perimeter 175772 1.93099e+09 62144.8\n\n### Cube\n\nLength = n\n Surface area 1.15859e+10 8.48534e+13 76111.5\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 131829 8.36142e+08 38055.8\n\n### Triangular Pyramid\n\nLength = n\n Surface area 3.34457e+09 1.00001e+13 35879.3\n\n## Cryptographic Hash Functions\n\nmd5 5b7fc89c42ee0f99a9db33e8baed5c52 7f1efd2ac85f5f9f17a4ee026712364a46b0ad2b 3fc6262266ed9abd399f388fc92b586a184b8f2ab20993fc1484a3ef07ff8403 967799371ac1456039d17b47577f7572a780f405bd3ca2e4fe440a5d1dee52599caeb565435e19611f24dce3be8588f7019dbc64d5a09e568e68a080c60f4dba 327859669824166eda14315749cc1fb9d34fdd08" ]

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[ "## What are the examples of discriminant?\n\nExamples. In our example, our quadratic equation gives us 1 for our letter a, 5 for letter b, and 4 for letter c. We take these values and plug them into their appropriate places in the discriminant formula, and we’ll find that our discriminant equals 9, a positive number.\n\nWhat are the 3 types of discriminant?\n\nThe discriminant can be positive, zero, or negative, and this determines how many solutions there are to the given quadratic equation.\n\n• A positive discriminant indicates that the quadratic has two distinct real number solutions.\n• A discriminant of zero indicates that the quadratic has a repeated real number solution.\n\nWhat is the discriminant of 3x 2 10x =- 2?\n\nThe discriminant of is 76.\n\n### How do you find the discriminant of two variables?\n\nIn two variables, the general quadratic equation is ax2 + bxy + cy2 + dx + ey + f = 0, in which a, b, c, d, e, and f are arbitrary constants and a, c ≠ 0. The discriminant (symbolized by the Greek letter delta, Δ) and the invariant (b2 − 4ac) together provide information as to the shape of the curve.\n\nWhat is a discriminant in standard form?\n\nThe discriminant is the expression b2 – 4ac, which is defined for any quadratic equation ax2 + bx + c = 0. Based upon the sign of the expression, you can determine how many real number solutions the quadratic equation has. Here’s how the discriminant works.\n\nIs the discriminant negative or positive?\n\nThe discriminant is the term underneath the square root in the quadratic formula and tells us the number of solutions to a quadratic equation. If the discriminant is positive, we know that we have 2 solutions. If it is negative, there are no solutions and if the discriminant is equal to zero, we have one solution.\n\n## What are real and complex solutions?\n\nYou can use a graph to check the number of real solutions of an equation. two solutions given by the Quadratic Formula. 1) If the discriminant is less than zero, the equation has two complex solution(s). 2) If the discriminant is equal to zero, the equation has one repeated real solution(s).\n\nWhat is the discriminant of 9×2 +2 10x?\n\nWe do this by subtracting 10x both sides (to get it on the other side).. 9x^2 + 2 = 10x 9x^2 – 10x + 2 = 10x – 10x 9x^2 – 10x + 2 = 0 Now the discriminant is: b^2 – 4ac The quadratic equation is Ax^2 + Bx + C = 0 Here, A = 9, B = -10 and C = 2 so plug them in…\n\nWhy does a discriminant of zero give you exactly one real zero?\n\nWhen the discriminant is equal to 0, there is exactly one real root. When the discriminant is less than zero, there are no real roots, but there are exactly two distinct imaginary roots. In this case, there is exactly one real root. This value of x is the one distinct real root of the given equation.\n\n### Which is an example of an onomatopoeia letter?\n\nCommon Onomatopoeia Words & Letter Combinations 1 bloop 2 splash 3 spray 4 sprinkle 5 squirt 6 dribble 7 drip 8 drizzle\n\nWhich is an example of a negative discriminant?\n\nThe discriminant can be either positive or negative or zero. Here is a discriminant example. The discriminant of the given equation is: Thus, the discriminant of the given equation is: The roots of a quadratic equation ax2 +bx+c = 0 a x 2 + b x + c = 0 are the values of x x that satisfy the equation.\n\nWhen do you use the discriminant in math?\n\nThe discriminant in Math is used to find the nature of the roots of a quadratic equation. The value of the discriminant will determine if the roots of the quadratic equation are real or imaginary, equal or unequal. The discriminant of a polynomial is a function of its coefficients which gives an idea about the nature of its roots.\n\n## How is the discriminant of a polynomial determined?\n\nThe discriminant of a polynomial of degree n is homogeneous of degree 2n − 2 in the coefficients. The discriminant value helps to determine the nature of the roots of the quadratic equation. The relationship between the discriminant value and the nature of roots are as follows:" ]

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[ "## Functions Modeling Change: A Preparation for Calculus, 5th Edition\n\n$$x\\leq -3$$\nWe write an expression that matches the shaded area on the graph. Note, if the dot on the edge of the shaded region is filled, then we included that point. Thus, we find: $$x\\leq -3$$" ]

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[ "", null, "# Best Path Calculation Methods - OSPF and EIGRP Metric Calculation\n\nLast Updated : November 16, 2022", null, "OSPF metric calculation is used to find the shortest path by using cost as a metric and determining the cost of paths to know the best path to the destination. Before understanding the OSPF and EIGRP Metric calculation, you need to know that:\n\nThe automatic and dynamic exchange of routing information between routers is made possible by the use of routing protocols. Numerous routing protocols exist, each with its own set of advantages and disadvantages due to being tailored to perform best in a certain kind of network deployment. Open Shortest Path First (OSPF) and Enhanced Interior Gateway Routing Protocol (EIGRP) are two of the most widely deployed routing protocols. In this article, we'll explain and comprehend the two best-path-calculating routing protocols.\n\n## OSPF Metric Calculation\n\nAs we know, OSPF is a link state protocol, so the router learns all the paths and their costs to the destination and selects the routes with the lowest costs to the destination. This is known as OSPF Metric calculation.\n\nMetric – It is a parameter that OSPF uses to choose its best path. OSPF metrics are calculated using a cost-based algorithm.\n\nFor any given interface, the cost is always inversely proportional to the bandwidth.\n\n• A higher bandwidth leads to a lower cost.\n• A lower bandwidth leads to a higher cost.\n\nSo, the path with the lowest cost will be the best path for OSPF.\n\nNow, as we know, OSPF utilizes link-state advertisements for network destinations, and it is known as a link-state routing protocol. For this reason, the (shortest path first) SPF algorithm is used for routing calculations based on link-state information. With OSPF, all routers share metrics and link-state information about their connected interfaces with each other.\n\nLet's move on to know more about the shortest-path-first algorithm.\n\n## SPF Algorithm\n\nSpecifically, OSPF employs a (shortest-path-first) SPF algorithm to determine and construct the shortest route to all known destinations.\n\n• All routers in OSPF exchange link states by flooding LSAs. Every router that receives an LSA will store a copy of its link-state database and then forward the LSA to other routers.\n• Once the database of every router is synced, it will calculate the shortest path to all possible destinations, which we call the shortest path tree (SPT).\n• The algorithm puts every router in a tree and determines the shortest path to each destination by taking into account the total cost of getting there.\n• Lastly, when the router builds the SPT, it starts creating the routing table.\n\nWe hope you have a good overview of the SPF Algorithm. Let's understand the OSPF cost formula now.\n\n## OSPF Cost Formula\n\nThe formula used by OSPF to determine cost is as follows:\n\nCost = Reference bandwidth/Interface bandwidth\n\nIn OSPF's documentation (RFC 2338), reference bandwidth was given a completely random number. Each manufacturer must determine its reference bandwidth. When referring to bandwidth, Cisco uses 100 Mbps (108) as reference bandwidth. With this, the equation would be:\n\n### Cost = 108/interface bandwidth in bps\n\nSome of the key points to follow while calculating cost is discussed below.\n\nKey points -\n\n• Cost is a positive integer value.\n• All decimal values will be rounded to the nearest positive integer.\n• Any value in decimal or less than 1 will be considered 1.\n\nNow that we have the formula let's perform the arithmetic and get the base cost of the necessary interfaces.\n\nDefault cost of essential interfaces.\n\nThe image below shows the default OSPF Cost values for various interface bandwidths.\n\n#### How to check the cost of a link?\n\n#show ip ospf interface fa0/0 | include cost\n\n#### How to change the reference bandwidth?\n\nRouter(config)#Router ospf 1\n\nRouter(config)#auto-cost reference-bandwidth 10000 (in Mbps)\n\n#### How to check reference bandwidth?\n\n#Show ip ospf | include reference\n\n#### How to check the bandwidth of a link?\n\n#Show ip ospf interface fa0/0 | include BW\n\nThis is everything you need to learn about OSPF Metric Calculation. Moving on to the EIGRP Section.\n\n## EIGRP Metric Calculation\n\nJust like OSPF, EIGRP also has a metric calculation. The Metric of EIGRP depends on five parameters which are considered as K-Values.\n\nThese parameters are:\n\n• Bandwidth - In computing, it is the quantity of information that can be sent through a connection in a certain length of time. Bandwidth is measured in Kilobits.\n• Load – It is defined as the traffic passing through the interface. It is measured on a scale of 255 where 1 represents that an interface is empty and 255 represents that an interface is fully utilized.\n• Delay – The time that has been passed in processing a particular packet. It is generally calculated in microseconds.\n• Reliability - It is expressed on a scale of 0 to 255. Where 255 expresses 100% reliability, and 0 represents 0% reliability.\n• MTU - stands for Maximum Transmission Unit.\n\nSo, EIGRP uses these five factors (Bandwidth, Delay, Reliability, Load, and MTU) to calculate its metric, i.e., cost.\n\nCISCO technically calls these five factors K Values which are represented as shown in the image below -\n\nBecause load and reliability are dynamic variables, EIGRP's default metric calculation values are (K1 & K3). However, the EIGRP procedure will become a more over-headed protocol if these values are constantly updated.\n\n### EIGRP Metric calculation formula\n\nFormula used for EIGRP Metric = (10^7/BW + Delay/10)} *256\n\nNow that we have learned about EIGRP Metric Calculation, it is time to under the EIGRP dual-diffused update algorithm.\n\n### EIGRP DUAL–DIFFUSED UPDATE ALGORITHM\n\nDUAL Algorithm uses four parameters –\n\n• Feasible distance (FD): Total cost from the local router to the destination.\n• Advertised distance/Reported distance (AD/RD): Cost from the next-hop router to the destination.\n• Successor: The best path of EIGRP to reach a destination network. It is Stored in the routing table as well as the topology table.\n• Feasible successor: A backup path that is stored in the topology table only.\n\nThe Feasibility Condition states that a route will not be accepted if the Reported Distance exceeds the best path's Feasible Distance.\n\nConclusion\n\nOpen Shortest Path First (OSPF) and the extended Interior Gateway Routing Protocol (EIGRP) are two of the most widely deployed IGPs today. Both are commonly used in various business networks and have shown to be stable and reliable options throughout the years. The OSPF metric calculation depends on a metric known as cost; the lower the cost, the more suitable path will be. At the same time, EIGRP Metric Calculation is dependent on five factors.\n\nIf you are eager to learn more about EIGRP and OSPF Metric Calculation, you should join PyNet Labs' CCNP ENCOR Training. The CCNP ENCOR training will help you comprehend all the advanced networking skills, including OSPF and EIGRP. If you are a network engineer who wants to grow in your career, then you should check out PyNet Labs' Job Guarantee Courses. These courses will help you upskill and get your dream networking job." ]

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[ "", null, "", null, "### Home > CCG > Chapter 10 > Lesson 10.1.1 > Problem10-7\n\n10-7.\n\nIn the diagram below, $\\overline{AD}$ is a diameter of $⊙B$.", null, "1. If $m∠A = 35°$, what is $m∠CBD$?\n\n$∠A$ is the inscribed angle and $∠CBD$ is the central angle intercepting the same arc.\nHow are these angles related?\n\n$70°$\n\n2. If $m∠CBD = 100°$, what is $m∠A$?\n\n$50°$\n\n3. If $m∠A = x$, what is $m∠CBD$?\n\nUse the same method you used to solve part (a)." ]

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", 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[ "Cm To Inches\n\n# 5840 cm to in5840 Centimeters to Inches\n\ncm\n=\nin\n\n## How to convert 5840 centimeters to inches?\n\n 5840 cm * 0.3937007874 in = 2299.21259843 in 1 cm\nA common question is How many centimeter in 5840 inch? And the answer is 14833.6 cm in 5840 in. Likewise the question how many inch in 5840 centimeter has the answer of 2299.21259843 in in 5840 cm.\n\n## How much are 5840 centimeters in inches?\n\n5840 centimeters equal 2299.21259843 inches (5840cm = 2299.21259843in). Converting 5840 cm to in is easy. Simply use our calculator above, or apply the formula to change the length 5840 cm to in.\n\n## Convert 5840 cm to common lengths\n\nUnitLengths\nNanometer58400000000.0 nm\nMicrometer58400000.0 µm\nMillimeter58400.0 mm\nCentimeter5840.0 cm\nInch2299.21259843 in\nFoot191.601049869 ft\nYard63.8670166229 yd\nMeter58.4 m\nKilometer0.0584 km\nMile0.0362880776 mi\nNautical mile0.0315334773 nmi\n\n## What is 5840 centimeters in in?\n\nTo convert 5840 cm to in multiply the length in centimeters by 0.3937007874. The 5840 cm in in formula is [in] = 5840 * 0.3937007874. Thus, for 5840 centimeters in inch we get 2299.21259843 in.\n\n## 5840 Centimeter Conversion Table", null, "## Alternative spelling\n\n5840 cm to Inches, 5840 cm in Inches, 5840 Centimeters to in, 5840 Centimeters in in, 5840 Centimeters to Inches, 5840 Centimeters in Inches, 5840 Centimeter to in, 5840 Centimeter in in, 5840 Centimeter to Inches, 5840 Centimeter in Inches, 5840 Centimeters to Inch, 5840 Centimeters in Inch, 5840 cm to Inch, 5840 cm in Inch" ]

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[ " Mixed Age Year 3 and 4 Properties of Shape Step 7 Resource Pack – Classroom Secrets | Classroom Secrets\nAll › Mixed Age Year 3 and 4 Properties of Shape Step 7 Resource Pack\n\n# Mixed Age Year 3 and 4 Properties of Shape Step 7 Resource Pack", null, "## Step 7: Mixed Age Year 3 and 4 Properties of Shape Step 7\n\nMixed Age Year 3 and 4 Properties of Shape Step 7 Resource Pack includes a teaching PowerPoint and differentiated varied fluency and reasoning and problem solving resources for this step which covers Year 3 Recognise and Describe 2D Shapes & Year 4 Lines of Symmetry for Summer Block 4.", null, "", null, "", null, "### What's included in the Pack?\n\nThis Mixed Age Year 3 and 4 Properties of Shape Step 7 pack includes:\n\n• Mixed Age Year 3 and 4 Properties of Shape Step 7 Teaching PowerPoint with examples.\n• Year 3 Recognise and Describe 2D Shapes Varied Fluency with answers.\n• Year 3 Recognise and Describe 2D Shapes Reasoning and Problem Solving with answers.\n• Year 4 Lines of Symmetry Varied Fluency with answers.\n• Year 4 Lines of Symmetry Reasoning and Problem Solving with answers.\n\n#### National Curriculum Objectives\n\nMathematics Year 3: (3G3a) Draw 2-D shapes\n\nMathematics Year 3: (3G4a) Recognise angles as a property of shape or a description of a turn\n\nMathematics Year 3: (3G2) Identify horizontal and vertical lines and pairs of perpendicular and parallel lines\n\nMathematics Year 4: (4G2b) Identify lines of symmetry in 2-D shapes presented in different orientations\n\nDifferentiation for Year 3 Recognise and Describe 2D Shapes:\n\nVaried Fluency\nDeveloping Questions to support recognising common 2D shapes and basic polygons, and describing 2D shapes using the number and length of sides and lines of symmetry.\nExpected Questions to support recognising common 2D shapes, quadrilaterals and regular polygons using the number and length of sides, lines of symmetry, types of angles and types of lines.\nGreater Depth Questions to support recognising regular and irregular polygons and quadrilaterals, including within compound shapes, using the number and length of sides, lines of symmetry, types of angles and types of lines.\n\nReasoning and Problem Solving\nQuestions 1, 4 and 7 (Problem Solving)\nDeveloping Draw 3 shapes which could represent each letter in 2 sorting circles. Includes recognising and describing common 2D shapes and basic polygons using the number and length of sides and lines of symmetry.\nExpected Draw 4 shapes which could represent each letter in a Venn diagram. Includes recognising and describing common 2D shapes, quadrilaterals and regular polygons using the number and length of sides, lines of symmetry, types of angles and types of lines.\nGreater Depth Draw 6 shapes which could represent each letter in a Venn diagram with 3 circles. Includes recognising and describing regular and irregular polygons, including within compound shapes, using the number and length of sides, lines of symmetry, types of angles and types of lines.\n\nQuestions 2, 5 and 8 (Reasoning)\nDeveloping Explain and prove whether the statement is always, sometimes or never true. Statements to include shapes fitting the criteria given for Question 1.\nExpected Explain and prove whether the statement is always, sometimes or never true. Statements to include shapes fitting the criteria given for Question 4.\nGreater Depth Explain and prove whether the statement is always, sometimes or never true. Statements to include shapes fitting the criteria given for Question 7.\n\nQuestions 3, 6 and 9 (Reasoning)\nDeveloping Explain the similarities between the shapes referenced for Question 1.\nExpected Explain the similarities between the shapes referenced for Question 4.\nGreater Depth Explain the similarities between the shapes referenced for Question 7.\n\nDifferentiation for Year 4 Lines of Symmetry:\n\nVaried Fluency\nDeveloping Questions to support finding and identifying lines of symmetry using simple regular polygons with up to 2 lines of symmetry. All shapes in ‘standard’ orientation.\nExpected Questions to support finding and identifying lines of symmetry using regular polygons with up to 8 lines of symmetry. All shapes in the same ‘non-standard’ orientation in each question.\nGreater Depth Questions to support finding and identifying lines of symmetry using irregular polygons with any number of lines of symmetry. All shapes in unique orientations.\n\nReasoning and Problem Solving\nQuestions 1, 4 and 7 (Reasoning)\nDeveloping Find and explain the mistake that has been made when sorting shapes into a Venn diagram. 3 simple regular polygons with up to 2 lines of symmetry used. All shapes in ‘standard’ orientation.\nExpected Find and explain the mistake that has been made when sorting shapes into a Venn diagram. 4 regular polygons with up to 8 lines of symmetry used. All shapes in the same ‘non-standard’ orientation in each question.\nGreater Depth Find and explain the mistake that has been made when sorting shapes into a Venn diagram. 5 irregular polygons with any number of lines of symmetry used. All shapes in unique orientations.\n\nQuestions 2, 5 and 8 (Problem Solving)\nDeveloping Find all the lines of symmetry (up to 2) in an image made up of multiple shapes. Up to 4 simple regular polygons used, all in ‘standard’ orientation.\nExpected Find all the lines of symmetry (up to 8) in an image made up of multiple shapes. Up to 8 regular polygons used, in a mix of ‘standard’ and ‘non-standard’ orientations.\nGreater Depth Find all the lines of symmetry (any number) in an image made up of multiple shapes. Up to 12 irregular polygons used, all in ‘non-standard’ orientations.\n\nQuestions 3, 6 and 9 (Reasoning)\nDeveloping Find the symmetrical shape from a set of 2 shapes and explain what makes it symmetrical. Simple shapes with up to 4 sides used, with one obviously asymmetrical.\nExpected Find the 2 asymmetrical shapes from a set of 3 shapes and explain what makes each of them asymmetrical. More complex shapes with up to 8 sides used, with more subtle reasons for asymmetry.\nGreater Depth Find the 3 asymmetrical shapes from a set of 4 shapes and explain what makes each of them asymmetrical. Very complex shapes with any number of sides used, with very subtle reasons for asymmetry." ]

