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https://www.excelcampus.com/functions/unique-non-adjacent-columns/
[ "", null, "# UNIQUE Formula for Non-Adjacent Columns\n\nBottom Line: Learn how to create a formula that returns unique combinations of values from non-adjacent columns.\n\nSkill Level: Intermediate\n\n## Watch the Tutorial\n\nIf you want to follow along, you can use this workbook. It's the same file I use in the video.\n\n## Pulling Unique Lists from Columns That Aren't Adjacent\n\nWe received a great question from Irene, who is an Elevate Excel member. She's curious to know if the UNIQUE function can be used to pull entries from columns that are not side by side or next to each other.\n\nThe answer to Irene's question is: on it's own, the UNIQUE function can only find unique combinations from adjacent columns. However, we can write a formula to work around that. This formula will essentially create an array of the non-adjacent columns that you want to reference in the UNIQUE function.\n\nIt's a similar process to what I explained in this tutorial on the FILTER Formula to Return Non-Adjacent Columns in Any Order.\n\nHere's how to write the formula we need:\n\n## Use INDEX to Create an Array for Non-Adjacent Columns\n\nBelow is an example where we want to use the UNIQUE function, which is a dynamic array function, to return a list of unique combinations of Customer Names and Product Names. As you can see, those two columns are not next to each other.\n\nSince the UNIQUE function does not handle that on its own, we have to use use another formula to help. So we're going to use INDEX.\n\nFor the INDEX function, the first argument is the array. We will select the entire table for the array.\n\nThe next argument is row number (row_num). To define the row numbers we want, we'll first create a list of row numbers using the SEQUENCE function.\n\nThe first argument (and the only one we need) in SEQUENCE is rows. To define rows, we will actually use the ROWS function. The only argument for the ROWS function is array, and for that, we will again reference the entire table. This will return every row from the table to our new array.\n\nTo define the column number argument (column_num), we can reference column numbers as an array in curly brackets. In our case we want to pull from columns 1 and 3. Those column numbers are specific to the table we're working with, not the sheet overall.\n\nSo, all together, our formula to return an array of non-adjacent columns looks like this:\n\n`=INDEX(tblUnique,SEQUENCE(ROWS([tblUnique]),{1,3})`\n\n## Wrap the Formula with UNIQUE\n\nNow that we've created a spill range that shows the columns we want, we simply have to wrap our existing formula with the UNIQUE function.\n\nThe UNIQUE function has other arguments, but the only one we need to use is array. Our array is defined by the formula we already created. So the final formula looks like this:\n\n`=UNIQUE(INDEX(tblUnique,SEQUENCE(ROWS([tblUnique]),{1,3}))`\n\nWith our duplicates filtered out, our spill range only shows unique entries from the two columns we wanted to pull from.\n\n## Conclusion\n\nYou can then sort the list of unique values by wrapping the formula in the SORT or SORBY function. Or you could further filter down the results with the FILTER function.\n\nThe possibilities are endless in terms of what you can do with dynamic arrays to create flexible, interactive reports. Checkout our video & post on Dynamic Array Formulas & Spill Ranges to learn more.\n\n•", null, "Paul Mwangi says:\n\nThis good insight never thought of this. Thanks\n\n•", null, "Jon Acampora says:\n\nThanks Paul!\n\n•", null, "Dan says:\n\nThis is really helpful! Could you take it 1 step further and look for counts of each of these unique combinations as well?\n\nThanks!\n\n•", null, "Jon Acampora says:\n\nHi Dan,\nGreat question! Yes, it is possible and there are several ways to go about it. With formulas you could use a COUNTIFS function.\n\n=COUNTIFS(tblUnique[Customer Name],F4,tblUnique[Product Name],G4)\n\nYou could also use a pivot table to create the summary report.\n\nI hope that helps. Thanks again and have a nice day!\n\n•", null, "Rahul Chheda says:\n\nThank you Jon for this Video. I have a question. If I want to get the dates sequence , which is only Saturday& Sunday for 3 months. How can I do this?\n\n•", null, "Jon Acampora says:\n\nHi Rahul,\nGreat question! This is also possible with dynamic array formulas. Here is a formula that uses SEQUENCE and FILTER. It assumes the start date is in cell M5 and end date in M6.\n\n=FILTER(SEQUENCE(M6-M5,,M5),WEEKDAY(SEQUENCE(M6-M5,,M5),2)>5)\n\nWe’ll add this formula to our list for future videos. The SEQUENCE function has a lot of uses with dates.\n\nThanks!\n\n•", null, "Austin says:\n\nHi, thank you so much for this! How could i take it step further and add a filter . For example, return these UNIQUE columns only when column C= “apples”\n\n•", null, "Noorol Jannah says:\n\nHi Jon,\nwhy i dont have SEQUENCE and UNIQUE function?\n\n•", null, "Neale Blackwood says:\n\nHi Jon\n\nThanks for sharing you can also use the CHOOSE function to join ranges together like this.\n\n=UNIQUE(CHOOSE({1,2},A2:A6,D2:D6))\n\n•", null, "Aashish Ilwadhi says:\n\nI can’t find Sequence function in Excel 2019. Is it replaced by some other function in this version of excel?\n\n•", null, "Maood says:\n\nThis is very helpful. If i want to sort in descending order, then what should be the formula. Thanks\n\n•", null, "Roy says:\n\nYou just reverse the values in the arrays. So where Jon used “{1,3}” to get columns A and C in that order, you’d use “{3,1}” to get them in the reverse order (C, then A).\n\nWorks the same for the rows. Just make the array in reverse order to have the bottom row at the top and so on.\n\nIf using SEQUENCE instead of writing the arrays out, do it in the last two of its four arguments. For instance, the following would create the array {1,3,5,7,9}:\n\nSEQUENCE(5,1,1,2)\n\nbut this would get the reverse ( {9,7,5,3,1} ):\n\nSEQUENCE(5,1,9,-2)\n\nOlder ways of creating these arrays used things like “ROW(1:3)” and that took a different technique. One said, OK, that’s 3 rows, so if I take the array ( {1,2,3} ) it creates and subtract it from one value higher (“4”), then I’d have the array {3,2,1} and INDEX() would turn the table upside down. Then a VLOOKUP() or similar thing would get the latest record for a match, not the earliest one. (Can’t do “ROW(3:1)” because Excel changes it, no matter what, to “ROW(1:3)”… so… thanks, Excel.\n\n•", null, "Serg says:\n\nDoesn’t work in Google Sheets. Use QUERY instead. For ex. UNIQUE(QUERY(A1:C10;”select A, C”))\n\n•", null, "pgSystemTester says:\n\nYou don’t need to use QUERY to do this in google sheets, just two ranges within another array {brackets}\n=UNIQUE({A1:A10,C1:C10})\n\nHopefully, Excel will develop something like this one of these days.\n\n•", null, "Charles says:\n\nHi, how can I make the 1 and 4 in this formula below (which are in “{ }” refer to cell inputs? I can only get it to work with “{1,4}” actually typed in the INDEX formula. Ideally I’d like the 1 and 4 to be MATCH() results.\n\n=INDEX(A1:D10,SEQUENCE(ROWS(A1:D10)),{1,4})\n\nI’ve tried “{“&1&”,”&4&”}” – with 1 and 4 being cell results, but can’t get it to return the 2 arrays that the typed {1,4} gives.\n\nI’m using this to create a single array from 2 non-adjacent arrays to be used in LINEST as the independent variables (allowing multiple regression) against a given dependent variable array.\n\nThanks\n\n•", null, "Roy says:\n\nExcel does not permit building an array constant from string materials as you are trying.\n\nIf you can figure a way to generate such an array constant from a function it would work. Excel creates many internal-to-the-calculation array constants when evaluating formulas and that would just be “one more.” But since it would really be an array constant, not a string-building construct, it would do the job.\n\nI’m not saying whatever function you use will be naturally intuitive or even seem “natural” in any way, but it will work. For example, say you are working with the formula you present, and you use XMATCH (or MATCH, both will produce an array constant if set up like this):\n\n=INDEX(A1:D10,SEQUENCE(ROWS(A1:D10)), XMATCH({1,4},{1,0,0,4},0))\n\nYou’d have to type it and get that typing right, but you have to do that to build a string anyway so the effort’s a wash. For most, you might even be able to “automate” the internal array constant that is typed here (the “{1,0,0,4}”) the same way-ish that ROWS “automates” the SEQUENCE function used in the formula.\n\nFor that, and a more complex array constant, one might just enter a formula like:\n\n=IF(a1=”x”,COLUMN(),0)\n\nin row 2 of a sheet, for however many columns that there are in your INDEX’ed table/range. In A3, place the formula:\n\n=ARRAYTOTEXT(A2:(whatever column)2).\n\nIn row 1 place an “x” in any column you need.\n\nNotice the formula in A3 is the exact value you need for your XMATCH from above. Copy it, “Paste|Special|Values” in an easy cell, edit that cell (F2) and copy it all to the Clipboard. Paste that string of the array constant into your XMATCH and you built it rather than typed it.\n\nAnd get rid of the helper stuff, of course. (Or maybe keep it in a “Useful Things” spreadsheet.)\n\n•", null, "Martin says:\n\nIs there a similar way to do it with non-adjacent rows?\n\nIt doesn’t seem to work, as the sequence function has a mandatory row argument, but I need only the column argument. I want the unique headings of two difference tables (which are themselves results of data-based filtering).\n\n•", null, "Roy says:\n\nUse “1” for the row argument.\n\nFor example, if A1:C9 has data in the odd rows and odd columns and you wish to have a tight little table with no blank rows (the even ones in this case) or columns (so, losing column B in this case), the following would do the trick:\n\n=INDEX(A1:C9,SEQUENCE(5,1,1,2),SEQUENCE(1,2,1,2))\n\nIn it, the row argument SEQUENCE() will return 5 rows starting with row 1 and then adding 2 each iteration to get 3, 5, 7, and 9. That’s the “1,2” at the end. The 5 rows is obvious: the “5” it starts with. Seems perfectly natural for the column argument to be a “1” doesn’t it? After all, you are just wanting to specify rows in this argument so a simple “1” in the columns argument of SEQUENCE() makes it plain vanilla.\n\nSame idea when working with the columns SEQUENCE() function: except here, it’s the rows argument that needs to have no impact on things so you use a “1” for it, then a real value for the columns (“2” since you want the 2 columns that have data), and finish it off with the start value (“1” for column 1) and the jump value (“2” so it gets column 3, not column 2).\n\nSo if you use SEQUENCE() on a row related argument, in INDEX() or anything else, you use a “1” in the columns value for it so the only variable has to do with rows. And the other way around if creating an array relating to columns (use the “1” in the rows argument so only the columns argument of SEQUENCE() is doing anything special).\n\n(That’s just inside functions needing row and column related arrays. Lots of other times, you’d be happy to use “non-1” values in both!)\n\n•", null, "Haresh P says:\n\nhow to add sumifs formula in below formula\n\n=UNIQUE(INDEX(tblUnique,SEQUENCE(ROWS([tblUnique]),{1,3}))\n\n•", null, "Haresh says:\n\nhow can we add Sumifs from different sheet\n\n=UNIQUE(INDEX(tblUnique,SEQUENCE(ROWS([tblUnique]),{1,3}))\n\n•", null, "Tarek says:\n\nHi, I have tried both methods (index and choose) but I only have 1 column of data. Do you know why?\n\n•", null, "pgSystemTester says:\n\nFYI the formula you have in your post is incorrect in that it doesn’t close the sequence function. It’s correct in your screenshot.\n\nYou have:\n=INDEX(tblUnique,SEQUENCE(ROWS([tblUnique]),{1,3})\n\nShould be:\n=INDEX(tblUnique,SEQUENCE(ROWS([tblUnique])),{1,3})\n\nEverything else is great, thanks for posting.\n\nGeneric filters\nExact matches only\n\n#### Excel Shortcuts List", null, "Learn over 270 Excel keyboard & mouse shortcuts for Windows & Mac.\n\nExcel Shortcuts List" ]
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https://www.quizzes.cc/calculator/weight/milligrams/1350
[ "### How much is 1350 milligrams?\n\nConvert 1350 milligrams. How much does 1350 milligrams weigh? What is 1350 milligrams in other units? How big is 1350 milligrams? Convert 1350 milligrams to lbs, kg, mg, oz, grams, and stone. To calculate, enter your desired inputs, then click calculate. Some units are rounded.\n\n### Summary\n\nConvert 1350 milligrams to lbs, kg, mg, oz, grams, and stone.\n\n#### 1350 milligrams to Other Units\n\n 1350 milligrams equals 1.35 grams 1350 milligrams equals 0.00135 kg 1350 milligrams equals 1350 mg\n 1350 milligrams equals 0.04761988748 oz 1350 milligrams equals 0.002976242967 lbs 1350 milligrams equals 0.0002125887164 stone" ]
[ null ]
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https://wiki.ivoa.net/internal/IVOA/IvoaVOQLHistoric/ADQ-Core-v0.2.xsd
[ "The only permitted root element of a query, the SELECT element The SELECT part of a query Represents the TOP part of a query abstract for Selection List List of items to be selected in the Query Represents an aggregate function count(*) Represent all columns as in Select * query The base type for alised selection item Used to select an expression as a new alias column Represents the From part of the query The base type for all tables used in the From clause of the query Represents a table, its name (required) and its alias name (required) Represents the Where part of the query The base type for searches in Where clauses of the query Represents expressions like A And B Represents expressions like A Or B The Like expression of a query Represents expressions like (A) Represents the Comparison of two expressions The Comparison operators such as Less-than or More-than, etc Represents the Between expression of a query Represents SQL IN expression The base type for selection set in a SQL IN expression Represents a list of constants provided for a SQL IN expression Represents SQL IS NULL expression The base type for a scalar expression Represents an expression inside a bracket such as (a) Represents a binary expression such as a+b Used for expressing operations like A+B Represents an unary expression such as -(a.ra) Operators for expressing a single element operation Represents a column. Table attribute represents a table alias name. The base type for all literals The base type for all numbers Represents a real number Represents an integer Represents a literal of non-number type, such as string, timestamp, boolean and so on Enumeration of allowed aggregate functions Represent a function Enumeration of allowed math functions The base type for a boolean function abstract for Region Search Represents the Regions such as circle in Where clause Restricted STC space frame type Restricted STC double1Type Equivalent to Restricted STC double2Type Restricted STC shapeType Equivalent to Restricted STC circleType Equivalent to Restricted STC circleType" ]
[ null ]
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https://investeringarovfa.web.app/8518/39729.html
[ "# Slå upp mätskala på Psykologiguiden i Natur & Kulturs\n\nMetoder för evalueringar av arbetsmarknadspolitik\n\nDuring this period, it has generated considerable controversy among stat-isticians. (nominal, ordinal, interval, and ratio) are best understood with example, as you’ll see below. Nominal Let’s start with the easiest one to understand. Nominal scales are used for labeling variables, without any quantitative value. “Nominal” scales could simply be called “labels.” Here are some examples, below. 2017-12-08 2020-12-16 Start studying nominal, ordinal, interval, ratio.\n\ntypes of markers (washable, permanent, etc.) nominal 6. time it takes to sing the National Anthem ratio 7. total annual income for statistics students ratio 8. Meetniveaus: Nominaal, Ordinaal, Interval en Ratio Wanneer je onderzoek doet heb je vaak variabelen die je hierin moet verwerken. Variabelen zijn elementen uit een onderzoek die verschillende waarden kunnen aannemen. Most mathematical operations work well on ratio values, but when interval, ordinal, or nominal values are multiplied, divided, or evaluated for the square root, the results are typically meaningless.\n\n## Statistics Learning for Data Science - vital terms – Appar på\n\n(Nominal, Ordinal, Interval, Ratio) 1. Cars described as compact, midsize, and full-size. ordinal 2. Colors of M&M candies.\n\n### ENGELSK - SVENSK - Department of Mathematics KTH", null, "Dates themselves are interval, but I could see cases where they could be any of those four.\n\nNominal Scale; Ordinal Scale; Interval Scale; Ratio Scale; Nominal Scale.\nFamilj bergengren", null, "The interval level of measurement includes all the properties of the nominal and ordinal level of measurement but it has an additional property that the difference (interval) between the values is known and constant size. In this measurement 0 is used as an arbitrary point. Nakatulong ba sa'yo ang video na 'to? You can support the channel in producing better educational content for both students and teachers. You can buy me a co (nominal, ordinal, interval, and ratio) are best understood with example, as you’ll see below.\n\nVilka är de categorical data types? Ordinal and Nominal.\nAretha franklin låtar", null, "strömsholm hästtävlingar\naktiebolaget stockholms bryggerier\nfibromyalgi triggerpunkter bild\ngul parkeringsskylt\ne after decimal\n\n### Dejt algutstorp - Eiy\n\n95 percent confidence interval:. Nominal, ordinal and ratio scales were used in the questionnaires. Variables performed during a shorter interval might improve health (14). av S HOLMBERG · 2004 · Citerat av 7 — the chi-square test for ordinal and nominal data.\n\nJoe jones mma\nvilja lied\n\n### Mellanmålsvanor i Sverige - SLU\n\nNominal scale: A scale used to label variables that have no quantitative values. Nominal Scale; Ordinal Scale; Interval Scale; Ratio Scale; Nominal Scale. A nominal scale is the 1 st level of measurement scale in which the numbers serve as “tags” or “labels” to classify or identify the objects. A nominal scale usually deals with the non-numeric variables or the numbers that do not have any value. Characteristics of Levels of measurement: Nominal, ordinal, interval, ratio.\n\n## Pin på Dagens skratt : Humor och jobb. - Pinterest\n\nNominal Let’s start with the easiest one to understand. Nominal scales are used for labeling variables, without any quantitative value. “Nominal” scales could simply be called “labels.” Here are some examples, below. Se hela listan på formpl.us 2020-04-22 · There are four basic levels: nominal, ordinal, interval, and ratio. A variable measured on a \"nominal\" scale is a variable that does not really have any evaluative distinction. One value is really not any greater than another.\n\n2019-10-03 · In the 1940s, Stanley Smith Stevens introduced four scales of measurement: nominal, ordinal, The interval level of measurement includes all the properties of the nominal and ordinal level of measurement but it has an additional property that the difference (interval) between the values is known and constant size." ]
[ null, "https://picsum.photos/800/624", null, "https://picsum.photos/800/620", null, "https://picsum.photos/800/616", null ]
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https://mihaibojin.com/coding-puzzles/arrays/sum-of-two-elements
[ "", null, "Photo by Geran de Klerk on Unsplash\n\n## The two-sum puzzle\n\nGiven an array of integers and an integer target, return indices of the two numbers such that they add up to the target.\n\nSumming up two elements to add to a specified integer is a classic (and basic) interview problem.\n\nA good interviewer will discuss a few variants for this problem explain time-complexity and space complexity implications, identify edge cases, write a few test cases and write working code.\n\nLet's look at an input and its expected output:\n\n``````// input\nint[] data = {1, 3, -1, 5, 8, 10};\nint sum = 9;\n\n// output\nint[] solution = {0, 4};``````\n\nThe two elements, at indexes `0` and `4`, sum up to the searched value `9`.\n\nLet's see some code.\n\n## Simplest solution\n\nThe simplest solution would be to iterate over each element in the array twice and stop when we found a solution.\n\n``````public static int[] findSumTwo(int[] data, int sum) {\n// iterate until the next to last element\nfor (int i = 0; i < data.length-1; i++) {\n// iterate to the last element, starting from the second element\nfor (int j = 1; i < data.length; j++) {\n// (the double loop will start from the\n// {0, 1} index pair and stop iterating at {n-1, n})\n\n// if the elements add up, return their indices\nif (data[i]+data[j] == sum) return new int[]{i,j};\n}\n}\n\nreturn new int[]{-1};\n}``````\n\nThis implementation has terrible performance (`O(N^2)`) since we could be looking for `18`, and the result would be the last two numbers in the array.\n\nIt works, but I wouldn't use it in an interview.\n\nWe can do better!\n\n## A fast but costly solution\n\nSince we're only looking at two changing values, we can precompute differences to the sum we've already seen.\n\nGoing through our initial array (`{1, 3, -1, 5, 8, 10}`), we can compute and store the difference to the sum for each element; given the desired value of `9`, we can add the following to a set:\n\n• `9-1 = 8`\n• `9-3 = 6`\n• `9-(-1) = 10`\n• `9-5 = 4`\n• `9-8`...\n\nWhen we reach `8` at position 4, we can determine that it forms a solution pair since it already exists in our set (position 0).\n\nWe can now stop and return.\n\nBut since the problem is asking for indexes, we also need to store the position of the elements making up the solution. We can do this by using a `HashMap<Integer, Integer>` data structure or by using an extra `int[]` array.\n\nLet's see some code:\n\n``````public static int[] findSumTwo(int[] data, int sum) {\n// value->index map\nMap<Integer, Integer> seen = new HashMap<>();\n\n// iterate through the array once (O(N))\nfor (int pos=0; i < data.length; i++) {\n// assigning the element for convenience\nint element = data[i];\n\n// if we've already seen the current element\nif (seen.containsKey(element)) {\n// form a solution\nreturn new int[]{seen.get(element), pos};\n}\n\n// otherwise mark its difference as 'seen'\nseen.put(sum-element, i);\n}\n\nreturn new int[]{-1};\n}``````\n\nThis looks pretty clean, but what's the cost?\n\nWe're only iterating through the input once, so time-complexity-wise it's O(N), but we're also storing all the differences in the `seen` map.\n\nWe are accruing a space-complexity code of at least O(N).\n\nWith a sufficiently large input, this would take up a lot of extra heap, or it wouldn't work at all.\n\nCan we do better?\n\n## No extra space-complexity\n\nYou might have noticed that the input array in the example above is already sorted.\n\nThis may be a trap, don't fall for it! Don't assume things and ask clarifying questions to determine what the interviewer is interested in. (I will be writing more about interview best practices in a later article.)\n\nIf the input is sorted, we can rely on a bit of math to find a more optimal solution. If we chose two starting indices, one at the beginning of the array (`left`) and one at the end (`right`), we could be sure that incrementing the left index or decrementing the right index would result in either a larger sum or a lower sum, respectively.\n\nGiven `left + right = found`, by comparing `found` to `sum`, we would know that the searched element needed to be larger, smaller, or the one we were looking for.\n\nHere's how the implementation looks like:\n\n``````public static int[] findSumTwo(int[] data, int sum) {\nint left=0;\nint right=data.length-1;\n\n// only iterate until the positions are valid\n// O(N)\nwhile (left < right) {\nint found = data[left] + data[right];\n\nif (found == sum) {\n// we have found a solution\nreturn new int[]{left, right};\n} else if (found < sum) {\n// we need a larger number\nleft++;\n} else {\n// or a lower one\nright--;\n}\n}\n\nreturn new int[]{-1};\n}``````\n\nSince we can't assume the input is already sorted, we would also need to sort it:\n\n``````// most sorting algorithms run in O(N*lgN) time\nArrays.sort(data);``````\n\nAnd there you have it - Another solution to this problem.\n\nAt this point, you may be congratulating yourself for a job well done. Don't...\n\n## Edge cases\n\nNever call a coding interview complete before talking about edge-cases! Figuring out how your code can break is a crucial trait of a talented software engineer!\n\nLet's consider some inputs:\n\n``````// empty input\nint[] data = {};\nint sum = /*irrelevant*/;\n\n// null input\nint[] data = null;\nint sum = /*irrelevant*/;\n\n// overflow\nint[] data = {Integer.MAX_VALUE, Integer.MAX_VALUE};\nint sum = -2; // doesn't seem correct, but it is given the Integer input\n\n// not enough elements\nint[] data = {9};\nint sum = 9; // we need at least two elements for a valid solution\n\n// equal elements\nint[] data = {8,8};\nint sum = 16; // this would cause the first solution to wrongly return {0,0}\n\n// unsorted input\nint[] data = {3,1};\nint sum = 4; // the correct solution is {0,1}\n\n// no solution\nint[] data = {1,1};\nint sum = 3; // no pair can sum up to 3``````\n\nA few tips to keep in mind:\n\n• always assume the input can be tainted unless explicitly told otherwise\n• think of null values and how they can break the code\n• do not assume the input is sorted\n• clarify what the output should look like\n• avoid printing outputs - it's hard to test and results in messy code\n• understand the consequences of making a choice (e.g., \"if I need to sort the input, the time-cost is at least N*lgN)\n• explicitly state assumptions (e.g., \"I assume the input is sorted\", or \"I assume the input has a valid solution\")\n• show a good understanding of big-O notation and algorithmic performance\n\nThat's all for now!\n\nI'll leave you with some food for thought: How would you solve this problem if the input could not fit in the memory of a single instance?" ]
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http://www.expertsmind.com/questions/determine-series-is-convergent-or-divergent-by-root-test-30153528.aspx
[ "## Determine series is convergent or divergent by root test, Mathematics\n\nAssignment Help:\n\nFind out if the following series is convergent or divergent.", null, "Solution\n\nThere really is not very much to these problems another than calculating the limit and then using the root test.  Here is the boundary for this problem.", null, "Thus, by the Root Test this series is divergent.\n\n#### Statistics Assignment, I need help in assignment of stats? Please give me a...\n\nI need help in assignment of stats? Please give me assist in my stats exam.\n\n#### Lance has 70 cents margaret has 3/4 who has the most money, Lance has 70 ce...\n\nLance has 70 cents, Margaret has three-fourths of a dollar, Guy has two quarters and a dime, and Bill has six dimes. Who has the most money? Lance has 70 cents. Three-fourths o\n\n#### Proportional Relationships, Carmen bought 3 pounds of bananas for \\$1.08. Ju...\n\nCarmen bought 3 pounds of bananas for \\$1.08. June paid for her purchase of bananas. If they paid the same price per pound, how many pounds did June buy?\n\n#### Math help until tuesday, I need help with pre algebra in 5th grade intermid...\n\nI need help with pre algebra in 5th grade intermidate school math until Tuesday afternoon please\n\n#### Assignment Marketing Mix, How to do assignment Marketing Mix\n\nHow to do assignment Marketing Mix\n\n(19 + 7 i)\n\n#### Prove that ac2 =ab2 + bc2+2bcxbd, If ABC is an obtuse angled triangle, obtu...\n\nIf ABC is an obtuse angled triangle, obtuse angled at B and if AD⊥CB Prove that AC 2 =AB 2 + BC 2 +2BCxBD Ans:    AC 2 = AD 2 + CD 2 = AD 2 + (BC + BD) 2 = A\n\n#### Multiplyig, why is multiplying inportent in our lifes\n\nwhy is multiplying inportent in our lifes\n\n#### Mensuration, In an equilateral triangle 3 coins of radius 1cm each are kept...\n\nIn an equilateral triangle 3 coins of radius 1cm each are kept along such that they touch each other and also the side of the triangle. Determine the side and area of the triangle.\n\n#### Differentiate functions f ( x ) = 15x100 - 3x12 + 5x - 46, Differentiate f...\n\nDifferentiate following functions. (a) f ( x ) = 15x 100 - 3x 12 + 5x - 46 (b) h ( x ) = x π   - x √2  Solution (a)    f ( x ) = 15x 100 - 3x 12 + 5x - 46 I", null, "", null, "" ]
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https://math.stackexchange.com/questions/3150392/find-the-solution-to-the-recurrence-relation-a-n-3a-n%E2%88%921-4a-n%E2%88%922-with-i
[ "# Find the solution to the recurrence relation $a_{n} = 3a_{n−1} +4a_{n−2}$ with initial terms $a_{0}=5$ and $a_{1} =8$\n\nFor the question\n\nFind the solution to the recurrence relation $$a_{n} = 3a_{n−1} +4a_{n−2}$$ with initial terms $$a_{0}=5$$ and $$a_{1} =8$$\n\nI think the way to solve this is the Characteristic Root Technique since the recurrence relation is a combination of two previous terms.\n\nSo the characteristic polynomial is $$x^2 + 3x + 4 =0$$. In the example solutions I have seen the two distinct characteristic roots should be easy to find, but for this question, I don't know how to move forward. If I do completing the square then I get $$(x+\\frac{3}{2})^2 + \\frac{7}{4}=0$$.\n\nI am stuck here. How do I find the solution?\n\n• Do you remember the quadratic formula from Algebra I? – amd Mar 16 '19 at 18:17\n\nFor the given recurrence relation $$a_{n} -3a_{n−1} -4a_{n−2}=0$$, the characteristic polynomial is $$x^2 - 3x - 4 =0$$, so we have to solve $$x^2 - 3x - 4=\\left(x-\\frac{3}{2}\\right)^2 - \\left(\\frac{5}{2}\\right)^2=(x+1)(x-4)=0.$$ Therefore $$a_n=C_1 \\cdot 4^n+ C_2\\cdot (-1)^n$$ where $$C_1$$ and $$C_2$$ are constants to be determined. Now find such constants by using the initial terms $$a_{0}=5$$ and $$a_{1} =8$$." ]
[ null ]
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https://learn.mindset.africa/resources/information-technology/grade-10/data-and-information-management/data-representation/conversions-using-binary-and-decimal-no-systems
[ "# Conversions using Binary and Decimal No. Systems\n\n.\n182 | 1 | 0\nIn this lesson we use a pattern of numbers to generate a formula for working out the number of combinations for n bits. We also demonstrate how to convert decimal numbers to binary and using division to convert from binary to decimal numbers.\nLearner Video" ]
[ null ]
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https://www.colorhexa.com/00de08
[ "# #00de08 Color Information\n\nIn a RGB color space, hex #00de08 is composed of 0% red, 87.1% green and 3.1% blue. Whereas in a CMYK color space, it is composed of 100% cyan, 0% magenta, 96.4% yellow and 12.9% black. It has a hue angle of 122.2 degrees, a saturation of 100% and a lightness of 43.5%. #00de08 color hex could be obtained by blending #00ff10 with #00bd00. Closest websafe color is: #00cc00.\n\n• R 0\n• G 87\n• B 3\nRGB color chart\n• C 100\n• M 0\n• Y 96\n• K 13\nCMYK color chart\n\n#00de08 color description : Pure (or mostly pure) lime green.\n\n# #00de08 Color Conversion\n\nThe hexadecimal color #00de08 has RGB values of R:0, G:222, B:8 and CMYK values of C:1, M:0, Y:0.96, K:0.13. Its decimal value is 56840.\n\nHex triplet RGB Decimal 00de08 `#00de08` 0, 222, 8 `rgb(0,222,8)` 0, 87.1, 3.1 `rgb(0%,87.1%,3.1%)` 100, 0, 96, 13 122.2°, 100, 43.5 `hsl(122.2,100%,43.5%)` 122.2°, 100, 87.1 00cc00 `#00cc00`\nCIE-LAB 77.434, -77.48, 74.175 26.164, 52.257, 8.937 0.299, 0.598, 52.257 77.434, 107.261, 136.248 77.434, -73.263, 94.303 72.289, -61.901, 43.272 00000000, 11011110, 00001000\n\n# Color Schemes with #00de08\n\n• #00de08\n``#00de08` `rgb(0,222,8)``\n• #de00d6\n``#de00d6` `rgb(222,0,214)``\nComplementary Color\n• #67de00\n``#67de00` `rgb(103,222,0)``\n• #00de08\n``#00de08` `rgb(0,222,8)``\n• #00de77\n``#00de77` `rgb(0,222,119)``\nAnalogous Color\n• #de0067\n``#de0067` `rgb(222,0,103)``\n• #00de08\n``#00de08` `rgb(0,222,8)``\n• #7700de\n``#7700de` `rgb(119,0,222)``\nSplit Complementary Color\n• #de0800\n``#de0800` `rgb(222,8,0)``\n• #00de08\n``#00de08` `rgb(0,222,8)``\n• #0800de\n``#0800de` `rgb(8,0,222)``\n• #d6de00\n``#d6de00` `rgb(214,222,0)``\n• #00de08\n``#00de08` `rgb(0,222,8)``\n• #0800de\n``#0800de` `rgb(8,0,222)``\n• #de00d6\n``#de00d6` `rgb(222,0,214)``\n• #009205\n``#009205` `rgb(0,146,5)``\n• #00ab06\n``#00ab06` `rgb(0,171,6)``\n• #00c507\n``#00c507` `rgb(0,197,7)``\n• #00de08\n``#00de08` `rgb(0,222,8)``\n• #00f809\n``#00f809` `rgb(0,248,9)``\n• #12ff1b\n``#12ff1b` `rgb(18,255,27)``\n• #2cff33\n``#2cff33` `rgb(44,255,51)``\nMonochromatic Color\n\n# Alternatives to #00de08\n\nBelow, you can see some colors close to #00de08. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #30de00\n``#30de00` `rgb(48,222,0)``\n• #1dde00\n``#1dde00` `rgb(29,222,0)``\n• #0bde00\n``#0bde00` `rgb(11,222,0)``\n• #00de08\n``#00de08` `rgb(0,222,8)``\n• #00de1b\n``#00de1b` `rgb(0,222,27)``\n• #00de2d\n``#00de2d` `rgb(0,222,45)``\n• #00de3f\n``#00de3f` `rgb(0,222,63)``\nSimilar Colors\n\n# #00de08 Preview\n\nThis text has a font color of #00de08.\n\n``<span style=\"color:#00de08;\">Text here</span>``\n#00de08 background color\n\nThis paragraph has a background color of #00de08.\n\n``<p style=\"background-color:#00de08;\">Content here</p>``\n#00de08 border color\n\nThis element has a border color of #00de08.\n\n``<div style=\"border:1px solid #00de08;\">Content here</div>``\nCSS codes\n``.text {color:#00de08;}``\n``.background {background-color:#00de08;}``\n``.border {border:1px solid #00de08;}``\n\n# Shades and Tints of #00de08\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, #000600 is the darkest color, while #f2fff2 is the lightest one.\n\n• #000600\n``#000600` `rgb(0,6,0)``\n• #001a01\n``#001a01` `rgb(0,26,1)``\n• #002d02\n``#002d02` `rgb(0,45,2)``\n• #004102\n``#004102` `rgb(0,65,2)``\n• #005503\n``#005503` `rgb(0,85,3)``\n• #006804\n``#006804` `rgb(0,104,4)``\n• #007c04\n``#007c04` `rgb(0,124,4)``\n• #009005\n``#009005` `rgb(0,144,5)``\n• #00a306\n``#00a306` `rgb(0,163,6)``\n• #00b707\n``#00b707` `rgb(0,183,7)``\n• #00ca07\n``#00ca07` `rgb(0,202,7)``\n• #00de08\n``#00de08` `rgb(0,222,8)``\n• #00f209\n``#00f209` `rgb(0,242,9)``\n• #06ff0f\n``#06ff0f` `rgb(6,255,15)``\n• #1aff22\n``#1aff22` `rgb(26,255,34)``\n• #2dff35\n``#2dff35` `rgb(45,255,53)``\n• #41ff48\n``#41ff48` `rgb(65,255,72)``\n• #55ff5b\n``#55ff5b` `rgb(85,255,91)``\n• #68ff6e\n``#68ff6e` `rgb(104,255,110)``\n• #7cff81\n``#7cff81` `rgb(124,255,129)``\n• #90ff94\n``#90ff94` `rgb(144,255,148)``\n• #a3ffa6\n``#a3ffa6` `rgb(163,255,166)``\n• #b7ffb9\n``#b7ffb9` `rgb(183,255,185)``\n• #caffcc\n``#caffcc` `rgb(202,255,204)``\n• #deffdf\n``#deffdf` `rgb(222,255,223)``\n• #f2fff2\n``#f2fff2` `rgb(242,255,242)``\nTint Color Variation\n\n# Tones of #00de08\n\nA tone is produced by adding gray to any pure hue. In this case, #667867 is the less saturated color, while #00de08 is the most saturated one.\n\n• #667867\n``#667867` `rgb(102,120,103)``\n• #5e805f\n``#5e805f` `rgb(94,128,95)``\n• #558957\n``#558957` `rgb(85,137,87)``\n• #4d914f\n``#4d914f` `rgb(77,145,79)``\n• #449a47\n``#449a47` `rgb(68,154,71)``\n• #3ca23f\n``#3ca23f` `rgb(60,162,63)``\n• #33ab38\n``#33ab38` `rgb(51,171,56)``\n• #2bb330\n``#2bb330` `rgb(43,179,48)``\n• #22bc28\n``#22bc28` `rgb(34,188,40)``\n• #1ac420\n``#1ac420` `rgb(26,196,32)``\n• #11cd18\n``#11cd18` `rgb(17,205,24)``\n• #09d510\n``#09d510` `rgb(9,213,16)``\n• #00de08\n``#00de08` `rgb(0,222,8)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #00de08 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 ]
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https://academia.stackexchange.com/questions/98580/how-to-cite-an-equation-that-is-common-knowledge-but-i-use-others-notation
[ "# how to cite an equation that is common knowledge but I use others notation\n\nI am writing my masters thesis on an econometrics topic. It is very closely related to another paper by NW. Since our papers are very closely related, I am using their notation in order to draw comparisons (which I state before introducing my model).\n\nNW motivate their model using known findings from their field. They demonstrate that the standard model does not behave well under their assumptions by (mathematically) decomposing the standard model's objective function. This approach is very common, however, every author formats their arguments to fit their purpose. In my paper, I want to build on NWs argument. I.e. I want to make the same point and extend it. Do I need to cite their equation (the decomposition of the objective function) in any special way?\n\nTo paraphrase what I currently have:\n\n... NW illustrate this by deconstructing the expectation of the objective function into a signal term and a noise term\n\n$$E[Q(\\beta)] = Eg(\\beta)'Wg(\\beta) + tr(W Omega(\\beta))$$\n\n...\n\n(I copy their notation exactly.)\n\nThe reason I am unsure is that NWs argumentation is not unique to their paper but how they frame it is." ]
[ null ]
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http://mantascode.com/java-lol-gamification-of-a-dataset-d/
[ "# Java: LOL!: Gamification of a DataSET :D\n\nRandomized multiple choice quiz of the relationship sets of Fallout 2 Characters and their Locations.\n\n(Character (is from/is related to) Location)\n(Concept 1 -> Relationship -> Concept 2)\nExample:\n( Sulik from Klamath)", null, "Now we want to ask a random question, so we get a random number between 1 and 13 , and pick that row in the data set.\n\nWe roll the dice and get “1” so we take the first data row and ask\n\n“Where is Sulik Found?”\n\nThis quiz is going to be multiple choice of 4, so we need 3 random incorrect answers, and we need to randomize the correct answer’s location somewhere within the set of incorrect answers.\n\nMake a List of 3 unique/distinct incorrect answers out of the remaining values within the hashtable.  Again, exclude the correct answer  and handle any duplicates.  So in Sulik‘s case, the distinct set of potential incorrect answers looks like this.", null, "Pick another random number and pick 3 distinct values from this list.  Then, append the 3 incorrect values with the correct one.  Pick another random random number 1-4 and display the randomized answers 4 times for “a.)” through “d.)“.\n\ncode for now…\n\n```import java.util.Enumeration; import java.util.Hashtable; import java.util.LinkedList; import java.util.ListIterator; import java.util.Random;   public class QuizStructure { public static void main(String[] args) { LinkedList EveryPotentialQuestion_AnswerSet = new LinkedList();   //HOW TO MAKE A Randomized QUIZ off OF a DataSet   //Gamification of a DataSET// LOL :D   System.out.println(\"indeed.\"); Hashtable Character_to_Location = new Hashtable();   //\"FALLOUT 2 Black Isle Character to Location\" Relationship// //( Character - Is_From - Location )// //( Concept - Relationship - Concept ) Character_to_Location.put(\"Sulik\", \"Klamath\"); Character_to_Location.put(\"Vic\", \"Den\"); Character_to_Location.put(\"Miria\", \"Modoc\"); Character_to_Location.put(\"Davin\", \"Modoc\"); Character_to_Location.put(\"Cassidy\", \"Vault City\"); Character_to_Location.put(\"Lenny\", \"Gecko\"); Character_to_Location.put(\"Marcus\", \"Broken Hills\"); Character_to_Location.put(\"Myron\", \"New Reno Stables\"); Character_to_Location.put(\"Skynet Robot\", \"Sierra Army Depot\"); Character_to_Location.put(\"K-9\", \"Navarro\"); Character_to_Location.put(\"Goris\", \"Vault 13\"); Character_to_Location.put(\"Cyber Dog\", \"NCR\"); Character_to_Location.put(\"Dogmeat\", \"Cafe of Broken Dreams\"); System.out.println(Character_to_Location.toString()); Enumeration keys = Character_to_Location.keys(); while( keys.hasMoreElements() ) { Object key = keys.nextElement(); Object value = Character_to_Location.get(key); System.out.println(\"Question/Answer Preview: Where is \"+key+\" from in Fallout 2? Answer : \"+value); } Random randomGenerator = new Random(); System.out.println(\"size: \"+Character_to_Location.size()); System.out.println(\"___________________________________\"); //ask Ten randomized questions with 4 randomized answers for( int i = 0 ; i &lt; 10 ; i++) { int randomInt = randomGenerator.nextInt(Character_to_Location.size()); randomInt++; int iStop = 0; Enumeration morekeys = Character_to_Location.keys();   while( morekeys.hasMoreElements() ) { Object key = morekeys.nextElement(); Object value = Character_to_Location.get(key); if ( iStop == randomInt) { //System.out.println(\"index:\" + iStop + \" \"+randomInt); System.out.println(); System.out.println(\"Where is \"+key+\" from?\"); //Generate Multiple Choice //we know the correct answer is System.out.println(\" Correct Answer : \" + value); //now we need to generate other \"incorrect\" Choices //we need 3, Multiple Choice is Typically 4 questions LinkedList wrongAnswerList = new LinkedList(); //while wrongAnswerList is not unique //we want to remove the possibility of duplicates //we want to not include correct answer boolean is_wronglist_unique = false;   while(!is_wronglist_unique) { int randomWrong = randomGenerator.nextInt(Character_to_Location.size()); randomWrong++; Enumeration wrongkeys = Character_to_Location.keys(); int iStopInner =0; while( wrongkeys.hasMoreElements() ) { Object key2 = wrongkeys.nextElement(); Object value2 = Character_to_Location.get(key2); if ( iStopInner == randomWrong) { if ( !wrongAnswerList.contains(value2)) { if(!value2.equals(value)) wrongAnswerList.add(value2); }   } iStopInner++; } if (wrongAnswerList.size() ==3) {is_wronglist_unique = true;} } System.out.println(\" Incorrect Answers : \"+wrongAnswerList.toString()); System.out.println(); wrongAnswerList.add(value); System.out.println(); String [] AnswerSet = (String[]) wrongAnswerList.toArray(new String); //get 4 random ints :D //0-3 LinkedList numberset = new LinkedList(); boolean isnumbersetunique = false; int deerpCount = 0; while(!isnumbersetunique) { int randomfinal4 = randomGenerator.nextInt(4); if (!numberset.contains(randomfinal4)) { numberset.add(randomfinal4); deerpCount++; } if(deerpCount == 4) { isnumbersetunique = true; String individualquestion =\"\"; System.out.println(\"** FINAL QUESTION *** Where is \"+key+\" from?\"); individualquestion += \"** FINAL QUESTION *** Where is \"+key+\" from?\"; ListIterator itr = numberset.listIterator(); String[] abcd = {\"a.)\",\"b.)\",\"c.)\",\"d.)\"}; int ix = 0; while(itr.hasNext()) { int get =(int) itr.next(); System.out.println(\" \" +abcd[ix]+\" \"+ AnswerSet[get]); individualquestion += \" \" +abcd[ix]+\" \"+ AnswerSet[get]; ix++; }   if ( !EveryPotentialQuestion_AnswerSet.contains(individualquestion)) { EveryPotentialQuestion_AnswerSet.add(individualquestion); //how do I know when this ^ is full ?????? ??????? //What is the most optimal way to answer this ^ without doing it manually? } } } wrongAnswerList.clear(); } iStop++; } } //the more questions you ask the more answers you'll get } }```\n\nOutput Example:", null, "What is the total number of distinct Questions and Possible combinations of Answers?  How many unique questions could be asked? Submit your answers below :D" ]
[ null, "http://mantascode.com/wp-content/uploads/2012/04/f2ds.png", null, "http://mantascode.com/wp-content/uploads/2012/04/uniquiewrong.png", null, "http://mantascode.com/wp-content/uploads/2012/04/outputgamificationDataset.png", null ]
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https://blog.tonycube.com/2015/01/laravel-15-query-builder.html
[ "# Laravel 學習筆記(15) - 資料庫之 Query Builder", null, "Query Builder\n\nLaravel 提供了方便易用的資料庫查詢機制,基於 PDO 參數綁定(parameter binding),保護應用程式免於 SQL資料隱碼 (SQL injection) 攻擊。\n\n### Selects(查詢)\n\n#### 取得整個資料表\n\n``````\\$posts = DB::table('posts')->get();\n\nforeach (\\$posts as \\$post) {\nvar_dump(\\$post->title);\n}\n``````\nforeach 中的 \\$post 代表一個資料列,`\\$post->title` 為這個列的名稱為 title 欄位的值。\n\n#### 取得單一資料列\n\n``````\\$post = DB::table('posts')->where('id', '=', 1)->first();\nvar_dump(\\$post->title);\n``````\n\n#### 取得單一資料列中的單一欄位\n\n``````\\$title = DB::table('posts')->where('id', '=', 1)->pluck('title');\nvar_dump(\\$title);\n``````\npluck 英文為摘取的意思,這裡就是從資料列中摘取 title 這個欄位的值。查看原始碼,pluck 已經做了 first() 的動作,所以只會取單一資料列。\n\n#### 取得多資料列中的單一欄位\n\n``````\\$titles = DB::table('posts')->lists('title');\nvar_dump(\\$titles);\n``````\n\n#### 查詢子句\n\n``````\\$posts = DB::table('posts')->select('id', 'title')->get();\n\\$posts = DB::table('posts')->distinct()->get();\n\\$posts = DB::table('posts')->select('title as subject')->get();\n``````\n\n#### 在查詢結果中,再加入查詢\n\n``````\\$query = DB::table('posts')->select('title');\n``````\n\n#### where 語句\n\n``````\\$posts = DB::table('posts')->where('id', '=', 1)->get();\n``````\nwhere 有 3 個參數,(欄位名稱, 運算子, 值),前面個參數有單引號包住,運算子可以用 '=', '<', '>', '>=', '<=', '<>' 等等,第 3 個參數是要比對的值,要注意的是,數值不加引號,字串、日期等則要加。\n\n##### or 多個條件:\n``````\\$posts = DB::table('posts')->where('id', '=', 1)\n->orWhere('title', '111')\n->get();\n``````\n\n``````SELECT * FROM posts WHERE (id = 1) or (title = '111');\n``````\n\n##### between 範圍:\n``````\\$posts = DB::table('posts')->whereBetween('id', [2,4])->get();\n``````\n\n``````SELECT * FROM posts WHERE id BETWEEN 2 AND 4;\n``````\n\n``````\\$posts = DB::table('posts')->whereNotBetween('id', [2,4])->get();\n``````\n\n``````SELECT * FROM posts WHERE id NOT BETWEEN 2 AND 4;\n``````\n\n#### In 某個(或多個)值\n\n``````\\$posts = DB::table('posts')->whereIn('id', [1,3,5])->get();\n``````\n\n``````SELECT * FROM posts WHERE id IN (1,3,5);\n``````\n\n``````\\$posts = DB::table('posts')->whereNotIn('id', [1,3,5])->get();\n``````\n\n``````SELECT * FROM posts WHERE id NOT IN (1,3,5);\n``````\n\n#### Order By 排序\n\n``````\\$posts = DB::table('posts')->orderBy('id', 'desc')->get();\n``````\n\n``````SELECT * FROM posts ORDER BY id DESC;\n``````\n\n#### Group By 群組及 Having 條件\n\n``````\\$posts = DB::table('posts')->groupBy('tag')->having('words','>', 100)->get();\n``````\n\nhaving 功能和 where 相同,只是 having 可以用在運算結果(這裡指群組化,有時候可能會是加總之類的)之後,而 where 不能。\n\n``````SELECT * FROM posts GROUP BY tag HAVING words > 100;\n``````\n\n#### Offset & Limit\n\n``````\\$posts = DB::table('posts')->skip(5)->take(3)->get();\n``````\nskip(5)表示,前面 5 筆資料不要,從第 6 筆開始取,而且 take(3) 只取 3 筆。也就是 6, 7, 8 筆。\n\n``````\\$posts = DB::table('posts')->take(1)->get();\n``````\n\n#### Joins 結合資料表\n\n##### 基本結合\n\n``````\\$posts = DB::table('posts')\n->get();\n``````\njoin() 方法的第 1 個參數表示要結合的另一個表格的名稱;第 2 到 4 個參數,則是這兩個表格要比較的欄位。\n\n``````SELECT posts.id, posts.title, comments.content FROM posts, comments WHERE posts.id = comments.post_id;\n``````\n\n##### LEFT JOIN 左外部連接\n``````\\$posts = DB::table('posts')\n->get();\n``````\n\n``````SELECT posts.id, posts.title, comments.content FROM posts LEFT JOIN comments ON posts.id = comments.post_id;\n``````\n\n#### 進階 Where 條件\n\n``````\\$posts = DB::table('posts')\n->where('title', '=', '111')\n->orWhere(function(\\$query)\n{\n\\$query->where('words', '>', 100)\n->where('tag', '=', 'php');\n})\n->get();\n``````\n\n``````SELECT * FROM posts WHERE title = '111' or (words > 100 and tag = 'php');\n``````\n\n#### Aggregates 集合方法(函數)\n\n``````//計算資料筆(列)數\n\\$postCount = DB::table('posts')->count();\n//SELECT count(*) FROM posts;\n\n//取 words 欄位最大值\n\\$maxWords = DB::table('posts')->max('words');\n//SELECT max(words) FROM posts;\n\n//取 words 欄位最小值\n\\$minWords = DB::table('posts')->min('words');\n//SELECT min(words) FROM posts;\n\n//計算 words 欄位平均值\n\\$avgWords = DB::table('posts')->avg('words');\n//SELECT avg(words) FROM posts;\n\n//計算 words 欄位總計值\n\\$sumWords = DB::table('posts')->sum('words');\n//SELECT sum(words) FROM posts;\n``````\n\n#### Raw Expressions 原生表達式\n\n``````\\$posts = DB::table('posts')\n->select(DB::raw('count(*) as post_count'))\n->get();\n``````\n\n``````SELECT count(*) as post_count FROM posts;\n``````\n\n### Inserts (新增)\n\n``````DB::table('posts')->insert(\n['title'=>'Hello!', 'content'=>'Laravel Demo~']\n);\n``````\n\n``````INSERT INTO posts (title, content) VALUES ('Hello!', 'Laravel Demo~');\n``````\n\n``````\\$id = DB::table('posts')->insertGetId(\n['title'=>'Hello 2!', 'content'=>'Laravel Demo~']\n);\n\nvar_dump(\\$id);\n``````\n\n``````DB::table('posts')->insert([\n['title'=>'Hello! 1', 'content'=>'Laravel Demo~'],\n['title'=>'Hello! 2', 'content'=>'Laravel Demo~'],\n['title'=>'Hello! 3', 'content'=>'Laravel Demo~']\n]);\n``````\n\n``````DB::table('posts')\n->where('id', 1)\n->update(['title'=>'Hi!']);\n``````\n\n``````UPDATE posts SET title = 'Hi!' WHERE id = 1;\n``````\n\n``````DB::table('posts')->increment('words');\nDB::table('posts')->increment('words', 10);\nDB::table('posts')->decrement('words');\nDB::table('posts')->decrement('words', 10);\n``````\n\n``````UPDATE posts SET words = words + 1;\nUPDATE posts SET words = words + 10;\nUPDATE posts SET words = words - 1;\nUPDATE posts SET words = words - 10;\n``````\n\n``````DB::table('posts')\n->where('id',1)\n->increment('words', 1);\n``````\n\n``````UPDATE posts SET words = words + 1 WHERE id = 1;\n``````\n\n### Deletes (刪除)\n\n``````DB::table('posts')\n->where('words', '>', 200)\n->delete();\n``````\n\n``````DELETE FROM posts WHERE words > 200;\n``````\n\n``````DB::table('posts')->delete();\n``````\n\n``````DELETE FROM posts;\n``````\n\n``````DB::table('posts')->truncate();\n``````\n\n``````TRUNCATE TABLE posts;\n``````\n\nTony Blog 撰寫,請勿全文複製,轉載時請註明出處及連結,謝謝 😀\n\n#### 2 則留言\n\n1.", null, "作者已經移除這則留言。\n\n2.", null, "您好,首先謝謝您的系列教學讓我收穫很多!\n\n有些問題想請教一下,我目前使用最新版的laravel,安裝 VS Code、Mac OS\n\nLaravels內大部分的函數都可以透過cmd + 滑鼠右鍵連到原始檔看程式碼,唯獨DB操作相關的皆無法連到原始檔看程式碼? 我有試過使用 PHPStorm也有一樣的問題。\n\n舉例來說:\nDB::insert(.......); 我想查看DB和 insert的原始碼、函數的用途、要放哪些參數,完全看不到。\n\nXXModel::orderBy(......)->get(); 我想查看orderBy, get的原始碼和函數的用途、參數,也是都看不到。" ]
[ null, "https://3.bp.blogspot.com/-WKOjA0lADyw/VK45AGnioaI/AAAAAAAAEPs/g84WlizZWGE/s1600/laravel.png", null, "https://lh5.googleusercontent.com/-nRKQAC7iy_E/AAAAAAAAAAI/AAAAAAAAAD8/qu9TNoKAnaY/s35-c/photo.jpg", null, "https://lh5.googleusercontent.com/-nRKQAC7iy_E/AAAAAAAAAAI/AAAAAAAAAD8/qu9TNoKAnaY/s35-c/photo.jpg", null ]
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https://downloads.hindawi.com/journals/jam/2012/257140.xml
