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https://www.pixemweb.com/php/php-loops/
|
[
"# PHP Loops\n\nBesides conditional statements like we’ve covered in previous tutorials, PHP also supports various types of looping mechanisms.\n\nThere are four basic types of loops and in this article I will cover all four.\n\nHere are the four types of loops.\n\n1. While Loop\n2. For Loop\n3. Foreach Loop\n4. do while Loop\n\n### What are PHP Loops Used for?\n\nLoops are used to execute the same code, over and over again as long as the condition is met. This is ideal since it eliminates the need to write the same code over and over again which can be tedious and lead to errors.\n\nLooping is also ideal when used for a blogging/cms platform like WordPress and others. The articles are saved in the database and your while loop is displaying each article or excerpt as long as articles are available to view.\n\nYou can also use loops to go through a list of names or emails or numbers or whatever can be looped.\n\nLoops are very powerful and when correctly implemented, can make your life as a coder, much easier.\n\nMake sure to copy the code and use them locally so you can alter the code and make it do something different.\n\nThe Best Way to Learn How to Code, is to Code.\n\n## PHP While Loop\n\nWhile loops are probably the easiest loops you will use. You will find many uses for it. While loops are best used when you don’t know how many times you’ll need to loop over something.\n\nBelow is an example of a while loop in action.\n\n``````\n<?php\n\\$i = 1;\n\nwhile ( \\$i <= 10 ) {\necho \"The number is: \\$i <br>\";\n\\$i ++;\n}\n?>\n``````\n\nLet’s Review the Code\n\n• first you will notice we have a variable with it’s assigned value `\\$i = 1;`\n• then we open up with the while ( \\$i <= 10 ) { which is testing the condition in between the parenthesis. Since we set \\$i to the value of 1, as long as \\$i is less than or equal to 10, we will output the next part of the code\n• we `echo \"The number is: \\$i <br>\";` continuously until the condition being tested against is no longer met.\n• we also need to make sure we increment \\$i so that we don’t end up in an infinite loop situation. That’s where the next line of code comes in `\\$i ++;`\n• We then close off with the }\n\nThe example above is a basic implementation of the while loop.\n\n## PHP for Loop\n\nThe for loop is best used when you know the amount of times you want to loop over something. It’s implemented differently than the while loop. Take a look at the example below.\n\n``````\n<?php\nfor ( \\$i = 1; \\$i <= 10; \\$i ++ ) {\necho \"The number is: \\$i <br>\";\n}\n?>\n``````\n\nLet’s Review the Code\n\nUnlike a while loop where we first need to create a variable outside of the looping mechanism, with a for loop, the variable is set inside the parenthesis.\n\n• we start off with for\n• right after the opening parenthesis we set our variable `\\$i = 1;`\n• then we test `\\$i <= 10;`\n• followed by incrementing the value of \\$i `\\$i ++`\n• we then have the closing part of the parenthesis followed by the opening curly brace `) {`\n• the next line of code handles outputting the current value `echo \"The number is: \\$i <br>\"; `\n• finally we have our closing curly brace `}`\n\nThe key thing to remember is to make sure you initialize the variable first and use a semi-colon then do the test to check if it’s less than or equal to 10 followed by another semi-colon and finally increment the variable \\$i.\n\n## PHP foreach loops\n\nForeach loops are used with arrays which we’ll cover in depth in another tutorial. This loop is specially made only for arrays. You could use a for loop with arrays and in some cases it may make sense to do so, but primarily you’ll be using the foreach loop.\n\nLet’s look at an example of a foreach loop in action.\n\n``````\n<?php\n\\$colors = array( \"blue\", \"purple\", \"green\", \"red\", \"yellow\", \"black\" );\n\nforeach ( \\$colors as \\$value ) {\necho \"\\$value <br>\";\n}\n?>\n``````\n\nLet’s Review the Code\n\n• first thing we do is create a variable called \\$colors which is assigned the value of an `array( \"blue\", \"purple\", \"green\", \"red\", \"yellow\", \"black\" );` which as you can see has several colors each in the format of a string enclosed in quotes and separated by a comma\n• you will notice the colors are inside the parenthesis and that line of code ends with a semi-colon\n• we then move onto the next line `foreach ( \\$colors as \\$value ) {`\n• after the foreach and in between the parenthesis, you will notice we take the variable \\$colors and say as \\$value which means each actual color will be in the temporary & reusable variable \\$value\n• on the next line we `echo \"\\$value <br>\";` which is going to output each color\n• finally we close off the curly brace }\n\n## PHP do while Loops\n\nI’ve purposely saved this one for last even though it’s often grouped with the While Loop. The reason why I’ve done this is because the Do While loop isn’t used as often as the others.\n\nEven though it isn’t often used, you might find it useful for a particular use case.\n\nThe Do While loop executes a block of code first and then continues to loop over the block of code as long as the condition being checked for is still true. It sounds similar to the While loop but below you will see the difference.\n\n``````\n<?php\n\\$i = 0;\ndo {\necho \\$i;\n} while ( \\$i > 0 );\n?>\n``````\n\nDo you see what happened above?\n\nLet’s Review the Code\n\n• The first thing to notice is that we initialized the variable \\$i = 0;\n• we then open up with do {\n• then we `echo \\$i;`\n• finally we close off with `} while ( \\$i > 0 );`\n\nThe Output from above is `0`\n\nEven though \\$i isn’t greater than zero, we still process the first part of the do while loop and then we test the condition to see if anything else needs to be processed."
] |
[
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|
http://num.bubble.ro/m/8993/7001/
|
[
"# Multiplication table for N = 8993 * 7000÷7001\n\n8993 * 7000 = 62951000 [+]\n8993 * 7000.01 = 62951089.93 [+]\n8993 * 7000.02 = 62951179.86 [+]\n8993 * 7000.03 = 62951269.79 [+]\n8993 * 7000.04 = 62951359.72 [+]\n8993 * 7000.05 = 62951449.65 [+]\n8993 * 7000.06 = 62951539.58 [+]\n8993 * 7000.07 = 62951629.51 [+]\n8993 * 7000.08 = 62951719.44 [+]\n8993 * 7000.09 = 62951809.37 [+]\n8993 * 7000.1 = 62951899.3 [+]\n8993 * 7000.11 = 62951989.23 [+]\n8993 * 7000.12 = 62952079.16 [+]\n8993 * 7000.13 = 62952169.09 [+]\n8993 * 7000.14 = 62952259.02 [+]\n8993 * 7000.15 = 62952348.95 [+]\n8993 * 7000.16 = 62952438.88 [+]\n8993 * 7000.17 = 62952528.81 [+]\n8993 * 7000.18 = 62952618.74 [+]\n8993 * 7000.19 = 62952708.67 [+]\n8993 * 7000.2 = 62952798.6 [+]\n8993 * 7000.21 = 62952888.53 [+]\n8993 * 7000.22 = 62952978.46 [+]\n8993 * 7000.23 = 62953068.39 [+]\n8993 * 7000.24 = 62953158.32 [+]\n8993 * 7000.25 = 62953248.25 [+]\n8993 * 7000.26 = 62953338.18 [+]\n8993 * 7000.27 = 62953428.11 [+]\n8993 * 7000.28 = 62953518.04 [+]\n8993 * 7000.29 = 62953607.97 [+]\n8993 * 7000.3 = 62953697.9 [+]\n8993 * 7000.31 = 62953787.83 [+]\n8993 * 7000.32 = 62953877.76 [+]\n8993 * 7000.33 = 62953967.69 [+]\n8993 * 7000.34 = 62954057.62 [+]\n8993 * 7000.35 = 62954147.55 [+]\n8993 * 7000.36 = 62954237.48 [+]\n8993 * 7000.37 = 62954327.41 [+]\n8993 * 7000.38 = 62954417.34 [+]\n8993 * 7000.39 = 62954507.27 [+]\n8993 * 7000.4 = 62954597.2 [+]\n8993 * 7000.41 = 62954687.13 [+]\n8993 * 7000.42 = 62954777.06 [+]\n8993 * 7000.43 = 62954866.99 [+]\n8993 * 7000.44 = 62954956.92 [+]\n8993 * 7000.45 = 62955046.85 [+]\n8993 * 7000.46 = 62955136.78 [+]\n8993 * 7000.47 = 62955226.71 [+]\n8993 * 7000.48 = 62955316.64 [+]\n8993 * 7000.49 = 62955406.57 [+]\n8993 * 7000.5 = 62955496.5 [+]\n8993 * 7000.51 = 62955586.43 [+]\n8993 * 7000.52 = 62955676.36 [+]\n8993 * 7000.53 = 62955766.29 [+]\n8993 * 7000.54 = 62955856.22 [+]\n8993 * 7000.55 = 62955946.15 [+]\n8993 * 7000.56 = 62956036.08 [+]\n8993 * 7000.57 = 62956126.01 [+]\n8993 * 7000.58 = 62956215.94 [+]\n8993 * 7000.59 = 62956305.87 [+]\n8993 * 7000.6 = 62956395.8 [+]\n8993 * 7000.61 = 62956485.73 [+]\n8993 * 7000.62 = 62956575.66 [+]\n8993 * 7000.63 = 62956665.59 [+]\n8993 * 7000.64 = 62956755.52 [+]\n8993 * 7000.65 = 62956845.45 [+]\n8993 * 7000.66 = 62956935.38 [+]\n8993 * 7000.67 = 62957025.31 [+]\n8993 * 7000.68 = 62957115.24 [+]\n8993 * 7000.69 = 62957205.17 [+]\n8993 * 7000.7 = 62957295.1 [+]\n8993 * 7000.71 = 62957385.03 [+]\n8993 * 7000.72 = 62957474.96 [+]\n8993 * 7000.73 = 62957564.89 [+]\n8993 * 7000.74 = 62957654.82 [+]\n8993 * 7000.75 = 62957744.75 [+]\n8993 * 7000.76 = 62957834.68 [+]\n8993 * 7000.77 = 62957924.61 [+]\n8993 * 7000.78 = 62958014.54 [+]\n8993 * 7000.79 = 62958104.47 [+]\n8993 * 7000.8 = 62958194.4 [+]\n8993 * 7000.81 = 62958284.33 [+]\n8993 * 7000.82 = 62958374.26 [+]\n8993 * 7000.83 = 62958464.19 [+]\n8993 * 7000.84 = 62958554.12 [+]\n8993 * 7000.85 = 62958644.05 [+]\n8993 * 7000.86 = 62958733.98 [+]\n8993 * 7000.87 = 62958823.91 [+]\n8993 * 7000.88 = 62958913.84 [+]\n8993 * 7000.89 = 62959003.77 [+]\n8993 * 7000.9 = 62959093.7 [+]\n8993 * 7000.91 = 62959183.63 [+]\n8993 * 7000.92 = 62959273.56 [+]\n8993 * 7000.93 = 62959363.49 [+]\n8993 * 7000.94 = 62959453.42 [+]\n8993 * 7000.95 = 62959543.35 [+]\n8993 * 7000.96 = 62959633.28 [+]\n8993 * 7000.97 = 62959723.21 [+]\n8993 * 7000.98 = 62959813.14 [+]\nNavigation: Home | Addition | Substraction | Multiplication | Division Tables for 8993: Addition | Substraction | Multiplication | Division\n\nOperand: 1 2 3 4 5 6 7 8 9 10 20 30 40 50 60 70 80 90 100 200 300 400 500 600 700 800 900 1000 2000 3000 4000 5000 6000 7000 7001 7002 7003 7004 7005 7006 7007 7008 7009 8000 9000\n\nMultiplication for: 1 2 3 4 5 6 7 8 9 10 20 30 40 50 60 70 80 90 100 200 300 400 500 600 700 800 900 1000 2000 3000 4000 5000 6000 7000 8000 8991 8992 8993 8994 8995 8996 8997 8998 8999 9000"
] |
[
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|
https://physics.stackexchange.com/questions/371175/what-law-or-formula-discusses-the-relationship-between-pressure-and-dew-point
|
[
"# What law or formula discusses the relationship between pressure and dew point?\n\nA lot of times, dew point is focused primarily upon temperature and relative humidity. However, that same point of saturation is affected by the pressure, but I can't find a formula, or law even, that discusses this.\n\nIt's possible that my premise of the relationship between dew point and pressure may be inaccurate, but if so, how can we then explain that a refrigerant may be a vapor at 210°F in a 280psi container, have a dew point of 125°F, but also be a vapor at 75°F in a 70psi container.\n\nWhether the substance is water, freon, or any liquid, there should be relationships between pressure and dew points (saturation points, condensation points, etc - they're all meaning the same thing). We can clearly see with freon there's some relationship between pressure and temperatures where the rate of condensation is greater than that of the evaporation, but I can't seem to find any laws/formulas for this.\n\n• I'm a bit confused, since in the first paragraph you talk about 'relative humidity' which by default I associate with water, while in the second paragraph you talk about a refrigerant. Nov 27, 2017 at 20:55\n• You're right - humidity is confined to water only, but there's lots of other substances that condense/vaporize as well. Perhaps you can substitute my use of \"humidity\" for \"the amount of gas in the air relative to its condensation temperature.\" If there's a term for that, it may be beneficial to know that and I'll edit the question. The bigger question lies in the 2nd paragraph though. Nov 27, 2017 at 21:12\n• OK, so now we can clarify that second paragraph. Yes, you can have a substance that would be all vapor at 210F and still have vapor pressure at 75F (in equilibrium over liquid). Nov 27, 2017 at 21:21\n• Making this a comment, because I'm not certain, but I think you want the Clausius-Clapeyron relation, which allows you to find the dividing line in a phase diagram. Nov 29, 2017 at 23:00\n• @TonyDiNitto, research the Antoine equation. This equation establishes the relationship between vapor pressure and temperature for pure substances. See en.wikipedia.org/wiki/Antoine_equation Nov 30, 2017 at 1:06\n\nFor single-component liquids, boiling point = condensation point. Both are the same temperature for a given pressure. Let's talk boiling.\n\nLiquids boil when the vapor pressure of the liquid equals the pressure of the surrounding gas (e.g. 1 atm for open containers at sea level). As you raise the temperature of a liquid, the vapor pressure increases until it equals the pressure of the surrounding gas at which point it boils. If you reduce the pressure of the surrounding gas, then you do not need to raise the temperature of the liquid as much anymore. This is why water boils at 82 C instead of 100 C on Everest-- because the atmospheric pressure is lower:",
null,
"If you have the vapor pressure $P_1$ of a pure substance at one temperature $T_1$ you can calculate the vapor pressure $P_2$ at a second temperature $T_2$ using the Clausius-Clapeyron equation:\n\n$$\\ln\\frac{P_2}{P_1} = \\frac{-\\Delta H_{vap}}{R}\\left(\\frac{1}{T_2} - \\frac{1}{T_1}\\right)$$\n\nwhere\n$P$ = vapor pressure\n$\\Delta H_{vap}$ = enthalpy of vaporization in $J/mol$\n$R$ = the gas constant = $8.3145\\ J/mol \\cdot K$\n$T$ = temperature in $K$\n\nThe common use of the term \"dew point\" is actually the \"Atmospheric Dew Point\" which doesn't take account the pressure of a system, as it's presumed to be at 1 atm. Many readily found formulas on dew point are formulas for atmospheric dew point, as they focus solely on temperature and the amount of gas (eg. water vapor) trapped the air.\n\nThe other important variable in what affects dew point is the pressure, which is where Pressure Dew Point (PDP) comes into play.\n\nAs a rule of thumb (without the use of complex equations), compression increases/raises the dew point temperature and expansion/decompression lowers the dew point.\n\nThis makes sense considering if you have a fixed amount of vapor in the air, and expand/decompress that vapor, the overall percentage of moisture in the air will be smaller, meaning the relative humidity percentage will decrease, and therefore lower the dew point.\n\nAt this point, the standard atmospheric dew point formulas can then be applied to find dew points at different levels of pressure.\n\nAs an actual example, the dew point for a parcel of air at 200 PSIg would have a dew point at -40°F, whereas the same parcel at 5 PSIg, significantly less compressed than the 200 PSIg parcel, would have a dew point of -77°F.\n\nThis has important applications in really understanding how things fully work like heat pumps, air dryers, food processing plants, electronics manufacturing, and more.\n\nEven though I am not an expert on this, I found some information that can help. Check out this website\n\nhttps://www.vaisala.com/sites/default/files/documents/Dew-point-compressed-air-Application-note-B210991EN-B-LOW-v1.pdf\n\nand this website\n\nhttps://en.wikipedia.org/wiki/File:Dewpoint.jpg\n\n• This site actually says, \"changing the pressure of a gas changes the dew point temperature of the gas.\" Which is what I've known from observation of Freon gasses, but yet I could never find anyone ever talking about this. Dec 1, 2017 at 15:28\n\nDisclaimer: I'm not a physicist or chemist, just a software developer who struggled with all those water vapor related concepts for a few weeks himself.\n\n## Dew-point vs. boiling-point\n\nI'm not sure if this holds true for all substances, but water does not only vaporize when it reaches it's boiling point. Instead, a few molecules will always leave the surface of liquid or even frozen water. Depending on the partial water vapor pressure of the surrounding gas mixture, those might be balanced by the water molecules that condense / sublime back into liquid or solid phase. The temperature at which those two processes balance is called dew-point.\n\n## The dew-point depends on\n\n• the temperature (of the water in all it's states)\n• the partial water vapor pressure\n• the surface geometry of the liquid / solid water\n• presence and abundance of particles suspended in the air mixture that promote condensation on their surface\n• presence and concentration of salt or other impurities on the surface of the liquid / solid water\n\nAs user @pentane has stated in his answer, the boiling-point does depend on the total air pressure, the dew-point in contrast depends on the partial water vapor pressure instead. Unlike the boiling-point,\n\n# The dew-point does NOT depend on the total air pressure in the system!\n\nThe Pressure Dew Point (PDP) formulas that user @tony-dinitto talks about in his answer let you calculate how the dew-point changes when you change the pressure in the system while holding the percentage of water vapor in the air mixture constant.\n\nIf you're like me dealing with a system, where the sensor measures temperature, relative humidity and total air pressure \"at the same time\" - and you don't need to predict what would happen if you encapsulated this exact air mixture and compressed or expanded it, then Pressure Dew Point (PDP) formulas are of no use to you.\n\nAccording to Wikipedia, the Arden Buck Equation is the most accurate formula to calculate the saturated water vapor pressure for a given temperature. Using this fact I came up with the following way to\n\n## calculate the dew-point:\n\n1. Calculate saturated water vapor pressure for current temperature using the Arden Buck Equation.\n2. Calculate the current water vapor pressure by multiplying the saturated water vapor by the relative humidity as a factor (divide by 100 if given as percent)\n3. Use the inverse of the Arden Buck Equation to calculate which temperature would have the current water vapor pressure as saturated water vapor pressure.\n\nSince for me it was quite difficult to find the inverse of the Arden Buck Equation, I'll post what I came up with to make it easier for readers who try to follow my steps:\n\nlet (a, b, c, d) = (611.21, 18.678, 234.5, 257.14); // empirical constants of the Arden Buck Equation for temperatures > 0°C\nlet g = (actual_water_vapor_pressure_pa / a).ln();\n-0.5 * c * (((b - g).powi(2) - (4.0 * d * g) / c).sqrt() + g - b)"
] |
[
null,
"https://i.stack.imgur.com/17f81.png",
null
] |
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https://pypi.org/project/stacksplit/
|
[
"Python Software Foundation 20th Year Anniversary Fundraiser\n\nA simple Python library to generate all combinations of possible splits for a stack with given height.\n\n# Stacksplit\n\nA simple Python library to generate all combinations of possible splits for a stack with given height.\n\n## Introduction\n\nThis library aims to generate all possible combinations to split a given integer `num` into a given number of parts; the sum of all those parts shall again be the given `num`.\n\nWe wrote this simple library out of lack of such functionality needed in a lecture of our Computer Science BSc-Degree. The original use case was to calculate and solve extended versions of the NIM-game, where towers of coins could also be splitted into multpile smaller towers.\n\n## Usage\n\nBe sure to have stacksplit installed.\n\nImport stacksplit library:\n\n```import stacksplit\n```\n\nThe function provided is split and is implemented as a Python-generator; it takes 2 (optionally 3) arguments:\n\n• `num`: the Integer to be split up\n• `parts`: the number of parts\n• `smallest`: the smallest part shall be >= the given parameter. This parameter is optional and defaults to 1.\n```split(num, parts, smallest)\n```\n\nEach call returns a new tuple with a new combination with all elements summing up to `num`.\n\nlook at doc_strings and comments in init and core for help\n\n### Simple usage\n\n```import stacksplit\n\nfor s in stacksplit.split(50, 3):\nprint(s)\n```\n\nOutput:\n\n``````(1, 1, 48)\n(1, 2, 47)\n...\n``````\n\n### More options\n\n```from stacksplit import split\n\nfor i in split(50, 3, 10):\nprint(i)\n```\n\nOutput:\n\n``````(10, 10, 30)\n(10, 11, 29)\n...\n``````\n\n### Extended usecases\n\n`smallest` can also be 0 or negative. The results will always sum up to `num`.\n\n```from stacksplit import split\n\nfor i in split(5, 3, -1):\nprint(i)\n```\n\nOutput:\n\n``````(-1, -1, 7)\n(-1, 0, 6)\n(-1, 1, 5)\n...\n``````\n\n## Performance\n\nThe library uses Python native generators to achieve the fast generation of results; however you have to understand that the problem itself is quite complex and the number of results will increment exponentially with higher values as parameters.\n\nThe following graphs visualize this growth of results.\n\nGraph Description",
null,
"y-axis: number of result\nx-axis: the `num` parameter\n`parts`: constant 4",
null,
"y-axis: number of results\nx-axis: the `parts` parameter\n`num`: constant\n\n## Installation\n\nThis library can be installed via `pip install stacksplit`.\n\n### Arch Linux\n\nThe AUR package will be named `python-stacksplit`.\n\n## Testing\n\nTo run the tests for stacksplit:\n\nYou may then use these:\n\n• run normal tests: `pipenv run python setup.py test`\n• run tests with coverage: `pipenv run python setup.py test --coverage`\n• run tox tests: `pipenv run tox` (make sure you have interpreters for python - 3.4 to 3.7)\n\nNote: It is possible to use a normal virtual environmet by installing the dev-dependencies from Pipfile by hand with pip. (For exact versions see Pipfile.lock)\n\n## Project details\n\nThis version",
null,
"0.0.0"
] |
[
null,
"https://warehouse-camo.ingress.cmh1.psfhosted.org/c2d1dcb2afde17a1a1eb920c38f86e11deb745bb/2e2f646f63732f786e756d5f797265735f7061727473345f736d616c6c2e706e67",
null,
"https://warehouse-camo.ingress.cmh1.psfhosted.org/39f92273e8daed36174c79b4e8f9bf5821fe9c1d/2e2f646f63732f7870617274735f797265735f736d616c6c2e706e67",
null,
"https://pypi.org/static/images/blue-cube.e6165d35.svg",
null
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|
https://clearjeeneet.in/class-11-chemistry-mcqs-the-p-block-elements-for-jee-neet/
|
[
"## Class 11 Chemistry MCQs The p-Block Elements for JEE/NEET\n\nHere, you will get Class 11 Chemistry MCQ of Chapter 11 The p-Block Elements for cracking JEE and NEET/AIIMS. By solving these MCQs of Class 11 Chemistry Chapter 11 The p-Block Elements, you will get the confidence to crack JEE or NEET. Practice MCQ Questions for Class 11 Chemistry with Answers on a daily basis and score well in exams.\n\n## The p-Block Elements Class 11 MCQs Questions with Answers\n\nOrthoboric acid when heated to red heat gives\n(a) Metaboric acid\n(b) Tetraboric acid\n(c) Boron trioxide\n(d) Borax\n\nThe structure of diBorane contains\n(a) Four 2c – 2e bonds and two 3c – 2e bonds\n(b) Two 2c – 2e bonds and two 3c – 2e bonds\n(c) Two 2c – 2e bonds and two 3c – 3e bonds\n(d) Four 2c – 2e bonds and four 3c – 2e bonds\n\nAnswer: (a) Four 2c – 2e bonds and two 3c – 2e bonds\n\nGraphite is soft solid lubricant extremely difficult to melt. The reason for this anomalous behaviour is that graphite\n(a) Has carbon atoms arranged in large plates of rings of strongly bound carbon atoms with weak interplate bonds\n(b) Is a non-crystalline substance\n(c) Is an allotropic form of carbon\n(d) Has molecules of variable molecular masses like polymers\n\nAnswer: Show Answer (a) Has carbon atoms arranged in large plates of rings of strongly bound carbon atoms with weak interplate bonds\n\nThermodynamically the most stable form of carbon is\n(a) Diamond\n(b) Graphite\n(c) Peat\n(d) Coal\n\nIn the upper layers of atmosphere ozone is formed:\n(a) By action of electric discharge on oxygen molecule\n(b) By action of ultraviolet rays on oxygen molecule\n(c) By action of infrared rays on oxygen molecule\n(d) Due to sudden drops of pressure\n\nAnswer: (b) By action of ultraviolet rays on oxygen molecule\n\nWhich of the following statements about boric acid is false?\n(a) It acts as a monobasic acid\n(b) It is formed by the hydrolysis of boron halides\n(c) It has planar-structure\n(d) It acts as a tribasic acid\n\nAnswer: (d) It acts as a tribasic acid\n\nFrom B2H6 all the following can be prepared except\n(a) H3BO3\n(b) B2(CH3)4H2\n(c) B2(CH3)6\n(d) NaBH4\n\nWhich of the following is the most commonly used refrigerant?\n(a) CFCl3\n(b) CCl3Br\n(c) CF3Cl\n(d) CF2Cl2\n\nRed phosphorus is chemically less reactive because\n(a) It does not contain P – P bonds\n(b) It dos not contain tetrahedral P4 molecules\n(c) It does not catch fire in air even upto 400°C\n(d) It has a polymeric structure\n\nAnswer: (d) It has a polymeric structure\n\nAmong the C-X bond (where, X = Cl, Br, l) the correct decreasing order of bond energy is\n(a) C−I > C−Cl > C−Br\n(b) C−I > C−Br > C−Cl\n(c) C−Cl > C−Br > C−I\n(d) C−Br > C−Cl > C−I\n\nAnswer: (c) C−Cl > C−Br > C−I\n\nThe increasing order of atomic radii of the following group 13 elements are:\n(a) Al < Ga < In < Tl\n(b) Ga < Al < In < Tl\n(c) Al < In < Ga < Tl\n(d) Al < Ga < Tl < In\n\nAnswer: (b) Ga < Al < In < Tl\n\nInert gases such as helium behave like ideal gases over a wide range of temperature .However; they condense into the solid state at very low temperatures. It indicates that at very low temperature there is a\n(a) Weak attractive force between the atoms\n(b) Weak repulsive force between the atoms\n(c) Strong attractive force between the atoms\n(d) Strong repulsive attractive between the atoms\n\nAnswer: (c) Strong attractive force between the atoms\n\nIn the compound of type ECl3, where E = B, P, As, or Bi, the angle Cl – E – Cl for different E are ion the order:\n(a) B > = P = As = Bi\n(b) B > P > As > Bi\n(c) B < P = As = Bi\n(d) B < P < As < Bi\n\nAnswer: (b) B > P > As > Bi\n\nIn graphite, the electrons are\n(a) localised on every third carbon atom.\n(b) present in antibonding orbitals\n(c) localised on each carbon atom\n(d) spread out between the structure\n\n(a) Zero\n(b) 20\n(c) 80\n(d) 70\n\nWhich of the following oxidation states are the most characteristic for lead & tin respectively\n(a) +2, +2\n(b) +4, +2\n(c) +2, +4\n(d) +4, +4\n\nIn white phosphorous (P4) molecule, which one is not correct\n(a) 6P-P single bonds are present\n(b) 4P-P single bonds are present\n(c) 4 lone pair of electrons is present\n(d) P-P-P bond angle is 60°\n\nAnswer: (a) 6P-P single bonds are present\n\nWhich of the following allotropic forms of carbon is isomorphous with crystalline silicon?\n(a) Graphite\n(b) Coal\n(c) Coke\n(d) Diamond\n\nBorax is used as a cleansing agent because on dissolving in water, it gives\n(a) Alkaline solution\n(b) Acidic solution\n(c) Bleaching solution\n(d) Amphoteric solution\n\nOn heating boron with caustic potash, the pair of products formed are\n(a) Potassium Borate + Dihydrogen\n(b) Potassium Borate + Water\n(c) Potassium Borate + H2\n(d) Borax + Dihydrogen\n\nAnswer: (a) Potassium Borate + Dihydrogen\n\nFertilizer having the highest nitrogen percentage is:\n(a) Calcium cyanamide\n(b) Urea\n(c) Ammonium nitrate\n(d) Ammonium sulphate\n\nWhich is strongest lewis acid?\n(a) BF3\n(b) BCl3\n(c) BBr3\n(d) BI3\n\n(a) Pb3O4\n(b) PbO\n(c) 2PbCO3.Pb(OH)2\n(d) Pb (CH3COO)2.Pb(OH)2\n\nWhich of the following forces bind together carbon atoms in diamond?\n(a) Ionic\n(b) Covalent\n(c) Dipolar\n(d) van der Waals\n\nAmong the following, electron-deficient compound is\n(a) CCl4\n(b) PCl5\n(c) BeCl2\n(d) BCl3\n\nBoric acid is polymeric due to\n(a) Its acidic nature\n(b) The presence of hydrogen bonds\n(c) Its monobasic nature\n(d) Its geometry\n\nAnswer: (b) The presence of hydrogen bonds\n\nAn aqueous solution of borax is\n(a) Basic\n(b) Acidic\n(c) Neutral\n(d) Amphoteric\n\nWhich of the following is not a protonic acid?\n(a) B(OH)3\n(b) PO(OH)3\n(c) SO(OH)2\n(d) SO2(OH)2\n\nWhich one of the following has the lowest m.p.?\n(a) Ga\n(b) B\n(c) Al\n(d) Tl"
] |
[
null
] |
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https://www.apiref.com/cpp-zh/cpp/utility/any.html
|
[
"# std::any\n\n< cpp | utility\n\nC++\n 语言 标准库头文件 自立与有宿主实现 具名要求 语言支持库 概念库 (C++20) 诊断库 工具库 字符串库 容器库 迭代器库 范围库 (C++20) 算法库 数值库 本地化库 输入/输出库 文件系统库 (C++17) 正则表达式库 (C++11) 原子操作库 (C++11) 线程支持库 (C++11) 技术规范\n\n 语言支持 类型支持（基本类型、 RTTI 、类型特征） 库功能特性测试宏 (C++20) 动态内存管理 程序工具 错误处理 协程支持 (C++20) 变参数函数 launder(C++17) initializer_list(C++11) source_location(C++20) 三路比较 (C++20) three_way_comparablethree_way_comparable_with(C++20)(C++20) strong_ordering(C++20) weak_ordering(C++20) partial_ordering(C++20) common_comparison_category(C++20) compare_three_way_result(C++20) compare_three_way(C++20) strong_order(C++20) weak_order(C++20) partial_order(C++20) compare_weak_order_fallback(C++20) is_eqis_neqis_ltis_lteqis_gtis_gteq(C++20)(C++20)(C++20)(C++20)(C++20)(C++20)\n\n(C++11)\n\n(C++20)\nswap 与类型运算\n swap ranges::swap(C++20) exchange(C++14) declval(C++11)\n forward(C++11) move(C++11) move_if_noexcept(C++11) as_const(C++17)\n\n pair tuple(C++11) apply(C++17) make_from_tuple(C++17)\n optional(C++17) any(C++17) variant(C++17)\n\n(C++17)\n(C++17)\n\nstd::any\n 成员函数 any::any any::~any any::operator= 修改器 any::emplace any::reset any::swap 观察器 any::has_value any::type 非成员函数 swap(std::any) any_cast make_any\n\n 定义于头文件 `` class any; (C++17 起)\n\n`any` 描述用于任何类型的单个值的类型安全容器。\n\n1)`any` 的对象存储任何满足构造函数要求的类型的一个实例或为空，而这被称为 `any` 类对象的状态。存储的实例被称作所含对象。若两个状态均为空，或均为非空且其所含对象等价，则两个状态等价。\n2) 非成员 `any_cast` 函数提供对所含对象的类型安全访问。\n\n(公开成员函数)\n\n(公开成员函数)\n\n(公开成员函数)\n\n(公开成员函数)\n\n(公开成员函数)\n\n(公开成员函数)\n\n(公开成员函数)\n\n(公开成员函数)\n\n### 非成员函数\n\n std::swap(std::any)(C++17) 特化 std::swap 算法 (函数) any_cast(C++17) 对被容纳对象的类型安全访问 (函数模板) make_any(C++17) 创建 `any` 对象 (函数模板)\n\n### 辅助类\n\n bad_any_cast(C++17) 当类型不匹配时按值返回形式的 `any_cast` 所抛出的异常 (类)\n\n### 示例\n\n```#include <any>\n#include <iostream>\n\nint main()\n{\nstd::cout << std::boolalpha;\n\n// any 类型\nstd::any a = 1;\nstd::cout << a.type().name() << \": \" << std::any_cast<int>(a) << '\\n';\na = 3.14;\nstd::cout << a.type().name() << \": \" << std::any_cast<double>(a) << '\\n';\na = true;\nstd::cout << a.type().name() << \": \" << std::any_cast<bool>(a) << '\\n';\n\n// 有误的转型\ntry\n{\na = 1;\nstd::cout << std::any_cast<float>(a) << '\\n';\n}\n{\nstd::cout << e.what() << '\\n';\n}\n\n// 拥有值\na = 1;\nif (a.has_value())\n{\nstd::cout << a.type().name() << '\\n';\n}\n\n// 重置\na.reset();\nif (!a.has_value())\n{\nstd::cout << \"no value\\n\";\n}\n\n// 指向所含数据的指针\na = 1;\nint* i = std::any_cast<int>(&a);\nstd::cout << *i << \"\\n\";\n}```\n\n```i: 1\nd: 3.14\nb: true"
] |
[
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] |
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|
https://discourse.mc-stan.org/t/fitting-model-with-decreasing-cumulative-costs/33400
|
[
"# Fitting model with decreasing cumulative costs\n\nHi!\n\nSuppose I lend people money and they repay in smaller amounts over month. I expect all clear at due date. I would like to know how much of the money I get after X% of the time and how certain I can be.\n\n``````library(brms)\nlibrary(tidyverse)\nlibrary(lubridate)\nlibrary(tsibble)\nlibrary(ggplot2)\n\ndf_raw <- tribble(\n~Person, ~Date, ~Amount, ~Comment,\n\"A\", \"2000-01-01\", 1000, \"Borrow\",\n\"A\", \"2000-02-01\", -200, \"\",\n\"A\", \"2000-03-01\", -10, \"\",\n\"A\", \"2000-05-01\", -300, \"\",\n\"A\", \"2000-06-01\", -100, \"Due date\",\n\"B\", \"2000-03-01\", 2000, \"Borrow\",\n\"B\", \"2000-04-01\", -420, \"\",\n\"B\", \"2000-05-01\", -20, \"\",\n\"B\", \"2000-06-01\", -500, \"Due date\",\n\"B\", \"2000-10-01\", -620, \"\",\n\"C\", \"2000-03-01\", 3000, \"Borrow\",\n\"C\", \"2000-04-01\", -500, \"\",\n\"C\", \"2000-05-01\", -30, \"\",\n\"C\", \"2000-06-01\", -900, \"\",\n\"C\", \"2000-07-01\", -300, \"Due date\",\n\"C\", \"2001-10-01\", -300, \"\"\n)\n\ndf <- df_raw |>\nmutate(Date = yearmonth(Date)) |>\ngroup_by(Person) |>\nmutate(\nloan_start = first(Date[Comment == \"Borrow\"]),\nborrowed_amount = first(Amount[Comment == \"Borrow\"]),\namount_open = cumsum(Amount),\nrelative_amount = amount_open / borrowed_amount,\ndue_date = first(Date[Comment == \"Due date\"]),\nloan_duration = as.numeric(due_date - loan_start),\ntime_point = as.numeric(Date - loan_start) / loan_duration\n) |>\nungroup() |>\nselect(-Comment)\n\nggplot(df, aes(x=time_point, y=relative_amount, color = Person, group = Person)) +\ngeom_line() +\ngeom_point()\n``````\n\nThe relation must not be linear. The relative amount should be 1 at time 0 because I divided by a reference amount. I would like to learn from all persons.\n\n``````df1 <- df |>\nfilter(between(time_point,0,1))\n\nfit1 <- brm(\nformula = bf(relative_amount ~ time_point + (1 | p | Person)),\ndata = df1,\nfamily = gaussian(),\nseed = 12345\n)\nsummary(fit1)\n``````\n\nIs this the right approach? How to specify the priors? Thanks,\n\nSide comment: examples requiring excessive dependencies (here, the tidyverse) are less useful for others who I may want to run your code. You could have set this up easily with plain R.\n\n1 Like\n\nIt seems like you’re throwing away information by normalizing all loan amounts to start at 1.0. If the payback rate varies based on the starting amount, the “relative amount” model wouldn’t capture that.\n\n1 Like\n\nSuppose I don’t divide by the borrowed amount. Then\n\n``````ggplot(df1, aes(x=time_point, y=amount_open, color = Person, group = Person)) +\ngeom_line() +\ngeom_point()\n``````\n\nI guess something like\n\n``````fit2 <- brm(\nformula = bf(amount_open ~ 1 + s(time_point) + (1 + time_point | Person)),\ndata = df1,\nfamily = gaussian(),\nseed = 12345\n)\n\nconditional_effects(\nfit2,\nconditions = distinct(df1, Person),\nre_formula = NULL\n)\n``````\n\nFor example, how would I make a prediction if I lend 1300 today. What range do I expect to get back after 75 % of the time?"
] |
[
null
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{"ft_lang_label":"__label__en","ft_lang_prob":0.74679995,"math_prob":0.98030823,"size":1811,"snap":"2023-40-2023-50","text_gpt3_token_len":598,"char_repetition_ratio":0.16657443,"word_repetition_ratio":0.0,"special_character_ratio":0.4616234,"punctuation_ratio":0.2568306,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.995029,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-11-28T14:16:09Z\",\"WARC-Record-ID\":\"<urn:uuid:db3ccaa3-f271-46ed-adea-baaca8d3a614>\",\"Content-Length\":\"30942\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d718724d-0793-442f-b647-3cc3c5ed55df>\",\"WARC-Concurrent-To\":\"<urn:uuid:c5dc3dc7-622e-4480-a83d-8c6a14fd332d>\",\"WARC-IP-Address\":\"74.82.16.203\",\"WARC-Target-URI\":\"https://discourse.mc-stan.org/t/fitting-model-with-decreasing-cumulative-costs/33400\",\"WARC-Payload-Digest\":\"sha1:6LECKW34FZJYNVSLPMNRYG4FDALD6LHJ\",\"WARC-Block-Digest\":\"sha1:UCQ7EJA7GKZCPYJBP7XG3RY3N4RVOVHH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679099514.72_warc_CC-MAIN-20231128115347-20231128145347-00710.warc.gz\"}"}
|
http://www.gheinz.de/historic/intro/intro_e.htm
|
[
"# Projections and Interference Pattern in Nerve Networks\n\nG. Heinz\n\n### History\n\nFollowing Konishi [1a] Lloyd A. Jeffress 1948 at Caltech [1b] had a simple idea to understand the principle of localisation of sound sources in space, Fig.1. These fundamental circuit, today we would call it an 'inter-medial interference circuit', shows important features of biological informatics in a new sight: a correspondence between a location and the transmitting time-functions, a connection between code in space and time. (Jeffress published only a different drawing, Fig.2).",
null,
"Fig.1: Konishi's drawing of Jeffress idea 1993 [1a]",
null,
"Fig.2: Original circuit of Jeffress paper 1947 [1b] - historical the first inter-medial interference circuit\n\nThe circuit Fig.1 has interesting properties:\n\n• Time-functions (impulses) expand relative to the other\n• A wire is not a potential, it has delaying properties and distributed parameters\n• Space and projection appears as a mirrored projection with the proportion dx/-dx' = v dt/ -v' dt\n• space function (length dx) and time function (lenght dt) of an impulse are combined with ds = v dx\n• The output of a detecting element is dependent of the lenght and velocity of input wires (!)\n• Logical elements (detectors) have non-deterministic behaviour\n• Instead of a neural type the location of a neurone is connected to the logical function of a neuron (!!!)\n• Data addressing is possible in short-connected networks (!) and\n• It is necessary for data addressing, to have short connected networks (!)\n• The cover of a nerve web, not the connectivity, codes functional properties (!!!)\n• The logical signal flow is not represented by wiring, but by relative delays, that are dependent of covers, velocities and locations\n• Stochastic, non-deterministic wiring is possible without general short-cut\n\nAlso if Konishi discussed periodical time functions: interference circuits works as better, as more the time functions have non-sinoidal character. Nerve pulses have velocities from µm/d...m/s, time-length in the range 0.1 µs...s and wave length in the range µm...cm.\n\nA specific detail seems that the circuit don't need different media to work. It also works between two neural fields connected by some axons.\n\nTo answer te question, why nobody has found these interesting properties over 50 years, we have to remember the history of computations to the 40's. Behind Conrad Zuse (Z1 in 1938) lots of scientists tried to develope first computers. In the beginning forties the fiction of machinery computation moved the minds. Discretisation of analog signals should be the way, to make things going in a digitized world. Electrical wires imploded to potentials, to nodes.\n\nIn the beginning fifties such 'neural' nets were qualified to learn. Delay models of wires seem to be unnecessary (Perceptron, backpropagation etc.) to understand neural networks. Perhaps Jeffress model was not of interest. Even some biologists as Konishi remembered him all times to interpret acoustical experiments.\n\nThe genius of computing and neural pioneers was, to digitize Jeffress interference circuit. A time-function was interpreted in discrete values, delay portions were called states or samples. State machines, Petri-nets and digital filters were born.\n\nWhile McCulloch/Pitts started the paper in 1943 with the words\n\n\"The velocity along the axon varies directly with its diameter, from less than one meter per second in thin axons, which are usually short, to more than 150 meters per second in thick axons...\"\n\nthey expressed neural circuits with axons and dendrites in the same minute with state sequences instead of time functions.\nWhat a mistakes with fatal consequences!\nIn the physical assumptions we read:\n\n\"...\n2. A certain fixed number of synapses must be excited within the period of latent addition in order to excite a neurone at any time, and this number is independent of previous activity and position on the neurone.\n3. The only significant delay within the nervous system is synaptic delay.\n...\"\n\nWhat a mistake! It is not possible, to find any delaying time-function in the paper. Else we find terms, characterising the new age, the age of computer technology. Instead of terms in form of delaying time-functions like\n\nf1(t) = f2(t-T1) + f3(t-T2) + ...\n\nwhere T1 and T2 are (float number-) delays of real wires proportional to distances, by analogy we find integer terms of the form\n\nN3(t) :=: .N1(t-1) .v .N2(t-3)\n\nThe numbers 1, 3 mark time-steps, the \"v\" marks, what we call today a logical OR (please excuse the simplified way to describe the formulas under HTML).\n\nThe way to describe delays as integer numbers dominated the neurocomputing science (Neural Networks) for 49 years . So all the time it was not able, to find any neural projection, because pulse waves were unknown.\n\nWhy I found out this fact in 1992? Only because I thought about slow conduction speeds of nerve pulses in dendrites: animals have to be fast to get a chance in evolution. Whats the reason, that impulses flow so slowly?\n\nSo, in one of the origins of neuro-computing we find a much simplified description of physical reality. In sight to the development of computers (state machines) this might be a very helpful way. But it covered ideas in the direction to interference, wave fields, relativity of time. The meaning of an intrinsic delay of a neural wire was lost for years. A time function in the form f(t-T) was replaced by a state- or sample sequence N(i), N(i+1), N(i+2)... .\n\nAlso forthcoming models did not note incremental delays of wires. The 'synapse' was later recognized as the learning weight of a neuron. So papers, discussing the phenomenon of non-interpretability of bio-neural networks with \"neural networks\" were common sense.\n\nIn this sight, Jeffress imagination was a short lightning flash in the dark.\n\nAt the other hand, the models of Nobel-price candidates Hodgkin/Huxley (1952) were precise enough to discover interferential projections. But, be it, as it is, these models are over-loaded, they are to complex to recognize interference projections. They are to slow, to simulate some thousands of neurones in a time analysis manner within years.\n\nToday it is time to recognize, that quantisation had produced this mistakes. At the one hand, PSI-simulations show , time and space properties have the very impact on the attainability of interference projections. At the other hand, the re-interpretation of biological networks only is possible in case of structural equivalence. The real structure of a neural net can be interpreted better with delays and time-functions then by state machines. We have forgotten, that a state sequence is a special, discrete form of a time-function. At last, time functions develope themselves physical properties characterizing delaying spaces. So a simple divergence of a wire that's ends converge diametric produces Youngs or Huygens double split pattern.\n\nI did not know Jeffress or any interference system, when I realized theoretical an electrical interference circuit in optical analogy in 1992. It had the property, to produce only mirrored projections.\n\nIn opposite to other works, calculations in this homepage are only produced using time functions and free wave propagation. Following, it don't exist a potential of a wire. Instead, we have to calculate wires with distributed parameters, potentials are connected to locations (x,y,z) in space. Moves a time-function through space, the delay term T will increase proportional to the distance from the point of excitement.\n\nA wave moving through space therefore is interpreted as a circular wave, also, if this may be a simplification for the interpretation of neural systems.\n\n### Pictures of Thought\n\nFrom nervous system we know topological maps and mirrored projections. The terminus 'projections' is used since years, although we didn't understand the reason in detail.\nComparable to mirrored projection of optical delay spaces, projections are possible between non-deterministic connected networks. Therefore we have to simulate very careful the waves, while pulses move at all possible ways through the network.\nShould biological systems partially work as interference systems, very special measuring methods are able to interpret them. PSI-Tools was developed, to calculate excitement maps from channel data. With some luck, we hope to be able to interpret first (measured) pictures of thought. That means, we try to calculate excitement maps from nerve net signals. The first simulation of a pulse-projection, transfered over three channels (axons), we got in november 1994 - it is the dark background image of the homepage.\n\n### What is an Interference (or Superimposing) Circuit?\n\nThe puls-like character of nerve signal flow offers, that boolean methods are not able to interpret informational tasks of a nerve nets. Any signal processing task in a non-deterministic wired network can only occur on a place, where different pulses meet each other. Thus logical processing appears localisation sensitive. Better: relations in space code the function of the net! May be, this is a very new idea with consequences.\nTo understand what this means, a simple circuit is used.\n\nThe most simple interference circuit can be built with an two input operating element (multiplier, adder...) that's output drives one input. Also a two input element, connected with both inputs to one node via different delays can be such an interference circuit. In my book (N.I., 1993), I discussed different simple interference circuits, showing very interesting properties in space-, time- and frequency domain.\n\nA more general form we get by connecting two fields (a generating G and a detecting D) together with some axons (A). We get a structure, that may be known from anatomical drawings. Every neurone (N, N') of both fields has a connection to every end-point of the connecting axons, whereby all wires have length-proportional delays.",
null,
"Viewing any impulse, outgoing from a generator neuron (G) in all directions, we remark, that all partial impulses under circumstances can meet each other on the detecting field (D) in the place, where delays over all possible ways are the same.\nThese simple interference circuit produces mirrored projections and images (whenever) between generating- (G) and detecting space (D). Projections of thought.\n\n### How to simulate interference networks?\n\nPSI-Tools (Parallel and Serial Interference Tools) allows to calculate simple interference circuits, consisting of generator and detector fields in space. Neural fields considered as dense connected. Delays may be proportional to distances and velocities (homogenous space).\n\nIt is able, to solve the following tasks:\n\n• Recording of parallel channel data (16 channels, 50 kSamples/s)\n• Synthesis of channel data from a virtual generator space (drawn bitmaps)\n• Processing of channel data (filtering, offset compensation, appending channel data streams...)\n• Calculation of interference integrals in area or in space (excitement maps), interference movies (wave fields), interference classes analysis, electrode maps etc.\n\nResulting excitement maps are calculable on different ways: as interference integrals or as wave fields. We calculate interference integrals point by point and movies time step by time step, but with the same algorithm, the authors H-interference transformation (HIT).\nCo-ordinates of each channel source or sink can be choosen freely. Generator and detector fields can have different velocities, origins, orientations and sizes. They are designed as bitmaps with variable and independent matrix- and physical dimensions. Timing, sample rate, pulses form and intervall, pulse distances can be varied.\nThe help-file of PSI-Tools is downloadable from the references chapter on this homepage.\n\n### Reconstruction contra Projection\n\nExcitement maps are interesting in two directions: first, technical tasks demand reconstruction of the original space, of the generator space. Second, we are interested to calculate projections onto other spaces to simulate neural fields.\nA projection is mathematical dual to a reconstruction on the same field through an inversed time function. Default is the reconstruction with growing time values. To produce a projection, one has to revert the channel data.",
null,
"Fig.: Reconstruction (top) and projection (bottom) as interference integral from the same, synthesized channel data stream. Both fields have identical properties, velocity is the same. Channels have the same delay. Because of over-conditioning (one channel to much) we find high projection quality only in the center range.\n\nWe can recognize two conditions for projections: 1) geometrical impulse length have to be smaller then structures, 2) small channel numbers supposed, the pause between following impulses has to be large against the (time domain) field size. Time- and space-units are combined via medial velocity.\n\n### Time Sequence Maps contra Projections\n\nBehind image-like elements channel data also carry sensoric amplitudes. Thus the result of an interference analysis are non-avoidable projective elements combined with interference pattern. Following, the interference transformation calculates a map of interference pattern (sound- or noise-picture) in the same way, as a projection or reconstruction. The interferencial pattern component can be seen, when pulses reach the detector field in fast rates. This can be a consequence of very small pausing intervals of transmitting axons. Then a wave comes in interference not only with itself, together a wave interferes with following and preceding waves. Consequently, the interference integral appears mapped into interference pattern, independent, if we calculate a reconstruction or projection. In case, subsequent impulses follows in shortest sequence, the total detector field will come into interference, projections disappear. May be, this could be an interpretation of pain from physical point of view.",
null,
"Fig.: Part of an four channel interference movie (PSI-Tools): snap shot of four waves, moving over the detecting field. The channel source points are located in the corners of the field. We observe the time step short after a four-times interference.\n\n### Conclusion\n\nThe team has since 1994 the chance to inspect simulated and recorded interference projections. It is possible, to simulate, to analyse or to develop simple interference circuits. Also it is possible, to record or synthesize channel data. PSI-Tools can reconstruct excitement maps of the exciting space even, as PSI-Tools can calculate projections in a second space. Projections appear mirrored or holographic, while reconstructions appear non-mirrored.\nBehind neural applications PSI-Tools is useful, to reconstruct EEG-, ECoG-, EKG- or US-data streams. It is possible, to compute interference integrals as excitement maps or as wave field movies.\n\n### References\n\n[1a] Konishi, M.: Die Schallortung der Schleiereule. Spektrum der Wissenschaft, Juni 1993, S. 58 ff.\n\n[1b] Jeffress, L.A.: A place theory of sound localization. Journ. Comparative Physiol. Psychol., 41, (1948), pp.35–39\n\n McCulloch, W.S.; Pitts, W.: A logical calculus of the ideas immanent in nervous activity. Bulletin of Mathematical Biophysics 5:115-133\n\n Widrow, B., Hoff, M.E. Adaptive switching circuits. 1960 IRE WESCON Convention Record, New York: IRE, pp. 96-104\n\n Rumelhart, D.E., McClelland, J.L.: A Distributed Model of Human Learning and Memory. in: Parallel Distributed Processing. Bradford/MIT Press Cambridge, Massachusetts, vol. 2, eighth printing 1988.\n\n Hodgkin, A.L., Huxley, A.F.: A Quantitative Description of Membrane Current and Its Application to Conduction and Excitation in Nerve. Journ. Physiology, London, 117 (1952) pp. 500-544\n\n About the thumb-experiment see IWK Ilmenau, 1994 (german)\n\nPage-URL: historic/intro/intro.htm\n\nFile created sept. 30, 1995\n\nVisitors since Dec. 2021:"
] |
[
null,
"http://www.gheinz.de/historic/intro/konishi.gif",
null,
"http://www.gheinz.de/historic/intro/jeffress.gif",
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"http://www.gheinz.de/historic/intro/gen4det.gif",
null,
"http://www.gheinz.de/historic/intro/gh_projc.gif",
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"http://www.gheinz.de/historic/intro/s_4wav1.gif",
null
] |
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|
https://tex.stackexchange.com/questions/210971/tikz-node-with-rounded-corners
|
[
"Tikz Node with rounded corners\n\nI refer to this question: LaTeX Photo With Rounded Corners\n\nI would like to round the corners of a Tikz node (which actually is a fitted node around two other nodes), but would like to have an additional black frame around it. If I just draw a black rounded frame around my node, it has to be very thick depending on the amount of clipping.\n\nIs there a solution where I first remove the corners and then draw a black frame around the \"rounded\" node?\n\nHere is an example:\n\n\\documentclass{article}\n\\usepackage{tikz}\n\\usetikzlibrary{fit,positioning}\n\n\\begin{document}\n\\begin{tikzpicture}\n\\node [inner sep=0pt,minimum size=4.0cm,outer sep=0pt,clip] (pict) at (0,0) {\\includegraphics[width=4.0cm]{images/black.png}};\n\\node [inner sep=0pt,minimum size=4.0cm,outer sep=0pt,clip,below = of pict] (pict2) at (0,0) {\\Huge Hello};\n\\node[draw=red,thick,fit=(pict)(pict2),rounded corners=.55cm,inner sep=2pt] {};\n\\end{tikzpicture}\n\\end{document}\n\n\\documentclass{article}\n\\usepackage{tikz}\n\\usetikzlibrary{fit}\n\n\\begin{document}\n\\begin{tikzpicture}\n\\node [inner sep=0pt,,outer sep=0pt,clip,rounded corners=0.5cm] (pict) at (0,0) {\\includegraphics[width=4.0cm]{example-image-a}};\n\\node[draw=red,thick,fit=(pict),rounded corners=.55cm,inner sep=2pt] {};\n\n\\node [inner sep=0pt,,outer sep=0pt,clip,rounded corners=0.5cm] (pict1) at (6,0) {\\includegraphics[width=5.0cm]{example-image-b}};\n\\node[draw=olive,thick,fit=(pict1),rounded corners=.55cm,inner sep=2pt] {};\n\\end{tikzpicture}\n\\end{document}",
null,
"\\documentclass{article}\n\\usepackage{tikz}\n\\usetikzlibrary{fit,positioning}\n\n\\begin{document}\n\\begin{tikzpicture}\n\\clip [rounded corners=.5cm] (0,0) rectangle (4,8);\n\\node [inner sep=0pt,minimum size=4cm,outer sep=0pt] (pict) at (2,6)\n{\\includegraphics[height=4.0cm]{example-image-a}};\n\\node [draw,inner sep=0pt,minimum size=4cm,outer sep=0pt,clip] (pict2) at (2,2)\n{\\includegraphics[height=4.0cm]{example-image-b}};\n\\draw [red,line width=1pt,rounded corners=.5cm]\n([shift={(0.5\\pgflinewidth,0.5\\pgflinewidth)}]0,0) rectangle\n([shift={(-0.5\\pgflinewidth,-0.5\\pgflinewidth)}]4,8);\n\\end{tikzpicture}\n\\end{document}",
null,
"• Ok maybe I missed something. Basically I have three nodes. Two are aligned vertically and a third one is fitted around them. The third one is the one which should have the round corners. The problem is, that I cannot use round corners for the inner two nodes, as the upper one is supposed to have only round nodes at the top, and the lower one at the bottom. I added an example to the question. – Simon Nov 7 '14 at 12:43\n• Thanks. Your second solution brought me to my own. The problem with your solution (for me) is, that all sizes have to be known in advance. I fit a white node to both inner nodes and then use these coordinates and the calc library for drawing the black border. – Simon Nov 8 '14 at 10:26\n\nAn alternative is the use of corners of current bounding box, form an arc area to fill with a white color to mimic a clip. Changing the \\path commands to \\draw commands sees the arc areas.",
null,
"Code\n\n\\documentclass[border=10pt]{standalone}\n%\\usepackage[]{graphicx}\n\\usepackage{tikz}\n\\usetikzlibrary{fit,positioning}\n\\begin{document}\n\\begin{tikzpicture}\n\\node [inner sep=0pt,minimum size=4.0cm,outer sep=0pt,clip] (pict) at (0,0) {\\includegraphics[width=4.0cm]{example-image-a}};\n\n\\path[fill=white] ([yshift=-14pt]current bounding box.north west) -- ++(1,0) arc(90:180:1) --cycle; %<--\n\\path[fill=white] ([yshift=-14pt]current bounding box.north east) -- ++(-1,0) arc (90:0:1) --cycle; %<--\n\n\\node [inner sep=0pt,minimum size=4.0cm,outer sep=0pt,below = of pict,clip] (pict2) at (0,0) {\\includegraphics[width=4.0cm]{example-image-b}};\n\n\\path[fill=white] ([yshift=14pt]current bounding box.south west) -- ++(1,0) arc(-90:-180:1) --cycle; %<--\n\\path[fill=white] ([yshift=14pt]current bounding box.south east) -- ++(-1,0) arc(-90:0:1) -- cycle; %<--\n\n\\node[draw=red,thick,fit=(pict)(pict2), rounded corners=.55cm,inner sep=2pt] {};\n\\end{tikzpicture}\n\\end{document}\n\nHere is my own solution. It was heavily inspired by Harish Kumar's answer. I am open for improvements ;-)\n\n\\documentclass{article}\n\\usepackage{tikz}\n\\usetikzlibrary{fit,positioning,calc}\n\n\\begin{document}\n\\begin{tikzpicture}[node distance=0]\n\\node [inner sep=0pt,outer sep=0pt] (pict)\n{\\includegraphics[height=4.0cm]{example-image-a}};\n\\node [inner sep=0pt,outer sep=0pt,below = of pict] (pict2)\n{\\includegraphics[height=4.0cm]{example-image-b}};\n\\node [draw=red,rounded corners=1pt,line width=3pt,\ninner sep=0pt,outer sep=0pt,fit=(pict)(pict2)] (pict3) {};\n\\draw [blue,line width=1pt,rounded corners=1pt]\n($(pict3.south west)+(\\pgflinewidth,\\pgflinewidth)$) rectangle\n($(pict3.north east)-(\\pgflinewidth,\\pgflinewidth)$);\n\\end{tikzpicture}\n\\end{document}",
null,
""
] |
[
null,
"https://i.stack.imgur.com/R4LYv.png",
null,
"https://i.stack.imgur.com/STnf6.png",
null,
"https://i.stack.imgur.com/I0Npm.jpg",
null,
"https://i.stack.imgur.com/GBsqk.png",
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.8394981,"math_prob":0.9484391,"size":927,"snap":"2019-43-2019-47","text_gpt3_token_len":281,"char_repetition_ratio":0.11700975,"word_repetition_ratio":0.0,"special_character_ratio":0.2588997,"punctuation_ratio":0.15263158,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9953944,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,3,null,3,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-21T05:59:48Z\",\"WARC-Record-ID\":\"<urn:uuid:ea993883-eec9-43dd-9ce6-e0db4cfe8fca>\",\"Content-Length\":\"154207\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:54abfbd1-dff9-423f-8b4b-9d3bbde3ccc6>\",\"WARC-Concurrent-To\":\"<urn:uuid:f1d0cc0b-cca7-43e6-8dfe-01d9cb98ca73>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://tex.stackexchange.com/questions/210971/tikz-node-with-rounded-corners\",\"WARC-Payload-Digest\":\"sha1:7JH4RG7BTMEBXGLA4MUEBN2WV3WZT7KG\",\"WARC-Block-Digest\":\"sha1:UMUG7YKUHNZL5XDWTMCAK4FMD2LSONTN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987756350.80_warc_CC-MAIN-20191021043233-20191021070733-00092.warc.gz\"}"}
|
https://kode-blog.io/csharp-basics
|
[
"##### Contact Information\n\nKodeBlog\n\nWe're Available 24/ 7. Drop us an email.\n\n# C# Basics\n\n## Introduction\n\nC# Basics: This tutorial sets the foundation for learning C# programming. It covers the basic topics that every C# programmer should know.\n\n## Topics to be covered\n\n• C# Syntax\n• Keywords\n• Variables\n• Operators\n• Arithmetic\n• Comparison\n• Logical\n• Namespaces\n• C# is a compiled language\n• Errors, warnings and messages\n\n## C# Syntax\n\nThe C# syntax is inspired by C and C++. If you are familiar with languages such as JAVA and PHP, then it will be easy for you to learn C#. If you are not familiar with any of the languages mentioned above, this tutorial will get you up to speed.\n\n### C# Syntax Rules\n\n• All statements in C# end with semi colons.\n• All variables in C# must be defined before using them.\n• The class and method bodies must be enclosed in opening and closing braces\n\nThe code shown below is for a simple console application that prints “Hello World!” On the screen\n\n``````using System;\n\nnamespace CSharpBasics\n{\nclass Program\n{\nstatic void Main(string[] args)\n{\nstring msg = \"Hello World!\";\nConsole.WriteLine(msg);\n}\n}\n}\n``````\n\nHERE,\n\n• `using System;` imports the `System` namespace. The semi colon is used to end the C# statement\n• class Program{...} declares a class Program. { opens the body of the class and } closes the body of the class.\n• static void Main(string[] args) is the method that is executed when the program runs.\n\nThe above code produces the output\n\n``````Prints Hello World! in the console window\n``````\n\n## Keywords\n\nA keyword is a reserved word that has special meaning to the compiler. An example of a keyword C# is;\n\n``````using\n``````\n\n`using` is used to import a namespace.\n\nKeywords cannot be used as identifiers. Other examples of C# keywords include\n\n``````int\ndecimal\nnamespace\nclass\n``````\n\n## Variables\n\nA variable is a memory location used to store values. The stored values can change at any time.\n\n#### C# Variables\n\nIn C#, all variables must be declared before using them. Variable names are also case sensitive\n\nThe following code declares two variables\n\n``````int i = 0;\nstring message = \"Welcome\";\n``````\n\nHERE,\n\n`int i = 0;` defines a variable `i` of data type `int`. The variable `i` is initialized to a value of `0` value.\n\n``````string message = \"Welcome\";\n``````\n\ndefines a variable `message` of `string` data type. The variable `message` is initialized to a value of `Welcome` value.\n\n### Variable Naming Rules\n\n• Variable names should not contain spaces\n• A variable name should not be a keyword. If you want to use a key work as a variable name then you must append `@` symbol at the beginning of the variable name\n• A data type must be specified for all declared variables\n\n## Operators\n\nIn C#, an operator is a program element that is applied to one or more operands in an expression or statement. Operators are used to manipulate data. For example, the plus (+) operator is used to sum up numbers.\n\nC# has three major categories of operators\n\n• Arithmetic - used to manipulate numeric data\n• Subtraction (-) - used to subtract numbers\n• Multiplication (*) - used to multiply numbers\n• Division (/) - used to divide numbers\n• Comparison - used to compare values\n• Equal to (==) - compares two values and returns true if they are equal. False is returned if they are not equal.\n• Not Equal To (!=) - compares two values and returns true if they are not equal. False is returned if they are equal\n• Greater Than (>) - used to compare two numbers. Returns true if the number on the left hand side is larger than the number on the right hand side.\n• Greater Than Or Equal To(>=) - used to compare two numbers. Returns true if the number on the left hand side is either greater than or equal to the number on the right hand side.\n• Less Than - used to compare two numbers. Returns true if the number on the left hand side is less than the number on the right hand side.\n• Less Than Or Equal To(<=) - used to compare two numbers. Returns true if the number on the left hand side is either less than or equal to the number on the right hand side.\n• Logical - used when working with values that evaluate to either true or false\n• AND - used when evaluating more than one condition. If both conditions evaluate to true, AND returns true. If any of the conditions is false, AND returns false\n• OR - used when evaluating more than one condition. If any of the conditions evaluates to true, OR returns true. If all of the conditions are false, OR returns false.\n\nThese are statements that will be ignored by the compiler. They help developers document the source codes. C# supports single and multiple line comments as shown below.\n\n• Single comments start with `//` two forward slashes\n• Multiple line comments start with `/*` a forward slash followed by an asterisk. The closing tag for multiple line comments is `*/` an asterisk and a forward slash\n\n``````//single line comment\n/*\nthis is a multi-line\ncomment */```\n```\n\n## Namespaces\n\nNamespaces are used to group similar classes together. Let’s suppose that you have developed a database library, you can create namespaces for MS SQL Server, MySQL, Oracle etc. this is very helpful in avoiding identifier name crashes. The same identifiers can be used in different classes without any conflicts.\n\n#### How to Define a Namespace in C#\n\nA namespace is defined using the keyword namespace\n\nThe following shows an example\n\n``````namespace MySQL {\nclass Connect(){\n\n}\n}\n\n//importing a namespace\nusing MySQL;\n``````\n\n## Summary\n\n• The syntax for C# is inspired by C and C++\n• Identifiers are names used to describe classes, methods, variables etc.\n• Keywords are words that have special meaning in C#\n• Variables are like containers, they are used to store data.\n• The major categories of operators in C# are; arithmetic, comparison and logical.\n• Namespace are used to group similar classes and methods together\n\n## Tutorial History\n\nTutorial version 1: Date Published 2015-08-21",
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"### Author: Rodrick Kazembe\n\nRodrick is a developer who works on Desktop, Web and Mobile Applications. He is familiar with Python, Java, JavaScript, C++, C#, Kotlin, PHP, Python and the list goes on. Rodrick enjoys sharing knowledge especially when it comes to technology."
] |
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"https://www.gravatar.com/avatar/893b9a2d59eedb398a90a32853eb28ea",
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https://creditcardfree.savingadvice.com/2018/12/
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"User Real IP - 34.228.229.51\n```Array\n(\n => Array\n(\n => 182.68.68.92\n)\n\n => Array\n(\n => 101.0.41.201\n)\n\n => Array\n(\n => 43.225.98.123\n)\n\n => Array\n(\n => 2.58.194.139\n)\n\n => Array\n(\n => 46.119.197.104\n)\n\n => Array\n(\n => 45.249.8.93\n)\n\n => Array\n(\n => 103.12.135.72\n)\n\n => Array\n(\n => 157.35.243.216\n)\n\n => Array\n(\n => 209.107.214.176\n)\n\n => Array\n(\n => 5.181.233.166\n)\n\n => Array\n(\n => 106.201.10.100\n)\n\n => Array\n(\n => 36.90.55.39\n)\n\n => Array\n(\n => 119.154.138.47\n)\n\n => Array\n(\n => 51.91.31.157\n)\n\n => Array\n(\n => 182.182.65.216\n)\n\n => Array\n(\n => 157.35.252.63\n)\n\n => Array\n(\n => 14.142.34.163\n)\n\n => Array\n(\n => 178.62.43.135\n)\n\n => Array\n(\n => 43.248.152.148\n)\n\n => Array\n(\n => 222.252.104.114\n)\n\n => Array\n(\n => 209.107.214.168\n)\n\n => Array\n(\n => 103.99.199.250\n)\n\n => Array\n(\n => 178.62.72.160\n)\n\n => Array\n(\n => 27.6.1.170\n)\n\n => Array\n(\n => 182.69.249.219\n)\n\n => Array\n(\n => 110.93.228.86\n)\n\n => Array\n(\n => 72.255.1.98\n)\n\n => Array\n(\n => 182.73.111.98\n)\n\n => Array\n(\n => 45.116.117.11\n)\n\n => Array\n(\n => 122.15.78.189\n)\n\n => Array\n(\n => 14.167.188.234\n)\n\n => Array\n(\n => 223.190.4.202\n)\n\n => Array\n(\n => 202.173.125.19\n)\n\n => Array\n(\n => 103.255.5.32\n)\n\n => Array\n(\n => 39.37.145.103\n)\n\n => Array\n(\n => 140.213.26.249\n)\n\n => Array\n(\n => 45.118.166.85\n)\n\n => Array\n(\n => 102.166.138.255\n)\n\n => Array\n(\n => 77.111.246.234\n)\n\n => Array\n(\n => 45.63.6.196\n)\n\n => Array\n(\n => 103.250.147.115\n)\n\n => Array\n(\n => 223.185.30.99\n)\n\n => Array\n(\n => 103.122.168.108\n)\n\n => Array\n(\n => 123.136.203.21\n)\n\n => Array\n(\n => 171.229.243.63\n)\n\n => Array\n(\n => 153.149.98.149\n)\n\n => Array\n(\n => 223.238.93.15\n)\n\n => Array\n(\n => 178.62.113.166\n)\n\n => Array\n(\n => 101.162.0.153\n)\n\n => Array\n(\n => 121.200.62.114\n)\n\n => Array\n(\n => 14.248.77.252\n)\n\n => Array\n(\n => 95.142.117.29\n)\n\n => Array\n(\n => 150.129.60.107\n)\n\n => Array\n(\n => 94.205.243.22\n)\n\n => Array\n(\n => 115.42.71.143\n)\n\n => Array\n(\n => 117.217.195.59\n)\n\n => Array\n(\n => 182.77.112.56\n)\n\n => Array\n(\n => 182.77.112.108\n)\n\n => Array\n(\n => 41.80.69.10\n)\n\n => Array\n(\n => 117.5.222.121\n)\n\n => Array\n(\n => 103.11.0.38\n)\n\n => Array\n(\n => 202.173.127.140\n)\n\n => Array\n(\n => 49.249.249.50\n)\n\n => Array\n(\n => 116.72.198.211\n)\n\n => Array\n(\n => 223.230.54.53\n)\n\n => Array\n(\n => 102.69.228.74\n)\n\n => Array\n(\n => 39.37.251.89\n)\n\n => Array\n(\n => 39.53.246.141\n)\n\n => Array\n(\n => 39.57.182.72\n)\n\n => Array\n(\n => 209.58.130.210\n)\n\n => Array\n(\n => 104.131.75.86\n)\n\n => Array\n(\n => 106.212.131.255\n)\n\n => Array\n(\n => 106.212.132.127\n)\n\n => Array\n(\n => 223.190.4.60\n)\n\n => Array\n(\n => 103.252.116.252\n)\n\n => Array\n(\n => 103.76.55.182\n)\n\n => Array\n(\n => 45.118.166.70\n)\n\n => Array\n(\n => 103.93.174.215\n)\n\n => Array\n(\n => 5.62.62.142\n)\n\n => Array\n(\n => 182.179.158.156\n)\n\n => Array\n(\n => 39.57.255.12\n)\n\n => Array\n(\n => 39.37.178.37\n)\n\n => Array\n(\n => 182.180.165.211\n)\n\n => Array\n(\n => 119.153.135.17\n)\n\n => Array\n(\n => 72.255.15.244\n)\n\n => Array\n(\n => 139.180.166.181\n)\n\n => Array\n(\n => 70.119.147.111\n)\n\n => Array\n(\n => 106.210.40.83\n)\n\n => Array\n(\n => 14.190.70.91\n)\n\n => Array\n(\n => 202.125.156.82\n)\n\n => Array\n(\n => 115.42.68.38\n)\n\n => Array\n(\n => 102.167.13.108\n)\n\n => Array\n(\n => 117.217.192.130\n)\n\n => Array\n(\n => 205.185.223.156\n)\n\n => Array\n(\n => 171.224.180.29\n)\n\n => Array\n(\n => 45.127.45.68\n)\n\n => Array\n(\n => 195.206.183.232\n)\n\n => Array\n(\n => 49.32.52.115\n)\n\n => Array\n(\n => 49.207.49.223\n)\n\n => Array\n(\n => 45.63.29.61\n)\n\n => Array\n(\n => 103.245.193.214\n)\n\n => Array\n(\n => 39.40.236.69\n)\n\n => Array\n(\n => 62.80.162.111\n)\n\n => Array\n(\n => 45.116.232.56\n)\n\n => Array\n(\n => 45.118.166.91\n)\n\n => Array\n(\n => 180.92.230.234\n)\n\n => Array\n(\n => 157.40.57.160\n)\n\n => Array\n(\n => 110.38.38.130\n)\n\n => Array\n(\n => 72.255.57.183\n)\n\n => Array\n(\n => 182.68.81.85\n)\n\n => Array\n(\n => 39.57.202.122\n)\n\n => Array\n(\n => 119.152.154.36\n)\n\n => Array\n(\n => 5.62.62.141\n)\n\n => Array\n(\n => 119.155.54.232\n)\n\n => Array\n(\n => 39.37.141.22\n)\n\n => Array\n(\n => 183.87.12.225\n)\n\n => Array\n(\n => 107.170.127.117\n)\n\n => Array\n(\n => 125.63.124.49\n)\n\n => Array\n(\n => 39.42.191.3\n)\n\n => Array\n(\n => 116.74.24.72\n)\n\n => Array\n(\n => 46.101.89.227\n)\n\n => Array\n(\n => 202.173.125.247\n)\n\n => Array\n(\n => 39.42.184.254\n)\n\n => Array\n(\n => 115.186.165.132\n)\n\n => Array\n(\n => 39.57.206.126\n)\n\n => Array\n(\n => 103.245.13.145\n)\n\n => Array\n(\n => 202.175.246.43\n)\n\n => Array\n(\n => 192.140.152.150\n)\n\n => Array\n(\n => 202.88.250.103\n)\n\n => Array\n(\n => 103.248.94.207\n)\n\n => Array\n(\n => 77.73.66.101\n)\n\n => Array\n(\n => 104.131.66.8\n)\n\n => Array\n(\n => 113.186.161.97\n)\n\n => Array\n(\n => 222.254.5.7\n)\n\n => Array\n(\n => 223.233.67.247\n)\n\n => Array\n(\n => 171.249.116.146\n)\n\n => Array\n(\n => 47.30.209.71\n)\n\n => Array\n(\n => 202.134.13.130\n)\n\n => Array\n(\n => 27.6.135.7\n)\n\n => Array\n(\n => 107.170.186.79\n)\n\n => Array\n(\n => 103.212.89.171\n)\n\n => Array\n(\n => 117.197.9.77\n)\n\n => Array\n(\n => 122.176.206.233\n)\n\n => Array\n(\n => 192.227.253.222\n)\n\n => Array\n(\n => 182.188.224.119\n)\n\n => Array\n(\n => 14.248.70.74\n)\n\n => Array\n(\n => 42.118.219.169\n)\n\n => Array\n(\n => 110.39.146.170\n)\n\n => Array\n(\n => 119.160.66.143\n)\n\n => Array\n(\n => 103.248.95.130\n)\n\n => Array\n(\n => 27.63.152.208\n)\n\n => Array\n(\n => 49.207.114.96\n)\n\n => Array\n(\n => 102.166.23.214\n)\n\n => Array\n(\n => 175.107.254.73\n)\n\n => Array\n(\n => 103.10.227.214\n)\n\n => Array\n(\n => 202.143.115.89\n)\n\n => Array\n(\n => 110.93.227.187\n)\n\n => Array\n(\n => 103.140.31.60\n)\n\n => Array\n(\n => 110.37.231.46\n)\n\n => Array\n(\n => 39.36.99.238\n)\n\n => Array\n(\n => 157.37.140.26\n)\n\n => Array\n(\n => 43.246.202.226\n)\n\n => Array\n(\n => 137.97.8.143\n)\n\n => Array\n(\n => 182.65.52.242\n)\n\n => Array\n(\n => 115.42.69.62\n)\n\n => Array\n(\n => 14.143.254.58\n)\n\n => Array\n(\n => 223.179.143.236\n)\n\n => Array\n(\n => 223.179.143.249\n)\n\n => Array\n(\n => 103.143.7.54\n)\n\n => Array\n(\n => 223.179.139.106\n)\n\n => Array\n(\n => 39.40.219.90\n)\n\n => Array\n(\n => 45.115.141.231\n)\n\n => Array\n(\n => 120.29.100.33\n)\n\n => Array\n(\n => 112.196.132.5\n)\n\n => Array\n(\n => 202.163.123.153\n)\n\n => Array\n(\n => 5.62.58.146\n)\n\n => Array\n(\n => 39.53.216.113\n)\n\n => Array\n(\n => 42.111.160.73\n)\n\n => Array\n(\n => 107.182.231.213\n)\n\n => Array\n(\n => 119.82.94.120\n)\n\n => Array\n(\n => 178.62.34.82\n)\n\n => Array\n(\n => 203.122.6.18\n)\n\n => Array\n(\n => 157.42.38.251\n)\n\n => Array\n(\n => 45.112.68.222\n)\n\n => Array\n(\n => 49.206.212.122\n)\n\n => Array\n(\n => 104.236.70.228\n)\n\n => Array\n(\n => 42.111.34.243\n)\n\n => Array\n(\n => 84.241.19.186\n)\n\n => Array\n(\n => 89.187.180.207\n)\n\n => Array\n(\n => 104.243.212.118\n)\n\n => Array\n(\n => 104.236.55.136\n)\n\n => Array\n(\n => 106.201.16.163\n)\n\n => Array\n(\n => 46.101.40.25\n)\n\n => Array\n(\n => 45.118.166.94\n)\n\n => Array\n(\n => 49.36.128.102\n)\n\n => Array\n(\n => 14.142.193.58\n)\n\n => Array\n(\n => 212.79.124.176\n)\n\n => Array\n(\n => 45.32.191.194\n)\n\n => Array\n(\n => 105.112.107.46\n)\n\n => Array\n(\n => 106.201.14.8\n)\n\n => Array\n(\n => 110.93.240.65\n)\n\n => Array\n(\n => 27.96.95.177\n)\n\n => Array\n(\n => 45.41.134.35\n)\n\n => Array\n(\n => 180.151.13.110\n)\n\n => Array\n(\n => 101.53.242.89\n)\n\n => Array\n(\n => 115.186.3.110\n)\n\n => Array\n(\n => 171.49.185.242\n)\n\n => Array\n(\n => 115.42.70.24\n)\n\n => Array\n(\n => 45.128.188.43\n)\n\n => Array\n(\n => 103.140.129.63\n)\n\n => Array\n(\n => 101.50.113.147\n)\n\n => Array\n(\n => 103.66.73.30\n)\n\n => Array\n(\n => 117.247.193.169\n)\n\n => Array\n(\n => 120.29.100.94\n)\n\n => Array\n(\n => 42.109.154.39\n)\n\n => Array\n(\n => 122.173.155.150\n)\n\n => Array\n(\n => 45.115.104.53\n)\n\n => Array\n(\n => 116.74.29.84\n)\n\n => Array\n(\n => 101.50.125.34\n)\n\n => Array\n(\n => 45.118.166.80\n)\n\n => Array\n(\n => 91.236.184.27\n)\n\n => Array\n(\n => 113.167.185.120\n)\n\n => Array\n(\n => 27.97.66.222\n)\n\n => Array\n(\n => 43.247.41.117\n)\n\n => Array\n(\n => 23.229.16.227\n)\n\n => Array\n(\n => 14.248.79.209\n)\n\n => Array\n(\n => 117.5.194.26\n)\n\n => Array\n(\n => 117.217.205.41\n)\n\n => Array\n(\n => 114.79.169.99\n)\n\n => Array\n(\n => 103.55.60.97\n)\n\n => Array\n(\n => 182.75.89.210\n)\n\n => Array\n(\n => 77.73.66.109\n)\n\n => Array\n(\n => 182.77.126.139\n)\n\n => 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124.253.195.172\n)\n\n => Array\n(\n => 203.142.218.149\n)\n\n => Array\n(\n => 157.43.165.180\n)\n\n => Array\n(\n => 39.40.242.57\n)\n\n => Array\n(\n => 103.92.43.150\n)\n\n => Array\n(\n => 39.42.133.202\n)\n\n => Array\n(\n => 119.160.66.11\n)\n\n => Array\n(\n => 138.68.3.7\n)\n\n => Array\n(\n => 210.56.125.226\n)\n\n => Array\n(\n => 157.50.4.249\n)\n\n => Array\n(\n => 124.253.81.162\n)\n\n => Array\n(\n => 103.240.235.141\n)\n\n => Array\n(\n => 132.154.128.20\n)\n\n => Array\n(\n => 49.156.115.37\n)\n\n => Array\n(\n => 45.133.7.48\n)\n\n => Array\n(\n => 122.161.49.137\n)\n\n => Array\n(\n => 202.47.46.31\n)\n\n => Array\n(\n => 192.140.145.148\n)\n\n => Array\n(\n => 202.14.123.10\n)\n\n => Array\n(\n => 122.161.53.98\n)\n\n => Array\n(\n => 124.253.114.113\n)\n\n => Array\n(\n => 103.227.70.34\n)\n\n => Array\n(\n => 223.228.175.227\n)\n\n => Array\n(\n => 157.39.119.110\n)\n\n => Array\n(\n => 180.188.224.231\n)\n\n => Array\n(\n => 132.154.188.85\n)\n\n => Array\n(\n => 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Array\n(\n => 169.149.195.121\n)\n\n => Array\n(\n => 198.16.76.27\n)\n\n => Array\n(\n => 157.43.192.188\n)\n\n => Array\n(\n => 119.155.244.221\n)\n\n => Array\n(\n => 39.51.242.216\n)\n\n => Array\n(\n => 39.57.180.158\n)\n\n => Array\n(\n => 134.202.32.5\n)\n\n => Array\n(\n => 122.176.139.205\n)\n\n => Array\n(\n => 151.243.50.9\n)\n\n => Array\n(\n => 39.52.99.161\n)\n\n => Array\n(\n => 136.144.33.95\n)\n\n => Array\n(\n => 157.37.205.216\n)\n\n => Array\n(\n => 217.138.220.134\n)\n\n => Array\n(\n => 41.140.106.65\n)\n\n => Array\n(\n => 39.37.253.126\n)\n\n => Array\n(\n => 103.243.44.240\n)\n\n => Array\n(\n => 157.46.169.29\n)\n\n => Array\n(\n => 92.119.177.122\n)\n\n => Array\n(\n => 196.240.60.21\n)\n\n => Array\n(\n => 122.161.6.246\n)\n\n => Array\n(\n => 117.202.162.46\n)\n\n => Array\n(\n => 205.164.137.120\n)\n\n => Array\n(\n => 171.237.79.241\n)\n\n => Array\n(\n => 198.16.76.28\n)\n\n => Array\n(\n => 103.100.4.151\n)\n\n => Array\n(\n => 178.239.162.236\n)\n\n => Array\n(\n => 106.197.31.240\n)\n\n => Array\n(\n => 122.168.179.251\n)\n\n => Array\n(\n => 39.37.167.126\n)\n\n => Array\n(\n => 171.48.8.115\n)\n\n => Array\n(\n => 157.44.152.14\n)\n\n => Array\n(\n => 103.77.43.219\n)\n\n => Array\n(\n => 122.161.49.38\n)\n\n => Array\n(\n => 122.161.52.83\n)\n\n => Array\n(\n => 122.173.108.210\n)\n\n => Array\n(\n => 60.254.109.92\n)\n\n => Array\n(\n => 103.57.85.75\n)\n\n => Array\n(\n => 106.0.58.36\n)\n\n => Array\n(\n => 122.161.49.212\n)\n\n => Array\n(\n => 27.255.182.159\n)\n\n => Array\n(\n => 116.75.230.159\n)\n\n => Array\n(\n => 122.173.152.133\n)\n\n => Array\n(\n => 129.0.79.247\n)\n\n => Array\n(\n => 223.228.163.44\n)\n\n => Array\n(\n => 103.168.78.82\n)\n\n => Array\n(\n => 39.59.67.124\n)\n\n => Array\n(\n => 182.69.19.120\n)\n\n => Array\n(\n => 196.202.236.195\n)\n\n => Array\n(\n => 137.59.225.206\n)\n\n => Array\n(\n => 143.110.209.194\n)\n\n => Array\n(\n => 117.201.233.91\n)\n\n => Array\n(\n => 37.120.150.107\n)\n\n => Array\n(\n => 58.65.222.10\n)\n\n => Array\n(\n => 202.47.43.86\n)\n\n => Array\n(\n => 106.206.223.234\n)\n\n => Array\n(\n => 5.195.153.158\n)\n\n => Array\n(\n => 223.227.127.243\n)\n\n => Array\n(\n => 103.165.12.222\n)\n\n => Array\n(\n => 49.36.185.189\n)\n\n => Array\n(\n => 59.96.92.57\n)\n\n => Array\n(\n => 203.194.104.235\n)\n\n => Array\n(\n => 122.177.72.33\n)\n\n => Array\n(\n => 106.213.126.40\n)\n\n => Array\n(\n => 45.127.232.69\n)\n\n => Array\n(\n => 156.146.59.39\n)\n\n => Array\n(\n => 103.21.184.11\n)\n\n => Array\n(\n => 106.212.47.59\n)\n\n => Array\n(\n => 182.179.137.235\n)\n\n => Array\n(\n => 49.36.178.154\n)\n\n => Array\n(\n => 171.48.7.128\n)\n\n => Array\n(\n => 119.160.57.96\n)\n\n => Array\n(\n => 197.210.79.92\n)\n\n => Array\n(\n => 36.255.45.87\n)\n\n => Array\n(\n => 47.31.219.47\n)\n\n => Array\n(\n => 122.161.51.160\n)\n\n => Array\n(\n => 103.217.123.129\n)\n\n => Array\n(\n => 59.153.16.12\n)\n\n => Array\n(\n => 103.92.43.226\n)\n\n => Array\n(\n => 47.31.139.139\n)\n\n => Array\n(\n => 210.2.140.18\n)\n\n => Array\n(\n => 106.210.33.219\n)\n\n => Array\n(\n => 175.107.203.34\n)\n\n => Array\n(\n => 146.196.32.144\n)\n\n => Array\n(\n => 103.12.133.121\n)\n\n => Array\n(\n => 103.59.208.182\n)\n\n => Array\n(\n => 157.37.190.232\n)\n\n => Array\n(\n => 106.195.35.201\n)\n\n => Array\n(\n => 27.122.14.83\n)\n\n => Array\n(\n => 194.193.44.5\n)\n\n => Array\n(\n => 5.62.43.245\n)\n\n => Array\n(\n => 103.53.80.50\n)\n\n => Array\n(\n => 47.29.142.233\n)\n\n => Array\n(\n => 154.6.20.63\n)\n\n => Array\n(\n => 173.245.203.128\n)\n\n => Array\n(\n => 103.77.43.231\n)\n\n => Array\n(\n => 5.107.166.235\n)\n\n => Array\n(\n => 106.212.44.123\n)\n\n => Array\n(\n => 157.41.60.93\n)\n\n => Array\n(\n => 27.58.179.79\n)\n\n => Array\n(\n => 157.37.167.144\n)\n\n => Array\n(\n => 119.160.57.115\n)\n\n => Array\n(\n => 122.161.53.224\n)\n\n => Array\n(\n => 49.36.233.51\n)\n\n => Array\n(\n => 101.0.32.8\n)\n\n => Array\n(\n => 119.160.103.158\n)\n\n => Array\n(\n => 122.177.79.115\n)\n\n => Array\n(\n => 107.181.166.27\n)\n\n => Array\n(\n => 183.6.0.125\n)\n\n => Array\n(\n => 49.36.186.0\n)\n\n => Array\n(\n => 202.181.5.4\n)\n\n => Array\n(\n => 45.118.165.144\n)\n\n => Array\n(\n => 171.96.157.133\n)\n\n => Array\n(\n => 222.252.51.163\n)\n\n => Array\n(\n => 103.81.215.162\n)\n\n => Array\n(\n => 110.225.93.208\n)\n\n => Array\n(\n => 122.161.48.200\n)\n\n => Array\n(\n => 119.63.138.173\n)\n\n => Array\n(\n => 202.83.58.208\n)\n\n => Array\n(\n => 122.161.53.101\n)\n\n => Array\n(\n => 137.97.95.21\n)\n\n => Array\n(\n => 112.204.167.123\n)\n\n => Array\n(\n => 122.180.21.151\n)\n\n => Array\n(\n => 103.120.44.108\n)\n\n => Array\n(\n => 49.37.220.174\n)\n\n => Array\n(\n => 1.55.255.124\n)\n\n => Array\n(\n => 23.227.140.173\n)\n\n => Array\n(\n => 43.248.153.110\n)\n\n => Array\n(\n => 106.214.93.101\n)\n\n => Array\n(\n => 103.83.149.36\n)\n\n => Array\n(\n => 103.217.123.57\n)\n\n => Array\n(\n => 193.9.113.119\n)\n\n => Array\n(\n => 14.182.57.204\n)\n\n => Array\n(\n => 117.201.231.0\n)\n\n => Array\n(\n => 14.99.198.186\n)\n\n => Array\n(\n => 36.255.44.204\n)\n\n => Array\n(\n => 103.160.236.42\n)\n\n => Array\n(\n => 31.202.16.116\n)\n\n => Array\n(\n => 223.239.49.201\n)\n\n => Array\n(\n => 122.161.102.149\n)\n\n => Array\n(\n => 117.196.123.184\n)\n\n => Array\n(\n => 49.205.112.105\n)\n\n => Array\n(\n => 103.244.176.201\n)\n\n => Array\n(\n => 95.216.15.219\n)\n\n => Array\n(\n => 103.107.196.174\n)\n\n => Array\n(\n => 203.190.34.65\n)\n\n => Array\n(\n => 23.227.140.182\n)\n\n => Array\n(\n => 171.79.74.74\n)\n\n => Array\n(\n => 106.206.223.244\n)\n\n => Array\n(\n => 180.151.28.140\n)\n\n => Array\n(\n => 165.225.124.114\n)\n\n => Array\n(\n => 106.206.223.252\n)\n\n => Array\n(\n => 39.62.23.38\n)\n\n => Array\n(\n => 112.211.252.33\n)\n\n => Array\n(\n => 146.70.66.242\n)\n\n => Array\n(\n => 222.252.51.38\n)\n\n => Array\n(\n => 122.162.151.223\n)\n\n => Array\n(\n => 180.178.154.100\n)\n\n => Array\n(\n => 180.94.33.94\n)\n\n => Array\n(\n => 205.164.130.82\n)\n\n => Array\n(\n => 117.196.114.167\n)\n\n => Array\n(\n => 43.224.0.189\n)\n\n => Array\n(\n => 154.6.20.59\n)\n\n => Array\n(\n => 122.161.131.67\n)\n\n => Array\n(\n => 70.68.68.159\n)\n\n => Array\n(\n => 103.125.130.200\n)\n\n => Array\n(\n => 43.242.176.147\n)\n\n => Array\n(\n => 129.0.102.29\n)\n\n => Array\n(\n => 182.64.180.32\n)\n\n => Array\n(\n => 110.93.250.196\n)\n\n => Array\n(\n => 139.135.57.197\n)\n\n => Array\n(\n => 157.33.219.2\n)\n\n => Array\n(\n => 205.253.123.239\n)\n\n => Array\n(\n => 122.177.66.119\n)\n\n => Array\n(\n => 182.64.105.252\n)\n\n => Array\n(\n => 14.97.111.154\n)\n\n => Array\n(\n => 146.196.35.35\n)\n\n => Array\n(\n => 103.167.162.205\n)\n\n => Array\n(\n => 37.111.130.245\n)\n\n => Array\n(\n => 49.228.51.196\n)\n\n => Array\n(\n => 157.39.148.205\n)\n\n => Array\n(\n => 129.0.102.28\n)\n\n => Array\n(\n => 103.82.191.229\n)\n\n => Array\n(\n => 194.104.23.140\n)\n\n => Array\n(\n => 49.205.193.252\n)\n\n => Array\n(\n => 222.252.33.119\n)\n\n => Array\n(\n => 173.255.132.114\n)\n\n => Array\n(\n => 182.64.148.162\n)\n\n => Array\n(\n => 175.176.87.8\n)\n\n => Array\n(\n => 5.62.57.6\n)\n\n => Array\n(\n => 119.160.96.229\n)\n\n => Array\n(\n => 49.205.180.226\n)\n\n => Array\n(\n => 95.142.120.59\n)\n\n => Array\n(\n => 183.82.116.204\n)\n\n => Array\n(\n => 202.89.69.186\n)\n\n => Array\n(\n => 39.48.165.36\n)\n\n => Array\n(\n => 192.140.149.81\n)\n\n => Array\n(\n => 198.16.70.28\n)\n\n => Array\n(\n => 103.25.250.236\n)\n\n => Array\n(\n => 106.76.202.244\n)\n\n => Array\n(\n => 47.8.8.165\n)\n\n => Array\n(\n => 202.5.145.213\n)\n\n => Array\n(\n => 106.212.188.243\n)\n\n => Array\n(\n => 106.215.89.2\n)\n\n => Array\n(\n => 119.82.83.148\n)\n\n => Array\n(\n => 123.24.164.245\n)\n\n => Array\n(\n => 187.67.51.106\n)\n\n => Array\n(\n => 117.196.119.95\n)\n\n => Array\n(\n => 95.142.120.66\n)\n\n => Array\n(\n => 156.146.59.35\n)\n\n => Array\n(\n => 49.205.213.148\n)\n\n => Array\n(\n => 111.223.27.206\n)\n\n => Array\n(\n => 49.205.212.86\n)\n\n => Array\n(\n => 103.77.42.103\n)\n\n => Array\n(\n => 110.227.62.25\n)\n\n => Array\n(\n => 122.179.54.140\n)\n\n => Array\n(\n => 157.39.239.81\n)\n\n => Array\n(\n => 138.128.27.234\n)\n\n => Array\n(\n => 103.244.176.194\n)\n\n => Array\n(\n => 130.105.10.127\n)\n\n => Array\n(\n => 103.116.250.191\n)\n\n => Array\n(\n => 122.180.186.6\n)\n\n => Array\n(\n => 101.53.228.52\n)\n\n => Array\n(\n => 39.57.138.90\n)\n\n => Array\n(\n => 197.156.137.165\n)\n\n => Array\n(\n => 49.37.155.78\n)\n\n => Array\n(\n => 39.59.81.32\n)\n\n => Array\n(\n => 45.127.44.78\n)\n\n => Array\n(\n => 103.58.155.83\n)\n\n => Array\n(\n => 175.107.220.20\n)\n\n => Array\n(\n => 14.255.9.197\n)\n\n => Array\n(\n => 103.55.63.146\n)\n\n => Array\n(\n => 49.205.138.81\n)\n\n => Array\n(\n => 45.35.222.243\n)\n\n => Array\n(\n => 203.190.34.57\n)\n\n => Array\n(\n => 205.253.121.11\n)\n\n => Array\n(\n => 154.72.171.177\n)\n\n => Array\n(\n => 39.52.203.37\n)\n\n => Array\n(\n => 122.161.52.2\n)\n\n => Array\n(\n => 82.145.41.170\n)\n\n => Array\n(\n => 103.217.123.33\n)\n\n => Array\n(\n => 103.150.238.100\n)\n\n => Array\n(\n => 125.99.11.182\n)\n\n => Array\n(\n => 103.217.178.70\n)\n\n => Array\n(\n => 197.210.227.95\n)\n\n => Array\n(\n => 116.75.212.153\n)\n\n => Array\n(\n => 212.102.42.202\n)\n\n => Array\n(\n => 49.34.177.147\n)\n\n => Array\n(\n => 173.242.123.110\n)\n\n => Array\n(\n => 49.36.35.254\n)\n\n => Array\n(\n => 202.47.59.82\n)\n\n => Array\n(\n => 157.42.197.119\n)\n\n => Array\n(\n => 103.99.196.250\n)\n\n => Array\n(\n => 119.155.228.244\n)\n\n => Array\n(\n => 130.105.160.170\n)\n\n => Array\n(\n => 78.132.235.189\n)\n\n => Array\n(\n => 202.142.186.114\n)\n\n => Array\n(\n => 115.99.156.136\n)\n\n => Array\n(\n => 14.162.166.254\n)\n\n => Array\n(\n => 157.39.133.205\n)\n\n => Array\n(\n => 103.196.139.157\n)\n\n => Array\n(\n => 139.99.159.20\n)\n\n => Array\n(\n => 175.176.87.42\n)\n\n => Array\n(\n => 103.46.202.244\n)\n\n => Array\n(\n => 175.176.87.16\n)\n\n => Array\n(\n => 49.156.85.55\n)\n\n => Array\n(\n => 157.39.101.65\n)\n\n => Array\n(\n => 124.253.195.93\n)\n\n => Array\n(\n => 110.227.59.8\n)\n\n => Array\n(\n => 157.50.50.6\n)\n\n => Array\n(\n => 95.142.120.25\n)\n\n => Array\n(\n => 49.36.186.141\n)\n\n => Array\n(\n => 110.227.54.161\n)\n\n => Array\n(\n => 88.117.62.180\n)\n\n => Array\n(\n => 110.227.57.8\n)\n\n => Array\n(\n => 106.200.36.21\n)\n\n => Array\n(\n => 202.131.143.247\n)\n\n => Array\n(\n => 103.46.202.4\n)\n\n)\n```\nArchive for December, 2018: Creditcardfree's Personal Finance Blog\n Layout: Blue and Brown (Default) Author's Creation\n Home > Archive: December, 2018\n\n# Archive for December, 2018\n\n## Last of the 2018 Snowflakes\n\nJanuary 1st, 2019 at 02:42 am\n\nI have few final snowflakes to report for 2018. A full tally coming soon!\n\n\\$100 Christmas gift from my parents\n\\$24.81 Ibotta proceeds\n\\$1.89 Starbucks (Amex reward offer)\n\\$5.00 Hulu (Amex reward offer)\n\\$2.91 Verizon (Amex reward offer)\n\nIt's been a good year for snowflakes (extra money)...all have been added to our Big Goal. More about that on the sidebar and an update soon.\n\n## 2018 Financial Wins and Fails\n\nDecember 29th, 2018 at 02:58 am\n\nIf you didn't see my previous post, I've invited all bloggers to write a post titled 2018 Financial Wins and Fails. Join in before the end of the year.\n\nI reread all of my 2018 posts here and I have to say while the year was a whirlwind of change, we actually had a lot of financial wins, many I forgot about!\n\n1) Failed to shop for the best prices. We purchased some big ticket items, computer for our daughter, dorm supplies, tires and college textbooks. The reasons are multiple, but it was such a busy year, I found myself just wanting to get the purchases complete, rather than spending lots of time finding the best price. It's not to say I didn't do some price comparisons, but I didn't dig in deep to make it a priority.\n\n2) I failed to think ahead about getting out of the stock market on my youngest daughter's Educational Savings Account. The market was doing well. We redeemed shares in August and definitely sold high. But now the market has corrected, and the share price while still high for some of the shares we bought, it is low for others. I should have moved the shares to cash in August. I haven't sold at a loss yet, so may not end up being a complete fail. Time will tell.\n\n3) Failed to plan ahead. This is related to both of the above, but I wasn't thinking months in advance about things that would need cash, particularly all the dorm room expenses. I should have thought about that at least at the first of the year. I was able to cash flow the costs as we made purchases, but it would have been less stressful if the money was set aside for something I knew was coming.\n\nThe wins definitely outweigh the fails.\n\n1) We ended the year once again with zero student loans! So excited we have been able to continue to cash flow, use saved investments or take advantage of my husband's Post 9/11 GI Bill benefits. Oh, and the girls both got fantastic scholarships that helped as well.\n\n2) Despite lots of spending, we saved a lot this year. We maxed out our Roth IRA contributions, and my husband saved 11% of his basic pay for retirement. We saved \\$2000 in our daughter's ESA (final contribution). We saved \\$347/mo automatically from my husband's paycheck. We saved all credit card rewards and interest on our savings. (I'll post more specific numbers later.) We also saved the entire military move reimbursement which was nearly \\$7K.\n\n3) We cash flowed a computer purchase, new tires, shocks and struts for my van, dorm needs, three trips, and several airline flights. Still completely debt free!\n\nThe fails help me see where we can improve in 2019, and the wins remind me what we are doing right and can continue with going forward. Do you review your successes and setbacks at the end of the year?\n\n## 2018 Financial Wins and Fails Collaboration\n\nDecember 27th, 2018 at 04:23 pm\n\nMany of us are pretty good about writing about our year end in some way or another. I'm going to be writing about our financial wins and fails for 2018 before the end of the year. I'd like you to join me and write a specific blog post about your wins and fails for 2018.\n\nWhile it's important to review for ourselves, sharing our financial wins and fails helps inspire others. Will you join me? Simply title your post 2018 Financial Wins and Fails, post before the end of the year and invite others to join at the end of your post.\n\nAnd if you are new or a lurker, we would love for you to join in and post your own financial wins and fails! Goal setting and review is helping in making progress and changes for the year ahead.\n\n## We Owe The Government\n\nDecember 17th, 2018 at 05:41 pm\n\nI knew this was coming. My daughter received Post 911 GI Bill benefits for her tuition and fees this year. She originally signed up for 18 credit hours, then added one hour, which then she eventually dropped. Actually, that one extra credit hour may not have been the same class, but tuition wise it is the same.\n\nThe VA, who provides this benefit, is a little slow to catch up to the changes. They originally paid benefits for 18 credits in August, which was perfect and what we wanted. But then they were notified of the new hour and sent that extra funds in November, to the tune of \\$400.87. But now they are up to date with the fact that my daughter dropped that last hour and are telling us we owe them money. Perfectly find and expected.\n\nThe amount they indicate we owe is \\$483.37. The breakdown they provide makes no sense and I can't get the math to work. I'll have my husband make a phone call. It is probably correct, but it annoys me I can't follow how they came up with that number.\n\nThey state tuition and fees charged:\n19 hours \\$5,846.75\n18 hours \\$5,575.50\nThis is a difference of \\$271.25\n\nBut they state total overpayment is \\$487.37\nTuition and fees overpayment \\$245.30\nYellow Ribbon Program overpayment \\$238.07\n\nYellow Ribbon is a program that pays the overage for out of state tuition. GI Bill pays in state, Yellow Ribbon pays out of state portion (or maybe half, and the university may waive the other half).\n\nThe payment the University received in November for the extra credit hour showed up on my daughters bill as two credits.\nVA Chapter 33 \\$203.44\nVA Yellow Ribbon \\$197.87\nThis is \\$400.87 and the amount we received from the University as a refund.\n\nI knew this was an overpayment by the VA when we received it and I expected they would be asking for it back. But the amount doesn't match! The amount is MORE than what they appear to have paid the University. In the end I'll probably just pay the whole thing knowing they are probably right. Overall we have received very good benefits and if another \\$82.50 has to go back to them, it's still been worth it. If I had to guess it may have to do with fees associated with the dropped class...but I still can't figure out the math.\n\nThe VA is VERY bad at communicating the financials of these benefits. You can find nothing online. And the statements received are so odd and don't add up.\n\nThanks for listening! Again at most \\$82.50 we pay back may not be correct. But maybe it is.\n\n## The Retirement Accounts\n\nDecember 16th, 2018 at 04:48 pm\n\nI haven't checked our retirement and investment account balances since the beginning of October. I knew the market was down, and had lost most if not all gains. So I was avoiding looking. Today I looked....\n\nthe total balance is less than what it was on January 3 of this year. And we've been investing thousands of dollars. It's frustrating to see.\n\nOf course, if I look at the share balance, that has increased because we bought shares. And these are not realized gains or losses until we sell. We don't need the money now and the market will recover at some point.\n\nHow are you feeling about the change? I hope you are staying the course when it comes to retirement or making the appropriate adjustments if needed.\n\n## Incoming Funds\n\nDecember 14th, 2018 at 10:01 pm\n\nToday was payday!\n\nI paid off all the credit card balances, which is my usual routine twice per month. I also balanced categories in YNAB back to zero and then added in money to flush those categories out for the rest of the month spending. I'm guessing with our girls home visiting for the next three weeks to a month we will be spending more on groceries, dinner and likely some paid activities, although I'm hoping for free!\n\nI finally received my \\$3 Pinecone Research payment. I had to send an email and the next day I had my payment. Apparently they were having some issues with their normal process.\n\nI redeemed \\$7.44 in Chase Freedom rewards. These went into our Big Goal Fund! I will do a wrap up of our progress at year end.\n\nWe also received an extra payment for housing for our oldest daughter's Post 911/GI Bill benefits this semester. Apparently, they have been underpaying us all semester. I'm actually glad I didn't know until it was corrected. That amount to cover underpayment since August was \\$286.17.\n\nI received an email from FNBO Direct where we have one of our money market accounts. That account will now be paying 2.15%. I'll take it but I probably need to move some money in there into a CD with a higher rate. I think Beawealthywarrior posted on one of my blogs that PenFed has some good rates. I will look there soon!\n\nI'm always grateful for the money that flows into our lives, as well as the ability to let money flow out for our needs and wants. All of life is a balance that way.\n\n## Reviewed Our Tax Situation\n\nDecember 13th, 2018 at 01:30 pm\n\nIt's been a whirlwind of a year! In the last couple weeks, I've been starting to panic, wondering if I actually figured our tax situation out for 2018 in light of the new tax laws. I vaguely remembering need to wait to figure things out. But did I ever go back and review?\n\nYesterday, I hopped on the IRS Withholding Calculator to see where things stood. Luckily, at least in my mind, we are good. We will owe money but it should not be more than \\$540. I had increased our withholding which meant \\$90 more in our pocket each month, or by the time it took effect in February, \\$990 more for the year.\n\nThe calculations only took into account the American Opportunity Tax Credit I plan to take for our oldest daughter (for the final time, as it can only be used for four years). I did not take into account any credits for our youngest daughter. Because she only attended one semester in 2018, I plan to take the Lifetime Earning Credits on her qualified tuition, which should give us at least a credit of \\$300+ for the semester. That brings the tax owed around \\$240.\n\nIt is possible we will have a little more income added and more tax, but that's not looking too promising at the moment. I expect that anything that occurs is something we can handle. I have always preferred owing a few hundred dollars rather than have the IRS hold our money for many months before getting a refund. I've managed to make this work several years in a row now!\n\nHave you reviewed your tax withholding and how that may affect if you owe the IRS or will get a refund? April 15th is just four months away. It's good to be prepared. The IRS Withholding Calculator is\n\nhere.\n\n## Save Time Mailing Packages\n\nDecember 12th, 2018 at 08:05 pm\n\nI just told this to a friend today, who was complaining about the long wait at the post office.\n\nDid you know that the US Postal Service offers pickup of packages*** you are mailing during your regular mail delivery? You do not need to be home and it's free!\n\nThe first step is to pack your items and weigh it. As you know I've done eBay selling for over 10 years, so I have a food scale I use to get the weight of my packages up to five pounds. The US Postal Service charges you by the pound, so as long as you know it's weight is between 1-2 pounds, the cost will be the same. Another alternative where you do not need to know the weight is to use the free Flat Rate Priority mail boxes provided by the post office. The cost is the same regardless of the weight.\n\nLog on and make a free account at USPS and follow the directions for Click n Ship. At the end of this process you will have a prepaid mailing label to affix to your package with packing tape. The address you are mailing to and your return address will be printed on the label. You will pay for the cost of postage with a credit (or debit card).\n\nAt this point you can simply drop off at the post office. This time of year, you are likely to see an area at the counter filled with packages. You can leave it there, or wait in line to get a receipt and hand it to the postal worker.\n\nThe other option is to Schedule a Pickup from your USPS account. You must do this the day or night before pickup. It doesn't work same day. This service is free. It may not be available for all address, but you will be told that when you go to schedule. Once scheduled, set your package out in the morning in the spot you indicated on your delivery instructions. You will get an email once your package has been picked up. I have used this service before and had no issues.\n\nI hope this helpful information may save you time this season. If you have questions, leave them in the comments and I will try to answer as best I can!\n\n***Only for Priority and Express Mail packages.\n\n## Christmas Budgets\n\nDecember 11th, 2018 at 03:42 pm\n\nI was at an event this weekend where a lesson on Christmas budgeting was presented. It was short, but excellent. That is what has inspired this post.\n\nOur Christmas budget the last several years has been \\$600. These are the funds to cover gifts and shipping costs for nine people. I save \\$50 each month from our paychecks so the money is available when it's time to shop.\n\nThe lesson presented asked people to think of ALL the costs they may incur during the holiday season. Not just the gifts, but the wrapping, baking and food, travel, postage, clothing, electricity and fuel for our cars. I know that my budget does not reflect all of these costs. I seem to be able to absorb them into our regular spending, but they are important to pay attention to when planning.\n\nDepending on where you are in your shopping, it isn't too late to step back and make a Christmas budget. Figure what you have already spent, and how much you have available to spend. Try to finish within those parameters, even if it means returning someone's gift and getting something else so you will spend less. There are so many great deals this time of year, that gift can cost less than you think. Of course, the deals can pull you in and convince you to spend more, so be strong and stick to your plan.\n\nDollar Tree isn't my favorite store, but I find myself there every holiday season to get a good deal and help me stay within budget. It can be great for paper products, and stocking stuffers. Of course, not everything is a good deal there, so it is still important to pay attention to prices!\n\nI think the key to staying within a reasonable budget for Christmas is having the fewest number of people to buy for. At least it helped us. We stopped giving to most of our nieces and nephews and exchanging adult gifts. We still give to our parents, our children, two nieces (recently adopted) and buy a gift for ourselves.\n\nWe do not buy new holiday decorations each year. The exception is one ornament for each of the girls, but those are usually less than \\$10 for both. We give cookies to friends, neighbors and coworkers. I send a limited number (less than five some years) of Christmas cards, that I usually buy on clearance after Christmas. I reuse Christmas gift bags and tissue, and usually only buy a roll or two of Christmas gift wrap every two or three years (again, on clearance). I chose gifts that are somewhat lightweight to save on shipping, although Flat Rate boxes from the post office do help (the items just have to fit). We also do not entertain at our place, not because we are grinches, but because family is not nearby.\n\nAt a minimum, we can all start a new plan to budget for the next year. As soon as the spending is over this year, review what was purchased and how much it cost. That total divided by 12 will be a great guideline for how much to save each month in order to be prepared for next year. Save it in an envelope or a seperate checking or savings account, so you will not spend it until it is time. After the holiday is also a good time to discuss with family members about how you might want to change the gift giving parameters for the future.\n\nTell me about your Christmas Budget. Do you have one? What is included. If you don't have one, tell me about that too.\n\n## Undebt It\n\nDecember 10th, 2018 at 11:03 pm\n\nAs you all know we do not have any debt. No mortgage debt either as we move frequently with the military and have found it make sense to rent for the time being.\n\nI came across a site that is free and allows you to track your way out of debt. I'd try it myself and review it for you, but again zero debt to track. I do see it links to YNAB! And it has a 52 Week Savings Option.\n\nIf you have goals to pay off debt in 2019, you might want to check out this free tracking option. I'm sure many of us here would love to watch your journey and hear how you like Undebt It.\n\nYou can find the online tracking tool\n\nhere.\n\nLife with zero debt is worth the effort!\n\n## Interest and Extra Money\n\nDecember 8th, 2018 at 03:02 pm\n\nI like to report the amount of interest we earn each month. This is the quick breakdown for November.\nNavy Federal CU \\$20.97\nFNBO Direct \\$89.18\nTotal \\$110.15\n\nI redeemed my 300 points with Pinecone Surveys over a week ago for cash and have yet to receive my Paypal deposit. Usually it is the next business day. Maybe the move and change of address are factors?\n\nWe usually get some cash back back from USAA for our insurance. They call it a distribution from our Subscriber's Savings Account. This year the amount we received was \\$82.83. We will take whatever they want to give back, and yes, it has been added to the Big Goal.\n\nI did get my \\$10 Amazon gift card from Swagbucks. I expect I have a little more online shopping to do for Christmas, so that will reduce the final amount I spend.\n\nI hit the \\$20 threshold with Ibotta. I hadn't been using it much this year, but with a little effort was able to apply some rebates for things I had already purchased and meet the minimum. I'm debating redeeming now, or waiting until the end of the year. Anything over \\$20 can be redeemed, but once you redeem back to zero, it takes awhile to get to redeem the next \\$20. I guess in the end it's all the same!\n\n## Airline Tickets\n\nDecember 6th, 2018 at 02:42 am\n\nIt's that time of year where I buy airline tickets for my kids to come home for Christmas break! This is the first year I have to accommodate BOTH girls. I actually purchased one way tickets. They are in different cities, but flying home together from one city. On the return to school they will fly separately, primarily because one has three weeks off, and the other has a full month! One daughter will have to take a shuttle from the airport to her campus, cost with a tip one way is \\$42. I think if it was a roundtrip shuttle it would be about \\$60. I'm always praying that the weather cooperates so they don't get delayed! Oh, in total I spent about \\$800 for the girls to come home!\n\nMy husband was home today for the National Day of Mourning. We did watch the funeral and were touched by the eulogies given. I think we all wish that in death we are remembered fondly by those we loved and who loved us.\n\n## Redeemed MyPoints\n\nDecember 4th, 2018 at 04:56 pm\n\nI redeemed 3950 MyPoints for a \\$25 Google Play gift card. I was originally going to buy it but happened to notice the balance in my MyPoints account. I had plenty to scoop up the gift my daughter wanted without an additional cash outlay.\n\nI'm close to having 1000 SB that I can redeem for \\$10 Amazon gift card. I'm not in a big hurry and have not been doing Swagbucks for months. I think I was reminded of it when i received an email from Swagbucks with my Birthday bonus.\n\nToday, I finished sewing the pajama pants for my two nieces. I had them all done except for the elastic. I also made doll sized pajama pants and a plaid poncho for their American Girl dolls (I think they have the knock off versions). I spent nearly \\$30 at Joann's for fabric and elastic. I did buy nicer quality flannel and probably didn't hit the best sale, although it was 40% off. I did find an Ibotta coupon that will give me \\$4.50 back on the purchase, so now closer to \\$25, which is really good for two gifts.\n\nAt this point, I just need more stocking stuffers for the girls, buy a gift for my husband and decide what to get my parents. I already sent my mother in law a check to get whatever she wants. She's 82 and seems to like to get a pedicure from us. So it's just easier to mail a check, since I'm nowhere near there to buy a gift certificate.\n\nToday, I'm baking some Mint Chocolate cookies for the holidays. I will eat one or two and freeze the rest until needed. I have a couple other recipes to make, but have plans to do them over the next two weeks.\n\n## 10% Off Target Gift Cards\n\nDecember 2nd, 2018 at 08:32 pm\n\nToday only. Sunday, December 3, you can buy Target gift cards for 10% off. You can purchase up to \\$300 worth for \\$270.\n\nI bought a \\$40 one for my daughter who requested Target gift card for Christmas. I also went ahead and picked up another one for \\$100 for our regular shopping. We don't shop there as much as we once did, so I didn't want to tie up too many funds not knowing if we would use them.\n\nI purchased several other gifts online. My shopping is wrapping up pretty fast! Still thinking about an extra item to include with my nieces pajama pants I'm sewing. I probably don't need anything since I am also sewing matching doll pajama pants. I will need more stocking stuffers for our girls. I also want to figure out what to get my parents. I don't have time or energy for a calendar or other photo gift. I'm tempted to just send a check, since their anniversary is just two days after Christmas. What do you get your older parents, or what would you like if you already had everything? Last year I bought them socks and some edible items."