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[ "Generative Adversarial Networks are great for generating something from essentially nothing and there are some interesting uses for them. Most uses are some sort of image processing. Nvidia’s GauGAN is an impressive example, giving the user an MS paint-like interface and generating landscapes. You can give the beta a shot here.\n\nI wanted to take a step back and use a small example to understand the basics and build some intuition behind GAN’s. There’s a tonne of information out there on how to fit a GAN to generate new hand drawn numbers, faces or Pokemon to varying success (the jury is still out as to whether or not the output can pass as Pokemon, but anyway). This isn’t the focus for this post. Instead, I’m going to simplify things further and use a GAN to generate sine waves.\n\n## R + Python with {reticulate}\n\nKeras and Tensorflow are used for this analysis. While there are R libraries I personally find it easier using Python via reticulate for deep learning tasks. You can find the code on Github. All the code bits in this post refer to functions from this repo.\n\nFor this to run correctly you’ll need Ananconda, Python 3.6-3.7 installed with keras and Tensorflow, as well as the standard libraries, numpy, pandas, etc. Along with python you’ll need reticulate installed and configured to use the appropriate version on Python. In short, run py_config() to initialise python for the session and py_available() to check if it’s all good to go. This can be tricky to set up and relies on how you’ve installed Python. If you have trouble refer to the reticulate cheat sheet and documentation.\n\n## Example data\n\nFor training data I’m going to use two wave functions,", null, "with random noise", null, "added to throw in some minor variation.\n\nget_training_data <- function(n = 200, m = 250, shape = list(a = c(1, 3), p = c(2, 10))) {\nmat <- matrix(NA, nrow = n, ncol = m)\nn_waves <- length(shape$a) for(k in 1:n){ ak <- shape$a[(k-1) %% n_waves + 1]\npk <- shape$p[(k-1) %% n_waves + 1] mat[k,] <- ak*sin(2*pi*seq(0, 1, length = m)*pk) + rnorm(m, 0, 0.05) } mat } train <- get_training_data() plot_waves(train) Nothing too complicated, just two distinct waves. Because we are generating training data using these two wave functions we only need to generate a handful of observations. ## Model There are two main components, the generator and discriminator. The generator generates new waves from random input, in this case a standard normal distribution. The discriminator sorts the real from the fake data. During training it will switch between training the discriminator and the generator. At each iteration both components perform better – the generator gets better at generating real observations and the discriminator gets better at determining whether or not the observation is real or fake. Like any neural network determining the number of hidden layers and sizes is more a process of experimentation than it is a science. For this example what I found worked well is, All are dense layers with LeakyReLU activation and 20% dropout. Given the input data is distinct it seems like overkill however I found this worked best. I’m sure other network configurations also work. The input data is time series data, in which case it is appropriate to use recurrent layers for the generator and discriminator. I actually found this to not be very successful. Either it takes far longer to train or just has trouble converging to a good solution, not saying it can’t be done though. For more challenging problems you’ll need more advanced models. Fortunately for this example we can keep it simple. ## Training To train the GAN on these two functions we simply run. gan(train, epochs = 2000) This will call the Python script with the GAN code, run it in Python for 2000 epochs and return the results. The training is saved in the global environment as x_train which is then able to be imported into the Python environment with r.x_train. A log file is created within the working directory and records the progress every 100 epochs. ## Output Once training has finished, view the output by py$gan_r$output. At each", null, "iteration set by trace a set of waves are generated. The plot_waves function will plot a set from the final iteration. Recall, we only fed the GAN two sine functions, which makes the output below pretty cool. Here we see 12 randomly generated waves from the final iteration of the GAN. plot_waves(py$gan_r$output, seq(1, 24, 2)) With a single draw from a random normal distribution the GAN iteratively adjusted it’s weights until it learnt to generate and identify each wave. But also it learnt to generate new waves. What stands out here is the new waves appear to be some combination of the input waves. What we’ve done is found a really, really expensive and not particularly accurate way to estimate this…", null, "where", null, "is between 0 and 1. This can be seen in the plot below where 12 waves have been plotted for different values of", null, ". Without explicitly telling the GAN what the functions were, it managed to learn them and explore the space between. While it estimated the frequency well it didn’t quite explore the whole range of amplitudes. They tend to range between 1.5-2.5 rather than 1-3. With more training it would probably get there. This took a few goes as training the GAN tends to converge to one of the input functions. By generating only one of the waves with high accuracy it can trick the discriminator into thinking it’s real every time. It’s a solution to the problem just not a very exciting one. With tuning we can get the desired result. Each starting value will correspond to some kind of wave. Out of the 12 random waves, 4 are very similar, right down to the two little spikes at the top of the second crest (see the waves in the third column). This suggests this wave is mapped to a set of values that may be drawn with a higher probability. ## Thoughts This isn’t as sexy as generating new landscape images using Paint but it’s helpful to understand what is going on within the guts of the GAN. It attempts to identify the key features in the observations making it distinct from random noise, pass as a real observation and map the space between. The same process is essentially happening at scale for more complex tasks. In this example it’s very inefficient to get the result we were looking for, but as the problem becomes more and more complex the trade-off makes sense. With an image example such as faces, the GAN will identify what a nose looks like and the range of noses that exist in the population, what an eye looks like and that they generally come in pairs at a slightly varying distance apart and so on. It will then generate eyes and noses along some space of possible eyes and noses, in the same way it generated a wave along some space of possible waves given the input. What’s interesting, the GAN only maps the space between. There are no examples where it generated a wave with a frequency greater than 10 or less than 2. Nor did it generate a wave with an amplitude greater than 3. The input waves essentially formed the boundaries of what could be considered real. There may be GAN variations that allow for this exploration. ## Code bits The easiest way to get the code for this example is from Github. Either clone or install. devtools::install_github(\"doehm/rgan\") The Python component is quite long and don’t think there is much to gain pasting it here. As mentioned the trickiest part is going to be configuring Python, keras and tensorflow but the R bits should work. If you want to explore the output of the GAN the data is found at inst/extdata/gan-output.Rdata in the Github repo. This will show you how the GAN improved with each iteration. Each element of the list is a sample of generated waves at iteration 100, 200, …, 2000. This data is the basis of the output waves above. e.g. plot_waves(gan_output[], id = 1:12, nrow = 3) The code below created the title graphic. It is an area chart of waves using 48 values for", null, ". Perhaps worthy of an accidental aRt submission. shape <- list(a = c(1, 3), p = c(2, 10)) m <- 250 ln <- 48 pal <- c(\"#4E364B\", \"#8D4E80\", \"#D86C15\", \"#F3C925\", \"#48B2A8\") map_dfr(seq(0, 1, length = ln), ~data.frame( x = seq(0, 1, length = m), y = .x*shape$a*sin(2*pi*seq(0, 1, length = m)*shape$p) + (1-.x)*shape$a*sin(2*pi*seq(0, 1, length = m)*shape\\$p),\na = round(.x, 2))) %>% filter(x < 0.5) %>%\nggplot(aes(x = x, y = y, colour = a, fill = a, group = as.factor(a))) +\ngeom_area() +\ntheme_void() +\ntheme(legend.position = \"none\") +\nscale_colour_gradientn(colors = colorRampPalette(pal)(200))" ]