[ "JAM Journal of Applied Mathematics 1687-0042 1110-757X Hindawi Publishing Corporation 257140 10.1155/2012/257140 257140 Research Article The Global Existence of Nonlinear Evolutionary Equation with Small Delay Yin Xunwu Yao Yonghong School of Science Tianjin Polytechnic University Tianjin 300387 China tjpu.edu.cn 2012 10 8 2012 2012 11 04 2012 09 05 2012 2012 Copyright © 2012 Xunwu Yin. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.\n\nWe investigate the global existence of the delayed nonlinear evolutionary equation tu+Au=f(u(t),u(tτ)). Our work space is the fractional powers space Xα. Under the fundamental theorem on sectorial operators, we make use of the fixed-point principle to prove the local existence and uniqueness theorem. Then, the global existence is obtained by Gronwall’s inequality.\n\n1. Introduction\n\nOn the existence for solutions of evolutionary equations, there are many works and methods . For example, the fixed principle [1, 35, 7] and Galerkin approximations [2, 6]. They are very classical methods to prove existence and uniqueness. Generally speaking, there are four solution concepts. That is, weak solution, mild solution, strong solution, and classical solution. We can obtain different types for different conditions. For instance , consider the following inhomogeneous initial value problem: (1.1)u(t)+Au=f(t),t>0,u(0)=xX, where X is Banach space. If the nonlinearity fL1(0,T;X), the initial value problem has a unique mild solution. If the nonlinearity f is differentiable a.e. on [0,T] and fL1(0,T;X), then for every xD(A) the initial value problem has a unique strong solution. Furthermore, if the nonlinearity fL1(0,T;X) is locally Hölder continuous, then the initial value problem has a unique classical solution.\n\nIn the article , the author considered scalar reaction-diffusion equations with small delay (1.2)u(t)-Δu=f(u(t),u(t-τ)). There the nonlinearity is assumed to be locally lipschitz and to satisfy the one-sided growth estimates (1.3)f(u,v)(u+1)γ(v),u0,f(u,v)-(|u|+1)γ(v),u0, for some continuous γ. To prove existence, he treated the equation stepwise as a nonautonomous undelayed parabolic partial differential equation on the time intervals [(j-1)τ,jτ] by regarding the delayed values as fixed. His strategy was to mimic the results of Henry [3, Theorem 3.3.3 and Corollary 3.3.5], but with his assumption of Hölder continuity in replaced by p-integrability. Many authors had investigated the nondelayed one in .\n\nIn this paper, we consider the following nonlinear evolutionary equation with small delay: (1.4)u(t)+Au=f(u(t),u(t-τ)),t>0,u[-τ,0]=φ(t). Under the hypothesis of (A1), (A2), and (A3) (see Section 2), we firstly make use of the fixed principle to prove the local existence and uniqueness theorem. Then we obtain the global existence and uniqueness by Gronwall inequality. In the whole paper, our work space is fractional powers space Xα. Its definition can be referred to [1, 3, 4].\n\n2. Preliminaries\n\nIn this section, we will give some basic notions and facts. Firstly, basic assumptions are listed.\n\nLet A be a positive, sectorial operator on a Banach Space X. e-At is an analytic semigroup generated by -A. Fractional powers operator Aα is well defined. Fractional powers space Xα=D(Aα) with the graph norm uα=AαuX. For simplicity, we will denote ·X as ·.\n\nFor some 0<α<1, the nonlinearity f:Xα×XαX is locally Lipschitz in (u,v). More precisely, there exists a neighborhood U such that for ui,viU and some constant L(2.1)f(u1,v1)-f(u2,v2)L(u1-u2α+v1-v2α).\n\nThe initial value φ(t) is Hölder continuous from [-τ,0] to Xα.\n\nDefinition 2.1.\n\nLet I be an interval. A function u is called a (classical) solution of (1.4) in the space Xα provided that u:IXα is continuously differentiable on I with tuC(I,X) and satisfies (1.4) everywhere in I.\n\nObviously the (classical) solution of (1.4) can be expressed by the variation of constant formula (2.2)u(t)=e-Atx+0te-A(t-s)f(u(s),u(s-τ))ds,for  t0, where we let φ(0)=x. Next we come to the main theorem on analytic semigroup which is extremely important in the study of the dynamics of nonlinear evolutionary equations .\n\nTheorem 2.2 (fundamental theorem on sectorial operators).\n\nLet A be a positive, sectorial operator on a Banach Space X and e-At be the analytic semigroup generated by -A. Then the following statements hold.\n\nFor any α0, there is a constant Cα>0 such that for all t>0(2.3)Aαe-AtL(X)Cαt-αe-at,(a>0).\n\nFor 0<α1, there is a constant Cα>0 such that for t0 and xD(Aα)(2.4)e-Atx-xCαtαAαx.\n\nFor every α0, there is a constant Cα>0 such that for all t>0 and xX(2.5)(e-A(t+h)-e-At)xαCα|h|t-(1+α)x.\n\nLemma 2.3 (Gronwall’s equality, [<xref ref-type=\"bibr\" rid=\"B2\">2</xref>–<xref ref-type=\"bibr\" rid=\"B4\">4</xref>]).\n\nLet v(t)0 and be continuous on [t0,T]. If there exists positive constants a,b,α(α<1) such that for t[t0,T](2.6)v(t)a+bt0t(t-s)α-1v(s)d, then there exists positive constant M such that for t[t0,T](2.7)v(t)Ma.\n\n3. Main Results Theorem 3.1.\n\nSuppose (A1), (A2), and (A3) hold. Then there exists a sufficiently small T>0 such that (1.4) has a unique solution on [-τ,T].\n\nProof.\n\nFor convenience, we still denote φ(0)=x. Select δ>0 and construct set (3.1)V={(u,v)u-xαδ,v-xαδ}. Let B=f(x,x), choose sufficient small T<τ such that (3.2)(e-At-I)xαδ2,0t<T,(3.3)Cα(B+2Lδ)0Tu-αe-auduδ2. Let Y be the Banach space C([-τ,T];X) with the usual supremum norm which we denote by ·Y. Let S be the nonempty closed and bounded subset of Y defined by (3.4)S={y:yY,y(t)-Aαxδ}. On S we define a mapping F by (3.5)Fy(t)={e-tAAαx+0tAαe-(t-s)Af(A-αy(s),A-αy(s-τ))ds,0<t<T,Aαφ(t),-τt0. Next we will utilize the contraction mapping theorem to prove the existence of fixed point. In order to complete this work, we need to verify that F maps S into itself and F is a contraction mapping on S with the contraction constant ≤1/2.\n\nIt is easy to see from (3.4) and (3.5) that for -τt0, F:SS. For 0<t<T, considering (2.1), (2.3), (3.2), and (3.3), we obtain (3.6)Fy(t)-Aαxe-tAAαx-Aαx+0tAαe-(t-s)Af(A-αy(s),A-αy(s-τ))dse-tAAαx-Aαx+0tAαe-(t-s)Af(A-αy(s),A-αy(s-τ))-f(x,x)ds+0tAαe-(t-s)Af(x,x)dsδ2+Cα(2Lδ+B)0t(t-s)-αe-a(t-s)dsδ2+Cα(2Lδ+B)0Tu-αe-auduδ. Therefore F:SS. Furthermore if y1,y2S then from (3.3) and (3.5) (3.7)Fy1(t)-Fy2(t)0tAαe-(t-s)A[f(A-αy1(s),A-αy1(s-τ))-f(A-αy2(s),A-αy2(s-τ))]ds0tAαe-(t-s)A[f(A-αy1(s),A-αy1(s-τ))-f(A-αy2(s),A-αy2(s-τ))]ds0tCα(t-s)-αe-a(t-s)Ly1(s)-y2(s)dsCαL0Tu-αe-auds(y1-y2)Y12(y1-y2)Y, which implies that (3.8)Fy1(t)-Fy2(t)Y12(y1-y2)Y. By the contraction mapping theorem the mapping F has a unique fixed point yS. This fixed point satisfies the following: (3.9)y(t)=e-tAAαx+0tAαe-(t-s)Af(A-αy(s),A-αy(s-τ))ds,0<t<T,(3.10)y(t)=Aαφ(t),-τt0.\n\nFrom (2.1) and the continuity of y it follows that tf(A-αy(t),A-αy(t-τ)) is continuous on [0,T] and a fortiori bounded on this interval. Let (3.11)f(A-αy(t),A-αy(t-τ))N. Next we want to show that tf(A-αy(t),A-αy(t-τ)) is locally Hölder continuous on (0,T). To this end, we show first that the solution y of (3.9) is locally Hölder continuous on (0,T).\n\nSelect [t0,t1](0,T), t0t<t+ht1 such that (3.12)y(t+h)-y(t)(e-hA-I)Aαe-tAx+0t(e-hA-I)Aαe-(t-s)Af(A-αy(s),A-αy(s-τ))ds+tt+hAαe-(t+h-s)Af(A-αy(s),A-αy(s-τ))ds=I1+I2+I3. Considering (2.3) and (2.4), we select β(0,1-α) such that (3.13)I1CβhβAα+βe-tAxCβhβCα+βt-(α+β)xM1hβ,I2NCβhβ0tAα+βe-(t-s)AdsNChβ0t(t-s)-(α+β)dsM2hβ,I3NCαtt+h(t+h-s)-αds=NCα1-αh1-αM3hβ. Synthesizing (3.12) and (3.13), we get (3.14)y(t+h)-y(t)Chβ,t[t0,t1](0,T).\n\nSo we proved the solution y of (3.9) is locally Hölder continuous on (0,T). Furthermore, in view of (2.1) we have (3.15)f[A-αy(t+h),A-αy(t+h-τ)]-f[A-αy(t),A-αy(t-τ)]L(y(t+h)-y(t)+y(t+h-τ)-y(t-τ))LChβ+Aαφ(t+h-τ)-Aαφ(t-τ)Mhγ.\n\nLet y be the solution of (3.9) and (3.10) and f~(t)=f(A-αy(t),A-αy(t-τ)). In view of locally Hölder continuous on (0,T) of f~(t), consider the inhomogeneous initial value problem (3.16)u(t)+Au=f~(t),0<t<T,u(0)=x. By Corollary 4.3.3 in , this problem has a unique solution and the solution is given by (3.17)u(t)=e-tAx+0te-(t-s)Af(A-αy(s),A-αy(s-τ))ds. Each term of (3.17) is in D(A) and a fortiori in D(Aα). Operating on both sides of (3.17) with Aα we find (3.18)Aαu(t)=e-tAAαx+0tAαe-(t-s)Af(A-αy(s),A-αy(s-τ))ds. By (3.9) the right-hand side of (3.18) equals y(t) and therefore u(t)=A-αy(t). So for 0<t<T, by (3.17) we have (3.19)u(t)=e-tAx+0te-(t-s)Af(u(s),u(s-τ))ds. So u is a uC1(0,T;X) solution of (1.4). The uniqueness of u follows readily from the uniqueness of the solutions of (3.9) and (3.16), and the proof is complete.\n\nBefore giving our global existence theorem, we should first prove extended theorem of solution.\n\nTheorem 3.2 (extended theorem).\n\nAssume that (A1), (A2), and (A3) hold. And also assume that for every closed bounded set BU, the image f(B) is bounded in X. If u is a solution of (1.1) on [-τ,Tmax), then either Tmax=+ or there exists a sequence tnTmax as n+ such that u(tn)U. (If U is unbounded, the point at infinity is included in U.)\n\nProof.\n\nSuppose Tmax<+, there exists a closed bounded B subset of U and τ0<Tmax such that for τ0t<Tmaxu(t)B. We prove there exists x*B such that (3.20)limtTmax-u(t)=x* in Xα, which implies the solution may be extended beyond time Tmax.\n\nNow let (3.21)C=sup{f(u,v),(u,v)B}. We show firstly that u(t)β remains bounded as tTmax- for any β[α,1).\n\nObserve that if αβ<1, τ0t<Tmax, in view of (2.3) and (3.19) we have (3.22)u(t)βAβ-αe-tAxα+0tAβe-(t-s)Af(u(s),u(s-τ))dsCβ-αt-(β-α)xα+CβC0t(t-s)-βds=Cβ-αt-(β-α)xα+CβC1-βt1-βM,0<τ0t<Tmax.\n\nSecondly, suppose τ0t1<t<Tmax, so (3.23)u(t)-u(t1)=(e-(t-t1)A-I)u(t1)+t1te-(t-s)Af(u(s),u(s-τ))ds. From (2.3) and (2.4) we get (3.24)u(t)-u(t1)α(e-(t-t1)A-I)Aαu(t1)+Ct1tAαe-(t-s)AdsCβ-α(t-t1)β-αAβ-α+αu(t1)+CCαt1t(t-s)-αds=Cβ-α(t-t1)β-αu(t1)β+CCα1-α(t-t1)1-αdsC0(t-t1)β-α. Thus (3.20) holds, and the proof is completed.\n\nTheorem 3.3 (global existence and uniqueness).\n\nAssume that (A1), (A2), and (A3) hold. And for all (u,v)Xα×Xα, f satisfies (3.25)f(u,v)L(uα+vα). Then, the unique solution of (1.4) exists for all t-τ.\n\nProof.\n\nWe need to verify that u(t)α is bounded when tTmax-. As for 0t<Tmax(3.26)u(t)=e-tAx+0te-(t-s)Af(u(s),u(s-τ))ds. Considering (3.25), we can obtain (3.27)u(t)α=Aαu(t)e-tAAαx+0tAαe-(t-s)AL(u(s)α+u(s-τ)α)dsC1xα+LCα0t(t-s)-αu(s)αds+LCα0t(t-s)-αu(s-τ)αds, For (3.28)0t(t-s)-αu(s-τ)ds=s-τ=w-τt-τ(t-τ-w)-αu(w)dw.\n\nCase 1. If Tmaxτ. Because u(t)=φ(t) for -τt0 and φ is Hölder continuous from [-τ,0] to Xα. Let (3.29)M=maxt[-τ,0]φ(t),(3.30)0t(t-s)-αu(s-τ)αds=-τt-τ(t-τ-w)-αφ(w)αdwM-τ0(t-τ-w)-αdwM1-αt1-αM1,t[0,Tmax). From (3.27), we immediately get (3.31)u(t)αa+b0t(t-s)-αu(s)αds. From Lemma 2.3, that is, Gronwall’s inequality, we find u(t)αC.\n\nCase 2. If Tmax>τ, still let (3.32)M=maxt[-τ,0]φ(t), because (3.33)0t(t-s)-αu(s-τ)αds=-τt-τ(t-τ-w)-αu(w)αdw=-τ0(t-τ-w)-αφ(w)αdw+0t-τ(t-τ-w)-αu(w)αdwM-τ0(t-τ-w)-αdw+0t(t-s)-αu(s)αdsM0+0t(t-s)-αu(s)αds. From (3.27) again, we obtain (3.34)u(t)αa0+b00t(t-s)-αu(s)αds. By Gronwall’s inequality again, we get u(t)αC. This completes the proof of this theorem.\n\nAcknowledgments\n\nThis work is supported by NNSF of China (11071185) and NSF of Tianjin (09JCYBJC01800).\n\nPazy A. Semigroups of Linear Operators and Applications to Partial Differential Equations 1983 44 New York, NY, USA Springer Applied Mathematical Sciences 710486 10.1007/978-1-4612-5561-1 ZBL0528.47030 Evans L. C. Partial Differential Equation 1998 19 Proceedings of the American Mathematical Society Graduate Studies in Mathematics Henry D. Geometric Theory of Semilinear Parabolic Equations 1981 840 Berlin, Germany Springer Lecture Notes in Mathematics 610244 ZBL1215.78002 Sell G. R. You Y. Dynamics of Evolutionary Equations 2002 44 New York, NY, USA Springer Applied Mathematical Sciences 1873467 Friesecke G. Convergence to equilibrium for delay-diffusion equations with small delay Journal of Dynamics and Differential Equations 1993 5 1 89 103 10.1007/BF01063736 1205455 Yin X. On asymptotic behavior for reaction diffusion equation with small time delay Abstract and Applied Analysis 2011 2011 13 142128 10.1155/2011/142128 2861497 Deimling K. Nonlinear Functional Analysis 1985 Berlin, Germany Springer 787404 Hale J. K. Asymptotic Behavior of Dissipative Systems 1988 25 Providence, RI, USA American Mathematical Society Mathematical Surveys and Monographs 941371 Matano H. Asymptotic behavior and stability of solutions of semilinear diffusion equations Publications of the Research Institute for Mathematical Sciences 1979 15 2 401 454 10.2977/prims/1195188180 555661 ZBL0445.35063 Fitzgibbon W. E. Semilinear functional differential equations in Banach space Journal of Differential Equations 1978 29 1 1 14 0492663 10.1016/0022-0396(78)90037-2 ZBL0392.34041" ]
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http://quadr-prog.ru/english.html
[ " Quadratic programming, common case на русский\n\n### Arbitrary matrix of the quadratic form\n\nWe consider the problem of finding the largest or smallest values of the objective function F = (X,AX) + (L,X) + C in case of an arbitrary matrix \"A\" of the quadratic form. Domain of the arguments \"X\" defined by a system of linear constraints-equality and linear constraints-inequalities. Here is not imposed simplicity demand of positive definite quadratic form in the case minimization of the function or negative definite quadratic form in the case maximization of the function. Using quadratic functions can be approximated an arbitrary nonlinear function (solving the problem of mathematical programming), herewith as a rule are not expected the constant sign of the quadratic form. In this case it is necessary to solve the optimization problem with an arbitrary matrix of the quadratic form.\n\n### The exact solution\n\nIt is proposed the principle of exact solution not based on any modification of the gradient methods or other descent methods. The basis of finding the global extremum of the function is taken the well-known algorithm for computing the largest or smallest values of the function on a compact - selection of the global extremum of the set from all local extrema inside the feasible region and on the boundary. Of course, not always a domain defined by a system of restrictions, is compact. For an unbounded domain in the proposed method introduces additional conditions, cutting off the compact part. Then taking the limit as the shift of additional restrictions to infinity, we obtain an exact solution for a given unbounded domain.\n\n### Free of charge testing this method\n\nAll those wishing to offer free opportunity to test above specified idea at concrete examples. For the purity of the experiment you create the example yourself and offer to solve it to author. After receiving solution, you can analyze it thoroughly to make sure working capacity of the algorithm or abandon it.\nSoftware testing. In order to avoid confusion in the description of examples of problems, is designed a program. You can download it here. This self-extracting archive file is containing four files: qpu.exe, qpu.aux, qpu.t0 and qpu.t1. All four files must be placed in one folder, executable file - qpu.exe. The program qpu.exe allows you to specify numerical parameters the quadratic programming problem and write them in a special format file. This file can be sent to the author [email protected], which undertakes to solve any problem that was sent or declare of inability to obtain the result with the claimed algorithm. In any case, response will be sent to your e-mail.\nTasks to test online. Numerical parameters of the quadratic programming problem small dimension, in standard form can be entered directly: Structure of the task -> Numerical data. In this case, the answer will appear on this site in a few days by page Solving quadratic programming problems (author visits the site 1-2 times a week).\n\n### Examples\n\nExample 1   Example 2   Example 3\n\n<\n\n#### Autor\n\nGennadij Bulanov, 1948 yr birth, Ph.D. phys.-math. sciences, docent department of higher mathematics Donbass State Engineering Academy, Kramatorsk, Donetsk region, Ukraine.\n\[email protected]\n\n#### From the methodological developments G.Bulanov for educational purposes\n\nAutomated generator problems on functional analysis (in Russian)" ]
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https://cs.stackexchange.com/questions/19755/most-common-subset-of-size-k
[ "# Most common subset of size $k$\n\nI'm trying to write an algorithm that detects the most common subset of at least size $k$, from a collection of sets. If there are ties for the most common subset, I want the one of them whose size is as large as possible.\n\nFor example if I have:\n\ns1 = {A, B, C }\ns2 = {A, B, C, D}\ns3 = { B, C, D}\n\n\nThen the most common subset of size $\\ge k=2$ is {B, C}. As another example, if I have:\n\ns1 = {A, B, C D}\ns2 = {A, B, C, D}\ns3 = { B, C, D}\n\n\nThen the most common subset of size $\\ge k=2$ is {B, C, D}. It's important that in this instance the algorithm would give me {B, C, D} and not {B, C}, {B, D} etc. Note that I'm not interested in the longest common subset (a different problem), I'm interested in the longest most common subset if you will. I also don't care about enumerating all the different subsets, I just want to find the most common.\n\nIs there an efficient algorithm for this problem?\n\nI have an algorithm for this problem, but I don't think it's very efficient. For $k=2$ I enumerate all subsets of size 2 and count how many times each one appears in the collection. If the most-frequently occurring pair is more frequently occurring than any other pair then that must be the most common subset. If there is more than one with the same (maximum) frequency then I look at the sets they are contained in. If these overlap exactly then I take the union of the pairs and that gives me the most common subset (with size > 2).\n\nI think this could be related to the maximum clique problem but I'm not certain.\n\nNote that just taking the intersection does not give the correct answer. For instance, if I have\n\ns1 = {A, B }\ns2 = { C, D}\ns3 = {A, C, D}\n\n\nthen the intersection is the empty set, but the most common subset is {C, D}." ]
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https://oncologyleader.info/?post=2344
[ "# What are possible random variables\n\n## Random variables and distributions\n\nIn this article, we'll explain everything about random variables and distributions. We discuss the following sub-topics in detail:\n\nOur Mathe-Abi'21 learning booklets Explanations ✔ Examples ✔ Free learning videos ✔\n\nNew!\n\n### What is a random variable?\n\nA random variable (\\$ X \\$) is a function that assigns real numbers to the results of a random experiment:\n\n\\ begin {align *}\nX: \\ Omega \\ rightarrow \\ mathbb {R}, \\ quad X: \\ omega \\ rightarrow X (\\ omega) = x\n\\ end {align *}\n\n### example 1\n\nRoll the dice twice with the random variable \"\\$ X \\$ = total\"\n\n\\$\n\\ begin {array} {c | c | c | c | c | c | c}\ne_i & (1 | 1) & (1 | 2) & (2 | 1) & (2 | 2) & \\ dots & (6 | 6) \\ \\ hline\nX (e_i) = x_i & 2 & 3 & 3 & 4 & \\ dots & 12\n\\ end {array} \\$", null, "Note: The realization (rolled number) can be achieved by different throw combinations. So one can achieve the realization “4” with the throws (2 | 2), (1 | 3) and (3 | 1). The probability 3/36 results from the number of natural events.\n\n### Example 2\n\nA coin is tossed twice with the random variable \"\\$ X \\$ = number of number throws\"\n\nWhat are the options? What does the result space look like? Knowing this, we can create the following table:\n\n\\$\n\\ begin {array} {c | c | c | c}\ne_i & (K | K) & (Z | K), (K | Z) & (Z | Z) \\ \\ hline\nX (e_i) = x_i & 0 & 1 & 2 \\ \\ hline\nP (x_i) & 1/4 & 1/2 & 1/4\n\\ end {array}\n\\$\n\nDaniel explains again what a random variable is.\n\nRandom size and probability distribution, basics with example | Math by Daniel Jung\n\nMath Abi'21 study booklet incl. Exercise collection\n\nNew!\n\n### Discrete random variables\n\nA random variable is called discreet, if it at last or countable infinite values \\$ X_1 \\$, \\$ X_2 \\$, \\$ X_3 \\$, \\$ \\ dots \\$, \\$ X_n \\$.\n\nA random variable \\$ (X) \\$, which only has a finite or countably infinite number of values, is always discrete.\n\nExamples: number of pips when rolling the dice, tossing a coin\n\n### Carrier of a discrete random variable\n\n• The carrier \\$ T_X \\$ of a discrete \\$ ZV \\$ \\$ X \\$ is the set of all values ​​that \\$ X \\$ assumes with a positive probability.\n• Usually the carrier of a discrete random variable is a subset of the natural numbers \\$ (1,2,3, \\ dots, n) \\$.\n• The carrier \\$ T_X \\$ can also be understood as an event space.\n• Important: First of all, all carriers of the random variables must always be determined in order to then determine the probability.\n\n### Examples of discrete random variables\n\n• Number of pips when rolling the dice once (event space \\$ \\ Omega = \\ {1,2,3,4,5,6 \\}) \\$.\n• Number of times a normal deck of cards was drawn without replacing until the 7 of diamonds is drawn (\\$ \\ Omega = \\ {1,2,3, \\ dots, 32 \\} \\$).\n\nOur Mathe-Abi'21 learning booklets Explanations ✔ Examples ✔ Free learning videos ✔\n\nNew!\n\n### Probability function of a discrete random variable\n\nA probability distribution indicates how the probabilities are distributed among the possible values ​​of a random variable and is only defined for discrete random variables.\n\nDefinition:\n\n\\ begin {align *}\nf: \\ mathbb {R} \\ rightarrow [0; 1], \\ quad f: x \\ rightarrow f (x) = P (X = x) = p\n\\ end {align *}\n\n\\$ P (X = x) \\$ indicates the probability that the ZV \\$ X \\$ assumes the value \\$ x \\$. It might seem complicated at first, but let's look at a small example:\n\nA normal die is thrown. As already described, the carriers of the random variables must first be determined, which can also be understood as an event space. So the event space is\n\n\\ begin {align *}\n\\ Omega = \\ {1,2,3,4,5,6 \\}.\n\\ end {align *}\n\nThe probabilities are then calculated using the Laplace probability, analogous to the previous probabilities.", null, "Our Mathe-Abi'21 learning booklets Explanations ✔ Examples ✔ Free learning videos ✔\n\nNew!\n\n### Distribution function of a discrete random variable\n\nThe distribution function is a tool for describing a discrete (or continuous) probability distribution. A function \\$ F \\$ that assigns exactly one probability \\$ P (X \\ leq x) \\$ to each \\$ x \\$ of a random variable \\$ X \\$ is called a distribution function.\n\n\\ begin {align *}\nF: \\ mathbb {R} \\ rightarrow [0,1], \\ quad\nF: x \\ rightarrow \\ F (x) = P (X≤x)\n\\ end {align *}\n\nAh yes, and what does that mean? Interpretation:\n\nThe distribution function measures the probability that the random variable \\$ X \\$ takes on at most the value \\$ x \\$: \\$ F (X) = P (X \\ leq x) \\$ = “Probability that \\$ X \\$ is less than or equal to a certain value \\$ x \\$ Has.\"\n\nProperties:\n\n• The distribution function of a discrete random variable is a step function.\n• \\$ F (x) \\$ is defined for each \\$ x \\$ and takes values ​​from 0 to 1.\n• Used in hypothesis tests! Signal word: At most.\n\nNote:\n\n\\ begin {align *}\nP (X \\ geq x) = 1-P (X x) = 1-P (X \\ leq x)\n\\ end {align *}\n\n### example\n\nA coin is tossed twice with the random variable \\$ X \\$ = number of \"number throws\"!\n\n\\$\n\\ begin {array} {c | c | c | c}\ne_i & (K | K) & (Z | K), (K | Z) & (Z | Z) \\ \\ hline\nX (e_i) = x_i & 0 & 1 & 2 \\ \\ hline\nP (x_i) & 1/4 & 1/2 & 1/4 \\ \\ hline\nF (X) & 0.25 & 0.75 & 1 \\\n\\ end {array}\n\\$", null, "1) How high is the probability that a “number” will be thrown at least once in the above random experiment?\n\nBasically, we express the operator at least as \\$ \\ geq \\$ and write what is sought from the task. It follows:\n\n\\ begin {align *}\nP (X \\ geq 1)\n\\ end {align *}\n\nIn the second step it can be seen from the task that the carriers of the random variables are \\$ \\ Omega = \\ {0,1,2 \\} \\$.\n\nAt this point you have to ask yourself what is really greater than or equal to one. From this it follows that only \\$ P (X = 1) \\$ and \\$ P (X = 2) \\$ are really greater than or equal to one and the probabilities for both random variables are known.\n\n\\ begin {align *}\nP (X \\ geq 1) = P (X = 1) + P (X = 2) = 0.5 +0.25 = 0.75\n\\ end {align *}\n\nAlternatively, you can proceed as follows: Ask yourself what is really less than one. Only \\$ P (X = 0) \\$ is really less than one, and the probability of this is known. It follows\n\n\\ begin {align *}\nP (X \\ geq 1) = 1- P (X <1) = 1 - P (X = 0) = 1 - 0.25 = 0.75\n\\ end {align *}\n\n2) What is the probability that a “number” will be thrown at most once in a random experiment?\n\nBasically, we express the operator at most as \\$ \\ leq \\$ and write out what is sought from the task. It follows:\n\n\\ begin {align *}\nP (X \\ leq 1)\n\\ end {align *}\n\nAt this point you have to ask yourself what is really less than or equal to one. From this it follows that only \\$ P (X = 0) \\$ and \\$ P (X = 1) \\$ are really less than or equal to one and the probabilities for both random variables are known.\n\n\\ begin {align *}\nP (X \\ leq 1) = P (X = 0) + P (X = 1) = 0.25 +0.5 = 0.75\n\\ end {align *}\n\n> You can also make use of the distribution function when answering, since the distribution function measures the probability that the random variable \\$ X \\$ at most (\\$ \\ leq \\$) takes on the value \\$ x \\$! It follows\n\n\\ begin {align *}\nP (X \\ leq 1) = F (X = 1) = 0.75.\n\\ end {align *}\n\nTake a closer look at Daniel's video tutorial on the topic of distribution functions.\n\nDistribution function, cumulative, stochastics, probability theory, math by Daniel Jung\n\nMath Abi'21 study booklet incl. Exercise collection\n\nNew!\n\n### Distribution parameter of a discrete random variable\n\nDistribution parameters are quantities that characterize certain aspects of a distribution, such as the position, scatter or skewness of a distribution.\n\nImportant parameters are:\n\n### Expected value (location parameter):\n\n• The expected value is the center of gravity of the distribution and describes the number that the random variable assumes on average.\n• The expected value \\$ E (X) \\$ is also often referred to as \\$ \\ mu \\$.\n\n\\ begin {align *}\n\\ mu = E (X) = \\ sum_ {i = 1} ^ k x_i \\ cdot \\ underbrace {P (X = x_i)} _ {= p_i} = x_1 \\ cdot p_1 + x_2 \\ cdot p_2 + \\ dots + x_k \\ cdot p_k\n\\ end {align *}\n\n### Variance (dispersion parameter):\n\n• Variance describes the spread of a random variable and does not depend on chance.\n• The variance of the random variable \\$ X \\$ is the expected value of the squared deviation from its expected value.\n• Often \\$ \\ sigma ^ 2 \\$ is written instead of \\$ V (X) \\$.\n\n\\ begin {align *}\n\\ sigma ^ 2 = V (X) = \\ sum_ {i = 1} ^ k (x_i- \\ mu) ^ 2 \\ cdot p_i\n\\ end {align *}\n\n• The shift theorem \\$ \\ sigma ^ 2 = \\ sum_ {i = 1} ^ k x_i ^ 2 \\ cdot p_i - \\ mu ^ 2 \\$ usually simplifies the calculation of the variance.\n\n### Standard deviation (variance parameter):\n\n• The standard deviation is the positive square root of the variance and indicates the spread of the values ​​around the mean.\n• This means that the standard deviation is also a measure of the spread, only that it increases a little more slowly than the variance. If you know the variance, it can easily be converted into the standard deviation (and vice versa).\n\n\\ begin {align *}\n\\ sigma = \\ sqrt {V (X)} = \\ sqrt {\\ sigma ^ 2} \\ notag\n\\ end {align *}\n\n• Calculate and interpret expected value and standard deviation\n• Calculate the probability that the random variable will take on values ​​that deviate from the expected value by specified values.\n\n### example\n\nA teacher wants to know how his students are doing and how much the good or bad students deviate from the average. The following table shows the distribution of grades\n\n\\$\n\\ begin {array} {c | c | c | c | c | c | c}\n\\ text {Note} \\ x_j & 1 & 2 & 3 & 4 & 5 & 6 \\ \\ hline\np_j & 0.1 & 0.15 & 0.5 & 0.2 & 0 & 0.05\n\\ end {array}\n\\$\n\nThe expected value is calculated as follows:\n\n\\ begin {align *}\n\\ mu = E (x) = 1 \\ times 0.1 + 2 \\ times 0.15 + 3 \\ times 0.5 + 4 \\ times 0.2 + 5 \\ times 0 + 6 \\ times 0.05 = 3 \\ notag\n\\ end {align *}\n\nInterpretation: The grade point average is \\$ 3 \\$.\n\nWe calculate the standard deviation using the variance:\n\n\\ begin {align *}\n\\ sigma ^ 2 & = (1-3) ^ 2 \\ times 0.1 + (2-3) ^ 2 \\ times 0.15 + (3-3) ^ 2 \\ times 0.5 + (4-3) ^ 2 \\ times 0.2 \\ notag \\ & + (5-3) ^ 2 \\ times 0 + (6-3) ^ 2 \\ times 0.05 = 1.2 \\ notag \\\n\\ Rightarrow \\ quad \\ sigma & = \\ sqrt {1,2} \\ notag\n\\ end {align *}\n\nThe spread around the grade point average is low. A stick diagram would make this further clear. With such tasks you often have to compare several scenarios and say which scenario spreads more.\n\nOur Mathe-Abi'21 learning booklets Explanations ✔ Examples ✔ Free learning videos ✔\n\nNew!" ]
[ null, "https://www.studyhelp.de/online-lernen/wp-content/uploads/2015/03/bil_verteilung-300x176.png", null, "https://www.studyhelp.de/online-lernen/wp-content/uploads/2015/03/bil_diskreteverteilung1.png", null, "https://www.studyhelp.de/online-lernen/wp-content/uploads/2015/03/bil_diskreteverteilung_1.png", null ]
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https://forums.fast.ai/t/how-to-load-learner-for-new-data/98650
[ "# How to load learner for new data?\n\nI have tried doing transfer learning for my model when I try to load the previously model for my new data, it gives me this error:\n\n``````/usr/local/lib/python3.7/dist-packages/fastai/basic_train.py in load(self, file, device, strict, with_opt, purge, remove_module)\n271 model_state = state['model']\n272 if remove_module: model_state = remove_module_load(model_state)\n--> 273 get_model(self.model).load_state_dict(model_state, strict=strict)\n274 if ifnone(with_opt,True):\n275 if not hasattr(self, 'opt'): self.create_opt(defaults.lr, self.wd)\n\n/usr/local/lib/python3.7/dist-packages/torch/nn/modules/module.py in load_state_dict(self, state_dict, strict)\n1496 if len(error_msgs) > 0:\n1497 raise RuntimeError('Error(s) in loading state_dict for {}:\\n\\t{}'.format(\n-> 1498 self.__class__.__name__, \"\\n\\t\".join(error_msgs)))\n1499 return _IncompatibleKeys(missing_keys, unexpected_keys)\n1500\n\nRuntimeError: Error(s) in loading state_dict for RetinaNet:\nsize mismatch for classifier.3.weight: copying a param with shape torch.Size([3, 128, 3, 3]) from checkpoint, the shape in current model is torch.Size([2, 128, 3, 3]).\nsize mismatch for classifier.3.bias: copying a param with shape torch.Size() from checkpoint, the shape in current model is torch.Size().\n``````\n\nThe images in data loader are not binary they are 3 channel images, but still the learn. load throws this error. The new data has 2 slide images whereas the model was trained on 100 slides. I have loaded the new data like this:\n\n``````batch_size=20\ndo_flip = True\nflip_vert = True\nmax_rotate = 90\nmax_zoom = 1.1\nmax_lighting = 0.2\nmax_warp = 0.2\np_affine = 0.75\np_lighting = 0.75\n\ntfms = get_transforms(do_flip=do_flip,\nflip_vert=flip_vert,\nmax_rotate=max_rotate,\nmax_zoom=max_zoom,\nmax_lighting=max_lighting,\nmax_warp=max_warp,\np_affine=p_affine,\np_lighting=p_lighting)\ntrain, valid = ObjectItemListSlide(train_images) ,ObjectItemListSlide(valid_images)\nitem_list = ItemLists(\".\", train, valid)\nlls = item_list.label_from_func(lambda x: x.y, label_cls=SlideObjectCategoryList)\nlls = lls.transform(tfms, tfm_y=True, size=patch_size)\ndata = lls.databunch(bs=batch_size, collate_fn=bb_pad_collate,num_workers=0).normalize()\n``````\n\nI have tried using `learn=load_learner(\"path/pklfile\")` also but it has empty dataloader so I am not able to perform `show_results ` or `show_results_side_by_side(learn, anchors, detect_thresh=detect_thresh, nms_thresh=nms_thresh, image_count=image_count)`\nDoes anyone know how one can solve this issue, and what needs to be changed?" ]
[ null ]
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https://vceela.com/shop/category/gifts-gifts-by-person-gifts-for-kids-117
[ "", null, "x\nClear all x\n\nRs. 12,500\nRs. 10,000\nRs. 3,000\nRs. 700\nRs. 1,499\nRs. 850\nRs. 700\nRs. 1,150\nRs. 250\nRs. 150\nRs. 2,000\nRs. 900\nRs. 870\nRs. 870\nRs. 999\nRs. 999\nRs. 400\nRs. 999\nRs. 999\nRs. 350\nRs. 350\nRs. 25,000\nRs. 11,000" ]
[ null, "https://vceela.com/web/image/645/shop.jpeg", null ]
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https://www.javacodeexamples.com/java-print-array-example/658
[ "# Java Print Array Example\n\nJava print array example shows how to print an array in Java in various ways as well as print 2d array (two dimensional) or multidimensional array.\n\n## How to print an array in Java?\n\nThere are various ways using which you can print an array in Java as given below.\n\n### 1) Using while loop\n\nWe can use a while loop to print the array. The below-given code example prints an array of String.\n\nOutput\n\n### 2) Using for loop\n\nProbably the simplest way to print an array is to print it using a for loop as given below. The array index starts at 0 and go up to array length – 1 index.\n\nOutput\n\n### 3) Using enhanced for loop\n\nIf you are using Java 1.5 or a later version, you can use the enhanced for loop instead of the for loop as given below.\n\nOutput\n\n### 4) Using Arrays class\n\nThe Arrays class provides many useful utility methods and the `toString` method is one of them. It converts array to string which can then be printed as given below.\n\nOutput\n\nIf you want to remove square brackets and comma from the output, you can remove them using a regular expression as given below.\n\nOutput\n\n### 5) Using Java 8 Stream\n\nIf you are using Java 8 or a later version, you can use the stream to print the array as given below.\n\nOutput\n\n### 6) Using Java 8 String join\n\nIf you are using Java 8, you can use the `join` method of the String class to join array elements by delimiter as given below.\n\nOutput\n\n### 7) Using Apache Commons library\n\nFinally, if you are using the Apache Commons library, you can use the `join` method of StringUtils class to join the array elements and print them as given below.\n\nOutput\n\nThe `join` method is overloaded for byte array, char array, double array, float array, int array, long array, and an Object array.\n\n## How to print a 2D array (two-dimensional array)?\n\nIf you have a two-dimensional array, you can print it using the below-given approaches.\n\n### a) Using for loop\n\nYou can use for loop to print a two-dimensional array as given below.\n\nOutput\n\n### b) Using Arrays class\n\nYou can also use the `deepToString` method of the Arrays class to print two-dimensional array as given below.\n\nOutput\n\nThis example is a part of the Java Array tutorial with examples." ]
[ null ]
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https://acm.timus.ru/problem.aspx?space=28&num=3
[ "ENG  RUS Timus Online Judge", null, "Online Judge\nProblems\nAuthors\nOnline contests\nSite news\nWebboard\nProblem set\nSubmit solution\nJudge status\nGuide\nRegister\nAuthors ranklist\nCurrent contest\nScheduled contests\nPast contests\nRules\n\nContest is over\n\n## C. Nontrivial Numbers\n\nTime limit: 2.0 second\nMemory limit: 64 MB\nSpecialists of SKB Kontur have developed a unique cryptographic algorithm for needs of information protection while transmitting data over the Internet. The main advantage of the algorithm is that you needn't use big numbers as keys; you may easily do with natural numbers not exceeding a million. However, in order to strengthen endurance of the cryptographic system it is recommended to use special numbers - those that psychologically seem least \"natural\". We introduce a notion of triviality in order to define and emphasize those numbers.\nTriviality of a natural number N is the ratio of the sum of all its proper divisors to the number itself. Thus, for example, triviality of the natural number 10 is equal to 0.8 = (1 + 2 + 5) / 10 and triviality of the number 20 is equal to 1.1 = (1 + 2 + 4 + 5 + 10) / 20. Recall that a proper divisor of a natural number is the divisor that is strictly less than the number.\nThus, it is recommended to use as nontrivial numbers as possible in the cryptographic protection system of SKB Kontur. You are to write a program that will find the less trivial number in a given range.\n\n### Input\n\nThe only line contains two integers I and J, 1 ≤ IJ ≤ 106, separated with a space.\n\n### Output\n\nOutput the only integer N satisfying the following conditions:\n1. INJ;\n2. N is the least trivial number among the ones that obey the first condition.\n\n### Sample\n\ninputoutput\n```24 28\n```\n```25\n```\nProblem Author: Leonid Volkov\nProblem Source: USU Open Collegiate Programming Contest October'2001 Junior Session\nTo submit the solution for this problem go to the Problem set: 1118. Nontrivial Numbers" ]
[ null, "https://acm.timus.ru/images/usu-summer.jpg", null ]
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https://www.zadmei.com/mdjjz.html
[ "# MATLAB 对角矩阵\n\n## 在 MATLAB 中使用 `diag()` 函数制作对角矩阵\n\n``````vector = [1 2 3 4 5];\ndiagonal = diag(vector)\n``````\n\n``````diagonal =\n1 0 0 0 0\n0 2 0 0 0\n0 0 3 0 0\n0 0 0 4 0\n0 0 0 0 5\n``````\n\n``````vector = [1 2 3 4 5];\ndiagonal = diag(vector,1)\n``````\n\n``````diagonal =\n0 1 0 0 0 0\n0 0 2 0 0 0\n0 0 0 3 0 0\n0 0 0 0 4 0\n0 0 0 0 0 5\n0 0 0 0 0 0\n``````\n\n``````vector = [1 2 3;4 5 6;7 8 9]\ndiagonal = diag(vector)\n``````\n\n``````vector =\n1 2 3\n4 5 6\n7 8 9\ndiagonal =\n1\n5\n9\n``````\n\n## 在 MATLAB 中使用 `spdiags()` 函数制作对角矩阵\n\n``````v1 = [1 2 3 4 5].';\nv2 = [2 2 2 2 2].';\nv3 = [3 3 3 3 3].';\ndiagonal = spdiags([v2 v1 v3],-1:1,5,5);\nmatrix = full(diagonal)\n``````\n\n``````matrix =\n1 3 0 0 0\n2 2 3 0 0\n0 2 3 3 0\n0 0 2 4 3\n0 0 0 2 5\n``````\n\n``````v1 = [1 2 3 4 5].';\nv2 = [2 2 2 2 2].';\nv3 = [3 3 3 3 3].';\ndiagonal = spdiags([v2 v1 v3],-1:1,5,5);\nmatrix1 = full(diagonal)\nv4 = [9 9 9 9 9].';\ndiagonal = spdiags(v4,0,diagonal);\nmatrix2 = full(diagonal)\n``````\n\n``````matrix1 =\n1 3 0 0 0\n2 2 3 0 0\n0 2 3 3 0\n0 0 2 4 3\n0 0 0 2 5\nmatrix2 =\n9 3 0 0 0\n2 9 3 0 0\n0 2 9 3 0\n0 0 2 9 3\n0 0 0 2 9\n``````\n\n``````v1 = [1 2 3 4 5].';\nv2 = [2 2 2 2 2].';\nv3 = [3 3 3 3 3].';\ndiagonal = spdiags([v2 v1 v3],-1:1,5,5);\nmatrix = full(diagonal)\ndiag_Values = spdiags(matrix)\n``````\n\n``````matrix =\n1 3 0 0 0\n2 2 3 0 0\n0 2 3 3 0\n0 0 2 4 3\n0 0 0 2 5\ndiag_Values =\n2 1 0\n2 2 3\n2 3 3\n2 4 3\n0 5 3\n``````" ]
[ null ]
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https://bestbtcxmsjlkwf.netlify.app/gochal49759bo/flat-interest-rate-to-effective-interest-rate-calculator-not.html
[ "## Flat interest rate to effective interest rate calculator\n\n5 Apr 2013 Credit Card Interest Rate – Why You May Be Suffering Because of It How to calculate monthly flat rate interest? the compounding effect, the actual interest rate you're being charged (also known as “effective interest rate”)  30 Jun 2010 Effective Interest Rate Calculator - Free download as Excel Spreadsheet (.xls), PDF For \"Flat\" Interest Rate (Personal Loan, Hire Purchase).\n\nThe effective interest rate is the interest rate on a loan or financial product restated from the nominal interest rate as an interest rate with annual compound interest payable in arrears. It is used to compare the annual interest between loans with different compounding terms (daily, monthly, quarterly, semi-annually, annually, or other). Effective interest rate calculation. Effective period interest rate calculation The effective period interest rate is equal to the nominal annual interest rate divided by the number of periods per year n: Effective Period Rate = Nominal Annual Rate / n Calculation of the effective interest rate on loan in Excel. The effective rate of interest on the loan (as with almost on any other financial instrument) – this is the expression of all future cash payments (incomes from a financial instrument), which are included in the treaty provision of the contract, in the figure annual interest. If you are shopping around for a personal loan, you have no doubt seen banks advertise two different interest rates: Annual Flat Rate and Effective Interest Rate (EIR). If you are confused by how these are different and what you should care about, you are not alone. Banks do not quite explain or demonstrate why and how these two rates are different, or how to even calculate them. Not only that The very simple formula to calculate Flat Rate Interest. Say for example, you’re taking out a personal loan of RM100,000 with a flat rate interest of 5.5% over 10 years. This would be your flat rate interest per instalment calculation: (RM100,000 x 10 x 5.5%) ÷ 120 = RM458\n\n## Therefore, the entire monthly payment would amount to \\$105. This would result in a total interest payment of \\$60 for the entire year. Diminishing Rate Loans. The\n\n10 Jan 2018 The simple interest rate is the interest rate that the bank charges you for It is also commonly known as the flat rate, nominal rate or advertised rate. for a good comparison as they can be calculated in various methods. 31 Oct 2018 The difference between flat and effective interest rate is that, the rates under former is calculated on the entire loan principal over the course of  5 Sep 2018 Why do loans have an effective interest rate, or EIR, in addition to the advertised interest rate? Here's a guide to understanding and calculating  13 Aug 2018 Flat interest rate is a system of interest stipulation imposed by banking institutions and financing institutions which is calculated based on the\n\n### The effective interest rate (EIR), effective annual interest rate, annual equivalent rate (AER) or simply effective rate is the interest rate on a loan or financial\n\nThe effective interest rate is the interest rate on a loan or financial product restated from the nominal interest rate as an interest rate with annual compound interest payable in arrears. It is used to compare the annual interest between loans with different compounding terms (daily, monthly, quarterly, semi-annually, annually, or other). How to Calculate Flat Rate Loan. Let's be honest - sometimes the best flat rate loan calculator is the one that is easy to use and doesn't require us to even know what the flat rate loan formula is in the first place! But if you want to know the exact formula for calculating flat rate loan then please check out the \"Formula\" box above. Compound interest is calculated based on the principal amount but also includes all the accrued interest of previous periods of a loan or investment. It can, therefore, be called as ‘interest on interest’ and can enormously grow the sum at a speedy rate than how it goes with a stated interest rate which is calculated by principal amount only. The flat interest rate schedule calculates interest at 1% on the opening principal balance of 3,000, whereas the APR interest rate schedule calculates interest at 1.5875% on the reducing balance at the start of each month. Effective rates vary from 7.07% p.a. to 7.15% p.a. for tenures from 2 – 5 years. Based on the internet search, flat interest rate is based on the total loan amount, while effective interest rate is based on the remaining loan amount. May I know to convert the flat rate to effective rate? For example, I loan 100k for 4 years tenures.\n\n### Learn about flat and monthly rest rates, and how they affect interest calculations. Flat rate With a flat rate, interest payments are calculated based on the original loan amount. The monthly interest stays the same throughout, even though your outstanding loan reduces over time. A flat rate is commonly used for car loans and personal term loans.\n\nEquivalent flat interest rate is calculated based on a front-end add-on calculation method and is for reference purposes only. 2 Effective interest rate is inclusive\n\n## The flat interest rate schedule calculates interest at 1% on the opening principal balance of 3,000, whereas the APR interest rate schedule calculates interest at 1.5875% on the reducing balance at the start of each month.\n\nThe flat interest rate is mostly used for personal and car loans. A flat interest rate is always a fixed percentage. For example: Imagine you applied for a personal loan of RM100,000 at a flat interest rate of 5% p.a. with a tenure of 10 years. Calculator Use. Calculate the effective interest rate per period given the nominal interest rate per period and the number of compounding intervals per period.. Commonly the effective interest rate is in terms of yearly periods and stated such as the effective annual rate, effective annual interest rate, annual equivalent rate (AER), or annual percentage yield (APY), however, the formula is in\n\nInterest Rate (% p.a. flat):. (%). Hiring Period (in Years). Calculate. Reset. Calculated Interest Charges (RM):. 0.00. Total Payment (RM) : 0.00. Monthly  Car Loan Calculator. Computation Results. Interest Rate. 2.48% p.a.. Effective Rate. 5.38% p.a. For flat rate and payment in advance. Rates are only  1.99% flat per month (equivalent to effective interest 3.44% - 3.46% per month, depending on the loan period) How to calculate installments / month. Installment / month = Loan + (Flat interest rate / month x Loan Term x Loan) / Time period. Note that the effect of this method of calculation is that the interest rate has the same effect as if a Using the amortised loan formula to calculate the monthly investment required to generate paid and compounded monthly, be equivalent to an effective annual rate of 3%. flat rate of 1% per month (compounded monthly)." ]
[ null ]
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https://www.statisticshowto.com/efficient-estimator-efficiency/
[ "", null, "# Efficiency / Relative Efficiency and the Efficient Estimator\n\nShare on\n\nSampling in Statistics > Efficiency & Efficient Estimator\n\n## What is Efficiency?\n\nThe meaning of “efficient” in statistics isn’t much different from the dictionary definition of efficient:\n\n“…capable of producing desired results without wasting materials, time, or energy. ~ Miriam Webster.\n\nIn other words, an efficient procedure produces results that maximize your use of materials, time and energy. You’ll use less energy if you have smaller sample sizes, for example. So a procedure that can work with a smaller sample is usually more efficient than one that requires a larger sample.\n\nEfficiency can refer to any procedure you want to optimize. For example, an efficient experimental design is one that produces your desired experimental results with the minimum amount of resources (e.g. time and money). An efficient hypothesis testing procedure also conserves resources in the same way. Generally, the most efficient hypothesis test, experimental design or estimator is going to be the one with the fewest observations.\n\n## What is Relative Efficiency?\n\nRelative efficiency has a couple of meanings, depending on the context:\n\n• As another name for relative precision in sampling. If you use simple random sampling to find the mean of a large population, relative precision and relative efficiency are equal. In other cases, they may not be equal.\n• As a reference to a gold standard — a “best possible” procedure. It is the ratio of the potential procedure against the gold standard. for example, 75/100 indicates the prospective procedure is 75% as efficient as the theoretically best possible procedure.\n\n## What is an Efficient Estimator?\n\nAn estimator is a statistic that estimates some fact about the population. for example, the X̄(the sample mean) is an estimator for the population mean, μ. Your options for finding X̄ are limitless: you could have a sample of ten, fifty of three hundred and one. You could use different classes, ages, or heights (depending on what you are trying to estimate). You could use a simple random sampling technique, or a more complex one like stratified sampling. Out of all these possible scenarios, an efficient estimator is one that has small variances (the estimator with the smallest possible variance is also called the “best” estimator). In other words, the estimator deviates as little as possible from the “true” value you are trying to estimate.\n\nThe more formal definition of an efficient estimator is covered in the next article (What is the Cramer-Rao Lower Bound?):\n\nAn unbiased estimator is efficient if the variance of Θ equals the Cramer-Rao Lower Bound\n\nCITE THIS AS:\nStephanie Glen. \"Efficiency / Relative Efficiency and the Efficient Estimator\" From StatisticsHowTo.com: Elementary Statistics for the rest of us! https://www.statisticshowto.com/efficient-estimator-efficiency/\n---------------------------------------------------------------------------", null, "", null, "Need help with a homework or test question? With Chegg Study, you can get step-by-step solutions to your questions from an expert in the field. Your first 30 minutes with a Chegg tutor is free!" ]
[ null, "https://www.statisticshowto.com/wp-content/uploads/2013/10/cropped-banner-21.jpg", null, "https://www.statisticshowto.com/14422-1200563/", null, "https://imp.pxf.io/i/3128517/1200563/14422", null ]
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https://tug.org/pipermail/tugindia/2003-September/002169.html
[ "# [Tugindia] font size inside the math environment\n\n#NAVEEN.N# naveen at pmail.ntu.edu.sg\nSat Sep 13 22:54:48 CEST 2003\n\nHi all,\n\n1) How do I change the font size inside the displaymath environment.\nI tried the following\n\n${\\large A} = [a_{ij}] = % I tried {\\large A} here \\begin{bmatrix} a_{11} & a_{12} & \\ldots & a_{1n} \\\\ a_{21} & a_{22} & \\ldots & a_{2n} \\\\ \\vdots \\\\ a_{m1} & a_{m2} & \\ldots & a_{mn} \\\\ \\end{bmatrix}$\n\n2) In the final output of the above code the size of the number\nsubscripts is\nseen to be bigger than the size of the alphabet subscripts. How do I\nmake both\nof them the same size?\n\nIllustration: In a_{m1} the size of m is smaller than the size of 1.\n\nNaveen" ]
[ null ]
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https://study-astrophysics.com/marion-01-03/
[ "# Thornton & Marion (Fifth Edition), Chapter 01, Exercise Problem 03 Solution\n\n## Chapter 01. Matrices, Vectors, and Vector Calculus\n\n### Problem 03. Transformation Matrix\n\n##### The problem asks you to\nfind a transformation matrix that satisfying some condition.\n##### This problem gives (or assumes)\nThis matrix rotates a rectangular coordinate through an angle of 120 degrees about an axis making equal angles with the original one.\n1) Direction cosine", null, "$\\lambda_{ij} = cos(x'_i, x_j)$", null, "$\\lambda_{ij} = cos(x'_i, x_j)$\n2) Transformation matrix\n##### Solution", null, "", null, "We can see the relation between the rotated and the original coordinates system. This picture shows that", null, "$\\vec{e_1}' = \\vec{e_2}$", null, "$\\vec{e_2}' = \\vec{e_3}$", null, "$\\vec{e_3}' = \\vec{e_1}$\n\nSo, the transformation matrix is", null, "$\\lambda = \\left( \\begin{array}{ccc} \\lambda_{11} & \\lambda_{12} & \\lambda_{13} \\\\ \\lambda_{21} & \\lambda_{22} & \\lambda_{23} \\\\ \\lambda_{31} & \\lambda_{32} & \\lambda_{33} \\\\ \\end{array} \\right)$", null, "$\\lambda = \\left( \\begin{array}{ccc} \\cos 90^\\circ & \\cos 0^\\circ & \\cos 90^\\circ \\\\ \\cos 90^\\circ & \\cos 90^\\circ & \\cos 0^\\circ \\\\ \\cos 0^\\circ & \\cos 90^\\circ & \\cos 90^\\circ \\\\ \\end{array} \\right)$\n\nTherefore,", null, "$\\lambda = \\left( \\begin{array}{ccc} 0 & 1 & 0 \\\\ 0 & 0 & 1 \\\\ 1 & 0 & 0 \\\\ \\end{array} \\right)$\n\n##### Reference\nhttps://math.stackexchange.com/questions/1599561/determining-the-transformation-matrix-r?newreg=f85754c5968d4b7fae383aabe7bfd2a5" ]
[ null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://i1.wp.com/study-astrophysics.com/wp-content/uploads/2018/02/marion-01-03.jpg", null, "https://i1.wp.com/study-astrophysics.com/wp-content/uploads/2018/02/marion-01-03.jpg", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null ]
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https://www.semanticscholar.org/paper/Classical-field-theories-from-Hamiltonian-and-laws-Zatloukal/272e8360e894bb1d9cbdb9b0dcfe9cdf8c34ed90
[ "• Corpus ID: 119137777\n\n# Classical field theories from Hamiltonian constraint: Symmetries and conservation laws\n\n```@article{Zatloukal2016ClassicalFT,\ntitle={Classical field theories from Hamiltonian constraint: Symmetries and conservation laws},\nauthor={V{\\'a}clav Zatloukal},\njournal={arXiv: Mathematical Physics},\nyear={2016}\n}```\n• V. Zatloukal\n• Published 13 April 2016\n• Mathematics\n• arXiv: Mathematical Physics\nWe discuss the relation between symmetries and conservation laws in the realm of classical field theories based on the Hamiltonian constraint. In this approach, spacetime positions and field values are treated on equal footing, and a generalized multivector-valued momentum is introduced. We derive a field-theoretic Hamiltonian version of the Noether theorem, and identify generalized Noether currents with the momentum contracted with symmetry-generating vector fields. Their relation to the…\n4 Citations\nWe consider the Hamiltonian constraint formulation of classical field theories, which treats spacetime and the space of fields symmetrically, and utilizes the concept of momentum multivector. The\n• V. Zatloukal\n• Physics" ]
[ null ]
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https://www.hackmath.net/en/math-problem/2100
[ "# Cube and water\n\nHow many liters of water can fit into a cube with an edge length of 0.11 m?\n\nx =  1.33 l\n\n### Step-by-step explanation:", null, "Did you find an error or inaccuracy? Feel free to write us. Thank you!", null, "Tips to related online calculators\nDo you know the volume and unit volume, and want to convert volume units?\n\n## Related math problems and questions:\n\n• Cubes", null, "How many cubes with an edge length of 5 cm may fit in the cube with an inner edge of 0.4 m maximally?\n• Cube 6", null, "The surface area of one wall cube is 1600 cm square. How many liters of water can fit into the cube?\n• Water well", null, "Drilled well has a depth of 20 meters and a 0.1 meters radius. How many liters of water can fit into the well?\n• Hectoliter into cube", null, "Can 1hl of water fit into a half-meter cube?\n• Cuboids in cube", null, "How many cuboids with dimensions of 6 cm, 8 cm and 12 cm can fit into a cube with side 96 centimeters?\n• Cube containers", null, "Two containers shaped of cube with edges of 0.7 m and 0.9 m replace a single cube so that it has the same volume as the original two together. What is the length of the edges of the new cube?\n• Water lake", null, "The length of the lake water is 8 meters width 7 meters and depth 120 centimeters. How many liters of water can fit into the water lake?\n• Container", null, "The container has a cylindrical shape the base diameter 0.8 m and the area of the base is equal to the area of the wall. How many liters of water can we pour into the container?\n• Collect rain water", null, "The garden water tank has a cylindrical shape with a diameter of 80 cm and a height of 12 dm. How many liters of water will fit into the tank?\n• Rainfall", null, "The annual average rainfall in India was in Cherrapunji in the year 1981 26 461 mm. How many hectoliters of water fell on 1 m2? Would fit this amount of water into a cube of three meters?\n• Cube into cylinder", null, "If we dip a wooden cube into a barrel with a 40cm radius, the water will rise 10 cm. What is the size of the cube edge?\n• Water", null, "On the lawn which has area 914 m2 rained 3 mm of water. How many liters of water rained?\n• Cube 9", null, "What was the cube's original edge length after cutting 39 small cubes with an edge of 2 dm left 200 dm3?\n• Water level", null, "How many cm will the water level in a cube-shaped tank with an edge of 3 m drop if we discharge 189 hl of water?\n• Cube basics", null, "How long is the edge length of a cube with volume 15 m3?\n• Snails", null, "How many liters of water will fit in an aquarium with bottom dimensions of 30 cm and 25 cm and a height of 60 cm, if we pour water up to a height of 58 cm? How many most snails can we keep in an aquarium, if we know that snails need 600 cm ^ 3 of water fo\n• For thinkings", null, "The glass cube dive into the aquarium, which has a length of 25 cm, width 20 cm and height of 30 cm. Aquarium water rises by 2 cm. a) What is the volume of a cube? b) How many centimeters measure its edge?" ]
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https://nhsjs.com/2012/an-analysis-of-spatial-changes-due-to-special-relativity/