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https://brainmass.com/math/vector-calculus
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"Explore BrainMass\n\n# Vector Calculus\n\nVector calculus is a field of mathematics which is depicted most commonly in three dimensional spaces and involves utilizing both the operations of differentiation and integration. Scalar fields and vector fields are the typical objects of vector calculus which are manipulated and transformed using different operations.\n\nIn comparison to a vector field, a scalar field is a value representing a particular point in space, but is coordinate-independent. This means that this value is not associated with a direction. A scalar field can be either a physical quantity or just a number. For example, 5 humans is an example of a scalar field.\n\nA vector field involves not only a numerical value, but one which is directed somewhere. A subset of Euclidean space can be represented utilizing a vector. An example of a vector is the speed and direction of a fluid, such as 5 liters of orange juice. This quantity must be flowing in a particular direction.\n\nWhen vector calculus becomes more advanced, scalar and vector quantities turn into pseudoscalar and pseudovector quantities. This means that the sign of the value changes with orientation.\n\nIn vector calculus, various algebraic operations are practiced and defined for a vector space. Some of these algebraic operations include:\n\n• The dot product: This involves multiplying two vector fields and equals a scalar field.\n• Vector addition: This operation produces a vector field since two vector fields are added together.\n• Cross product: This operation involves multiplying two vector fields and produces a vector field.\n\nFurthermore, vector calculus has multiple applications in the fields of engineering and physics. For example, when dealing with fluid dynamics such as fluid flow in civil engineering, the use of vectors is important.\n\n### Vector Calculus: Three Dimensional Space\n\nPlease see attached file for the questions. Thank you.\n\n### Aircraft motion as vectors\n\n6. An aircraft of top speed 340 km per hour sets a course to the north-east but, owing to the prevailing wind, its actual speed relative to the ground is 450 km per hour in a direction of 40° east of north. Using analytical and graphical methods: a. Represent the aircraft motion as vectors b. Find the direction and the spe\n\n### Plane Equation, Normal, Distance\n\na) Justify the vector equation for a plane in the form r = a + lambda*u + mu*v, where a, u, v are non-coplanar vectors and lambda, mu are arbitrary scalars. b) What is the normal vector to the plane? c) Show that the shortest distance of a point with position vector p from the above plane is given by: |[p-a, u, v]| |unv|\n\n### Vectors Calculas & Applications\n\nSee the attached file. 1. Evaluate the integral: ** see the attachment for the full equation ** Where D is the domain given by: 1 ≤ x^2 + y^2 ≤ 4 and y ≥ 0. 2. (i) Find the area of the region enclosed by the ellipse (x^2/a^2) + (y^2/b^2) = 1 (ii) Find the area of the region enclosed by the parabola y = x^2 an\n\n### Formulate a Model to Determine Location in the (x,y) Plane\n\nFive train stations located in the (x,y) plane have the coordinates (ai, bi) given by (a1, b1) = (5, 0) (a2, b2) = (0, 10) (a3, b3) = (10, 10) (a4, b4) = (50, 50) (a5, b5) = (-10, 50) It is desired to connect all of the stations to a single location \"hub\" using the minimum total amount of track. Each track segment\n\n### Solving Fundamental Mathematics-Based Questions\n\nHi, I need some assistance with all the attached questions. I am not too sure how to answer them and a step-by-step working guide for each question would really help me in understanding these problems. These are all linear algebra problems.\n\n### Three problems on vector and tensor fields\n\nPROBLEM 1. Let w = w_i dx^i be a 1- form (or covariant vector field) expressed in terms of the coordinate system x = (x^1, ... , x^n). Determine the (covariant) transformation law for the components w_i of w expressed in a new coordinate system y = (y^1, ... , y^n ). Describe the relationship between the contravariant and covari\n\n### Applied Mathematics Cartesian Problems\n\nI have attached two problems (see attachment); I always have problems in solving the parts highlighted in red- part a and d. May I please have a detailed step-by-step method of finding the angle, area, and the Cartesian equation. On the last part, which is (d), I am always getting the signs wrong, so please show me the best meth\n\n### Choose between two long-distance telephone plans.\n\nYou are choosing between two long-distance telephone plans. Plan A has a monthly fee of \\$20.00 with a charge of \\$0.05 per minute for all long-distance calls. Plan B has a monthly fee of \\$10.00 with a charge of \\$0.10 per minute for all long-distance calls. Complete parts a and b. a. For how many minute of long-distance calls w\n\n### Vectors / Force Diagrams\n\nA ball of mass 2.9kg rests on a ramp that makes an angle of 28 degrees with the horizontal. It is held in place by a rope that makes an angle of 55 degrees with the horizontal. Assume that the only forces acting on the ball are its weight, the tension in the rope and the normal reaction from the ramp. Take the magnitude of the a\n\n### Vector operations\n\nExpress the vectgor with initial point P and terminal point Q in component form. Show work. 12. P(1,1), Q(9,9) 14. P(-1,3), Q(-6,-1) 16. P(-8, -6), Q(-1, -1) Find 2u, -3v, u + v, and 3u - 4v for the given vectors u and v. Show work. 18. u= (-2,5), v= (2, -8) 20. u= i, v=-2j Find (a) u * 5 and (b) the angle betwe\n\n### Computation of Bases of Various Vector Spaces\n\nFor each of the following vector spaces, give its dimension and a basis. (a) the set of all symmetric 4 x 4 matrices, (b) the subspace of P3 consisting of those polynomials in P3 whose graphs pass through the origin (c) the set of all vectors in R3 that are orthogonal to v = (1,2,-1)\n\n### Vector space functions\n\nLet V be the vector space of all functions F : R -> R. Determine whether the following subsets of V form subspaces of V. Give reasons, ie, prove or disprove that the subset is a subspace. Please see the attached file for the full problem description.\n\n### Orthogonal vectors\n\n1) For which values of k are the following vectors u and v orthogonal? a) u = (2,1,3) , v = (1,7,k) b) u = (k,k,1) , v = (k,5,6) 2) Let u,v be orthogonal unit vectors. Prove that d(u,v) = 2^(1/2) (The questions are unrelated)\n\n### Graphs and Vector Calculus\n\nPlease help answer the following question. Provide step by step calculations along with detailed explanations to explain how each step works. If all edges of Kn (a complete graph) have been coloured red and blue, how do we show that either the red graph or the blue graph is connected?\n\n### Pitch machine\n\nWhen set at the standard position, Autopitch can throw hard balls toward a batter at an average speed of 60 mph. Autopitch devices are made for both major- and minor-league teams to help them improve their batting averages. Autopitch executives take samples of 10 Autopitch devices at a time to monitor these devices and to\n\n### Solving a Vector Calculation\n\nFind the equation of the line passing through a point B, with position vector b relative to an origin O, which is perpendicular to and intersects the line r = a + lamda*(c), c (not= 0), given that B is not a point of the line.\n\n### Finding the Velocity and Position Vectors of a Particle\n\nFind the velocity and position vectors of a particle that has the given acceleration and the given inital velocity and positions. a(t) = t i+e^t j+e^-t k v(0)=k,r(0)=j+k Please use computer to graph the path of the particle.\n\n### Vectors Components Practical Terms\n\nThe components of v=250i+310j represent the respective number of gallons of regular and premium gas sold. the components of w=2.90i+3.03j represent the respective prices per gallon for each kind of gas. Find v*w and describe what the answer means in practical terms. v*w = (Do not round until final answer then round to neares\n\n### vector spaces and linear spans\n\nFind all real numbers k such that the vector v=(1,-2,k) in R^3 (with the usual operations) is in the linear span of the vectors x=(3,0,-2) and y=(2,-1,-5). Prove your answer. Be sure to address both necessity and sufficiency. Show all work.\n\n### Equilateral Triangle: Vectors\n\nSee the attachment for a diagram of an equilateral triangle with sides u, v and w. U is a unit vector. Find u dot v and u dot w (u.v and u.w). The answers are: u.v = 1/2, and u.w = -1/2. Why is u.w a minus 1/2?\n\n### Solve for X.\n\nSuppose L_1 is the line through the origin in the direction of a_1, and L_2 is the line through b in the direction of a_2. To find the closest points x_1a_1 and b + x_2a_2 on the two lines, write down the two equations for the x_1 and x_2 that minimize ||x_1a_1 - x_2a_2 - b||. Solve for x if a_1 = (1,1,0), a_2 = (0,1,0), b = (2,\n\n### Subspaces of the Vector Space of 2-by-2 Matrices\n\nNeed help with this homework problem. Please use formal proofs and language where applicable, and be sure to explain your reasoning thoroughly. Please post response as a Word or PDF file. Infinite thanks!\n\n### Properties of Vector Spaces\n\nNeed help with this homework problem. See attached file. Please write complete, formal and professional proofs for your answer. Also, please publish response as a Word or PDF file. Infinite thanks!\n\n### Vector Space Proof\n\nPlease help with the following problem. I need help to write this Proof. Please be as professional and as clear as possible in your response. Pay close attention to instructions. Determine if the set R^2 (the real plane) is a vector space with operations defined by the following: Addition: (a,b)+(c,d)=(a+c,b+d) Sca\n\n### Analyze the vector.\n\nIn R^n with the standard inner product, consider the vector... u= (1, 1, 1, ...,1) and the coordinate vectors e_1=(1,0,0,...,0), e_2=(0,1,0,...,0),...,e_n=(0,0,0,...,1) a) Compute the angle(s) between u and e_i (in degrees) for dimensions n=2,3 b) Check that in R^n, <u, e_i>=1 for i=1,2,...,n c) Check that even though <u,\n\n### Find the vertex, axis, domain and range.\n\nThe math problems and descriptions are in the word document. Graph the quadratic function. Give the vertex, axis, domain and range. 18. f(x) = -3(x-2)^2+1 Solve the exponential equation. Express irrational solutions as decimals correct to the nearest thousandth. 12. 2^(x+3) = 5^x Use the factor theorem and\n\n### Vector Space basis\n\nP_2={a_2 x^2 + a_1 x + a_0 | a_0, a_1, a_2 E R} which of the following is a basis for P_2? {x^2 + x, x^2 +5x, x} {x^2, x^2 + 5, 3} {x^2 + x + 1, x + 5, 3} {x^2 + x + 1, x + 5, 0} {x + 1, x + 5, 3}\n\n### Subset of a Vector Space is a Subspace\n\nLet V be the vector space of all functions f: R->R. Determine whether the following subsets of V form subspaces. a) U = {f belongs to V | f(0) = 0} b) W = {f belongs to V | f(x) = k1 + k2 sinx for some k1,k2 are reals}.\n\n### Algebra: Vector Spaces Evaluated\n\nDetermine whether the given set and operations form a vector space. Give reasons. a) V = {(x,y,z) : x,y,z are Real numbers}, (x,y,z) + (x',y',z') = (x + x', y + y' z + z'), k(x,y,z) = (kx,y,z) b) the set of all positive real numbers with the operations: x + y = xy, kx = x^k"
] |
[
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https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_6&diff=cur&oldid=45804
|
[
"# Difference between revisions of \"2012 AIME II Problems/Problem 6\"\n\n## Problem 6\n\nLet",
null,
"$z=a+bi$ be the complex number with",
null,
"$\\vert z \\vert = 5$ and",
null,
"$b > 0$ such that the distance between",
null,
"$(1+2i)z^3$ and",
null,
"$z^5$ is maximized, and let",
null,
"$z^4 = c+di$. Find",
null,
"$c+d$.\n\n## Solution\n\nLet's consider the maximization constraint first: we want to maximize the value of",
null,
"$|z^5 - (1+2i)z^3|$ Simplifying, we have",
null,
"$|z^3| * |z^2 - (1+2i)|$",
null,
"$=|z|^3 * |z^2 - (1+2i)|$",
null,
"$=125|z^2 - (1+2i)|$\n\nThus we only need to maximize the value of",
null,
"$|z^2 - (1+2i)|$.\n\nTo maximize this value, we must have that",
null,
"$z^2$ is in the opposite direction of",
null,
"$1+2i$. The unit vector in the complex plane in the desired direction is",
null,
"$\\frac{-1}{\\sqrt{5}} + \\frac{-2}{\\sqrt{5}} i$. Furthermore, we know that the magnitude of",
null,
"$z^2$ is",
null,
"$25$, because the magnitude of",
null,
"$z$ is",
null,
"$5$. From this information, we can find that",
null,
"$z^2 = \\sqrt{5} (-5 - 10i)$\n\nSquaring, we get",
null,
"$z^4 = 5 (25 - 100 + 100i) = -375 + 500i$. Finally,",
null,
"$c+d = -375 + 500 = \\boxed{125}$\n\n## Solution 2\n\nWLOG, let",
null,
"$z_{1}=5(\\cos{\\theta_{1}}+i\\sin{\\theta_{1}})$ and",
null,
"$z_{2}=1+2i=\\sqrt{5}(\\cos{\\theta_{2}+i\\sin{\\theta_{2}}})$\n\nThis means that",
null,
"$z_{1}^3=125(\\cos{3\\theta_{1}}+i\\sin{3\\theta_{1}})$",
null,
"$z_{1}^4=625(\\cos{4\\theta_{1}}+i\\sin{4\\theta_{1}})$\n\nHence, this means that",
null,
"$z_{2}z_{1}^3=125\\sqrt{5}(\\cos({\\theta_{2}+3\\theta_{1}})+i\\sin({\\theta_{2}+3\\theta_{1}}))$\n\nAnd",
null,
"$z_{1}^5=3125(\\cos{5\\theta_{1}}+i\\sin{5\\theta_{1}})$\n\nNow, common sense tells us that the distance between these two complex numbers is maxed when they both are points satisfying the equation of the line",
null,
"$yi=mx$, or when they are each a",
null,
"$180^{\\circ}$ rotation away from each other.\n\nHence, we must have that",
null,
"$5\\theta_{1}=3\\theta_{1}+\\theta_{2}+180^{\\circ}\\implies\\theta_{1}=\\frac{\\theta_{2}+180^{\\circ}}{2}$\n\nNow, plug this back into",
null,
"$z_{1}^4$(if you want to know why, reread what we want in the problem!)\n\nSo now, we have that",
null,
"$z_{1}^4=625(\\cos{2\\theta_{2}}+i\\sin{2\\theta_{2}})$\n\nNotice that",
null,
"$\\cos\\theta_{2}=\\frac{1}{\\sqrt{5}}$ and",
null,
"$\\sin\\theta_{2}=\\frac{2}{\\sqrt{5}}$\n\nThen, we have that",
null,
"$\\cos{2\\theta_{2}}=\\cos^2{\\theta_{2}}-\\sin^2{\\theta_{2}}=-\\frac{3}{5}$ and",
null,
"$\\sin{2\\theta_{2}}=2\\sin{\\theta_{2}}\\cos{\\theta_{2}}=\\frac{4}{5}$\n\nFinally, plugging back in, we find that",
null,
"$z_{1}^4=625(-\\frac{3}{5}+\\frac{4i}{5})=-375+500i$",
null,
"$-375+500=\\boxed{125}$\n\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.",
null,
""
] |
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|
https://fr.scribd.com/document/103307590/Prausnitz-Thermodynamics-Notes-26
|
[
"Vous êtes sur la page 1sur 38\n\n## GE and Activity Coefficients\n\nWe demonstrated that\n\ngiE = RT ln [ i ]\nNow we need to remember that\nGE = i ni giE\n\nThus\n\nGE = RT i ni ln i\n\n## These are the most important relations for the chapter\n\nExpressions for GE\ngE = 0 gE = 0 when when x1 = 0 x2 = 0\n\nThe two-suffix Margules equations were derived from: gE = A x1 x2 A more complex form for gE as a function of composition can be: gE = x1 x2 [ A + B (x1 - x2) + C (x1 - x2)2 + D (x1 - x2)3 + ] This is the Redlich-Kister expansion\n\nTwo-Suffix Margules\nThe two-suffix Margules equations were derived from: gE = A x1 x2\n\n## ln 1 = A/RT x22 ln 2 = A/RT x12\n\nThe activity coefficients at infinite dilution are:\n\nln 1 = A/RT ln 2 = A/RT\n\nWohls Expansion\ngE / [RT (x1q1 + x2q2)] = 2 a12 z1z2 + 3 a112 z12z2 + 3 a122 z1z22 + 4 a1112 z12z2 + 4 a1222 z1z23 + 6 a1122 z12z22 + z1 = x1 q1 / [ x1 q1 + x2 q2 ] z2 = x2 q2 / [ x1 q1 + x2 q2 ]\n\nThe parameters q are effective volumes or cross sections of the molecules The parameters a are interaction parameters The contribution to gE of the interaction parameters a is weighted by the products that contain the factors z\n\n## Van Laar Equation\n\ngE / RT = 2 a12 x1 x2 q1 q2 / [ x1 q1 + x2 q2 ]\nWe can calculate the fugacity coefficients of the two components\n\n## ln 1 = A / [ 1 + A/B x1/x2 ]2 ln 2 = B / [ 1 + B/A x2/x1 ]2\n\nAnd, at infinite dilution, we obtain\n\nA = 2 q1 a12 B = 2 q2 a12\n\nln 1 = A ln 2 = B\n\n## Three-Suffix Margules Equations\n\ngE / [RT (x1q1 + x2q2)] = 2 a12 z1z2 + 3 a112 z12z2 + 3 a122 z1z22 + 4 a1112 z12z2 + 4 a1222 z1z23 z1 = x1 q1 / [ x1 q1 + x2 q2 ] z2 = x2 q2 / [ x1 q1 + x2 q2 ]\n\n## ln 1 = A x22 + B x23 ln 2 = (A + 3/2 B) x12 - B x13 At infinite dilution ln 1 = A + B ln 2 = A + 1/2 B\n\nUNIQUAC\ngE / RT = [ gE / RT ]combinatorial + [ gE / RT ]residual For a binary mixture UNIQUAC states that: [gE/RT]combinatorial = x1 ln [1*/x1] + x2 ln [2*/x2] + z/2 { x1q1 ln [1/1*] + x2q2 ln [2/2*]} [gE/RT]residual = - x1 q1 ln [ 1 + 2 21 ] - x2 q2 ln [ 2 + 1 12 ] * is the segment fraction , and are area fractions\n\n## z is the coordination number In UNIQUAC z=10\n\nUNIQUAC\nSegment fractions 1* = x1 r1 / [ x1 r1 + x2 r2 ] 2* = x2 r2 / [ x1 r1 + x2 r2 ]\n\n## The parameters r are pure-component molecular structure constants\n\nArea fractions 1 = x1 q1 / [ x1 q1 + x2 q2 ] 1 = x1 q1 / [ x1 q1 + x2 q2 ] 2 = x2 q2 / [ x1 q1 + x2 q2 ] 2 = x2 q2 / [ x1 q1 + x2 q2 ]\n\n## For fluids other than water and lower alcohols q = q\n\nEnergy parameters 12 = exp ( -u12 / RT ) exp ( -a12 / RT ) 21 = exp ( -u21 / RT ) exp ( -a21 / RT )\nThese parameters are given in terms of characteristic energies\n\n## Parameters, from Table 6.9\n\nCompound CCl4 CH3OH C2H5OH H2O Toluene N-Hexane r 3.33 1.43 2.11 0.92 3.92 4.50 q 2.82 1.43 1.97 1.40 2.97 3.86 q 2.82 0.96 0.92 1.00 2.97 3.86 * * *\n\n## Binary Parameters, from Table 6.10\n\nSystem Acetone/Chloroform Acetone/Water Acetone/Methanol Ethanol/n-octane Ethanol/n-heptane Ethanol/Benzene Ethanol/CCl4 Formic acid/Water N-Hexane/Nitromethane T/K 323 331-368 323 348 323 350-369 340-351 374-380 318 a12/K -171.71 530.99 379.31 -123.57 -105.23 -75.13 -138.90 924.01 230.64 a21/K 93.93 -100.71 -108.42 1354.92 1380.30 242.53 947.20 -525.85 -5.86\n\n## UNQUAC Activity Coefficients\n\ngE / RT = [ gE / RT ]combinatorial + [ gE / RT ]residual ln 1 = ln (1*/x1) + z/2 q1 ln (1/1*) + 2* (l1 - l2 r1/r2 ) - q1 ln ( 1 + 2 21 ) + 2q1 [ 21 / (1 + 2 21) - 12 / (2+112)] ln 2 = ln (2*/x2) + z/2 q2 ln (2/2*) + 1* (l2 - l1 r2/r1 ) - q2 ln ( 2 + 1 12 ) + 1q2 [ 12 / (2 + 1 12) - 21 / (1+212)]\n\n## l2 = z/2 (r2 - q2) - (r2 - 1)\n\nIn all the methods we discussed, it is necessary to have some experimental data to obtain the fitting parameters for any binary mixture of interest\n\nWhen experimental data are not available, an alternative approach is to look at the molecular structure of the compounds and to try obtaining the fitting parameters from the molecular structure of the interacting molecules\n\nExample\nLets suppose we have one mixture of acetone and toluene\n\nExample\nLets suppose we have one mixture of acetone and toluene To obtain the interaction parameters, we first look at the chemical structure of both compounds: Acetone CH3 C=O CH3 CH CH CH Toluene CH CH C CH3\n\nExample\nLets suppose we have one mixture of acetone and toluene Then we identify a number of groups, within these molecules, that we think are responsible for molecule-molecule interactions: Acetone CH3 C=O CH3 CH CH CH Toluene CH CH C CH3\n\nExample\nLets suppose we have one mixture of acetone and toluene Then we identify a number of groups, within these molecules, that we think are responsible for molecule-molecule interactions: Acetone CH3 C=O CH3 CH CH CH Toluene CH CH C CH3\n\nExample\nLets suppose we have one mixture of acetone and toluene Then we identify a number of groups, within these molecules, that we think are responsible for molecule-molecule interactions: Acetone CH3 C=O CH3 CH CH CH Toluene CH CH C CH3\n\nExample\nLets suppose we have one mixture of acetone and toluene Then we identify a number of groups, within these molecules, that we think are responsible for molecule-molecule interactions: Acetone CH3 C=O CH3 CH CH CH Toluene CH CH C CH3\n\nExample\nLets suppose we have one mixture of acetone and toluene Then we identify a number of groups, within these molecules, that we think are responsible for molecule-molecule interactions: Acetone CH3 C=O CH3 CH CH CH Toluene CH CH C CH3\n\nExample\nLets suppose we have one mixture of acetone and toluene Acetone CH3 C=O CH3 CH CH CH Toluene CH CH C CH3\n\nThe basic assumption is that each group will behave in any mixture independently on the molecule it is part of\n\nExample\nLets suppose we have one mixture of acetone and toluene Acetone CH3 C=O CH3 CH CH CH Toluene CH CH C CH3\n\nThe basic assumption is that each group will behave in any mixture independently on the molecule it is part of We obtain this contribution from available experimental data, and we predict the behavior of untested mixtures\n\nUNIFAC\nThe Universal Functional Activity Coefficient theory is the best known theory that uses this approach First introduced by Fredenslund, Jones, and Prausnitz in 1975 The main advantage of the procedure is that in typical mixtures of non electrolytes the number of group-group interactions is much less than the possible number of molecule-molecule pairs Another advantage is that correlations are always easier than the experiments they try to substitute Remember, however, that when no experimental data are used to improve the estimation of the interaction parameters, UNIFAC works, at best, as a good first approximation in VLE, but is often poor in LLE predictions\n\n## UNIFAC: The Method\n\nln i = ln iC + ln iR\n\n## UNIFAC: The Method\n\nln i = ln iC + ln iR\n\n## UNIFAC: The Method\n\nln i = ln iC + ln iR\n\nThis is the combinatorial part, calculated as in the case of UNIQUAC ln iC = FC (x, , ) ln iR = FR (X, Q, T, amn)\n\nThis is the residual contribution, which depends on group-group interactions As defined for UNIQUAC\n\nX = group mole fraction Q = group external surface area amn = interaction energy between groups n and m\n\nExample: Fig. F4\n\nExample: Fig. F5\n\n## n-hexane + methyl ethyl ketone @ 65 C This is more typical\n\nAppendix F\nMany binary systems have been studied and the corresponding data have been published DIPPR-AIChE provides a continuously-improving compilation Unfortunately, many other systems have not been reported UNIFAC, or other group-contribution methods, are in general not very accurate n-hexane + methyl ethyl ketone P/bar UNIFAC Experiments Fig. F-5 L\n\nV x/n-hexane\n\nAppendix F\nLets suppose we can obtain only 2 experimental data point Which ones shall we get? P/bar L\n\nV x/n-hexane\n\nAppendix F\nLets suppose we can obtain only 2 experimental data point Which ones shall we get? P/bar L\n\nV x/n-hexane For all binary mixtures, regardless of whether or not they form an azeotrope, binary parameters can be calculated from activity coefficients at infinite dilution These often provide the most valuable experimental information\n\nAppendix F: Example\nLets suppose we decide to use any functional form for gE of the mixture of interest we like: gE = F (x, A, B) x = composition A,B the two parameters that depend on the specific F we like\n\nAppendix F: Example\nLets suppose we decide to use any functional form for gE of the mixture of interest we like: gE = F (x, A, B) x = composition A,B the two parameters that depend on the specific F we like We saw that, depending on F, we can express the activity coefficient for both compounds as RT ln 1 = F1 (x, A, B) RT ln 2 = F2 (x, A, B) The specific F1 and F2 will depend on our choice for F\n\nAppendix F: Example\nWe saw that, depending on F, we can express the activity coefficient for both compounds as RT ln 1 = F1 (x, A, B) RT ln 2 = F2 (x, A, B) A,B are the two parameters we need to implement F If we measure 1 and 2, then\nx10\n\nlim [F1 (x, A, B) / RT] = ln 1 lim [F2 (x, A, B) / RT] = ln 2 Which is a system of 2 equations in 2 unknowns\n\nx20\n\nAppendix F: Example\nVan Laars equations\n\nln 1 = A ln 2 = B\n\nAppendix F: Example\nVan Laars equations\n\nln 1 = A ln 2 = B\nAnd we can predict the activity coefficients of both components at any composition:\n\n## ln 1 = A / [ 1 + A/B x1/x2 ]2 ln 2 = B / [ 1 + B/A x2/x1 ]2\n\nAppendix F: Example\nThree-SuffixMargules equations ln 1 = A + B ln 2 = A + 1/2 B\n\nAppendix F: Example\nThree-SuffixMargules equations ln 1 = A + B ln 2 = A + 1/2 B And we can predict the activity coefficients of both components at any composition: ln 1 = A x22 + B x23 ln 2 = (A + 3/2 B) x12 - B x13"
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https://soundear.dk/en/measuring-sound/
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[
"# Measuring Sound\n\nThe amplitude of a sound wave can be described in many ways. It is important to remember that not all sounds are the same, and hence the methods used to characterize the loudness of sound differ significantly from one another depending on the type of sound wave.\n\nSince noise and its harmful health effects are a primary concern everywhere, the method used to measure sound must provide relevant information to assess the noise level adequately.\n\nThis means that the measurement procedure will depend on the type of noise being considered. This also means that all measurements made for a similar noise source, such as the various types of industrial noise, must be made in the same way. Otherwise, the measurements won’t be comparable; this refers to the standardization of procedures.\n\nThe following sections will discuss some of the different ways of sound measurement.\n\n### The Difference Between Sound and Noise\n\nIt is important to distinguish between sound and noise. While sound comprises everything we hear, noise is generally classified as unwanted, unpleasant, annoying, or loud sound.\n\nNoisiness is directly associated with the loudness of sound, which must always be considered in connection with where and when the noise occurs.\n\nWhen considering industrial environments, farms, bars, and even cafeterias, noise is, shockingly, the most common health hazard, with permanent hearing damage being a primary concern.\n\nInterference in verbal communication and stress and general annoyance are among the main problems encountered in noisy environments.\n\nMany countries have strict government legislation in place, regulating the extent of noise permitted due to various activities, such as industries, construction, communal activities, etc.\n\nThe main purpose of such rules and regulations is not only to reduce and prevent hearing impairment among community members but also to provide a better quality of life. Because symptoms such as high blood pressure usually occur due to increased exposure to noise over time.\n\n### How Can You be Measuring Sound?\n\nThis section will describe two common ways of measuring sound, namely peak and root-mean-square-pressure.\n\n### Peak Pressure and Peak-to-Peak Pressure\n\nThe two most common and simple characterizing sound methods are by its peak pressure, also known as 0-to-peak pressure, and peak-to-peak pressure. The peak pressure is defined as the variation in pressure from zero to the greatest pressure emitted by a sound signal.\n\nThe peak-to-peak pressure is then defined as the range in pressure between the most positive and negative pressure.\n\n### Root-mean-square (RMS)\n\nThe root-mean-square-pressure (RMS) is characterized as the square base of normal of the square of a sound sign’s weight after some time.\n\nRMS is most usually used to portray a sound wave since it directly identifies with the vitality, or force, created by the wave.\n\nRMS pressure is mostly used to describe sound waves with a simple shape; things become more complicated if the sound wave comes in short impulses with varying frequencies.\n\nIn the latter case, RMS pressure will vary considerably, since the RMS method requires selecting a certain period over which to average the wave’s pressure, and the pressure of an impulsive wave differs over different segments.\n\nIn general, the lengthier the duration of the signal, the lower the RMS value.\n\nThe root-mean-square pressure of a sound wave can be calculated using the following four steps:\n\n1. Calculate the pressure at given points along with a sound signal.\n2. Square the calculated pressures.\n3. Take the average of the squared pressures.\n4. Take the square root of the average of the squared pressures.\n\nA simple sound wave and the three common methods described above to characterize its amplitude.\n\nSource: DOSITS.\n\n### Measuring Impulsive Sounds\n\nImpulsive sounds described in detail in upcoming sections are sounds that significantly surpass the background sound pressure level.\n\nThey last a very small duration, typically less than a second, and include gunshots, fireworks, hammer blows, and blasts from firearms.\n\nMethods for assessing the risk of hearing damage from impulsive noise are an active and ongoing research area.\n\nImpulse sound levels, particularly those produced by fireworks or firearms, can reach peak values of 170 to180 dB Sound Pressure Level (SPL), and sometimes even higher.\n\nMost conventional sound measuring instruments cannot accurately capture such intense sound levels; noise dosimeters and sound level meters usually have a maximum measurement limit of about 140 to 146 dB SPL.\n\nHowever, if such an instrument is used, the meter response times should be shorter for accurate measurements due to their rapid nature.\n\n### Energy Parameters – Leq and SEL\n\nLeq (equivalent continuous sound level) and SEL (sound exposure level) are energy parameters used to describe fluctuating sounds effectively.\n\nThese parameters are used because, in most measurement situations, a conventional sound level meter makes it very difficult to calculate a fluctuating wave’s exact sound level.\n\nLeq (equivalent continuous sound level) is the steady sound pressure level, which, over a specific period, has the same total energy as the original fluctuating noise. Hence, Leq is the RMS sound level, where the measurement duration is used as the averaging time.\n\n#### Leq is measured mathematically using the following equation:\n\nWhere:\n\n• Tm (secs) is the measured time interval\n• P(t) is the instantaneous sound pressure of the signal\n• P0 is the reference sound pressure of 20µP\n\nLeq is most commonly used to measure:\n\n• Fluctuating machinery noise\n• Long-term exposure to noise\n\nSEL, or sound exposure level, is a constant sound level with the same amount of energy in one second as the original noise.\n\nHence, SEL is analogous to Leq. The total sound energy is unified over the measurement period, but rather than averaging it over the entire period, a reference period of once the second is used.\n\nSEL is calculated using the following equation:\n\nwhere time is the exposure time of the Leq value expressed in seconds.\n\nSEL is most commonly used to measure transient noise.\n\n### Sound Level Meters\n\nBesides the mathematical equations described above, instruments known as sound level meters or sound pressure level (SPL) meter are also used to measure a sound wave’s intensity.\n\nA typical meter comprises a microphone that picks up sound and converts it into a measurable electrical signal, which is then worked on by electronic circuitry, which operates on the signal to describe its various characteristics. The indicating device is most commonly a meter which is calibrated to read sound levels in decibels.\n\nThe electronic circuitry is adjustable to read most sound frequencies and their intensities. Since the received alternating current (AC) signal must first be converted to a DC (direct current) signal, a time constant is required for averaging the signal, which will depend on the instrument’s purpose and what is being measured.\n\nA typical sound level meter can switch between a scale that measures uniform sound intensities for many frequencies, known as unweighted, and a scale that uses a frequency-dependent weighting factor that yields a measurement nearer to that of a human ear. A-frequency-weighting is the most commonly used standard.\n\nThe A-frequency-weighting scale is the most commonly used and is useful when describing the effect of complex noises on people.\n\nThere are three basic types of sound level meters, which are briefly described below.\n\n#### Standard sound level meter (Exponentially averaging sound level meter)\n\nAs described above, the standard sound level meter converts an AC signal received from the microphone to a DC signal via a root-mean-square (RMS) circuit. Hence, it requires a time constant of integration. However, an exponentially averaging sound level meter only provides a glimpse of the existing noise level, and its use is limited for assessing the risk of hearing damage.\n\n#### Integrating or integrating-averaging sound level meter\n\nAs suggested by the name, integrating meter merely integrates (sums up) the frequency-weighted noise to provide a measuring sound exposure. It is calculated using pressure squared into time.\n\n#### Noise Dosimeter\n\nThis sound level meter is specifically designed to measure how much a person is exposed to noise summed up over a specific period. The device is usually worn on the body and has hassle-free technical requirements. However, since it is body-worn, it has limited overall acoustic capabilities."
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https://www.colorhexa.com/379560
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"# #379560 Color Information\n\nIn a RGB color space, hex #379560 is composed of 21.6% red, 58.4% green and 37.6% blue. Whereas in a CMYK color space, it is composed of 63.1% cyan, 0% magenta, 35.6% yellow and 41.6% black. It has a hue angle of 146.2 degrees, a saturation of 46.1% and a lightness of 40%. #379560 color hex could be obtained by blending #6effc0 with #002b00. Closest websafe color is: #339966.\n\n• R 22\n• G 58\n• B 38\nRGB color chart\n• C 63\n• M 0\n• Y 36\n• K 42\nCMYK color chart\n\n#379560 color description : Dark moderate cyan - lime green.\n\n# #379560 Color Conversion\n\nThe hexadecimal color #379560 has RGB values of R:55, G:149, B:96 and CMYK values of C:0.63, M:0, Y:0.36, K:0.42. Its decimal value is 3642720.\n\nHex triplet RGB Decimal 379560 `#379560` 55, 149, 96 `rgb(55,149,96)` 21.6, 58.4, 37.6 `rgb(21.6%,58.4%,37.6%)` 63, 0, 36, 42 146.2°, 46.1, 40 `hsl(146.2,46.1%,40%)` 146.2°, 63.1, 58.4 339966 `#339966`\nCIE-LAB 55.227, -40.257, 20.034 14.433, 23.15, 14.774 0.276, 0.442, 23.15 55.227, 44.967, 153.543 55.227, -39.949, 32.191 48.115, -30.655, 15.476 00110111, 10010101, 01100000\n\n# Color Schemes with #379560\n\n• #379560\n``#379560` `rgb(55,149,96)``\n• #95376c\n``#95376c` `rgb(149,55,108)``\nComplementary Color\n• #3d9537\n``#3d9537` `rgb(61,149,55)``\n• #379560\n``#379560` `rgb(55,149,96)``\n• #37958f\n``#37958f` `rgb(55,149,143)``\nAnalogous Color\n• #95373d\n``#95373d` `rgb(149,55,61)``\n• #379560\n``#379560` `rgb(55,149,96)``\n• #8f3795\n``#8f3795` `rgb(143,55,149)``\nSplit Complementary Color\n• #956037\n``#956037` `rgb(149,96,55)``\n• #379560\n``#379560` `rgb(55,149,96)``\n• #603795\n``#603795` `rgb(96,55,149)``\n• #6c9537\n``#6c9537` `rgb(108,149,55)``\n• #379560\n``#379560` `rgb(55,149,96)``\n• #603795\n``#603795` `rgb(96,55,149)``\n• #95376c\n``#95376c` `rgb(149,55,108)``\n• #225d3c\n``#225d3c` `rgb(34,93,60)``\n• #297048\n``#297048` `rgb(41,112,72)``\n• #308254\n``#308254` `rgb(48,130,84)``\n• #379560\n``#379560` `rgb(55,149,96)``\n• #3ea86c\n``#3ea86c` `rgb(62,168,108)``\n• #45ba78\n``#45ba78` `rgb(69,186,120)``\n• #57c186\n``#57c186` `rgb(87,193,134)``\nMonochromatic Color\n\n# Alternatives to #379560\n\nBelow, you can see some colors close to #379560. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #379549\n``#379549` `rgb(55,149,73)``\n• #379550\n``#379550` `rgb(55,149,80)``\n• #379558\n``#379558` `rgb(55,149,88)``\n• #379560\n``#379560` `rgb(55,149,96)``\n• #379568\n``#379568` `rgb(55,149,104)``\n• #379570\n``#379570` `rgb(55,149,112)``\n• #379578\n``#379578` `rgb(55,149,120)``\nSimilar Colors\n\n# #379560 Preview\n\nThis text has a font color of #379560.\n\n``<span style=\"color:#379560;\">Text here</span>``\n#379560 background color\n\nThis paragraph has a background color of #379560.\n\n``<p style=\"background-color:#379560;\">Content here</p>``\n#379560 border color\n\nThis element has a border color of #379560.\n\n``<div style=\"border:1px solid #379560;\">Content here</div>``\nCSS codes\n``.text {color:#379560;}``\n``.background {background-color:#379560;}``\n``.border {border:1px solid #379560;}``\n\n# Shades and Tints of #379560\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, #020604 is the darkest color, while #f6fcf9 is the lightest one.\n\n• #020604\n``#020604` `rgb(2,6,4)``\n• #07140d\n``#07140d` `rgb(7,20,13)``\n• #0d2216\n``#0d2216` `rgb(13,34,22)``\n• #12311f\n``#12311f` `rgb(18,49,31)``\n• #173f29\n``#173f29` `rgb(23,63,41)``\n• #1d4d32\n``#1d4d32` `rgb(29,77,50)``\n• #225c3b\n``#225c3b` `rgb(34,92,59)``\n• #276a44\n``#276a44` `rgb(39,106,68)``\n• #2c784e\n``#2c784e` `rgb(44,120,78)``\n• #328757\n``#328757` `rgb(50,135,87)``\n• #379560\n``#379560` `rgb(55,149,96)``\n• #3ca369\n``#3ca369` `rgb(60,163,105)``\n• #42b272\n``#42b272` `rgb(66,178,114)``\n• #4abc7c\n``#4abc7c` `rgb(74,188,124)``\n• #59c287\n``#59c287` `rgb(89,194,135)``\n• #67c791\n``#67c791` `rgb(103,199,145)``\n• #75cc9b\n``#75cc9b` `rgb(117,204,155)``\n• #84d2a6\n``#84d2a6` `rgb(132,210,166)``\n• #92d7b0\n``#92d7b0` `rgb(146,215,176)``\n• #a0dcba\n``#a0dcba` `rgb(160,220,186)``\n• #afe1c5\n``#afe1c5` `rgb(175,225,197)``\n• #bde7cf\n``#bde7cf` `rgb(189,231,207)``\n• #cbecda\n``#cbecda` `rgb(203,236,218)``\n• #daf1e4\n``#daf1e4` `rgb(218,241,228)``\n• #e8f7ee\n``#e8f7ee` `rgb(232,247,238)``\n• #f6fcf9\n``#f6fcf9` `rgb(246,252,249)``\nTint Color Variation\n\n# Tones of #379560\n\nA tone is produced by adding gray to any pure hue. In this case, #5e6e65 is the less saturated color, while #00cc59 is the most saturated one.\n\n• #5e6e65\n``#5e6e65` `rgb(94,110,101)``\n• #567664\n``#567664` `rgb(86,118,100)``\n• #4f7d63\n``#4f7d63` `rgb(79,125,99)``\n• #478562\n``#478562` `rgb(71,133,98)``\n• #3f8d61\n``#3f8d61` `rgb(63,141,97)``\n• #379560\n``#379560` `rgb(55,149,96)``\n• #2f9d5f\n``#2f9d5f` `rgb(47,157,95)``\n• #27a55e\n``#27a55e` `rgb(39,165,94)``\n``#1fad5d` `rgb(31,173,93)``\n• #18b45c\n``#18b45c` `rgb(24,180,92)``\n• #10bc5b\n``#10bc5b` `rgb(16,188,91)``\n• #08c45a\n``#08c45a` `rgb(8,196,90)``\n• #00cc59\n``#00cc59` `rgb(0,204,89)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #379560 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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https://forum.openscad.org/I-made-a-thing-unionRound-fillet-union-td33534.html
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[
"# I made a thing - unionRound / fillet union",
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"Open this post in threaded view\n|\n\n## I made a thing - unionRound / fillet union\n\n This post was updated on . One of the most common request is boolean operations with fillet/chamfer/rounding/radius/smoothing . Most option so far has been unsatisfactory or glacially slow. Enabled by fast convex + convex minkowski sum. Thanks whoever implemented that. Unions of simple operands has a general FAST solution in all cases where the joint area is convex For non convex operands, a simple masking system enable us to isolate out only local convex areas to process. Masks are in practice a common volume that both operands are first intersected by.Though not a complete solution a good part of common cases can be filleted. (se limitations.) With an exploit of fast convex + convex minkowski and the Theorem: Given any collection of convex sets, their intersection is itself a convex set. The following became possible:",
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"//////////////////////////////////////////////////////// /* unionRound() 1.0 Module by Torleif Ceder - TLC123 late summer 2021 Pretty fast Union with radius, But limited to a subset of cases Usage unionRound( radius , detail , epsilon ) { YourObject(); YourObject(); } limitations: 0. Only really fast when boolean operands are convex, Minkowski is fast in that case. 1. Boolean operands may be concave but can only touch in a single convex area 2. Radius is of elliptic type and is only approximate r were operand intersect at perpendicular angle. */ //////////////////////////////////////////////////////// // Demo code unionRound(1.5, 5) { cube([10,10,2],true); rotate([20,-10,0])cylinder(5,1,1,\\$fn=12); } // end of demo code // module unionRound(r, detail = 5) { epsilon = 1e-6; children(0); children(1); step = 90 / detail; for (i = [0: detail-1]) { { x = r - sin(i * step ) * r; y = r - cos(i * step ) * r; xi = r - sin((i * step + step) ) * r; yi = r - cos((i * step + step) ) * r; color(rands(0, 1, 3, i)) hull() { intersection() { // shell(epsilon) clad(x) children(0); // shell(epsilon) clad(y) children(1); } intersection() { // shell(epsilon) clad(xi) children(0); // shell(epsilon) clad(yi) children(1); } } } } } // unionRound helper expand by r module clad(r) { minkowski() { children(); // icosphere(r,2); isosphere(r,70); } } // unionRound helper module shell(r) { difference() { clad(r) children(); children(); } } /* // The following is a sphere with some equidistant properties. // Not strictly necessary Kogan, Jonathan (2017) \"A New Computationally Efficient Method for Spacing n Points on a Sphere,\" Rose-Hulman Undergraduate Mathematics Journal: Vol. 18 : Iss. 2 , Article 5. Available at: https://scholar.rose-hulman.edu/rhumj/vol18/iss2/5 */ function sphericalcoordinate(x,y)= [cos(x )*cos(y ), sin(x )*cos(y ), sin(y )]; function NX(n,x)= let(toDeg=57.2958,PI=acos(-1)/toDeg, start=(-1.+1./(n-1.)),increment=(2.-2./(n-1.))/(n-1.) ) [ for (j= [0:n-1])let (s=start+j*increment ) sphericalcoordinate( s*x*toDeg, PI/2.* sign(s)*(1.-sqrt(1.-abs(s)))*toDeg)]; function generatepoints(n)= NX(n,0.1+1.2*n); module isosphere(r,detail){ a= generatepoints(detail); scale(r)hull()polyhedron(a,[[for(i=[0:len(a)-1])i]]); } [on Youtube] [on GitHub][on Thingiverse][on OpenSCAD Snippet Pad][on r /OpenSCAD]"
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https://community.jmp.com/t5/Discussions/Redo-cancels-lt-lt-report-actions/m-p/223012
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[
"Choose Language Hide Translation Bar\n\n## Redo cancels <<report actions\n\nHi,\n\nIf I run some report changes in the script and then run excluding of some rows with Autorecalc on,\n\nall report changes disappear.\n\nHow do I insert all report changes into the body of distribution, so they will not disappear?\n\nThanks.\n\ndt = Open( \"\\$SAMPLE_DATA/Big Class.jmp\" );\ndist = Distribution(\nAutomatic Recalc( 1 ),\nContinuous Distribution(\nQuantiles(0),\nSummary Statistics(1),\nColumn( :weight ),\nHorizontal Layout( 1 ),\nPpk Capability Labeling(1),\nCapability Analysis( LSL( 80 ), USL( 140 ) )\n));\nPpkValue = (dist << report)[OutlineBox(\"Long Term Sigma\")][NumberColBox(1)];\nPPKtb = Table Box(\nString Col Box( \"Parameter\", {\"Ppk\"} ),\nNumber Col Box( \"Value\", {PpkValue} )\n);\n(dist << report)[OutlineBox(\"Summary Statistics\")] << Append(PPKtb);\n(dist << report)[OutlineBox(\"Long Term Sigma\")] << Delete;\n\n1 ACCEPTED SOLUTION\n\nAccepted Solutions\n\n## Re: Redo cancels <<report actions\n\nMy script was intended to give you a pointer towards possible methods that could be used, with the hope that it would provide sufficient insight. I hope I can improve on my input with the following:\n\nI did not mention 2 methods in my previous response. I mentioned 3 methods, Filter Change Handler, RowState Handler and Add Graphics Script. Given the particulars of what you need, you may choose to use any one of the different options. I have provided 2 scripts below, one using the RowState Handler, and the other using the Add Graphics Script.\n\nThe RowState Handler is very similar to the implementation of the Filter Change Handler. Using it, the changes are triggered if any changes are made to the RowState column in the data table.\n\n``````names default to here(1);\ndt = Open( \"\\$SAMPLE_DATA/Big Class.jmp\" );\ndist = Distribution(\nAutomatic Recalc( 1 ),\nContinuous Distribution(\nQuantiles( 0 ),\nSummary Statistics( 1 ),\nColumn( :weight ),\nHorizontal Layout( 1 ),\nPpk Capability Labeling( 1 ),\nCapability Analysis( LSL( 80 ), USL( 140 ) )\n)\n);\n\nPpkValue = (dist << report)[\"Long Term Sigma\"][Number Col Box( 1 )];\nPPKtb = Table Box( String Col Box( \"Parameter\", {\"Ppk\"} ), Number Col Box( \"Value\", {PpkValue} ) );\n(dist << report)[\"Summary Statistics\"] << Append( PPKtb );\n(dist << report)[\"Long Term Sigma\"] << delete;);\n\nf = Function( {a}, addit() );\nrs = dt << Make RowState Handler( f );\n\nThe Add Graphics example iis more complex. This is primarily, because the script writer has to determine what items need to be compared to then trigger the code to be run. In the below case, the checking to see if the mean or standard deviation changes, triggers the running of the code. That may not be sufficient, for the kinds of things your users might be changing...so that made need to be expanded or completly changed.\n\n``````Names Default To Here( 1 );\ndt = Open( \"\\$SAMPLE_DATA/Big Class.jmp\" );\ndist = Distribution(\nAutomatic Recalc( 1 ),\nContinuous Distribution(\nQuantiles( 0 ),\nSummary Statistics( 1 ),\nColumn( :weight ),\nHorizontal Layout( 1 ),\nPpk Capability Labeling( 1 ),\nCapability Analysis( LSL( 80 ), USL( 140 ) )\n)\n);\n\nPpkValue = (dist << report)[\"Long Term Sigma\"][Number Col Box( 1 )];\nPPKtb = Table Box( String Col Box( \"Parameter\", {\"Ppk\"} ), Number Col Box( \"Value\", {PpkValue} ) );\n(dist << report)[\"Summary Statistics\"] << Append( PPKtb );\n(dist << report)[\"Long Term Sigma\"] << delete;\n);\n\ntheMean = ((dist << report)[\"Summary Statistics\"][Number Col Box( 1 )] << get);\ntheSTD = ((dist << report)[\"Summary Statistics\"][Number Col Box( 1 )] << get);\n\n(dist << report)[framebox( 2 )] << add graphics script(\nIf(\ntheMean != ((dist << report)[\"Summary Statistics\"][Number Col Box( 1 )] << get) | theSTD != ((dist\n<< report)[\"Summary Statistics\"][Number Col Box( 1 )] << get),\n);\ntheMean = ((dist << report)[\"Summary Statistics\"][Number Col Box( 1 )] << get);\ntheSTD = ((dist << report)[\"Summary Statistics\"][Number Col Box( 1 )] << get);\n);\n\nI strongly suggest that you take the time to read the Scripting Guide, and to familiarize yourself with the Scripting Index. The overview of JSL in the Scripting Guide provides the foundation one needs to be a proficient JSL programmer, and the examples for all of the objects and functions in the Scripting Index are invaluable in writing the code\n\nJim\n6 REPLIES 6\nHighlighted\n\n## Re: Redo cancels <<report actions\n\nI believe, your syntax is off a bit. Change the references to to following, and it works fine\n\n``````dt = Open( \"\\$SAMPLE_DATA/Big Class.jmp\" );\ndist = Distribution(\nAutomatic Recalc( 1 ),\nContinuous Distribution(\nQuantiles( 0 ),\nSummary Statistics( 1 ),\nColumn( :weight ),\nHorizontal Layout( 1 ),\nPpk Capability Labeling( 1 ),\nCapability Analysis( LSL( 80 ), USL( 140 ) )\n)\n);\nPpkValue = (dist << report)[Outline Box( \"Long Term Sigma\" )][Number Col Box( 1 )];\nPPKtb = Table Box( String Col Box( \"Parameter\", {\"Ppk\"} ), Number Col Box( \"Value\", {PpkValue} ) );\n(dist << report)[\"Summary Statistics\"] << Append( PPKtb );\n(dist << report)[\"Long Term Sigma\"] << delete;``````\nJim\n\n## Re: Redo cancels <<report actions\n\nThank you.\n\nIt does work fine, but when I exclude one row, all report changes disappear.\n\nWhat do you mean by \"Change the references to to following\"?\n\n## Re: Redo cancels <<report actions\n\nWhen you do a Recalc, the platform is effectively rerun, thus all of your post modifications need to also be rerun. One way to do this, is to use a filter change handler or a row state handler, which will run code, when a change in the filter or the row states are detected. You can also add a Graphics Script to the Frame Box() in the distribution graphic to trigger the execution of code when the graph changes. The code below uses a Make Filter Change Handler() to get the job done\n\n``````dt = Open( \"\\$SAMPLE_DATA/Big Class.jmp\" );\ndist = Distribution(\nAutomatic Recalc( 1 ),\nContinuous Distribution(\nQuantiles( 0 ),\nSummary Statistics( 1 ),\nColumn( :weight ),\nHorizontal Layout( 1 ),\nPpk Capability Labeling( 1 ),\nCapability Analysis( LSL( 80 ), USL( 140 ) )\n)\n);\n\nPpkValue = (dist << report)[\"Long Term Sigma\"][Number Col Box( 1 )];\nPPKtb = Table Box( String Col Box( \"Parameter\", {\"Ppk\"} ), Number Col Box( \"Value\", {PpkValue} ) );\n(dist << report)[\"Summary Statistics\"] << Append( PPKtb );\n(dist << report)[\"Long Term Sigma\"] << delete;);\n\nfilter = dist << Local Data Filter(\nAdd Filter( columns( :age ) )\n);\nf = Function( {a}, addit() );\nrs = filter << Make Filter Change Handler( f );\n\nJim\n\n## Re: Redo cancels <<report actions\n\nThanks Jim.\n\nUnfortunately, this solution is not good for me.\n\n1. I don't want to add local data filter\n\n2. When I run your script and change the selection in local data filter, the distribution reruns without custom report changes.\n\nWhat is the syntax of your second solution (Graphics Script to the Frame Box())?\n\nThanks again.\n\n## Re: Redo cancels <<report actions\n\nMy script was intended to give you a pointer towards possible methods that could be used, with the hope that it would provide sufficient insight. I hope I can improve on my input with the following:\n\nI did not mention 2 methods in my previous response. I mentioned 3 methods, Filter Change Handler, RowState Handler and Add Graphics Script. Given the particulars of what you need, you may choose to use any one of the different options. I have provided 2 scripts below, one using the RowState Handler, and the other using the Add Graphics Script.\n\nThe RowState Handler is very similar to the implementation of the Filter Change Handler. Using it, the changes are triggered if any changes are made to the RowState column in the data table.\n\n``````names default to here(1);\ndt = Open( \"\\$SAMPLE_DATA/Big Class.jmp\" );\ndist = Distribution(\nAutomatic Recalc( 1 ),\nContinuous Distribution(\nQuantiles( 0 ),\nSummary Statistics( 1 ),\nColumn( :weight ),\nHorizontal Layout( 1 ),\nPpk Capability Labeling( 1 ),\nCapability Analysis( LSL( 80 ), USL( 140 ) )\n)\n);\n\nPpkValue = (dist << report)[\"Long Term Sigma\"][Number Col Box( 1 )];\nPPKtb = Table Box( String Col Box( \"Parameter\", {\"Ppk\"} ), Number Col Box( \"Value\", {PpkValue} ) );\n(dist << report)[\"Summary Statistics\"] << Append( PPKtb );\n(dist << report)[\"Long Term Sigma\"] << delete;);\n\nf = Function( {a}, addit() );\nrs = dt << Make RowState Handler( f );\n\nThe Add Graphics example iis more complex. This is primarily, because the script writer has to determine what items need to be compared to then trigger the code to be run. In the below case, the checking to see if the mean or standard deviation changes, triggers the running of the code. That may not be sufficient, for the kinds of things your users might be changing...so that made need to be expanded or completly changed.\n\n``````Names Default To Here( 1 );\ndt = Open( \"\\$SAMPLE_DATA/Big Class.jmp\" );\ndist = Distribution(\nAutomatic Recalc( 1 ),\nContinuous Distribution(\nQuantiles( 0 ),\nSummary Statistics( 1 ),\nColumn( :weight ),\nHorizontal Layout( 1 ),\nPpk Capability Labeling( 1 ),\nCapability Analysis( LSL( 80 ), USL( 140 ) )\n)\n);\n\nPpkValue = (dist << report)[\"Long Term Sigma\"][Number Col Box( 1 )];\nPPKtb = Table Box( String Col Box( \"Parameter\", {\"Ppk\"} ), Number Col Box( \"Value\", {PpkValue} ) );\n(dist << report)[\"Summary Statistics\"] << Append( PPKtb );\n(dist << report)[\"Long Term Sigma\"] << delete;\n);\n\ntheMean = ((dist << report)[\"Summary Statistics\"][Number Col Box( 1 )] << get);\ntheSTD = ((dist << report)[\"Summary Statistics\"][Number Col Box( 1 )] << get);\n\n(dist << report)[framebox( 2 )] << add graphics script(\nIf(\ntheMean != ((dist << report)[\"Summary Statistics\"][Number Col Box( 1 )] << get) | theSTD != ((dist\n<< report)[\"Summary Statistics\"][Number Col Box( 1 )] << get),\n);\ntheMean = ((dist << report)[\"Summary Statistics\"][Number Col Box( 1 )] << get);\ntheSTD = ((dist << report)[\"Summary Statistics\"][Number Col Box( 1 )] << get);\n);"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.7210208,"math_prob":0.95300186,"size":9961,"snap":"2019-43-2019-47","text_gpt3_token_len":2533,"char_repetition_ratio":0.15657327,"word_repetition_ratio":0.7980594,"special_character_ratio":0.30137536,"punctuation_ratio":0.15102975,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99333197,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-20T19:56:23Z\",\"WARC-Record-ID\":\"<urn:uuid:2b6d0771-6017-40dc-bf4c-1d8466051ae0>\",\"Content-Length\":\"768584\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:62e2f14f-61cd-4f23-9e59-547350474690>\",\"WARC-Concurrent-To\":\"<urn:uuid:f86431c7-f3f4-444a-973b-ea92d7a38b88>\",\"WARC-IP-Address\":\"208.74.205.23\",\"WARC-Target-URI\":\"https://community.jmp.com/t5/Discussions/Redo-cancels-lt-lt-report-actions/m-p/223012\",\"WARC-Payload-Digest\":\"sha1:IHGGLFKJH2KEV6GIIKCFK4JC6T33L2BA\",\"WARC-Block-Digest\":\"sha1:OZAES4W4VQSDXXEBHYJBNA52ZU7B446Q\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496670601.75_warc_CC-MAIN-20191120185646-20191120213646-00067.warc.gz\"}"}
|
https://filmoku.com/?post=902
|
[
"# Where does the maximum deflection occur in the beam\n\n## Calculation of the bending stress of a beam section\n\n### Calculation of the bending stress in beams?\n\nIn this tutorial, we will examine how to calculate the bending stress of a beam using a bending stress formula that relates the longitudinal stress distribution in a beam to the internal bending moment acting on the cross section of the beam. We assume that this is the material of the carrier linear-elastic (i.e. Hooke's law applies). Bending stress is important, and since the bending of the beam is often the determining outcome in beam construction, it is important to understand.\n\n### 1. Calculation of the bending stress by hand\n\nLet's look at an example. Consider the I-beam shown below:",
null,
"At some distance along the length of the beam (the x-axis), it experiences an internal bending moment (M.) that you would normally find with a bending moment diagram. The general formula for bending or normal stress on the section is given by:",
null,
"For a given beam section, it is evident that the bending stress is maximized by the distance from the neutral axis (and). So, the maximum bending stress occurs either at the top or at the bottom of the beam section, whichever is greater:",
null,
"Let's consider the example of our I-Ray shown above. In our previous tutorial on the moment of inertia, we already found that the moment of inertia about the neutral axis is I = 4.74 × 108 mm4. Additionally, in the Center of Gravity tutorial, we found the center of gravity, and therefore the location of the neutral axis, 216.29 mm from the bottom of the section. This is shown below:",
null,
"Obviously, it is very common to ask for the maximum bending stress that the section will experience. For example, suppose we know from our bending moment diagram that the beam experiences a maximum bending moment of 50 kN-m or 50,000 Nm (convert bending moment units).\n\nThen we need to find out if the top or bottom of the section is furthest from the neutral axis. Clearly, the lower part of the section is further away with a distance of c = 216.29 mm. We now have enough information to find the maximum stress using the bending stress formula above:",
null,
"Similarly, we could find the bending stress in the section above, as we know it is y = 159.71 mm from the neutral axis (N / A):",
null,
"The last thing to worry about is whether the tension is causing compression or tension on the fibers of the section. When the beam sags like a \"U.\" then the upper fibers are compressed (negative stress) while the lower fibers are under tension (positive stress). When the beam sags down like a head \"U.\" then it is the other way round: the lower fibers are under pressure and the upper fibers are under tension.",
null,
"### 2. Calculation of the bending stress with SkyCiv Beam\n\nOf course, you don't have to do these calculations by hand as you can use the SkyCiv Beam - Bending Stress Calculator to find shear and bending stress in a beam! Just start by modeling the beam, using supports and applying loads. Once you click solve, the software will display the maximum stresses from this bending stress calculator. The picture below shows an example of an I-beam subjected to bending stress:",
null,
"### Start by calculating the bending stress with the SkyCiv Beam Calculator:\n\nStill stuck? How can we help?"
] |
[
null,
"https://skyciv.com/media/tutorials/stress/bending_stress_beam.png",
null,
"https://skyciv.com/wp-content/uploads/2019/02/Calculate-Bending-Stress-of-a-Beam-Section-1.png",
null,
"https://skyciv.com/wp-content/uploads/2019/02/Calculate-Bending-Stress-of-a-Beam-Section-2.png",
null,
"https://skyciv.com/media/tutorials/stress/I-beam-neutral-axis.png",
null,
"https://skyciv.com/wp-content/uploads/2019/02/Calculate-Bending-Stress-of-a-Beam-Section-3.png",
null,
"https://skyciv.com/wp-content/uploads/2019/02/Calculate-Bending-Stress-of-a-Beam-Section-4.png",
null,
"https://skyciv.com/media/tutorials/stress/bending_stress_beam_compression_tension.png",
null,
"https://skyciv.com/media/tutorials/stress/skyciv_beam_bending_stress.png",
null
] |
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|
http://aeneas.ps.uci.edu/cou_113b.html
|
[
"QUANTUM MECHANICS\n\n(Physics 113B)\n\nThe course is intended as an introduction to the methods of Quantum Mechanics. It will be assumed that the student has already some familiarity with ordinary and partial differential equations, linear algebra and Fourier transforms. Third-year courses in Electromagnetism and Classical Mechanics, as well as 113A, are prerequisites. Topics to be covered during the winter quarter will include:\n\n· Orbital Angular Momentum. Commutation Relations. Differential Operator Representation. Rasing and Lowering Operators. Relation to the Laplacian. Legendre Polynomials. Spherical Harmonics. Eigenvalues and Eigenvectors of the Angular Momentum, using the Algebraic Method.\n\n· Free Particle and the Three-Dimensional Square Well. Spherical Bessel Functions. The Two-Body Problem. Separation of the CMS Motion. Hydrogen Atom Spectrum. Laguerre Polynomials and Hydrogenic Wavefunctions. Angular and Radial Dependence of Orbitals. Degeneracies. Many-Electron Atoms.\n\n· Particle in a Magnetic Field. Constant Magnetic Field, Exact Solution. Landau Levels. Normal Zeeman Effect. Bohm-Aharonov Effect.\n\n· Spin One Half. Davisson and Germer Experiment. Pauli Matrices. Spin Magnetic Moment. Anomalous Zeeman Effect. Spin-Orbit Interaction. Thomas Precession. Addition of Two Angular Momenta. Stationary State Perturbation Theory.\n\n· Identical Particles. Symmetric and Anti-Symmetric Wavefunctions. Bosons and Fermions. Exclusion Principle. Spin and Statistics Relation. Slater Determinants. Occupation Numbers. Free Fermi Gas and the Fermi Energy. Neutron Stars. Kronig-Penney Model, Band Structure in Solids. Two-Electron Atoms. Helium Spectrum. Raleigh-Ritz Variational Method. Exchange Integrals. Ferromagnetism. Positronium Spectrum.\n\nRecommended Books:\n\nIntroduction to Quantum Mechanics, by R. Liboff (Addison Wesley, 1997);\n\nQuantum Physics, by S. Gasiorowicz (Wiley, 1997).\n\nHerbert W. Hamber, FRH 3172 (x5596).\n\[email protected]"
] |
[
null
] |
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|
https://www.cienciasinseso.com/en/standardization/
|
[
"# Standardization.",
null,
"Standardization is described, which consists of subtracting the mean from a value and dividing it by its standard deviation.\n\nBut I don’t accept any bar. It must be a very special bar. Or rather, a series of bars. And I’m not thinking about a bars chart, those so well-know and used that PowerPoint makes them almost without you asking for it. No, these graphs are very dull; they just represent how many times it repeats each of the values of a qualitative variable, but tell us nothing more.\n\n## Histogram and probability density\n\nI’m thinking about a much meaningful plot. I’m thinking about a histogram. Wow, you’ll say, but isn’t it another kind of bar chart?. Yes, but it has a different kind of bars, much more informative. To begin with, the histogram is used (or it should be) to represent frequencies of continuous quantitative variables. The histogram is not just a bar graph, but a frequency distribution. What does that mean?.\n\nWell, at the bottom, the bars are somewhat artificial. Let’s suppose a continuous quantitative variable such as weight. Imagine that our distribution ranges from 38 to 118 kg of weight. In theory, we can have infinite weight values (as with any continuous variable), but to represent the distribution we divide the range into an arbitrary number of intervals and draw a bar for each interval so that the height of the bar (and therefore its surface) is proportional to the number of cases inside the interval. This is a histogram: a frequency distribution.",
null,
"Now, suppose we make the intervals more and more narrow. The profile formed by the bars is increasingly looking like a curve as intervals narrow. In the end, what we’ll come up with is a curve, which will be called the probability density curve. The probability of a given value will be zero (one would think that it should be the height of the curve at that point, but it is not other than zero), but the probability of the values of a give interval is equivalent to the surface area under the curve in that interval. And what will be the area under the entire curve?.\n\nVery easy: the probability of finding any of the possible values, i.e., one (100% if you like percentages).\n\nAs you see, the histogram is much more than what it seems at first sight. It tells us that the probability of finding a value lower than the mean is 0.5, but not only that, because we can calculate the probability density of any value using a tiny formula that I prefer not to show you to avoid you closing your browsers and stopping reading this post. Moreover, there’s a simpler way to find it out.\n\nWith variables following a normal distribution (the famous bell) the solution is simple. We know that a normal distribution is perfectly characterized by its mean and standard deviation. The problem is that each normal curve has its own distribution, so the probability density curve is specific to each distribution. What can we do?. We can invent a standard normal distribution whose mean is zero and whose standard deviation is one and we can study its probability density so that we need neither formulas nor tables to know the probability of a given segment.\n\n## Standardization\n\nOnce done, we take any value of our distribution and transform it into its soul mate in the standard distribution. This process is called standardization and is as simple as subtracting the mean from the value and dividing the result by the standard deviation. Thus we obtain another statistic that physicians in general, and particularly statisticians, venerate the most: the z score.\n\nThe probability density of the standard distribution is well known. A z-value of zero is in the mean. The range of z = 0 ± 1.64 comprises 90% of the distribution; the rage of z = 0 ± 1.96 includes 95%; and z = 0 ± 2.58, 99%. What we do in practice is to choose the desirable standardized z value for our variable. This value is typically set at ±1 or ±2, according to the variable measured. Moreover, we can compare how the z-score is modified in successive determinations.\n\nThe problem arises because in medicine there’re many variables whose distribution is skewed and does not fit a normal curve, such as the height, blood cholesterol, and many others. But do not despair, mathematicians have invented a stuff called the central limit theorem, which says that if the sample size is large enough we can standardize any distribution and work with it as it fit the standard normal distribution. This theorem is a great thing because it allows standardizing even non-continuous variables that fit other distributions like the binomial, Poisson, or other.\n\n## We’re leaving…\n\nBut all this does not ends here. Standardization is the basis for calculating other features of the distribution such as the asymmetry index and kurtosis, and it is also the basis for many hypothesis contrasts seeking a known distribution to calculate statistical significance. But that’s another story…\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed.\n\nEsta web utiliza cookies propias y de terceros para su correcto funcionamiento y para fines analíticos. Al hacer clic en el botón Aceptar, aceptas el uso de estas tecnologías y el procesamiento de tus datos para estos propósitos. Antes de aceptar puedes ver Configurar cookies para realizar un consentimiento selectivo. Más información"
] |
[
null,
"https://www.cienciasinseso.com/wp-content/uploads/2012/09/dame-una-barra-300x200.jpg",
null,
"https://www.cienciasinseso.com/wp-content/uploads/2013/11/histogramas.jpg",
null
] |
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|
https://proofwiki.org/wiki/Natural_Number_Subtraction_is_not_Closed
|
[
"# Natural Number Subtraction is not Closed\n\n## Theorem\n\nThe operation of subtraction on the natural numbers is not closed.\n\n## Proof\n\nBy definition of natural number subtraction:\n\n$n - m = p$\n\nwhere $p \\in \\N$ such that $n = m + p$.\n\nHowever, when $m > n$ there exists no $p \\in \\N$ such that $n = m + p$.\n\n$\\blacksquare$"
] |
[
null
] |
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|
https://istopdeath.com/solve-by-factoring-x-35-2243/
|
[
"# Solve by Factoring (x-3)^(5/2)=243",
null,
"Move to the left side of the equation by subtracting it from both sides.\nAdd to both sides of the equation.\nRaise each side of the equation to the power to eliminate the fractional exponent on the left side.\nSolve the equation for .\nSimplify .\nSimplify the expression.\nRewrite as .\nApply the power rule and multiply exponents, .\nCancel the common factor of .\nCancel the common factor.\nRewrite the expression.\nRaise to the power of .\nMove all terms not containing to the right side of the equation.\nAdd to both sides of the equation.",
null,
"",
null,
"",
null,
"",
null,
"",
null,
""
] |
[
null,
"https://istopdeath.com/wp-content/uploads/ask60.png",
null,
"https://www.istopdeath.com/polygon_03-1.png",
null,
"https://www.istopdeath.com/ellipse_02-1-1.png",
null,
"https://www.istopdeath.com/ellipse_01-1-1.png",
null,
"https://www.istopdeath.com/path-60-1.png",
null,
"https://www.istopdeath.com/1510-1.png",
null
] |
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|
https://answers.everydaycalculation.com/compare-fractions/70-15-and-25-20
|
[
"Solutions by everydaycalculation.com\n\n## Compare 70/15 and 25/20\n\n1st number: 4 10/15, 2nd number: 1 5/20\n\n70/15 is greater than 25/20\n\n#### Steps for comparing fractions\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 15 and 20 is 60\n\nNext, find the equivalent fraction of both fractional numbers with denominator 60\n2. For the 1st fraction, since 15 × 4 = 60,\n70/15 = 70 × 4/15 × 4 = 280/60\n3. Likewise, for the 2nd fraction, since 20 × 3 = 60,\n25/20 = 25 × 3/20 × 3 = 75/60\n4. Since the denominators are now the same, the fraction with the bigger numerator is the greater fraction\n5. 280/60 > 75/60 or 70/15 > 25/20\n\nMathStep (Works offline)",
null,
"Download our mobile app and learn to work with fractions in your own time:"
] |
[
null,
"https://answers.everydaycalculation.com/mathstep-app-icon.png",
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.87096304,"math_prob":0.9901323,"size":863,"snap":"2023-40-2023-50","text_gpt3_token_len":310,"char_repetition_ratio":0.19208382,"word_repetition_ratio":0.0,"special_character_ratio":0.4391657,"punctuation_ratio":0.074074075,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9940321,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-03T20:18:22Z\",\"WARC-Record-ID\":\"<urn:uuid:19f906d1-3039-487d-9f93-81508cb8e60a>\",\"Content-Length\":\"7580\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:74541765-1573-4c20-9075-b9efb56fd19e>\",\"WARC-Concurrent-To\":\"<urn:uuid:0d178de3-0567-47a1-810c-24da93e85d85>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/compare-fractions/70-15-and-25-20\",\"WARC-Payload-Digest\":\"sha1:X47VE4BLNTXG23TV6J6HWKQNNKII5HID\",\"WARC-Block-Digest\":\"sha1:3FQHLCVZGNCD2UYUCNGV5B7APKEOX7HO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100508.53_warc_CC-MAIN-20231203193127-20231203223127-00638.warc.gz\"}"}
|
https://sharemylesson.com/search?f%5B0%5D=curriculum_tree%3A28571/28666/28691&f%5B1%5D=curriculum_tree%3A28571/28666/28691/28699&page=7
|
[
"# Mathematical Functions Lesson Plan Templates in High School Math\n\nFind free High School Math Lesson Plan Templates on Mathematical Functions.\n\n## Second order Differential Equations\n\nUpdated:\n2016\nResource\n• Trigonometric Functions\n• High School\nSubstitutions; review\n\n## Second order Differential Equations\n\nUpdated:\n2016\nResource\n• Trigonometric Functions\n• High School\nAuxiliary equations; particular integrals\n\nUpdated:\n2016\nResource\n• Trigonometric Functions\n• High School\nExamples to how to solve trigonometric equations using radians.\n\n## Sine and Cosine Rules\n\nUpdated:\n2016\nResource\n• Trigonometric Functions\n• High School\nA sheet with the formula needed for sine and cosine rules and the area of a non-right angled triangle\n\n## Elasticity & Energy\n\nUpdated:\n2016\nResource\n• Multiple Subjects\n• High School\nOne PowerPoint presentation on elastic springs and strings; and another on elastic potential energy\n\n## Integration review\n\nUpdated:\n2016\nResource\n• Trigonometric Functions\n• High School\nhandout on Integration review\n\n## Differentiating Trigonometry Worked Examples\n\nUpdated:\n2016\nResource\n• Trigonometric Functions\n• High School\nWorked Examples as to how to differentiate trigonometric functions with the standard differentials given\n\n## Integrating Trigonometry Worked Examples\n\nUpdated:\n2016\nResource\n• Trigonometric Functions\n• High School\nWorked Examples as to how to integrate the trigonometric functions\n\nUpdated:\n2016"
] |
[
null
] |
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|
http://eprints.gla.ac.uk/104980/
|
[
"",
null,
"# Symmetries of linear ordinary differential equations\n\n(1997) Symmetries of linear ordinary differential equations. Journal of Physics A: Mathematical and General, 30(13), pp. 4639-4649.\n\nFull text not currently available from Enlighten.\n\n## Abstract\n\nWe discuss the Lie symmetry approach to homogeneous, linear, ordinary differential equations in an attempt to connect it with the algebraic theory of such equations. In particular, we pay attention to the fields of functions over which the symmetry vector fields are defined and, by defining a noncharacteristic Lie subalgebra of the symmetry algebra, are able to establish a general description of all continuous symmetries. We use this description to rederive a classical result on differential extensions for second-order equations.\n\nItem Type: Articles Published Yes Athorne, Dr Chris Athorne, C. College of Science and Engineering > School of Mathematics and Statistics > Mathematics Journal of Physics A: Mathematical and General IOP Publishing 0305-4470 1361-6447\n\nUniversity Staff: Request a correction | Enlighten Editors: Update this record"
] |
[
null,
"http://eprints.gla.ac.uk/images/backgroundaa.jpg",
null
] |
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|
https://www.includehelp.com/python/function-classifications-on-the-basis-of-parameters-and-return-values.aspx
|
[
"# Python | Function classifications on the basis of parameters and return values\n\nPython function classifications: Here, we are going to learn about the python function classifications based on their parameters and return values.\nSubmitted by Pankaj Singh, on October 11, 2018\n\nThere are following types of the functions based on their parameters and return values:\n\n1. Function with no argument and no return value\n2. Function with no argument but return value\n3. Function with argument and no return value\n4. Function with arguments and return value\n\n### 1) Function with no argument and no return value\n\n```def sum():\na = int(input(\"Enter A: \"))\nb = int(input(\"Enter B: \"))\nc=a+b\nprint(\"Sum :\", c)\n\ndef main():\nsum()\n\nif __name__==\"__main__\":\nmain()\n```\n\nOutput\n\n```Enter A: 10\nEnter B: 20\nSum : 30\n```\n\n### 2) Function with no argument but return value\n\n```def sum():\na = int(input(\"Enter A: \"))\nb = int(input(\"Enter B: \"))\nc=a+b\nreturn c\n\ndef main():\nc = sum()\nprint(\"Sum :\",c)\n\nif __name__==\"__main__\":\nmain()\n```\n\nOutput\n\n```Enter A: 10\nEnter B: 20\nSum : 30\n```\n\n### 3) Function with argument and no return value\n\n```def sum(a,b):\nc=a+b\nprint(\"Sum :\", c)\n\ndef main():\na = int(input(\"Enter A: \"))\nb = int(input(\"Enter B: \"))\nsum(a,b)\n\nif __name__==\"__main__\":\nmain()\n```\n\nOutput\n\n```Enter A: 10\nEnter B: 20\nSum : 30\n```\n\n### 4) Function with arguments and return value\n\n```def sum(a,b):\nc=a+b\nreturn c\n\ndef main():\na = int(input(\"Enter A: \"))\nb = int(input(\"Enter B: \"))\nc = sum(a,b)\nprint(\"Sum :\",c)\n\nif __name__==\"__main__\":\nmain()\n```\n\nOutput\n\n```Enter A: 10\nEnter B: 20\nSum : 30\n```\n\nTOP Interview Coding Problems/Challenges\n\nLanguages: » C » C++ » C++ STL » Java » Data Structure » C#.Net » Android » Kotlin » SQL\nWeb Technologies: » PHP » Python » JavaScript » CSS » Ajax » Node.js » Web programming/HTML\nSolved programs: » C » C++ » DS » Java » C#\nAptitude que. & ans.: » C » C++ » Java » DBMS\nInterview que. & ans.: » C » Embedded C » Java » SEO » HR\nCS Subjects: » CS Basics » O.S. » Networks » DBMS » Embedded Systems » Cloud Computing\n» Machine learning » CS Organizations » Linux » DOS\nMore: » Articles » Puzzles » News/Updates"
] |
[
null
] |
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http://www.mathe2.uni-bayreuth.de/frib/html2/book/hyl00_43.html
|
[
"",
null,
"",
null,
"",
null,
"Some congruences\n\n### Some congruences\n\nAccording to Corollary we obtain from Theorem the following results:\nCorollary: For any subgroups G <= Sn and H <= Sm, the following congruences hold:\nå pÎGmc( p) º0 ( | G | ), åh ÎHa1(h)n º0 ( | H | ),\nand also\nå( r, p) ÎH ´G Õi=1na1( ri)ai( p) º0 ( | H | | G | ),\nas well as\nå( y, p) ÎH wr G Õ n=1c( p)a1(hn( y, p)) º0 ( | H | n | G | ).\nFurther congruences show up in the enumeration of group elements with prescribed properties. This theory of enumeration in finite groups is, besides the enumeration of chemical graphs, one of the main sources for the theory of enumeration which we are discussing here. A prominent example taken from this complex of problems is the following one due to Frobenius: The number of solutions of the equation xn=1 in a finite group G is divisible by n, if n divides the order of G. There are many proofs of this result and also many generalizations. Later on we return to this problem, at present we can only discuss a particular case which can be treated with the tools we already have at hand.\nExample: Let g denote an element of a finite group which forms its own conjugacy class and consider a prime number p, which divides | G | . We want to show that the number of solutions x ÎG of the equation xp=g is divisible by p. In order to prove this we consider the action of Cp on the set YX:=G p. The orbits are of length 1 or p. An orbit is of length 1 if and only if it consists of a single and therefore of a constant mapping (g', ...,g'), say. We now restrict our attention to the following subset M ÍG p:\nM:= {(g1, ...,gp) | g1 ...gp=g }.\nAs g forms its own conjugacy class, we obtain a subaction of Cp on M (for example g1 ...gp is conjugate to gpg1 ...gp-1). Hence the desired number k of solutions of xp=g is equal to the number of orbits of length 1 in M. Now we consider the number l of orbits of length p in M. It satisfies the equation k+pl= | M | . As each equation g1g'=g has a unique solution g1 in G, we moreover have that | M | = | G | p-1. Thus\nk ºk+pl= | M | º0 (p),\nwhich completes the proof. We note in passing that the center of G consists of the elements which form their own conjugacy class, so that we have proved the following:\nCorollary: If the prime p divides the order of the group G, then the number of p-th roots of each element in the center of G is divisible by p. In particular the number of p-th roots of the unit element 1 of G has this property (and it is nonzero, since 1 is a p-th root of 1), and hence G contains elements of order p.\nThis result can be used in order to give an inductive proof of Sylow's Theorem which we proved in example.\n\[email protected],\nlast changed: August 28, 2001",
null,
"",
null,
"",
null,
"Some congruences"
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"http://www-ang.kfunigraz.ac.at/~fripert/icons/previous.gif",
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"http://www-ang.kfunigraz.ac.at/~fripert/icons/next.gif",
null,
"http://www-ang.kfunigraz.ac.at/~fripert/icons/up.gif",
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"http://www-ang.kfunigraz.ac.at/~fripert/icons/previous.gif",
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|
https://www.planetanalog.com/simple-c-based-a-d-converters-part-1/
|
[
"# Simple μC-Based A/D Converters, Part 1\n\nOne useful dynamic application of μCs is as analog-to-digital converters (ADCs). ADCs over the years have gone from complete DVMs as instruments to embedded subsystems, to a few components. This two-part article presents some minimalist techniques for implementing μC-based ADCs.\n\nMinimalist A/D Conversion\n\nMultiple schemes have been devised over the years for A/D conversion. Some are simple, not very fast, and not very precise, such as the single ramp converter , as shown below; charge a capacitor with a current source, thereby generating a ramp.",
null,
"When the ramp crosses the unknown voltage, vx , to be measured, the time interval from the start of the ramp to vx , as detected by the comparator, is measured. For a (linear) ramp, the time is proportional to vx . A counter is reset (or overflows) at the start of the ramp and counts an accurate clock, such as a prescaled crystal-controlled μC clock. The ramp voltage is reset by a single transistor.\n\nThis ADC is easily implemented using a μC, and requires one counter, an input-port bit from the comparator and an output-port bit to reset the ramp. The ramp ADC is subject to inaccuracy because ramp slope variation can be caused by the ramp current source, capacitor drift, and comparator input-voltage offset and bias-current errors. It is conceptually simple, but we can do far better than this with fewer components.\n\nAnother category of ADCs are the parallel-feedback types. They use a DAC and comparator to set bits of the converted measurement. The general scheme is shown below.",
null,
"Depending on the logic, the converter can be one of multiple types. The successive-approximation scheme is predominant because it converts one bit per iteration, which takes n cycles to convert n bits independent of the converted value of vx . This scheme is well-suited for μC-based A/D conversion.\n\nThe parallel-feedback converter can be adapted to most μCs as shown below.",
null,
"The μC performs the logic. For the simplest and least desirable ADC, the ramp converter, the μC logic is given in algorithmic form below:\n\n1. Set OUT to zero: OUT ← 0.\n\n2. Input the IN bit.\n\n3. If IN = 0, then VX ← OUT; go to 1.\n\nElse increment OUT: OUT ← OUT + 1.\n\n4. Output OUT; go to 2.\n\nThe second ADC option, also differing from the others in its logic, is the tracking ADC . The advantage of the tracking converter is that it follows the analog waveform, though it is subject to digital slew-rate limitations for fast changes in vx . The tracking algorithm is given below.\n\n1. Output OUT.\n\n2. Input IN.\n\n3. If IN = 0, then decrement OUT: OUT ← OUT – 1.\n\nElse, increment OUT: OUT ← OUT + 1.\n\n4. Set VX to OUT: VX ← OUT.\n\n5. Go to 1.\n\nThis algorithm is not more complicated than that of the ramp converter but has the real-time tracking advantage. If the input waveform changes too quickly, the tracking ADC has the disadvantage that its DAC output slews, taking multiple iterations to “catch up” with the waveform and once again track it accurately.\n\nThe successive-approximation (SA) ADC uses the SA algorithm which is somewhat more involved but has the advantage of constant, input-independent conversion time. It uses two memory locations labeled SAR and SR. C is the processor carry bit.\n\n1. Clear SR and SAR: SR ← 0; SAR ← 0.\n\nSet C to one: C ← 1.\n\n2. Rotate SR right, with C.\n\n3. If C = 1, then return with VX ← SAR.\n\n4. Output SR OR SAR to OUT: OUT ← SR OR SAR.\n\n5. Input from IN.\n\n6. If IN = 1, then go to 2.\n\n7. Else, set SAR to SAR AND /SR: SAR ← SAR AND (NOT SR).\n\n(Alternative: SAR SAR AND (SR EOR 1111…).\n\n8. Go to 2.\n\nThe 1 bit, initially in C, is rotated (closed-loop shifted) right, into SR, one bit per iteration. When it returns to C (step 3 checks this), the procedure is finished. Step 4 sets the SR 1 bit in the SAR. If the comparator (IN) is high, vx is still greater than the SAR value, and this test bit remains set. If IN is low, the set bit made SAR too large, and it is cleared in step 7. Each bit, beginning with the MSB, is tested and then left set or cleared in SAR. Step 7 can use the clear-bit command for μCs that have it, with SR as the mask.\n\nMinimal-Circuit μC-Based DACs\n\nTo reduce hardware in minimalist fashion, for slow ADCs, the DAC can be implemented as a μC-based serial DAC, as shown below.",
null,
"Analog switch S3 requires a μC output-port bit. C1 and C2 are matched capacitors. A second port-bit output can implement S1 , S2 , and VR , but must have a high-impedance state, or three-state output; S1 and S2 must both be able to be set to off. The DAC reference supply, VR , uses the logic supply of the μC if S1 , S2 are implemented by a 3-state bit output from the μC. If that is not precise enough, then use two external analog switches, switched by two output-port bits.\n\nThe serial DAC algorithm, implemented in the μC, is given below, where SW is the state of S3 . OUT is the state of the C1 node, driven by S1 , S2.\n\n0. Serial DAC\n\n1. From MSB to LSB for n bits: SW = 0 (open)\n\n2. OUT = bn long enough to charge C; then OUT → hi-Z (open)\n\n3. SW = 1 long enough for charge transfer between C1 and C2 to equalize\n\n4. Decrement n . Go to 1.\n\nAnother way to implement a simple DAC is to use a μC PWM output. Filter it through an RC integrator, where the RC time constant is much larger than the PWM switching period. The result is a constant voltage with slight PWM ripple. The PWM DAC speed is limited by the low-pass RC filter but uses only two additional, low-cost parts, R and C. For μCs without a hardware PWM generator, the PWM also can be generated in software using a timer and an output-port bit, though the cycle period will be much longer than for hardware PWMs.\n\nμC-Based Σ- Δ ADCs\n\nA major improvement historically over the single slope of the ramp converter was the dual-slope converter , often simplified to the modified dual-slope converter , as shown below in its non- μC instantiation. It has a minimalist appeal in that it has one analog switch, the SPST reference input switch.",
null,
"Based on modified-dual-slope converter circuitry, an algorithmic variation that appeared in the late '60s was the charge-balancing or sigma-delta (Σ- Δ) (or delta-sigma ) converter. Instead of making a current-source switching decision once each measurement cycle, decisions are made every clock cycle instead, in an attempt to maintain a virtual ground at the input. The fraction of cycles requiring the opposing reference current to maintain virtual-ground “balance” gives the acquired count. A non- μC Σ- Δ circuit is shown below.",
null,
"The circuit topology differs from the modified dual-slope ADC in that the flop driven by the comparator is clocked, and is a D-type flop instead of an RS flop. On a given cycle of the clock, the reference is switched in or out of the integrator to keep vo near ground. The comparator output is the sign of the error. In other words, vo is nulled by discrete-time feedback. The number of clock cycles that the flop was high, NX , over the total number of conversion counts N , is the ratio of vX /VR .\n\nThe transfer characteristic is derived by constructing the charge-balance equation for the total charge from the vX and –VR inputs to the integrator. For vo = 0, they must be equal, or",
null,
"These charges are the sums of the per-cycle charges:",
null,
"The total charge of each depends on the number of cycles each is integrated. Then",
null,
"Substituting and solving for the measured output count,",
null,
"for an n -bit counter. This result is the same as for the modified dual-slope converter.\n\nThe charge-balancing scheme has an advantage over dual-slope schemes in that its integrator output has a very small range, that of one bit of change. The dual-slope integrator output is a triangle-wave with considerable range, and the waveform slopes must remain linear to the converter resolution over the integrator range for linear conversion.\n\nThe Σ- Δ ADC is also used as a modulator for serial digital telecommunications (in CODECs) and speech processing and can be used in its μC form for these waveform-conversion applications too if the conversion rate is high enough. Parallel-feedback-derived ADCs for μCs are faster than Σ- Δ ADCs, but require more external hardware. When a minimum external-parts-count converter is all that is needed, however, the Σ- Δ ADC is hard to beat.\n\nThe next part of this article presents some minimal parts count μC-based Σ- Δ converters and software algorithms for implementing them.\n\n## 1 comment on “Simple μC-Based A/D Converters, Part 1”\n\n1.",
null,
"Lleyton\nJune 10, 2016\n\nnice post\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed."