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[ "## Educational Data Mining - Semantic Scholar\n\nTiffany Barnes, John Stamper. Difficulties in inferring ...... 14: p. 1-33. 10. Murphy, K., Brief Introduction to Graphical Models and Bayesian Networks. 1998:.\nWorkshop of Educational Data Mining\n\nEducational Data Mining\n\nCecily Heiner (Co-chair) Neil Heffernan (Co-chair) Tiffany Barnes (Co-chair)\n\nSupplementary Proceedings of the 13th International Conference of Artificial Intelligence in Education. Marina del Rey, CA. USA. July 2007\n\nWorkshop of Educational Data Mining\n\nEducational Data Mining Workshop (http://www.educationaldatamining.org/) Cecily Heiner; University of Utah; [email protected] (Co-chair) Neil Heffernan; Worcester Polytechnic Institute; [email protected] (Co-chair) Tiffany Barnes; University of North Carolina at Charlotte, [email protected] (Co-chair)\n\nIntroduction Educational data mining is the process of converting raw data from educational systems to useful information that can be used to inform design decisions and answer research questions. Data mining encompasses a wide range of research techniques that includes more traditional options such as database queries and simple automatic logging as well as more recent developments in machine learning and language technology. Educational data mining techniques are now being used in ITS and AIED research worldwide. For example, researchers have used educational data mining to: o Detect affect and disengagement o Detect attempts to circumvent learning called \"gaming the system\" o Guide student learning efforts o Develop or refine student models o Measure the effect of individual interventions o Improved teaching support o Predict student performance and behavior However, these techniques could achieve greater use and bring wider benefits to the ITS and AIED communities. We need to develop standard data formats, so that researchers can more easily share data and conduct meta-analysis across tutoring systems, and we need to determine which data mining techniques are most appropriate for the specific features of educational data, and how these techniques can be used on a wide scale. The workshop will provide a forum to present preliminary but promising results that can advance our knowledge of how to appropriately conduct educational data mining and extend the field in new directions. Topics They include, but are not limited to: o What new discoveries does data mining enable? o What techniques are especially useful for data mining? o How can we integrate data mining and existing educational theories? o How can data mining improve teacher support? o How can data mining build better student models? o How can data mining dynamically alter instruction more effectively? o How can data mining improve systems and interventions evaluations? o How do these evaluations lead to system and intervention improvements? o How is data mining fundamentally different from other research methods?\n\nWorkshop of Educational Data Mining\n\nTABLE OF CONTENTS Evaluating problem difficulty rankings using sparse student data………….…………1 Ari Bader-Natal, Jordan Pollack Toward the extraction of production rules for solving logic proofs…………………...11 Tiffany Barnes, John Stamper Difficulties in inferring student knowledge from observations (and why you should care)………………………………………………………………………………………...21 Joseph E. Beck What’s in a word? Extending learning factors analysis to modeling reading transfer………………………………………………………………………..…………….31 James M. Leszczenski, Joseph E. Beck Predicting student engagement in intelligent tutoring systems using teacher expert knowledge ……….………………………………………………………………………...40 Nicholas Lloyd, Neil Heffernan, Carolina Ruiz Analyzing fine-grained skill models using Bayesian and mixed effects methods ……………………………………………………………………………………………….50 Zachary Pardos, Mingyu Feng, Neil Heffernan, Cristina Heffernan, Carolina Ruiz Mining learners’ traces from an online collaboration tool……………………………..60 Dilhan Perera, Judy Kay, Kalina Yacef, Irena Koprinska Mining on-line discussions: Assessing technical quality for student scaffolding and classifying messages for participation profiling………………………………………. .70 Sujith Ravi, Jihie Kim, Erin Shaw\n\nAll in the (word) family: Using learning decomposition to estimate transfer between skills in a reading tutor that listens………………………………………………..……..80 Xiaonan Zhang, Jack Mostow, Joseph Beck\n\nWorkshop of Educational Data Mining\n\nEvaluating Problem Difficulty Rankings Using Sparse Student Response Data Ari BADER-NATAL 1 , Jordan POLLACK DEMO Lab, Brandeis University Abstract. Problem difficulty estimates play important roles in a wide variety of educational systems, including determining the sequence of problems presented to students and the interpretation of the resulting responses. The accuracy of these metrics are therefore important, as they can determine the relevance of an educational experience. For systems that record large quantities of raw data, these observations can be used to test the predictive accuracy of an existing difficulty metric. In this paper, we examine how well one rigorously developed – but potentially outdated – difficulty scale for American-English spelling fits the data collected from seventeen thousand students using our SpellBEE peer-tutoring system. We then attempt to construct alternate metrics that use collected data to achieve a better fit. The domain-independent techniques presented here are applicable when the matrix of available student-response data is sparsely populated or non-randomly sampled. We find that while the original metric fits the data relatively well, the datadriven metrics provide approximately 10% improvement in predictive accuracy. Using these techniques, a difficulty metric can be periodically or continuously recalibrated to ensure the relevance of the educational experience for the student.\n\n1. Introduction Estimates of student proficiency and problem difficulty play central roles in Item Response Theory (IRT) . Several current educational systems make use of this theory, including our own BEEweb peer-tutoring activities [2,8,9,13]. IRT-based analysis often focuses on estimating student proficiency in the task domain, but the challenge of estimating problem difficulty should not be overlooked. While student proficiency estimates can inform assessment, problem difficulty estimates can be used to refine instruction: these metrics can affect the selection and ordering of problems posed and can influence the interpretation of the resulting responses . It is therefore important to choose a good difficulty metric initially and to periodically evaluate the accuracy of a chosen metric with respect to available student data. In this paper, we examine how accurately one rigorously developed – but potentially outdated – difficulty scale for the domain of American-English spelling predicts the data collected from students using our SpellBEE system . The defining challenge in providing this assessment lies in the nature of the data. As SpellBEE is a peer-tutoring system, the challenges posed to students are determined by other students, resulting in data that is neither random nor complete. In this 1 Correspondence to: Ari Bader-Natal, Brandeis University, Computer Science Department – MS 018. Waltham, MA 02454. USA. Tel.: +1 781 736 3366; Fax: +1 781 736 2741; E-mail: [email protected].\n\n1\n\nWorkshop of Educational Data Mining\n\nwork, we rely on a pairwise comparison technique designed to be robust to data with these characteristics. After assessing the relevance of this existing metric (in terms of predictive accuracy), we will examine some related techniques for initially constructing a difficulty metric based on non-random, incomplete samples of observed student data.\n\n2. American-English spelling: A sample task domain The educational system examined here, SpellBEE, was designed to address the task domain of American-English spelling . SpellBEE is the oldest of a growing suite of web-based reciprocal tutoring systems using the Teacher’s Dilemma as a motivational mechanism . For the purposes of this paper, however, the mechanisms for motivation and interaction can be ignored, and the SpellBEE system and the difficulty metric used by it can be specifically re-characterized for an educational data mining audience. 2.1. Relevant characteristics of the SpellBEE system Students access SpellBEE online at SpellBEE.org from their homes or schools. As of May 2007, over 17,000 students have actively participated. After creating a user account, a student is able to log in, choose a partner, and begin the activity.2 During the activity, students take turns posing and solving spelling problems. When posing a problem, the student selects from a short list of words randomly drawn from the database of wordchallenges. This database is comprised of 3,129 words drawn from Greene’s New Iowa Spelling Scale (NISS), which will be discussed in the next section .3 When responding to a problem, the student types in what they believe to be the correct spelling of the challenge word. The accuracy of the response is assessed to be either correct or incorrect. Figure 1 presents a list of the relevant data stored in the SpellBEE server logs. To date, we have observed over 64,000 unique (case-insensitive) responses to the challenges posed,4 distributed across over 22,000 completed games consisting of seven questions attempted per student. Student participation, measured in games completed, has not been uniform, however. Of the challenges in the space, most students have only attempted a very small fraction. In fact, when examining the response matrix of every student by every challenge, less than 1% of the matrix data is known. An important characteristic of the SpellBEE data, then, is that the response matrix is remarkably sparse. Given that the students acting as tutors are able to – and systemically motivated to – express their preferences and hunches through the problems that they select, another important characteristic of the SpellBEE data is that the data present in the studentchallenge response matrix is also biased. The effects of this bias can be found in the following example: 16% of student attempts to spell the word “file” were correct, while 66% of attempts to spell the word “official” were correct. The average grade level among the first set of students was 3.9, while for the second set it was 6.4. In Section 3.2 we 2 In the newer BEEweb activities, if no one else is present, a student can practice alone on problems randomly drawn from the database of challenges posed in the past. 3 In SpellBEE, the word-challenges are presented in the context of a sentence, and so of the words in Greene’s list, we only use those found in the seven public-domain books that we parsed for sentences. 4 Of these, 17,391 were observed more than once. In this paper, we restrict the set of responses that we consider to this subset. See Footnote 7 for the rationale behind this.\n\n2\n\nWorkshop of Educational Data Mining\n\nFigure 1. The SpellBEE server logs data about each turn taken by each student, as shown in the first list. The data in the first list is sufficient to generate the data included in the second list.\n\n1. 2. 3. 4. 5. 6.\n\ntime : a time-stamp allows responses to be ordered game : identifies the game in which this turn occurred tutor : identifies the student acting as the tutor in this turn tutee : identifies the student acting as the tutee in this turn challenge : identifies the challenge posed by the tutor response : identifies the response offered by the tutee\n\n1. difficulty : the difficulty rating of the challenge posed by the tutor 2. accuracy : the accuracy rating of the response offered by the tutee will present techniques designed to draw more meaningful difficulty information from this type of data. 2.2. Origin, use, and application of the problem difficulty metric When trying to define a measure of problem difficulty for the well-studied domain of American-English spelling, we were able to benefit from earlier research in the field. Greene’s “New Iowa Spelling Scale” provides a rich source of data on word spelling difficulty, drawn from a vast study published in 1954. Greene first developed a methodology for selecting words for his list (5,507 were eventually used.) Approximately 230,000 students from 8,800 classrooms (grades 2 through 8) around the United States participated in the study, totally over 23 million spelling responses . From these, Greene calculated the percentage of correct responses for each word for each grade. This table of success rates is used in SpellBEE to calculate the difficulty of each spelling problem for students, whose grade level is known.\n\n3. Techniques for assessing relative challenge difficulty The research questions addressed in this paper focus on the fit of the difficulty model based on the NISS data to the observed SpellBEE student data. Two different techniques are involved in the calculating this fit. The first converts the graded NISS data to a linear scale. The second identifies from the observed student data a difficulty ordering over pairs of problems, in a manner appropriate for a sparse and biased data matrix. Both will be employed to address the research questions in the following sections. 3.1. Linearly ordering challenges using the difficulty metric Many subsequent studies have explored various aspects of Greene’s study and the data that it produced. Cahen, Craun, and Johnson and, later, Wilson and Bock explore the degree to which various combinations of domain-specific predictors could account for Greene’s data. Initially starting with 20 predictors, Wilson and Bock work down to a regression model with an adjusted R2 value of 0.854.5 Here, we not interested in 5 The\n\nmost influential of which being the length of the word.\n\n3\n\nWorkshop of Educational Data Mining\n\nFigure 2. Difficulty data for two words from the NISS study are plotted, and the I50 statistics are calculated.\n\nI50(\"acknowledge\")=7.875\n\n\"above\" \"acknowledge\"\n\n50 % I50(\"above\")=3.7\n\nPercentage of students with correct responses\n\n100 %\n\n0% 2\n\n3\n\n4 5 6 Student grade level\n\n7\n\n8\n\npredicting the NISS results, but instead are interested in assessing the fit (or predictive power) of the 1954 NISS results to observations made of students using SpellBEE over 50 years later. We will drawn upon one statistic used by Wilson and Bock: the onedimensional flattening of the seven-graded NISS data. This statistic, which they refer to as the “location” of the word, is the (fractional) grade level at which 50% of the NISS students correctly spell word w.6 We denote this as I50 (w). Figure 2 illustrates how the graded difficulty data that is used to derive this statistic for two different words. The value of this statistic is that it provides a single grade-independent difficulty value for a word that can be compared directly to that of other words. 3.2. Identifying pairwise difficulty orderings using observed student data Given the characteristics of the data collected from the SpellBEE system, identifying the more difficult of a pair of problems based on this data is not trivial. The percentage of correct responses to a challenge, the calculation used to generate the NISS data, is not appropriate here, as the assignment of challenges to students was done in a biased, nonrandom manner (recall the “file”/“official” example from Section 2.1.) Tutors, in fact, are motivated to base their challenge selection on the response accuracies that they anticipate. A more appropriate measure, rooted in several different literatures, is to assess pairwise problem difficulties on distinctions indirectly indicated by the students. In the statistics literature, McNemar’s test provides a statistic based on this concept , in the IRT literature, this is used as a data reduction strategy for Rasch model parameter estimation , and the Machine Learning literature includes various approaches to learning 6 Wilson and Bock calculate the 50% threshold based on a logistic model fit to the discrete grade-level data, while we calculate the threshold slightly differently, based on a linear interpolation of the grade-level data.\n\n4\n\nWorkshop of Educational Data Mining\n\nTable 1. While the I50 metric flattens the grade-specific NISS data to a single dimension, the relative difficulty ordering of most word-pairs based on the graded NISS data is the same as when based on the I50 scale. In this table, we quantify the amount of agreement between I50 and each set of grade-specific NISS data using Spearman’s rank correlation coefficient. The strong correlations observed suggest that the unidimensional scale sufficiently captures the relative difficulty information from the original NISS dataset. (The number of words, N, varies by grade, as the NISS study did not show several of the harder words to the younger students.) Grade\n\nN\n\nSpearman’s ρ\n\n2 3 4\n\n2218 3059 3126\n\n0.751 0.933 0.977\n\n5 6 7 8\n\n3129 3129 3129 3129\n\n0.974 0.960 0.935 0.915\n\nrankings based on pairwise preferences . Assume that for some specific pair of problems, such as the spelling of the words “about” and “acknowledge”, we first identify all students in the SpellBEE database who have attempted both words. Given that response accuracy is dichotomous, there are only four possible configurations of a student’s response accuracy to the pair of challenges. In the cases where the student responds to both correctly or incorrectly, no distinction is made between the pair. But in the cases where the student correctly responds to one but incorrectly to the other, we classify this as a distinction indicating a difficulty ordering between the two problems.7 It is also worth stating that in this study, we assume a “static student” model, so we are not concerned with the order of these two responses. At the cost of some data loss, one could instead assume a “learning student” model, for which only a correct response on one problem followed by an incorrect response on the other would define a distinction. Had the incorrect response been observed first, we could not rule out the possibility that the difference was due to a change in the student’s abilities over time, and not necessarily an indication of difference in problem difficulties.8 An example may clarify. If counting the number of both directional distinctions made by all students (e.g. 12 students in SpellBEE spelled “about” correctly and “acknowledge” incorrectly, while 2 students spelled “about” incorrectly and “acknowledge” correctly), we have a strong indication of relative problem difficulty. McNemar’s test assigns a significance to this pair of distinction counts. In this work, we more closely follow the IRT approach, relying only the relative size of the two counts (and not the significance.) Thus, since 12 distinctions were found in one direction and only 2 in the other, we say that we observed the word “about” to be easier than the word “acknowledge” based on collected SpellBEE student data. If distinctions were available for every problem pair, 7 We recognize that some distinctions are spurious, for which the incorrect response was not reflective of the student’s abilities. Here we take a simplistic approach of identifying and ignoring non-responses (in which the student typed nothing) and globally-unique responses (which no other student ever responded, to any challenge.) Globally-unique responses encompass responses from students who don’t yet understand the activity, responses from students who did not hear the audio recording, responses from student attempting to use the response field as a chat interface, and responses from students making no effort to engage in the activity. 8 Another possible model is a “dynamic student” model, for which student abilities may get better or worse over time. Under this model, no distinctions can be definitively attributed to difference in problem difficulty.\n\n5\n\nWorkshop of Educational Data Mining\n\na total of 3,129 × 3,128 = 9,787,512 pairwise problem orderings could be expressed. In our collected data so far, we have 3,349,602 of these problem pairs for which we have distinctions recorded. In the subsequent sections, we measure the fitness of a predictive model (like I50 ) based on how many of these pairwise orderings are satisfied.9\n\n4. Assessing the fit of the NISS-based I50 model to the SpellBEE student data Given the NISS-based I50 difficulty model of problem difficulty and the data-driven technique for turning observed distinctions recorded in the SpellBEE database into pairwise difficulty orderings, we can now explore various methods to assess the applicability of the model to the data. 4.1. Assessing fit with a regression model The first method is to construct a regression model that uses I50 to predict observed difficulty. Since observed difficulty is currently available only in pairwise form, this requires an additional step in which we flatten these pairwise orderings into one total ordering over all problems. As this is a highly non-trivial step, the results should be interpreted tentatively. Here, we accomplish a flattening by calculating, for each challenge, the percentage of available pairwise orderings for which the given challenge was the more difficult of the pair. So if 100 pairwise orderings involve the challenge word “acknowledge”, and 72 of these found “acknowledge” to be the harder of the pair, we would mark “acknowledge” as harder than 72% of other words. A regression model was then built on this, using I50 as a predictor of the pairwise-derived percentage. The model, after filtering out data points causing ceiling and floor effects (i.e. I50 (w) = 2.0 or I50 (w) = 8.0), had an adjusted R2 value of 0.337 (p < 0.001 for the model). The corresponding scatterplot is shown in Figure 3.10 The relatively low adjusted R2 value is likely at least partially a result of the flattening step (rather than solely due to poor fit.) Had we flattened the data differently, this value would clearly change. In order to obtain a more reliable measure of model fitness, we seek to avoid any unnecessary processing of the mined data. 4.2. Assessing fit with as the percentage of agreements on pairwise difficulties The second method that we explore provides a more direct comparison, without any further flattening of the student data. Here, we simply calculate the percentage of observed pairwise difficulty orderings (across all challenges) for which the I50 model correctly predicts the observed pairwise difficulty ordering. When we do this across all of the 3,349,602 difficulty orderings that we have constructed from the student data, we find that the I50 model correctly predicts 2,534,228 of these pairwise orderings, providing a 75.66% agreement with known pairwise orderings from the mined data. Remarkably, we found that the predictive accuracy of the I50 model did not significantly change as the 9 Note\n\nthat it is not be possible to achieve a 100% fit, as some cycles exist among these pairwise orderings.\n\n10 The outliers in this plot mark the problems that are ranked most differently by the two measures. The word\n\n“arithmetic”, for example, was found to be difficult by SpellBEE students, but was not found to be particularly difficult for the students in the NISS study. Variations like this one may reflect changes in the teaching or in the frequency of usage since the NISS study was performed 50 years ago.\n\n6\n\nWorkshop of Educational Data Mining\n\nFigure 3. Words are plotted by their difficulty on the I50 scale and by the percentage of other words for which the observed pairwise orderings found the word to be the harder of the pair. An adjusted R2 value of 0.490 was calculated for this model. (When ignoring the words affected by a ceiling or floor effect in either variable, the adjusted R2 value drops to 0.377.)\n\nPercentage of comparisons finding word harder\n\n100 %\n\n50 %\n\n0% 2\n\n3\n\n4\n\n5 6 I50 Difficulty Value\n\n7\n\n8\n\nquantity of student data used for the distinction varied. 75.1% of predictions based on one distinction were accurate, while 74.7% of predictions based on 25 distinctions were accurate (intermediate values ranged from 71.0% to 77.6%). This flat relationship suggests that pairwise difficulty orderings constructed from a minimal amount of observed data may be just as accurate, in the aggregate, as those orderings constructed when additional data is available.\n\n5. Incorporating SpellBEE student data in a revised difficulty model We now know that there is a 75.66% agreement in pairwise difficulty orderings between the I50 difficulty metric derived from the NISS data and the observed pairwise preferences mined from the SpellBEE database. Can we improve upon this? We will present an approach that iteratively updates the I50 problem difficulty estimates using the mined data and a logistic regression model. Rather than producing a single predictive model, we construct one logistic model for each challenge, and use these fitted model to update our estimates of the problem difficulty. Applied iteratively, we hope to converge on problem difficulty metric that better fits the observed data. This process is inspired by the parameter estimation procedures for Rasch models , which may not be directly applicable due to the large size of our problem space. For a given challenge c1 (e.g. “acknowledge”), we can first generate the list of all other challenges for which SpellBEE students have expressed distinctions (in either direction.) In Section 3.2, we chose to censor these distinctions in order to generate a bi-\n\n7\n\nWorkshop of Educational Data Mining\n\nFigure 4. A logistic regression model is used to estimate the difficulty of the word “abandon.” At left, the first estimate is based on the original I50 difficulty values. At right, the third iteration of the estimate is constructed based on data from the previous best estimate. The point estimate dropped from 8.0 (from I50 ) to 7.06 (from iteration 1) to 6.81 (from iteration 3.) 100 % Percentage of distinctions finding word harder\n\nPercentage of distinctions finding word harder\n\n100 %\n\n50 %\n\n0% 2\n\n3\n\n4\n\n5 6 Difficulty Estimate\n\n7\n\n8\n\n50 %\n\n0% 2\n\n3\n\n4\n\n5 6 Difficulty Estimate\n\n7\n\n8\n\nnary value representing the difficulty ordering. Here we will make use of the actual distinction counts in each direction. For each challenge with which pairwise distinctions for c1 are available, we note our current-best estimate of the difficulty of c2 (initially, using I50 values), and note the number of distinctions indicating that c1 is the more difficult challenge. We can then regress the grouped distinction data on the problem difficulty estimate data to construct a logistic model relating the two. For some c1 , if the relationship is statistically significant, we can use it to generate a revised estimate for the difficulty of that challenge. By solving the regression equation for the c2 problem difficulty value for which 50% of distinctions find c1 harder, we can calculate the difficulty of a problem for which relative-difficulty distinctions are equally likely in either direction. This provides a revised estimate for the difficulty of the original problem, c1 . We use this procedure to calculate revised estimates for every challenge in the space (unless the resulting logistic regression model is statistically not significant, in which case we retain our previous difficulty estimate.) This process can be iteratively repeated, using the revised difficulty estimates as the basis of the new regression models. Figure 4 plots this data for one word, using the difficulty estimates resulting from the third iteration of the estimation. A second approach towards incorporating observed distinction data into a unified problem difficulty scale is briefly introduced and compared to the other metrics. Here, we recast the estimation problem as a sorting problem, and use a probabilistic variant of the bubble-sort algorithm to reorder consecutive challenges based on available distinction data. Initially ordering the challenge words alphabetically, we repeatedly step through the list, reordering challenges at indices i and i + 1 with a probability based on the proportion of distinctions finding the first challenge harder than the second.11 After “bubbling” through the ordered list of challenges 200,000 times, we interpret the rankorder of each challenge as a difficulty index. These indices provide a metric of difficulty (which we refer to as P robBubble), and a means for predicting the relative difficulty of any pair of challenges (based on index ordering.) 11 If distinctions have been observed in both directions, the challenges are reordered with a probability determined by the proportion of distinctions in that direction. If no distinctions in either direction have been observed, the challenges are reordered with a probability of p = 0.5. If distinctions have been observed in one direction but not the other, the challenges are reordered with a fixed minimal probability (p = 0.1).\n\n8\n\nWorkshop of Educational Data Mining\n\nTable 2. Summary table for the predictive accuracy of various difficulty metrics. For each metric, the percentage of accurate predictions of pairwise difficulty orderings is noted. The accuracy of the I50 metric is measured against all of the 3,349,602 pairwise orderings identified by student distinctions. The accuracy of the datadriven metrics (I50 rev.1 and P robBubble) are based on the average results from a 5-fold cross-validation, in which the metrics are constructed or trained on a subset of the pairwise distinction data and are evaluated on a different set of pairwise data (the remaining portion.) Difficulty Model\n\nPredictive Accuracy\n\nI50 I50 rev.1 P robBubble\n\n75.66% 84.79% 84.98%\n\nTable 3. Spearman’s rank correlation coefficient between pairs of problem difficulty rank-orderings (N = 3129, p < 0.01, two-tailed.) Metric 1\n\nMetric 2\n\nSpearman’s ρ\n\nI50 I50 I50 rev.3\n\nI50 rev.3 P robBubble P robBubble\n\n0.677 0.673 0.908\n\nGiven the pairwise technique used in Section 4.2 for analyzing the fit of a difficulty metric for a set of pairwise difficulty orderings, we can examine how these two data-driven models compare to the original I50 difficulty metric. Table 2 summarizes our findings. Here we observe that the data-driven approaches provide an improvement of almost 10% accuracy with regard to the prediction of pairwise difficulty orderings. As was noted earlier, cycles in the observed pairwise difficulty orderings prevent any linear metric from achieving 100% prediction accuracy, and the maximum achievable accuracy for the SpellBEE student data is not know. We do note that two different data-driven approaches, logistic regression-based iterative estimation and the probabilistic sorting, arrived at very similar levels of predictive accuracy. Table 3 uses Spearman’s rank correlation coefficient as a tool to quantitatively compare the three metrics. One notable finding here is the extremely high rank correlation between the P robBubble and I50 rev.3 data-driven metrics.\n\n6. Conclusion The findings from the research questions posed here are both reassuring and revealing. Although the NISS study was done over 50 years ago, much of its value seems to have been retained. The NISS-based I50 difficulty metric was observed to correctly predict 76% of the pairwise difficulty orderings mined from SpellBEE student data. Many of the challenges for which the difficulty metric achieved low predictive accuracies corresponded with words whose cultural relevance or prominence has changed over the past few decades. The data-driven techniques presented in Section 5 offers a means for incorporating these changes back into a difficulty metric. After doing so, we found the predictive accuracy increased approximately 10%, to the 85% agreement level. The key technique used here to enable the assessment and improvement of problem difficulty estimates works even when not all students have attempted all challenges or\n\n9\n\nWorkshop of Educational Data Mining\n\nwhen the selection of challenges for students is highly biased. It is data-driven, based on identifying and counting pairwise distinctions indicated indirectly through observations of student behavior over the duration of use of an education system. The pairwise distinction-based techniques for estimating problem difficulty information explored here is a part of a larger campaign to develop methods for constructing educational systems that require a minimal amount of expert domain knowledge and model-building. Our BEEweb model is but one such approach, the Q-matrix method is another , and most the IRT-based systems discussed in the introduction are, also. Designing BEEweb activities only requires domain knowledge in the form of a problem difficulty function and a response accuracy function. The latter can usually be created without expertise, and the former can now be approached, even when collected data is sparse and biased, using the techniques discussed in this paper.\n\nReferences \n\n\n\n \n\n\n\n \n\n\n\nAri Bader-Natal and Jordan B. Pollack. Motivating appropriate challenges in a reciprocal tutoring system. In C.-K. Looi, G. McCalla, B. Bredeweg, and J. Breuker, editors, Proceedings of the 12th International Conference on Artificial Intelligence in Education (AIED-2005), pages 49–56, Amsterdam, July 2005. IOS Press. Ari Bader-Natal and Jordan B. Pollack. BEEweb: A multi-domain platform for reciprocal peer-driven tutoring systems. In M. Ikeda, K. Ashley, and T.-W. Chan, editors, Proceedings of the 8th International Conference on Intelligent Tutoring Systems (ITS-2006), pages 698–700. Springer-Verlag, June 2006. Tiffany Barnes. The q-matrix method: Mining student response data for knowledge. Technical Report WS-05-02, AAAI-05 Workshop on Educational Data Mining, Pittsburgh, 2005. Klaus Brinker, Johannes Fürnkranz, and Eyke Hüllermeier. Label ranking by learning pairwise preferences. Journal of Machine Learning Research, 2005. Leonard S. Cahen, Marlys J. Craun, and Susan K. Johnson. Spelling difficulty – a survey of the research. Review of Educational Research, 41(4):281–301, October 1971. Chih-Ming Chen, Chao-Yu Liu, and Mei-Hui Chang. Personalized curriculum sequencing utilizing modified item response theory for web-based instruction. Expert Systems with Applications, 30, 2006. Bruce Choppin. A fully conditional estimation procedure for rasch model parameters. CSE Report 196, Center for the Study of Evaluation, University of California, Los Angeles, 1983. Ricardo Conejo, Eduardo Guzmán, Eva Millán, Mónica Trella, José Luis Pérez-De-La-Cruz, and Antonia Ríos. Siette: A web-based tool for adaptive testing. International Journal of Artificial Intelligence in Education, 14:29–61, 2004. Michel C. Desmarais, Shunkai Fu, and Xiaoming Pu. Tradeoff analysis between knowledge assessment approaches. In C.-K. Looi, G. McCalla, B. Bredeweg, and J. Breuker, editors, Proceedings of the 12th International Conference on Artificial Intelligence in Education (AIED-2005). IOS Press, 2005. B. S. Everitt. The Analysis of Contingency Tables. Chapman and Hall, 1977. Gerhard H. Fischer and Ivo W. Molenaar, editors. Rasch Models: Foundations, Recent Developments, and Applications. Springer-Verlag, New York, 1995. Harry A. Greene. New Iowa Spelling Scale. State University of Iowa, Iowa City, 1954. Jeff Johns, Sridhar Mahadevan, and Beverly Woolf. Estimating student proficiency using an item response theory model. In M. Ikeda, K. Ashley, and T.-W. Chan, editors, Proceedings of the 8th International Conference on Intelligent Tutoring Systems (ITS-2006), pages 473–480, 2006. Mark Wilson and R. Darrell Bock. Spellability: A linearly ordered content domain. American Educational Research Journal, 22(2):297–307, Summer 1985.\n\n10\n\nWorkshop of Educational Data Mining\n\nToward the extraction of production rules for solving logic proofs Tiffany Barnes, John Stamper Department of Computer Science, University of North Carolina at Charlotte [email protected], [email protected]\n\nAbstract: In building intelligent tutoring systems, it is critical to be able to understand and diagnose student responses in interactive problem solving. However, building this understanding into the tutor is a time-intensive process usually conducted by subject experts. Much of this time is spent in building production rules that model all the ways a student might solve a problem. We propose a novel application of Markov decision processes (MDPs), a reinforcement learning technique, to automatically extract production rules for an intelligent tutor that learns. We demonstrate the feasibility of this approach by extracting MDPs from student solutions in a logic proof tutor, and using these to analyze and visualize student work. Our results indicate that extracted MDPs contain many production rules generated by domain experts and reveal errors that experts do not always predict. These MDPs also help us identify areas for improvement in the tutor. Keywords: educational data mining, Markov decision processes\n\n1. Introduction According to the ACM computing curriculum, discrete mathematics is a core course in computer science, and an important topic in this course is solving formal logic proofs. However, this topic is of particular difficulty for students, who are unfamiliar with logic rules and manipulating symbols. To allow students extra practice and help in writing logic proofs, we are building an intelligent tutoring system on top of our existing proof verifying program. Our experience in teaching discrete math, and in student surveys, indicate that students particularly need feedback when they get stuck. The problem of offering individualized help and feedback is not unique to logic proofs. Through adaptation to individual learners, intelligent tutoring systems (ITS) can have significant effects on learning . However, building one hour of adaptive instruction takes between 100-1000 hours of work of subject experts, instructional designers, and programmers , and a large part of this time is used in developing production rules that are used to model student behavior and progress. A variety of approaches have been used to reduce the development time for ITSs, including ITS authoring tools (such as ASSERT and CTAT), or building constraint-based student models instead of production rule systems. ASSERT is an ITS authoring system that uses theory refinement to learn student models from an existing knowledge base and student data . Constraint-based tutors, which look for violations of problem constraints, require less time to construct and have been favorably compared to cognitive tutors, particularly for problems that may not be heavily procedural .\n\n11\n\nWorkshop of Educational Data Mining\n\nSome systems, including RIDES, DIAG, and CTAT use teacher-authored or demonstrated examples to develop ITS production rules. RIDES is a “Tutor in a Box” system used to build training systems for military equipment usage, while DIAG was built as an expert diagnostic system that generates context-specific feedback for students . These systems cannot be easily generalized, however, to learn from student data. CTAT has been used to develop “pseudo-tutors” for subjects including genetics, Java, and truth tables . This system has also been used with data to build initial models for an ITS, in an approach called Bootstrapping Novice Data (BND) . Similar to the goal of BND, we seek to use student data to directly create student models for an ITS. However, instead of feeding student behavior data into CTAT to build a production rule system, we propose to generate Markov Decision Processes that represent all student approaches to a particular problem, and use these MDPs directly to generate feedback. We believe one of the most important contributions of this work is the ability to generate feedback based on frequent, low-error student solutions. We propose a method of automatically generating production rules using previous student data to reduce the expert knowledge needed to generate intelligent, contextdependent feedback. The system we propose is capable of continued refinement as new data is provided. We illustrate our approach by applying MDPs to analyze student work in solving formal logic proofs. This example is meant to demonstrate the applicability of using MDPs to collect and model student behavior and generate a graph of student responses that can be used as the basis for a production rule system. 2. Background and Proofs Tutorial Context Several computer-based teaching systems, including Deep Thought , CPT and the Logic-ITA have been built to support teaching and learning of logic proofs. Of these, the Logic-ITA is the most intelligent, verifying proof statements as a student enters them, and providing feedback after the proof is complete on student performance. Logic-ITA also has facilities for considerable logging and teacher feedback to support exploration of student performance , but does not offer students help in planning their work. In this research, we propose to augment our own existing Proofs Tutorial, with a cognitive architecture derived using educational data mining, that can provide students feedback to avoid error-prone solutions, find optimal solutions, and inform students of other student approaches. In , the first author has applied educational data mining to analyze completed formal proof solutions for automatic feedback generation. However, this work did not take into account student errors, and could only provide general indications of student approaches, as opposed to feedback tailored to a student’s current progress. In this work, we explore all student attempts at proof solutions, including partial proofs and incorrect rule applications, and use visualization tools to learn how this work can be extended to automatically extract a production rule system to add to our logic proof tutorial. In , the second author performed a pilot study to extract Markov decision processes for a simple proof from three semesters of student data from Deep Thought, and verified that the rules extracted by the MDP conformed with expert-derived rules and generated buggy rules that surprised experts. In this work, we apply the technique and extend it with visualization tools to new data from the Proofs Tutorial. The Proofs Tutorial is a computer-aided learning tool implemented on NovaNET (http://www.pearsondigital.com/novanet/). This program has been used for practice and feedback in writing proofs in university discrete mathematics courses taught by the\n\n12\n\nWorkshop of Educational Data Mining\n\nfirst author and others at North Carolina State University since 2002. In the Proofs Tutorial, students are assigned a set of 10 problems that range from simpler logical equivalence applications to more complex inference proofs. (The tutorial can check arbitrary proofs, but we use it for a standard set of exercises). In the tutorial, students type in consecutive lines of a proof, which consist of 4 parts: the statement, reference lines, the axiom used, and the substitutions which allow the axiom to be applied. After the student enters these 4 parts to a line, the statement, reference lines, axiom, and substitutions are verified. If any of these conditions does not hold, a warning messageis shown, and the line is deleted (but saved for later analysis). In this work, we examine student solutions to Proof 1. Table 1 lists an example student solution. Figure 2 is a graphical representation of this proof, with givens as white circles, errors as orange circles, and premises as rectangles. Table 1: Sample Proof 1 Solution (red lines are errors) Statement 1. a → b 2. c → d 3. ¬ (a → d) ¬avd 4. a ^ ¬ d 5. a b b 6. b 7. ¬ d 8. ¬c 9. b ^ ¬c\n\nLine\n\n3 3 4 4 1 1,5 4 2,7 6,8\n\nReason Given Given Given rule IM (error) rule IM implication rule S simplification rule MP (error) rule MP (error) rule MP modus ponens rule S simplification rule MT modus tollens rule CJ conjunction\n\nFigure 2: Graph of Proof 1 Solution (Red lines are errors)\n\n3.\n\nMarkov Decision Processes and ACT-R\n\nA Markov decision process (MDP) is defined by its state set S, action set A, transition probabilities P, and a reward function R . On executing action a in state s the probability of transitioning to state s´ is denoted P(s´ | s, a) and the expected reward associated with that transition is denoted R(s´|s, a). For a particular point in a student’s proof, our method takes the current premises and the conclusion as the state, and the student’s input as the action. Therefore, each proof attempt can be seen as a graph with a sequence of states (each describing the solution up to the current point), connected by actions. We combine all student solution graphs into a single graph, by taking the union of all states and actions, and mapping identical states to one another. Once this graph is constructed, it represents all of the paths students have taken in working a proof. Typically, at this step reinforcement learning is used to find an optimal solution to the MDP. We propose instead, to create multiple functions to deliver different types of feedback, such as functions that could:\n\n13\n\nWorkshop of Educational Data Mining\n\n1.\n\n2.\n\n3.\n\nFind expert-like paths. To derive this policy, we would assign a high reward to the goal state, negative rewards to error states and smaller negative rewards for taking an action. This function would return an optimal (short and correct) choice, and hence expert feedback, at every point in the MDP. Find a typical path to the goal state. We could derive this policy by assigning high rewards to successful paths that many students have taken. Vygotsky’s theory of the zone of proximal development states that students are able to learn new things that are closest to what they already know. Presumably, frequent actions could be those that more students feel fluent using. Therefore, paths based on typical student behavior may be more helpful than optimal solutions, which may be above a student’s current ability to understand. Find paths with low probabilities of errors. It is possible that some approaches could be much less error-prone than others. We could this policy by assigning large penalties to error states. Students often need to learn a simple method that is easily understood, rather than an elegant but complex solution.\n\nThese MDPs are also intended to serve as the basis for learning a production rule system that will perform model tracing, as in a cognitive tutor, while a student is solving a problem. ACT-R Theory is a cognitive architecture that has been successfully applied in the creation of cognitive tutors, and consists of declarative and procedural components. The procedural module contains a production rules system, and creating its production rules is the most time consuming part of developing an ITS based on ACT-R. A production rule system consists of the current problem state (called working memory in ACT-R), a set of production rules, and a rule interpreter. A production rule can be seen as a condition-action pair such as “if A then C” with associated probability x. An example production rule for solving proofs is “if the goal is to prove b^-c, then set as subgoal to prove b”. The rule interpreter performs model tracing to find a sequence of productions that produce the actions a student has taken. This allows for individualized help, tracks student mastery (using the correctness of the productions being applied), and can provide feedback on recognized errors. Production rules in ACT-R are of a general nature to allow them to apply to multiple problems. However, they could be written very specifically to apply to a particular problem. An MDP extracted for a particular problem could be seen as a set of production rules that describe the actions a student might take from each problem state. In this case, MDP state-action pairs are production rules, where the full current problem state is the condition and the action is the applying a particular axiom. By building an MDP using data from several semesters, we generate a significant number of rules and record probabilities that reflect different student approaches. We can use this MDP as it is to perform model tracing as a student is working a proof. If a student is working the problem in a way that has already been encountered, we can use the MDP to provide feedback. If he or she asks for help, we can direct that student to optimal, frequent, or simply away from error-prone paths, based on that particular student and/or path. Similarly to constraint-based tutors , if a student is solving a proof in a way that is not already present in our MDP, we simply add their steps to our model. If such a student does commit an error, only default feedback will be given. As a first step, MDPs created from student data can be used to add intelligent feedback to every problem. This would require storing the MDP, adding a process to the end of the statement checking to match the student’s state to the MDP, and if it is found, adding a hint option that would reveal the optimal next choice in the MDP. If a student’s state is not found in the MDP, we can add it (storing its correctness), and periodically run reinforcement learning to update the reward function values.\n\n14\n\nWorkshop of Educational Data Mining\n\nOur next step will be to apply machine learning to the MDP to learn more general rules for building a true production rule system. The list of axioms itself can be used as general production rules, and represent most valid student actions. For example, Modus Ponens can be expressed as the production rule: “If A and AB, then apply Modus Ponens to get B”. However, there are no priorities assigned to the axioms themselves. We can cycle through the states of the MDP, examining all cases where the premises hold. We can use the MDP reward functions or the frequency of student use, or a combination of these, to set priorities for each axiom. Then when more than one axiom applies, we know which one was most frequently used with success (and therefore which one should fire). Alternatively, we could apply machine learning techniques to our MDPs to find common subgraphs in student approached, or build hierarchical MDPs to group substrategies in proof solutions. We believe this approach could be done in a general way so that it could be applied to other domains. Toward that end, we have generated visualizations of our MDPs to investigate the structure of student solutions, what errors they commit, and where they occur most frequently. This process can provide insight into where machine learning could provide the most benefit. In addition, this analysis can help us discover areas for improvement in the ITS. 4. Method This experiment uses data from the four fall semesters of 2003-2006, where an average of 120 students take the discrete math course at NC State University each fall. Students in this course are typically engineering and computer science students in their second or third year of college, but most have not been exposed to a course in logic. Students attend several lectures on propositional logic and complete an online homework where students complete truth tables and fill in the blanks in partially-completed proofs. Students then use the Proofs Tutorial to solve 10 proofs as homework, directly or using proof by contradiction. The majority of students used direct proof to solve proof 1. We extracted 429 of students’ first attempts at direct solutions to proof 1 from the Proofs Tutorial. We then removed invalid data (such as proofs with only one step to reach the conclusion), resulting in 416 student proofs. Of these, 283 (70%) were complete and 133 (30%) were partial proofs. Due to storage limitations, a few (6) of these proofs may have been completed by students but not fully recorded. The average lengths were 13 and 10 lines, respectively, for completed and partial proofs. This indicates that students did attempt to complete the proof. After cleaning the data, we load the proofs into a database and build an MDP for the data. We then set a large reward for the goal state (100) and penalties for incorrect states (10) and a cost for taking each action (1). Setting a non-zero cost on actions causes the MDP to penalize longer solutions (but we set this at 1/10 the cost of taking an incorrect step). These values may need to be adjusted for different sizes of MDPs. We apply the value iteration Bellman backup reinforcement learning technique to assign reward values to all states in the MDP. The equation for calculating values V(s) for each state s, where R(s) is the reward for the state, γ is the discount factor (set to 1), and Pa(s,s’) is the probability that action a will take state s to state s’:\n\nV(s) := R(s) + \" max a\n\n# P (s,s') V(s') a\n\ns'\n\nFor value iteration, V is calculated for each state until there is little change in the value function over the entire state space. Once this is complete, the optimal solution\n\n!\n\n15\n\nWorkshop of Educational Data Mining\n\nin the MDP corresponds to taking a greedy traversal approach in the MDP . The rewards for each state then indicate how close to the goal a state is, while probabilities of each transition reveal the frequency of taking a certain action in a certain state. For the purposes of this paper, just one step of value iteration was performed to cascade goal values through the MDP. We then ran the MDP for each semester of data, and for the aggregate. This resulted in a set of states, reward values, and actions for each MDP. On average, the four semesters yielded 158 states (ranging from 95-226 states in a semester). The most frequent approaches to problems were very similar across semesters, and the most frequent errors were also repeated. Therefore, in the remainder of this paper, we examine the aggregate MDP. Using Excel®, we assigned labels to each state in the MDP (just using the latest premise added), colors for errors, state values, and action frequencies, and prepared the data for display. We used GraphViz to display the output and convert into pictures. Table 2 shows the legend for nodes and edges. After graphing each MDP, we continually refined the data being displayed to explore questions about the student data. We present our findings in the following section. Table 2: Legend for MDP edges and nodes Edges (Values=Frequency)\n\nNodes (Values=Rewards)\n\n5. Results The aggregate MDP run on all four semesters of data has a total of 547 states (each semester alone generated between 95-226 unique states). Since it is hard to view statically, we omit it here. From interactive exploration, we found that 90% of all student errors related to explaining their actions, and a great majority of these were on the simplification rule. We plan to improve the interface for this explanation. We also found that students commit a great number of errors in the first step but less as they progress. To assist in visualizing student behavior, we plan to build a tool for teacher visualization that will allow for pruning nodes below a certain reward or frequency, and also for highlighting all the correct or incorrect applications of a particular action. We demonstrate some of these views in Figures 3-4. Figure 3 shows the aggregate MDP restricted to only valid states and frequent state-action pairs. The graph shows only one frequent successful path – indicating an optimal solution that students can perform. This path also corresponds to an expert solution. On this path, it seems that errors are occurring in going from state (a^-d) to (a), demonstrating our observation that applying simplification is difficult for students. We note many errors at the start, indicating that students have trouble getting started. Following the shaded path, second from the lowest in Figure 4, we observe that several (20-49) students apply IM in various ways; this path is very error-prone and does not (frequently) lead to a solution. This indicates a need to help students plan proof strategies. We can detect this type of path in the MDP and offer hints to help avoid it.\n\n16\n\nWorkshop of Educational Data Mining\n\nFigure 3: View of MDP restricted to valid states and frequent actions\n\nTo further examine student approaches, we expand this graph to include error states in Figure 4. In most errors, students find the correct premise (e.g. –d, which is correct) but have trouble explaining how the rule was applied to get the premise. For example, the rule for S (simplification) states (p^q)p, so to obtain (a^-d)-d the student must show that we substitute p=-d and q=a. We conclude that we need a better interface to help students show how a rule applies. For example, in Deep Thought, students are not required to perform substitutions if the program can detect them itself, and for simplification, the program asks: Right or Left? Anecdotally and intuitively, students seem to have less trouble with this approach. All but two of the remaining error states (indicated with darker shading and a double border) are due to substitution errors. (Start)  (–(a^-d)) is an incorrect application of IM (it’s missing a not), while Contrapositive (CP) was frequently applied to obtain –c from c>d and –d (which needs Modus Tollens, MT). These findings indicate that more practice might be needed with these rules in the context of negated variables.\n\nFigure 4: View of MDP restricted to frequent states and actions\n\nThese visualizations are useful in predicting what the MDP will find when generating feedback based on frequent responses. However, this does not yield insight into a more general application of MDPs to proofs and what types of production rules might be generated for general problems. To make more general production rules for\n\n17\n\nWorkshop of Educational Data Mining\n\nproof problems, we will take MDPs from several problems and attempt to learn general rules. For instance, a production rule often applied by experts is, “if (p  q) and (p) are premises then apply MP to obtain the new premise (q)”. To learn this, we need to break MDP states down into their constituent premises. We created Figure 5, (which is a graph, not an MDP), by mapping all correct states with a common last premise into a single node, which corresponds to grouping unique action/premise pairs. We then eliminated all low-frequency actions and the simplification rule (since we plan to change the interface to improve usage of this rule). This new visualization of the paths of student solutions allows us to track frequent solution attempts regardless of order. In other words, our previous MDP views correspond to unique paths, while this graph shows us relationships between consecutive steps in the proof regardless of what was done before. In Figure 5, there are some dead-end paths, meaning that several students started proofs in these directions but few students were able to reach the solution this way. These dead-end paths can be used to derive feedback to students that their approach may not be productive. We can also use Figure 5 to derive most frequent orderings of student solutions. To do so, we start at the (top) start node and choose a frequent (wide) edge, and repeat without visiting nodes twice, until we reach the solution, as in Figure 6.\n\nFigure 5: Graph showing inferences, unique premises and frequent (>9) actions\n\nFigure 6: Most frequent proof solution sequences, derived from Figure 5\n\nSome secondary approaches are shown in Figure 7, demonstrating how a teacher could use Figure 5 by following particular paths but not repeating any nodes visited, to find unique solutions to the proof. These approaches demonstrate students’ frequent preference to use Disjunctive Syllogism (DS), even though these solutions are longer.\n\n18\n\nWorkshop of Educational Data Mining\n\nFigure 7: Secondary proof approaches derived from Figure 5\n\nIn both Figures 6 and 7, we see that there are errors (on the double edges) leading to and from states containing negated variables. This observation reflects our experience that students need more practice in using rules when variables are of negated, and in applying rules in sequence to achieve a goal. 6. Conclusions and Future Work We have proposed a general approach to mining Markov decision processes from student work to automatically generate production rules and discussed how this approach can be applied to a particular CBT. This approach differs from prior work in authoring tutoring systems by mining actual student data, rather than relying on teachers to add examples the system can learn from. We have explored visualizations of the Markov decision processes extracted from solutions to a formal logic proof to determine how to improve the tutor and how we might proceed in building useful production rules and feedback. We believe that the process we have applied to creating problem visualizations can be useful in learning about other problem-solving processes from student data, instead of creating expert systems by hand. From these visualizations, we have learned: 1.\n\nAlthough we hypothesized that students need help in planning, this did not seem to be the case. Instead, students needed help on getting started. 2. As we expected, students need more practice with negation. 3. The overwhelming majority of student errors were in explaining rule applications. We plan to add a better interface for this step. We have also concluded that the extracted MDPs will be useful in generating student feedback. The extracted MDP does contain a frequent expert-like path and contains a significant number of correct paths and student errors. Our tutor can already classify many errors students make. Adding the MDP to this tutor will enable it to model student mastery, provide hints, and feedback on errors. This MDP can constantly learn from new student data. We note that on cold start for a new problem that has no student data, the system will still act as a problem-solving environment, but after even one semester of data is collected, a limited amount of feedback can be generated. As more data are added, more automated feedback can be generated. In our future work we plan to implement this system in our tutor and in Deep Thought, a logic tutor created by Marvin Croy . Once implemented, we will test the feedback generated based on the MDP, and we will continue to explore ways to learn general rules to build production rule systems with greater coverage and robustness. A novel contribution of this work is the idea of providing feedback on the most frequent approach to a problem. It is often the case that students are not ready to apply optimal solutions to a problem. However, detecting this readiness is a challenging user modeling problem. It may be possible that student feedback based on frequency rather\n\n19\n\nWorkshop of Educational Data Mining\n\nthan optimality can provide most students with the help they need. This type of feedback may also encourage student reflection, when the tutor is no longer allknowing, but has a memory that students tap into instead. In our future work we plan to investigate the impact of this type of feedback on student learning. 7. References \n\nAnderson, J., Gluck, K. 2001. What role do cognitive architectures play in intelligent tutoring systems? In D. Klahr & S. Carver (Eds.) Cognition & Instruction: 25 years of progress, 227-262. Erlbaum.\n\n\n\nMurray, Tom. 1999. Authoring intelligent tutoring systems: An analysis of the state of the art. Intl. J. Artificial Intelligence in Education, 10: 98-129.\n\n Baffes, P. & Mooney, R.J. 1996. A novel application of theory refinement to student modeling. Proc. AAAI-96, pp. 403-408, Portland, OR, August 1996. \n\nMitrovic, A., Koedinger, K. & Martin, B. 2003. A comparative analysis of cognitive tutoring and constraint-based modeling. User Modeling 2003: 313-322.\n\n\n\nKoedinger, K. R., Aleven, V., Heffernan. T., McLaren, B. & Hockenberry, M. 2004. Opening the door to non-programmers: Authoring intelligent tutor behavior by demonstration. Proc. 7th Intelligent Tutoring Systems Conference, Maceio, Brazil, pp. 162-173.\n\n\n\nMcLaren, B., Koedinger, K., Schneider, M., Harrer, A., & Bollen, L. 2004. Bootstrapping Novice Data: Semi-automated tutor authoring using student log files, In Proc. Workshop on Analyzing Student-Tutor Interaction Logs to Improve Educational Outcomes, 7th Intl. Conf. Intelligent Tutoring Systems (ITS2004), Maceió, Brazil, August 30, 2004.\n\n\n\nCroy, M. 2000. \"Problem Solving, Working Backwards, and Graphic Proof Representation,\" Teaching Philosophy, 23(2), 169-187.\n\n\n\nScheines, R., & Sieg, W. 1994. Computer environments for proof construction. Interactive Learning Environments, 4(2), 159-169.\n\n\n\nMerceron, A., & Yacef, K. 2005. Educational data mining: a case study. In: C.-K. Looi, G. McCalla, B. Bredeweg & J. Breuker (eds.), (Proc. 12th Intl. Conf. Artificial Intelligence), pp. 467–474. Amsterdam: IOS Press.\n\n Barnes, T. 2006. Evaluation of the q-matrix method in understanding student logic proofs. Proc. 19th Intl. Florida AI Research Society Conference (FLAIRS 2006), Melbourne Beach, FL, May 11-13, 2006. Stamper. J. 2006. Automating the Generation of Production Rules for Intelligent Tutoring Systems. Proc. Intl. Conf. Computer-Aided Learning (ICL2006). Austria, Sep. 27-29, 2006. Richard S. Sutton and Andrew G. Barto. 1998. Reinforcement Learning: An Introduction. The MIT Press, Cambridge, MA. Vygotsky, L. (1986). Thought and language. Cambridge, MA: MIT Press.\n\n20\n\nWorkshop of Educational Data Mining\n\nDifficulties in inferring student knowledge from observations (and why you should care) Joseph E. Beck [email protected] Machine Learning Department Carnegie Mellon University 5000 Forbes Avenue Pittsburgh, PA USA Abstract. Student modeling has a long history in the field of intelligent educational software and is the basis for many tutorial decisions. Furthermore, the task of assessing a student’s level of knowledge is a basic building block in the educational data mining process. If we cannot estimate what students know, it is difficult to perform fine-grained analyses to see if a system’s teaching actions are having a positive effect. In this paper, we demonstrate that there are several unaddressed problems with student model construction that negatively affect the inferences we can make. We present two partial solutions to these problems, using Expectation Maximization to estimate parameters and using Dirichlet priors to bias the model fit procedure. Aside from reliably improving model fit in predictive accuracy, these approaches might result in model parameters that are more plausible. Although parameter plausibility is difficult to quantify, we discuss some guidelines and propose a derived measure of predicted number of trials until mastery as a method for evaluating model parameters. Keywords: student modeling, mastery learning, parameter estimation, Bayesian networks, Dirichlet priors\n\n1\n\nIntroduction\n\nThe focus of this paper is on the difficulties in mapping observable student performance to estimate his level of knowledge about underlying skills. This task, better known as student modeling, has been around for at least three decades . Given the long history, it is fair to ask whether there are large, fundamental problems to be overcome. We discuss several problems resulting from the existence of local-maxima as well as multiple global-maxima in the parameter search space for a common student modeling approach. These problems may seem esoteric, but they result in seemingly similar models that make very different claims about the student’s knowledge. There are two direct impacts of the results presented in this paper on Educational Data Mining (EDM). First, since estimating student knowledge is relatively well understood, it should be one of the easiest tasks for us to analyze. Given the surprising complexity and difficulty in evaluating models, we should be cautious about our analyses of more exotic phenomena. Second, finegrained estimates of student knowledge are a useful lever in attacking other data mining tasks. For example, it is difficult to “measure the effect of individual interventions” (from the workshop’s call for papers) if we cannot track student knowledge. The goal of student modeling is to take observations of a student’s performance and use those to estimate the student’s knowledge, goals, preferences, and other latent characteristics. The key aspect of the problem is that the characteristics of interest are not directly observable. Thus, some means of mapping the observations to characteristics is required. Knowledge tracing is an example of such a technique, and is the one we will focus on in the explorations in this paper. Knowledge tracing, shown in Figure 1, is an approach for taking student observations and using those to estimate the student’s level of knowledge. It assumes that students either know a skill or do not; similarly they either respond correctly to an item or not. There are two parameters, slip and guess, that mediate student knowledge and student performance. These two parameters are called the performance parameters in the model. An assumption of the model is that even if a student knows a skill, there is a chance he might still respond incorrectly to a question that utilizes that skill. This probability is the slip parameter. There are a variety of reasons for an incorrect response, the student could have made a simple typo (e.g. typed ‘12’ instead of ‘21’ for “7 x 3”), he could have been frustrated with the 21\n\nWorkshop of Educational Data Mining\n\nsystem and behaving randomly, or he may not have felt like solving the problem and would rather have the tutor tell him how to solve it. Doesn’t know Skill\n\nKnows Skill\n\nP(slip)\n\nP(guess)\n\n1 – P(slip)\n\n1 – P(guess)\n\nObserve correct\n\nObserve incorrect\n\nFigure 1. Diagram of knowledge tracing Conversely, a student who does not know the skill might still be able to generate a correct response. This probability is referred to as the guess parameter. A guess could occur either through blind chance (e.g. in a 4choice multiple choice test there is a ¼ chance of getting a question right even if one does not understand it), the student being able to utilize a weaker version of the correct rule that only applies in certain circumstances, or through errors in the tutor’s scoring mechanism. For example, experiments using knowledge tracing with speech recognition had a very high guess parameter because the speech recognizer had a hard time noticing when students made mistakes . In addition to the two performance parameters, there are two learning parameters. The first is K0, the likelihood the student knows the skill when he first uses the tutor. Initial knowledge can come from a variety of places including material from a prior course or declarative instruction delivered by the tutor before procedural lessons (e.g. ). The second learning parameter is t, the probability a student will acquire a skill as a result of an opportunity to practice it. Every skill to be tracked has these four parameters, slip, guess, K0, and t, associated with it. To train a knowledge tracing model, historical data of student performance is provided to model fitting software. There exist a variety of software packages to perform the optimization such as the curve fitting approach used by some of the cognitive tutors1 and the Bayes Net Toolkit for Student Modeling (BNT-SM, ). To use a knowledge tracing model, when a student begins using an ITS his initial knowledge estimates are set to the K0 estimates for each skill. When a student responds to an item, Bayesian inference is used to update his internal knowledge on the basis of whether he responded correctly or not, and on the skill’s slip and guess parameters. The student is then credited with an opportunity to learn the skill via the t parameter. This work also assumes the tutor’s designers already know which skill will be modeled. Although the construction and refinement of domain models from data is an interesting topic (e.g. [6-8]), it is beyond the scope of this paper. This paper uses the knowledge tracing model as a platform for experimentation. We are not asserting knowledge tracing is correct, and in fact would argue the opposite; for one thing, it does not acknowledge that students can forget what they have learned. It is simply an approach that is the closest the AIED community has to a standard. Furthermore, it is the simplest modeling approach that has the promise to be analytically interesting. If we are able to uncover and study basic problems with a model as simple as knowledge tracing, it is probable that more complex modeling schemes will suffer similar drawbacks.\n\n2\n\nDifficulties\n\nKnowledge tracing has two main flaws. The first is that the parameter estimation task is subject to local maxima. That means that given a set of observations, the parameters returned by model fitting software might not the best possible fit to the data. The second problem, less well known, is that there are multiple local maxima. That is, there can be more than one set of parameters that fit the data equally well.\n\n1\n\nSource code is courtesy of Albert http://www.cs.cmu.edu/~rsbaker/curvefit.tar.gz\n\nCorbett\n\nand\n\nRyan\n\nBaker\n\nand\n\nis\n\navailable\n\nat 22\n\nWorkshop of Educational Data Mining\n\n2.1" ]