[ "# An Analysis of Spatial Changes Due to Special Relativity\n\n1\n1346", null, "### Abstract:\n\nThis thought experiment addresses loss of volume that occurs when a three-dimensional object is contracted due to relativistic processes. Specifically we investigate the dependence of contraction due to orientation relative to direction of motion. Length contraction states that relative constant velocity will lead to a shortened distance in the direction of motion, when viewed from a separate inertial reference frame. When applied to two dimensional shapes, this results in a reduced area. The problem initially focuses on a square pyramid with the height perpendicular to the direction of motion. Using geometry, it is found that the area of the base is invariant under rotation. The problem is then expanded to general shapes. Any shape can be approximated by inscribing circles within the perimeter, coming arbitrarily close to the shape’s actual area. Since circles are fundamentally invariant under rotation, the change in area of a circle is independent of orientation.  This argument is used to postulate that any two dimensional shape, when contracted in one direction, will have a reduced area that is invariant under rotation. Therefore the contracted area and subsequent reduced volume depend only on the Lorentz factor $$\\gamma$$, and not on relative orientation of the object. A key next step is to determine if this can be extrapolated to three dimensions, using spheres instead of circles.\n\nA PDF version of this paper is available here.\n\n### Introduction:\n\nAccording to Einstein’s work on special relativity, transformations between inertial reference frames are governed by the following two postulates: First, all laws of physics must remain consistent in all inertial reference frames. Second, the speed of light is constant in all inertial reference frames.\n\nRelativistic processes dictate how systems in our universe behave when traveling close to the speed of light. Length contraction, a relativistic process, is a necessary outcome due to time dilation, in order to preserve the measured relative velocity between two inertial reference frames. The time dilation equation is stated as\n\n\\begin{equation}\n\\Delta{}t^{‘}=\\ \\gamma{}\\Delta{}t\n\\end{equation}\n\nThe contracted length divided by the proper time must yield the same value as the proper length over the dilated time.\n\n\\begin{equation}\nL=\\ \\frac{L_o}{\\gamma{}}\n\\end{equation}\n\nwith the Lorentz factor, $$\\gamma$$, given as\n\n\\begin{equation}\n\\gamma{}=\\frac{1}{\\sqrt{1-\\ {\\beta{}}^2}}\n\\end{equation}\n\nWhere $$\\Large \\beta{}=\\ \\frac{v}{c}$$\n\nHere L is the length that is perceived from the external reference frame while Lo, the proper length, is the measured length at rest within a given reference frame. This property of length contraction will lead to a reduction in area of two-dimensional figures and a subsequent reduction in volume of three dimensional shapes when measured from an external inertial reference frame.\n\nThe question is posed as to whether the orientation of three dimensional objects has an effect on the degree of contraction. Specifically, a square pyramid is considered with the base situated in the plane of motion and the height perpendicular to the direction of motion. See figure 1.\n\n### Geometric Analysis of a Pyramid Under Length Contraction:\n\nThe volume of a square pyramid is given as\n\n\\begin{equation}\n\\large\nV=\\frac{1}{3}A_bH\n%eq3\n\\end{equation}\n\nwith Ab the area of the base and H the height.\n\nAs stated, dimensions perpendicular to the direction of motion are not contracted, leading to unilateral effects of special relativity. If the pyramid is oriented with one side of the base perpendicular to the direction of motion, $$\\theta$$ = 0.0 degrees, the base will contract as shown in figure 1. The resulting contracted base area is found by simply reducing one side by 1/$$\\gamma$$.\n\n\\begin{equation}\\large\nA^{‘}=\\ w\\bullet{}w^{‘}=\\ \\frac{w^2}{\\gamma{}}=\\ \\frac{A_o}{\\gamma{}}\n\\end{equation}", null, "Figure 1. A square pyramid with base sides of length w and height H. The top down view shows the contracted dimension of the pyramid base with one side perpendicular to the direction of motion, $$\\theta$$ = 0.0 degrees.\n\nDue to symmetry, the meaningful range of orientation is $$\\theta$$ = [0o, 45o]. At $$\\theta$$ = 45o, the analysis is only slightly more involved. See figure 2. One diagonal dimension is contracted resulting in a parallelogram. The resulting contracted area is found by adding the two resulting equilateral triangles of base $$w\\sqrt{2}$$ and height $$\\Large \\frac{w\\sqrt{2}}{2\\gamma{}}$$\n\n\\begin{equation}\\large\nA^{‘}=\\ w\\sqrt{2}\\ \\frac{w\\sqrt{2}}{2\\gamma{}}=\\ \\frac{w^2}{\\gamma{}}=\\ \\frac{A_o}{\\gamma{}}\n\\end{equation}", null, "Figure 2. Top down view of the pyramid base oriented with the diagonal parallel to the direction of motion, $$\\theta$$ = 45 degrees.\n\nWhen the angle of orientation lies between zero and 45 degrees, the contracted area no longer has equal lengths and therefore involves considerably more math. Equations 6 – 13 give the expressions used to calculate the final contracted area. Again symmetry is used to reduce the problem to matching triangles. See figure 3. Sides L1’ and L2’ are calculated straightforward using length contraction and right angle trigonometry. The remaining side, L3’, is found using the law of cosines. Finally Heron’s formula yields the reduced area. Figure 4 is a plot of the area as a function of orientation using the geometrical approach just described. Included in figure 4 are the lengths of one of the triangle pairs. Note that the solution found here reduces to the limiting cases of zero and forty-five degrees. Also, with $$\\gamma$$ = 1, the proper area is regained. The contracted area remains unchanged through the full range of rotation from zero to forty-five degrees.", null, "Figure 3. Top down view of the pyramid base oriented between zero and 45 degrees to the direction of motion.\n\n\\begin{equation} \\large\n{L_1}^{‘}=w\\ \\sqrt{\\frac{{sin}^2(\\theta{})}{{\\gamma{}}^2}+\\\n{cos}^2(\\theta{})}\\\n\\end{equation}\n\n\\begin{equation} \\large\n{L_2}^{‘}=w\\ \\sqrt{\\frac{cos^2(\\theta{})}{{\\gamma{}}^2}+\\\n{sin}^2(\\theta{})}\n\\end{equation}\n\n\\begin{equation}\n\\large\n{\\theta ^{\\rm{‘}}} = {\\sin ^{ – 1}}\\left( {\\frac{{\\gamma \\sin \\theta }}{{\\sqrt {{{\\cos }^2}\\theta + {\\gamma ^2}{{\\sin }^2}\\theta } }}} \\right)\n\\end{equation}\n\n\\begin{equation}\n\\large\n{\\varphi ^{\\rm{‘}}} = {\\sin ^{ – 1}}\\left( {\\frac{{\\gamma \\cos \\theta }}{{\\sqrt {{\\gamma ^2}{{\\cos }^2}\\theta + {{\\sin }^2}\\theta } }}} \\right)\n\\end{equation}\n\n\\begin{equation}\n\\large\n\\left( {L_3^{\\rm{‘}}} \\right) = \\frac{w}{\\gamma }\\sqrt {\\left( {1 + {\\gamma ^2}} \\right) – 2\\left( {\\cos \\theta \\sin \\theta \\left( {1 – {\\gamma ^2}} \\right)} \\right)}\n\\end{equation}\n\n\\begin{equation}\n\\large\nA^{‘}=\\frac{1}{2}\\sqrt{{{{(L}_1^{‘}}^2+{L_2^{‘}}^2+{L_3^{‘}}^2)}^2+{L_1^{‘}}^4+{L_2^{‘}}^4+{L_3^{‘}}^4}= \\frac{{{A_o}}}{\\gamma }\n\\end{equation}", null, "Figure 4. Plot of contracted area as a function of base orientation with respect to direction of motion. $$\\beta$$ = 0.9/$$\\gamma$$ = 2.3. The proper area is 16 (arb units). The plotted lengths represent one triangle and therefore half of the pyramid base.\n\n### Analysis For General Shapes:\n\nIt is proposed that any two dimensional closed shape can be comprised of an arbitrarily large number of smaller and smaller inscribed circles. If the number of circles inscribed begins with the largest possible circle, the area of the shape can be described as the sum of the areas of all the circles with i = 1 being the largest. However, ordering the size of the circles is not necessary and in fact may cause loss of precision depending on the shape.\n\n\\begin{equation}\n\\large\nA\\left( C \\right) = \\pi \\mathop \\sum \\limits_{i = 1}^\\infty  {\\left( {{r_i}} \\right)^2}\n\\end{equation}\n\nIf the shape is contracted at relativistic speeds, as with the pyramid base, it follows that the circles will become ellipses. See figure 5. The semi-minor axis a represents one half of the contracted dimension of the circle while the semi-major axis b remains the original radius of the circle since it is orthogonal to the direction of motion. Once the shape is contracted, all of the inscribed circles will become ellipses with the new area equation being the sum of all the ellipses.\n\n\\begin{equation}\n\\large\n{\\left( {\\frac{x}{a}} \\right)^2} + {\\left( {\\frac{y}{b}} \\right)^2} = 1\n\\end{equation}\n\n\\begin{equation}\n\\large\nA\\left(E\\right)=\\ \\pi{}\\sum_{i=1}^{\\infty{}}r_ia_i\n\\end{equation}\n\nwith ri, the original radius, becoming the semi-major axis.", null, "Figure 5. A circle before and after contraction due to traveling at relativistic speeds\n\nThen using length contraction for the minor axis, $$\\large a_i = r_i/\\gamma$$\n\n\\begin{equation}\n\\large\nA\\left(E\\right)=\\\n\\pi{}\\sum_{i=1}^{\\infty{}}r_i\\frac{r_i}{\\gamma{}}\n\\end{equation}\n\n\\begin{equation}\n\\large\nA\\left(E\\right)=\\\n\\frac{\\pi{}}{\\gamma{}}\\sum_{i=1}^{\\infty{}}{\\left(r_i\\right)}^2=A_o/\\gamma\n\\end{equation}\n\nAgain, the contracted area reduces to the proper area divided by the Lorentz factor. Due to perfect symmetry of the circle, any rotation of the entire shape, relative to the direction of motion, will only change the direction of contraction while leaving the area of each new ellipse unchanged. It follows that the summation of all of the ellipses will remain invariant under rotation as well.\n\nFigure 6 shows results of a simulation for a square pyramid base at $$\\theta$$ = 0.0 degrees as well as rotated 30 degrees relative to the direction of motion. The contracted areas are simulated for both orientations and compared to the known values as well as the proper area. The simulation is run by subtracting the area of an inscribed circle (non-contracted) or an ellipse (contracted) from the original shape. This process reiterates, scaling the new area appropriately. The rotated case takes longer to converge to the correct area since each newly inscribed ellipse takes up less of the remaining region. However after fourteen iterations the simulation data is within 0.10 percent and continues to converge.", null, "Figure 6. Simulation results of the area of a square pyramid with base oriented at 30 degrees relative to the direction of motion. $$\\beta$$ = 0.40, $$\\gamma$$ ~ 1.10. After fourteen iterations, the simulated and actual contracted areas differ by ~ 0.10%.", null, "Figure 7. The contraction of a circle inscribed inside a square. The original square and circle are dashed lines and the contracted forms are solid lines.\n\n### Discussion:\n\nThe contraction will only occur in the direction of motion. It was found that the reduced area of a pyramid’s base will depend on $$\\gamma$$ and not the orientation of the base relative to the direction of motion. The problem was expanded to any shape and a more elegant solution is proposed using inscribed circles to estimate the area of any two dimensional closed region. Circles were chosen due to their perfect symmetry and resulting invariance under rotation.\n\nIn the future, this problem could be expanded to three dimensions and a similar argument using spheres to replace circles could eliminate the necessity for the shape’s top to be orthogonal to the direction of motion.\n\n### Acknowledgments:\n\nTo Mr. Paul McClernon, we owe him our deepest gratitude.  Without his support and guidance over the entire group’s work, we could not have made it this far.  Thank you for helping us no matter how often our dedication seemed to waver.\n\nTo Mr. Robert Witz, thank you for helping us with all the ins and outs of the English language.\n\n### Citations:\n\nEinstein, Albert. (1905) “On the Electrodynamics of Moving Bodies.” Translated by John Walker. (1999) Retrieved from http://www.fourmilab.ch/etexts/einstein/specrel/www/\n\nTipler, Paul Allen, and Ralph A. Llewellyn. Modern Physics. 5th ed. New York, NY: W.H. Freeman, 2008. Print.\n\nA PDF version of this paper is available here." ]
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https://linen.growthbook.io/t/2400605/hi-team-we-re-running-an-experiment-where-the-growthbook-ana
[ "n\n\nnarrow-psychiatrist-15437\n\n08/17/2022, 10:10 AM\nHi, team! We’re running an experiment where the Growthbook analysis does not find users allocated to\n``variant_id = 0``\n. Looking at the raw data from BigQuery, the variants (\n``0``\nand\n``1``\n) are equally distributed (see screenshot), but variant\n``0``\nusers do not show up in the experiment results. I’ve also verified that the there are no `userId`s exposed to both variants. Here’s the query from the analysis:\nCopy code\n``````-- Activation (binomial)\nWITH __rawExperiment as (\nSELECT\nuser_id,\nexperiment_id,\nvariation_id\nFROM\n`rudderstack_data.experiment`\n),\n__experiment as (\n-- Viewed Experiment\nSELECT\ne.user_id as user_id,\ncast(e.variation_id as string) as variation,\nCAST(e.timestamp as DATETIME) as conversion_start,\nDATETIME_ADD(CAST(e.timestamp as DATETIME), INTERVAL 336 HOUR) as conversion_end\nFROM\n__rawExperiment e\nWHERE\ne.experiment_id = 'hello-sanity_template'\nAND CAST(e.timestamp as DATETIME) >= DATETIME(\"2022-08-15 19:00:00\")\n),\n__metric as (\n-- Metric (Activation)\nSELECT\nuser_id as user_id,\n1 as value,\nCAST(m.timestamp as DATETIME) as conversion_start,\nCAST(m.timestamp as DATETIME) as conversion_end\nFROM\n(\nSELECT\nuser_id,\nCAST(activation_date AS TIMESTAMP) AS timestamp\nFROM\n`dbt_production.dim_user`\nWHERE\nis_activated = TRUE\n) m\nWHERE\nCAST(m.timestamp as DATETIME) >= DATETIME(\"2022-08-15 19:00:00\")\n),\n__distinctUsers as (\n-- One row per user/dimension\nSELECT\ne.user_id,\ncast('All' as string) as dimension,\n(\nCASE\nWHEN count(distinct e.variation) > 1 THEN '__multiple__'\nELSE max(e.variation) END\n) as variation,\nMIN(e.conversion_start) as conversion_start,\nMIN(e.conversion_end) as conversion_end\nFROM\n__experiment e\nGROUP BY\ne.user_id\n),\n__userMetric as (\n-- Add in the aggregate metric value for each user\nSELECT\nd.variation,\nd.dimension,\n1 as value\nFROM\n__distinctUsers d\nJOIN __metric m ON (m.user_id = d.user_id)\nWHERE\nm.conversion_start >= d.conversion_start\nAND m.conversion_start <= d.conversion_end\nGROUP BY\nvariation,\ndimension,\nd.user_id\n),\n__overallUsers as (\n-- Number of users in each variation\nSELECT\nvariation,\ndimension,\nCOUNT(*) as users\nFROM\n__distinctUsers\nGROUP BY\nvariation,\ndimension\n),\n__stats as (\n-- Sum all user metrics together to get a total per variation/dimension\nSELECT\nvariation,\ndimension,\nCOUNT(*) as count,\nAVG(value) as mean,\nSTDDEV(value) as stddev\nFROM\n__userMetric\nGROUP BY\nvariation,\ndimension\n)\nSELECT\ns.variation,\ns.dimension,\ns.count,\ns.mean,\ns.stddev,\nu.users\nFROM\n__stats s\nJOIN __overallUsers u ON (\ns.variation = u.variation\nAND s.dimension = u.dimension\n)``````\nAre you able to spot what’s wrong here?\nf\n\nfuture-teacher-7046\n\n08/17/2022, 12:03 PM\nIf you view the queries in GrowthBook, what do you see as the returned rows for that metric?\nOh, actually I think I know what's going on. We use an inner join in our queries right now so nothing is returned unless there is at least 1 conversion. We're planning to change that to a left join so you can still see the user count.\nn\n\nnarrow-psychiatrist-15437\n\n08/17/2022, 2:13 PM\nOhh, yeah, that could make sense!\nI’ll give it some time until we have more conversions\nUpdate: everything works fine now, after we got more conversion on variant\n``0``\n!\n19 Views" ]
[ null ]
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https://www.shaalaa.com/question-bank-solutions/if-one-of-the-lines-given-by-ax2-2hxy-by2-0-bisect-an-angle-between-the-coordinate-axes-then-show-that-a-b-2-4h2-homogeneous-equation-of-degree-two_142030
[ "# If one of the lines given by ax2 + 2hxy + by2 = 0 bisect an angle between the coordinate axes, then show that (a + b)2 = 4h2 . - Mathematics and Statistics\n\nSum\n\nIf one of the lines given by ax2 + 2hxy + by2 = 0 bisect an angle between the coordinate axes, then show that (a + b)2 = 4h2 .\n\n#### Solution\n\nThe auxiliary equation of the lines given by ax2 + 2hxy + by2 = 0 is bm2 + 2hm + a = 0.\n\nSince one of the lines bisects an angle between the coordinate axes, that line makes an angle of 45° or 135° with the positive direction of X-axis.\n\n∴ the slope of that line = tan 45° or tan 135°\n\n∴ m = tan 45° = 1\n\nor m = tan 135° = tan (180° - 45°)\n\n= - tan 45° = - 1\n\n∴ m = ± 1 are the roots of the auxiliary equation bm2 + 2hm + a = 0.\n\n∴ b(±1)2 + 2h(±1) + a = 0\n\n∴ b ± 2h + a = 0\n\n∴ a + b = ± 2h\n\n∴ (a + b)2 = 4h2\n\nThis is the required condition.\n\nConcept: Homogeneous Equation of Degree Two\nIs there an error in this question or solution?" ]
[ null ]
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http://encyclopedia.kids.net.au/page/st/String_field_theory
[ "", null, "", null, "## Encyclopedia > String field theory\n\nArticle Content\n\n# String theory\n\nRedirected from String field theory\n\nA string theory is a physical model whose fundamental building blocks are extended objects (strings, membranes and higher-dimensional objects) rather than points. String theories are able to avoid problems, such as infinite energy density, associated with the presence of mathematical points in a physical theory.\n\nThe term 'string theory' properly refers to both the 26-dimensional bosonic string theories and to the 10-dimensional superstring theories discovered by adding supersymmetry to bosonic string theory.\nNowadays, 'string theory' usually refers to the supersymmetric variant while the earlier is given its full name 'bosonic string theory'.\nThe different superstring theories were discovered to be different limits of an unknown 11-dimensional theory called M-theory proposed by Horava and Witten in the 1990s.\n\nA central consequence of string theory is that the observed physics of the known universe can be stated as arising from two seemingly incompatible geometries: one being 'as large as we see', and the other being smaller by far than the Planck length. Since both geometries lead to the same observed physics, but the small scale phenomena are beyond human investigation, there has been considerable discussion of the impact of string theory on both the philosophy of science (is it 'science' if no experiment can be run to disprove it?) and the philosophy of mathematics (if two geometries lead to the same physics, should geometry itself not reflect the same universe its discoverers and codifiers live in, and prove that there is a single common phenomenom that the two geometries reflect?).\nOn a more practical level, string theory has led to advances in the mathematics of folding, knots and Calabi-Yau spaces. Because much of this mathematics is new, the uncertainty has been increased somewhat, as very few people can understand either the physics or the mathematics on which it depends.\n\n### Problems with string theory\n\nString theory suffers from two major problems. The first problem is that, in the words of Wolfgang Pauli, it is not even wrong. In other words, it does not make any predictions that may be subject to experimental verification. It can be neither proven nor disproven. That is a serious problem for any theory of physics.\nThe second problem is that it assumes, as did Newtonian mechanics and special relativity, a fixed space-time background. Ultimately, a theory subsuming quantum mechanics is needed which is independent of any fixed space-time and thus consistent with general relativity. M-theory has been hypothesized to overcome the latter problem. Loop quantum gravity overcomes both.\n\n### bosonic string theory\n\nto be written\n\nAll Wikipedia text is available under the terms of the GNU Free Documentation License\n\nSearch Encyclopedia\n Search over one million articles, find something about almost anything!\n\nFeatured Article", null, "", null, "" ]
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https://mcrlabs.com/grease-monkey-3/
[ "", null, "22.6%\n\nTotal Cannabinoids\n\n-\n\nTotal Terpenes\nTested 06/28/2019 for Potency\n\nPOTENCY PROFILES\n\nCannabinoid Profile\n\n22.6%\nTotal Cannabinoids\n19.88%\nMax THC\n\nCannabinoids are quantified using high-performance liquid chromatography.\n\nTotal Cannabinoids is the sum of all detected cannabinoids.\n\nMax THC THCa is converted to THC by heat. Max THC is the maximum amount of THC that can be derived from this product. Use the following formula to find the Max THC: Max THC = THC + (THCa * 0.877)\n\nMax CBD CBDa is converted to CBD by heat. Max CBD is the maximum amount of CBD that can be derived from this product. Use the following formula to find the Max CBD: Max CBD = CBD + (CBDa * 0.877)\n\nOther* This represents materials other than cannabinoids in the product.\n\nTo determine how much Grease Monkey to consume to achieve your desired dose, use the:\n\nmg" ]
[ null, "https://mcrlabs.com/wp-content/uploads/2019/07/S19-20263-1-1.jpg", null ]
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https://discuss.codechef.com/questions/145292/editorial-for-chef-and-secret-ingredients-chefing
[ "You are not logged in. Please login at www.codechef.com to post your questions!\n\n×\n\n# Editorial for Chef and Secret Ingredients [ CHEFING ]\n\n1. Actually this problem is very simple if we use set() function of respective language.\n\n2. You just need to know the very basics of set theory.\n\n3. Here we have to find those ingredients (actually the characters) which are common in all of the given dishes (actually the strings) .\n\n4. So actually we need to find the intersection of all the inputs considering the characters in the input dish(actually a string) as the elements of the set.\n\n5. And then we need to find the length of the set.\n\n## Python Code for the given method\n\n#iterate through testcases\nfor _ in range(int(input())):\n#input N <- number of dishes\nN=int(input())\n\n#input the first dish as a set and name it arr\narr=set(input())\n\n#iterate for N-1 times for getting the other dishes\nfor i in range(N-1):\n# take other dishes as set arr2 in each iteration\narr2=set(input())\n\n#save in arr the value of arr (intersection) arr2\narr=arr.intersection(arr2)\n\n# finally after all iterations output the number of elements in set arr\nprint(len(arr))\n\n\nasked 13 Feb, 17:25", null, "112\naccept rate: 0%\n\n 0 actually I used a smaller code check my submission - Python Solution answered 13 Feb, 17:59", null, "11●2 accept rate: 0%\n 0 Instead we can use a character map and keep updating the character map for each new string and it's alphabet encountered. -- C++ Solution answered 15 Feb, 13:23", null, "1●1 accept rate: 0%\n toggle preview community wiki:\nPreview\n\n### Follow this question\n\nBy Email:\n\nOnce you sign in you will be able to subscribe for any updates here\n\nMarkdown Basics\n\n• *italic* or _italic_\n• **bold** or __bold__\n• image?![alt text](/path/img.jpg \"title\")\n• numbered list: 1. Foo 2. Bar\n• to add a line break simply add two spaces to where you would like the new line to be.\n• basic HTML tags are also supported\n• mathemetical formulas in Latex between \\$ symbol\n\nQuestion tags:\n\n×15,852\n×427\n×189\n×13\n×8\n\nquestion asked: 13 Feb, 17:25\n\nquestion was seen: 322 times\n\nlast updated: 15 Feb, 13:23" ]
[ null, "https://www.codechef.com/sites/default/files/uploads/pictures/f9165c51ef985940d8514751ed722331.jpg", null, "https://www.codechef.com/sites/default/files/uploads/pictures/f9165c51ef985940d8514751ed722331.jpg", null, "https://www.codechef.com/sites/default/files/uploads/pictures/53f5cb93ba55b990972d74eab41b6e8a.jpg", null ]
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https://www.cut-the-knot.org/m/Geometry/BMO2014_3.shtml
[ "BMO 2014, Problem 3\n\nProblem\n\nLet $ABCD$ be a trapezium inscribed in a circle $\\Gamma$ with diameter $AB.$ Let $E$ be the intersection point of the diagonals $AC$ and $BD.$ The circle with center $B$ and radius $BE$ meets $\\Gamma$ at the points $K$ and $L,$ where $K$ is on the same side of $AB$ as $C.$ The line perpendicular to $BD$ at $E$ intersects $CD$ at $M.$", null, "Prove that $KM$ is perpendicular to $DL.$\n\nSolution 1\n\nSince $AB\\parallel CD,$ we have that $ABCD$ is isosceles trapezoid. Let $O$ be the center of $\\Gamma$ and $EM$ meets $AB$ at point $Q.$ Then, from the right angled triangle $BEQ,$ we have $BE^{2} = BO\\cdot BQ.$ Since $BE = BK,$ we get\n\n(1)\n\n$BK^{2} = BO\\cdot BQ.$\n\nSuppose that $KL$ meets $AB$ at $P.$ Then, from the right angled triangle $BAK,$ we have\n\n(2)\n\n$BK^{2} = BP\\cdot BA.$", null, "From (1) and (2) we get $\\displaystyle \\frac{BP}{BQ} =\\frac{BO}{BA} =\\frac{1}{2},$ and therefore $P$ is the midpoint of $BQ:$\n\n(3)\n\n$BP=PQ.$\n\nHowever, $DM\\parallel AQ$ and $MQ\\parallel AD$ (both are perpendicular to $DB).$ Hence, $AQMD$ is parallelogram and thus $MQ = AD = BC.$ We conclude that $QBCM$ is isosceles trapezoid. It follows from (3) that $KL$ is the perpendicular bisector of $BQ$ and $CM,$ that is, $M$ is symmetric to $C$ with respect to $KL.$ Finally, we get that $M$ is the orthocenter of $\\Delta DLK$ by using the well-known result that the reflection of the orthocenter of a triangle to every side belongs to the circumcircle of the triangle and vise versa.\n\nSolution 2\n\nLet $\\Gamma$ be described by the equation $x^2+y^2=1$ and assume the vertices are given as $A(-1,0),$ $B(1,0),$ $C(\\cos 2t, \\sin 2t)$ and $D(-\\cos 2t, \\sin 2t),$ $t\\in(0,\\pi /4).$ This, in particular, means that $\\angle BOC=2t,$ whereas $\\angle BAC=t.$ From here, the slope of $AC$ equals $\\tan t$ such that $AC$ is defined by the equation $y=\\tan t\\cdot (x+1),$ which for $x=0,$ gives $y=\\tan t,$ implying $E=(0,\\tan t).$\n\nThe square of the distance between $B$ and $E$ is\n\n$\\displaystyle BE^{2}=1^{2}+\\tan^{2}t=\\frac{1}{\\cos^{2}t}=\\frac{2}{1+\\cos 2t}.$\n\nThis gives the equation of circle $C(B,E)$ as $(x-1)^2+y^2=2/(1+\\cos 2t),$ wherefrom\n\n$\\displaystyle K=\\bigg(\\frac{\\cos 2t}{1+\\cos 2t}, \\frac{\\sqrt{1+2\\cos 2t}}{1+\\cos 2t}\\bigg)$ and $\\displaystyle L=\\bigg(\\frac{\\cos 2t}{1+\\cos 2t}, -\\frac{\\sqrt{1+2\\cos 2t}}{1+\\cos 2t}\\bigg).$\n\nThe equation of the tangent at $E$ to $C(B,E)$ is $y-\\tan t=x \\cot t.$ It intersects $CD$ at $\\displaystyle M=\\bigg(\\cos 2t \\frac{1-\\cos 2t}{1+\\cos 2t},\\sin2t\\bigg).$ Now it is easy to verify that the product of slopes $MK$ and $DL$ is $-1.$\n\nAcknowledgment\n\nThis is Problem 3 from the XXXI Balkan Mathematical Olympiad. I am grateful to Leo Giugiuc for communicating the problem and the second solution.", null, "" ]
[ null, "https://www.cut-the-knot.org/m/Geometry/BMO2014_3P.jpg", null, "https://www.cut-the-knot.org/m/Geometry/BMO2014_3S.jpg", null, "https://www.cut-the-knot.org/gifs/tbow_sh.gif", null ]
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https://www.programmingr.com/r-error-messages/unexpected-symbol-in-r/
[ "# Fixing the R Error: unexpected symbol in r\n\nThe unexpected symbol in r error message occurs in r programming when a function receives unexpected input. This unexpected input is normally the result of a syntax error resulting from careless typing. The way to fix this problem is to fix any syntactic errors in your code. While this results from a common mistake, there are many things you can do to cause it, making it tricky to diagnose and fix.\n\n### Description of the R error\n\nThe unexpected symbol error usually results from syntactic errors in your code, such as forgetting a closing bracket, forgetting to include commas, and many other typographical mistakes. It can occur when creating or processing an object, but it is not a problem with a variable name or its values. This error Is easy to cause but tricky to diagnose and fix because there are many ways that you can accidentally cause it even within the same function. The good news is, that once you find the source of the problem, it is typically easy to figure out how to fix it. The tricky part is finding out exactly what needs to be fixed.\n\n### Explanation of the error\n\nHere are seven examples of r code that produce our error message. Most of them are simple syntax errors that are easy to catch as you are writing the code.\n\n> x = c(1,2,3,4,5,6,7,8)\n> 2 x\nError: unexpected symbol in “2 x”\n\nIn this example, the operator symbol was omitted in the second line. As a result, it produces our message.\n\n> x = 4\n> if x > 1 {x^2}\nError: unexpected symbol in “if x”\n\nIn this example, the parentheses are omitted in the if statement. As a result, it produces our message.\n\n> c(2 5)\nError: unexpected numeric constant in “c(2 5”\n\nIn this example, the comma was omitted in the first line. As a result, it produces our message.\n\n> x = 4\n> if (x = 4) {x^2}\nError: unexpected ‘=’ in “if (x =”\n\nIn this example, the second equal sign was omitted from the if statement. As a result, it produces our message.\n\n> df = data.frame(“x y” = 1:5)\n> df\\$x y\nError: unexpected symbol in “df\\$x y”\n\nIn this example, the column name has an unusual format that causes it to produce our message.\n\n> data.frame(x = c(1,2,3,4,5,6,7,8)\n+ y = c(2,3,4,5,6,7,8,9))\nError: unexpected symbol in:\n“data frame(x = c(1,2,3,4,5,6,7,8)\ny”\n\nIn this example, a comma has been omitted in creating this data frame. As a result, it produces our message.\n\n> df = data.frame(x = c(1,2,3,4,5,6,7,8),\n+ y = c(2,3,4,5,6,7,8,9)\n+ df\nError: unexpected symbol in:\n” y = c(2,3,4,5,6,7,8,9)\ndf”\nIn this example, the ending parenthesis has been omitted in creating this data frame. As a result, it produces our message.\n\n### How to fix the R error\n\nHere are seven examples of code that fixes our error message as produced in the previous section.\n\n> x = c(1,2,3,4,5,6,7,8)\n> 2*x\n 2 4 6 8 10 12 14 16\n\nIn this example, we added a multiplication sign to fix the problem. It could have been other signs as well.\n\n> x = 4\n> if (x > 1) {x^2}\n 16\n\nIn this example, we added parentheses to fix the problem.\n\n> c(2, 5)\n 2 5\n\nIn this example, we are in a comma to fix the problem.\n\n> x = 4\n> if (x == 4) {x^2}\n 16\n\nIn this example, we added a second equal sign to fix the problem by creating a proper logical statement.\n\n> df = data.frame(“x_y” = 1:5)\n> df\\$x_y\n 1 2 3 4 5\n\nIn this example, we added an underscore to fix the problem by eliminating the space. Another solution is placing x y in single quotes after the dollar sign.\n\n> data.frame(x = c(1,2,3,4,5,6,7,8),\n+ y = c(2,3,4,5,6,7,8,9))\n\nIn this example, we added a comma at the end of the first line to fix the problem.\n\n> df = data.frame(x = c(1,2,3,4,5,6,7,8),\n+ y = c(2,3,4,5,6,7,8,9))\n> df\n\nIn this example, we added the ending parenthesis to fix the problem.\n\nThis error message results from the common mistake of making syntactic errors. This is an easy error to make because there are several ways that it can be produced. However, once you figure out the cause, fixing the problem is quite easy, the challenge is in finding the cause.\n\nScroll to top" ]
[ null ]
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https://technologyformindfulness.com/podcasts/ep33-rohan-gunatilake-buddhify-mindfulness-everywhere/
[ "", null, "## Podcast", null, "### Ep33: Rohan Gunatilake, Creator of ‘Buddhify’ & ‘Mindfulness Everywhere’\n\n``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ```\n\nWhether writing, talking or making products, Rohan Gunatilake is one of the most original and creative voices in modern mindfulness and meditation. Through his company Mindfulness Everywhere, he is the creator of Kara, Sleepfulness, Designing Mindfulness and the best-selling hit app Buddhify. Rohan’s first book, Modern Mindfulness, in out now. Based in Glasgow, Rohan is a trustee of the British Council and was named by Wired magazine in their Smart List of 50 people who will change the world. Rohan created Buddhify as an expression of his love both of meditation and of design & innovation. He started training in mindfulness meditation in 2003 and now many years later considers it a privilege & a delight to be doing the work he does. He is the author of Modern Mindfulness. Find more at https://rohangunatillake.com/.\n\n``` `````` `````` `````` `````` `````` `````` `````` `````` `````` `````` `````` `````` `````` `````` `````` `````` `````` `````` `````` `````` `````` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ```\n\n---> Listen via YouTube at https://youtu.be/OOQ8achZA6k <---\n``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ```\n\n``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ```On each episode of the Technology For Mindfulness Podcast, Robert Plotkin, co-creator of the “Hack Your Mind” series at MIT, explores the intersection between the practice of mindfulness & the use of technology in the modern age. Show notes can be found at TechnologyForMindfulness.com/. Come back often & feel free to subscribe in iTunes or add the Technology For Mindfulness Podcast to your favorite podcast application.\n\n``` `````` `````` `````` `````` `````` `````` `````` `````` `````` `````` `````` `````` `````` `````` `````` `````` `````` `````` `````` `````` `````` `````` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ```\n\n``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` https://Twitter.com/TechForMindful\n``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` https://Facebook.com/TechnologyForMindfulness\n\n``` `````` `````` `````` `````` `````` `````` `````` `````` `````` `````` `````` `````` `````` `````` `````` `````` `````` `````` `````` `````` `````` `````` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` 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``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ```\n\nMusic courtesy of Tobu - Colors [NCS Release]\n``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` 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``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` www.youtube.com/tobuofficial", null, "## The average cellphone user touches their phone 2,617 times a day.\n\nSign up to receive a free, 5 minute guided meditation that helps you gain control over your smartphone, instead of being controlled by it." ]
[ null, "https://technologyformindfulness.com/wp-content/themes/tfm/images/innerbanners2.jpg", null, "https://technologyformindfulness.com/wp-content/uploads/2018/08/Rohan-Gunatilake-1600x1600.jpg", null, "https://technologyformindfulness.com/wp-content/plugins/bloom/images/premade-image-10.png", null ]
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https://www.coolstuffshub.com/weight/convert/pennyweights-to-us-gallons-(dry)/
[ "# Convert Pennyweights to Us gallons (dry) (dwt to US gal dry Conversion)\n\n1 pennyweight is equal to 0.000146708 US gallons (dry).\n\n1 dwt = 0.000146708 US gal dry\n\n## How to convert pennyweights to US gallons (dry)?\n\nTo convert pennyweights to US gallons (dry), multiply the value in pennyweights by 0.000146708.\n\nYou can use the conversion formula :\nUS gallons (dry) = pennyweights × 0.000146708\n\nTo calculate, you can also use our pennyweights to US gallons (dry) converter, which is a much faster and easier option as compared to calculating manually.\n\n## How many US gallons (dry) are in a pennyweight?\n\nThere are 0.000146708 US gallons (dry) in a pennyweight.\n\n• 1 pennyweight = 0.000146708 US gallons (dry)\n• 2 pennyweights = 0.000293416 US gallons (dry)\n• 3 pennyweights = 0.000440124 US gallons (dry)\n• 4 pennyweights = 0.000586832 US gallons (dry)\n• 5 pennyweights = 0.00073354 US gallons (dry)\n• 10 pennyweights = 0.00146708 US gallons (dry)\n• 100 pennyweights = 0.0146708 US gallons (dry)\n\n## Examples to convert dwt to US gal dry\n\nExample 1:\nConvert 50 dwt to US gal dry.\n\nSolution:\nConverting from pennyweights to us gallons (dry) is very easy.\nWe know that 1 dwt = 0.000146708 US gal dry.\n\nSo, to convert 50 dwt to US gal dry, multiply 50 dwt by 0.000146708 US gal dry.\n\n50 dwt = 50 × 0.000146708 US gal dry\n50 dwt = 0.0073354 US gal dry\n\nTherefore, 50 pennyweights converted to US gallons (dry) is equal to 0.0073354 US gal dry.\n\nExample 2:\nConvert 125 dwt to US gal dry.\n\nSolution:\n1 dwt = 0.000146708 US gal dry\n\nSo, 125 dwt = 125 × 0.000146708 US gal dry\n125 dwt = 0.0183385 US gal dry\n\nTherefore, 125 dwt converted to US gal dry is equal to 0.0183385 US gal dry.\n\nFor faster calculations, you can simply use our dwt to US gal dry converter.\n\n## Pennyweights to US gallons (dry) conversion table\n\nPennyweights US gallons (dry)\n0.001 dwt 1.46708 × 10-7 US gal dry\n0.01 dwt 1.46708 × 10-6 US gal dry\n0.1 dwt 1.46708 × 10-5 US gal dry\n1 dwt 0.000146708 US gal dry\n2 dwt 0.000293416 US gal dry\n3 dwt 0.000440124 US gal dry\n4 dwt 0.000586832 US gal dry\n5 dwt 0.00073354 US gal dry\n6 dwt 0.000880248 US gal dry\n7 dwt 0.001026956 US gal dry\n8 dwt 0.001173664 US gal dry\n9 dwt 0.001320372 US gal dry\n10 dwt 0.00146708 US gal dry\n20 dwt 0.00293416 US gal dry\n30 dwt 0.00440124 US gal dry\n40 dwt 0.00586832 US gal dry\n50 dwt 0.0073354 US gal dry\n60 dwt 0.00880248 US gal dry\n70 dwt 0.01026956 US gal dry\n80 dwt 0.01173664 US gal dry\n90 dwt 0.01320372 US gal dry\n100 dwt 0.0146708 US gal dry" ]
[ null ]
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https://metanumbers.com/13261390
[ "## 13261390\n\n13,261,390 (thirteen million two hundred sixty-one thousand three hundred ninety) is an even eight-digits composite number following 13261389 and preceding 13261391. In scientific notation, it is written as 1.326139 × 107. The sum of its digits is 25. It has a total of 4 prime factors and 16 positive divisors. There are 5,280,480 positive integers (up to 13261390) that are relatively prime to 13261390.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Even\n• Number length 8\n• Sum of Digits 25\n• Digital Root 7\n\n## Name\n\nShort name 13 million 261 thousand 390 thirteen million two hundred sixty-one thousand three hundred ninety\n\n## Notation\n\nScientific notation 1.326139 × 107 13.26139 × 106\n\n## Prime Factorization of 13261390\n\nPrime Factorization 2 × 5 × 229 × 5791\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 13261390 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 13,261,390 is 2 × 5 × 229 × 5791. Since it has a total of 4 prime factors, 13,261,390 is a composite number.\n\n## Divisors of 13261390\n\n16 divisors\n\n Even divisors 8 8 4 4\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 16 Total number of the positive divisors of n σ(n) 2.39789e+07 Sum of all the positive divisors of n s(n) 1.07175e+07 Sum of the proper positive divisors of n A(n) 1.49868e+06 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 3641.62 Returns the nth root of the product of n divisors H(n) 8.84871 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 13,261,390 can be divided by 16 positive divisors (out of which 8 are even, and 8 are odd). The sum of these divisors (counting 13,261,390) is 23,978,880, the average is 1,498,680.\n\n## Other Arithmetic Functions (n = 13261390)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 5280480 Total number of positive integers not greater than n that are coprime to n λ(n) 220020 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 863678 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 5,280,480 positive integers (less than 13,261,390) that are coprime with 13,261,390. And there are approximately 863,678 prime numbers less than or equal to 13,261,390.\n\n## Divisibility of 13261390\n\n m n mod m 2 3 4 5 6 7 8 9 0 1 2 0 4 2 6 7\n\nThe number 13,261,390 is divisible by 2 and 5.\n\n• Arithmetic\n• Deficient\n\n• Polite\n\n• Square Free\n\n## Base conversion (13261390)\n\nBase System Value\n2 Binary 110010100101101001001110\n3 Ternary 220221202012121\n4 Quaternary 302211221032\n5 Quinary 11343331030\n6 Senary 1152123154\n8 Octal 62455116\n10 Decimal 13261390\n12 Duodecimal 45364ba\n20 Vigesimal 42hd9a\n36 Base36 7w8jy\n\n## Basic calculations (n = 13261390)\n\n### Multiplication\n\nn×i\n n×2 26522780 39784170 53045560 66306950\n\n### Division\n\nni\n n⁄2 6.6307e+06 4.42046e+06 3.31535e+06 2.65228e+06\n\n### Exponentiation\n\nni\n n2 175864464732100 2332207253953623619000 30928309955508044724770410000 410152380360874829232623067469900000\n\n### Nth Root\n\ni√n\n 2√n 3641.62 236.699 60.3458 26.5777\n\n## 13261390 as geometric shapes\n\n### Circle\n\n Diameter 2.65228e+07 8.33238e+07 5.52495e+14\n\n### Sphere\n\n Volume 9.76913e+21 2.20998e+15 8.33238e+07\n\n### Square\n\nLength = n\n Perimeter 5.30456e+07 1.75864e+14 1.87544e+07\n\n### Cube\n\nLength = n\n Surface area 1.05519e+15 2.33221e+21 2.29694e+07\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 3.97842e+07 7.61515e+13 1.14847e+07\n\n### Triangular Pyramid\n\nLength = n\n Surface area 3.04606e+14 2.74853e+20 1.08279e+07" ]
[ null ]
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http://www.rodsmith.org.uk/alfred-korzybski/science-sanity%20-%200831.htm
[ "# SCIENCE AND SANITY - online book\n\n### An Introduction To Non-aristotelian Systems And General Semantics.\n\n ON GEOMETRY 631 One of the main difficulties is that the structure of this world is such, that it is made up of absolute individuals, each with unique relationship toward environment (in the broadest sense); and we have to speak about it in terms of generalities. 'Laws'. , formulated in the old two-valued ways, can never account adequately for the facts at hand, being only approximations. The mathematical methods which have already been explained give us at once a great advantage. We have seen that if we have a function, y =/(x), let us say, and take the graph of this function, to every point of the graph there corresponds a pair of values x and y. We have seen also that each of the four quadrants I, II, III, IV has a characteristic pair of signs. In quadrant I, both x and y are positive; in II, * is                           Y negative and y positive; in III, both * and y are negative; and finally, in IV, * is positive and y negative. We can easily see that the value of the variables may be thought of as variable conditions different for each individual, and that definite localizations correspond to them. In our example we had to do with a function of one independent variable, and we had a one-dimensional line, curved in two dimensions. When we had a function of two independent variables we had a surface, which in general was curved in a third dimension. By analogy we may pass to any number of dimensions, where by dimension we do not mean anything mysterious, but roughly the number of variables involved in the problem. We see that if we think of the activity of the nervous system in terms of a mathematical function with an enormous number of variables, we shall not only have place for the uniqueness of each individual, determined by the value of the variables and the character of the function, but that this would also imply a localization, which is permanent in a given individual at a given 'time'; which again implies the totality of 'circumstances',. Our function would be N\"f(xu x2,xt, . . .xn). In fact it is hard to see how it is possible to analyse the activities of the nervous system in any other way. The facts are, that every organism is an individual, distinct and different from others, and so we must have means to take this individuality into account. Different values for different variables take care of this point. Similarities are accounted for by the general structural character of the functions. For instance, any quadratic equation with two unknowns gives us a conic section. An equation of the type yi =ax represents a parabola, the graph of any equation of the form xy =o represents a hyperbola ,. For every definite set of values of our variables the implied localization is also definite, which corresponds to the fact that in a given individual at a given 'time'. , the localization is definite. One value for the whole function can be", null, "" ]
[ null, "http://www.rodsmith.org.uk/alfred-korzybski/science-sanity - 0831-1.png", null ]
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https://www.arxiv-vanity.com/papers/astro-ph/0611645/
[ "astro-ph/yymmnnn\n\nLarge Non-Gaussianities in Single Field Inflation\n\nXingang Chen, Richard Easther and Eugene A. Lim\n\n[.3in] Newman Laboratory, Cornell University, Ithaca, NY 14853\n\nDepartment of Physics, Yale University, New Haven, CT 06511\n\n[0.3in]\n\nAbstract\n\nWe compute the 3-point correlation function for a general model of inflation driven by a single, minimally coupled scalar field. Our approach is based on the numerical evaluation of both the perturbation equations and the integrals which contribute to the 3-point function. Consequently, we can analyze models where the potential has a “feature”, in the vicinity of which the slow roll parameters may take on large, transient values. This introduces both scale and shape dependent non-Gaussianities into the primordial perturbations. As an example of our methodology, we examine the “step” potentials which have been invoked to improve the fit to the glitch in the for , present in both the one and three year WMAP data sets. We show that for the typical parameter values, the non-Gaussianities associated with the step are far larger than those in standard slow roll inflation, and may even be within reach of a next generation CMB experiment such as Planck. More generally, we use this example to explain that while adding features to potential can improve the fit to the 2-point function, these are generically associated with a greatly enhanced signal at the 3-point level. Moreover, this 3-point signal will have a very nontrivial shape and scale dependence, which is correlated with the form of the 2-point function, and may thus lead to a consistency check on models of inflation with non-smooth potentials.\n\n## 1 Introduction\n\nThe Cosmic Microwave Background is a treasure trove of information about the primordial universe. An exciting set of current and future experiments [1, 2, 3] will yield an even richer data set, allowing us to make precise tests of inflationary models via their predictions for the form of the primordial perturbations. Typically, these predictions are expressed in terms of the power spectrum, or 2-point function. For a general set of initial inhomogeneities, further information about the perturbations might be gleaned from their higher order correlation functions. If the perturbations are exactly Gaussian, then the N-point functions vanish exactly when is odd, and is fully specified in terms of the 2-point function for even .\n\nAll inflationary models predict some level of non-Gaussianity. Moreover, non-linear corrections will generate non-Gaussianities in the CMB even if the primordial spectrum is purely Gaussian . Thus any observed non-Gaussianities will be a combination of the primordial non-Gaussianity laid down during inflation, and that arising from weak second-order couplings between modes. A rough measure of the non-Gaussianities is provided by the parameter ,\n\n Φ(x)=ΦL(x)+fNL(ΦL(x)2−⟨ΦL(x)2⟩) (1.1)\n\nwhere is the gravitational potential which sources the temperature anisotropies in the CMB . Generically, second order couplings between modes yield , while it has been shown in great detail that for standard single field slow roll inflation, , where is the usual slow roll parameter . The two contributions are cumulative, and the primordial 3-point function is thus swamped by the evolutionary contribution. Moreover, cosmic variance ensures that even a perfect CMB dataset will not allow a conclusive detection of the 3-point function if is of order unity. In addition, the presence of foreground contamination [2, 7] further complicates observational efforts. Current estimates suggest that Planck can achieve a limit of . In this paper paper we will show that this may permit the detection of interesting features in the 3-point statistics for some classes of scale-dependent inflationary models.\n\nThe 3-point correlation function is a separate and independent statistic. Therefore it potentially discriminates between models of inflation with degenerate power spectra. More optimistically, if the amplitude of the 3-point function is large enough for us to map its dependence on both the scale and shape of the momenta triangle, this will be an enormous boon to early universe cosmology, since the 3-point function encodes information in both these properties. However, to realize this possibility, we need to compute the 3-point correlation function for a given inflationary model, and compare these predictions to data. Here we show how to compute the 3-point function for a general model of inflation driven by a single, minimally coupled scalar field. Crucially, we do not assume slow roll, and our methodology applies to both simple models with smooth potentials, and more complicated scenarios where the potential has an isolated “feature”. Since the 3-point function can have both shape and scale dependence, our calculations will underscore the limitations of a single scalar statistic such as . Our analysis is purely theoretical, and we do not compare our computed 3-point function to data, but this issue is being actively addressed by others .\n\nAs cosmological data improves, we will frequently be faced with the problem of interpreting “glitches” in the observed power spectra, and determining whether these represent genuine departures from some concordance cosmology. For instance, the derived from the temperature anisotropies in both the 1 Year and 3 Year WMAP datasets show a noticeable departure from the best fit LCDM spectra around , and the fit can be improved by adding a small “step” to the inflationary potential. The best-fit parameters for this step were extracted from the 1 Year data set in , and from the 3 Year dataset in , and the two analyses find broadly similar values. Of course, any proposal that adds free parameters to the inflationary potential is likely to produce a better fit to data. The question then turns to whether this improvement is good enough to justify the additional parameters. If we focus solely on the 2-point function, this question is addressed via the resulting change in the per degree of freedom, or by Bayesian evidence. However, we will see that a potential with a sharp, localized feature induces a massive enhancement of the primordial 3-point function, relative to our expectations from slow-roll. This enhancement is sufficiently dramatic to boost the non-Gaussianities to the point where they may be detected experimentally. Consequently, this provides a stringent check on any proposal for a nontrivial inflationary potential, since the specific form of the non-Gaussianities will be heavily correlated with the 2-point function. Such correlations will also extend to higher order statistics such as the trispectrum and beyond, though we do not consider them here [12, 13, 14, 15, 16, 17, 18].\n\nIn models with a non-trivial potentials, an accurate computation of the power spectrum must be performed numerically. We begin our examination of the 3-point function by reproducing existing semi-analytical results that assume slow roll. We then analyze a specific model where the inflaton potential has a small, sharp “step” to illustrate our methodology in a non-trivial setting. Since this model has already been used to improve the fit between LCDM cosmology and the observed power spectrum, we can take the best-fit parameter values for this step and compute the resulting 3-point function. Interestingly, although such a feature only causes a small correction in the 2-point correlation function , it amplifies the non-Gaussianity by a factor around 1000. Roughly speaking111As we will discuss later, is not a good statistic for such non-trivial non-Gaussianities, although it remains useful as a rough guide to the amplitude of the 3-point function., is boosted from to , making the non-Gaussianity a potentially useful constraint. The leading term in the cubic Lagrangian responsible for this boost differs from those used in the previous calculations in Ref. [6, 20, 21].\n\nThe distinctive feature of this non-Gaussianity is its dependence on the scale of the momenta triangle . Firstly, it has a characteristic “ringing” behavior; as we will see, a suitably defined generalization of oscillates between positive and negative values. Secondly, the impact on the 3-point function is localized around the wavenumbers that are most affected by any feature, and dies away over several e-folds. These features distinguish a step model from other single field theories with large non-Gaussianities, such as DBI inflation [23, 24] or k-inflation models , where the non-Gaussianities are present at all scales and the running is very slow [26, 22, 21, 27, 28].\n\nThis paper is divided into the following sections. Section 2 details the step potential model that we are considering; we lay out our conventions in this section. Section 3 details the analytic forms of the 3-pt correlation functions, which we will proceed to integrate numerically in Section 4. We discuss our results in Section 5 and conclude in Section 6.\n\n## 2 Slow roll model with a feature\n\nFor single field inflation driven by a minimally coupled scalar, the action is\n\n S=∫dx4√g[M2p2R+12(∂ϕ)2−V(ϕ)] (2.2)\n\nwhere the potential is designed such that the inflaton field is slowly rolling for long enough to drive inflation. We define the slow roll parameters by222Note these are related to the potential slow roll parameters via , , by , .\n\n ϵ = ˙ϕ22M2pH2 , (2.3) η = ˙ϵϵH. (2.4)\n\nFor the quadratic potential, , . Assuming transplanckian field values, or , both slow roll parameters and their higher derivative cousins are sub-unity.\n\nWe add a step into the slow roll potential, with the form proposed by \n\n V(ϕ)=12m2ϕ2[1+ctanh(ϕ−ϕsd)], (2.5)\n\nwhere the step is at , with size and gradient respectively. While the presence of the step at just the “right” place would require a tuning, one can imagine potentials with many steps, so that the odds are high that at least one of them would fall inside the range of relevant to the cosmological perturbations.\n\nAs discussed in [30, 19] this step induces an oscillatory ringing in the power spectrum. When the inflaton rolls through the step, it undergoes a strong momentary acceleration. In realistic models the step is typically less than 1% of the overall height of the potential. In this case, at all times, and . On the other hand, and its higher derivatives can grow dramatically as the inflaton crosses the step, often to the point where they exceed unity, as shown in Fig. (1).\n\nWe begin with an order of magnitude estimate of the values of the slow roll parameters. The step has a depth and a width (setting ). When the inflaton falls down the step, of the potential energy is converted to kinetic energy, resulting in an increase in on the order of , within a time interval . Using these quantities, we can estimate\n\n η=˙ϵHϵ≈7c3/2dϵ (2.6)\n\nand\n\n ˙η = H(¨ϵH2ϵ+ϵη−η2) (2.7) ≈ H10c2d2ϵ(1+0.7dϵ√c−4.5cϵ) ≈ H10c2d2ϵ .\n\nThese values have the same order of magnitude as the peaks obtained numerically in Fig. 1.\n\nAfter this acceleration, the inflaton is damped by the friction term in\n\n ¨ϕ+3H˙ϕ+dVdϕ=0 (2.8)\n\nand relaxes to the attractor solution. This relaxation time is determined by the first two terms in Eq. (2.8), and is of order . This corresponds to the decay width of the peaks in Fig. 1. It is easy to see that and during the relaxation are of order and , respectively. Our numerical results were obtained with , – the central values of the step corresponding to the low- glitch analyzed by Covi et al. . Thus , and get , consistent with Fig. 1. Note that is chosen to be around 1 in the plot, is the conformal time, and prime is the derivative with respect to .\n\nThe non-linear couplings of the perturbations are proportional to powers of the slow roll parameters which characterize the background evolution, even when the slow roll parameters are large. Consequently, a deviation from slow roll results in a large mixing (“interaction”) of modes, and large non-Gaussianities. The amplification of the non-Gaussianity relative to a smooth potential is roughly characterized by . From our previous discussion we know that the during relaxation after traversing the feature is of order . During acceleration this quantity will depend on the firm of the feature, and is greater than unity for the step potential. Thus any sharp feature will greatly boost the non-Gaussianity.", null, "Figure 1: The η′ (left) and ϵ (right) evolution over the step for the model c=0.0018, d=0.022, with its amplitude momentarily η′≈10c2/(ϵd2). This is the primary source of the large non-Gaussianities in the step potential, as the leading term in the 3 point expansion is of order η′ϵ. On the other hand, since the height of the step is small, so the ratio of the kinetic energy to the potential energy remains tiny and ϵ≪1.\n\n## 3 3-point correlation functions\n\nIn this section, we sketch out the derivation of the 3-point correlation function, following [6, 20, 21], writing in terms of integrals over conformal time , with the integrand being a finite expansion in slow roll parameters. Although we work with the step potential, the analysis is applicable to a generic feature. We begin by perturbing the fields in the ADM metric \n\n ds2=−N2dt2+hij(dxi+Nidt)(dxj+Njdt) (3.9)\n\nusing the comoving gauge [32, 6]\n\n hij=a2e2ζδij , (3.10)\n\nwhere and are Lagrangian multipliers, and is the scalar perturbation. Note that in this gauge vanishes. The power spectrum is given by the 2-point correlation function of the curvature perturbation\n\n ⟨ζ(x)ζ(x)⟩ = ∫d3k1(2π)3d3k2(2π)3⟨ζ(k1)ζ(k2)⟩ei(k1+k2)⋅x (3.11) = ∫dkkPζ ,\n\nwhere is the power spectrum.