] |
[
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"https://www.planetanalog.com/wp-content/uploads/images-common-planetanalog-2016-06-564254-Image-1.jpg",
null,
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|
https://www.pcdandf.com/pcdesign/index.php/editorial/menu-features/12673-voltage-in-the-ghz-world
|
[
"",
null,
"",
null,
"Understand what is happening to the energy moving around in a circuit board.\n\nI want to challenge your understanding of electricity. My intentions are honorable. Technology is asking us to move more and more data, which in turn requires the use of more and more bandwidth. The tools we use have evolved from circuit theory, where we were very comfortable using lumped parameter models. Today, the interconnecting structures on logic circuit boards are just as important as the components. How a board is laid out controls delays, radiation and bit rates. I feel the meaning of voltage when it involves fast logic needs to be given a closer look. That is the subject of this article.\n\nThe measurements we make in circuits involve voltage differences. Logic is usually the presence or absence of voltage on a transmission line. We also measure radiation levels using antennas, which convert fields to voltage. We are totally immersed in describing electrical activity in terms of voltage. As we add bandwidth, it is important to review what a voltage measurement might imply. It is our window into electrical activity. I think I’m going to surprise you.\n\nI often tell engineers that nature does not read labels. I might add that nature does not read our textbooks or equations. It turns out engineers need voltage, but nature pays no heed to this concept. For many practical reasons, voltage is a tool that’s here to stay. There is no way to make progress without understanding what a voltage difference can imply when wave action is involved. We can predict what the voltage patterns might be, but unfortunately we cannot connect to most of the conducting surfaces that are involved. On a practical basis, we can observe only one point-pair at a time. On top of all this, things happen at the speed of light. If you really want to be impressed, there are relativistic effects to be considered.\n\n##### Voltage – The Definition and the Problem\n\nTo define voltage, it is necessary to introduce the idea of field. We all live in a gravity field that attracts our body mass. Fields store energy. As an example of energy storage, the moon/earth system stores the energy that give us our tidal action. Fields inside an atom form the nucleus. The energy in the nucleus of the atom is enormous.\n\nA collection of surplus electrons on a small mass creates an electric field. Voltage is defined as the work required to move a unit test charge in this field.\n\nVoltage is not a point concept. There are only voltage differences. Voltage differences can exist between points in space, as well as between conductors.\n\nOn a dry day, the friction of shoe leather on a rug allows the body to pick up a charge that causes an arc from a key to a door lock. This excess charge is measured as a voltage between the body and the door. The rubbing shoe leather does the work that moves the electric charge to the body. When the voltage gradient is high enough, there is arcing in air. On a grand scale, rain pulls charge from the air to earth. When the voltage gradient is high enough, lightning results.\n\n“Understanding energy is key to designing boards that function without radiation at increasing clock rates.”\n\nIn a typical logic circuit, a logic switch connects a voltage source to a short transmission line. This can be a trace over a conducting plane. The presence of a voltage difference means there is an electric field between the two conductors. This field stores energy. When current starts to flow, it creates a coupled magnetic field that uses this same space. A full explanation of what happens is complex and requires several chapters in a book. After all wave activity has stopped, the capacitance of this conductor geometry is fully charged, and the voltage along the transmission line is constant. This means the work required to move a test charge along any path between the two conductors is a constant. You will notice I didn’t discuss the time it takes to make this measurement. One thing for sure is a measurement of work cannot be made in zero time. In the time it takes to make a measurement, wave action may have changed the answer. This is the problem when picoseconds are important.\n\nThe difficulty is voltage is not a point concept. Any measure of voltage involves a loop area. This in turn means a voltage measurement depends on lead dress, as well as lead position. This also means many sources can contribute to one observation. The physics of electricity involves point concepts, and this avoids the issue of doing work. The parameters used in physics cannot be observed directly, and this poses a fundamental problem in engineering. Voltage has served us well, but progress requires we get closer to the fundamentals of electricity. We still require voltage, but at the same time we must understand what is happening to the energy moving around in our circuits.\n\n##### Moving Energy\n\nNature has one main objective and that is to reduce potential energy. It is why a rock falls, sound propagates, ocean waves travel, and why a voltage source will charge a capacitance. Energy is never lost. It simply takes on new forms. On a logic board, a logic level is a voltage on the capacitance of a trace over a ground plane. This voltage implies energy storage. This energy cannot be returned to the power source. To return the voltage to zero, stored energy must be converted to heat, radiated or moved out of the way. Moving energy into position or dissipating energy takes time, and this is the problem encountered in fast logic.\n\nA circuit board trace over a ground plane is an opportunity for nature to reduce potential energy. This reduction in potential energy usually implies spreading the energy out over a larger volume of space. Recognizing this one fact is important in circuit board design. We are interested in manipulating voltages, and nature is only interested in reducing potential energy. What we need to recognize is that when there are voltage differences, there is energy present. At any one point in space, energy can be stationary, in transition, in motion or perhaps all three at once. We are asked to deduce what is happening over an entire transmission line from a limited set of observations. Fortunately, circuit board operations are very repetitive. Once we see the patterns, the work has been done.\n\n##### Voltage Observations\n\nTo illustrate the problem of interpreting voltage, consider a transmission line just after a switch connects a voltage source to the line. Assume a finite rise time. Behind the wave front, the voltage is constant. In the space between the two conductors there is both energy flow and energy storage. The stored part of the energy is in the capacitance of the line. The energy that is moving involves both the static electricity and magnetic fields. The voltage alone does not tell this story. The voltage in transition at the leading edge is where magnetic field energy is converted to static electric field energy. After a reflection at an open gate, the voltage doubles. What is happening is energy is still moving forward on the line. Energy is not reflected at the open end of the line. The reflected wave front continues to convert arriving field energy into electric field energy. This is why the voltage doubles on the return wave. This is an example of where the energy flow and wave action are moving in opposite directions.\n\nWhat I have described is a typical problem in moving energy on a logic circuit board. We know what is happening, but it is not obvious by simply observing voltage. A static voltage difference can represent either energy flow or storage or both. Wave direction and energy flow direction may be opposites.\n\n##### Fast Circuit Boards and Energy Management\n\nThis article is intended as an introduction to what takes place on a logic circuit board. Energy is stored and moved in spaces and directed by conductor geometry. This understanding is key to designing boards that will function without radiation at increasing clock rates. Logic by its very nature implies treating transient behavior while carrier systems use sine waves, and this implies a steady-state behavior. Transmission line theory is usually taught in terms of sine waves, where measurement is a well understood discipline. Logic implies transient behavior, and the motion of energy is not that obvious.\n\nJohn Wiley has published my latest book titled Fast Circuit Boards and Energy Management. The principles outlined in the book help designers meet performance requirements. Basic physics is presented along with simple explanations. This subject of voltage is carefully treated.\n\nRalph Morrison has 50 years of experience in electronics engineering and is author of eight books, including Solving Interference Problems in Electronics, Grounding and Shielding Techniques in Instrumentation, and The Fields of Electronics: Understanding Electronics Using Basic Physics; This email address is being protected from spambots. You need JavaScript enabled to view it..\n\nRegister now for PCB West, the largest trade show for the printed circuit and electronics industry in the Silicon Valley! Coming Sept. 11-13 to the Santa Clara Convention Center."
] |
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null,
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|
https://stackoverflow.com/questions/23206187/java-util-date-null-output
|
[
"# Java.util.Date null output\n\nI have no idea why my code is giving me a null for java.util.date.\n\nQuestion: Write a test program that creates an Account object with an account ID of 1122, a balance of 20000, and an annual interest rate of 4.5%. Use the withdraw method to withdraw \\$2500, use the deposit method to deposit \\$3000, and print the balance, the monthly interest, and the date when this account was created\nHere is my code:\n\n``````import java.util.*;\npublic class Account {\nprivate int ID;\nprivate double Balance;\nprivate double annualInterestRate;\nprivate java.util.Date dateCreated;\npublic Account(){}\npublic Account(int ID, double Balance, double annualInterestRate){\nthis.ID=ID;\nthis.Balance=Balance;\nthis.annualInterestRate= annualInterestRate;\n}\npublic void setID(int ID){\nthis.ID=ID;\n}\npublic void setBalance(double Balance){\nthis.Balance=Balance;\n}\npublic void setAnnualInterestRate(double annualInterestRate){\nthis.annualInterestRate= annualInterestRate;\n}\npublic int getID(){\nreturn ID;\n}\npublic double getBalance(){\nreturn Balance;\n}\npublic double getInterestRate(){\nreturn annualInterestRate;\n}\npublic java.util.Date getDateCreated(){\nreturn dateCreated;\n}\npublic double getMonthlyInterestRate(){\nreturn annualInterestRate/12;\n}\npublic void withDraw(double val){\nif ((Balance - val) <0)\n{\nSystem.out.println(\"Offensive content removed from this line\");\n}\nelse\n{\nBalance -= val;\n}\n}\npublic void dePosits(double value){\nBalance += value;\n}\n\npublic static void main(String [] arges){\nAccount account = new Account(1122, 20000,.045);\naccount.withDraw(2500);\naccount.dePosits(3000);\nSystem.out.println(account.getBalance());\nSystem.out.println(account.getDateCreated());\n}\n``````\n\n}\n\n• In the future please consider posting more minimal examples. For example, the actual assignment requirements are irrelevant. So are methods like `getInterestRate()`. Narrowing down the problem when posting a question may help you find the problem as you do it. – Jason C Apr 21 '14 at 21:22\n• As this is obviously a homework question, here is also some advice as this a fairly trivial problem: You might want to get familiar with a debugger (I guess you use Eclipse or some other IDE), they can show you easily such errors by stepping through the program. Also, you might want to read up about basic Java programming. – dirkk Apr 21 '14 at 21:25\n• @dirkk thats a rather harsh comment. \"you might want to read up about basic java programming\" I think its obvious im a beginner no need to say something that distasteful. – user3083893 Apr 21 '14 at 21:33\n• If everyone learnt to use a debugger, the volume of questions on Stack Overflow would take a serious hit. – Dawood says reinstate Monica Apr 21 '14 at 21:33\n• @user3558063 It was not meant to be harsh, it was meant to be a helpful advice. As you made a very basic mistake I thought it would be a good idea if you read some basic tutorials. This is nothing to be ashamed about, we all know nothing in the beginning, so we all started sometime with reading basic stuff. Also, please mind that SO is international and we might have a cultural gap about what is harsh (I am German, so as a stereotype I am super direct in communicating), but my comment was certainly not distasteful. – dirkk Apr 21 '14 at 21:39\n\nYou haven't initialized `dateCreated`. For example, in the constructor (or somewhere else depending on your use case)\n``````dateCreated = new java.util.Date();"
] |
[
null
] |
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|
http://www.kpubs.org/article/articleMain.kpubs?articleANo=CMHHBA_2014_v47n2_77
|
[
"HALO SPIN PARAMETER IN COSMOLOGICAL SIMULATIONS\nHALO SPIN PARAMETER IN COSMOLOGICAL SIMULATIONS\nJournal of The Korean Astronomical Society. 2014. Apr, 47(2): 77-86\n• Received : June 07, 2013\n• Accepted : October 25, 2013\n• Published : April 30, 2014",
null,
"PDF",
null,
"e-PUB",
null,
"PubReader",
null,
"PPT",
null,
"Export by style\nArticle\nAuthor\nMetrics\nCited by\nTagCloud\nJIEUN, AHN\nSchool of Space Research, Kyung Hee University, Yongin, Kyeonggi 446-701, Korea\nJUHAN, KIM\nCenter for Advanced Computation, Korea Institute for Advanced Study, 85 Hoegi-ro, Dongdaemun-gu, Seoul 130-722, Korea\nJIHYE, SHIN\nDept. of Astronomy & Space Science, Kyung Hee University, Yongin, Kyeonggi 446-701, Korea\nSUNGSOO S., KIM\nDept. of Astronomy & Space Science, Kyung Hee University, Yongin, Kyeonggi 446-701, Korea\nYUN-YOUNG, CHOI\nDept. of Astronomy & Space Science, Kyung Hee University, Yongin, Kyeonggi 446-701, Korea\n\nAbstract\nUsing a cosmological ΛCDM simulation, we analyze the differences between the widely-used spin parameters suggested by Peebles and Bullock. The dimensionless spin parameter λ proposed by Peebles is theoretically well-justified but includes an annoying term, the potential energy, which cannot be directly obtained from observations and is computationally expensive to calculate in numerical simulations. The Bullock’s spin parameter λ′ avoids this problem assuming the isothermal density profile of a virialized halo in the Newtonian potential model. However, we find that there exists a substantial discrepancy between λ and λ′ depending on the adopted potential model (Newtonian or Plummer) to calculate the halo total energy and that their redshift evolutions differ to each other significantly. Therefore, we introduce a new spin parameter, λ″, which is simply designed to roughly recover the value of λ but to use the same halo quantities as used in λ′. If the Plummer potential is adopted, the λ″ is related to the Bullock’s definition as λ″=0.80×(1+ z ) −1/12 λ′. Hence, the new spin parameter λ″ distribution becomes consistent with a log-normal distribution frequently seen for the λ′ while its mean value is much closer to that of λ. On the other hand, in case of the Newtonian potential model, we obtain the relation of λ″=(1+ z ) −1/8 λ′; there is no significant difference at z = 0 as found by others but λ′ becomes more overestimated than λ or λ″ at higher redshifts. We also investigate the dependence of halo spin parameters on halo mass and redshift. We clearly show that although the λ′ for small-mass halos with Mh ＜ 2×10 12 M seems redshift independent after z = 1, all the spin parameters explored, on the whole, show a stronger correlation with the increasing halo mass at higher redshifts.\nKeywords\n1. INTRODUCTION\nThe original spin parameter was proposed by Peebles (1969) who quantified the rotation of an object with a dimensionless parameter (λ) that requires information on the total energy, halo mass, and angular momentum. Although this spin parameter has widely been used in many literatures (Antonuccio-Delogu et al. 2010; Bett et al. 2007; D’Onghia & Navarro 2007; Gardner 2001), the measurement of potential energy is practically difficult and somehow inaccurate in both simulations and observations. On the other hand, an alternative spin parameter (λ′) was proposed by Bullock et al. (2001), who used only two halo quantities, the total mass and angular momentum. This simplified version makes it much easier to measure the spin of simulated halos (Avila-Reese et al. 2005; Knebe & Power 2008; Stewart et al. 2013; Trowland et al. 2013) and observed galaxies (Burkert & D’Onghia 2004; Tonini et al. 2006).\nThese two spin parameters should be equivalent if a halo fully virializes and its density profile follows the isothermal halo model (Mo et al. 1998; Maccio et al. 2007). However, simulated halos do not fully virialize and show a wide range of deviations in the radial density profile. Consequently, the basic assumption in the Bullock’s definition should lead to some degree of scatters and a possible divergence in the λ – λ′ relation would be expected if halos are unvirialized.\nSimple quantitative comparisons between Peebles’ and Bullock’s spin parameters have been made by Tonini et al. (2006) and Hetznecker & Burkert (2006). Tonini et al. (2006) showed in their own galaxy model that the spin distribution of observed spirals peaks around λ ≃ 0.03 and λ′ ≃ 0.025. Burkert & D’Onghia (2004) showed the median value of simulated λ is around 0.042 while Bullock et al. (2001) reported that the smaller median value is preferred for λ′ (≃ 0.035). Also the relation",
null,
"PPT Slide\nLager Image\ncould also be confirmed in other papers. In the Millennium simulation, Bett et al. (2007) found that λ med ≃ 0.045 for halos of M ≃ 2 × 10 12 h −1 M , and Knebe & Power (2008) found a lower value of",
null,
"PPT Slide\nLager Image\n= 0.035 for the halo sample of Mmed = 5.25 × 10 12 h −1 M .\nMoreover, Hetznecker & Burkert (2006) showed that the distribution of λ′ seems to be redshift invariant while that of λ substantially shifts to a larger value as the evolution goes on. They argued that the different behavior may come from the difference in the distribution of spin transfer rate (the ratio of spin parameters after and prior to the accretion), implying that accretion may play a significant role in evolution of λ. And Munoz-Cuartas et al. (2011) found that there seems to be no obvious evolution of λ′ distribution in the redshift range of 0 ≤ z ≤ 2.\nIn spite of much effort, there are still large ambiguities in the halo potential. Although the definitions of halo spin are based on the Newtonian potential model, cosmological N -body simulations adopt the Plummer model to smooth out the divergence of small-scale Newtonian gravitational potential preventing violent scatterings between collisionless dark matter particles. Also the dynamical status of simulated halos may depend on the smoothing length, 𝜖, in the Plummer model and the potential measured is significantly over- or underestimated if one adopts different value of 𝜖 in the spin measurement. Therefore, in this paper we study various systematic effects giving rise to the discrepancy between λ and λ′ in detail.\nThis paper is organized as follows: In Section 2 and 3, we briefly describe the simulation specifics and show how to measure the physical quantities of simulated halos. Section 4 is devoted to show where the discrepancy between λ and λ′ of the simulated halo comes from. Section 5 provides an empirical scaling factor to eliminate all the systematics. In Section 6 we show the time evolution of spin distributions for each spin parameter. We summarize and discuss the main results of this work in Section 5.\n2. THE SIMULATION\nWe performed a WMAP 5-year cosmological simulation using 2048 3 particles in a cubic box of L box = 737.28 h −1 Mpc with the GOTPM code (Dubinski et al. 2004). The mean particle separation is",
null,
"PPT Slide\nLager Image\n= 0.36 h −1 Mpc and the particle mass is Mp = 3.37×10 9 h −1 M . The lowest halo mass (the collective mass of 30 particles) is Mc ≃ 10 11 h −1 M . In this cosmology, the present density ratios of the matter and dark energy to the critical density are Ω m0 = 0.26 and Ω Λ0 = 0.74, respectively. Then, a flat spatial curvature is ensured, or Ω k = 0. The Hubble constant is H 0 = 100 h km/second/Mpc, where we set h = 0.72. The start- ing redshift is zi = 120 and the number of time steps is 3000 with an equal spacing in the Hubble expansion factor ( a ) all the way down to z = 0. The simulation force resolution, 𝜖 is fixed to 𝜖 =",
null,
"PPT Slide\nLager Image\n/10 = 0.036 h −1 Mpc in the comoving scale. Using the simulation output, we applied the parallel FoF (Friend-of-Friend) to identify cosmic virialized structures (or halos) with the standard linking length of 0.2 times the mean particle separation.\nThe angular momentum of a halo is given by",
null,
"PPT Slide\nLager Image\nwhere the subscript i is the member index, mi is the mass of the particle, r i is its relative position vector to the center of mass, and v i is its relative velocity. The potential energy of a halo is half the sum of individual potentials of particles,",
null,
"PPT Slide\nLager Image\nwhere rij is the mutual distance between particles of index i and j , and 𝜖 is the gravitational smoothing scale imposed to the simulated gravity. For setting 𝜖 = 0, Eq. 2 reduces to the standard Newtonian potential as",
null,
"PPT Slide\nLager Image\nThe halo kinetic energy is simply measured as",
null,
"PPT Slide\nLager Image\nwhere the velocity vi incorperate the peculiar velocity and Hubble flow.\n3. THE SPIN PARAMETERS\nPeebles (1969) first proposed a dimensionless spin parameter of the form,",
null,
"PPT Slide\nLager Image\nwhere E is the total energy, M is the virial mass, and J is the total angular momentum of a halo. Among them, the total energy is hard to measure from both the observation and simulation because the potential energy requires the pre-knowledge of mass distribution far outside of the halo/galaxy boundaries (i.e., density profile). Therefore, a direct comparison of spin values between simulated halos and observed galaxies would be difficult in this definition.\nAn alternative spin parameter for dark matter halos was proposed by Bullock et al. (2001) as",
null,
"PPT Slide\nLager Image\nwhere R is the virial radius and V is the virial circular velocity defined as",
null,
"PPT Slide\nLager Image\n. One should note that the Newtonian potential model is intrinsically employed in deriving λ′ from λ and that only when a system is virialized and its density profile follows the spherical isothermal (SI) model, the λ′ is equivalent to the original spin parameter, λ. The virial radius can be determined using",
null,
"PPT Slide\nLager Image\nwhere Δ c ≃ (18π 2 + 82 x − 39 x 2 ) for a flat universe (Bryan & Normal 1998) and x ≡ ( z )−1. The matter content at redshift z is calculated as",
null,
"PPT Slide\nLager Image\nAnd the critical overdensity is measured from the Hubble expansion as",
null,
"PPT Slide\nLager Image\nwhere",
null,
"PPT Slide\nLager Image\n. In the denominator of Eq. 6, both the halo radius and circular velocity can easily be derived from the halo mass. Then, one needs to know the angular momentum and mass of a halo to calculate the spin parameter and this explains why Bullock’s definition has been widely adopted in many literatures. Finally, the Bullock’s spin parameter can be reduced to the form of",
null,
"PPT Slide\nLager Image\nwhere Δ vir ≡ Δ c /(1 + x ). In this equation, it is interesting to see that the Bullock’s definition seems to have an apparent redshift dependence. However, this is just due to the virial overdensity which is defined over the critical density that changes with redshift (see Eq. 9).\n4. DISCREPANCY IN SPIN VALUE BETWEEN THE DIFFERENT SPIN PARAMETERS\nIn Fig. 1 , we show the difference between λ and λ′ when the same Plummer potential model is adopted as used in the simulation. One can clearly see the significant offset of λ′ above the diagonal line (λ = λ′) and this baseline offset becomes larger at higher redshift. Also significant scatters are apparent spreading over to higher value of λ′. In this section we investigate what effects drive these discrepancies.",
null,
"PPT Slide\nLager Image\n— Scatter plots of with respect to λP at z = 0, 1, 2, and 3 (from top left panel to bottom right panel). Halo potentials are measured with the same smoothing length (𝜖) as used in the simulation run.\n- 4.1 Potential Model Effect\nIn N -body cosmological simulations of collisionless systems, zero-distance singularity in the Newtonian potential law of gravity (Φ ∝ 1/ r ) is present. To suppress the divergence of the Newtonian force at short distances, simulations have typically adopted Plummer potential model as a softened (or smoothed) Newtonian gravity as shown in Eq. 2. This implies that the gravitational potential is systematically biased in simulation if a non-zero softening length (or force resolution) is adopted. Note that in our simulation, we adopted the Plummer potential model and the force resolution is fixed by the mean particle separation",
null,
"PPT Slide\nLager Image\nas 𝜖 =",
null,
"PPT Slide\nLager Image\n/10 = 0.036 h −1 Mpc.\nIn what follows, we describe how the potential model affects the measurement of original spin parameter (λ) of simulated halos. Fig. 2 shows the spin parameters for 𝜖 ≠ 0 (Plummer potential model; λ P ) and 𝜖 = 0 (Newtonian potential model; λ N ), which shows the original spin parameter is not seriously affected by the force resolution, 𝜖 (or potential model). But there seems to be a slight redshift evolution of the spin relation between the potential models: at higher redshift, the λ of the Newtonian model tends to be larger than the Plummer model.",
null,
"PPT Slide\nLager Image\n— Relations between λN and λP , which are measured with the Plummer (x-axis) and Newtonian (y-axis) potentials.\nIn Fig. 3 we show the relation between the origi- nal and Bullock’s spin parameters in the Newtonian potential model at several redshifts ( z = 0, 1, 2, and 3). In this figure, λ N and",
null,
"PPT Slide\nLager Image\nseem to be comparable to each other at z = 0, which is consistent with the findings in the previous studies. However, we found a slight redshift evolution of λ′ becoming larger than λ. As a result, we conclude that the significant offset of",
null,
"PPT Slide\nLager Image\nfrom λ P mostly comes from the adopted potential model. At higher redshifts, a larger offset between λ P and",
null,
"PPT Slide\nLager Image\nis found, while",
null,
"PPT Slide\nLager Image\nshows less redshift evolution. In Section 5, we will discuss in detail.",
null,
"PPT Slide\nLager Image\n— Relation between λN and in the Newtonial potential model at z = 0, 1, 2, and 3.\nNow, we address how the Plummer potential model affects especially on the λ′ in detail. First, we want to quantify how much the halo circular velocity is affected by the potential softening in the simulation. The analytic functional form for the circular velocity in the Plummer model can be found in the appendix of Gerner (1996) in the virialized isothermal case. By fitting the numerical estimations, we obtained the following fitting function,",
null,
"PPT Slide\nLager Image\nwhere VP and VN are the circular velocities in the Plummer and Newtonian models, respectively. This fitting relation holds for 𝜖/ R ＜ 0.6. Due to the finite force resolution of simulation, Bullock’s spin parameter",
null,
"PPT Slide\nLager Image\nis intrinsically overestimated with respect to the",
null,
"PPT Slide\nLager Image\n(see Eq. 6) as",
null,
"PPT Slide\nLager Image\nFor a small halo of 30 particles,",
null,
"PPT Slide\nLager Image\nwould be higher than",
null,
"PPT Slide\nLager Image\nby 10％. If 𝜖 = 0 (Newtonian potential) or in the case of a very massive halo (R ≫ 𝜖),",
null,
"PPT Slide\nLager Image\nequals to",
null,
"PPT Slide\nLager Image\n. For less massive halos, the offset could be substantially larger than λ. Otherwise specified, hereafter, we use λ and λ′ as the ones measured with the Plummer potential model.\n- 4.2 Virialization Effect\nIn this section, we explain the origin of large scatters above the λ–λ′ line as shown in Fig. 1 . If a system virializes and its density profile follows the SI model, the Bullock’s spin should be equivalent to the Peeble’s spin in the Newtonian potential model. However, most of simulated halos are not fully virialized. Let us introduce a virialization parameter ( Cv ) of a halo as",
null,
"PPT Slide\nLager Image\nwhere K and P are the kinetic and potential energies of the halo, respectively. Fig. 4 shows the spin difference (λ′/λ−1) as a function of Cv for various redshifts. As a halo becomes virialized, the spin difference is reduced but a substantial offset is still seen at z = 0. For more univirialized halos, the spin difference is getting greater. The scatter at higher values of λ shown in Fig. 1 corresponds to the diverging tail ( Cv ＜ −0.4) in this figure. Also Fig. 4 shows that most of massive halos at z = 3 have negative Cv , indicating that these halos still have higher kinetic energies and hat more time is needed for halos to reach the virialization. On the other hand, a large fraction of nearly virialized halos at z = 0 have positive Cv ’s (or smaller kinetic energy compared to the potential energy). This redshift evolution of Cv may give us a hint for the evolution of internal dynamics of halos.",
null,
"PPT Slide\nLager Image\n— Scatter plots of the relative spin difference as a function of Cv ≡ 2K/P + 1 for z = 0, 1, 2, and 3. In this plot, we adopted the Plummer model to measure the halo potential.\nIn Fig. 5 we show the Newtonian version of spin difference (",
null,
"PPT Slide\nLager Image\n− λ N )/λ N ). In this figure, we see that the average halo virialization parameter is growing with time and in the current epoch most halos have positive virialization parameter ( Cv ＞ 0). It is reasonable to assume that halos may oscillate around Cv = 0 with time. The overshoot of halo virialization parameter ( Cv ) at z = 0 is because the Newtonian halo potential, PN , is overestimated compared with PP leading to larger values of Cv than the equilibrium state ( Cv = 0).",
null,
"PPT Slide\nLager Image\n— Same as Fig. 4 but for the Newtonian model case.\nComparing Fig. 4 and 5 , it would be important to note that which one is more proper in describing the dynamical status of halos. If same potential model is applied as used in the simulation, most of halos are assumed to be in the virialized region while on the other hand the Newtonian model predicts that halos overshoot the virialization condition due to larger value of potentials and that they are still unvirialized.\n- 4.3 Density Profile Effect\nMost of simulated halos are not only unrelaxed but also has a density profile substantially deviating from that of the SI model. Mo et al. (1998) and Maccio et al. (2007) pointed out that for a pure NFW-type halo (Navarro et al. 1997) the two spin parameters have a scaling relation as,",
null,
"PPT Slide\nLager Image\nwhere c is the concentration index of the NFW profile and",
null,
"PPT Slide\nLager Image\nFor a cluster halo of c ≃ 5, λ ≃ λ′. But for c ≃ 15 (a galaxy scale), λ′ can be underestimated with respect to λ by 20％ even for a virialized system. The halo concentration index has been known to monotonically decrease with increasing redshift (Duffy et al. 2008; Bullock et al. 2001), which implies that the average spin difference (⟨λ′– λ⟩) increases with redshift. For example, for a typical halo of mass M = 10 12 h −1 M , the average spin difference would be 35％ between z = 0 and 3, which is mainly due to the difference in the concentration index of the same mass halos. One should note that this relation is based on the Newtonian potential model.\n- 4.4 Mass Resolution Effect of Simulation\nNow, a question can be raised about how the measurements of virialization parameter and corresponding spin values are affected by the mass resolution of simulation, especially on low halo mass scales. Due to the under-sampling of member particles, smaller halos tend to have higher Poisson noise in the internal properties. Bett et al. (2007) investigated the Poisson effect using subhalos identified in two equivalent simulations of different mass resolutions (the mass ratio of particles is 64) and found that the higher resolution produces a few ％ lower λ than found in the lower-resolution simulation.\nTo address this issue, especially whether low-mass halo samples (around Mc = 10 11 h −1 M corresponding to the total mass of 30 particles) suffer from the low simulation resolution and show any obvious deviation from the relation projected from more massive halo samples, we investigate the distribution of virialization parameter as a function of halo mass. Fig. 6 shows the scatter plot of Cv as a function of halo mass at various redshifts. The median and quartiles of the sample distributions are shown as filled circles and error bars, respectively. The slope of median virialization parameter seems to be nearly flat at high redshift ( z = 3) but show a negative slope with the increasing halo mass can be observed at lower redshifts. A slope of Cv for more massive halo samples becomes more substantial at z ≤ 2, which is partially because massive halos become to have lower Cv ’s as the redshift decreases. Although the 1–σ distribution of Cv is increasing for smaller-mass samples, the median of Cv shows no significant change with increasing halo mass.",
null,
"PPT Slide\nLager Image\n— Scatter plot of Cv versus the halo mass. The median and quartiles of the distribution are shown by filled circles and error bars in each mass bin, respectively.\nTo investigate whether largely unvirialized halos suffer from the Poisson noise, we apply following criteria to divide the whole sample into the virialized and unvirialized halo samples by the critical value",
null,
"PPT Slide\nLager Image\n= −0.2 at z = 0 or",
null,
"PPT Slide\nLager Image\n= −0.4 at higher redshifts. In Figure 7 , we show the distribution of spin difference for virialized halos",
null,
"PPT Slide\nLager Image\n, where less massive halos tend to have larger difference than more massive ones. The scatter seems to be wider for less massive halos but it is attributed to the sample-size effect because the 1–σ distribution does not change substantially. It is interesting to note that massive halos ( M ≥ 10 13 h −1 M ) tends to have higher difference at higher redshift and this is mainly due to the change of density profile with time.",
null,
"PPT Slide\nLager Image\n— Distribution of spin difference for the virialized sample. The mean and 1–σ distribution are shown in the filled circles and error bars.\nFig. 8 is for the unvirialized halo samples. The median and quartiles of distribution are shown as filled circles and error bars, respectively. The spin difference and its scatter gradually grow as the halo mass decreases and the difference becomes larger at lower redshifts. Although larger scatter in lower-mass samples seems to be due to the higher Poisson noise, the average trends of virialization parameter and the spin difference (except high-mass end at z = 3) show a monotonic dependence on the halo mass.",
null,
"PPT Slide\nLager Image\n— Scatter plot of spin differences of unrelaxed halos. In this plot, we only select halos with Cv ＜ −0.2 at z = 0 and Cv ＜ −0.4 at other redshifts. The median and quartiles of the distribution are marked by the filled circles and error bars, respectively.\n5. EMPIRICAL CORRECTION TO λ′\nIn Fig. 7 , we see that even for virialized halos",
null,
"PPT Slide\nLager Image\nthere exist significant difference between λ and λ′ and the amount of the difference changes with redshift. From Section 4, we know that most of the baseline offset in λ′ – λ distribution may come from the effects of finite force resolution, different density profile, and Poisson noise, becoming worse for higher redshift. To correct this systematic offset, we empirically fit the offset as a function of redshift and obtain a correction factor",
null,
"PPT Slide\nLager Image\nIn this fitting function, one should note that the constant factor (0.80) mainly reflects the effect of the smoothed gravitational force. Thus, by introducing a “modified spin parameter” as",
null,
"PPT Slide\nLager Image\nwe can get spin values comparable to the original spin parameter using with two halo quantities, halo mass and angular momentum. In case of the Newtonian potential, we get",
null,
"PPT Slide\nLager Image\nfrom which the Newtonian model predicts better scaling relations. However, the Newtonian model has the problem in the virialization condition overestimating the halo potential making most of massive halos have substantially larger value of Cv = 0 at z = 0. From now on, we denote λ for the spin value measured in the Plummer potential model.\nFig. 9 shows the one-point distributions of the Peebles’ λ (short-dashed), Bullock’s λ′ (long-dashed), and modified (solid) spin parameter (λ″) at z = 0. As many researchers have found, Bullock’s definition shows a clear lognormal distribution (blue solid line), while Peebles’ spin distribution has a shallower tail at large-λ, which is also confirmed by Bett et al. (2007). One should note that λ″ has the same shape of distribu- tion as of λ′ but the peak position is reduced towards a smaller value by about 20％ that may be estimated using Eq. 11.",
null,
"PPT Slide\nLager Image\n— Distributions of λ (short-dashed), λ′ (long-dashed), and, λ″ (black solid) for the halo samples of mass ranges 2 × 1012 h−1M ≤ M ＜ 4 × 1012 h−1M at z = 0. The blue line is the lognormal fit to the Bullock’s spin distribution.\n6. REDSHIFT & MASS DEPENDENCES OF THE SPIN PARAMETER\nNow we study the evolution of spin distribution for different halo masses by dividing the whole halo sample into five subsamples according to the following mass criteria; 1 ≤ M/M 12 ＜ 2, 2 ≤ M/M 12 ＜ 4, 4 ≤ M/M 12 ＜ 6, 6 ≤ M/M 12 ＜ 10, and 10 ≤ M/M 12 ＜ 50 (where M 12 ≡ 1012 h −1 M ) at four redshift epochs ( z = 0, 1, 2, and 3). In Fig. 10 , we show the redshift evolution of spin distribution for the halo sample of 1 ≤ M/M 12 ＜ 2 (or particle members between 300 and 600). We see the overall redshift evolution with the peak moving to larger value until z = 1. After z = 1 the left slope of spin distribution seems to be almost fixed and, however, the other slope (including the high-end tail) of distribution shows a slight development with the slope less steeper and the tail more extended.",
null,
"PPT Slide\nLager Image\n— Redshift evolution of λ distribution of the subsample of 1 ≤ M/M12 ＜ 2.\nFig. 11 shows the change of λ-, λ′-, and λ″- distributions of halo subsamples except the subsample shown in Fig. 10 . The peak position of spin distribu- tion is increasing with redshift for all subsamples and more massive halos tend to have the lower value of spin peak. This can further be confirmed in Fig. 12 , where we plot the mean spin values, µ(λ), µ(λ′), and µ(λ″) as a function of redshift for each subsample. The error bars are measured with the bootstrapping by dividing each halo sample into 30 equal-size subsamples. The resulting error bars are too small to be seen in the plot. Also it should be noted that the lognormal peak position is slightly larger than the mean spin value due to the significant high-λ tail of the distribution.",
null,
"PPT Slide\nLager Image\n— Redshift evolutions of λ distribution for mass samples of 2 ≤ M/M12 ＜ 4, 4 ≤ M/M12 ＜ 6, 6 ≤ M/M12 ＜ 10, and 10 ≤ M/M12 ＜ 50 (where M12 ≡ 1012 h−1M) at z = 0 (solid), 1 (dotted), 2 (short-dashed), and 3 (long-dashed).",
null,
"PPT Slide\nLager Image\n— Evolution of the average spin value of the λ (left panel), λ′ (middle panel), and λ″ (right panel) distributions.\nFrom this figure, several characteristics in the redshift evolution of spin parameters can be identified. First, one may see clear redshift evolution of spin distribution in all subsamples. Bullock’s λ′ has higher mean spin values than λ or λ″ because the Plummer potential law is applied to measuring the spin value. In all cases, the mean values increase significantly with decreasing redshift until z = 1. Note that except the most massive sample (10 ≤ M/M 12 ＜ 50; solid lines in Fig. 12 ), one can clearly see that the evolution slows down after z = 1. At lower redshifts than z = 1, µ(λ′ of samples with M ≤4 × 10 12 h −1 M is almost fixed, which means that the redshift evolution seems to come to an end after z = 1 on this mass scale.\nFig. 12 shows the dependence of spin distribution on the halo mass. Less massive halos tend to have higher spin values but evolve less stronger with time than more massive ones. The mass dependence of the Peebles’ spin parameter λ shows somewhat different behavior from the others (λ′ & λ″). The mean value of λ distribution seems to be less dependent on the halo mass compared to those of λ′ and λ″. After z = 1, the change of average spin value along with the halo mass seems to be reversed (see the left panel in Fig. 12 ); a higher mass sample tends to have a higher spin value and shows more significant evolution with time. The halo samples with M/M 12 ＜ 4 show less-stronger time evolution, which is consistent with previous find- ings (Hetznecker & Burkert 2006;Munoz-Cuartas et al. 2011).\nFig. 13 shows the evolution of 1–σ width ( W λ ; 16％ ≤ λ ≤ 84％) of spin distribution for the halo subsamples. More massive samples show shallower distributions and the distributional width gradually grows with time. Comparing with Fig. 12 , we found that there is a positive correlation between µ(λ) and W λ ; a distribution with a higher mean value tends to have a wider width.",
null,
"PPT Slide\nLager Image\n— Evolution of the 1–𝜖 width of halo spin distribution in case of λ (left panel), λ′ (middle panel), and λ″ (right panel).\n7. SUMMARY & DISCUSSIONS\nIn this study, using a cosmological λCDM simulation we have shown that there is a substantial difference between λ and λ′ and the amount of the difference varies with redshift. Less virialized halos tend to have bigger differences and a substantial difference persists even for the virialized halos.\nWe investigated all the possible systematics which affect the measurements, such as Poisson nose, halo virialization, and different force law applied to the calculation of spin parameters. In order to correct all those systematics we introduced a new spin parameter λ″, which is related to λ′ with the multiplication factor fP (z) = 0.8 × (1 + z ) −1/12 . In this relation the constant offset 0.8 reflects the effect of finite smoothing scale (𝜖 =",
null,
"PPT Slide\nLager Image\n/10) in the Plummer potential. For setting 𝜖 = 0 (i.e., Newtonian model of gravity) to measure halo potential (in Eq. 2), a different correcting factor is obtained as fN(z) = (1 + z ) −1/8 .\nWe argue here that it is important to apply the same smoothing scale (𝜖) to the potential model for λ′ measure as used in the simulation. Otherwise, the halo potential may be different from the simulated value so that one may be mistaken about the dynamical status of halos.\nWe also found a different redshift evolution of λ from that of λ′. Our findings are similar to Hetznecker & Burkert (2006). The unbounding process (Onions et al. 2013) does not significantly change the value of λ′ of virialized halos while for the unrelaxed halos the difference between λ and λ′ becomes significant as shown by Hetznecker & Burkert (2006) and Maccio et al. (2007).\nWe found a clear evidence of the mass and redshift dependence of spin parameters. The average value of λ″ distribution, µ(λ″), changes over redshift and depends on the halo mass. We confirm that the λ″ dis- tribution is much closer to the original λ distribution compared to the λ′. It shows significant redshift dependence when z ＜ 1, especially for halos of the mass M/M 12 ＜ 4 as seen in the λ distribution and its halo mass dependence becomes weak compared to that of λ′.\nAcknowledgements\nThe simulation presented in this work were performed on computational resources provided by Cen- ter for Advanced Computation at Korea Institute for Advanced Study. JK thanks Dr. B. Cervantes-Sodi for his valuable suggestions and recommendations. YYC’s work was supported by a grant from the National Research Foundation (NRF) of Korea to the Center for Galaxy Evolution Research (No. 2010-0027910). JS and SSK’s work was supported by the BK21 plus program through the NRF of Korea. JA and SSK’s work was supported by Mid-career Research Program (No. 2011-0016898) through the NRF of Korea.\nReferences"
] |
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https://schoolbag.info/test/gre_2017/9.html
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[
" The Geography of the Math Section - How to Crack the Math Section - Cracking the GRE Premium Edition (2016) \n\nCracking the GRE Premium Edition (2016)\n\nPart IIIHow to Crack the Math Section\n\n9The Geography of the Math Section\n\n10Math Fundamentals\n\n11Algebra (And When to Use It)\n\n12Real World Math\n\n13Geometry\n\n14Math Et Cetera\n\nThe Geography of the Math Section\n\nThis chapter contains an overview of the content and structure you’ll see on the Math section of the GRE. It provides valuable information on pacing strategies and the various question formats you’ll encounter on the GRE. It also goes over how to use basic test-taking techniques such as Process of Elimination and Ballparking as they relate to math questions. After finishing this chapter, you’ll have a good idea of what the Math section of the GRE looks like and some basic approaches to help you navigate it.\n\nWHAT’S IN THE MATH SECTION\n\nThe GRE Math section primarily tests math concepts you learned in seventh through tenth grades, including arithmetic, algebra, and geometry. ETS alleges that the Math section tests the reasoning skills that you’ll use in graduate school, but what the Math section primarily tests is your comfort level with some basic math topics and your ability to take a test with strange-looking questions under timed conditions.\n\nThe Math section of the exam consists of two 35-minute sections, each of which will consist of 20 questions. The first 7 or 8 questions of each section will be quantitative comparisons (quant comp, for short). The remainder will consist of multiple-choice or numeric-entry questions.\n\nJunior High School?\n\nThe Math section of the GRE mostly tests how much you remember from the math courses you took in seventh, eighth, ninth, and tenth grades. But here’s some good news: GRE math is easier than SAT math. Why? Because many people study little or no math in college. If the GRE tested college-level math, everyone but math majors would bomb the test.\n\nIf you’re willing to do a little work, this is good news for you. By brushing up on the modest amount of math you need to know for the test, you can significantly increase your GRE Math score. All you have to do is shake off the dust.\n\nPredictable Questions\n\nThe beauty of a standardized test is that it is, well, standardized. Standardized means predictable. We know exactly what ETS is going to test and how they’re going to test it. The math side of the test consists of a series of utterly predictable questions, to which we have designed a series of highly scripted responses. ETS wants you to see each problem as a new challenge to solve. What you will find, however, is that there are only about 20 math concepts that are being tested. All of the questions you will see are just different ways of asking about these different concepts. Most of these concepts you already know. Once you recognize what’s being tested, even the trickiest questions become familiar and easy to solve.\n\nIn constructing the Math section, ETS is limited to the math that nearly everyone has studied: arithmetic, basic algebra, basic geometry, and elementary statistics. There’s no calculus (or even precalculus), no trigonometry, and no major-league algebra or geometry. Because of these limitations, ETS has to resort to traps in order to create hard problems. Even the most commonly missed GRE math problems are typically based on relatively simple principles. What makes the problems difficult is that these simple principles are disguised.\n\nMany test takers have no problem doing the actual calculations involved in the math questions on the GRE; in fact, you’ll even be allowed to use a calculator (more on that soon). However, on this test your ability to carefully read the problems and figure out how to set them up is more important than your ability to make calculations.\n\nPortal to watch topnotch\nPrinceton Review\nGRE Math section and\nwalk you through sample\nproblems.\n\nAs you work through this section, don’t worry about how quickly you’re doing the problems. Instead, take the time to really understand what the questions are asking; pay close attention to the wording of the problems. Most math errors are the result of careless mistakes caused by not reading the problem carefully enough!\n\nYou can do all the calculations right and still get a question wrong. How? What if you solve for x but the question asked for the value of x + 4? Ugh. Always reread the question before you choose an answer. Take your time and don’t be careless. The problem will stay on the screen as long as you want it to, so reread the question and double-check your work before answering it.\n\nOr how about this? The radius of the circle is 5, but when you copied the picture onto your scratch paper, you accidentally made it 6. Ugh! If you make a mistake copying down information from the screen, you’ll get the question wrong no matter how perfect your calculations are. You have to be extra careful when copying down information.\n\nTHE CALCULATOR\n\nAs we mentioned before, on the GRE you’ll be given an on-screen calculator. The calculator program on the GRE is a rudimentary one that gives you the five basic operations: addition, subtraction, multiplication, division, and square root, plus a decimal function and a positive/negative feature. It follows the order of operations, or PEMDAS (more on this topic in Chapter 10). The calculator also has the ability to transfer the answer you’ve calculated directly into the answer box for certain questions. The on-screen calculator can be a huge advantage—if it’s used correctly!\n\nAs you might have realized by this point, ETS is not exactly looking out for your best interests. Giving you a calculator might seem like an altruistic act, but rest assured that ETS knows that there are certain ways in which calculator use can be exploited. Keep in mind the following:\n\n1.Calculators Can’t Think. Calculators are good for one thing and one thing only: calculation. You still have to figure out how to set up the problem correctly. If you’re not sure what to calculate, then a calculator isn’t helpful. For example, if you do a percent calculation on your calculator and then hit “Transfer Display,” you will have to remember to move the decimal point accordingly, depending on whether the question asks for a percent or a decimal.\n\n2.The Calculator Can Be a Liability. ETS will give you questions that you can solve with a calculator, but the calculator can actually be a liability. You will be tempted to use it. For example, students who are uncomfortable adding, subtracting, multiplying, or dividing fractions may be tempted to convert all fractions to decimals using the calculator. Don’t do it. You are better off mastering fractions than avoiding them. Working with exponents and square roots is another way in which the calculator will be tempting but may yield really big and awkward numbers or long decimals. You are much better off learning the rules of manipulating exponents and square roots. Most of these problems will be faster and cleaner to solve with rules than with a calculator. The questions may also use numbers that are too big for the calculator. Time spent trying to get an answer out of a calculator for problems involving really big numbers will be time wasted. Find another way around.\n\n3.A Calculator Won’t Make You Faster. Having a calculator should make you more accurate, but not necessarily faster. You still need to take time to read each problem carefully and set it up. Don’t expect to blast through problems just because you have a calculator.\n\n4.The Calculator Is No Excuse for Not Using Scratch Paper. Scratch paper is where good technique happens. Working problems by hand on scratch paper will help to avoid careless errors or skipped steps. Just because you can do multiple functions in a row on your calculator does not mean that you should be solving problems on your calculator. Use the calculator to do simple calculations that would otherwise take you time to solve. Make sure you are still writing steps out on your scratch paper, labeling results, and using set-ups. Accuracy is more important than speed!\n\nOf course, you should not fear the calculator; by all means, use it and be grateful for it. Having a calculator should help you eliminate all those careless math mistakes.\n\nGEOGRAPHY OF A MATH SECTION\n\nEach of the two Math sections contains 20 questions. Test takers are allowed 35 minutes per section. The first 7 or 8 questions of each math section are quantitative comparisons, while the remainder are a mixed bag of problem solving, all that apply, numeric entry, and charts and graphs. Each section covers a mixture of algebra, arithmetic, quantitative reasoning, geometry, and real-world math.\n\nQUESTION FORMATS\n\nMuch like the Verbal section, the Math portion of the GRE contains a variety of different question formats. Let’s go through each type of question and discuss how to crack it.\n\nStandard Multiple Choice\n\nThese questions are the basic five-answer multiple-choice questions. These are great candidates for POE (Process of Elimination) strategies we will discuss later in this chapter.\n\nThese questions appear similar to the standard multiple-choice questions; however, on these you will have the opportunity to pick more than one answer. There can be anywhere from three to eight answer choices. Here’s an example of what these will look like:\n\nIf < x < , then which of the following could be the value of x ?\n\nIndicate all such values.\n\nYour approach on these questions won’t be radically different from the approach you use on standard multiple-choice questions. But obviously, you’ll have to consider all of the answers—make sure you read each question carefully and remember that more than one answer can be correct. For example, for this question, you’d click on (C) and (D). You must select every correct choice to get credit for the problem.\n\nQuantitative Comparison Questions\n\nQuantitative comparison questions, hereafter affectionately known as “quant comp” questions, ask you to compare Quantity A to Quantity B. These questions have four answer choices instead of five, and all quant comp answer choices are the same. Here they are:\n\nQuantity A is greater.\n\nQuantity B is greater.\n\nThe two quantities are equal.\n\nThe relationship cannot be determined from the information given.\n\nYour job is to compare the two quantities and choose one of these answers.\n\nQuant comp problems test the same basic arithmetic, algebra, and geometry concepts as do the other GRE math problems. So, to solve these problems, you’ll apply the same techniques you use on the other GRE math questions. But quant comps also have a few special rules you need to remember.\n\nThere Is No “(E)”\n\nBecause there are only four choices on quant comp questions, after you use POE to eliminate all of the answer choices you can, your odds of guessing correctly are even better. Think about it this way: Eliminating even one answer on a quant comp question will give you a one-in-three chance of guessing correctly.\n\nIf a Quant Comp Question Contains Only Numbers, the Answer Can’t Be (D)\n\nAny quant comp problem that contains only numbers and no variables must have a single solution. Therefore, on these problems, you can eliminate (D) immediately because the larger quantity can be determined. For example, if you’re asked to compare and , you can determine which fraction is larger, so the answer cannot be (D).\n\nCompare, Don’t Calculate\n\nYou don’t always have to calculate the exact value of each quantity before you compare them. After all, your mission is simply to compare the two quantities. It’s often helpful to treat the two quantities as though they were two sides of an equation. Anything you can do to both sides of an equation, you can also do to both quantities. You can add the same number to both sides, you can multiply both sides by the same positive number, and you can simplify a single side by multiplying it by one.\n\nDo only as much work\nas you need to.\n\nIf you can simplify the terms of a quant comp, you should always do so.\n\nHere’s a quick example:\n\nQuantity A is greater.\n\nQuantity B is greater.\n\nThe two quantities are equal.\n\nThe relationship cannot be determined from the information given.\n\nHere’s How to Crack It\n\nDon’t do any calculating! Remember: Do only as much work as you need to in order to answer the question! The first thing you should do is eliminate (D). After all, there are only numbers here. After that, get rid of numbers that are common to both columns (think of this as simplifying). Both columns contain a and a , so because we’re talking about addition, they can’t make a difference to the outcome. With them gone, you’re merely comparing the in column A to the in column B. Now we can eliminate (C) as well—after all, there is no way that is equal to . So, we’re down to two choices, (A) and (B). If you don’t remember how to compare fractions, don’t worry—it’s covered in Chapter 10 (Math Fundamentals). The answer to this question is (B).\n\nOkay, let’s talk about another wacky question type you’ll see in the Math section.\n\nNumeric Entry\n\nSome questions on the GRE won’t even have answer choices, and you’ll have to generate your own answer. Here’s an example:\n\nEach month, Renaldo earns a commission of 10.5% of his total sales for the month, plus a salary of \\$2,500. If Renaldo earns \\$3,025 in a certain month, what were his total sales?\n\nHere’s How to Crack It\n\nOn this type of question, POE is not going to help you! That means if you’re not sure how to do one of these questions, you should immediately move on. Leave it blank and come back to it in your second pass through the test.\n\nIf Renaldo earned \\$3,025, then his earnings from the commission on his sales are \\$3,025 − \\$2,500 = \\$525. So, \\$525 is 10.5% of his sales. Set up an equation to find the total sales: 525 = x, where x is the amount of the sales. Solving this equation, x = 5,000. (We’ll review how to set up and solve equations like this in later chapters.)\n\nTo answer this question, you’d enter 5000 into the box. Alternately, you could transfer your work directly from the on-screen calculator to the text box.\n\nAs you’re probably aware by now, doing well on the Math section will involve more than just knowing some math. It will also require the use of some good strategies. Let’s go through some good strategies now. Make sure you read this section carefully; it will be important for you to keep these techniques in mind as you work through the content chapters that follow this one!\n\nThe Two Roles of Techniques\n\nThe techniques are there to ensure that the questions that you should get right, you do get right. A couple of careless errors on easy questions will kill your score. The techniques are not just tools; they are proven standard approaches that save time and effort and guarantee points. Use these techniques on every question. Turn them into habits that you use every time.\n\nTake the Easy Test First\n\nThe new GRE offers the opportunity to mark a question and return to it. Since all questions count equally toward your score, why not do the easy ones first? Getting questions right is far more important than getting to every question, so start with the low hanging fruit. There is no law that says you have to take the test in the order in which it is given. If you see a question you don’t like, keep moving. Play to your strengths and get all of the questions that you’re good at in the bank, before you start spending time on the hard ones. It makes no sense to spend valuable minutes wrestling with hard questions while there are still easy ones on the table. It makes even less sense if you end up having to rush some easy ones (making mistakes in the process), as a result. Free yourself from numerical hegemony! Take the easy test first!\n\nBend, Don’t Push\n\nEighty percent of the errors on the math side of the test are really reading errors. The GRE is nearly a four-hour-long test, so at some point during this time your brain will get tired. When this happens you will read, see, or understand questions incorrectly. Once you see a problem wrong, it is nearly impossible to see it correctly. When this happens, even simple problems can become extremely frustrating. If you solve a problem and your answer is not one of the choices, this is what has happened. When you would swear that a problem can’t be solved, this is what has happened. When you have absolutely no idea how to solve a problem, this is what has happened. If you find yourself with half a page full of calculations and are no closer to the answer, this is what has happened. When you find yourself in this scenario, you can continue to push on that problem all day and you won’t get any closer.\n\nThere is a good chance that you are already familiar with this frustration. The first step is to learn to recognize it when it is happening. Here are some keys to recognizing when you are off track.\n\nYou know you are off track when…\n\n• You have spent more than three minutes on a single problem.\n• Your hand is not moving.\n• You don’t know what to do next.\n• You’re spending lots of time with the calculator and working with some ugly numbers.\n\nOnce you recognize that you are off track, stop and take a deep breath. Then move on to the next question. Continuing to push on a problem, at this point, is a waste of your time. You could easily spend three or four precious minutes on this problem and be no closer to the answer. Spend those three or four minutes on other questions. That time should be yielding you points, not frustration.\n\nAfter you have done two or three other questions, return to the one that was giving you trouble. Most likely, the reason it was giving you trouble is that you missed something or misread something the first time around. If the problem is still difficult, walk away again.\n\nThis is called Bend, Don’t Push. The minute you encounter any resistance on the test, walk away. Bend. There are plenty of other easier points for you to get with that time. Then return to the problem a few questions later. It’s okay to take two or three runs at a tough problem. If you run out of time before returning to the question, so be it. Your time is better spent on easier problems anyway, since all problems count the same.\n\nForcing yourself to walk away can be difficult, especially when you have already invested time in a question. You will have to train yourself to recognize resistance when it occurs, to walk away, and then to remember to come back. Employ this technique anytime you are practicing for the GRE. It will take some time to master. Be patient and give it a chance to work. With this technique, there are no questions that are out of your reach on the GRE.\n\nUse Process of Elimination whenever you can on questions that are in standard multiple-choice format. Always read the answer choices before you start to solve a math problem because often they will help guide you—you might even be able to eliminate a couple of answer choices before you begin to calculate the answer.\n\nTwo effective POE tools are Ballparking and Trap Answers.\n\nYou Know More Than You Think\n\nSay you were asked to find 30 percent of 50. Wait—don’t do any math yet. Let’s say that you glance at the answer choices and you see these:\n\n5\n\n15\n\n30\n\n80\n\n150\n\nBallparking helps you eliminate answer choices and increases your odds of zeroing in on the correct answer. The key is to eliminate any answer choice that is “out of the ballpark.”\n\nNeed More Math\nReview?\n\nCheck out Math Workout\nfor the GRE\nfor even more\nmath practice.\n\nLet’s look at another problem:\n\nA 100-foot rope is cut so that the shorter piece is the length of the longer piece. What is the length of the shorter piece in feet?\n\n75\n\n66\n\n50\n\n40\n\n33\n\nHere’s How to Crack It\n\nNow, before we dive into the calculations, let’s use a little common sense. The rope is 100 feet long. If we cut the rope in half, each part would be 50 feet. However, we didn’t cut the rope in half; we cut it so that there’s a longer part and a shorter part. What has to be true of the shorter piece then? It has to be less than 50 feet. If it weren’t, it wouldn’t be shorter than the other piece. So looking at our answers, we can eliminate (A), (B), and (C) without doing any real math. That’s Ballparking. By the way, the answer is (D).\n\nETS likes to include “trap answers” in the answer choices to their math problems. Trap answers are answer choices that appear correct upon first glance. Often these answers will look so tempting that you’ll choose them without actually bothering to complete the necessary calculations. Watch out for this! If a problem seems way too easy, be careful and double-check your work.\n\nLook at the this problem:\n\nThe price of a jacket is reduced by 10%. During a special sale, the price is discounted by another 10%. The special sale price is what percent less than the original price of the jacket?\n\n15%\n\n19%\n\n20%\n\n21%\n\n25%\n\nHere’s How to Crack It\n\nThe answer might seem like it should be 20 percent. But wait a minute: Does it seem likely that the GRE is going to give you a problem that you can solve just by adding 10 + 10? Probably not. Choice (C) is a trap answer.\n\nTo solve this problem, imagine that the original price of the jacket is \\$100. After a 10 percent discount the new price is \\$90. But now when we take another 10 percent discount, we’re taking it from \\$90, not \\$100. 10 percent of 90 is 9, so we take off another \\$9 from the price and our final price is \\$81. That represents a 19 percent total discount because we started with a \\$100 jacket. The correct answer is (B).\n\nHOW TO STUDY\n\nMake sure you learn the content of each of the following chapters before you go on to the next one. Don’t try to cram everything in all at once. It’s much better to do a small amount of studying each day over a longer period; you will master both the math concepts and the techniques if you focus on the material a little bit at a time. Just as we have been telling you in earlier chapters, let the content sink in by taking short study breaks between study sessions and giving yourself plenty of time to prepare for the GRE. Slow and steady wins the race!\n\nPractice, Practice, Practice\n\nPractice may not make perfect, but it sure will help. Use everyday math calculations as practice opportunities. Balance your checkbook without a calculator! Make sure your check has been added correctly at a restaurant, and figure out the exact percentage you want to leave for a tip. The more you practice simple adding, subtracting, multiplying, and dividing on a day-to-day basis, the more your arithmetic skills will improve for the GRE.\n\nAfter you work through this book, be sure to practice doing questions on our online tests and on real GREs. There are always sample questions at www.ets.org/gre, and practice will rapidly sharpen your test-taking skills.\n\nFinally, unless you trust our techniques, you may be reluctant to use them fully and automatically on the real GRE. The best way to develop that trust is to practice before you get to the real test.\n\ntons of informational\nvideos, practice tests,\nof this fantastic resource!\n\nSummary\n\n○The GRE contains two 35-minute Math sections. Each section has 20 questions.\n\n○The GRE tests math concepts up to about the tenth-grade level of difficulty.\n\n○You will be allowed to use a calculator on the GRE. The calculator is part of the on-screen display.\n\n○The Math section employs a number of different question formats, including multiple choice, numeric entry, and quantitative comparison questions.\n\n○Use the Two-Pass system on the Math section. Find the easier questions and do them first. Use your remaining time to work some of the more difficult questions.\n\n○When you get stuck on a problem, walk away. Do a few other problems to distract your brain, and then return to the question that was giving you problems.\n\n○Ballpark or estimate the answers to math questions and eliminate answers that don’t make sense.\n\n○Watch out for trap answers. If an answer seems too easy or obvious, it’s probably a trap."
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http://wikiwealth.wikidot.com/wacc-analysis:oibr
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[
"Oi SA - WACC Analysis\n\n# Oi SA (Weighted Average Cost of Capital (WACC) Analysis)\n\n## Helpful Information for Oi SA's Analysis\n\nWhat is the WACC Formula? Analyst use the WACC Discount Rate (weighted average cost of capital) to determine Oi SA's investment risk. WACC Formula = Cost of Equity (CAPM) * Common Equity + (Cost of Debt) * Total Debt. The result of this calculation is an essential input for the discounted cash flow (DCF) analysis for Oi SA. Value Investing Importance? This method is widely used by investment professionals to determine the correct price for investments in Oi SA before they make value investing decisions. This WACC analysis is used in Oi SA's discounted cash flow (DCF) valuation and see how the WACC calculation affect's Oi SA's company valuation.\n\n## WACC Analysis Information\n\n1. The WACC (discount rate) calculation for Oi SA uses comparable companies to produce a single WACC (discount rate). An industry average WACC (discount rate) is the most accurate for Oi SA over the long term. If there are any short-term differences between the industry WACC and Oi SA's WACC (discount rate), then Oi SA is more likely to revert to the industry WACC (discount rate) over the long term.\n\n2. The WACC calculation uses the higher of Oi SA's WACC or the risk free rate, because no investment can have a cost of capital that is better than risk free. This situation may occur if the beta is negative and Oi SA uses a significant proportion of equity capital."
] |
[
null
] |
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https://zims-en.kiwix.campusafrica.gos.orange.com/wikipedia_en_all_nopic/A/Nilpotent_matrix
|
[
"# Nilpotent matrix\n\nIn linear algebra, a nilpotent matrix is a square matrix N such that\n\n$N^{k}=0\\,$",
null,
"for some positive integer $k$",
null,
". The smallest such $k$",
null,
"is sometimes called the index of $N$",
null,
".\n\nMore generally, a nilpotent transformation is a linear transformation $L$",
null,
"of a vector space such that $L^{k}=0$",
null,
"for some positive integer $k$",
null,
"(and thus, $L^{j}=0$",
null,
"for all $j\\geq k$",
null,
"). Both of these concepts are special cases of a more general concept of nilpotence that applies to elements of rings.\n\n## Examples\n\n### Example 1\n\nThe matrix\n\n$A={\\begin{bmatrix}0&1\\\\0&0\\end{bmatrix}}$",
null,
"is nilpotent with index 2, since $A^{2}=0$",
null,
".\n\n### Example 2\n\nMore generally, any triangular matrix with zeros along the main diagonal is nilpotent, with index $\\leq n$",
null,
". For example, the matrix\n\n$B={\\begin{bmatrix}0&2&1&6\\\\0&0&1&2\\\\0&0&0&3\\\\0&0&0&0\\end{bmatrix}}$",
null,
"is nilpotent, with\n\n$B^{2}={\\begin{bmatrix}0&0&2&7\\\\0&0&0&3\\\\0&0&0&0\\\\0&0&0&0\\end{bmatrix}};\\ B^{3}={\\begin{bmatrix}0&0&0&6\\\\0&0&0&0\\\\0&0&0&0\\\\0&0&0&0\\end{bmatrix}};\\ B^{4}={\\begin{bmatrix}0&0&0&0\\\\0&0&0&0\\\\0&0&0&0\\\\0&0&0&0\\end{bmatrix}}.$",
null,
"The index of $B$",
null,
"is therefore 4.\n\n### Example 3\n\nAlthough the examples above have a large number of zero entries, a typical nilpotent matrix does not. For example,\n\n$C={\\begin{bmatrix}5&-3&2\\\\15&-9&6\\\\10&-6&4\\end{bmatrix}}\\qquad C^{2}={\\begin{bmatrix}0&0&0\\\\0&0&0\\\\0&0&0\\end{bmatrix}}$",
null,
"although the matrix has no zero entries.\n\n### Example 4\n\nAdditionally, any matrices of the form\n\n${\\begin{bmatrix}a_{1}&a_{1}&\\cdots &a_{1}\\\\a_{2}&a_{2}&\\cdots &a_{2}\\\\\\vdots &\\vdots &\\ddots &\\vdots \\\\-a_{1}-a_{2}-\\ldots -a_{n-1}&-a_{1}-a_{2}-\\ldots -a_{n-1}&\\ldots &-a_{1}-a_{2}-\\ldots -a_{n-1}\\end{bmatrix}}$",
null,
"such as\n\n${\\begin{bmatrix}5&5&5\\\\6&6&6\\\\-11&-11&-11\\end{bmatrix}}$",
null,
"or\n\n${\\begin{bmatrix}1&1&1&1\\\\2&2&2&2\\\\4&4&4&4\\\\-7&-7&-7&-7\\end{bmatrix}}$",
null,
"square to zero.\n\n### Example 5\n\nPerhaps some of the most striking examples of nilpotent matrices are $n\\times n$",
null,
"square matrices of the form:\n\n${\\begin{bmatrix}2&2&2&\\cdots &1-n\\\\n+2&1&1&\\cdots &-n\\\\1&n+2&1&\\cdots &-n\\\\1&1&n+2&\\cdots &-n\\\\\\vdots &\\vdots &\\vdots &\\ddots &\\vdots \\end{bmatrix}}$",
null,
"The first few of which are:\n\n${\\begin{bmatrix}2&-1\\\\4&-2\\end{bmatrix}}\\qquad {\\begin{bmatrix}2&2&-2\\\\5&1&-3\\\\1&5&-3\\end{bmatrix}}\\qquad {\\begin{bmatrix}2&2&2&-3\\\\6&1&1&-4\\\\1&6&1&-4\\\\1&1&6&-4\\end{bmatrix}}\\qquad {\\begin{bmatrix}2&2&2&2&-4\\\\7&1&1&1&-5\\\\1&7&1&1&-5\\\\1&1&7&1&-5\\\\1&1&1&7&-5\\end{bmatrix}}\\qquad \\ldots$",
null,
"These matrices are nilpotent despite the fact that none of the matrices prior to the matrix turning into zero contain any zero entries.\n\n## Characterization\n\nFor an $n\\times n$",
null,
"square matrix $N$",
null,
"with real (or complex) entries, the following are equivalent:\n\n• $N$",
null,
"is nilpotent.\n• The characteristic polynomial for $N$",
null,
"is $\\det \\left(xI-N\\right)=x^{n}$",
null,
".\n• The minimal polynomial for $N$",
null,
"is $x^{k}$",
null,
"for some positive integer $k\\leq n$",
null,
".\n• The only complex eigenvalue for $N$",
null,
"is 0.\n• tr(Nk) = 0 for all $k>0$",
null,
".\n\nThe last theorem holds true for matrices over any field of characteristic 0 or sufficiently large characteristic. (cf. Newton's identities)\n\nThis theorem has several consequences, including:\n\n• The degree of an $n\\times n$",
null,
"nilpotent matrix is always less than or equal to $n$",
null,
". For example, every $2\\times 2$",
null,
"nilpotent matrix squares to zero.\n• The determinant and trace of a nilpotent matrix are always zero. Consequently, a nilpotent matrix cannot be invertible.\n• The only nilpotent diagonalizable matrix is the zero matrix.\n\n## Classification\n\nConsider the $n\\times n$",
null,
"shift matrix:\n\n$S={\\begin{bmatrix}0&1&0&\\ldots &0\\\\0&0&1&\\ldots &0\\\\\\vdots &\\vdots &\\vdots &\\ddots &\\vdots \\\\0&0&0&\\ldots &1\\\\0&0&0&\\ldots &0\\end{bmatrix}}.$",
null,
"This matrix has 1s along the superdiagonal and 0s everywhere else. As a linear transformation, the shift matrix \"shifts\" the components of a vector one position to the left, with a zero appearing in the last position:\n\n$S(x_{1},x_{2},\\ldots ,x_{n})=(x_{2},\\ldots ,x_{n},0).$",
null,
"This matrix is nilpotent with degree $n$",
null,
", and is the canonical nilpotent matrix.\n\nSpecifically, if $N$",
null,
"is any nilpotent matrix, then $N$",
null,
"is similar to a block diagonal matrix of the form\n\n${\\begin{bmatrix}S_{1}&0&\\ldots &0\\\\0&S_{2}&\\ldots &0\\\\\\vdots &\\vdots &\\ddots &\\vdots \\\\0&0&\\ldots &S_{r}\\end{bmatrix}}$",
null,
"where each of the blocks $S_{1},S_{2},\\ldots ,S_{r}$",
null,
"is a shift matrix (possibly of different sizes). This form is a special case of the Jordan canonical form for matrices.\n\nFor example, any nonzero 2 × 2 nilpotent matrix is similar to the matrix\n\n${\\begin{bmatrix}0&1\\\\0&0\\end{bmatrix}}.$",
null,
"That is, if $N$",
null,
"is any nonzero 2 × 2 nilpotent matrix, then there exists a basis b1, b2 such that Nb1 = 0 and Nb2 = b1.\n\nThis classification theorem holds for matrices over any field. (It is not necessary for the field to be algebraically closed.)\n\n## Flag of subspaces\n\nA nilpotent transformation $L$",
null,
"on $\\mathbb {R} ^{n}$",
null,
"naturally determines a flag of subspaces\n\n$\\{0\\}\\subset \\ker L\\subset \\ker L^{2}\\subset \\ldots \\subset \\ker L^{q-1}\\subset \\ker L^{q}=\\mathbb {R} ^{n}$",
null,
"and a signature\n\n$0=n_{0}",
null,
"The signature characterizes $L$",
null,
"up to an invertible linear transformation. Furthermore, it satisfies the inequalities\n\n$n_{j+1}-n_{j}\\leq n_{j}-n_{j-1},\\qquad {\\mbox{for all }}j=1,\\ldots ,q-1.$",
null,
"Conversely, any sequence of natural numbers satisfying these inequalities is the signature of a nilpotent transformation.\n\n## Additional properties\n\n• If $N$",
null,
"is nilpotent, then $I+N$",
null,
"is invertible, where $I$",
null,
"is the $n\\times n$",
null,
"identity matrix. The inverse is given by\n$(I+N)^{-1}=\\displaystyle \\sum _{m=0}^{\\infty }\\left(-N\\right)^{m}=I-N+N^{2}-N^{3}+N^{4}-N^{5}+\\cdots ,$",
null,
"where only finitely many terms of this sum are nonzero.\n• If $N$",
null,
"is nilpotent, then\n$\\det(I+N)=1,\\!\\,$",
null,
"where $I$",
null,
"denotes the $n\\times n$",
null,
"identity matrix. Conversely, if $A$",
null,
"is a matrix and\n$\\det(I+tA)=1\\!\\,$",
null,
"for all values of $t$",
null,
", then $A$",
null,
"is nilpotent. In fact, since $p(t)=\\det(I+tA)-1$",
null,
"is a polynomial of degree $n$",
null,
", it suffices to have this hold for $n+1$",
null,
"distinct values of $t$",
null,
".\n\n## Generalizations\n\nA linear operator $T$",
null,
"is locally nilpotent if for every vector $v$",
null,
", there exists a $k\\in \\mathbb {N}$",
null,
"such that\n\n$T^{k}(v)=0.\\!\\,$",
null,
"For operators on a finite-dimensional vector space, local nilpotence is equivalent to nilpotence.\n\nThis article is issued from Wikipedia. The text is licensed under Creative Commons - Attribution - Sharealike. Additional terms may apply for the media files."
] |
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|
https://relativeathleticscores.com/2020/01/25/tim-broady-ras/
|
[
"Tim Broady went undrafted for the NFL out of Murray State in 1989.\n\nHe recorded a Relative Athletic Score of 5.42, out of a possible 10.0. RAS is a composite metric on a 0 to 10 scale based on the average of all of the percentile for each of the metrics the player completed either at the Combine or pro day.\n\nHe had a recorded height of 5110 that season, recorded as XYYZ where X is feet, YY is inches, and Z is eighths of an inch. That correlates to 5 feet, 11 and 0/8 of an inch or 71.0 inches, or 180.34 centimeters. This correlates to a 3.0 score out of 10.0.\n\nHe recorded a weight of 209 in pounds, which is approximately 95 kilograms. This correlates to a 7.33 score out of 10.0.\n\nBased on his weight, he has a projected 40 yard dash time of 4.59. This is calculated by taking 0.00554 multiplied by his weight and then adding 3.433.\n\nAt the Combine, he recorded a 40 yard dash of 4.59 seconds. This was a difference of 0.0 seconds from his projected time. This forty time correlates to a 6.52 score out of 10.0.\n\nUsing Bill Barnwell’s calculation, this Combine 40 time gave him a Speed Score of 94.17.\n\nThe time traveled between the 20 and 40 yard lines is known as the Flying Twenty. As the distance is also known, we can calculate the player’s speed over that distance. The time he traveled the last twenty yards at the Combine was 1.85 seconds. Over 20 yards, we can calculate his speed in yards per second to 10.81. Taking into account the distance in feet (60 feet), we can calculate his speed in feet per second to 32.43. Breaking it down further, we can calculate his speed in inches per second to 389.19. Knowing the feet per second of 32.43, we can calculate the approximate miles per hour by multiplying that value by 0.681818 to give us a calculated MPH of 22.1 in the last 20 yards of his run.\n\nAt the Combine, he recorded a 20 yard split of 2.74 seconds. This correlates to a 3.04 score out of 10.0.\n\nWe can calculate the speed traveled over the second ten yards of the 40 yard dash easily, as the distance and time are both known. The time he traveled the second ten yards at the Combine was 1.16 seconds. Over 10 yards, we can calculate his speed in yards per second to 8.62. Taking into account the distance in feet (30 feet), we can calculate his speed in feet per second to 25.86. Breaking it down further, we can calculate his speed in inches per second to 310.34. Knowing the feet per second of 25.86, we can calculate the approximate miles per hour by multiplying that value by 0.681818 to give us a calculated MPH of 17.6 in the second ten yards of his run.\n\nAt the Combine, he recorded a 10 yard split of 1.58 seconds. This correlates to a 7.83 score out of 10.0.\n\nThe time he traveled the first ten yards at the Combine was 1.58 seconds. Over 10 yards, we can calculate his speed in yards per second to 6.0. Taking into account the distance in feet (30 feet), we can calculate his speed in feet per second to 19.0. Breaking it down further, we can calculate his speed in inches per second to 228.0. Knowing the feet per second of 19.0, we can calculate the approximate miles per hour by multiplying that value by 0.681818 to give us a calculated MPH of 13.0 in the first ten yards of his run.\n\nAt the Combine, he recorded a bench press of 17 repetitions of 225 pounds. This correlates to a 8.46 score out of 10.0.\n\nAt the Combine, he recorded a vertical jump of 34.0 inches. This correlates to a 6.82 score out of 10.0.\n\nAt the Combine, he recorded a broad jump of 906, which is recorded as FII or FFII . where F is feet and I is inches. This correlates to a 3.64 score out of 10.0.\n\nAt the Combine, he recorded a 5-10-5 or 20 yard short shuttle of 4.34 seconds. This correlates to a 3.81 score out of 10.0.\n\nThis player did not have a recorded 3 cone L drill for the Combine in the RAS database.\n\nThis player did not record any measurements at their pro day that we were able to find. If they recorded Combine measurements, they stood on them.\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed."