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[ "## 18425\n\n18,425 (eighteen thousand four hundred twenty-five) is an odd five-digits composite number following 18424 and preceding 18426. In scientific notation, it is written as 1.8425 × 104. The sum of its digits is 20. It has a total of 4 prime factors and 12 positive divisors. There are 13,200 positive integers (up to 18425) that are relatively prime to 18425.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Odd\n• Number length 5\n• Sum of Digits 20\n• Digital Root 2\n\n## Name\n\nShort name 18 thousand 425 eighteen thousand four hundred twenty-five\n\n## Notation\n\nScientific notation 1.8425 × 104 18.425 × 103\n\n## Prime Factorization of 18425\n\nPrime Factorization 52 × 11 × 67\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 3685 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 18,425 is 52 × 11 × 67. Since it has a total of 4 prime factors, 18,425 is a composite number.\n\n## Divisors of 18425\n\n1, 5, 11, 25, 55, 67, 275, 335, 737, 1675, 3685, 18425\n\n12 divisors\n\n Even divisors 0 12 6 6\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 12 Total number of the positive divisors of n σ(n) 25296 Sum of all the positive divisors of n s(n) 6871 Sum of the proper positive divisors of n A(n) 2108 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 135.739 Returns the nth root of the product of n divisors H(n) 8.74051 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 18,425 can be divided by 12 positive divisors (out of which 0 are even, and 12 are odd). The sum of these divisors (counting 18,425) is 25,296, the average is 2,108.\n\n## Other Arithmetic Functions (n = 18425)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 13200 Total number of positive integers not greater than n that are coprime to n λ(n) 660 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 2107 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares\n\nThere are 13,200 positive integers (less than 18,425) that are coprime with 18,425. And there are approximately 2,107 prime numbers less than or equal to 18,425.\n\n## Divisibility of 18425\n\n m n mod m 2 3 4 5 6 7 8 9 1 2 1 0 5 1 1 2\n\nThe number 18,425 is divisible by 5.\n\n• Arithmetic\n• Deficient\n\n• Polite\n\n## Base conversion (18425)\n\nBase System Value\n2 Binary 100011111111001\n3 Ternary 221021102\n4 Quaternary 10133321\n5 Quinary 1042200\n6 Senary 221145\n8 Octal 43771\n10 Decimal 18425\n12 Duodecimal a7b5\n16 Hexadecimal 47f9\n20 Vigesimal 2615\n36 Base36 e7t\n\n## Basic calculations (n = 18425)\n\n### Multiplication\n\nn×i\n n×2 36850 55275 73700 92125\n\n### Division\n\nni\n n⁄2 9212.5 6141.67 4606.25 3685\n\n### Exponentiation\n\nni\n n2 339480625 6254930515625 115247094750390625 2123427720775947265625\n\n### Nth Root\n\ni√n\n 2√n 135.739 26.4121 11.6507 7.12987\n\n## 18425 as geometric shapes\n\n### Circle\n\nRadius = n\n Diameter 36850 115768 1.06651e+09\n\n### Sphere\n\nRadius = n\n Volume 2.62006e+13 4.26604e+09 115768\n\n### Square\n\nLength = n\n Perimeter 73700 3.39481e+08 26056.9\n\n### Cube\n\nLength = n\n Surface area 2.03688e+09 6.25493e+12 31913\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 55275 1.46999e+08 15956.5\n\n### Triangular Pyramid\n\nLength = n\n Surface area 5.87998e+08 7.37151e+11 15043.9\n\n## Cryptographic Hash Functions\n\nmd5 fc60f5d7abc2e080599bb6dc465db54d 512e15e3675f90121af56453af8c1f26a6b26d55 6d21361cf8735de1aa833ac0f97d2afcc1e3c357d921303cb33761bd8441ae79 76353fee61aa8df98dc7641d6ebf26d8a00da59e87f76d6eb8cf84e5fb3b8eac87769f57327f043c13c8610b78e33a50d90a62a85d361b408197557418c1e9b2 59bffc5f37f2f4e2bf33ee3153a8745c600488c2" ]