\n\nThe 3-point correlation function is substantially more complicated. As non-Gaussianities arise from departures from nonlinear couplings, we must compute the action (2.2) to cubic order in the perturbation:\n\n S3 = ∫dtd3x{a3ϵ2ζ˙ζ2+aϵ2ζ(∂ζ)2−2aϵ˙ζ(∂ζ)(∂χ) (3.12) +a3ϵ2dηdtζ2˙ζ+ϵ2a(∂ζ)(∂χ)∂2χ+ϵ4a(∂2ζ)(∂χ)2+2f(ζ)δLδζ|1} ,\n\nwhere\n\n χ = a2ϵ∂−2˙ζ , (3.13) δLδζ∣1 = a(d∂2χdt+H∂2χ−ϵ∂2ζ) , (3.14) f(ζ) = η4ζ2+terms with derivatives on ζ . (3.15)\n\nHere is the inverse Laplacian and is the variation of the quadratic action with respect to the perturbation . The cubic action (3.12) is exact for arbitrary and . Note that the highest power of slow-roll parameters which appears is of , when we recall that contains a multiplicative factor of . The fact that the series expansion in slow-roll parameters is finite is important since, although remains small in the presence of a feature, and may become large. The last term in (3.12) can be absorbed by a field redefinition of ,\n\n ζ→ζn+f(ζn) . (3.16)\n\nAfter this redefinition, the interaction Hamiltonian in conformal time is\n\n Hint(τ) = −∫d3x{aϵ2ζζ′2+aϵ2ζ(∂ζ)2−2ϵζ′(∂ζ)(∂χ) (3.17) +a2ϵη′ζ2ζ′+ϵ2a(∂ζ)(∂χ)(∂2χ)+ϵ4a(∂2ζ)(∂χ)2} .\n\nIt is more convenient to work in Fourier space, so\n\n ⟨ζ(x)ζ(x)ζ(x)⟩=∫d3k1(2π)3d3k2(2π)3d3k3(2π)3⟨ζ(k1)ζ(k2)ζ(k3)⟩ei(k1+k2+k3)⋅x. (3.18)\n\nThe field redefinition equation (3.16) introduces some extra terms in the 3-point correlation function,\n\n ⟨ζ(k1)ζ(k2)ζ(k3)⟩ = ⟨ζn(k1)ζn(k2)ζn(k3)⟩ (3.19) + η⟨ζ2n(k1)ζn(k2)ζn(k3)⟩+sym+O(η2(Pζk)3) ,\n\nwhere denotes the Fourier transform of – note that this is not the square of . In (3.19), the slow roll parameters are evaluated at the end of inflation. Although we can no longer perturbatively expand in terms of if is large, the terms that we neglected are of higher order in . Since , we can neglect these unless becomes ridiculously large, which it never does in any case of interest to us. We have only considered the first term in (3.15), since all other terms involve at least one derivative of , and vanish when evaluated outside the horizon.\n\nThe 3-point correlation function at some time after horizon exit is then the vacuum expectation value of the three point function in the interaction vacuum\n\n ⟨ζ(τ,k1)ζ(τ,k2)ζ(τ,% k3)⟩=−i∫ττ0dτ′ a ⟨[ζ(τ,k1)ζ(τ,k2)ζ(τ,k3),Hint(τ′)]⟩ . (3.20)\n\nBefore we proceed, let us pause and point out a subtlety hidden inside equation (3.20): the 3-point on the left hand side is evaluated with the interaction vacuum while the the right hand side is evaluated at the “true” vacuum. The interaction Hamiltonian evolves the true vacuum to the interaction vacuum at the time we evaluate the 3-point function. Neglecting this will lead to errors as first pointed out by Maldacena [6, 33, 34].\n\nThe three terms in the second line of (3.17) are of higher order in slow-roll parameters, and were properly neglected in Refs. [6, 20, 21], which assume that and are always small. In our calculation, the terms are small, but the term may be large and in fact dominant. the step.\n\nWe now decompose into its Fourier modes and quantize it by writing\n\n ζ(τ,x)=∫d3p(2π)3ζ(τ,k)eip⋅x , (3.21)\n\nwith associated operators and mode functions\n\n ζ(τ,k)=u(τ,k)a(k)+u∗(τ,−k)a†(−k) , (3.22)\n\nwhere and satisfy the commutation relation . The “true” vacuum is annihilated by the lowering operator . The power spectrum is then\n\n Pζ≡k32π2|uk|2 . (3.23)\n\nwith is evaluated after each mode crosses the horizon.\n\nMeanwhile is the solution of the linear equation of motion of the quadratic action,\n\n v′′k+k2vk−z′′zvk=0 , (3.24)\n\nwhere we have used the definitions\n\n vk≡zuk ,    z≡a√2ϵ . (3.25)\n\nOur choice of vacuum implies that the initial condition for the mode function is given by the Bunch-Davies vacuum\n\n vk(τ0) = √12k v′k(τ0) = −i√k2 (3.26)\n\nwhere we have neglected an irrelevant phase, since (3.24) is rotationally invariant.\n\nTo compute the 3-point correlation function, one simply substitutes these solutions into equation (3.17), and integrates the mode functions from through to the end of inflation. This integral can be done semi-analytically for simple models, provided the slow roll parameters are small and relatively constant. Any departures from this serene picture will render the semi-analytic approach intractable – including the step potential we consider here.\n\nInspecting equations (3.17) and (3.20) and recalling that , it is clear that the 3-point correlation function consists of a sum of integrals of the form\n\n Iϵ2∝R[∏iui(τend)∫τendτ0dτϵ2a2ξ1(τ)ξ2(τ)ξ3(τ)+O(ϵ3)] (3.27)\n Iϵη′∝R[∏iui(τend)∫τendτ0dτϵη′a2ξ1(τ)ξ2(τ)ξ3(τ)] (3.28)\n\nwhere is either or . In a single field model after Hubble crossing as it freezes out, while oscillates rapidly at early times, so its contribution to the integral tends to cancel. Thus the integral is dominated by the range of during which the modes leave the horizon.\n\nNow let us consider the potential (2.5). We expect , making (3.27) negligible. However, and can clearly be very large as changes rapidly over a very short time. This flagrant violation of slow roll means that we expect a large contribution from the term in the integral, of the order , where is the acceleration time for the inflaton by the feature. From the qualitative estimation in Sec. 2, we know this is . The best parameters from Covi et al. for a step that matches the low- glitch seen in current CMB data are and . Thus we expect a boost in the 3-point statistic of at scales affected by the step. Given that the baseline 3-point function is proportional to , we see that this pushes the expected value up to from .\n\nWe now focus on the term, and relegate the discussion of the subleading terms to an appendix (where a field redefinition term is also subleading since the effect of the feature dies away towards the boundary). So our leading term is\n\n i(∏iui(τend))∫τend−∞dτa2ϵη′(u∗1(τ)u∗2(τ)ddτu∗3(τ)+two perm)(2π)3δ3(∑iki)+c.c. ,\n\nwhere the “two perm” stands for two other terms that are symmetric under permutations of the indices 1, 2 and 3, where 1, 2, 3 are shorthand for , and .\n\nWe must numerically solve the equations of motion to get , and then evaluate (3). The contributions from the other five terms in the interaction Hamiltonian (3.17) in the appendix can be calculated in a similar fashion.\n\n## 4 Numerical Method\n\nThe equations of motion for the background fields and the perturbations are\n\n H(τ) = a′a2, (4.29) d2adτ2 = a6M2p(−ϕ′2+4a2V(ϕ)), (4.30) d2ϕdτ2 = −2a′aϕ′−a2dVdϕ, (4.31) d2vkdτ2 = −(k2−z′′z)vk, (4.32)\n\nwhere is our time parameter. We choose initial conditions and units for such that the step occurs around , and the mode with comoving wavenumber crosses the horizon at the same point. The initial conditions of the perturbations are given by (3.26).\n\nWe need to evaluate (3) and equations (A.47) to (A). They all have the form (3.27) and (3.28). At early times the integrand is highly oscillatory and the net contribution per period is thus very small. However the early time limit of the integral must be handled with care, since a sharp cutoff imposed by a finite initial time will introduce a spurious contribution of into the result.\n\nAnalytically, the standard procedure is to Wick rotate the integral slightly into the imaginary plane [6, 20, 21] in order to eliminate the oscillatory terms. Unfortunately, it is not straightforward to implement this approach numerically. Instead we introduce a “damping” factor into the integrand\n\n Iϵ2∝∫τendτearlydτa2ϵ2ξ1ξ2ξ3×eβ(k1+k2+k3)(τ−τend), (4.33)\n\nand similarly for the term. This is approximately equivalent to rotating the contour integral from the real axis slightly into the imaginary plane . The relative error introduced by this numerical procedure can be estimated by taking the difference between the integration over and and is thus . If we ensure\n\n |βkτcrossing| ≪ 1 , (4.34) |βkτearly| ≫ 1 , (4.35)\n\nthen this contribution is negligible, recalling that .\n\nWe can adjust both and . Choosing an appropriate combination will damp out the oscillatory contributions at early times while having no effect on the integrand during Hubble crossing, which provides the dominant contribution to the 3-point function. We tested this scheme against the known analytical results for 3-point functions, which we match to within a few percent. However, and are set by hand in our code, so while this approach is suitable for an initial survey, we are developing a more flexible scheme333Since finishing this paper, we have developed a more robust numerical methods which regulates the integral at early times without using a damping factor . The regularization is prone to suppressing the 3-point function since pushing to large negative values make the numerical integrations very time consuming. In practice the results shown here for the step potential may be lower than their exact values, with a greater suppression at larger values of .. As an aside, the imaginary component of the integral goes to infinity, scaling as . However, as long as we stop the integration a few efolds after horizon crossing, the integration is stable.\n\n## 5 Results and Discussion\n\n### 5.1 Shape and Scale\n\nEvery 3-point correlation function has two main attributes: shape and scale. Unlike previous treatments, we cannot assume that the 3-point function is scale invariant. Before we compare the analytical and numerical results, let us define a more useful parameter than the rather unwieldy raw 3-point function,\n\n ⟨ζ(k1)ζ(k2)ζ(k3)⟩=(2π)7Nδ3(k1+k2+k3)1Πik3iG(k1,k2,k3).s (5.36)\n\nWe have factored out the product , since during inflation and there are 6 factors of in the 3-point function. The function depends on the details of the integrals listed in Sect. 3 and is some constant that we will come back to later. If the slow roll parameters stay small and constant we can factor the power spectrum out of [6, 20, 21]\n\n ⟨ζ(k1)ζ(k2)ζ(k3)⟩=(2π)7δ3(k1+k2+k3)(Pζk)21Πik3iA(k1,k2,k3) , (5.37)\n\nwhere is a quantity of and the convention here follows Ref. . In terms of ,\n\n NG=(Pζk)2A. (5.38)\n\nSince the power spectrum is statistically isotropic , we assume that this is also true in the bispectrum and impose rotational symmetry so that, combined with the conserved momentum constraint, the 3-point correlation function depends only on the amplitudes of the ’s.\n\nRecall the definition of , equation (1.1), which is often used when computing the non-Gaussianities in the CMB sky [37, 38, 1, 8]. Current constraints on , from 3 years of WMAP data are \n\n −36\n\nAs pointed out in , this ansatz assumes that the 3-point correlation function has the following local form\n\n ⟨ζ(k1)ζ(k2)ζ(k3)⟩local=(2π)7δ3(k1+k2+k3)(−310fNL(Pζk)2)Σik3iΠik3i, (5.40)\n\nor, in terms of\n\n Alocal=−310fNLΣik3i. (5.41)\n\nIn other words, the constraint (5.39) is applicable only if the primordial non-Gaussianities have the local form (5.40).\n\nSince equation (1.1) does not contain an explicit scale, this form is scale-invariant (neglecting the correction terms)\n\n ⟨ζ(k1)ζ(k2)ζ(k3)⟩local=λ9⟨ζ(λk1)ζ(λk2)ζ(λk3)⟩local. (5.42)\n\nNote that there is a small violation of scale-invariance in the 3-point correlation function stemming from the power spectrum term which we have ignored in the proceeding argument. In fact, equation (5.40) itself hides a further ambiguity: at what value of are we evaluating the power spectrum? This omission is justified if we assume that the power spectrum is almost scale-invariant, which is true for most “standard” slow roll single scalar field models. However, for the step potential we are considering, the power spectrum itself undergoes drastic oscillations of . We will come back to this point later.\n\nWe first reproduce known results for a simple slow roll inflation model ,\n\n V(ϕ)=12m2ϕ2 (5.43)\n\nwhere .\n\nFollowing , we can write down the explicit form of\n\n ASR=ϵ(−18Σik3i+18Σi≠jkik2j+1k1+k2+k3Σi>jk2ik2j)+η(18Σik3i)+O(ϵ2) , (5.44)\n\nwhere terms include . Comparing this to the local form equation (5.41) we can see that for the equilateral case, where the second equality comes from the fact that for the potential (5.43). In the other extreme of the squeezed triangle case, . This result was first derived by Maldacena, who pointed out that for single field slow roll inflationary models, and is thus unobservable. We compared our results to equation (5.44), and as Figure 2 shows, our code is accurate to within a few percent, which is within the error expected from the analytical estimate itself.", null, "Figure 2: Comparison of A/k3 between the analytical slow roll results of equation (5.44) and numerical results from our code, for the equilateral case where k1=k2=k3=k run from 0.5\n\nSince the potential has no features, the 3-point correlation function is essentially “scale independent”. By this, we mean that triangles which exhibit scaling symmetry posses identical values for their 3-point functions.\n\n### 5.3 Step potential\n\nConfident that our code is robust, we now compute the 3-point correlation functions for the step potential (2.5). We focus on the best fit model of Covi et al. for a step which affects modes at large angular scales, namely where . We choose the same inflaton mass used with the model, so that we can compare the results.", null, "Figure 3: The power spectrum for the standard m2ϕ2 model and the step potential (2.5) for a decade of k. In this plot, m=10−6Mp for the m2ϕ2 model while for the step potential we have used the parameters (c,ϕs,d)=(0.0018,14.81Mp,0.022).\n\nConsider the power spectrum, Figure (3). As first described in , the violation of slow roll induces a temporary growing/decaying behavior in the evolution of the modes before horizon crossing, resulting in a “ringing” of the power spectrum. Thus we cannot simply factor out the power spectrum from as we have done in equation (5.37); doing so will introduce spurious contributions to any statistic that we might obtain.444For example, dividing it by [40, 41] will entangle the ringing in the power spectrum with the ringing in the bispectrum. The 3-pt correlation function is, in principle, a completely independent statistic from the 2-pt correlation function. In our definition, we make it clear that the only content of the division by is simply to state that the 3-pt amplitude is roughly the product of 2-pt amplitude. In the future, we envisage that the 3-pt will be part of the set of cosmological parameter space that we can constrain from observations of the CMB and large scale structure, and this definition completely disentangle the 2-pt from the 3-pt and thus avoid any unphysical cross-correlations between the two statistics. On the other hand, we expect the 3-point correlation to be of order , so if we set in equation (5.36) where (i.e. the observed value of the power spectrum), then the dimensionless quantity\n\n G(k1,k2,k3)k1k2k3=1δ3(k1+k2+k3)(k1k2k3)2~P2(2π)7⟨ζ(k1)ζ(k2)ζ(k3)⟩ (5.45)\n\nis what we want to plot. This quantity has the great advantage that the strong running from the scaling of the 3-point correlation function (see for example equation (5.42)) is factored out, so we are left with the scaling that is generated by from the non-linear evolution of the modes.\n\nConsider the equilateral case, Figure (4). Analogously to the power spectrum, the presence of the step induces a ringing of the equilateral bispectrum. The ringing frequency in the 3-point function is roughly 1.5 times that in power spectrum, due to the presence of three factors of in the integrand instead of two.", null, "Figure 4: The running of non-Gaussianity G/k1k2k3 (in the equilateral case k1=k2=k3≡k) for the large scale step potential model of (c,ϕs,d)=(0.0018,14.81Mp,0.022). For comparison, the standard slow roll model will yield G/k1k2k3≈O(ϵ).\n\nThe plots in Figures 5 illustrate the complicated landscape of the non-Gaussianities. Compared to the amplitude of the model, the non-Gaussianities are magnified several hundredfold and can be larger for other viable step parameters (Figure 7), since we are taking the mid-point values from Covi et al.. As we explained in Sec. 3, the dominant contribution to the non-Gaussianities comes from the term (Figure 1). As we estimated, is of order since the contribution from the term is . This is confirmed by our numerical results; the shape of the non-gaussianity remains constant but its amplitude is rescaled as we vary and together (Figure (7)).\n\nIn Figure 6, we plot the integrand with respect to for the modes which cross the Hubble horizon around the step. We have suppressed the early time oscillations via our “damping trick” in order to isolate the actual contribution to the integral. The step also occurs around this time. Comparing this with the plot Figure (1), we see that the oscillatory nature of modulates the single hump of , yielding complicated structures. Depending on the combined phases of the modes, the modulation can be both constructive and destructive, leading to the “ringing” of the primordial bispectrum. The non-linear interactions between the modes have become dominant due to the large coupling term around crossing, mixing up the perturbations in non-trivial manner and generating large departures from Gaussianity. This is in contrast to the ringing of the power spectrum which is caused by the change in the effective mass of the modes as they cross the horizon.", null, "Figure 5: The shape of non-Gaussianities G/k1k2k3 for the large scale step potential model of (c,ϕs,d)=(0.0018,14.81Mp,0.022), with k1 ranging from 6.5 to 2.5 going from top left to bottom right. We have set the forbidden triangle regions outside of ki≤kj+kl , i≠j,l to −10 for visualization purposes.", null, "Figure 6: The real component of the integrand (k1k2k3)2a2ϵη′u1(0)u2(0)u3(0)u∗1u∗2u∗3′ plotted for the modes (k1,k2,k3), with the step occurring around τ=−1 and inflation ending at τ=0. Note that the early time oscillations have been suppressed by the inclusion of a damping factor β in the integrand.", null, "Figure 7: Variation of G/k3 in response to changing the parameters of the step for the equilateral triangle case. The left (right) plot describes the variation of c (d) from the smallest to biggest amplitude in the order c=0.0009,0.0018,0.0036,0.0054 fixing d=0.022 (d=0.044,0.022,0.011,0.0074 fixing c=0.0018). We keep the location of the step fixed at ϕs=14.81Mp.\n\nTo bound this specific non-Gaussian signal with current WMAP data, we would have to redo the full sky analysis using the data for our complicated scale and shape dependent predictions of the bispectrum. This is not a trivial task and requires new methods to be developed before we can approach the problem. A brute force computation of the likelihood is computationally daunting since the bispectrum here is clearly unfactorizable into a product of separate integrals over the ’s555A recent paper has begun to address this important question.. However, we can make some guesses. We have previously alluded to the fact that the statistic (5.45) is similar to . By comparing equation (5.41) to equation (5.45), we see that\n\n fNL∼−10k1k2k33Σik3iG(k1,k2,k3)k1k2k3. (5.46)\n\nSince the factor is roughly of then which is within sights of the next generation CMB experiments . However, in our case only a subset of all possible triangles we could draw on the sky have a large 3-point function, which increases the risk that cosmic variance will swamp any signal. On the other hand, we have a specific template for the 3-point function, which can then be cross-correlated with the 2-point function associated with any putative step.\n\n## 6 Conclusion\n\nWhile simple models of slow roll inflation do not exhibit observable non-Gaussianities, the space of models of inflation that can produce a power spectrum which fits the CMB data is infinite. To discriminate between these degenerate models, some other independent statistic derived from the CMB must be used and the bispectrum is becoming the most promising candidate for this role. Unlike the power spectrum, the computation of the bispectrum for a given inflationary model is a messy affair. In this paper, we have developed a robust numerical code which can compute the 3-point correlation function for a non-slow roll single field model.\n\nUsing this method, we numerically integrated the 3-point correlation functions for a class of inflationary models where slow roll is violated for a brief moment. We show that the addition of this step breaks the scale invariance of the bispectrum in a non-trivial way, leading to large non-Gaussianities. This numerical technique can be extended to solve more complicated problems such as multifield models of inflation [42, 43] and to compute higher statistics such as the trispectrum .\n\nMirroring the behavior of the power spectrum, a temporary violation of slow roll introduces a ringing into the bispectrum, resulting in a structure much like the transient vibrations of a rug being beaten. These large non-Gaussianities are generated when the non-linear coupling between the modes, , becomes temporarily large during this period of non-slow roll behavior. In order to check whether we can constrain such models with data, a full sky analysis must be undertaken with either current or future data, a task which is beyond the scope of this paper. However, we argue that at least for the models considered in this paper, the non-Gaussianities may be large enough to be within reach of the Planck satellite.\n\nIn the near future we can expect even better quality CMB data to become available, opening up the possibility of cross-correlating higher order statistics such as the bispectrum with the power spectrum. This should provide a powerful consistency check on any models of inflation with a sharp feature in the potential, even if this improves the fit to the 2-point function. Consequently, we have laid the groundwork for an analog of the “consistency relationship” of simple models of slow roll inflation, where the amplitude and slope of the scalar and tensor spectra are expressed as functions of just three independent parameters - the observed amplitude of the scalar spectrum, and the first two slow roll parameters, and . In this case considered here, the parameters that describe any step or other near-discontinuity would fix both the 2- and 3-point functions, yielding a consistency test for a very broad class of inflationary models. In practice we could fit separately to the 2- and 3-point functions and then check that the results matched. Alternatively, one could simultaneously include information from both the power spectrum and bispectrum into the cosmological parameter estimation process. In this case, we could fit directly to the parameters , and (along with the “average” slow roll parameters), generalizing the approach of [45, 46]. Finally, if it ever becomes possible to determine the 3-point function from observations of large scale structure, this information would further tighten the constraints on models where slow roll is briefly violated.\n\n## 7 Acknowledgments\n\nWe thank Wayne Hu, Hiranya Peiris, Kendrick Smith, Henry Tye and an anonymous referee for a number of useful discussions. XC thanks the physics department of Yale university and KITP for their hospitality. XC is supported in part by the National Science Foundation under grant PHY-0355005. RE and EAL are supported in part by the United States Department of Energy, grant DE-FG02-92ER-40704.\n\n## Appendix A Other terms\n\nIn Section 3, we have given the expressions for the leading non-Gaussianity in the case when is large but remains small. In this appendix, we give the expressions for the subleading terms for this case from the Hamiltonian (3.17).\n\nThe subleading contributions come from the first line of (3.17). Contribution from term:\n\n 2i∫τend−∞dτ a2ϵ2(∏iui(τend))(u∗1du∗2dτdu∗3dτ+two perm)(2π)3δ3(∑iki)+c.c. . (A.47)\n\nContribution from term:\n\n −2i∫τend−∞dτ a2ϵ2(∏iui(τend)u∗i(τ))(k1⋅k2+two perm)(2π)3δ3(∑iki)+c.c. . (A.48)\n\nContribution from term:\n\n −2i∫τend−∞dτ a2ϵ2(∏iui(τend))(u∗1du∗2dτdu∗3dτk1⋅k2k22+five perm)(2π)3δ3(∑iki)+c.c. .\n\nThe redefinition (3.15) contributes\n\n η2 |u2|2|u3|2∣∣∣τ→τend(2π)3δ3(∑iki)+two perm . (A.49)\n\nNote that in Ref. [6, 20, 21], where sharp features are absent, these four terms give leading contributions to non-Gaussianities.\n\nThe last two terms in (3.17) are further suppressed by , but for completeness we list their contribution in the following. Contribution from term:\n\nContribution from term:\n\n i2∫τend−∞dτ a2ϵ3(∏iui(τend))(u∗1du∗2dτdu∗3dτk21k2⋅k3k22k23+two perm)(2π)3δ3(∑iki)+c.c. ." ]
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https://groupprops.subwiki.org/wiki/Contrasting_symmetric_groups_of_various_degrees
[ "# Contrasting symmetric groups of various degrees\n\nThis is a survey article related to:symmetric group\nView other survey articles about symmetric group\n\nThe symmetric group on a set is defined as the group of all permutations on that set under composition. A bijection between two sets induces an isomorphism of the corresponding symmetric groups -- in particular, the isomorphism type of a symmetric group is completely determined by the cardinality of the set it acts on. Further, except for the case of sets of size zero and one, sets of distinct cardinalities have non-isomorphic symmetric groups. We shall use the term symmetric group of degree", null, "$n$ for a symmetric group on a set with", null, "$n$ elements, which for convenience we take to be the set", null, "$\\{ 1,2, \\dots, n \\}$.\n\nThis article contrasts the properties and behavior of symmetric groups of small degrees, specifically the symmetric groups of degree", null, "$n$ for", null, "$n = 0,1,2,3,4,5,6,7$, compared with higher values. We shall use", null, "$S_n$ to denote the symmetric group of degree", null, "$n$.\n\n## Order and basic information\n\n### Some easy firsts involving orders\n\nLooking at the table, we see the following:\n\n•", null, "$n = 2$ is the first case of a nontrivial symmetric group.\n•", null, "$n = 3$ is the first case of a non-abelian, as well as non-nilpotent symmetric group. All", null, "$S_n, n \\ge 3$ are not nilpotent. In particular, there is no element whose order equals the order of the whole group.\n•", null, "$n = 4$ is the first case of a symmetric group whose exponent is not equal to the order of the group. The order of the group is", null, "$4! = 24$, while the exponent is the lcm of", null, "$1,2,3,4$, which is", null, "$12$. All", null, "$n \\ge 4$ have the property that the exponent of", null, "$S_n$ is strictly less than its order.\n\n## Composition factors and simplicity\n\n### The initial cases\n\n• The cases", null, "$n = 0,1$ are unusual, in that", null, "$S_n$ is trivial, so its composition series has length", null, "$0$.\n•", null, "$n = 2$ is the only case where", null, "$S_n$ is a simple group, since", null, "$A_n$ is trivial and has index two.\n\n### The special cases of", null, "$n = 3$ and", null, "$n = 4$\n\n•", null, "$n = 3$ is the only case where", null, "$S_n$ is a non-abelian metacyclic group (and hence also the only case of a non-abelian supersolvable group).", null, "$S_3$ has a cyclic normal subgroup (which is, in fact, a characteristic subgroup) of order three and a quotient of order two -- hence, it has a composition series of length two with both factors being cyclic groups.\n•", null, "$n = 4$ is the only case where", null, "$S_n$ is a solvable group that is not supersolvable.", null, "$S_4$ has a composition series of length four, with three composition factors being cyclic groups of order two and one composition factor being cyclic of order three. However, its unique chief series does not comprise only cyclic groups: the chief series has the Klein four-group of double transpositions and the alternating group. The chief factors are the Klein four-group, the cyclic group of order three, and the cyclic group of order two.", null, "$S_4$ is also the only symmetric group whose chief series is not a composition series.\n\n### The cases", null, "$n \\ge 5$\n\nFurther information: Alternating groups are simple", null, "$A_n$ is simple non-abelian for", null, "$n \\ge 5$, so the composition series of", null, "$S_n$ coincides with its unique chief series and has length two, with", null, "$A_n$ being the intermediate subgroup. The composition factors are", null, "$A_n$ and a cyclic group of order two.\n\n## Inner automorphisms and outer automorphisms; normal and characteristic subgroups\n\n### Completeness\n\nFurther information: Automorphism group of alternating group equals symmetric group, Symmetric groups are complete\n\n• For", null, "$n = 2$, the group", null, "$S_2$ has the property that every automorphism is inner, but it is abelian (in fact,", null, "$S_2$ is, up to isomorphism, the only nontrivial group with a trivial automorphism group). Hence, it is not a complete group. Further information: Trivial automorphism group implies trivial or cyclic of order two\n• For", null, "$n = 3$ and", null, "$n = 4$,", null, "$S_n$ is the automorphism group of the group", null, "$A_n$. Also,", null, "$S_n$ is a complete solvable group.\n• For", null, "$n = 5$ or", null, "$n \\ge 7$,", null, "$S_n$ is the automorphism group of the simple non-abelian group", null, "$A_n$. Hence, it is a complete group, and is also an almost simple group.\n\n### General facts\n\nFor all", null, "$n$, the following are true about", null, "$S_n$ and", null, "$A_n$. Note that since for", null, "$n \\ge 5$,", null, "$A_n$ is simple, most of these statements can be proved simply by checking them in all the small cases:\n\n• For all", null, "$S_n$ and all", null, "$A_n$, every normal subgroup is characteristic.\n• Any two normal subgroups of", null, "$S_n$ are comparable, so", null, "$S_n$ is a normal-comparable group. The same is true for", null, "$A_n$.\n• With the exception of", null, "$n = 4$, every", null, "$S_n$ and every", null, "$A_n$ is a T-group: a normal subgroup of a normal subgroup is normal.\n• For every", null, "$S_n$ and", null, "$A_n$ except", null, "$n = 4$, there is a unique composition series that also equals the unique chief series.\n• For every", null, "$S_n$ and", null, "$A_n$, there is a unique chief series. Although there may be multiple composition series, the order of appearance of composition factors is the same in all." ]
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https://en.citizendium.org/wiki/Prime_number/Citable_Version
[ "# Prime number/Citable Version", null, "", null, "Main Article Discussion Related Articles  [?] Bibliography  [?] External Links  [?] Citable Version  [?] This version approved either by the Approvals Committee, or an Editor from the listed workgroup. The Mathematics Workgroup is responsible for this citable version. While we have done conscientious work, we cannot guarantee that this version is wholly free of mistakes. See here (not History) for authorship. Help improve this work further on the editable Main Article!", null, "The prime number 11 illustrated with square tiles. 12 squares can be arranged into a rectangle with sides of length 3 and 4, so 12 is not a prime number. There is no way to form a full rectangle more than one square wide with 11 squares, so 11 is a prime number.\n\nA prime number is a number that can be evenly divided by exactly two positive whole numbers, namely 1 and itself. The first few prime numbers are 2, 3, 5, 7, 11, 13, and 17. A prime number $p$", null, "cannot be factored as the product of two numbers $p=a\\times b,$", null, "except for the trivial factorizations $p=1\\times p=p\\times 1$", null, "that all numbers possess.\n\nWith the exception of 2, all prime numbers are odd numbers, but not every odd number is prime. For example, $9=3\\times 3$", null, "and $15=3\\times 5,$", null, "so neither 9 nor 15 is prime. By the strict mathematical definition, the number 1 is not considered to be prime (although this is a matter of definition, and mathematicians in the past often did consider 1 to be a prime).\n\nThe importance of prime numbers in arithmetic comes in large part from the unique factorization of numbers. Every number $N>1$", null, "can be written as a product of prime factors, and any two such expressions for $N$", null, "will differ only in the order of the factors. For example, we can write $5040=7\\times 5\\times 3\\times 3\\times 2\\times 2\\times 2\\times 2$", null, ", where all of the factors in the product are prime numbers; there is no other way of writing 5040 as a product of prime numbers except by rearranging the prime factors we already have, such as $5040=2\\times 3\\times 2\\times 7\\times 2\\times 3\\times 2\\times 5$", null, ". Because of the unique factorization of numbers into prime numbers, an analogy can be made between the role prime numbers play in arithmetic and the role atoms play in chemistry.\n\nThe study of prime numbers has a long history, going back to ancient times, and it remains an active part of number theory today. Although the study of prime numbers used to be an interesting but not terribly useful area of mathematical research, today it has important applications. Understanding properties of prime numbers and their generalizations is essential to modern cryptography—in particular to public key ciphers that are crucial to Internet commerce, wireless networks, and military applications. Less well-known is that other computer algorithms also depend on properties of prime numbers.\n\n## There are infinitely many primes\n\nOne basic fact about the prime numbers is that there are infinitely many of them. In other words, the list of prime numbers 2, 3, 5, 7, 11, 13, 17, ... never stops. There are a number of ways of showing that there are infinitely many primes, but one of the oldest and most familiar proofs goes back go Euclid.\n\nEuclid proved that for any finite set of prime numbers, there is always another prime number which is not in that set. Choose any finite set of prime numbers $\\{p_{1},p_{2},p_{3},\\ldots ,p_{n}\\}$", null, ". Then the number\n\n$N=p_{1}p_{2}\\cdots p_{n}+1$", null, "presents a problem. On the one hand, the number $N-1=p_{1}p_{2}\\cdots p_{n}$", null, "is a multiple of $p_{1}$", null, ", and multiples of the same prime number are never next to each other, so N itself can't be a multiple of $p_{1}$", null, ". The same argument actually shows that $N$", null, "itself cannot be a multiple of any of the prime numbers $p_{1},p_{2},p_{3},\\dots ,p_{n}$", null, ". On the other hand, every number $N>1$", null, "is divisible by some prime. (For example, its smallest divisor greater than 1 must be a prime.)\n\nThis shows that there is at least one prime number (namely, the smallest divisor of N greater than 1) that is excluded from our initial finite set. Since any finite set of prime numbers can thus be extended to a larger finite set of prime numbers, we conclude that there are infinitely many prime numbers.\n\nAlthough most other proofs of the infinitude of prime numbers are more complicated, they can also provide more information. One mathematical milestone known as the Prime Number Theorem estimates how many of the numbers between 1 and x are prime numbers (see below). Another such proof is Euler's demonstration that the sum of the reciprocals of the primes diverges to ∞.\n\n## Locating primes\n\nHow can we tell which numbers are prime and which are not? It is sometimes possible to tell that a number is not prime by looking at its digits: for example, any number whose last digit is even is divisble by 2, and any number ending with 5 or 0 is divisible by 5. Therefore, except for the prime numbers 2 and 5, any prime number must end with the digit 1, 3, 7, or 9. This check can be used to rule out the possibility of a randomly chosen number being prime more than half of the time, but numbers that end with 1, 3, 7, or 9 often have divisors that are harder to spot.\n\nTo find large prime numbers, we must use a systematic procedure — an algorithm. Nowadays, prime-finding calculations are performed by computers, often using very complicated algorithms, but there are simple algorithms that can be carried out by hand if the numbers are small. In fact, the simplest methods for locating prime numbers are some of the oldest algorithms, known since antiquity. Two classical algorithms are called trial division and the sieve of Eratosthenes.\n\n### Trial division\n\nTrial division consists of systematically searching the list of numbers $2,3,\\dots ,n-1$", null, "for a divisor; if none is found, the number is prime. If n has a small divisor, we can quit as soon as we've found it, but in the worst case — if n is prime — we have to test all $n-2$", null, "numbers to be sure. This algorithm can be improved by realizing the following: if n has a divisor a that is larger than ${\\sqrt {n}}$", null, ", there must be another divisor b that is smaller than ${\\sqrt {n}}$", null, ". Thus, it is sufficient to look for a divisor up to ${\\sqrt {n}}$", null, ". This makes a significant difference: for example, we only need to try dividing by 2, 3, ..., 31 to verify that 997 is prime, rather than all the numbers 2, 3, ..., 996. Trial division might be described as follows using pseudocode:\n\nAlgorithm: trial division\n\nGiven $n$", null, ",\nFor each $i$", null, "= 2, 3, ... less than or equal to ${\\sqrt {n}}$", null, ":\nIf $i$", null, "divides $n$", null, ":\nReturn \"$n$", null, "is not prime\"\nElse:\nContinue with the next $i$", null, "When all $i$", null, "have been checked:\nReturn \"$n$", null, "is prime\"\n\n### Sieve of Eratosthenes\n\nThe Sieve of Eratosthenes not only provides a method for testing a number to see if it is prime, but also for enumerating the (infinite) set of prime numbers. The idea of the method to write down a list of numbers starting from 2 ranging up to some limit, say:\n\n2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20\n\nThe first number (2) is prime, so we mark it, and cross out all of its multiples:\n\n2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20\n\nThe smallest unmarked number is 3, so we mark it and cross out all its multiples (some of which may already have been crossed out):\n\n2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20\n\nThe smallest unmarked number (5) is the next prime, so we mark it and cross out all of its multiples:\n\n2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20\n\nNotice that there are no multiples of 5 $(\\leq 20)$", null, "that have not already been crossed out, but that doesn't matter at this stage. Proceeding as before, we add 7, 11, 13, 17 and 19 to our list of primes:\n\n2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20\n\nWe have now found all prime numbers up to 20.\n\n## Distribution of prime numbers\n\nIt is evident that the prime numbers are randomly distributed but, unfortunately, we don't know what 'random' means. - R. C. Vaughan\n\nThe list of prime numbers suggests that they thin out the further you go: 44% of the one-digit numbers are prime, but only 23% of the two-digit numbers and 16% of the three-digit numbers. The trial division method explained above provides an intuitive explanation. To test whether a number $n$", null, "is prime, you have to try whether it can be divided by all numbers between 2 and ${\\sqrt {n}}$", null, ". Large numbers have to undergo more tests, so fewer of them will be prime.\n\nThe Prime Number Theorem explains how fast the prime numbers thin out. It says that if you are looking around the number $n$", null, ", about one in every $\\log \\,n$", null, "numbers is prime (here, $\\log \\,n$", null, "denotes the natural logarithm of $n$", null, "). The formal statement of the prime number theorem is\n\n$\\lim _{x\\to \\infty }{\\frac {\\pi (x)\\log x}{x}}=1$", null, "where $\\pi (x)$", null, "is the number of primes $\\leq x.$", null, "## Some unsolved problems\n\nThere are many unsolved problems concerning prime numbers. A few such problems (posed as conjectures) are:\n\n### The twin prime conjecture\n\nTwin primes are pairs of prime numbers differing by 2. Examples of twin primes include 5 and 7, 11 and 13, and 41 and 43. The Twin Prime Conjecture states that there are an infinite number of these pairs. It remains unproven.\n\n### The Goldbach conjecture\n\nThe Goldbach conjecture is that every even number greater than 2 can be expressed as the sum of two primes. For example, if you choose the even number 48, you can find $48=41+7$", null, "where 41 and 7 are prime numbers.\n\n### Primes of special forms\n\n• It is not known whether there are infinitely many primes of the form $2^{n}-1$", null, "(called Mersenne primes). These primes arise in the study of perfect numbers, and factors of numbers of the form $2^{n}-1$", null, "(sometimes called 'Mersenne numbers') are a fruitful source of large prime numbers.\n• It is not known whether there are infinitely many primes of the form $2^{n}+1$", null, "(called 'Fermat primes'). Fermat primes arise in elementary geometry because if $p$", null, "is a Fermat prime, it is possible to construct a regular $p$", null, "-gon with a ruler and compass. In particular, it is possible to construct a regular 17-sided polygon (or 17-gon, for short) with a ruler and compass.\n• It is not known whether there are infinitely many primes of the form $n^{2}+1$", null, ".\n\n## Alternative definition\n\nA prime number is usually defined as a positive integer other than 1 that is (evenly) divisible only by 1 and itself.\n\nThere is another way of defining prime numbers, and that is that a number is prime if whenever it divides the product of two numbers, it must divide one of those numbers. A nonexample (if you will) is that 4 divides 12 (i.e. is a factor of 12), but 4 does not divide 2 and 4 does not divide 6 even though 12 is 2 times 6. This means that 4 is not a prime number. We may express this second possible definition in mathematical notation as follows: A number $p\\in \\mathbb {N}$", null, "(natural number) is prime if for any $a,b\\in \\mathbb {N}$", null, "such that $p|ab$", null, "(read p divides ab), either $p|a$", null, "or $p|b$", null, ".\n\nIf the first characterization of prime numbers is taken as the definition, the second is derived from it as a theorem (known as Euclid's lemma), and vice versa. The equivalence of these two definitions (in the integers $\\mathbb {Z}$", null, ") is not immediately obvious. In fact, it is a significant result.\n\n## References and notes\n\n1. The number N has at least one divisor greater than 1, because N is a divisor of itself. Let q denote the smallest divisor of N greater than 1. To see why q must be prime: Any divisor of q is also a divisor of N, and since q is the smallest, then q has no divisors smaller than itself except 1, therefore it is prime.\n2. The Euclidean algorithm may be used to show that $\\mathbb {Z}$", null, "is a principal ideal domain, and this implies that irreducibles are prime." ]
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https://reverseengineering.stackexchange.com/questions/20210/reversing-pe32-executable-ctf-question
[ "# Reversing PE32 executable - CTF Question\n\nSo we were playing CTF , and we found this interesting RE challenge . And when we did a statical analysis to the file we found an interesting for loop .\n\n``````for (var_84 = 0x0; var_84 < 0x13; var_84 = var_84 + 0x1) {\nedx = var_84;\n*(int8_t * )(var_84 + \"Catch Me If You Can\") = sign_extend_32( * (int8_t * )(var_84 + \"Catch Me If You Can\")) ^ * (ebp + (edx * 0x4 - 0xd8));\n\n}\n``````\n\nCan anyone explain the code above ?\n\nIt is byte-byte string xor in specific order. \"Catch me if you can\" is a string, or array of bytes.\n\nvar_84 (and edx register) is an index in it. I don't know what exactly is a stack layout and can not say with what exactly it is XORed.\n\nIn more readable C it will look like as follows:\n\n``````array = &\"Catch me if you can\";\nfor (var_84 = 0x0; var_84 < 0x13; var_84 = var_84 + 0x1) {\nedx = var_84;\narray[var_84] = sign_extend_32( array[var_84]) ^ * (ebp + (edx * 0x4 - 0xd8));\n\n}\n``````" ]
[ null ]
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https://brilliant.org/discussions/thread/fun-with-ms-paint/
[ "# Fun with MS Paint\n\nI was wondering about MS paint and stumbled upon certain interesting problems. Here they are.\n\nProblem 1: Can you draw an equilateral triangle in MS Paint?\n\nProblem 2: Can you draw a regular hexagon in MS Paint?\n\nProblem 3: Can you draw a regular polygon having odd number of sides in MS Paint?\n\nRules: You cannot use any aid to make measurements of any kind (length or angle). You can only use MS Paint (obviously). You are not supposed to use the co-ordinates to make measurements of any kind (length or angle). However, you can use the co-ordinates for precise placement of your mouse pointer.", null, "Note by Bruce Wayne\n6 years, 6 months ago\n\nThis discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.\n\nWhen posting on Brilliant:\n\n• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .\n• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting \"I don't understand!\" doesn't help anyone.\n• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n• bulleted\n• list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.\n2 \\times 3 $2 \\times 3$\n2^{34} $2^{34}$\na_{i-1} $a_{i-1}$\n\\frac{2}{3} $\\frac{2}{3}$\n\\sqrt{2} $\\sqrt{2}$\n\\sum_{i=1}^3 $\\sum_{i=1}^3$\n\\sin \\theta $\\sin \\theta$\n\\boxed{123} $\\boxed{123}$\n\nSort by:\n\nIt's impossible; drawing on MS Paint requires the vertices to be lattice points, and it's a well-known result that a regular polygon that is not a square cannot have their vertices on lattice points (using that if $a,b,c$ are rationals and $\\cos a\\pi = b, \\sin a\\pi = c$, then $2a$ is an integer).\n\nOf course, assuming that MS Paint's canvas is infinitely divisible (so it's basically a plane of reals times reals), then it's the usual compass and straightedge problem.\n\n- 6 years, 6 months ago\n\nRead my comment below. The equilateral triangle and regular hexagon can be constructed (and should looks okay if the screen resolution and the size of the computer screen is 1:1). Only the pentagon is impossible to construct.\n\n- 6 years, 6 months ago\n\nIn Windows 7, you can make a hexagon and a pentagon. Plus, you didn't say a REGULAR hexagon. So, problem 2 and 3 have been solved. But I don't know how to make equilateral triangle.\n\n- 6 years, 6 months ago\n\nProblems 2 and 3 have been revised. Try them out now :P\n\n- 6 years, 6 months ago\n\nHeh. I knew it you meant that way. But in Windows 7, I just realized that I can solve those three puzzles easily. Okay, let's say that you do it in Windows XP. I'll try solving these problems. Because I might got it. Might. Hehe.\n\n- 6 years, 6 months ago\n\nCan you provide your Win7 MS Paint solution for the three problems?\n\n- 6 years, 6 months ago\n\nIn windows 7, there are three tools. You can make a triangle, a hexagon and a pentagon by using these tools.\n\nAfter choosing a tool, let's say a pentagon, while holding the shift key, you can make a regular pentagon. That will also work on other tools.\n\nI think making a pentagon in Windows XP paint is impossible.\n\n- 6 years, 6 months ago\n\nI think I got it. picture\n\n- 6 years, 6 months ago\n\nerr how do you draw the circles so precisely?\n\n- 6 years, 6 months ago\n\nThis video shows you how to make a 60 degree angle, an equilateral triangle and a hexagon. Youtube video I hope I didn't break any rules. Plus, constructing a regular pentagon is impossible in Windows XP paint.\n\n- 6 years, 6 months ago" ]
[ null, "https://ds055uzetaobb.cloudfront.net/site_media/images/default-avatar-globe.png", null ]
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http://www.benfordonline.net/fullreference/647
[ "## View Complete Reference\n\n### Chaitin-Chatelin, F (1996)\n\n#### Is finite precision arithmetic useful for physics?\n\nJournal of Universal Computer Science 2(5), pp. 380-395.\n\nISSN/ISBN: Not available at this time. DOI: Not available at this time.\n\nAbstract: Both empirical sciences and computations are fundamentally restricted to measurements/computations involving a finite amount of information. These activities deal with the FINITE - some finite precision numbers, coming out from measurements, or from calculations run for some finite amount of time. By way of contrast, as Leibniz expressed it, mathematics is the science of the INFINITE, which contains the concept of continuum. The related concepts of limit points, derivatives and Cantor sets also belong to mathematics, the realm of the infinite, and not to the world of the finite. One is then lead to wonder about the basis for the \"unreasonable effectiveness of mathematics in the natural sciences\" (Wigner (1960). This puzzling situation gave birth, over the centuries, to a very lively philosophical discussion between mathematicians and physicists. We intend to throw into the debate a few simple examples drawn from practice in numerical analysis as well as in finite precision computations. By means of these examples, we illustrate some aspects of the subtle interplay between the discrete and the continuous, which takes place in Scientific Computing, when solving some equations of Physics. Is Nature better described by discrete or continuous models at its most intimate level, that is below the atomic level? With the theory of quantum physics, it seems that the question has received a significant push towards a discrete space. However one can argue equally that the time variable in Schrödinger's equation is continuous. We will not get involved in the scholarly dispute between the continuous and the discrete. Instead, we will show on simple examples taken from Scientific Computing, the subtlety of the interplay between the continuous and the discrete, which can take place in computations, be it with finite precision or exact arithmetic.\n\nBibtex:\n```@Article{, author = \"Françoise Chaitin-Chatelin\", title = \"Is Finite Precision Arithmetic Useful For Physics ?\", journal = j-jucs, year = \"1996\", volume = \"2\", number = \"5\", pages = \"380--395\", date = \"1996-05-28\", month = \"may\", note = \"\\url|http://www.jucs.org/is_finite_precision_arithmetic_useful_for_physics|\"} ```\n\nReference Type: Journal Article\n\nSubject Area(s): Computer Science" ]
[ null ]
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https://ncatlab.org/nlab/show/term+elimination
[ "# Contents\n\n## Idea\n\nIn type theory, term elimination are the natural deduction rules for how to use terms of a given type.\n\nFor example, the term elimination rules for the sum type are as follows:\n\n$\\frac{\\Gamma, z:A + B \\vdash C \\; \\mathrm{type} \\quad \\Gamma, x:A \\vdash c:C[\\mathrm{inl}(x)/z] \\quad \\Gamma, y:B \\vdash d:C[\\mathrm{inr}(y)/z] \\quad \\Gamma \\vdash e:A + B}{\\Gamma \\vdash \\mathrm{ind}_{A + B}(z.C, x.c, y.d, e):C[e/z]}$\n\n## Contextual term elimination\n\nSimilar to the conversion rules for types, there are also contextual term elimination rules for types. These differ from the usual term elimination rules in that in that there is an additional context member $\\Delta$ attached to the end of the context $\\Gamma, x:A$ so that the full context becomes $\\Gamma, x:A, \\Delta$. By definition, $\\Delta$ is dependent upon $x:A$, and the conclusion usually involves substituting $x:A$ by some given term $a:A$ in the context, becoming $\\Gamma, \\Delta[a/x]$. For example, the contextual term elimination rules for the sum type are given by:\n\n$\\frac{\\Gamma, z:A + B, \\Delta \\vdash C \\; \\mathrm{type} \\quad \\Gamma, x:A, \\Delta[\\mathrm{inl}(x)/z] \\vdash c:C[\\mathrm{inl}(x)/z] \\quad \\Gamma, y:B, \\Delta[\\mathrm{inr}(y)/z] \\vdash d:C[\\mathrm{inr}(y)/z] \\quad \\Gamma \\vdash e:A + B}{\\Gamma, \\Delta[e/z] \\vdash \\mathrm{ind}_{A + B}(z.C, x.c, y.d, e):C[e/z]}$\n\nAnd in first order logic over type theory, the contextual term elimination rules for the sum types are given by:\n\n$\\frac{\\Gamma, z:A + B, \\Delta \\vert \\Phi \\vdash C \\; \\mathrm{type} \\quad \\Gamma, x:A, \\Delta[\\mathrm{inl}(x)/z] \\vert \\Phi[\\mathrm{inl}(x)/z] \\vdash c:C[\\mathrm{inl}(x)/z] \\quad \\Gamma, y:B, \\Delta[\\mathrm{inr}(y)/z] \\vert \\Phi[\\mathrm{inr}(y)/z] \\vdash d:C[\\mathrm{inr}(y)/z] \\quad \\Gamma \\vdash e:A + B}{\\Gamma, \\Delta[e/z] \\vert \\Phi[e/z] \\vdash \\mathrm{ind}_{A + B}(z.C, x.c, y.d, e):C[e/z]}$\n\nContextual dependent product types and contextual identity types are defined in the appendix of:\n\nwhere the term elimination rules are contextual term elimination rules." ]
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https://tw.dictionary.search.yahoo.com/search?p=subtract&ei=UTF-8&context=gsmcontext%3A%3Alimlangpair%3A%3Aen_en
[ "# Yahoo奇摩字典 網頁搜尋\n\n### subtract\n\n• IPA[səbˈtrakt]\n\n美式\n\n• take away (a number or amount) from another to calculate the difference;take away (something) from something else so as to decrease the size, number, or amount\n\n• 釋義\n\n### 動詞\n\n• 1. take away (a number or amount) from another to calculate the difference subtract 43 from 60\n• take away (something) from something else so as to decrease the size, number, or amount programs were added and subtracted as called for\n• 更多解釋\n\n### subtract\n\n• IPA[səbˈtrakt]\n\n英式\n\n• take away (a number or amount) from another to calculate the difference: subtract 43 from 60\n\n2. ### 知識+\n\n• #### deduct 和 subtract在用法上有什麼不同嗎?\n\n冰封玫瑰為您解答... deduct 和 subtract 的意思一樣 使用方法有些顯微的差別 在描述數學的「減法」時 兩個字都能...\n\n• #### 英文單字,請問什麼是unusable box?\n\nSubtract 50 from the total -- I just found an unusable box. 我在美國查原書比較...在講「庫存」 inventory,所舉的例子也都與倉庫、存貨有關。這個例句是單字 subtract 的例句,意思是: 「從總數扣掉 50 個,我剛發現一箱不能用。」 an unusable...\n\n• #### 有幾首幼稚園的英文歌需要英文歌詞\n\n...nine, eight, seven, six, five, four, three, two, one, zero 3. Jumping to add and subtract Jumping is how you can add and subtract It sounds funny..." ]
[ null ]
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https://uk.mathworks.com/matlabcentral/cody/problems/605-whether-the-input-is-vector/solutions/2002316
[ "Cody\n\n# Problem 605. Whether the input is vector?\n\nSolution 2002316\n\nSubmitted on 4 Nov 2019 by Son Pham\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1   Pass\nx = [1 2 3 1]; y_correct = 1; assert(isequal(checkvector(x),y_correct))\n\ny = 1\n\n2   Pass\nx = [1 2 7; 6 3 1]; y_correct = 0; assert(isequal(checkvector(x),y_correct))\n\ny = 0\n\n3   Pass\nx = [1;2;0]; y_correct = 1; assert(isequal(checkvector(x),y_correct))\n\ny = 1\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!" ]
[ null ]
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https://nrich.maths.org/parabolicpatterns