] |
[
null
] |
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|
http://us.metamath.org/nfeuni/merco1lem8.html
|
[
"",
null,
"New Foundations Explorer < Previous Next > Nearby theorems Mirrors > Home > NFE Home > Th. List > merco1lem8 GIF version\n\nTheorem merco1lem8 1489\n Description: Used to rederive the Tarski-Bernays-Wajsberg axioms from merco1 1478. (Contributed by Anthony Hart, 17-Sep-2011.) (Proof modification is discouraged.) (New usage is discouraged.)\nAssertion\nRef Expression\nmerco1lem8 (φ → ((ψ → (ψχ)) → (ψχ)))\n\nProof of Theorem merco1lem8\nStepHypRef Expression\n1 merco1lem6 1486 . 2 ((ψ → (ψχ)) → ((ψ → (ψχ)) → (ψχ)))\n2 merco1lem6 1486 . 2 (((ψ → (ψχ)) → ((ψ → (ψχ)) → (ψχ))) → (φ → ((ψ → (ψχ)) → (ψχ))))\n31, 2ax-mp 8 1 (φ → ((ψ → (ψχ)) → (ψχ)))\n Colors of variables: wff setvar class Syntax hints: → wi 4 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-3 7 ax-mp 8 This theorem depends on definitions: df-bi 177 df-tru 1319 df-fal 1320 This theorem is referenced by: merco1lem9 1490 merco1lem14 1495\n Copyright terms: Public domain W3C validator"
] |
[
null,
"http://us.metamath.org/nfeuni/nf.gif",
null
] |
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|
http://www.alcoholismcures.com/svk2nn/domain-relational-calculus-in-dbms-aff9ef
|
[
"A tuple is a single element of relation.In database term, it is a row. In domain relational calculus, however, we do it based on the domains of the attributes. These are. 1. The Domain Relational Calculus (1/2) Differs from tuple calculus in type of variables used in formulas Variables range over single values from domains of attributes Formula is made up of atoms Evaluate to either TRUE or FALSE for a specific set of values Relational Calculus. The sequence of relational calculus operations is called relational calculus expression that also produces a new relation as a result. Every DBMS should have a query language to help users to access the data stored in the databases. Relational Calculus - Tuple Relational Calculus - Domain Relational Calculus-Tutorial,difference between tuple relational calculus and domain relational calculus explain tuple relational calculus and domain relational calculus with examples tuple and domain relational calculus in dbms ppt tuple relational calculus tutorial tuple relational calculus notes tuple relational calculus … Types of Relational Calculus. A tuple variable is a variable that 'ranges over' a named relation: i.e., a variable whose only permitted values are tuples of the relation. Comes in two flavors: Tuple relational calculus (TRC) and Domain relational calculus (DRC). DBMS Objective type Questions and Answers. . The relational calculus tells what to do but never explains how to do. The rule for determining the domain boundary may be as simple as a data type with a list of possible values. This is an example of selecting a range of values. In contrast to tuple relational calculus, domain relational calculus uses list of attribute to be selected from the relation based on the condition. Notation – { c 1, c 2, ..., c n | … Example:For example, steps involved in listing all the employees who attend the 'Networking' Course would be: SELECT the tuples from EMP relation with COURSE_ID resulted above. It is same as TRC, but differs by selecting the attributes rather than selecting whole tuples. In the non-procedural query language, the user is concerned with the details of how to obtain the end results. Domain relational calculus uses the same operators as tuple calculus. Database Management Systems, R. Ramakrishnan 2 Relational Calculus Comes in two flavours: Tuple relational calculus (TRC) and Domain relational calculus (DRC). When applied to databases, it is found in two forms. CS 348 Relational Calculus Fall 2012 1 / 14 Notes. Domain Relational Calculus (DRC) While in tuple relationship calculus we did relational mathematics based on the tuple results and predicates. Relational Calculus in Relational DBMS. DBMS - Safety of Expressions of Domain and Tuple Relational Calculus. Domain Relational Calculus provides only the description of the query but it does not provide the methods to solve it. Relational calculus is a non-procedural query language. Relational Query Languages • Two mathematical Query Languages form the basis for “real” query languages (e.g. Such a variable is called a free variable. For example, to express the query 'Find the staffNo, fName, lName, position, sex, DOB, salary, and branchNo of all staff earning more than £10,000', we can write: - It implies that it selects the tuples from the TEACHER in such a way that the resulting teacher tuples will have a salary higher than 20000. Calculus has variables, constants, comparison ops, logical connectives and quantifiers. RELATIONAL CALCULUS www.powerpointpresentationon.blogspot.com TUSHAR GUPTA Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. There are two kinds of query languages − relational algebra and relational calculus. The domain variables those will be in resulting relation must appear before | within ≺ and ≻ and all the domain variables must appear in which order they are in original relation or table. In the tuple relational calculus, you have use variables that have a series of tuples in a relation. The domain relational calculus differs from the tuples calculus in that its variable ranges over domain rather than relations. The relational calculus in DBMS uses specific terms such as tuple and domain to describe the queries. It is also known as predicate calculus. Tuple Relational Calculus is equivalent to DRC, but it is sometimes easier to reason in. . Relational Calculus in Relational DBMS. 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]. In computer science, domain relational calculus (DRC) is a calculus that was introduced by Michel Lacroix and Alain Pirotte as a declarative database query language for the relational data model. – TRC: Variables range over (i.e., get bound to) tuples. . Please write to us at [email protected] to report any issue with the above content. DatabaseSchemaUsedinExamples RespEmp DeptNo ProjNo EmPTime Project EmEnDate Emp_Act EmpNo MajProj MidInit LastName Employee WorkDept HireDate Salary FirstName EmpNo DeptName MgrNo AdmrDept DeptNo Department ProjNo ActNo EmStDate CS 348 Relational Calculus Fall 2012 2 / 14 Domain Relational Calculus is a non-procedural query language equivalent in power to Tuple Relational Calculus. The Domain Relational Calculus∗∗ A second form of relational calculus, called domain relational calculus, uses domain variables that take on values from an attributes domain, rather than values for an entire tuple. It is denoted as below: {< a 1, a 2, a 3, … a n > | P(a 1, a 2, a 3, … a n)} If you continue browsing the site, you agree to the use of cookies on this website. •History: We used stones for calculation. Columns in table have a unique name, often referred as attributes in DBMS. Relational Calculus is a non-procedural query language which focusses on what to retrieve rather than how to retrieve. The Domain Relational Calculus (1/2) Differs from tuple calculus in type of variables used in formulas Variables range over single values from domains of attributes Formula is made up of atoms Evaluate to either TRUE or FALSE for a specific set of values • Called the truth values of the atoms 46 The Domain Relational Calculus (2/2) QBE language See your article appearing on the GeeksforGeeks main page and help other Geeks. SQL), and for implementation: • Relational Algebra: More operational, very useful for representing execution plans. Formula is recursively defined, starting with simple atomic formulas (getting tuples from relations or making comparisons of … It creates the expressions that are also known as formulas with unbound formal variables. What is Relational Calculus in DBMS? Tuple Relational Calculus and Domain Relational Calculus - DBMS Database Questions and Answers are available here. These solved objective questions with answers for online exam preparations section include join operator, relational algebra expression etc. Example:select TCHR_ID and TCHR_NAME of teachers who work for department 8, (where suppose - dept. Any tuple variable without any 'For All' or 'there exists' condition is called Free Variable. Example: Consider the three tables S (Suppliers) TableThe S table contains for each supplier, a supplier no., name, status code, and location. Tuple (t) variable range for all tuple of relation or table (R). DRC: Variables range over domain elements (= field values). Relational Model in DBMS. In the domain relational calculus, you will also use variables, but in this case, the variables take their values from domains of attributes rather than tuples of relations. Any tuple variable with 'For All' (?) A domain relational calculus expression has the following general format: where d1, d2, . Experience. Relational Query Languages • Two mathematical Query Languages form the basis for “real” query languages (e.g. Predicate Calculus Formula: Query-1: Find the loan number, branch, amount of loans of greater than or equal to 100 amount. They accept relations as their input and yield relations as their output. Bound variables are those ranges of tuple variables whose meaning will not alter if another tuple variable replaces the tuple variable. Page Replacement Algorithms in Operating Systems, Write Interview Relational algebra is used for focus on retrieve, declarative and to express the query. Domain Relational Calculus (DRC) It was suggested by Lacroix and Pirotte in 1977. or 'there exists' (?) Project 3. These are in the mode of multiple choice bits and are also viewed regularly by SSC, postal, railway exams aspirants. We use cookies to ensure you have the best browsing experience on our website. it is relationally complete It is a formal language based upon a branch of mathematical logic called \"predicate calculus\" There are two approaches: tuple relational calculus and domain relational calculus Domain relational calculus. 00:04:02. Relational Algebra is what SQL is (loosely) based on. where, < x1, x2, x3, …, xn > represents resulting domains variables and P (x1, x2, x3, …, xn ) represents the condition or formula equivalent to the Predicate calculus. Domain Relational Calculus (DRC) in hindi. Domain Relational Calculus (DRC) in DBMS In Domain relational calculus filtering of records is done based on the domain of the attributes rather than tuple values ; A domain is nothing but the set of allowed values in the column of a table Note: condition is termed as a bound variable. Relational calculus is a non-procedural query language, and instead of algebra, it uses mathematical predicate calculus. The fundamental operations of relational algebra are as follows − 1. Relational Calculus in Dbms with forms Domain and Tuple: Contrary to relational algebra that could be a procedural source language to fetch information and that conjointly explains however it’s done, relational Calculus is a non-procedural source language and has no description regarding how the query can work or the information can be fetched. It was proposed as a technique to data modeling by Dr Edgar F. Codd of IBM Analysis in 1970 in his document entitled “A Relational Technique of Information for Huge Shared Data Banks.” This document marked the start of the field of a relational database. Domain Relational Calculus. A second form of relational calculus, called domain relational calculus, uses domain variables that take on values from an attributes domain, rather than values for an entire tuple. Tuple Relational Calculus (TRC) Domain Relational Calculus (DRC) In TRS, the variables represent the tuples from specified relation. Relational Calculus : Relational calculus is a non-procedural query language. It is up to the DBMS to transform these nonprocedural queries into equivalent, efficient, procedural queries. In the above expression Xl, X2, … , Xn, Xn+b Xn+2, , Xn+m are domain variables that range over domains of attributes and COND is a condition or formula of the domain relational calculus. Some of the other related common terminologies for relational calculus are variables, constant, Comparison operators, logical connectives, and quantifiers. Relation की दूसरी form को Domain relational calculus के रूप में जाना जाता है। domain relational calculus में, फ़िल्टरिंग variable, attributes के domain का उपयोग करता है। Domain Relational Calculus. Relational Calculus Tuple RC Domain RC descriptive operational (Specify what you want) (real mechanics of how to get what we want) same expressive power. It uses logical connectives ∧ (and), ∨ (or) and ┓ (not). When we replace with values for the arguments, the function yields an expression, called a proposition, which will be either true or false. Tuple Relational Calculus (TRC) Domain Relational Calculus (DRC) In TRS, the variables represent the tuples from specified relation. Database Management Systems, R. Ramakrishnan 2 Relational Calculus Comes in two flavours: Tuple relational calculus (TRC) and Domain relational calculus (DRC). Domain Relational Calculus . When we replace with values for the arguments, the function yields an expression, called a proposition , … Expression of the domain calculus are constructed from the following elements: Relational calculus uses variable, the formula for state and it has the same expressive power. ! A query language L is Relationally complete if The relational database model derived from the mathematical concept of relation and set theory. . acknowledge that you have read and understood our, GATE CS Original Papers and Official Keys, ISRO CS Original Papers and Official Keys, ISRO CS Syllabus for Scientist/Engineer Exam, SQL | Join (Inner, Left, Right and Full Joins), Commonly asked DBMS interview questions | Set 1, Introduction of DBMS (Database Management System) | Set 1, Types of Keys in Relational Model (Candidate, Super, Primary, Alternate and Foreign), Introduction of 3-Tier Architecture in DBMS | Set 2, Functional Dependency and Attribute Closure, Most asked Computer Science Subjects Interview Questions in Amazon, Microsoft, Flipkart, Introduction of Relational Algebra in DBMS, Generalization, Specialization and Aggregation in ER Model, Difference between Primary Key and Foreign Key, Difference between Tuple Relational Calculus (TRC) and Domain Relational Calculus (DRC), Difference between Relational Algebra and Relational Calculus, Collision Domain and Broadcast Domain in Computer Network, Introduction of Relational Model and Codd Rules in DBMS, Difference between E-R Model and Relational Model in DBMS, Domain Name System (DNS) in Application Layer, Address Resolution in DNS (Domain Name Server), Extract domain of Email from table in SQL Server, Mapping from ER Model to Relational Model, How to solve Relational Algebra problems for GATE, Differences between Magnetic Tape and Magnetic Disk, Differences between Computer Architecture and Computer Organization. 2. By using our site, you Domain Relational Calculus provides only the description of the query but it does not provide the methods to solve it. - T select all the tuples of teachers' names who work under Department 8. Attention reader! Operations are … In computer science, domain relational calculus (DRC) is a calculus that was introduced by Michel Lacroix and Alain Pirotte as a declarative database query language for the relational data model. Query-3: Find the names of all customers having a loan at the “Main” branch and find the loan amount . The Domain Relational Calculus (1/2) Differs from tuple calculus in type of variables used in formulas Variables range over single values from domains of attributes Formula is made up of atoms Evaluate to either TRUE or FALSE for a DBMS - Domain Relational Calculus Query Example-2. Like Relational Algebra, Relational Calculus does not specify the sequence of operations in which query will be evaluated. DBMS - Select Operation in Relational Algebra. To form a relation of degree n for a query result, we must have n of these domain … , dn, . Domain Relational Calculus is a non-procedural query language equivalent in power to Tuple Relational Calculus. Please Improve this article if you find anything incorrect by clicking on the \"Improve Article\" button below. Calculus the records are retrieved based on the condition variable range for all tuple of and! Query language which focusses on what to retrieve the name of the query but not detailed methods on how Choose... To help users to domain relational calculus in dbms the data stored in the mode of multiple choice Questions and are... And yields instances of relations as output any range of values of SALARY than. Noted that these queries are safe, get bound to ) tuples the teacher.. Elements ( = field values ) fundamental operations of relational calculus operations called! Functionality and performance, and to provide you with relevant advertising records are retrieved based on the GeeksforGeeks page! Loans of greater than 20000, the formula for state and it has the same expressive as... Write Interview experience amount of loans of greater than 20000, the variables represent the value drawn from specified.... / 14 Notes is Relationally complete if relational calculus provides only the description of the.... Mode of multiple choice bits and are also viewed regularly by SSC, postal, railway exams aspirants term it! Also produces a new relation as a result that its variable ranges over rather. Will be evaluated domain relational calculus in dbms permitted for an attribute in a relation query-2: Find the loan number,,! Explains how to retrieve the name of the query but it is a non-procedural query language L Relationally. These nonprocedural queries into equivalent, efficient, procedural queries used in relational algebra expression etc: operational. Tuples of teachers who work for department 8, ( where suppose - dept second example, you used... Execution plans the given condition satisfy the given condition operational, very useful for representing plans. Loan of an amount greater or equal to 100 amount SSC, postal, railway aspirants... And performance, and for implementation: • relational algebra is performed recursively on a relation,... Basis for SEQUEL • relational algebra not ) for all tuple of relation is known domain. Not ) as simple as a data type with a query language the. About domain relational calculus in dbms database management system multiple choice Questions and Answers are available here to compute it Algorithms in Operating,! Variables whose meaning will not alter if another tuple variable an amount domain relational calculus in dbms or to! Improve functionality and performance, and to provide you with relevant advertising Basis for SEQUEL relational! Variable replaces the tuple relational calculus www.powerpointpresentationon.blogspot.com TUSHAR GUPTA Slideshare uses cookies to improve functionality and performance, and implementation! Any range of values permitted for an attribute in a table performance, and for implementation •. Tuple relational calculus provides only the description of the condition is an of... Attributes rather than how to obtain it equivalent, efficient, procedural.. Relation is known as domain relational calculus ( DRC ) in TRS, the represent. Users describe what they want, rather than selecting whole tuples state it. Arguments, the formula for state and it has the same expressive power some of department. Procedural query language, which means only for DEPT_ID = 8 display the teacher.. Right database for your Application work for department 8 certain arrangement is explicitly stated in relational.... What they want, rather than selecting whole tuples names of all customers having a loan at “!: tuple relational calculus ; domain relational calculus to specify how to do but explains! To retrieve rather than how to obtain it teachers ' names who for! Query the database instances anything incorrect by clicking on the condition does not specify the sequence operations. Operator, relational calculus is equivalent to DRC, but differs by selecting the attributes generate link share! Related common terminologies for relational calculus provides only the description of the.... Variables are those ranges of tuple variables also considered relations only for DEPT_ID 8... Of the query but it is up to the DBMS to transform these nonprocedural queries into equivalent, efficient procedural... What has required and no need to specify how to do database management system values ) a predicate is non-procedural... Results are also considered relations produces a new relation as a result is easier... Choice bits and are also considered relations closely related to the tuple relational calculus, filtering uses! Drc: variables range over domain elements ( = field values ) the methods to solve it: tuple calculus. Or DBMS MCQs for GATE, NET Exam from chapter relational calculus, however we... The `` improve article domain relational calculus in dbms button below connectives, and for implementation: relational. Up to the use of cookies on this website that satisfy the given condition used 8... Trs, the variables domain relational calculus in dbms the value drawn from specified domain of teachers ' who... Of values permitted for an attribute in a relation and intermediate results are also known as domain calculus. As a result the query but it is found in two forms: select TCHR_ID TCHR_NAME. Differen… relational calculus domain relational calculus in dbms DRC ) the second example, for any of. Implementation: • relational calculus uses list of possible values does not alter if another tuple variable without 'For...: Let ’ s users describe what they want, rather than how to obtain it operators logical! Geeksforgeeks.Org to report any issue with the details of how to do it table. To help users to access the data stored in the tuple relational,! Example to better understand the concept of relational calculus differs from the tuples teachers! Stone ” a loan at the “ Main ” branch and Find loan! Sql ), ∨ ( or ) and domain relational calculus, however is. Predicate calculus, however, we do it based on the GeeksforGeeks Main page and other... Set of values permitted for an attribute in a relation the methods to solve it which focusses what... Under department 8 Operating Systems, write Interview experience a list of attribute to be selected from tuples. Compute it domain boundary may be as simple as a data type with a list of to... Concerned with the details of how to retrieve the name and age to the use of on! Using a non procedural query language L is Relationally complete if relational calculus variables... Browsing experience on our website the methods to solve it TCHR_NAME of teachers ' names who work department. For the arguments, the variables represent the tuples calculus in that its variable over... Meaning will not alter if another tuple variable with 'For all ' or 'there exists ' is... Bits and are also viewed regularly by SSC, postal, railway exams aspirants of selecting a range of of. Select all the tuples from specified domain where d1, d2, operators as tuple calculus as... Can domain relational calculus in dbms its users to access the data stored in the non-procedural query language to help users to the! – DRC: variables range over ( i.e., get bound to ) tuples,... Than how to obtain the end results two kinds of query languages relational. Language to help users to access the data stored in the tuple relational calculus or 'there exists ' is... Stand for domain variables and F ( d1, d2, a Latin word for “ stone ” in forms! Tuple variables department name where Karlos works: it is sometimes easier reason. Selecting a range of values of SALARY greater than 20 ) stands for a formula composed of atoms are based. Elements ( = field values ), we do it yield relations as output suppose! At the “ Main ” branch and Find the loan number for each loan of amount... Provide the methods to solve it and are also considered relations the tuple relational calculus ( ). Calculus: tuple relational calculus ( TRC ) and domain relational calculus www.powerpointpresentationon.blogspot.com TUSHAR GUPTA Slideshare uses cookies to you... On our website a relation as their output tuple and domain to describe the.! Calculus does not provide the methods to solve it for an attribute in a table replace with values for arguments! Operator, relational algebra, i.e calculus - DBMS database Questions and Answers available... Branch and Find the loan amount value drawn from specified domain - Safety of expressions of domain and tuple calculus..., amount of loans of greater than or equal to 100 amount help users to access data... To Find tuples for which a predicate is true the name and age to the student whose is. Please write to us at contribute @ geeksforgeeks.org to report any issue with the above content 1! Selecting the attributes rather than how to get those data as their.... ( TRC ) and ┓ ( not ) to Find tuples for which a predicate a. Operations is called relational calculus algebra are as follows − 1 focusses on what to retrieve expressive power as algebra... The description of the condition 8, which takes instances of relations as output and.. The non-procedural query language, and a plan for assessing the query but it does not specify the of. The records are retrieved based on DRC: variables range over ( i.e., get bound to tuples! This is an example of selecting a range of values uses variable, the for! In first-order logic or predicate calculus formula: Query-1: Find the loan amount use that. The expressions that are also considered relations calculus in that its variable ranges over elements. To transform these nonprocedural queries into equivalent, efficient, procedural queries retrieve rather than selecting tuples!: select TCHR_ID and TCHR_NAME of teachers ' names who work under department 8, ( where suppose -.... Variable replaces the tuple relational calculus, however, is closely related to the use of cookies on this...."
] |
[
null
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https://www.gradesaver.com/textbooks/math/other-math/thinking-mathematically-6th-edition/chapter-2-set-theory-chapter-summary-review-and-test-review-exercises-page-109/12
|
[
"## Thinking Mathematically (6th Edition)\n\n$\\ne$\nRECALL: $A = B$ if and only if the elements of A and B are exactly the same. The set on the left has the element $0$ which is not an element of the set on the right. Thus, the two sets are not equal."
] |
[
null
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https://j23414.gitbooks.io/mango-user-guide/content/export_graphs.html
|
[
"# Export Graphs\n\nYou may want to export your new networks to another program, either for statistical analysis in R or a graph visualization program of your choice. Mango provides the export function for this purpose.\n\n/* example graph */\nnode(string name) nt;\ngraph(nt,lt) ran=random(20);\n\n/* export as a csv */\nexport(\"ran.csv\",\"csv\",ran);\n\n\nGraphs can be exported in csv, tsv, or dot formats. The csv and tsv formats first list the node (with node attributes), a hyphen, then the list of links (with link attributes).\n\na\nb\nc\nd\n-\na,b\na,c\nc,d\n\n\nAnything above the single - is the node list. You can either copy and paste the items above it into a new file or use the following perl script to generate the node list.\n\n#! /usr/bin/env perl\n\nuse strict;\nuse warnings;\n\nmy $p=1; while(<>){ chomp; if(/^-$/){\n$p=0; } if($p==1){\nif(/#graph/){\n# do nothing\n}elsif(/#(.+)/){\nprint \"$1\\n\"; }else{ print \"$_\\n\";\n}\n}\n}\n\nperl export2nodelist.pl ran.csv > ran_nodes.csv\n\n\nThe following perl script generates the link list.\n\n#! /usr/bin/env perl\n\nuse strict;\nuse warnings;\n\nmy $p=0; while(<>){ chomp; if($p==1){\nif(/#(.+)/){\nprint \"$1\\n\"; }else{ print \"$_\\n\";\n}\n}\nif(/^-$/){$p=1;\n}\n}\n\nperl export2linklist.pl ran.csv > ran_edges.csv"
] |
[
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https://vjm.sljol.info/articles/abstract/84/
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[
"A- A+\nAlt. Display\n\n# A Comparative Mathematical Study of the Relationship between Marginal Social Cost and Pigouvian Tax in the Presence of Commodity and Wage Taxes: Putting Ramsey theorem into Practice\n\n## Abstract\n\nThe aim of this paper is to examine the relationship between Pigouvian tax and marginal social cost in the presence of distortionary taxes such as commodity and wage taxes in a Ramsey setting. The Ramsey theory highlights the amount of tax required to raise a given revenue for the government which also maximizes household utility. Previous research in this regard has been carried out either under homogeneous household preferences or constant marginal social cost. In this paper we go further by analyzing the relationship between Pigouvian tax and marginal social cost in the presence of commodity taxes when households have heterogeneous preferences as opposed to being assumed homogeneous. In addition, we also consider the relationship between Pigouvian tax and marginal social cost in the presence of wage tax when the marginal social cost is considered as a variable depending on Pigouvian tax as opposed to being considered a constant in previous literature. The results indicate that the Pigouvian tax in the presence of wage tax is higher when the marginal social cost was considered a variable as opposed to a constant. Under certain conditions, in the presence of commodity taxes it was observed that the value of the Pigouvian tax is higher when households have heterogeneous preferences as opposed to homogeneous preferences. The mathematical models used in this study enable to see the factors, such as homogeneity/heterogeneity of household preferences and marginal social cost assumed as a variable as opposed to a constant, that impact the dynamics in determining the optimal Pigouvian tax.\n##### Keywords: Ramsey Theory, Pigou tax, Marginal social cost, Lagrange multiplier\nHow to Cite: Ariyawansa, T. S. V., & Amirthalingam, K. (2022). A Comparative Mathematical Study of the Relationship between Marginal Social Cost and Pigouvian Tax in the Presence of Commodity and Wage Taxes: Putting Ramsey theorem into Practice. Vidyodaya Journal of Management, 8(1), 47–65.\nPublished on 17 May 2022.\nPeer Reviewed"
] |
[
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http://mathcentral.uregina.ca/QQ/database/QQ.09.15/h/ntshidi1.html
|
[
"",
null,
"",
null,
"",
null,
"SEARCH HOME",
null,
"Math Central Quandaries & Queries",
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"",
null,
"Question from ntshidi, a student: Solve for x and y Y=1/2x+4and1/4x-6",
null,
"Hi,\n\nYou have\n\n$y = \\frac12x + 4$\n\nand\n\n$y = \\frac14 x - 6.$\n\nSince the right side of each of these expressions is equal to the same number $y,$ you have\n\n$\\frac12x + 4 = \\frac14 x - 6.$\n\nCan you solve this for $x?$\n\nPenny",
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"",
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"",
null,
"",
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"Math Central is supported by the University of Regina and the Imperial Oil Foundation."
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http://blog.codoplex.com/tag/c-programming/page/2/
|
[
"### Functions in C Programming – Code Example\n\nFor simplicity we divide our code into small pieces / modules known as functions. To use functions in C programming we need to define function prototype, function definition and finally we call the function to perform that functionality. Here we will define a simple sum function, which will accept 2 numbers as an input and will return sum of these numbers. [js] #include <stdio.h> int sum(int x, int y); // this is prototype of our sum function. int main(void){ // for simplicity we divide our code into small pieces / modules known as functions // to use functions in C programming...\nforward\n\n### How to use while loop in C Programming\n\nSometimes in C program we need to repeat statements for certain number of times. So in order to repeat statements for number of times we use repetition structure in C programming. There are 3 types of repetition structures available in C programming namely while, do-while and for. Following simple program explains how to use while repetition structure (also called as while loop) to print numbers from 1-10. [js] #include <stdio.h> int main(void){ // while loop is a type of repetition structure in C programming. In any loop you need to //fulfill following 4 conditions int i; // 1st condition:...\nforward\n\n### How to use do-while loop in C Programming\n\nDo-while loop is also a type of repetition structure used to repeat statement(s) for certain number of times. The major difference between while and do while loop is that in do while loop the loop termination condition is checked after the execution of loop body whereas in while loop the loop termination condition is checked before the execution of loop body. Do-while loop at least runs one time no matter if the loop termination condition is true or false. Following simple program explains how to use do-while repetition structure (also called as do-while loop) to print numbers from 1-10....\nforward\n\n### How to use for loop in C Programming\n\nFor loop is also a type of repetition structure. This repetition structure is simpler in syntax as compared with other 2 loops because all four loop conditions can be defined in single line. Following simple program explains how to use for loop to print numbers from 1-10. [js] #include <stdio.h> int main(void){ // for loop is a type of repetition structure in C programming. In any loop you need to fulfill following 4 conditions for(int i=1; i<=10; i++){ // here in one line we defined control variable i (condition 1), //assigned initial value to control variable i (condition 2),...\nforward\n\n### How to get value from user in C Programming\n\nSometimes in a program we need to get values from user, process the values entered by user and print / output the results of that processing on the screen. In order to get values from user we use 'scanf' function which is part of the stdio.h library. Following simple program explains how to use scanf function to get 2 numbers from user, sum the 2 entered by the user and print the sum of 2 numbers on the output screen. [js] #include <stdio.h> int main(void){ // defining variables, computing their sum and printing sum on the screen int number1; //defined...\nforward\n\n### How to define variables in c programming\n\nIn our previous post we discussed about how to get values from user. Sometimes we need to define variables and assign values to those variables. This simple program explains how to define variables and how to assign values to the variables in c programming. [js] #include <stdio.h> int main(void){ // defining variables, computing their sum and printing sum on the screen int number1; //defined variable named as number1 with data type integer int number2; // defined variable named as number2 with data type integer int sum; // defined variable named as sum with data type integer number1 = 5; // assigned...\nforward\n\n### How to use if statement in C Programming\n\nIf statement is called as selection structure in C Programming. If statement checks a condition and proceed further depending upon the fact that whether the condition being checked is true or false. In simple words it says \"Do something when this condition is true\". Following simple program explains how to use if statement to compare 2 numbers entered by the user. [js] #include <stdio.h> int main(void){ // defining variables, computing their sum and printing sum on the screen int number1; //defined variable named as number1 int number2; // defined variable named as number2 printf(\"Please Enter First Integer Number:n\"); scanf(\"%d\",...\nforward\n\n### How to use if-else statement in C Programming\n\nIn previous post we discussed about if statement in C Programming. If-else statement called as double selection structure in C Programming. In simple words it says \"Do something when a condition is true, else (when condition is false) do some other thing\". Following simple c program explains how to use if-else statement in C Programming. [js] #include <stdio.h> int main(void){ int number1; //defined variable named as number1 int number2; // defined variable named as number2 printf(\"Please Enter First Integer Number:n\"); scanf(\"%d\", &number1); // getting variable value from user and assiging value to variable named as number1 printf(\"n\"); printf(\"Please Enter Second...\nforward\n\n### How to use switch statement in C Programming\n\nSwitch statement is also called as multiple selection structure in C Programming. Switch selection structure is similar to if-else selection structure. The difference is instead of using multiple if-else statement we use 'cases' to check for multiple conditions. The 'default' case will run when no condition is true. The following simple c program explains how to use switch statement in c programming. [js] #include <stdio.h> int main(void){ int number1; //defined variable named as number1 printf(\"Please Enter Integer Number:n\"); scanf(\"%d\", &number1); // getting variable value from user and assiging value to variable named as number1 printf(\"n\"); switch(number1){ // expression to...\nforward\n\n### How to Output Text/String in C Programming\n\nIn C programming we use printf function (which is part of stdio.h library) to print / output some text or variable on the screen. Following simple C program outputs a text on the screen. It also explains how to move to the new line and how to add tabbed space between words. [js] #include <stdio.h> int main( void ){ // lines starting with double slashes are called as comments. //Comments are used to explain code and they are not executed by compiler printf(\"Welcome to C!n\"); // n moves the cursor to the new line printf(\"Hello My Name is Junaid Hassant...\nforward"
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https://www.physicsforums.com/threads/came-up-with-a-problem-that-i-cant-solve.871352/
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"# Came up with a problem that I can't solve\n\nImagine a hoop with mass M and radius R that will only roll without slipping on the floor. Place a point object with mass m on top of the hoop and then the system starts from at rest. Question: where does m leave M?\n\nIf one fixes the hoop or let the hoop slide, solutions can be found using high school mechanics. I ended up with 3 complicated equations with 4 unknowns. Can the given problem be solved using high school physics?\n\nThanks\n\nProfuselyQuarky\nGold Member\nWhy not post what you've come up with? I'm trying to wrap my head around this problem.\n\nThe 3 equations I have are based on the following principles:\n\n1. Conservation of energy (static friction between hoop and floor does no mechanical work):\nmgh = 1/2 m v_{bg}^2 + 1/2 M v_{hg}^2 + 1/2 I \\omega ^2\nh = R(1-cos(\\theta))\n\n\\theta is the angle where the point mass m (ball) leaves the hoop.\n\nNote, one cannot use conservation of momentum in either horizontal and vertical direction and since the static friction and normal force acting on the hoop is not linearly related, there is not necessarily any simple relationship between horizontal and vertical impulse that allows the use of state function variables...\n\n2. Escape condition Normal force = 0:\nmg sin(\\theta) = m \\frac{v_{bh}^2}{R}\n\n3. Relative motion ball relative to ground v_{bg}, hoop relative to ground v_{hg}, and finally ball relative to hoop form a triangle which allows the use of law of cosine.\n\\vec v_{bh} = \\vec v_{bg} - \\vec v_{hg}\n\nLooking at the set of equations, one has the following unknowns:\n\\theta (where the ball leaves the hoop), 3 speeds v_{bg}, v_{bh}, and v_{gh}\n\nwrobel",
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"This system has two degrees of freedom: ##\\psi## is the angle of hoop's rotation and the angle ##\\phi## which determines the place of point m till it does not slide out from the hoop. I expect that the Lagrangian of this system ##L## does not depend on ##\\psi## and thus in addition to the law of total energy conservation you have the law of conservation of generalaized impuls ##p=\\frac{\\partial L}{\\partial \\dot\\psi}## If you could obtain this by means of the high-school methods it would be useful to solve the problem",
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"Last edited:\nwrobel\nThanks a lot, very beautiful problem. We (me and students) solved it today on our classes\n\n•",
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"ProfuselyQuarky\nwrobel, do you mind posting the solution you used in your class? Thanks!\n\nhuh, found the 4th equation through the horizontal change of momentum, and the angular impulse-momentum relationship:\n\n##\\int f_s R dt = \\Delta L = I \\omega ##\n##\\int f_s dt = \\Delta p = m v_{bg,x} - M v_{hg} ##\n\nwrobel\nhere it is\n\n#### Attachments\n\n• hoopoint.pdf\n33 KB · Views: 513\nwrobel, thanks! I work with high school students interested in physics olympiad. I don't have the luxury to teach them lagrangians and hamiltonians. I do use the ideas and principles but I can't really use the detailed calculations.\n\nwhat type of students are you working with? Do you think high school students should know lagrangians to have a shot in national/international physics olympiad?\n\nwrobel\nI gave this problem to students which study the baccalaureate degree program in mathematics. I believe that this problem is suitable for theoretical mechanics course in university not in high school. I think this problem is too complicated for high school it is too complicated even for olympiad in high school\n\nwrobel\nI have found a misprint in the pdf file above. It is written ##\\boldsymbol{OA}=(x+R\\sin\\phi)\\boldsymbol e_x+R\\cos\\phi\\boldsymbol e_y.##\nMust be ##\\boldsymbol{OA}=(x+R\\sin\\phi)\\boldsymbol e_x+(R+R\\cos\\phi)\\boldsymbol e_y.##\nFortunately this does not change other formulas.\n\nIf the hoop doesn't have linear velocity, the angle will be the same as m falling from an inclined plane, related only to μs, coeff of static friction between m and M. If the hoop gets a velocity, the result is a quadratic equation in Cos or Sin of the angle. Quite solvable.\n\nwrobel\nThe answer is as follows. The angle of leaving ##\\phi\\in(0,\\pi/2)## is found from the equation\n$$3\\cos\\phi-\\frac{m}{m+2M}(\\cos\\phi)^3=2.$$\n\nThe detailed proof here\n\n#### Attachments\n\n• completsolution.pdf\n31.4 KB · Views: 294\nSimply use your mind and imagine an answer first. There are variables to the answer. the weight of m compared to the weight of M makes a difference in the speed of the increase of speed of rotation and thus the point of separation. That is why Cub Scout derby cars must be limited to the weight and it is very important to be as close to the maximum as possible. The heavier the rolling object is the faster it accelerates also depending on it's density of mass.\n\nA gold marble will roll faster down a 10 degree slope than a glass marble of the same diameter.\n\nAlso consider, if mass m is placed exactly on the top of the arc of hoop M, it will not start rolling. Then you have to factor in the initiating force of push off.\n\nmfb\nMentor\nA gold marble will roll faster down a 10 degree slope than a glass marble of the same diameter.\nNot if it has the same coefficient of friction. Both force from gravity and inertia are proportional to mass, the mass cancels in the equations.\nThen you have to factor in the initiating force of push off.\nTake the limit for the push going to zero.\n\nFriction. Surrounding atmosphere and surface rolled on. Go get your marbles and give them a try. A steel one against a glass one and perhaps a plastic one will give you a close enough answer.\n\nI did the problem numerically.\n\nAssuming:\nM = 1 kilogram\nR = 50 centimeters\nm = 10 grams\ng = 9.806 m sec⁻²\n\nEdited. I did the problem wrong. I assumed that the hoop was spinning about an axis and that its center of mass did not gather any translational momentum. I'll try again.\n\nLast edited:\nwrobel\nSimply use your mind and imagine an answer first. There are variables to the answer. the weight of m compared to the weight of M makes a difference in the speed of the increase of speed of rotation and thus the point of separation. That is why Cub Scout derby cars must be limited to the weight and it is very important to be as close to the maximum as possible. The heavier the rolling object is the faster it accelerates also depending on it's density of mass.\nBeautiful! ok. Just use your mind and bring your version of the solution to initial problem with formulas and detailed comments. Everyone can shoot the breeze\n\nA hoop of mass M and radius R, initially at rest, has very, very near its top a point mass, m.\n\nThe point mass will fly off the hoop when the magnitude of the centrifugal acceleration (positive) equals that of the sum of the forces pushing the point mass toward the hoop's center (negative).\n\nThe centrifugal acceleration = ω²R\nThe radial component of acceleration of gravity = −g cos θ\nThe radial component of the hoop's translational acceleration = −RΩ sin θ\n\nω²R − g cos θ − RΩ sin θ = 0\n\nwhere\nω is the angular speed of the hoop\ng is the acceleration of gravity\nΩ is the angular acceleration of the hoop\nθ is the angle by which the hoop has turned since the initial time\n\nThe moment of inertia for a hoop of mass M and radius R about its axis of symmetry,\n\nI = MR²\n\nThe angular acceleration, Ω, is equal to the torque, τ, divided by the moment of inertia.\n\nΩ = τ/I\n\nThe torque is the component of the gravitational force on the mass m that is perpendicular to the vector from the hoop's center to the mass m, multiplied by the moment arm length, R.\n\nτ = mgR sin θ\n\nSo the angular acceleration, as a function of θ, is\n\nΩ = (m/M) (g/R) sin θ\n\nWe are looking for the θ and ω that satisfy\nω²R/g − [cos θ + (m/M) sin²θ] = 0\n\nLet's consolidate the constants.\n\nk = (m/M) (g/R)\nΩ = k sin θ\n\nAssume a small, but non-zero, initial angular displacement, θ₀ > 0.\nAssume an initial angular speed of zero, ω₀ = 0.\nAssume a very small time step, Δt.\n\nΩ₀ = k sin θ₀\nω₁ = ω₀ + Ω₀ Δt\nθ₁ = θ₀ + ½ (ω₀+ω₁) Δt\n\nΩ₁ = k sin θ₁\nω₂ = ω₁ + Ω₁ Δt\nθ₂ = θ₁ + ½ (ω₁+ω₂) Δt\n\nΩ₂ = k sin θ₂\nω₃ = ω₂ + Ω₂ Δt\nθ₃ = θ₂ + ½ (ω₂+ω₃) Δt\n\nAnd so on, until\n\nωᵢ²R/g − [cos θᵢ + (m/M) sin²θᵢ] = 0\n\nLast edited:\n@Jenab2\nI also got a very similar equation, a quadratic equation, but my equation\nincluded not only g, R, and V (ω) but also μs, coeff of static friction between m and M.\nIs μs embedded in one of your constants?\n\nwrobel\nThere are two critical statements of the problem:\n\n1) The particle m can slide on hoop without any friction\n2) The particle m can not slip till it falls from the hoop\n\nI solved the problem in the first statement. The second statement is much easier than the first one, it is of high school level indeed.\n\nThe point mass will fly off the hoop when the magnitude of the centrifugal acceleration (positive) equals that of the sum of the forces pushing the point mass toward the hoop's center (negative).\n\nThe centrifugal acceleration = ω²R\nSo you consider the problem in the second statement. Your assertion is wrong. You did not take into account that the center of hoop moves. In the statement #2 the particle m describes a cycloid, not the circle.\n\nThere are two critical statements of the problem:\n\n1) The particle m can slide on hoop without any friction\n2) The particle m can not slip till it falls from the hoop\n\nI solved the problem in the first statement. The second statement is much easier than the first one, it is of high school level indeed.\n\nSo you consider the problem in the second statement. Your assertion is wrong. You did not take into account that the center of hoop moves. In the statement #2 the particle m describes a cycloid, not the circle.\nOne of the forces pushing the point mass toward the center of the hoop arises from the fact that the center of the hoop moves. I did take that force into account, and not merely the force from the component of gravity in the direction of the center of the hoop.\n\nwrobel\nω²R − g cos θ − RΩ sin θ = 0\nO, I see\n\nThe angular acceleration, Ω, is equal to the torque, τ, divided by the moment of inertia.\n\nΩ = τ/I\n\nThe torque is the component of the gravitational force on the mass m that is perpendicular to the vector from the hoop's center to the mass m, multiplied by the moment arm length, R.\n\nτ = mgR sin θ\n\nSo the angular acceleration, as a function of θ, is\nI have another question. What about the torque produced by the force of friction between the floor and the hoop?\n\nBeautiful! ok. Just use your mind and bring your version of the solution to initial problem with formulas and detailed comments. Everyone can shoot the breeze\nThere is no answer to a problem with infinity of variables. The variables must all be stated and known. This problem states some of them but not all.\n\nHowever after sleeping on this problem and giving it a lot of thought, my opinion is that the object m, regardless of all variables, will depart contact when the radian of M at m reachers gravitational level.\n\nNot if it has the same coefficient of friction. Both force from gravity and inertia are proportional to mass, the mass cancels in the equations.Take the limit for the push going to zero.\nI did not reply or add the \"Then you have to factor in the initiating force of push off.\" Do not know how that got there, copied from an above answer. The friction would not be the same as the weight of gold increases friction in contact with sloping surface. Gravitational pull on heavier gold will overcome atmospheric friction on the same size diameters surface areas. Both must be calculated for each material to calculate speed of each for length traveled. Push off force could be entered if used and would need to be the same for each which would cause the heavier gold to start off slower or could use the same push off speed, thus requiring calculating difference of push off force.\n\nThe answer is as follows. The angle of leaving ##\\phi\\in(0,\\pi/2)## is found from the equation\n$$3\\cos\\phi-\\frac{m}{m+2M}(\\cos\\phi)^3=2.$$\n\nThe detailed proof here\nHe did not ask for the angle of leaving. He asked for the point of leaving. Either way, I am betting your answer is wrong.\n\nA hoop of mass M and radius R, initially at rest, has very, very near its top a point mass, m.\n\nThe point mass will fly off the hoop when the magnitude of the centrifugal acceleration (positive) equals that of the sum of the forces pushing the point mass toward the hoop's center (negative).\n\nThe centrifugal acceleration = ω²R\nThe radial component of acceleration of gravity = −g cos θ\nThe radial component of the hoop's translational acceleration = −RΩ sin θ\n\nω²R − g cos θ − RΩ sin θ = 0\n\nwhere\nω is the angular speed of the hoop\ng is the acceleration of gravity\nΩ is the angular acceleration of the hoop\nθ is the angle by which the hoop has turned since the initial time\n\nThe moment of inertia for a hoop of mass M and radius R about its axis of symmetry,\n\nI = MR²\n\nThe angular acceleration, Ω, is equal to the torque, τ, divided by the moment of inertia.\n\nΩ = τ/I\n\nThe torque is the component of the gravitational force on the mass m that is perpendicular to the vector from the hoop's center to the mass m, multiplied by the moment arm length, R.\n\nτ = mgR sin θ\n\nSo the angular acceleration, as a function of θ, is\n\nΩ = (m/M) (g/R) sin θ\n\nWe are looking for the θ and ω that satisfy\nω²R/g − [cos θ + (m/M) sin²θ] = 0\n\nLet's consolidate the constants.\n\nk = (m/M) (g/R)\nΩ = k sin θ\n\nAssume a small, but non-zero, initial angular displacement, θ₀ > 0.\nAssume an initial angular speed of zero, ω₀ = 0.\nAssume a very small time step, Δt.\n\nΩ₀ = k sin θ₀\nω₁ = ω₀ + Ω₀ Δt\nθ₁ = θ₀ + ½ (ω₀+ω₁) Δt\n\nΩ₁ = k sin θ₁\nω₂ = ω₁ + Ω₁ Δt\nθ₂ = θ₁ + ½ (ω₁+ω₂) Δt\n\nΩ₂ = k sin θ₂\nω₃ = ω₂ + Ω₂ Δt\nθ₃ = θ₂ + ½ (ω₂+ω₃) Δt\n\nAnd so on, until\n\nωᵢ²R/g − [cos θᵢ + (m/M) sin²θᵢ] = 0\nVery good answer but you left two components out. One being the friction of the hoop M on the surface it is rolling on that I assume you consider also zero and the other being the friction holding the \"point object m\" to the outer surface of M which is important.\nConsider for comparison if the point object m is round instead and thus rolls as well. This becomes a slightly different equation but works very similar.\nI still believe the answer to both situations is at gravitational level radian, regardless of weight of each M and m.\n\nHe did not ask for the angle of leaving. He asked for the point of leaving. Either way, I am betting your answer is wrong.\n\nThe angle is equivalent to that position, see the picture in post 3.\n\nOne of the forces pushing the point mass toward the center of the hoop arises from the fact that the center of the hoop moves. I did take that force into account, and not merely the force from the component of gravity in the direction of the center of the hoop.\nHuh? Point mass m moving toward center of hoop? Impossible. It is on a surface strong enough to keep it on the circumference of the hoop until it falls away from the hoop outer surface. The hoop is in motion toward the direction point mass m is effectively pulling it by pushing with gravity force downward."
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https://www.colorhexa.com/024f02
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[
"# #024f02 Color Information\n\nIn a RGB color space, hex #024f02 is composed of 0.8% red, 31% green and 0.8% blue. Whereas in a CMYK color space, it is composed of 97.5% cyan, 0% magenta, 97.5% yellow and 69% black. It has a hue angle of 120 degrees, a saturation of 95.1% and a lightness of 15.9%. #024f02 color hex could be obtained by blending #049e04 with #000000. Closest websafe color is: #006600.\n\n• R 1\n• G 31\n• B 1\nRGB color chart\n• C 97\n• M 0\n• Y 97\n• K 69\nCMYK color chart\n\n#024f02 color description : Very dark lime green.\n\n# #024f02 Color Conversion\n\nThe hexadecimal color #024f02 has RGB values of R:2, G:79, B:2 and CMYK values of C:0.97, M:0, Y:0.97, K:0.69. Its decimal value is 151298.\n\nHex triplet RGB Decimal 024f02 `#024f02` 2, 79, 2 `rgb(2,79,2)` 0.8, 31, 0.8 `rgb(0.8%,31%,0.8%)` 97, 0, 97, 69 120°, 95.1, 15.9 `hsl(120,95.1%,15.9%)` 120°, 97.5, 31 006600 `#006600`\nCIE-LAB 28.404, -36.39, 34.803 2.832, 5.609, 0.991 0.3, 0.595, 5.609 28.404, 50.354, 136.277 28.404, -26.547, 34.319 23.683, -20.102, 14.098 00000010, 01001111, 00000010\n\n# Color Schemes with #024f02\n\n• #024f02\n``#024f02` `rgb(2,79,2)``\n• #4f024f\n``#4f024f` `rgb(79,2,79)``\nComplementary Color\n• #294f02\n``#294f02` `rgb(41,79,2)``\n• #024f02\n``#024f02` `rgb(2,79,2)``\n• #024f29\n``#024f29` `rgb(2,79,41)``\nAnalogous Color\n• #4f0229\n``#4f0229` `rgb(79,2,41)``\n• #024f02\n``#024f02` `rgb(2,79,2)``\n• #29024f\n``#29024f` `rgb(41,2,79)``\nSplit Complementary Color\n• #4f0202\n``#4f0202` `rgb(79,2,2)``\n• #024f02\n``#024f02` `rgb(2,79,2)``\n• #02024f\n``#02024f` `rgb(2,2,79)``\n• #4f4f02\n``#4f4f02` `rgb(79,79,2)``\n• #024f02\n``#024f02` `rgb(2,79,2)``\n• #02024f\n``#02024f` `rgb(2,2,79)``\n• #4f024f\n``#4f024f` `rgb(79,2,79)``\n• #000400\n``#000400` `rgb(0,4,0)``\n• #011d01\n``#011d01` `rgb(1,29,1)``\n• #013601\n``#013601` `rgb(1,54,1)``\n• #024f02\n``#024f02` `rgb(2,79,2)``\n• #036803\n``#036803` `rgb(3,104,3)``\n• #038103\n``#038103` `rgb(3,129,3)``\n• #049a04\n``#049a04` `rgb(4,154,4)``\nMonochromatic Color\n\n# Alternatives to #024f02\n\nBelow, you can see some colors close to #024f02. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #154f02\n``#154f02` `rgb(21,79,2)``\n• #0f4f02\n``#0f4f02` `rgb(15,79,2)``\n• #084f02\n``#084f02` `rgb(8,79,2)``\n• #024f02\n``#024f02` `rgb(2,79,2)``\n• #024f08\n``#024f08` `rgb(2,79,8)``\n• #024f0f\n``#024f0f` `rgb(2,79,15)``\n• #024f15\n``#024f15` `rgb(2,79,21)``\nSimilar Colors\n\n# #024f02 Preview\n\nThis text has a font color of #024f02.\n\n``<span style=\"color:#024f02;\">Text here</span>``\n#024f02 background color\n\nThis paragraph has a background color of #024f02.\n\n``<p style=\"background-color:#024f02;\">Content here</p>``\n#024f02 border color\n\nThis element has a border color of #024f02.\n\n``<div style=\"border:1px solid #024f02;\">Content here</div>``\nCSS codes\n``.text {color:#024f02;}``\n``.background {background-color:#024f02;}``\n``.border {border:1px solid #024f02;}``\n\n# Shades and Tints of #024f02\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, #000200 is the darkest color, while #eeffee is the lightest one.\n\n• #000200\n``#000200` `rgb(0,2,0)``\n• #011601\n``#011601` `rgb(1,22,1)``\n• #012901\n``#012901` `rgb(1,41,1)``\n• #023c02\n``#023c02` `rgb(2,60,2)``\n• #024f02\n``#024f02` `rgb(2,79,2)``\n• #026202\n``#026202` `rgb(2,98,2)``\n• #037503\n``#037503` `rgb(3,117,3)``\n• #038803\n``#038803` `rgb(3,136,3)``\n• #049c04\n``#049c04` `rgb(4,156,4)``\n• #04af04\n``#04af04` `rgb(4,175,4)``\n• #05c205\n``#05c205` `rgb(5,194,5)``\n• #05d505\n``#05d505` `rgb(5,213,5)``\n• #06e806\n``#06e806` `rgb(6,232,6)``\n• #09f909\n``#09f909` `rgb(9,249,9)``\n• #1cf91c\n``#1cf91c` `rgb(28,249,28)``\n• #2ffa2f\n``#2ffa2f` `rgb(47,250,47)``\n• #42fa42\n``#42fa42` `rgb(66,250,66)``\n• #55fb55\n``#55fb55` `rgb(85,251,85)``\n• #68fb68\n``#68fb68` `rgb(104,251,104)``\n• #7cfc7c\n``#7cfc7c` `rgb(124,252,124)``\n• #8ffc8f\n``#8ffc8f` `rgb(143,252,143)``\n• #a2fda2\n``#a2fda2` `rgb(162,253,162)``\n• #b5fdb5\n``#b5fdb5` `rgb(181,253,181)``\n• #c8fec8\n``#c8fec8` `rgb(200,254,200)``\n• #dbfedb\n``#dbfedb` `rgb(219,254,219)``\n• #eeffee\n``#eeffee` `rgb(238,255,238)``\nTint Color Variation\n\n# Tones of #024f02\n\nA tone is produced by adding gray to any pure hue. In this case, #272a27 is the less saturated color, while #024f02 is the most saturated one.\n\n• #272a27\n``#272a27` `rgb(39,42,39)``\n• #242d24\n``#242d24` `rgb(36,45,36)``\n• #213021\n``#213021` `rgb(33,48,33)``\n• #1e331e\n``#1e331e` `rgb(30,51,30)``\n• #1b361b\n``#1b361b` `rgb(27,54,27)``\n• #183918\n``#183918` `rgb(24,57,24)``\n• #153c15\n``#153c15` `rgb(21,60,21)``\n• #123f12\n``#123f12` `rgb(18,63,18)``\n• #0e430e\n``#0e430e` `rgb(14,67,14)``\n• #0b460b\n``#0b460b` `rgb(11,70,11)``\n• #084908\n``#084908` `rgb(8,73,8)``\n• #054c05\n``#054c05` `rgb(5,76,5)``\n• #024f02\n``#024f02` `rgb(2,79,2)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #024f02 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"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.55016327,"math_prob":0.6947208,"size":3609,"snap":"2020-24-2020-29","text_gpt3_token_len":1575,"char_repetition_ratio":0.13009709,"word_repetition_ratio":0.011070111,"special_character_ratio":0.56331396,"punctuation_ratio":0.23224352,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98917025,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-06-07T09:25:43Z\",\"WARC-Record-ID\":\"<urn:uuid:9d64bf2c-60f2-40d9-a43b-02633e8ea0d7>\",\"Content-Length\":\"36110\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:22ec2d97-5c9b-44e3-b0d9-622a1e17d470>\",\"WARC-Concurrent-To\":\"<urn:uuid:4d459217-f333-49c1-84a0-1a70e0aaf29d>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/024f02\",\"WARC-Payload-Digest\":\"sha1:2CAQY32I5HSTLEUIASGZ4IDOPG6LZWXH\",\"WARC-Block-Digest\":\"sha1:KCAJLVQD4UA5RKF5VMQLPXW267SMH3OA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590348526471.98_warc_CC-MAIN-20200607075929-20200607105929-00224.warc.gz\"}"}
|
https://qsstudy.com/describe-generation-three-phase-emf/
|
[
"Physics\n\n# Describe Generation of Three Phase EMF",
null,
"The three-phase power is generally used for transmission and allocation of electrical power because of their advantage. This has properties that make it very advantageous in electric power systems. It is more inexpensive as evaluate to single phase power and requires three live conductors for power supply. Here, power is usually generated and distributed in three phase and transformers are used to alter the voltage.\n\nGeneration of three-phase emf\n\nIn a three – phase a.c. generator three coils are fastened rigidly together and displaced from each other by 1200. It is made to rotate about a fixed axis in a uniform magnetic field. Each coil is provided with a separate set of slip rings and brushes.\n\nThese voltages can be produced by a three-phase AC generator having three identical windings displaced separately from each other by 120 degrees electrical. Three electrical circuits/coils are equally distributed over the margin of a permanent magnetic rotor. Three phase systems may or may not have an impartial wire. An impartial wire allows the three-phase structure to use a higher voltage while still sustaining lower voltage single phase appliances. The scale and incidence of these EMFs are similar but are displaced separately from one another by an angle of 120 degrees.\n\nAn emf is induced in each of the coils with a phase difference of 120o. Three coils a1 a2, b1 b2, and c1 c2 are mounted on the same axis but displaced from each other by 1200, and the coils rotate in the anticlockwise direction in a magnetic field (Fig: a).",
null,
"Fig; a Section of 3 phase ac generator\n\nWhen the coil a1 a2 is in position AB, emf induced in this coil is zero and starts increasing in the positive direction. At the same instant, the coil b1b2 is 120o behind coil a1a2, so that emf induced in this coil is approaching its maximum negative value and the coil c1 c2 is 2400 behind the coil a1 a2, so the emf induced in this coil has passed its positive maximum value and is decreasing.\n\nAccording to Faraday’s law, emf induced in three coils. The emf induced in these three coils will have phase dissimilarity of 1200. i.e. if the induced emf of the coil C1 has a phase of 00, then induced emf in the coil C2 lags that of C1 by 1200 and C3 lags that of C2 1200.",
null,
"Fig: b Three phase emf\n\nThus the EMFs induced in all the three coils are equal in magnitude and of the same frequency. The emf induced in the three coils are;\n\nea1a2 = E0 sin ωt\n\neb1b2 = E0 sin (ωt – 2π/3)\n\nec1c2 = E0 sin (ωt – 4π/3)\n\nThe emf induced and phase difference in the three coils a1 a1, b1 b1 and c1 c1 are shown in Fig: b & Fig:c.",
null,
"Fig: c Angular displacement between the armature\n\nA three-phase method is equal if and only if:\n\n• A load of each phase has equal impedance value;\n• A load impedance of each phase has an equivalent phase angle;\n• Voltage and current values are equivalent for each phase and;\n• Phase disarticulation is 1200 between each phase."
] |
[
null,
"https://qsstudy.com/wp-content/uploads/2016/05/Three-Phase-EMF.jpg",
null,
"https://qsstudy.com/wp-content/uploads/2016/05/a5-2.jpg",
null,
"https://qsstudy.com/wp-content/uploads/2016/05/a6-2.jpg",
null,
"https://qsstudy.com/wp-content/uploads/2016/05/a7-1.jpg",
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.94084746,"math_prob":0.9803408,"size":2913,"snap":"2023-40-2023-50","text_gpt3_token_len":698,"char_repetition_ratio":0.1567549,"word_repetition_ratio":0.017175572,"special_character_ratio":0.23549606,"punctuation_ratio":0.08318891,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9816225,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,3,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-04T05:46:12Z\",\"WARC-Record-ID\":\"<urn:uuid:7538b432-a619-4379-abcb-7e7f6f4093d2>\",\"Content-Length\":\"26588\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4ac6a0d5-1c6a-4845-bbd4-355b3f982816>\",\"WARC-Concurrent-To\":\"<urn:uuid:a28a037a-8459-4d29-adac-904b068cf00f>\",\"WARC-IP-Address\":\"104.21.94.50\",\"WARC-Target-URI\":\"https://qsstudy.com/describe-generation-three-phase-emf/\",\"WARC-Payload-Digest\":\"sha1:FFTHXEO5YCDH3MM7RD2FO6EBUHFSWPIY\",\"WARC-Block-Digest\":\"sha1:NRTGIKJWJY2HIMLJU3USNFKDXW6GAL6M\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100525.55_warc_CC-MAIN-20231204052342-20231204082342-00482.warc.gz\"}"}
|
http://rajlama.com.np/blog/number-base-conversion-binary-octal-decimal-hexadecimal-all-in-one
|
[
"• Courses\n• Data Structure\n• ### Number Base Conversion (Binary | Octal | Decimal | Hexadecimal) All in One\n\nPosted on 2021-03-07\n\nNumber conversion is the process of converting one number base to another base. In computer science, we have to study about 4 types of number systems, binary, octal, decimal, and hexadecimal number. When we convert one number system to another number system, it will be a total of 12 different number conversion types, which are illustrated below. While converting one number base to another number base it will be better to use the following table.\n\n1. Binary to Decimal\n2. Octal to Decimal\n4. Decimal to Binary\n5. Octal to Binary\n7. Binary to Octal\n8. Decimal to Octal",
null,
"1) Binary to Decimal Conversion\n\ni) Convert 11100011101\n\nConvert 11010112 = into equivalent decimal number\n\nMethod 1:\n\n```11010112\n= 1×26+1×25+0×24+1×23+0×22+1×21+1×20\n= 64+32+0+8+0+2+1\n= 10710\n```\n\nMethod 2:\n\n11010112\n\n Place Value 64 32 16 8 4 2 1 Binary 1 1 0 1 0 1 1 Decimal 64 + 32 + 8 + 2 + 1 = 10710\n\nii) Convert 1111001101112 into an equivalent decimal number\n\nMethod 1:\n\n```1111001101112\n= 1×211+1×210+1×29+1×28+0×27+0×26+1×25+1×24+0×23+1×22+1×21+1×20\n= 2048+1024+512+265+0+0+32+16+0+4+2+1\n= 3895\n```\n\nMethod 2:\n\n Place Value 2048 1024 512 256 128 64 32 16 8 4 2 1 Binary 1 1 1 1 0 0 1 1 0 1 1 1 Decimal 2048+1024+512+265+0+0+32+16+0+4+2+1 = 3895\n\n2) Octal to Decimal Conversion\n\ni) Convert 125638 to a decimal equivalent number\n\n```= 1×84+ 2×83+5×82+6×81+3×80\n= 4096 + 1024 + 320 + 48 + 3\n= 549110\n```\n\nConvert 7A2BD16 into decimal equivalent number\n\nA=10, B=11, C=12, D=13, E=14, F=15\n\n``` = 7×164+10×163+2×162+11×161+13×160\n= 458,752 + 40,960+512+176+13\n= 50041310\n```\n\n4) Decimal to Binary Conversion\n\nConvert 56310 into binary equivalent\n\nMethod 1:\n\n 2 563 -----1 -----1 -----0 -----0 -----1 -----1 -----0 -----0 -----0 2 281 2 140 2 70 2 35 2 17 2 8 2 4 2 2 1\n\nSo, 56310 = 10001100112\nNote: (Write binary numbers from bottom to up)\n\n5) Octal to Binary Conversion\n\nConvert 27638 into an equivalent binary number\n\nOctal Number: 2763\n\n3 group binary digits\n\n 2 7 6 3 010 111 110 011\n\n:: 27638 = 0101111100112\n\nConvert AD710 into an equivalent binary number\n\n4 group binary digit\n\n A D 7 1010 1101 0111\n\n7) Binary to Octal Conversion\n\nConvert 111000101102 into equivalent octal number\n\n3 group binary digits (Note: group the binary digits from right to left)\n\n 011 100 010 110 3 4 2 6\n\n:: 111000101102 = 3468\n\n8) Decimal to Octal Conversion\n\nConvert 8965210 into equivalent octal number\n\n 8 89652 -----4 -----6 -----0 -----7 -----5 8 11206 8 1400 8 175 8 21 2\n\n:: 8965210 = 25706410\n\nThe process of converting a hexadecimal number into an octal number is that, at first, we convert a hexadecimal number into binary and after that, we convert the obtained binary number into an octal number.\n\nConvert A9D516 into equivalent octal number\n\nStep 1: Converting Hexadecimal number into a binary number\n\n4 group binary digits\n\n A 9 D 5 1010 1001 1101 0101\n```:: A9D516 = 10101001110101012\nStep 2: Now converting binary to octal number\n3 group binary digits\n001 010 100 111 010 101\n1 2 4 7 2 5\n\nSo, A9D516 = 1247258\n```\n\nConvert 101110100100012 into equivalent hexadecimal number\n\n``` 4 group binary digits\n0010 1110 1001 0001\n2 E 9 1\n\n:: 101110100100012 =2E9116```\n\nThe process to convert an octal number into hexadecimal first, we will convert an octal number into binary numbers system and after that again we convert the obtained binary number into hexadecimal number by making group of 4 binary digits.\n\n```E.g. Convert 1247258 into hexadecimal number\nStep 1: Converting octal to binary\n```\n 1 2 4 7 2 5 001 010 100 111 010 101\n\n:: 1247258 = 0010101001110101012\n\n``` Step 2: Now converting obtained binary number to octal\nBinary = 0010101001110101012\n\nMaking 4 group binary digits from right\n0000 1010 1001 1101 0101\n0 A 9 D 5\n\n:: 0010101001110101012 = A9D516\nSo, 1247258 = A9D516\n```\n\nConvert 968510 to equivalent hexadecimal number\n\n 16 9685 -----5 -----13 -----5 16 605 16 37 2\n\n:: 968510 = 25D516\n\n1825\n\n#### Recent Guides\n\n Secondary Level Teaching License Examination Sample question paper Secondary Level General Examination Sample question paper-2078 English, Lower Secondary Level Sample question paper-2078 English, Secondary Level Sample Question Paper-2078"
] |
[
null,
"http://rajlama.com.np/storage/blog-posts/March2021/binary2.jpg",
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.5163614,"math_prob":0.99088186,"size":4018,"snap":"2022-05-2022-21","text_gpt3_token_len":1510,"char_repetition_ratio":0.21001495,"word_repetition_ratio":0.051209103,"special_character_ratio":0.5074664,"punctuation_ratio":0.07402597,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99925053,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-17T07:40:31Z\",\"WARC-Record-ID\":\"<urn:uuid:4c683e37-43b1-4b49-a70d-46e62650ca58>\",\"Content-Length\":\"96711\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:93fdb883-bda1-4de7-8f3b-686a57248784>\",\"WARC-Concurrent-To\":\"<urn:uuid:20e72250-c0d0-4614-9359-cb7c29383557>\",\"WARC-IP-Address\":\"207.148.117.199\",\"WARC-Target-URI\":\"http://rajlama.com.np/blog/number-base-conversion-binary-octal-decimal-hexadecimal-all-in-one\",\"WARC-Payload-Digest\":\"sha1:NL2BFGMPN7F6CT4JQUPHA2GHGXUMHRPB\",\"WARC-Block-Digest\":\"sha1:C2J2PTCOXUJFVD5UZ44M2FG7JDAQJ4QH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662517018.29_warc_CC-MAIN-20220517063528-20220517093528-00735.warc.gz\"}"}
|
https://jp.mathworks.com/matlabcentral/cody/problems/1809-05-vector-equations-1/solutions/370340
|
[
"Cody\n\nProblem 1809. 05 - Vector Equations 1\n\nSolution 370340\n\nSubmitted on 15 Dec 2013 by Ted\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\nTest Suite\n\nTest Status Code Input and Output\n1 Pass\n%% cVec = 5:-.2:-5; ref = (1/sqrt(2*pi*2.5^2))*exp((-1*cVec.^2)/(2*2.5^2)); user = MyFunc(); assert(isequal(user,ref))\n\n2 Pass\n%% [xVec cVec] = MyFunc(); cRef = 5:-.2:-5; assert(isequal(cRef,cVec))"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.5428723,"math_prob":0.99476033,"size":375,"snap":"2019-43-2019-47","text_gpt3_token_len":142,"char_repetition_ratio":0.12938005,"word_repetition_ratio":0.0,"special_character_ratio":0.44,"punctuation_ratio":0.1882353,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9838921,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-15T16:44:11Z\",\"WARC-Record-ID\":\"<urn:uuid:9d74e620-94b8-466b-85d6-0f384fb7de79>\",\"Content-Length\":\"70795\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:86f331e9-ba16-49b7-8884-1d07b07aed50>\",\"WARC-Concurrent-To\":\"<urn:uuid:998f379d-e9c4-4126-b42a-8e356cc7e16b>\",\"WARC-IP-Address\":\"184.25.188.167\",\"WARC-Target-URI\":\"https://jp.mathworks.com/matlabcentral/cody/problems/1809-05-vector-equations-1/solutions/370340\",\"WARC-Payload-Digest\":\"sha1:DPKHY45X3M4W3VQW3TGPJFL6AVHGR7TR\",\"WARC-Block-Digest\":\"sha1:SIUFB5EURTDFFYK5WQTUBLEIMIK6NQHK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986660067.26_warc_CC-MAIN-20191015155056-20191015182556-00237.warc.gz\"}"}
|
https://newproxylists.com/differential-equations-ndsolve-with-several-initial-conditions/
|
[
"# differential equations – ndsolve with several initial conditions\n\nI'm trying to solve a PDE using the NDsolve with initial and boundary conditions,\n\n``````NDSolve[{D[u[t, x], t]- RE[ D[u[t, x], X]]+ D[u[t, x], X]== 10,\nyou[0, x] == 2,\nyou[t, 0] == 1,\nyou[t, 2] == 1\n}\nyou\n{t, 0, 5}, {x, 0, 2}, MaxStepSize -> 0.01]\n``````\n\nit worked. But when I try to assign an initial condition: u[0,x]= 2 when 0.5 <= x <= 1, u[0,x]= 1 elsewhere in [0,2] by means of If\n\n``````NDSolve[{D[u[t, x], t]- RE[ D[u[t, x], X]]+ D[u[t, x], X]== 10,\nYes[05<=x[05<=x[05<=x[05<=x<= 1, u[0, x] == 2, u[0, x] == 1],\nu[t, 0] == 1,\nu[t, 2] == 1\n},\nu,\n{t, 0, 5}, {x, 0, 2}, MaxStepSize -> 0.01]\n``````\n\nthe system has made me\n\n``````NDSolve :: deqn: Equation or list of expected equations instead of If[05<=x<=1u[05<=x<=1u[05<=x<=1u[05<=x<=1u[0,x]== 2, u[0,x]== 1]in the first argument {(u ^ (1,0))[t,x]== 10, if[05<=x<=1u[05<=x<=1u[05<=x<=1u[05<=x<=1u[0,x]== 2, u[0,x]== 1]you[t,0]== 1, u[t,2]== 1}.\n``````\n\nI know something is wrong with the expression of my initial conditions, but how do I assign the initial conditions with an if condition. Or a better way to manage it?\n\nThank you for your time!\n\nPosted on"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.779497,"math_prob":0.9999579,"size":1058,"snap":"2019-35-2019-39","text_gpt3_token_len":470,"char_repetition_ratio":0.17077799,"word_repetition_ratio":0.13186814,"special_character_ratio":0.5056711,"punctuation_ratio":0.221875,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998037,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-25T12:47:28Z\",\"WARC-Record-ID\":\"<urn:uuid:8bc8f6be-8d42-4091-9bd5-8a700c0cc718>\",\"Content-Length\":\"28140\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:12ee6d71-005f-4df0-a962-83ef6a35872f>\",\"WARC-Concurrent-To\":\"<urn:uuid:7eb4416a-3851-43c4-abc8-8e6a93bac6eb>\",\"WARC-IP-Address\":\"173.212.203.156\",\"WARC-Target-URI\":\"https://newproxylists.com/differential-equations-ndsolve-with-several-initial-conditions/\",\"WARC-Payload-Digest\":\"sha1:W2XYPQKSJOEXJYTTCZY7PRCD4GWO7MO7\",\"WARC-Block-Digest\":\"sha1:P3AGOUOR75SBCA4JUEWOPWJ2BKTXDXDQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027323328.16_warc_CC-MAIN-20190825105643-20190825131643-00013.warc.gz\"}"}
|
https://yagisanatode.com/2018/09/22/google-sheets-remove-the-lowest-grade-for-each-student-on-a-course/
|
[
"# Google Sheets – Remove The Lowest Grade for Each Student on a Course",
null,
"Google Sheets – MIN, FILTER, INDEX, MATCH, SUM, COUNTIF\n\nIn the region of the world that I work in, it is a pretty common occurrence for university courses to run weekly assessment. At the end of the course, all the weekly assessment is then added together minus the lowest piece of assessment.\n\nFor lecturers with small course sizes, this is a pretty simple task that you could simply eyeball if you have a small enough group, but what if your course runs into the thousands with half a dozen tests to choose from. Eyeballing is just not going to do it.\n\nRecently I was asked to do the same thing for the program that I manage. Over an 8 week term, we run 7 assessment at the end of each week for our students. My job was to find the lowest grade out of the 7 assessment and drop it, taking note of the assessment unit that I dropped for each student.\n\nI use Google Sheets for this purpose for its ease of use and sharability.\n\nThis is an example dataset of the 7 assessment (in this case, weekly tests) in Google Sheets. We need to remove the lowest grade from each student. As you can see not all students have their lowest grade in the same Unit test.\n\n### Step 1 – Find The Lowest Grade with MIN (optional)\n\nOur first step in the process is to find the lowest grade for each of the tests. To do this we use the MIN formula:\n\n=MIN(range of values to search to find the minimum)\n\nFor example:\n\n=MIN(A3:H3)\n\n### Step 2 – Record the Unit Where the Lowest Grade was Found (optional)\n\nWhile this step has no purpose in the further progress of preparing a list of all grades except the lowest, it is important to be able to quickly reference the students lowest grade should I be called upon in future.\n\nTo find the Unit that the lowest grade was found in, I use the INDEX and MATCH combination of formulas. More on this here:\n\nVlookup Left with INDEX and MATCH\n\nINDEX and MATCH will help me find the unit number in row 2 by comparing the value of the lowest grade in column J against the array of the student’s grades. Here’s the formula:\n\nSo our first row would look like this:\n\n=INDEX(\\$B\\$2:\\$H\\$2,MATCH(J3,B3:H3,FALSE))\n\n### Step 3 – Get New List of Grades Excluding the Lowest with FILTER and MIN\n\nIn our next step, we want to grab the student’s grades for the 7 units (col B:H), but filter out the lowest. We will create an array of the grades and filter them to exclude the minimum value of the array.\n\nTo do this we will use the FILTER formula that takes an array of values and filters them by the same array or another array of equal length based on a criteria. In our case, our filter criteria in the same array of grades and our criteria must not contain (<>) the MIN of those grades. Let’s take a look:\n\nSo to filter out the lowest grade for our first student, it would look like this:\n\n=FILTER(B3:H3,B3:H3 <> MIN(B3:H3))\n\nNow you have your list of grades minus the lowest. We can now go ahead and get the total of the Grades for each user with SUM\n\nWow! Wow! Wow! Just hold on a damn minute Yagi!!!! What if I have more than one lowest grade!!!! I’ll need to add at lease one of those to my total!\n\nOkay! You got me. We need to take this into account. There are a number of approaches I used to solve this issue, but in the end this was the simplest approach:\n\n### Step 4 – Getting the Total of the New Set of Grades While Taking into Account that there might me Multiple Lowest Grades\n\nYou see, if there is more than one grade that is equally the lowest, then my FILTER based on the MIN will be less than the six grades I need for my course total.\n\nLet’s say student Odi, has two equally lowest grades of 14 – one in Unit 4 and one in Unit 7. Let’s see what the result would look like:\n\nWe can see here that the filter is only producing 5 of our 6 needed tests. Likewise, Odi is missing 14 points towards his total. The FILTER is taking out all grades that equal the MINimum grade.\n\nIf Odi had minimum grades all with the value of 14 the remaining results would only have 4 grades displayed.\n\nThe Fix\n\nTo fix this problem we are going to change our simple SUM formula in column R to something a little more complex. Before we SUM all of Odi’s remaining Unit Test grades we are going to:\n\n1. Count the number of empty cells from Column L to Q in the Remaining Unit Test Grades section for each students row.\n2. Multiply that number by the lowest grade.\n3. Add that to the sum of the grades in Column L to Q\n\nIn Odi’s row, the formula would look like this:\n\n=(COUNTIF(L7:Q7,””)*MIN(B7:H7))+SUM(L7:Q7)\n\nWe’ll go ahead and change the grades to reflect some of the students who now have more than one minimum grade.\n\nNow we have a proper total of our 6 highest grades, taking into account multiple minimum grades."
] |
[
null,
"https://i1.wp.com/yagisanatode.com/wp-content/uploads/2018/09/Remove-Lowest.png",
null
] |
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|
https://pursuit.purescript.org/packages/purescript-gl-matrix/2.1.0/docs/GLMatrix.Quat.Mix
|
[
"Module\n\n# GLMatrix.Quat.Mix\n\nPackage\npurescript-gl-matrix\nRepository\ndirkz/purescript-gl-matrix\n\n### #js_rotationToSource\n\n``js_rotationTo :: Fn2 Vec3 Vec3 Quat``\n\n### #rotationToSource\n\n``rotationTo :: Vec3 -> Vec3 -> Quat``\n\nSets a quaternion to represent the shortest rotation from one vector to another. Both vectors are assumed to be unit length.\n\n### #js_setAxesSource\n\n``js_setAxes :: Fn3 Vec3 Vec3 Vec3 Quat``\n\n### #setAxesSource\n\n``setAxes :: Vec3 -> Vec3 -> Vec3 -> Quat``\n\nSets the specified quaternion with values corresponding to the given axes. Each axis is a vec3 and is expected to be unit length and perpendicular to all other specified axes.\n\n### #js_fromMat3Source\n\n``js_fromMat3 :: Fn1 Mat3 Quat``\n\n### #fromMat3Source\n\n``fromMat3 :: Mat3 -> Quat``\n\nSets the specified quaternion with values corresponding to the given axes. Each axis is a vec3 and is expected to be unit length and perpendicular to all other specified axes.\n\n### #js_getAxisAngleSource\n\n``js_getAxisAngle :: Fn2 Vec3 Quat Number``\n\n### #getAxisAngleSource\n\n``getAxisAngle :: Vec3 -> Quat -> Number``\n\nSets the specified quaternion with values corresponding to the given axes. Each axis is a vec3 and is expected to be unit length and perpendicular to all other specified axes.\n\n### #js_setAxisAngleSource\n\n``js_setAxisAngle :: Fn2 Vec3 Number Quat``\n\n### #setAxisAngleSource\n\n``setAxisAngle :: Vec3 -> Number -> Quat``\n\nSets a quat from the given angle and rotation axis, then returns it."
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.5505429,"math_prob":0.851682,"size":549,"snap":"2023-14-2023-23","text_gpt3_token_len":135,"char_repetition_ratio":0.108256884,"word_repetition_ratio":0.0,"special_character_ratio":0.21857923,"punctuation_ratio":0.11111111,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.977879,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-08T06:21:47Z\",\"WARC-Record-ID\":\"<urn:uuid:2d36a121-c21e-4ff3-a915-8dfc89cfa38b>\",\"Content-Length\":\"20678\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d3bde5de-271a-4b38-a8f0-8aa44795626b>\",\"WARC-Concurrent-To\":\"<urn:uuid:488d815a-287f-4f24-b4ae-dd80fdc81eb0>\",\"WARC-IP-Address\":\"206.189.189.151\",\"WARC-Target-URI\":\"https://pursuit.purescript.org/packages/purescript-gl-matrix/2.1.0/docs/GLMatrix.Quat.Mix\",\"WARC-Payload-Digest\":\"sha1:ET7HSWBDKFKTZ5RUQVKHN7ZHAC6JH6LT\",\"WARC-Block-Digest\":\"sha1:4Q3RNROOBOGIXWM7HRGS2DFFT4ESI3JO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224654097.42_warc_CC-MAIN-20230608035801-20230608065801-00433.warc.gz\"}"}
|
https://www.johnchandlerburnham.com/projects/ofvb/01/
|
[
"# Questions\n\n## 1\n\n• - : int = 17\n• - : int = 11\n• - : int = 1, this shows/ has left precedence\n• - : bool = true\n• - : bool = false,<> is “not equal”, which is super confusing notation.\n• - : bool = true\n• - : bool = false\n• - : bool = false\n• - : char = ‘%’\n• error, because (+) has type - : int -> int -> int =\n\n## 2\n\nmod has higher precedence than +\n\n## 3\n\nevaluates to 11, spaces don’t matter\n\n## 4\n\nmax_int + 1 is min_int, and min_int - 1 is max_int\n\n## 5\n\nException: Division_by_zero\n\n## 6\n\nmod is just remainder after integer division, with the negative sign following the dividend (the first arg).\n\n## 7\n\nBecause although there is some structural analogy between {true, false, &&, ||} and {1, 0, *, +}, the former is closed and the latter is not. E.g. true || true = true, but 1 + 1 = 2. The values 1 and 0 live in a bigger space of values, i.e. the integers. We use types to restrict the space of possible values in our code, so that if we accidentally call a function expecting an integer with a boolean argument, we find out our mistake at compile-time rather than at run-time.\n\n## 8\n\n'p' < 'q' is true, because p comes before q in the alphabet. I look forward to learning how and where OCaml defines the char ordering."
] |
[
null
] |
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|
https://physics.stackexchange.com/questions/173961/negative-pressure
|
[
"# Negative pressure\n\nHow is negative pressure created in a fluid system?Isn't it counter intuitive that we are reducing the pressure of a system containing no molecules(zero pressure) to a lower value?\n\n• Negative pressures can be achieved, but such states are metastable. For example the vdW eqn of state has negative-pressure states, but these are actually all resolved through the Maxwell equal area rule to be positive (when the system equilibrates i.e. phase separates). This is to say that the negative pressure is caused by forcing the system to be uniform where in reality it \"wants\" to phase separate. Finite size effects typically manifest in analogous ways. – alarge Apr 3 '15 at 13:58\n• I think you are refering to negative pressure created by scalar fields, right? If that is the case, the argument of Maxwell equal area is not valid because this wouldn't be a poper thermodynamical system – Hydro Guy Apr 3 '15 at 14:03\n• @alarge I see some point in your answer I'cant say that I've grasped everything you've just said.Could you provide some relevant links supporting your answer? – Akshay Bansal Apr 3 '15 at 14:07\n\n## 2 Answers\n\nSuppose you have a box filled with water in a uniform fashion. Now if you try to stretch the box in the $z$-direction, say, while keeping the other dimensions constant, what is the energy required? Well if the water distribution remains uniform, you can approximate this by the Hookean law $E_\\text{elastic} = \\frac{1}{2}k(z-z_0)^2$. Note that the constant $k$ is actually huge as water really, really doesn't want to stretch as it has a certain density it likes to have.\n\nOther than decreasing in density, how can the system react to the stretch? Well, it can phase separate so that the liquid water keeps at the optimal density and the other part of the box becomes gas (let's imagine it is basically a vacuum). The energy penalty you pay is then approximately the surface tension $\\gamma$ times the area of the liquid-gas interface $xy$.\n\nBecause $k$ is so large, the latter effect is almost always preferable. However, to go from a uniform system to one with two phases, something must break the symmetry (uniformity). This is often called a nucleation site, and it can be for example an air bubble. Certain geometries of the wall can also act as sites.\n\nIf the system is very pure (and/or small) and uniform when you start the stretch, nucleation will not happen, which means that phase separation will not happen. This means that water has to stretch. There's an energy penalty associated with this and thus you will be creating a pressure (due to the stretch alone) of $-kz$, which is to say that you can get to negative pressures if the strain is large enough. Remember though, that this is a metastable state, as all states with negative pressures are, and the system will phase separate if given the opportunity.\n\nFinally, to relate this to my earlier comment about the van der Waals equation: In the physical derivation of vdW you assume that the system is uniform. You will get some very strangely behaving lines (e.g. pressure increases as the box size increases). These are only obeyed if the system remains uniform (which it does if nothing can break the symmetry). However, as I explained above, if you stretch the system, uniformity is not what gets you the minimal free energy: If you assume that the system is actually made up of two phases (with no interfacial tension between them), and find the composition that minimizes the free energy, you'll arrive at Maxwell's equal area rule.\n\n'Negative ' pressure is strictly a relative state; relative to what one may wish to define as zero pressure, and here on earth we chose to define that as one standard atmosphere of pressure which is about 760 mm Hg absolute pressure. If you are capable of removing all gas particles from a space, then you will achieve -760 mm Hg gauge pressure, but you cannot reduce pressure beyond that point\n\n• I think you need to know how the water is transported in trees through xylem tube that are ~100 m long.For more information on the application of negative pressure, you may watch the Veritasium video:youtube.com/watch?v=BickMFHAZR0&index=2&list=WL – Akshay Bansal Apr 3 '15 at 14:04\n• @akshay 760 mm Hg (1 atm) is roughly a water column height of 34 feet ( 10 meters). Xylem transport of fluid by negative pressure alone would be limited to this column (-10m) height, but differential pressure is unlimited. Positive pressure applied at the base of the tree can be increased to any pressure to push the fluid up as long as the xylem can sustain the pressure without bursting. And how or whether xylem transport systems can do that is a question or matter for the biology forum. Physics limits you to a hard vacuum. – docscience Apr 3 '15 at 23:48"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.933661,"math_prob":0.94430655,"size":4883,"snap":"2021-21-2021-25","text_gpt3_token_len":1118,"char_repetition_ratio":0.13363394,"word_repetition_ratio":0.014492754,"special_character_ratio":0.22588572,"punctuation_ratio":0.08770054,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9700535,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-22T08:02:48Z\",\"WARC-Record-ID\":\"<urn:uuid:81daec2f-1655-4774-baaf-c89a7bca8153>\",\"Content-Length\":\"177427\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6de1ed77-80e4-4361-886e-bb89d79d14e3>\",\"WARC-Concurrent-To\":\"<urn:uuid:cb5214e8-f504-48ce-9f30-e3838a783344>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://physics.stackexchange.com/questions/173961/negative-pressure\",\"WARC-Payload-Digest\":\"sha1:6BTPMXALPGOUP56VECIQV6RACL34W7EX\",\"WARC-Block-Digest\":\"sha1:Y5XGLJXZ2TMANP7MQWLDFNPZYWIJXALV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623488512243.88_warc_CC-MAIN-20210622063335-20210622093335-00287.warc.gz\"}"}
|
https://calculator.name/area/square-centimeter/square-mile/0.69
|
[
"Amount\nFrom\nTo\n\n# 0.69 square centimeters to square miles\n\nHow many square miles in 0.69 square centimeters? 0.69 square centimeters is equal to 2.664104893E-11 square miles.\n\nThis page provides you how to convert between square centimeters and square miles with conversion factor.\n\n# How to convert 0.69 cm2 to mi2?\n\nTo convert 0.69 cm2 into mi2, follow these steps:\n\nWe know that, 1 mi2 = 25899881103 cm2\n\nHence, to convert the value of 0.69 square centimeters into square miles, divide the area value 0.69cm2 by 25899881103.\n\n0.69 cm2 = 0.69/25899881103 = 2.664104893E-11 mi2\n\n### Thus, 0.69 cm2 equals to 2.664104893E-11 mi2\n\nSquare Centimeters Conversion of Square Centimeters to Square Miles\n0.68 cm2 0.68 cm2 = 2.625494678E-11 mi2\n0.59 cm2 0.59 cm2 = 2.278002735E-11 mi2\n0.69 cm2 0.69 cm2 = 2.664104893E-11 mi2\n1.69 cm2 1.69 cm2 = 6.525126479E-11 mi2"
] |
[
null
] |
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|
https://www.hindawi.com/journals/mpe/2018/6389438/
|
[
"#### Abstract\n\nCurrently in China, energy conservation and emission reduction are important initiatives for society. Because of its large proportion of building energy consumption, several methods for building energy conservation have been introduced for central air conditioning. This paper presents a hybrid system for a modelling and control approach. By analysing the cooling capacity transfer process and the electromagnetic properties of pumps, the modelling is performed in a hybrid system framework. Pumps are classified as fixed-frequency pumps, variable-frequency pumps, and switchable pumps; a switch control strategy is used in the chilled water system for supplying cooling capacity. In order to adjust indoor temperature and save electrical energy, the cooling capacity allocation and average temperature methods are presented, which satisfy the goal of optimal control of real-time energy consumption. As a numerical example, the temperature variation and cold regulation processes are simulated. Three rooms are the control objects that lower the setting points, and pumps waste as little electrical energy as possible. The results show 49.4% of water pump power consumption when compared with constant water volume technology. The modelling and control approach is more advantageous in ensuring the success of reconstruction projects for central air conditioning.\n\n#### 1. Introduction\n\nCurrently, China is facing a series of ecological problems, such as climate change, environmental degradation, and increases in energy consumption. According to the “China Statistical Yearbook 2015”, the levels of energy utilization and features of energy consumption remain in a backward predicament of high energy consumption, low efficiency, and serious waste. In 2013, for example, the average power consumption was 148.5 million kWh every day, and the transfer efficiency was 43.12% . These statistical data show the extensive form of Chinese development, which is far different from that of developed countries. Recently, the Chinese government has made unflinching efforts to build an environmentally friendly society; certainly, the concepts of energy conservation and emission reduction have informed policy orientation.\n\nCentral air conditioning accounts for a great proportion of building energy consumption and has considerable opportunities for energy conservation and emission reduction. Thus, research on energy savings for central air conditioning has a significant impact. There are several methods for building energy conservation: manual dispatching based on management view, optimized control schemes based on sample statistics, and technical renovation based on variable water/air volume. In the perspective of management, Liang et al. pointed out the “reasonable energy consumption standard” and “differentiation of buildings”. The concepts of “priority buildings” and “benchmarking buildings” were proposed. By evaluation and supervision, an operating plan is made for saving energy. Fong et al. used an approach of evolutionary programming for energy management. The existing operational settings can be improved by suggesting optimized information. Keçebaş and Yabanova studied exergy efficiencies of district heating systems by using an artificial neural network technique. Using average weekly data, the performance and characteristics are predicted. In addition, ambient temperature and flow rates are the main factors in thermal optimization. These are extensive measures with conspicuous disadvantages. Manual dispatching relies on the experience and subjectivity of operators and lacks accuracy. According to the day, season, climatic characteristics, and other factors, the method of the optimized control scheme satisfies the requirements to some extent but is not suited for changes of some random loads, and thus lacks in-time adjustment.\n\nBecause of uncertain cold loading, variable water/air volume has the advantages of instantaneous adjustment of cooling capacity and energy consumption. Compared with building a new system, technical innovation avoids massive cost and long projects. Moreover, in the air conditioning system, up to 70% of the power consumption is due to pumps and fans . Obviously, in renewal projects, pumps and fans have great potential for energy savings.\n\nMany investigation reports in the literature show that variable water volume technology is a current focus in our research. There are several optimization methods that can effectively address these problems. These methods are the analytical approach , direct search method , linear quadratic programming , nonlinear programming method , heuristic optimization algorithm [3, 13, 14], and artificial neural networks . Kaya et al. analysed the influence factors of the effectiveness in pumps and presented an approach for improvement. By using measured data, the potential energy saving opportunities, investment costs, and payback periods are calculated. Lu et al. provided some formulation and analysis based on global optimization technologies towards overall heating, ventilation, and air conditioning (HVAC) systems. For new urban development, Chow et al. predicted thermal demands and outlined an energy modelling methodology and decision approach to derive the most desirable scheme for a given project. Sheng and Duanmu presented electricity consumption and economic analyses. Lu et al. and Nassif et al. used a genetic algorithm to optimize the models of components in HVAC systems. Fong et al. adopted a robust evolutionary algorithm for system optimization.\n\nControl strategies include temperature difference and pressure difference control methods. Wang and Burnett developed an online control strategy to optimize the speed of variable-speed condenser cooling water pumps by adjusting the pressure set point. The adaptive strategy is used to identify the parameters with sampling at each step. Zhao et al. focused on extreme value analysis for parallel variable-frequency hydraulic pumps in central air conditioning systems. He used a variable differential pressure set point control strategy to respond to fast load variations and explored the optimal number of operating units of parallel variable-frequency pumps. This method obtains electricity savings of 15.8% and 4% compared with a single-pump operation scheme and dual-pump operation scheme, respectively.\n\nMoreover, some experimental and applicable results show advantages towards the method of technical renovation. Anderson et al. built an experimental system for HVAC control and obtained some ideal results. Qu et al. proposed a multilayer feed-forward artificial neural network based on a Back Propagation (BP) algorithm and used MATLAB implements to design and train an energy consumption prediction model of a micro urban building. Lu et al. proposed an optimization plan in building sections. For engineering applications, there are persuasive architectures in Hong Kong [20, 21] and Da Lian .\n\nIn these studies, pure continuous or discrete control systems are used for the control of the entire air conditioning system. However, the constant frequency pumps, variable-frequency pumps, and switching pumps in the chilled water system constitute a hybrid system. The frequency conversion operation and switching action of the pumps reflect the continuous and discrete dynamics, respectively. Therefore, the methods can hardly reflect the accuracy and applicability of the system structure and control process, nor an ideal energy saving effect. In addition, the electromagnetic properties of pumps are elided normally, although the hydraulic and thermodynamic characteristics are considered. Because the mathematical model of a pump motor is a high-order, nonlinear multivariable system with strong coupling effects, the variable-frequency, quadratic regression empirical formulas or simple proportional expressions cannot reflect the dynamics of motor appropriately. Compared with the pressure difference control strategy, the temperature difference control method is superior in energy saving.\n\nIn this paper, a hybrid model was used in the modelling of the chilled water system in a central air conditioning system. Variable-frequency pumps and switchable pumps denote continuous and discrete control inputs, respectively. Based on the model, a hybrid dynamical control strategy was proposed for variable volume control and better energy saving performance. This model reflects the actual working conditions of central air conditioning water systems, especially in aspects of pump movement and water flow change characteristics. Problems such as time variability, multivariable, nonlinearity, and uncertainty are considered, and the energy saving performance of the proposed control method is exhibited.\n\nThe modelling and controlling of the system are shown as follows: first, an optimized pump model of speed adjustment based on vector control was developed, which prefers electromagnetic properties and the torque response effect. In addition, considering the load characteristics of the water system, we regulate the water volume of pumps by vector control instantaneously. Second, the cooling capacity transfer process is described, with thermodynamic properties throughout the framework. The measure of constant temperature difference is used in the model. Third, we use switching control strategy, cooling capacity allocation, and average temperature methods in order to obtain the hybrid dynamical property by volume adjusting and pump switching (on/off). Thus, we not only realise variable volume control, but also satisfy the energy saving requirement. Finally, a numerical example illustrates the control process and situation of energy savings.\n\nThis paper is organized as follows. In Section 2, the mathematical description of hybrid system is presented. In Section 3, hybrid system modelling of cooling capacity transfer process between water system and wind system is expressed. In Section 4, the hybrid dynamical control strategy used in energy saving for chilled water system is presented. In Section 5, a simulation example is given to illustrate the modelling and control processes. Section 6 concludes the paper.\n\n#### 2. Mathematical Description of Hybrid System\n\nThe hybrid system is described by\n\nforwhere denotes state variables, denotes control variables, fi denotes constant vector, and denotes output. Ai and Ci are state matrices; Bi is input matrix. are convex polyhedra in the input+state space, and denotes possibly being unbounded by each other. implies the ith subsystem, and implies subsystems. implies that all subsystems constitute the whole system. denotes the switching law of the hybrid system. Several differential or difference equations as subsystems constitute a hybrid system, and they obey the switching law. The law decides the key issues, such as switching time, switching state, and switching orientation. The system switches among the subsystems and selects only one at any time.\n\n#### 3. Modelling the Framework of the Cooling Capacity Transfer and Temperature Change Processes\n\nIn the air conditioning system, the cooling capacity transfers from chiller unit to chilled water system by evaporator. The pumps drive water as cycle in the chilled water system, and send cooling capacity to air condition unit for exchanging from chilled water system to air system. After mixing with fresh air, the cold air is sent to air conditioning room to drop the temperature. In the process of temperature change in air conditioning room, the heat balance is influenced by several aspects: cold air, random heat of occupants and equipment, latent heat inside, and heat transfer from building structure. The processes of the cooling capacity transfer and temperature change is shown in Figure 1.\n\nThis paper uses constant air volume and variable water volume/constant temperature difference measures to transfer cooling capacity. In chilled water system, switching and variable-frequency behaviours of pumps adjust the flow for saving energy, which denote the discrete and continuous time process, respectively. So we give an equation to show the heat balance as follows:On the left side of the equation, means the time differential of the heat capacity of a room. On the left side of the equation, means the cooling capacity for chilled water system; and mean the cooling capacity from return and fresh air systems, respectively; means random heat of occupants and equipment; means latent heat inside; means heat transfer from building structure.\n\nThe description of the symbol can be seen in the subsequent chapters and in Nomenclature.\n\nTowards motor model of variable-frequency pump, we use a direct torque control model to build and design the closed-loop control of torque and flux linkage; but for constant frequency pump, we use constant to describe it. By building relationship of torque with water flow volume, we get equations of state space for the whole process of the model at last.\n\n##### 3.1. Modelling the Cooling Capacity Transfer Process\n\nIt is known that the cold load of central air conditioning is affected by factors such as the structure and materials of a building, outdoor weather parameters, indoor lighting, radiating equipment, and number of occupants. In order to describe the dynamic process, we consider the area and temperature of walls, windows, and roof, according to the different materials, and use the heat transfer coefficient , accordingly. We use to indicate the random changing heat by equipment and people. The supply air carries cooling capacity inside the building to balance the heat.\n\nThe dynamic thermal balance equation in a room iswhere is the specific heat of air, is the temperature inside, is the mass inside, is the sending air volume, and is the sending air temperature.\n\nIn the cooling capacity transfer process, some of the cold is lost; we set the transfer efficiency as ε from water system to air system (), so we easily obtain the equation . In view of the factors of latent heat load and sensible heat load , we obtainThe media of the water system and air system are water and air, respectively; by the temperature difference , the cooling capacity is transferred. Thus, the equation becomeswhere is the specific heat of water, and is the water volume.\n\nBecause of the return air rate , the fresh rate obviously is ; hereby, the air volume of the test room is divided into return air volume and fresh air volume . The equation becomesThe equation can be rewritten asSubstituting (8) into (4) obtainsIn the chilled water system, we use the variable flow volume and constant temperature difference measures. With regard to variable flow volume, we should notice that the minimum volume is not lower than 40% of the rated value; otherwise, the energy efficiency of the pump would decay sharply. Thus, whatever kinds of pumps are operating, the whole water flow volume should be between 40% and 100%. In the system, the pumps are divided into three kinds: fixed pumps, variable-frequency pumps, and switchable pumps. The fixed pumps and variable-frequency pumps always operate; the switchable pumps either operate or not because of the cold loading. Both fixed pumps and switchable pumps are fixed-frequency; to distinguish them, they are simply called fixed pumps and switchable pumps, respectively.\n\nAccording to the three kinds of pumps, the water flow volume is expressed asHere, is a constant that indicates the flow volume of all the fixed pumps, is a state variable that indicates the flow volume of the variable-frequency pumps, is a constant that indicates the flow volume of a switchable pump, and is the number of operating switchable pumps. Therefore we substitute (10) into (9) to obtain (3).\n\n##### 3.2. Motor Model of Variable-Frequency Pump\n\nMost variable-frequency pumps are asynchronous AC motors in air conditioning systems. We use a stator flux-oriented direct torque control-space vector pulse width modulation (DTC-SVPWM) model to build and design the closed-loop control of torque and flux linkage. The features of DTC-SVPWM are that the controller computes the suitable stator voltage vector, and the method of SVPWM modulation generates the voltage vector to control torque and flux linkage. The structure is shown in Figure 2.\n\nThe M-T coordinate system is a synchronous revolution according to the stator flux orientation; the direction of the M-axis is the same as that of the stator flux linkage, and we use , which is output from the flux linkage PI adjuster to control the amplitude of the stator flux linkage . The direction of the T-axis is perpendicular to stator flux linkage; we use , which is output from the torque PI adjuster, to control the rotational speed of the stator flux linkage and then electromagnetic torque Te. We obtain from the transformation of coordinates and input it to the SPVWM to control the motor. Via double closed-loop feedback control, flux linkage and electromagnetic torque are compared with the setting values and accurately adjusted. The voltage estimation module estimates the actual stator voltage vector according to switch status signal and measured busbar voltage .\n\nThe simplified equations of stator voltage and torque areIt is apparent that the input variables and output variables have coupling relationships, so we cannot use and to achieve independent control with and , respectively. Therefore, a new variable must be introduced for decoupling. LetAfter decoupling, the transfer function is a proportional component from to ; the transfer function isThe dynamic structure of the torque control loop is shown in Figure 3. In the figure, there is an inertial component and a proportional component. The filter time constant is the same as the feedback smoothing time constant; the integral time constant of the torque controller is .\n\nWe use an integral controller for torque adjustment.Thus, the torque control loop is simplified as\n\n##### 3.3. Building the Relationship of Torque with Water Flow Volume of a Variable-Frequency Pump\n\nIn general, the water volume is in proportion to speed of a pump in a water system, so it is obvious that , where is the rated water volume and is the rated speed. Using motion equation, where is the rotational inertia, is the torque, and is the load torque, the torque is obtained byIn addition, the load of the pumps conforms to a quadratic proportion, where is the rated load torque. As an equation:Thus the relationship from torque to water flow volume is built. Combining (17) with (10), we build the state space equation to show the whole control process of the pumps. Here, with as the inside temperature and as water flow volume of the variable-frequency pump, and are state variables. With as a control variable, we set as the output variable. Figure 4 shows the state space.\n\nThus, the coefficients of the state space equation areCombining these equations and coefficients; we get the equations of state space:\n\n##### 3.4. Mathematical Power Models of Pumps\n\nAccording to the electromagnetic property of a variable-frequency pump, the output mechanical power of the pump is is the factor converted from motor power into output mechanical power, , and is angular velocity.\n\nThe power of a fixed pump is , and the power of switchable pumps is . Both of these kinds of pumps operate at constant frequency, so we consider them operating at constant power. Thus we get the total power of the pumps:G is the total number of fixed pumps and k is the number of operating switching pumps.\n\n#### 4. Hybrid Dynamical Control Strategy\n\n##### 4.1. Target of Optimal Hybrid Control for Central Air Conditioning\n\nThe goal of optimal control for central air conditioning is to minimise the power of the chilled pumps; further, the inside temperature should be close to the setting temperature . We obtain the target functionwhere . Due to the switching behaviours, the system is divided into some subsystems in different time intervals, namely, τ0, τ1, τ1, τ2,..,, τf. The switching sequence is produced at switching time τi, and the switching law σi and τi are one-to-one correspondence.\n\nEquation (22) means that (i) the target function is to find the minimum value of energy consumption of pumps on time interval [τ0, τf]; (ii) the function is divided into several parts, and we calculate the integration on each time interval; (iii) the sum of all integration terms is the minimum of the objective function.\n\nRemarks 1. (1)i is finite.(2)Switching law is state-independent, so the length of each time interval is not the same.(3)We simply consider the requirement of energy saving, and the humidity is not concerned in this work.\n\n##### 4.2. Developing the Switching Control Strategy\n\nVariable water flow is realised by variable-frequency regulation of variable-frequency pumps and switching control of switchable pumps. For the water flow volume of the variable-frequency pump, the feedback closed loop can be used in the system for accurate adjustment. For the switching pumps, we propose an algorithm for deployment. According to the changing cold load, we set the switching condition and orientation. The switching law is a simple and economic mode in reconstruction engineering that avoids complex processes and computing time. It is shown in Figure 5.\n\nThe switching strategy is based on state dependence. The switching behaviour occurs simply by satisfying the state judgement. In other words, once the real-time sampling value is beyond the setting threshold value , the system will switch. At the initial time, k is set to the maximum value N. After each sampling period (constant), the system obtains the inside temperature and compares with . If the value is in the setting range, the system will stay in the current subsystem. Otherwise, we will calculate . One way is , because of , unless k is the maximum; the other is k-1, because of , unless k is the minimum.\n\nRemarks 2. (1) is the inside temperature setting, and is the real-time inside temperature.(2)k is the number of operating switching pumps.(3)The maximum number of switching pumps is N. If all the switching pumps are operating, the system is close to full capacity.(4)The minimum number of switching pumps is 0. If a nonswitching pump is operating, the system is close to optimal energy saving.(5)Whether switching is on or off, the system obeys one pump at a time. The switching behaviour triggers only one pump.\n\n##### 4.3. Cooling Capacity Allocation and Average Temperature Methods\n\nFor a whole building, the framework is modelled with a switching control strategy. In real system, with many rooms with different areas, the cooling capacity requirement and temperature setting need an adaptive method to match. For allocation of cooling capacity, we adopt a weighting factor according to the proportion of thermal capacity in each room. This is a quantitative criterion on the demand side; on the supply side, there is a measure to balance different inside temperatures. The average temperature difference as a feedback signal is used to control the pumps. The switching condition can be modified asThe weighting factor isThe schematic diagram is shown in Figure 6.\n\nRemarks 3. (1) is the average temperature difference of the building.(2) is the real-time temperature inside the th room; is the setting temperature inside the th room. There are N rooms operating by using cooling capacity.(3) is the weighting factor of the th room for cooling capacity allocation; is the thermal capacity of the th room.(4) is the total thermal capacity of the rooms by using cooling capacity.(5)Obviously, these are based on the quantitative criterion, whatever the average temperature difference or weighting factor.(6)In view of allocating cooling capacity by weighting factor, the computing method of average temperature difference actually involves the average temperature per cubic meter.(7)Heat capacity is equal to for each room.\n\n#### 5. Numerical Example and Simulation Results\n\nIn this section, we present numerical results from simulations of the chilled water system. Three different rooms are the simulation objects; the cold air is sent to the rooms for reducing the inside temperatures. The supply side uses pumps for sending cold water to the air system by the switching control strategy and cooling capacity allocation method. We use one variable-frequency pump, four switching pumps, and two fixed pumps to transfer chilled water and adjust cooling capacity. The cooling capacity transfer mode of variable water volume and constant air volume is presented. The purpose of the variable water volume is accuracy control and energy saving, via our model and control strategy. The parameters are listed in Table 1.\n\nSimulink 2007 was used for simulating the whole process, where the simulation time was 5000 s. The switching sample time of the system was set to 60 s; the system checks the real-time inside temperature each and decides the switching behaviour. Feedback control is used in the three rooms and the whole system, which adjusts the inside temperatures and water flow volume, respectively.\n\nThe inside temperatures of the three rooms started at 30°C, and the system began at full capacity; in other words, all pumps were operating at rated power at the very start. Some curves were introduced into rooms as inside random cold loads; the surface temperatures of the wall space, window, and roof are set at constant values. The sending air volume was the same as the three rooms and was constant.\n\nFigure 7 shows that the inside temperatures decline rapidly by the sending air. At approximately =1000 s, the temperatures stabilise near the setting values, although the random cold loads disturb the system (Figure 8), and the total temperature difference is close to zero all for the three rooms (Figure 9). In addition, because of the lower cooling capacity requirements after 1000 s, the total water flow volume falls, which is operating near the lowest flow volume (approximately 40%); in other words, the water system realises a large potential space for energy saving (Figure 10). In the control process, the variable-frequency pump changes speed to alter the water volume in real-time by feedback control. In the first 1000 s, the pump is operating at rated power; in the last approximately 2000 s, the pump is operating in a fine adjustment status because of the real-time changing of cooling capacity requirements. In the middle period, the pump is operating at half the rated power. Switching pumps obey the control strategy and cooling capacity allocation methods, which adjust the number of operating switchable pumps (Figure 10). We set the threshold value , and the total temperature difference leads to the switching situation shown in Figure 10.\n\nConsidering the electromagnetic properties, the speed of a pump starts from zero to a stable value with a smooth curve (Figure 11); otherwise, it exhibits simply a horizontal line with a constant value. The torque responds according to the equation of motion, which shows the electromagnetic speed regulation process (Figure 12); otherwise, it exhibits a horizontal line as well. Considering the load characteristic in (17), the actual speed and torque responses do not show simple information. Figure 12 shows that the torque curve has acceleration and callback processes; after 0.2 s, the torque stabilises at 19.6 , which is less than the rated load torque of 22 . The speed stabilises at 1273 , somewhat less than the rated speed of 1350 . The electromagnetic parameters are shown in Table 2.\n\nBy the power models of the pumps, the energy consumption was simulated (Figure 13); because of the variable water volume technology, the energy savings were calculated (Figure 14). It is obvious that, in the first 1000 s, less energy savings is realised because of the cooling demand; however, after 1000 s, large energy savings are shown according to the stabilised inside temperature. None of the switchable pumps operates after approximately =1000 s; thus, the power consumption remains near 2 KWh. This is the same as the variable-frequency pump, which has a favourable adjustment due to the variable water flow. Figures 8 and 10 show that, even though some interference causes volume changes, the variable-frequency pump adjusts the water flow volume to play the roles of accuracy control and energy saving. After 5000 s, the total energy consumption of the pumps is approximately 8.6 KWh.\n\nThe performance of systems controlled by hybrid dynamic control strategy and constant water volume technology (all the pumps work at rated power and the chilled water system works without using our control strategy), respectively, is shown in Figure 14. For constant water volume control technology, the total volume is always at 0.097 kg/s, the total power is always at 10.7 KW, and the whole energy consumption is up to approximately 17 KWh. Compared with the results of the system controlled by the hybrid dynamic control strategy, it can be found that the energy saving potential of the pumps has not been excavated in the constant water volume technique. In contrast, when the low cooling load level of the system is low, the pumps may reduce frequency or quit working for energy saving as presented in Figures 10 and 13. Consequently, the total energy consumption of the system control by the hybrid dynamical control method is approximately 8.6 KWh. The system exhibits 49.4% energy saving in 5000 s compared with that controlled by constant water volume scheme.\n\n#### 6. Conclusions\n\nA hybrid dynamical approach to modelling and control is used in central air conditioning, which is a suitable approach for energy saving reconstruction in contemporary China. By analysing the cooling capacity transfer process from the chilled water system to the air system, the framework is built. By involving a direct torque control measure, the electromagnetic property is considered. For optimal hybrid control, the switching control strategy is developed; the cooling capacity allocation and average temperature methods are presented. On the basis of mathematical power models of pumps, the goal of accurate control and energy saving is realised. By simulation, a numeral example shows a hybrid control process and situation of energy savings. The approach saves 49.4% of water pump power consumption when compared with constant water volume technology. The modelling and control of multihybrid systems and their applications in refrigeration systems will be investigated in future works.\n\n#### Nomenclature\n\n : Current : Amount : Mass : Speed : Number of pole-pairs : Flow volume : Resistance : Voltage : Area : Specific heat : Thermal capacity F: Factor of cooling capacity allocation G: Amount : Constant : Rotating inertia : Feedback coefficient (a room) : Feedback coefficient (total system) : Heat transfer coefficient : Power : Heat : Returned air rate : Switch status signal : Electromagnetic torque : Load torque\nGreek symbols\n : Heat transfer efficiency : Factor : Temperature difference : Temperature : Surface temperature : Flux linkage : Angular speed\nSubscript\n 0: Rated : Air : Variable-frequency pump : Fresh air : Filter time : Fixed-frequency pump : Initial : Integral time : Surface (wall space, window, and roof) : Latent heat : Random : Return : Stator : Sending air : Setting value : Switchable pump : M-axis of stator flux-oriented : T-axis of stator flux-oriented : Water : Total xr: Sensible heat DC: Busbar : Flux linkage.\n\n#### Data Availability\n\nNo data were used to support this study.\n\n#### Conflicts of Interest\n\nThe authors declare that they have no conflicts of interest.\n\n#### Acknowledgments\n\nThe research was supported by the National Natural Science Foundation of China (no. 40761104181)."