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[ "# Estimate a Regression Model with ARIMA Errors\n\nThis example shows how to estimate the sensitivity of the US Gross Domestic Product (GDP) to changes in the Consumer Price Index (CPI) using estimate.\n\nLoad the US macroeconomic data set, Data_USEconModel. Plot the GDP and CPI.\n\nload Data_USEconModel gdp = DataTable.GDP; cpi = DataTable.CPIAUCSL; figure plot(dates,gdp) title('{\\bf US Gross Domestic Product, Q1 in 1947 to Q1 in 2009}') datetick axis tight", null, "figure plot(dates,cpi) title('{\\bf US Consumer Price Index, Q1 in 1947 to Q1 in 2009}') datetick axis tight", null, "gdp and cpi seem to increase exponentially.\n\nRegress gdp onto cpi. Plot the residuals.\n\nXDes = [ones(length(cpi),1) cpi]; % Design matrix beta = XDes\\gdp; u = gdp - XDes*beta; % Residuals figure plot(u) h1 = gca; hold on plot(h1.XLim,[0 0],'r:') title('{\\bf Residual Plot}') hold off", null, "The pattern of the residuals suggests that the standard linear model assumption of uncorrelated errors is violated. The residuals appear autocorrelated.\n\nPlot correlograms for the residuals.\n\nfigure subplot(2,1,1) autocorr(u) subplot(2,1,2) parcorr(u)", null, "The autocorrelation function suggests that the residuals are a nonstationary process.\n\nApply the first difference to the logged series to stabilize the residuals.\n\ndlGDP = diff(log(gdp)); dlCPI = diff(log(cpi)); dlXDes = [ones(length(dlCPI),1) dlCPI]; beta = dlXDes\\dlGDP; u = dlGDP - dlXDes*beta; figure plot(u); h2 = gca; hold on plot(h2.XLim,[0 0],'r:') title('{\\bf Residual Plot, Transformed Series}') hold off", null, "figure subplot(2,1,1) autocorr(u) subplot(2,1,2) parcorr(u)", null, "The residual plot from the transformed data suggests stabilized, albeit heteroscedastic, unconditional disturbances. The correlograms suggest that the unconditional disturbances follow an AR(1) process.\n\nSpecify the regression model with AR(1) errors:\n\n$\\begin{array}{c}\\text{dlGDP}=\\text{Intercept}+\\text{dlCPI}\\beta +{u}_{t}\\\\ {u}_{t}=\\varphi {u}_{t-1}+{\\epsilon }_{t}.\\end{array}$\n\nMdl = regARIMA('ARLags',1);\n\nestimate estimates any parameter having a value of NaN.\n\nFit Mdl to the data.\n\nEstMdl = estimate(Mdl,dlGDP,'X',dlCPI,'Display','params');\n Regression with ARMA(1,0) Error Model (Gaussian Distribution): Value StandardError TStatistic PValue __________ _____________ __________ __________ Intercept 0.012762 0.0013472 9.4734 2.7098e-21 AR{1} 0.38245 0.052494 7.2856 3.2031e-13 Beta(1) 0.3989 0.077286 5.1614 2.4516e-07 Variance 9.0101e-05 5.947e-06 15.151 7.5075e-52 \n\nAlternatively, estimate the regression coefficients and Newey-West standard errors using hac.\n\nhac(dlCPI,dlGDP,'intercept',true,'display','full');\nEstimator type: HAC Estimation method: BT Bandwidth: 4.1963 Whitening order: 0 Effective sample size: 248 Small sample correction: on Coefficient Estimates: | Coeff SE ------------------------ Const | 0.0115 0.0012 x1 | 0.5421 0.1005 Coefficient Covariances: | Const x1 -------------------------- Const | 0.0000 -0.0001 x1 | -0.0001 0.0101 \n\nThe intercept estimates are close, but the regression coefficient estimates corresponding to dlCPI are not. This is because regARIMA explicitly models for the autocorrelation of the disturbances. hac estimates the coefficients using ordinary least squares, and returns standard errors that are robust to the residual autocorrelation and heteroscedasticity.\n\nAssuming that the model is correct, the results suggest that an increase of one point in the CPI rate increases the GDP growth rate by 0.399 points. This effect is significant according to the t statistic.\n\nFrom here, you can use forecast or simulate to obtain forecasts and forecast intervals for the GDP rate. You can also compare several models by computing their AIC statistics using aicbic." ]

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[ "The Annals of Probability\n\nOn unbounded invariant measures of stochastic dynamical systems\n\nAbstract\n\nWe consider stochastic dynamical systems on $\\mathbb{R}$, that is, random processes defined by $X_{n}^{x}=\\Psi_{n}(X_{n-1}^{x})$, $X_{0}^{x}=x$, where $\\Psi_{n}$ are i.i.d. random continuous transformations of some unbounded closed subset of $\\mathbb{R}$. We assume here that $\\Psi_{n}$ behaves asymptotically like $A_{n}x$, for some random positive number $A_{n}$ [the main example is the affine stochastic recursion $\\Psi_{n}(x)=A_{n}x+B_{n}$]. Our aim is to describe invariant Radon measures of the process $X_{n}^{x}$ in the critical case, when $\\mathbb{E}\\log A_{1}=0$. We prove that those measures behave at infinity like $\\frac{dx}{x}$. We study also the problem of uniqueness of the invariant measure. We improve previous results known for the affine recursions and generalize them to a larger class of stochastic dynamical systems which include, for instance, reflected random walks, stochastic dynamical systems on the unit interval $[0,1]$, additive Markov processes and a variant of the Galton–Watson process.\n\nArticle information\n\nSource\nAnn. Probab., Volume 43, Number 3 (2015), 1456-1492.\n\nDates\nFirst available in Project Euclid: 5 May 2015\n\nPermanent link to this document\nhttps://projecteuclid.org/euclid.aop/1430830287\n\nDigital Object Identifier\ndoi:10.1214/13-AOP903\n\nMathematical Reviews number (MathSciNet)\nMR3342668\n\nZentralblatt MATH identifier\n1352.37155\n\nCitation\n\nBrofferio, Sara; Buraczewski, Dariusz. On unbounded invariant measures of stochastic dynamical systems. Ann. Probab. 43 (2015), no. 3, 1456--1492. doi:10.1214/13-AOP903. https://projecteuclid.org/euclid.aop/1430830287" ]

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[ "Instruction\n1\nTo find out the cost of servicing the loan portfolio, calculate the weighted average interest rate for all loans. Calculate the total amount of expenses for payment of percent for the year by multiplying the loan amount by the interest rate on each contract separately and adding the values. Divide the final value on the index credit volume of the company and multiply the quotient by 100.\n2\nTo calculate the weighted average interest rates on loans and deposits use the formula proposed by the Central Bank of the Russian Federation:\nPav = (V1 x P1 + V2 x P2 + ... + Pn x Vn):(V1 + V2 + ... + Vn), where\nV1, V2, ..., Vn – the volume of loans or deposits,\nP1, P2, ..., PN is the nominal interest rate under the contract.\n3\nFor the loans in different banks, and you have a large number of contracts for convenience, make calculations using the spreadsheet: in column A enter the loan amount, in column b – interest rate in the column, specify the formula for calculating the amount of annual interest (A x B), and in the lower part of the table or the formula to calculate totals for columns. In a separate cell, select the algorithm of calculating weighted average rates:\n(Total column C / Total of column A) x 100.\n4\nIf you do not know the interest rate in terms of contracts, but there is the total amount of expenses for payment of percent on credits, divide it by the total volume of credit mass and multiply by 100 - you will receive the average rate.\n5\nIn addition, loan and Deposit transactions may be issued on terms of variable interest rates. In this case, its mean value is need to calculate taking into account the change of its value during the entire period of the contract. To do this, multiply the loan amount by the interest rate, divide by the number of days in the year (365 or 366), and multiply by the number of days in which it was applied. Calculate and add the interest charges for each rate, and then divide the total amount by the loan amount and multiply the result by 100." ]