[ "#### You may also like", null, "### Cubic Spin\n\nProve that the graph of f(x) = x^3 - 6x^2 +9x +1 has rotational symmetry. Do graphs of all cubics have rotational symmetry?", null, "### Sine Problem\n\nIn this 'mesh' of sine graphs, one of the graphs is the graph of the sine function. Find the equations of the other graphs to reproduce the pattern.", null, "### More Parabolic Patterns\n\nThe illustration shows the graphs of twelve functions. Three of them have equations y=x^2, x=y^2 and x=-y^2+2. Find the equations of all the other graphs.\n\n# Parabolic Patterns\n\n##### Age 14 to 18Challenge Level\nThe illustration shows the graphs of fifteen functions. Two of them have equations\n\n$y = x^2$\n$y = - (x - 4)^2$", null, "Can you find the equations of the other parabolas in the picture?\nYou may wish to use a graphical calculator or software such as Desmos to recreate the pattern for yourself.\n\nCan you find the equations of these parabolas?", null, "NOTES AND BACKGROUND\nThis sort of challenge is sometimes called an inverse problem because the question is posed the opposite way round to what might have been expected. This is almost like saying: 'here is the answer, what was the question?' Instead of giving the equations of some functions and asking you to sketch the graphs, this challenge gives the graphs and asks you to find their equations.\n\nYou are being asked to sketch a family of graphs. What makes this a family? All the graphs are obtained by transformations such as reflections and translations of other graphs in the family. The key is to find the simplest function and then to find transformations of the graph of that function which give the other graphs in the family.\n\nIf you have access to a graphic calculator, or to graph drawing software, it will not give you the answers. You will have to think for yourself what the equations should be and then the software will enable you to test your own theories and see if you were right." ]
[ null, "https://nrich.maths.org/content/02/09/15plus1/icon.jpg", null, "https://nrich.maths.org/content/02/10/15plus1/icon.png", null, "https://nrich.maths.org/content/01/05/six4/icon.jpg", null, "https://nrich.maths.org/content/01/04/six6/parabolicpatterns16.png", null, "https://nrich.maths.org/content/01/04/six6/Parabolic%20Patterns%202.png", null ]
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http://blog.hudongdong.com/c/187.html
[ "如果文章对您有用,麻烦顺手点一下文章的广告吧~\n\nC++的运算符\nC++的运算符十分丰富,使得C++的运算十分灵活方便。例如把赋值号(=)也作为运算符处理,这样,a=b=c=4就是...", null, "13\n2015/07\n\n# C++的运算符\n\nC++的运算符十分丰富,使得C++的运算十分灵活方便。例如把赋值号(=)也作为运算符处理,这样,a=b=c=4就是合法的表达式,这是与其他语言不同的。C++提供了以下运算符:\n\n+(加)  -(减)  *(乘)  /(除)  %(整除求余)  ++(自加)  --(自减)\n\n>(大于)  <(小于)   ==(等于)  >=(大于或等于)  <=(小于或等于)  !=(不等于)\n\n&&(逻辑与)  ||(逻辑或)   !(逻辑非)\n\n<<(按位左移)  >>(按位右移)  &(按位与)  |(按位或)   ^(按位异或)  ~(按位取反)", null, "Last modification:January 1st, 1970 at 08:00 am", null, "" ]
[ null, "http://blog.hudongdong.com/usr/themes/handsome/libs/GetCode.php", null, "http://blog.hudongdong.com/img/weixin.png", null, "https://cdn.v2ex.com/gravatar/d41d8cd98f00b204e9800998ecf8427e", null ]
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https://getcalc.com/howmany-ways-lettersofword-hi-canbe-arranged.htm
[ "# How Many Ways are There to Order the Letters of Word HI?\n\nThe 2 letters word HI can be arranged in 2 distinct ways. The below detailed information shows how to find how many ways are there to order the letters HI and how it is being calculated in the real world problems.\n\n Enter word :\n\nDistinguishable Ways to Arrange the Word HI\nThe below step by step work generated by the word permutations calculator shows how to find how many different ways can the letters of the word HI be arranged.\n\nObjective:\nFind how many distinguishable ways are there to order the letters in the word HI.\n\nStep by step workout:\nstep 1 Address the formula, input parameters and values to find how many ways are there to order the letters HI.\nFormula:\nnPr =n!/(n1! n2! . . . nr!)\n\nInput parameters and values:\nTotal number of letters in HI:\nn = 2\n\nDistinct subsets:\nSubsets : H = 1; I = 1;\nSubsets' count:\nn1(H) = 1, n2(I) = 1\n\nstep 2 Apply the values extracted from the word HI in the (nPr) permutations equation\nnPr = 2!/(1! 1! )\n\n= 1 x 2/{(1) (1)}\n\n= 2/1\n\n= 2\nnPr of word HI = 2\n\nHence,\nThe letters of the word HI can be arranged in 2 distinct ways.\n\nApart from the word HI, you may try different words with various lengths with or without repetition of letters to observe how it affects the nPr word permutation calculation to find how many ways the letters in the given word can be arranged.", null, "" ]
[ null, "https://getcalc.com/cdn/graphics/getcalc-logo.png", null ]
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http://carolinacashfast.com/blog/how-to-calculate-loan-repayments/
[ "If you’re considering applying for some sort of loan, then it may be helpful to first know what the payment amounts with interest might be. Not only will calculating loan repayments give you the confidence in your ability to make your payments, but you will also be able to gain trust with your lender by showing her your research.\n\n## Easily Calculate Loan Repayment Amounts\n\nThere are several ways to calculate a loan repayment. Whichever method you choose, you must first understand the variables that are involved in loan repayment calculation and their symbols.\n\nSome methods will ask for the symbols while others will spell out exactly what they are asking for. Referring to the descriptions below can make the process go a little more smoothly.\n\n• M is the Payment Amount\n• P is the Principal, or the Amount of Money Borrowed\n• J is the Effective Interest Rate\n• To calculate the Effective Interest Rate, follow these steps:\n• Divide the annual interest rate by 100 in order to get it into a decimal form.\n• Then, divide the decimal form by the number of payments that will be made.\n• N is the Total Number of Payments\n\nNow that you are familiar with the common variables involved in calculating loan payment amounts you will be able to more easily understand how to arrive at a solution.\n\nHow to Calculate Loan Payments With a Formula\n\nIf you would like to calculate your loan payments manually, then you can use the loan payments formula:\n\n## M = P * (J/1 – (1 + J) – N))\n\nIn order to solve this formula, you need all of the necessary information. It is also essential to use a graphing calculator because a standard calculator opens you up to the risk of significant rounding errors.\n\nThough just looking at the formula can be intimidating, it can be completed by taking it one step at a time. However, if math is not your strongest asset, then consider one of the two following options.\n\nHow to Calculate Loan Payments Using Online Calculators\n\nPerforming a simple Google search for online loan payment calculators brings up hundreds of results. Some of the advanced ones require users to set up an account, and others are just not user friendly. In order to save you some time, we’ve rounded up several of the best loan repayment calculators.\n\nUsing any of these online loan payment calculators is a quick and easy way to determine your payment amounts.\n\nHow to Calculate Loan Payments in Excel\n\nOne of the most popular methods for calculating loan repayment amounts is the use of Excel. Not only is it easy to input information, but you’ll be able to save the document for reference later. It only takes a few steps to calculate a loan repayment in Excel and they are best explained in this video.\n\nIf you don’t have Excel, then don’t worry. Google Sheets works very similarly to Excel. The application is free, but you must register for a Google account in order to use it.\n\nRepaying loans on time is extremely important. Whether you’re interested in a payday loan, student loans, or a car title loan, calculating loan repayment amounts is a great first step toward financial responsibility.\n\nCalculating your loan repayments by hand, through an online loan repayment calculator, or by plugging in all your information into Excel or Google Sheets is a great way to get a realistic idea of your payments. When you have that down, the path to debt-free living is much easier to follow." ]
[ null ]
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https://edoc.unibas.ch/69975/
[ "# The Moser-Trudinger inequality and its extremals on a disk via energy estimates\n\nMancini, Gabriele and Martinazzi, Luca. (2016) The Moser-Trudinger inequality and its extremals on a disk via energy estimates. Preprints Fachbereich Mathematik, 2016 (17).", null, "PDF - Published Version 854Kb\n\nOfficial URL: https://edoc.unibas.ch/69975/\n\nWe study the Dirichlet energy of non-negative radially symmetric critical points $u_\\mu$ of the Moser-Trudinger inequality on the unit disc in $\\mathbb{R}^2$, and prove that it expands as\n$4\\pi + \\frac{4\\pi}{\\mu^4}+o(\\mu^{-4}) \\le \\int_{B_1} |\\grad u_\\mu|^2 dx \\le 4\\pi + \\frac{6\\pi}{\\mu^4}+o(\\mu^{-4}), as \\mu \\to \\infty,$\nwhere $\\mu = u_\\mu (0)$ is the maximum of $u_\\mu$. As a consequence, we obtain a new proof of the Moser-Trudinger inequality, of the Carleson-Chang result about the existence of extremals, and of the Struwe and Lamm-Robert-Struwe multiplicity result in the supercritical regime (only in the case of the unit disk)." ]
[ null, "https://edoc.unibas.ch/style/images/fileicons/application_pdf.png", null ]
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http://tomee.apache.org/microprofile-2.0/javadoc/org/eclipse/microprofile/metrics/Timer.html
[ "org.eclipse.microprofile.metrics\n\n## Interface Timer\n\n• All Superinterfaces:\nCounting, Metered, Metric, Sampling\n\n```public interface Timer\nextends Metered, Sampling```\nA timer metric which aggregates timing durations and provides duration statistics, plus throughput statistics via `Meter`. The timer measures duration in nanoseconds.\n• ### Nested Class Summary\n\nNested Classes\nModifier and Type Interface and Description\n`static interface ` `Timer.Context`\nA timing context.\n• ### Method Summary\n\nAll Methods\nModifier and Type Method and Description\n`long` `getCount()`\nReturns the number of events which have been marked.\n`double` `getFifteenMinuteRate()`\nReturns the fifteen-minute exponentially-weighted moving average rate at which events have occurred since the meter was created.\n`double` `getFiveMinuteRate()`\nReturns the five-minute exponentially-weighted moving average rate at which events have occurred since the meter was created.\n`double` `getMeanRate()`\nReturns the mean rate at which events have occurred since the meter was created.\n`double` `getOneMinuteRate()`\nReturns the one-minute exponentially-weighted moving average rate at which events have occurred since the meter was created.\n`Snapshot` `getSnapshot()`\nReturns a snapshot of the values.\n`Timer.Context` `time()`\n`<T> T` `time(java.util.concurrent.Callable<T> event)`\nTimes and records the duration of event.\n`void` `time(java.lang.Runnable event)`\nTimes and records the duration of event.\n`void` ```update(long duration, java.util.concurrent.TimeUnit unit)```\nAdds a recorded duration.\n• ### Method Detail\n\n• #### update\n\n```void update(long duration,\njava.util.concurrent.TimeUnit unit)```\nAdds a recorded duration.\nParameters:\n`duration` - the length of the duration\n`unit` - the scale unit of `duration`\n• #### time\n\n```<T> T time(java.util.concurrent.Callable<T> event)\nthrows java.lang.Exception```\nTimes and records the duration of event.\nType Parameters:\n`T` - the type of the value returned by `event`\nParameters:\n`event` - a `Callable` whose `Callable.call()` method implements a process whose duration should be timed\nReturns:\nthe value returned by `event`\nThrows:\n`java.lang.Exception` - if `event` throws an `Exception`\n• #### time\n\n`void time(java.lang.Runnable event)`\nTimes and records the duration of event.\nParameters:\n`event` - a `Runnable` whose `Runnable.run()` method implements a process whose duration should be timed\n• #### getCount\n\n`long getCount()`\nDescription copied from interface: `Metered`\nReturns the number of events which have been marked.\nSpecified by:\n`getCount` in interface `Counting`\nSpecified by:\n`getCount` in interface `Metered`\nReturns:\nthe number of events which have been marked\n• #### getFifteenMinuteRate\n\n`double getFifteenMinuteRate()`\nDescription copied from interface: `Metered`\nReturns the fifteen-minute exponentially-weighted moving average rate at which events have occurred since the meter was created. This rate has the same exponential decay factor as the fifteen-minute load average in the `top` Unix command.\nSpecified by:\n`getFifteenMinuteRate` in interface `Metered`\nReturns:\nthe fifteen-minute exponentially-weighted moving average rate at which events have occurred since the meter was created\n• #### getFiveMinuteRate\n\n`double getFiveMinuteRate()`\nDescription copied from interface: `Metered`\nReturns the five-minute exponentially-weighted moving average rate at which events have occurred since the meter was created. This rate has the same exponential decay factor as the five-minute load average in the `top` Unix command.\nSpecified by:\n`getFiveMinuteRate` in interface `Metered`\nReturns:\nthe five-minute exponentially-weighted moving average rate at which events have occurred since the meter was created\n• #### getMeanRate\n\n`double getMeanRate()`\nDescription copied from interface: `Metered`\nReturns the mean rate at which events have occurred since the meter was created.\nSpecified by:\n`getMeanRate` in interface `Metered`\nReturns:\nthe mean rate at which events have occurred since the meter was created\n• #### getOneMinuteRate\n\n`double getOneMinuteRate()`\nDescription copied from interface: `Metered`\nReturns the one-minute exponentially-weighted moving average rate at which events have occurred since the meter was created. This rate has the same exponential decay factor as the one-minute load average in the `top` Unix command.\nSpecified by:\n`getOneMinuteRate` in interface `Metered`\nReturns:\nthe one-minute exponentially-weighted moving average rate at which events have occurred since the meter was created\n• #### getSnapshot\n\n`Snapshot getSnapshot()`\nDescription copied from interface: `Sampling`\nReturns a snapshot of the values.\nSpecified by:\n`getSnapshot` in interface `Sampling`\nReturns:\na snapshot of the values" ]
[ null ]
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http://interactivewebtech.com/uahp8h/transformations-calculator.html
[ "# Transformations calculator", null, "Transformations calculator\n\nCorrectness guaranteed. To find the transformation, compare the two functions and check to see if there is a horizontal or vertical shift, Free absolute value equation calculator - solve absolute value equations with all the steps. Which says \"all the x and y coordinates become x+30 and y+40\" Matrix Transformation Calculators: Feel free to use all the matrix calculators in this collection. Back substitution of Gauss-Jordan calculator reduces matrix to reduced row echelon form. SAT Math Test Prep Online Crash Course Algebra & Geometry Study Guide Review, Functions,Youtube - Duration: 2:28:48. Fourier Series Calculator is a Fourier Series on line utility, simply enter your function if piecewise, introduces each of the parts and calculates the Fourier coefficients may also represent up to 20 coefficients. Matrix Representations of Linear Transformations and Changes of Coordinates 0. Now add the. Note that we may need to use several points from the graph and “transform” them, to make With help of this calculator you can: find the matrix determinant, the rank, raise the matrix to a power, find the sum and the multiplication of matrices, calculate the inverse matrix. I personally find it is easier to separate the two, so the view transformation can be modified independently of the model matrix. To add the calculator. Our Transformations Worksheets are free to download, easy to use, and very flexible. Laplace transform is the most commonly used transform in calculus to solve Differential equations. Transformations of f(x)=-¼√(-x)+9. By Sharon K. BAZI calculator is free for everyone. *xyz file that you have already created. Data transformations are an important tool for the proper statistical analysis of biological data. This calculator will save you time, energy and frustration. This exploration is about recognizing what happens to the graph of the logarithmic function when you change one or more of the coefficients a, b, c, and d. Enter the sine value, select degrees (°) or radians (rad) and press the = button. The Estimate Geometric Transformation block supports Nonreflective similarity, Affine, and Projective transformation types, which are described in this section. This Linear Algebra Toolkit is composed of the modules listed below. For example, a calculator configure for WGS84 to ED50 Transformations can be accessed by clicking the following calculator button (login required): After an overview of coordinate notation, students explore transformations including translation, reflection, rotation, and dilation in a coordinate plane. The Fisher Z-Transformation is a way to transform the sampling distribution of Pearson’s r (i. Thus, we can take linear combinations of linear transformations, where the domain and target are two Fvector spaces V and Wrespectively. In particular, part 3 of the beer sales regression example illustrates an application of the log transformation in modeling the effect of price on demand, including how to use the EXP (exponential) function to “un-log” the forecasts and confidence limits to convert them back into the units of the original data. Notice in these two functions, I've replaced the the x and root x by something else. To find the transformation, compare the two functions and check to see if there is a horizontal or vertical shift, reflection about the x-axis, and if there is a vertical  Free graphing calculator instantly graphs your math problems. The image created is This calculator is online sandbox for playing with Discrete Fourier Transform (DFT). Millions of people around the world use GeoGebra to learn  This point will help you when describing the transformation that has occurred. You can use this Laplace calculator software to calculate Laplace or inverse Laplace transform. Also note that you can't just back-transform the confidence interval and add or subtract that from the back-transformed mean; you can't take 10 0. Use this fraction calculator for adding, subtracting, multiplying and dividing fractions. Transforming a variable involves using a mathematical operation to change its Also note that you can't just back-transform the confidence interval and add or subtract that from the back-transformed mean; you can't take 10 0. Loading Parent Function Transformations Free Laplace Transform calculator - Find the Laplace and inverse Laplace transforms of functions step-by-step Free functions and graphing calculator - analyze and graph line equations and functions step-by-step The calculator will find the Laplace Transform of the given function. What is the energy transformation during operation of a solar calculator? The energy transformation during the operation of a solar calculator is radiant energy to mechanical energy. To Find A Logarithm. You will learn how to perform the transformations, and how to map one figure into another using these transformations. What energy transformations in calculator? 3 4 5. Then he dropped 150 pounds in the same amount of time. It's destined for BaZi consultants as a helpful tool for BaZi readings. The funtool app is a visual function calculator that manipulates and displays functions of one variable. Embed this widget » Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. To the left zooms in, to the right zooms out. 0 to perform three-dimensional (latitude, longitude, ellipsoid height) coordinate transformations for a wide range of datums and regions in the National Spatial Reference System. com version of the graphing calculator to your web site copy and paste the following code where ever you want the calculator to appear. The basic parent function of any exponential function is f(x) = b x, where b is the base. changes the x-values). Dilation is the transformation which is an extreme, radical change in appearance. Asked in Solar Power Jul 27, 2019- Gravity Transformation - Macro Calculator. The author, Samuel Chukwuemeka aka Samdom For Peace gives credit to Our Lord, Jesus Christ. Jeor equation, to help NASA/IPAC EXTRAGALACTIC DATABASE Coordinate Transformation & Galactic Extinction Calculator Help | Comment | NED Home Geographic Calculator supports a wide range of file formats and is built on the foundation of the largest geodetic parameter database available anywhere. We are experts in geometry transformation calculators. with a≠0, where we will be concerned with three general types of transformations in the variables x and y. See what this looks like with   Line straight ell is graphed in the xy-plane below. You can click-and-drag to move the graph around. changes the y-values) or horizontally (i. Natural logarithm calculator. Enter the time domain data in the Time Domain Data box below with each sample on a new line. Enter the input number x and press the = button: * Use e for scientific notation. Examples of Horizontal Stretches and Shrinks wxMaxima is one of the best free Laplace transform calculator for Windows. It uses real DFT, that is, the version of Discrete Fourier Transform which uses real numbers to represent the input and output signals. CPM Student Tutorials CPM Core Connections eTools & Videos CC Course 3 eTools General eTools Rigid Transformations eTool (CPM) Rigid Transformations eTool (CPM) This eTool will record the steps you create showing translation, rotation, and reflection. In other words, if f (x) = 0 for some value of x, then k f (x) = 0 for the same value of x. By the way, these substitutions for horizontal stretch and compression are also counterintuitive to a lot of students. The first, flipping upside down, is found by taking the negative of the original function; that is, the rule for this transformation is –f (x). (1964) An analysis of transformations. Calculates transpose, determinant, trace, rank, inverse, pseudoinverse, eigenvalues and eigenvectors. Input proper or improper fractions, select the math sign and click Calculate. We are fortunate to live in an era of technology that we can now access such incredible resources that were never at the palm of our hands like they are today. Coordinate Transformation & Galactic Extinction Calculator. Vertical Translations: [ Interactive Graph] If k is any positive real number then, Matrix Calculator. The Organic Chemistry Tutor 1,214,051 views Use our online point reflection calculator to know the point reflection for the given coordinates. the correlation coefficient) so that it becomes normally distributed. After any of those transformations (turn, flip or slide), the shape still has the same size, area, angles and line lengths. Leave extra cells empty to enter non-square matrices. Enter a value for Step, the step size. In this section, we cover a method used to quickly sketch graphs related to some basic functions. Play alternates as players transform their triangles and collect bugs. ector to transform x into kxky, kyk= 1. Calculator Use. For more about how to use the Integral Calculator, go to \"Help\" or take a look at the examples. Wolfram|Alpha has the ability to compute the transformation matrix for a specific 2D or 3D transformation activity or to return a general transformation calculator for rotations, reflections and shears. rotation, translation, re ection page 10 Transformations Worksheet Calculate the Fourier Transform of your data, graph the frequency domain spectrum from the Fast Fourier Transform (FFT), Inverse Fourier Transform with the IFFT, and much more. The calculator does source transformations and presents the new circuits with the new values. Loading Rotation about a Point @rhodes_math Loved teaching parallel and perpendicular lines using @desmos on iPads @sl_academy #innovation #12joysofteaching @kirk_humphreys 7th graders used @desmos today to calculate linear/quad. Matrices (singular matrix) are rectangular arrays of mathematical elements, like numbers or variables. In this video, I introduce the idea of a linear transformation of vectors from one space to another. Box, G. That means that to simulate a camera transformation, you actually have to transform the world with the inverse of that transformation. If line straight ell is translated up 5  Please have a look at the online calculators on the page Computation of Effect  Translate different types of function graphs using sliders. Free graphing calculator instantly graphs your math problems. . Feb 25, 2018 Statistics Definitions > Transformations are when you literally \"transform\" your data into something slightly different. For details on using the Transformation Calculator, see  One fun way to think about functions is to imagine that they literally move the points from the input space over to the output space. Sometimes, it is necessary to apply a linear transformation to a random variable. If an input is given then it can easily show the result for the given number. Using the graphing calculator to examine transformations:  Free function shift calculator - find phase and vertical shift of periodic functions step-by-step. The Rref calculator is used to transform any matrix into the reduced row echelon form. In addition to showing the original and transformed curves, it displays an individual movable point on the original curve and the image of the point on the transformed curve. riding a bike after lunch answer: chemical energy transforms to mechanical energy. Your answer noted above can be got by the \"rotation of points\" Calculator. From Medical Weight Loss Treatments, to Botox, to Fraxel Laser Treatemnts, to our incredible EmSculpt Machine for Fat Reduction and Muscle building. Transformations with a Quadratic Function. Byju's Laplace Transform Calculator is a tool which makes calculations very simple and interesting. In the Output Coordinate System area (right), click “Use coordinate system selected below” and select the desired, or output, coordinate system. Each module is designed to help a linear algebra student learn and practice a basic linear algebra procedure, such as Gauss-Jordan reduction, calculating the determinant, or checking for linear independence. Transformation Efficiency: Transformants / μg DNA or: Transformants / ng of DNA * This calculation assumes that no futher dilutions of the transformation reaction are made after DNA, competent cells, and media are added. Not sure what to avoid on the ketogenic diet? Here's a quick list of 7 foods you absolutely can not eat on the keto diet. This allows us to use linear regression techniques more effectively with nonlinear data. The figure presents a xy- coordinate plane with the graph of a line and 2. Normality is an important assumption for many statistical techniques; if your data isn’t normal, applying a Box-Cox means that you are able to run a broader number of tests. Enlargements (Dilations) Enlargement, sometimes called scaling or dilation, is a kind of transformation that changes the size of an object. Also, a vertical stretch/shrink by a factor of k means that the point (x, y) on the graph of f (x) is transformed to the point (x, ky) on the graph of g(x). reshish. chemical go in and mechanical cause you are moving your fingers. Field A, B, and C: Enter the value(s) for executing the specified calculator function in the Calculation field. . Once a token is used in an action, it is removed from your bank. In the “Coordinate Transformation” area, click “Apply coordinate transformation”. DFT is part of Fourier analysis, which is a set of math techniques based on decomposing signals into sinusoids. Transformation efficiency ( transformants/µg) is calculated as follows: # colonies on plate/ng of DNA plated X 1000  This simple calculator allows you to calculate a standardized z-score for any raw value of X. Base 2, base e, base 10. In fact, the set L(V;W) of all linear transformations T: V 21. The natural logarithm of x is: ln x = log e x = y. 10. Graph your equations with MathPapa! This graphing calculator will show you how to graph your problems. Pearson product moment correlation coefficient is also referred as Pearson's r or bivariate correlation. In this section let c be a positive real number. 3 Transformations of Graphs MATH 1330 Precalculus 87 Looking for a Pattern – When Does the Order of Transformations Matter? When deciding whether the order of the transformations matters, it helps to think about whether a transformation affects the graph vertically (i. Let V and Wbe vector spaces over F. This lesson explains how to make a linear transformation and how to compute the mean and variance of the result. At the click of a button, for example, funtool draws a graph representing the sum, product, difference, or ratio of two functions that you specify. When you let go of the slider it goes back to the middle so you can zoom more. When and have the same dimension, it is possible for to be invertible, meaning there exists a such that . Linear Transformation (Geometric transformation) calculator in 2D, including, rotation, reflection, shearing, projection, scaling (dilation). However matrices can be not only two-dimensional, but also one-dimensional (vectors), so that you can multiply vectors, vector by matrix and vice versa. Check the Sigmas. The “z” in Fisher Z stands for a z-score. Graphical Transformations of Functions In this section we will discuss how the graph of a function may be transformed either by shifting, stretching or compressing, or reflection. So, for example, you could use this test to find out whether people Loading Loading First, we need a little terminology/notation out of the way. You can also remove transformations from the action bar by dragging the tokens back to your bank. It holds calculators like N x N Rank of Matrix Calculator, Transpose of a Matrix Calculator, Rank of Matrix, Matrix Inverse (determinant, adjoint), 4x4 Matrix Inverse Calculator, Matrix Inverse Calculator, Moore-Penrose Pseudo Inverse Calculator etc. Then any linear combination of linear transformations with domain V and target Wis also linear. So there is a lot to see here so look around or better yet give us a call at 1-833-DIET-MDs. Just enter your raw score, population mean and standard deviation, and hit \"Calculate Z\". Calculator for Matrices. How do you get scattergram on T83 calculator, lcm worksheet ks3, algebra 2 curriculum. Just type matrix elements and click the button. Includes all the functions and options you might need. A horizontal stretch or shrink by a factor of 1/k means that the point (x, y) on the graph of f(x) is transformed to the point (x/k, y) on the graph of g(x). A common way of describing this situation is to say that as an object approaches the speed of light, its mass increases and more force must be exerted to produce a given acceleration. You can also check your answers! Interactive graphs/plots help visualize and better understand the functions. Transformations. Label each parabola with its equation. Also you can compute a number of solutions in a system of linear equations (analyse the compatibility) using Rouché–Capelli theorem. Square root function, up 9, vertical compression, vertical and horizontal reflection (graph falls down and to the left) To do an FFT. Try and follow what happens each time. For complex (I and Q) data, the real and imaginary components should be on the same line saparated by a comma. We have translation, rotation, and reflection worksheets for your use. maximum angle explained; test1; LC calculus-slope of a tangent; Mapping 3; Linear Inequalities in 2 Variables You can always \"cheat\", by the way, especially if you have a graphing calculator, by quickly graphing y = (x – 2) 4 + 1 and verifying that it matches what you've drawn. A linear transformation may or may not be injective or surjective. Download free on Google Play. *xyz file. Answers are fractions in lowest terms or mixed numbers in reduced form. If you don't choose any base, the default is base 10. Math solver for factoring, math investment problems, aegbra problem solver. What is a Box Cox Transformation? A Box Cox transformation is a way to transform non-normal dependent variables into a normal shape. Function Transformations. It makes the lives of people who use matrices easier. Tracking your nutrition for just a few weeks can provide a wake-up call that makes a huge difference in your results for years to come. Wiki User 12/12/2010. For math, science, nutrition, history Loading Transforming the Graph of a Quadratic Function Transformation Matrices. We call the equations that define the change of variables a transformation. transformations to introduce subdiagonal zeros in Gaussian elimination. All you have to do is enter your details, select your goals and retrieve your macros. For these cards, it will tell you an object and then tell the energy transformation, ex. Graphs up to five functions, exploring transformations (shifting In geometry and complex analysis, a Möbius transformation of the plane is a rational function of one complex variable. Xlist is the name of the list where the x-coordinates will be found. Type in any equation to get the solution, steps and graph Transformation Graphing allows you to investigate only one variable at a time, so only one equal sign will be highlighted. Article ID: 819 | Rating: Unrated | Last Updated: Fri, Sep 4, 2015 at 9:17 PM. Online calculator which allows you to separate the variable to one side of the algebra equation and everything else to the other side,for solving the equation easily. r to z' Computes transformations in both directions. Source: Math on Call - A Mathematics Handbook Great Source Education   Easily graph functions and equations, find special points of functions, save and share your results. re ection, dilation, translation C. To continue calculating with the result, click Result to A or Result to B. When transformations have to be correct, consistent and certifiable, GIS professionals around the world choose Geographic Calculator. After you enter Transformation Graphing, you are returned to the calculator’s Home screen, and it looks like the application is not running. Use a standard window as shown in figure 1, to graph the function f ⁢ x = x 2 f x x 2 using a graphing calculator. Related Questions. Remember that x-intercepts do not move under vertical stretches and shrinks. (figure 2) z' z' r The Fisher Z-Transformation is a way to transform the sampling distribution of Pearson’s r (i. A Möbius transformation can be obtained by first performing stereographic projection from the plane to the unit two-sphere, rotating and moving the sphere to a new location and orientation in space, and then performing stereographic projection (from the new position of the Transformations in Function Notation (based on Graph and/or Points) You may also be asked to perform a transformation of a function using a graph and individual points; in this case, you’ll probably be given the transformation in function notation. Plane strain transformation calculator is used to calculate normal strains and shear strain at a specific point for plane strain state (εz=γzx=γzy=0) after the  Coordinate Transformations with Geographic Calculator. Transformations in math occur when there is a change in position, shape, or size. 3) Select the number base. You can insert value pairs to the text area labeled as \"Input coordinate pairs\" - also by using copy/paste even from MS Excell or similar programs. This calculator solves Systems of Linear Equations using Gaussian Elimination Method, Inverse Matrix Method, or Cramer's rule. Transforming a variable involves using a mathematical operation to change its Linear Transformation. Minitab determines an optimal power transformation. The graph of a quadratic equation is in Welcome to Transformations Weight Loss & Skin Clinic. Performs LU, Cholesky, QR, Singular value Start your transformation and realize your future as a digital organization. Each of these moves is a transformation of the puzzle piece. Transforming the Graph of a Quadratic Function. Print “Find constellations” so that the match is searched for automatically. e. Calculator Introduction: Given the strains at a space point in the body, this calculator computes the strains of the same space point in a rotated coordinate system. 2) Enter the number to the right of the \"Logarithm of\" box. Interactive, free online graphing calculator from GeoGebra: graph functions, plot data, drag sliders, and much more! The author, Samuel Chukwuemeka aka Samdom For Peace gives credit to Our Lord, Jesus Christ. Linear Transformations , Example 1, Part 2 Example: to say the shape gets moved 30 Units in the \"X\" direction, and 40 Units in the \"Y\" direction, we can write:. The tool will display the respective transformation in the output field. This calculator allows one to transform a delta (Δ) network to a wye (Y), and wye (Y) to delta (Δ) network, thus solving for unknown resistor values. In 2016, she founded TransformNation. Graph the function. A quadratic equation is an equation of the form y = ax^2 + bx + c, where a, b and c are constants. Inverse sine calculator. 5 Coordinate Transformation of Vector Components Very often in practical problems, the components of a vector are known in one coordinate system but it is necessary to find them in some other coordinate system. But you do need to know how to do function transformations, because there are ways to ask the questions that don't allow you to cheat, as we'll see in the next section. Laplace transform calculator is the online tool which can easily reduce any given differential equation into an algebraic expression as the answer. Select a transformation if needed by double clicking in the blue transformation box. This calculator helps you to find the point reflection A, for the given coordinates of A(x,y). But practically it is more convenient to eliminate all elements below and above at once when using Gauss-Jordan elimination calculator. Vertical Translations . P. O’Kelley . For example, you can  Basic Calculator Use. The macro calculator takes the guess work out of dieting. Matrix Calculator. 344 and add or subtract that. Through her own struggle of hitting rock bottom, she realized her calling was to help others regain control over their lives, starting with fitness. 1. If they are close to zero, you can create the alignment. Here, it is calculated with matrix A and B, the result is given in the result matrix. 2017/10/21 06:32 Male/Under 20 years old/Elementary school/ Junior high-school student/Very/ Purpose of use Homework Comment/Request My only compliant is it doesn''t tell you whether it is clockwise or counter-clockwise. With this calculator, you can try out as many examples as you want. Discover Resources. BYJU’S online transformation calculator is simple and easy to use. It's often the opposite of what you might expect. All the basic matrix operations as well as methods for solving systems of simultaneous linear equations are implemented on this site. IBAN Calculator: lets you convert a national account number into an IBAN, validate an IBAN, find bank information. One way to graph functions is to simply plot points. Share a link to this widget: More. To help you determine where you stand, we've provided a total daily energy expenditure (TDEE) calculator, which is based on the Mifflin-St. Complex FFT calculator, IFFT calculator, online FFT calculator SciStatCalc (FFT) or an Inverse Fast Fourier Transform (IFFT) on a complex input, dependent on the For coordinate transformations, NCAT uses NADCON 5. Jul 8, 2003 The objective of this module is for students to explore the transformations of functions using a graphing calculator. Easy to use and 100% Free! Find area calculators and other resources to complete your project. The higher your BMI, the higher your risk of developing such conditions as heart disease, high blood pressure, sleep apnea, and type 2 diabetes. You use energy no matter what you're doing, even when sleeping. A linear transformation between two vector spaces and is a map such that the following hold: 1. Import the. Not only does this technology provide intuitive interfaces for setting up complicated transformations, late binding datum transformations, direct connections to the OGP's EPSG registry, survey seismic formats, batch coordinate transformations and all types of file format translation, but the Geographic Calculator combined with the Blue Marble desktop allows GIS managers and Geodesists to A matrix is a rectangular array of numbers. For coordinate transformations, NCAT uses NADCON 5. Help | Comment | NED Home  Feb 1, 2014 Transformations with the Desmos Graphing Calculator. To find the image of a point, we multiply the transformation matrix by a column vector that represents the point's coordinate. Generalized Procrustes analysis , which compares two shapes in Factor Analysis , uses geometric transformations (i. a. When you have finished building your transformation, click Go! to perform the transformation(s). E. The size of a matrix is its dimension, namely the number of rows and columns of the matrix. com is the most convenient free online Matrix Calculator. This simple calculator allows you to calculate a standardized z-score for any raw value of X. Answer. For example, one might know that the force f acting “in the x1 direction” has a certain Transformations of Graphs . Vertical Translations A shift may be referred to as a translation. Jul 27, 2019- Gravity Transformation - Macro Calculator. This Demonstration allows you to investigate the transformation of the graph of a function to for various values of the parameters , , , and . Combinations and Transformations of Functions. Loading Graph Transformations Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. Anti-logarithm calculator. Fisher Z Transformation Calculator . and Cox, D. 0 matrix. The macro calculator is first flexible dieting tool of its kind. Learn how to graph quadratic equations in vertex form. Transformations of the Sine and Cosine Graph – An Exploration. Matrix Multiplication Calculator Here you can perform matrix multiplication with complex numbers online for free. What is a Vertical Translation? Vertically translating a graph is equivalent to shifting the base graph up or down Using the Fisher r-to-z transformation, this page will calculate a value of z that can be applied to assess the significance of the difference between two correlation coefficients, r a and r b, found in two independent samples. You have 4 choices: base 'e', base '10', base '2' and \"Other\". Choose a DNA, RNA, genome editing, qPCR calculator from NEB, a leader in production and supply of reagents for the life science industry. Try It Yourself. Improve your math knowledge with free questions in \"Transformations of functions\" and thousands of other math skills. This is a fraction calculator with steps shown in the solution. Choosing the right transformation. Otherwise, it tries different substitutions and transformations until either the integral  Create an expression using the Columns/Variables and Functions tabs, and the calculator buttons. Enter the calculator function to use in the transformation. Use transformations in combination to explore powerful concepts, create beautiful artwork, and more! The IF calculator is first Intermittent Fasting calculator of its kind. We developed it to be the most comprehensive and easy to use fitness calculator for people following the diet. graph of rational parent function with asymptotes. In geometry and complex analysis, a Möbius transformation of the plane is a rational function of one complex variable. r to P Calculator If the true correlation between X and Y within the general population is rho=0, and if the size of the sample, N, on which an observed value of r is based is equal to or greater than 6, then the quantity The macro calculator is first flexible dieting tool of its kind. Teach me algebra video, algebra help step by step answers, algebra math cal, www. 2. It has four degrees of freedom and requires two pairs of points. matrix. You now know what a transformation is, so let's introduce a special kind of transformation called a linear transformation. More powerful than a graphing calculator! @GraemeCampbell Go to Registration and select the transformation calculator. But when addiction and depression caused the weight to boomerang back, he was faced with a second transformation, and attacked it with wisdom. Proposition 6. A simple calculator taking expressions as input. The graphing calculator uses the list editor and functions with lists including the augment command and line graphs of familiar objects, a broken heart (half a heart) and a flower petal or feather. regression as well as plotting residuals. After an overview of coordinate notation, students explore transformations including translation, reflection, rotation, and dilation in a coordinate plane. Provide the number of inputs, point value, and center of dilation to find the dilation point(s) using this online center of dilation calculator. The calculator provides accurate calculations after submission. Using the x and y values from Sign in to come back to your work later: Sign in with Google. This lesson will define and give examples of each of the four common transformations and end with a quiz to BaZi calculator features: the bazi calculator counts stem and branch interactions, symbolic stars, Na Yin, DHHS, Qi phases, day master strength, draws charts. Rref Calculator for the problem solvers. When: b y = x. Just like Transformations in Geometry, we can move and resize the graphs of functions: Let us start with a function, in this case it is f(x) This calculator is the \"rotation of axes\" Calculator. Using our powerful and blazingly-fast math engine, the calculator can instantly plot any equation, from lines and parabolas up through derivatives and Fourier series. Three transformations will be performed on triangle ABC. 1. Mathway. Transformation on the Graphing Calculator TI-84+, Rotations Using the graphing calculator to examine transformations: To see Reflections (in the x -axis, y -axis, line y = x , origin), Translations , and Dilations , Purplemath. Then the base b logarithm of a number x: log b x = y. Sketch the graphs on the same axes. Modernize with IT infrastructure that takes you to the next level and give your workforce the power to perform their best. Interactive lesson on how to perform a composition of reflections The Integral Calculator supports definite and indefinite integrals (antiderivatives) as well as integrating functions with many variables. To achieve this vision, we’ve started by building the next generation of the graphing calculator. Here at Transformations we are dedicated to changing lives for the better. The most sophisticated and comprehensive graphing calculator online. Your Body Mass Index (BMI) is an estimate of your body fat, based on your height and weight. For instance, the graph for y = x 2 + 3 looks like this: c as Speed Limit The speed of light c is said to be the speed limit of the universe because nothing can be accelerated to the speed of light with respect to you. g: 5e3, 4e-8, 1. The Mohr's Circle calculator provides an intuitive way of visualizing the state of stress at a point in a loaded material. Compositions of reflections in math. In order to calculate log-1 (y) on the calculator, enter the base b (10 is the default value, enter e for e constant), enter the logarithm value y and press the = or calculate button: To add the original graphing calculator, written by Richard Ye, to your web site go to: GitHub and download the code from there. , established in 1987, is an advanced medical weight loss company. vector size (in bp): vector amount (in ng): . Journal of the   Logarithm calculator online. Studentized Range Distribution: Use for range tests such as the Tukey hsd test. Above all, they are used to display linear transformations. Transform coordinates Online convertor for lat & long coordinates, geodetic datums and projected systems Natural Logarithm Calculator. Just select an axis from the drop-down and enter the coordinates, the point reflection calculator will show the result. This is an exploration for Advanced Algebra or Precalculus teachers who have introduced their students to the basic sine and cosine graphs and now want their students to explore how changes to the equations affect the graphs. This on-line tool allows you to insert value pairs of geographic coordinates and transform them to different coordinate system or cartographic projection. b. The last two easy transformations involve flipping functions upside down (flipping them around the x-axis), and mirroring them in the y-axis. As soon as it is changed into the reduced row echelon form the use of it in linear algebra is much easier and can be really convenient for mostly mathematicians. Transformations of Functions with a Graphing Calculator. for any vectors and in , and 2. A description of each function is in the Calculator Functions List. It only makes sense that we have something called a linear transformation because we're studying linear algebra. NEBioCalculator v1. r to P Calculator If the true correlation between X and Y within the general population is rho=0, and if the size of the sample, N, on which an observed value of r is based is equal to or greater than 6, then the quantity Take your geometric explorations to the next level with transformations! Start by defining a reflection, translation, rotation, or dilation. higher molar ratios can be used for blunt end ligations) Transformations A Brush With Death Motivated Noah To Change His Life Noah Kingery gained over 200 pounds in under a year. Linear transformation, sometimes called linear mapping, is a special case of a vector transformation. Matrix Multiplication, Addition and Subtraction Calculator; Matrix Inverse, Determinant and Adjoint Calculator z' z' r Currently, the calculator can be accessed by clicking on the calculator button (machine cog icon) which appears in the headers of Projected CRS pages and Transformation pages (see graphic below). Using Geographic  Compositions of reflections in math. The Laplace Transform Calculator an online tool which shows Laplace Transform for the given input. Which set of transformations will always produce a congruent triangle? A. alzebra 10th qishanes, college algebra test, Pre-Algebra Formula Rules. The only thing that changes is it is now in parallel for a current source transformation. Nonreflective similarity transformation supports translation, rotation, and isotropic scaling. dilation, rotation, translation B. This calculator is online sandbox for playing with Discrete Fourier Transform (DFT). To perform a Box-Cox transformation, choose Stat > Control Charts > Box-Cox Transformation. The above transformations (rotation, reflection, scaling, and shearing) can be represented by matrices. A function transformation takes whatever is the basic function f (x) and then \"transforms\" it (or \"translates\" it), which is a fancy way of saying that you change the formula a bit and thereby move the graph around. To apply these transformations directly to your data in the worksheet, use the Minitab Calculator. Z Score Calculator. rotation, re ection, dilation D. Transformation is termed ROTN, where N is shift  This free online software (calculator) computes the Box-Cox Normality Plot. Here x+4, here x-1. Value type This calculator can compute logs for any number base. Download free on iTunes. For instance, the graph for y = x 2 + 3 looks like this: Graphing an exponential function is helpful when you want to visually analyze the function. Asked in Solar Power The ModelView matrix combined the model and view transformations into one. Use this calculator to find out how much protein you need to transform your body or maintain your size. The symbols to the left of the functions in this screen indicate that the Transformation Graphing application is running. Resizing The other important Transformation is Resizing (also called dilation, contraction, compression, enlargement or even expansion ). Free quadratic equation calculator - Solve quadratic equations using factoring, complete the square and the quadratic formula step-by-step When you will be needing assistance with math and in particular with Algebra Calculator Transforming Formulas or graphing linear come visit us at Algebrahomework. Abby Pollock is an engineer turned transformation specialist who has personally experienced both weight loss and muscle gain transformations. Transform your coordinates online easily with epsg. 1 De nitions A subspace V of Rnis a subset of Rnthat contains the zero element and is closed under addition The Transformations Worksheets are randomly created and will never repeat so you have an endless supply of quality Transformations Worksheets to use in the classroom or at home. Step-by-Step Examples. The ModelView matrix combined the model and view transformations into one. Transformation calculator is a free online tool that gives the transformations of the given input function. Transformation on the Graphing Calculator TI-84+. logo. Forward elimination of Gauss-Jordan calculator reduces matrix to row echelon form. insert size (in bp): Please enter the molar vector : insert ratio: (normally a vector to insert ratio of 1 to 3 is used of cohesive end ligations. Describing Transformations Rotation - centre, angle & direction (clockwise or anti-clockwise) Reflection - the line of reflection Enlargement - centre & scale factor Translation - vector Pearson Correlation Coefficient Calculator. Go to StatPlot - #1 Plot - highlight On - highlight second icon. Let us first look specifically at the basic monic quadratic equation for a parabola with vertex at the origin, (0,0): y = x². When a residual plot reveals a data set to be nonlinear, it is often possible to \"transform\" the raw data to make it more linear. Note that we may need to use several points from the graph and “transform” them, to make Power Calculator: Computes power for a two-sample t-test. Bacteria Transformation Efficiency Calculator. io. Laplace transforms are used to reduce differential equations into algebraic expressions. Each time you tell the calculator to graph a transformation, it will increment the chosen variable by this specified step size and graph the resulting function. Furniture Transformations lets you paint your  Wolfram|Alpha has the ability to compute the transformation matrix for a specific 2D or 3D transformation activity or to return a general transformation calculator  NASA/IPAC EXTRAGALACTIC DATABASE. In particular for each linear geometric transformation, there is one unique real matrix representation. Recall that the Laplace transform of a function is F(s)=L(f(t))=\\int_0^{\\infty} Transformation on the Graphing Calculator TI-84+ 2. I want to talk about the transformation y equals f of x minus h. You can always \"cheat\", by the way, especially if you have a graphing calculator, by quickly graphing y = (x – 2) 4 + 1 and verifying that it matches what you've drawn. For math, science, nutrition, history To zoom, use the zoom slider. Transformations of Exponential Functions: The basic graph of an exponential function in the form The calculator shows us the following graph for this function. The inverse is calculated using Gauss-Jordan . Free functions and graphing calculator - analyze and graph line equations and functions step-by-step. In this topic you will learn about the most useful math concept for creating video game graphics: geometric transformations, specifically translations, rotations, reflections, and dilations. The BMR Calculator will calculate your Basal Metabolic Rate (BMR); the number of calories you'd burn if you stayed in bed all day. 1 Subspaces and Bases 0. Then apply your transformation to any object—just once, or over and over. Our calculator uses this method. The Integral Calculator supports definite and indefinite integrals (antiderivatives) as well as integrating functions with many variables. Sliders make it a breeze to demonstrate function transformations. This leads us to the following algorithm to compute the QR decomposition: function [Q,R] = lec16hqr1(A) % Compute the QR decomposition of an m-by-n matrix A using % Householder transformations. If you just click-and-release (without moving), then the spot you clicked on will be the new Free Geometry calculator - Calculate properties of planes, coordinates and 3d shapes step-by-step Algebra Examples. Online arcsin(x) calculator. It is a measure of linear correlation between two variables x and y and its represented with the symbol 'r'. What happens to the graph of a   Frameworks and Algorithms for the Analysis and Transformation of Scientific the Uniform Library, the Omega Library, and Omega calculator, please use the  The Integral Calculator lets you calculate integrals and antiderivatives of functions Our calculator allows you to check your solutions to calculus exercises. Use basic keys on a calculator (add, subtract, multiply, divide). In this lesson you will investigate transformations of functions using a graphing calculator. f(x) = a ln(b (x - c)) + d. 1) Click on the logarithm button. Linear Transformations of Random Variables. Calculator Introduction: Given the stresses at a space point in the body, s x, s y, and t xy, this calculator computes the stresses of the same space point in a rotated coordinate system, s x', s y', and t x'y'. Visit Mathway on the web. Press [Y=], and you see a screen similar to the following. ln(x) calculator. In order to use this software for Laplace transform calculation, you need to follow these steps: What energy transformations in calculator? 3 4 5. The program is a great tool! Not only does it give you the answers but it also shows you how and why you come up with those answers. R. Built By Science: Protein Calculator Protein is a key nutrient for gaining muscle strength and size, losing fat, and smashing hunger. Here we focus on rigid transformations, that is, transformations that do not change the shape of the graph. Furniture Transformations Color Tool. rescaling, reflection, rotation, or translation) of matrices to compare the sets of data. When you are playing with a jigsaw puzzle, you could move a puzzle piece by sliding it, flipping it, or turning it. For operations of matrices, please use the two calculators below. Linear Transformations , Example 1, Part 1 of 2. [m,n] = size(A); Day 2: Investigating Transformations of Quadratic Relations Chapter 4: Quadratic Relations 2 Investigation A: Compare y x2 and 2 k Use a graphing calculator to graph the quadratic functions on the same set of axis and complete the following table. Transposition Equations Solving Calculator. We already had linear combinations so we might as well have a Transformations International, Inc. 45e12. Be careful of horizontal transformations (shifting left and right, stretching/compressing). Note: If you already know the value of z, and want to calculate p, this calculator will do the job. It provides medical weight loss services to patients desiring a healthy, effective weight loss and management program as well as the advantages of medical assistance. t Distribution: Computes areas of the t distribution. Description and tutorials transformation calculator function. System of linear equations calculator. org. The Y–Δ Transform, also known as \"delta–star\", and \"delta–wye\", is a mathematical process used in the field of electronic engineering to simplify complex resistor networks. Just enter your raw score, population mean and standard deviation,  Caesar cipher is one of the oldest known encryption methods. Stress Transformations & Mohr's Circle. Get our free online math tools for graphing, geometry, 3D, and more! Transposition Equations Solving Calculator. Combinations and Transformations of Functions Graphs up to 5 functions simultaneously in rectangular coordinates, allowing definitions of one function in terms of others. for any scalar. Performs LU, Cholesky, QR, Singular value Transformations > Box Cox Transformation. You will never find a fasting calculator so sophisticated in the whole wide web especially one dedicated solely for The Leangains & OMAD Protocols. The BAZI calculator also enables to manually set transformations. Turn on the Connected Graph icon under StatPlots. If c is added to the function, where the Transformations in math. Also, we will typically start out with a region, $$R$$, in $$xy$$-coordinates and transform it into a region in $$uv$$-coordinates. In geometry, transformation refers to the movement of objects in the coordinate plane. Use the Filter field to search for a specific function. Notice that the right side doesn't get compressed as much. I recommend the Algebrator to students who need help with fractions, equations and algebra. And to understand what kind of transformation this gives us, let's look at an example where I graph three functions that are all related by this transformation. The Pearson correlation coefficient is used to measure the strength of a linear association between two variables, where the value r = 1 means a perfect positive correlation and the value r = -1 means a perfect negataive correlation. Here you can calculate inverse matrix with complex numbers online for free with a very detailed solution. You can rotate different shapes (point-by-point) by an angle, around a center point below. or The Y–Δ Transform, also known as \"delta–star\", and \"delta–wye\", is a mathematical process used in the field of electronic engineering to simplify complex resistor networks. Enter any function in the input field. Doing so allows you to really see the growth or decay of what you’re dealing with. Square root function, up 9, vertical compression, vertical and horizontal reflection (graph falls down and to the left) SECTION 1. Try out our calculator below. Interactive lesson on how to perform a composition of reflections. It is very simple - it is just shifting an alphabet. transformations calculator\n\nfwe, rnutcomx, 11cfj, pm8vzb, vsb, g7u8d, cff, kvm2a1, tbfx, b2a1, i7," ]