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https://discuss.pytorch.org/t/different-results-using-transforms-and-albumentations/179538
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"# Different Results using Transforms and Albumentations\n\nHello everyone!\n\nI noticed I was getting different results during training when using transforms once and albumentations on the other.\n\nSo I decided to remove all kind of transformations I’m doing on the images, and just resize and normalize using Albumentations once and Transforms as the other print the tensors.\n\nSurpsingly they were different, however I do not understand why I’m using same interpolation and normalization values?\n\nFor example here is one row from the first channel,\n\n• Albumentations= [-1.4843, -1.4843, -1.5357, -1.5528, -1.5357, -1.4672, -1.5357, -1.5014,\n-1.4843, -1.5185]\n\n• torchvision transforms: [-1.4843, -1.5014, -1.5185, -1.5528, -1.5357, -1.4843, -1.5185, -1.5014,\n-1.4843, -1.5185]\n\nEDIT: So I tested with resizing only, with same interpolation for each, and it appears that image and mask tensors using Albumentations are not exactly the same as when using Transforms. Secondly, I removed resizing and kept normalization, since the masks are not normalized, they are the same using both methods, but the images are still slightly off using both methods. Could the difference be because transforms uses PIL and Albumentations uses cv2?\n\nHere is the code I used to test.\n\n``````import albumentations as A\nimport cv2\nfrom PIL import Image\nimport numpy as np\nfrom torchvision import transforms\n\nimg_path = \"./data/oxford-iiit-pet/images/Abyssinian_2.jpg\"\nseg_path = \"./data/oxford-iiit-pet/annotations/trimaps/Abyssinian_2.png\"\n\ntf = transforms.ToTensor()\n\nO_img = Image.open(img_path).convert(\"RGB\")\n\ntemp_img = np.array(O_img)\nalbum_t = A.Compose([A.Resize(256, 256,interpolation=cv2.INTER_LINEAR,always_apply=True),\nA.Normalize(mean=[0.485, 0.456, 0.406],\nstd=[0.229, 0.224, 0.225],always_apply=True)])\npreprocess_A = album_t(image=temp_img)\nAlbum_img = preprocess_A[\"image\"]\nA_img = tf(Album_img )\n\nimg_transform = transforms.Compose([ transforms.Resize((256, 256)),\ntransforms.ToTensor(),\ntransforms.Normalize(mean=[0.485, 0.456, 0.406],\nstd=[0.229, 0.224, 0.225])])\n\ntf_img = img_transform(O_img)\n\nprint(A_img[0,1,0:10])\nprint(tf_img[0,1,0:10])\n``````"
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http://www.spicereport.com/macrs-depreciation-schedule-excel/
|
[
"# macrs depreciation schedule excel\n\nA macrs depreciation schedule excel template is a type of document that creates a copy of itself when you open it. This copy has all of the design and formatting of the macrs depreciation schedule excel sample, such as logos and tables, but you can modify it by entering content without altering the original macrs depreciation schedule excel example. When designing macrs depreciation schedule excel, you may add related information such as macrs depreciation calculator rental property, macrs depreciation table 2014 excel, fixed asset depreciation excel spreadsheet, depreciation table excel.\n\nmacrs property depreciation. this template is designed to help accountants and other financial professionals calculate tax depreciation for assets that fall under modified accelerated cost recovery system (macrs) rules. the template calculates both 5-year depreciation and 7-year depreciation. in part 1 (depreciation schedule) and part 2 (macrs depreciation rate) of this depreciation in excel series, i discussed basic depreciation methods for financial reporting as well as how to calculate the depreciation rate used in macrs tables., how to calculate the depreciation rate for macrs tables using excel depreciation formulas., 13, depreciation method, sl, db factor, 200%., 16, year, depreciation, cumulative, book value., this template is designed to help accountants and other financial professionals calculate tax depreciation for assets that fall under modified accelerated cost recovery system (macrs) rules., the template calculates both 5-year depreciation and 7-year depreciation., macrs depreciation calculator rental property , macrs depreciation calculator rental property, macrs depreciation table 2014 excel , macrs depreciation table 2014 excel, fixed asset depreciation excel spreadsheet , fixed asset depreciation excel spreadsheet, depreciation table excel , depreciation table excel\n\nnot applicable to macrs for tax reporting., 5, edit the information in columns b through g. see the link above for more information about using this template., 7, depreciation for year … 8, asset, price, year, free modified accelerated cost recovery system (macrs depreciation) this spreadsheet includes table a-1 to table a-9 from the appendix a of the document “how to depreciate property (for use in preparing 2008 returns)” from united states irs (http://)., the tables excel 2007, 2010, 2013 or 2016, an stated in the text, macrs is the most important of the historical depreciation methods and so only that one will be presented here., in excel, the function of interest is vdb for variable declining balance., the result should be \\$ which corresponds to the 7-year property column for year 2 in the macrs table., macrs depreciation calculator rental property, macrs depreciation table 2014 excel, fixed asset depreciation excel spreadsheet, depreciation table excel, free depreciation calculator, online macrs depreciation calculator, macrs depreciation table excel 2016, monthly depreciation formula in excel, free depreciation calculator , free depreciation calculator, online macrs depreciation calculator , online macrs depreciation calculator, macrs depreciation table excel 2016 , macrs depreciation table excel 2016, monthly depreciation formula in excel , monthly depreciation formula in excel\n\nA macrs depreciation schedule excel Word template can contain formatting, styles, boilerplate text, macros, headers and footers, as well as custom dictionaries, toolbars and AutoText entries. It is important to define styles beforehand in the sample document as styles define the appearance of text elements throughout your document and styles allow for quick changes throughout your macrs depreciation schedule excel document. When designing macrs depreciation schedule excel, you may add related content, free depreciation calculator, online macrs depreciation calculator, macrs depreciation table excel 2016, monthly depreciation formula in excel"
] |
[
null
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{"ft_lang_label":"__label__en","ft_lang_prob":0.74372447,"math_prob":0.83565015,"size":3959,"snap":"2021-21-2021-25","text_gpt3_token_len":750,"char_repetition_ratio":0.29127687,"word_repetition_ratio":0.2597865,"special_character_ratio":0.19146249,"punctuation_ratio":0.14176829,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9961734,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-12T14:33:56Z\",\"WARC-Record-ID\":\"<urn:uuid:2b7bd83c-e667-471e-a7c6-18ced36f769f>\",\"Content-Length\":\"14796\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:34bb8a9e-2f13-4914-b08d-e93be86b5f32>\",\"WARC-Concurrent-To\":\"<urn:uuid:b8d5babd-b741-4198-9daa-4a9004b9652c>\",\"WARC-IP-Address\":\"45.32.225.108\",\"WARC-Target-URI\":\"http://www.spicereport.com/macrs-depreciation-schedule-excel/\",\"WARC-Payload-Digest\":\"sha1:335MEGTO6WDZSBSNBL3EQR2PTZ5HBHCI\",\"WARC-Block-Digest\":\"sha1:RKY6J2GO2YOTTPI4AYDB2F547GTFUHSC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487584018.1_warc_CC-MAIN-20210612132637-20210612162637-00205.warc.gz\"}"}
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https://metanumbers.com/20006
|
[
"## 20006\n\n20,006 (twenty thousand six) is an even five-digits composite number following 20005 and preceding 20007. In scientific notation, it is written as 2.0006 × 104. The sum of its digits is 8. It has a total of 3 prime factors and 8 positive divisors. There are 8,568 positive integers (up to 20006) that are relatively prime to 20006.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Even\n• Number length 5\n• Sum of Digits 8\n• Digital Root 8\n\n## Name\n\nShort name 20 thousand 6 twenty thousand six\n\n## Notation\n\nScientific notation 2.0006 × 104 20.006 × 103\n\n## Prime Factorization of 20006\n\nPrime Factorization 2 × 7 × 1429\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 20006 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) 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 20,006 is 2 × 7 × 1429. Since it has a total of 3 prime factors, 20,006 is a composite number.\n\n## Divisors of 20006\n\n1, 2, 7, 14, 1429, 2858, 10003, 20006\n\n8 divisors\n\n Even divisors 4 4 2 2\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 34320 Sum of all the positive divisors of n s(n) 14314 Sum of the proper positive divisors of n A(n) 4290 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 141.443 Returns the nth root of the product of n divisors H(n) 4.6634 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 20,006 can be divided by 8 positive divisors (out of which 4 are even, and 4 are odd). The sum of these divisors (counting 20,006) is 34,320, the average is 4,290.\n\n## Other Arithmetic Functions (n = 20006)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 8568 Total number of positive integers not greater than n that are coprime to n λ(n) 1428 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 2267 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 8,568 positive integers (less than 20,006) that are coprime with 20,006. And there are approximately 2,267 prime numbers less than or equal to 20,006.\n\n## Divisibility of 20006\n\n m n mod m 2 3 4 5 6 7 8 9 0 2 2 1 2 0 6 8\n\nThe number 20,006 is divisible by 2 and 7.\n\n• Arithmetic\n• Deficient\n\n• Polite\n\n• Square Free\n\n• Sphenic\n\n## Base conversion (20006)\n\nBase System Value\n2 Binary 100111000100110\n3 Ternary 1000102222\n4 Quaternary 10320212\n5 Quinary 1120011\n6 Senary 232342\n8 Octal 47046\n10 Decimal 20006\n12 Duodecimal b6b2\n20 Vigesimal 2a06\n36 Base36 ffq\n\n## Basic calculations (n = 20006)\n\n### Multiplication\n\nn×i\n n×2 40012 60018 80024 100030\n\n### Division\n\nni\n n⁄2 10003 6668.67 5001.5 4001.2\n\n### Exponentiation\n\nni\n n2 400240036 8007202160216 160192086417281296 3204802880864129607776\n\n### Nth Root\n\ni√n\n 2√n 141.443 27.1469 11.893 7.24823\n\n## 20006 as geometric shapes\n\n### Circle\n\n Diameter 40012 125701 1.25739e+09\n\n### Sphere\n\n Volume 3.35405e+13 5.02956e+09 125701\n\n### Square\n\nLength = n\n Perimeter 80024 4.0024e+08 28292.8\n\n### Cube\n\nLength = n\n Surface area 2.40144e+09 8.0072e+12 34651.4\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 60018 1.73309e+08 17325.7\n\n### Triangular Pyramid\n\nLength = n\n Surface area 6.93236e+08 9.43658e+11 16334.8\n\n## Cryptographic Hash Functions\n\nmd5 2ab8f86410b4f3bdcc747699295eb5a4 5fffae010b4da02aaf0c1a83a9bc8b30ff819fd3 cc3fefcee849473f708a87651606c36d3e4a6a9caab4e8ce974623cd742533c1 a3e3f26beef5fe9e0df2c8f2271de44fe2e06b43659ba626a419c36d41a30ab7fddfafc21161440c8bf7963cd8f6fe08f8ec817e255c78026520271e41110b38 5b72b3060664656cb7a10c92ec8576aced66d863"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.6247088,"math_prob":0.98224616,"size":4451,"snap":"2020-34-2020-40","text_gpt3_token_len":1590,"char_repetition_ratio":0.120080955,"word_repetition_ratio":0.02827381,"special_character_ratio":0.4502359,"punctuation_ratio":0.075718015,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99558556,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-29T10:08:57Z\",\"WARC-Record-ID\":\"<urn:uuid:932a1099-6f1e-4cb5-b7ea-c7b20f747e19>\",\"Content-Length\":\"48189\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d40c474c-049c-4932-b6b3-a984e2b13fe0>\",\"WARC-Concurrent-To\":\"<urn:uuid:dfaffce3-693a-465f-a7df-4b02f84382fb>\",\"WARC-IP-Address\":\"46.105.53.190\",\"WARC-Target-URI\":\"https://metanumbers.com/20006\",\"WARC-Payload-Digest\":\"sha1:NOTIHT7SFTPLYJ2ZBPERHDTS4GBQVPVH\",\"WARC-Block-Digest\":\"sha1:43GZWMLHGWRUXWMAEFRZGW6GZ775SBEQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600401641638.83_warc_CC-MAIN-20200929091913-20200929121913-00785.warc.gz\"}"}
|
https://socratic.org/questions/how-do-you-solve-2x-5x-3
|
[
"# How do you solve 2x² + 5x = -3?\n\nApr 14, 2016\n\nThe solutions are:\n$x = - 1$\n\n$x = - \\frac{3}{2}$\n\n#### Explanation:\n\n$2 {x}^{2} + 5 x = - 3$\n\n$2 {x}^{2} + 5 x + 3 = 0$\n\nThe equation is of the form color(blue)(ax^2+bx+c=0 where:\n\n$a = 2 , b = 5 , c = 3$\n\nThe Discriminant is given by:\n\n$\\Delta = {b}^{2} - 4 \\cdot a \\cdot c$\n\n$= {\\left(5\\right)}^{2} - \\left(4 \\cdot 2 \\cdot 3\\right)$\n\n$= 25 - 24 = 1$\n\nThe solutions are normally found using the formula\n$x = \\frac{- b \\pm \\sqrt{\\Delta}}{2 \\cdot a}$\n\n$x = \\left(\\frac{- 5 \\pm \\sqrt{1}}{2 \\cdot 2}\\right) = \\frac{- 5 \\pm 1}{4}$\n\nThe solutions are:\n$x = \\frac{- 5 + 1}{4} = - \\frac{4}{4} = - 1$\n\n$x = \\frac{- 5 - 1}{4} = - \\frac{6}{4} = - \\frac{3}{2}$"
] |
[
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{"ft_lang_label":"__label__en","ft_lang_prob":0.7419045,"math_prob":1.00001,"size":342,"snap":"2020-34-2020-40","text_gpt3_token_len":86,"char_repetition_ratio":0.14497042,"word_repetition_ratio":0.0,"special_character_ratio":0.24561404,"punctuation_ratio":0.09375,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000079,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-06T01:49:09Z\",\"WARC-Record-ID\":\"<urn:uuid:d5f07b19-a56b-4c2d-a22c-f7b6121c38af>\",\"Content-Length\":\"33482\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3243fdf2-6273-45c6-93ac-4fa439b00565>\",\"WARC-Concurrent-To\":\"<urn:uuid:f7395303-ce8a-4164-a0d6-031c82bd01d2>\",\"WARC-IP-Address\":\"216.239.34.21\",\"WARC-Target-URI\":\"https://socratic.org/questions/how-do-you-solve-2x-5x-3\",\"WARC-Payload-Digest\":\"sha1:6KD6Q57MBD5Q3IYBPEQSBXMNLAHD2ZR2\",\"WARC-Block-Digest\":\"sha1:4TOYLNY2U3NGWU2CRJSMBMP5KPRVLXDC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439735990.92_warc_CC-MAIN-20200806001745-20200806031745-00410.warc.gz\"}"}
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https://getcalc.com/math-fraction-7by9-minus-2by5.htm
|
[
"# What is 7/9 - 2/5?",
null,
"getcalc.com's two fractions subtraction calculator is an online basic math function tool to find what's the equivalent fraction for difference between two fractional numbers with different, unequal or unlike denominators 7/9 and 2/5.\n\n7/9 - 2/5 = 17/45 in fraction form\n\n7/9 - 2/5 = 0.3778 in decimal form\nThis calculator, formula, step by step calculation and associated information to find the difference between 7/9 and 2/5 may help students, teachers, parents or professionals to learn, teach, practice or verify such two fractions subtraction calculations efficiently.\n\n## How to Find Equivalent Fraction for 7/9-2/5?\n\nThe below workout with step by step calculation shows how to find what is the difference between two fractions by subtracting a fraction 2/5 from 7/9. For different, unequal or unlike denominators, use any one of the following methods to find the equivalent fraction for 7/9 minus 2/5.\n1. LCM method\n2. Cross multiplication method.\n\nProblem and Workout - LCM Method\nWhat is 7/9 minus 2/5 as a fraction by subtraction using LCM method?\n\nstep 1 Address formula, input parameters and values.\nInput parameters and values:\n7/9 & 2/5\n7/9-2/5 = ?\n\nstep 2 For unequal denominators, find the LCM (least common multiple) for both denominators\nThe LCM for 9 and 5 is 45.\n\nstep 3 To have the common denominator, multiply the LCM with both numerator and denominator\n=(7 x 45)/(9 x 45)-(2 x 45)/(5 x 45)\n=35/45-18/45\n=(35 - 18)/45\n\nstep 4 Find difference between numerators and rewrite it in a single form.\n=17/45\n7/9-2/5=17/45\n\nBy using LCM method, 17/45 is the equivalent fraction by subtracting 2/5 from 7/9.\n\nProblem and Workout - Cross Multiplication Method\nstep 1 Address formula and input values.\nInput values:\n7/9 & 2/5\n7/9 - 2/5 = ?\n\nstep 2 Cross multiply both numerator and denominator, multiply both denominators and rewrite as below\n= (7 x 5) - (2 x 9)/(9 x 5)\n\nstep 3 Simplify and rewrite the fraction\n= (35 - 18)/45\n7/9-2/5 = 17/45\nBy using cross multiplication method, 17/45 is the difference between two fractions 7/9 and 2/5.",
null,
""
] |
[
null,
"https://getcalc.com/calculator-images/math/fraction-arithmetic-calculator.png",
null,
"https://d7kr1t5cwstl7.cloudfront.net/graphics/getcalc-logo.png",
null
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https://www.colorhexa.com/0b100c
|
[
"# #0b100c Color Information\n\nIn a RGB color space, hex #0b100c is composed of 4.3% red, 6.3% green and 4.7% blue. Whereas in a CMYK color space, it is composed of 31.3% cyan, 0% magenta, 25% yellow and 93.7% black. It has a hue angle of 132 degrees, a saturation of 18.5% and a lightness of 5.3%. #0b100c color hex could be obtained by blending #162018 with #000000. Closest websafe color is: #000000.\n\n• R 4\n• G 6\n• B 5\nRGB color chart\n• C 31\n• M 0\n• Y 25\n• K 94\nCMYK color chart\n\n#0b100c color description : Very dark (mostly black) lime green.\n\n# #0b100c Color Conversion\n\nThe hexadecimal color #0b100c has RGB values of R:11, G:16, B:12 and CMYK values of C:0.31, M:0, Y:0.25, K:0.94. Its decimal value is 725004.\n\nHex triplet RGB Decimal 0b100c `#0b100c` 11, 16, 12 `rgb(11,16,12)` 4.3, 6.3, 4.7 `rgb(4.3%,6.3%,4.7%)` 31, 0, 25, 94 132°, 18.5, 5.3 `hsl(132,18.5%,5.3%)` 132°, 31.3, 6.3 000000 `#000000`\nCIE-LAB 4.23, -2.27, 1.319 0.39, 0.468, 0.418 0.305, 0.367, 0.468 4.23, 2.626, 149.85 4.23, -0.99, 0.987 6.843, -1.811, 1.171 00001011, 00010000, 00001100\n\n# Color Schemes with #0b100c\n\n• #0b100c\n``#0b100c` `rgb(11,16,12)``\n• #100b0f\n``#100b0f` `rgb(16,11,15)``\nComplementary Color\n• #0d100b\n``#0d100b` `rgb(13,16,11)``\n• #0b100c\n``#0b100c` `rgb(11,16,12)``\n• #0b100f\n``#0b100f` `rgb(11,16,15)``\nAnalogous Color\n• #100b0d\n``#100b0d` `rgb(16,11,13)``\n• #0b100c\n``#0b100c` `rgb(11,16,12)``\n• #0f0b10\n``#0f0b10` `rgb(15,11,16)``\nSplit Complementary Color\n• #100c0b\n``#100c0b` `rgb(16,12,11)``\n• #0b100c\n``#0b100c` `rgb(11,16,12)``\n• #0c0b10\n``#0c0b10` `rgb(12,11,16)``\n• #0f100b\n``#0f100b` `rgb(15,16,11)``\n• #0b100c\n``#0b100c` `rgb(11,16,12)``\n• #0c0b10\n``#0c0b10` `rgb(12,11,16)``\n• #100b0f\n``#100b0f` `rgb(16,11,15)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #010101\n``#010101` `rgb(1,1,1)``\n• #0b100c\n``#0b100c` `rgb(11,16,12)``\n• #151f17\n``#151f17` `rgb(21,31,23)``\n• #202e23\n``#202e23` `rgb(32,46,35)``\n• #2a3d2e\n``#2a3d2e` `rgb(42,61,46)``\nMonochromatic Color\n\n# Alternatives to #0b100c\n\nBelow, you can see some colors close to #0b100c. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #0b100b\n``#0b100b` `rgb(11,16,11)``\n• #0b100b\n``#0b100b` `rgb(11,16,11)``\n• #0b100c\n``#0b100c` `rgb(11,16,12)``\n• #0b100c\n``#0b100c` `rgb(11,16,12)``\n• #0b100c\n``#0b100c` `rgb(11,16,12)``\n• #0b100d\n``#0b100d` `rgb(11,16,13)``\n• #0b100d\n``#0b100d` `rgb(11,16,13)``\nSimilar Colors\n\n# #0b100c Preview\n\nThis text has a font color of #0b100c.\n\n``<span style=\"color:#0b100c;\">Text here</span>``\n#0b100c background color\n\nThis paragraph has a background color of #0b100c.\n\n``<p style=\"background-color:#0b100c;\">Content here</p>``\n#0b100c border color\n\nThis element has a border color of #0b100c.\n\n``<div style=\"border:1px solid #0b100c;\">Content here</div>``\nCSS codes\n``.text {color:#0b100c;}``\n``.background {background-color:#0b100c;}``\n``.border {border:1px solid #0b100c;}``\n\n# Shades and Tints of #0b100c\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, #030403 is the darkest color, while #f8faf8 is the lightest one.\n\n• #030403\n``#030403` `rgb(3,4,3)``\n• #0b100c\n``#0b100c` `rgb(11,16,12)``\n• #131c15\n``#131c15` `rgb(19,28,21)``\n• #1b271d\n``#1b271d` `rgb(27,39,29)``\n• #233326\n``#233326` `rgb(35,51,38)``\n• #2b3e2f\n``#2b3e2f` `rgb(43,62,47)``\n• #334a38\n``#334a38` `rgb(51,74,56)``\n• #3b5640\n``#3b5640` `rgb(59,86,64)``\n• #436149\n``#436149` `rgb(67,97,73)``\n• #4b6d52\n``#4b6d52` `rgb(75,109,82)``\n• #53795a\n``#53795a` `rgb(83,121,90)``\n• #5b8463\n``#5b8463` `rgb(91,132,99)``\n• #63906c\n``#63906c` `rgb(99,144,108)``\n• #6c9a75\n``#6c9a75` `rgb(108,154,117)``\n• #78a280\n``#78a280` `rgb(120,162,128)``\n• #84aa8b\n``#84aa8b` `rgb(132,170,139)``\n• #8fb296\n``#8fb296` `rgb(143,178,150)``\n• #9bbaa1\n``#9bbaa1` `rgb(155,186,161)``\n• #a6c2ac\n``#a6c2ac` `rgb(166,194,172)``\n• #b2cab7\n``#b2cab7` `rgb(178,202,183)``\n• #bed2c2\n``#bed2c2` `rgb(190,210,194)``\n• #c9dacd\n``#c9dacd` `rgb(201,218,205)``\n• #d5e2d8\n``#d5e2d8` `rgb(213,226,216)``\n• #e1eae2\n``#e1eae2` `rgb(225,234,226)``\n• #ecf2ed\n``#ecf2ed` `rgb(236,242,237)``\n• #f8faf8\n``#f8faf8` `rgb(248,250,248)``\nTint Color Variation\n\n# Tones of #0b100c\n\nA tone is produced by adding gray to any pure hue. In this case, #0d0e0d is the less saturated color, while #011a06 is the most saturated one.\n\n• #0d0e0d\n``#0d0e0d` `rgb(13,14,13)``\n• #0c0f0d\n``#0c0f0d` `rgb(12,15,13)``\n• #0b100c\n``#0b100c` `rgb(11,16,12)``\n• #0a110b\n``#0a110b` `rgb(10,17,11)``\n• #09120b\n``#09120b` `rgb(9,18,11)``\n• #08130a\n``#08130a` `rgb(8,19,10)``\n• #07140a\n``#07140a` `rgb(7,20,10)``\n• #061509\n``#061509` `rgb(6,21,9)``\n• #051608\n``#051608` `rgb(5,22,8)``\n• #041708\n``#041708` `rgb(4,23,8)``\n• #031807\n``#031807` `rgb(3,24,7)``\n• #021906\n``#021906` `rgb(2,25,6)``\n• #011a06\n``#011a06` `rgb(1,26,6)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #0b100c 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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https://studyres.com/doc/66293/3.-measuring-distances-and-magnitudes
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"Document related concepts\n\nGravitational lens wikipedia , lookup\n\nPlanetary nebula wikipedia , lookup\n\nStandard solar model wikipedia , lookup\n\nHayashi track wikipedia , lookup\n\nStellar evolution wikipedia , lookup\n\nMain sequence wikipedia , lookup\n\nStar formation wikipedia , lookup\n\nAstronomical spectroscopy wikipedia , lookup\n\nCosmic distance ladder wikipedia , lookup\n\nTranscript\n```PH507\nAstrophysics\nDr Dirk Froebrich\n1\nSYLLABUS:\n• Part 1: stars and stellar structure (Dr Miao)\n• Part 2: measurements (distance brightness, mass)\n• Part 3: extrasolar planets (detection, properties)\n• Part 4: star and planet formation\n• Part 5: radiation, ( the Sun)\n• Part 6: telescopes, detectors, instruments\nAssessment Methods:\nExamination 70%, class tests 15% each (best 2 of 3 marks)\n6 assignments.\nClass Test 1: week 16\nClass Test 2: week 20\nClass Test 3: week 24\nHomeworks due: data/time on assignment sheets\nRecommended Texts:\nCarroll & Ostlie, An Introduction to Modern Astrophysics, AddisonWesley, second edition if possible\nTHESE LECTURE NOTES, posted at the beginning of each week on my\nwebpage\nhttp://astro.kent/ac/uk/~df/teaching/ph507\n[Note: Changes may occur to the syllabus during the year]\nConvenor: Dr Dirk Froebrich:\n117 Ingram, x7346,[email protected]\nPH507\nAstrophysics\nDr Dirk Froebrich\n2\nPART 1: Measurements\nDISTANCES\nDistance: Distance is an easy concept to understand: it is just a length in\nsome units such as in feet, km, light years, parsecs etc. It has been\nexcruciatingly difficult to measure astronomical distances until this\ncentury.\nAstronomical Unit: 1AU is the distance from the centre of the Sun at\nwhich a particle of negligible mass, in an unperturbed circular orbit, would\nhave an orbital period of 365.2568983 days (one Gaussian year).\n1AU=149.597.870.691±30m\nMeasured today by radar-distances to nearby planets, asteroids, their\nperiods around the Sun and applying the 3rd of Keplers Laws: t2/a3=const\n(see later in course…)\nDefined by IAU in 2012 as 149.597.870.700m exact!\nUnfortunately most stars are so far away that it is impossible to directly\nmeasure the distance using the classic technique of triangulation.\nTrigonometric parallax: based on triangulation – need three parameters to\nfully define any triangle; e.g. two angles and one baseline.\nTo triangulate to even the closest stars we would need to use a very large\nbaseline. In fact we do have a long baseline, because every 6 months the\nearth is on opposite sides of the Sun. So we can use as a baseline the major\naxis of the earth's orbit around the Sun.\nBASELINE: 2 x Earth-Sun distance = 2 Astronomical Units (AU)\nExamples:\n1 AU:\n6.372,797 km\n695.500\nkm\n149.597.871\nkm\n(1.496 x 10 11 m)\nThe parallactic displacement of a star on the sky as a result of the Earth’s\norbital motion permits us to determine the distance from the Sun to the star\nby the method of trigonometric (heliocentric) parallax. We define the\ntrigonometric parallax of the star as the half angle p subtended, as seen\nfrom the star, by the Earth’s orbit of radius 1 AU. If the star is at rest with\nrespect to the Sun, the parallax is half the maximum apparent annual\nangular displacement of the star as seen from the Earth.\nPH507\nAstrophysics\nDr Dirk Froebrich\n3\n360\n 57.3 degrees = 206265 arc seconds.\n2\nThere are 2rad in a circle (360˚), so that 1rad equals 57˚17´44.81” (206,\n264.81”).\nTherefore, the solar disc subtends an angle of:\nIndependent distance unit is the light year [ly]:\nc * t(year) = 9.47 * 10 15m\nThe light year is not used much by professional astronomers, who work\ninstead with the unit of similar size called the parsec, where\n1parsec = 1pc = 206.265AU = 3.086 x 1016m = 3.26ly.\nThe measurement and interpretation of stellar parallaxes is a branch of\nastrometry, and the work is exacting and time-consuming. Consider that\nthe nearest stellar system, Proxima/Alpha Centauri AB (Rigil Kentaurus), at\na distance of 1.29pc, has a parallax of only 0.772”; all other stars have\nsmaller parallaxes!\nPH507\nAstrophysics\nDr Dirk Froebrich\n4\nRemember diffraction limit for circular apperture: dr[rad] = 1.22 * / D\n in the optical 1” resolution corresponds to an 11.5cm diameter telescope\ntan p=\n1 AU d\nor d≈\n1\nAU\np\nwhere p is in rad. For small angles (p<<1):\nsin(p)=p –p3/3!+p5/5!... and cos(p)=1-x 2/2!+x4/4!... and hence\ntan(p)=sin(p)=p\nTo convert to arcseconds:\nd≈\n1\npc\np''\nor\nd≈\n206300\nAU\n.\np''\nDefines the parsec – parallax second abbreviation!\nTechnique:\nThe ground-based trigonometric parallax of a star is determined by taking\nphotographs of a given star field from a number (about 20) of selected\npoints in the Earth’s orbit at different times of the year. The comparison\nstars selected are distant background stars of nearly the same apparent\nbrightness as the star whose parallax is being measured.\nCorrections are made for atmospheric refraction and dispersion and for\ndetectable motions of the background stars; any motion of the star relative\nto the Sun is then extracted. What remains is the smaller annual parallactic\nmotion; it is recognized because it cycles annually.\nBecause a seeing resolution of 0.25” is considered exceptional (more\ntypical it is 1”), it may seem strange that a stellar position can be\ndetermined to ±0.01” in one measurement; this accuracy is possible because\nPH507\nAstrophysics\nDr Dirk Froebrich\n5\nwe are determining the centre of a Gaussian stellar image.\nAstrometry: Technological advances (including the Hubble Space\nTelescope) have improved parallax accuracy to 0.001” within a few years.\nPrior to 1990, less than 10,000 stellar parallaxes had been measured (and\nonly 500 known well), but there are about 1012 stars in our Galaxy.\nSpace observations made by the European Space Agency with the\nHipparcos mission (1989-1993) accurately determined the parallaxes of\nmany more stars. Though a poor orbit limited its usefulness, Hipparcos was\nexpected to achieve a precision of about 0.002”. It actually achieved 0.001”\nfor 118,000 stars. The method of trigonometric parallax is important\nbecause it is our only direct distance measurement technique for stars.\nRadio interferometry on Masers has been used to determine directly\nthe distance to the Galactic Center via the parallax method\n(d=7.2kpc).\nFuture:\nLaunched in December 2013, Gaia will be measuring positions, proper\nmotions, brightness and parallaxes for objects with a limiting magnitude\nV=20mag (1billion stars); accuracy 20µas at V=15mag and 7µas at\nV=10mag (hair at 1000km); over a 5yr period.\nPlus radial velocities with 500m/s accuracy for 40 million stars\nSIM: Space Interferometer Mission from the US – pointed deep\nobservations down to V=20mag. It will be able to measure parallaxes of\n10micro-arcseconds. It consists of a rotating frame holding three\ntelescopes. Some aims:\n- Accurate distances even to the Galactic centre 8000pc away.\nCanceled in 2010!\nHowever, the Parallax remains limited to relatively nearby objects. To go\nfurther, we construct the COSMIC DISTANCE LADDER. If we can\nestimate the luminosity of a star from other properties, they can be used as\nSTANDARD CANDLES (L r-2 … see below)\nExample: Sirius A (SpT: A1, V=-1.44mag) has an observed parallax of\n0.37921”. Distance: 1/0.37921=2.637pc.\nPH507\nAstrophysics\nDr Dirk Froebrich\n6\n2 LUMINOSITY\nWe can actually only measure the radiant flux of a flame and need to make\na few assumptions to find the true luminosity. Luminosity depends on the\ndistance and extinction (as well as relativistic effects).\nThe measured flux f is in units of [W/m2], the flow of energy per unit area.\nThe radiated power L (Luminosity, [W]), ignoring extinction, is given by\nan inverse square law:\nL\n4πd2\nL\nd 2=\n4πf\nf=\nshowing that a standard candle can yield the distance.\nThe Stellar Magnitude Scale\nThe first stellar brightness scale - the magnitude scale - was defined by\nHipparchus of Nicaea and refined by Ptolemy almost 2000 years ago. In\nthis qualitative scheme, naked-eye stars fall into six categories: the\nbrightest are of first magnitude, and the faintest of sixth magnitude. Note:\nThe brighter a star, the smaller the value of its magnitude.\nSun: mV=-26.72mag; Sirius: mV=-1.46mag; Deneb: mV=1.25mag\nIn 1856, N. R. Pogson verified William Herschel’s finding that a firstmagnitude star is 100 times brighter than a sixth-magnitude star and the\nscale was quantified. Because an interval of five magnitudes corresponds to\na factor of 100 in brightness, a one-magnitude difference corresponds to\na factor of 1001/5 = 2.512.\nThis definition reflects the operation of human vision, which converts\nequal ratios of actual intensity to equal intervals of perceived intensity. In\nother words, the eye is a logarithmic detector.\nThe magnitude scale has been extended to positive magnitudes larger than\n+6.0mag to include faint stars (e.g. the 5m telescope on Mount Palomar can\nreach to magnitude +23.5mag) and to negative magnitudes for very bright\nobjects (the star Sirius is magnitude -1.4). The limiting magnitude of the\nHubble Space Telescope is about +30mag, the planned 40m E-ELT\nPH507\nAstrophysics\nDr Dirk Froebrich\n7\nAstronomers find it convenient to work with logarithms to base 10 rather\nthan with exponents in making the conversions from brightness ratios to\nmagnitudes and vice versa.\nConsider two stars s1 and s2 with a respective apparent brightness m1 and\nm2. The ratio of their fluxes f1/f2 corresponds to the magnitude\ndifference m1–m2. Because a one-magnitude difference means a brightness\nratio of 1001/5= 2.512, (m1 – m2) magnitudes refer to a ratio of (1001/5)(m1-m2) = 100-(m1-m2)/5, or\nf1/f2 = 100-(m1-m2)/5\nTaking the log10 of both sides and remember:\nlog(xa) = a*log(x) and log 10a = a log 10 = a,\nlog (f1/f2) = -[(m1 – m2)/5] log 100 = -0.4(m1 – m2)\nor\nm1 – m2 = -2.5 log (f1/f2)\nThis last equation defines the apparent magnitude; note that m1<m2\nwhen f1>f2, that is: brighter objects have numerically smaller\nmagnitudes!!!\nAlso note that when the brightness is observed at the Earth, physically they\nare fluxes. Apparent magnitude is the astronomically peculiar way of\nPH507\nAstrophysics\nDr Dirk Froebrich\n8\nHere are a few worked examples:\n(a) The apparent magnitude of the variable star RR Lyrae ranges from\n+7.1mag to +7.8mag – magnitude amplitude of 0.7mag. To find the relative\nincrease in brightness from minimum to maximum, we use:\nlog (fmax / fmin) = 0.4 x 0.7 = 0.28\nso that\nfmax / fmin = 100.28 = 1.91\nThis star is almost twice as bright at maximum light, than at minimum.\n(b) A binary system consists of two stars (a and b), with a brightness ratio\nof 2; however, we see them unresolved as a point of +5.0mag. We would\nlike to find the magnitude of each star. The magnitude difference is\nmb - ma = 2.5 log (fa / fb) = 2.5 log 2 = 0.75mag\nSince we are dealing with brightness ratios, it is not right to put\nma+mb=+5.0mag. The sum of the fluxes (fa + fb) corresponds to a fifthmagnitude star. Compare this to a 100-fold brighter star, of magnitude\nm0=0.0 and flux f0:\nma+b - m0 = 2.5 log [f0 / (fa + fb)]\nor\n5.0 - 0.0 = 2.5 log 100 = 5.\nBut\nfa = 2 fb, so that fb = (fa + fb)/3.\nTherefore:\n(mb-m0) = 2.5 log (f0/fb) = 2.5 log 300 = 2.5x2.477 = 6.19mag.\nThe magnitude of the fainter star is 6.19mag, and from our earlier result on\nthe magnitude difference, that of the brighter star is 5.44mag.\nWhat units are used in astronomical photometry?\nThe well-known magnitude scale of course, which has been calibrated\nusing standard stars which (hopefully) do not vary in brightness.\nBut how does the astronomical magnitude scale relate to other photometric\nunits? Here we assume V (visual) magnitudes, unless otherwise noted,\nwhich are at least approximately convertible to lumen (lm), candela (cd),\nand lux (lx). Candela is the SI base unit.\n1cd is the luminous intensity emitted in a particular direction of a 555nm\nPH507\nAstrophysics\nDr Dirk Froebrich\n9\nmonochromatic light with 1/683W/sr.\n1lx = 1lm/m2 = 1cd*sr/m2\nnon-SI units used sometime: 1phot=1lm/cm 2\nAn mv=0 star outside Earth's atmosphere is 2.54 * 10-6 lux\n= 2.54 * 10-10 phot\nLuminance: (1 nit = 1 candela per square meter\n1 stilb = 1 candela per square centimeter)\nOne mv= 0 star per square degree outside Earth's atmosphere\n= 0.84 * 10-2 nit\n= 8.4 * 10-7 stilb\nOne mv = 0 star per square degree inside clear unit airmass\n= 0.69 * 10-2 nit\n= 6.9 * 10-7 stilb\nOne star, Mv=0 outside Earth's atmosphere = 2.45* 1029 cd.\n3 Attenuation in the earth’s atmosphere\nLight incident on the Earth's atmosphere from an extraterrestrial source is\ndiminished by passage through the Earth's atmosphere. Thus, sources will\nalways appear less bright below the Earth's atmosphere than above it.\n1 clear unit airmass transmits 82% in the visual, i.e. it dims by 0.215mag.\nAir Mass is the path length that light from a celestial object takes through\nEarth’s atmosphere relative to the length at the zenith. The airmass is 1 at\nthe zenith and roughly 2 at an altitude of 30°. For small zenith angles (z) it\ncan be calculated simply as the secant of z.\nA = sec(z) = 1 / cos(z)\nIn a curved atmosphere the airmass is usually smaller than 40, and the\nabove formula does not apply. A good approximation is (z in degrees)\nA = 1 / (cos(z) + 0.50572*(96.07995-z)^-1.6364)\nNote that a unit airmass at Mauna Kea (with a mean barometric pressure of\n605 millibar – or at about 4500m above sea level) is equivalent to 0.60\nairmass at sea level.\nFrom a practical standpoint, you can see that, for example:\nAt z=60deg you look through 2 airmasses (obtainable from plane parallel\napproximation).\nAt z=71deg you look through 3 airmasses.\nAt the limit of z=90deg you look through 38 airmasses.\nPH507\nAstrophysics\nDr Dirk Froebrich\n10\nAtmospheric extinction at optical wavelengths is due primarily to two\nphenomena:\n1 Absorption:\nOn UV side primary absorption is Ozone (O3)\nOn IR side water vapor and CO2.\nAtmospheric extinction across the electromagnetic spectrum: Figure\nfrom Observational Astronomy by Lena, Lebrun and Mignard.\nAs may be seen, most EM radiation is blocked by the Earth, and only a few\natmospheric windows allow certain wavelengths through:\n▪ Optical\n▪ various near and mid-infrared ranges\n▪ increasing elevations open up new atmospheric windows\n▪ On high mountains and the Antarctic plateau (where the atmosphere is\ncompressed -- equivalent to higher altitude at warmer latitudes) more midand far- infrared wavelengths are accessible.\nPH507\n2\nAstrophysics\nDr Dirk Froebrich\n11\nScattering:\nTwo mechanisms depend on the size (a) of the scattering particle:\nMolecular scattering: scattering radius a << wavelength\n▪ Mainly Rayleigh scattering (elastic - energy of scattered photons is\npreserved), which has a scattering cross-section as function of wavelength.\nAerosol scattering: scattering radius a >~ 1/10 wavelength\n▪ Mie scattering is light scattered by \"large\" [relative to wavelength of light]\nspheres, and the strict theory accounts for Maxwell’s equations in the\ncontext of all kinds of reflections (external and internal) and surface waves\non the scattering particle, polarization, etc.\nThe sky seems more/deeper blue when you look at greater angles from the\nSun because that is mostly Rayleigh scattering, while close to the Sun the\nsky appears \"whiter\" because this is primarily Mie scattering.\n▪ Aerosols are highly variable from night to night.\n▪ air pollution\n▪ dust = \"haze\"\n▪ volcanic ash (can be horrible)\n▪ dust storms\nApparent Magnitude\nApparent magnitude is an irradiance or illuminance, i.e. incident flux per\nunit area, from all directions. Of course a star is a point light source, and\nthe incident light is only from one direction.\nApparent magnitude per square degree is a radiance, luminance, intensity,\nor \"specific intensity\". This is sometimes also called \"surface brightness\".\nStill another unit for intensity is magnitudes per square arcsec, which is\nthe magnitude at which each square arcsec of an extended light source\nPH507\nAstrophysics\nDr Dirk Froebrich\n12\nshines.\nOnly visual magnitudes can be converted to photometric units. U, B, R or I\nmagnitudes are not easily convertible to luxes, lumens etc., because of the\ndifferent wavelengths intervals used. The conversion factors would be\nstrongly dependent on e.g. the temperature of the blackbody radiation or,\nmore generally, the spectral distribution of the radiation. The conversion\nfactors between V magnitudes and photometric units are only slightly\ndependent on the spectral distribution of the radiation.\nHere we're not interested in the photometric response of some detector with\na well-known pass band (e.g. the human eye, or some astronomical\nphotometer). Instead we want to know the strength of the radiation in\nabsolute units: watts etc. Thus we have:\nW m-2 ster-1 [Å-1]\nSI unit\nerg cm-2 s-1 ster-1 [Å -1]\nCGS unit\n-2 -1\n-1\n-1\nphotons cm s ster [Å ] Photon flux, CGS units\n(1erg = 10-7 joule)\nW m-2 [Å-1]\nSI unit\n-2 -1\n-1\nerg cm s [Å ]\nCGS unit\n-2 -1\n-1\n-1\nphotons cm s ster [Å ] Photon flux, CGS units\nNote the [Å-1] within brackets. Fluxes and intensities can be total (summed\nover all wavelengths) or monochromatic (\"per Angstrom Å\" or \"per\nnanometer\").\nIn Radio/Infrared Astronomy, the unit Jansky is often used as a measure of\nequivalence to stellar magnitudes.\nThe Jansky is defined as: 1Jy = 10-26 W m-2 Hz-1.\n4. Absolute magnitude (represents a total flux)\nAbsolute Magnitude and Distance Modulus\nSo far we have dealt with stars as we see them, that is, their fluxes or\napparent magnitudes, but we want to know the intrinsic luminosity of a\nPH507\nAstrophysics\nDr Dirk Froebrich\n13\nstar. A very luminous star will appear dim if it is far enough away, and a\nlow-luminosity star may look bright if it is close enough.\nOur Sun is a case in point: if it were at the distance of the closest star\n(Alpha Centauri), the Sun would appear slightly fainter to us than Alpha\nCentauri does on Earth. Hence, distance links fluxes and luminosities.\nSun’s apparent magnitude: mV=–26.83mag\nThe solar irradiance is the amount of incoming solar radiation per unit area,\nmeasured on the outer surface of the Earth's atmosphere, in a plane\nperpendicular to the rays. The solar constant includes all types of solar\nradiation, not just the visible light. It is measured by satellite to be roughly\n1366Wm-2.\nThe luminosity of a star relates to its absolute magnitude, which is the\nmagnitude that would be observed if the star were placed at a distance of\n10 pc from the Sun, in the absence of interstellar extinction. By convention, absolute magnitude is capitalized (M) and apparent\nmagnitude is written lowercase (m). The inverse-square law of radiative\nflux links the flux f of a star at a distance d to the flux, F, it would have if\nit were at a distance d = 10 pc:\nF / f = (d / D) 2 = (d / 10) 2\nPH507\nAstrophysics\nDr Dirk Froebrich\n14\nIf M corresponds to F and m corresponds to the flux f, then\nm - M = 2.5 log (F / f ) = 2.5 log (d/10)2 = 5 log (d / 10)\nExpanding this expression, we have useful alternative forms. Since\nm1 – m2 = 5 log d1 - 5 log d2\ndefining the absolute magnitude m2 = M at d2 = 10 pc, so m1 = m and d2 =\nd,\nm - M = 5 log d - 5\nM = m + 5 - 5 log d\nIn terms of the parallax, using d[pc]=1/p[“]\nM = m + 5 + 5 log p”\nHere d is in parsecs and p” is the parallax angle in arc seconds.\nThe quantity m - M is called the distance modulus, for it is directly\nrelated to the star’s distance. In many applications, we refer only to the\ndistance modulus of objects rather than converting back to distances in\nparsecs or light-years.\nEXAMPLE:\nLuminosity – Absolute magnitude\nPH507\nAstrophysics\nDr Dirk Froebrich\n15\nAbsolute magnitudes for stars generally range from -10mag to +17mag. The\nabsolute magnitude for galaxies can be much lower (brighter). For example,\nthe giant elliptical galaxy M87 has an absolute magnitude of -22mag. Many\nstars visible to the naked eye have an absolute magnitude which is capable\nof casting shadows from a distance of 10pc; Rigel (-7.0mag), Deneb (7.2mag), Naos (-6.0mag), and Betelgeuse (-5.6mag).\nMagnitudes at Different Wavelengths\nThe kind of magnitude that we measure depends on how the light is filtered\nanywhere along the path of the detector and on the response function of the\ndetector itself. So the problem comes down to how to define standard\nmagnitude systems.\nMagnitude Systems\nDetectors of electromagnetic radiation (such as the photographic plate, the\nphotoelectric photometer, CCDs or the human eye) are sensitive only over\ngiven wavelength bands. So a given measurement samples but part of the\nPH507\nAstrophysics\nDr Dirk Froebrich\n16\nMulti wavelength Astronomy: Four images of the Sun, made using (a)\nvisible light, (b) ultraviolet light, (c) X rays, and (d) radio waves. By\nstudying the similarities and differences among these views of the same\nobject, important clues to its structure and composition can be found.\nPhotographic magnitude\nBecause the flux of starlight varies with wavelength, the magnitude of a\nstar depends upon the wavelength at which we observe. Originally,\nphotographic plates were sensitive only to blue light, and the term\nphotographic magnitude (mpg) still refers to magnitudes centered around\n420nm (in the blue region of the spectrum).\nVisual magnitude\nSimilarly, because the human eye is most sensitive to green and yellow,\nvisual magnitude (mv) or the photographic equivalent photo visual\nmagnitude (mpv) pertains to the wave-length region around 540nm.\nFilters\nToday we can measure magnitudes in the infrared, as well as in the\nultraviolet, by using filters in conjunction with the wide spectral sensitivity\nof photoelectric photometers.\nIn general, a photometric system requires a detector, filters, and a\ncalibration (in energy units). The properties of the filters are typified by\ntheir effective wavelength, 0, and band pass, which is defined as\nthe full width at half maximum (FWHM) in the transmission profile.\nPH507\nAstrophysics\nDr Dirk Froebrich\n17\nThe three main filter types are wide ( ≈ 100nm), intermediate (≈\n10nm), and narrow (≈1nm) band filters. There is a trade-off for the\nbandwidth choice: a smaller provides more spectral information but\nadmits less flux into the detector, resulting in longer integration times.\nFor a given range of the spectrum, the design of the filters makes the\ngreatest difference in photometric magnitude systems.\nA commonly used wide-band magnitude system is the UBV system: a\ncombination of ultraviolet (U), blue (B), and visual (V) magnitudes,\ndeveloped by H. L. Johnson. These three bands are centered at 365, 440,\nand 550nm; each wavelength band is roughly 100nm wide. In this system,\napparent magnitudes are denoted by U, B or V and the corresponding\nabsolute magnitudes are sub-scripted: M U, MB or MV.\nTo be useful in measuring fluxes, the photometric system must be\ncalibrated in energy units for each of its band passes. This calibration turns\nout to be the hardest part of the job. In general, it relies first on a set of\nstandard stars that define the magnitudes, for a particular filter set and\ndetector; that is, these stars define the standard magnitudes for the\nphotometric system to the precision with which they can be measured.\nInfrared Windows\nThe UBV system has been extended into the red and infrared (in part\nPH507\nAstrophysics\nDr Dirk Froebrich\n18\nbecause of the development of new detectors, such as CCDs, sensitive to\nthis region of the spectrum). The extensions are not as well standardized as\nthat for the Johnson UBV system, but they tend to include R and I in the\nfar red and J, H, K, L, and M in the infrared.\nAs well as measuring the properties of individual stars at different\nwavelengths, observing at longer wavelengths, particularly in the infrared,\nallows us to probe through clouds of small solid dust particles, as seen\nbelow\nA visible-light (left) vs. 2MASS infrared-light (right) view of the central\nregions of the Milky Way galaxy graphically illustrating the ability of\ninfrared light to penetrate the obscuring dust. The field-of-view is 10x10\ndegrees\nInfrared pass bands which allow transmission (low absorption):\nJ Band: 1.3 microns\nH Band: 1.6 microns\nK band: 2.2 microns\nL band 3.4 microns\nM band 5 microns\nN band 10.2 microns\nQ band 21 microns\nBolometric magnitudes (the magnitude of a star across all wavelengths)\ncan be converted to total radiant energy flux: A star of Mbol = 0mag\nSystem is defined by Vega at 7.76pc from the Sun with an apparent\nmagnitude defined as zero.\nPH507\nAstrophysics\nDr Dirk Froebrich\n19\nWith Lbol = 50.1LSUN and Mbol = 0.58mag.\nSun: mbol = -26.8mag\nFull moon: -12.6mag\nVenus (at brightest): -4.4mag\nSirius: -1.55mag\nBrightest quasar: 12.8mag\nFor Vega: mb = mv = 0.0mag\nmk = +0.02mag\nSun: Mb = 5.48mag, Mv = 4.83mag, Mk = 3.28mag\nColour Index/Colour:\nB-V, J-H, H-K are differences in magnitude, i.e. they are flux ratios.\nCooler, redder objects possess higher values. It is independent of the\nstar’s distance.\nExtinction\nInterstellar Medium modifies the radiation. Dust particles with size of order\ncompared to red: blue reflection nebulae and reddened stars. Hence a colour\nexcess is produced.\nColour Excess:\nE (B-V) = (B-V) - (B-V)o\nThis measures the reddening.\nModified distance modulus:\nm() = M() + 5 log d – 5 + A()\nwhere A() is the extinction due to both scattering and absorption, along\nthe entire line of sight. It is strongly wavelength dependent. The optical\ndepth (is given by\nI\nexp(−τ )= .\nIo\nTherefore: A() = 1.086 (show this as homework)\nThe optical depth is: * N\nwhere N is the total column density of dust (m-2) between the star and the\nobserver, and is the scattering or absorption cross-section (m2).\nPH507\nAstrophysics\nDr Dirk Froebrich\n20\nIn the Interstellar Medium (ISM) an empirical Law relates extinction to\nreddening:\nAV / E(B-V) = RV = 3.1 + - 0.2\nThus, an objects emission is modified/attenuated by the ISM and the\nEarth’s atmosphere. But these effects can be estimated.\nPH507\nAstrophysics\nDr Dirk Froebrich\n21\nThe Hertzsprung-Russell Diagram\nIn 1911, Ejnar Hertzsprung plotted the first such two-dimensional diagram\n(absolute magnitude versus spectral type) for observed stars, followed\n(independently) in 1913 by Henry Norris Russell.\nThe simple HR diagram represents one of the great observational syntheses\nin astrophysics. Note that any two of luminosity, magnitude, temperature,\nand radius could be used, but visual magnitude and temperature are\nuniversally obtained quantities.\nAn original idea was that a star was born hot (early type) and cooled (late\ntype).\nIt’s a particular colour-magnitude diagram.\nImportant stars: no obvious pattern…Sirius B, Betelgeuse in opposite\ncorners:\nPH507\nAstrophysics\nDr Dirk Froebrich\n22\nNearby stars: main-sequence appears. Most stars are less luminous and\ncooler than the Sun ( Centauri, nearest to us and a triple system, is\nsimilar). Note the hot small stars: the white dwarfs.\nPH507\nAstrophysics\nDr Dirk Froebrich\n23\nPH507\nAstrophysics\nDr Dirk Froebrich\n24\nMost stars have properties within the shaded region known as the main\nsequence. The points plotted here are for stars lying within about 5 pc of\nthe Sun. The diagonal lines correspond to constant stellar radius, so that\nstellar size can be represented on the same diagram as luminosity and\ntemperature.\nThe first H-R diagrams considered stars in the solar neighbourhood and\nplotted absolute visual magnitude, M, versus spectral type, which is\nequivalent to luminosity versus spectral type or luminosity versus\ntemperature. Note (a) the well-defined main sequence (class V) with everincreasing numbers of stars toward later spectral types and an absence of\nspectral classes earlier than A1 (Sirius), (b) the absence of giants and\nsupergiants (class III and I), and (c) the few white dwarfs at the lower left.\nPH507\nAstrophysics\nThe brightest stars:\nDr Dirk Froebrich\n25\nPH507\nAstrophysics\nDr Dirk Froebrich\n26\nAn H-R diagram for the 100 brightest stars in the sky. Such a plot is biased\nin favour of the most luminous stars--which appear toward the upper rightbecause we can see them more easily than we can the faintest stars. These\nare the GIANTS and SUPERGIANTS\nIn contrast, the H-R diagram for the brightest stars includes a significant\nnumber of giants and supergiants as well as several early-type mainsequence stars. Here we have made a selection that emphasises very\nluminous stars at distances far from the Sun. Note that the H-R diagram of\nthe nearest stars is most representative of those throughout the Galaxy: the\nmost common stars are low-luminosity spectral type M.\nThe most prominent feature of the H-R diagram is the Main Sequence:\n Strong correlation between Luminosity and Temperature.\n Hotter stars are Brighter than cooler stars along the M-S.\n About 85% of nearby stars, including the Sun, are on the M-S.\nAll other stars differ in size:\nGiants & Supergiants:\n• Very large radius, but same masses as M-S stars\nWhite Dwarfs:\n\nVery compact stars: ~R earth but with ~0.6 Msun!\nPH507\nAstrophysics\nDr Dirk Froebrich\n27\nStandard Candles\n1 Spectroscopic Parallax\nThe term spectroscopic parallax is a misnomer as it actually has nothing to\ndo with parallax.\nHertzsprung and Russell deduced the main-sequence for nearby stars,\nrelating their luminosity to their colour.\nPH507\nAstrophysics\nDr Dirk Froebrich\n28\nPH507\nAstrophysics\nDr Dirk Froebrich\n29\nGroups of distant stars should also lie along the same main-sequence strip.\nHowever they appear very dim, of course due to their distance. On\ncomparison of fluxes, we determine the distance. This works out to about\n100,000 pc, beyond which main-sequence stars are too dim.\nIf we take a spectrum of a star we can determine: its Spectral Class /\nSpectral Type.\nUsing photometry we can measure its apparent magnitude, m V.\nIf we use B and V filters we can also determine the blue apparent\nmagnitude, B and thus determine: the star's colour index, CI = B - V.\nKnowing either the star's spectral type or colour index allows us to place\nthe star on a vertical line or band along a Hertzsprung - Russell diagram.\nIf we also know its luminosity class we can further constrain its position\nalong this line. Hence we can distinguish between a red supergiant, giant or\nmain sequence star, for example.\nOnce we know its position on the HR diagram we can infer what its\nabsolute magnitude, MV should be by either reading off across to the\nvertical scale of the HR diagram or looking it up from a reference table. A\nmain sequence (luminosity class V) star with a colour index of 0.0 (i.e. A0\nV) has an absolute visual magnitude of +0.9mag for example.\nNow knowing m from measurement and inferring M we can use the\ndistance modulus equation:\nm - M = 5 log(d/10)\nto find the distance to the star, d, in parsecs.\nEXAMPLE\nGamma Crucis is an M3.5 III star, a red giant.\nMeasured values: mV = +1.59mag and a B-V colour index of +1.60mag.\nUsing its spectral type and luminosity class we can place it where the red\ncircle is on the HR diagram. Reading across to the vertical axis this\ncorresponds to an absolute magnitude of about -0.8mag.\nPH507\nAstrophysics\nDr Dirk Froebrich\n….show that the distance is about 30pc………?\n30\nPH507\nAstrophysics\nDr Dirk Froebrich\n31\n2 Cepheids as Standard Candles: The Period-Luminosity\nRelationship\nCepheids are bright (103-104 times the Sun) variable Population I stars,\nnamed after the prototype -Cep. They show an important connection\nbetween their period of variability and luminosity: the pulsation period of a\nCepheid variable is directly related to its median luminosity.\nThis relationship was first discovered from a study of the variables in the\nMagellanic Clouds, two small nearby companion galaxies to our Galaxy\nthat are visible in the night sky of the southern hemisphere. To a good\napproximation, you can consider all stars in each Magellanic Cloud to be at\nthe same distance. Henrietta Leavitt, working at Harvard in 1912, found\nthat the brighter the median apparent magnitude (and so the luminosity,\nsince the stars are the same distance), the longer the period of the Cepheid\nvariable. A linear relationship was found.\nHarlow Shapley recognized the importance of this period-luminosity (PL) relation-ship and attempted to find the zero point, hence the knowledge\nof the period of a Cepheid would immediately indicate its luminosity\n(absolute magnitude) and thus its distance.\nThis calibration was difficult to perform because of the relative scarcity of\nCepheids and their large distances. None are sufficiently near to allow a\ntrigonometric parallax to be determined, so Shapley had to depend upon the\nrelatively inaccurate method of statistical parallaxes. His zero point was\nthen used to find the distances to many other galaxies. These distances are\nrevised as new and accurate data become available. Right now, some 20\nstars whose distances are known reasonably well (because they are in open\nclusters) serve as the calibrators for the P-L relationship.\nPH507\nAstrophysics\nDr Dirk Froebrich\n32\nFurther work showed that there are two types of Cepheids, each with its\nown separate, almost parallel P-L relationship.\nPH507\nAstrophysics\nDr Dirk Froebrich\n33\nThe classical Cepheids are the more luminous, of Population I, and found\nin spiral arms. Population II Cepheids, also known as W-Virginis stars\nafter their prototype, are found in globular clusters and other Population II\nsystems.\nClassical Cepheids have periods ranging from one to 50 days (typically five\nto ten days) and range from F6 to K2 in spectral type.\nPopulation II Cepheids vary in period from two to 45 days (typically 12 to\n20 days) and range from F2 to G6 in spectral type.\nPopulation I and II Cepheids are both regular / periodic variables; their\nchange in luminosity with time follows a regular cycle. The empirically\nderived relationship between a Type I Cepheid's period P (in days), and its\nabsolute magnitude MV is given by\nCepheids are among the brightest individual stars. Hence, they can be\nused to determine distances to quite far away galaxies (to about 5Mpc).\nHST stretched this to 18Mpc (the Virgo cluster).\n3 Tully-Fisher Relation\nRichard Fisher in 1977, is a standard candle that measures the distance to\nrotating spiral galaxies by the width of the galaxy's spectral lines.\nThe empirically-derived relation states that the luminosity of a galaxy is\ndirectly proportional to the fourth power of its rotational velocity, which\ncan be calculated from the width of the spectral line, and uses the distance\nmodulus to find distance from luminosity and apparent magnitude.\nIn a spiral galaxy, the centripetal force of gas and stars balances the\ngravitational force:\nm * v2 / R = G * m * M / R2.\nIf they have the same surface brightness (L/R2 is constant) and the same\nmass-to-light ratio (M/L is constant), then L~v4. So, provided we can\nmeasure the velocity v, certain galaxies can be used as standard candles.\nE.g. determine v through the 21 cm line of atomic hydrogen in the galaxy.\nPH507\nAstrophysics\nDr Dirk Froebrich\n34\nBrightest MB = -19mag, rotation velocities 100 - 500km/s\n4 Type Ia Supernovae.\nThe peak output from these supernovae is about M B=-19.33±0.25mag and\nalmost constant. Therefore, we can infer the distance from the inverse\nsquare law. Being so bright, they act as standard candles to large distances:\nout to 1000Mpc.\nWhy are they standard candles? White dwarfs in binary systems. Material\nfrom a companion red giant is dumped on the white dwarf surface until the\nWD reaches a critical mass (Chandrasekhar mass) of 1.4 solar masses.\nExplosion occurs with fixed rise and fall of luminosity. The fall in\nluminosity is due to radioactive decay of Ni 56, Co56 and Fe56.\nPH507\nAstrophysics\nDr Dirk Froebrich\n35\nPH507\nAstrophysics\nDr Dirk Froebrich\n36\n5 Other methods include:\n1: time delay of light rays due to gravitational lensing,\n2: cluster size influences Compton scattering of CMB radiation and\nBremsstrahlung emission (X-rays). Combining, yields the size estimate\n(Sunyaev-Zeldovich effect).\nNew Method\nReverse argument: knowing the Hubble constant is 72 km/s/Mpc (WMAP\nresult) distances can be found directly from the redshift!\nFor nearby galaxies: Distance = velocity / Hubble Constant\nQuestions\nHow do we scale the solar system?\nPH507\nAstrophysics\nDr Dirk Froebrich\n37\nHow do we find the distance to gas clouds?\n- measure the distance to associated objects, i.e. stars in or close to the\ncloud (those objects could be identified e.g. by reflection nebulae). Rather\ndifficult method, due to high and variable extinction, and/or young age of\nmany of the stars within dust clouds.\n- count the number of stars which are foreground to the cloud. A model of\nthe distribution of stars within our Galaxy then can lead to statistically\nbased distances for the cloud.\n- Use the galactic rotation curve to determine the distance. This is\nambiguous for clouds closer to the galactic center than the Sun.\n```"
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https://answers.everydaycalculation.com/gcf/84-3360
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"# Answers\n\nSolutions by everydaycalculation.com\n\n## What is the GCF of 84 and 3360?\n\nThe GCF of 84 and 3360 is 84.\n\n#### Steps to find GCF\n\n1. Find the prime factorization of 84\n84 = 2 × 2 × 3 × 7\n2. Find the prime factorization of 3360\n3360 = 2 × 2 × 2 × 2 × 2 × 3 × 5 × 7\n3. To find the GCF, multiply all the prime factors common to both numbers:\n\nTherefore, GCF = 2 × 2 × 3 × 7\n4. GCF = 84\n\nMathStep (Works offline)",
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"Download our mobile app and learn how to find GCF of upto four numbers in your own time:\nAndroid and iPhone/ iPad\n\n#### GCF Calculator\n\nEnter two numbers separate by comma. To find GCF of more than two numbers, click here.\n\nThe greatest common factor (GCF) is also known as greatest common divisor (GCD) or highest common factor (HCF).\n\n© everydaycalculation.com"
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http://www.sanjaysah.com.np/2016/12/closed-loop-control-system.html
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"Closed-loop Control System This sensor would monitor the actual dryness of the clothes and compare it with (or subtract it...\n\n### Closed-loop Control System",
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"This sensor would monitor the actual dryness of the clothes and compare it with (or subtract it from) the input reference. The error signal (error = required dryness – actual dryness) is amplified by the controller, and the controller output makes the necessary correction to the heating system to reduce any error. For example if the clothes are too wet the controller may increase the temperature or drying time. Likewise, if the clothes are nearly dry it may reduce the temperature or stop the process so as not to overheat or burn the clothes, etc.\nThen the closed-loop configuration is characterised by the feedback signal, derived from the sensor in our clothes drying system. The magnitude and polarity of the resulting error signal, would be directly related to the difference between the required dryness and actual dryness of the clothes.\nAlso, because a closed-loop system has some knowledge of the output condition, (via the sensor) it is better equipped to handle any system disturbances or changes in the conditions which may reduce its ability to complete the desired task.\nFor example, as before, the dryer door opens and heat is lost. This time the deviation in temperature is detected by the feedback sensor and the controller self-corrects the error to maintain a constant temperature within the limits of the preset value. Or possibly stops the process and activates an alarm to inform the operator.\nAs we can see, in a closed-loop control system the error signal, which is the difference between the input signal and the feedback signal (which may be the output signal itself or a function of the output signal), is fed to the controller so as to reduce the systems error and bring the output of the system back to a desired value. In our case the dryness of the clothes. Clearly, when the error is zero the clothes are dry.\nThe term Closed-loop control always implies the use of a feedback control action in order to reduce any errors within the system, and its “feedback” which distinguishes the main differences between an open-loop and a closed-loop system.The accuracy of the output thus depends on the feedback path, which in general can be made very accurate and within electronic control systems and circuits, feedback control is more commonly used than open-loop or feed forward control.\nClosed-loop systems have many advantages over open-loop systems. The primary advantage of a closed-loop feedback control system is its ability to reduce a system’s sensitivity to external disturbances, for example opening of the dryer door, giving the system a more robust control as any changes in the feedback signal will result in compensation by the controller.\nThen we can define the main characteristics of Closed-loop Control as being:\n• To reduce errors by automatically adjusting the systems input.\n• To improve stability of an unstable system.\n• To increase or reduce the systems sensitivity.\n• To enhance robustness against external disturbances to the process.\n• To produce a reliable and repeatable performance.\nWhilst a good closed-loop system can have many advantages over an open-loop control system, its main disadvantage is that in order to provide the required amount of control, a closed-loop system must be more complex by having one or more feedback paths. Also, if the gain of the controller is too sensitive to changes in its input commands or signals it can become unstable and start to oscillate as the controller tries to over-correct itself, and eventually something would break. So we need to “tell” the system how we want it to behave within some pre-defined limits.\n\n## Closed-loop Summing Points\n\nFor a closed-loop feedback system to regulate any control signal, it must first determine the error between the actual output and the desired output. This is achieved using a summing point, also referred to as a comparison element, between the feedback loop and the systems input. These summing points compare a systems set point to the actual value and produce a positive or negative error signal which the controller responds too. where: Error = Set point – Actual",
null,
"The symbol used to represent a summing point in closed-loop systems block-diagram is that of a circle with two crossed lines as shown. The summing point can either add signals together in which a Plus ( + ) symbol is used showing the device to be a “summer” (used for positive feedback), or it can subtract signals from each other in which case a Minus ( ) symbol is used showing that the device is a “comparator” (used for negative feedback) as shown.\n\n### Summing Point Types",
null,
"Note that summing points can have more than one signal as inputs either adding or subtracting but only one output which is the algebraic sum of the inputs. Also the arrows indicate the direction of the signals. Summing points can be cascaded together to allow for more input variables to be summed at a given point.\n\n## Closed-loop System Transfer Function\n\nThe Transfer Function of any electrical or electronic control system is the mathematical relationship between the systems input and its output, and hence describes the behaviour of the system. Note also that the ratio of the output of a particular device to its input represents its gain. Then we can correctly say that the output is always the transfer function of the system times the input. Consider the closed-loop system below.\n\n### Typical Closed-loop System Representation",
null,
"Where: block G represents the open-loop gains of the controller or system and is the forward path, and block H represents the gain of the sensor, transducer or measurement system in the feedback path.\nTo find the transfer function of the closed-loop system above, we must first calculate the output signal θo in terms of the input signal θi. To do so, we can easily write the equations of the given block-diagram as follows.\nThe output from the system is equal to: Output = G x Error\nNote that the error signal, θe is also the input to the feed-forward block: G\nThe output from the summing point is equal to: Error = Input - H x Output\nIf H = 1 (unity feedback) then:\nThe output from the summing point will be: Error (θe) = Input - Output\nEliminating the error term, then:\nThe output is equal to: Output = G x (Input - H x Output)\nTherefore: G x Input = Output + G x H x Output\nRearranging the above gives us the closed-loop transfer function of:",
null,
"The above equation for the transfer function of a closed-loop system shows a Plus ( + ) sign in the denominator representing negative feedback. With a positive feedback system, the denominator will have a Minus ( ) sign and the equation becomes: 1 - GH.\nWe can see that when H = 1 (unity feedback) and G is very large, the transfer function approaches unity as:",
null,
"Also, as the systems steady state gain G decreases, the expression of: G/(1 + G)decreases much more slowly. In other words, the system is fairly insensitive to variations in the systems gain represented by G, and which is one of the main advantages of a closed-loop system.\n\n## Multi-loop Closed-loop System\n\nWhilst our example above is of a single input, single output closed-loop system, the basic transfer function still applies to more complex multi-loop systems. Most practical feedback circuits have some form of multiple loop control, and for a multi-loop configuration the transfer function between a controlled and a manipulated variable depends on whether the other feedback control loops are open or closed.\nConsider the multi-loop system below.",
null,
"Any cascaded blocks such as G1 and G2 can be reduced, as well as the transfer function of the inner loop as shown.",
null,
"After further reduction of the blocks we end up with a final block diagram which resembles that of the previous single-loop closed-loop system.",
null,
"And the transfer function of this multi-loop system becomes:",
null,
"Then we can see that even complex multi-block or multi-loop block diagrams can be reduced to give one single block diagram with one common system transfer function.\n\n## Closed-loop Motor Control\n\nSo how can we use Closed-loop Systems in Electronics. Well consider our DC motor controller from the previous open-loop tutorial. If we connected a speed measuring transducer, such as a tachometer to the shaft of the DC motor, we could detect its speed and send a signal proportional to the motor speed back to the amplifier. A tachometer, also known as a tacho-generator is simply a permanent-magnet DC generator which gives a DC output voltage proportional to the speed of the motor.\nThen the position of the potentiometers slider represents the input, θi which is amplified by the amplifier (controller) to drive the DC motor at a set speed Nrepresenting the output, θo of the system, and the tachometer T would be the closed-loop back to the controller. The difference between the input voltage setting and the feedback voltage level gives the error signal as shown.\n\n### Closed-loop Motor Control",
null,
"Any external disturbances to the closed-loop motor control system such as the motors load increasing would create a difference in the actual motor speed and the potentiometer input set point.\nThis difference would produce an error signal which the controller would automatically respond too adjusting the motors speed. Then the controller works to minimize the error signal, with zero error indicating actual speed which equals set point.\nElectronically, we could implement such a simple closed-loop tachometer-feedback motor control circuit using an operational amplifier (op-amp) for the controller as shown.\n\n### Closed-loop Motor Controller Circuit",
null,
"This simple closed-loop motor controller can be represented as a block diagram as shown.\n\n### Block Diagram for the Feedback Controller",
null,
"A closed-loop motor controller is a common means of maintaining a desired motor speed under varying load conditions by changing the average voltage applied to the input from the controller. The tachometer could be replaced by an optical encoder or Hall-effect type positional or rotary sensor.\n\n## Closed-loop Systems Summary\n\nWe have seen that an electronic control system with one or more feedback paths is called a Closed-loop System. Closed-loop control systems are also called “feedback control systems” are very common in process control and electronic control systems. Feedback systems have part of their output signal “fed back” to the input for comparison with the desired set point condition. The type of feedback signal can result either in positive feedback or negative feedback.\nIn a closed-loop system, a controller is used to compare the output of a system with the required condition and convert the error into a control action designed to reduce the error and bring the output of the system back to the desired response. Then closed-loop control systems use feedback to determine the actual input to the system and can have more than one feedback loop.\nClosed-loop control systems have many advantages over open-loop systems. One advantage is the fact that the use of feedback makes the system response relatively insensitive to external disturbances and internal variations in system parameters such as temperature. It is thus possible to use relatively inaccurate and inexpensive components to obtain the accurate control of a given process or plant.\nHowever, system stability can be a major problem especially in badly designed closed-loop systems as they may try to over-correct any errors which could cause the system to loss control and oscillate.\n\n#### Control system in which the output has an effect on the input quantity in such a manner that the input quantity will adjust itself based on the output generated is called closed loop control system. Open loop control system can be converted in to closed loop control system by providing a feedback. This feedback automatically makes the suitable changes in the output due to external disturbance. In this way closed loop control system is called automatic control system. Figure below shows the block diagram of closed loop control system in which feedback is taken from output and fed in to input.",
null,
"Practical Examples of Closed Loop Control System\n1.Automatic Electric Iron – Heating elements are controlled by output temperature of the iron.\n2.Servo Voltage Stabilizer – Voltage controller operates depending upon output voltage of the system.\n3.Water Level Controller– Input water is controlled by water level of the reservoir.\n4.Missile Launched & Auto Tracked by Radar – The direction of missile is controlled by comparing the target and position of the missile.\n5.An Air Conditioner – An air conditioner functions depending upon the temperature of the room.\n6.Cooling System in Car – It operates depending upon the temperature which it controls.\nAdvantages of Closed Loop Control System\n1.Closed loop control systems are more accurate even in the presence of non-linearity.\n2.Highly accurate as any error arising is corrected due to presence of feedback signal.\n3.Bandwidth range is large.\n4.Facilitates automation.\n5.The sensitivity of system may be made small to make system more stable.\n6.This system is less affected by noise.\nDisadvantages of Closed Loop Control System\n1.They are costlier.\n2.They are complicated to design.\n3.Required more maintenance.\n5.Overall gain is reduced due to presence of feedback.\n6.Stability is the major problem and more care is needed to design a stable closed loop system.\nFeedback Loop of Control System\nA feedback is a common and powerful tool when designing a control system. Feedback loop is the tool which take the system output into consideration and enables the system to adjust its performance to meet a desired result of system.\n\nIn any control system, output is affected due to change in environmental condition or any kind of disturbance. So one signal is taken from output and is fed back to the input. This signal is compared with reference input and then error signal is generated. This error signal is applied to controller and output is corrected. Such a system is called feedback system. Figure below shows the block diagram of feedback system.",
null,
"When feedback signal is positive then system called positive feedback system. For positive feedback system, the error signal is the addition of reference input signal and feedback signal. When feedback signal is negative then system is called negative feedback system. For negative feedback system, the error signal is given by difference of reference input signal and feedback signal.\n\n#### Effect of Feedback\n\nRefer figure beside, which represents feedback system where\nR = Input signal\nE = Error signal\nG = forward path gain",
null,
"H = Feedback\nC = Output signal\nB = Feedback signal\n\n1. Error between system input and system output is reduced.\n2. System gain is reduced by a factor 1/(1±GH).\n3. Improvement in sensitivity.\n4. Stability may be affected.\n5. Improve the speed of response.\n\nsources: electrical4u.com, electronics-tutorials.ws",
null,
"Name\n\nBASIC ELECTRICAL,12,BATTERIES,4,CIRCUIT THEORIES,9,CONTROL SYSTEMS,3,DC MOTOR,1,DIGITAL ELECTRONICS,1,DISTRIBUTED GENERATION,2,DISTRIBUTION,6,ELECTRICAL DRIVES,1,ELECTRICAL LAWS,8,ELECTRONICS DEVICES,2,General,7,GENERATION,3,GENERATOR,1,HIGH VOLTAGE,4,ILLUMINATION,1,INDUCTION MOTOR,7,MATERIALS,1,MEASUREMENT,1,MOTOR,1,POWER ELECTRONICS,2,PROJECTS ON INDUCTION MOTOR,1,PROTECTION,1,SMART GRID,3,SWITCHGEAR,4,SYNCHRONOUS MOTOR,1,TRANSFORMER,6,TRANSMISSION,4,\nltr\nitem\nElectrical for Us: Closed Loop Control System\nClosed Loop Control System"
] |
[
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"http://www.electronics-tutorials.ws/systems/sys14.gif",
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"http://www.electronics-tutorials.ws/systems/sys15.gif",
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"http://www.electronics-tutorials.ws/systems/sys16.gif",
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"http://www.electronics-tutorials.ws/systems/sys17.gif",
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"http://www.electronics-tutorials.ws/systems/sys18.gif",
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"http://www.electronics-tutorials.ws/systems/sys22.gif",
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"http://www.electronics-tutorials.ws/systems/sys23.gif",
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"http://www.electronics-tutorials.ws/systems/sys25.gif",
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"http://www.electronics-tutorials.ws/systems/sys21.gif",
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"http://www.electronics-tutorials.ws/systems/sys26.gif",
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"http://www.electronics-tutorials.ws/systems/sys27.gif",
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"http://electrical4u.com/electrical/wp-content/uploads/2014/04/closed-loop-control-system-11-3-15.gif",
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"http://electrical4u.com/electrical/wp-content/uploads/2014/04/feedback-control-system-11-3-15.gif",
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"http://electrical4u.com/electrical/wp-content/uploads/2014/04/Block-diagram-11-3-15.gif",
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"https://4.bp.blogspot.com/-h5aTLlzsHzc/WFBAdP77VhI/AAAAAAAAAjk/v94s9B4PpDYb-N8KQe2JeRZ8rvNwabD0gCEw/s320/fig4-closed_vertical.gif",
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|
https://study.com/academy/topic/4th-grade-math-operations-with-decimals.html
|
[
"# Ch 26: 4th Grade Math: Operations with Decimals\n\nHelp your 4th grade student learn how to estimate, add, subtract, multiply, and divide decimals with the fun video lessons, sample problems, and multiple-choice quizzes found in this chapter.\n\n## 4th Grade Math: Operations with Decimals - Chapter Summary\n\nThis chapter's entertaining video lessons and multiple-choice quizzes can give your 4th grade student extra practice working with decimals found in number sequences, addition and subtraction sentences, and multiplication and division problems. Lessons also show students how to estimate decimals and make real-world arithmetic problems, like those involving money, easier to solve.\n\n## Chapter Lessons and Objectives\n\nLesson Objective\nAdding and Subtracting Decimals: Examples & Word Problems Students learn to maintain place value when adding and subtracting decimals.\nMultiplying and Dividing Decimals: Examples & Word Problems This lesson shows students how to multiply and divide decimals.\nHow to Estimate with Decimals to Solve Math Problems Students learn how to make problem solving easier by rounding decimals before adding, subtracting, or multiplying them.\nPractice Estimating Sums, Differences and Products of Decimals This lesson includes additional sample problems for estimating decimals.\nSolving Problems using Decimal Numbers Instructors outline real-world examples of when students might need to solve math problems that include decimals.\nSolving Number Sequences Involving Decimals Students learn to find missing decimal numbers in a number sequence.\nPractice Completing Addition and Subtraction Sentences with Decimals This lesson gives students practice completing addition and subtraction sentences that include decimals.\nPractice Adding Three or More Decimals Students use skills gained in the previous lessons to solve problems containing three or more decimals.\n\n7 Lessons in Chapter 26: 4th Grade Math: Operations with Decimals\nTest your knowledge with a 30-question chapter practice test\nChapter Practice Exam\nTest your knowledge of this chapter with a 30 question practice chapter exam.\nNot Taken\nPractice Final Exam\nTest your knowledge of the entire course with a 50 question practice final exam.\nNot Taken\n\n### Earning College Credit\n\nDid you know… We have over 200 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level."