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[ "• Research\n• Open Access\n\n# Nonconvex composite multiobjective nonsmooth fractional programming\n\nJournal of Inequalities and Applications20132013:508\n\nhttps://doi.org/10.1186/1029-242X-2013-508\n\n• Received: 4 April 2013\n• Accepted: 25 September 2013\n• Published:\n\n## Abstract\n\nWe consider nonsmooth multiobjective programs where the objective function is a fractional composition of invex functions and locally Lipschitz and Gâteaux differentiable functions. Kuhn-Tucker necessary and sufficient optimality conditions for weakly efficient solutions are presented. We formulate dual problems and establish weak, strong and converse duality theorems for a weakly efficient solution.\n\nMSC:90C46, 90C29, 90C32.\n\n## Keywords\n\n• composite functions\n• fractional programming\n• nonsmooth programming\n• multiobjective problems\n• necessary and sufficient conditions\n• duality\n\n## 1 Introduction\n\nRecently there has been an increasing interest in developing optimality conditions and duality relations for nonsmooth multiobjective programming problems involving locally Lipschitz functions. Many authors have studied under kinds of generalized convexity, and some results have been obtained. Schaible and Bector et al. derived some Kuhn-Tucker necessary and sufficient optimality conditions for the multiobjective fractional programming. By using ρ-invexity of a fractional function, Kim obtained necessary and sufficient optimality conditions and duality theorems for nonsmooth multiobjective fractional programming problems. Lai and Ho established sufficient optimality conditions for multiobjective fractional programming problems involving exponential V-r-invex Lipschitz functions. In , Kim and Schaible considered nonsmooth multiobjective programming problems with inequality and equality constraints involving locally Lipschitz functions and presented several sufficient optimality conditions under various invexity assumptions and regularity conditions. Soghra Nobakhtian obtained optimality conditions and a mixed dual model for nonsmooth fractional multiobjective programming problems. Jeyakumar and Yang considered nonsmooth constrained multiobjective optimization problems where the objective function and the constraints are compositions of convex functions and locally Lipschitz and Gâteaux differentiable functions. Lagrangian necessary conditions and new sufficient optimality conditions for efficient and properly efficient solutions were presented. Mishra and Mukherjee extended the work of Jeyakumar and Yang and the constraints are compositions of V-invex functions.\n\nThe present article begins with an extension of the results in [7, 8] from the nonfractional to the fractional case. We consider nonsmooth multiobjective programs where the objective functions are fractional compositions of invex functions and locally Lipschitz and Gâteaux differentiable functions. Kuhn-Tucker necessary conditions and sufficient optimality conditions for weakly efficient solutions are presented. We formulate dual problems and establish weak, strong and converse duality theorems for a weakly efficient solution.\n\n## 2 Preliminaries\n\nLet ${\\mathbb{R}}^{n}$ be the n-dimensional Euclidean space and ${\\mathbb{R}}_{+}^{n}$ be its nonnegative orthant. Throughout the paper, the following convention for inequalities will be used for $x,y\\in {\\mathbb{R}}^{n}$:\nThe real-valued function $f:{\\mathbb{R}}^{n}\\to \\mathbb{R}$ is said to be locally Lipschitz if for any $z\\in {\\mathbb{R}}^{n}$ there exists a positive constant K and a neighbourhood N of z such that, for each $x,y\\in N$,\n$|f\\left(x\\right)-f\\left(y\\right)|\\leqq K\\parallel x-y\\parallel .$\nThe Clarke generalized directional derivative of a locally Lipschitz function f at x in the direction d denoted by ${f}^{\\circ }\\left(x;d\\right)$ (see, e.g., Clarke ) is as follows:\n${f}^{\\circ }\\left(x;d\\right)=\\underset{t↓0}{\\underset{y\\to x}{lim sup}}{t}^{-1}\\left(f\\left(y+td\\right)-f\\left(y\\right)\\right).$\nThe Clarke generalized subgradient of f at x is denoted by\n\nProposition 2.1 \n\nLet f, h be Lipschitz near x, and suppose $h\\left(x\\right)\\ne 0$. Then $\\frac{f}{h}$ is Lipschitz near x, and one has\n$\\partial \\left(\\frac{f}{h}\\right)\\left(x\\right)\\subset \\frac{h\\left(x\\right)\\partial f\\left(x\\right)-f\\left(x\\right)\\partial h\\left(x\\right)}{{h}^{2}\\left(x\\right)}.$\n\nIf, in addition, $f\\left(x\\right)\\geqq 0$, $h\\left(x\\right)>0$ and if f andh are regular at x, then equality holds and $\\frac{f}{h}$ is regular at x.\n\nIn this paper, we consider the following composite multiobjective fractional programming problem:\n$\\begin{array}{lll}\\left(\\text{P}\\right)& \\text{Minimize}& \\left(\\frac{{f}_{1}\\left({F}_{1}\\left(x\\right)\\right)}{{h}_{1}\\left({F}_{1}\\left(x\\right)\\right)},\\dots ,\\frac{{f}_{p}\\left({F}_{p}\\left(x\\right)\\right)}{{h}_{p}\\left({F}_{p}\\left(x\\right)\\right)}\\right)\\\\ \\text{subject to}& {g}_{j}\\left({G}_{j}\\left(x\\right)\\right)\\leqq 0,\\phantom{\\rule{1em}{0ex}}j=1,2,\\dots ,m,x\\in C,\\end{array}$\nwhere\n1. (1)\n\nC is an open convex subset of a Banach space X,\n\n2. (2)\n\n${f}_{i}$, ${h}_{i}$, $i=1,2,\\dots ,p$, and ${g}_{j}$, $j=1,2,\\dots ,m$, are real-valued locally Lipschitz functions on ${\\mathbb{R}}^{n}$, and ${F}_{i}$ and ${G}_{j}$ are locally Lipschitz and Gâteaux differentiable functions from X into ${\\mathbb{R}}^{n}$ with Gâteaux derivatives ${F}_{i}^{\\prime }\\left(\\cdot \\right)$ and ${G}_{j}^{\\prime }\\left(\\cdot \\right)$, respectively, but are not necessarily continuously Fréchet differentiable or strictly differentiable ,\n\n3. (3)\n\n${f}_{i}\\left(x\\right)\\geqq 0$, ${h}_{i}\\left(x\\right)>0$, $i=1,2,\\dots ,p$,\n\n4. (4)\n\n${f}_{i}\\left(x\\right)$ and $-{h}_{i}\\left(x\\right)$ are regular.\n\nDefinition 2.1 A feasible point ${x}_{0}$ is said to be a weakly efficient solution for (P) if there exists no feasible point x for which\n$\\frac{{f}_{i}\\left({F}_{i}\\left(x\\right)\\right)}{{h}_{i}\\left({F}_{i}\\left(x\\right)\\right)}<\\frac{{f}_{i}\\left({F}_{i}\\left({x}_{0}\\right)\\right)}{{h}_{i}\\left({F}_{i}\\left({x}_{0}\\right)\\right)},\\phantom{\\rule{1em}{0ex}}\\mathrm{\\forall }i=1,2,\\dots ,p.$\n\nDefinition 2.2 \n\nA function f is invex on ${X}_{0}\\subset {\\mathbb{R}}^{n}$ if for $x,u\\in {X}_{0}$ there exists a function $\\eta \\left(x,u\\right):{X}_{0}×{X}_{0}\\to {\\mathbb{R}}^{n}$ such that\n${f}_{i}\\left(x\\right)-{f}_{i}\\left(u\\right)\\geqq {\\xi }_{i}^{T}\\eta \\left(x,u\\right),\\phantom{\\rule{1em}{0ex}}\\mathrm{\\forall }{\\xi }_{i}\\in \\partial {f}_{i}\\left(u\\right).$\n\nDefinition 2.3 \n\nA function $f:{X}_{0}\\to {\\mathbb{R}}^{n}$ is V-invex on ${X}_{0}\\subset {\\mathbb{R}}^{n}$ if for $x,u\\in {X}_{0}$ there exist functions $\\eta \\left(x,u\\right):{X}_{0}×{X}_{0}\\to {\\mathbb{R}}^{n}$ and ${\\alpha }_{i}\\left(x,u\\right):{X}_{0}×{X}_{0}\\to {\\mathbb{R}}_{+}\\setminus \\left\\{0\\right\\}$ such that\n${f}_{i}\\left(x\\right)-{f}_{i}\\left(u\\right)\\geqq {\\alpha }_{i}\\left(x,u\\right){\\xi }_{i}^{T}\\eta \\left(x,u\\right),\\phantom{\\rule{1em}{0ex}}\\mathrm{\\forall }{\\xi }_{i}\\in \\partial f\\left(u\\right).$\n\nThe following lemma is needed in necessary optimality conditions, weak duality and converse duality.\n\nLemma 2.1 \n\nIf ${f}_{i}\\geqq 0$, ${h}_{i}>0$, ${f}_{i}$ and $-{h}_{i}$ are invex at u with respect to $\\eta \\left(x,u\\right)$, and ${f}_{i}$ and $-{h}_{i}$ are regular at u, then $\\frac{{f}_{i}}{{h}_{i}}$ is V-invex at u with respect to $\\overline{\\eta }$, where $\\overline{\\eta }\\left(x,u\\right)=\\frac{{h}_{i}\\left(u\\right)}{{h}_{i}\\left(x\\right)}\\eta \\left(x,u\\right)$.\n\n## 3 Optimality conditions\n\nNote that if $F:X\\to {\\mathbb{R}}^{n}$ is locally Lipschitz near a point $x\\in X$ and Gâteaux differentiable at x and if $f:{\\mathbb{R}}^{n}\\to \\mathbb{R}$ is locally Lipschitz near $F\\left(x\\right)$, then the continuous sublinear function, defined by\n${\\pi }_{x}\\left(h\\right):=max\\left\\{\\sum _{k=1}^{n}{w}_{k}{F}_{k}^{\\prime }\\left(x\\right)h|w\\in \\partial f\\left(F\\left(x\\right)\\right)\\right\\},$\nsatisfies the inequality\n${\\left(f\\circ F\\right)}_{+}^{\\prime }\\left(x,h\\right)\\leqq {\\pi }_{x}\\left(h\\right),\\phantom{\\rule{1em}{0ex}}\\mathrm{\\forall }h\\in X.$\n(3.1)\n\nRecall that ${q}_{+}^{\\prime }\\left(x,h\\right)={lim}_{\\lambda ↓0}sup{\\lambda }^{-1}\\left(q\\left(x+\\lambda h\\right)-q\\left(x\\right)\\right)$is the upper Dini-directional derivative of $q:X\\to \\mathbb{R}$ at x in the direction of h, and $\\partial f\\left(F\\left(x\\right)\\right)$ is the Clarke subdifferential of f at $F\\left(x\\right)$. The function ${\\pi }_{x}\\left(\\cdot \\right)$ in (3.1) is called upper convex approximation of $f\\circ F$ at x, see [11, 12].\n\nNote that for a set C, int C denotes the interior of C, and ${C}^{+}=\\left\\{v\\in {X}^{\\prime }|v\\left(x\\right)\\geqq 0,\\mathrm{\\forall }x\\in C\\right\\}$, denotes the dual cone of C, where ${X}^{\\prime }$ is the topological dual space of X. It is also worth noting that for a convex set C, the closure of the cone generated by the set C at a point a, cl $cone\\left(C-a\\right)$, is the tangent cone of C at a, and the dual $cone-{\\left(C-a\\right)}^{+}$ is the normal cone of C at a, see [9, 13].\n\nTheorem 3.1 (Necessary optimality conditions)\n\nSuppose that ${f}_{i}$, ${h}_{i}$ and ${g}_{j}$ are locally Lipschitz functions, and that ${F}_{i}$ and ${G}_{j}$ are locally Lipschitz and Gâteaux differentiable functions. If $a\\in C$ is a weakly efficient solution for (P), then there exist Lagrange multipliers ${\\lambda }_{i}\\geqq 0$, $i=1,2,\\dots ,p$, and ${\\mu }_{j}\\geqq 0$, $j=1,2,\\dots ,m$, not all zero, satisfying\n$\\begin{array}{c}0\\in \\sum _{i=1}^{p}{\\lambda }_{i}{T}_{i}\\left(a\\right){F}_{i}^{\\prime }\\left(a\\right)+\\sum _{j=1}^{m}{\\mu }_{j}\\partial {g}_{j}\\left({G}_{j}\\left(a\\right)\\right){G}_{j}^{\\prime }\\left(a\\right)-{\\left(C-a\\right)}^{+},\\hfill \\\\ {\\mu }_{j}{g}_{j}\\left({G}_{j}\\left(a\\right)\\right)=0,\\phantom{\\rule{1em}{0ex}}j=1,2,\\dots ,m,\\hfill \\\\ {T}_{i}\\left(a\\right)=\\frac{\\partial {f}_{i}\\left({F}_{i}\\left(a\\right)\\right)-{\\varphi }_{i}\\left(a\\right)\\partial {h}_{i}\\left({F}_{i}\\left(a\\right)\\right)}{{h}_{i}\\left({F}_{i}\\left(a\\right)\\right)},\\phantom{\\rule{1em}{0ex}}{\\varphi }_{i}\\left(a\\right)=\\frac{{f}_{i}\\left({F}_{i}\\left(a\\right)\\right)}{{h}_{i}\\left({F}_{i}\\left(a\\right)\\right)}.\\hfill \\end{array}$\n\nProof Let $I=\\left\\{1,2,\\dots ,p\\right\\}$, ${J}_{p}=\\left\\{p+j|j=1,2,\\dots ,m\\right\\}$, ${J}_{p}\\left(a\\right)=\\left\\{p+j|{g}_{j}\\left({G}_{j}\\left(a\\right)\\right)=0,j\\in \\left\\{1,2,\\dots ,m\\right\\}\\right\\}$.\n\nFor convenience, we define\n${l}_{k}\\left(x\\right)=\\left\\{\\begin{array}{cc}\\left(\\frac{{f}_{k}}{{h}_{k}}\\circ {F}_{k}\\right)\\left(x\\right),\\hfill & k=1,2,\\dots ,p,\\hfill \\\\ \\left({g}_{k-p}\\circ {G}_{k-p}\\left(x\\right),\\hfill & k=p+1,\\dots ,p+m.\\hfill \\end{array}$\nSuppose that the following system has a solution:\n$d\\in cone\\left(C-a\\right),\\phantom{\\rule{2em}{0ex}}{\\pi }_{a}^{k}\\left(d\\right)<0,\\phantom{\\rule{2em}{0ex}}k\\in I\\cup {J}_{p}\\left(a\\right),$\n(3.2)\nwhere ${\\pi }_{a}^{k}\\left(d\\right)$ is given by\n${\\pi }_{a}^{k}\\left(d\\right)=\\left\\{\\begin{array}{cc}max\\left\\{{\\sum }_{k=1}^{p}{\\nu }_{k}{F}_{k}^{\\prime }\\left(a\\right)d|\\nu \\in {T}_{k}\\left(a\\right)\\right\\},\\hfill & k\\in I,\\hfill \\\\ max\\left\\{{\\sum }_{k-p=1}^{m}{w}_{k-p}{G}_{k-p}^{\\prime }\\left(a\\right)d|w\\in \\partial {g}_{k-p}\\left({G}_{k-p}\\left(a\\right)\\right)\\right\\},\\hfill & k\\in {J}_{p}\\left(a\\right).\\hfill \\end{array}$\nThen the system\n$d\\in cone\\left(C-a\\right),\\phantom{\\rule{2em}{0ex}}{\\left({l}_{k}\\right)}_{+}^{\\prime }\\left(a;d\\right)<0,\\phantom{\\rule{2em}{0ex}}k\\in I\\cup {J}_{p}\\left(a\\right)$\n\nhas a solution. So, there exists ${\\alpha }_{1}>0$ such that $a+\\alpha d\\in C$, ${l}_{k}\\left(a+\\alpha d\\right)<{l}_{k}\\left(a\\right)$, $k\\in I\\cup {J}_{p}\\left(a\\right)$, whenever $0<\\alpha \\leqq {\\alpha }_{1}$. Since ${l}_{k}\\left(a\\right)<0$ for $k\\in {J}_{p}\\setminus {J}_{p}\\left(a\\right)$ and ${l}_{k}$ is continuous in a neighbourhood of a, there exists ${\\alpha }_{2}>0$ such that ${l}_{k}\\left(a+\\alpha d\\right)<0$, whenever $0<\\alpha \\leqq {\\alpha }_{2}$, $k\\in {J}_{p}\\setminus {J}_{p}\\left(a\\right)$. Let ${\\alpha }^{\\ast }=min\\left\\{{\\alpha }_{1},{\\alpha }_{2}\\right\\}$. Then $a+\\alpha d$ is a feasible solution for (P) and ${l}_{k}\\left(a+\\alpha d\\right)<{l}_{k}\\left(a\\right)$, $k\\in I$ for sufficiently small α such that $0<\\alpha \\leqq {\\alpha }^{\\ast }$.\n\nThis contradicts the fact that a is a weakly efficient solution for (P). Hence (3.2) has no solution.\n\nSince, for each k, ${\\pi }_{a}^{k}\\left(\\cdot \\right)$ is sublinear and $cone\\left(C-a\\right)$ is convex, it follows from a separation theorem [12, 14] that there exist ${\\lambda }_{i}\\geqq 0$, $i=1,\\dots ,p$, ${\\mu }_{j}\\geqq 0$, $j\\in {J}_{p}\\left(a\\right)$, not all zero, such that\n$\\sum _{i=1}^{p}{\\lambda }_{i}{\\pi }_{a}^{i}\\left(x\\right)+\\sum _{j\\in {J}_{p}\\left(a\\right)}{\\mu }_{j}{\\pi }_{a}^{j}\\left(x\\right)\\geqq 0,\\phantom{\\rule{1em}{0ex}}\\mathrm{\\forall }x\\in cone\\left(C-a\\right).$\nThen, by applying standard arguments of convex analysis (see [15, 16]) and choosing ${\\mu }_{j}=0$ whenever $j\\in {J}_{p}\\setminus {J}_{p}\\left(a\\right)$, we have\n$0\\in \\sum _{i=1}^{p}{\\lambda }_{i}\\partial {\\pi }_{a}^{i}\\left(0\\right)+\\sum _{j=1}^{m}{\\mu }_{j}\\partial {\\pi }_{a}^{j+p}\\left(0\\right)-{\\left(C-a\\right)}^{+}.$\nSo, there exist ${\\nu }_{i}\\in {T}_{i}\\left(a\\right)$, ${w}_{j}\\in \\partial {g}_{j}\\left({G}_{j}\\left(a\\right)\\right)$ satisfying\n$\\sum _{i=1}^{p}{\\lambda }_{i}{\\nu }_{i}^{T}{F}_{i}^{\\prime }\\left(a\\right)+\\sum _{j=1}^{m}{\\mu }_{j}{w}_{j}^{T}{G}_{j}^{\\prime }\\left(a\\right)\\in {\\left(C-a\\right)}^{+}.$\n\nHence, the conclusion holds. □\n\nUnder the following generalized Slater condition, we do the following:\n$\\mathrm{\\exists }{x}_{0}\\in cone\\left(C-a\\right),\\phantom{\\rule{1em}{0ex}}{\\mu }^{T}{G}_{j}^{\\prime }\\left(a\\right){x}_{0}<0,\\phantom{\\rule{1em}{0ex}}\\mathrm{\\forall }\\mu \\in \\partial {g}_{j}\\left({G}_{j}\\left(a\\right)\\right),\\mathrm{\\forall }j\\in J\\left(a\\right),$\n\nwhere $J\\left(a\\right)=\\left\\{j|{g}_{j}\\left({G}_{j}\\left(a\\right)\\right)=0,j=1,\\dots ,m\\right\\}$.\n\nChoosing $q\\in {\\mathbb{R}}^{p}$, $q>0$ with ${\\lambda }^{T}q=1$ and defining $\\mathrm{\\Lambda }=q{q}^{T}$, we can select the multipliers $\\overline{\\lambda }=\\mathrm{\\Lambda }\\lambda =q{q}^{T}\\lambda =q>0$ and $\\overline{\\mu }=\\mathrm{\\Lambda }\\mu =q{q}^{T}\\mu \\geqq 0$. Hence, the following Kuhn-Tucker type optimality conditions (KT) for (P) are obtained:\n$\\begin{array}{ll}\\left(\\text{KT}\\right)& \\overline{\\lambda }\\in {\\mathbb{R}}^{p},{\\overline{\\lambda }}_{i}>0,\\overline{\\mu }\\in {\\mathbb{R}}^{m},{\\overline{\\mu }}_{j}\\geqq 0,{\\overline{\\mu }}_{j}{g}_{j}\\left({G}_{j}\\left(a\\right)\\right)=0,\\\\ 0\\in \\sum _{i=1}^{p}{\\overline{\\lambda }}_{i}{T}_{i}\\left(a\\right){F}_{i}^{\\prime }\\left(a\\right)+\\sum _{j=1}^{m}{\\overline{\\mu }}_{j}\\partial {g}_{j}\\left({G}_{j}\\left(a\\right)\\right){G}_{j}^{\\prime }\\left(a\\right)-{\\left(C-a\\right)}^{+},\\\\ {T}_{i}\\left(a\\right)=\\frac{\\partial {f}_{i}\\left({F}_{i}\\left(a\\right)\\right)-{\\varphi }_{i}\\left(a\\right)\\partial {h}_{i}\\left({F}_{i}\\left(a\\right)\\right)}{{h}_{i}\\left({F}_{i}\\left(a\\right)\\right)},\\phantom{\\rule{1em}{0ex}}{\\varphi }_{i}\\left(a\\right)=\\frac{{f}_{i}\\left({F}_{i}\\left(a\\right)\\right)}{{h}_{i}\\left({F}_{i}\\left(a\\right)\\right)}.\\end{array}$\n\nWe present new conditions under which the optimality conditions (KT) become sufficient for weakly efficient solutions.\n\nThe following null space condition is as in :\n\nLet $x,a\\in X$. Define $K:X\\to {\\mathbb{R}}^{n\\left(p+m\\right)}:=\\pi {\\mathbb{R}}^{n}$ by $K\\left(x\\right)=\\left({F}_{1}\\left(x\\right),\\dots ,{F}_{p}\\left(x\\right),{G}_{1}\\left(x\\right),\\dots ,{G}_{m}\\left(x\\right)\\right)$. For each $x,a\\in X$, the linear mapping ${A}_{x,a}:X\\to {\\mathbb{R}}^{n\\left(p+m\\right)}$ is given by\n${A}_{x,a}\\left(y\\right)=\\left({\\alpha }_{1}\\left(x,a\\right){F}_{1}^{\\prime }\\left(a\\right)y,\\dots ,{\\alpha }_{p}\\left(x,a\\right){F}_{p}^{\\prime }\\left(a\\right)y,{\\beta }_{1}\\left(x,a\\right){G}_{1}^{\\prime }\\left(a\\right)y,\\dots ,{\\beta }_{m}\\left(x,a\\right){G}_{m}^{\\prime }\\left(a\\right)y\\right),$\n\nwhere ${\\alpha }_{i}\\left(x,a\\right)$, $i=1,2,\\dots ,p$ and ${\\beta }_{j}\\left(x,a\\right)$, $j=1,2,\\dots ,m$, are real positive constants. Let us denote the null space of a function H by $N\\left[H\\right]$.\n\nRecall, from the generalized Farkas lemma , that $K\\left(x\\right)-K\\left(a\\right)\\in {A}_{x,a}\\left(X\\right)$ if and only if ${A}_{x,a}^{T}\\left(u\\right)=0⇒{u}^{T}\\left(K\\left(x\\right)-K\\left(a\\right)\\right)=0$. This observation prompts us to define the following general null space condition:\n\nFor each $x,a\\in X$, there exist real constants ${\\alpha }_{i}\\left(x,a\\right)>0$, $i=1,2,\\dots ,p$, and ${\\beta }_{j}\\left(x,a\\right)>0$, $j=1,2,\\dots ,m$, such that\n$N\\left[{A}_{x,a}\\right]\\subset N\\left[K\\left(x\\right)-K\\left(a\\right)\\right],$\n(NC)\nwhere\n${A}_{x,a}\\left(y\\right)=\\left({\\alpha }_{1}\\left(x,a\\right){F}_{1}^{\\prime }\\left(a\\right)y,\\dots ,{\\alpha }_{p}\\left(x,a\\right){F}_{p}^{\\prime }\\left(a\\right)y,{\\beta }_{1}\\left(x,a\\right){G}_{1}^{\\prime }\\left(a\\right)y,\\dots ,{\\beta }_{m}\\left(x,a\\right){G}_{m}^{\\prime }\\left(a\\right)y\\right).$\n\nEquivalently, the null space condition means that for each $x,a\\in X$, there exist real constants ${\\alpha }_{i}\\left(x,a\\right)>0$, $i=1,2,\\dots ,p$, and ${\\beta }_{j}\\left(x,a\\right)>0$, $i=1,2,\\dots ,m$, and $\\zeta \\left(x,a\\right)\\in X$ such that ${F}_{i}\\left(x\\right)-{F}_{i}\\left(a\\right)={\\alpha }_{i}\\left(x,a\\right){F}_{i}^{\\prime }\\left(a\\right)\\zeta \\left(x,a\\right)$ and ${G}_{j}\\left(x\\right)-{G}_{j}\\left(a\\right)={\\beta }_{j}\\left(x,a\\right){G}_{j}^{\\prime }\\left(a\\right)\\zeta \\left(x,a\\right)$. For our problem (P), we assume the following generalized null space condition for invex function (GNCI):\n\nFor each $x,a\\in C$, there exist real constants ${\\alpha }_{i}\\left(x,a\\right)>0$, $i=1,2,\\dots ,p$, and ${\\beta }_{j}\\left(x,a\\right)>0$, $j=1,2,\\dots ,m$, and $\\zeta \\left(x,a\\right)\\in \\left(C-a\\right)$ such that $\\eta \\left({F}_{i}\\left(x\\right),{F}_{i}\\left(a\\right)\\right)={\\alpha }_{i}\\left(x,a\\right){F}_{i}^{\\prime }\\left(a\\right)\\zeta \\left(x,a\\right)$ and $\\eta \\left({G}_{j}\\left(x\\right),{G}_{j}\\left(a\\right)\\right)={\\beta }_{j}\\left(x,a\\right){G}_{j}^{\\prime }\\left(a\\right)\\zeta \\left(x,a\\right)$.\n\nNote that when $C=X$ and $\\eta \\left({F}_{i}\\left(x\\right),{F}_{i}\\left(a\\right)\\right)={F}_{i}\\left(x\\right)-{F}_{i}\\left(a\\right)$ and $\\eta \\left({G}_{j}\\left(x\\right),{G}_{j}\\left(a\\right)\\right)={G}_{j}\\left(x\\right)-{G}_{j}\\left(a\\right)$, the generalized null space condition for invex function (GNCI) reduces to (NC).\n\nTheorem 3.2 (Sufficient optimality conditions)\n\nFor the problem (P), assume that ${f}_{i}$, $-{h}_{i}$ and ${g}_{j}$ are invex functions and ${F}_{i}$ and ${G}_{j}$ are locally Lipschitz and Gâteaux differentiable functions. Let u be feasible for (P). Suppose that the optimality conditions (KT) hold at u. If (GNCI) holds at each feasible point x of (P), then u is a weakly efficient solution of (P).\n\nProof From the optimality conditions (KT), there exist ${\\nu }_{i}\\in {T}_{i}\\left(u\\right)$, ${w}_{j}\\in \\partial {g}_{j}\\left({G}_{j}\\left(u\\right)\\right)$ such that\n$\\sum _{i=1}^{p}{\\lambda }_{i}{\\nu }_{i}^{T}{F}_{i}^{\\prime }\\left(u\\right)+\\sum _{j=1}^{m}{\\mu }_{j}{w}_{j}^{T}{G}_{j}^{\\prime }\\left(u\\right)\\in {\\left(C-u\\right)}^{+},\\phantom{\\rule{2em}{0ex}}{\\mu }_{j}{g}_{j}\\left({G}_{j}\\left(u\\right)\\right)=0.$\nSuppose that u is not a weakly efficient solution of (P). Then there exists a feasible $x\\in C$ for (P) with\n$\\frac{{f}_{i}\\left({F}_{i}\\left(x\\right)\\right)}{{h}_{i}\\left({F}_{i}\\left(x\\right)}<\\frac{{f}_{i}\\left({F}_{i}\\left(u\\right)\\right)}{{h}_{i}\\left({F}_{i}\\left(u\\right)\\right)},\\phantom{\\rule{1em}{0ex}}i=1,2,\\dots ,p.$\nBy (GNCI), there exists $\\zeta \\left(x,u\\right)\\in \\left(C-u\\right)$, same for each ${F}_{i}$ and ${G}_{j}$, such that $\\eta \\left({F}_{i}\\left(x\\right),{F}_{i}\\left(u\\right)\\right)={\\alpha }_{i}\\left(x,u\\right){F}_{i}^{\\prime }\\left(u\\right)\\zeta \\left(x,u\\right)$, $i=1,2,\\dots ,p$, and $\\eta \\left({G}_{j}\\left(x\\right),{G}_{j}\\left(u\\right)\\right)={\\beta }_{j}\\left(x,u\\right){G}_{j}^{\\prime }\\left(u\\right)\\zeta \\left(x,u\\right)$, $j=1,2,\\dots ,m$. Hence\n$\\begin{array}{rcl}0& \\geqq & \\sum _{j=1}^{m}\\frac{{\\mu }_{j}}{{\\beta }_{j}\\left(x,u\\right)}\\left({g}_{j}\\left({G}_{j}\\left(x\\right)\\right)-{g}_{j}\\left({G}_{j}\\left(u\\right)\\right)\\right)\\phantom{\\rule{1em}{0ex}}\\text{(by feasibility)}\\\\ \\geqq & \\sum _{j=1}^{m}\\frac{{\\mu }_{j}}{{\\beta }_{j}\\left(x,u\\right)}{w}_{j}^{T}\\eta \\left({G}_{j}\\left(x\\right),{G}_{j}\\left(u\\right)\\right)\\phantom{\\rule{1em}{0ex}}\\text{(by subdifferentiability)}\\\\ =& \\sum _{j=1}^{m}{\\mu }_{j}{w}_{j}^{T}{G}_{j}^{\\prime }\\left(u\\right)\\zeta \\left(x,u\\right)\\phantom{\\rule{1em}{0ex}}\\text{(by (GNCI))}\\\\ \\geqq & -\\sum _{j=1}^{m}{\\lambda }_{i}{\\nu }_{i}^{T}{F}_{i}^{\\prime }\\left(u\\right)\\zeta \\left(x,u\\right)\\phantom{\\rule{1em}{0ex}}\\text{(by a hypothesis)}\\\\ =& -\\sum _{i=1}^{p}\\frac{{\\lambda }_{i}}{{\\alpha }_{i}\\left(x,u\\right)}{\\nu }_{i}^{T}\\eta \\left({F}_{i}\\left(x\\right),{F}_{i}\\left(u\\right)\\right)\\phantom{\\rule{1em}{0ex}}\\text{(by (GNCI))}\\\\ =& -\\sum _{i=1}^{p}\\frac{{\\lambda }_{i}}{{\\alpha }_{i}\\left(x,u\\right)}\\left(\\frac{{h}_{i}\\left({F}_{i}\\left(x\\right)\\right)}{{h}_{i}\\left({F}_{i}\\left(u\\right)\\right)}\\right)\\left(\\frac{{f}_{i}\\left({F}_{i}\\left(x\\right)\\right)}{{h}_{i}\\left({F}_{i}\\left(x\\right)\\right)}-\\frac{{f}_{i}\\left({F}_{i}\\left(u\\right)\\right)}{{h}_{i}\\left({F}_{i}\\left(u\\right)\\right)}\\right)\\phantom{\\rule{1em}{0ex}}\\text{(by subdifferentiability)}\\\\ >& 0.\\end{array}$\n\nThis is a contradiction and hence u is a weakly efficient solution for (P). □\n\n## 4 Duality theorems\n\nIn this section, we introduce a dual programming problem and establish weak, strong and converse duality theorems. Now we propose the following dual (D) to (P).\n$\\begin{array}{lll}\\left(\\text{D}\\right)& \\text{Maximize}& \\left(\\frac{{f}_{1}\\left({F}_{1}\\left(u\\right)\\right)}{{h}_{1}\\left({F}_{1}\\left(u\\right)\\right)},\\dots ,\\frac{{f}_{p}\\left({F}_{p}\\left(u\\right)\\right)}{{h}_{p}\\left({F}_{p}\\left(u\\right)\\right)}\\right)\\\\ \\text{subject to}& 0\\in \\sum _{i=1}^{p}{\\lambda }_{i}{\\nu }_{i}^{T}{F}_{i}^{\\prime }\\left(u\\right)+\\sum _{j=1}^{m}{\\mu }_{j}{w}_{j}^{T}{G}_{j}^{\\prime }\\left(u\\right)-{\\left(C-u\\right)}^{+},\\\\ {\\mu }_{j}{g}_{j}\\left({G}_{j}\\left(u\\right)\\right)\\geqq 0,\\phantom{\\rule{1em}{0ex}}j=1,2,\\dots ,m,\\\\ u\\in C,\\lambda \\in {\\mathbb{R}}^{p},{\\lambda }_{i}>0,{\\mu }_{j}\\in {\\mathbb{R}}^{m},{\\mu }_{j}\\geqq 0.\\end{array}$\n\nTheorem 4.1 (Weak duality)\n\nLet x be feasible for (P), and let $\\left(u,\\lambda ,\\mu \\right)$ be feasible for (D). Assume that (GNCI) holds with ${\\alpha }_{i}\\left(x,u\\right)={\\beta }_{j}\\left(x,u\\right)=1$. Moreover, ${f}_{i}$, $-{h}_{i}$ and ${g}_{j}$ are invex functions and ${F}_{i}$ and ${G}_{j}$ are locally Lipschitz and Gâteaux differentiable functions. Then\n${\\left(\\frac{{f}_{1}\\left({F}_{1}\\left(x\\right)\\right)}{{h}_{1}\\left({F}_{1}\\left(x\\right)\\right)},\\dots ,\\frac{{f}_{p}\\left({F}_{p}\\left(x\\right)\\right)}{{h}_{p}\\left({F}_{p}\\left(x\\right)\\right)}\\right)}^{T}-{\\left(\\frac{{f}_{1}\\left({F}_{1}\\left(u\\right)\\right)}{{h}_{1}\\left({F}_{1}\\left(u\\right)\\right)},\\dots ,\\frac{{f}_{p}\\left({F}_{p}\\left(u\\right)\\right)}{{h}_{p}\\left({F}_{p}\\left(u\\right)\\right)}\\right)}^{T}\\notin -{{\\mathbb{R}}^{p}}_{+}\\setminus \\left\\{0\\right\\}.$\nProof Since $\\left(u,\\lambda ,\\mu \\right)$ is feasible for (D), there exist ${\\lambda }_{i}>0$, ${\\mu }_{j}\\geqq 0$, ${\\nu }_{i}\\in {T}_{i}\\left(u\\right)$, $i=1,2,\\dots ,p$, ${w}_{j}\\in \\partial {g}_{j}\\left({G}_{j}\\left(u\\right)\\right)$, $j=1,2,\\dots ,m$, satisfying ${\\mu }_{j}{g}_{j}\\left({G}_{j}\\left(u\\right)\\right)\\geqq 0$ for $j=1,2,\\dots ,m$ and\n$\\sum _{i=1}^{p}{\\lambda }_{i}{\\nu }_{i}^{T}{F}_{i}^{\\prime }\\left(u\\right)+\\sum _{j=1}^{m}{\\mu }_{j}{w}_{j}^{T}{G}_{j}^{\\prime }\\left(u\\right)\\in {\\left(C-u\\right)}^{+}.$\nSuppose that $x\\ne u$ and\n${\\left(\\frac{{f}_{1}\\left({F}_{1}\\left(x\\right)\\right)}{{h}_{1}\\left({F}_{1}\\left(x\\right)\\right)},\\dots ,\\frac{{f}_{p}\\left({F}_{p}\\left(x\\right)\\right)}{{h}_{p}\\left({F}_{p}\\left(x\\right)\\right)}\\right)}^{T}-{\\left(\\frac{{f}_{1}\\left({F}_{1}\\left(u\\right)\\right)}{{h}_{1}\\left({F}_{1}\\left(u\\right)\\right)},\\dots ,\\frac{{f}_{p}\\left({F}_{p}\\left(u\\right)\\right)}{{h}_{p}\\left({F}_{p}\\left(u\\right)\\right)}\\right)}^{T}\\in -{{\\mathbb{R}}^{p}}_{+}\\setminus \\left\\{0\\right\\}.$\nThen\n$0>\\frac{{f}_{i}\\left({F}_{i}\\left(x\\right)\\right)}{{h}_{i}\\left({F}_{i}\\left(x\\right)\\right)}-\\frac{{f}_{i}\\left({F}_{i}\\left(u\\right)\\right)}{{h}_{i}\\left({F}_{i}\\left(u\\right)\\right)}.$\nBy the invexity of ${f}_{i}$ and $-{h}_{i}$, we have\nsince $\\frac{{h}_{i}\\left({F}_{i}\\left(u\\right)\\right)}{{h}_{i}\\left({F}_{i}\\left(x\\right)\\right)}>0$ and ${\\lambda }_{i}>0$, then\n$\\sum _{i=1}^{p}{\\lambda }_{i}{\\nu }_{i}^{T}{F}_{i}^{\\prime }\\left(u\\right)\\zeta \\left(x,u\\right)<0.$\n(4.1)\nFrom the feasibility conditions, we get ${\\mu }_{j}{g}_{j}\\left({G}_{j}\\left(x\\right)\\right)\\leqq 0$, ${\\mu }_{j}{g}_{j}\\left({G}_{j}\\left(u\\right)\\right)\\geqq 0$, and so\n$\\sum _{j=1}^{m}\\frac{{\\mu }_{j}}{{\\beta }_{j}\\left(x,u\\right)}\\left({g}_{j}\\left({G}_{j}\\left(x\\right)\\right)-{g}_{j}\\left({G}_{j}\\left(u\\right)\\right)\\right)\\leqq 0.$\nSimilarly, by the invexity of ${g}_{j}$, positivity of ${\\beta }_{j}\\left(x,u\\right)$ and by (GNCI), we have\n$\\sum _{j=1}^{m}{\\mu }_{j}{w}_{j}^{T}{G}_{j}^{\\prime }\\left(u\\right)\\zeta \\left(x,u\\right)\\leqq 0.$\n(4.2)\nBy (4.1) and (4.2), we get\n$\\left[\\sum _{i=1}^{p}{\\lambda }_{i}{\\nu }_{i}^{T}{F}_{i}^{\\prime }\\left(u\\right)+\\sum _{j=1}^{m}{\\mu }_{j}{w}_{j}^{T}{G}_{j}^{\\prime }\\left(u\\right)\\right]\\zeta \\left(x,u\\right)<0.$\n\nThis is a contradiction. The proof is completed by noting that when $x=u$ the conclusion trivially holds. □\n\nTheorem 4.2 (Strong duality)\n\nFor the problem (P), assume that the generalized Slater constraint qualification holds. If u is a weakly efficient solution for (P), then there exist $\\lambda \\in {\\mathbb{R}}^{p}$, ${\\lambda }_{i}>0$, $\\mu \\in {\\mathbb{R}}^{m}$, ${\\mu }_{j}\\geqq 0$ such that $\\left(u,\\lambda ,\\mu \\right)$ is a weakly efficient solution for (D).\n\nProof It follows from Theorem 3.1 that there exist $\\lambda \\in {\\mathbb{R}}^{p}$, ${\\lambda }_{i}>0$, $\\mu \\in {\\mathbb{R}}^{m}$, ${\\mu }_{j}\\geqq 0$ such that\n$\\begin{array}{c}0\\in \\sum _{i=1}^{p}{\\lambda }_{i}{T}_{i}\\left(u\\right){F}_{i}^{\\prime }\\left(u\\right)+\\sum _{j=1}^{m}{\\mu }_{j}\\partial {g}_{j}\\left({G}_{j}\\left(u\\right)\\right){G}_{j}^{\\prime }\\left(u\\right)-{\\left(C-u\\right)}^{+},\\hfill \\\\ {\\mu }_{j}{g}_{j}\\left({G}_{j}\\left(u\\right)\\right)=0,\\phantom{\\rule{1em}{0ex}}j=1,2,\\dots ,m.\\hfill \\end{array}$\nThen $\\left(u,\\lambda ,\\mu \\right)$ is a feasible solution for (D). By weak duality,\n${\\left(\\frac{{f}_{1}\\left({F}_{1}\\left(x\\right)\\right)}{{h}_{1}\\left({F}_{1}\\left(x\\right)\\right)},\\dots ,\\frac{{f}_{p}\\left({F}_{p}\\left(x\\right)\\right)}{{h}_{p}\\left({F}_{p}\\left(x\\right)\\right)}\\right)}^{T}-{\\left(\\frac{{f}_{1}\\left({F}_{1}\\left(u\\right)\\right)}{{h}_{1}\\left({F}_{1}\\left(u\\right)\\right)},\\dots ,\\frac{{f}_{p}\\left({F}_{p}\\left(u\\right)\\right)}{{h}_{p}\\left({F}_{p}\\left(u\\right)\\right)}\\right)}^{T}\\notin -{{\\mathbb{R}}^{p}}_{+}\\setminus \\left\\{0\\right\\}.$\n\nSince $\\left(u,\\lambda ,\\mu \\right)$ is a feasible solution for (D), $\\left(u,\\lambda ,\\mu \\right)$ is a weakly efficient solution for (D). Hence the result holds. □\n\nTheorem 4.3 (Converse duality)\n\nLet $\\left(u,\\lambda ,\\mu \\right)$ be a weakly efficient solution of (D), and let a be a feasible solution of (P). Assume that ${f}_{i}$, $-{h}_{i}$ and ${g}_{j}$ are invex functions and ${F}_{i}$ and ${G}_{j}$ are locally Lipschitz and Gâteaux differentiable functions. Moreover, (GNCI) holds with ${\\alpha }_{i}\\left(x,u\\right)={\\beta }_{j}\\left(x,u\\right)=1$. Then u is a weakly efficient solution of (P).\n\nProof Suppose, contrary to the result, that u is not a weakly efficient solution of (P). Then there exists $x\\in D$ such that\n$\\frac{{f}_{i}\\left({F}_{i}\\left(x\\right)\\right)}{{h}_{i}\\left({F}_{i}\\left(x\\right)\\right)}<\\frac{{f}_{i}\\left({F}_{i}\\left(u\\right)\\right)}{{h}_{i}\\left({F}_{i}\\left(u\\right)\\right)}.$\nSince ${f}_{i}$, $-{h}_{i}$ are invex functions, for each ${\\nu }_{i}\\in {T}_{i}\\left(x\\right)$, we have\n$0>\\frac{{h}_{i}\\left({F}_{i}\\left(u\\right)\\right)}{{h}_{i}\\left({F}_{i}\\left(x\\right)\\right)}{\\nu }_{i}^{T}\\eta \\left({F}_{i}\\left(x\\right),{F}_{i}\\left(u\\right)\\right).$\nSince $\\left(u,\\lambda ,\\mu \\right)$ are feasible for (P), we get\n(4.3)\nFrom the hypothesis ${\\mu }_{j}{g}_{j}\\left({G}_{j}\\left(x\\right)\\right)\\leqq {\\mu }_{j}{g}_{j}\\left({G}_{j}\\left(u\\right)\\right)$, ${g}_{j}$ is an invex function and for each ${w}_{j}\\in \\partial {g}_{j}\\left({G}_{j}\\left(x\\right)\\right)$, it follows that\nand ${\\mu }_{j}\\geqq 0$, $j=1,2,\\dots ,m$, then we have\n$\\sum _{j=1}^{m}{\\mu }_{j}{w}_{j}^{T}{G}_{j}^{\\prime }\\left(u\\right)\\zeta \\left(x,u\\right)\\leqq 0.$\n(4.4)\nFrom (4.3) and (4.4), we get\n$\\left[\\sum _{i=1}^{p}{\\lambda }_{i}{\\nu }_{i}^{T}{F}_{i}^{\\prime }\\left(u\\right)+\\sum _{j=1}^{m}{\\mu }_{j}{w}_{j}^{T}{G}_{j}^{\\prime }\\left(u\\right)\\right]\\zeta \\left(x,u\\right)<0.$\n\nThis is a contradiction. □\n\n## Declarations\n\n### Acknowledgements\n\nThis work was supported by a research grant of Pukyong National University (2013). The authors wish to thank the anonymous referees for their suggestions and comments.\n\n## Authors’ Affiliations\n\n(1)\nDepartment of Applied Mathematics, Pukyong National University, Busan, 608-737, Republic of Korea\n\n## References", null, "" ]