[ null, "http://interactivewebtech.com/uahp8h/transformations-calculator.html", null ]
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https://www.litscape.com/word_analysis/toprail
[ "# toprail in Scrabble®\n\nThe word toprail is playable in Scrabble®, no blanks required.\n\nTOPRAIL\n(86 = 36 + 50)\nTOPRAIL\n(86 = 36 + 50)\n\ntoprail\n\nTOPRAIL\n(86 = 36 + 50)\nTOPRAIL\n(86 = 36 + 50)\nTOPRAIL\n(80 = 30 + 50)\nTOPRAIL\n(80 = 30 + 50)\nTOPRAIL\n(80 = 30 + 50)\nTOPRAIL\n(80 = 30 + 50)\nTOPRAIL\n(80 = 30 + 50)\nTOPRAIL\n(80 = 30 + 50)\nTOPRAIL\n(80 = 30 + 50)\nTOPRAIL\n(80 = 30 + 50)\nTOPRAIL\n(77 = 27 + 50)\nTOPRAIL\n(76 = 26 + 50)\nTOPRAIL\n(74 = 24 + 50)\nTOPRAIL\n(72 = 22 + 50)\nTOPRAIL\n(72 = 22 + 50)\nTOPRAIL\n(72 = 22 + 50)\nTOPRAIL\n(72 = 22 + 50)\nTOPRAIL\n(70 = 20 + 50)\nTOPRAIL\n(70 = 20 + 50)\nTOPRAIL\n(70 = 20 + 50)\nTOPRAIL\n(70 = 20 + 50)\nTOPRAIL\n(70 = 20 + 50)\nTOPRAIL\n(70 = 20 + 50)\nTOPRAIL\n(70 = 20 + 50)\nTOPRAIL\n(68 = 18 + 50)\nTOPRAIL\n(68 = 18 + 50)\nTOPRAIL\n(68 = 18 + 50)\nTOPRAIL\n(68 = 18 + 50)\nTOPRAIL\n(68 = 18 + 50)\nTOPRAIL\n(67 = 17 + 50)\nTOPRAIL\n(64 = 14 + 50)\nTOPRAIL\n(63 = 13 + 50)\nTOPRAIL\n(63 = 13 + 50)\nTOPRAIL\n(63 = 13 + 50)\nTOPRAIL\n(63 = 13 + 50)\nTOPRAIL\n(62 = 12 + 50)\nTOPRAIL\n(61 = 11 + 50)\nTOPRAIL\n(61 = 11 + 50)\nTOPRAIL\n(61 = 11 + 50)\nTOPRAIL\n(61 = 11 + 50)\nTOPRAIL\n(61 = 11 + 50)\nTOPRAIL\n(60 = 10 + 50)\n\nTOPRAIL\n(86 = 36 + 50)\nTOPRAIL\n(86 = 36 + 50)\nTOPRAIL\n(80 = 30 + 50)\nTOPRAIL\n(80 = 30 + 50)\nTOPRAIL\n(80 = 30 + 50)\nTOPRAIL\n(80 = 30 + 50)\nTOPRAIL\n(80 = 30 + 50)\nTOPRAIL\n(80 = 30 + 50)\nTOPRAIL\n(80 = 30 + 50)\nTOPRAIL\n(80 = 30 + 50)\nTOPRAIL\n(77 = 27 + 50)\nTOPRAIL\n(76 = 26 + 50)\nTOPRAIL\n(74 = 24 + 50)\nTOPRAIL\n(72 = 22 + 50)\nTOPRAIL\n(72 = 22 + 50)\nTOPRAIL\n(72 = 22 + 50)\nTOPRAIL\n(72 = 22 + 50)\nTOPRAIL\n(70 = 20 + 50)\nTOPRAIL\n(70 = 20 + 50)\nTOPRAIL\n(70 = 20 + 50)\nTOPRAIL\n(70 = 20 + 50)\nTOPRAIL\n(70 = 20 + 50)\nTOPRAIL\n(70 = 20 + 50)\nTOPRAIL\n(70 = 20 + 50)\nTOPRAIL\n(68 = 18 + 50)\nTOPRAIL\n(68 = 18 + 50)\nTOPRAIL\n(68 = 18 + 50)\nTOPRAIL\n(68 = 18 + 50)\nTOPRAIL\n(68 = 18 + 50)\nTOPRAIL\n(67 = 17 + 50)\nTOPRAIL\n(64 = 14 + 50)\nTOPRAIL\n(63 = 13 + 50)\nTOPRAIL\n(63 = 13 + 50)\nTOPRAIL\n(63 = 13 + 50)\nTOPRAIL\n(63 = 13 + 50)\nTOPRAIL\n(62 = 12 + 50)\nTOPRAIL\n(61 = 11 + 50)\nTOPRAIL\n(61 = 11 + 50)\nTOPRAIL\n(61 = 11 + 50)\nTOPRAIL\n(61 = 11 + 50)\nTOPRAIL\n(61 = 11 + 50)\nTOPRAIL\n(60 = 10 + 50)\nPORTAL\n(33)\nPATROL\n(33)\nPATIO\n(30)\nAPRIL\n(30)\nPOLAR\n(30)\nPARTI\n(30)\nPILOT\n(30)\nPLAIT\n(30)\nPATROL\n(28)\nPORTAL\n(28)\nPITA\n(27)\nPATROL\n(27)\nPORTAL\n(27)\nPATROL\n(27)\nPATROL\n(27)\nPATROL\n(27)\nPATROL\n(27)\nPORTAL\n(27)\nTRIP\n(27)\nTRAP\n(27)\nPAIR\n(27)\nPORTAL\n(27)\nPART\n(27)\nPORTAL\n(27)\nPORTAL\n(27)\nPAIL\n(27)\nATOP\n(27)\nTARP\n(27)\nPORT\n(27)\nPLOT\n(27)\nPOLAR\n(26)\nPLAIT\n(26)\nPARTI\n(26)\nPILOT\n(26)\nPATIO\n(26)\nPILOT\n(24)\nTAPIR\n(24)\nPILOT\n(24)\nTAPIR\n(24)\nPATROL\n(24)\nPARTI\n(24)\nPATIO\n(24)\nPATROL\n(24)\nPATIO\n(24)\nPATIO\n(24)\nTAPIR\n(24)\nPARTI\n(24)\nPARTI\n(24)\nTAPIR\n(24)\nPILOT\n(24)\nAPRIL\n(24)\nPORTAL\n(24)\nPOLAR\n(24)\nPOLAR\n(24)\nPOLAR\n(24)\nAPRIL\n(24)\nPLAIT\n(24)\nPLAIT\n(24)\nPLAIT\n(24)\nPORTAL\n(24)\nAPRIL\n(24)\nPORTAL\n(22)\nPATROL\n(22)\nPATROL\n(22)\nPORTAL\n(22)\nPILOT\n(21)\nOPAL\n(21)\nAPRIL\n(21)\nOPAL\n(21)\nPITA\n(21)\nAPRIL\n(21)\nPLAIT\n(21)\nPLAIT\n(21)\nPATIO\n(21)\nPATIO\n(21)\nPATIO\n(21)\nPAIL\n(21)\nATOP\n(21)\nTRIP\n(21)\nPAIR\n(21)\nRAPT\n(21)\nRAPT\n(21)\nPLOT\n(21)\nPOLAR\n(21)\nPOLAR\n(21)\nPOLAR\n(21)\nPART\n(21)\nPARTI\n(21)\nPARTI\n(21)\nPARTI\n(21)\nTAILOR\n(21)\nPLAIT\n(21)\nTAILOR\n(21)\nTAPIR\n(21)\nTAILOR\n(21)\nTAILOR\n(21)\nTAILOR\n(21)\nPILOT\n(21)\nAPRIL\n(21)\nPILOT\n(21)\nPORT\n(21)\nTRAP\n(21)\nTAPIR\n(21)\nTAPIR\n(21)\nTAILOR\n(21)\nTARP\n(21)\nPORTAL\n(20)\nPOLAR\n(20)\nPILOT\n(20)\nPILOT\n(20)\nPARTI\n(20)\nPOLAR\n(20)\nPATROL\n(20)\nPLAIT\n(20)\nPLAIT\n(20)\nPARTI\n(20)\nPORTAL\n(20)\nPATROL\n(20)\nPATIO\n(20)\nPATROL\n(20)\nPATIO\n(20)\nPORTAL\n(20)\nRAPT\n(18)\nPART\n(18)\nPART\n(18)\nRAPT\n(18)\nTAILOR\n(18)\nTRIP\n(18)\nPART\n(18)\nTARP\n(18)\nPART\n(18)\nPORT\n(18)\nPART\n(18)\nTAPIR\n(18)\nTAPIR\n(18)\nTRIP\n(18)\nPORT\n(18)\nRAPT\n(18)\nTARP\n(18)\nRAPT\n(18)\nPAIL\n(18)\nTAILOR\n(18)\nOPAL\n(18)\nOPAL\n(18)\nOPAL\n(18)\nTARP\n(18)\nOPAL\n(18)\nTARP\n(18)\nRATIO\n(18)\nRATIO\n(18)\nRATIO\n(18)\nPAIL\n(18)\nTARP\n(18)\nRATIO\n(18)\nPAIL\n(18)\nPAIL\n(18)\nPORT\n(18)\nPAIL\n(18)\n\n# toprail in Words With Friends™\n\nThe word toprail is playable in Words With Friends™, no blanks required.\n\nTOPRAIL\n(98 = 63 + 35)\n\ntoprail\n\nTOPRAIL\n(98 = 63 + 35)\nTOPRAIL\n(86 = 51 + 35)\nTOPRAIL\n(80 = 45 + 35)\nTOPRAIL\n(80 = 45 + 35)\nTOPRAIL\n(80 = 45 + 35)\nTOPRAIL\n(79 = 44 + 35)\nTOPRAIL\n(79 = 44 + 35)\nTOPRAIL\n(79 = 44 + 35)\nTOPRAIL\n(74 = 39 + 35)\nTOPRAIL\n(74 = 39 + 35)\nTOPRAIL\n(74 = 39 + 35)\nTOPRAIL\n(74 = 39 + 35)\nTOPRAIL\n(74 = 39 + 35)\nTOPRAIL\n(73 = 38 + 35)\nTOPRAIL\n(65 = 30 + 35)\nTOPRAIL\n(65 = 30 + 35)\nTOPRAIL\n(61 = 26 + 35)\nTOPRAIL\n(61 = 26 + 35)\nTOPRAIL\n(61 = 26 + 35)\nTOPRAIL\n(61 = 26 + 35)\nTOPRAIL\n(61 = 26 + 35)\nTOPRAIL\n(59 = 24 + 35)\nTOPRAIL\n(59 = 24 + 35)\nTOPRAIL\n(59 = 24 + 35)\nTOPRAIL\n(59 = 24 + 35)\nTOPRAIL\n(58 = 23 + 35)\nTOPRAIL\n(57 = 22 + 35)\nTOPRAIL\n(57 = 22 + 35)\nTOPRAIL\n(57 = 22 + 35)\nTOPRAIL\n(57 = 22 + 35)\nTOPRAIL\n(57 = 22 + 35)\nTOPRAIL\n(57 = 22 + 35)\nTOPRAIL\n(57 = 22 + 35)\nTOPRAIL\n(56 = 21 + 35)\nTOPRAIL\n(54 = 19 + 35)\nTOPRAIL\n(53 = 18 + 35)\nTOPRAIL\n(52 = 17 + 35)\nTOPRAIL\n(51 = 16 + 35)\nTOPRAIL\n(51 = 16 + 35)\nTOPRAIL\n(50 = 15 + 35)\nTOPRAIL\n(50 = 15 + 35)\nTOPRAIL\n(50 = 15 + 35)\nTOPRAIL\n(50 = 15 + 35)\nTOPRAIL\n(49 = 14 + 35)\nTOPRAIL\n(49 = 14 + 35)\nTOPRAIL\n(49 = 14 + 35)\nTOPRAIL\n(48 = 13 + 35)\nTOPRAIL\n(48 = 13 + 35)\nTOPRAIL\n(48 = 13 + 35)\nTOPRAIL\n(48 = 13 + 35)\nTOPRAIL\n(48 = 13 + 35)\nTOPRAIL\n(48 = 13 + 35)\nTOPRAIL\n(48 = 13 + 35)\nTOPRAIL\n(47 = 12 + 35)\nTOPRAIL\n(47 = 12 + 35)\nTOPRAIL\n(47 = 12 + 35)\nTOPRAIL\n(47 = 12 + 35)\nTOPRAIL\n(46 = 11 + 35)\n\nTOPRAIL\n(98 = 63 + 35)\nTOPRAIL\n(86 = 51 + 35)\nTOPRAIL\n(80 = 45 + 35)\nTOPRAIL\n(80 = 45 + 35)\nTOPRAIL\n(80 = 45 + 35)\nTOPRAIL\n(79 = 44 + 35)\nTOPRAIL\n(79 = 44 + 35)\nTOPRAIL\n(79 = 44 + 35)\nTOPRAIL\n(74 = 39 + 35)\nTOPRAIL\n(74 = 39 + 35)\nTOPRAIL\n(74 = 39 + 35)\nTOPRAIL\n(74 = 39 + 35)\nTOPRAIL\n(74 = 39 + 35)\nTOPRAIL\n(73 = 38 + 35)\nTOPRAIL\n(65 = 30 + 35)\nTOPRAIL\n(65 = 30 + 35)\nTOPRAIL\n(61 = 26 + 35)\nTOPRAIL\n(61 = 26 + 35)\nTOPRAIL\n(61 = 26 + 35)\nTOPRAIL\n(61 = 26 + 35)\nTOPRAIL\n(61 = 26 + 35)\nPORTAL\n(60)\nPATROL\n(60)\nTOPRAIL\n(59 = 24 + 35)\nTOPRAIL\n(59 = 24 + 35)\nTOPRAIL\n(59 = 24 + 35)\nTOPRAIL\n(59 = 24 + 35)\nTOPRAIL\n(58 = 23 + 35)\nTOPRAIL\n(57 = 22 + 35)\nTOPRAIL\n(57 = 22 + 35)\nTOPRAIL\n(57 = 22 + 35)\nTOPRAIL\n(57 = 22 + 35)\nTOPRAIL\n(57 = 22 + 35)\nTOPRAIL\n(57 = 22 + 35)\nTOPRAIL\n(57 = 22 + 35)\nTOPRAIL\n(56 = 21 + 35)\nPATROL\n(54)\nPORTAL\n(54)\nTOPRAIL\n(54 = 19 + 35)\nTOPRAIL\n(53 = 18 + 35)\nTOPRAIL\n(52 = 17 + 35)\nPILOT\n(51)\nPLAIT\n(51)\nAPRIL\n(51)\nPOLAR\n(51)\nTOPRAIL\n(51 = 16 + 35)\nTOPRAIL\n(51 = 16 + 35)\nTOPRAIL\n(50 = 15 + 35)\nTOPRAIL\n(50 = 15 + 35)\nTOPRAIL\n(50 = 15 + 35)\nTOPRAIL\n(50 = 15 + 35)\nTOPRAIL\n(49 = 14 + 35)\nTOPRAIL\n(49 = 14 + 35)\nTOPRAIL\n(49 = 14 + 35)\nPARTI\n(48)\nPORTAL\n(48)\nTOPRAIL\n(48 = 13 + 35)\nTOPRAIL\n(48 = 13 + 35)\nTOPRAIL\n(48 = 13 + 35)\nTOPRAIL\n(48 = 13 + 35)\nTOPRAIL\n(48 = 13 + 35)\nTOPRAIL\n(48 = 13 + 35)\nPLOT\n(48)\nPAIL\n(48)\nPATROL\n(48)\nTOPRAIL\n(48 = 13 + 35)\nPATIO\n(48)\nTOPRAIL\n(47 = 12 + 35)\nTOPRAIL\n(47 = 12 + 35)\nTOPRAIL\n(47 = 12 + 35)\nTOPRAIL\n(47 = 12 + 35)\nTOPRAIL\n(46 = 11 + 35)\nPART\n(45)\nPAIR\n(45)\nATOP\n(45)\nPITA\n(45)\nTARP\n(45)\nTRAP\n(45)\nTRIP\n(45)\nPORT\n(45)\nPATROL\n(42)\nPORTAL\n(42)\nPORTAL\n(40)\nPATROL\n(40)\nPATROL\n(40)\nPORTAL\n(40)\nPLAIT\n(39)\nTAILOR\n(39)\nAPRIL\n(39)\nAPRIL\n(36)\nPOLAR\n(36)\nPILOT\n(36)\nPLAIT\n(36)\nPAIL\n(36)\nPATROL\n(36)\nPATROL\n(36)\nPORTAL\n(36)\nPORTAL\n(36)\nOPAL\n(36)\nPATROL\n(36)\nPATROL\n(36)\nPORTAL\n(36)\nPORTAL\n(36)\nPORTAL\n(36)\nPATROL\n(36)\nPILOT\n(34)\nPOLAR\n(34)\nPLAIT\n(34)\nTAILOR\n(33)\nAPRIL\n(33)\nPLAIT\n(33)\nTAILOR\n(33)\nAPRIL\n(33)\nPOLAR\n(33)\nPILOT\n(33)\nPOLAR\n(33)\nPILOT\n(33)\nPLAIT\n(33)\nPILOT\n(33)\nPOLAR\n(33)\nPATIO\n(32)\nPARTI\n(32)\nTAPIR\n(32)\nPATIO\n(32)\nPARTI\n(32)\nPARTI\n(30)\nTRAIL\n(30)\nTRIAL\n(30)\nPATROL\n(30)\nPORTAL\n(30)\nPATIO\n(30)\nPATROL\n(30)\nPATIO\n(30)\nPARTI\n(30)\nPORTAL\n(30)\nPARTI\n(30)\nPATIO\n(30)\nTAPIR\n(30)\nTAPIR\n(30)\nTAPIR\n(30)\nTAPIR\n(30)\nOPAL\n(30)\nPLOT\n(30)\nTAILOR\n(28)\nPORTAL\n(28)\nTAILOR\n(28)\nPATROL\n(28)\nPORTAL\n(28)\nPATROL\n(28)\nTAIL\n(27)\nLIAR\n(27)\nPOLAR\n(27)\nAPRIL\n(27)\nPART\n(27)\nPOLAR\n(27)\nPILOT\n(27)\nPILOT\n(27)\nLARI\n(27)\nTRIP\n(27)\nLAIR\n(27)\nPILOT\n(27)\nPAIR\n(27)\nAPRIL\n(27)\nPLAIT\n(27)\nRAPT\n(27)\nATOP\n(27)\nPORT\n(27)\nPOLAR\n(27)\nRAPT\n(27)\nROIL\n(27)\nORAL\n(27)\nPLAIT\n(27)\nPLAIT\n(27)\nTAILOR\n(27)\nRIAL\n(27)\nPITA\n(27)\nTAILOR\n(27)\nTAILOR\n(27)\nTARP\n(27)\nTAILOR\n(27)\nRAIL\n(27)\nARIL\n(27)\nTRAP\n(27)\nAPRIL\n(27)\nTOIL\n(27)\nTAILOR\n(27)\nLIRA\n(27)\nLOTI\n(27)\nPOLAR\n(26)\nAPRIL\n(26)\nPILOT\n(26)\nPLAIT\n(26)\nAPRIL\n(26)\nPARTI\n(24)\nTRIAL\n(24)\nTRIAL\n(24)\nPATROL\n(24)\nPARTI\n(24)\nTRIAL\n(24)\nPORTAL\n(24)\n\nai ail rail\n\nto top\n\ntoprails" ]
[ null ]
{"ft_lang_label":"__label__en","ft_lang_prob":0.7178014,"math_prob":1.0000063,"size":359,"snap":"2023-40-2023-50","text_gpt3_token_len":109,"char_repetition_ratio":0.3774648,"word_repetition_ratio":0.20408164,"special_character_ratio":0.18941504,"punctuation_ratio":0.1,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9985776,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-11T16:09:21Z\",\"WARC-Record-ID\":\"<urn:uuid:15710a52-23f7-4b55-b54a-cd083d20a374>\",\"Content-Length\":\"134284\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:af57ce65-0912-46fd-8844-1b075d555bc0>\",\"WARC-Concurrent-To\":\"<urn:uuid:6dcdb5c9-7150-4886-bfae-ab248c9104d7>\",\"WARC-IP-Address\":\"104.21.48.183\",\"WARC-Target-URI\":\"https://www.litscape.com/word_analysis/toprail\",\"WARC-Payload-Digest\":\"sha1:ASRUQIN2MLRA2R23ZO45Z6FEKJU2ELTF\",\"WARC-Block-Digest\":\"sha1:TYIN7DZOZGLV6ACY6SIBAXDBJRDTMOQ5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679515260.97_warc_CC-MAIN-20231211143258-20231211173258-00535.warc.gz\"}"}
https://blog.biswa.co/js-coding-questions-ckfcd1a9d06tm2zs10s6k8xsh?guid=none&deviceId=a4736d60-b633-45f9-82a1-6a3a871c7cd2
[ "# JS Coding Questions\n\nSubscribe to my newsletter and never miss my upcoming articles\n\nQ. Write a program to check for duplicates in an array, it the distance between two duplicate items it less than a specified amount.\n\n``````const arr1 = [3, 1, 5, 7, 9, 5];\n\nfunction checkDuplicates(arr, dis) {\nconst map = {};\n\nfor(let i = 0; i < arr.length; i++) {\nif(map[arr[i]]) {\nif(Math.abs(map[arr[i]] - (i + 1)) <= dis) {\nreturn true;\n}\n}\n\nmap[arr[i]] = i + 1;\n}\n\nreturn false;\n}\n\nconsole.log('Result: ', checkDuplicates(arr1, 3));\n``````\n\nQ. Write a curried version of a sum function which gives results like the 4 use cases mentioned below:\n\n``````sum(1, 2, 3).sumOf(); // 6\nsum(2, 3)(2).sumOf(); // 7\nsum(1)(2)(3)(4).sumOf(); // 10\nsum(2)(4, 1)(2).sumOf(); // 9\n``````\n\nExplanation: This is a curried sum function, which returns a new sum function. This new function must have a property called sumOf which is a function that returns the sum of all elements passed as arguments to the function till the point is is called.\n\nTo achieve this, you need to store all the passed argument continuously in an array and return a sumOf function that can reduce the array to its sum.\n\n``````function sum(...args) {\nlet elements = [];\n\nif(args && args.length) {\nelements = [...elements, ...args];\n}\n\nconst newFn = function(...args) {\nif(args && args.length) {\nelements = [...elements, ...args];\n}\nreturn newFn;\n}\n\nnewFn.sumOf = () => {\nreturn elements.reduce((acc, obj) => acc + obj, 0);\n}\n\nreturn newFn;\n}\n``````" ]
[ null ]
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https://socratic.org/questions/how-do-you-write-the-inequality-and-solve-eight-less-than-a-number-is-no-more-th
[ "# How do you write the inequality and solve \"eight less than a number is no more than 14 and no less than 5#?\n\nApr 14, 2017\n\nSee the entire solution process below:\n\n#### Explanation:\n\nFirst, let's call \"a number\": $n$\n\nThen. \"eight less than a number\" can be written as:\n\n$n - 8$\n\nThis expression \"is no more than 14\" can be written as:\n\n$n - 8 \\le 14$\n\nThis expression \"and no less than 5\" can be written as:\n\n$n - 8 \\ge 5$\n\nOr\n\n$5 \\le n - 8$\n\nWe can combine these two inequalities as:\n\n$5 \\le n - 8 \\le 14$\n\nTo solve we can add $\\textcolor{red}{8}$ to each segment of the inequality to solve for $n$ while keeping the system of inequalities balanced:\n\n$5 + \\textcolor{red}{8} \\le n - 8 + \\textcolor{red}{8} \\le 14 + \\textcolor{red}{8}$\n\n$13 \\le n - 0 \\le 22$\n\n$13 \\le n \\le 22$\n\nOr\n\n$n \\ge 13$ and $n \\le 22$\n\nOr\n\n$\\left[13 , 22\\right]$" ]
[ null ]
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https://tug.org/pipermail/texhax/2005-May/004096.html
[ "# [texhax] inserting text in an enumeration list\n\nE. Krishnan ekmath at asianetindia.com\nFri May 27 06:09:59 CEST 2005\n\nOn Thu, 26 May 2005, Zbigniew Nitecki wrote:\n\n> I am thinking in terms\n> of a subgrouping the exercises at the end of each section of a long\n> book into (in my case) Practical Exercises, Theoretical Exercises,\n> and Challenge Problems.\n\nPlease see whether the example below meets your needs:\n\n\\documentclass[a4paper]{article}\n\\usepackage[defblank]{paralist}\n\n\\newenvironment{problist}{%\n\\renewcommand{\\theenumii}{\\arabic{enumii}}\n\\begin{asparablank}}%\n{\\end{asparablank}}\n\n\\begin{document}\n\nWe give below some exercises to test your understanding of the\nconcepts discussed.\n\\begin{problist}\n\\item \\noindent\\textbf{Practical Exercises}\\\\[2pt]\nThese problems suggest various applications of the concepts\n\\begin{enumerate}\n\\item The first problem\n\\item The second problem\n\\end{enumerate}\n\\item \\noindent\\textbf{Theoretical Exercises}\\\\[2pt]\nThese problems indicate theoretical extensions of the concepts\n\\begin{enumerate}\n\\item The first problem\n\\item The second problem\n\\end{enumerate}\n\\end{problist}\n\n--\nKrishnan\n\n\n\nMore information about the texhax mailing list" ]
[ null ]
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https://gitlab.haskell.org/nineonine/ghc/-/commit/ad0e3c1e2b5edc0b95252acd1c615faeec8b99dc
[ "### Massive patch for the first months work adding System FC to GHC #3\n\nFri Aug 4 15:21:36 EDT 2006 Manuel M T Chakravarty <[email protected]>\n* Massive patch for the first months work adding System FC to GHC #3\n\nBroken up massive patch -=chak\nOriginal log message:\nThis is (sadly) all done in one patch to avoid Darcs bugs.\nIt's not complete work... more FC stuff to come. A compiler\nusing just this patch will fail dismally.\nparent d5bba9ee\n ... ... @@ -157,6 +157,7 @@ expr_fvs (Lit lit) = noVars expr_fvs (Note _ expr) = expr_fvs expr expr_fvs (App fun arg) = expr_fvs fun union expr_fvs arg expr_fvs (Lam bndr body) = addBndr bndr (expr_fvs body) expr_fvs (Cast expr co) = expr_fvs expr union someVars (tyVarsOfType co) expr_fvs (Case scrut bndr ty alts) = expr_fvs scrut union someVars (tyVarsOfType ty) union addBndr bndr ... ... @@ -217,7 +218,8 @@ exprFreeNames e go (Type ty) = tyClsNamesOfType ty -- Don't need free tyvars go (App e1 e2) = go e1 unionNameSets go e2 go (Lam v e) = go e delFromNameSet varName v go (Note n e) = go e go (Note n e) = go e go (Cast e co) = go e unionNameSets tyClsNamesOfType co go (Let (NonRec b r) e) = go e unionNameSets go r go (Let (Rec prs) e) = exprsFreeNames (map snd prs) unionNameSets go e go (Case e b ty as) = go e unionNameSets tyClsNamesOfType ty ... ... @@ -404,13 +406,12 @@ freeVars (Let (Rec binds) body) body2 = freeVars body body_fvs = freeVarsOf body2 freeVars (Note (Coerce to_ty from_ty) expr) = (freeVarsOf expr2 unionFVs tfvs1 unionFVs tfvs2, AnnNote (Coerce to_ty from_ty) expr2) freeVars (Cast expr co) = (freeVarsOf expr2 unionFVs cfvs, AnnCast expr2 co) where expr2 = freeVars expr tfvs1 = tyVarsOfType from_ty tfvs2 = tyVarsOfType to_ty expr2 = freeVars expr cfvs = tyVarsOfType co freeVars (Note other_note expr) = (freeVarsOf expr2, AnnNote other_note expr2) ... ...\n ... ... @@ -6,7 +6,7 @@ \\begin{code} module CoreUtils ( -- Construction mkInlineMe, mkSCC, mkCoerce, mkCoerce2, mkInlineMe, mkSCC, mkCoerce, bindNonRec, needsCaseBinding, mkIfThenElse, mkAltExpr, mkPiType, mkPiTypes, ... ... @@ -42,7 +42,7 @@ import GLAEXTS -- For xori import CoreSyn import CoreFVs ( exprFreeVars ) import PprCore ( pprCoreExpr ) import Var ( Var ) import Var ( Var, TyVar ) import VarSet ( unionVarSet ) import VarEnv import Name ( hashName ) ... ... @@ -51,8 +51,9 @@ import Packages ( isDllName ) #endif import Literal ( hashLiteral, literalType, litIsDupable, litIsTrivial, isZeroLit, Literal( MachLabel ) ) import DataCon ( DataCon, dataConRepArity, dataConInstArgTys, isVanillaDataCon, dataConTyCon ) import DataCon ( DataCon, dataConRepArity, isVanillaDataCon, dataConTyCon, dataConRepArgTys, dataConUnivTyVars ) import PrimOp ( PrimOp(..), primOpOkForSpeculation, primOpIsCheap ) import Id ( Id, idType, globalIdDetails, idNewStrictness, mkWildId, idArity, idName, idUnfolding, idInfo, ... ... @@ -65,8 +66,12 @@ import Type ( Type, mkFunTy, mkForAllTy, splitFunTy_maybe, splitFunTy, tcEqTypeX, applyTys, isUnLiftedType, seqType, mkTyVarTy, splitForAllTy_maybe, isForAllTy, splitRecNewType_maybe, splitTyConApp_maybe, coreEqType, funResultTy, applyTy splitTyConApp_maybe, coreEqType, funResultTy, applyTy, substTyWith ) import Coercion ( Coercion, mkTransCoercion, coercionKind, splitRecNewTypeCo_maybe, mkSymCoercion, mkLeftCoercion, mkRightCoercion, decomposeCo, coercionKindTyConApp ) import TyCon ( tyConArity ) import TysWiredIn ( boolTy, trueDataCon, falseDataCon ) import CostCentre ( CostCentre ) ... ... @@ -93,7 +98,8 @@ exprType (Var var) = idType var exprType (Lit lit) = literalType lit exprType (Let _ body) = exprType body exprType (Case _ _ ty alts) = ty exprType (Note (Coerce ty _) e) = ty -- **! should take usage from e exprType (Cast e co) = let (_, ty) = coercionKind co in ty exprType (Note other_note e) = exprType e exprType (Lam binder expr) = mkPiType binder (exprType expr) exprType e@(App _ _) ... ... @@ -145,7 +151,7 @@ applyTypeToArgs e op_ty (Type ty : args) applyTypeToArgs e op_ty (other_arg : args) = case (splitFunTy_maybe op_ty) of Just (_, res_ty) -> applyTypeToArgs e res_ty args Nothing -> pprPanic \"applyTypeToArgs\" (pprCoreExpr e) Nothing -> pprPanic \"applyTypeToArgs\" (pprCoreExpr e $$ppr op_ty) \\end{code} ... ... @@ -161,7 +167,6 @@ mkNote removes redundant coercions, and SCCs where possible \\begin{code} #ifdef UNUSED mkNote :: Note -> CoreExpr -> CoreExpr mkNote (Coerce to_ty from_ty) expr = mkCoerce2 to_ty from_ty expr mkNote (SCC cc) expr = mkSCC cc expr mkNote InlineMe expr = mkInlineMe expr mkNote note expr = Note note expr ... ... @@ -197,18 +202,20 @@ mkInlineMe e = Note InlineMe e \\begin{code} mkCoerce :: Type -> CoreExpr -> CoreExpr mkCoerce to_ty expr = mkCoerce2 to_ty (exprType expr) expr mkCoerce2 :: Type -> Type -> CoreExpr -> CoreExpr mkCoerce2 to_ty from_ty (Note (Coerce to_ty2 from_ty2) expr) = ASSERT( from_ty coreEqType to_ty2 ) mkCoerce2 to_ty from_ty2 expr mkCoerce2 to_ty from_ty expr | to_ty coreEqType from_ty = expr | otherwise = ASSERT( from_ty coreEqType exprType expr ) Note (Coerce to_ty from_ty) expr mkCoerce :: Coercion -> CoreExpr -> CoreExpr mkCoerce co (Cast expr co2) = ASSERT(let { (from_ty, to_ty) = coercionKind co; (from_ty2, to_ty2) = coercionKind co2} in from_ty coreEqType to_ty2 ) mkCoerce (mkTransCoercion co2 co) expr mkCoerce co expr = let (from_ty, to_ty) = coercionKind co in -- if to_ty coreEqType from_ty -- then expr -- else ASSERT2(from_ty coreEqType (exprType expr), text \"Trying to coerce\" <+> text \"(\" <> ppr expr$$ text \"::\" <+> ppr (exprType expr) <> text \")\" $$ppr co$$ ppr (coercionKindTyConApp co)) (Cast expr co) \\end{code} \\begin{code} ... ... @@ -219,6 +226,7 @@ mkSCC cc (Lit lit) = Lit lit mkSCC cc (Lam x e) = Lam x (mkSCC cc e) -- Move _scc_ inside lambda mkSCC cc (Note (SCC cc') e) = Note (SCC cc) (Note (SCC cc') e) mkSCC cc (Note n e) = Note n (mkSCC cc e) -- Move _scc_ inside notes mkSCC cc (Cast e co) = Cast (mkSCC cc e) co -- Move _scc_ inside cast mkSCC cc expr = Note (SCC cc) expr \\end{code} ... ... @@ -256,7 +264,7 @@ mkAltExpr :: AltCon -> [CoreBndr] -> [Type] -> CoreExpr -- This guy constructs the value that the scrutinee must have -- when you are in one particular branch of a case mkAltExpr (DataAlt con) args inst_tys = mkConApp con (map Type inst_tys ++ map varToCoreExpr args) = mkConApp con (map Type inst_tys ++ varsToCoreExprs args) mkAltExpr (LitAlt lit) [] [] = Lit lit ... ... @@ -353,6 +361,7 @@ exprIsTrivial (Lit lit) = litIsTrivial lit exprIsTrivial (App e arg) = not (isRuntimeArg arg) && exprIsTrivial e exprIsTrivial (Note (SCC _) e) = False -- See notes above exprIsTrivial (Note _ e) = exprIsTrivial e exprIsTrivial (Cast e co) = exprIsTrivial e exprIsTrivial (Lam b body) = not (isRuntimeVar b) && exprIsTrivial body exprIsTrivial other = False \\end{code} ... ... @@ -375,6 +384,7 @@ exprIsDupable (Var v) = True exprIsDupable (Lit lit) = litIsDupable lit exprIsDupable (Note InlineMe e) = True exprIsDupable (Note _ e) = exprIsDupable e exprIsDupable (Cast e co) = exprIsDupable e exprIsDupable expr = go expr 0 where ... ... @@ -423,6 +433,7 @@ exprIsCheap (Type _) = True exprIsCheap (Var _) = True exprIsCheap (Note InlineMe e) = True exprIsCheap (Note _ e) = exprIsCheap e exprIsCheap (Cast e co) = exprIsCheap e exprIsCheap (Lam x e) = isRuntimeVar x || exprIsCheap e exprIsCheap (Case e _ _ alts) = exprIsCheap e && and [exprIsCheap rhs | (_,_,rhs) <- alts] ... ... @@ -513,10 +524,11 @@ side effects, and can't diverge or raise an exception. \\begin{code} exprOkForSpeculation :: CoreExpr -> Bool exprOkForSpeculation (Lit _) = True exprOkForSpeculation (Type _) = True exprOkForSpeculation (Var v) = isUnLiftedType (idType v) exprOkForSpeculation (Note _ e) = exprOkForSpeculation e exprOkForSpeculation (Lit _) = True exprOkForSpeculation (Type _) = True exprOkForSpeculation (Var v) = isUnLiftedType (idType v) exprOkForSpeculation (Note _ e) = exprOkForSpeculation e exprOkForSpeculation (Cast e co) = exprOkForSpeculation e exprOkForSpeculation other_expr = case collectArgs other_expr of (Var f, args) -> spec_ok (globalIdDetails f) args ... ... @@ -567,6 +579,7 @@ exprIsBottom e = go 0 e where -- n is the number of args go n (Note _ e) = go n e go n (Cast e co) = go n e go n (Let _ e) = go n e go n (Case e _ _ _) = go 0 e -- Just check the scrut go n (App e _) = go (n+1) e ... ... @@ -618,6 +631,7 @@ exprIsHNF (Type ty) = True -- Types are honorary Values; -- we don't mind copying them exprIsHNF (Lam b e) = isRuntimeVar b || exprIsHNF e exprIsHNF (Note _ e) = exprIsHNF e exprIsHNF (Cast e co) = exprIsHNF e exprIsHNF (App e (Type _)) = exprIsHNF e exprIsHNF (App e a) = app_is_value e [a] exprIsHNF other = False ... ... @@ -643,8 +657,27 @@ check_args fun_ty (arg : args) \\end{code} \\begin{code} -- deep applies a TyConApp coercion as a substitution to a reflexive coercion -- deepCast t [a1,...,an] co corresponds to deep(t, [a1,...,an], co) from -- FC paper deepCast :: Type -> [TyVar] -> Coercion -> Coercion deepCast ty tyVars co = ASSERT( let {(lty, rty) = coercionKind co; Just (tc1, lArgs) = splitTyConApp_maybe lty; Just (tc2, rArgs) = splitTyConApp_maybe rty} in tc1 == tc2 && length lArgs == length rArgs && length lArgs == length tyVars ) substTyWith tyVars coArgs ty where -- coArgs = [right (left (left co)), right (left co), right co] coArgs = decomposeCo (length tyVars) co exprIsConApp_maybe :: CoreExpr -> Maybe (DataCon, [CoreExpr]) exprIsConApp_maybe (Note (Coerce to_ty from_ty) expr) -- Returns (Just (dc, [x1..xn])) if the argument expression is -- a constructor application of the form (dc x1 .. xn) exprIsConApp_maybe (Cast expr co) = -- Maybe this is over the top, but here we try to turn -- coerce (S,T) ( x, y ) -- effectively into ... ... @@ -654,6 +687,7 @@ exprIsConApp_maybe (Note (Coerce to_ty from_ty) expr) -- (# r, s #) -> ... -- where the memcpy is in the IO monad, but the call is in -- the (ST s) monad let (from_ty, to_ty) = coercionKind co in case exprIsConApp_maybe expr of { Nothing -> Nothing ; Just (dc, args) -> ... ... @@ -666,14 +700,15 @@ exprIsConApp_maybe (Note (Coerce to_ty from_ty) expr) -- Type constructor must match -- We knock out existentials to keep matters simple(r) let arity = tyConArity tc val_args = drop arity args to_arg_tys = dataConInstArgTys dc tc_arg_tys mk_coerce ty arg = mkCoerce ty arg new_val_args = zipWith mk_coerce to_arg_tys val_args arity = tyConArity tc val_args = drop arity args arg_tys = dataConRepArgTys dc dc_tyvars = dataConUnivTyVars dc deep arg_ty = deepCast arg_ty dc_tyvars co new_val_args = zipWith mkCoerce (map deep arg_tys) val_args in ASSERT( all isTypeArg (take arity args) ) ASSERT( equalLength val_args to_arg_tys ) ASSERT( equalLength val_args arg_tys ) Just (dc, map Type tc_arg_tys ++ new_val_args) }} ... ... @@ -823,6 +858,8 @@ arityType dflags (Note n e) = arityType dflags e -- | ok_note n = arityType dflags e -- | otherwise = ATop arityType dflags (Cast e co) = arityType dflags e arityType dflags (Var v) = mk (idArity v) (arg_tys (idType v)) where ... ... @@ -933,7 +970,8 @@ etaExpand :: Arity -- Result should have this number of value args etaExpand n us expr ty | manifestArity expr >= n = expr -- The no-op case | otherwise = eta_expand n us expr ty | otherwise = eta_expand n us expr ty where -- manifestArity sees how many leading value lambdas there are ... ... @@ -941,6 +979,7 @@ manifestArity :: CoreExpr -> Arity manifestArity (Lam v e) | isId v = 1 + manifestArity e | otherwise = manifestArity e manifestArity (Note _ e) = manifestArity e manifestArity (Cast e _) = manifestArity e manifestArity e = 0 -- etaExpand deals with for-alls. For example: ... ... @@ -987,7 +1026,8 @@ eta_expand n us (Lam v body) ty -- and round we go eta_expand n us expr ty = case splitForAllTy_maybe ty of { = ASSERT2 (exprType expr coreEqType ty, ppr (exprType expr) ppr ty) case splitForAllTy_maybe ty of { Just (tv,ty') -> Lam tv (eta_expand n us (App expr (Type (mkTyVarTy tv))) ty') ; Nothing -> ... ... @@ -1006,11 +1046,10 @@ eta_expand n us expr ty -- eta_expand 1 e T -- We want to get -- coerce T (\\x::[T] -> (coerce ([T]->Int) e) x) -- Only try this for recursive newtypes; the non-recursive kind -- are transparent anyway case splitRecNewType_maybe ty of { Just ty' -> mkCoerce2 ty ty' (eta_expand n us (mkCoerce2 ty' ty expr) ty') ; case splitRecNewTypeCo_maybe ty of { Just(ty1,co) -> mkCoerce co (eta_expand n us (mkCoerce (mkSymCoercion co) expr) ty1) ; Nothing -> -- We have an expression of arity > 0, but its type isn't a function ... ... @@ -1053,6 +1092,7 @@ exprArity e = go e go (Lam x e) | isId x = go e + 1 | otherwise = go e go (Note n e) = go e go (Cast e _) = go e go (App e (Type t)) = go e go (App f a) | exprIsCheap a = (go f - 1) max 0 -- NB: exprIsCheap a! ... ... @@ -1129,13 +1169,13 @@ tcEqExprX env (Case e1 v1 t1 a1) env' = rnBndr2 env v1 v2 tcEqExprX env (Note n1 e1) (Note n2 e2) = eq_note env n1 n2 && tcEqExprX env e1 e2 tcEqExprX env (Cast e1 co1) (Cast e2 co2) = tcEqTypeX env co1 co2 && tcEqExprX env e1 e2 tcEqExprX env (Type t1) (Type t2) = tcEqTypeX env t1 t2 tcEqExprX env e1 e2 = False eq_alt env (c1,vs1,r1) (c2,vs2,r2) = c1==c2 && tcEqExprX (rnBndrs2 env vs1 vs2) r1 r2 eq_note env (SCC cc1) (SCC cc2) = cc1 == cc2 eq_note env (Coerce t1 f1) (Coerce t2 f2) = tcEqTypeX env t1 t2 && tcEqTypeX env f1 f2 eq_note env (CoreNote s1) (CoreNote s2) = s1 == s2 eq_note env other1 other2 = False \\end{code} ... ... @@ -1160,11 +1200,11 @@ exprSize (App f a) = exprSize f + exprSize a exprSize (Lam b e) = varSize b + exprSize e exprSize (Let b e) = bindSize b + exprSize e exprSize (Case e b t as) = seqType t seq exprSize e + varSize b + 1 + foldr ((+) . altSize) 0 as exprSize (Cast e co) = (seqType co seq 1) + exprSize e exprSize (Note n e) = noteSize n + exprSize e exprSize (Type t) = seqType t seq 1 noteSize (SCC cc) = cc seq 1 noteSize (Coerce t1 t2) = seqType t1 seq seqType t2 seq 1 noteSize InlineMe = 1 noteSize (CoreNote s) = s seq 1 -- hdaume: core annotations ... ... @@ -1193,12 +1233,21 @@ altSize (c,bs,e) = c seq varsSize bs + exprSize e \\begin{code} hashExpr :: CoreExpr -> Int -- Two expressions that hash to the same Int may be equal (but may not be) -- Two expressions that hash to the different Ints are definitely unequal -- -- But \"unequal\" here means \"not identical\"; two alpha-equivalent -- expressions may hash to the different Ints -- -- The emphasis is on a crude, fast hash, rather than on high precision hashExpr e | hash < 0 = 77 -- Just in case we hit -maxInt | otherwise = hash where hash = abs (hash_expr e) -- Negative numbers kill UniqFM hash_expr (Note _ e) = hash_expr e hash_expr (Cast e co) = hash_expr e hash_expr (Let (NonRec b r) e) = hashId b hash_expr (Let (Rec ((b,r):_)) e) = hashId b hash_expr (Case _ b _ _) = hashId b ... ... @@ -1304,6 +1353,7 @@ rhsIsStatic this_pkg rhs = is_static False rhs is_static in_arg (Note (SCC _) e) = False is_static in_arg (Note _ e) = is_static in_arg e is_static in_arg (Cast e co) = is_static in_arg e is_static in_arg (Lit lit) = case lit of ... ... @@ -1346,6 +1396,7 @@ rhsIsStatic this_pkg rhs = is_static False rhs go (Note (SCC _) f) n_val_args = False go (Note _ f) n_val_args = go f n_val_args go (Cast e co) n_val_args = go e n_val_args go other n_val_args = False ... ...\nSupports Markdown\n0% or .\nYou are about to add 0 people to the discussion. Proceed with caution.\nFinish editing this message first!" ]
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https://www.preservearticles.com/chemistry/essential-characteristics-of-an-electron/499
[ "Electron is a negatively charged particles having very small mass. Mass and charge of an electron are given below.\n\n(i) Mass of an electron (me): The mass of a electron is about 1/1840 times that of a hydrogen atom. Its absolute mass is ,\n\nme= 9.108 × 10-31kg=9.108 × 10-28g\n\nIt is a very light particle, and therefore, it makes very little contribution to the total mass of the atom in which it is contained.\n\n(ii) Charge on an electron (e): an electron possess one unit negative charge, it has been found to be the smallest negative charge that any particle can carry.\n\nCharge on an electron, e = – 1.602× 10-19 Coulombs.\n\nHome››Chemistry››" ]
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https://www.mathworks.com/help/optim/ug/problem-based-workflow.html
[ "## Problem-Based Optimization Workflow\n\nNote\n\nOptimization Toolbox™ provides two approaches for solving single-objective optimization problems. This topic describes the problem-based approach. Solver-Based Optimization Problem Setup describes the solver-based approach.\n\nTo solve an optimization problem, perform the following steps.\n\n• Create an optimization problem object by using `optimproblem`. A problem object is a container in which you define an objective expression and constraints. The optimization problem object defines the problem and any bounds that exist in the problem variables.\n\nFor example, create a maximization problem.\n\n`prob = optimproblem('ObjectiveSense','maximize');`\n• Create named variables by using `optimvar`. An optimization variable is a symbolic variable that you use to describe the problem objective and constraints. Include any bounds in the variable definitions.\n\nFor example, create a 15-by-3 array of binary variables named `'x'`.\n\n`x = optimvar('x',15,3,'Type','integer','LowerBound',0,'UpperBound',1);`\n• Define the objective function in the problem object as an expression in the named variables.\n\nNote\n\nIf you have a nonlinear function that is not composed of polynomials, rational expressions, and elementary functions such as `exp`, then convert the function to an optimization expression by using `fcn2optimexpr`. See Convert Nonlinear Function to Optimization Expression and Supported Operations for Optimization Variables and Expressions.\n\nIf necessary, include extra parameters in your expression as workspace variables; see Pass Extra Parameters in Problem-Based Approach.\n\nFor example, assume that you have a real matrix `f` of the same size as a matrix of variables `x`, and the objective is the sum of the entries in `f` times the corresponding variables `x`.\n\n`prob.Objective = sum(sum(f.*x));`\n• Define constraints for optimization problems as either comparisons in the named variables or as comparisons of expressions.\n\nNote\n\nIf you have a nonlinear function that is not composed of polynomials, rational expressions, and elementary functions such as `exp`, then convert the function to an optimization expression by using `fcn2optimexpr`. See Convert Nonlinear Function to Optimization Expression and Supported Operations for Optimization Variables and Expressions.\n\nFor example, assume that the sum of the variables in each row of `x` must be one, and the sum of the variables in each column must be no more than one.\n\n```onesum = sum(x,2) == 1; vertsum = sum(x,1) <= 1; prob.Constraints.onesum = onesum; prob.Constraints.vertsum = vertsum;```\n• For nonlinear problems, set an initial point as a structure whose fields are the optimization variable names. For example:\n\n```x0.x = randn(size(x)); x0.y = eye(4); % Assumes y is a 4-by-4 variable```\n• Solve the problem by using `solve`.\n\n```sol = solve(prob); % Or, for nonlinear problems, sol = solve(prob,x0)```\n\nIn addition to these basic steps, you can review the problem definition before solving the problem by using `show` or `write`. Set options for `solve` by using `optimoptions`, as explained in Change Default Solver or Options.\n\nWarning\n\nThe problem-based approach does not support complex values in an objective function, nonlinear equalities, or nonlinear inequalities. If a function calculation has a complex value, even as an intermediate value, the final result might be incorrect.\n\nNote\n\nAll names in an optimization problem must be unique. Specifically, all variable names, objective function names, and constraint function names must be different.\n\nFor a basic mixed-integer linear programming example, see Mixed-Integer Linear Programming Basics: Problem-Based or the video version Solve a Mixed-Integer Linear Programming Problem Using Optimization Modeling. For a nonlinear example, see Solve a Constrained Nonlinear Problem, Problem-Based. For more extensive examples, see Problem-Based Nonlinear Optimization, Linear Programming and Mixed-Integer Linear Programming, or Quadratic Programming and Cone Programming." ]
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http://kth.diva-portal.org/smash/resultList.jsf?p=551&fs=false&language=en&searchType=SIMPLE&query=&af=%5B%5D&aq=%5B%5B%7B%22organisationId%22%3A%226139%22%7D%5D%5D&aq2=%5B%5B%5D%5D&aqe=%5B%5D&noOfRows=50&sortOrder=author_sort_asc&sortOrder2=title_sort_asc&onlyFullText=false&sf=all
[ "Change search\nRefine search result\n9101112131415 551 - 600 of 764\nCite\nCitation style\n• apa\n• harvard1\n• ieee\n• modern-language-association-8th-edition\n• vancouver\n• Other style\nMore styles\nLanguage\n• de-DE\n• en-GB\n• en-US\n• fi-FI\n• nn-NO\n• nn-NB\n• sv-SE\n• Other locale\nMore languages\nOutput format\n• html\n• text\n• asciidoc\n• rtf\nRows per page\n• 5\n• 10\n• 20\n• 50\n• 100\n• 250\nSort\n• Standard (Relevance)\n• Author A-Ö\n• Author Ö-A\n• Title A-Ö\n• Title Ö-A\n• Publication type A-Ö\n• Publication type Ö-A\n• Issued (Oldest first)\n• Created (Oldest first)\n• Last updated (Oldest first)\n• Disputation date (earliest first)\n• Disputation date (latest first)\n• Standard (Relevance)\n• Author A-Ö\n• Author Ö-A\n• Title A-Ö\n• Title Ö-A\n• Publication type A-Ö\n• Publication type Ö-A\n• Issued (Oldest first)\n• Created (Oldest first)\n• Last updated (Oldest first)\n• Disputation date (earliest first)\n• Disputation date (latest first)\nSelect\nThe maximal number of hits you can export is 250. When you want to export more records please use the Create feeds function.\n• 551.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics. KTH, School of Engineering Sciences (SCI), Theoretical Physics.\nMonte Carlo Simulations of a Lattice Gas with Logarithmic InteractionsManuscript (Other academic)\n• 552.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics. Department of Physics, Luleå University of Technology.\nCritical scaling properties at the superfluid transition of He-4 in aerogel2006In: Physical Review Letters, ISSN 0031-9007, E-ISSN 1079-7114, Vol. 97, no 22, p. 225702-Article in journal (Refereed)\n\nWe study the superfluid transition of He-4 in aerogel by Monte Carlo simulations and finite size scaling analysis. Aerogel is a highly porous silica glass, which we model by a diffusion limited cluster aggregation model. The superfluid is modeled by a three dimensional XY model, with excluded bonds to sites on the aerogel cluster. We obtain the correlation length exponent nu=0.73 +/- 0.02, in reasonable agreement with experiments and with previous simulations. For the heat capacity exponent alpha, both experiments and previous simulations suggest deviations from the Josephson hyperscaling relation alpha=2-d nu. In contrast, our Monte Carlo results support hyperscaling with alpha=-0.2 +/- 0.05. We suggest a reinterpretation of the experiments, which avoids scaling violations and is consistent with our simulation results.\n\n• 553.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics.\nMonte Carlo Simulation Study of 1D Josephson Junction Arrays2016Independent thesis Advanced level (degree of Master (Two Years)), 20 credits / 30 HE creditsStudent thesis\n• 554.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics.\nThe Klein-Gordon Equation and Pionic Atoms2014Independent thesis Basic level (degree of Bachelor), 10 HE creditsStudent thesis\n\nThis bachelor thesis introduces the Klein-Gordon equation and pionic atoms through\n\na historical review. It discusses properties of the equation and its continuity equation\n\nbased on a comparison with the Schrödinger equation and its continuity equation. The\n\nconclusion that the Klein-Gordon equation is a relativistic extension of the Schrödinger\n\nequation for spin-0 particles is drawn. This makes it possible to derive a Klein-Gordon\n\nequation and Coulomb potential based model for pionic atoms. The transition energies\n\nand the charge density for a pionic atom are derived. The model and a Schrödinger equation\n\nmodel are used to draw conclusions on the differences between regular and pionic\n\natoms. Numerical predictions of the models are compared to experimental data and the\n\naccuracy of the models is discussed. Properties of the pionic atom are discussed based\n\n• 555.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics.\nMagnetic Monopoles in Spin Ice2014Independent thesis Advanced level (degree of Master (Two Years)), 20 credits / 30 HE creditsStudent thesis\n• 556.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics.\n– A study of the Langevin equation in a thermal ratchet model of the myosin V motor protein2012Independent thesis Basic level (degree of Bachelor), 10 credits / 15 HE creditsStudent thesis\n• 557. Nussinov, Z.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Condensed Matter Theory.\nExact ground states of extended t-J(z) models on a square lattice2007In: Philosophical Magazine Letters, ISSN 0950-0839, E-ISSN 1362-3036, Vol. 87, no 7, p. 515-526Article in journal (Refereed)\n\nWe examine special extended spin S = 1/2 fermionic and hard core bosonic t - J(z) models with nearest neighbour and next nearest neighbour interactions to find exact ground states. Some of these models display an exponentially large degeneracy with diagonal stripe-like patterns.\n\n• 558. Nyblom, Maria\nKTH, School of Engineering Sciences (SCI), Theoretical Physics. KTH, Centres, SeRC - Swedish e-Science Research Centre. KTH, Centres, Science for Life Laboratory, SciLifeLab. KTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical & Computational Biophysics. KTH, Centres, SeRC - Swedish e-Science Research Centre. KTH, Centres, Science for Life Laboratory, SciLifeLab.\nCrystal Structure of Na+, K+-ATPase in the Na+-Bound State2013In: Science, ISSN 0036-8075, E-ISSN 1095-9203, Vol. 342, no 6154, p. 123-127Article in journal (Refereed)\n\nThe Na+, K+-adenosine triphosphatase (ATPase) maintains the electrochemical gradients of Na+ and K+ across the plasma membrane-a prerequisite for electrical excitability and secondary transport. Hitherto, structural information has been limited to K+-bound or ouabain-blocked forms. We present the crystal structure of a Na+-bound Na+, K+-ATPase as determined at 4.3 angstrom resolution. Compared with the K+-bound form, large conformational changes are observed in the a subunit whereas the beta and gamma subunit structures are maintained. The locations of the three Na+ sites are indicated with the unique site III at the recently suggested IIIb, as further supported by electrophysiological studies on leak currents. Extracellular release of the third Na+ from IIIb through IIIa, followed by exchange of Na+ for K+ at sites I and II, is suggested.\n\n• 559. Nys, Mieke\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical & Computational Biophysics. KTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical & Computational Biophysics. KTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical & Computational Biophysics.\nThe Crystal Structure of ELIC in Complex with Chlorpromazine Unexpectedly Unveils an Allosteric Binding Site in the Ligand-Binding Domain2016In: Biophysical Journal, ISSN 0006-3495, E-ISSN 1542-0086, Vol. 110, no 3, p. 457A-457AArticle in journal (Other academic)\n• 560. Nys, Mieke\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical & Computational Biophysics. KTH, Centres, Science for Life Laboratory, SciLifeLab. KTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical & Computational Biophysics. KTH, Centres, Science for Life Laboratory, SciLifeLab. KTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical & Computational Biophysics. KTH, Centres, Science for Life Laboratory, SciLifeLab. Stockholm University, Sweden.\nAllosteric binding site in a Cys-loop receptor ligand-binding domain unveiled in the crystal structure of ELIC in complex with chlorpromazine2016In: Proceedings of the National Academy of Sciences of the United States of America, ISSN 0027-8424, E-ISSN 1091-6490, Vol. 113, no 43, p. E6696-E6703Article in journal (Refereed)\n\nPentameric ligand-gated ion channels or Cys-loop receptors are responsible for fast inhibitory or excitatory synaptic transmission. The antipsychotic compound chlorpromazine is a widely used tool to probe the ion channel pore of the nicotinic acetylcholine receptor, which is a prototypical Cys-loop receptor. In this study, we determine the molecular determinants of chlorpromazine binding in the Erwinia ligand-gated ion channel (ELIC). We report the X-ray crystal structures of ELIC in complex with chlorpromazine or its brominated derivative bromopromazine. Unexpectedly, we do not find a chlorpromazine molecule in the channel pore of ELIC, but behind the beta 8-beta 9 loop in the extracellular ligand-binding domain. The beta 8-beta 9 loop is localized downstream from the neurotransmitter binding site and plays an important role in coupling of ligand binding to channel opening. In combination with electrophysiological recordings from ELIC cysteine mutants and a thiol-reactive derivative of chlorpromazine, we demonstrate that chlorpromazine binding at the beta 8-beta 9 loop is responsible for receptor inhibition. We further use molecular-dynamics simulations to support the X-ray data and mutagenesis experiments. Together, these data unveil an allosteric binding site in the extracellular ligand-binding domain of ELIC. Our results extend on previous observations and further substantiate our understanding of a multisite model for allosteric modulation of Cys-loop receptors.\n\n• 561.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics.\nConsequences of Violation of Special Relativity and Causality2012Independent thesis Basic level (degree of Bachelor), 10 credits / 15 HE creditsStudent thesis\n• 562.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics.\nA brief status of non-standard neutrino interactions2013In: NOW 2012: Proceedings of the Neutrino Oscillation Workshop / [ed] Paolo Bernardini, Gianluigi Fogli, Eligio Lisi, Elsevier, 2013, p. 301-307Conference paper (Refereed)\n\nIn this plenary talk, we review the status of non-standard neutrino interactions (NSIs). First, we give a brief introduction to neutrino flavor transitions with NSIs based on the standard paradigm of neutrino oscillations. Then, we discuss alternative scenarios for neutrino flavor transitions such as neutrino decoherence, neutrino decay, and NSIs. Second, we investigate NSIs with three neutrino flavors. In general, we introduce production and detection NSIs, including the so-called zero-distance effect, and matter NSIs. In addition, we study mappings and approximate formulas for NSIs. Third, we present a brief account of theoretical models for NSIs. Fourth and most important, we investigate in detail the phenomenology of NSIs based on different types of data from neutrino experiments. Fifth, we give some phenomenological bounds on both matter and production/detection NSIs as well as we present sensitivity and discovery reach of NSIs at future experiments. Finally, we present a summary and state our conclusions.\n\n• 563.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics.\nAfter Higgs2012In: New scientist (1971), ISSN 0262-4079, Vol. 215, no 2881, p. 29-29Article in journal (Other (popular science, discussion, etc.))\n• 564.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics.\nAnother collider is not the way forward2013In: Nature, ISSN 0028-0836, E-ISSN 1476-4687, Vol. 494, no 7435, p. 35-35Article in journal (Other academic)\n• 565.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics.\nBimaximal fermion mixing from the quark and leptonic mixing matrices2005In: Physics Letters B, ISSN 0370-2693, E-ISSN 1873-2445, Vol. 622, no 02-jan, p. 159-164", null, "Article in journal (Refereed)\n\nIn this Letter. we show how the mixing angles of the standard parameterization add when multiplying the quark and leptonic mixing matrices. i.e., we derive explicit sum rules for the quark and leptonic mixing angles. In this connection, we also discuss other recently proposed sum rules for the mixing angles assuming bimaximal fermion mixing. In addition, we find that the present experimental and phenomenological data of the mixing angles naturally fulfill our sum rules, and thus, give rise to bilarge or bimaximal fermion mixing.\n\n• 566.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics.\nCutting with Occam's razor2012In: Physical Review D, ISSN 1550-7998, E-ISSN 1550-2368, Vol. 86, no 9, p. 097301-Article in journal (Refereed)\n\nWe comment positively on the recent article \"Seesaw mechanism with Occam's razor\" [K. Harigaya et al. Phys. Rev. D 86, 013002 (2012).] by Harigaya, Ibe, and Yanagida. In this article, the authors are using the principle of Occam's razor in neutrino physics-a principle that should be applied more often in science in general. At the end, we also discuss the Bayesian method in statistics, in which Occam's razor is naturally built in.\n\n• 567.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics.\nDon't let furore over neutrinos blur results2012In: Nature, ISSN 0028-0836, E-ISSN 1476-4687, Vol. 485, no 7398, p. 309-309Article in journal (Refereed)\n• 568.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics.\nEffects of Non-Standard Interactions in the MINOS Experiment2008In: Neutrino Factories, Superbeams And Betabeams / [ed] Osamu Yasuda, Naba Mondal, Chihiro Ohmori, 2008, p. 213-215Conference paper (Refereed)\n\nIn this talk, we investigate the effects of non-standard interactions (NSI) in the MINOS experiment based on a full three-flavor neutrino oscillation framework simulation. The simulation was performed using the GLoBES software. The results indicate that the allowed region in the sin(2)(2 theta(23))-Delta m(31)(2) plane is extended due to NSI effects, that there is a degeneracy between the leptonic mixing angle theta(13) and the NSI parameter epsilon(e tau), and that MINOS can put a lower upper bound on sin(2)(2 theta(13)) than CHOOZ only for small values of vertical bar epsilon(e tau)vertical bar.