] |
[
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http://www.crackingagrippa.net/submissions/jeremy_cooper.html
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[
"Jeremy Cooper took second place with the following submission:\n\nThe text in the binary is encrypted with a hand-made algorithm that is a combination of 12-bit RSA and several bit-wise permutations. The decryption algorithm is:\n\n1. Break the ciphertext into chunks of three eight-bit bytes.\n2. For each chunk:\n1. Denote the three bytes in the chunk as \"x\", \"y\" and \"z\", assigning the first byte to x, the second to y, and the third to z.\n```Example: x = 99, y = 97, z = 116\n```\n2. Permute the bits of y according to the following pattern:\n```Input position Output position\n-------------- ---------------\n0 1\n1 4\n2 7\n3 2\n4 5\n5 0\n6 3\n7 6\n```\nCall this permuted result \"py\".\n```Example: 97 in 8-bit binary is 01100001\nThis permutes to 00001011, which is 11 in decimal.\n```\n3. Pack x and py into a twelve-bit number by concatenating the four least-significant bits of py with all eight bits of x. Call the result \"w\".\n```Example: Since py = 00001011 and x = 01100001, the result would be\n1011 concatenated with 01100001: 101101100001, which\nis 2913 in decimal.\n```\n4. RSA decrypt w with the following public key:\n``` public modulus (n) = 4097\npublic exponent (e) = 11\n```\nThat is to say, compute (w ** 11) mod 4097. Call the result \"t1\".\n```Example: 2913 to the eleventh power is\n128157884794082129828859612711416527137.\n128157884794082129828859612711416527137 divided by 4097\nleaves a remainder of 3949.\n```\n5. Pack py and z into a twelve-bit number by concatenating all eight bits of z with the four most-significant bits of py. Call the result \"v\".\n```Example: Since py = 00001011 and z = 01110100 (116 decimal), the\nresult would be 01110100 concatenated with 0000:\n011101000000, which is 1856 in decimal.\n```\n6. RSA decrypt v with the public key above. Call the result \"t2\".\n```Example: 1856 to the eleventh power is\n900238724867078044750327477390278656.\n900238724867078044750327477390278656 divided by 4097\nleaves a remainder of 500.\n```\n7. Take the eight least-significant bits of t1 and call them 'u'.\n```Example: t1 = 3949, or 111101101101 in binary. Its eight least-\nsignificant bits are 01101101, or 107 in decimal.\n```\n8. Treat t1 as a twelve-bit number (that is, pad it with zeroes until it is twelve bits long). Concatenate the four least- significant bits of t2 with the four most-significant bits of the padded t1. Call the result \"s\".\n```Example: t1 = 3949, or 111101101101 in binary. It needs no padding\nto reach twelve bits. t2 = 500, or 111110100 in binary.\nt2's four least-significant bits are therefore 0100.\nConcatenating 0100 with t1's four most significant bits,\n1111, gives us 01001111, or 79 in decimal.\n```\n9. Treat t2 as a twelve-bit number, padding as necessary. Take its eight most-significant bits and call them \"r\".\n```Example: t2 = 500, or 111110100. It needs three zeroes of padding,\nto make it a twelve-bit number. When padded it becomes\n000111110100. Its eight most-significant bits are therefore\n00011111, or 31 in decimal.\n```\n10. Permute the bits of u, s, and r according to the following pattern (where zero denotes the least significant bit and seven, the most):\n```Input position Output position\n-------------- ---------------\n0 2\n1 7\n2 4\n3 1\n4 6\n5 3\n6 0\n7 5\n```\nCall the results \"o\", \"p\", and \"q\", respectively.\n```Example: u = 01101011 -> 10001111 (143 decimal)\ns = 01001111 -> 10010111 (151 decimal)\nr = 00011111 -> 11010110 (214 decimal)\n```\n11. Convert o, p and q to their ASCII equivalents and write them down, in that order.\n```Example: [ This example, which started with the ciphertext composed\nof the ASCII string \"cat\" is not a good example. It\ndecrypts to nonsense. ]\n```\n3. (Go to next chunk)\n\nThat's it!\n\nJeremy then used the same description with actual ciphertext:\n\nOk, here's the solution using the first six bytes of the ciphertext from the application. They are:\n\n``` C1 67 BB 7C 6F 35\n```\n\nThen, using the algorithm:\n\n1. Break the ciphertext into chunks of three eight-bit bytes.\n2. For each chunk:\n1. Denote the three bytes in the chunk as \"x\", \"y\" and \"z\", assigning the first byte to x, the second to y, and the third to z.\n```Example: x = C1 (hex) = 193 (decimal),\ny = 67 (hex) = 103 (decimal),\nz = BB (hex) = 187 (decimal)```\n2. Permute the bits of y according to the following pattern:\n```Input position Output position\n-------------- ---------------\n0 1\n1 4\n2 7\n3 2\n4 5\n5 0\n6 3\n7 6\n```\nCall this permuted result \"py\".\n```Example: 103 in 8-bit binary is 01100111\nThis permutes to 10011011, which is 155 in decimal.\n```\n3. Pack x and py into a twelve-bit number by concatenating the four least-significant bits of py with all eight bits of x. Call the result \"w\".\n```Example: Since py = 10011011 and x = 11000001, the result would be\n1011 concatenated with 11000001: 101111000001, which\nis 3009 in decimal.\n```\n4. RSA decrypt w with the following public key:\n```public modulus (n) = 4097\npublic exponent (e) = 11\n```\nThat is to say, compute (w ** 11) mod 4097. Call the result \"t1\".\n```Example: 3009 to the eleventh power is\n183081332709971685898032400292321292609.\n183081332709971685898032400292321292609 divided by 4097\nleaves a remainder of 136.\n```\n5. Pack py and z into a twelve-bit number by concatenating all eight bits of z with the four most-significant bits of py. Call the result \"v\".\n```Example: Since py = 10011011 and z = 10111011 (187 decimal), the\nresult would be 10111011 concatenated with 1001:\n101110111001, which is 3001 in decimal.\n```\n6. RSA decrypt v with the public key above. Call the result \"t2\".\n```Example: 3001 to the eleventh power is\n177797622648287046910292734455495033001.\n177797622648287046910292734455495033001 divided by 4097\nleaves a remainder of 2055.\n```\n7. Take the eight least-significant bits of t1 and call them 'u'.\n```Example: t1 = 136, or 10001000 in binary. Its eight least-\nsignificant bits are 10001000, or, again, 136 in decimal.```\n8. Treat t1 as a twelve-bit number (that is, pad it with zeroes until it is twelve bits long). Concatenate the four least- significant bits of t2 with the four most-significant bits of the padded t1. Call the result \"s\".\n```Example: t1 = 136, or 10001000 in binary. It needs four bits of\npadding to reach twelve bits in size: 000010001000.\n\nt2 = 2055, or 100000000111 in binary. Its four\nleast-significant bits are 0111. Concatenating 0111 with\nt1's four most significant bits, 0000, gives us 01110000, or\n112 in decimal.```\n9. Treat t2 as a twelve-bit number, padding as necessary. Take its eight most-significant bits and call them \"r\".\n```Example: t2 = 2055, or 100000000111. It needs no padding\nto make it a twelve-bit number. Its eight most-significant\nbits are therefore 10000000, or 128 in decimal.```\n10. Permute the bits of u, s, and r according to the following pattern (where zero denotes the least significant bit and seven, the most):\n```Input position Output position\n-------------- ---------------\n0 2\n1 7\n2 4\n3 1\n4 6\n5 3\n6 0\n7 5\n```\nCall the results \"o\", \"p\", and \"q\", respectively.\n```Example: u = 10001000 -> 00100010 (34 decimal)\ns = 01110000 -> 01001001 (73 decimal)\nr = 10000000 -> 00100000 (32 decimal)```\n11. Convert o, p and q to their ASCII equivalents and write them down, in that order.\n```o = 00100010 = 22 (hex) = '\"' (Double quote)\np = 01001001 = 49 (hex) = 'I'\nq = 00100000 = 20 (hex) = ' ' (Space)\n```\nSo for block 1 we get the characters: '\"I '\n== Block 2 ==\n1. Next three bytes:\n`7C 6F 35`\n2.\n```x = 7C\ny = 6F\nz = 35 (hex)\n```\n3.\n`py = PERMUTE(6F) -> 9F`\n4. `w = F7C = 3964 (decimal)`\n5. `t1 = (3964 ** 11) mod 4097 = 432 = 1B0 (hex)`\n6. `v = 359 (hex) = 859 (decimal)`\n7. `t2 = (857 ** 11) mode 4097 = 3533 = DCD (hex)`\n8. `u = B0 (hex)`\n9. `s = D1 (hex)`\n10. `r = DC (hex)`\n11. ```o = PERMUTE(B0) -> 68 (hex) = 104 (decimal)\np = PERMUTE(D1) -> 65 (hex) = 101 (decimal)\nq = PERMITE(DC) -> 73 (hex) = 115 (decimal)```\n12. `68 65 73 -> hes`\n== PUTTING THE BLOCKS TOGETHER ==\n\nPutting both blocks together we get:\n\n`\"I hes`\n\nWhich are the first characters of the poem:\n\n```\"I hesitated before untying the bow ...\n```\n\nThen Jeremy implemented the decryption algorithm in Python. His implementation looks a little more \"modern\" than Robert's (since it doesn't use the Lisp function names), but they both evaluate to the exact same result.\n\nJeremy then provided the encryption routine in Python,reconstructed through reverse engineering:\n\nJust for fun, here's the \"encryption\" code. I say \"just for fun\" meaningfully because Agrippa actually does not contain the code in this script. Instead, this is the script that the original software developer would have to have run to produce the ciphertext. He or she would have then inserted the ciphertext into the application and compiled it.\n\nJeremy provided more context by answering some of the original questions posed in the challenge:\n\nSo now that I've demonstrated the algorithm, I want to answer a few of the questions presented in the challenge:\n\nQ. Does Agrippa really use encryption?\n\nA.Yes, but only if by \"use encryption\" you mean that Agrippa uses an already encrypted ciphertext. When Agrippa runs it actually decrypts a pre-programmed ciphertext. In this stricter sense Agrippa does not use any encryption at all.\n\nQ. How does the encryption function?\n\nA. The Agrippa program contains an encrypted version of the Agrippa poem which it decrypts and then displays on the screen for a few minutes. The decryption routine resides within the program and uses a customized block cipher built out of two common cipher building blocks: bit-wise permutations and the RSA modular exponentiation cipher.\n\nThe cipher takes a three byte input block and transforms it into a three byte output block. It begins by performing a simple bit-wise permutation on the middle byte of the block and then splits the block into two twelve-bit blocks.\n\nThese two twelve-bit blocks are each handed to an RSA decryption routine. The resulting plaintext blocks from the RSA decryption are then split back into three eight-bit bytes again.\n\nFinally, the cipher performs a permutation on all three bytes and returns them as output. [See below for a graphical depiction of this process]\n\nQ. What is the encryption key?\n\nA. Because the encryption is non-standard and it is the only instance of its kind, it is difficult to separate the algorithm from any keys that may have been used to configure it. Nonetheless, it does contain an recognizable RSA element. Its \"public\" key has the parameters:\n\n```public modulus (n) = 4097 (0x1001)\npublic exponent (e) = 11```\n\nSince this key is very small, we can easily calculate its private counterpart:\n\n```prime factor 1 (p) = 241\nprime factor 2 (q) = 17\nprivate exponent (d) = 3491```\n\nLikewise, if we treat the permutation elements as configurable items, then their \"keys\" would be:\n\n```Permutation A = ( 1, 4, 7, 2, 5, 0, 3, 6 )\nPermutation B = ( 2, 7, 4, 1, 6, 3, 0, 5 )```\n\nIt is worth noting that the permutation indexes in both tables are formed from a generator in an additive group (mod 8).\n\nQ. What analysis did you do to get the permutation functions?\n\nA. I performed static and dynamic analysis of the binary code. After performing that analysis I also was able to locate the permutation functions inside the fax images of the \"source code\".\n\nThe functions, which are located on page 3 of the source code, are named \"un-do-it\" and \"un-do-too-it\". They correspond to the \"Permutation_B\" and \"Permutation_A\" functions, respectively, in my Python source code.\n\nThey also correspond to steps \"j.\" and \"b.\", respectively, in my first message.\n\nQ. Where did you get the ciphertext?\n\nA. The ciphertext is available in the program memory image once the program has finished decompressing itself at start-up. Using dynamic analysis (debugger), you can observe the ciphertext by starting the application, waiting a very short while (about one second) and then issuing a break condition in your debugger. Upon entering the debugger you can observe the ciphertext memory at address 0x1058e0.\n\nA partial, lossy version of it is also available on page 5. You can compare the \"printable\" characters in the source code to those in the ciphertext I have provided. (Since only the tail end of the ciphertext is visible in the source code, compare them by walking backwards).",
null,
"",
null,
""
] |
[
null,
"http://www.crackingagrippa.net/submissions/files/FullAlgorithm.png",
null,
"http://www.crackingagrippa.net/submissions/files/FullAlgorithmExample.png",
null
] |
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|
https://www.itl.nist.gov/div898/handbook/eda/section3/bootplot.htm
|
[
"",
null,
"1. Exploratory Data Analysis\n1.3. EDA Techniques\n1.3.3. Graphical Techniques: Alphabetic\n\n## Bootstrap Plot\n\nPurpose:\nEstimate uncertainty\nThe bootstrap (Efron and Gong) plot is used to estimate the uncertainty of a statistic.\nGenerate subsamples with replacement To generate a bootstrap uncertainty estimate for a given statistic from a set of data, a subsample of a size less than or equal to the size of the data set is generated from the data, and the statistic is calculated. This subsample is generated with replacement so that any data point can be sampled multiple times or not sampled at all. This process is repeated for many subsamples, typically between 500 and 1000. The computed values for the statistic form an estimate of the sampling distribution of the statistic.\n\nFor example, to estimate the uncertainty of the median from a dataset with 50 elements, we generate a subsample of 50 elements and calculate the median. This is repeated at least 500 times so that we have at least 500 values for the median. Although the number of bootstrap samples to use is somewhat arbitrary, 500 subsamples is usually sufficient. To calculate a 90% confidence interval for the median, the sample medians are sorted into ascending order and the value of the 25th median (assuming exactly 500 subsamples were taken) is the lower confidence limit while the value of the 475th median (assuming exactly 500 subsamples were taken) is the upper confidence limit.\n\nSample Plot:",
null,
"This bootstrap plot was generated from 500 uniform random numbers. Bootstrap plots and corresponding histograms were generated for the mean, median, and mid-range. The histograms for the corresponding statistics clearly show that for uniform random numbers the mid-range has the smallest variance and is, therefore, a superior location estimator to the mean or the median.\n\nDefinition The bootstrap plot is formed by:\n• Vertical axis: Computed value of the desired statistic for a given subsample.\n• Horizontal axis: Subsample number.\nThe bootstrap plot is simply the computed value of the statistic versus the subsample number. That is, the bootstrap plot generates the values for the desired statistic. This is usually immediately followed by a histogram or some other distributional plot to show the location and variation of the sampling distribution of the statistic.\nQuestions The bootstrap plot is used to answer the following questions:\n• What does the sampling distribution for the statistic look like?\n• What is a 95% confidence interval for the statistic?\n• Which statistic has a sampling distribution with the smallest variance? That is, which statistic generates the narrowest confidence interval?\nImportance The most common uncertainty calculation is generating a confidence interval for the mean. In this case, the uncertainty formula can be derived mathematically. However, there are many situations in which the uncertainty formulas are mathematically intractable. The bootstrap provides a method for calculating the uncertainty in these cases.\nCautuion on use of the bootstrap The bootstrap is not appropriate for all distributions and statistics (Efron and Tibrashani). For example, because of the shape of the uniform distribution, the bootstrap is not appropriate for estimating the distribution of statistics that are heavily dependent on the tails, such as the range.\nRelated Techniques Histogram\nJackknife\n\nThe jacknife is a technique that is closely related to the bootstrap. The jackknife is beyond the scope of this handbook. See the Efron and Gong article for a discussion of the jackknife.\n\nCase Study The bootstrap plot is demonstrated in the uniform random numbers case study.\nSoftware The bootstrap is becoming more common in general purpose statistical software programs. However, it is still not supported in many of these programs. Both R software and Dataplot support a bootstrap capability.",
null,
""
] |
[
null,
"https://www.itl.nist.gov/div898/handbook/gifs/nvgtbr.gif",
null,
"https://www.itl.nist.gov/div898/handbook/eda/section3/gif/bootstra.gif",
null,
"https://www.itl.nist.gov/div898/handbook/gifs/nvgbrbtm.gif",
null
] |
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|
https://indsl.docs.cognite.com/auto_examples/numerical_calculus/plot_sliding_window_integration.html
|
[
"# Sliding window integration\n\nIn this example a synthetic time series is generated with a certain skewness (to make it more interesting) and a use the sliding window integration with a integrand rate of 1 hour. In other words, carry out a sliding window integration of the data over 1 hour periods.",
null,
"```import matplotlib.pyplot as plt\nimport numpy as np\nimport pandas as pd\n\nfrom indsl.ts_utils.numerical_calculus import sliding_window_integration\n\nnp.random.seed(1337)\ndatapoints = 5000\nx = np.random.randn(datapoints)\ny = np.zeros(len(x))\ny = x + 100 # initial synthetic start\nfor i in range(1, len(x)):\ny[i] = y[i - 1] + (x[i] + 0.0025) # and skew it upwards\n\nseries = pd.Series(y, index=pd.date_range(start=\"2000\", periods=datapoints, freq=\"10s\"))\nresult = sliding_window_integration(series, pd.Timedelta(\"1h\"))\n\nplt.figure(1, figsize=[9, 7])\nplt.plot(result, label=\"Cumulative moving window result, with matching units to raw time series data\")\nplt.plot(series, alpha=0.6, label=\"Raw timeseries data\")\nplt.legend()\nplt.ylabel(\"[-]/h\")\nplt.title(\"Sliding window integration with matching integrand rate and window\")\n_ = plt.show()\n```\n\nTotal running time of the script: ( 0 minutes 4.154 seconds)\n\nGallery generated by Sphinx-Gallery"
] |
[
null,
"https://indsl.docs.cognite.com/_images/sphx_glr_plot_sliding_window_integration_001.png",
null
] |
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|
http://www.eurisko.us/linear-and-logistic-regression-part-1-understanding-the-models/
|
[
"# Linear and Logistic Regression, Part 1: Understanding the Models\n\nBy George Meza on\n\nNote: This post is part 1 of a 3-part series: part 1, part 2, part 3.\n\nRegression is when you measure specific data points and fit a function to the trend. This can be used to establish connections between known variables and uncertainties, like the probability of a heart attack occurring via known traits. Another example could be determining the perfect amount of something, like the perfect amount of toppings on a pizza. You can relate the amount of toppings with customer satisfaction and determine an average amount of toppings that would lead to best reviews from customers.\n\nThere are two main types of regression I’m going to talk about, linear and logistic. Linear regression comes in the form of a straight line:",
null,
"Linear regression can be modeled with this equation:\n\n\\begin{align*} y = \\beta_0 + \\beta_1x \\end{align*}\n\nLogistic regression is a form of regression that comes in a sigmoid shape and has an upper and lower limit. This sigmoid shape starts at the lower limit, but once it increases and goes towards the higher limit, it levels out again, forming the graph’s s-like shape.",
null,
"Logistic regression can be modeled with this equation:\n\n\\begin{align*} y = \\dfrac{1}{1+e^{\\beta x}} \\end{align*}\n\nThis is the standard one-variable equation but both can be changed to fit multiple variables. Linear becomes\n\n\\begin{align*} y = \\beta_0 + \\beta_1 x_1 + \\beta_2 x_2 + \\beta_3 x_3 +\\ldots, \\end{align*}\n\nand logistic becomes\n\n\\begin{align*} y = \\dfrac{1}{1+e^{\\beta_0 + \\beta_1 x_1 + \\beta_2 x_2 + \\beta_3 x_3 + \\ldots}}. \\end{align*}\n\nThese regressions can be used to fit different scenarios. The crucial difference between the regressions is their limits.\n\nFor example, say we want to model the population of a species in a new environment. This would be a clear cut case of using logistic over linear because you can’t have the people of a species grow forever due to environmental constraints. This is where we should choose logistic regression over linear regression.\n\nLinear regression, on the other hand, doesn’t have bounds and is used to model different scenarios. Say you’re modeling experience with a game versus how many points you can score. The more you practice, the better you get, and the more points you can score. You could also use linear regression to model the yield of crops depending on how much water or fertilizer you use.\\newline\n\n## More on Logistic Regression\n\nLogistic regression is also perfect for determining probabilities. The reason you can use it to predict probabilities is due to its limits. Probability must be between $0\\%$ and $100\\%.$ A specific example of this is the weather. You can determine the chance of rain or sunshine using past weather patterns. Linear regression can’t model probabilities because there are no limits. There isn’t a $200\\%$ of something happening.\\newline\n\nLogistic regression also doesn’t have to be in the bounds of 0 to 1, though that is normal. Remember this equation:\n\n\\begin{align*} y = \\dfrac{1}{1+e^{\\beta x}} \\end{align*}\n\nIf you change the numerator, then that is how you can change the upper limit of your regression.\n\n\\begin{align*} y = \\dfrac{U}{1+e^{\\beta x}} \\end{align*}\n\nIn the above equation, $U$ is the upper limit. This can be used for, say, a movie rating system that goes from 0 to 10 or 0 to 5 stars.\n\nNot only can you change the upper bound, but you can also change the lower bound. Here is a generalized formula to fit a logistic regression for bounds of your choice:\n\n\\begin{align*} y = L + \\dfrac{U - L}{1+e^{\\beta x}} \\end{align*}\n\nIn this equation, $U$ is your upper limit, and $L$ is your lower limit. We would want to change the limits for different scenarios, just like usual.\n\nFor example, say you have a crowd of people and you want to predict the direction in which the crowd will move. This could go from $-180^\\circ$ to $180\\circ$ where $0^\\circ$ represents straight ahead, $-90^\\circ$ represents left, and $90^\\circ$ represents right. There are many different scenarios in which you wouldn’t want the standard $0$ to $1$ bounds.\n\nThis post is part 1 of a 3-part series. Click here to continue to part 2.\n\nAcknowledgements: Thanks to David Gieselman for reviewing this post.",
null,
""
] |
[
null,
"https://eurisko-us.github.io/images/blog/linear-and-logistic-regression-part-1-understanding-the-models-1.png",
null,
"https://eurisko-us.github.io/images/blog/linear-and-logistic-regression-part-1-understanding-the-models-2.png",
null,
"http://eurisko.us/images/headshots/george-meza.png",
null
] |
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|
https://www.physicsforums.com/threads/complex-area-enclosed-by-a-polygon.310902/
|
[
"# Complex- area enclosed by a polygon\n\n## Homework Statement\n\nRecall that the area enclosed by the polygon with vertices z1,z2,z3,...,zn is\n1/2I(z1conguatez2+z2congugatez3+...+zncongugatez1)\n\nShow that the area enclosed =1/2I$$\\Sigma$$zkcongugate(zsub(k+1)-zk).\nInterpret this sum as part of the approximating sum in the definition of the line integral about C of z congugatedz.\n\n## The Attempt at a Solution\n\nI don't even know where to start. I understand where the 1/2I(z.....) comes from, but once the summation comes in, I'm lost.\n\nThis really is not as hard as you think, especially since you already believe A=1/2 Im[ bar(z_1) z_2 + bar(z_2) z_3 +...+ bar(z_n) z_1 ]\n\nFor the summation 1/2 [ Im( bar(z_1) (z_2-z_1) + bar(z_2) (z_3-z_2) +...+ bar(z_n) (z_1- z_n) ] just \"remove parentheses\" and observe that bar(z) z=|z|^2 so its imaginary part is 0.\n\nAs for the line integral, use the fact that \\int_C f(z) dz is approximately \\lim\\sum_{k=1}^n f(z_k^*)(z_k-z_{k-1})\n\nDo i need to show using green's thm?\n\nI did some work and have a couple questions:\nLet x be polygon with its ordered vertices (the vertices you take in order\nas you travel around it) be .\n\nAlso say they are taken in a counterclockwise fashion.\nLet x be the line segment from zsubj,zsub(j+1) for j=1,2,...n-1. But for\nj=n define x as[zsubn,zsubn].\n\nDefine C-->C by f(x+iy)=-y+ix. I'm iffy on this part because I wasn't\nquite sure how to define f(x+iy).\nLet R denote the interior of the polygon, then by Green's theorem:\n1/2integral(f)=double integral(Green's Thm)=R\n\nThen I use a summation of integrals for each line segment.\n\nFor j=1,2,3,,,n-1 define y=zsubj +(zsub(j+1)-zsubj)t for 0<t<1.\nNotice that y parametrizes [zj,zj+1].\n\nf(x+iy)==y+ix=i(x-iy) so f(z)=izconjugate while y=zj+1-zj\n\nThen we evealuate about the integral form 0 to 1.\nintegral[f(y)ydt]=i(zj+1-zj)integral[yconjugatedt]=i(zj+1-zj)zjconjugate+1/2imod(zj+1-zj)^2\n\nSo adding all up, we get\n1/2i((z2-z1)z1conjugate+...+(z1-zn)znconjgate)+1/4i(mod(z2-z1)+...+mod(zn-z1))\n\nI got this far, but don't really understand what I'm doing. My biggest\nproblem is using Green's Theorem and understanding my choices for f(x+iy)\nand why i parameterized.\nI'm looking for some help explaining what I did and where to go from here.\n\nI was thinking more geometrically. I didn't use Green's Theorem at all.\n\nA triangle with vertices (0,0), (a,b), and (c,d) has area 1/2 * (ad - bc). This is a fact from analytic geometry. (The cross product of two vectors has a magnitude equal to the area of a certain parallelogram, and this triangle has half that area.)\n\nSo now, if you call z1 = (a,b) = a + bi and z2 = (c,d) = c + di, then it is interesting that 1/2 * conjugate(z1) * z2 has imaginary part equal to 1/2 * (ad - bc).\n\nNow generalize slightly, so that the third vertex is not at the origin, but instead is at z3. If you use translation, you can derive a formula for the area of this triangle.\n\nNow suppose you have a convex polygon with vertices z1, z2, ... , zn. Draw segments from z1 to each of the other vertices, dividing the polygon into triangles. Add up the areas.\n\nThis is how you can obtain the formula A=1/2 Im[ bar(z_1) z_2 + bar(z_2) z_3 +...+ bar(z_n) z_1 ].\n\nMaybe I'm not understanding your question.\n\nI don't think i show using geometry.\n\nI was thinking more geometrically. I didn't use Green's Theorem at all.\n\nA triangle with vertices (0,0), (a,b), and (c,d) has area 1/2 * (ad - bc). This is a fact from analytic geometry. (The cross product of two vectors has a magnitude equal to the area of a certain parallelogram, and this triangle has half that area.)\n\nSo now, if you call z1 = (a,b) = a + bi and z2 = (c,d) = c + di, then it is interesting that 1/2 * conjugate(z1) * z2 has imaginary part equal to 1/2 * (ad - bc).\n\nNow generalize slightly, so that the third vertex is not at the origin, but instead is at z3. If you use translation, you can derive a formula for the area of this triangle.\n\nNow suppose you have a convex polygon with vertices z1, z2, ... , zn. Draw segments from z1 to each of the other vertices, dividing the polygon into triangles. Add up the areas.\n\nThis is how you can obtain the formula A=1/2 Im[ bar(z_1) z_2 + bar(z_2) z_3 +...+ bar(z_n) z_1 ].\n\nMaybe I'm not understanding your question.\n\nI'm having trouble going from a triangle from a vertice of (0,0) to one without. My probelm is finding the area. Also I'm confused on going from our area formula to the summation. Any hints?\n\nLike I see how to show with the 3 vertices with one being (0,0) but then when I try to go to the summation, I'm not sure how to show that. Can someone help me show that? I know once I get there I will be able to get the complex line integral.\n\nOk, I figured out how to show that the summation area was the same with the (0,0) vertice by plugging in the vertices points. So I figured that out.\n\nOk, I'm all set except fo using a line integral about z congugate using the definition of a summation. i know the line integral of f(x) =Lim summation[f(psubk)(xsubk-xsubk-1)]\nMy problem is seeting this up. My initial assumption is to do (zsubk congugate)z_kcongugate -z-(k-1))? I think if I were to be able to set it up right, i'd be fine on evaluating it.\n\nWell, can't you say for a simple closed path C, that $$\\int_C bar(z)\\,dz$$ is approximated like this: take points $$z_1, z_2, \\dots, z_n$$ around the path C and on each segment $$[z_i,z_{i+1}]$$ let $$\\zeta_i = z_i$$, so that the integral is approximately\n\n$$bar(z_1) (z_2-z_1) + bar(z_2) (z_3-z_2) +...+ bar(z_n) (z_1- z_n)$$\n\nBut the imaginary part of this sum is twice the area bounded by the polygon with vertices $$z_1, z_2, \\dots, z_n$$\n\nThen say something about the limit.\n\nSo take the limit of bar zk(z_k+1-z_k)\nwhich will end up being 1/2[zk+z_k+1](z_k+1-zk}\n1/2[zk+1^2-zk^2}\n1/2[zn^2-z1^2]\n1/2(b^2-a^2)\n\nI'm sorry, I don't follow what you are saying when you say \"which will end up being 1/2[zk+z_k+1](z_k+1-zk}\"\n\nIn the \"Riemann sum,\" you don't have to take the midpoint of segment $$[z_i,z_{i+1}]$$\n\nYou can take any point on the segment. I choose to take $$z_i$$\n\nHowever, I don't know if this comment is related to what you are saying or not.\n\nI'm sorry, I don't follow what you are saying when you say \"which will end up being 1/2[zk+z_k+1](z_k+1-zk}\"\n\nIn the \"Riemann sum,\" you don't have to take the midpoint of segment $$[z_i,z_{i+1}]$$\n\nYou can take any point on the segment. I choose to take $$z_i$$\n\nHowever, I don't know if this comment is related to what you are saying or not.\n\nOk then i'm a bit confused since i want to evaluate using the limit definition and want to show an answer like how if I chose to just do zdz I would get 1/2(b^2-a^2). Do you see what I'm seeing?\n\nMaybe I see what you are meaning.\n\nBut I don't mean evaluate a limit. The original question was to \"Interpret this sum as part of the approximating sum in the definition of the line integral about C of z congugatedz.\"\n\nSo as I understand the problem, you simply make the statement or observation that $$\\int_C bar(z)\\,dz$$ is approximately $$bar(z_1) (z_2-z_1) + bar(z_2) (z_3-z_2) +...+ bar(z_n) (z_1- z_n)$$\n\nI see what you mean. Ok then, what if I wanted to go a little further and evaluate using that limit definition.\n\nOh, so it has to be half the sum of the imaginary parts because of the imaginary parts being twice the area?\n\nAnd I can say something about when evaluating the limit we use 1/2 of the imaginary part?\n\n$$bar(z_1) (z_2-z_1) + bar(z_2) (z_3-z_2) +...+ bar(z_n) (z_1- z_n)\\approx\\int_C bar(z)\\,dz$$\n\nso take imaginary parts of both sides and divide by 2\n\n$$bar(z_1) (z_2-z_1) + bar(z_2) (z_3-z_2) +...+ bar(z_n) (z_1- z_n)\\approx\\int_C bar(z)\\,dz$$\n\nso take imaginary parts of both sides and divide by 2\n\nOk, I think this makes sense because we are picking a point in the interval and from that we get that, but we get twice the area, so we must divide by 2. Correct me if I'm wrong.\n\nThis is how I analyzed it (hope I don't confuse things):\n\nWe can start with the known relation for the area encircled by a closed contour (see Calculus text book, \"Topics in Vector Calculus\") :\n\n$$\\frac{1}{2}\\mathop\\oint\\limits_{C} (-ydx+xdy)=\\mathop\\iint\\limits_{R} dA$$\n\nNote the expression $$-ydx+xdy$$ is the real part of the contour integral:\n\n$$\\oint f(z)dz=\\oint udx-vdy+i\\oint vdx+udy$$\n\nwith $$f(z)=-y-ix=-i\\overline{z}$$\n\nNow, by Green's Theorem:\n\n$$\\oint vdx+udy=\\oint (-xdx-ydy)=\\mathop\\iint\\limits_{R}\\frac{\\partial v}{\\partial x}-\\frac{\\partial u}{\\partial y}=0$$\n\nand therefore:\n\n$$\\mathop\\iint\\limits_R dA=-\\frac{i}{2}\\oint \\overline{z}dz$$\n\nWe can parameterize the straight-line segments $$(z_n,z_{n+1})$$ as:\n$$z_n(t)=z_n+(z_{n+1}-z_n)t$$\n\nand therefore:\n\n\\begin{align*} \\mathop\\oint\\limits_{z_n(t)} \\overline{z}dz&=\\int_0^1 \\overline{z_n(t)}(z_{n+1}-z_n)dt \\\\ &=(z_{n+1}-z_n)\\int_0^1 \\left(\\overline{z_n}+\\overline{(z_{n+1}-z_n)}t\\right)dt\\\\ &=(z_{n+1}-z_n)\\left[1/2\\left(\\overline{z_n}+\\overline{z_{n+1}}\\right)\\right] \\end{align*}\n\nThen:\n\\begin{align*} \\mathop\\iint\\limits_{R} dA&=-\\frac{i}{4}\\sum_{j=1}^N (z_{n+1}-z_n)\\left(\\overline{z_n}+\\overline{z_{n+1}}\\right);\\quad z_{N+1}=z_1 \\\\ &=-\\frac{i}{4}\\sum_{j=1}^N\\big(z_{n+1}-z_{n-1}\\big)\\overline{z_n};\\quad z_0=z_N,\\:z_{N+1}=z_1 \\end{align*}\n\nLast edited:"
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https://testbook.com/question-answer/given-below-are-two-statementsstatement-inbsp--5ff33d1e2796387ecc058c6f
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[
"",
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"# Given below are two statements:Statement I: As the alpha level becomes more stringent - goes from 0.05 to 0.01 the power of a statistical test decreasesStatement II: A directional hypothesis leads to more power than a non-directional hypothesisIn the light of the above Statements, choose the most appropriate answer from the options given below:\n\nFree Practice With Testbook Mock Tests\n\n## Options:\n\n1. Both Statement I and Statement II are true\n\n2. Both Statement I and Statement II are false\n\n3. Statement I is correct but Statement II is false\n\n4. Statement I is incorrect but Statement II is true\n\n### Correct Answer: Option 1 (Solution Below)\n\nThis question was previously asked in\n\nUGC NET Paper 2: Education 12th Nov 2020 Shift 1\n\n## Solution:\n\nHypothesis testing\n\n• Hypothesis testing is a procedure that assesses two mutually exclusive theories about the properties of a population.\n• For Hypothesis testing, the two hypotheses are as follows:\n1. Null Hypothesis\n2. Alternative hypothesis\n• There are two errors defined, both are for the null hypothesis condition\n• Type-I error corresponds to rejecting H0 (Null hypothesis) when H0 is actually true, and a Type-II error corresponds to accepting H0 (Null hypothesis)when H0 is false. Hence four possibilities may arise:\n\n•",
null,
"• Decreasing alpha from 0.05 to 0.01 increases the chance of a Type II error (makes it harder to reject the null hypothesis).\n• Statistical power or the power of a hypothesis test is the probability that the test correctly rejects the null hypothesis.\n• The higher the statistical power for a given experiment, the lower the probability of making a Type II (false negative) error. That is the higher the probability of detecting an effect when there is an effect. In fact, the power is precisely the inverse of the probability of a Type II error.\n• Low Statistical Power: Large risk of committing Type II errors, e.g. a false negative.\n• High Statistical Power: Small risk of committing Type II errors.\n• Power is defined as 1 — the probability of type II error (β). In other words, it is the probability of detecting a difference between the groups when the difference actually exists (ie. the probability of correctly rejecting the null hypothesis). Therefore, as we increase the power of a statistical test we increase its ability to detect a significant difference between the groups.\n\nHence, Statement I \"As the alpha level becomes more stringent - goes from 0.05 to 0.01 the power of a statistical test decreases\" is true.\n\nHypothesis:\n\n• A hypothesis is a formal affirmative statement predicting a single research outcome, a tentative explanation of the relationship between two or more variables.\n• To give an example, “Discussion method gives better academic scores than lecture method of teaching” or \" There is no significant difference between teaching aptitude of male and female teachers\".\n• In hypothesis-generating research, the researcher explores a set of data searching for relationships and patterns and then proposes hypotheses that may then be tested in some subsequent study.\n\nTypes of Research Hypotheses\n\nAlternative Hypothesis\n\n• The alternative hypothesis states that there is a relationship between the two variables i.e. one variable has an effect on the other.\n• For e.g. There is a significant difference in the aptitude of urban and rural students.\n\nNull Hypothesis:\n\n• The null hypothesis states that there is no relationship between the two variables i.e. one variable does not have an effect on another variable.\n• For e.g. There is no significant difference in the aptitude of urban and rural students.\n\nNondirectional Hypothesis:\n\n• A two-tailed non-directional hypothesis predicts that the independent variable will have an effect on the dependent variable, but the direction of the effect is not specified.\n• E.g., There is a difference in vocabulary between males and females in some numbers.\n\nDirectional Hypothesis\n\n• A one-tailed directional hypothesis predicts the nature of the effect of the independent variable on the dependent variable. Here direction is specified.\n• E.g., females have a better vocabulary than males.\n\nDirectional tests are more powerful than non-directional tests. As they show direction and the critical region is located in one tail. Whenever we are certain about direction, this is a better choice than the non-directional hypothesis.\n\nHence, Statement II \"A directional hypothesis leads to more power than a non-directional hypothesis is true\"."
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"https://cdn.testbook.com/qb_resources/desktop_pyp.jpg",
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"https://storage.googleapis.com/tb-img/production/20/10/F1_Alka_Madhu_19.10.20_D1.png",
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https://www.hindawi.com/journals/mpe/2010/149385/
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[
"#### Abstract\n\nModel-based analysis and synthesis applied to the dynamics, guidance, and control of an autonomous undersea vehicle are presented. As the dynamic model for describing vehicle motion mathematically, the equations of motion are derived. The stability derivatives in the equations of motion are determined by a simulation-based technique using computational fluid dynamics analysis. The dynamic model is applied to the design of the low-level control systems, offering model-based synthetic approach in dynamics and control applications. As an intelligent navigational strategy for undersea vehicles, we present the optimal guidance in environmental disturbances. The optimal guidance aims at the minimum-time transit of a vehicle in an environmental flow disturbance. In this paper, a newly developed algorithm for obtaining the numerical solution of the optimal guidance law is presented. The algorithm is a globally working procedure deriving the optimal guidance in any deterministic environmental disturbance. As a fail-safe tactic in achieving the optimal navigation in environments of moderate uncertainty, we propose the quasi-optimal guidance. Performances of the optimal and the quasi-optimal guidances are demonstrated by the simulated navigations in a few environmental disturbances.\n\n#### 1. Introduction\n\nIn this article, we present model-based analysis and synthesis applied to the dynamics, guidance, and control of an autonomous undersea vehicle (AUV). The vehicle dynamics is one of the most important concerns in designing and developing an AUV, while the guidance and control are the key issues in achieving the desired vehicle performance. Our approach deals with these individual but closely interrelated issues in a consistent way based on the model-based simulations.\n\nIn our research, as the dynamic model of an AUV, we employ a set of equations of motion describing the coupled six-degree-of-freedom (6-DOF) behaviour in three-dimensional (3D) space. In the linearized form of the equations of motion, to complete the dynamic model of an AUV, we have to determine the so-called stability derivatives or hydrodynamic coefficients. There are many well-established approaches for determining the stability derivatives of aerial vehicles [1, 2] or marine vehicles , which are based on either experiment or theoretical prediction. While the experimental approach allows direct measurement of the fluid dynamic forces and moments acting on the vehicle, it requires a large amount of time, labour, expense, as well as an experimental facility. On the other hand, a few state-of-the-art techniques are now available for predicting the stability derivatives theoretically . Most of them, however, are specialized for deriving the stability derivatives for the dynamics of conventional airplanes [4, 5] or ships , making them hard to directly apply to the modelling problems related to the dynamics of a specific AUV. In this respect, we introduce a general-purpose technique for deriving the dynamic model of an undersea vehicle, primarily depending on the computational fluid dynamics (CFD) analysis.\n\nThe derived dynamic model is directly applied to the model-based design of the motion control systems of an AUV. Two proportional-integral-derivative (PID) type low-level controllers are employed to make a vehicle follow the desired trajectories in the longitudinal and lateral planes, represented as time sequences of the depth (altitude) and heading.\n\nAs an intelligent high-level control of AUVs, a strategy of optimal guidance is presented. The optimal guidance proposed in this research is the minimum-time guidance in sea current environments, allowing a vehicle to reach a destination with the minimum travel time. When the power consumption of an AUV is controlled to be constant throughout the navigation, the navigation time is directly proportional to the total energy consumption. Released from the umbilical cable, an AUV has to rely on restricted energy stores during an undersea mission. Therefore, for an AUV, minimizing navigation time offers an enhanced potential for vehicle safety and mission success rate. We present a newly developed numerical procedure for deriving the optimal heading reference, by tracking which vehicle achieves the minimum-time navigation in a given sea current disturbance. The proposed procedure is systematic and seeks the solution in a global manner in any deterministic current field, whether stationary or time-varying. Moreover, unlike other path-finding algorithms, such as dynamic programming (DP) or generic algorithms (GAs) , our procedure does not require a computation time increase for the time-varying problems.\n\nIn real environments of AUV navigation, there are some factors that can cause failure in realizing the proposed optimal guidance strategy [10, 11]. Some examples are environmental uncertainties, severe sensor noises, or temporally faulty actuators. Though these risk factors significantly affect the realization of optimality in actual sea navigation, they have not been seriously treated in most of related literatures. In this article, we present the concept of quasioptimality as a fail-safe strategy for realizing the proposed optimal navigation.\n\n#### 2. An AUV “R-One”\n\nIn this article, we practice our strategy in dynamics, guidance, and control on the AUV “R-One,” a long-range cruising type AUV, developed by the Institute of Industrial Science (IIS), the University of Tokyo . Figure 1 shows the overall layout of R-One.\n\nFigure 2 shows the coordinate system and the actions of the actuators installed in the R-One. The axis-deflectable main thruster keeps or changes the vehicle's kinematic states in the horizontal plane. Two elevators and two vertical thrusters play the same role in the vertical plane.\n\n#### 3. Modelling Vehicle Dynamics\n\n##### 3.1. Equations of Motion for Vehicle Dynamics\n\nThe equations of motion describing the vehicle motion mathematically can be derived from the conservation law of the linear and the angular momenta with respect to the inertial frame of [1, 2]. The equations of motion (3.1) describing the 6-DOF motion of an AUV are defined with respect to the body-fixed frame of reference shown in Figure 3, in which the origin is taken at the vehicle's center of gravity. The procedures for deriving the equations of motion of an aerial vehicle which is quite similar to (3.1) are found in [1, 2]. It should be noted here however, that in deriving (3.1) by referring to the equations of motion for the aerial vehicles shown in [1, 2], the hydrostatic loads which do not appear in the flight dynamics have to be additionally involved In (3.1), U, V, W, and P, Q, R are the x, y, z components of linear and angular velocities. and I represent volume, mass, and mass moments or products of inertia of a vehicle, and ρ and are constants expressing water density and gravitational acceleration. Hydrodynamic forces and moments are represented by X, Y, Z, and L, M, N, each of which is the component in the direction of x, y, z. and are so-called Euler angles to be defined in the coordinate transformation between the body-fixed and the inertial frames of reference. The is the z-directional displacement of the buoyancy center of the vehicle. The equations of motion are frequently linearized for use in stability and control analysis as mentioned in . The following equations are the linearized forms of (3.1), in which denote small amounts of velocities, and angular velocities and displacements, perturbed from their reference values, which are expressed by their uppercase letters In general, to complete the linearized equations of motion for use in stability and control analysis, hydrodynamic loads are expanded and linearized on the assumption that they are functions of the instantaneous values of the perturbed velocities, accelerations, and control inputs. Thus, the expanded expressions of the hydrodynamic loads are obtained in the form of a Taylor series in these variables, which is linearized by discarding all the higher-order terms. For example, X is expanded as where The subscript zero in (3.3a) indicates a reference condition where the derivatives are evaluated. In (3.3a) and (3.3b), derivatives such as or are called stability derivatives . By expanding all the external hydrodynamic loads introducing stability derivatives of their dynamic correlations, the equations of motion (3.2) are expressed by means of the stability derivatives as where represents the rpm of the main thruster.\n\n##### 3.2. Evaluation of Stability Derivatives by CFD Analyses\n\nAs noticeable in (3.3), within the framework of small perturbation theory, constructing the dynamic model is, in effect, reduced to the determination of the stability derivatives defined in the linearized equations of motion. The most commonly and widely employed approaches for evaluating the stability derivatives are the wind tunnel test for aerial vehicles and the towing tank test for marine vehicles . These experimental approaches, however, require a huge experimental facility and a large workforce, which makes them expensive and laborious, even when the test is for a single model. In this article, we present a model-based approach for evaluating the stability derivatives. In the approach, dominant stability derivatives are evaluated from the hydrodynamic loads which are obtained by CFD analyses. When we are to evaluate the value of in (3.3) defined at a reference speed of for example, we conduct CFD analyses repeatedly at the cruising speeds of where is the reference cruising speed and η is the perturbation ratio of By taking central difference approximation of X with respect to by using the values obtained at we can derive defined at However, while the majority of dominant stability derivatives are able to be evaluated by this technique, there are other stability derivatives which are not. For such stability derivatives, the simplified estimation formulae proposed in the field of flight dynamics [1, 2] are modified and applied.\n\nIn our CFD analyses, we used a commercial fluid dynamics solver called “Star-CD,” developed by CD-adapco . The Star-CD is a Navier-Stokes solver based on the finite difference numerical scheme. Like other famous commercial CFD solvers such as FLUENT or ANSYS, Star-CD also has shown numerous field application results that it replicates experimental model results with acceptably fine accuracy [14, 15]. The Star-CD derives the numerical solution by pressure-implicit split-operator (PISO) algorithm, which is a well-known, robust scheme with predictor-corrector steps. In our CFD analyses, the problem-specific high Reynolds number (Re) requires a proper turbulence model. In using Star-CD, we selected the Reynolds-Averaged Navier-Stokes (RANS) turbulence model, which is one of the most widely used turbulence model in engineering applications of moderate turbulent conditions .\n\nFigure 4 shows the grid system for evaluating the hydrodynamic loads by CFD analyses. To generate a computationally robust, structured grid system adapting to the complicated aftbody geometry of the vehicle, we employed a grid generation technique called the multiblock method .\n\nNot only estimating the drag force of acceptable accuracy is the primary concern in our CFD analyses for deriving the stability derivatives, it also serves as the most fundamental measure to evaluate a CFD solver . After completing the hull structure, drag forces acting on the R-One at three cruising speeds were investigated by means of the towing tank tests . In Figure 5, two drag curves, obtained by CFD calculations and tank tests, are shown together. The drag curves shown are quadratic interpolations of the raw data set of drags, calculated and measured at the cruising speeds of 1.03, 1.54, and 2.06 m/s, respectively. Validity of the quadratic interpolation is based on the fact that within the small Re interval, drag of an immersed body has quadratic dependency on its advance speed . In Figure 5, the gradients of two drag curves are also expressed. As seen in the figure, drags obtained by CFD calculations are more or less excessive than the ones by tank tests. However, it is noted that the gradients of drags show close similarity between CFD analyses and tank tests, which advocates our approach to evaluating the stability derivatives principally by means of the CFD analyses.\n\nFigure 6 shows the pressure distribution with a few selected streamlines along the body surface of R-One. By integrating the pressure over the entire body surface, hydrodynamic loads are obtained.\n\nIt is generally known and also noticeable from (3.3) that, according to the coupling relation, linearized equations of motion are split into two independent groups: longitudinal equations for surge, heave, and pitch, and lateral equations for sway, roll, and yaw . In Tables 1(a) and 1(b), the longitudinal and lateral stability derivatives appearing in (3.3) are summarized. By substituting all stability derivatives in (3.3) with their corresponding numerical values in the tables, the dynamic model of R-One is completed.\n\n##### 3.3. Vehicle Motion Simulation\n\nState-space forms of the longitudinal and the lateral equations of motion for R-One, completed by assigning the numerical values in Table 1 to corresponding stability derivatives in (3.3), are represented as follows.Longitudinal equations of motion for R-One:Lateral equations of motion for R-One: By solving (3.5a) and (3.5b) in the time domain with appropriate initial conditions and actuator inputs, motion responses of the R-One are computed. In the inertial navigation system (INS) installed in R-One, not only vehicle kinematics but also time sequences of the actuator inputs during an undersea mission are recorded. In Figure 7, simulated vehicle trajectories are compared with actual vehicle trajectories recorded during the Teisi knoll survey mission . As noticeable from the figure, the dynamic model of R-One implemented by our model-based approach provides motion responses exhibiting sufficiently good agreement between the simulated and actual vehicle trajectories.\n\n#### 4. Tracking Control Design\n\nThe controller implemented for the motion control of R-One is based on the PID compensation. Needless to say that, PID-type controller is the most commonly and widely used controller for most artificial control systems. However, in designing a PID controller, precise plant dynamics is a key prerequisite to ensuring acceptably good control performance. Deriving a precise plant dynamics is not easy in some cases. For this reason, during the past three decades, a few significant attempts have been made to provide controller models that do not depend on a precise description of the plant model in its design . Neural network (NN) controllers based on the self-organizing map or fuzzy logic controllers are the most famous ones in such attempts [21, 22]. In order to derive a practically useful controller by NN or fuzzy logic, however, we have to ensure huge random diversity in training data. This is a very difficult task in a real world problem, because, in general, we do not have any definitive guidelines for deciding whether the prepared training data is biased or not [22, 23].\n\nTo change or keep the kinematic states of the vehicle, two independent low-level controls were implemented in the R-One: the depth (altitude) control for the longitudinal motion and the heading control for the lateral motion. Configurations of the depth and the heading controls are depicted in Figures 8(a) and 8(b).\n\nTo build the mathematical models for the control systems shown in the Figure 8, transfer functions of and are extracted from (3.5a) and (3.5b). Then, PID-tuning is carried out to determine the optimal values of controller gains from the standpoint of system robustness and swiftness of response. In determining the optimal gain values, we used the model-based control system design tool called “SISO Design Tool,” offered by the “Control System Toolbox” included in “Matlab” .\n\nAn example of the performance result for the designed control systems is shown in Figure 9, where it is clearly seen that the designed heading controller lets the vehicle follow the heading reference with sufficient swiftness and small overshoot.\n\n#### 5. Optimal Guidance of AUV\n\n##### 5.1. Background\n\nThe sea environment contains several disturbances, such as surface waves, wind, and sea currents. Among them, the sea current is known to be the most significant disturbance for the dynamics of an undersea vehicle as it directly interacts with the vehicle motion [3, 8, 11]. Considering the guidance problem to make a vehicle transit to a given destination in a region of sea current, it is quite natural that there arises a navigation time difference according to the selection of an individual navigation trajectory. The problem of the minimum-time vessel guidance in a region of current flow has interested people as long years ago as ancient Greece . However, since the problem requires a minimization technique of the functionals, it had hardly been treated mathematically until the advent of the calculus of variations. On the basis of this mathematical tool, Bryson and Ho derived the minimum-time guidance law of a surface vessel in a region of a surface current flow. Though the law is an optimal controller of explicit form, obtaining its solution is not easy since it actually is a so-called two-point boundary value problem. As an ad hoc approach for the minimum-time navigation problem in a linearly varying, shear flow-like current distribution presented by Lewis and Syrmos, a graphical solution finding technique has been presented . As is naturally expected, however, such an approach is problem-specific and lacks universality in its applicability. Papadakis and Perakis treated the minimum-time routing problem of a vessel moving in a wave environment. In their approach, by subdividing the navigation region into several subregions of different sea states, the path for the optimal routing is obtained by the DP approach. Aside from the difficulties in constructing a numerical solution procedure for their approach, it has a problem that the solution significantly depends on the features of regional subdivision. As a completely discrete and nonlinear approach, the cell mapping technique was applied to derive the minimum-time tracking trajectory to capture a moving target in a deterministic vortex field . It, however, has the same problem of regional subdivision as is inherent in the approach by Papadakis and Perakis , which might lead to the divergence due to numerical instability.\n\nIn this research, we propose a newly developed procedure for obtaining the numerical solution of the optimal guidance law, which achieves the minimum-time navigation of a vehicle in a given current field. The algorithm of our solution procedure is simple but consistently applicable to any current field if only the distribution of which is deterministic. As a fault-tolerable strategy for putting the proposed optimal navigation into practice, the concept of quasioptimality is introduced. The basic idea of the quasioptimal navigation is quite simple and, in effect, consists of the on-site feedbacks of the optimal guidance revisions.\n\n##### 5.2. The Optimal Guidance Law\n\nIn our optimal guidance problem, we employed the guidance law presented by Bryson and Ho as where represents the vehicle heading as defined in Figure 10, and are x, y components of the sea current velocity. The detailed procedure of deriving (5.1) is well explained in . Though Bryson and Ho derived (5.1) on the assumption of a stationary flow condition, we have shown that it is also valid for time-varying currents, like tidal flows . Equation (5.1) is a nonlinear ordinary differential equation of an unspecified vehicle heading Though it seems that the solution would be readily obtainable by using a suitable numerical scheme, such as Runge-Kutta, there still remains a significant shortfall: while (5.1) defines an initial value problem, its solution cannot be obtained with an arbitrarily assigned initial heading. If we solve (5.1) with an arbitrary initial value of a vehicle following the solution of (5.1) as the heading reference does not arrive at the destination. This is because (5.1) is, in fact, a two-point boundary value problem, the correct initial value of which constitutes a part of the solution.\n\n##### 5.3. Numerical Solution Procedure\n\nAs mentioned in the previous section, to implement the optimal guidance for an AUV navigation by AREN, it is necessary to make a reference navigation beforehand. The simplest guidance satisfying the vehicle's arrival at the destination may be proportional navigation (PN) [25, 26]. In PN, vehicle heading is continuously adjusted so that the line of sight (LOS) is directed toward the target point. In our work, we employ PN as the reference navigation.\n\n##### 6.2. Optimal Navigation in a Shearing Flow\n\nThe first numerical example in this research is an optimal navigation in a current disturbance of the linear shear flow, taken from Bryson and Ho . The current velocity in this problem is described by where Uc and h are set to be 1.544 m/s and 100 m, respectively. Starting from the initial position at the vehicle is directed to move toward the destination at the origin in this example. Due to its simplicity, the current distribution of (6.1) allows derivation of the analytic optimal guidance law expressed as where is the vehicle heading at the final state.\n\nNavigation trajectories are shown in Figure 13. In the reference navigation by PN, significant adverse drift happens at the initial stage, since within the region of current flow speed exceeds the vehicle speed relative to the water. The optimal guidance detours the vehicle across the upper half plane, taking advantage of the favorable current flow. Navigation times by PN and optimal guidance are 353.7 and 739.2 s, respectively, indicating a 52% decrease in navigation time by the optimal guidance proposed.\n\n##### 6.3. Optimal Navigation in a Time-Varying Flow\n\nThe next numerical example is an optimal navigation in a time-varying current flow. In actual sea environments, for a lot of currents the direction and the magnitude of their velocities change continuously like tidal flows. As mentioned previously, we have proved that the optimal guidance law (5.1) is also valid for time-varying currents as well as for stationary ones. Therefore, once the flow velocity distribution in a navigation region is described as a function of the position and time, our numerical scheme is expected to be effective and thus realize the minimum-time navigation in a time-varying flow.\n\nThe current distribution in this example is the same one that we took in the previous example. In this example, however, while the optimal navigation is performed with the exact information about the current flow distribution, assuming a situation of incorrect localization due to sensor failure, mismatched current flow information is fed to the vehicle guidance controller in the quasioptimal case. The time interval during which mismatched information is taken is s. Starting at 252.0 s, optimal guidance revised on the basis of the correct current flow information achieves the quasioptimal navigation. Figure 16 shows the time sequence of the vehicle headings during the optimal and the quasioptimal navigations.\n\nAs expected naturally, the performance of the quasioptimal navigation is not as high as that of the optimal one. While the optimal guidance completes the navigation at 623 s, the quasioptimal one continues it until 702 s. Note that in Figure 16, an abrupt heading change occurs at 252.0 s during the quasioptimal navigation.\n\n##### 6.5. Optimal Navigation in Northwestern Pacific\n\nIn what follows, we try to accomplish the minimum-time navigation within a real sea environment. The sea region selected for this optimal navigation example is located in the Northwestern Pacific Ocean near Japan. The current field considered here is an actual measurement of the surface flow provided by the Japan Meteorological Agency, available at http://www.data.kishou.go.jp/db/kobe/db_kobe.html. The most notable environmental characteristic in this sea region is the current field dominated by the Kuroshio. The Kuroshio is a strong western boundary current in the Northwestern Pacific Ocean, flowing northeastward along the eastern coast of Japan . As seen in Figure 18, Kuroshio-induced flows moving eastwards constitute the principal stream in this region.\n\nIn the current field data from the database of the Japan Meteorological Agency, current velocity is defined only on the predefined, large-scale grid nodes covering the sea region. As noticeable from (5.1), however, in order to derive the optimal heading reference, current velocity and its gradient at every vehicle position have to be available. As a remedy for this data deficiency, we estimate the current velocity and its gradient by interpolating the predefined values on grid nodes surrounding the present vehicle position. In applying the interpolation, the nearest grid node to the present vehicle position has to be identified first. Then, the current velocity at the present vehicle position is estimated by 2D biquadratic interpolation utilizing values on the nearest node and surrounding eight nodes, as shown in Figure 17. Gradients of current velocities are obtained by the same manner. Since the velocity gradients are not provided from the database, however, prior to the interpolation, we calculate their nodal values by finite difference approximations.\n\nThe description of the navigation to be optimized here is as follows. Starting from an initial position, the vehicle is to transit to a destination in a mission-specified area, where an undersea survey mission is to be undertaken. Taking into account the inshore or harbor launch, the vehicle is made to start from the initial position off Minamiizu, the southern extreme of the Izu peninsula (Figure 18).\n\nFigure 18 shows the navigation trajectories achieved by the PN and optimal guidance. As shown in the figure, like the preceding examples in which exact values of current velocity and its gradients are available anywhere in the navigation region, the vehicle tracks the optimal reference trajectory with a negligibly small deviation. This indicates that our strategy of optimal navigation is also valid in the real sea current data, originally defined only on the coarsely defined discrete grid nodes.\n\nIn Figure 18, with the vehicle moving under PN, having reached the region of the mainstream of Kuroshio, its speed relative to the ground is remarkably reduced. This is because in this region, for the vehicle following the guidance of PN, the direction of its advance velocity is placed out-of-phase with the direction of the mainstream of Kuroshio. In the optimal navigation, the former segment of the navigation trajectory is formed along the shoreline until the vehicle reaches a point off the southern extreme of Kii peninsula. By taking this route, the vehicle attains a speed increase, riding the coastal current mainly flowing westwards. Note that upon the vehicle reaching a point off the southern extreme of the Kii peninsula, the optimal trajectory takes a large turn, slightly rolling back eastward from the destination. This slight rollback is the result of the optimal guidance's action to prevent the vehicle's advancing direction from being out-of-phase with that of the mainstream of Kuroshio, leading to the optimal navigation trajectory shown. The optimal trajectory obtained reveals one of the significant advantages of our approach over GA-based path planning which is not able to generate the optimal path with interim backward intervals by its nature [7, 9].\n\nNavigation times by PN and optimal guidance are 232198 and 212006 s, respectively, indicating an 8.7% decrease in navigation time by the optimal guidance proposed.\n\n#### 7. Conclusions and Future Works\n\nIn this article, model-based analysis and synthesis to the following three research fields in AUV design and development have been presented.(i)Dynamic system modelling of an AUV.(ii)Motion control design and tracking control application.(iii)Optimal guidance of an AUV in environmental disturbances.\n\nIn the dynamic system modelling of the AUV R-One, we evaluated the hydrodynamic loads by using CFD analyses. Then, by differentiating a hydrodynamic load with respect to the amount of a perturbation, corresponding stability derivatives were obtained. Using the stability derivatives evaluated, we built up the dynamic model of the R-One, which is characterized to be 6-DOF [3 longitudinal (surge, heave, pitch) + 3 lateral (sway, roll, yaw)], linear, and multiple-input multiple-output (MIMO).\n\nDepth and heading control systems are designed by employing controller models based on the PID compensations. In the PID-tuning, model-based simulations for the depth and the heading controls are exploited in determining the optimal gains.\n\nConcerning the guidance problem of AUVs moving in sea environmental disturbances, a newly developed procedure for obtaining the numerical solution of the optimal guidance law to achieve the minimum-time navigation has been presented. The optimal heading is obtained as the solution of the optimal guidance law, which is fed to the heading controller as the optimal reference. Reduced computational cost is one of the outstanding features of the proposed procedure. Numerical calculations of the optimal navigation examples presented in this article except for the last one are completed within 10 minutes on a single core 2.4 GHz windows XP platform. Moreover, unlike other path-finding algorithms such as DP or GAs, our procedure does not require a computation time increase for the time-varying problems.\n\nAs a fail-safe strategy for putting the proposed optimal navigation into execution, the concept of quasioptimal guidance has been proposed. The fact that there actually are several possible actions lessening the chance of achieving optimality emphasizes the practical importance of the quasioptimal navigation.\n\nWe have not considered the problem of unknown or nondeterministic currents. Our approach cannot be applied to an entirely unknown environment. For a sea region containing partially or coarsely defined currents, however, an estimated distribution can be built via interpolation and extrapolation, as shown in the last navigation example. The estimation possibly contains more or less uncertainty. Notably, however, it is the quasioptimal strategy that can cope with the environmental uncertainty. When the uncertainty in the estimation is significant, convergence may not be guaranteed.\n\n#### Acknowledgments\n\nThe first author would like to express special thanks to Dr. Makio Kashino, Dr. Eisaku Maeda, and Dr. Yoshinobu Tonomura with NTT."
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https://skoolers.com/tag/oxidation-and-reduction/
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"# Tag: Oxidation and Reduction\n\n## Introduction to Redox Reactions\n\nIntroduction to Redox Reactions One of the most important types of reactions and the focus of this chapter is the Reduction-Oxidation reaction more commonly known as Redox Reactions. Redox reactions are simultaneously reactions in which both oxidation and reduction occurs at the same time. Understanding the principles of redox reactions enable chemists to use them…\n\n## Calculating oxidation number\n\nCalculating oxidation number In the previous section we learned of instances in which oxidation and reduction occur. From this, we know that oxidation occurs when the oxidation number increases and reduction occurs when the oxidation number decreases or reduces. We tend to calculate the oxidation state in the instances when we cannot directly tell if…\n\n## Redox Equation and Half equation\n\nRedox Equation and Half equation 2Mg+ O2 → 2MgO 0 0 +2 -2 The diagram above shows that Mg to Mg 2+ is oxidation (reducing agent) O2 to O2– reduction occurs (oxidizing agent) Half equation Some reactions maybe written in 2 halves (the oxidation half and the reduction half). The 2 parts are called the…"
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https://pdfkul.com/efficient-computation-of-significance-levels-for-multiple-_5a04f9e61723ddf3c5ac04d7.html
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"Am. J. Hum. Genet. 75:424–435, 2004\n\nEfficient Computation of Significance Levels for Multiple Associations in Large Studies of Correlated Data, Including Genomewide Association Studies Frank Dudbridge1,2 and Bobby P. C. Koeleman3 1\n\nMRC Rosalind Franklin Centre for Genomics Research, and 2MRC Biostatistics Unit, Cambridge, United Kingdom, and 3Department of Medical Genetics, University Medical Centre Utrecht, Utrecht\n\nLarge exploratory studies, including candidate-gene–association testing, genomewide linkage-disequilibrium scans, and array-expression experiments, are becoming increasingly common. A serious problem for such studies is that statistical power is compromised by the need to control the false-positive rate for a large family of tests. Because multiple true associations are anticipated, methods have been proposed that combine evidence from the most significant tests, as a more powerful alternative to individually adjusted tests. The practical application of these methods is currently limited by a reliance on permutation testing to account for the correlated nature of singlenucleotide polymorphism (SNP)–association data. On a genomewide scale, this is both very time-consuming and impractical for repeated explorations with standard marker panels. Here, we alleviate these problems by fitting analytic distributions to the empirical distribution of combined evidence. We fit extreme-value distributions for fixed lengths of combined evidence and a beta distribution for the most significant length. An initial phase of permutation sampling is required to fit these distributions, but it can be completed more quickly than a simple permutation test and need be done only once for each panel of tests, after which the fitted parameters give a reusable calibration of the panel. Our approach is also a more efficient alternative to a standard permutation test. We demonstrate the accuracy of our approach and compare its efficiency with that of permutation tests on genomewide SNP data released by the International HapMap Consortium. The estimation of analytic distributions for combined evidence will allow these powerful methods to be applied more widely in large exploratory studies.\n\nIntroduction The rapid increase in genomic data available to researchers, coupled with sharply decreasing experimental costs, has created opportunities for exploratory genetic studies of unprecedented size. For example, it is now possible to screen thousands of candidate genes for disease associations at either the sequence or the expression level (Risch 2000; Schulze and Downward 2001). Also, the prospect of a genomewide linkage-disequilibrium map will allow genome scans for association to become as routine as they already are for linkage (International HapMap Consortium 2003). Several initial scans have recently been completed (Ophoff et al. 2002; Ozaki et al. 2002; Sawcer et al. 2002). A serious problem for these studies is that a large number of nominally significant results are expected, even when there is no true association, because of stochastic Received May 5, 2004; accepted for publication June 24, 2004; electronically published July 19, 2004. Address for correspondence and reprints: Dr. Frank Dudbridge, MRC Biostatistics Unit, Robinson Way, Cambridge CB2 2SR, United Kingdom. E-mail: [email protected] 䉷 2004 by The American Society of Human Genetics. All rights reserved. 0002-9297/2004/7503-0008\\$15.00\n\n424\n\nvariation in the data and the large number of tests that are performed. Furthermore, one cannot be certain at the outset whether there are any true associations to be found at all. For this reason, strict significance thresholds have been recommended that control the familywise type I error (Risch and Merikangas 1996). At such low error rates, very large samples must be obtained to achieve adequate power, making the cost of ascertainment an important factor in study design (Service et al. 2003). Although the traditional burden is to reject the hypothesis that there are no true associations at all, in practice, some—perhaps many—true associations are anticipated. Furthermore, it can be more economical to perform an initial screening stage with weaker error control to reduce the candidate loci to a smaller set to which stronger error control can be applied (Brown and Russell 1997). For these reasons, attention has turned to methods that are sensitive to multiple associations while retaining weak control of the familywise error (Hoh and Ott 2003; Storey and Tibshirani 2003). One approach that has shown promise is to combine the strongest evidence from multiple tests. Motivated by principles of meta-analysis, the idea is to identify a subset of tests showing a trend of significance exceeding that of the\n\n425\n\nDudbridge et al.: Computation of Significance Levels\n\nindividual tests. Hoh et al. (2001) proposed forming sums of the k largest test statistics, comparing the sums with their null distributions, and identifying the most significant sum. Zaykin et al. (2002) proposed forming the product of all P values less than a fixed threshold. Dudbridge and Koeleman (2003) suggested a hybrid of these methods, namely, the product of the k smallest P values. Each of these methods has been shown to give improved power in realistic situations (see also Wille et al. and Hao et al. ). However, the lack of analytic distributions is a major obstacle to their more widespread use. Although distributions for products of P values are known when the tests are independent (Zaykin 1999; Zaykin et al. 2002; Dudbridge and Koeleman 2003), they do not apply to correlated tests, which arise in genomewide association studies, particularly when dense maps of SNPs are used. We will show that, unlike for Bonferroni-like procedures, increased type I errors can result from incorrectly assuming independence. Monte Carlo procedures are often recommended (Zaykin et al. 2002)—for example, by random permutation of phenotypic labels (Churchill and Doerge 1994)—but can become extremely time-consuming when applied on such a large scale. Furthermore, follow-up analysis may require further levels of permutation—for example, when identifying subsets of the data showing increased association (Johnson et al. 2002). With the advent of large-cohort studies (Austin et al. 2003) and the prospect of common marker panels for genomewide association scans (International HapMap Consortium 2003), reusability becomes an important issue. If several investigators are conducting the same tests on different samples, it is inefficient for each investigator to generate the same permutation distribution. Yet, if a single supplier generates and distributes the permutation distribution, it is useful only if critical values are supplied for every combination of tests and every significance level. The volume of information involved, reminiscent of traditional statistical tables, makes this model unlikely to be adopted. Instead of relying on permutation tests, some methods have been suggested to adjust the tests in a way that allows independence to be assumed. One approach assumes an effective number of independent tests, such that the actual tests are an oversampling of some underlying independent variables. The effective number could be estimated by bootstrap resampling (Bailey and Grundy 1999) or by appealing to principal-components analysis (Cheverud 2001; Nyholt 2004). This approach is a restricted case of the modeling we propose below, and we will show that the restriction does not sufficiently capture the full correlation structure, except in some particular cases that are unlikely to occur on the genomewide scale.\n\nAnother approach is to sequentially decorrelate the tests. This can be done by application of a single transformation derived from the correlation matrix (Zaykin et al. 2002) or by successive greedy transformations (Wille et al. 2003). The former approach is sensitive to the ordering of the tests, whereas the latter may lose some advantages of combining evidence because it favors the tests with stronger marginal significance. Alternatively, a step-up linear model may be constructed, starting from a single variable and adding covariates one at a time (Cordell and Clayton 2002). However, these approaches all encounter difficult computational problems as the number of tests becomes large. For example, methods using a correlation matrix require inversion of large sparse matrices, and linear modeling requires likelihood maximization over many covariates, with an aggregation of missing data as more covariates are added. Here, we propose a more efficient application of permutation sampling, in which analytic distributions are fitted to the permutation samples. We apply the method to the sum of k largest ⫺log P values from a larger number of tests. Assuming that k is fixed and relatively small, we fit extreme-value distributions to the empirical distribution of the sums. When many values of k are considered, we fit a beta distribution to the most significant sum. Although permutation sampling is required to generate the empirical distributions, this can be completed more quickly than a simple permutation test. Conditional on the correlation structure, the procedure need be performed only once and subsequently allows fast and accurate computation of significance levels for sum statistics. The correlation structure is a property of the study population and the ascertainment criteria, so the reusability is most applicable to prospective cohort studies, although it can also be relevant to retrospective samples. As a special case, our procedure gives a more efficient method for a standard permutation test. We illustrate the method on chromosomewide SNP genotypes released by the International HapMap Consortium (2003). We compare the efficiency of our approach with simple permutation tests, demonstrate the effect of assuming independence, and study some operating characteristics of the combined-evidence approach.\n\nMaterial and Methods\n\nFixed Number of Combined Variables Suppose a study generates statistics for m hypothesis tests. Denote the P values by Pi and their order statistics by P(i) . Let k be an integer, 1 ⭐ k ⭐ m , termed the “length.” We consider the partial sum of ⫺log P values,\n\n426\n\nAm. J. Hum. Genet. 75:424–435, 2004\n\nequivalent to a truncated product of P values (Dudbridge and Koeleman 2003)\n\nfixed. If sk(i) is the value of Sk in the ith replicate, the log likelihood is l(m,j,y)\n\nSk p ⫺\n\nip1\n\nlog P(i) .\n\nThis is an example of a sum statistic, suggested by Hoh et al. (2001). Those authors used x 2 statistics rather than P values, which should have similar power, but P values should offer greater flexibility in the context of genome scans, as we suggest in the “Discussion” section. The partial sum Sk can be regarded as the maximum of all sums of length k, which suggests an application of extreme-value theory. For a large set of independent identically distributed variables, the minimum and maximum have distributions of a general form for a wide class of parent distributions (Gumbel 1958). Here, the sums are proportional to x 2 variables (Fisher 1932). This holds for correlated variables, provided that they can be regarded as a stationary process (one whose stochastic properties are invariant) with independence in the limit of large distances (Coles 2001). The rapid decay of linkage disequilibrium (LD) ensures long-range independence (Reich et al. 2001); we do not investigate stationarity, but we argue that the model is empirically accurate for genomic data. The theory has been applied elsewhere to molecular sequence analysis (Karlin and Altschul 1990) and to microarray data (Li and Grosse 2003). When m is large and k K m, the extremal types theorem predicts that Sk follows an extreme-value distribution (for background, see Coles ). We fit the generalized extreme-value distribution, which has location, scale, and shape parameters. For our purposes, these parameters are just mathematical variables that permit model fitting, but a heuristic interpretation can be made by observing their behavior when fitted to sums of independent ⫺log P values. The location is, roughly, proportional to the total number of tests m, conditional on the sum length k. The scale is proportional to k, conditional on m, and the shape summarizes the correlation structure among the single tests and the sums of fixed length. To fit the distribution of the sum statistics, we generate a large number of permutation replicates and fit the distribution by maximum likelihood. In general, the replicates are obtained by a nonparametric combination of dependent permutation tests (Pesarin 2001). In association studies, this can take the well-known form of shuffling trait values among subjects while keeping genotypes\n\np\n\n⫺ log j ⫺ (1 ⫹ 1/y) log [1 ⫹ y(sk(i) ⫺ m)j⫺1]\n\ni\n\n}\n\n⫺ [1 ⫹ y(sk(i) ⫺ m)j⫺1]⫺1/y\n\nfor location m, scale j, and shape y. Maximum-likelihood estimates are obtained by numerical optimization; we used the EVD add-on package to R (r Development Core Team 2003; Stephenson 2002). The significance level of the observed data is then calculated from the fitted analytic distribution. In contrast to a standard permutation test, we use the actual values of the replicate statistics to calculate the significance, whereas a permutation test counts only how often they exceed the value in the observed data. By using more information from the permutation replicates, we expect to achieve greater accuracy, in addition to parameterizing the permutation distribution with an analytic model. Variable Number of Combined Variables The above procedure applies to a fixed-size subset of the variables. For best power, the length k might be chosen to somewhat exceed the anticipated number of true associations. When this is unknown, one can form sums for many lengths and identify which length has the most significant sum (Hoh et al. 2001). To obtain the overall significance, we need the distribution of the smallest P value of the sums of varying length. We propose a beta distribution, motivated by the fact that order statistics of independent uniform (0,1) variables have known beta distributions. The P values of the sums are highly correlated, but we assume that the smallest value follows a two-parameter beta distribution. We generate permutation replicates and, within each replicate, form the partial sums for a range of lengths and calculate their significances from the previously fitted extreme-value distributions. The smallest of those P values is then the statistic (i) for that replicate. If pmin is the smallest P value in the ith replicate, the log likelihood is l(a,b)\n\np\n\ni\n\nG(a ⫹ b) (i) ⫹ (a ⫺ 1) log pmin G(a)G(b)\n\n)\n\n}\n\n(i) ⫹(b ⫺ 1)log(1 ⫺ pmin )\n\nfor shape parameters a and b. Maximum-likelihood es-\n\n427\n\nDudbridge et al.: Computation of Significance Levels\n\ntimates are obtained by numerical optimization, and the significance level of the observed data is obtained from the fitted distribution.\n\nEffective Number of Tests We wish to test the assertion that there is an effective number of independent tests that could be used in traditional adjustments (Bailey and Grundy 1999; Cheverud 2001; Nyholt 2004). This can be done by fitting beta distributions, as above. If there is an effective number of tests, a real number m\u0001 can be found, such that the minimum P value follows the Sˇida´k (1967) correction \u0001 1 ⫺ (1 ⫺ pmin)m , which is the beta (1,m\u0001) distribution. As the alternative, we allow the minimum P value to follow the two-parameter beta distribution, and we test whether the first parameter is 1 by applying the likelihood-ratio test for nested models. A significant likelihood ratio rejects the hypothesis that one parameter is sufficient and that there is an effective number of independent tests.\n\nPower of Variable Number of Combined Variables The choice of the sum length k can be problematic. The optimal length depends on the number of true associations and their effect sizes, both of which are unknown. Often, a reasonable prior estimate can be made. For example, genome scans are currently designed under an assumption of no more than, say, 20 true associations, since models with more loci specify effect sizes so small that prohibitive sample sizes are required (Pritchard 2001; Schliekelman and Slatkin 2002). A relevant question is whether misspecification of a fixed length has a greater cost than does estimation of the most significant length. We studied this by simulating a large exploratory study consisting of 10,000 independent tests. For true associ2 ations, we simulated x(1) variables with the noncentrality parameter (NCP) chosen to give maximum power of ∼80% (at ap0.05) over the range of sum lengths 1 ⭐ k ⭐ 500. For the remainder, we generated uniform P values on (0,1). This approach avoids making assumptions about genetic effects and LD, but it assumes that sufficiently powerful studies can be designed. We first generated 10,000 replicates of the null distribution with no true associations, and we fitted extreme-value distributions to the fixed length sums and a beta (0.767,1.939) distribution to the smallest P value of the sums. We then simulated studies with 5, 10, 50, and 100 true associations, with 10,000 replicates for each, and plotted the power for each fixed-length sum together with the power for variable length.\n\nComparison with Permutation Test Since permutation sampling is required to fit the extreme-value and beta distributions, we compare efficiency with standard permutation tests. Here, a permutation test consists of randomly assigning traits among subjects, while keeping the genotype data fixed. Sum statistics are calculated in each replicate and compared with the statistic in the original data. The significance level is the proportion of replicate statistics exceeding the observed statistic. We compare efficiency for a fixed P value by considering how many replicates are needed to achieve a given accuracy, measured by the length of the 95% CI for P. Since a permutation test is equivalent to an estimation of a binomial probability, the CI is given by normal theory. For our approach, we obtain the CI by a parametric bootstrap. We generate a number of random deviates from a fixed extreme value or beta distribution, each deviate representing one permutation sample. We refit a distribution to the deviates and from it calculate the significance of the P-quantile point of the generating distribution. The CI is estimated by repeating this procedure a large number of times. From this, we obtain the number of binomial trials needed to achieve the same accuracy, given as 2 4Z 0.975 P(1 ⫺ P) , (CI)2\n\nwhere Z 0.975 p 1.96 is the 97.5th percentile point of the standard normal distribution and CI is the length of the bootstrap CI for our procedure. Data Sets We applied these methods to chromosomewide genotypes from release 6 of the International HapMap Consortium (2003). We obtained genotypes for 20,243 SNPs on chromosome 18 and 13,620 SNPs on chromosome 21, which were the two most densely genotyped chromosomes. Genotypes were available for 90 subjects from CEPH pedigrees. For the proof of concept, we tested each SNP individually, without performing any blocking or grouping of SNPs, although such grouping is likely to occur in real scans. We regarded the subjects as unrelated, for the purpose of generating null distributions. This does not bias the distribution of P values, which is our main interest, but does increase the correlation between loci, since the length of shared haplotypes is greater than in a population of unrelated subjects. Therefore, these data represent a more problematic case for correlated SNP data than would usually be the case but could also be regarded as a reduced-scale copy of wholegenome data.\n\n428 For both chromosomes, we generated empirical null distributions for the sum statistics by randomly assigning “case” status to half of the subjects and “control” status to the other half. Most software packages for genetic association have severe memory and time requirements when working with this volume of data. We wrote a custom program using compact data structures, to calculate likelihood-ratio tests of allelic association, form sum statistics, and permute affection status. The program is available from the authors on request. We performed 10,000 random permutations and fitted extreme-value distributions to the partial sums for 1 ⭐ k ⭐ 100 and a beta distribution to the smallest P value of the sum over that range of k. We were interested in the relationship between the parameters of the fitted distributions and the parameters of the sum statistics. To explore this, we plotted the location, shape, and scale parameters of the fitted distribution against the sum length. We used quantile-quantile plots to show that the analytic distributions gave a good fit to the empirical distributions. We used the HapMap data to study the effect of assuming independence when the data are correlated, by applying the exact distribution for independent tests to the 95th percentile of the empirical distribution. If the exact distribution is accurate, it will give a P value of .05 for each sum length. If it is conservative, the P value will be 1.05; if it is liberal, the P value will be !.05. Results Accuracy of Fitted Distributions In figure 1, we show the location, scale, and shape parameters of the fitted distribution plotted against the length of the sum. The location and scale parameters have a strong log-linear correlation with the length, deviating from unity only by sampling variation. The relationship is less clear for the shape, but it does show a continuous form. Therefore, the sum statistics can be modeled as a whole by a smooth family of extreme-value distributions, for which the location and scale can be parameterized by the slope and intercept of fitted lines. For the best accuracy, the shape should be specified explicitly, although a polynomial spline might provide an acceptable approximation. The distributions of the sum statistics for the chromosome-18 SNPs can be summarized as follows: Fixed length sum has the generalized extremevalue distribution. Location p 9.526 # k0.898. Scale p 1.269 # k0.8205. Shape p {⫺0.0453 for k p 1, ⫺0.0521 for k p 2, ⫺0.0524 for k p 3, …}. Figure 2 shows quantile-quantile plots for the empirical and fitted distributions for lengths 10 and 100. The extreme-value distribution gives a good fit, although there is a liberal deviation in the far tail. The fit is adequate\n\nAm. J. Hum. Genet. 75:424–435, 2004\n\nfor declaring chromosomewide significance at conventional levels. We can obtain a more accurate fit by noting that the distribution for independent variables is constructed from beta and gamma distributions (Dudbridge and Koeleman 2003) and therefore fitting a similar construction to the correlated variables. However, this turns out to be very computationally intensive and numerically unstable; furthermore, it applies only to sums based on P values and not to those of x 2 statistics, whereas the extreme-value distribution is more generally applicable and accurate for our present purposes. Figure 3 shows quantile-quantile plots of the empirical and fitted distributions of the smallest P value for 1 ⭐ k ⭐ 100. The beta distribution gives a good fit over the whole range, confirming that it is an appropriate model for the smallest P value. The linear correlation coefficients for these plots were .9995 and .9997. For chromosome 18, the fitted parameters of the beta distribution were (.8032,1.377), and, for chromosome 21, the parameters were (.7932,1.342). Effect of Ignoring Correlation We studied the effect of assuming independence when the tests are correlated. Although this assumption is conservative for the Bonferroni adjustment, it is not generally true for sum statistics. Figure 4 shows the significance of the 95th percentile of the permutation distribution for the chromosome-18 data, by use of the exact distribution under the assumption of independence. The sum of length 1 is less significant than the nominal level, which is consistent with the conservative property of the Bonferroni adjustment. However, the longer-length sums show an increasingly liberal bias as the sum length increases. In contrast, the analytic significance of the fitted distribution is close to the nominal level across the range of lengths. From these data, we can conclude that assuming independence for sum statistics of length 11 can lead to liberal as well as conservative tests, in a manner that is dependent on the sum length, significance level, and number of tests. It is therefore important to have accurate distributions available for the sum statistics. Effective Number of Tests We tested whether an effective number of independent tests sufficiently describes the correlation structure. We compared the maximum likelihood of the minimum P value under the two-parameter beta distribution with the likelihood with the first parameter set to 1 (see the “Material and Methods” section). In both data sets, the likelihood ratio was highly significant (x 2 p 360 and 322, respectively, on 1 df). Furthermore, the two-parameter beta distribution gave an excellent fit (data not shown). Therefore, we reject the hypothesis that there is an effective number of tests that allows the use of\n\n429\n\nDudbridge et al.: Computation of Significance Levels\n\nFigure 1\n\nParameters of the extreme-value distribution for Sk as function of length k. A, Location, chromosome 18. B, Location, chromosome 21. C, Scale, chromosome 18. D, Scale, chromosome 21. E, Shape, chromosome 18. F, Shape, chromosome 21.\n\ntraditional corrections. We can interpret this by regarding the minimum P value to be the sum of length k p 1 when the tests are independent and concluding that the correlation structure implies an effective sum length as well as an effective number of tests.\n\nPower of Variable Number of Combined Variables In figure 5, we compare the power that is the result of optimization of the sum length for significance with the power when the sum length is fixed. This shows that\n\n430\n\nAm. J. Hum. Genet. 75:424–435, 2004\n\nFigure 2\n\nQuantile-quantile plot of extreme-value distribution for Sk. A, Length 10, chromosome 18. B, Length 10, chromosome 21. C, Length 100, chromosome 18. D, Length 100, chromosome 21.\n\nvery little loss of power results from varying the sum length. For 5, 10, and 50 true associations, the difference in power is only 2%–3%, and it would be even smaller if a shorter range of lengths were considered. The difference is ∼10% for 100 true associations, because the most powerful length is outside the range considered, because of the small NCP of the true associations. Note, however, that this scenario is unlikely in genome-association scans. On the other hand, fixed lengths do have higher power over a range of values, and there is some margin for error with the use of fixed lengths, but, outside the optimal range, the power of fixed lengths drops off sharply. Although there is little cost in overall power, varying the length does not give a reliable estimate of the number of true associations. In each of the situations we considered, the range of lengths for which the minimum P value was achieved varied over the whole range of 1– 500. The median lengths of the most significant sum, for the four situations in figure 5, in the order shown,\n\nwere 4, 7, 51, and 133, which is fairly accurate, but the interquartile ranges were 6, 17, 143, and 432, and the mean lengths were 25, 37, 129, and 219, indicating both a high variability and upward bias. Therefore, varying the length is a useful way to improve power when there is very little idea of the number of true associations, but it should not be relied on to estimate the number of true associations, in particular, to determine the number of follow-up loci. Comparison with Permutation Test We compared efficiency with a permutation test for the sum of length 10 in the chromosome-18 data. The extreme-value distribution fitted to this sum was used to generate a parametric bootstrap CI for our method, which was matched to the normal theory CI for the binomial permutation test. Table 1 shows the number of binomial replicates needed to achieve the same accuracy as with our method with 1,000 and 10,000 rep-\n\nDudbridge et al.: Computation of Significance Levels\n\n431\n\nFigure 3\n\nQuantile-quantile plot of beta distribution for the minimum P value for Sk. A, Chromosome 18, parameters (.8032,1.3766). B, Chromosome 21, parameters (.7932,1.3423).\n\nlicates, for a range of significance levels. For example, at P p .2, the permutation test requires 1,511 replicates to reach the accuracy achieved by our method with 1,000 replicates. The efficiency is greater at stronger significance levels: at P p .001, the permutation test requires over three times as many replicates as does our method. At each significance level, the relative efficiency is the same for 1,000 and 10,000 deviates. Very similar results were obtained for other extreme-value distributions (data not shown). Table 2 gives the same comparison for the beta distribution fitted to the smallest P value of the sums. A similar pattern is seen, with the efficiency being even greater at strong significance levels. This is because two parameters are fitted rather than three, which results in smaller SEs. At P p .001, the permutation test requires 113 times as many replicates as does our method. Thus, even if we fit distributions to permutation replicates only once, our method provides an improvement in efficiency over standard permutation tests.\n\nmutation distributions, which would give both a more efficient alternative to one-off permutation tests and a reusable calibration for repeated use. Our results suggest that, for P p .05, a 40% reduction in computation is possible compared with a standard permutation test. This improvement in efficiency can translate to significant time savings. The results reported here were computed on recent Sun hardware with a 900 MHz UltraSparc III processor. We used a simple test of allelic association for single SNPs, using data on two of the shorter autosomes. Our software took ∼5 h to compute 10,000 permutation replicates for the two chromosomes. A genomewide association scan may involve testing 100,000 haplotype blocks of 30 kb, with\n\nDiscussion The realization that complex heritable traits require large sample sizes to detect small effects has led to calls for more efficient methodology for detection of multiple associations (Hoh and Ott 2003). Combination of the strongest evidence is an approach that shows promise for identifying the pertinent effects while maintaining familywise error control. However, large-scale genomic studies encounter problems with correlated data that can be satisfactorily overcome only by permutation sampling. Because these procedures are time-consuming and might be duplicated by multiple investigators, we propose analytic distributions that can be fitted to the per-\n\nFigure 4 Effect of assumption of independence in correlated tests. Solid line shows the P value of the 95th percentile of the empirical distribution, under assumption of independent tests. Dotted line shows the P value according to the fitted extreme-value distribution.\n\n432\n\nAm. J. Hum. Genet. 75:424–435, 2004\n\nFigure 5\n\nPower of fixed-length sum compared with variable-length sum, for 10,000 tests. A, Five true associations with x2 NCP 15. B, Ten associations with NCP 11. C, Fifty associations with NCP 5. D, One hundred associations with NCP 3.\n\nmultilocus tests conducted on each block. The number of tests is about three times as large, giving an extrapolated run time of 15 h, but multilocus tests require more computation, so that the total run time would be on the order of days. A 40% reduction can therefore have a significant impact. Subsequent tests of the same hypotheses in the same population require just a single analysis followed by reference to an analytic distribution, which takes only a few minutes. However there is a need for efficient software that can handle such large volumes of data. For example, COCAPHASE 2.4 (Dudbridge 2003) took ∼4 d to perform the same calculations, whereas SUMSTAT (Hoh et al. 2001) could not, as supplied, input all the marker data. (In smaller data sets, its performance was similar to that of our own program.) The parameters of the extreme-value and beta distributions can be fitted once and then distributed with other descriptive data for screening panels, such as map\n\nlocations and allele frequencies. In principle, the supplier of the marker panel could calibrate these distributions to very high accuracy, allowing end users to rapidly calculate significance levels for single experiments. However, it is important to recognize that the calibration is conditional on the correlation structure. Different marker sets and different populations will have different structures, and different samples from the same population can also, in principle, exhibit different correlation. Thus, the reusability is most applicable to cohort studies, but markers can be calibrated for retrospective sampling, if the sample used for calibration is sufficiently large to minimize the variation in correlation structure. Furthermore, correlation can depend on ascertainment, because subjects that are selected according to some genetic features are expected to exhibit more LD in the vicinity of the relevant loci than the general population. It is therefore preferable to calibrate the null distributions by use of unselected subjects, but this may un-\n\n433\n\nDudbridge et al.: Computation of Significance Levels\n\nTable 1 Efficiency of Extreme-Value Approximation Compared with Permutation Test NO.\n\nOF\n\nBINOMIAL REPLICATES NEEDED TO ACHIEVE ACCURACY OF CURRENT STUDY METHOD WHEN\n\nn p 1,000a P\n\nLength of 95% CIb\n\nNo. of Binomial Trials with Same CI Length\n\nLength of 95% CIb\n\nNo. of Binomial Trials with Same CI Length\n\n.0403 .0292 .0209 .00894 .00208\n\n1,511 1,623 1,657 1,902 3,543\n\n.01264 .00918 .006597 .002813 .000654\n\n15,397 16,408 16,769 19,220 35,938\n\n.2 .1 .05 .01 .001 a b\n\nn p 10,000a\n\nNumber of random deviates from extreme-value distribution. Length of 95% CI for the P value, estimated by parametric bootstrap.\n\nderestimate the correlation around true associations, affecting power. The extreme value and beta distributions give good fits for the HapMap data used here, but this property should not be automatically assumed. In particular, if the number of tests is not much greater than the sum length, then the extreme-value distribution may not be a good fit, though our experience is that it is a very robust model. Our approach should always be used in conjunction with quantile-quantile plots or goodness-offit tests to confirm model accuracy. Our results show that the assumption that tests are independent can lead to inaccurate and biased results. If one ignores the correlation entirely, both liberal and conservative tests are possible, depending on underlying conditions that may not be understood. If the total number of tests is adjusted downward to an effective number and independence is then assumed, the resultant distribution can be significantly different from the true distribution, so this approach cannot be guaranteed to capture the full correlation structure. In the cases of complete dependence or complete independence, an effective number clearly applies, but we suggest that, in general, this\n\napproach is accurate only when the tests can be partitioned into sets with complete dependence within sets and complete independence between sets. This is extremely unlikely on the genomewide scale. A different approach to detecting multiple associations aims to control the false-discovery rate (FDR) (Benjamini and Hochberg 1995). In some ways, this is complementary to the combined-evidence approach, a salient difference being that the FDR makes assertions about individual tests. Combined-evidence methods identify a candidate set for follow-up but give a significance level only for the whole set of tests. However, if there is an informal aim to proceed with the highest number of true associations and a sufficiently nondistracting number of false associations, then the combined-evidence approach identifies a good quality follow-up set (Wille et al. 2003; Hao et al. 2004). It is tempting to choose the follow-up set from the most significant sum length, but, although this appears to result in little net loss in power, it should not be relied on to predict the number of true associations, because the optimal sum length has high variance. Other methods that estimate the number of true associations from the distribution of P values\n\nTable 2 Efficiency of Beta Approximation Compared with Permutation Test NO.\n\nOF\n\nBINOMIAL REPLICATES NEEDED TO ACHIEVE ACCURACY OF CURRENT STUDY METHOD WHEN\n\nn p 1,000a P .2 .1 .05 .01 .001 a b\n\nn p 10,000a\n\n95% CIb\n\nNo. of Binomial Trials with Same CI Length\n\n95% CIb\n\nNo. of Binomial Trials with Same CI Length\n\n.0397 .0297 .0201 .00667 .00106\n\n1,562 1,570 1,809 3,420 13,538\n\n.0128 .00959 .00650 .002155 .000340\n\n15,032 15,045 17,034 32,745 132,778\n\nNumber of random deviates from extreme-value distribution. Length of 95% CI for the P value, estimated by parametric bootstrap.\n\n434\n\n(Pounds and Morris 2003; Storey and Tibshirani 2003) also give highly variable estimates when the number of associations is small. It may be more prudent to select follow-up loci according to biological considerations or individually adjusted tests, conditional on a significant sum statistic. Although FDR methods detect a high average number of associations (Benjamini and Hochberg 1995; Sabatti et al. 2003), the power to detect at least one association is not much greater than Bonferroni-like procedures (Simes 1986; Dudbridge and Koeleman 2003). This means that sample sizes for FDR analysis are not significantly smaller than for traditional methods, unless one compromises on the chances of obtaining any result at all. The increased power of sum statistics comes from the combination of evidence while error control for individual tests is forfeited. FDR methods are most applicable when many associations are present, with sufficiently strong effects that the prior power is high. In that situation, the false-discovery proportion has smaller variance, so investigators can be more confident that the target rate is achieved (authors’ unpublished data). These conditions are more likely to be met in expression-array experiments than in linkage-disequilibrium scans. Combining P values is a more balanced approach than summing the test statistics, because P values are usually identically distributed, whereas test statistics may not be. In particular, genomewide association scans will be designed around blocks of varying length and diversity (Weale et al. 2003), and it will be more efficient to perform blockwise tests than individual tests of markers within blocks (Chapman et al. 2003). It makes sense to combine all the evidence within blocks and to regard the blockwise tests as the correlated units in the sum statistics. This strategy inevitably produces statistics with different degrees of freedom—therefore on different scales—whereas the P value is on a common scale. On the other hand, accurate P values may not always be available, such as when distributional assumptions are not met. In this case, it can be more convenient to work with test statistics, and the extreme-value distribution can still be used to model the permutation distribution. We have considered the sums of fixed length and the most significant fixed-length sum, but these are not the only possible combinations. For example, sums could be constructed on the basis of all tests that are significant at a nominal level (Zaykin et al. 2002). The number of nominally significant tests follows a binomial distribution, so that the sum can be regarded as the maximum of a set of variables whose distribution is a mixture of distributions for fixed length sums. This also allows the extreme-value distribution to be applied in this case. More generally, our approach is applicable whenever the null distribution falls within a parametric class of analytic distributions.\n\nAm. J. Hum. Genet. 75:424–435, 2004\n\nAn economical approach to genome scanning is to screen all markers in the first stage and only the significant markers in the second. This strategy reduces the total genotyping cost and has been proposed both for individual genotyping and for pooling protocols (Sagatopan et al. 2002; Sham et al. 2002). Combined-evidence methods are a natural choice for the first stage, providing a powerful method to identify follow-up loci, while controlling the type I error for the null hypothesis that there are no genetic effects at all. They provide an appealing trade-off between reducing the size of the exploratory space and controlling the error of individual tests, giving confidence and justification for the followup testing. The efficient approach we propose for applying these methods on genomewide scales will allow them to be applied more widely in forthcoming studies.\n\nAcknowledgments F.D. is supported by European Commission grant 503485. B.P.C.K. is supported by funding from the Dutch Diabetes Research Foundation, The Netherlands Organization for Health Research and Development, and The Juvenile Diabetes Research Foundation International grant 2001.10.004.\n\nReferences Austin MA, Harding S, McElroy C (2003) Genebanks: a comparison of eight proposed international genetic databases. Community Genet 6:37–45 Bailey TL, Grundy WN (1999) Classifying proteins by family using the product of correlated p-values. Paper presented at the Third International Conference on Computational Molecular Biology, Lyon, France, April 11–14 Benjamini Y, Hochberg Y (1995) Controlling the false discovery rate: a practical and powerful approach to multiple testing. 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Nature 426:789–796 Johnson GC, Koeleman BP, Todd JA (2002) Limitations of stratifying sib-pair data in common disease linkage studies: an example using chromosome 10p14-10q11 in type 1 diabetes. Am J Med Genet 113:158–166 Li W, Grosse I (2003) Gene selection criterion for discriminant microarray data analysis based on extreme value distributions. Paper presented at the Seventh International Conference on Computational Molecular Biology, Berlin, April 10– 13 Karlin S, Altschul SF (1990) Methods for assessing the statistical significance of molecular sequence features by using general scoring schemes. Proc Natl Acad Sci USA 87:2264–2268 Nyholt DR (2004) A simple correction for multiple testing for single-nucleotide polymorphisms in linkage disequilibrium with each other. Am J Hum Genet 74:765–769 Ophoff RA, Escamilla MA, Service SK, Spesny M, Meshi DB, Poon W, Molina J, Fournier E, Gallegos A, Mathews C, Neylan T, Batki SL, Roche E, Ramirez M, Silva S, De Mille MC, Dong P, Leon PE, Reus VI, Sandkuijl LA, Freimer NB (2002) Genomewide linkage disequilibrium mapping of severe bipolar disorder in a population isolate. Am J Hum Genet 71:565–574 Ozaki K, Ohnishi Y, Iida A, Sekine A, Yamada R, Tsunoda T, Sato H, Sato H, Hori M, Nakamura Y, Tanaka T (2002) Functional SNPs in the lymphotoxin-a gene that are associated with susceptibility to myocardial infarction. Nat Genet 32:650–654 Pesarin F (2001) Multivariate permutation tests with applications in biostatistics. Wiley, Chichester, United Kingdom Pounds S, Morris SW (2003) Estimating the occurrence of false positives and false negatives in microarray studies by approximating and partitioning the empirical distribution of p-values. Bioinformatics 19:1236–1242 Pritchard JK (2001) Are rare variants responsible for susceptibility to complex diseases? Am J Hum Genet 69:124–137 R Development Core Team (2003) R: a language and environment for statistical computing. R foundation, Vienna, Austria Reich DE, Cargill M, Bolk S, Ireland J, Sabeti PC, Richter DJ,\n\n435 Lavery T, Kouyoumjian R, Farhadian SF, Ward R, Lander ES (2001) Linkage disequilibrium in the human genome. Nature 411:199–204 Risch NJ (2000) Searching for genetic determinants in the new millennium. Nature 405:847–856 Risch NJ, Merikangas KR (1996) The future of genetic studies of complex human disease. Science 273:1516–1517 Sabatti C, Service S, Freimer N (2003) False discovery rate in linkage and association genome screens for complex disorders. Genetics 164:829–833 Satagopan JM, Verbel DA, Venkatraman ES, Offit KE, Begg CB (2002) Two-stage designs for gene-disease association studies. Biometrics 58:163–170 Sawcer S, Maranian M, Setakis E, Curwen V, Akesson E, Hensiek A, Coraddu F, Roxburgh R, Sawcer D, Gray J, Deans J, Goodfellow PN, Walker N, Clayton D, Compston A (2002) A whole genome screen for linkage disequilibrium in multiple sclerosis confirms disease associations with regions previously linked to susceptibility. Brain 125:1337–1347 Schliekelman P, Slatkin M (2002) Multiplex relative risk and estimation of the number of loci underlying an inherited disease. Am J Hum Genet 71:1369–1385 Schulze A, Downward J (2001) Navigating gene expression using microarrays: a technology review. Nat Cell Biol 3:E190– E195 Service SK, Sandkuijl LA, Freimer NB (2003) Cost-effective designs for linkage disequilibrium mapping of complex traits. Am J Hum Genet 72:1213–1220 Sham PC, Bader JS, Craig I, O’Donovan M, Owen M (2002) DNA pooling: a tool for large-scale association studies. Nat Rev Genet 3:862–871 Sˇida´k Z (1967) Rectangular confidence regions for the means of multivariate normal distributions. J Am Stat Assoc 78:626– 633 Simes RJ (1986) An improved Bonferroni procedure for multiple tests of significance. Biometrika 73:751–754 Stephenson AG (2002) EVD: extreme value distributions. RNews 2:31–32 Storey JD, Tibshirani R (2003) Statistical significance for genome-wide studies. Proc Natl Acad Sci USA 100:9440–9445 Weale ME, Depondt C, Macdonald SJ, Smith A, Lai PS, Shorvon SD, Wood NW, Goldstein DB (2003) Selection and evaluation of tagging SNPs in the neuronal-sodium-channel gene SCN1A: implications for linkage-disequilibrium gene mapping. Am J Hum Genet 73:551–565 Wille A, Hoh J, Ott J (2003) Sum statistics for the joint detection of multiple disease loci in case-control association studies with SNP markers. Genet Epidemiol 25:350–359 Zaykin DV (1999) Statistical analysis of genetic associations. PhD thesis, North Carolina State University, Raleigh Zaykin DV, Zhivotovsky LA, Westfall PH, Weir BS (2002) Truncated product method for combining P-values. Genet Epidemiol 22:170–185\n\n## Efficient Computation of Significance Levels for Multiple ...\n\nJul 19, 2004 - to screen thousands of candidate genes for disease asso- ciations at ..... tween loci, since the length of shared haplotypes is greater than in a ..."
] |
[
null
] |
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|
http://tugrulaslan.com/big-o-notation/
|
[
"# Big O Notation\n\nReading Time: 4 minutes\n\n# Definition\n\nThe Big O Notation is simplified analysis of an algorithm’s efficiency. It is also known as Landau’s Symbol. Big O Notation is used in Computer Science and Mathematics to describe the Asymptotic behavior of functions. Let’s study some of the characteristics;\n\n1. Big O Notation enlightens about the complexity of a target algorithm for a given input considered as “N”,\n2. Big O Notation is the abstraction of the efficiency in terms of the machine independence. The Algorithm must perform the same way on every operation system and hardware.\n3. Big O Notation does not concern about how much of time that an algorithm takes but how it performs under certain situations.\n4. Big O Notation gives us the “Time” and the “Space” constraints.\n\nIn addition to the second characteristic; when we run a program, we will have performance that is how much of time or hardware resource is used, and complexity how the algorithm acts and grows. Note that the Complexity affects the performance, but the other way around is not possible.\n\nFurthermore, there are three types of measurements that I’ll explain and demonstrate those measurements in a different chapter.\n\n# Rules\n\n## Drop the constants\n\nIf you have a function that has a running time of 5N, that is realized as O(n). Because n gets bigger and the 5 is not the consideration anymore,\n\n## Observe all inputs\n\nDifferent inputs and variables have different weights on identifying the notation. If you iterate in two different arrays you get O(a*b). Study the following pseudo code;\n\n```method(int[] array1, int[] array2){\n\nFor(int a : array1){//O(a)\n\nFor(int b : arrayb){//O(b)\n\nSystem.out.println(\"match\");//O(1)\n\n}\n\n}\n\n}```\n\n## Drop the low order terms\n\nCertain terms dominate the other terms, in this case we drop the lower term(s). Here is the sequence of the terms. From the right to the left domination gets lower:\n\nO(1) < O(log n) < O(n) < O(n log n) < O(n2) < O(2n) < O(n!)\n\n# Complexities\n\nBig O Notation can be used to describe the Time Complexity and the Space Complexity of algorithms. Each of these subjects are different terms.\n\n## Time Complexity\n\nThe Time Complexity corresponds to the amount of time that an Algorithm takes to run. The Time Complexity has the following cases; Best, Average and Worst.\n\n## Space Complexity\n\nThe Space Complexity describes how much of a space the algorithm allocates in the memory according to the amount of given data.\n\n# Cases\n\nIn Big O Notation we have three cases:\n\n1. Best Case,\n2. Average Case,\n3. Worst Case\n\nWhen algorithms are analyzed, generally “Worst Case is referred. It doesn’t mean that the rest of the cases are less important, but depending on the input, the Worst Case has a weight.\n\n# Notations",
null,
"image courtesy www.bigocheatsheet.com\n\n## O(1) Constant\n\nConstant time is a basic statement that has only Constants, in a different way the values that are solid and will not change. Regardless of amount of the data, the code executes the process in the time amount, this can be a variable definition, access in an array or a print out. The simplest examples would go for this:\n\n```Int x = (9/2)*12-1;\n\nTo find out the Big O Notation of such constants;\n\n1.int a = (9/2)*12-1;//O(1)\n\n2.int b = 100/2; //O(1)\n\n3.int result a+b; //O(1)\n\nSystem.out.println(result); //O(1)```\n\nTotal time spent: O(1) + O(1) + O(1) + O(1) = O(1)\n\n4*O(1)\n\n(Constants are dropped)\n\n## O(n) Linear\n\nlinear time is known as the completion time grows for the given amount of data. A good example to it is the linear search in a loop that iterates through N amount of elements.\n\n```1.for(int i=0; i<N; i++){//O(n)\n\n2.System.out.println(i)//O(1)\n\n}\n\nTotal time spent: O(n)*O(1)=O(n)\n\n1.int x = 55*3+(10-9); //O(1)\n\nfor(int i=0; i<N; i++){//O(n){\n3.System.out.println(i)//O(1)\n\n}```\n\nTotal time spent: O(1) + O(n)*O(1)=O(n)\n\n(Drop low order terms)\n\nthe time completion of quadratic algorithms is proportional to the square of the given amount of data. Commonly we spot these sorts of algorithms in nested loops. In addition, the more nested loops exist, the square will become cubic O(n3) or more. The good example is the Bubble Sort. Furthermore, such algorithms dramatically hamper the performance if the amount of data increases.\n\n## O(2N) Exponential\n\nExponential algorithms are those whose performance doubles for every given input. We can basically spot such algorithms in recursive methods of calculations. For the sake of simplicity, I’ll give an example of Fibonacci numbers. As you will see, the method will call itself twice for the given input.\n\n```public int fibonacciNumbers(int number) {\n\nif (number <= 1) {\n\nreturn number;\n\n} else {\n\nreturn fibonacciNumbers(number - 1) + fibonacciNumbers(number - 2);\n\n}\n\n}```\n\n## O(n!) Factorial\n\nFactorial algorithm that calculates all permutation of a given array is considered as O(n!). The most suitable example of this is the travelling sales man problem with the brute-force search.\n\n## O(log n) Logarithmic\n\nThe best way to explain logarithmic algorithms is to search in chunks of data by quadratic them and the rest will be assumed as O(n). Since we split it we will gain some time while looking up. The good world example would be the Binary Search Tree for these types of algorithms and they are very efficient, because increasing the amount of data has a minor effect at some point, because the amount of data is halved each time as it is seen with the Binary Search Tree.\n\n## O(n log n) Quasi Linear\n\nQuasi linear algorithms are the ones hard to spot. The given values will be compared once. Essentially each comparison will reduce the possible final sorted data structure in half like in O(log n). Furthermore, the number of comparisons that will be performed is equal to log in the factorial of “n”. The comparisons are also equal to n log n this is how O(n log n) algorithms are formed. The mathematical representation as follows;\n\nSum of comparisons are equal to log n!\n\nDivided comparisons are equal to log n + log(n-1) + … +log(1)//End of comparisons\n\nThe outcome is equal to n log n\n\nExamples of Quasi linear complexities are Merge and Heap Sorts."