[ null, "https://journalofinequalitiesandapplications.springeropen.com/track/article/10.1186/1029-242X-2013-508", null ]

[ "# How to compare classification methods in terms of performance?\n\nI'd like to compare logistic regression to classification trees. In a first step, I compared the theoretical framework of the two classifiers. In a second step, I compared the performance using a rather unbalanced data set containing two classes. I therefore compared confusion-matrices, balanced-accuracy, sensitivity and specificity. Moreover, I compared the ROC curves and derived therefrom the AUC values. Are there any other value adding measures to compare the classifiers? How would you compare the efficiency in terms of running time?\n\nYou seem to be using a mish-mash of methods. Focus on getting predicted risks and using proper accuracy scores such as the Brier score or logarithmic scoring rule (log likelihood; related to pseudo $R^2$). Things started going south when you chose to use classifiers rather than predictors. And note that regression trees are highly unstable when the sample size is $<100,000$ subjects for example. That's why people use bagging, boosting, and random forests instead of single trees.\nProper accuracy scores are not destroyed by imbalanced $Y$.\n• The concordance probability (AUROC) is less sensitive. Think of a pair of predicted probabilities 0.2 and 0.8 corresponding to a non-event and an event. Then think of a pair 0.1 and 0.9. Both get one point in the concordance calculation even though the second goes out on a limb. You can also think about this from the standpoint that the concordance probability is essentially the Wilcoxon-Mann-Whitney statistic for comparing two groups ($Y=0, Y=1$). Comparing 2 $c$-indexes is essentially the same as subtracting one Wilcoxon stat from another, which no one does. – Frank Harrell Dec 31 '15 at 21:06" ]

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[ "# 1277344 (number)\n\n1,277,344 (one million two hundred seventy-seven thousand three hundred forty-four) is an even seven-digits composite number following 1277343 and preceding 1277345. In scientific notation, it is written as 1.277344 × 106. The sum of its digits is 28. It has a total of 7 prime factors and 24 positive divisors. There are 632,256 positive integers (up to 1277344) that are relatively prime to 1277344.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Even\n• Number length 7\n• Sum of Digits 28\n• Digital Root 1\n\n## Name\n\nShort name 1 million 277 thousand 344 one million two hundred seventy-seven thousand three hundred forty-four\n\n## Notation\n\nScientific notation 1.277344 × 106 1.277344 × 106\n\n## Prime Factorization of 1277344\n\nPrime Factorization 25 × 179 × 223\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 7 Total number of prime factors rad(n) 79834 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 1,277,344 is 25 × 179 × 223. Since it has a total of 7 prime factors, 1,277,344 is a composite number.\n\n## Divisors of 1277344\n\n24 divisors\n\n Even divisors 20 4 2 2\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 24 Total number of the positive divisors of n σ(n) 2.54016e+06 Sum of all the positive divisors of n s(n) 1.26282e+06 Sum of the proper positive divisors of n A(n) 105840 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1130.2 Returns the nth root of the product of n divisors H(n) 12.0686 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 1,277,344 can be divided by 24 positive divisors (out of which 20 are even, and 4 are odd). The sum of these divisors (counting 1,277,344) is 2,540,160, the average is 105,840.\n\n## Other Arithmetic Functions (n = 1277344)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 632256 Total number of positive integers not greater than n that are coprime to n λ(n) 158064 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 98187 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares\n\nThere are 632,256 positive integers (less than 1,277,344) that are coprime with 1,277,344. And there are approximately 98,187 prime numbers less than or equal to 1,277,344.\n\n## Divisibility of 1277344\n\n m n mod m 2 3 4 5 6 7 8 9 0 1 0 4 4 5 0 1\n\nThe number 1,277,344 is divisible by 2, 4 and 8.\n\n## Classification of 1277344\n\n• Arithmetic\n• Deficient\n\n### Expressible via specific sums\n\n• Polite\n• Non-hypotenuse\n\n## Base conversion (1277344)\n\nBase System Value\n2 Binary 100110111110110100000\n3 Ternary 2101220012001\n4 Quaternary 10313312200\n5 Quinary 311333334\n6 Senary 43213344\n8 Octal 4676640\n10 Decimal 1277344\n12 Duodecimal 517254\n20 Vigesimal 7jd74\n36 Base36 rdls\n\n## Basic calculations (n = 1277344)\n\n### Multiplication\n\nn×y\n n×2 2554688 3832032 5109376 6386720\n\n### Division\n\nn÷y\n n÷2 638672 425781 319336 255469\n\n### Exponentiation\n\nny\n n2 1631607694336 2084124298713923584 2662143668216438006480896 3400473241734257788950333620224\n\n### Nth Root\n\ny√n\n 2√n 1130.2 108.502 33.6184 16.6441\n\n## 1277344 as geometric shapes\n\n### Circle\n\n Diameter 2.55469e+06 8.02579e+06 5.12585e+12\n\n### Sphere\n\n Volume 8.72996e+18 2.05034e+13 8.02579e+06\n\n### Square\n\nLength = n\n Perimeter 5.10938e+06 1.63161e+12 1.80644e+06\n\n### Cube\n\nLength = n\n Surface area 9.78965e+12 2.08412e+18 2.21242e+06\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 3.83203e+06 7.06507e+11 1.10621e+06\n\n### Triangular Pyramid\n\nLength = n\n Surface area 2.82603e+12 2.45616e+17 1.04295e+06" ]

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[ "# Sides of Similar Triangles Worksheets\n\nWhat is Related Between Sides of Similar Triangles? There are two concepts in geometry that are pretty common and taught in early grades. They are congruency and similarity, and they can be applied to any shape. Congruency means that two shapes should be exactly equal in every aspect that are: shape, size, and structure. Whereas in similarity, only the shape needs to be similar. Both shapes could differ in size or dimensions if they are similar. If we talk about triangles, two similar triangles can assist in solving each other’s dimensions and angles. When we are talking about two similar triangles, each of their sides are similar to each other as well. It's just that they can differ in their measurements. Usually, the sides of similar triangles are a factor of each other, and they differ in a certain ratio. For example, if one triangle has a side of three centimeters, then that very side in the other triangle would be six or nine centimeters. So if we find out that differentiating factor between them, we can solve for the measurements of each side of both the triangles. Also, each angle in two similar triangles is equal to its corresponding angle. So, there is no need for actually finding the value of an angle in either of the triangles if its corresponding angle's value in the other triangle is known.\n\n• ### Basic Lesson\n\nIntroduces students to corresponding angles and proportions of similar triangles. Identical triangles are similar. Mathematically, similar triangles have both corresponding angles equal, while the lengths of the corresponding sides are in proportion.\n\n• ### Intermediate Lesson\n\nGuides students through writing equal ratios to corresponding sides of similar triangles.\n\n• ### Independent Practice 1\n\nA really great activity for allowing students to understand the concepts of similar triangles.\n\n• ### Independent Practice 2\n\nStudents tackle 20 similar triangles problems. The answers can be found below.\n\n• ### Homework Worksheet\n\nStudents are provided with 12 problems to understand the concepts of Sides of Similar Triangles.\n\n• ### Skill Quiz\n\nThis tests the students ability to understand similar figures." ]

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[ "# ANGULAR VELOCITY-DIFFERENCE BETWEEN W×R AND R×W\n\n## 1/12/20\n\n ANGULAR VELOCITY\nANGULAR VELOCITY is the Physical quantity derived from base physical quantity defined as - \" Angular Velocity is rate at which angular position of a body changes w.r.t chosen point centre \"\n\nBy convention positive angular velocity indicates counterclockwise rotation why negative is clock wise.\n\nConventionally angular velocity as specified by Right Hand Thumb Rule.\n\nThe angular velocity of particle with respect to origin is Defined by perpendicular component of velocity vector v.\n\nAngular velocity determined by formulae\n\nWhere , V prependicular is cross radial component of v .\n\nHence, According to Right hand thumb rule\n\nAccording to left Hand Thumb Rule" ]

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[ "# [Wolfram Mathematica] Using Newton's method to solve non-linear system\n\nHi. This is not actually not part of the homework; but it's something I'd like to do.\nI have to solve the following system using Newton-Raphson's method:\n\n$$\\begin{matrix} \\frac{X}{\\mu }+Y=1 \\\\ X=\\left( \\lambda -\\left( K-1 \\right)X \\right)Y \\\\ \\end{matrix}$$\n\nSurfing the internet I found this precious Wolfram Demonstration:\nhttp://demonstrations.wolfram.com/IterationsOfNewtonsMethodForTwoNonlinearEquations/\n\nwhich is an interactive plot where you can select the starting point and see how and where it converges.\n\nSo I took the code and put those functions of mine:\n\nCode:\nManipulate[\nQuiet@DynamicModule[{sol},\nsol = Prepend[\nReap[FindRoot[f, {vars, start}\\[Transpose],\nStepMonitor :> Sow[vars]]][[2, 1]], start];\nShow[g1, g2,\nGraphics[{PointSize[Medium], Point[sol], Thickness[Medium],\nLine[sol], Style[Text[start, {-20, 20}, {-1, 1}], 12]}]]],\n{{start, {3.7, 5.7}}, Locator},\nSaveDefinitions -> True,\nSynchronousInitialization -> False,\nTrackedSymbols -> True,\nInitialization :> {\nf = {x + y - 1, x - y*(0.99 - 4*x) },\nvars = {x, y},\nroot = vars /. FindRoot[f, {vars, {1, 1}}\\[Transpose]],\npp = {},\nDo[sol = vars /. FindRoot[f, {vars, {x0, y0}}\\[Transpose]];\nIf[Norm[sol - root] < 10^-7, pp = {pp, {x0, y0}}], {x0, -20,\n20, .12}, {y0, -20, 20, .12}],\ng1 = ListPlot[Partition[Flatten[pp], 2], AspectRatio -> Automatic,\nPlotStyle -> PointSize[Tiny],\nPlotRange -> {{-20, 20}, {-20, 20}},\nTicks -> {Range[-20, 20], Range[-20, 20]},\nImageSize -> {480, 480}],\ng2 = ContourPlot[\nEvaluate[Thread[f == 0]], {x, -20, 20}, {y, -20, 20},\nContourStyle -> {{Thick, Green}, {Thick, Orange}}]}]\nIt uses the FindRoot command which is supposed to use Newton's method but it actually does not purely. I checked this because I wrote the method on Octave and the region of convergence is the following:", null, "As you can see it's \"solid\" whereas in the Wolfram Demonstration you can see stripes so FindRoot clearly does not use Newton's method.\n\nSo now I want to replace that FindRoot command to match the existing code but using the pure Newton's method. I just don't know where to start because I barely understand that code.\nSo if someone is a pro in Wolfram I would really appreciate him/her !!\n\n#### Attachments\n\n• 15.7 KB Views: 1,646\n• 1.5 MB Views: 248\n\nRelated Engineering and Comp Sci Homework Help News on Phys.org\nI think what you want to replace is the\n\nCode:\nFindRoot[f, {vars, start}\\[Transpose], StepMonitor :> Sow[vars]]\nThe StepMonitor:>Sow[vars] is going to put the result of each successive iteration \"into the bag\" and when finished Reap[][[2,1]] around this will extract the contents of the bag for Show to then display.\n\nSo if you replace all of that FindRoot[] with your code that uses Newton and puts successive {x,y} points into the bag with Sow[vars] then you should be close to what you want.\n\nYou could start with something that just uses a For[Sow[{x,y}],{x,1,3},{y,1,4}]] instead of the FindRoot[] to put points in the bag to verify this much of the process. Then you could replace that with more of your Newton.\n\nThere are two additional FindRoot[] down further in the code. It looks like those are being used for graphics showing either the beginning or the end of the iterative process. It doesn't look like those are absolutely critical to what you are trying to do. You might try gently commenting those out of the code and see how the graphics change.\n\nThe Demonstrations project often provide something that looks cute, but all the stuff they have to shovel in to create the interactive graphical result almost always makes this very complicated and difficult to reverse engineer for even a slightly different purpose. This is very different from providing brilliantly simple examples to show how algorithms can be implemented and that you can easily see how to incorporate into what you need to do. But cute graphics are what many people seem to think they want to see.\n\nAlmost all the internal algorithms used for things like FindRoot[] are very complicated and not adequately documented for the user to be able to tell exactly how they work. Many of these supposedly take dozens or even hundreds of pages of code to implement. There isn't even documentation of the available arguments \"somemethod\" that can be used with Method->\"somemethod\" or even a rough description of what each somemethod might do, you have to guess from the name and do trial and error experimentation." ]

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[ "## Pages\n\n`“Life is like riding a bicycle. To keep your balance you must keep moving.”–Albert Einstein`\n\n## Thursday, March 27, 2008\n\n### AP Physics B – Answer to Free-Response Question (for practice) involving Compton Scattering", null, "A free response question involving Compton Scattering was posted on 24th March 2008 for your practice. This was the question:\n\nAn X-ray photon of energy 31000 eV proceeds in the positive X-direction. It collides with a free electron at rest. After the collision the electron moves in the positive X-direction with a momentum greater than that of the X-ray photon. In the process, the wave length of the X-ray photon is changed by an amount dλ = 0.0486 Ǻ.\n\n(a) What is the direction of motion of the photon after the collision? Justify your answer.\n\n(b) Will the wave length of the photon decrease or increase? Justify your answer.\n\n(c) Determine the wave length of the photon before collision.\n\n(d) Calculate the kinetic energy transferred (by the photon) to the electron.\n\n(a) The initial momentum of the photon electron system is the initial momentum of the photon which is in the positive X-direction. After the collision, the electron has a momentum in the X-direction. Since it is greater than the initial momentum of the photon, the final momentum of the photon must be in the negative X-direction so that the total final momentum of the system is equal to the initial momentum of the photon (in accordance with the law of conservation of momentum).\n\n(b) The energy of the photon is given by\n\nE = hc/λ where h is Planck’s constant, c is the speed of light in free space and λ is the wave length.\n\nThe wave length λ of the photon must be increased since the energy of the photon is decreased by transferring kinetic energy to the electron.\n\n(c) The product of the wave length in Angstrom and the energy in electron volt in the case of any photon is 12400. Therefore, wave length of the X-ray photon before collision is\n\nλ = 12400/31000 = 0.4 Ǻ\n\n[The wave length λ can be found also by using the expression for the photon energy, E =hc/λ, where h is Planck’s constant and c is the speed of light in free space. The energy E should be in joule: E = 31000×1.6×10–19 joule. The value of λ will then be in metre].\n\n(d) The initial energy of the photon = 31000 eV = 31000×1.6×10–19 joule = 4.96×10–15 J.\n\nThe final wave length of the photon is λ’ = (0.4 + 0.0486) Ǻ = 0.4486 Ǻ. Therefore, the final energy of the photon is hc/λ’ = (6.63×10–34×3×108)/(0.4486×10–10) = 4.4338×10–15 J.\n\nThe energy transferred by the photon to the electron = change in energy of the photon = (4.96×10–15 – 4.4338×10–15) J = 0.5262×10–15 J..\n\nI give you a multiple choice question related to Compton scattering. This question will be quite simple once you understand the answer to the free response qestion we discussed above. Here is the question:\n\nAn X-ray photon of wave length λ proceeding in the positive X-direction suffers a head-on collision with a free electron at rest and retraces its path. If the wave length of the photon increases to λ’ in the process, the momentum transferred (by the photon) to the electron is (h = Planck’s constant, c = speed f light in free space)\n\n(a) hc[(λ’ + λ)/ λλ]\n\n(b) hc[(λλ)/ λλ]\n\n(c) h[(λ’ + λ)/ λλ]\n\n(d) h[(λλ)/ λλ]\n\n(e) ) h[(λλ)\n\nThe initial momentum of the photon = h/λ\n\nThe final momentum of the photon = h/λ\n\nThe negative sign shows that the final momentum of the photon is in the negative X-direction.\n\nThe momentum transferred by the photon to the electron = change in momentum of the photon = [(h/λ – (–h/λ’) = h(1/λ + 1/λ’) = h[(λ’ + λ)/ λλ]" ]