\n\n• 569.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics.\nExtrinsic CPT Violation in Neutrino Oscillations2004In: NEUTRINO FACTORIES AND SUPERBEAMS: 5th International Workshop on Neutrino Factories and Superbeams; NuFact 03 / [ed] Adam Para, Fermi National Accelerator Laboratory, American Institute of Physics , 2004, p. 265-268Conference paper (Refereed)\n\nIn this talk, we investigate extrinsic CPT violation in neutrino oscillations in matter with three flavors. Note that extrinsic CPT violation is different from intrinsic CPT violation. Extrinsic CPT violation is one way of quantifying matter effects, whereas intrinsic CPT violation would mean that the CPT invariance theorem is not valid. We present analytical formulas for the extrinsic CPT probability differences and discuss their implications for long-baseline experiments and neutrino factory setups.\n\n• 570.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics.\nFour's a crowd2013In: New scientist (1971), ISSN 0262-4079, Vol. 219, no 2925, p. 32-32Article in journal (Refereed)\n• 571.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics.\nInternational Linear Collider: Another collider is not the way forward2013In: Nature, ISSN 0028-0836, E-ISSN 1476-4687, Vol. 494, no 7435Article in journal (Refereed)\n• 572.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics.\nLHC reality check2012In: New scientist (1971), ISSN 0262-4079, Vol. 216, no 2895, p. 36-36Article in journal (Other academic)\n• 573.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics.\nNeutrino extras2012In: New scientist (1971), ISSN 0262-4079, Vol. 216, no 2885, p. 31-31Article in journal (Other (popular science, discussion, etc.))\n• 574.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics.\nNeutrinos and the hunt for the last mixing angle2012In: Europhysics News, ISSN 0531-7479, Vol. 43, no 4, p. 22-25Article in journal (Refereed)\n\nIn summary, the measurement of Ξ 23 by the Super-Kamiokande experiment resulted in one of the first indications of physics beyond Standard Model, the measurement of Ξ 12 was the first precision measurement in neutrino physics, and the hunt for the value of the third and final mixing angle Ξ 13 was the beginning of the end of the measurements of the mixing angles for the neutrinos, but the beginning of the continuation of measurements of the remaining neutrino parameters.\n\n• 575.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics.\nNeutrinos from Kaluza-Klein Dark Matter Annihilations in the Sun2011In: Physics beyond the Standard Models of Particles, Cosmology and Astrophysics: Proceedings of the Fifth International Conference / [ed] H V Klapdor-Kleingrothaus, I V Krivosheina, R Viollier, World Scientific Publishing , 2011, p. 496-502Conference paper (Refereed)\n\nIn this talk, we compute fluxes of neutrinos from Kaluza–Klein dark matter annihilations in the Sun based on cross-sections from both five- and six-dimensional models. For our numerical calculations, we use WimpSim and DarkSUSY. In addition, we compare our results with the ones derived earlier in the literature.\n\n• 576.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics.\nNeutrinos from WIMP Annihilations2008In: Dark Matter in Astroparticle and Particle Physics - Dark 2007: Proceedings of the 6th International Heidelberg Conference / [ed] Hans Volker Klapdor-Kleingrothaus, Geraint F. Lewis, World Scientific Publishing , 2008, p. 51-59Conference paper (Refereed)\n\nIn this talk, we make an improved analysis on the production of neutrinos coming from WIMP annihilations inside the Sun as well as the Earth. We treat both neutrino interaction and oscillation effects in aconsistent three-flavor framework. Our numerical simulations are performed in an event-based setting, which is useful for both theoretical studies and creating neutrino telescope Monte Carlo simulations. We find that the yield of muon-type neutrinos is enhanced or suppressed depending on the dominant WIMP annihilation channel. In addition, we find that oscillations can significantly affect the neutrino yields from WIMP annihilations in the Sun. Effectively, the neutrino flavors are mixed. Finally, for the Earth, the oscillations have no large impact at energies for the new neutrino telescopes such as IceCube, ANTARES, and NESTOR.\n\n• 577.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics.\nNon-Hermitian neutrino oscillations in matter with PT symmetric Hamiltonians2016In: Europhysics letters, ISSN 0295-5075, E-ISSN 1286-4854, Vol. 113, no 6, article id 61001Article in journal (Refereed)\n\nWe introduce and develop a novel approach to extend the ordinary two-flavor neutrino oscillation formalism in matter using a non-Hermitian PT symmetric effective Hamiltonian. The condition of PT symmetry is weaker and less mathematical than that of hermicity, but more physical, and such an extension of the formalism can give rise to sub-leading effects in neutrino flavor transitions similar to the effects by so-called non-standard neutrino interactions. We derive the necessary conditions for the spectrum of the effective Hamiltonian to be real as well as the mappings between the fundamental and effective parameters. We find that the real spectrum of the effective Hamiltonian will depend on all new fundamental parameters introduced in the non-Hermitian PT symmetric extension of the usual neutrino oscillation formalism and that either i) the spectrum is exact and the effective leptonic mixing must always be maximal or ii) the spectrum is approximate and all new fundamental parameters must be small.\n\n• 578.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics.\nNon-standard neutrino interactions2009In: SIXTH INTERNATIONAL WORKSHOP ON NEUTRINO-NUCLEUS INTERACTIONS IN THE FEW-GEV REGION / [ed] Sanchez F; Sorel M; Alvarez-Ruso L; Cervera A; Vicente-Vacas M, American Institute of Physics , 2009, Vol. 1189, p. 16-23Conference paper (Refereed)\n\nIn this talk, I will review non-standard interactions in neutrino physics, especially I will emphazise the impact of non-standard interactions on neutrino oscillations. First, I will give a brief introduction about non-standard interactions and what they are. Then, I will present what has been performed in the literature, what I have done in the field, and what could be done in the future. Next, I will discuss how important non-standard interactions are for neutrino cross-sections. Finally, I will give a summary of the field.\n\n• 579.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics.\nPreprint servers: Follow arXiv's lead2012In: Nature, ISSN 0028-0836, E-ISSN 1476-4687, Vol. 489, no 7416, p. 367-Article in journal (Refereed)\n• 580.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics.\nRelativistic Quantum Physics: From Advanced Quantum Mechanics to Introductory Quantum Field Theory2011 (ed. 1)Book (Refereed)\n\nQuantum physics and special relativity theory were two of the greatest breakthroughs in physics during the twentieth century and contributed to paradigm shifts in physics. This book combines these two discoveries to provide a complete description of the fundamentals of relativistic quantum physics, guiding the reader effortlessly from relativistic quantum mechanics to basic quantum field theory. The book gives a thorough and detailed treatment of the subject, beginning with the classification of particles, the Klein–Gordon equation and the Dirac equation. It then moves on to the canonical quantization procedure of the Klein–Gordon, Dirac and electromagnetic fields. Classical Yang–Mills theory, the LSZ formalism, perturbation theory, elementary processes in QED are introduced, and regularization, renormalization and radiative corrections are explored. With exercises scattered through the text and problems at the end of most chapters, the book is ideal for advanced undergraduate and graduate students in theoretical physics.\n\n• 581.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics.\nSearches for hyperbolic extra dimensions at the LHC2008Other (Refereed)\n\nIn this poster, we present a model of large extra dimensions where the internal space has the geometry of a hyperbolic disc. Compared with the ADD model, this model provides a more satisfactory solution to the hierarchy problem between the electroweak scale and the Planck scale, and it also avoids constraints from astrophysics. Since there is no known analytic form of the Kaluza-Klein spectrum for our choice of geometry, we obtain a spectrum based on a combination of approximations and numerical computations. We study the possible signatures of our model for hadron colliders, especially the LHC, where the most important processes are the production of a graviton together with a hadronic jet or a photon. We find that for the case of hadronic jet production, it is possible to obtain relatively strong signals, while for the case of photon production, this is much more difficult.\n\n• 582.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics.\nStatus of non-standard neutrino interactions2013In: Reports on progress in physics (Print), ISSN 0034-4885, E-ISSN 1361-6633, Vol. 76, no 4, p. 044201-Article, review/survey (Refereed)\n\nThe phenomenon of neutrino oscillations has been established as the leading mechanism behind neutrino flavor transitions, providing solid experimental evidence that neutrinos are massive and lepton flavors are mixed. Here we review sub-leading effects in neutrino flavor transitions known as non-standard neutrino interactions (NSIs), which is currently the most explored description for effects beyond the standard paradigm of neutrino oscillations. In particular, we report on the phenomenology of NSIs and their experimental and phenomenological bounds as well as an outlook for future sensitivity and discovery reach.\n\n• 583.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics.\nSterile search2013In: New scientist (1971), ISSN 0262-4079, Vol. 218, no 2913, p. 32-32Article in journal (Refereed)\n• 584.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics.\nTesting CPT invariance with neutrinos2002In: Proceedings of ICHEP 2002 / [ed] S. Bentvelsen, P. de Jong, J. Koch, and E. Laenen, Amsterdam: Elsevier Science B.V. , 2002, p. 53-56Conference paper (Refereed)\n\nWe investigate possible tests of CPT invariance on the level of event rates at neutrino factories. We do not assume any specific model, but phenomenological differences in the neutrino-antineutrino masses and mixing angles in a Lorentz invariance preserving context, which could be induced by physics beyond the Standard Model. We especially focus on the muon neutrino and antineutrino disappearance channels in order to obtain constraints on the neutrino-antineutrino mass and mixing angle differences. In a typical neutrino factory setup simulation, we find, for example, that", null, ".\n\n• 585.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics.\nT-violating effects in three flavor neutrino oscillations in matter2001In: International Europhysics Conference on High Energy Physics: hep2001 / [ed] Horváth Dezsõ (chairman), Lévai Péter, Patkós András, SISSA , 2001, p. 195-Conference paper (Refereed)\n\nIn this talk, we consider the interplay of fundamental and matter-induced T-violating effects in neutrino oscillations in matter. We present a simple approximative analytical formula for the T-violating probability asymmetry for three flavor neutrino oscillations in matter with an arbitrary density profile. We also discuss some implications of the obtained results. Since there are no T-violating effects in two flavor neutrino case (in the limit of vanshing $\\theta_{13}$ or $\\Delta m_{21}^2$, the three flavor neutrino oscillations reduces to the two flavor ones), the T-violating probability asymmetry can, in principle, provide a way to measure $\\theta_{13}$ and $\\Delta m_{21}^2$.\n\n• 586.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics.\nBlennow, MattiasKTH, School of Engineering Sciences (SCI), Theoretical Physics.Badelek, BarbaraEdsjö, JoakimHällgren, TomasKonstandin, ThomasPearce, MarkKTH, School of Engineering Sciences (SCI), Physics, Particle and Astroparticle Physics.\nThe 2nd Scandinavian NeutrinO Workshop2006Conference proceedings (editor) (Refereed)\n\nThe first Scandinavian NeutrinO Workshop (SNOW) was held in Uppsala, Sweden, in February 2001. About five years passed until the next SNOW took place—this time in Stockholm, Sweden between 2 May 2006 and 6 May 2006. The aim of the workshop was to cover a variety of topics in neutrino physics with leading researchers in the field as speakers. The Royal Swedish Academy of Sciences (KVA) awarded SNOW 2006 a grant for inviting such speakers. The workshop was mainly directed towards phenomenology and theory with connections to experiments and gave an opportunity for theorists and experimentalists to work together, discuss the latest results, and combine the different branches of neutrino physics. The different topics discussed were: solar and atmospheric neutrinos, reactor and accelerator neutrinos, neutrinos in astrophysics and cosmology, phenomenology of neutrino data, neutrino oscillations, theory and model building, fundamental properties of neutrinos, neutrinoless double beta decay, and flavor physics.\n\nAround 70 scientists (spanning from graduate students to world-leading researchers) in the field of neutrino physics participated in SNOW 2006 and 44 talks were presented in plenary sessions. Out of the 44 talks, 37 have been contributed to these proceedings.\n\nThe talks of SNOW 2006 took place in the Oskar Klein Auditorium at the AlbaNova University Center in Stockholm. The AlbaNova University Center is a joint endeavour between the Royal Institute of Technology (KTH) and Stockholm University. The social program included a welcome reception at KVA, an excursion to the Royal Armoury at the Royal Palace in Stockholm as well as a boat trip in the archipelago of Stockholm, a reception at the City Hall of Stockholm arranged by the city, and finally, a workshop dinner at Häringe Castle south of Stockholm.\n\n• 587.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics. KTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics.\nNon-unitarity of the leptonic mixing matrix in the TeV-scale type-I seesaw model2010In: Physics Letters B, ISSN 0370-2693, E-ISSN 1873-2445, Vol. 692, no 4, p. 257-260", null, "Article in journal (Refereed)\n\nThe non-unitarity effects in leptonic flavor mixing are regarded as one of the generic features of the type-1 seesaw model. Therefore, we explore these effects in the TeV-scale type-1 seesaw model, and show that there exist non-trivial correlations among the non-unitarity parameters, stemming from the typical flavor structure of the low-scale seesaw model. In general, it follows from analytical discussions and numerical results that all the six non-unitarity parameters are related to three model parameters, while the widely studied parameters eta(e tau) and eta(mu tau) cannot be phenomenologically significant simultaneously.\n\n• 588.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics.\nRunning of neutrino parameters and the Higgs self-coupling in a six-dimensional UED model2013In: Physics Letters B, ISSN 0370-2693, E-ISSN 1873-2445, Vol. 718, no 3, p. 1002-1007", null, "Article in journal (Refereed)\n\nWe investigate a six-dimensional universal extra-dimensional model in the extension of an effective neutrino mass operator. We derive the β-functions and renormalization group equations for the Yukawa couplings, the Higgs self-coupling, and the effective neutrino mass operator in this model. Especially, we focus on the renormalization group running of physical parameters such as the Higgs self-coupling and the leptonic mixing angles. The recent measurements of the Higgs boson mass by the ATLAS and CMS Collaborations at the LHC as well as the current three-flavor global fits of neutrino oscillation data have been taken into account. We set a bound on the six-dimensional model, using the vacuum stability criterion, that allows five Kaluza-Klein modes only, which leads to a strong limit on the cutoff scale. Furthermore, we find that the leptonic mixing angle θ12 shows the most sizeable running, and that the running of the angles θ13 and θ23 are negligible. Finally, it turns out that the findings in this six-dimensional model are comparable with what is achieved in the corresponding five-dimensional model, but the cutoff scale is significantly smaller, which means that it could be detectable in a closer future.\n\n• 589.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics.\nNon-standard neutrino interactions in the Zee-Babu model2009In: Physics Letters B, ISSN 0370-2693, E-ISSN 1873-2445, Vol. 681, no 3, p. 269-275", null, "Article in journal (Refereed)\n\nWe investigate non-standard neutrino interactions (NSIs) in the Zee-Babu model. The size of NSIs predicted by this model is obtained from a full scan over the parameter space. taking into account constraints from low-energy experiments such as searches for lepton flavor violation (LFV) and the requirement to obtain a viable neutrino mass matrix. The dependence on the scale of new physics as well as on the type of the neutrino mass hierarchy is discussed. We find that NSIs at the source of a future neutrino factory may be at an observable level in the nu(e) -> nu(tau) and/or nu(mu) -> nu(tau) channels. In particular, if the doubly charged scalar of the model has a mass in reach of the LHC and if the neutrino mass hierarchy is inverted, a highly predictive scenario is obtained with observable signals at the LHC, in upcoming neutrino oscillation experiments, in LFV processes, and for NSIs at a neutrino factory.\n\n• 590.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics.\nNon-standard interaction effects at reactor neutrino experiments2009In: Physics Letters B, ISSN 0370-2693, E-ISSN 1873-2445, Vol. 671, no 1, p. 99-104", null, "Article in journal (Refereed)\n\nWe study non-standard interactions (NSIs) at reactor neutrino experiments, and in particular, the mimicking effects on theta(13). We present generic formulas for oscillation probabilities including NSIs from sources and detectors. Instructive mappings between the fundamental leptonic mixing parameters and the effective leptonic mixing parameters are established. In addition, NSI corrections to the mixing angles theta(13) and theta(12) are discussed in detailed. Finally, we show that, even for a vanishing theta(13), all Oscillation phenomenon may still be observed ill future short baseline reactor neutrino experiments, such as Double Chooz and Daya Bay, due to the existences of NSIs.\n\n• 591.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics.\nEffects of nonstandard neutrino interactions at PINGU2013In: Physical Review D, ISSN 1550-7998, E-ISSN 1550-2368, Vol. 88, no 1, p. 013001-Article in journal (Refereed)\n\nNeutrino oscillation experiments in the past decades have greatly improved our knowledge on neutrinos by measuring the fundamental neutrino parameters. The ongoing and upcoming neutrino oscillation experiments are intended to pin down the neutrino mass hierarchy and to discover the leptonic CP violation. By means of neutrino oscillograms, we analyze the impact of nonstandard neutrino interactions on neutrino oscillations in Earth matter. The standard neutrino oscillation probabilities may be significantly changed by nonstandard interaction parameters, and, in particular, the CP-violating effects in the energy range E = 1-20 GeV are greatly enhanced. In addition, the event rates of muon neutrinos in the proposed huge atmospheric neutrino experiment, PINGU at the South Pole, have been estimated in the presence of nonstandard neutrino interactions. It has been found that the PINGU experiment has very good sensitivities to the nonstandard neutrino interaction parameters.\n\n• 592.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics.\nNonstandard interaction effects on neutrino parameters at medium-baseline reactor antineutrino experiments2014In: Physics Letters B, ISSN 0370-2693, E-ISSN 1873-2445, Vol. 728, p. 148-155", null, "Article in journal (Refereed)\n\nPrecision measurements of leptonic mixing parameters and the determination of the neutrino mass hierarchy are the primary goals of the forthcoming medium-baseline reactor antineutrino experiments, such as JUNO and RENO-50. In this work, we investigate the impact of nonstandard neutrino interactions (NSIs) on the measurements of {sin(2) theta(12), Delta m(21)(2)} and {sin(2) theta(13), Delta m(31)(2)}. and on the sensitivity to the neutrino mass hierarchy, at the medium-baseline reactor experiments by assuming a typical experimental setup. It turns out that the true mixing parameter sin(2) theta(12) can be excluded at a more than 3 sigma level if the NSI parameter epsilon(e mu), or epsilon(e tau), is as large as 2% in the most optimistic case. However, the discovery reach of NSI effects has been found to be small, and depends crucially on the CP-violating phases. Finally, we show that NSI effects could enhance or reduce the discrimination power of the JUNO and RENO-50 experiments between the normal and inverted neutrino mass hierarchies.\n\n• 593.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics.\nProbing the leptonic Dirac CP-violating phase in neutrino oscillation experiments2013In: Physical Review D, ISSN 1550-7998, E-ISSN 1550-2368, Vol. 87, no 5, p. 053006-Article in journal (Refereed)\n\nThe discovery of leptonic CP violation is one of the primary goals of next-generation neutrino oscillation experiments, which is feasible due to the recent measurement of a relatively large leptonic mixing angle theta(13). We suggest two new working observables Delta A(alpha beta)(m) equivalent to max[A(alpha beta)(CP)(delta) - min[A(alpha beta)(CP)(delta)] and Delta A(alpha beta)(CP)(delta) equivalent to A(alpha beta)(CP)(delta) - A(alpha beta)(CP)(0) to describe the CP-violating effects in long-baseline and atmospheric neutrino oscillation experiments. The former signifies the experimental sensitivity to the leptonic Dirac CP-violating phase delta and can be used to optimize the experimental setup, while the latter measures the intrinsic leptonic CP violation and can be used to extract delta directly from the experimental observations. Both analytical and numerical analyses are carried out to illustrate their main features. It turns out that an intense neutrino beam with sub-GeV energies and a baseline of a few 100 km may serve as an optimal experimental setup for probing leptonic CP violation.\n\n• 594.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics.\nRadiative corrections to the leptonic Dirac CP-violating phase2013In: Physical Review D, ISSN 1550-7998, E-ISSN 1550-2368, Vol. 87, no 1, p. 013012-Article in journal (Refereed)\n\nSince the smallest leptonic mixing angle theta(13) has been measured to be relatively large, it is quite promising to constrain or determine the leptonic Dirac CP-violating phase delta in future neutrino oscillation experiments. Given some typical values of delta = pi/2, pi, and 3 pi/2 at the low energy scale, as well as current experimental results of the other neutrino parameters, we perform a systematic study of radiative corrections to delta by using the one-loop renormalization group equations in the minimal supersymmetric standard model and the universal extra-dimensional model. It turns out that delta is rather stable against radiative corrections in both models, except for the minimal supersymmetric standard model with a very large value of tan beta. Both cases of Majorana and Dirac neutrinos are discussed. In addition, we use the preliminary indication of delta = (1.08(-0.31)(+0.28))pi or delta = (1.67(-0.77)(+0.37))pi from the latest global-fit analyses of data from neutrino oscillation experiments to illustrate how it will be modified by radiative corrections.\n\n• 595.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics.\nExtrinsic and intrinsic CPT asymmetries in neutrino oscillations2015In: Nuclear Physics B, ISSN 0550-3213, E-ISSN 1873-1562, Vol. 893, p. 482-500", null, "Article in journal (Refereed)\n\nWe reconsider the extrinsic and possible intrinsic CPT violation in neutrino oscillations, and point out an identity, i.e., A(alpha beta)(CP) = A(beta alpha)(CPT) + A(alpha beta)(T), among the CP, T, and CPT asymmetries in oscillations. For three-flavor oscillations in matter of constant density, the extrinsic CPT asymmetries A(ee)(CPT), A(e mu)(CPT), A(mu e)(CPT), and A(mu mu)(CPT) caused by Earth matter effects have been calculated in the plane of different neutrino energies and baseline lengths. It is found that two analytical conditions can be implemented to describe the main structure of the contours of vanishing extrinsic CPT asymmetries. Finally, without assuming intrinsic CPT symmetry in the neutrino sector, we investigate the possibility to constrain the difference of the neutrino CF-violating phase and the antineutrino one (delta) over bar using a low-energy neutrino factory and the super-beam experiment ESS nu SB. We find that \\delta - (delta) over bar\\ less than or similar to 0.35 pi in the former case and \\delta - (delta) over bar\\ less than or similar to 0.7 pi in the latter case can be achieved at the 3 sigma confidence level if delta = (delta) over bar = pi/2 is assumed.\n\n• 596.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical Particle Physics. Chinese Acad Sci,China.\nRenormalization group running of neutrino parameters2014In: Nature Communications, ISSN 2041-1723, E-ISSN 2041-1723, Vol. 5, p. 5153-Article, review/survey (Refereed)\n\nNeutrinos are the most elusive particles in our Universe. They have masses at least one million times smaller than the electron mass, carry no electric charge and very weakly interact with other particles, meaning that they are rarely captured in terrestrial detectors. Tremendous efforts in the past two decades have revealed that neutrinos can transform from one type to another as a consequence of neutrino oscillations-a quantum mechanical effect over macroscopic distances-yet the origin of neutrino masses remains puzzling. The physical evolution of neutrino parameters with respect to energy scale may help elucidate the mechanism for their mass generation.\n\n• 597.\nTampere University of Technology.\nUniversity of Groningen. University of Groningen. KTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical & Computational Biophysics. Tampere University of Technology,. University of Groningen.\n3D pressure field in lipid membranes and membrane-protein complexes2009In: Physical Review Letters, ISSN 0031-9007, E-ISSN 1079-7114, Vol. 102, no 7, p. 078101-Article in journal (Refereed)\n\nWe calculate full 3D pressure fields for inhomogeneous nanoscale systems using molecular dynamics simulation data. The fields represent systems with increasing level of complexity, ranging from semivesicles and vesicles to membranes characterized by coexistence of two phases, including also a protein-membrane complex. We show that the 3D pressure field is distinctly different for curved and planar bilayers, the pressure field depends strongly on the phase of the membrane, and that an integral protein modulates the tension and elastic properties of the membrane.\n\n• 598. Olsen, Richard W.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical & Computational Biophysics.\nStructural Models of Ligand-Gated Ion Channels: Sites of Action for Anesthetics and Ethanol2014In: Alcoholism: Clinical and Experimental Research, ISSN 0145-6008, E-ISSN 1530-0277, Vol. 38, no 3, p. 595-603Article in journal (Refereed)\n\nThe molecular mechanism(s) of action of anesthetic, and especially, intoxicating doses of alcohol (ethanol [EtOH]) have been of interest even before the advent of the Research Society on Alcoholism. Recent physiological, genetic, and biochemical studies have pin-pointed molecular targets for anesthetics and EtOH in the brain as ligand-gated ion channel (LGIC) membrane proteins, especially the pentameric (5 subunit) Cys-loop superfamily of neurotransmitter receptors including nicotinic acetylcholine (nAChRs), GABA(A) (GABA(A)Rs), and glycine receptors (GlyRs). The ability to demonstrate molecular and structural elements of these proteins critical for the behavioral effects of these drugs on animals and humans provides convincing evidence for their role in the drugs' actions. Amino acid residues necessary for pharmacologically relevant allosteric modulation of LGIC function by anesthetics and EtOH have been identified in these channel proteins. Site-directed mutagenesis revealed potential allosteric modulatory sites in both the trans-membrane domain (TMD) and extracellular domain (ECD). Potential sites of action and binding have been deduced from homology modeling of other LGICs with structures known from crystallography and cryo-electron microscopy studies. Direct information about ligand binding in the TMD has been obtained by photoaffinity labeling, especially in GABA(A)Rs. Recent structural information from crystallized procaryotic (ELIC and GLIC) and eukaryotic (GluCl) LGICs allows refinement of the structural models including evaluation of possible sites of EtOH action.\n\n• 599. Orellana, Laura\nKTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical & Computational Biophysics. KTH, School of Engineering Sciences (SCI), Theoretical Physics, Theoretical & Computational Biophysics.\nPrediction and Validation of Protein Intermediate States from Structurally Rich Ensembles and Coarse-Grained Simulations2016In: Nature Communications, ISSN 2041-1723, E-ISSN 2041-1723, Vol. 7Article in journal (Refereed)\n\nProtein conformational changes are at the heart of cell functions, from signaling to ion transport. However, the transient nature of the intermediates along transition pathways hampers their experimental detection, making the underlying mechanisms elusive. Here, we retrieve dynamic information on the actual transition routes from Principal Component Analysis (PCA) of structurally-rich ensembles and, in combination with coarse-grained simulations, explore the conformational landscapes of five well-studied proteins. Modeling them as elastic networks in a hybrid Elastic-Network Brownian Dynamics simulation (eBDIMS), we generate trajectories connecting stable end-states that spontaneously sample the crystallographic motions, predicting the structures of known intermediates along thepaths. We also show that the explored non-linear routes can delimit the lowest energy passages between end-states sampled by atomistic molecular dynamics. The integrative methodology presented here provides a powerful framework to extract and expand dynamic pathway information from the Protein Data Bank, as well as to validate sampling methods in general.\n\n• 600.\nKTH, School of Engineering Sciences (SCI), Theoretical Physics.\nBacterial growth models2011Independent thesis Advanced level (degree of Master (Two Years)), 20 credits / 30 HE creditsStudent thesis\n\nIn this thesis the growth patterns of certain bacterial colonies are studied through numerical simulations of a continuous model, describing the growth of the colony. The objective was to construct a mathematical model capable of recreating observed growth patterns and thus try to gain some insight into the dynamics of bacterial growth. The model constructed in this thesis consists of a set of reaction-diffusion equations describing the evolution of the bacterial colony viewed as a continuous body rather than many individual bacteria. This approach was partially successful, in the sense that similar patterns to those observed in experiments were indeed observed in the simulations. On the other hand, the model was found inadequate in the sense that it could not satisfactory account for the initial rapid expansion of the colony in combination with the fact that the expansion is severely diminished after a few days. From the results it could be concluded that in order to get a satisfactory description of the complete evolution of the bacterial colonies studied here a more sophisticated theory for the diffusion processes is needed. The main conclusion is that the changes in the growth patterns are most likely due to genetic changes, or ’mutations’, in some bacteria causing the mutated bacteria to diffuse faster. There is an upper bound for the mutation frequency in order to get the patterns seen in experiments. The biological mechanism behind this phenomenon could be that the bacteria that mutate lose their pili enabling them to diffuse faster.\n\n9101112131415 551 - 600 of 764\nCite\nCitation style\n• apa\n• harvard1\n• ieee\n• modern-language-association-8th-edition\n• vancouver\n• Other style\nMore styles\nLanguage\n• de-DE\n• en-GB\n• en-US\n• fi-FI\n• nn-NO\n• nn-NB\n• sv-SE\n• Other locale\nMore languages\nOutput format\n• html\n• text\n• asciidoc\n• rtf" ]
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https://baymn.org/guides-to-learn-online-mathematics-easily/
[ "Math is one of the most popular subjects in school, but it can also be a challenge to learn on your own. However, there is no need to fear as this blog post will take you through every statistic you’ll ever need to know about. It’s going to teach you not just the basics but also get you through problems that might otherwise feel impossible. By the time you’re finished, you’ll have every single stat that you need to ace an exam, quiz, or test on mathematics online courses .\n\nThere are many ways to approach this subject but we’re going to go over the subject from a general perspective that applies to a wide range of topics. We’ll begin by going over the basics of math then dive into progression and regression, Next, we will talk about probability and how these relate to various games of chance. Finally, we will go over some basic statistics concepts so that you can understand how they apply in real-life situations.\n\n## Basics of Mathematics\n\nThe first thing you’ll want to learn is how to count. It’s also important to learn basic vocabulary. Each installment of our math series will go over more advanced information but we’re going to cover the basics first. There are four basic operations in math which you’ll want to memorize: addition, subtraction, multiplication, and division.\n\nAddition: Addition is one of the most basic operations in mathematics and you should know how to do it perfectly. The order in which you add numbers is extremely important because if two numbers are added incorrectly this can lead to inaccurate results. You should learn how to add correctly before moving on as you’ll see later on.\n\nSubtraction: Subtraction is very important when you’re learning mathematics with a focus on statistics. If you’re subtracting small numbers or even ones that are close together it’s best to do it with a calculator but some things are so simple that you can easily do it mentally.\n\nMultiplication: The multiplication operation is also very important in mathematics. There are four main types of multiplication: push down, horizontal, vertical, and cube. These are pretty complicated and the different ways to do it vary by subject so don’t worry if you forget how they work there’s nothing wrong with just doing it correctly rather than different ways.\n\nDivision: Division is very similar to multiplication. There are four different types of division: fractional, decimal, quotient, and remainders. Some people don’t realize that the order in which you divide numbers is important, since it can lead to inaccurate results. Make sure you know how to do division properly before moving on.\n\n## Progression and Regression\n\nProgression and regression are two things you should learn about when working with statistics in mathematics. Platforms that “sell courses online” where you can find the best educators for mathematics theory & formulas. about how numbers relate to one another and it’s extremely important for understanding more complicated topics down the road such as derivatives.\n\nProgression: Progression is a mathematical term used to describe something that moves forward, as in an increase. Most people use the word progression to refer to something moving upwards in value so that it doesn’t drop, like an investment. However, it can also move downwards and this is called regression. Both of them are very important when working with statistics so make sure you understand progression and regression perfectly before moving on.\n\nRegression: Regression is the opposite of progression and it’s easy to understand if you think about the fact that numbers don’t have to move upwards all the time. When they drop or when they decrease in value, this is known as regression and it’s the opposite of progression. Many people use the word regression to refer to the economy but they are actually referring to a situation where there has been a decrease in numbers." ]
[ null ]
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https://rdrr.io/cran/rmutil/man/MultBinom.html
[ "# MultBinom: Multiplicative Binomial Distribution In rmutil: Utilities for Nonlinear Regression and Repeated Measurements Models\n\n Multiplicative Binomial R Documentation\n\n## Multiplicative Binomial Distribution\n\n### Description\n\nThese functions provide information about the multiplicative binomial distribution with parameters `m` and `s`: density, cumulative distribution, quantiles, and random generation.\n\nThe multiplicative binomial distribution with total = n and `prob` = m has density\n\np(y) = c(n,m,s) Choose(n,y) m^y (1-m)^(n-y) s^(y(n-y))\n\nfor y = 0, …, n, where c(.) is a normalizing constant.\n\n### Usage\n\n```dmultbinom(y, size, m, s, log=FALSE)\npmultbinom(q, size, m, s)\nqmultbinom(p, size, m, s)\nrmultbinom(n, size, m, s)\n```\n\n### Arguments\n\n `y` vector of frequencies `q` vector of quantiles `p` vector of probabilities `n` number of values to generate `size` vector of totals `m` vector of probabilities of success `s` vector of overdispersion parameters `log` if TRUE, log probabilities are supplied.\n\n### Author(s)\n\nJ.K. Lindsey\n\n`dbinom` for the binomial, `ddoublebinom` for the double binomial, and `dbetabinom` for the beta binomial distribution.\n\n### Examples\n\n```# compute P(45 < y < 55) for y multiplicative binomial(100,0.5,1.1)\nsum(dmultbinom(46:54, 100, 0.5, 1.1))\npmultbinom(54, 100, 0.5, 1.1)-pmultbinom(45, 100, 0.5, 1.1)\npmultbinom(2,10,0.5,1.1)\nqmultbinom(0.025,10,0.5,1.1)\nrmultbinom(10,10,0.5,1.1)\n```\n\nrmutil documentation built on March 18, 2022, 6:55 p.m." ]
[ null ]
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http://www.ijpam.eu/contents/2012-78-5/14/index.html
[ "# IJPAM: Volume 78, No. 5 (2012)\n\nEXISTENCE AND UNIQUENESS OF SOLUTIONS OF\nSEMILINEAR EQUATIONS WITH ANALYTIC\nSEMIGROUPS IN BANACH SPACES\n\nRichard Olu Awonusika\nDepartment of Mathematics and Industrial Mathematics\nP.M.B. 001, Akungba Akoko, Ondo State, NIGERIA\n\nAbstract. This paper studies the existence and uniqueness of mild solutions of the semilinear equation", null, "where the linear operator", null, "is the infinitesimal generator of an analytic semigroup", null, "satisfying the exponential stability and the function", null, "is locally Hölder continuous in", null, "and locally Lipschitz continuous in", null, "The basic tool in this paper is the use of fractional power of operators.\n\nAMS Subject Classification: 35B08, 35B09, 35B10, 35B15, 35B65\n\nKey Words and Phrases: semilinear equations in Banach spaces, infinitesimal generator of analytic semigroup, fractional powers" ]
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https://www.colorhexa.com/001f8b
[ "# #001f8b Color Information\n\nIn a RGB color space, hex #001f8b is composed of 0% red, 12.2% green and 54.5% blue. Whereas in a CMYK color space, it is composed of 100% cyan, 77.7% magenta, 0% yellow and 45.5% black. It has a hue angle of 226.6 degrees, a saturation of 100% and a lightness of 27.3%. #001f8b color hex could be obtained by blending #003eff with #000017. Closest websafe color is: #003399.\n\n• R 0\n• G 12\n• B 55\nRGB color chart\n• C 100\n• M 78\n• Y 0\n• K 45\nCMYK color chart\n\n#001f8b color description : Dark blue.\n\n# #001f8b Color Conversion\n\nThe hexadecimal color #001f8b has RGB values of R:0, G:31, B:139 and CMYK values of C:1, M:0.78, Y:0, K:0.45. Its decimal value is 8075.\n\nHex triplet RGB Decimal 001f8b `#001f8b` 0, 31, 139 `rgb(0,31,139)` 0, 12.2, 54.5 `rgb(0%,12.2%,54.5%)` 100, 78, 0, 45 226.6°, 100, 27.3 `hsl(226.6,100%,27.3%)` 226.6°, 100, 54.5 003399 `#003399`\nCIE-LAB 19.406, 36.579, -60.934 5.149, 2.844, 24.702 0.157, 0.087, 2.844 19.406, 71.07, 300.976 19.406, -7.288, -65.192 16.863, 24.996, -75.048 00000000, 00011111, 10001011\n\n# Color Schemes with #001f8b\n\n• #001f8b\n``#001f8b` `rgb(0,31,139)``\n• #8b6c00\n``#8b6c00` `rgb(139,108,0)``\nComplementary Color\n• #00658b\n``#00658b` `rgb(0,101,139)``\n• #001f8b\n``#001f8b` `rgb(0,31,139)``\n• #26008b\n``#26008b` `rgb(38,0,139)``\nAnalogous Color\n• #658b00\n``#658b00` `rgb(101,139,0)``\n• #001f8b\n``#001f8b` `rgb(0,31,139)``\n• #8b2600\n``#8b2600` `rgb(139,38,0)``\nSplit Complementary Color\n• #1f8b00\n``#1f8b00` `rgb(31,139,0)``\n• #001f8b\n``#001f8b` `rgb(0,31,139)``\n• #8b001f\n``#8b001f` `rgb(139,0,31)``\n• #008b6c\n``#008b6c` `rgb(0,139,108)``\n• #001f8b\n``#001f8b` `rgb(0,31,139)``\n• #8b001f\n``#8b001f` `rgb(139,0,31)``\n• #8b6c00\n``#8b6c00` `rgb(139,108,0)``\n• #000e3f\n``#000e3f` `rgb(0,14,63)``\n• #001458\n``#001458` `rgb(0,20,88)``\n• #001972\n``#001972` `rgb(0,25,114)``\n• #001f8b\n``#001f8b` `rgb(0,31,139)``\n• #0025a5\n``#0025a5` `rgb(0,37,165)``\n• #002abe\n``#002abe` `rgb(0,42,190)``\n• #0030d8\n``#0030d8` `rgb(0,48,216)``\nMonochromatic Color\n\n# Alternatives to #001f8b\n\nBelow, you can see some colors close to #001f8b. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #00428b\n``#00428b` `rgb(0,66,139)``\n• #00368b\n``#00368b` `rgb(0,54,139)``\n• #002b8b\n``#002b8b` `rgb(0,43,139)``\n• #001f8b\n``#001f8b` `rgb(0,31,139)``\n• #00138b\n``#00138b` `rgb(0,19,139)``\n• #00088b\n``#00088b` `rgb(0,8,139)``\n• #04008b\n``#04008b` `rgb(4,0,139)``\nSimilar Colors\n\n# #001f8b Preview\n\nThis text has a font color of #001f8b.\n\n``<span style=\"color:#001f8b;\">Text here</span>``\n#001f8b background color\n\nThis paragraph has a background color of #001f8b.\n\n``<p style=\"background-color:#001f8b;\">Content here</p>``\n#001f8b border color\n\nThis element has a border color of #001f8b.\n\n``<div style=\"border:1px solid #001f8b;\">Content here</div>``\nCSS codes\n``.text {color:#001f8b;}``\n``.background {background-color:#001f8b;}``\n``.border {border:1px solid #001f8b;}``\n\n# Shades and Tints of #001f8b\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, #000002 is the darkest color, while #edf1ff is the lightest one.\n\n• #000002\n``#000002` `rgb(0,0,2)``\n• #000515\n``#000515` `rgb(0,5,21)``\n• #000929\n``#000929` `rgb(0,9,41)``\n• #000e3d\n``#000e3d` `rgb(0,14,61)``\n• #001250\n``#001250` `rgb(0,18,80)``\n• #001664\n``#001664` `rgb(0,22,100)``\n• #001b77\n``#001b77` `rgb(0,27,119)``\n• #001f8b\n``#001f8b` `rgb(0,31,139)``\n• #00239f\n``#00239f` `rgb(0,35,159)``\n• #0028b2\n``#0028b2` `rgb(0,40,178)``\n• #002cc6\n``#002cc6` `rgb(0,44,198)``\n• #0030d9\n``#0030d9` `rgb(0,48,217)``\n• #0035ed\n``#0035ed` `rgb(0,53,237)``\n• #023aff\n``#023aff` `rgb(2,58,255)``\n• #1549ff\n``#1549ff` `rgb(21,73,255)``\n• #2959ff\n``#2959ff` `rgb(41,89,255)``\n• #3d68ff\n``#3d68ff` `rgb(61,104,255)``\n• #5077ff\n``#5077ff` `rgb(80,119,255)``\n• #6486ff\n``#6486ff` `rgb(100,134,255)``\n• #7796ff\n``#7796ff` `rgb(119,150,255)``\n• #8ba5ff\n``#8ba5ff` `rgb(139,165,255)``\n• #9fb4ff\n``#9fb4ff` `rgb(159,180,255)``\n• #b2c3ff\n``#b2c3ff` `rgb(178,195,255)``\n• #c6d3ff\n``#c6d3ff` `rgb(198,211,255)``\n• #d9e2ff\n``#d9e2ff` `rgb(217,226,255)``\n• #edf1ff\n``#edf1ff` `rgb(237,241,255)``\nTint Color Variation\n\n# Tones of #001f8b\n\nA tone is produced by adding gray to any pure hue. In this case, #40434b is the less saturated color, while #001f8b is the most saturated one.\n\n• #40434b\n``#40434b` `rgb(64,67,75)``\n• #3b4050\n``#3b4050` `rgb(59,64,80)``\n• #353d56\n``#353d56` `rgb(53,61,86)``\n• #303a5b\n``#303a5b` `rgb(48,58,91)``\n• #2b3760\n``#2b3760` `rgb(43,55,96)``\n• #253466\n``#253466` `rgb(37,52,102)``\n• #20316b\n``#20316b` `rgb(32,49,107)``\n• #1b2e70\n``#1b2e70` `rgb(27,46,112)``\n• #152b76\n``#152b76` `rgb(21,43,118)``\n• #10287b\n``#10287b` `rgb(16,40,123)``\n• #0b2580\n``#0b2580` `rgb(11,37,128)``\n• #052286\n``#052286` `rgb(5,34,134)``\n• #001f8b\n``#001f8b` `rgb(0,31,139)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #001f8b 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":0.50431377,"math_prob":0.84336454,"size":3641,"snap":"2022-40-2023-06","text_gpt3_token_len":1648,"char_repetition_ratio":0.15067363,"word_repetition_ratio":0.0074074073,"special_character_ratio":0.56303215,"punctuation_ratio":0.23276836,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9924764,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-04T19:45:44Z\",\"WARC-Record-ID\":\"<urn:uuid:574792b4-4c15-47e1-8b2b-fec0d8444ccc>\",\"Content-Length\":\"36075\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9c5a66f8-f455-44f5-97c5-2c521766fbe8>\",\"WARC-Concurrent-To\":\"<urn:uuid:b06bc5a4-de84-4ac8-91c8-e1ad1a0313d5>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/001f8b\",\"WARC-Payload-Digest\":\"sha1:ENCORZ435LYYJQEKV7B52PVT4O62HC64\",\"WARC-Block-Digest\":\"sha1:W7A24MAE4YJTC5T4JHYHYMNZLZXKYWQU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500151.93_warc_CC-MAIN-20230204173912-20230204203912-00025.warc.gz\"}"}
https://visualfractions.com/calculator/half-fractions/what-is-half-of-2-235/
[ "# What is Half of 2/235?\n\nAre you looking to work out and calculate half of 2/235? In this really simple guide, we'll teach you exactly what half of 2/235 is and walk you through the step-by-process of how to calculate half of any fraction.\n\nAs always with our series of calculation posts, it's important to remember that the number above the fraction line is called the numerator and the number below the line is called the denomination.\n\nWant to quickly learn or refresh memory on how to calculate a half fraction play this quick and informative video now!\n\nSo what do we mean by half? Half means there are two of the thing, and so all you need to do is divide it by two. Half of 2/235 is just another way of saying 2/235 divided by 2:\n\n2 / 235 ÷ 2\n\nNow we know that \"half\" means to divide by 2, how do we half 2/235? Remember that a fraction a part of the whole, so the higher the denominator is, the smaller the piece. The answer is that the numerator stays the same and we multiply the denominator by 2:\n\n2 / 235 x 2 = 2 / 470\n\nThat's it! Working out half of 2/235 really is that easy. Hopefully you understood the process and can use the same techniques to halve other fractions as well. The complete answer is below (simplified to the lowest form):\n\n1/235\n\n## Convert Half of 2/235 to Decimal\n\nHere's a little bonus calculation for you to easily work out the decimal format of half 2/235. All you need to do is divide the numerator by the denominator and you can convert any fraction to decimal:\n\n2 / 470 = 0.0043" ]
[ null ]
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https://studylib.net/doc/7300807/security
[ "# Security", null, "```(1901758): Computer Protection and Security\nComputer Science Dept., Jordan University, 2007/2008\nCourse Description: Computer security and basic cryptography topics, introduction to the\nmathematical principles of data security, information security, various developments in security are\ndiscussed, with emphasis on public-key encryption, digital signature, the data encryption standard\n(DES), key-safeguarding schemes vulnerabilities, policy formation, control and protection methods,\nand Internet security. (This description is the same as appeared in the computer science dept.\ncurriculum).\nIntended Learning Outcomes:\nThe intended learning outcomes of this course are:\nA- Knowledge and Understanding: Students should …\nA1) Learn the basic concepts in computer security..\nA2) Understand the advance d features in public-key cryptosystems.\nB- Intellectual skills: with the ability to …\nB1) Compare and analyze algorithms as fundamental tools of security.\nB2) Apply mathematical tools to algorithm verification and analysis.\nB3) Analytically recognize the needed security schema in relevance to some application.\nC- Subject specific skills – with ability to …\nC1) Work on case studies to show how all the tools are used together to build a complete project and research.\nC2) Translate abstract ideas into practice.\nC3) Implement and handle security projects.\nD- Transferable skills – with ability to\nD1) Possess good security policy.\nD2) Develop advanced public key cryptanalysis.\nD3) Choose the appropriate methods for public key generation of RSA algorithm.\nCourse Contents:\n\nCourse Introduction: Digital Threats,\nSecurity needs, Security systems, Security Policy,\nSymmetric Cryptography, Seizer Cipher, One Time\n\nFinite Fields Groups: Fields, and Rings\nModular Arithmetic, Euclid’s Algorithm, Finite\nField of the form GF(p), Polynomial Arithmetic,\nFinite Fields of the form GF(2).\n\nSymmetric key Encryption: DES Encryption\nmodes. Triple DES , and AES.\n\nNumber Theory: Prime numbers, Fermat’s\ntheorem, Euler’s Totient Function, Euler’s\nTheorem. Testing for Primality, The Chinese\nremainder theorem, Discrete logarithms.\n\nPublic-Key Cryptography: Public-Key\ncryptosystems, Applications for public Key\ncryptosystems, Requirements for Public-Key\ncryptosystems.\n\nThe RSA algorithm: Description of the\nAlgorithm, Computational Aspects, RSA\nCryptanalysis and the security of RSA.\n\nOther Public-Key approaches: The DiffieHellman key exchange, Elliptic curve arithmetic,\nAbelian groups, Eleptic curves over real numbers,\nElliptic curves over Zp, Elliptic curves over\nGF(2^m), Elliptic curves cryptography, analog of\nDiffie-Hellman key exchange, Security of Elliptic\ncurve cryptography.\n\nMessage Authentication and Hash functions:\nAuthentication functions.\n\nDigital signatures\n\nAuthentication servers: Kerberos\n\nHash algorithms: MD5 and MD4, Secure\nHash algorithms.\n\n\nWeb security: PGP\nIntrusion Detection\nTeaching methods. Lectures, Demos, Researches and\nProjects, Seminars and Presentations, Invited Speakers.\nEvaluations:\n1- Mid Term Exam: 35\n2- Project, Technical Report Writing, and\nSeminars: 25\n3- Final Exam: 40.\nTextbooks;\nCryptography and Network Security, Third Edition. By William Stallings. from Prentice Hall.\nA good overview of computer security in plain English from one of the leading cryptographic\nexperts: Secrets &amp; Lies: Digital Security in a Networked World, by Bruce Schneier. ISBN:\n0471253111.\nReferences: Journal of Computer Security, IEEE Security and Privacy\n```" ]
[ null, "https://s3.studylib.net/store/data/007300807_1-33d05c2c06e8333436549041bbb0967e-768x994.png", null ]
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http://activepatience.com/two-way-tables-worksheet/two-way-tables-worksheet-pictogram-maths-worksheet-tables-worksheets-printable/
[ "# Two Way Tables Worksheet Pictogram Maths Worksheet Tables Worksheets Printable", null, "two way tables worksheet pictogram maths worksheet tables worksheets printable.\n\ntruth tables worksheets pdf time for 3rd grade creating frequency worksheet 1 answers awesome two way ratio,fun activities to practice two way relative frequency tables dodging worksheets for grade 2 periodic table worksheet 5th 1,3 times tables worksheets pdf division funding k s and worksheet answers along with two way input output for 5th grade,dodging tables worksheets for grade 5 time 3rd printable two way frequency worksheet the best image,division tables worksheets pdf dodging two way relative frequency video khan academy table grade 4,division tables worksheets printable and graphs for 3rd grade probability two way worksheet answers times 1,multiplication tables worksheet generator two way independent practice math truth worksheets pdf division,dodging tables worksheets for grade 2 3 two way worksheet the best image collection table of contents 5th,table of contents worksheet for grade 2 times tables worksheets year 1 two way frequency 5,two way tables worksheets for all download and share table of contents worksheet grade 2 practice pdf dodging.\n\nPosted on" ]
[ null, "http://activepatience.com/wp-content/uploads/2018/05/two-way-tables-worksheet-pictogram-maths-worksheet-tables-worksheets-printable.jpg", null ]
{"ft_lang_label":"__label__en","ft_lang_prob":0.72161746,"math_prob":0.8834756,"size":1229,"snap":"2019-13-2019-22","text_gpt3_token_len":235,"char_repetition_ratio":0.2946939,"word_repetition_ratio":0.0,"special_character_ratio":0.17005695,"punctuation_ratio":0.055,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9951867,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-19T20:58:26Z\",\"WARC-Record-ID\":\"<urn:uuid:827c8932-7305-4acc-bf96-24ddd083e442>\",\"Content-Length\":\"36918\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8393975b-c55e-44cb-a498-58ff2724a0a7>\",\"WARC-Concurrent-To\":\"<urn:uuid:c1279660-bcd0-4468-8575-3e837cf14f91>\",\"WARC-IP-Address\":\"104.28.29.198\",\"WARC-Target-URI\":\"http://activepatience.com/two-way-tables-worksheet/two-way-tables-worksheet-pictogram-maths-worksheet-tables-worksheets-printable/\",\"WARC-Payload-Digest\":\"sha1:PIHW2CCDC4ETE76VRL4K7MNQYJO3SXUC\",\"WARC-Block-Digest\":\"sha1:V55CYNGO7UJ6IO5L3LEFK2ZUTCAXLITS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232255165.2_warc_CC-MAIN-20190519201521-20190519223521-00556.warc.gz\"}"}
https://www.maa.org/press/maa-reviews/geometry-through-history-euclidean-hyperbolic-and-projective-geometries