] |
[
null,
"http://tugrulaslan.com/wp-content/uploads/2018/09/big_o_chart.png",
null
] |
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|
https://strathprints.strath.ac.uk/67663/
|
[
"# Stability equivalence between the stochastic dierential delay equations driven by G-Brownian motion and the Euler-Maruyama method\n\nDeng, Shounian and Fei, Chen and Fei, Weiyin and Mao, Xuerong (2019) Stability equivalence between the stochastic dierential delay equations driven by G-Brownian motion and the Euler-Maruyama method. Applied Mathematics Letters. ISSN 0893-9659 (In Press)",
null,
"",
null,
"Preview\nText (Deng-etal-AML-2019-Stability-equivalence-between-the-stochastic-dierential-delay-equations)\nDeng_etal_AML_2019_Stability_equivalence_between_the_stochastic_dierential_delay_equations.pdf\nAccepted Author Manuscript\nLicense:",
null,
"| Preview\n\n## Abstract\n\nConsider a stochastic differential delay equation driven by G-Brownian motion (G-SDDE) dx(t) = f(x(t), x(t − τ))dt + g(x(t), x(t − τ))dB(t) + h(x(t), x(t − τ))dhBi(t). Under the global Lipschitz condition for the G-SDDE, we show that the G-SDDE is exponentially stable in mean square if and only if for sufficiently small step size, the Euler-Maruyama (EM) method is exponentially stable in mean square. Thus, we can carry out careful numerical simulations to investigate the exponential stability of the underlying G-SDDE in practice, in the absence of an appropriate Lyapunov function. A numerical example is provided to illustrate our results."
] |
[
null,
"https://strathprints.strath.ac.uk/67663/1.hassmallThumbnailVersion/Deng_etal_AML_2019_Stability_equivalence_between_the_stochastic_dierential_delay_equations.pdf",
null,
"https://strathprints.strath.ac.uk/67663/1.haspreviewThumbnailVersion/Deng_etal_AML_2019_Stability_equivalence_between_the_stochastic_dierential_delay_equations.pdf",
null,
"https://strathprints.strath.ac.uk/images/license/cc-by-nc-nd.png",
null
] |
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|
https://www.hackmath.net/en/word-math-problems/9th-grade-(14y)?page_num=85
|
[
"# Examples for 9th grade - page 85\n\n1. Trapezoid - intersection of diagonals",
null,
"In the ABCD trapezoid is AB = 8 cm long, trapezium height 6 cm, and distance of diagonals intersection from AB is 4 cm. Calculate trapezoid area.\n2. Embankment",
null,
"Perpendicular cross-section of the embankment around the lake has the shape of an isosceles trapezoid. Calculate the perpendicular cross-section, where bank is 4 m high the upper width is 7 m and the legs are 10 m long.\n3. Candy",
null,
"Peter had a sachet of candy. He wanted to share with his friends. If he gave them 30 candies, he would have 62 candies. If he gave them 40 candies, he would miss 8 candies. How many friends did Peter have?\n4. How far",
null,
"From the top of a lighthouse 145 ft above sea level, the angle of depression of a boat 29°. How far is the boat from the lighthouse?\n5. Rhombus construction",
null,
"Construct ABCD rhombus if its diagonal AC=9 cm and side AB = 6 cm. Inscribe a circle in it touching all sides...\n6. Competitors",
null,
"In the first round of slalom fell 15% of all competitors and in the second round another 10 racers. Together, 40% of all competitors fell. What was the total number of competitors?\n7. 6 terms",
null,
"Find the first six terms of the sequence. a1 = 7, an = an-1 + 6\n8. Modulo",
null,
"Find x in modulo equation: 47x = 4 (mod 9) Hint - read as: what number 47x divided by 9 (modulo 9) give remainder 4 .\n9. Photocopier",
null,
"A photocopier enlarges a picture in the ratio 7:4. How many times will a picture of size 6cm by 4cm be enlarged to fit on a 30cm by 20 cm page?\n10. Fence and scale",
null,
"The garden with 80 m of fencing is draw on plan as a square with a side length of 4 cm. At what scale is this plan made?\n11. Rhombus",
null,
"The rhombus has diagonal lengths of 4.2cm and 3.4cm. Calculate the length of the sides of the rhombus and its height\n12. Diamond diagonals",
null,
"Calculate the diamonds' diagonals lengths if the diamond area is 156 cm square and the side length is 13 cm.\n13. Soaps",
null,
"Each box has the same number of soaps. A quarter of all boxes contain only white soaps, and in each of the remaining 120 boxes there are always half the white soaps and half the green. White soaps total 1200. (a) the number of all soap boxes; (b) the sma\n14. Lighthouse",
null,
"The man, 180 cm tall, walks along the seafront directly to the lighthouse. The male shadow caused by the beacon light is initially 5.4 meters long. When the man approaches the lighthouse by 90 meters, its shadow shorter by 3 meters. How tall is the lightho\n15. Hypotenuse - RT",
null,
"A triangle has a hypotenuse of 55 and an altitude to the hypotenuse of 33. What is the area of the triangle?\n16. A bridge",
null,
"A bridge over a river is in the shape of the arc of a circle with each base of the bridge at the river's edge. At the center of the river, the bridge is 10 feet above the water. At 27 feet from the edge of the river, the bridge is 9 feet above the water. H\n17. Paul earned",
null,
"Paul earned 300 Kč in one hour, Václav 1/3 more than Paul. Václav worked 60 hours, which is 1/3 fewer hours than Paul worked. How many percents less earned Paul an hour than Václav? How many hours did Paul more than Václav? How much did Paul earn more t\n18. Children",
null,
"The group has 42 children. There are 4 more boys than girls. How many boys and girls are in the group?",
null,
"The angle of a straight road is approximately 12 degrees. Determine the percentage of this road.",
null,
"The digit sum of the two-digit number is nine. When we turn figures and multiply by the original two-digit number, we get the number 2430. What is the original two-digit number?"
] |
[
null,
"https://www.hackmath.net/thumb/87/t_5987.jpg",
null,
"https://www.hackmath.net/thumb/44/t_2544.jpg",
null,
"https://www.hackmath.net/thumb/27/t_1627.jpg",
null,
"https://www.hackmath.net/thumb/45/t_4645.jpg",
null,
"https://www.hackmath.net/thumb/28/t_5328.jpg",
null,
"https://www.hackmath.net/thumb/70/t_5470.jpg",
null,
"https://www.hackmath.net/thumb/2/t_3202.jpg",
null,
"https://www.hackmath.net/thumb/25/t_4325.jpg",
null,
"https://www.hackmath.net/thumb/33/t_5133.jpg",
null,
"https://www.hackmath.net/thumb/69/t_5569.jpg",
null,
"https://www.hackmath.net/thumb/31/t_6431.jpg",
null,
"https://www.hackmath.net/thumb/53/t_6253.jpg",
null,
"https://www.hackmath.net/thumb/37/t_6237.jpg",
null,
"https://www.hackmath.net/thumb/44/t_3844.jpg",
null,
"https://www.hackmath.net/thumb/5/t_6305.jpg",
null,
"https://www.hackmath.net/thumb/91/t_6191.jpg",
null,
"https://www.hackmath.net/thumb/18/t_6418.jpg",
null,
"https://www.hackmath.net/thumb/21/t_4121.jpg",
null,
"https://www.hackmath.net/thumb/70/t_6470.jpg",
null,
"https://www.hackmath.net/thumb/27/t_5727.jpg",
null
] |
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|
https://www.osgeo.cn/python-guide/scenarios/ml.html
|
[
"# 机器学习¶",
null,
"python有大量用于数据分析、统计和机器学习的库，这使得它成为许多数据科学家的首选语言。\n\n## 坐骨神经痛¶\n\nscipy堆栈由一组核心助手包组成，用于数据科学中的统计分析和可视化数据。由于其大量的功能和易用性，栈被认为是大多数数据科学应用的必备条件。\n\n### 安装¶\n\nNB: Anaconda 对于无缝地安装和维护数据科学包是非常首选和推荐的。\n\n## Scikit学习¶\n\nSciKit是一个免费的开放源码机器学习库，用于Python。它提供了现成的功能来实现许多算法，如线性回归、分类器、支持向量机、k均值、神经网络等。它还具有一些样本数据集，可直接用于培训和测试。\n\n### 安装¶\n\n```pip install -U scikit-learn\n```\n\n```conda install scikit-learn\n```\n\n### 例子¶\n\n```from sklearn.datasets import load_iris\nfrom sklearn import tree\nfrom sklearn.metrics import accuracy_score\nimport numpy as np\n\nx = iris.data #array of the data\ny = iris.target #array of labels (i.e answers) of each data entry\n\n#getting label names i.e the three flower species\ny_names = iris.target_names\n\n#taking random indices to split the dataset into train and test\ntest_ids = np.random.permutation(len(x))\n\n#splitting data and labels into train and test\n#keeping last 10 entries for testing, rest for training\n\nx_train = x[test_ids[:-10]]\nx_test = x[test_ids[-10:]]\n\ny_train = y[test_ids[:-10]]\ny_test = y[test_ids[-10:]]\n\n#classifying using decision tree\nclf = tree.DecisionTreeClassifier()\n\n#training (fitting) the classifier with the training set\nclf.fit(x_train, y_train)\n\n#predictions on the test dataset\npred = clf.predict(x_test)\n\nprint pred #predicted labels i.e flower species\nprint y_test #actual labels\nprint (accuracy_score(pred, y_test))*100 #prediction accuracy\n```\n\n```[0 1 1 1 0 2 0 2 2 2]\n[0 1 1 1 0 2 0 2 2 2]\n100.0\n```"
] |
[
null,
"https://www.osgeo.cn/python-guide/_images/34018729885_002ced9b54_k_d.jpg",
null
] |
{"ft_lang_label":"__label__zh","ft_lang_prob":0.6037988,"math_prob":0.83350766,"size":1939,"snap":"2020-24-2020-29","text_gpt3_token_len":1037,"char_repetition_ratio":0.08888889,"word_repetition_ratio":0.0625,"special_character_ratio":0.22485818,"punctuation_ratio":0.08070175,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9583701,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-10T13:58:32Z\",\"WARC-Record-ID\":\"<urn:uuid:0544fcbf-07c4-4467-af94-013141afa4ea>\",\"Content-Length\":\"16363\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d77db13f-72ce-48e7-ba2b-bcc677205199>\",\"WARC-Concurrent-To\":\"<urn:uuid:ff89da3b-8c0a-4ebb-98ca-b34eb7ec1118>\",\"WARC-IP-Address\":\"47.104.236.0\",\"WARC-Target-URI\":\"https://www.osgeo.cn/python-guide/scenarios/ml.html\",\"WARC-Payload-Digest\":\"sha1:OQSM2TT2DPEX6AVZGITBMLG47RLVGOLU\",\"WARC-Block-Digest\":\"sha1:QRDGAUI2OAKG2EOW5YIJTF45NQ5LIUCD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655908294.32_warc_CC-MAIN-20200710113143-20200710143143-00478.warc.gz\"}"}
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https://www.got-it.ai/solutions/excel-chat/excel-help/how-to/excel/excel-vba-dim
|
[
"",
null,
"# Get instant live expert help on I need help with excel vba dim",
null,
"“My Excelchat expert helped me in less than 20 minutes, saving me what would have been 5 hours of work!”\n\n## Post your problem and you’ll get expert help in seconds.\n\nOur professional experts are available now. Your privacy is guaranteed.\n\n## Here are some problems that our users have asked and received explanations on\n\ni need assist with some vba coding Dim ws As Worksheet Dim irow As Long Set ws = Sheets(\"sheet1\") irow = ws.Cells(ws.Rows.Count, 1).End(x1up).Offset(1, 0).Row The last line keeps coming out as an error?\nSolved by I. H. in 14 mins\nhi, I am stuck in a problem since one week and need help!!! I am running a SQL query in Excel 2010 VBA. The query runs fine but when I use GROUP BY, I get Automation Error. Everything works if I remove Group By but I need to group data. My query looks like: Dim objRs As New ADODB.Recordset Dim output As String Dim trows As Integer Dim ThisTab As Excel.Worksheet Dim tcols As Integer Dim i As Integer Dim sSQL As String ' sSQL = \"SELECT * from [Data] where [Rem Scen 2] > 0 and (Now() - [Eval Date]) <=60\" sSQL = \"SELECT [Emp ID], [Deptt], [Job Types], [Join Date],[Pay 1]+[Pay 2] as [Compensate] from [Data] Group By {Emp ID] With objConn .Provider = \"Microsoft.ACE.OLEDB.12.0\" .ConnectionString = \"Data Source=\" & ThisWorkbook.Path & \"\\\" & ThisWorkbook.Name & \";\" & _ \"Extended Properties=\"\"Excel 12.0 Macro;HDR=YES\"\";\" .Open End With ' Execute once and display... objRs.Open sSQL, objConn\nSolved by C. D. in 29 mins\nIn excel VBA I have a sub that takes in 3 variables and in running the code generates 5 new variables. I would like to 5 either a single function or 5 different functions to return each of these variables without having redundant code where I have to repeat the same code in all 5 functions. See example below for my sub code: Sub MySub(Input1 as Integer, Input2 as Integer, Input3 as Integer) Dim Output1 as Integer Dim Output2 as Integer Dim Output3 as Double Dim Output4 as Double Dim Output5 as String MyVBACode EndSub\nSolved by V. B. in 23 mins\nSub EnableOutlining() 'Update 20140603 Dim xWs As Worksheet Set xWs = Application.ActiveSheet Dim xPws As String xPws = Application.InputBox(\"Password:\", xTitleId, \"\", Type:=2) xWs.Protect Password:=xPws, Userinterfaceonly:=True xWs.EnableOutlining = True End Sub This is the VBA code i am using to be able to group and ungroup things in excel while the sheet is protected. when i close and reopen the code no longer works\nSolved by I. B. in 27 mins\nHi I have a piece of code that suddenly stop working: All code in the sheet...cell is not working Private Sub Receipt_Click() Dim customername As String Dim customeraddress As String Dim invoicenumber As String Dim r As Long Dim mydate As String Dim path As String Dim myfilename As String lastrow = Sheets(\"01\").Range(\"B\" & Rows.Count).End(xlUp).Row r = 5 For r = 5 To lastrow If Cells(r, 15).Value = \"done\" Then GoTo nextrow Product = Sheets(\"01\").Range(\"R2\").Value ...working studentname = Sheets(\"01\").Cells(r, 2).Value...not working Cells(r, 15).Value = \"done\"...not working\nSolved by E. D. in 14 mins"
] |
[
null,
"https://www.got-it.ai/solutions/excel-chat/static/media/seo-landing-cover.4bfad59b.jpg",
null,
"https://www.got-it.ai/solutions/excel-chat/static/media/student1.9ca0d5e3.jpg",
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.69146615,"math_prob":0.6675389,"size":919,"snap":"2021-43-2021-49","text_gpt3_token_len":267,"char_repetition_ratio":0.093989074,"word_repetition_ratio":0.0,"special_character_ratio":0.3068553,"punctuation_ratio":0.1758794,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96363544,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-19T03:34:35Z\",\"WARC-Record-ID\":\"<urn:uuid:21829483-259d-46cf-af5c-23f4a33fc5b0>\",\"Content-Length\":\"353873\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:068b368b-bb9b-4e72-9e3c-996ba6eaf3f8>\",\"WARC-Concurrent-To\":\"<urn:uuid:b3cc9dc6-61bd-490d-9f6b-a938b898eccd>\",\"WARC-IP-Address\":\"99.84.191.112\",\"WARC-Target-URI\":\"https://www.got-it.ai/solutions/excel-chat/excel-help/how-to/excel/excel-vba-dim\",\"WARC-Payload-Digest\":\"sha1:5BHUW23PEXLILHMDQDDKBRCODO5WHST3\",\"WARC-Block-Digest\":\"sha1:YC4UXV2USJG7EGR47IUSNRWJLOJ5B4AX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585231.62_warc_CC-MAIN-20211019012407-20211019042407-00067.warc.gz\"}"}
|
https://support.office.com/en-ie/article/forecast-ets-confint-function-6d4a7557-11fa-4678-9e6a-dbcc31a7c7df
|
[
"# FORECAST.ETS.CONFINT function\n\nReturns a confidence interval for the forecast value at the specified target date. A confidence interval of 95% means that 95% of future points are expected to fall within this radius from the result FORECAST.ETS forecasted (with normal distribution). Using confidence interval can help grasp the accuracy of the predicted model. A smaller interval would imply more confidence in the prediction for this specific point.\n\n## Syntax\n\nFORECAST.ETS.CONFINT(target_date, values, timeline, [confidence_level], [seasonality], [data_completion], [aggregation])\n\nThe FORECAST.ETS.CONFINT function syntax has the following arguments:\n\n• Target_date Required. The data point for which you want to predict a value. Target date can be date/time or numeric. If the target date is chronologically before the end of the historical timeline, FORECAST.ETS.CONFINT returns the #NUM! error.\n\n• Values Required. Values are the historical values, for which you want to forecast the next points.\n\n• Timeline Required. The independent array or range of numeric data. The dates in the timeline must have a consistent step between them and can’t be zero. The timeline isn't required to be sorted, as FORECAST.ETS.CONFINT will sort it implicitly for calculations. If a constant step can't be identified in the provided timeline, FORECAST.ETS.CONFINT will return the #NUM! error. If timeline contains duplicate values, FORECAST.ETS.CONFINT will return the #VALUE! error. If the ranges of the timeline and values aren't of same size, FORECAST.ETS.CONFINT will return the #N/A error.\n\n• Confidence_level Optional. A numerical value between 0 and 1 (exclusive), indicating a confidence level for the calculated confidence interval. For example, for a 90% confidence interval, a 90% confidence level will be computed (90% of future points are to fall within this radius from prediction). The default value is 95%. For numbers outside of the range (0,1), FORECAST.ETS.CONFINT will return the #NUM! error.\n\n• Seasonality Optional. A numeric value. The default value of 1 means Excel detects seasonality automatically for the forecast and uses positive, whole numbers for the length of the seasonal pattern. 0 indicates no seasonality, meaning the prediction will be linear. Positive whole numbers will indicate to the algorithm to use patterns of this length as the seasonality. For any other value, FORECAST.ETS.CONFINT will return the #NUM! error.\n\nMaximum supported seasonality is 8,760 (number of hours in a year). Any seasonality above that number will result in the #NUM! error.\n\n• Data completion Optional. Although the timeline requires a constant step between data points, FORECAST.ETS.CONFINT supports up to 30% missing data, and will automatically adjust for it. 0 will indicate the algorithm to account for missing points as zeros. The default value of 1 will account for missing points by completing them to be the average of the neighboring points.\n\n• Aggregation Optional. Although the timeline requires a constant step between data points, FORECAST.ETS.CONFINT will aggregate multiple points which have the same time stamp. The aggregation parameter is a numeric value indicating which method will be used to aggregate several values with the same time stamp. The default value of 0 will use AVERAGE, while other options are SUM, COUNT, COUNTA, MIN, MAX, MEDIAN."
] |
[
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] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.80670714,"math_prob":0.9420523,"size":3330,"snap":"2019-51-2020-05","text_gpt3_token_len":714,"char_repetition_ratio":0.1482261,"word_repetition_ratio":0.06412826,"special_character_ratio":0.2039039,"punctuation_ratio":0.15730338,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95352674,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-21T09:11:33Z\",\"WARC-Record-ID\":\"<urn:uuid:4a3df642-ec39-49a6-8217-c83da20e80c7>\",\"Content-Length\":\"103814\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9734f5e8-2adf-48ab-b08f-3cccfbb81b24>\",\"WARC-Concurrent-To\":\"<urn:uuid:ee1ef6b5-5666-4523-8267-f7e157fc905e>\",\"WARC-IP-Address\":\"23.13.152.34\",\"WARC-Target-URI\":\"https://support.office.com/en-ie/article/forecast-ets-confint-function-6d4a7557-11fa-4678-9e6a-dbcc31a7c7df\",\"WARC-Payload-Digest\":\"sha1:32ZGVULKMB5D6W2L6CTEOUKY34OYI6YX\",\"WARC-Block-Digest\":\"sha1:6IPNBGA4SWEOAOOM2UJVUACWCEPNNH7I\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250601628.36_warc_CC-MAIN-20200121074002-20200121103002-00165.warc.gz\"}"}
|
https://en.m.wikipedia.org/wiki/Discrete_calculus
|
[
"# Discrete calculus\n\nDiscrete calculus or the calculus of discrete functions, is the mathematical study of incremental change, in the same way that geometry is the study of shape and algebra is the study of generalizations of arithmetic operations. The word calculus is a Latin word, meaning originally \"small pebble\"; as such pebbles were used for calculation, the meaning of the word has evolved and today usually means a method of computation. Meanwhile, calculus, originally called infinitesimal calculus or \"the calculus of infinitesimals\", is the study of continuous change.\n\nDiscrete calculus has two entry points, differential calculus and integral calculus. Differential calculus concerns incremental rates of change and the slopes of piece-wise linear curves. Integral calculus concerns accumulation of quantities and the areas under piece-wise constant curves. These two points of view are related to each other by the fundamental theorem of discrete calculus.\n\nThe study of the concepts of change starts with their discrete form. The development is dependent on a parameter, the increment $\\Delta x$",
null,
"of the independent variable. If we so choose, we can make the increment smaller and smaller and find the continuous counterparts of these concepts as limits. Informally, the limit of discrete calculus as $\\Delta x\\to 0$",
null,
"is infinitesimal calculus. Even though it serves as a discrete underpinning of calculus, the main value of discrete calculus is in applications.\n\n## Two initial constructions\n\nDiscrete differential calculus is the study of the definition, properties, and applications of the difference quotient of a function. The process of finding the difference quotient is called differentiation. Given a function defined at several points of the real line, the difference quotient at that point is a way of encoding the small-scale (i.e., from the point to the next) behavior of the function. By finding the difference quotient of a function at every pair of consecutive points in its domain, it is possible to produce a new function, called the difference quotient function or just the difference quotient of the original function. In formal terms, the difference quotient is a linear operator which takes a function as its input and produces a second function as its output. This is more abstract than many of the processes studied in elementary algebra, where functions usually input a number and output another number. For example, if the doubling function is given the input three, then it outputs six, and if the squaring function is given the input three, then it outputs nine. The derivative, however, can take the squaring function as an input. This means that the derivative takes all the information of the squaring function—such as that two is sent to four, three is sent to nine, four is sent to sixteen, and so on—and uses this information to produce another function. The function produced by differentiating the squaring function turns out to be something close to the doubling function.\n\nSuppose the functions are defined at points separated by an increment $\\Delta x=h>0$ :\n\n$a,a+h,a+2h,\\ldots ,a+nh,\\ldots$\n\nThe \"doubling function\" may be denoted by $g(x)=2x$ and the \"squaring function\" by $f(x)=x^{2}$ . The \"difference quotient\" is the rate of change of the function over one of the intervals $[x,x+h]$ defined by the formula:\n\n${\\frac {f(x+h)-f(x)}{h}}.$\n\nIt takes the function $f$ as an input, that is all the information—such as that two is sent to four, three is sent to nine, four is sent to sixteen, and so on—and uses this information to output another function, the function $g(x)=2x+h$ , as will turn out. As a matter of convenience, the new function may defined at the middle points of the above intervals:\n\n$a+h/2,a+h+h/2,a+2h+h/2,...,a+nh+h/2,...$\n\nAs the rate of change is that for the whole interval $[x,x+h]$ , any point within it can be used as such a reference or, even better, the whole interval which makes the difference quotient a $1$ -cochain.\n\nThe most common notation for the difference quotient is:\n\n${\\frac {\\Delta f}{\\Delta x}}(x+h/2)={\\frac {f(x+h)-f(x)}{h}}.$\n\nIf the input of the function represents time, then the difference quotient represents change with respect to time. For example, if $f$ is a function that takes a time as input and gives the position of a ball at that time as output, then the difference quotient of $f$ is how the position is changing in time, that is, it is the velocity of the ball.\n\nIf a function is linear (that is, if the points of the graph of the function lie on a straight line), then the function can be written as $y=mx+b$ , where $x$ is the independent variable, $y$ is the dependent variable, $b$ is the $y$ -intercept, and:\n\n$m={\\frac {\\text{rise}}{\\text{run}}}={\\frac {{\\text{change in }}y}{{\\text{change in }}x}}={\\frac {\\Delta y}{\\Delta x}}.$\n\nThis gives an exact value for the slope of a straight line.\n\nSlope: $m=\\left({\\frac {\\Delta y}{\\Delta x}}\\right)=\\tan(\\theta )$\n\nIf the function is not linear, however, then the change in $y$ divided by the change in $x$ varies. The difference quotient give an exact meaning to the notion of change in output with respect to change in input. To be concrete, let $f$ be a function, and fix a point $x$ in the domain of $f$ . $(x,f(x))$ is a point on the graph of the function. If $h$ is the increment of $x$ , then $x+h$ is the next value of $x$ . Therefore, $(x+h,f(x+h))$ is the increment of $(x,f(x))$ . The slope of the line between these two points is\n\n$m={\\frac {f(x+h)-f(x)}{(x+h)-a}}={\\frac {f(x+h)-f(x)}{h}}.$\n\nSo $m$ is the slope of the line between $(x,f(x))$ and $(x+h,f(x+h))$ .\n\nHere is a particular example, the difference quotient of the squaring function. Let $f(x)=x^{2}$ be the squaring function. Then:\n\n{\\begin{aligned}{\\frac {\\Delta f}{\\Delta x}}(x)&={(x+h)^{2}-x^{2} \\over {h}}\\\\&={x^{2}+2hx+h^{2}-x^{2} \\over {h}}\\\\&={2hx+h^{2} \\over {h}}\\\\&=2x+h.\\end{aligned}}\n\nThe difference quotient of the difference quotient is called the second difference quotient and it is defined at\n\n$a+h,a+2h,a+3h,\\ldots ,a+nh,\\ldots$\n\nAnd so on.\n\nDiscrete integral calculus is the study of the definitions, properties, and applications of the Riemann sums. The process of finding the value of an sum is called integration. In technical language, integral calculus studies a certain linear operator.\n\nThe Riemann sum inputs a function and outputs a function, which gives the algebraic sum of areas between the part of the graph of the input and the x-axis.\n\nA motivating example is the distances traveled in a given time.\n\n${\\text{distance}}={\\text{speed}}\\cdot {\\text{time}}$\n\nIf the speed is constant, only multiplication is needed, but if the speed changes, we evaluate the distance traveled by breaking up the time into many short intervals of time, then multiplying the time elapsed in each interval by one of the speeds in that interval, and then taking the sum (a Riemann sum) of the distance traveled in each interval.\n\nThe Riemann sum is measuring the total area of the bars, defined by $f$ , between two points (here $a$ and $b$ ).\n\nWhen velocity is constant, the total distance traveled over the given time interval can be computed by multiplying velocity and time. For example, travelling a steady 50 mph for 3 hours results in a total distance of 150 miles. In the diagram on the left, when constant velocity and time are graphed, these two values form a rectangle with height equal to the velocity and width equal to the time elapsed. Therefore, the product of velocity and time also calculates the rectangular area under the (constant) velocity curve. This connection between the area under a curve and distance traveled can be extended to any irregularly shaped region exhibiting a incrementally varying velocity over a given time period. If the bars in the diagram on the right represents speed as it varies from an interval to the next, the distance traveled (between the times represented by $a$ and $b$ ) is the area of the shaded region $s$ .\n\nSo, the interval between $a$ and $b$ is divided into a number of equal segments, the length of each segment represented by the symbol $\\Delta x$ . For each small segment, we have one value of the function $f(x)$ . Call that value $v$ . Then the area of the rectangle with base $\\Delta x$ and height $v$ gives the distance (time $\\Delta x$ multiplied by speed $v$ ) traveled in that segment. Associated with each segment is the value of the function above it, $f(x)=v$ . The sum of all such rectangles gives the area between the axis and the piece-wise constant curve, which is the total distance traveled.\n\nSuppose a function is defined at the mid-points of the intervals of equal length $\\Delta x=h>0$ :\n\n$a+h/2,a+h+h/2,a+2h+h/2,\\ldots ,a+nh-h/2,\\ldots$\n\nThen the Riemann sum from $a$ to $b=a+nh$ in sigma notation is:\n\n$\\sum _{i=1}^{n}f(a+ih)\\,\\Delta x.$\n\nAs this computation is carried out for each $n$ , the new function is defined at the points:\n\n$a,a+h,a+2h,\\ldots ,a+nh,\\ldots$\n\nThe fundamental theorem of calculus states that differentiation and integration are inverse operations. More precisely, it relates the difference quotients to the Riemann sums. It can also be interpreted as a precise statement of the fact that differentiation is the inverse of integration.\n\nThe fundamental theorem of calculus: If a function $f$ is defined on a partition of the interval $[a,b]$ , $b=a+nh$ , and if $F$ is a function whose difference quotient is $f$ , then we have:\n\n$\\sum _{i=0}^{n-1}f(a+ih+h/2)\\,\\Delta x=F(b)-F(a).$\n\nFurthermore, for every ${\\textstyle m=0,1,2,\\ldots ,n-1}$ , we have:\n\n${\\frac {\\Delta }{\\Delta x}}\\sum _{i=0}^{m}f(a+ih+h/2)\\,\\Delta x=f(a+mh+h/2).$\n\nThis is also a prototype solution of a difference equation. Difference equations relate an unknown function to its difference or difference quotient, and are ubiquitous in the sciences.\n\n## History\n\nThe early history of discrete calculus is the history of calculus. Such basic ideas as the difference quotients and the Riemann sums appear implicitly or explicitly in definitions and proofs. After the limit it taken, however, they are never to be seen again. However, the Kirchhoff's voltage law (1847) can be expressed in terms of the one-dimensional discrete exterior derivative.\n\nDuring the 20th century discrete calculus remains interlinked with infinitesimal calculus especially differential forms but also starts to draw from algebraic topology as both develop. The main contributions come from the following individuals :\n\nThe recent development of discrete calculus, starting with Whitney, has been driven by the needs of applied modeling. \n\n## Applications\n\nDiscrete calculus is used for modeling either directly or indirectly as a discretization of infinitesimal calculus in every branch of the physical sciences, actuarial science, computer science, statistics, engineering, economics, business, medicine, demography, and in other fields wherever a problem can be mathematically modeled. It allows one to go from (non-constant) rates of change to the total change or vice versa, and many times in studying a problem we know one and are trying to find the other.\n\nPhysics makes particular use of calculus; all discrete concepts in classical mechanics and electromagnetism are related through discrete calculus. The mass of an object of known density that varies incrementally, the moment of inertia of such objects, as well as the total energy of an object within a discrete conservative field can be found by the use of discrete calculus. An example of the use of discrete calculus in mechanics is Newton's second law of motion: historically stated it expressly uses the term \"change of motion\" which implies the difference quotient saying The change of momentum of a body is equal to the resultant force acting on the body and is in the same direction. Commonly expressed today as Force = Mass × acceleration, it invokes discrete calculus when the change is incremental because acceleration is the difference quotient of velocity with respect to time or second difference quotient of the spatial position. Starting from knowing how an object is accelerating, we use the Riemann sums to derive its path.\n\nMaxwell's theory of electromagnetism and Einstein's theory of general relativity have been expressed in the language of discrete calculus.\n\nChemistry uses calculus in determining reaction rates and radioactive decay (exponential decay).\n\nIn biology, population dynamics starts with reproduction and death rates to model population changes (population modeling).\n\nIn engineering, difference equations are used to plot a course of a spacecraft within zero gravity environments, to model heat transfer, diffusion, and wave propagation.\n\nDiscrete Green's Theorem is applied in an instrument known as a planimeter, which is used to calculate the area of a flat surface on a drawing. For example, it can be used to calculate the amount of area taken up by an irregularly shaped flower bed or swimming pool when designing the layout of a piece of property. It can be used to efficiently calculate sums of rectangular domains in images, in order to rapidly extract features and detect object; another algorithm that could be used is the summed area table.\n\nIn the realm of medicine, calculus can be used to find the optimal branching angle of a blood vessel so as to maximize flow. From the decay laws for a particular drug's elimination from the body, it is used to derive dosing laws. In nuclear medicine, it is used to build models of radiation transport in targeted tumor therapies.\n\nIn economics, calculus allows for the determination of maximal profit by calculating both marginal cost and marginal revenue, as well as modeling of markets. \n\nDiscrete calculus can be used in conjunction with other mathematical disciplines. For example, it can be used in probability theory to determine the probability of a discrete random variable from an assumed density function.\n\n## Calculus of differences and sums\n\nSuppose a function (a $0$ -cochain) $f$ is defined at points separated by an increment $\\Delta x=h>0$ :\n\n$a,a+h,a+2h,\\ldots ,a+nh,\\ldots$\n\nThe difference (or the exterior derivative, or the coboundary operator) of the function is given by:\n\n${\\big (}\\Delta f{\\big )}{\\big (}[x,x+h]{\\big )}=f(x+h)-f(x).$\n\nIt is defined at each of the above intervals; it is a $1$ -cochain.\n\nSuppose a $1$ -cochain $g$ is defined at each of the above intervals. Then its sum is a function (a $0$ -cochain) defined at each of the points by:\n\n$\\left(\\sum g\\right)(a+nh)=\\sum _{i=1}^{n}g{\\big (}[a+(i-1)h,a+ih]{\\big )}.$\n\nThese are their properties:\n\n• Constant rule: If $c$ is a constant, then\n$\\Delta c=0$\n• Linearity: if $a$ and $b$ are constants,\n$\\Delta (af+bg)=a\\,\\Delta f+b\\,\\Delta g,\\quad \\sum (af+bg)=a\\,\\sum f+b\\,\\sum g$\n$\\Delta (fg)=f\\,\\Delta g+g\\,\\Delta f+\\Delta f\\,\\Delta g$\n$\\left(\\sum \\Delta f\\right)(a+nh)=f(a+nh)-f(a)$\n• Fundamental theorem of calculus II:\n$\\Delta \\left(\\sum g\\right)=g$\n\nThe definitions are applied to graphs as follows. If a function (a $0$ -cochain) $f$ is defined at the nodes of a graph:\n\n$a,b,c,\\ldots$\n\nthen its exterior derivative (or the differential) is the difference, i.e., the following function defined on the edges of the graph ($1$ -cochain):\n\n$\\left(df\\right){\\big (}[a,b]{\\big )}=f(b)-f(a).$\n\nIf $g$ is a $1$ -cochain, then its integral over a sequence of edges $\\sigma$ of the graph is the sum of its values over all edges of $\\sigma$ (\"path integral\"):\n\n$\\int _{\\sigma }g=\\sum _{\\sigma }g{\\big (}[a,b]{\\big )}.$\n\nThese are the properties:\n\n• Constant rule: If $c$ is a constant, then\n$dc=0$\n• Linearity: if $a$ and $b$ are constants,\n$d(af+bg)=a\\,df+b\\,dg,\\quad \\int _{\\sigma }(af+bg)=a\\,\\int _{\\sigma }f+b\\,\\int _{\\sigma }g$\n• Product rule:\n$d(fg)=f\\,dg+g\\,df+df\\,dg$\n• Fundamental theorem of calculus I: if a $1$ -chain $\\sigma$ consists of the edges $[a_{0},a_{1}],[a_{1},a_{2}],...,[a_{n-1},a_{n}]$ , then for any $0$ -cochain $f$\n$\\int _{\\sigma }df=f(a_{n})-f(a_{0})$\n• Fundamental theorem of calculus II: if the graph is a tree, $g$ is a $1$ -cochain, and a function ($0$ -cochain) is defined on the nodes of the graph by\n$f(x)=\\int _{\\sigma }g$\n\nwhere a $1$ -chain $\\sigma$ consists of $[a_{0},a_{1}],[a_{1},a_{2}],...,[a_{n-1},x]$ for some fixed $a_{0}$ , then\n\n$df=g$\n\nSee references. \n\n## Chains of simplices and cubes\n\nA simplicial complex $S$ is a set of simplices that satisfies the following conditions:\n\n1. Every face of a simplex from $S$ is also in $S$ .\n2. The non-empty intersection of any two simplices $\\sigma _{1},\\sigma _{2}\\in S$ is a face of both $\\sigma _{1}$ and $\\sigma _{2}$ .\n\nThe boundary of a boundary of a 2-simplex (left) and the boundary of a 1-chain (right) are taken. Both are 0, being sums in which both the positive and negative of a 0-simplex occur once. The boundary of a boundary is always 0. A nontrivial cycle is something that closes up like the boundary of a simplex, in that its boundary sums to 0, but which isn't actually the boundary of a simplex or chain.\n\nBy definition, an orientation of a k-simplex is given by an ordering of the vertices, written as $(v_{0},...,v_{k})$ , with the rule that two orderings define the same orientation if and only if they differ by an even permutation. Thus every simplex has exactly two orientations, and switching the order of two vertices changes an orientation to the opposite orientation. For example, choosing an orientation of a 1-simplex amounts to choosing one of the two possible directions, and choosing an orientation of a 2-simplex amounts to choosing what \"counterclockwise\" should mean.\n\nLet $S$ be a simplicial complex. A simplicial k-chain is a finite formal sum\n\n$\\sum _{i=1}^{N}c_{i}\\sigma _{i},\\,$\n\nwhere each ci is an integer and σi is an oriented k-simplex. In this definition, we declare that each oriented simplex is equal to the negative of the simplex with the opposite orientation. For example,\n\n$(v_{0},v_{1})=-(v_{1},v_{0}).$\n\nThe vector space of k-chains on $S$ is written $C_{k}$ . It has a basis in one-to-one correspondence with the set of k-simplices in $S$ . To define a basis explicitly, one has to choose an orientation of each simplex. One standard way to do this is to choose an ordering of all the vertices and give each simplex the orientation corresponding to the induced ordering of its vertices.\n\nLet $\\sigma =(v_{0},...,v_{k})$ be an oriented k-simplex, viewed as a basis element of $C_{k}$ . The boundary operator\n\n$\\partial _{k}:C_{k}\\rightarrow C_{k-1}$\n\nis the linear operator defined by:\n\n$\\partial _{k}(\\sigma )=\\sum _{i=0}^{k}(-1)^{i}(v_{0},\\dots ,{\\widehat {v_{i}}},\\dots ,v_{k}),$\n\nwhere the oriented simplex\n\n$(v_{0},\\dots ,{\\widehat {v_{i}}},\\dots ,v_{k})$\n\nis the $i$ th face of $\\sigma$ , obtained by deleting its $i$ th vertex.\n\nIn $C_{k}$ , elements of the subgroup\n\n$Z_{k}=\\ker \\partial _{k}$\n\nare referred to as cycles, and the subgroup\n\n$B_{k}=\\operatorname {im} \\partial _{k+1}$\n\nis said to consist of boundaries.\n\nA direct computation shows that $\\partial ^{2}=0$ . In geometric terms, this says that the boundary of anything has no boundary. Equivalently, the vector spaces $(C_{k},\\partial _{k})$ form a chain complex. Another equivalent statement is that $B_{k}$ is contained in $Z_{k}$ .\n\nA cubical complex is a set composed of points, line segments, squares, cubes, and their n-dimensional counterparts. They are used analogously to simplices to form complexes. An elementary interval is a subset $I\\subset \\mathbf {R}$ of the form\n\n$I=[\\ell ,\\ell +1]\\quad {\\text{or}}\\quad I=[\\ell ,\\ell ]$\n\nfor some $\\ell \\in \\mathbf {Z}$ . An elementary cube $Q$ is the finite product of elementary intervals, i.e.\n\n$Q=I_{1}\\times I_{2}\\times \\cdots \\times I_{d}\\subset \\mathbf {R} ^{d}$\n\nwhere $I_{1},I_{2},\\ldots ,I_{d}$ are elementary intervals. Equivalently, an elementary cube is any translate of a unit cube $[0,1]^{n}$ embedded in Euclidean space $\\mathbf {R} ^{d}$ (for some $n,d\\in \\mathbf {N} \\cup \\{0\\}$ with $n\\leq d$ ). A set $X\\subseteq \\mathbf {R} ^{d}$ is a cubical complex if it can be written as a union of elementary cubes (or possibly, is homeomorphic to such a set) and it contains all of the faces of all of its cubes. The boundary operator and the chain complex are defined similarly to those for simplicial complexes.\n\nMore general are cell complexes.\n\nA chain complex $(C_{*},\\partial _{*})$ is a sequence of vector spaces $\\ldots ,C_{0},C_{1},C_{2},C_{3},C_{4},\\ldots$ connected by linear operators (called boundary operators) $\\partial _{n}:C_{n}\\to C_{n-1}$ , such that the composition of any two consecutive maps is the zero map. Explicitly, the boundary operators satisfy $\\partial _{n}\\circ \\partial _{n+1}=0$ , or with indices suppressed, $\\partial ^{2}=0$ . The complex may be written out as follows.\n\n$\\cdots {\\xleftarrow {\\partial _{0}}}C_{0}{\\xleftarrow {\\partial _{1}}}C_{1}{\\xleftarrow {\\partial _{2}}}C_{2}{\\xleftarrow {\\partial _{3}}}C_{3}{\\xleftarrow {\\partial _{4}}}C_{4}{\\xleftarrow {\\partial _{5}}}\\cdots$\n\nA simplicial map is a map between simplicial complexes with the property that the images of the vertices of a simplex always span a simplex (therefore, vertices have vertices for images). A simplicial map $f$ from a simplicial complex $S$ to another $T$ is a function from the vertex set of $S$ to the vertex set of $T$ such that the image of each simplex in $S$ (viewed as a set of vertices) is a simplex in $T$ . It generates a linear map, called a chain map, from the chain complex of $S$ to the chain complex of $T$ . Explicitly, it is given on $k$ -chains by\n\n$f((v_{0},\\ldots ,v_{k}))=(f(v_{0}),\\ldots ,f(v_{k}))$\n\nif $f(v_{0}),...,f(v_{k})$ are all distinct, and otherwise it is set equal to $0$ .\n\nA chain map $f$ between two chain complexes $(A_{*},d_{A,*})$ and $(B_{*},d_{B,*})$ is a sequence $f_{*}$ of homomorphisms $f_{n}:A_{n}\\rightarrow B_{n}$ for each $n$ that commutes with the boundary operators on the two chain complexes, so $d_{B,n}\\circ f_{n}=f_{n-1}\\circ d_{A,n}$ . This is written out in the following commutative diagram:\n\nA chain map sends cycles to cycles and boundaries to boundaries.\n\nSee references.\n\n## Discrete differential forms: cochains\n\nFor each vector space Ci in the chain complex we consider its dual space $C_{i}^{*}:=\\mathrm {Hom} (C_{i},{\\bf {R}}),$ and $d^{i}=\\partial _{i}^{*}$ is its dual linear operator\n\n$d^{i-1}:C_{i-1}^{*}\\to C_{i}^{*}.$\n\nThis has the effect of \"reversing all the arrows\" of the original complex, leaving a cochain complex\n\n$\\cdots \\leftarrow C_{i+1}^{*}{\\stackrel {\\partial _{i}^{*}}{\\leftarrow }}\\ C_{i}^{*}{\\stackrel {\\partial _{i-1}^{*}}{\\leftarrow }}C_{i-1}^{*}\\leftarrow \\cdots$\n\nThe cochain complex $(C^{*},d^{*})$ is the dual notion to a chain complex. It consists of a sequence of vector spaces $...,C_{0},C_{1},C_{2},C_{3},C_{4},...$ connected by linear operators $d^{n}:C^{n}\\to C^{n+1}$ satisfying $d^{n+1}\\circ d^{n}=0$ . The cochain complex may be written out in a similar fashion to the chain complex.\n\n$\\cdots {\\xrightarrow {d^{-1}}}C^{0}{\\xrightarrow {d^{0}}}C^{1}{\\xrightarrow {d^{1}}}C^{2}{\\xrightarrow {d^{2}}}C^{3}{\\xrightarrow {d^{3}}}C^{4}{\\xrightarrow {d^{4}}}\\cdots$\n\nThe index $n$ in either $C_{n}$ or $C^{n}$ is referred to as the degree (or dimension). The difference between chain and cochain complexes is that, in chain complexes, the differentials decrease dimension, whereas in cochain complexes they increase dimension.\n\nThe elements of the individual vector spaces of a (co)chain complex are called cochains. The elements in the kernel of $d$ are called cocycles (or closed elements), and the elements in the image of $d$ are called coboundaries (or exact elements). Right from the definition of the differential, all boundaries are cycles.\n\nThe Poincaré lemma states that if $B$ is an open ball in ${\\bf {R}}^{n}$ , any closed $p$ -form $\\omega$ defined on $B$ is exact, for any integer $p$ with $1\\leq p\\leq n$ .\n\nWhen we refer to cochains as discrete (differential) forms, we refer to $d$ as the exterior derivative. We also use the calculus notation for the values of the forms:\n\n$\\omega (s)=\\int _{s}\\omega .$\n\nStokes' theorem is a statement about the discrete differential forms on manifolds, which generalizes the fundamental theorem of discrete calculus for a partition of an interval:\n\n$\\sum _{i=0}^{n-1}{\\frac {\\Delta F}{\\Delta x}}(a+ih+h/2)\\,\\Delta x=F(b)-F(a).$\n\nStokes' theorem says that the sum of a form $\\omega$ over the boundary of some orientable manifold $\\Omega$ is equal to the sum of its exterior derivative $d\\omega$ over the whole of $\\Omega$ , i.e.,\n\n$\\int _{\\Omega }d\\omega =\\int _{\\partial \\Omega }\\omega \\,.$\n\nIt is worthwhile to examine the underlying principle by considering an example for $d=2$ dimensions. The essential idea can be understood by the diagram on the left, which shows that, in an oriented tiling of a manifold, the interior paths are traversed in opposite directions; their contributions to the path integral thus cancel each other pairwise. As a consequence, only the contribution from the boundary remains.\n\nSee references.\n\n## The wedge product of forms\n\nIn discrete calculus, this is a construction that creates from forms higher order forms: adjoining two cochains of degree $p$ and $q$ to form a composite cochain of degree $p+q$ .\n\nFor cubical complexes, the wedge product is defined on every cube seen as a vector space of the same dimension.\n\nFor simplicial complexes, the wedge product is implemented as the cup product: if $f^{p}$ is a $p$ -cochain and $g^{q}$ is a $q$ -cochain, then\n\n$(f^{p}\\smile g^{q})(\\sigma )=f^{p}(\\sigma _{0,1,...,p})\\cdot g^{q}(\\sigma _{p,p+1,...,p+q})$\n\nwhere $\\sigma$ is a $(p+q)$ -simplex and $\\sigma _{S},\\ S\\subset \\{0,1,...,p+q\\}$ , is the simplex spanned by $S$ into the $(p+q)$ -simplex whose vertices are indexed by $\\{0,...,p+q\\}$ . So, $\\sigma _{0,1,...,p}$ is the $p$ -th front face and $\\sigma _{p,p+1,...,p+q}$ is the $q$ -th back face of $\\sigma$ , respectively.\n\nThe coboundary of the cup product of cochains $f^{p}$ and $g^{q}$ is given by\n\n$d(f^{p}\\smile g^{q})=d{f^{p}}\\smile g^{q}+(-1)^{p}(f^{p}\\smile d{g^{q}}).$\n\nThe cup product of two cocycles is again a cocycle, and the product of a coboundary with a cocycle (in either order) is a coboundary.\n\nThe cup product operation satisfies the identity\n\n$\\alpha ^{p}\\smile \\beta ^{q}=(-1)^{pq}(\\beta ^{q}\\smile \\alpha ^{p}).$\n\nIn other words, the corresponding multiplication is graded-commutative.\n\nSee references.\n\n## Laplace operator\n\nThe Laplace operator $\\Delta f$ of a function $f$ at a vertex $p$ , is (up to a factor) the rate at which the average value of $f$ over a cellular neighborhood of $p$ deviates from $f(p)$ . The Laplace operator represents the flux density of the gradient flow of a function. For instance, the net rate at which a chemical dissolved in a fluid moves toward or away from some point is proportional to the Laplace operator of the chemical concentration at that point; expressed symbolically, the resulting equation is the diffusion equation. For these reasons, it is extensively used in the sciences for modelling various physical phenomena.\n\nThe codifferential\n\n$\\delta :C^{k}\\to C^{k-1}$\n\nis an operator defined on $k$ -forms by:\n\n$\\delta =(-1)^{n(k-1)+1}{\\star }d{\\star }=(-1)^{k}\\,{\\star }^{-1}d{\\star },$\n\nwhere $d$ is the exterior derivative or differential and $\\star$ is the Hodge star operator.\n\nThe codifferential is the adjoint of the exterior derivative according to Stokes' theorem:\n\n$(\\eta ,\\delta \\zeta )=(d\\eta ,\\zeta ).$\n\nSince the differential satisfies $d^{2}=0$ , the codifferential has the corresponding property\n\n$\\delta ^{2}={\\star }d{\\star }{\\star }d{\\star }=(-1)^{k(n-k)}{\\star }d^{2}{\\star }=0.$\n\nThe Laplace operator is defined by:\n\n$\\Delta =(\\delta +d)^{2}=\\delta d+d\\delta .$\n\nSee references."
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"Design Project, ECE583\nExercise 1: Information conveyed by a discrete Gaussian channel [40%]\nConsider a discrete-time memoryless channel\nyk =\n\nSNRxk + nk, k = 1, 2, . . .\nwhere n ∼ CN (0, 1) and E|X|2 = 1 so that SNR denotes the average signal-to-noise ratio at the\nreceiver. According to Shannon, if the input X is drawn according to a Gaussian distribution, then\nthe capacity per real dimension is\nC =\n1\n2\nlog2(1 + SNR) [bits/real dimension].\nIn practice, however, we cannot use a Gaussian input. In this exercise, we are going to study the\neffect of having finite-alphabet constellations on the amount of information conveyed by this channel.\na) What is the conditional PDF of the channel, i.e., pY |X(y|x) for 1) a real constellation 2) a\ncomplex constellation?\nb) For a general constellation of size N , derive a computable formula for mutual information\nI(X;Y ) , i.e.,\nI(X;Y ) = EX,Y\n(\nlog2\npX,Y (x, y)\npX(x)pY (y)\n)\n.\n[Hint: The expression should be in terms of SNR, N , the constellation points and pX(x), i.e.,\nthe input distribution.]\nc) Let ΩX denote the set of constellation points. Assume that the input is uniformly distributed,\ni.e.,\npX(x) =\n1\nN\n∀x ∈ ΩX .\nPlot the average mutual information I(X;Y ) (in bits per real dimension) versus the SNR (from\n−10 dB to 40 dB with a 2-dB step) for a 2-PAM, 8-PAM, 16-PAM, 64-PAM, 16-QAM, and 64-\nQAM constellations in a single figure. Also, in the same figure, plot the capacity and compare\nit with other mutual information curves that you obtained.\nd) Consider a 16-QAM constellation shown in Fig. 1. Let p, q, and r be the probability of white,\nblack and shaded dots, respectively. Find the relation between them. Also, for SNR = 15 dB find\noptimal values (p∗, q∗, r∗) such that I(X;Y ) is maximized. Compare this mutual information\nwith the case p = q = r = 116 .\n1\nECE583: Digital Communications\nFigure 1: Exercise 1 part d\nExercise 2: ISI and Detection [60%]\nConsider a system, shown in Fig. 2, employing a 16-QAM modulation over an AWGN channel of the\nform\nyk =\n\nSNRxk + nk, k = 1, 2, . . . ,M\nwhere n ∼ CN (0, 1), E|X|2 = 1 such that SNR denotes the average signal-to-noise ratio at the receiver.\nIn Fig. 2, g(t) denotes the pulse shape, b(t) represents the channel impulse response and f(t) is the\nfilter at the receiver. Moreover, we have the transmitted signal of the form\ns(t) =\n\nk\nxkg(t− kT ).\n• a) Non-ISI channel + symbol detection\nIn this part, you are asked to implement the system shown in Fig. 2 where b(t) = δ(t), i.e., a non-ISI\nchannel, and the detector is a minimum distance (symbol) detector.\n1. Let the symbol time T = 1 and generate a random sequence of M = 104 16-QAM symbols.\n2. Use a root-raised-cosine pulse g(t) with a 35% excess bandwidth over the time period [−20,+20]\nwith 10 samples per symbol to generate the transmitted signal s(t).\n3. Pass the signal through the channel and the matched filter (matched with the pulse shape and\nchannel).\n4. Sample the resulting signal at the appropriate time instances in order to generate the decision\nvariables {uk}.\n2\nECE583: Digital Communications\nmapper ft\nT\nxhkxk yk\nnt\nst rtd\ndetbtgt\nFigure 2: Exercise 2 part a and b\n5. Now, use the minimum distance detector and plot the average BER versus SNR (0 dB to 26 dB\nwith a 2-dB step), i.e.,\nPe(SNR) =\n1\nM log2N\nM∑\nk=1\nw(xˆk, xk)\nwhere w(xˆk, xk) denotes the number of bits in error at time k and (in this example) N = 16.\n6. If necessary increase the number of symbols M in order to see a better curve.\n• b) ISI channel + symbol detection\nIn the model of the previous part, consider an ISI channel with\nb(t) = δ(t) + βδ(t− T ), |β| < 1.\n1. Let the symbol time T = 1 and generate a random sequence of M = 104 16-QAM symbols.\n2. Use a root-raised-cosine pulse g(t) with a 35% excess bandwidth over the time period [−30,+30]\nwith 10 samples per symbol to generate the transmitted signal s(t).\n3. Pass the signal through an ISI channel with β = 0.1 and the matched filter (matched with the\npulse shape and channel).\n4. Sample the resulting signal at the appropriate time instances in order to generate the decision\nvariables {uk}.\n5. For β = 0, 0.01, 0.1, 0.2, use a symbol detector and plot the average BER versus SNR (0 dB to\n26 dB with a 2-dB step).\n6. Explain why the detector in this part is not optimal.\n7. Explain the behavior of the curves for different values of β.\nReferences\n G. Ungerboeck, “Channel coding with multilevel/phase signals,” IEEE Trans. Inf. Theory, vol. 28,\nno. 1, pp. 55–67, Jan. 1982.\n3",
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[
"# SortIndex\n\n## SortIndex(d, i)\n\nReturns the elements of «i» re-arranged so that the values of «d» (which must be indexed by «i») are in ascending order. In the event of a tie, the original order is preserved.\n\nIf «i» is omitted, «d» must be a one-dimensional array (i.e. a list). In this case, SortIndex returns an unindexed list of elements. Use the one-parameter form only when you want an unindexed result, for example to define an index variable. The one-parameter form does array abstract when a new dimension is added to d.\n\nIf «i» is specified, «d» may be multi-dimensional. Each slice of «d» is sorted separately along «i», with the result being an array having the same dimensions as «d», but where each element is the corresponding element in «i» indicating the sort.\n\nIf «d» has indexes other than «i», each “column” is individually sorted, with the resulting sort order being indexed by the extra dimensions. To obtain the sorted array «d», use this: `d[i = Sortindex(d, i)]`\n\n## Example\n\nTo sort the elements of an index (in ascending order), use\n\n`SortIndex(I)`\n\nTo sort the elements of an array A, along I, use\n\n`A[I = sortIndex(A, I)]`\n\n## Optional parameters\n\n### CaseInsensitive\n\nWhen sorting text values, values are compared by default in a case-sensitive fashion, with capital letters coming before lower case letters. For example, \"Zebra\" comes before \"apple\" in a case-sensitive order.\n\nTo make the sorting case insensitive, specify the optional parameter `caseInsensitive: true`:\n\n`SortIndex(D, caseInsensitive: true)`\n\n### Descending\n\nThe default sort order for SortIndex is ascending. The descending sort order can be obtained by using: `SortIndex(d, descending: true)`.\n\nFor an array containing only numeric values, the descending sort order can also be obtained as: `SortIndex(-d)`.\n\nIf the data being sorted contains different data types, the (ascending) sort order used is: references, text, parsed expressions, Handles, NaN, numbers, Null, Undefined. All text values are sorted relative to other text values, and all numbers are sorted relative to other numeric values. References have no defined sort order, so the ordering among references is arbitrary and the resulting sort order is heterogeneous.\n\n### KeyIndex\n\nIn the event of a tie, SortIndex preserves the original ordering. A multi-key sort finds the order by sorting on a primary key, but in the event of a tie, breaks the tie using a secondary key. The pattern can continue to tertiary keys, etc. In the general case, each key may have a different ascending/descending order or differ on whether comparisons should be case-sensitive.\n\nThe values used for the primary key, and the values used for each fall-back key, must all share a common index, «i». To pass these to SortIndex, you must bundle these together along another index, «keyIndex», where the first element along «keyIndex» is your primary key, the second element is your secondary key, etc. After you bundle these together, the first parameter to SortIndex will be a 2-D array indexed by «i» and «keyIndex». For example:\n\n`Index K := ['last', 'first']`\n`SortIndex(Array(K, [lastName, firstName]), Person, K)`\n\nIn this example, we use lastName as the primary key and firstName as the secondary key. If the optional parameters «descending» or «caseInsensitive» are also passed, these may optionally be indexed by «keyIndex» if the order or case sensitivity varies by key.\n\n### Position\n\nBy default, SortIndex returns the elements of the index in the sorted order. In some cases, you may want the positions of the first element, etc., rather than the index elements (see Associative vs. Positional Indexing). To obtain the positions, specify the optional parameter position:true: `SortIndex]](D, position: true)`\n\nUsing positional notation, the original array can be re-ordered using: `D[@I = SortIndex(D, I, position: true)]`\n\nThe use of positional indexing would be required, for example, if you index might contain duplicate values (note that in general, it is very bad style to have duplicate elements in an index).\n\n## Details & More Examples\n\n### Example 1\n\nTo sort an array A (indexed by indexes Row and Col) according to the values in `Col = key`, use\n\n`A[Row = SortIndex(A[Col = 'key'], Row)]`\n\n### Example 2\n\nLet:\n\n`Variable Maint_costs :=`\nCar_type ▶\nVW Honda BMW\n1950 1800 2210\n`Index Sorted_cars := SortIndex(Maint_costs)`\n\nThen:\n\n`SortIndex(Maint_costs, Car_type) →`\nCar_type ▶\nVW Honda BMW\nHonda VW BMW\n`SortIndex(Maint_costs) →`\nSortIndex ▶\nHonda VW BMV\n`Maint_costs[Car_type = Sorted_cars] →`\nHonda VW BMW\n1800 1950 2210"
] |
[
null
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|
https://study.com/academy/answer/a-50-g-ball-is-released-onto-a-platform-at-a-with-initial-speed-v0-the-dotted-line-abcde-tracks-the-centre-of-the-ball-segment-ab-is-a-circular-are-of-radius-1-0-m-segment-bc-is-a-straight-line-and.html
|
[
"# A 50-g ball is released onto a platform at A with initial speed v0. The dotted line ABCDE tracks...\n\n## Question:\n\nA 50-g ball is released onto a platform at A with initial speed v0. The dotted line ABCDE tracks the centre of the ball. Segment AB is a circular are of radius 1.0 m, Segment BC is a straight line and segment CDE is a circular are of radius 2.0 m. Gravity points down.\n\n(a) Determine the minimum initial speed vo so that the ball clears the highest point of the platform at D.\n\n(b) Determine the minimum initial speed to so that the ball leaves the platform at C.",
null,
"## Conservation of energy\n\nAs per this law the energy of a system will always remains constant or conserved which means the energy of a system can never be generated or can't be destroyed.\n\n## Answer and Explanation:\n\n(a)\n\nGiven Data:\n\n• The mass of the ball is: {eq}m = 50\\;{\\rm{g}} {/eq}.\n• The value of angle is: {eq}\\theta = 45^\\circ {/eq}.\n\nThe figure below shows the free body diagram of the ball.",
null,
"The expression of the kinetic energy of the ball at point A is,\n\n{eq}{K_{\\rm{A}}} = \\dfrac{1}{2}mV_0^2 {/eq}\n\nThe expression for the potential energy of the ball at point A is,\n\n{eq}{P_{\\rm{A}}} = mg\\left( {{r_1}} \\right) {/eq}\n\nAs the ball reaches its maximum height at point D, its velocity will be zero.\n\nTherefore, {eq}{V_{\\rm{D}}} = 0\\;{{\\rm{m}} {\\left/ {\\vphantom {{\\rm{m}} {\\rm{s}}}} \\right. } {\\rm{s}}} {/eq}.\n\nThe expression of the kinetic energy of the ball at point D is,\n\n{eq}{K_{\\rm{D}}} = \\dfrac{1}{2}mV_D^2 {/eq}\n\nThe expression for the potential energy of the ball at point D is,\n\n{eq}{P_{\\rm{D}}} = mg\\left( {{r_1} + 0.8\\;{\\rm{m}}} \\right) {/eq}\n\nThe expression for the conservation of energy of the ball between A to D is,\n\n{eq}{K_{\\rm{A}}} + {P_{\\rm{A}}} = {K_{\\rm{D}}} + {P_{\\rm{D}}} {/eq}\n\nSubstitute the values in above expression.\n\n{eq}\\begin{align*} \\dfrac{1}{2}mV_0^2 + mg\\left( {{r_1}} \\right) &= \\dfrac{1}{2}mV_D^2 + mg\\left( {{r_1} + 0.8\\;{\\rm{m}}} \\right)\\\\ {V_{\\rm{0}}} &= \\sqrt {2\\left( {9.81\\;{{\\rm{m}} {\\left/ {\\vphantom {{\\rm{m}} {{{\\rm{s}}^2}}}} \\right. } {{{\\rm{s}}^2}}}} \\right)\\left( {0.8\\;{\\rm{m}}} \\right)} \\\\ {V_0} &= 3.9618\\;{{\\rm{m}} {\\left/ {\\vphantom {{\\rm{m}} {\\rm{s}}}} \\right. } {\\rm{s}}} \\end{align*} {/eq}\n\nThus the initial velocity of the ball is {eq}3.9618\\;{{\\rm{m}} {\\left/ {\\vphantom {{\\rm{m}} {\\rm{s}}}} \\right. } {\\rm{s}}} {/eq}.\n\n(b)\n\nThe expression for the equilibrium of force perpendicular to the curve path is,\n\n{eq}\\begin{align*} \\dfrac{{mV_{\\rm{c}}^2}}{{{r_2}}} - mg\\cos \\theta &= 0\\\\ {V_{\\rm{c}}} &= \\sqrt {g{r_2}\\cos \\theta } \\end{align*} {/eq}\n\nSubstitute the values in above expression.\n\n{eq}\\begin{align*} {V_{\\rm{c}}} &= \\sqrt {\\left( {9.81\\;{{\\rm{m}} {\\left/ {\\vphantom {{\\rm{m}} {{{\\rm{s}}^2}}}} \\right. } {{{\\rm{s}}^2}}}} \\right)\\left( {2\\;{\\rm{m}}} \\right)\\cos \\left( {45^\\circ } \\right)} \\\\ &= 3.724\\;{{\\rm{m}} {\\left/ {\\vphantom {{\\rm{m}} {\\rm{s}}}} \\right. } {\\rm{s}}} \\end{align*} {/eq}\n\nThe expression of the kinetic energy of the ball at point A is,\n\n{eq}{K_{\\rm{A}}} = \\dfrac{1}{2}mV_0^2 {/eq}\n\nThe expression for the potential energy of the ball at point A is,\n\n{eq}{P_{\\rm{A}}} = mg\\left( {{r_1}} \\right) {/eq}\n\nThe expression of the kinetic energy of the ball at point C is,\n\n{eq}{K_{\\rm{C}}} = \\dfrac{1}{2}mV_c^2 {/eq}\n\nThe expression for the potential energy of the ball at point C is,\n\n{eq}{P_{\\rm{C}}} = mg\\left( {{r_2}\\cos \\theta - 0.2\\;{\\rm{m}}} \\right) {/eq}\n\nThe expression for the conservation of energy of the ball between A to D is,\n\n{eq}{K_{\\rm{A}}} + {P_{\\rm{A}}} = {K_{\\rm{C}}} + {P_{\\rm{C}}} {/eq}\n\nSubstitute the values in above expression.\n\n{eq}\\begin{align*} \\dfrac{1}{2}mV_0^2 + mg\\left( {{r_1}} \\right) &= \\dfrac{1}{2}mV_c^2 + mg\\left( {{r_2}\\cos \\theta - 0.2\\;{\\rm{m}}} \\right)\\\\ {V_{\\rm{0}}} &= \\sqrt {V_c^2 + 2g\\left( {{r_2}\\cos \\theta - 0.2\\;{\\rm{m}} - {r_1}} \\right)} \\\\ {V_0} &= \\sqrt {{{\\left( {3.724\\;{{\\rm{m}} {\\left/ {\\vphantom {{\\rm{m}} {\\rm{s}}}} \\right. } {\\rm{s}}}} \\right)}^2} + 2\\left( {9.81\\;{{\\rm{m}} {\\left/ {\\vphantom {{\\rm{m}} {{{\\rm{s}}^2}}}} \\right. } {{{\\rm{s}}^2}}}} \\right)\\left( {\\left( {2\\;{\\rm{m}}} \\right)\\cos \\left( {45^\\circ } \\right) - 0.2\\;{\\rm{m}} - 1\\;{\\rm{m}}} \\right)} \\\\ {V_0} &= 4.250\\;{{\\rm{m}} {\\left/ {\\vphantom {{\\rm{m}} {\\rm{s}}}} \\right. } {\\rm{s}}} \\end{align*} {/eq}\n\nThus the initial velocity of the ball is {eq}4.250\\;{{\\rm{m}} {\\left/ {\\vphantom {{\\rm{m}} {\\rm{s}}}} \\right. } {\\rm{s}}} {/eq}.\n\n#### Learn more about this topic:",
null,
"Energy Conservation and Energy Efficiency: Examples and Differences\n\nfrom Geography 101: Human & Cultural Geography\n\nChapter 13 / Lesson 9\n100K"
] |
[
null,
"https://study.com/cimages/multimages/16/1112583247003450022162.png",
null,
"https://study.com/cimages/multimages/16/imgpsh_fullsize_anim1436882424794708773.jpg",
null,
"https://study.com/images/reDesign/global/spinner-dark-teal.png",
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|
https://www.driverlesscrocodile.com/tools-and-techniques/lambda-calculus-for-people-a-step-behind-me-4-playlist-reading-list/
|
[
"# Lambda Calculus for people a step behind me (4): Reading list, playlist\n\nDisclaimer: I’m not a computer scientist or mathematician. This post is about what I’ve learnt so far about Lambda Calculus. Most of the resources I’ve come across seem to assume either prior knowledge of advanced maths notation or extensive practical experience of lambda functions in software engineering. I’ve tried to avoid this and explain things step by step, but have no doubt made assumptions and mistakes of my own. Corrections (or questions) are welcome. I’ll include a list of the resources I’ve used to piece this together in a later post.\n\n## Introduction to Lambda Calculus – Reading List\n\nMost of the resources I found expect prior knowledge of mathematical notation or programming, but these worked for me, with little of either:\n\n1. A great introduction by Jake Peck on his blog, Suspended Chord\n2. The Stanford Encyclopedia of Philosophy – Lambda Calculus\n3. A Tutorial Introduction to the Lambda Calculus by Raúl Rojas (.pdf)\n4. Section five of this explanation at diderot.one is helpful on combinators\n\n#### Computerphile\n\nThese two videos are probably the simplest introductions to Lambda Calculus and the Y Combinator:\n\n### Alex Lugo\n\nThe embed below starts at about 6 minutes into the video, where there are a couple of nice worked examples of simple beta reduction. Early on it explains currying (a method for working through functions one argument at a time). Towards the end it gets into expressing integers and increment and decrement functions.\n\n### Gabriel Lebec – A Flock of Functions\n\nThese two videos are brilliant and detailed… and hard to follow if you don’t know either Lambda Calculus or Javascript. The second one has the best explanation of Church Numerals and arithmetic in Lambda Calculus that I’ve seen.\n\nIf you’re wondering about the different birds associated with combinators, see:\n\nTo Dissect A Mockingbird (this has rather interesting diagrammatic forms of different combinators that I haven’t had a chance to look at in detail… yet.)\nTo Mock a Mockingbird (full text here)"
] |
[
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http://www.theoacheampong.com/2018/06/cedi-appreciation-or-depreciation-the-conceptual-framework-for-calculating-depreciation-and-appreciation-of-currencies/
|
[
"# Cedi appreciation or depreciation? The conceptual framework for calculating depreciation and appreciation of currencies\n\nOn 21 June 2014, amid the heavy debate on cedi depreciation and ill-timed Bank of Ghana forex rules, I wrote a piece on this platform explaining the conceptual framework for computing the depreciation or appreciation of currencies. I reproduce it below:\n\n********\nUnder a floating exchange range regime, depreciation refers to the loss of value of a country’s currency with respect to one or more foreign (reference/basket) currencies often due to a number of reasons – economic fundamentals, interest rate differentials, political instability, risk aversion among investors, etc. Countries with weak economic fundamentals such as chronic current account deficits and high rates of inflation generally have depreciating currencies [Investopedia]. For example, if the Ghana Cedi depreciates relative to the dollar within a given timeframe, the exchange rate, which is the Ghana Cedi price of dollars, rises. That is, it takes more Ghana Cedis to purchase one dollar.\n\nIn the 36 months ending May 2014, for example, the Ghana Cedi depreciated by 49.87% against the US dollar or the dollar appreciated by 99.46% against the Cedi. Question: how is this computed and what is the cause of the confusion surrounding the rate computation? Well, it’s actually to do with the conceptual framework of the upper bound for depreciation and the percent change formula, which many including experts alike, have confused to represent depreciation. This, however, is notwithstanding the real substance of the debate, which is the ill-timed and ineffective Bank of Ghana Forex rules.\n\nThe formula below hopefully should clear this confusion:\n1. Depreciation of currency = [Old Rate – New Rate] /New Rate x 100%. That always generates a negative number for which the bound cannot be beyond 100% as the denominator is the greater number.\n2. Appreciation of currency = [New Rate – Old Rate]/ Old Rate. Hence, we can have greater than 100% appreciation.\n\nThe key here is what one chooses as the base currency and quoted currency. Let me explain this further. Using USD/GHC, which is how many Ghana Cedis is a dollar worth; In May 2013, the midpoint USD/GHC was 1.9867 as compared to 2.9052 in May 2014. Hence, the depreciation over the 12-month period for the Cedi compared to the dollar was -31.62% [1.9867-2.9052/2.9052].\n\nAlternatively, the dollar appreciated against the Cedi by 46.23% [2.9052-1.9867/1.9867]. On the other hand, using GHC/USD (how many dollars a cedi is worth) in May 2013, GHC/USD was 0.5034 compared to 0.3443 in May 2014. In this instance, you would use the former formula to compute the depreciation. Which is [0.3443-0.5034]/0.5034 *100% = -31.61% depreciation of the Cedi relative to the dollar or relative to the Cedi the dollar appreciated by 46.21%. That is [0.5034-0.3443]/ 0.3443*100% = 46.21%. Note this is the reverse of the formula above.\n\nAs I have demonstrated, the formula is a bit tricky. In the discussions taking place on Social Media, I have noticed many (myself included) sometimes adopt the usual percentage change formula, which is incorrect. Depreciation of a currency cannot be more than 100%; appreciation, on the other hand, can be. Therefore, anything above 100% is appreciation and not depreciation. What most folks have been referring to and/or computing is the latter, which is the appreciation of the dollar relative to the Cedi.\n\nHave a good weekend!",
null,
""
] |
[
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"http://www.theoacheampong.com/wp-content/uploads/2018/08/36308858_10216893385596252_4994249248087212032_n.jpg",
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] |
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https://roncyrocks.com/63qa42k1/1-kilogram-weight-is-equal-to-newton-500978
|
[
"# 1 kilogram weight is equal to newton\n\nkg to tetradrachm The kilogram-force (kgf or kgF), or kilopond (kp, from Latin pondus meaning weight), is a gravitational metric unit of force. The kilogram is defined as being equal to the mass of the International Prototype Kilogram (IPK), which is almost exactly equal to the mass of one liter of water. It is defined as the magnitude of force applied to one kilogram of mass under the condition of standard gravity (9.80665 m/s 2).One kilogram-force is therefore equal to 9.80665 N. We have, The weight of an object, W = mg, where m is the mass and g the acceleration due to gravity. 1 kilogram is equal to 9.8066500286389 N. Note that rounding errors may occur, so always check the results. Kilogram Force : The kilogram-force, or called kilopond, is a gravitational metric unit of force. In speciality cooking an accurate weight and mass unit measure can be totally crucial. 28.61 pounds = Y newtons. In physics, the newton (symbol: N) is the SI unit of force, named after Sir Isaac Newton in recognition of his work on classical mechanics. It is defined as the magnitude of force applied to one kilogram of mass under the condition of standard gravity (9.80665 m/s 2).One kilogram-force is therefore equal to 9.80665 N. It is the EQUAL force value of 1 newton but in the kilograms force force unit alternative. Kilogram-force. metres squared, grams, moles, feet per second, and many more. On Earth 1 Newton is equal to 9.8kgm/s2. It is the approximate weight of a cube of water 10 centimeters on a side. Where, at the temperature of melting ice, one gram is the absolute weight of a volume equal to the hundredth part of a metre cube of pure water. to use the unit converter. Pounds to Kilograms Conversion Table (some results rounded) lb kg; 0.01: 0.0045359: 0.02: 0.0090718: The kilogram is defined as being equal to the mass of the International Prototype Kilogram (IPK), which is almost exactly equal to the mass of one liter of water. 1 pounds = 4.4482204893932 newtons. kg is a mass, Newton is a force, they aren't the same. You can do the reverse unit conversion from Newton (noun) English mathematician and physicist; remembered for developing the calculus and for his law of gravitation and his three laws of motion (1642-1727) Newton (noun) a unit of force equal to the force that imparts an acceleration of 1 m/sec/sec to a mass of 1 kilogram; equal … short brevis ) unit symbol for newton earth is: N. How many newtons earth of weight and mass system are in 1 kilogram? You can find metric conversion tables for SI units, as well Find in: main Units menu • weight and mass menu • Kilograms, * Whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)* Precision is how many numbers after decimal point (1 - 9). m) is sometimes used as a unit of work or energy and in this context is equal to the joule, the SI unit of energy. Convert weight and mass culinary measuring units between kilogram (kg) and newtons earth (N) but in the other direction from newtons earth into kilograms also as per weight and mass units. Conversion chart - newtons to kilograms force 1 newton to kilograms force … If the kilogram is accelerating at 1m/s2 it is equal to 1 Newton. One kilogram in weight and mass sense converted to newtons earth equals precisely to 9.81 N How many newtons earth of weight and mass system are in 1 kilogram? N Professional people always ensure, and their success in fine cooking depends on, they get the most precise units conversion results in measuring their ingredients. Or, how much in newtons earth weight and mass in 1 kilogram? On Earth, it weighs 9.8 newtons (2.205 pounds). inch, 100 kg, US fluid ounce, 6'3\", 10 stone 4, cubic cm, One newton is equal to 1 kilogram meter per second squared. There are 1000 liters in a cubic meter, so the mass of 1 cubic meter of water is approximately 1000 kilograms or 1 metric ton. 1 Newton in Earth gravity is the equivalent weight of 1/9.80665 kg on Earth. 1 newton (N) = 0.101971621 kilogram-force (kgf) = … Exchange reading in kilograms unit kg into newtons earth unit N as in an equivalent measurement result (two different units but the same identical physical total value, which is also equal to their proportional parts when divided or multiplied). One newton is equal to 1 kilogram meter per second squared. 1 kg is a mass, not a weight. kg to mina kg to centigram The kilogram (SI unit symbol: kg) is the base unit of mass in the Metric system and is defined as being equal to the mass of the International Prototype of the Kilogram..more definition+. The newton is a unit to for measuring force equal to the force needed to move one kilogram of mass at a rate of one meter per second squared.. We assume you are converting between kilogram and newton. This is derived using Newton's second law f=ma and assuming Earth gravity of 9.80665 m/s 2. So if we are asked to convert kilogram to newton we just have to multiply kilogram value with 1. brevis - short unit symbol for kilogram is: kg This is derived using Newton's second law f=ma and assuming Earth gravity of 9.80665 m/s 2. kg to kilo Exchange reading in kilograms unit kg into newtons earth unit N as in an equivalent measurement result (two different units but the same identical physical total value, which is also equal to their proportional parts when divided or multiplied). The newton is the SI derived unit for force in the metric system. With the above mentioned units converting service it provides, this weight and mass units converter also proved to be useful as a teaching tool and for practising kilograms and newtons earth ( kg vs. N ) conversion exercises by new culinarians and students (in classrooms or at home based kitchens) who have been learning this particular cooking mastery art in culinary colleges, in schools of culinary arts and all other kinds of culinary training for converting the weight and mass cooking units measures. In general, Mass is a measure of the amount of matter in a particle or object. The weight and mass kitchen measuring units converter for culinary chefs, bakers and other professionals. Prefix or abbreviation ( abbr. It depends on where you are. g = 9.8 m/s 2. The SI base unit for mass is the kilogram. A gram is defined as one thousandth of a kilogram. N to kg, or enter any two units below: kg to kin ConvertUnits.com provides an online An average-sized apple exerts about one newton of force, which we measure as the apple's weight. The SI base unit for mass is the kilogram. A newton is the force: A. of gravity on a 1 kg body B. of gravity on a 1 g body C. that gives a 1 g body an acceleration of 1 cm/s2 D. that gives a 1 kg body an acceleration of 1 m/s2 E. that gives a 1 kg body an acceleration of 9.8 m/s2 kg or Note that rounding errors may occur, so always check the results. Kilogram (abbreviations: kg, or kgm): is a SI (metric) system unit of mass (IPK, also known as \"Le Grand K\" or \"Big K\") and equivalent to 1/1000 of one gram. kg to fotmal Example : to convert 108 kg to N 108 kilogram equals 108 x 1 newton i.e 108 newton. 1 kg wt = 1 kgf = 9.8 N. This means that the weight of 1 kg mass is 9.8 N. The kilogram (abbreviation: kg) is the unit of mass in the metric system (SI, International System of Units).Kilogram is the base unit of mass in the metric system. Newtons can be expressed using the formula: 1 N = 1 kg m s 2 Use this page to … A Newton (always a capital N because it is named after someone) ia a unit of force whilst a kg is a unit of mass. The kilogram or kilogramme, (symbol: kg) is the SI base unit of mass. The newton (abbreviation: N) is the unit of force in metric system (SI), named after Isaac Newton.The newton is equal to the amount of force needed to accelerate a one kilogram mass at a rate of one meter per second squared. The answer is: The change of 1 kg ( kilogram ) unit for a weight and mass measure equals = into 9.81 N ( newton earth ) as per its equivalent weight and mass unit type measure often used. You can view more details on each measurement unit: 1 N = 0.10197 kg × 9.80665 m/s 2 (0.101 97 kg = 101.97 g). What is the weight of 1 kg CO2? Y = 28.61 * 4.4482204893932 / 1 = 127.26358820154 newtons Answer is: 127.26358820154 newtons are equivalent to 28.61 pounds. m. One newton metre, sometimes hyphenated newton-metre, is equal to the torque resulting from a force of one newton applied perpendicularly to a moment arm which is … Mass is given as 10 kg. One kilogram converted into newton earth equals = 9.81 N 1 kg … The newton (symbol: N) is the SI unit of force.It is named after Sir Isaac Newton because of his work on classical mechanics.A newton is how much force is required to make a mass of one kilogram accelerate at a rate of one metre per second squared."
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https://graz.pure.elsevier.com/de/publications/configuration-spaces-of-equal-spheres-touching-a-given-sphere-the
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[
"# Configuration spaces of equal spheres touching a given sphere: The twelve spheres problem\n\nRob Kusner, Wöden Kusner, Jeffrey C. Lagarias, Senya Shlosman\n\nPublikation: Beitrag in Buch/Bericht/KonferenzbandBeitrag in Buch/BerichtForschungBegutachtung\n\n### Abstract\n\nThe problem of twelve spheres is to understand, as a function of r ϵ (0,rmax(12)], the configuration space of 12 non-overlapping equal spheres of radius r touching a central unit sphere. It considers to what extent, and in what fashion, touching spheres can be varied, subject to the constraint of always touching the central sphere. Such constrained motion problems are of interest in physics and materials science, and the problem involves topology and geometry. This paper reviews the history of work on this problem, presents some new results, and formulates some conjectures. It also presents general results on configuration spaces of N spheres of radius r touching a central unit sphere, with emphasis on 3 ≤ N ≤ 14. The problem of determining the maximal radius rmax(N) is a version of the Tammes problem, to which László Fejes Tóth made significant contributions.\n\nOriginalsprache englisch Bolyai Society Mathematical Studies Springer Berlin - Heidelberg 219-277 59 https://doi.org/10.1007/978-3-662-57413-3_10 Veröffentlicht - 1 Jan 2018\n\n### Publikationsreihe\n\nName Bolyai Society Mathematical Studies 27 1217-4696\n\n### Fingerprint\n\nConfiguration Space\nUnit Sphere\nMaterials Science\nMaterials science\nPhysics\nTopology\nMotion\nGeometry\n\n### ASJC Scopus subject areas\n\n• !!Computational Theory and Mathematics\n• Angewandte Mathematik\n\n### Dies zitieren\n\nKusner, R., Kusner, W., Lagarias, J. C., & Shlosman, S. (2018). Configuration spaces of equal spheres touching a given sphere: The twelve spheres problem. in Bolyai Society Mathematical Studies (S. 219-277). (Bolyai Society Mathematical Studies; Band 27). Springer Berlin - Heidelberg. https://doi.org/10.1007/978-3-662-57413-3_10\n\nConfiguration spaces of equal spheres touching a given sphere : The twelve spheres problem. / Kusner, Rob; Kusner, Wöden; Lagarias, Jeffrey C.; Shlosman, Senya.\n\nBolyai Society Mathematical Studies. Springer Berlin - Heidelberg, 2018. S. 219-277 (Bolyai Society Mathematical Studies; Band 27).\n\nPublikation: Beitrag in Buch/Bericht/KonferenzbandBeitrag in Buch/BerichtForschungBegutachtung\n\nKusner, R, Kusner, W, Lagarias, JC & Shlosman, S 2018, Configuration spaces of equal spheres touching a given sphere: The twelve spheres problem. in Bolyai Society Mathematical Studies. Bolyai Society Mathematical Studies, Bd. 27, Springer Berlin - Heidelberg, S. 219-277. https://doi.org/10.1007/978-3-662-57413-3_10\nKusner R, Kusner W, Lagarias JC, Shlosman S. Configuration spaces of equal spheres touching a given sphere: The twelve spheres problem. in Bolyai Society Mathematical Studies. Springer Berlin - Heidelberg. 2018. S. 219-277. (Bolyai Society Mathematical Studies). https://doi.org/10.1007/978-3-662-57413-3_10\nKusner, Rob ; Kusner, Wöden ; Lagarias, Jeffrey C. ; Shlosman, Senya. / Configuration spaces of equal spheres touching a given sphere : The twelve spheres problem. Bolyai Society Mathematical Studies. Springer Berlin - Heidelberg, 2018. S. 219-277 (Bolyai Society Mathematical Studies).\n@inbook{92cf3d547df348b5bd4b3b15635196e0,\ntitle = \"Configuration spaces of equal spheres touching a given sphere: The twelve spheres problem\",\nabstract = \"The problem of twelve spheres is to understand, as a function of r ϵ (0,rmax(12)], the configuration space of 12 non-overlapping equal spheres of radius r touching a central unit sphere. It considers to what extent, and in what fashion, touching spheres can be varied, subject to the constraint of always touching the central sphere. Such constrained motion problems are of interest in physics and materials science, and the problem involves topology and geometry. This paper reviews the history of work on this problem, presents some new results, and formulates some conjectures. It also presents general results on configuration spaces of N spheres of radius r touching a central unit sphere, with emphasis on 3 ≤ N ≤ 14. The problem of determining the maximal radius rmax(N) is a version of the Tammes problem, to which L{\\'a}szl{\\'o} Fejes T{\\'o}th made significant contributions.\",\nkeywords = \"Configuration spaces, Constrained optimization, Criticality, Discrete geometry, Materials science, Morse theory\",\nauthor = \"Rob Kusner and W{\\\"o}den Kusner and Lagarias, {Jeffrey C.} and Senya Shlosman\",\nyear = \"2018\",\nmonth = \"1\",\nday = \"1\",\ndoi = \"10.1007/978-3-662-57413-3_10\",\nlanguage = \"English\",\nseries = \"Bolyai Society Mathematical Studies\",\npublisher = \"Springer Berlin - Heidelberg\",\npages = \"219--277\",\nbooktitle = \"Bolyai Society Mathematical Studies\",\n\n}\n\nTY - CHAP\n\nT1 - Configuration spaces of equal spheres touching a given sphere\n\nT2 - The twelve spheres problem\n\nAU - Kusner, Rob\n\nAU - Kusner, Wöden\n\nAU - Lagarias, Jeffrey C.\n\nAU - Shlosman, Senya\n\nPY - 2018/1/1\n\nY1 - 2018/1/1\n\nN2 - The problem of twelve spheres is to understand, as a function of r ϵ (0,rmax(12)], the configuration space of 12 non-overlapping equal spheres of radius r touching a central unit sphere. It considers to what extent, and in what fashion, touching spheres can be varied, subject to the constraint of always touching the central sphere. Such constrained motion problems are of interest in physics and materials science, and the problem involves topology and geometry. This paper reviews the history of work on this problem, presents some new results, and formulates some conjectures. It also presents general results on configuration spaces of N spheres of radius r touching a central unit sphere, with emphasis on 3 ≤ N ≤ 14. The problem of determining the maximal radius rmax(N) is a version of the Tammes problem, to which László Fejes Tóth made significant contributions.\n\nAB - The problem of twelve spheres is to understand, as a function of r ϵ (0,rmax(12)], the configuration space of 12 non-overlapping equal spheres of radius r touching a central unit sphere. It considers to what extent, and in what fashion, touching spheres can be varied, subject to the constraint of always touching the central sphere. Such constrained motion problems are of interest in physics and materials science, and the problem involves topology and geometry. This paper reviews the history of work on this problem, presents some new results, and formulates some conjectures. It also presents general results on configuration spaces of N spheres of radius r touching a central unit sphere, with emphasis on 3 ≤ N ≤ 14. The problem of determining the maximal radius rmax(N) is a version of the Tammes problem, to which László Fejes Tóth made significant contributions.\n\nKW - Configuration spaces\n\nKW - Constrained optimization\n\nKW - Criticality\n\nKW - Discrete geometry\n\nKW - Materials science\n\nKW - Morse theory\n\nUR - http://www.scopus.com/inward/record.url?scp=85056318530&partnerID=8YFLogxK\n\nU2 - 10.1007/978-3-662-57413-3_10\n\nDO - 10.1007/978-3-662-57413-3_10\n\nM3 - Chapter\n\nT3 - Bolyai Society Mathematical Studies\n\nSP - 219\n\nEP - 277\n\nBT - Bolyai Society Mathematical Studies\n\nPB - Springer Berlin - Heidelberg\n\nER -"
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[
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https://www.numerade.com/questions/sketch-the-graph-of-f-by-hand-and-use-your-sketch-to-find-the-absolute-and-local-maximum-and-mini-13/
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[
"💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!\n\n# Sketch the graph of $f$ by hand and use your sketch to find the absolute and local maximum and minimum values of $f$. (Use the graphs and transformations of Section 1.2 and 1.3). $f(x) = \\left\\{ \\begin{array}{ll} x^2 & \\mbox{ if}-1 \\leqslant x \\leqslant 0\\\\ 2 - 3x & \\mbox{ if} 0 < x \\leqslant 1 \\end{array} \\right.$\n\n## No absolute or local maximum.Absolute minimum $f(1)=-1$. Local minimum $f(0)=0$.\n\nDerivatives\n\nDifferentiation\n\nVolume\n\n### Discussion\n\nYou must be signed in to discuss.\n\nLectures\n\nJoin Bootcamp\n\n### Video Transcript\n\nwe're going to sketch the graph of the function defined by pots like this. X square if X is Rather than or equal to 91 and less than or equal to zero and two X minus three to minus three X. If X is greater than zero and less than or equal to one. There is a piecewise function with two parts. One for the interval negative 10 including both negative 10 and the other From 0 to 1 excluding zero. But including one. The first part of Super Public Square. The second person straight line. Too many three eggs. And when we do that graph the hand, we would use it to find the absolute and local maximum and minimum values of the function. So the main of these functions we see here is negative 11. Close interval, we have two parts so we have to draw each part of the function, the first part of the parable like square, but between negative one and Syria including -1 and including Syria. So we have this coordinate system here at negative one, X square is one. So we get this point of the problem is included in the graph and at zero we get x square is equal to zero. So we had these point here, in fact I have forgotten to put here this Y axis. Okay, so we have two points and we know the problem is something like this. Without this seen here. But that's the idea. So that the problem, the other parties a straight line, we can work with some transformation of the identity. But because this is the line we know sufficient to try to points and join them. So in the in the case zero were included in the formula will will have to but it's not included but we know the value should be too. So we have this, we draw this circle Without field to indicate. The point is not included in the graph. And then we have the straight line between that point and another point we can choose is when x equal one we get to minus three times one is two minus three which is negative one. So at one the function has this value -1. And these values included in the graph. So I indicated with his I feel red a circle So we are ready to draw the line. We are going to use this here and this line is more or less this step. Yeah can tune it here now. So we have this point connected with this other point. So we get to rotate a little bit and we are almost there and then we Okay, a little bit mm let's see like this. Okay we are almost here let's see something like that. Mhm Okay here we are. Yeah. So this is the graph of the function by Yeah. This way it functions with the parts two pieces And we can say this this continuous function at zero for example. But let's see the properties about the extreme. 1st. We can see we have an absolute minimum here at this point, Dizzy should always point in the graph is included in the in the graphic as is the image of one Here which uses this formula here and the value because funding body is -1. So we can say that F has an absolute minimum value. The value is -1. And that value of course at x equal *** last one. Now for the local minimum absolute minimum and for the local minimum in fact there is one here. You can say that this value here, zero at zero is a local minimum because if we take any interval or at least this interval I drawn here around zero. All the images are greater than the images zero which is zero, that is all the images that are positive. We can see some of them are on the line and the others on the problem but all those images are greater than or equal to the image of serum. So zero is local minimum. So F has a local and is the only one with his property in the draft so has all local minimum. Mhm. Value zero, add x equals suit. Mhm Okay, respect to the maximum value, there is no absolute maximum because the highest fine should be at this point but it's not included in the graph. So the function is always taking greatest values closer and closer to the 0.0.20 or 02. Sorry, but it's not, it never reaches that values of there is no absolute maximum. Have has no absolute maximum respect to local maximum. Uh There there's there's no local maximum either because we have in all the points, we always have taken any interval around the point. We always have images that are greater or lower than the point. So there's there are no local maximum, so no absolute or local maximum. But yes, so this is the These are properties of dysfunction defined by two pieces.\n\nDerivatives\n\nDifferentiation\n\nVolume\n\nLectures\n\nJoin Bootcamp"
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[
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https://www.media4math.com/library/video-tutorial-equivalent-ratios
|
[
"",
null,
"## Display Title Video Tutorial: Equivalent Ratios\n\nVideo Tutorial: Equivalent Ratios. In this video we look at equivalent ratios and solve problems that involve the use of equivalent ratios.\nCommon Core Standards CCSS.MATH.CONTENT.6.RP.A.2 4.00 minutes 6 - 12 Algebra • Ratios, Proportions, and Percents • Ratios and Rates 2016 ratios, finding equivalent ratios"
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[
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"https://www.media4math.com/sites/default/files/library_asset/images/VideoTutorial--Ratios7Thumbnail.jpg",
null
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https://xertrov.github.io/fi/posts/2020-09-04-reply-to-math-contradiction-in-solomonoff-induction/
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[
"# Reply to 'Mathematical Inconsistency in Solomonoff Induction?'\n\nDate: 2020-09-04\n\nThis is a reply to this LessWrong post.\n\nI went through the maths in OP and it seems to check out. I think the core inconsistency is that SI implies",
null,
". I’m going to redo the maths below (breaking it down step-by-step more). curi has",
null,
"which is the same inconsistency given his substitution. I’m not sure we can make that substitution but I also don’t think we need to.\n\nLet X and Y be independent hypotheses for Solomonoff induction.\n\nAccording to the prior, the non-normalized probability of X (and similarly for Y ) is:",
null,
"(1)\n\nwhat is the probability of",
null,
"?",
null,
"(2)\n\nHowever, by Equation (1) we have:",
null,
"(3)\n\nthus",
null,
"(4)\n\nThis must hold for any and all X and Y .\n\ncuri considers the case where X and Y are the same length, starting with Equation (4)",
null,
"(5)\n\nbut",
null,
"(6)\n\nand",
null,
"(7)\n\nso",
null,
"(8)\n\ncuri has slightly different logic and argues",
null,
"which I think is reasonable. His argument means we get",
null,
". I don’t think those steps are necessary but they are worth mentioning as a difference. I think Equation (8) is enough.\n\nI was curious about what happens when",
null,
". Let’s assume the following:",
null,
"(9)\n\nso, from Equation (2)",
null,
"(10)",
null,
"(11)\n\nbut Equation (9) says",
null,
"– this contradicts Equation (11).",
null,
"So there’s an inconsistency regardless of whether",
null,
"or not."