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[ "# Net Of A Rectangular Prism\n\nby -1 views\n\nAll lengths are in feet 10 a Find the following side lengths for the net. Online calculators and formulas for a prism and other geometry problems.", null, "Rectangular Prism Opened Into Net Rectangular Prism Activities Bar Chart\n\n### 2 X 5.", null, "Net of a rectangular prism. A net of a polyhedron is a two-dimensional flattened out version of it. Net rectangular prism Title. Like other prisms two bases are congruent and parallel.\n\nThe net of a solid figure is formed when a solid figure is unfolded along its edges and its faces are laid out in a pattern in two dimensions. Geometric Nets For 3D Shapes. Nets of Solids – cubes cuboids rectangular solids prisms cylinders spheres cones pyramids net of solids What is meant by the net of a solid net of cylinder activities and demonstrations How to use nets to find surface area and volumes Interactive animations for nets of solids in video lessons with examples and step-by-step solutions.\n\nThe prism has 5 faces 9 sides and 6 vertices. In a rectangular prism any two pairs of congruent rectangles that are connected by a height can be used as the bases. Print out the pages below – ideally onto thicker paper or onto card – and cut out the shapes.\n\nA triangular prism is a polyhedron having two triangular bases and three rectangular sides. Use this printable graph paper generator to create grids that can help students draw their own nets. Nets of Triangular Prism.\n\nStart it with a sheet and watch it form into a neatly displayed rectangular prism in this 4th grade 5th grade and 6th grade pdf. It is also a cuboid. So this area right over here is going to be one half times the base so times 12 times the height times eight.\n\nFind the surface area Calculus. Each of the ten nets have square centimeters printed on each surface so students may quickly grasp the concept of surface area as the total number of units covered on all six surfaces. All of its angles are right angles.\n\nA rectangular prism and its net are shown below. Well in the net that corresponds to this area its a triangle it has a base of 12 and height of eight. A rectangular prism or cuboid is formed by folding a net as shown.\n\nBecause of its cross-section along the length it is said to be a prism. It can also be called a cuboid. A rectangular prism is a polyhedron with two congruent and parallel bases.\n\nNets of a Rectangular Prism. Use these to form 3-dimensional shapes by folding along the lines. The height is 3 ft.\n\nWhat is the formula for finding the surface area of a right rectangular prism. So this is the same thing as six times eight which is equal to 48 whatever units or square units. Geometry Nets – Rectangular Prisms A rectangular is a three dimensional shape with six rectangular shaped sides.\n\nSurface Area of a Rectangular Prism using Nets. We make the net by cutting the polyhedron along one or more of the edges until we can lay out the whole thing flat in a plane. 9122003 10311 PM.\n\nMicrosoft Word – net_rect_prismdoc Author. It has six faces and all the faces are in a rectangle shape and have twelve edges. Using a net to find the surface area of a rectangular prism.\n\nThere are different nets. Nets of rectangular prisms are made up of rectangles and squares. We can see from the net that there are two rectangles with dimensions 3 cm by 6 cm two rectangles with dimensions 2 cm by 6 cm and two rectangles with dimensions 2 cm by 3 cm.\n\nFoldable Net of a Rectangular Prism Snip the shape along the lines and glue the tabs one by one. In geometry a rectangular prism known by a number of names including cuboid is a convex polyhedron bounded by six quadrilateral faces whose polyhedral graph is the same as that of a cube. Explore the concept of surface area by folding the nets of cubes and rectangular prisms into 3-dimensional solids.\n\nThe base of a right rectangular prism is 10 ft by 7 ft. Ft D b Use the net to find the surface area of the prism. Calculator online for a rectangular prism.\n\nCreating shapes would never be this fun. Calculate the unknown defining surface areas lengths widths heights and volume of a rectangular prism with any 3 known variables. What is the surface area in square feet of a rectangular prism with dimensions of 2 feet by 3 feet by 4 feet as shown in the net.", null, "3d Shape Nets Teacherspayteachers Com 3d Shape 3d Shapes Nets Math Projects", null, "3d Shapes For Kids Shapes For Kids 3d Shapes For Kids Rectangular Prism", null, "Nets Of Rectangular Prism And Square Pyramid From Cpalms Site Mathematics Images Shapes Worksheets Rectangular Prism", null, "Pin By Leslie Marrie Scroggs Lasater On Math Measurement Geometry And Data Teaching Mathematics Math Resources Math Foldables", null, "Solid 3d Shapes Worksheets 3d Shapes Worksheets Shapes Worksheets 3d Shapes Nets", null, "Cuboid Net Teaching Geometry 3d Shapes Worksheets Geometry", null, "3d Geometric Shapes Nets 3d Geometric Shapes Box Template Printable Geometric Shapes", null, "Matching Rectangular Prisms To Their Nets 2 Rectangular Prisms Math For 6th Graders Middle School Geometry", null, "Net Rectangular Prism Cuboid Rectangular Prism Rectangular Prisms Rectangular", null, "Net Twisted Rectangular Prism Rectangular Prism Prism Rectangular", null, "8 X 4 X 6 Net Rectangular Prism Basic Geometry Rectangular Prisms", null, "Matching Rectangular Prisms To Their Nets Rectangular Prisms Rectangular Prism Surface Area", null, "Rectangular Prism Nets Foldables Rectangular Prism Rectangular Prisms Everyday Math", null, "Paper Model Of A Rectangular Prism Cuboid Rectangular Prism Math Models Rectangular Prisms", null, "Triangular Prism And Net Diagram Of Prism With Faces Labeled A B C D And E Length 18 In Width 10 In Height 12 In Math Geometry Triangular Prism Face E", null, "Cuboid Net Teaching Geometry 3d Shapes Worksheets Geometry", null, "Paper Model Of An Oblique Rectangular Prism Rectangular Prism Paper Models Cardboard Sculpture", null, "3 D Shapes Truncated Square Based Pyramid Tabs Paper Models Paper Christmas Ornaments Free Paper Models\n\nREAD:   Do Book Titles Go In Quotes" ]

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[ "# GPE Hamiltonian\n\n## Description\n\nThe GPE Hamiltonian node defines the total energy of the system modelled by the Gross-Pitaevskii equation describing the mean field of Bose-Einstein condensates. It is the first building block to get the energy spectrum of a system.", null, "## Inputs\n\nThe node has the following inputs:\n\n• Potental (V): A potential function defined in the Potential node\n• Self-interaction ($\\beta$): A scalar quantity denoting the interaction strength of the condensate\n\n## Content\n\nThe content in the node shows what is being calculated. It is the sum of the kinetic and potential energy and a non-linear term describing the self-interaction strength.\n\n## Output\n\nThe node outputs the Hamiltonian based on the GPE\n\n## Example\n\nIn the example below, the GPE Hamiltonian has a harmonic oscillator potential as the potential function and the self-interaction strength of the system is equal to 1.", null, "" ]

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[ "#", null, "Two-dimensional example of the Atiyah-Singer index theorem for the Dirac operator\n\n+ 2 like - 0 dislike\n230 views\n\nI am trying to generalize the example of the Atiyah-Singer index theorem for the Dirac operator given in the appendix of http://xxx.lanl.gov/pdf/0802.0634v1.pdf.\n\nPlease consider the generalized two-dimensional $U(1)$ instanton configuration given by\n\n$$A={\\frac {m{n}^{2}{k}^{2}y{\\it dx}}{{x}^{2}{k}^{2}+{y}^{2}{n}^{2}+{c}^{2}{n}^{2}{k}^{2}}}-{\\frac {m{n}^{2}{k}^{2}x{\\it dy}}{{x}^{2}{k}^{2}+{y}^{2}{n}^{2}+{c}^{2}{n}^{2}{k}^{2}}}$$\n\nwhere $m$, $n$ and $k$ are integers.\n\nThe corresponding Yang-Mills field takes the form\n\n$$F= dA = -2\\,{\\frac {m{n}^{4}{k}^{4}{c}^{2}{\\it dx}\\wedge {\\it dy}}{ \\left( {x}^{2}{k}^{2}+{y}^{2}{n}^{2}+{c}^{2}{n}^{2}{k}^{2} \\right) ^{2}}}$$\n\nThen,  given that $F = \\frac{1}{2}F_{ij}dx^i \\wedge dx^j$ we have that\n\n$$F = \\frac{1}{2}F_{ij}dx^i \\wedge dx^j = \\frac{1}{2}F_{12}dx \\wedge dy +\\frac{1}{2}F_{21}dy \\wedge dx$$\n\n$$F = \\frac{1}{2}F_{12}dx \\wedge dy -\\frac{1}{2}F_{12}dy \\wedge dx$$\n\n$$F = \\frac{1}{2}F_{12}dx \\wedge dy +\\frac{1}{2}F_{12}dx \\wedge dy$$\n\n$$F = F_{12}dx \\wedge dy$$\n\nand then we obtain\n\n$$F_{12}dx \\wedge dy = -2\\,{\\frac {m{n}^{4}{k}^{4}{c}^{2}{\\it dx}\\wedge {\\it dy}}{ \\left( {x}^{2}{k}^{2}+{y}^{2}{n}^{2}+{c}^{2}{n}^{2}{k}^{2} \\right) ^{2}}}$$\n\nwhich implies that\n\n$$F_{12} = -2\\,{\\frac {m{n}^{4}{k}^{4}{c}^{2}}{ \\left( {x}^{2}{k}^{2}+{y}^{2}{n}^ {2}+{c}^{2}{n}^{2}{k}^{2} \\right) ^{2}}}$$\n\nThe instantonic number for such configuration is\n\n$$-\\frac{1}{2\\pi}\\int_{\\mathbb{R}^2}F=-\\frac{1}{2\\pi}\\int_{\\mathbb{R}^2}F_{12}dxdy$$\n\nwhich is reduced to\n\n$$-\\frac{1}{2\\pi}\\int_{\\mathbb{R}^2}F=-\\frac{1}{2\\pi}\\int _{-\\infty }^{\\infty }\\int _{-\\infty }^{\\infty } -2\\,{\\frac {m{n}^{4}{k}^{4}{c}^{2}}{ \\left( {x}^{2}{k}^{2}+{y}^{2}{n}^{2}+{c}^{2}{n}^{2}{k}^{2} \\right) ^{2}}}dxdy$$\n\nand the computation gives\n\n$$-\\frac{1}{2\\pi}\\int_{\\mathbb{R}^2}F= mkn$$\n\nNow, the Dirac operator in such background gauge field is given by\n\n$$\\not{D}=\\left[ \\begin {array}{cc} 0&\\partial_{{x}}+i{\\partial}_{{y}}-{\\frac {im{n}^{2}{k}^{2}y+m{n}^{2}{k}^{2}x}{{x}^{2}{k}^{2}+{y}^{2}{n}^{2}+{c}^{ 2}{n}^{2}{k}^{2}}}\\\\\\partial_{{x}}-i{\\partial}_{{y}}- {\\frac {im{n}^{2}{k}^{2}y-m{n}^{2}{k}^{2}x}{{x}^{2}{k}^{2}+{y}^{2}{n}^ {2}+{c}^{2}{n}^{2}{k}^{2}}}&0\\end {array} \\right]$$\n\nA positive chirality zero-mode satisfies\n\n$$\\not{D}\\left[ \\begin {array}{c} \\chi \\left( x,y \\right) \\\\0\\end {array} \\right]=0$$\n\nit is to say", null, "with the explicit form", null, "These solutions are square normalizable and then there are $mnk$ linearly independent zero-modes of positive chirality.\n\nOn the other hand, a negative chirality zero-mode satisfies\n\n$$\\not{D}\\left[ \\begin {array}{c} 0 \\\\ \\eta \\left( x,y \\right) \\end {array} \\right]=0$$\n\nit is to say", null, "with the explicit form", null, "These solutions are not square normalizable and then there are $0$ linearly independent zero-modes of negative chirality.\n\nThen, we obtain that the index of $\\not{D}$ , which is the number of linearly independent normalizable\npositive chirality zero-modes minus the number of linearly independent normalizable negative chirality zero-modes is  $mnk-0 = mnk$.  Formally, the index theorem in this case reads:\n\n$$index(i\\not{D}) = mnk = -\\frac{1}{2\\pi}\\int_{\\mathbb{R}^2}F= -\\frac{1}{4\\pi}\\int_{\\mathbb{R}^2}\\epsilon^{\\mu\\nu}F_{\\mu\\nu}$$\n\nMy questions are:\n\n1.  Do you agree with the computations in this  example?\n\n2.  Do you know other example in two-dimensions?\n\n Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the \"link\" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\\varnothing$ in the following word:p$\\hbar$ysi$\\varnothing$sOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register." ]

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[ "## 530 - Binomial Showdown\n\nModerator: Board moderators\n\nRaiyan Kamal\nExperienced poster\nPosts: 106\nJoined: Thu Jan 29, 2004 12:07 pm\nContact:\nPerhaps you are consuming too much time by multiplying numbers one after another and in determining the GCD. Think how can you do it by using addition instead of multiplication\n\nA1\nExperienced poster\nPosts: 173\nJoined: Wed Jan 28, 2004 3:34 pm\n\nHedge\nNew poster\nPosts: 11\nJoined: Thu Jul 29, 2004 8:59 pm\n\n### Re: WA in 0.000 s\n\nHi, I think i know what might be the problem of your program. You declare 'p' as an unsigned long integer, but you print it with \"%ld\". This converts it to an singed long int, so it might end up as a negative value if it is above 2^31. You should print unsigned long int with \"%lu\", to be sure that values above 2^31 are positive.\nedit: Sorry, i just read the specification. The result will always be below 2^31. So what I mentioned wont affect you for this problem. But nevertheless you should always print unsinged integers with %u not %d.\nActually i used unsigned long long integer for the intermediate results. Maybe you should take a close look at the binom function, and if the values there might overflow.\n\nWR\nExperienced poster\nPosts: 145\nJoined: Thu Nov 27, 2003 9:46 am\nThanks for the post.\n\nI've solved this problem shortly after the previous post (using double, by the way). Of course my program caused the WA, but at that time I just wasn't sure if the OJ returns a WA immediately after a wrong answer or if it first reads all records and complains afterwards.\n\nleonardooo\nNew poster\nPosts: 8\nJoined: Sun Nov 28, 2004 9:26 am\nLocation: Campina Grande - PB / Brazil\n\n### 530 Binomial Showdown\n\nI dont know why my code is wrong, anybody has special test cases ?\n\n[java]\n//Problem 530\n\nimport java.util.*;\nclass Main {\n\npublic static void main(String[] args) {\n\nwhile(input != null && !input.equals(\"0 0\")) {\n\nStringTokenizer st = new StringTokenizer(input);\nint n = Integer.parseInt(st.nextToken());\nint k = Integer.parseInt(st.nextToken());\n\nif(k > n) {\nSystem.out.println(\"0\");\ncontinue;\n}\n\nif(k == n) {\nSystem.out.println(\"1\");\ncontinue;\n}\n\nif(k == 1) {\nSystem.out.println(n);\ncontinue;\n}\n\nlong result;\nlong divfat;\n\nif(n-k > k) {\ndivfat = fat(n-k+1,n);\nresult = divfat / fat(1,k);\n} else {\ndivfat = fat(k+1,n);\nresult = divfat / fat(1,n-k);\n}\n\nSystem.out.println(result);\n}\n\n}\n\nstatic long fat(int ini, int fim) {\nlong result = 1;\nfor(int i = ini; i <= fim; i++) result *= i;\nreturn result;\n}\n\nString newLine = System.getProperty(\"line.separator\");\nStringBuffer buffer = new StringBuffer();\nint car = -1;\ntry {\nwhile ((car > 0) && (car != newLine.charAt(0))) {\nbuffer.append((char)car);\n}\nif (car == newLine.charAt(0))\nSystem.in.skip(newLine.length() - 1);\n} catch (java.io.IOException e) { return (null);}\nif ((car < 0) && (buffer.length() == 0)) return (null);\nreturn (buffer.toString()).trim();\n}\n\n}\n[/java]", null, "thx\nEu sou foda? N\n\nMing Han\nLearning poster\nPosts: 77\nJoined: Thu Jun 06, 2002 7:10 pm\nLocation: Singapore\nContact:\n:: HanWorks ::\n\nRocky\nExperienced poster\nPosts: 124\nJoined: Thu Oct 14, 2004 9:05 am\nContact:\nI am not expert in java so i can't understand you'r code\nBUT I suggest that BY simplification the main equation become very\nsimple and check that Whether the mod of two number is ZERO\nif ZERO then dived them and multiply the result of divide with result1 OTHERWISE multiply them directly with result1 and result2 AND printf the result of result1/result2\n.....\n\nMing Han\nLearning poster\nPosts: 77\nJoined: Thu Jun 06, 2002 7:10 pm\nLocation: Singapore\nContact:\nhttp://mathworld.wolfram.com/PascalsTriangle.html\n\nQuite a good site to understand more....\n:: HanWorks ::\n\nleonardooo\nNew poster\nPosts: 8\nJoined: Sun Nov 28, 2004 9:26 am\nLocation: Campina Grande - PB / Brazil\nthanks Guys, now I got AC !!!", null, "I stored all numbers in a int array: 10! = [1,2,3,4,5,6,7,8,9,10] 5! = [1,2,3,4,5]\n\nI used this algorithm: 10! / 5!\n\n[java]\nfor(i = 0; i < 10; i++)\nfor(j = 0; j < 5; j++)\nif( array1 % array2[j] == 0)\narray1 /= array2[j];\narray2[j] = 1;\n[/java]", null, "", null, "", null, "", null, "", null, "Eu sou foda? N\n\nmurkho\nNew poster\nPosts: 33\nJoined: Mon Mar 28, 2005 6:41 pm\n\n### 530\n\ncan anybody tell me the reason to get wa?????\n\nCode: Select all\n\n``````\n#include<stdio.h>\n//#include<math.h>\nint main()\n{\nlong int n,ans;long int index,i,j,k;\nlong int n_a,k_a;\n//freopen(\"F:\\\\530.txt\",\"r\",stdin);\nwhile(scanf(\"%ld %ld\",&n,&k)==2 )\n{\nif (n == 0 && k == 0)\nbreak;\nif(k>n/2)\nk = (long)n -k;\nindex = -1;\nans = 1;\nfor(i = (long) n;i > n-k;i--)\nn_a[++index] =i;\n\nindex = -1;\nfor(i = k;i>1;i--)\nk_a[++index] = i;\n\nfor(i = 0;i<k;i++)\nfor(j=0;j<k-1;j++)\n{\nif(k_a[j]!=1 && (n_a[i]%k_a[j]==0))\n{\nn_a[i] = n_a[i] /k_a[j];\nk_a[j] = 1;\n\n}\n\n}\nfor(i = 0;i<k;i++)\nans = ans*n_a[i];\n\nprintf(\"%.0ld\\n\",ans);\n\n}\n\nreturn 0;\n}\n``````\n\nstubbscroll\nExperienced poster\nPosts: 151\nJoined: Tue Nov 16, 2004 7:23 pm\nLocation: Norway\nContact:\nYou have an overflow in your intermediate calculations, as n_a can be over 2^31. Remember that both long int and int is 32 bits wide. If you need a 64-bit datatype, use long long.\n\nDerekG\nNew poster\nPosts: 2\nJoined: Mon Apr 18, 2005 12:26 am\n\n### 530 WA\n\ncan anyone please help me on this my code works fine on my machine .. maybe im misunderstanding something but any help is appreciated Thx.\n\n#include <cstdlib>\n#include <iostream>\n#include <math.h>\n\nusing namespace std;\nint factorial(int n);\nint gcd(int a, int b);\nint main(int argc, char *argv[])\n{\nwhile(1){\nlong long n, d, sum=1, nminusd=0, temp=0;\ncin>>n>>d;\nif(n==0) return 0;\nif(n==d || d==0){\ncout<<\"1\"<<endl;\nreturn 0;\n}\nif(d>n){\ncout<<\"0\"<<endl;\nreturn 0;\n}\nif(d>=(n-d)){\nnminusd=factorial(n-d);\nfor(int i=n; i>d; i--){\nsum*=i;\ntemp=sum;\nsum/=gcd(sum, nminusd);\nnminusd/=gcd(temp, nminusd);\n}\ncout<<sum<<endl;\n}\nelse{\nnminusd=factorial(d);\nfor(int i=n; i>(n-d); i--){\nsum*=i;\ntemp=sum;\nsum/=gcd(sum, nminusd);\nnminusd/=gcd(temp, nminusd);\n}\ncout<<sum<<endl;\n}\n}\nreturn 0;\n}\nint factorial(int n)\n{\nint sum=1;\nfor(int i=n; i>1; i--)\nsum*=i;\nreturn sum;\n}\n\nint gcd(int a, int b)\n{\nlong r;\nif(a < 0) a = -a;\nif(b < 0) b = -b;\n\nif(a == 0) return(b);\nif(b == 0) return(a);\n\nwhile(b > 0)\n{\nr = a % b;\na = b;\nb = r;\n}\nreturn(a);\n}\n\nExperienced poster\nPosts: 154\nJoined: Sat Apr 17, 2004 9:34 am\nLocation: EEE, BUET\nTry using long double...\nYou should never take more than you give in the circle of life.\n\nDerekG\nNew poster\nPosts: 2\nJoined: Mon Apr 18, 2005 12:26 am\ni tried that, changing as much to long double as i could wihle still having it work and that still failed... any other ideas? my code works perfectly on my stuff, does my readin look ok for how they run their test cases? any idea how large there test cases go i thought this was supposed to fit into a regular int? if so, all this long and double stuff shouldnt be needed by reducing the sum by the gcd each iteration...\n\n59557RC\nNew poster\nPosts: 26\nJoined: Sun Mar 20, 2005 9:28 pm\nContact:\n\n### 530:confused WA?\n\ni used long double n dividebygcd but why WA:\n\n#include<stdio.h>\n#include<math.h>\n\nlong double gcd(long double a,long double b)\n{\nif(fmod(a,b)==0) return b; else return gcd(b,fmod(a,b));\n}\n\nint main()\n{\n\nlong double g,i,j,k,n,mul,div;\ndouble up,down;\nscanf(\"%Lf %Lf\",&n,&k);\nwhile(n!=0 && k!=0){\nup=1;down=1;\nif(k>n/2) k=n-k;\nfor(i=k;i>0;i--){\nj=n-i+1;\nmul=j;div=i;\ng=gcd(mul,div);mul/=g;div/=g;\ng=gcd(mul,down);mul/=g;down/=g;\ng=gcd(up,div);div/=g;up/=g;\n\nup*=mul;down*=div;\n}\nprintf(\"%.0Lf\\n\",up/down);\nscanf(\"%Lf %Lf\",&n,&k);\n}\nreturn 0;\n\n}\naaa" ]

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[ "# Python: Check list is empty or not\n\nGiven a Python list, we have to check whether it is an empty or not.\nSubmitted by Nijakat Khan, on July 15, 2022\n\nThere are multiple approaches to check the given list is empty or not. Here, we are discussing some of them.\n\n## Method 1: Using len() method\n\nThe len() method is used to check the length of a list, tuple, dictionary, string, etc.\n\nSyntax:\n\n```A=len(list)\n```\n\nThe statement will return the length of list and stores it in variable A.\n\nExample 1:\n\n```# Defining a list\nl = []\n\n# Checking the list is empty or not\nif len(l)==0:\nprint(\"list is empty\")\nelse:\nprint(\"list is not empty\")\n```\n\nOutput:\n\n```list is empty\n```\n\nBelow is another same example, but here the list is not empty.\n\nExample 2:\n\n```# Defining a list\nl = [\"a\", \"b\", \"c\"]\n\n# Checking the list is empty or not\nif len(l)==0:\nprint(\"list is empty\")\nelse:\nprint(\"list is not empty\")\n```\n\nOutput:\n\n```list is not empty\n```\n\n## Method 2: Using bool() method\n\nThe bool() method is used to return the result in the form of true and false. If the condition is true it will return true else it will return false.\n\nSyntax:\n\n```l = bool(list)\n```\n\nIf the list is true (if there is some element in the list) it will return true and if the list is false (if the list is empty) it will return false.\n\nExample 3:\n\n```# Defining an empty list\nlist=[]\n\nif bool(list):\nprint (\"list is not empty\")\nelse:\nprint (\"list is empty\")\n```\n\nOutput:\n\n```list is empty\n```\n\n## Method 3: Using the if else condition\n\nThis statement is used to check the condition, if the if condition is true, it will execute the block code of the if statement otherwise it will execute the block code of the else statement.\n\nSyntax:\n\n```if statement:\n# code block\nelse:\n# code block\n```\n\nThere are different ways to check if the list is empty or not using the if else condition.\n\nExample 4:\n\n```# Defining an empty list\nl= []\n\n# Checking the list is empty or not\nif not l:\nprint (\"list is empty\")\nelse:\nprint (\"list is not empty\")\n```\n\nOutput:\n\n```list is empty\n```\n\nExample 5:\n\n```# Defining an empty list\nlist=[]\n\n# Checking whether list is empty or not\nif list==[]:\nprint (\"List is empty\")\nelse:\nprint (\"List is not empty\")\n```\n\nOutput:\n\n```List is empty\n```" ]

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