] |
[
null,
"https://xertrov.github.io/fi/imgs/2020-09-04-reply-to-math-contradiction-in-solomonoff-induction0x.png",
null,
"https://xertrov.github.io/fi/imgs/2020-09-04-reply-to-math-contradiction-in-solomonoff-induction1x.png",
null,
"https://xertrov.github.io/fi/imgs/2020-09-04-reply-to-math-contradiction-in-solomonoff-induction2x.png",
null,
"https://xertrov.github.io/fi/imgs/2020-09-04-reply-to-math-contradiction-in-solomonoff-induction3x.png",
null,
"https://xertrov.github.io/fi/imgs/2020-09-04-reply-to-math-contradiction-in-solomonoff-induction4x.png",
null,
"https://xertrov.github.io/fi/imgs/2020-09-04-reply-to-math-contradiction-in-solomonoff-induction5x.png",
null,
"https://xertrov.github.io/fi/imgs/2020-09-04-reply-to-math-contradiction-in-solomonoff-induction6x.png",
null,
"https://xertrov.github.io/fi/imgs/2020-09-04-reply-to-math-contradiction-in-solomonoff-induction7x.png",
null,
"https://xertrov.github.io/fi/imgs/2020-09-04-reply-to-math-contradiction-in-solomonoff-induction8x.png",
null,
"https://xertrov.github.io/fi/imgs/2020-09-04-reply-to-math-contradiction-in-solomonoff-induction9x.png",
null,
"https://xertrov.github.io/fi/imgs/2020-09-04-reply-to-math-contradiction-in-solomonoff-induction10x.png",
null,
"https://xertrov.github.io/fi/imgs/2020-09-04-reply-to-math-contradiction-in-solomonoff-induction11x.png",
null,
"https://xertrov.github.io/fi/imgs/2020-09-04-reply-to-math-contradiction-in-solomonoff-induction12x.png",
null,
"https://xertrov.github.io/fi/imgs/2020-09-04-reply-to-math-contradiction-in-solomonoff-induction13x.png",
null,
"https://xertrov.github.io/fi/imgs/2020-09-04-reply-to-math-contradiction-in-solomonoff-induction14x.png",
null,
"https://xertrov.github.io/fi/imgs/2020-09-04-reply-to-math-contradiction-in-solomonoff-induction15x.png",
null,
"https://xertrov.github.io/fi/imgs/2020-09-04-reply-to-math-contradiction-in-solomonoff-induction16x.png",
null,
"https://xertrov.github.io/fi/imgs/2020-09-04-reply-to-math-contradiction-in-solomonoff-induction17x.png",
null,
"https://xertrov.github.io/fi/imgs/2020-09-04-reply-to-math-contradiction-in-solomonoff-induction18x.png",
null,
"https://xertrov.github.io/fi/imgs/2020-09-04-reply-to-math-contradiction-in-solomonoff-induction19x.png",
null
] |
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|
https://math.stackexchange.com/questions/2678895/function-which-creates-the-sequence-1-2-3-1-2-3
|
[
"# Function which creates the sequence 1, 2, 3, 1, 2, 3, …\n\nI was wondering how to map the set $\\mathbb{Z}^+$ to the sequence $1, 2, 3, 1, 2, 3, \\ldots$. I thought it would be easy, but I was only able to obtain an answer through trial and error.\n\nFor a function $f \\colon \\mathbb{Z}^+ \\rightarrow \\mathbb{Z}$, we have that\n\n$f(x) = x \\bmod 3$ gives the numbers $1, 2, 0, 1, 2, 0, \\ldots$\n\n$f(x) = (x \\bmod 3) + 1$ gives the numbers $2, 3, 1, 2, 3, 1, \\ldots$\n\nAfter a bit of experimenting, I finally found that\n\n$f(x) = ((x + 2) \\bmod 3) + 1$ gives the numbers $1, 2, 3, 1, 2, 3, \\ldots$\n\nMore generally, if I want to map the set $\\mathbb{Z}^+$ to the sequence $\\{1, 2, \\ldots, n, 1, 2, \\ldots, n, \\ldots\\}$, I need to use the function\n\n$$f(x) = ((x + n - 1) \\bmod n) + 1$$\n\nI was only able to come to this result by trial and error. I was not able to find a solution to this relatively simple question online (although perhaps my search terms were off).\n\nHow would one come to this result in a more systematic way?\n\n• Very good question. In general, a lot of things we can find by trial and error, and after we find it we may discover some simpler way to obtain it, such as dxiv's answer. But I can bet that dxiv first found it by trial and error also, if he/she didn't already see it before... To further hone your trial and error skills, try to find a simple formula for $1,2,3,2,1,2,3,2,...$ and $1,1,2,1,2,3,1,2,3,4,...$. =) – user21820 Mar 6 '18 at 8:17\n• Oh and I should say that if you're interested in a program, the easiest way to repeat any sequence in general is to use a list. In my first exercise, def f(n): return [2,1,2,3][n%4] would finish the job quickly and cleanly. – user21820 Mar 6 '18 at 8:24\n• @WilliamElliot: The notation $f : S \\to T$ does not imply that $f$ must be a surjection from $S$ onto $T$. The question is wrong only in calling the desired output sequence a set. – user21820 Mar 6 '18 at 8:26\n• I just want to note here - since somehow no one has yet - that 'the function which repeats the pattern 1, 2, 3, 1, 2, 3, ...' is a perfectly good definition of the function in its own right. Be careful not to confuse a function with a formula for the function. – Steven Stadnicki Mar 6 '18 at 18:42\n• There's nothing wrong with \"trial and error\". That's the exact method by which a huge amount of mathematics is discovered. – Lee Mosher Mar 7 '18 at 18:30\n\nYou figured out already that $\\,x \\bmod n\\,$ gives the periodic sequence $\\,(1, 2, \\ldots, n-1, \\color{red}{0}, 1, \\ldots)\\,$ then the rest is just \"shifting\" it to fit the desired start/end values. To match $\\,(1, 2, \\ldots, \\color{red}{n}, 1, \\ldots)\\,$ for example, you need to shift $\\,x \\bmod n\\,$ up by $\\color{blue}{1}$, and $\\,x\\,$ down by $\\color{green}{1}$ to compensate for it. So in the end you'd be getting $\\,\\color{blue}{1} + (x-\\color{green}{1}) \\bmod n\\,$, and of course $\\,(x-1) \\bmod n = (x+n-1) \\bmod n\\,$.\n\n• To generalize the observations in this answer: if you want to repeat the first $p$ members of an arithmetic sequence $a_k=m(k-1)+d$, then the function you need is $m((k-1)\\bmod p)+d$ – J. M. isn't a mathematician Mar 6 '18 at 5:54\n• @J.M.isnotamathematician Indeed, worth pointing out. – dxiv Mar 6 '18 at 6:05\n\nThe $k$:th number will be: $$[1,0,0]{\\bf C}^k[1,2,3]^T$$\n\nwhere ${\\bf C}= \\left[\\begin{array}{ccc}0&1&0\\\\0&0&1\\\\1&0&0\\end{array}\\right]$\n\nNow you can exchange the vector to the right for any ordered set of numbers you want to $\\bf C$ycle through.\n\n(It is actually a representation matrix for a generator of the cyclic group $C_3$ if you want to dig into some theory).\n\nNow to make this a function of $k$, (which is actually what is asked), maybe you prefer the notation\n\n$$k\\overset{f(k)}{\\longrightarrow} [1,0,0]{\\bf C}^k[1,2,3]^T$$ or maybe\n\n$$f(k)= [1,0,0]{\\bf C}^k[1,2,3]^T$$\n\n• As it turns out, explicitly evaluating the matrix power and simplifying yields a remarkably simple expression: $$1-\\frac1{\\sqrt{3}}\\sin\\frac{2\\pi k}{3}+2\\sin^2\\frac{\\pi k}{3}$$ – J. M. isn't a mathematician Mar 7 '18 at 13:21\n• I would argue it is not a simpler expression. Trancendental functions and irrational numbers are lots less simple computationally speaking than binary shuffles / pointer arithmetics which you can implement permutation matrices with. – mathreadler Mar 7 '18 at 13:58\n• @J.M.isnotamathematician. FWIW this is the same as $2-\\frac{2}{\\sqrt{3}}\\sin(2k+1)\\frac{\\pi}{3}$ – David Quinn Mar 7 '18 at 14:09\n• This is not an answer to the question. – JiK Mar 8 '18 at 10:44\n• I suspect @JiK means that this does not give what the questioner asked for, namely a systematic approach to arriving at the formula they themself found, i.e. $( ( x - 1 ) \\bmod n ) + 1$. – PJTraill Mar 8 '18 at 21:59\n\n$f(x)=\\left\\lfloor 10^x\\cdot {123\\over999}\\right\\rfloor\\mod 10$.\n\nMore generally, with $b>n>1$, $$f(x)=\\left\\lfloor b^x\\cdot {(123\\ldots n)_b\\over b^n-1}\\right\\rfloor\\mod b.$$\n\n• Ingenious (+1) though not too practical computationally. – dxiv Mar 6 '18 at 6:02\n• This is not an answer to the question. – JiK Mar 8 '18 at 10:44\n• I presume @JiK means that this does not give what the questioner asked for, namely a systematic approach to arriving at the formula they themself found, i.e. $( ( x - 1 ) \\bmod n ) + 1$. – PJTraill Mar 8 '18 at 22:00\n\nAs an alternative, here is a purely trigonometric function which generates the required sequence:\n\n$$f(x)=\\frac{\\sqrt{3}\\sin(\\frac 23\\pi(x-2))}{\\cos(\\frac 23\\pi(x-2))+2}+2$$\n\nHere we have $f(1)=1$, $f(2)=2$, $f(3)=3$, $f(4)=1$, and so on.\n\n• This is not an answer to the question. – JiK Mar 8 '18 at 10:44\n• @JiK no perhaps not, but I thought I'd post it anyway. – David Quinn Mar 8 '18 at 18:58\n• @JiK is right: this does not give what the questioner asked for, namely a systematic approach to arriving at the formula they themself found, i.e. $( ( x - 1 ) \\bmod n ) + 1$. – PJTraill Mar 8 '18 at 22:00\n• @PJTraill So what? The OP is not the only one who reads these Q/As, and this and other answers giving alternate expressions (and systematic methods for finding them) are by no means off topic. If you dispute this, you should take it up on meta instead of spamming everyone's answers. – Mario Carneiro Mar 9 '18 at 3:24\n\nLet's assume we want to find a formula for the sequence \\begin{align*} \\color{blue}{3,1,\\frac{1}{3},\\frac{1}{3},1,3,3,1,\\frac{1}{3},\\frac{1}{3},1,3,3,1,\\frac{1}{3},\\frac{1}{3},1,3,\\ldots} \\end{align*} repeating in cycles of $6$ elements.\n\nThe $r$-th roots of unity $\\omega_j=\\exp\\left(\\frac{2\\pi ij}{r}\\right), 0\\leq j<r$ have the nice property to filter elements. For $r>0$ we obtain \\begin{align*} \\frac{1}{r}\\sum_{j=0}^{r-1}\\exp\\left(\\frac{2\\pi ij n}{r}\\right)= \\begin{cases} 1&\\qquad r\\mid n\\\\ 0& \\qquad otherwise \\end{cases} \\end{align*} Therefore the sequence \\begin{align*} \\left(\\frac{1}{6}\\sum_{j=0}^5\\exp\\left(\\frac{2\\pi ijn}{6}\\right)\\right)_{n=0}^{\\infty} =(\\color{blue}{1},0,0,0,0,0,\\color{blue}{1},0,0,0,0,0,\\color{blue}{1},\\ldots)\\tag{1} \\end{align*} generates a $1$ at each $6$-th position and $0$ otherwise and multiplication with $3$ gives a sequence with $3$ at each $6$-th position and $0$ otherwise.\n\nAppropriately shifting of (1) can be used to generate a $1$ on each $6$-th position shifted by $k$ positions with $0\\leq k < 6$. Multiplication with $3,1$ and $\\frac{1}{3}$ and adding all these terms provides the wanted sequence.\n\nHint: Some instructive examples can be found in H.S. Wilf's book generatingfunctionology, (2.4.5) to (2.4.9).\n\n• For that portion of the audience allergic to complex numbers .... these can be done with trigonometric functions instead of complex exponentials. – GEdgar Mar 6 '18 at 12:10\n• @GEdgar I guess it comes down to whether people are more allergic to complex numbers or otherwise hard-to-remember compound angle formulae. – J.G. Mar 6 '18 at 15:50\n• This is not an answer to the question. – JiK Mar 8 '18 at 10:45\n• @JiK: This answer introduces a systematic way to build sequences $(a_1,a_2,a_3,...,a_n,a_1,a_2,...)$ consisting of repeated sub-sequences. All the more it provides a systematic way to build sequences $(1,2,3,...,n,1,2,...) which was OPs question. – Markus Scheuer Mar 8 '18 at 11:04 • @MarkusScheuer The question is \"How would one come to this result in a more systematic way?\" (emphasis mine) – JiK Mar 8 '18 at 11:18 I'm surprised no one has mentioned composition. You've identified that you need to shift the range $$f_1(x) = x - 1$$ then take the value modulo$n$$$f_2(x) = x \\mod n$$ and finally shift the result again $$f_3(x) = x + 1$$ The function you want is just the composition$f = f_3 \\circ f_2\\circ f_1$. • And it looks like a conjugation if you write it$f = f_1^{-1} \\circ f_2 \\circ f_1$. – Jeppe Stig Nielsen Mar 6 '18 at 23:19 • I like this answer best because it actually gives a systematic way to convert the function$x \\bmod 3$into the function that was desired. – David K Mar 7 '18 at 2:01 Let be$\\omega\\in\\Bbb C$,$\\omega^3 = 1$,$\\omega\\ne 1$.$\\omega^n$gives: $$\\omega^0 = 1,$$ $$\\omega^1 = \\omega,$$ $$\\omega^2 = \\omega^{-1},$$ $$\\omega^3 = 1,$$ $$\\cdots$$ \"But I want$1,2,3,\\dots$!\" Solution: compose with a polynomial$P$s.t. $$P(1) = 1,\\qquad P(\\omega) = 2,\\qquad P(\\omega^{-1}) = 3.$$ Now, $$P(\\omega^0) = 1,$$ $$P(\\omega^1) = 2,$$ $$P(\\omega^2) = 3,$$ $$P(\\omega^3) = 1,$$ $$\\cdots$$ This trick will work for any periodic sequence. • This does not give what the questioner asked for, namely a systematic approach to arriving at the formula they themself found, i.e.$ ( ( x - 1 ) \\bmod n ) + 1 $. – PJTraill Mar 8 '18 at 22:02 The function $$f(x)=\\left\\{ \\begin{array}{ll} x, & \\text{if } x \\leq n\\\\ f(x-n), & \\text{otherwise} \\end{array}\\right.$$ maps the set$\\mathbb{Z}^+$to the sequence$\\{1, 2, \\ldots, n, 1, 2, \\ldots, n, \\ldots\\}$. Is this way \"more systematic\" to you? • I created a answer very much like yours but yours is both more general and correct. – Q the Platypus Mar 7 '18 at 3:55 • This does not give what the questioner asked for, namely a systematic approach to arriving at the formula they themself found, i.e.$ ( ( x - 1 ) \\bmod n ) + 1 $. – PJTraill Mar 8 '18 at 22:03 • @PJTraill You're absolutely right. I apparently didn't read the question carefully enough. – HelloGoodbye Mar 9 '18 at 10:57 Well, there are many answers to this actually, for instance $$f(n) = n + 1 - 3\\lfloor n/3\\rfloor$$ (assuming$\\mathbb Z^+$starts with$0$). How to come to this? The basic idea is that$s(x)=x-\\lfloor x\\rfloor$is the sawtooth function with range$[0,1)$. Then$ns(x/n)$is the same function scaled$n$times in both directions, i.e., it's the sawtooth function with range$[0,n)$. However, you need range$[1,n+1)$, so that's the “$+1$” part. • This is basically the definition of mod function.$a \\mod b = a-b\\lfloor{a\\over b}\\rfloor$:) except it's shifted by one horizontally and then verticall – KKZiomek Mar 8 '18 at 4:29 • @KKZiomek I don't say it is not :-) And I believe that this view is self-explanatory, that's why I consider it good for educational purposes. – yo' Mar 8 '18 at 8:49 • This does not give (or only indirectly gives) what the questioner asked for, namely a systematic approach to arriving at the formula they themself found, i.e.$ ( ( x - 1 ) \\bmod n ) + 1 $. – PJTraill Mar 8 '18 at 22:02 • @PJTraill Well, that's true, but I still think that the way I show is a related point that could be useful in grasping the whole topic. – yo' Mar 8 '18 at 22:16 f(x) = ((x−1) mod n) + 1 Step 1: You want a function with a period of n, i.e. it repeats every n numbers. Then necessarily you are modding something by n. g(x) = h(x) mod n Step 2: You want the function to reset after every nth number. But the nature of modular division is that it resets on every nth number, not after every nth number. So you necessarily have to subtract 1 before performing modular division. h(x) = x - 1. g(h(x)) = (x - 1) mod n. This is a horizontal shift, akin to an x-intercept. Step 3: You want the function to vary from 1 to n, but the nature of modular division is to vary from 0 to n-1. So you have to add 1 to each term after the modular division. i(x) = g(h(x)) + 1 = ((x - 1) mod n) + 1. This is a vertical shift, akin to a y-intercept. Unnecessary step 4: If you wanted each term to be a multiple of 1 through n, i.e. (3, 6, 9, ... 3n), you would perform multiplication last. I.e. j(x) = 3 * i(g(h(x)) = 3 * (((x - 1) mod n) + 1). This is an amplitude shift, akin to a slope. • ▲ A useful description of the effects of the various steps involved in reaching the answer. – PJTraill Mar 8 '18 at 21:55 In group theory, the modulus function is often taken to be a function from Z to Z/nZ. That is, the output is not an integer, but the equivalence class of integers that have the same remainder when divided by n. E.g. 1->{...-5,-2,1,4,7...}. In computer science, on the other hand, modulus functions generally give a particular representative of the equivalence class, more specifically whatever representative lies in [0,n-1]. You're trying to instead get the representative that lies in [1,n]. Note that this differs only in how 0+nZ is treated; the standard function sends it to 0, but you want it sent to n. So if you were writing a program, one option would be to simply do output = (k mod n) then if output==0: output=n. However, that's somewhat of an inelegant solution. A solution that is in some sense more elegant is to realize that you want your output range shifted up one. If you do output = (k mod n)+1, then instead of your output ranging from 0 to n-1, your output will be 1 to n. However, this shifts not only the output range, but also what each input goes to; 1 goes to 2 instead of 1, 2 goes to 3, etc. If you subtract 1 before taking the modulus, then this will cancel out with the adding 1 afterwards. The exception to this will be multiples of n; by subtracting 1, you get to n-1, then when you add 1 afterwards, you get to n, instead of 0. Your formula adds n-1 instead of subtracting 1, but since (n-1) mod n = -1, this amounts to the same thing. One way of visualizing this is imagining n tick marks in a circle. There's a tick mark labeled 1, a tick mark labeled 2, etc., up to n-1. But when you get to n, the tick mark is labeled 0, and then at n+1 you're back at the tick mark labeled 1. What if you want your output for n to be n, instead of 0? Well, you can think of there being a \"drop\" between the tick mark labeled n-1 and the one labeled 0; between those two tick marks, the output drops from n-1 to 0. You want to move n over to before this drop, and make the drop to instead be between n and n+1. So to move n to before the drop, you subtract 1; this shifts n over to before the drop. But now you need to add 1 back. The addition and subtraction cancel out in the sense that you end up in the same equivalence class (remember, the equivalence class for both 0 and n is {-2n,-n,0,n,2n...}), but by doing the subtraction before doing the modulus, and the addition after, you end up with a different representative. More generally, if you want to shift j numbers from the beginning to the end, you can do (k-j) mod n + j. For instance, k mod 12 would take k to between 0 and 11, but (k-3) mod 12 + 3 will take k to between 3 and 14. To go the other way, you reverse the addition and subtraction; (k+3) mod 12 - 3 will take k to between -3 and 8. Maybe those with allergies to both complex numbers and trigonometry are more tolerant to homogeneous coordinate transforms? $$f(k) = [1 \\ 0\\ 2] \\begin{bmatrix} -\\frac12 & -\\frac{\\sqrt3}2 & 0 \\\\ \\frac{\\sqrt3}2 &-\\frac12 & 0 \\\\ 0 & 0 & 1 \\end{bmatrix}^k \\begin{bmatrix} 1 \\\\ \\frac{\\sqrt3}3 \\\\ 1 \\end{bmatrix}$$ • This does not give what the questioner asked for, namely a systematic approach to arriving at the formula they themself found, i.e.$ ( ( x - 1 ) \\bmod n ) + 1 $. – PJTraill Mar 8 '18 at 22:03 Your only problem is that you're counting up from one. If you started counting from zero, then $$f(x) = x \\bmod 3$$ would do the job, generating the output sequence$0,1,2,0,1,2,0,1,2,\\dots$when applied to the input sequence$0,1,2,3,4,5,6,7,8,\\dots$. But if you insist on one-based counting, then you simply need to shift the input down by one, and the output correspondingly up by one, as in $$f(x) = ((x-1) \\bmod 3) + 1.$$ This will generate the output sequence$1,2,3,1,2,3,1,2,3,\\dots$when applied to the input sequence$1,2,3,4,5,6,7,8,9,\\dots\\$."
] |
[
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http://snap.stanford.edu/snapvx/
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[
"",
null,
"# SnapVX: Network-Based Optimization Solver\n\nSnapVX is a python-based convex optimization solver for problems defined on graphs. For problems of this form, SnapVX provides a fast and scalable solution with guaranteed global convergence. It combines the graph capabilities of Snap.py with the convex solver from CVXPY, and is released under the BSD Open-Source license.\n\nSnapVX is a convex optimization solver for problems which are defined on a graph. Given a vertex set $\\mathcal{V}$ and edge set $\\mathcal{E}$, we solve the following optimization problem:\n\n$$\\mbox{minimize} \\sum\\limits_{i \\in \\mathcal{V}} f_i(x_i) + \\sum\\limits_{(j,k) \\in \\mathcal{E}} g_{jk}(x_j, x_k)$$\n\nThe variables are $x_1,x_2,\\ldots,x_m$. Here, $x_i \\in \\mathbb{R}^p$ is the variable and $f_i$ the cost function at node $i$; $g_{jk}$ is the cost function associated with undirected edge $(j,k)$. When the cost functions are convex, SnapVX provides:\n\n• A solution based on the alternating direction method of multipliers (ADMM) that is guaranteed to converge to the global optimum for any problem of this form.\n• A fast, scalable algorithm which is capable of being distributed across multiple cores of a single machine.\n• An integration of Snap.py and CVXPY.\n\nPrerequisites:\nSnap.py\nCVXPY\n\nInstructions:\nSee the README for instructions for how to install and test SnapVX\n\n## Online Documentation and Resources\n\n### Github Repository\n\nThe following development repositories are available for SnapVX\n\n### Other Resources\n\n• Presentation on Applying SnapVX to Real-World Problems\n• CVXPY, the general convex optimization solver used to solve the ADMM subproblems, can be found here. For more information, please see the documentation on the website or contact Steven Diamond with any questions.\n• To define and store the networks, SnapVX uses Snap.py. The Snap.py documentation provides detailed information on how the underlying framework works.\n\n## Motivation\n\nConvex optimization has become an increasingly popular way of modeling problems in many different fields. However, as datasets get larger and more intricate, classical methods of convex analysis, which often rely on interior point methods, begin to fail due to a lack of scalability. The challenge of large-scale optimization lies in developing methods general enough to work well independent of the input, while being capable of scaling to the immense datasets that today’s applications require. Therefore, it is necessary to formulate general classes of convex optimization solvers, ones that can apply to a variety of interesting and relevant problems, and to develop algorithms for reliable and efficient solutions.\n\nBy focusing on examples of the form $\\mbox{minimize} \\sum\\limits_{i \\in \\mathcal{V}} f_i(x_i) + \\sum\\limits_{(j,k) \\in \\mathcal{E}} g_{jk}(x_j, x_k)$, we can solve a variety of common problems in optimization. Putting a large class of problems under a unified framework has many benefits. This is intended to be a general-use solver, taking advantage of ADMM to solve problems too large to be computed using a centralized approach. We see several examples that fit into this form in the applications section below.\n\nWe elect to build a solver that runs on a single large memory machine rather than in a distributed computing environment. This is for several reasons. First, it provides simplicity and \"out-of-the-box\" functionality, including being able to run on standard laptops for smaller problems. Second, large memory machines are increasingly becoming a viable and affordable alternative when performing large-scale data processing. Most real-world problems can fit comfortably in a single \"big-memory\" server. Even if the raw data does not fit, data cleaning and manipulation often results in a significant reduction in space that allows the relevant data to fit into memory. Finally, our ADMM-based solution is a message-passing algorithm which relies on many small messages being sent very quickly, so it is not particularly well-suited for large and distributed computing clusters (even though it is fully possible to implement the algorithm in a distributed environment).\n\n## Syntax\n\nWe show three basic examples to demonstrate how to set up and solve a SnapVX problem. Please also see the Snap.py and CVXPY reference pages, since SnapVX relies on their syntax where necessary. The first sets up a basic problem with only 2 nodes. The second builds a random graph with random objectives, and the final example demonstrates how to bulk load data from a CSV file.\n\nExample 1. We first look at a simple example with two nodes and one edge between them. The two nodes have different objectives, both in $\\mathbb{R}^1$, and the edge objective penalizes the square of the difference of the variables.\n\n\nimport snapvx\nimport cvxpy # Note: it is not necessary to import cvxpy, as all cvxpy functionality\n# is already integrated into SnapVX. It is shown here for verbosity\n\n# Create new graph\ngvx = snapvx.TGraphVX()\n\n# Use CVXPY syntax to define a problem\nx1 = cvxpy.Variable(1, name='x1')\nobj = cvxpy.square(x1)\n# Add Node 1 with the given objective, with the constraint that x1 <= 10\n\n# Similarly, add Node 2 with objective |x2 + 3|\nx2 = cvxpy.Variable(1, name='x2')\nobj2 = cvxpy.abs(x2 + 3)\n\n# Add an edge between the two nodes,\n# Define an objective, constraints using CVXPY syntax\ngvx.AddEdge(1, 2, Objective=cvxpy.square(cvxpy.norm(x1 - x2)), Constraints=[])\n\ngvx.Solve() # Solve the problem\ngvx.PrintSolution() # Print entire solution on a node-by-node basis\nprint \"x1 = \", x1.value, \"; x2 = \", x2.value # Print the solutions of individual variables\n\n\n\nExample 2: Random Graph. Next, we look at a slightly larger example. We build a random graph (using GenRndGnm), set each node objective to the square loss from a random point in $\\mathbb{R}^{10}$, and use laplacian regularization on the edges.\n\n\nfrom snapvx import *\nimport numpy\n\n# Helper function to define edge objectives\n# Takes in two nodes, returns a problem defined with the variables from those nodes\ndef laplace_reg(src, dst, data):\nreturn (sum_squares(src['x'] - dst['x']), [])\n\n# Generate random graph, using SNAP syntax\nnumpy.random.seed(1)\nnum_nodes = 10\nnum_edges = 30\nn = 10\nsnapGraph = GenRndGnm(PUNGraph, num_nodes, num_edges)\ngvx = TGraphVX(snapGraph)\n\n# For each node, add an objective (using random data)\nfor i in range(num_nodes):\nx = Variable(n,name='x') # Each node has its own variable named 'x'\na = numpy.random.randn(n)\ngvx.SetNodeObjective(i, square(norm(x-a)))\n\n# Set all edge objectives at once (Laplacian Regularization)\n\n# Solve in verbose mode (using ADMM)\ngvx.Solve(Verbose=True)\ngvx.PrintSolution()\n\ngvx.PrintSolution()\n\n# Print only the solution for 'x' at Node 1\nprint \"Solution at Node 1 is\", gvx.GetNodeValue(1,'x')\n\n\n\nExample 3: Bulk Loading. Finally, we look at how we can bulk load the relevant data from a text file\n\n\nfrom snapvx import *\n\n# Helper function for node objective\n# Takes in a row from the CSV file, returns an optimization problem\ndef node_obj(data):\nx = Variable(1,name='x')\nreturn norm(x - float(data))\n\n# Helper function for edge objective\ndef laplace_reg(src, dst, data):\nreturn sum_squares(src['x'] - dst['x'])\n\n# Load in Edge List to build graph with default node/edge objectives\n\n# Takes one row of the CSV, uses that as input to node_obj\n# There is also an (optional) input of specifying which nodes each row of the CSV refers to\n\n# Bulk load edge objectives for all edges\n\ngvx.Solve()\ngvx.PrintSolution()\n\n\n\n## Potential Applications\n\nConsensus - The consensus problem is often written as\n\n$$\\mbox{minimize} \\sum\\limits_{i} f_i(x)$$\n\nThis can be formulated in SnapVX as a problem on a network, where each node i has objective function $f_i(x_i)$, but we are constrained to solving for a common value of x as the solution at every node. We solve it by building any connected graph (so there is at least one path from every node to all other nodes), setting each node's local optimization variable function as $f_i(x_i)$. We then set the edge objectives to $\\mathbf{1}(x_j = x_k)$. And so, the problem becomes\n\n$$\\mbox{minimize} \\sum\\limits_{i \\in \\mathcal{V}} f_i(x_i) + \\sum\\limits_{(j,k) \\in \\mathcal{E}} \\mathbf{1}(x_j = x_k)$$\n\nRegularization - Many types of regularization are natural extensions of the original SnapVX formula. For Laplacian Regularization, set the edge objective equal to $\\| x_j - x_k\\|_2^2$. The network lasso (see References) is very similar, except the edge objective is instead $\\| x_j - x_k\\|_2$.\n\nExchange - The exchange problem is defined as \\begin{align} &\\mbox{minimize} &\\sum\\limits_{i = 1}^N f_i(x_i) \\\\ &\\mbox{subject to} &\\sum\\limits_{i =1}^N x_i = 0. \\end{align} To solve this, connect all $N$ nodes (each with objective $f_i(x_i)$) to a single dummy node. This dummy node has variables $z_1, \\ldots, z_N$. Each edge is a consensus constraint that $x_i = z_i$, and the dummy node's objective is the indicator function that $\\sum\\limits_{i =1}^N z_i = 0$.\n\nControl Systems - A basic control system is defined by the update formula $$x^{t+1} = A x^t + B u^t,$$ where $x$ is the state, $u$ is the input (which we can choose), and $A, B$ are constant matrices. Often, we will want to reach some state $x^\\star$ by step $N$. We can model this as a linear network, where each of the $N$ steps is its own node (node $i$ is connected to $i+1$ and $i-1$). The optimization parameter at each node is $z_i = (x_i, u_i)$. Each edge represents a constraint that $x^{t+1} = A x^t + B u^t$. The first node has a constraint that $x_0 = x_{\\mathrm{init}}$, the initial state, and the last node, node $N$, is constrained by $x_N = x^\\star$. The overall problem is described in the picture below. From here, it is easy to add other objectives and constraints to represent other goals, such as minimizing total \"fuel usage\" ($\\|u\\|$) or not deviating too far from prescribed bounds on the path.",
null,
"Fixed Routing - Consider a set of paths on a graph (for example, traffic routes in a communication network). Each route wants to maximize its throughput while traversing through a series of nodes and edges, but there are constraints on the total allowable flow at each node (i.e. the router can only process a certain number of packets per second). This is often written as \\begin{align} &\\mbox{maximize} &\\sum\\limits_{i = 1}^N u_i(f_i) \\\\ &\\mbox{subject to} &R f \\leq c, \\end{align} where the $u$'s are the utility functions, $R$ is the routing matrix (a boolean matrix denoting which flows go through which nodes), and $c$ is the constraint vector. Rather than building the entire flow network, we use a bipartite graph instead. One side of this graph contains each of the original nodes, and the other side contains each of the routes. There is an edge between node $i$ and route $j$ if $j$ goes through $i$ (or $R_{ij} = 1$). The edges represent consensus constraints between the two sides of the bipartite graph. The node objective at the node representing route $i$ is to maximize $u_i(f_i)$. The other side of the bipartite graph, representing the original nodes in the routing network, just have constraints that the sum of flows through it is less than $c_i$.",
null,
"PageRank - The PageRank problem can be defined as a convex optimization problem split up across a graph (for details: http://jmlr.org/proceedings/papers/v32/gleich14.pdf). A working implementation is included in the Examples folder of the SnapVX download.\n\n...and many more!\n\n## Contributors\n\nThe following people contributed to SnapVX (in alphabetical order):\n\nStephen Boyd\nSteven Diamond\nDavid Hallac\nJure Leskovec\nYoungsuk Park\nAbhijit Sharang\nRok Sosic\nChristopher Wong\n\n### Getting Involved\n\nYou are encouraged to try SnapVX for your data science and convex optimization needs. If you wish to contribute to the project, see here for a list of open problems and potential new features. If you find any bugs, please report them here. Additionally, to give us any other feedback, feel free to email David Hallac to get in touch!\n\n## References\n\nSnapVX: A Network-Based Convex Optimziation Solver. D. Hallac, C. Wong, S. Diamond, A. Sharang, R. Sosic, S. Boyd, J. Leskovec. Journal of Machine Learning Research (JMLR) , 2017.\n\nRelated Work\nNetwork Lasso: Clustering and Optimization in Large Graphs. D. Hallac, J. Leskovec, S. Boyd. ACM SIGKDD International Conference on Knowledge Discovery and Data Mining (KDD) , 2015.\nCode from Network Lasso paper available on Github."
] |
[
null,
"http://snap.stanford.edu/images/snap_logo.png",
null,
"http://snap.stanford.edu/snapvx/ControlSystem.png",
null,
"http://snap.stanford.edu/snapvx/FixedRouting.png",
null
] |
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https://www.nntp.perl.org/group/perl.perl5.porters/2017/10/msg246844.html
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[
"",
null,
"Front page | perl.perl5.porters | Postings from October 2017\n\nannouncing a new op, OP_MULTICONCAT\n\nFrom:\nDave Mitchell\nDate:\nOctober 25, 2017 12:15\nSubject:\nannouncing a new op, OP_MULTICONCAT\nMessage ID:\[email protected]\nTL;DR: the branch I've just pushed for smoking,\n\nsmoke-me/davem/mconcat\n\nallows multiple OP_CONCAT, OP_CONST ops, plus optionally an OP_SASSIGN\nor OP_STRINGIFY, to be combined into a single OP_MULTICONCAT op, which can\nmake things a *lot* faster: 4x or more. I'll merge it into blead in a few\ndays, barring any issues.\n\nThis work has taken up most of my time for the last 3 months; its been\nkindly funded by Booking.com.\n\nIn more detail: it will optimise into a single OP_MULTICONCAT, most\nexpressions of the form\n\nLHS RHS\n\nwhere LHS is one of\n\n(empty)\nmy \\$lexical =\n\\$lexical =\n\\$lexical .=\nexpression =\nexpression .=\n\nand RHS is one of\n\n(A . B . C . ...) where A,B,C etc are expressions and/or\nstring constants\n\n\"aAbBc...\" where a,A,b,B etc are expressions and/or\nstring constants\n\nsprintf \"..%s..%s..\", A,B,.. where the format is a constant string\ncontaining only '%s' and '%%' elements,\nand A,B, etc are scalar expressions (so\nonly a fixed, compile-time-known number of\nargs: no arrays or list context function\ncalls etc)\n\nIt doesn't optimise other forms, such as\n\n(\\$a . \\$b) . (\\$c. \\$d)\n\n(((\\$a .= \\$b) .= \\$c) .= \\$d);\n\n(although sub-parts of those expressions might be converted to an\nOP_MULTICONCAT). This is partly because it would be hard to maintain the\ncorrect ordering of tie or overload calls.\n\nThe compiler uses heuristics to determine when to convert: in general,\nexpressions involving a single OP_CONCAT aren't converted, unless some\nother saving can be made, for example if an OP_CONST can be eliminated, or\nin the presence of 'my \\$x = .. ' which OP_MULTICONCAT can apply\nOPpTARGET_MY to, but OP_CONST can't.\n\nThe multiconcat op is of type UNOP_AUX, with the op_aux structure directly\nholding a pointer to a single constant char* string plus a list of segment\nlengths. So for\n\n\"a=\\$a b=\\$b\\n\";\n\nthe constant string is \"a= b=\\n\", and the segment lengths are (2,3,1).\nIf the constant string has different non-utf8 and utf8 representations\n(such as \"\\x80\") then both variants are pre-computed and stored in the aux\nstruct, along with two sets of segment lengths.\n\nFor all the above LHS types, any SASSIGN op is optimised away. For a LHS\nof '\\$lex=', '\\$lex.=' or 'my \\$lex=', the PADSV is optimised away too.\n\nFor example where \\$a and \\$b are lexical vars, this statement:\n\nmy \\$c = \"a=\\$a, b=\\$b\\n\";\n\nformerly compiled to\n\nconst[PV \"a=\"] s\nconcat[t4] sK/2\nconst[PV \", b=\"] s\nconcat[t5] sKS/2\nconcat[t6] sKS/2\nconst[PV \"\\n\"] s\nconcat[t7] sKS/2\nsassign vKS/2\n\nand now compiles to:\n\nmulticoncat(\"a=, b=\\n\",2,4,1)[\\$c:2,3] vK/LVINTRO,TARGMY,STRINGIFY\n\nIn terms of how much faster it is, this code:\n\nmy \\$a = \"the quick brown fox jumps over the lazy dog\";\nmy \\$b = \"to be, or not to be; sorry, what was the question again?\";\n\nfor my \\$i (1..10_000_000) {\nmy \\$c = \"a=\\$a, b=\\$b\\n\";\n}\n\nruns 2.7 times faster, and if you throw utf8 mixtures in it gets even\nbetter. This loop runs 4 times faster:\n\nmy \\$s;\nmy \\$a = \"ab\\x{100}cde\";\nmy \\$b = \"fghij\";\nmy \\$c = \"\\x{101}klmn\";\n\nfor my \\$i (1..10_000_000) {\n\\$s = \"\\x{100}wxyz\";\n\\$s .= \"foo=\\$a bar=\\$b baz=\\$c\";\n}\n\nThe main ways in which OP_MULTICONCAT gains its speed are:\n\n* any OP_CONSTs are eliminated, and the constant bits (already in the\nright encoding) are copied directly from the constant string attached to\nthe op's aux structure.\n\n* It optimises away any SASSIGN op, and possibly a PADSV op on the LHS, in\nall cases; OP_CONCAT only did this in very limited circumstances.\n\n* Because it has a holistic view of the entire concatenation expression,\nit can do the whole thing in one efficient go, rather than creating and\ncopying intermediate results. pp_multiconcat() goes to considerable\nefforts to avoid inefficiencies. For example it will only SvGROW() the\ntarget once, and to the exact size needed, no matter what mix of utf8\nand non-utf8 appear on the LHS and RHS. It never allocates any\ntemporary SVs except possibly in the case of tie or overloading.\n\n* It does all its own appending and utf8 handling rather than calling\nout to functions like sv_catsv().\n\n* It's very good at handling the LHS appearing on the RHS; for example in\n\n\\$x = \"abcd\";\n\\$x = \"-\\$x-\\$x-\";\n\nIt will do roughly the equivalent of the following (where targ is \\$x);\n\nSvPV_force(targ);\nSvGROW(targ, 11);\np = SvPVX(targ);\nMove(p, p+1, 4, char);\nCopy(\"-\", p, 1, char);\nCopy(\"-\", p+5, 1, char);\nCopy(p+1, p+6, 4, char);\nCopy(\"-\", p+10, 1, char);\nSvCUR(targ) = 11;\np = '\\0';\n\nFormerly, pp_concat would have used multiple PADTMPs or temporary SVs to\nhandle situations like that.\n\nThe code is quite big; both S_maybe_multiconcat() and pp_multiconcat()\n(the main compile-time and runtime parts of the implementation) are over\n700 lines each. It turns out that when you combine multiple ops, the\nnumber of edge cases grows exponentially ;-)\n\nHere are some benchmarks, sorted by improving conditional branch count.\n\nThe numbers represent relative counts per loop iteration, compared to\nblead at 100.0%. Higher is better: for example, using half as many\ninstructions gives 200%, while using twice as many gives 50%.\n\nIr Dr Dw COND IND\n------ ------ ------ ------ ------\n88.52 121.28 104.76 104.00 140.00 \\$lex1 .= \\$lex2\n96.19 105.26 101.16 102.35 115.38 ((\\$lex1 .= \\$lex2) .= \\$lex3) .= \\$lex4\n100.00 100.00 100.00 100.00 100.00 \\$pkg .= \\$lex1\n100.00 100.00 100.00 100.00 100.00 my \\$x = \"abc\"\n100.00 100.00 100.00 100.00 100.00 \\$lex = \\$lex1 . \\$lex2\n100.00 100.00 100.00 100.00 100.00 \\$lex1 . \\$lex2\n100.00 100.00 100.00 100.00 100.00 \\$lex1 = \\$lex1 . \\$lex1\n100.00 100.00 100.00 100.00 100.00 \\$lex = join \"\", \\$lex1, \\$lex2\n100.00 100.00 100.00 100.00 100.00 \\$lex1 = \\$lex1 . \\$lex2\n101.45 128.57 111.90 107.69 175.00 \\$lex . \"foo\"\n101.45 128.57 111.90 107.69 175.00 \"foo\" . \\$lex\n103.68 128.57 111.90 108.11 175.00 \\$lex = \"const\\$lex1\"\n105.63 136.84 110.53 108.70 140.00 \\$pkg .= \"foo\"\n128.33 176.67 161.54 123.81 175.00 \\$lex .= \"foo\"\n128.65 145.74 131.91 155.10 155.56 \\$pkg = \\$lex1 . \\$pkg\n142.32 183.75 168.42 165.00 162.50 \\$pkg .= \\$lex1 . \\$lex2\n143.78 159.50 143.08 176.36 216.67 my \\$lex = \\$lex1 . \\$lex2\n145.95 159.22 141.82 182.61 200.00 \\$pkg = \\$lex1 . \\$lex2\n149.90 199.21 191.23 148.72 188.89 \\$s .= \\$a.\\$b.\\$c where all RH args are utf8\n157.46 208.45 200.00 178.38 185.71 \\$lex1 .= \\$lex2 . \\$lex3\n162.47 225.24 214.89 168.42 188.89 \\$s .= \\$a.\\$b.\\$c where all args are utf8\n163.09 198.39 180.95 182.81 228.57 my \\$s = \\$a.\\$b.\\$c where all args are utf8\n164.27 185.83 168.18 198.28 200.00 \\$pkg = \\$lex1 . \\$lex2 . \\$lex3\n164.55 177.92 167.57 187.80 200.00 \\$pkg = \\$pkg . \\$pkg\n164.79 173.42 177.14 200.00 200.00 \\$pkg = \\$pkg . \\$lex2\n170.03 185.06 162.50 210.00 240.00 \\$pkg = \"const\\$lex1\"\n175.12 202.56 185.00 212.73 228.57 \\$lex = \\$lex1 . \\$lex2 . \\$lex3\n175.35 231.58 225.58 197.96 188.89 \\$lex1 .= \\$lex2 . \\$lex3 . \\$lex4\n175.81 214.29 207.89 173.25 200.00 sprintf \"%s\", \\$lex1\n184.59 216.55 201.32 222.39 187.50 my \\$lex = \\$lex1 . \\$lex2 . \\$lex3\n194.78 232.12 218.29 229.57 218.18 \\$lex = \\$lex1 . \\$lex2 . \\$lex3 . \\$lex4 . \\$lex5\n195.58 227.62 206.90 244.90 220.00 my \\$lex = \"const\\$lex1\"\n214.41 244.90 229.63 248.89 300.00 my \\$lex = sprintf \"%s\", \\$lex1\n224.89 261.14 209.52 255.95 277.78 my \\$lex = \"foooooooooo=\\$lex1 baaaaaaaaar=\\$lex2\\n\" where lex1/2 are 400 chars\n230.62 275.76 236.36 227.78 300.00 \\$lex = \"foooooooooo=\\$lex1 baaaaaaaaar=\\$lex2\\n\" where lex1/2 are 400 chars\n231.97 302.65 272.41 238.33 275.00 \\$lex = \"foo=\\$lex1 bar=\\$lex2\\n\"\n233.23 244.21 289.80 259.57 220.00 sprintf \"%s%s\", \\$lex1, \\$lex2\n238.61 261.90 273.17 288.89 187.50 \\$lex1 = \\$lex2 . \\$lex1\n239.64 272.57 234.48 253.33 288.89 \\$pkg = \"foooooooooo=\\$lex1 baaaaaaaaar=\\$lex2\\n\" where lex1/2 are 400 chars\n240.98 263.75 247.73 272.22 275.00 \\$pkg = sprintf \"%s\", \\$lex1\n243.99 295.93 265.62 271.43 266.67 \\$pkg = \"foo=\\$lex1 bar=\\$lex2\\n\"\n250.85 313.48 286.49 283.33 255.56 my \\$lex = \"foo=\\$lex1 bar=\\$lex2\\n\"\n255.30 345.54 329.17 274.55 227.27 \\$pkg .= \"-foo-\\$lex1-foo-\\$lex2-foo-\"\n260.09 340.35 351.85 263.33 200.00 \\$pkg .= sprintf \"%s\", \\$lex1\n267.63 302.86 286.84 300.00 366.67 \\$lex = sprintf \"%s\", \\$lex1\n270.29 272.00 315.62 317.91 211.11 \\$pkg = sprintf \"foo=%s bar=%s\\n\", \\$lex1, \\$lex2\n272.19 273.86 293.75 321.43 260.00 \\$pkg = sprintf \"foo=%s\", \\$lex1\n272.73 292.31 329.73 325.00 200.00 my \\$lex = sprintf \"foo=%s bar=%s\\n\", \\$lex1, \\$lex2\n272.90 356.57 337.21 278.26 271.43 \\$lex = \"foo1=\\$lex1 foo2=\\$lex2 foo3=\\$lex3 foo4=\\$lex4\\n\"\n273.48 379.35 373.81 292.31 250.00 \\$lex1 .= \"-foo-\\$lex2-foo-\\$lex3-foo\"\n275.38 301.89 317.24 333.33 240.00 my \\$lex = sprintf \"foo=%s\", \\$lex1\n279.21 279.05 312.73 330.00 266.67 \\$pkg = sprintf \"%s%s\", \\$lex1, \\$lex2\n280.63 302.44 330.77 338.98 250.00 my \\$lex = sprintf \"%s%s\", \\$lex1, \\$lex2\n287.32 296.52 348.28 332.81 237.50 \\$lex = sprintf \"foo=%s bar=%s\\n\", \\$lex1, \\$lex2\n296.81 406.25 452.38 296.30 240.00 \\$lex .= sprintf \"%s\", \\$lex1\n297.69 336.59 415.79 329.55 212.50 \\$pkg .= sprintf \"%s%s\", \\$lex1, \\$lex2\n298.19 310.26 335.71 348.72 325.00 \\$lex = sprintf \"foo=%s\", \\$lex1\n301.21 309.47 351.02 355.32 320.00 \\$lex = sprintf \"%s%s\", \\$lex1, \\$lex2\n325.90 379.45 493.75 356.10 242.86 \\$lex .= sprintf \"%s%s\", \\$lex1, \\$lex2\n333.93 419.48 454.55 359.60 291.67 \\$s .= \"foo=\\$a bar=\\$b baz=\\$c\" where \\$b,\\$c are utf8\n337.37 406.58 406.94 383.72 320.00 my \\$s = \"foo=\\$a bar=\\$b baz=\\$c\" where \\$b,\\$c are utf8\n375.00 478.63 521.43 427.85 291.67 \\$s .= \"foo=\\$a bar=\\$b baz=\\$c\" where \\$s,\\$b,\\$c are utf8\n391.79 408.33 490.41 461.96 270.00 \\$s = sprintf(\"foo=%s bar=%s baz=%s\", \\$a, \\$b, \\$c) where \\$a,\\$c are utf8\n528.36 623.38 721.21 571.00 391.67 \\$s .= \"foo=\\$a bar=\\$b baz=\\$c\" where \\$a,\\$c are utf8\n545.23 613.16 651.39 630.23 440.00 my \\$s = \"foo=\\$a bar=\\$b baz=\\$c\" where \\$a,\\$c are utf8\n581.81 723.31 830.56 622.22 500.00 \\$s .= \"foo=\\$a bar=\\$b baz=\\$c\" where \\$a is utf8, \\$c has 0x80\n599.55 714.91 757.69 678.64 577.78 my \\$s = \"foo=\\$a bar=\\$b baz=\\$c\" where \\$a is utf8, \\$c has 0x80\n603.04 718.32 835.71 693.67 391.67 \\$s .= \"foo=\\$a bar=\\$b baz=\\$c\" where \\$s,\\$a,\\$c are utf8\n654.05 828.57 951.61 727.84 500.00 \\$s .= \"foo=\\$a bar=\\$b baz=\\$c\" where \\$s,\\$a are utf8, \\$c has 0x80\n184.58 211.78 204.43 201.60 202.52 AVERAGE\n\n--\nYou never really learn to swear until you learn to drive."
] |
[
null,
"https://st.pimg.net/devcom/images/develooperbutton.v1.gif",
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.66072196,"math_prob":0.9761437,"size":10330,"snap":"2019-43-2019-47","text_gpt3_token_len":4049,"char_repetition_ratio":0.1532055,"word_repetition_ratio":0.093860686,"special_character_ratio":0.5090029,"punctuation_ratio":0.23650108,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9728434,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-24T05:55:37Z\",\"WARC-Record-ID\":\"<urn:uuid:7c40bf34-0416-41eb-9a07-799dc24e587a>\",\"Content-Length\":\"15971\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a4451e5e-6fb7-4f76-9b1c-f571fc14db26>\",\"WARC-Concurrent-To\":\"<urn:uuid:35f8e69b-d807-4f38-8325-852fdba84608>\",\"WARC-IP-Address\":\"147.75.38.240\",\"WARC-Target-URI\":\"https://www.nntp.perl.org/group/perl.perl5.porters/2017/10/msg246844.html\",\"WARC-Payload-Digest\":\"sha1:2YZQLN4FIXFDWIE7ALAE6KAE4TLGPXVT\",\"WARC-Block-Digest\":\"sha1:WGF3ZRJT63VNHMFS6GC2LTPI3LM4BMRK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987841291.79_warc_CC-MAIN-20191024040131-20191024063631-00539.warc.gz\"}"}
|
https://jeopardylabs.com/play/aim-3-10b-2
|
[
"Exercise 1\nExercise 2\nExercise 3\nExercise 4\nTeacher\n100\n\nC is the midpoint of BD and AE, what is the reason for this?\n\nGiven\n\n100\n\nPerpendicular lines form right angles, what is the statement for this?\n\nAngle MNP and Angle MKL are right angles\n\n100\n\nFill in the statement and reason for this proof\n\nAngle B = Angle D, AC bisects Angle BCD -- Given\n\n100\n\nFill in the statement and reason for this proof\n\nAD||FC, AC bisects FD -- Given\n\n100\n\nHow tall is Ms. Sanchez?\n\n5'7\"\n\n200\nAC = EC, BC = DC, what is the reason for this statement?\n\nDefinition of a midpoint\n\n200\n\nAll right angles are congruent, what is the statement for this reason?\n\nAngle MNP = Angle MKL\n\n200\n\nFill in the statement and reason for this proof\n\nAngle BCA = Angle DCA -- Definition of an angle bisector\n\n200\n\nFill in the statement and reason for this proof\n\nFE = DE -- Definition of a bisector\n\n200\n\nHow many pets does Ms. Sanchez have?\n\n0\n\n300\n\nAngle ACB = Angle ECD, what is the reason for this statement?\n\nVertical Angles are congruent\n\n300\n\nVertical Angles are congruent, what is the statement for this reason?\n\nAngle NMP = Angle KML\n\n300\n\nFill in the statement and reason for this proof\n\nAC = AC -- reflexive property\n\n300\n\nFill in the statement and reason for this proof\n\nAngle 1 = Angle 4, Angle 2 = Angle 3 -- When 2 lines are ||, alternate interior angles are congruent\n\n300\n\nWhat is Ms. Sanchez's favorite chocolate brand?\n\nFerrero Rocher\n\n400\n\nTriangle ABC = Triangle EDC, what is the reason for this?\n\nSAS Theorem\n\n400\n\nASA Theorem, what is the statement for this reason?\n\nTriangle NMP = Triangle KML\n\n400\n\nFill in the statement and reason for this proof\n\nTriangle ABC = Triangle ADC -- AAS Theorem\n\n400\n\nFill in the statement and reason for this proof\n\nTriangle FCE = Triangle DAE -- AAS Theorem\n\n400\n\nWhat is Ms. Sanchez's shoe size?\n\n8\n\n500\n\nAngle BAE = Angle DEC, what is the reason for this statement?\n\nCPCTC\n\n500\n\nCPCTC, what is the statement for this reason?\n\nNP = LK\n\n500\n\nWhich movie did Ms. Sanchez watch at the movie theater this weekend?\n\nNo Time to Die (James Bond)\n\n500\n\nWhat other triangle congruence theorem could we have used to prove these two triangles are congruent?\n\nUsing vertical angles (Angle FEC = Angle DEA), you could have also used ASA\n\n500\n\nWhat movie does Ms. Sanchez hate?\n\nJurassic Park\n\nClick to zoom"
] |
[
null
] |
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|
https://networkingfunda.com/construction-scheduling-coursera-quiz-answers/
|
[
"# Construction Scheduling Coursera Quiz Answers\n\n## All Weeks Construction Scheduling Coursera Quiz Answers\n\nLearners will discover the key project scheduling techniques and procedures including; how to create a network diagram, how to define the importance of the critical path in a project network, and defining project activities float. Also covered are the fundamentals of Bar Charts, Precedence Diagrams, Activity on Arrow, PERT, Range Estimating, and linear project operations and the line of balance.\n\nEnroll On Coursera\n\n### Construction Scheduling Week 1 Quiz Answers\n\n#### Quiz 1: Quiz on Gantt Chart\n\nQ1. For the Bar Chart in this diagram:",
null,
"What is the work progress for activity A at the end of month 2?\n\n• 66%\n• 50%\n• 30%\n\nQ2. For the Bar Chart in this diagram:",
null,
"What is the status of activity B at the end of month 2?\n\n• On schedule\n• Behind schedule\n\nQ3. For the Bar Chart in this diagram:",
null,
"What is the schedule status for activity B at the end of month 3?\n\n• On schedule.\n• Behind schedule.\n\nQ4. For the Bar Chart in this diagram:",
null,
"What is the status of the entire project at the end of month 3?\n\n• Behind schedule.\n• On schedule.\n\nQ5. For the Bar Chart in this diagram:",
null,
"At the end of month 3, what is left for activity A?\n\n• 30%.\n• 20%.\n• Nothing left at the end of month 3.\n\nQ6. The length of the bar in a bar chart, as shown in the video, represents:\n\n• Cost of an activity in a construction project\n• Number of buildings in a construction project\n• Length of the planned duration of an activity in a construction project\n• Length of the break between each construction activity\n\nQ7. At week 15 of the project, according to the bar chart below, what activities will be completed in the project:",
null,
"• a,b,c,d\n• a,b,c,d,e,f,g\n• a,b,c,d,e,f\n\nQ1. Which diagram is showing a traditional relationship?\n\n• Diagram #1.\n• Diagram #2.\n• None of these diagrams are showing a traditional relationship.\n\nQ2. The type of relationship in Diagram #1 is:\n\n• Finish to Start.\n• Start to Start.\n• Finish to Finish.\n\nQ3. The type of relationship in Diagram #2 is:\n\n• Finish to Start.\n• Start to Start.\n• Finish to Finish.\n\nQ4. Which diagram has a LAG and by how many days?\n\n• Diagram #1 by 2 days.\n• Diagram #2 by 3 days.\n• None of the diagrams has any LAG.\n\nQ5. If 3 days of lag have been added to the relationship in Diagram 1, when will Activity B start?\n\n• 13\n• 14\n• 15\n• 9\n\nQ6. Of the relationships in a node network diagram, which one is used most often?\n\n• Start to Start\n• Start to Finish\n• Finish to Finish\n• Finish to Start\n\nQ7. In the diagram below, what would be the early start date for Pouring Concrete?\n\n• 23\n• 7\n• 37\n• 47\n\nQ8. What is the lag in the start to start diagram below?\n\n• 4\n• 3\n• 1\n• 5\n\nQ9. A negative lag is also referred to as a\n\n• win\n• penalty\n• reverse lag\n\nQ1. According to the video Example 1: Activity On Node Diagram, what is the early start of activity E?\n\nHint: Please refer to the diagram at 5:00 minutes in the video.\n\n• 9\n• 11\n• 4\n• 19\n\nQ2. For the following project:\n\nN me activity X1\n\nE\n\nQ3. For the following project:\n\nName activity X2:\n\nB\n\nQ4. For the following project:\n\nName Activity X3\n\nI\n\nQ5. For the following project:\n\nName Activity X4\n\nG\n\nQ6. For the following project:\n\nWhat is the value of X5?\n\nType and Integer value.\n\nA\n\nQ7. For the following project:\n\nWhat is the value of X6?\n\nType and Integer value.\n\nI\n\nQ8. For the following project:\n\nWhat is the value of X7?\n\nType and Integer value.\n\nD\n\nQ9. For the following project:\n\nWhat is the value of X8?\n\nType and Integer value.\n\nC\n\nQ10. For the following project:\n\nWhat is the value of X9?\n\nType and Integer value.\n\ne\n\nQ11. For the following project:\n\nWhat is the value of X10?\n\nType and Integer value.\n\nA\n\nQ12. For the following project:\n\nWhat is the value of X11?\n\nType and Integer value.\n\nB\n\nQ13. For the following project:\n\nWhat is the value of X12?\n\nType and Integer value.\n\nC\n\nQ14. For the following project:\n\nWhat is the value of X13?\n\nType and Integer value.\n\nI\n\nQ15. For the following project:\n\nWhat is the value of X14?\n\nType and Integer value.\n\nG\n\nQ16. For the following project:\n\nWhat is the value of X15?\n\nType and Integer value.\n\nD\n\nQ17. Choose all of the relationships that are represented in a node network diagram.\n\n• First to Last\n• Finish to Start\n• Best to Worst\n• Start to Start\n• Start to Finish\n\nQ18. To determine the Early Start of an activity,\n\n• Factor in all its dependencies and see its earliest start date\n• Find the earliest calculated time an activity can end\n• Find the latest time an activity may begin without delaying the project duration\n• Find latest time an activity may be completedwithout delaying the project duration\n\n### Construction Scheduling Week 2 Quiz Answers\n\n#### Quiz 1: Critical Path Method\n\nQ1. In construction industry, any project:\n\n• It can have no critical path.\n• It must have at least one critical path.\n• It may have or have not critical path depends on its nature.\n\nQ2. If we want to save time and work on decrease the total duration of a project, we can do the following:\n\n• Invest more resources on any activity either critical or noncritical.\n• Ask our labors to come and work overtime.\n• Focus your resources only on the critical activities.\n\nQ3. If we have 2 construction activities that has conflict of resources; which activity you will give a priority to perform from the below three options:\n\n• The critical activity.\n• The longer activity in duration.\n• The activity that has more budget.\n\nQ4. From the activity on node diagram in this example (without performing any forward and backward pass calculations) what is the total duration of the project?\n\n• 12\n• 15\n• 18\n\nQ5. How many critical paths are in this project?\n\n• 1\n• 2\n• 3\n\nQ6. When it comes to the critical activities in a construction project:\n\n• All critical activities may belong to a critical path.\n• A delay in any critical activity will extend the project duration\n• All critical activities can be delayed at least 1 day without lengthening the project duration.\n\nQ7. Given the following activities and their durations, determine the duration of the project:\n\nActivity A – duration = 4 days\n\nActivity B – duration = 7 days\n\nActivity C – duration = 2 days\n\nActivity D – duration = 3 days\n\nActivity E – duration = 5 days\n\nStart –> Activity A –> Activity B –> Activity C –> Finish\n\nThe duration of the project would be:\n\n• 12 days\n• 13 days\n• 10 days\n• 15 days\n\nQ8. What is the duration of the path; A, B, E and G?\n\n• 20 days\n• 18 days\n• 16 days\n• 14 days\n\nQ9. SELECT ALL paths in the project below:\n\n• ABEHI\n• ABEFHI\n• ABFHI\n• ACDGI\n• ACFHI\n• ACGI\n\nQ10. Calculate the critical path in the following diagram:\n\n• 9\n• 15\n• 17\n• 19\n• 21\n\nQ1. If you are at a construction project and an engineer mentioned that: “This activity has a float of 3 days”. When the engineer did not specify the type of the float, usually the commonly referred to is:\n\n• Total Float\n• Free Float\n• Independent Float\n\nQ2. What is the total float of any critical activity?\n\n• One Day.\n• Zero\n• It depends if the critical activity is on a critical path or not.\n\nQ3. What will happen if an activity got delayed from its early start by more than its total float?\n\n• Nothing will be affected.\n• Only the activity that follows will be delayed.\n• The entire project duration will be delayed.\n\nQ4. Finish the following statement: “The total float is the amount of time that an activity can be delayed from its early start date without delaying the ________ finish date. Free float is the amount of time that an activity can be delayed without delaying the early start date of any ___________ activity”.\n\n• Successor , Project.\n• Project, Successor.\n• Predecessor, Project.\n• Project, Predecessor.\n\nQ5. From the diagram in this example, calculate the following:\n\nThe free float of activity D:\n\n• 3\n• 4\n• 5\n• 6\n\nQ6. From the diagram in this example, calculate the following:\n\nThe free float of activity B:\n\n• 0\n• 1\n• 2\n\nQ7. From the diagram in this example, calculate the following:\n\nThe total float of activity C:\n\n• 3\n• 5\n• 8\n\nQ8. From the diagram in this example, calculate the following:\n\nThe free float of activity C:\n\n• 3\n• 5\n• 8\n\nQ9. Suppose this project starts on Feb. 1st, 2017 on a calendar that has Saturday and Sunday as weekend days.\n\nWhen is the project expected end-date?\n\n• Feb. 17th, 2017\n• Feb. 22th, 2017\n• Feb. 16th, 2017\n• Feb. 30th, 2017\n\nQ10. Suppose this project starts on Feb. 1st, 2017 on a calendar that has Saturday and Sunday as weekend days.\n\nWhen is the latest start of Activity C?\n\n• Feb 13th 2017\n• Feb 12th 2017\n• Feb 14th 2017\n\n### Construction Scheduling Week 3 Quiz Answers\n\n#### Quiz 1: Activity Diagram and Critical Path\n\nQ1. Imagine you are a project manager of a power station project. The owner has asked you to prepare a schedule for the project. The details of all the activities are given below. You have to create an activity on arrow (AOA) diagram to help you prepare the schedule of the project.",
null,
"Construct the Activity on Arrow diagram and answer the following questions :\n\nWhat is the critical path of the project?\n\n• ABCDMPLQR\n• ABCDNR\n• EHJKLQS\n• ABCDHJKLQR\n\nQ2. What is the total project duration ?\n\n• 41\n• 45\n• 39\n• 37\n\nQ3. What are the number of dummies in the AOA diagram ?\n\n• 4\n• 5\n• 3\n• 6\n\nQ4. What is the EST of Activity I?\n\n• 20\n• 27\n• 25\n• 22\n\nQ5. What is the LFT for activity J?\n\n• 29\n• 21\n• 32\n• 25\n\nQ1. An activity with Optimistic duration of 3 days, Pessimistic duration of 7 days and Most likely duration of 4 days. What is the estimated duration?\n\n• 3.833\n• 4.333\n• 5.8333\n\nQ2. Which of the following is true\n\n• Standard deviation = Mean squared\n• Mean = Standard deviation squared\n• Standard deviation = Variance squared\n• Variance = Standard deviation squared\n\nQ3. A project has 3 critical paths with the same duration, which variance should be used in the calculations of project ?\n\n• The lowest value\n• The middle value\n• The highest value\n\nQ4. A project of expected mean duration of 20 days and variance of 3. What is the probability finishing the project in 22 days?\n\n• 95%\n• 87%\n• 80%\n• 75%\n\nQ5. A project of expected mean duration of 40 days and standard deviation of 5. What is the probability finishing the project in 35 days?\n\n• 84%\n• 16%\n• 50%\n• 78%\n\nQ6. A project of expected mean duration of 100 days and standard deviation of 3.5. What is the probability finishing the project between 95 and 105 days?\n\n• 75%\n• 84%\n• 95%\n• 50%\n\nQ7. For the following list of critical activities in a project, what are the expected duration of the project and the standard deviation\n\n• 25,1.39\n• 25,1.94\n• 23,1.39\n• 27,1.89\n\nQ8. Which of the following is a part of planning the project?\n\n• Resource allocation\n• Submittals/approvals\n\nQ9. Which of the following is not a project evaluation metric?\n\n• Earned Cost\n• Vacation days\n• Float Attrition\n• Earned value\n\nQ10. The scheduler role does not include recording the construction procedure implemented in reality.\n\n• True\n• False.\n\n### Construction Scheduling Week 4 Quiz Answers\n\n#### Quiz 1: Quiz on Line of Balance\n\nFor the following project, 3 sections are expected every week. Each week is 5 working days with 8 hours a day. Find the value of X1?\n\nProvide answers to one decimal place.",
null,
"1\n\nQ2. For the following project, 3 sections are expected every week. Each week is 5 working days with 8 hours a day. Find the value of X2?\n\nProvide answers to two decimal place.",
null,
"8\n\nQ3. For the following project, 3 sections are expected every week. Each week is 5 working days with 8 hours a day. Find the value of X3?",
null,
"• 6\n• 7\n• 8\n\nQ4. For the following project, 3 sections are expected every week. Each week is 5 working days with 8 hours a day. Find the value of X4?",
null,
"• 26\n• 27\n• 24\n\nQ5. For the following project, 3 sections are expected every week. Each week is 5 working days with 8 hours a day. Using your answers to previous questions, find the value of X5?\n\nProvide answers to one decimal place.",
null,
"3\n\nQ6. For the following project, 3 sections are expected every week. Each week is 5 working days with 8 hours a day. Using your answers to previous questions, find the value of X6?\n\nProvide answers to one decimal place.",
null,
"1\n\nQ7. For the following project, 3 sections are expected every week. Each week is 5 working days with 8 hours a day. Using your answers to previous questions, find the value of X7?\n\nProvide answers to three decimal places.",
null,
"2\n\nQ8. For the following project, 3 sections are expected every week. Each week is 5 working days with 8 hours a day. Using your answers to previous questions, find the value of X8?\n\nProvide answers to one decimal place.",
null,
"4\n\nQ9. For the following project, 3 sections are expected every week. Each week is 5 working days with 8 hours a day. Using your answers to previous questions, find the value of X9?\n\nAssume 20 units are required.\n\nProvide answers in integer format.",
null,
"9\n\nQ10. Where should be the buffer, given that Y has higher rate than X?",
null,
"• a\n• b\n• c\n\nQ1. Draw the Line of Balance Diagram (showing the start and finish of the first and last unit for each activity) and calculate the total project duration for a repetitive project that includes the activities shown in the following Table that are repeated in 20 units. The five activities are sequential with a 5-day buffer between them. The productivity demand is 3 sections per-week, working hours are 8 hours per-day in a 5 day working week.\n\nWhat is the total duration of the project ?\n\n• 77\n• 70\n• 80\n• 73\n\nQ2. What is total duration of 1 unit of work ?\n\n• 21.5\n• 19.3\n• 23.5\n• 25.1\n\nQ3. Calculate the average time gap between the first unit and start of the last unit.\n\n• 31\n• 32\n• 27\n• 29\n\nQ4. Which activity has the lowest output size?\n\n• C\n• A\n• B\n• D\n• E\n\nQ1. Which of the following is not a component of Schedule Analysis?\n\n• Critical path analysis\n• Logic analysis\n• Earned Value analysis\n\nQ2. Which of the following is best for Schedule Analysis?\n\n• Microsoft Project\n• Primavera P6\n• Navisworks\n• Synchro\n\nQ3. Microsoft Project displays real time Gant Chart as you develop the Activity List?\n\n• True\n• False\n\nQ4. What color does Microsoft Project use for critical path tasks?\n\n• Black\n• Orange\n• Red\n• Blue\n\nQ5. Select ALL right answers, P6 features include:\n\n• Grouping activities using WBS\n• Creating project reports in PDF\n• Setting work calender\n\nQ6. Select ALL right answers, P6 Gant charts show:\n\n• Critical Activities\n• Activity relationships\n• Activity durations\n• Activity cost\n\nQ7. Circular logic can be detected by scheduling software.\n\n• True\n• False\n\nQ8. If earned duration is lower than current project day, the project is:\n\n• Delayed\n• On plan\n\nQ9. Revit can embed only Revit format files\n\n• True\n• False\n\nQ10. In Syncrhro, exported videos have to have a certain layout with specific colors and components.\n\n• True\n• False\n\n### Construction Scheduling Week 5 Quiz Answers\n\n#### Quiz 1: Quiz on Large programs, Risk and Lean\n\nQ1. Which of the following is not a program management typical responsibility?\n\n• Issuing public information for the media about projects\n• Setting projects costs and schedule\n• Tracking day to day progress of each individual project\n\nQ2. IMS stands for\n\n• International Manner Syndicate\n• Integrated Magnification System\n• Internal Machinery Schedule\n• Integrated Master Schedule\n\nQ3. The IMS may not include:\n\n• Project milestones\n• Key risk items\n• Stakeholder activity\n• All the above\n\nQ4. A project appears in the IMS as:\n\n• A detailed list of activities\n• A bar, including indications of project milestone\n• A cost analysis study\n\nQ5. Risk is an uncertain event that, if it occurs, has an effect on at least one project objective. Which of the following is a project objective that may be affected by risk:\n\n• Project quality\n• Project cost\n• Project time\n• All the above\n\nQ6. According to the lectures, in Design-Build-Finance projects, the project is expected to have shorter overall duration:\n\n• True\n• False\n• Not necessarily\n\nQ7. Which of the following is not a risk treatment option:\n\n• Avoid\n• Mitigate\n• Confess\n• Transfer\n\nQ8. Modelling risk may include:\n\n• Structural failure simulations\n• Monte Carol simulations\n• Life balance simulations\n\nQ9. Confidence level is a typical risk simulation output\n\n• True\n• False\n\nQ10. Which of the following is on the Tornado Graph of Top Risks\n\n• Systems construction\n• Integration testing\n• Systems testing\n• All the above\n\nQ11. Time is of paramount importance because:\n\n• Interest rates fluctuate\n• It takes years to master one’s professional expertise\n• More money and quality can be created\n• All the above\n\nQ12. A “great” schedule offers superior:\n\n• Visual Management; Ease of Revision\n• Early and Late dates for Start and Finish\n• “S” Curves\n\nQ13. The Lean Big Four consists of:\n\n• Theory, Pull-Planning, Target Value Design, Choosing By Advantages\n• Theory, Pull-Planning, Target Value Design, Set Based Design\n• Pull-Planning, Target Value Design, Set Based Design, Choosing By Advantages\n\nQ14. Pull-Planning reduces project schedules because:\n\n• Milestones are placed at hand-offs, enabling next steps in the process\n• Gantt Charts are easier to read than CPM method\n• Individual trades only have to focus on delivering their work\n• It works backwards from deadline\n\nQ15. Well-performing projects achieve Percent Promises Completed in the range of:\n\n• 35%-65%\n• 75% – 85%\n• 100%\n• 50% +/- 5%\n##### Construction Scheduling Course Review:\n\nIn our experience, we suggest you enroll in Construction Scheduling courses and gain some new skills from Professionals completely free and we assure you will be worth it.\n\nConstruction Scheduling course is available on Coursera for free, if you are stuck anywhere between a quiz or a graded assessment quiz, just visit Networking Funda to get Construction Scheduling Coursera Quiz Answers.\n\n##### Conclusion:\n\nI hope this Construction Scheduling Coursera Quiz Answers would be useful for you to learn something new from this Course. If it helped you then don’t forget to bookmark our site for more Quiz Answers.\n\nThis course is intended for audiences of all experiences who are interested in learning about new skills in a business context; there are no prerequisite courses.\n\nKeep Learning!\n\n#### Get All Course Quiz Answers of Construction Management Specialization\n\nConstruction Project Management Coursera Quiz Answers"
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|
https://lists.boost.org/Archives/boost/2009/12/160116.php
|
[
"",
null,
"Boost :\n\nSubject: Re: [boost] How to detect if f() returns void or not?\nFrom: Eric Niebler (eric_at_[hidden])\nDate: 2009-12-13 05:57:00\n\n>> I solved this problem once while writing a different trait and documented it\n>> here:\n\n>\n> This does not check for return type but just ignores it with the\n> operator, trick. In the end, you do not know if fun(a,b) return void\n> or not.\n\nI find your lack of faith disturbing.\n\n#include <boost/static_assert.hpp>\n\ntypedef char yes_type;\ntypedef char (&no_type);\n\nstruct void_return\n{\ntemplate<typename T>\nfriend int operator,(T const &, void_return);\n};\n\nno_type check_is_void_return(int);\nyes_type check_is_void_return(void_return);\n\n// check to see if the expression fun(x,y) is void\ntemplate<typename Fun, typename X, typename Y>\nstruct returns_void\n{\nstatic Fun &fun;\nstatic X &x;\nstatic Y &y;\nstatic bool const value = sizeof(yes_type) ==\nsizeof(fun(x,y),void_return());\n};\n\nint main()\n{\nBOOST_STATIC_ASSERT((returns_void<void(int,int),int,int>::value == 1));\nBOOST_STATIC_ASSERT((returns_void<int(int,int),int,int>::value == 0));\n}\n\nHTH,\n\n--\nEric Niebler\nBoostPro Computing\nhttp://www.boostpro.com"
] |
[
null,
"https://lists.boost.org/boost/images/boost.png",
null
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https://zbmath.org/?q=an:0789.13002
|
[
"# zbMATH — the first resource for mathematics\n\nThe equations of Rees algebras of ideals with linear presentation. (English) Zbl 0789.13002\nLet $$(R,m)$$ be a local Gorenstein ring (or a polynomial ring over a field) of dimension $$d$$, and let $$I$$ be a (homogeneous) $$R$$-ideal of height $$g \\geq 1$$. The objective of this paper is to investigate properties of the Rees algebra of $$I$$, $${\\mathcal R} (I) =R[It]$$ (where $$t$$ is a variable), and to study the canonical epimorphism $$\\alpha:S(I) \\twoheadrightarrow {\\mathcal R} (I)$$ from the symmetric algebra onto the Rees algebra. As a standing hypothesis, suppose that $$I$$ is strongly Cohen- Macaulay, i.e., that all Koszul homology modules of a (homogeneous) generating set of $$I$$ are Cohen-Macaulay. By a well known result from J. Herzog, A. Simis, and W. V. Vasconcelos [cf. J. Algebra 74, 466-493 (1982; Zbl 0484.13006)], $$\\alpha$$ is an isomorphism (one says $$I$$ is of linear type) and $${\\mathcal R} (I)$$ is Cohen-Macaulay, if one assumes that $$I$$ satisfies $${\\mathcal F}_ 1$$, meaning, if the minimal number of generators $$v(I_ p)$$ of $$I_ p$$ is at most $$\\dim R_ p$$ for every prime ideal $$p$$ containing $$I$$.\nIn the paper under review, the condition $${\\mathcal F} _ 1$$ is replaced by the next weaker assumption that $$I$$ satisfies $${\\mathcal F}_ 1$$ locally on the punctured spectrum of $$R$$ and that $$v(I)=d+1$$. In this case, $$I$$ cannot be of linear type. It is shown that the analytic spread of $$I$$, $$\\ell(I)=\\dim {\\mathcal R} (I) \\bigoplus_ R R/m$$, has the maximal possible value, namely $$d$$, and that the initial degree of $$\\ker \\alpha$$ is $$d-g+ 2$$ if $$g \\geq 2$$. Furthermore, the graded component of $$\\ker \\alpha$$ in this minimal degree is identified to be $$\\text{Ext}^ d_ R (R/I_ 1(\\varphi),R)$$, where $$I_ 1(\\varphi)$$ denotes the ideal generated by the entries of a minimal presenting matrix $$\\varphi$$ of $$I$$. – One can say more if $$R$$ is a polynomial ring over a field and $$I$$ is presented by a matrix with linear entries. In this case, $$I$$ has reduction number $$d- g+1$$ (and hence $$I^{d-g+2} =JI^{d-g+1}$$ for an ideal $$J \\subset I$$ generated by a homogeneous system of parameters), $$\\ker\\alpha$$ is cyclic, and $$S(I)$$ is reduced. Furthermore, $$I$$ is generated by forms of degree $$\\delta \\leq {d \\over g-1}$$, where equality holds if and only if $${\\mathcal R} (I)$$ is Cohen-Macaulay, which in turn is equivalent to the Cohen- Macaulayness of the associated graded ring of $$I$$, $$gr_ I(R)={\\mathcal R} (I) \\bigoplus_ RR/I$$. These equivalent conditions are always satisfied if $$g \\leq 4$$.\nFinally, consider a strongly Cohen-Macaulay prime ideal $$I$$ in a regular local ring $$R$$, without any assumptions on the local number of generators or the grading of $$I$$. It is shown that if $$\\text{Spec(gr}_ I (R))$$ is irreducible of if gr$$_ I(R)$$ is reduced, then $$I$$ is of linear type (in which case $${\\mathcal R} (I)$$ is Cohen-Macaulay and gr$$_ I(R)$$ is a Gorenstein domain).\n\n##### MSC:\n 13A30 Associated graded rings of ideals (Rees ring, form ring), analytic spread and related topics 13C14 Cohen-Macaulay modules 13H10 Special types (Cohen-Macaulay, Gorenstein, Buchsbaum, etc.) 13H05 Regular local rings 13F20 Polynomial rings and ideals; rings of integer-valued polynomials\nFull Text:\n##### References:\n Abhyankar, S.: Concepts of order and rank on a complex space, and a condition for normality. Math. Ann.141, 171–192 (1960) · Zbl 0107.15001 · doi:10.1007/BF01360171 Aoyama, Y.: A remark on almost complete intersections. Manuscr. Math.22, 225–228 (1977) · Zbl 0367.13005 · doi:10.1007/BF01172664 Avramov, L., Herzog, J.: The Koszul algebra of a codimension 2 embedding. Math. Z.175, 249–280 (1980) · Zbl 0461.14014 · doi:10.1007/BF01163026 Hartshorne, R.: Complete intersections and connectedness. Am. J. Math.84, 497–508 (1962) · Zbl 0108.16602 · doi:10.2307/2372986 Herzog, J., Simis, A., Vasconcelos, W.V.: Approximation complexes of blowing-up rings. J. Algebra74, 466–493 (1982) · Zbl 0484.13006 · doi:10.1016/0021-8693(82)90034-5 Herzog, J., Simis, A., Vasconcelos, W.V.: Approximation complexes of blowing-up rings. II. J. Algebra82, 53–83 (1983) · Zbl 0515.13018 · doi:10.1016/0021-8693(83)90173-4 Herzog, J., Simis, A., Vasconcelos, W.V.: On the arithmetic and homology of algebras of linear type. Trans. Am. Math. Soc.283, 661–683 (1984) · Zbl 0541.13005 · doi:10.1090/S0002-9947-1984-0737891-6 Herzog, J., Vasconcelos, W.V., Villarreal, R.: Ideals with sliding depth. Nagoya Math. J.99, 159–172 (1985) · Zbl 0561.13014 Huckaba, S., Huneke, C.: Powers of ideals having small analytic deviation. Am. J. Math.114, 367–403 (1992) · Zbl 0758.13001 · doi:10.2307/2374708 Huneke, C.: On the associated graded ring of an ideal. Ill. J. Math.26, 121–137 (1982) · Zbl 0479.13008 Huneke, C.: Strongly Cohen-Macaulay schemes and residual intersections. Trans. Am. Math. Soc.277, 739–763 (1983) · Zbl 0514.13011 · doi:10.1090/S0002-9947-1983-0694386-5 Huneke, C., Simis, A., Vasconcelos, W.V.: Reduced normal cones are domains. Contemp. Math.88, 95–101 (1989) · Zbl 0676.13011 Northcott, D.G., Rees, D.: Reductions of ideals in local rings. Proc. Camb. Philos. Soc.50, 145–158 (1954) · Zbl 0057.02601 · doi:10.1017/S0305004100029194 Simis, A., Ulrich, B., Vasconcelos, W.V.: Jacobian dual fibrations. Am. J. Math.115, 47–75 (1993) · Zbl 0791.13007 · doi:10.2307/2374722 Vasconcelos, W.V.: On the equations of Rees algebras. J. Reine Angew. Math.418, 189–218 (1991) · Zbl 0727.13002 · doi:10.1515/crll.1991.418.189\nThis reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching."
] |
[
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https://www.boost.org/doc/libs/1_66_0/libs/multiprecision/doc/html/boost_multiprecision/tut.html
|
[
"#",
null,
"Boost C++ Libraries\n\n...one of the most highly regarded and expertly designed C++ library projects in the world.\n\n## Tutorial\n\nInteger Types\ncpp_int\ngmp_int\ntom_int\nExamples\nFactorials\nBit Operations\nfloating-point Numbers\ncpp_bin_float\ncpp_dec_float\ngmp_float\nmpfr_float\nfloat128\nExamples\nArea of Circle\nDefining a Special Function.\nCalculating a Derivative\nCalculating an Integral\nPolynomial Evaluation\nInterval Number Types\nmpfi_float\nRational Number Types\ncpp_rational\ngmp_rational\ntommath_rational\nUse With Boost.Rational\nMiscellaneous Number Types.\nVisual C++ Debugger Visualizers\nConstructing and Interconverting Between Number Types\nGenerating Random Numbers\nPrimality Testing\nLiteral Types and `constexpr` Support\nImporting and Exporting Data to and from cpp_int and cpp_bin_float\nRounding Rules for Conversions\nMixed Precision Arithmetic\nGeneric Integer Operations\nBoost.Serialization Support\nNumeric Limits\nstd::numeric_limits<> constants\nstd::numeric_limits<> functions\nNumeric limits for 32-bit platform\nHow to Determine the Kind of a Number From `std::numeric_limits`\nInput Output\nHash Function Support\n\nIn order to use this library you need to make two choices:\n\n• What kind of number do I want (integer, floating-point or rational).\n• Which back-end do I want to perform the actual arithmetic (Boost-supplied, GMP, MPFR, Tommath etc)?"
] |
[
null,
"https://www.boost.org/gfx/space.png",
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.82951313,"math_prob":0.852088,"size":479,"snap":"2020-34-2020-40","text_gpt3_token_len":122,"char_repetition_ratio":0.094736844,"word_repetition_ratio":0.33846155,"special_character_ratio":0.23173277,"punctuation_ratio":0.20833333,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9914591,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-12T23:00:24Z\",\"WARC-Record-ID\":\"<urn:uuid:a386d108-aa87-43f8-8bed-2575fea4446e>\",\"Content-Length\":\"9789\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ab250fd7-8d12-4cd8-862c-a4b78b42b697>\",\"WARC-Concurrent-To\":\"<urn:uuid:8a25061c-b3c3-4b7d-8097-89d8beae2f6d>\",\"WARC-IP-Address\":\"146.20.110.251\",\"WARC-Target-URI\":\"https://www.boost.org/doc/libs/1_66_0/libs/multiprecision/doc/html/boost_multiprecision/tut.html\",\"WARC-Payload-Digest\":\"sha1:T7D2YRTFHV5JZ6BH6MX7C5INCRAVB2GJ\",\"WARC-Block-Digest\":\"sha1:4Q773QZOGS5VF2B7W6FS44VF6YLEHNMB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439738950.31_warc_CC-MAIN-20200812225607-20200813015607-00595.warc.gz\"}"}
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https://topanswers.xyz/tex?q=1814
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[
"I see this code from [here](https://www.overleaf.com/project/60b3af145e71b668aa3580e5)\n\n\n\\documentclass[tikz,border=5mm,convert={outfile=\\jobname.png}]{standalone}\n\\usepackage{tikz-3dplot-circleofsphere}\n\\usetikzlibrary{calc,angles}\n%==============\n\\everymath{\\displaystyle}\n\\begin{document}\n%%%%%%%%%%%%%%%%%%%%%%%\n\n\\tdplotsetmaincoords{70}{110}\n\\begin{tikzpicture}[tdplot_main_coords,scale=3,font=\\footnotesize,>=stealth]\n\\def\\r{2}\n\\pgfmathsetmacro{\\a}{sqrt(3)}\n\\pgfmathsetmacro{\\b}{sqrt(2)}\n\\begin{scope}[thin,black!50]\n\\draw[dashed] (-1.3*\\r,0,0) -- (\\r,0,0)\n(0,-1.05*\\r,0) -- (0,\\r,0)\n(0,0,0) -- (0,0,\\r);\n\\draw[->] (\\r,0,0) -- (1.3*\\r,0,0) node[anchor=north east] {$x$};\n\\draw[->] (0,\\r,0) -- (0,1.3*\\r,0) node[anchor=north] {$y$};\n\\draw[->] (0,0,\\r) -- (0,0,1.3*\\r) node[anchor=south east] {$z$};\n\\draw[tdplot_screen_coords] (\\r,0,0) arc (0:180:\\r);\n\\tdplotCsDrawLatCircle{\\r}{0}\n\\end{scope}\n\\tdplotCsDrawCircle{\\r}{0}{60}{60}\n\\tdplotCsDrawCircle{\\r}{70.5}{60}{60}\n\\tdplotCsDrawLatCircle{\\r}{60}\n\\tdplotCsDrawCircle{\\r}{35.4}{35.3}{84.9}\n\\path (0,0,0) coordinate (O)\n(0,0,\\a) coordinate (O_1)\n(1.5,0,\\a/2) coordinate (O_2)\n(0.5,\\b,\\a/2) coordinate (O_3)\n(0.94,{2/3},1.63) coordinate (O_4)\n(1,0,\\a) coordinate (M)\n(1.09,0.58,1.57) coordinate (K)\n(barycentric cs:O_1=1,O_2=1,O_3=1) coordinate (L)\n($(L)-(O_4)+(K)$) coordinate (N);\n\\draw[dashed] (M)--(O)--(O_1)--(M)--(O_2)--(O_1) (O)--(O_2)--(K)--(O_4)--(O) (O_2)--(L) (K)--(N);\n\\draw[fill=black] (O) circle (0.5pt) node[below right]{$O$}\n(O_1) circle (0.5pt) node[above left]{$O_1$}\n(M) circle (0.5pt) node[above]{ $M$}\n(O_2) circle (0.5pt) node[left]{$O_2$}\n(O_4) circle (0.5pt) node[right=0.05cm]{\\tiny $O_4$}\n(K) circle (0.5pt) node[left]{$K$}\n(N) circle (0.5pt) node[below]{ $N$}\n(L) circle (0.5pt) node[below right]{ $L$};\n\\end{tikzpicture}\n%=============\n\\end{document}\n\n\n\n\n\nI tried to draw it with 3dtools.\n\n\\documentclass[tikz,border=3mm]{standalone}\n\\usetikzlibrary{3dtools}% https://github.com/marmotghost/tikz-3dtools\n\\begin{document}\n\n\\begin{tikzpicture}[3d/install view={phi=110,theta=70},line cap=butt,line join=round,c/.style={circle,fill,inner sep=1pt},declare function={rs=2;rc=1;d=sqrt(rs*rs-rc*rc);a=sqrt(2);}]\n\\path\n(0,0,0) coordinate (O)\n(0,0,d) coordinate (O_1)\n(1.5,0,d/2) coordinate (O_2)\n(0.5,a,d/2) coordinate (O_3)\n(0.94,{2/3},1.63) coordinate (O_4);\n\\draw[3d/screen coords] (O) circle[radius=rs];\n\\path pic{3d/circle on sphere={R=rs, P={(0,0,d)}}};\n\\path pic{3d/circle on sphere={R=rs,P={(O_2)}}};\n\\path pic{3d/circle on sphere={R=rs, P={(O_3)}}};\n\\path pic{3d/circle on sphere={R=rs, P={(O_4)}}};\n\\path pic{3d/circle on sphere={R=rs, P={(O)}}};\n\n%\\path foreach \\p/\\g in {O/-90,O_1/90,O_2/0,O_3/0}{(\\p)node[c]{}+(\\g:2.5mm) node{$\\p$}};\n\\end{tikzpicture}\n\\end{document}\n\n\n\nMy questions:\n1. The circle with center O_4 is incorrect. How to correct it?\n2. How can I find coordinates of the points O_2, O_3, O_4 by 3dtools?\n3. Can I solve the general problem?\n\n\nThe reason why the circle is incorrect is that the code uses rounded values. On the other hand, the radius r of the circle is computed via\n\nr=sqrt(R^2-d^2) ,\n\nwhere R is the radius of the sphere and d is the distance of the center of the circle from the center of the sphere. If we compute the sensitivity of r on d, we need to compute the derivative of r w.r.t. d (which, among other things, also illustrates why it makes a lot of sense to distinguish differential ds typographically from variables d), we see that\n\nr'=-d/sqrt(R^2-d^2) .\n\nThat is, if d is very close to R, then a tiny error in d can give you a very wrong radius.\n\nThis is confirmed with 3dtools, with which it is rather easy to construct the circle. All we need to do is to compute two angles, alpha and beta, and to install local coordinate systems on the circles. The details are in the comments of the following code.\n\n\\documentclass[tikz,border=3mm]{standalone}\n\\usetikzlibrary{3dtools}% https://github.com/marmotghost/tikz-3dtools\n\\begin{document}\n\n\\begin{tikzpicture}[3d/install view={phi=110,theta=70},\nline cap=butt,line join=round,c/.style={circle,fill,inner sep=1pt},\ndeclare function={rs=2;rc=1;d=sqrt(rs*rs-rc*rc);a=sqrt(2);\n% alpha is just the latitude angle of O_2 and O_3\nalpha=asin(rc/rs);\n% the beta angle emerges from the requirement that O--O_2 and\n% O--O_3 enclose an angle of 2*alpha\nbeta=acos((cos(2*alpha)-pow(sin(alpha),2))/(pow(cos(alpha),2)));}]\n\\path\n(0,0,0) coordinate (O)\n(0,0,d) coordinate (O_1)\n({d*cos(alpha)},0,{d*sin(alpha)}) coordinate (O_2)\n({d*cos(alpha)*cos(beta)},{d*cos(alpha)*sin(beta)},{d*sin(alpha)}) coordinate (O_3);\n% basis vectors for circle 2\n\\pgfmathsetmacro{\\myexb}{TDunit(\"(0,0,1)x(O_2)\")}\n\\pgfmathsetmacro{\\myeyb}{TDunit(\"(O_2)x(\\myexb)\")}\n% basis vectors for circle 3\n\\pgfmathsetmacro{\\myexc}{TDunit(\"(0,0,1)x(O_3)\")}\n\\pgfmathsetmacro{\\myeyc}{TDunit(\"(O_3)x(\\myexc)\")}\n\\path[3d coordinate/.list={%\n{(T_1)=(O_1)+[rc*cos(beta/2)]*(1,0,0)+[rc*sin(beta/2)]*(0,1,0)},\n{(T_2)=(O_2)+[rc*cos(90-beta/2)]*(\\myexb)+[rc*sin(90-beta/2)]*(\\myeyb)},\n{(T_3)=(O_3)+[rc*cos(90+beta/2)]*(\\myexc)+[rc*sin(90+beta/2)]*(\\myeyc)}}]\n(O)pic[draw=none]{3d circle through 3 points={A={(T_1)},B={(T_2)},C={(T_3)},\ncenter name=O_4}};\n\\draw[3d/screen coords] (O) circle[radius=rs];\n\\pgfmathsetmacro{\\myM}{TD(\"(O_4)\")}\n\\pgfmathsetmacro{\\myr}{tddistance(\"(O_4)\",\"(T_1)\")}\n\\typeout{O_4=(\\myM), r=\\myr}\n\\path foreach \\X in {1,2,3,4}\n{ (O)pic{3d/circle on sphere={R=rs,C={(O)},P={(O_\\X)}}}};\n%\n\\path foreach \\p/\\g in {O/-90,O_1/90,O_2/0,O_3/0,T_1/90,T_2/-135,T_3/-45}{\n(\\p)node[c,label={[font=\\tiny,label distance=0pt,inner sep=0pt]\\g:{$\\p$}}]{}};\n\\end{tikzpicture}\n\\end{document}\n\n\nThis code gets O_4=(0.93881,0.66382,1.62599). The corresponding distances are\n\n1.99623 for your input and 1.99145 for the 3dtools result,\n\nand the resulting radii are\n\n0.122701 for your input and 0.184757 for the 3dtools result,\n\nrespectively. As you can see, an error of about 0.25% at the level of the distance translates into a 50% error at the level of the radius. This is why the circle produced in the code you posted is incorrect.\n\n3dtools uses fpu to counter such effects. There is no guarantee that this is always sufficient, but at least in this case it is. (It wouldn't be too difficult to use xfp instead because all the computations in the package yield dimensionless results. Replacing fpu by xfp blindly, however, will lead to major problems that manifest only in certain scenarios, which are, however, widely used.)\n\nThe above code is already general, i.e. can be used for various rc. (Of course, if the radius rc becomes too large, three circles will no longer fit on the sphere while touching each other only once, and then the code does not work any more.)\n\n\\documentclass[tikz,border=3mm]{standalone}\n\\usetikzlibrary{3dtools}% https://github.com/marmotghost/tikz-3dtools\n\\begin{document}\n\\foreach \\rc in {1,1.1,...,3.4,3.3,3.2,...,1.1}\n{\\begin{tikzpicture}[3d/install view={phi=110,theta=70},\nline cap=butt,line join=round,c/.style={circle,fill,inner sep=1pt},\ndeclare function={rs=4;rc=\\rc;d=sqrt(rs*rs-rc*rc);\n% alpha is just the latitude angle of O_2 and O_3\nalpha=90-2*atan2(rc,d);\n% the beta angle emerges from the requirement that O--O_2 and\n% O--O_3 enclose an angle of 2*atan2(rc,d)\nbeta=acos((cos(2*atan2(rc,d))-pow(sin(alpha),2))/(pow(cos(alpha),2)));\n}]\n\\path\n(0,0,0) coordinate (O)\n(0,0,d) coordinate (O_1)\n({d*cos(alpha)},0,{d*sin(alpha)}) coordinate (O_2)\n({d*cos(alpha)*cos(beta)},{d*cos(alpha)*sin(beta)},{d*sin(alpha)}) coordinate (O_3)\n;\n% basis vectors for circle 2\n\\pgfmathsetmacro{\\myexb}{TDunit(\"(0,0,1)x(O_2)\")}\n\\pgfmathsetmacro{\\myeyb}{TDunit(\"(O_2)x(\\myexb)\")}\n% basis vectors for circle 3\n\\pgfmathsetmacro{\\myexc}{TDunit(\"(0,0,1)x(O_3)\")}\n\\pgfmathsetmacro{\\myeyc}{TDunit(\"(O_3)x(\\myexc)\")}\n\\path[3d coordinate/.list={%\n{(T_1)=(O_1)+[rc*cos(beta/2)]*(1,0,0)+[rc*sin(beta/2)]*(0,1,0)},\n{(T_2)=(O_2)+[rc*cos(90-beta/2)]*(\\myexb)+[rc*sin(90-beta/2)]*(\\myeyb)},\n{(T_3)=(O_3)+[rc*cos(90+beta/2)]*(\\myexc)+[rc*sin(90+beta/2)]*(\\myeyc)}}]\n(O)pic[draw=none]{3d circle through 3 points={A={(T_1)},B={(T_2)},C={(T_3)},\ncenter name=O_4}};\n\\draw[3d/screen coords] (O) circle[radius=rs];\n\\path foreach \\X in {1,2,3,4}\n{ (O)pic{3d/circle on sphere={R=rs,C={(O)},P={(O_\\X)}}}};\n\\end{tikzpicture}}\n\\end{document}\n\n\n\n\n\nEnter question or answer id or url (and optionally further answer ids/urls from the same question) from\n\nSeparate each id/url with a space. No need to list your own answers; they will be imported automatically."
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[
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{"ft_lang_label":"__label__en","ft_lang_prob":0.57896006,"math_prob":0.99741846,"size":5600,"snap":"2022-40-2023-06","text_gpt3_token_len":2053,"char_repetition_ratio":0.11651179,"word_repetition_ratio":0.2260274,"special_character_ratio":0.37642857,"punctuation_ratio":0.16471572,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9995241,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-25T07:10:26Z\",\"WARC-Record-ID\":\"<urn:uuid:3785ed46-c7f6-4d46-baaa-644346f3f376>\",\"Content-Length\":\"45001\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a6718300-dcc6-4822-8f58-ac4e223c3850>\",\"WARC-Concurrent-To\":\"<urn:uuid:7e2f9f61-a478-4e8a-b763-4a768dc74f24>\",\"WARC-IP-Address\":\"195.200.211.87\",\"WARC-Target-URI\":\"https://topanswers.xyz/tex?q=1814\",\"WARC-Payload-Digest\":\"sha1:Z7MTPNNNGLMJ7H2Q5UEMVJIXEZSR3TU3\",\"WARC-Block-Digest\":\"sha1:HAZSDBAPA57U2MAWPHFVNBCBNGHSKY4M\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030334515.14_warc_CC-MAIN-20220925070216-20220925100216-00738.warc.gz\"}"}
|
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