[ "# Geometry Through History: Euclidean, Hyperbolic, and Projective Geometries", null, "###### Meighan I. Dillon\nPublisher:\nSpringer\nPublication Date:\n2018\nNumber of Pages:\n350\nFormat:\nHardcover\nPrice:\n59.99\nISBN:\n9783319741345\nCategory:\nTextbook\n[Reviewed by\nMark Hunacek\n, on\n10/4/2018\n]\n\nThis book has its origins in a geometry course that was developed and taught by the author to prospective secondary school teachers, but which apparently came to be populated by other kinds of students (for example, physics and computer science majors) as well. The title may be a little misleading: although there is a lot of interesting geometry to be found here, and although the chapters follow an historical path and contain some historical content, the history is not developed as much as one might expect it to be, with a number of omissions that struck me as puzzling.\n\nThere are ten chapters. The first six comprise a course in what I would call the foundations of geometry. In more detail: the first chapter looks at the first book of Euclid’s Elements, reproducing proofs in somewhat more modern language, and also giving some indication of how Euclid made logical errors in reasoning that, by current standards of mathematical reasoning, do not stand up to scrutiny. The second then focuses on “neutral” geometry, which is the geometry we get if we assume all the axioms of Euclidean geometry except the famous fifth postulate; the idea here is to see what theorems of Euclidean geometry can be proved without the fifth postulate, and what changes to other theorems of Euclidean geometry are required. (For example, although it can no longer be proved that the sum of the angles of a triangle is 180 degrees, one can prove that the angle sum is less than or equal to this number.) In chapter 3, the assumption of neutrality regarding the parallel postulate is dropped and the hyperbolic parallel postulate, asserting that through a given point not on a given line, there are at least two lines parallel to the line, is adopted; the chapter then surveys some of the basic results of this non-Euclidean geometry.\n\nChapter 4 sketches a rigorous development of Euclidean (solid) geometry via the axioms used by Hilbert in his famous Foundations of Geometry. The exposition here is fairly streamlined (the entire chapter is about 25 pages long), and a number of results are left to the reader as exercises. Chapter 5 then returns to Euclidean geometry, going beyond Book I of the Elements to discuss topics discussed in later books, and also some topics not discussed in the Elements at all. One of the topics covered here is inversion in a circle, which is in turn used in chapter 6, which describes the Poincaré disc model of hyperbolic geometry. Other topics discussed in chapter 5 include the theorems of Ceva and Menelaus, and the nine-point circle.\n\nAlthough there is no chapter specifically devoted to axiom systems and their models, these topics are introduced “on the fly”, so to speak, especially in chapters 4 and 6. The discussion of these topics is fairly brief, perhaps because the author wished to avoid getting bogged down in these issues. Whether this approach is optimal is likely one of personal taste; I tend to think a more systematic discussion might have been advisable. Specifically, I think the use of models in chapter 4 would have enhanced that chapter, allowing the reader to see concrete interpretations of the various axioms discussed therein.\n\nThe remaining four chapters of the book shift gears somewhat, and cover a selection of topics in geometry that are all linked to abstract or linear algebra. Chapter 7 is on affine geometry, which, as done here, refers to the underlying geometry of a vector space: a point is an element of the space, and a line is a coset of a one-dimensional subspace. If we start, for example, with a two-dimensional space over the field $\\mathbb{R}$ of real numbers, we obtain the real affine plane, and can give elegant algebraic proofs of many theorems of Euclidean geometry that rely only on incidence alone, rather than on distance or angles. The author works in greater generality, however, not restricting herself to two-dimensional spaces, and also working not just over $\\mathbb{R}$ but over arbitrary fields.\n\nChapter 8 is on projective geometry. The approach to this subject is linear algebraic as well, and builds on the material of the preceding chapter on affine spaces defined by a field (or more precisely a vector space over that field). Dillon begins with affine three-space defined by a field F (i.e., a three-dimensional vector space over F), and adds “points at infinity” to form projective three-space. Then she shifts gears and gives some axioms for both projective space and the projective plane. Subsequent sections in the chapter touch on various additional topics, such as the theorems of Desargues and Pappus, harmonic sequences and homogenous coordinates for the projective plane defined by a field.\n\nChapter 9, on algebraic curves, gives an introduction to some of the ideas of algebraic geometry in affine and projective spaces, culminating in a statement of Bézout’s theorem, and a rough sketch of how it is proved.\n\nFinally, chapter 10 discusses rotations of the plane and three-dimensional Euclidean space. Quaternions are introduced late in the chapter and their relationship to rotations addressed.\n\nThis is quite a lot of material — more, I think, than can be covered in one semester. A dependence chart, or some indication of possible syllabi for courses, would have been helpful. The first six chapters seem to form, as previously mentioned, a fairly cohesive unit, but it would be helpful for a perspective reader to know just what material from these chapters is necessary for study of the remaining four.\n\nI thought that the last four chapters were somewhat less successful than the six that preceded them. For one thing, these chapters necessarily invest a fair amount of time in developing the relevant algebraic background. The chapter on affine geometry, for example, spends more time discussing background linear and abstract algebra than it does discussing geometry. Likewise, the chapter on algebraic curves spends a great deal of time discussing algebraic preliminaries (integral domains, UFDs, polynomials, resultants), and we only really get to geometry fairly late in the chapter. By the time the algebraic preliminaries in both of these chapters have been disposed of, the amount of time and space available to do geometry is limited; this all struck me as a lot of preliminary time spent leading up to an inadequate payoff.\n\nThe author’s chapter on projective geometry essentially makes it dependent on the previous chapter on affine geometry, and therefore makes it hard to introduce this subject directly after the first six chapters on Euclidean and hyperbolic geometry. Yet many instructors, having just discussed hyperbolic geometry (where there are lots of lines that are parallel to a given line and which pass through a given point not on that line) might wish to discuss a geometry where there are no such parallel lines.\n\nIn addition, the transition from the first six chapters to the last four seemed to me a bit abrupt, particularly since there are other topics that would fit in better. Particularly in a class with a number of prospective secondary school teachers in the audience, a more extended discussion of compass and straightedge constructions than the fairly brief account given here (particularly emphasizing the impossible ones) would be very valuable, as would a chapter on general geometric transformations (classifying plane isometries, for example, and using them to solve geometric problems, particularly with regard to symmetry; this topic would be of value to the physics majors in the class as well).\n\nI mentioned in the first paragraph of this review that I thought there were some puzzling omissions in the historical discussions. The lack of any treatment of the three unsolvable compass and straightedge constructions (trisecting the angle, doubling the cube, squaring the circle) is one of them, as is the omission of any discussion of the constructability of the regular n-gon, but there are others.\n\nFor one thing, geometry in this book begins with Euclid, which necessarily means that there is no discussion of pre-Greek geometry. This decision can be defended on the grounds that modern mathematics began with the Greeks, but it still struck me as odd that something like Plimpton 322 is not even mentioned, especially since the Pythagorean theorem is, and there is a connection between the two. And I find it even harder to believe that a book that contains the words “geometry” and “history” in its title would omit any reference to Thales.\n\nIn addition, given the fact that the Pythagorean theorem is discussed, I was surprised to see no discussion of the fact that $\\sqrt2$, the length of the hypotenuse of a right triangle with both legs having unit length, is irrational, and the significance of results like this for the ancient Greeks.\n\nAnother Greek mathematician that receives shorter shrift than one might expect is Archimedes. He is mentioned in connection with the Archimedean axiom, but there is little attention given to some of his geometrical accomplishments.\n\nThere is little discussion of the historical roots of projective geometry. The book’s introduction of projective planes and spaces in terms of arbitrary fields may lead a student to mistakenly assume that this was the way the subject arose historically, when in fact the term “projective plane” historically referred only to the extended Euclidean plane (i.e., the real number field, not arbitrary fields) and was based on problems in perspective drawing that arose during the Renaissance. The relationship between perspective drawing and projective geometry is only briefly mentioned in the book.\n\nThere is a beautiful connection between coordinatization and the theorems of Pappus and Desargues: a projective plane, defined axiomatically, can be coordinatized by a division ring if and only if Desargues’ theorem holds in that plane, and can be coordinatized by a field if and only if Pappus’ theorem does. Unfortunately, this topic is also not discussed in the chapter on projective geometry.\n\nI also thought that the author’s account of the history of hyperbolic geometry leaves many interesting things unsaid. Dillon summarizes the history in a few pages but omits much of the details that students find interesting. Although she says, for example, that attempts were made to prove Euclid’s Fifth Postulate but were ultimately all found to be faulty, she doesn’t give examples of these faulty proofs and explain why they were faulty. She also omits much of the human interest involved in the development of the subjects: the letters between Wolfgang and Janos Bolyai, for example, are fascinating, as is the correspondence between the elder Bolyai and Gauss, but they are not discussed or quoted here. Greenberg’s Euclidean and Non-Euclidean Geometries: Development and History, by contrast, gives a much more compelling account of how people gradually came to realize that hyperbolic geometry was a logically consistent area of mathematics in its own right.\n\nI should make it clear that I am not suggesting that any book on geometry must contain these topics. It seems to me, however, that when you title a book Geometry Through History, you raise certain expectations, and I am not sure these are fully met by this text. Books with similar titles (Geometry by Its History by Ostermann and Wanner and Revolutions of Geometry by O’Leary, for example), do seem to put more emphasis on history than does this text. And, of course, for a really detailed look at the history of geometry, there is 5000 Years of Geometry by Scriba and Schreiber and, for 19th century geometry, Worlds Out of Nothing by Jeremy Gray. Perhaps a title like “Topics in Geometry” or “Introduction to Geometry” would have been more accurate.\n\nBut there are worse things to say about a book than the fact that it may not have a fully accurate title. There is, in fact, a lot to like about this book. It discusses a lot of interesting topics in geometry, it is clearly written and should be accessible to its target audience of students who have “a year of calculus, an introductory course in linear algebra, and an interest in mathematics”, and it offers students an opportunity to see the relationship between geometry and other areas of mathematics, including abstract and linear algebra. If you are teaching a course in geometry at the college level and the topics discussed here are compatible with your course, this book certainly merits a serious look.\n\nMark Hunacek ([email protected]) teaches mathematics at Iowa State University.\n\n•", null, "•", null, "•", null, "•", null, "•", null, "" ]
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https://betterlesson.com/lesson/565909/time-for-my-catch?from=mtp_lesson
[ "# \"Time\" For My Catch\n\n3 teachers like this lesson\nPrint Lesson\n\n## Objective\n\nSWBAT name, notate, and tell time to the hour on a digital and analog clock. SWBAT use accurate measurement techniques.\n\n#### Big Idea\n\nThe day has come for students to create their own fish and to make sure that it is a keeper. The students will also be introduced to the clock face and the idea of hour notation. Hence the \"Time For My Catch\"\n\n## Warm Up\n\n5 minutes\n\nAdvanced preparation:  You will need to print out both sets of number cards from the resource section.  Cut out the 31-60 cards and put them into one envelope and do the same with the 61-90 and put them in their own envelope.\n\nI start this part of the lesson by asking the kids to sit in front of the classroom number line.\n\n\"Today we are going to change up our Start At/Stop At routine.  We are going to add the numbers 61-90.\"  I will pull one card out of the 31-60 envelope and we will use that as our start at number.  I will then pull out a card from our 61-90 envelope and use that as our Stop At number.  We will then count as a class from our starting number to our ending number.\"\n\nI will ask a student to point to each number as we count as a whole group.  I will continue this process as time allows.  I will also mix in counting backwards by starting at the higher number and counting to the lower number.  The Core Standards expect 1st graders to be able to count to 120, starting at any number, by the end of 1st grade.  This routine is the process in which I can assure that the students are continuously working toward that standard (CCSS.Math.Content.1.NBT.A.1).\n\n## Introducing Clocks and Telling Time to the Hour\n\n15 minutes\n\nI start this part of the lesson by asking all of the students to face the white board easel.\n\n\"We have been talking a lot about measuring objects, fish, and lengths of sides.  Mathematicians also like to measure time.  However, they don't use tiles to measure time.  What are some tools that people use to measure time?\"\n\nI then write their ideas on the easel (see picture in the resource section).\n\n\"Today we are going to look at clocks and learn how to tell time by the hour.  There are two types of clocks. There are two types of clocks and ways of notating times.  One is called an analog clock and the other is called a digital clock.\"\n\nI then write these two terms on the easel.\n\n\"Let's take a look at the analog clock. Does any one know what the hands are called?\"\n\nI then go over the hands and corresponding numbers.  I also have them count all of the minute marks around the perimeter of the clock (see video Looking at the Minutes on the Clock in section resource).\n\nI then draw the hands on an analog clock to show 2:00 and write the digital notation.\n\n\"This is what 2 o'clock looks like on an analog clock looks like and here is what it would look like on a digital clock (see visual created resource).\"\n\n\"If this is 2:00, then what time is this (I set the analog clock to another time)?  Who could come up and write this time with digital notation?\"  There is a video called Telling Time to the Hour, in the section resource, that models this part of the lesson.  I repeat this for several more hour times.  The Core Standards expect that 1st graders can tell and write time in hours and half-hours using analog and digital clocks (CCSS.Math.Content.1.MD.B.3).\n\n\"Now, I would like each of you to take out a pencil and a clipboard. Let's take a look at this sheet together.\"  The sheet is called Filling in the Hands on a Clock (section resource) and you will need to print enough for your class.  I hang a copy on the whiteboard easel for modeling purposes.  I go over the first few together and then ask the students to complete the form on their own.  Their is a video clip of them working on this and a finished sample in the resource section.\n\nThe students are using clocks to read times and placing hour and minute hands to represent set times. The core expects that mathematically proficient students to consider the available tools when solving a mathematical problem and that they are sufficiently familiar with tools appropriate for their grade or course to make sound decisions about when each of these tools might be helpful (CCSS.Math.Practice.MP5)\n\n## My Catch\n\n5 minutes\n\nThis next section is a continuation of the previous measurement activities and concepts that have been presented in this unit.  I want to continue to allow students the opportunity to perfect their accuracy and precision when using a unit to measure an object.\n\n\"Today you will have a chance to make your own fish or 'catch.'  You will need to decide how long for fish will have to be to be considered legal.  Your fish can be any size but must fit on this piece of paper (I give them legal sized paper).  What I would like you to do now is write your name on the paper and write who long your fish will have to be to be considered legal.\"\n\n\"When you get to this activity during station time, you can draw in your fish, measure it, record the length and then tell me weather or not it is a keeper.\"\n\nThere is a picture of one student's fish in this resource section.  There are also more examples of this activity in the Station Time section.\n\n## Station Time\n\n30 minutes\n\nMy Catch:  This activity was explained int he previous section.  There is also a student example of a completed poster and a video of a student working on this activity.\n\nMeasuring Fish:  This activity was explained in the previous lesson.  The resources and explanation is explained in the Measuring Fish Section of the linked lesson.\n\nIn this activity, students must be accurate with their measuring, make sure that their are no gaps or overlaps with their tiles, and record the total length of each fish.  The Core expects students to be able to express the length of an object as a whole number of length units, by laying multiple copies of a shorter object (the length unit) end to end; understand that the length measurement of an object is the number of same-size length units that span it with no gaps or overlaps. Limit to contexts where the object being measured is spanned by a whole number of length units with no gaps or overlaps (CCSS.Math.Content.1.MD.A.2 and CCSS.Math.Practice.MP6)\n\n## Exit Ticket and Continued Practice\n\n10 minutes\n\nI end the lesson with an informal assessment of students ability to look at an analog clock and write the time with digital notation.  I go over the sheet (see section resource) as a class and then have them go work not their own.\n\nAs they finish, I have ten complete the true or false equation sheet.  I want to continue to thread the concepts of the equal sign and understanding equations though out the year.  This is important in units like this because it is so focused on the measuring concepts and I don;t want to loose the operations knowledge that has been established.  By the end of the year, it is expected that students can understand the meaning of the equal sign, and determine if equations involving addition and subtraction are true or false. For example, which of the following equations are true and which are false? 6 = 6, 7 = 8 – 1, 5 + 2 = 2 + 5, 4 + 1 = 5 + 2 (CCSS.Math.Content.1.OA.D.7).\n\nThe sheets are in the section resource along with an example of the completed time sheet." ]
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https://webapps.stackexchange.com/questions/115113/conditional-formatting-depending-on-the-formula-that-the-cell-contains
[ "# Conditional formatting depending on the formula that the cell contains\n\nI have a spreadsheet that has 3 rows. Total received, Share A, and Share B. Share A will have a formula that is either `=A1*0.8` or `=A1*0.9`. Is there any way for me to set up a conditional format that will make the cell a different color depending on which mathematical formula I use?\n\nI'm trying to be able to easily tell whether Share A is getting 80% or 90%.\n\n• The value of total received will be different every time so I cannot use a number matching format – John C Mar 4 '18 at 3:55\n\nI don't think there is any built-in function that can access the formula in a cell.\n\nIf you want to go that route, you can use Apps Script function `getFormula()`.\n\nMy recommendation is that you adopt the following convention for yourself.\n\nAs much as possible,\n\nalways organize your data in two dimensions: each row is a new instance of your data type, each column is an attribute of your data type.\n\nIf you do that, you can easily apply Conditional Formatting, among many other things, based on column index. You can also apply Conditional Formatting based on the strings in the first row, ie. your header.\n\nIn your example, there should be 3 columns. Each column contains\n\n• ID/name of share;\n\n• share type (or a formula to output share based on the ID/name of your share);\n\n• and total received, which is a formula that computes your \"total received\" based on share type\n\nrespectively.\n\nYou can use `arrayformula` to apply your formula automatically and centralize your formulas in one cell. For example, in `C2`, enter\n\n``````=arrayformula(if(isblank(A2:A100,,ifs(exact(B2:B100,\"Share Type 1\"),A2:A100*0.8,\nexact(B2:B100,\"Share Type 2\"),A2:A100*0.9))\n``````\n\nUse `B2:B100` or some relatively small number instead of `B2:B` to reduce run time. The `isblank()` check will ensure that the formula is only applied to the rows you entered share information on.\n\nThe custom formula for Conditional Formatting would be\n\n``````=if(isblank(c2),false,exact(B2,\"Share Type 1\"))\n``````\n\nwith range `C2:C` for one color and duplicate the same thing for \"Share Type 2\" for another color.\n\nThere are a few things you can do in order to conveniently enter/output the type of the share. For example,\n\n• you can use Data Validation and then enter the type manually. Data Validation will help you enter the correct string every time and it also comes with a simple complete-what-you-type function;\n\n• or, if the type of a given share is fixed over time, you can have a list of all shares on a separate tab/sheet and each type of shares under a different column. Also enter the associated coefficients there using `arrayformula`. Then use `vlookup` on your main tab/sheet to retrieve those coefficients conveniently.\n\nCustom Formula:\n\n``````=AND(B1=\\$A1*0.9,ISNUMBER(B1))\n``````\n\nApply to:\n\n``````B:C\n``````" ]
[ null ]
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http://mathforcollege.com/nm/videos/youtube/01aae/taylor/0107_02_Taylor_Series_Part_Example.htm
[ "CHAPTER 01.07: TAYLOR SERIES REVISITED: Taylor Series: Example    In this segment we're going to take an example of Taylor series, and see how we can apply Taylor series to be able to get the value of the function at some other point. What I mean to say is that, let me write down the Taylor series again. So the general form of the Taylor series, or Taylor's Theorem, is that the value of the function can be calculated at some point x plus h, if we know the value of the function at x, if we know the value of the derivative of the function at x, we know the value of the second derivative of the function at x, and we know the third derivative of the value of the function at x, and so on and so forth, then I can give you the value of the function at some other point.   So in this example, let's suppose somebody tells you the following, that, hey, says you are given the value of the function at 4, some function, is given as 125, the value of the first derivative of the function is given as 74, the value of the second derivative at 4, same point, is given 30, the value of the third derivative at 4 is given as 6, and all other derivatives are 0. So you're given the value of the function at 4, the first derivative at 4, the second derivative at 4, and third derivative at 4, and then all the other derivatives are given as 0, that means the fourth derivative at 4, the fifth derivative at 4, the sixth derivative at 4, and all those other derivatives are all given to be 0.  So those are all 0, and what is asked for you is to find the value of the function at 6. Of course, it is already given to you, the fine print is already given to you that all these derivatives are continuous and exist between the point x equal to 4 to x equal to 6, that's the only way you can apply Taylor's Theorem.  So let's go ahead and see that, how we can find the value of the function at 6.  Now, again, I want to point out here is that nowhere are you given what the function itself is, because if you were given what the function itself is, then you could just substitute the value of x equal to 6, and you would get what the value of the function at 6 is.  So this is the beauty of Taylor's Theorem, that you are only given the value of the function at a particular point and all its derivatives at that particular point, and you are asked to find out what the value of the function at some other point is, in this case it is the value of x equal to 6.  So if I'm going to apply my Taylor series, my x is 4, my h is 2, because x plus h will then give me the value of 6, and that's where I'm interested in finding out the value of the function, right?  So what I'll get is that f of 6 is exactly equal to the value of the function at 4, plus the derivative of the function at 4 times h, which is 2, plus the second derivative of the function at 4 times h squared divided by factorial 2, h I have 2, plus the third derivative of the function at 4 divided by factorial 3 times h cubed, which is 2 cubed. And then the fourth derivative and so on and so forth, which are all 0. All the . . . all the fourth derivative, fifth derivative, sixth derivative, and so on and so forth, they're all 0, so I will not get any contribution from those terms. So this value of the function which I'm going to calculate at 6 will be an exact value, there is no approximation involved in it, because all the derivatives are given to me, and also that the derivatives which are after the fourth derivative, they're all 0. So f of 4, which was given to me as 125, plus f prime of 4, which is given to me as 74, plus f double prime of 4, which is given to me as 30, plus f triple prime of 4, which is the third derivative, which is given as 6. And this value here turns out to be 341. In fact, what I'm going to do is I'm going to write down the individual contributions, I get 125 from here, 148 from here, 60 from here, and 8 from there.  So, from the four terms which are available for the Taylor series, because the other ones are 0, I get 125 contribution from here, 148 from here, 60 from here, and 8 from there. So you can see that the contribution will decrease as you add more terms. It doesn't mean that it's always going to be decreasing, it could be increasing in certain cases, but if you choose h to be small enough, it's going to continue to decrease.  Here you are finding out the second term is contributing more than the first term, so please don't think that as you keep on using more terms in the Taylor series, that their contribution is smaller and smaller, term by term, and the value turns out to be 341, and this is the value of the function at 6.  And you were able to calculate the value of the function at 6 without knowing what the expression for the function is, but from the information that you knew what the initial value at x equal to 4 is, what the derivative of the function at 4 is, what the second derivative of the function at 4 is, the third derivative of the function at 4 is, and all the other derivatives, you're able to calculate what the value of the function at 6 is, in this case being 341, okay?  And that's the end of this segment." ]
[ null ]
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https://justaaa.com/finance/6921-at-5-percent-interest-how-long-does-it-take-to
[ "Question\n\n# At 5 percent interest, how long does it take to double your money? (Do not round...\n\nAt 5 percent interest, how long does it take to double your money? (Do not round intermediate calculations and round your answer to 2 decimal places, e.g., 32.16.)\n\nLength of time             years\n\nAt 5 percent interest, how long does it take to quadruple your money? (Do not round intermediate calculations and round your answer to 2 decimal places, e.g., 32.16.)\n\nLength of time             years\n\na.We use the formula:\nA=P(1+r/100)^n\nwhere\nA=future value(\\$2x)\nP=present value(\\$x say)\nr=rate of interest\nn=time period.\n\n2x=x*(1.05)^n\n\n2=(1.05)^n\n\nTaking log on both sides;\n\nlog 2=n*log 1.05\n\nn=log 2/log 1.05\n\nwhich is equal to\n\n=14.21 years(Approx).\n\nb.We use the formula:\n\nA=P(1+r/100)^n\nwhere\nA=future value(\\$4x)\nP=present value(\\$x say)\nr=rate of interest\nn=time period.\n\n4x=x*(1.05)^n\n\n4=(1.05)^n\n\nTaking log on both sides;\n\nlog 4=n*log 1.05\n\nn=log 4/log 1.05\n\nwhich is equal to\n\n=28.41 years(Approx).\n\n#### Earn Coins\n\nCoins can be redeemed for fabulous gifts." ]
[ null ]
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https://artofproblemsolving.com/wiki/index.php/De_Longchamps_point
[ "# De Longchamps point\n\nThe title of this article has been capitalized due to technical restrictions. The correct title should be de Longchamps point.", null, "$[asy] draw((0,0)--(44,60)--(44,-10)--cycle); draw((0,0)--(44,0),blue+dashed); draw((44,60)--(22,-5),blue+dashed); draw((44,-10)--(6.5,10),blue+dashed); label(\"H\",(24,0),(1,1)); dot((24,0)); draw((22,30)--(44,14),red); draw((22,-5)--(34,46),red); draw((44,25)--(18,25),red); dot((29,25)); label(\"C\",(29,25),(1,1)); draw(Circle((29,25),25),dashed); dot((34,50)); label(\"L\",(34,50),(1,1)); [/asy]$", null, "The de Longchamps point (", null, "$L$) is the the orthocenter (", null, "$H$) reflected through the circumcenter (", null, "$C$).\n\nThe de Longchamps point of a triangle is the reflection of the triangle's orthocenter through its circumcenter.\n\nThe point is collinear with the orthocenter and circumcenter." ]
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https://www.cnblogs.com/shine-lee/p/11908610.html
[ "# 权重初始化最佳实践", null, "", null, "• 因为对权重$w$的大小和正负缺乏先验,所以应初始化在0附近,但不能为全0或常数,所以要有一定的随机性,即数学期望$E(w)=0$\n\n• 因为梯度消失和梯度爆炸,权重不易过大或过小,所以要对权重的方差$Var(w)$有所控制\n\n• 深度神经网络的多层结构中,每个激活层的输出对后面的层而言都是输入,所以我们希望不同激活层输出的方差相同,即$Var(a^{[l]})=Var(a^{[l-1]})$,这也就意味不同激活层输入的方差相同,即$Var(z^{[l]})=Var(z^{[l-1]})$\n\n• 如果忽略激活函数,前向传播和反向传播可以看成是权重矩阵(转置)的连续相乘。数值太大,前向时可能陷入饱和区,反向时可能梯度爆炸,数值太小,反向时可能梯度消失。所以初始化时,权重的数值范围(方差)应考虑到前向和后向两个过程", null, "", null, "# 期望与方差的相关性质\n\n$Var(X) = E(X^2) - (E(X))^2$\n\n$Cov(X, Y) = 0$\n\n\\begin{align} Cov(X, Y) &= E((X-E(X))(Y-E(Y))) \\\\ &= E(XY)-E(X)E(Y) =0 \\end{align}\n\n\\begin{aligned} \\operatorname{Var}(X+Y) &=E\\left((X+Y)^{2}\\right)-(E(X+Y))^{2} \\\\ &=E\\left(X^{2}+Y^{2}+2 X Y\\right)-(E(X)+E(Y))^{2} \\\\ &=\\left(E\\left(X^{2}\\right)+E\\left(Y^{2}\\right)+2 E(X Y)\\right)-\\left((E(X))^{2}+(E(Y))^{2}+2 E(X) E(Y)\\right) \\\\ &=\\left(E\\left(X^{2}\\right)+E\\left(Y^{2}\\right)+2 E(X) E(Y)\\right)-\\left((E(X))^{2}+(E(Y))^{2}+2 E(X) E(Y)\\right) \\\\ &=E\\left(X^{2}\\right)-(E(X))^{2}+E\\left(Y^{2}\\right)-(E(Y))^{2} \\\\ &=\\operatorname{Var}(X)+\\operatorname{Var}(Y) \\end{aligned}\n\n\\begin{aligned} \\operatorname{Var}(X Y) &=E\\left((X Y)^{2}\\right)-(E(X Y))^{2} \\\\ &=E\\left(X^{2}\\right) E\\left(Y^{2}\\right)-(E(X) E(Y))^{2} \\\\ &=\\left(\\operatorname{Var}(X)+(E(X))^{2}\\right)\\left(\\operatorname{Var}(Y)+(E(Y))^{2}\\right)-(E(X))^{2}(E(Y))^{2} \\\\ &=\\operatorname{Var}(X) \\operatorname{Var}(Y)+(E(X))^{2} \\operatorname{Var}(Y)+\\operatorname{Var}(X)(E(Y))^{2} \\end{aligned}\n\n# 全连接层方差分析\n\n\\begin{align}a_i^{[l-1]} &= f(z_i^{[l-1]}) \\\\z_i^{[l]} &= \\sum_{j=1}^{fan\\_in} w_{ij}^{[l]} \\ a_j^{[l-1]}+b^{[l]} \\\\a_i^{[l]} &= f(z_i^{[l]})\\end{align}\n\n• 网络输入的每个元素$x_1,x_2,\\dots$独立同分布\n• 每层的权重随机初始化,同层的权重$w_{i1}, w_{i2}, \\dots$独立同分布,且期望$E(w)=0$\n• 每层的权重$w$和输入$a$随机初始化且相互独立,所以两者之积构成的随机变量$w_{i1}a_1, w_{i2}a_2, \\dots$亦相互独立,且同分布;\n• 根据上面的计算公式,同层的$z_1, z_2, \\dots$独立同分布,同层的$a_1, a_2, \\dots$也为独立同分布\n\n\\begin{align} Var(z) &=Var(\\sum_{j=1}^{fan\\_in} w_{ij} \\ a_j) \\\\ &= fan\\_in \\times (Var(wa)) \\\\ &= fan\\_in \\times (Var(w) \\ Var(a) + E(w)^2 Var(a) + Var(w) E(a)^2) \\\\ &= fan\\_in \\times (Var(w) \\ Var(a) + Var(w) E(a)^2) \\end{align}\n\n# tanh下的初始化方法\n\n$Var(a^{[l]}) = Var(z^{[l]}) = fan\\_in \\times Var(w) \\times Var(a^{[l-1]})$\n\n\\begin{align} Var(a^{L}) = Var(z^{[L]}) &= n^{[L-1]} Var(w^{[L]}) Var(a^{[L-1]}) \\\\ &=\\left[\\prod_{l=1}^{L} n^{[l-1]} Var(w^{[l]})\\right] {Var}(x) \\end{align}\n\n$\\prod^{t} n^{[l-1]} Var(w^{[l]}) \\\\ \\prod^{t} n^{[l]} Var(w^{[l]})$\n\n## Lecun 1998\n\nLecun 1998年的paper Efficient BackProp ,在输入Standardization以及采用tanh激活函数的情况下,令$n^{[l-1]}Var(w^{[l]})=1$,即在初始化阶段让前向传播过程每层方差保持不变,权重从如下高斯分布采样,其中第$l$层的$fan\\_in = n^{[l-1]}$\n\n$W \\sim N(0, \\frac{1}{fan\\_in})$\n\n## Xavier 2010\n\n$W \\sim N(0, \\frac{2}{fan\\_in + fan\\_out})$\n\n$Var(U(-n, n)) = \\frac{n^2}{3}$\n\n$n = \\frac{\\sqrt{6}}{\\sqrt{fan\\_in + fan\\_out}}$\n\n$W \\sim U(-\\frac{\\sqrt{6}}{\\sqrt{fan\\_in + fan\\_out}}, \\frac{\\sqrt{6}}{\\sqrt{fan\\_in + fan\\_out}})$", null, "# ReLU/PReLU下的初始化方法\n\n$Var(z)= fan\\_in \\times (Var(w) \\ Var(a) + Var(w) E(a)^2)$", null, "\\begin{align} Var(z) &= fan\\_in \\times (Var(w) \\ Var(a) + Var(w) E(a)^2) \\\\ &= fan\\_in \\times (Var(w) (E(a^2) - E(a)^2)+Var(w)E(a)^2) \\\\ &= fan\\_in \\times Var(w) \\times E(a^2) \\end{align}\n\n## He 2015 for ReLU\n\n$Var(z^{[l]}) = fan\\_in \\times Var(w^{[l]}) \\times E((a^{[l-1]})^2)$\n\n\\begin{align}Var(x) &= \\int_{-\\infty}^{+\\infty}(x-0)^2 p(x) dx \\\\&= 2 \\int_{0}^{+\\infty}x^2 p(x) dx \\\\&= 2 E(\\max(0, x)^2)\\end{align}\n\n\\begin {align}Var(z^{[l]}) &= fan\\_in \\times Var(w^{[l]}) \\times E((a^{[l-1]})^2) \\\\&= \\frac{1}{2} \\times fan\\_in \\times Var(w^{[l]}) \\times Var(z^{[l-1]}) \\end{align}\n\n$\\frac{1}{2} \\times fan\\_in \\times Var(w^{[l]}) = 1 \\\\ Var(w) = \\frac{2}{fan\\_in}$\n\n$W \\sim N(0, \\frac{2}{fan\\_in})$\n\n$W \\sim N(0, \\frac{2}{fan\\_out})$", null, "", null, "## He 2015 for PReLU", null, "$\\frac{1}{2} (1 + a^2) \\times fan\\_in \\times Var(w^{[l]}) = 1 \\\\Var(w) = \\frac{2}{(1 + a^2) fan\\_in} \\\\W \\sim N(0, \\frac{2}{(1 + a^2) fan\\_in}) \\\\W \\sim N(0, \\frac{2}{(1 + a^2) fan\\_out})$\n\n## caffe中的实现", null, "# 参考\n\nposted @ 2019-11-21 21:42  shine-lee  阅读(6038)  评论(1编辑  收藏  举报" ]
[ null, "https://s2.ax1x.com/2019/11/05/MpBFRU.png", null, "https://s2.ax1x.com/2019/11/05/MpBVsJ.png", null, "https://s2.ax1x.com/2019/11/20/MfIhi6.png", null, "https://s2.ax1x.com/2019/11/20/MW33zn.png", null, "https://s2.ax1x.com/2019/11/20/Mfi2z4.png", null, "https://s2.ax1x.com/2019/11/06/MPzLB6.png", null, "https://s2.ax1x.com/2019/11/20/Mf6E0U.png", null, "https://s2.ax1x.com/2019/11/20/Mf6YAe.png", null, "https://s2.ax1x.com/2019/11/20/Mf2jvn.png", null, "https://s2.ax1x.com/2019/11/20/Mf2Nh4.png", null ]
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https://haogroot.com/2020/11/29/bfs-leetcode/
[ "# BFS 廣度優先搜尋 – 陪你刷題\n\nLeetcode 邁向千題的關卡,想要把所有題目刷過一遍,對於一位上班族來說就跟寫出來的程式沒有 bug 一樣困難,因此想要將刷題常用的觀念技巧與其對應題目整理出來,一方面可以整理自己思緒,也可以做為未來複習用。\n\n# BFS 操作框架\n\n1. 透過 queue 的 FIFO 特性來紀錄走訪過且待處理的 vertex (頂點) 。\n2. `visited` 陣列來記錄各 vertex 是否被走訪過,避免 vertex 被重複訪問。\n• 如果圖是 binary tree ,就不會有重複造訪的情況,就不需要 visited\n• 如果 vertex 不是由固定範圍的 integer 所構成 ,那就用 unordered_set 來紀錄 vertex 是否走訪過 (以 C++ 來說)\n3. `prev` 來紀錄搜尋路徑。 `prev[s] = t` ,代表由 t 走到 s ,也就是說 `prev[s]` 表示頂點 s 是由哪個頂點走過來的 。\n\nBFS 操作順序如下,可以搭配下面的示意圖與程式碼來理解:\n\n1. 把起點放到 queue 裡,並將其在 visited 陣列中對應值寫成 1 表示拜訪過 。\n2. 從 queue 的前端拿出 vertex (最早放入的, queue 的 FIFO 特性 )。\n3. 檢查拿出的 vertex 的所有鄰近 vertices 中,將尚未走訪過的放進 queue 中,並執行以下動作:\n1. 將其在 `visited` 陣列中對應值寫成 1\n2. 將其在 `prev` 陣列中對應值寫成拿出的 vertex\n4. 移除 queue 前端的 vertex 。\n5. 不斷重複步驟 2~4 直到 queue 為空。\n6. 最終可以由 `prev` 陣列構建出 BFS 的走訪順序。", null, "``````int BFS (int start, int target)\n{\nqueue<int> qu;\nint shortest_step = 0;\nint visited[size] = {0};\nint prev[size] = {-1};\n\nqu.push (start);\nwhile (!qu.empty ())\n{\nint round = q.size ();\nfor (int i=0; i<round; i++)\n{\nint node = qu.front ();\nif (node == target)\nreturn shortest_step;\n// 將 vertex 所有的鄰居節點放入 queue\nfor (neighbor:neightborhood)\n{\nif (visited[neightbor] == 0)\n{\nqu.push (neighbor);\nvisited[neigbor] = 1;\nprev[neighbor] = node;\n}\n}\n// 從 queue 中拿出節點,處理完就 pop\nqu.pop ();\n}\nshortest_step++;\n}\n}``````\n\n## Leetcode #102 Binary Tree Level Order Traversal\n\n``````vector<vector<int>> levelOrder(TreeNode* root) {\nvector<vector <int>> result;\nqueue<TreeNode *> q;\nif (root != nullptr)\nq.push(root);\nwhile (!q.empty())\n{\nint qu_size = q.size();\nvector<int> level_vec;\nfor (int i=0; i<qu_size; i++)\n{\nTreeNode *top_node = q.front();\nlevel_vec.push_back (top_node->val);\nif (top_node->left != nullptr)\n{\nq.push (top_node->left);\n}\nif (top_node->right != nullptr)\n{\nq.push (top_node->right);\n}\nq.pop();\n}\nresult.push_back (level_vec);\n}\n\nreturn result;\n}``````\n\n1. 替換的字元不能跟未替換前一樣\n2. 替換後如果是目標,直接回傳結果\n3. 如果出現在 visited 內,直接忽略\n4. 確認是否屬於 wordList 內的單字,不屬於裏面的單字不需要處理\n\n### 優化\n\n``````// 將 wordList 轉換為 unordered_set\nunordered_set word_list (wordList.begin(), wordList.end());\n\n// 每當有一個新的單字 curr ,先檢查 word_list 內是否有這個單字\nif (word_list.count(curr))\n{\nqu.push (curr);\n// 走訪過的單字由 word_list 中移除,避免重複走訪\nword_list.erase(curr);\n}``````\n\n``````class Solution {\npublic:\nvector<string>& wordList)\n{\nunordered_set<string> path (wordList.begin (),\nwordList.end ());\nif (path.find (endWord) == path.end ())\nreturn 0;\n\nqueue<string> qu;\nqu.push (beginWord);\nint step = 0;\n\nwhile (!qu.empty ())\n{\nstep++;\nint todo = qu.size ();\nfor (int i=0; i<todo; i++)\n{\nstring curr = qu.front ();\nqu.pop ();\n\n// iterate each char of string\nfor (int idx=0; idx<curr.size (); idx++)\n{\nstring replaceCurr = curr;\n// change character from a to z\nfor (int ch='a'; ch<='z'; ch++)\n{\nreplaceCurr[idx] = ch;\nif (replaceCurr == endWord)\n{\nreturn step+1;\n}\nif (replaceCurr == curr)\ncontinue;\nif (path.find (replaceCurr)\n!= path.end())\n{\nqu.push (replaceCurr);\npath.erase (replaceCurr);\n}\n}\n}\n}\n}\nreturn 0;\n}\n};``````\n\n## 參考資料\n\nUpdated on 2021-02-07 21:23:04 星期日" ]
[ null, "https://haogroot.com/wp-content/uploads/2020/11/443C1A46-D227-4DAD-BEA9-1DAF28333767.jpeg", null ]
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https://creaticals.com/work-on-an-inclined-plane/
[ "# Work on an Inclined Plane: Explanation, Equation, and Problems With Solutions\n\nBy a Guy Who Teaches Physics for Fun", null, "Choose language:\nWork on an inclined plane is divided into 2 cases: a) when only conservative force like gravity applies and b) when non-conservative forces like friction or tension apply. The two cases are explained with different equation. Let’s discuss it one by one.\n\n## Work on inclined plane when only gravity force is acting on an object\n\nThis case is a simple case in which an object is descending an inclined plane and only affected by the force of gravity without friction or other non-conservative forces. We can calculate the work done on the object by using any of these equations written below.", null, "*You need to realize that the normal force is actually a non-conservative force, but because the normal force and the direction of displacement form a $90^o$ angle, the calculation of work done on the object by normal force would be zero which means it can be ignored.\nIn this case, the object’s potential energy converts to kinetic energy as the object descends down.\nWhen there is no friction or any other non-conservative force, an object descending an inclined plane experiences the same amount of work as an object in a free fall without friction(same mass and height).\nShare Button\n\n## Work on an inclined plane when non-conservative force applies\n\nThis case is a bit more complicated than the previous case. When friction force or any other non-conservative force applies on the object, we could calculate the work using the following equation.\n$W_{n c net}=\\Delta E_{p}+\\Delta E_{k}$\nDescription:\n\\begin{aligned} W_{nc}=&\\;\\text{net work by all non-consv forces}\\\\ \\Delta E_{p}=&mg(h_2-h_1)\\\\ \\Delta E_{k}=&\\frac{1}{2}m({v_2}^2-{v_1}^2)=\\sum F_x \\;s \\end{aligned}\nYou can use this equation for an object which is descending or climbing an inclined plane. You can see an example of using this equation on the slide below.\n\n## Work on inclined plane problems with solutions\n\nOf course, you can not fully understand without studying how we use the equation we have just discussed. Below we present some problems that you can study.\n\n### Problem 1(only conservative force applies)\n\nA block of ice is at the top of an inclined plane that has a height of 5 meters with a slope $30^o$ from horizontal. The block of ice is initially at rest and then slides down the plane. Assume because the block is made of ice, we can neglect the surface friction. Determine the work done on the ice block and determine the final velocity of the ice block.\n\n#### Solution:\n\nLet’s first sketch the forces the ice blocks experience. Please swipe the image above.\nThen we calculate the distance traveled by the ice block.\n\\begin{aligned} s=&\\; \\frac{\\text{tinggi bidang miring}}{\\sin 30^{\\circ}}\\\\ s=&\\; \\frac{5}{1/2}\\\\ s=&\\; 10\\;\\text{meter} \\end{aligned}\nWe can solve this problem with 3 different work equations. Let’s solve this problem with each of these three equations.\n##### Equation 1\n\\begin{aligned} W\\;=&\\;mg \\sin \\theta \\;s\\\\ =&(2)(9.81) \\sin 30^o \\;(10)\\\\ =&\\;98.1 \\;\\text{Joule} \\end{aligned}\n##### Equation 2\n\\begin{aligned} W\\;=&\\;-\\Delta Ep\\\\ W\\;=&\\;-mg(h_2-h_1)\\\\ =&-(2)(9.81)(0-5)\\\\ =&\\;98.1 \\;\\text{Joule} \\end{aligned}\n##### Equation 3\nWe have obtained the work done on the ice block by using equation 1 and 2. We could use equation 3 to find the final velocity of the ice block.\n\\begin{aligned} W=&\\;\\Delta Ek\\\\ W=&\\; \\frac{1}{2}m({v_2}^2-{v_1}^2)\\\\ 98.1=&\\;\\frac{1}{2}(2)({v_2}^2-0)\\\\ {v_2}=&\\sqrt{98.1}=\\;9.9 \\;m/s \\end{aligned}\nActually, in the case where only gravity force applies and the object is initially at rest, we can also use the equation $v=\\sqrt{2gh}$ to find the final velocity of the object. Please try it yourself.\n\n#### Discussion\n\nThus we have obtained the work done on the ice block. In this problem, we could see all the potential energy the ice block has is transformed into kinetic energy. You could see the energy diagram below.", null, "### Problem 2 (non-conservative force applies)\n\nNow the block is not made of ice but something solid that would make the surface friction significant. The coefficient of kinetic friction of the inclined plane with the block is 0.3. a) Calculate the work done by each force acting on the block. b) Also calculate the change in the kinetic energy of the block. c) Calculate the final velocity of the block.\n\n#### Solution:\n\nIn this problem, the non-conservative forces that apply are the frictional force and the normal force.\nFirst, let’s sketch every force that acts on the block. You can slide the image above to see it.\nWe know from problem 1 that the distance traveled by the block is $s=10\\; \\text{meter}$.\nThen we can calculate the work by each force as follows\n\n#### Work by each force\n\nThere are 3 forces that apply, namely gravity (w), friction ($f_k$), and normal(N).\nThe work done by the normal force is as follows.\n\\begin{aligned} W_{N}=&N \\cos 90 \\;s\\\\ =&N \\;(0) \\;(10)\\\\ =&0 \\end{aligned}\nThe work done by the frictional force is as follows.\n\\begin{aligned} W_{fk}=&f_{k} (\\cos 180^o) \\;s\\\\ =&mg(\\cos 30^o)\\mu_k\\;(-1)(10) \\\\ =&(2)(9.81)(\\frac{1}{2}\\sqrt{3})(0.3)(-1)(10)\\\\ =&-50.97 \\; \\text{Joule} \\end{aligned}\nThe work by the force of gravity is as follows.\n\\begin{aligned} W_{w}=&mg (\\sin 30^o) \\;s\\\\ =&(2)(9.81)(0,5)(10) \\\\ =&98.1\\;\\text{Joule} \\end{aligned}\n\n#### Change in the kinetic energy of the block\n\nWe could find the change in kinetic energy by using the following equation.\n\\begin{aligned} W_{nc}=&\\; \\Delta Ep \\;+\\; \\Delta Ek\\\\ \\end{aligned}\nWe will work the equation to get $\\Delta Ek$.\n\\begin{aligned} W_{fk} + W_N=&\\; \\Delta Ep\\;+\\;\\Delta Ep\\\\ -50.97 + 0=&mg(h_2-h_1)+\\Delta Ek\\\\ -50.97 =&(2)(9.81)(0-5)+\\Delta Ek\\\\ -50.97=&\\;-98.1+\\Delta Ek\\\\ \\Delta Ek=&\\;47.13\\;\\text{Joule} \\end{aligned}\n\n#### Block's final velocity\n\nWe can find the final velocity of the block by using the equation $\\Delta Ek=\\frac{1}{2}m({v_2}^2-{v_1}^2)$ and the operation is as follows.\n\\begin{aligned} \\Delta Ek=&\\frac{1}{2}m({v_2}^2-{v_1}^2)\\\\ 47.13=&\\frac{1}{2}(2)({v_2}^2-0)\\\\ v_2=&\\sqrt{47.13}\\\\ =\\; &6.87 \\;m/s \\end{aligned}\n\n#### Discussion\n\nIn contrast to problem 1, in this case, not all of the potential energy of the block is converted into kinetic energy. Nearly half of the potential energy of the block(50.97 joules) is released to the environment due to friction. So the kinetic energy of the block when it reaches the end of the inclined plane is only 47.13 Joules. You can see the energy chart of this case below.", null, "### Problem 3(non-conservative forces apply)", null, "A block(50 kg) is pulled by Jason through an inclined plane that has a height of 2.5 meters. The coefficient of kinetic friction of the inclined plane is 0.3. The inclined plane has an angle of inclination $30^o$ The force exerted by Jason’s muscles is 400N. (Assume $g=10m/s^2$)\n1. Calculate the work by each force acting on the block\n2. Determine the change in kinetic energy experienced by the block.\n3. Determine the block’s velocity when it reach the top of the inclined plane.\n\n#### Solution and Dicussion\n\nYou can see the solution and discussion of this problem in the article work by non-conservative force.\n\n### Problem 4(non-conservative force applies)", null, "An object climbs an inclined plane at a constant speed from a height of $h_1=2m$ meters to $h_2=6m$. The mass of the object is m=20 Kg. Calculate the work experienced by the object.\n\n#### Solution and discussion\n\nTo answer this problem, it is important to define the system and the environment.\nIf we isolate the object from the gravitational field so that the object’s kinetic energy is the only intrinsic energy the object has then the work that the object experiences is $W=\\Delta Ek=0$(remember its velocity is constant).\nHowever, if we don’t isolate the gravitational field from the object so that the potential energy is included as the object’s property, then the work is\n\\begin{aligned} W=&\\Delta Ep\\\\=&mg(h_2-h_1)\\\\=&(20)(9.81)( 6-2)\\\\=&784.8\\;\\text{Joule} \\end{aligned}\nThis means all the forces that act upon the object increase the potential energy of the object.\nShare Button\nRelated post", null, "", null, "", null, "", null, "" ]
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https://eurekamathanswers.com/slope-of-the-graph-of-y-equals-to-mx-plus-c/
[ "", null, "In the previous articles, we have discussed about slope and y-intercept, plotting points in the XY plane, independent variables and dependent variables, etc. Let us learn how to solve the slope of the graph y = mx + c from this page. In addition that you can find the problems on the slope of the graph y = mx + c in the below section.\n\nIntroduction to Slope-Intercept Form | y=mx+c Equation of Straight Line Slope\n\ny = mx + c is an equation of the line having a slope and y-intercept of the line in coordinate geometry. m represents the slope or gradient of the line and c is the y-intercept that cuts the point on the y-axis. The line cuts the y-axis at the point (0, c) which is the distance from the origin to the point c.\nm = (y – c)/(x – 0)\nm = (y – c)/x\nmx = y – c\nmx + c = y\ny = mx + c", null, "y = mx + c Straight Line Graphs Slope Examples\n\nExample 1.\nWhat is the slope of the equation 6x – 4y + 5 = 0?\nSolution:\nGiven that the equation is\n6x – 4y + 5 = 0\n4y = 6x + 5\ny = 6/4x + 5/4.\ny = 3/2x + 5/4\nComparing the equation with y = mx + c,\nwe have m = 3/2.\nTherefore, the slope of the line is 3/2", null, "Example 2.\nFind the equation of a line in the form of y = mx + c, having a slope of 4 units and an intercept of -6 units.\nSolution:\nGiven that\nThe slope of the line, m = 4 and The y-intercept of the line, c = -6.\nWe know that\nThe slope-intercept form of the equation of a line is y = mx + c.\nFrom the equation\ny = 4x – 6\nTherefore the required equation of the line is y = 4x – 6\n\nExample 3.\nConvert the equation 5x + 4y = 12 into y = mx + c and find its y-intercept.\nSolution:\nGiven that the equation is\n5x + 6y + 12 = 0\n6y = 5x + 12\ny = 5/6x + 12/6.\ny = 5/6x + 2\nComparing the equation with y = mx + c,\nwe have m = 5/6 and c = 2\nTherefore, the slope of the line is 5/6. y-intercept is c = 2\n\nExample 4.\nWhat is the slope of the equation 2x – 4y + 10 = 0?\nSolution:\nGiven that the equation is\n2x – 4y + 10 = 0\n4y = 2x + 10\ny = 2/4x + 10/4.\ny = 1/2x + 5/2\nComparing the equation with y = mx + c,\nwe have m = 1/2.\nTherefore, the slope of the line is 1/2.\n\nExample 5.\nWhat is the slope and y-intercept of the line 3x – 6y + 12 = 0?\nSolution:\nGiven that the equation is\n3x – 8y + 12 = 0\n8y = 3x + 12\ny = 3/8x + 12/8.\ny = 3/8x + 3/2\nComparing the equation with y = mx + c,\nwe have m = 3/8 and c = 3/2\nTherefore, the slope of the line is ⅜ and y intercept c = 3/2\n\nFAQs on Slope of the graph y = mx + c\n\n1. What is the equation of a straight line?\n\nThe equation of the straight line is y = mx +c\nwhere,\nm is the slope and c is the y-intercept.\n\n2. How do I find the slope in a graph?\n\nTake two points on the line and determine their coordinates. Determine the difference in y-coordinates of these two points. Find the difference in x-coordinates for these two points. Divide the difference in y-coordinates by the difference in x-coordinates.\n\n3. What is mx in the slope formula?\n\nIn the equation of a straight line y = mx + c, the slope is m which is multiplied by x and c is the y-intercept." ]
[ null, "https://eurekamathanswers.com/wp-content/uploads/2022/01/Slope-of-Graph-y-mxc.png", null, "https://eurekamathanswers.com/wp-content/uploads/2022/01/y-mxc.png", null, "https://eurekamathanswers.com/wp-content/uploads/2022/01/Slope-of-the-graph-y-mxc_1-300x293.png", null ]
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https://networkx.org/documentation/networkx-2.3/reference/algorithms/generated/networkx.algorithms.bipartite.redundancy.node_redundancy.html
[ "Warning\n\nThis documents an unmaintained version of NetworkX. Please upgrade to a maintained version and see the current NetworkX documentation.\n\n# networkx.algorithms.bipartite.redundancy.node_redundancy¶\n\nnode_redundancy(G, nodes=None)[source]\n\nComputes the node redundancy coefficients for the nodes in the bipartite graph G.\n\nThe redundancy coefficient of a node v is the fraction of pairs of neighbors of v that are both linked to other nodes. In a one-mode projection these nodes would be linked together even if v were not there.\n\nMore formally, for any vertex v, the redundancy coefficient of v is defined by\n\n$rc(v) = \\frac{|\\{\\{u, w\\} \\subseteq N(v), \\: \\exists v' \\neq v,\\: (v',u) \\in E\\: \\mathrm{and}\\: (v',w) \\in E\\}|}{ \\frac{|N(v)|(|N(v)|-1)}{2}},$\n\nwhere N(v) is the set of neighbors of v in G.\n\nParameters: G (graph) – A bipartite graph nodes (list or iterable (optional)) – Compute redundancy for these nodes. The default is all nodes in G. redundancy – A dictionary keyed by node with the node redundancy value. dictionary\n\nExamples\n\nCompute the redundancy coefficient of each node in a graph:\n\n>>> import networkx as nx\n>>> from networkx.algorithms import bipartite\n>>> G = nx.cycle_graph(4)\n>>> rc = bipartite.node_redundancy(G)\n>>> rc\n1.0\n\n\nCompute the average redundancy for the graph:\n\n>>> import networkx as nx\n>>> from networkx.algorithms import bipartite\n>>> G = nx.cycle_graph(4)\n>>> rc = bipartite.node_redundancy(G)\n>>> sum(rc.values()) / len(G)\n1.0\n\n\nCompute the average redundancy for a set of nodes:\n\n>>> import networkx as nx\n>>> from networkx.algorithms import bipartite\n>>> G = nx.cycle_graph(4)\n>>> rc = bipartite.node_redundancy(G)\n>>> nodes = [0, 2]\n>>> sum(rc[n] for n in nodes) / len(nodes)\n1.0\n\nRaises: NetworkXError – If any of the nodes in the graph (or in nodes, if specified) has (out-)degree less than two (which would result in division by zero, according to the definition of the redundancy coefficient).\n\nReferences\n\n Latapy, Matthieu, Clémence Magnien, and Nathalie Del Vecchio (2008). Basic notions for the analysis of large two-mode networks. Social Networks 30(1), 31–48." ]
[ null ]
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http://hltd.org/lib/exponential_growth.htm
[ "# exponential growth\n\nExponential Growth\n\nA model for growth of a quantity for which the rate of growth is directly proportional to the amount present. The equation for the model is A = A0bt (where b > 1 ) or A = A0ekt (where k is a positive number representing the rate of growth). In both formulas A0 is the original amount present at time t = 0.\n\nThis model is used for such phenomena as inflation or population growth. For example, A = 7000e0.05t is a model for the exponential growth of \\$7000 invested at 5% per year compounded continuously." ]
[ null ]
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https://www.tutorialspoint.com/how-to-find-the-number-of-times-a-variable-changes-its-sign-in-an-r-data-frame-column
[ "# How to find the number of times a variable changes its sign in an R data frame column?\n\nR ProgrammingServer Side ProgrammingProgramming\n\nTo find the number of times a variable changes its sign in an R data frame column, we can use sign function with diff and sum function.\n\nFor example, if we have a data frame called df that contains a column say C then, we can find the number of times C changes its sign by using the following command −\n\n## Output\n\nIf you execute all the above given snippets as a single program, it generates the following output −\n\n 9\n\n\n## Example 2\n\nFollowing snippet creates a sample data frame −\n\ny<-sample(-2:2,20,replace=TRUE)\ndf2<-data.frame(y)\ndf2\n\nThe following dataframe is created −\n\n    y\n1  -1\n2   0\n3  -1\n4   2\n5   2\n6   0\n7   1\n8   1\n9   1\n10  2\n11  0\n12 -2\n13 -1\n14 -2\n15 -1\n16  2\n17  0\n18  1\n19  1\n20 -2\n\nTo find the number of times y changes its sign, add the following code to the above snippet −\n\ny<-sample(-2:2,20,replace=TRUE)\ndf2<-data.frame(y)\n\n## Output\n\nIf you execute all the above given snippets as a single program, it generates the following output −\n\n 8" ]
[ null ]
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https://au.mathworks.com/matlabcentral/cody/problems/1807-04-scalar-equations-2/solutions/489963
[ "Cody\n\n# Problem 1807. 04 - Scalar Equations 2\n\nSolution 489963\n\nSubmitted on 24 Aug 2014 by bainhome\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1   Pass\n%% a=10; b=2.5e23; ref = (sqrt(a)+b^(1/21))^pi; user = MyFunc(); assert(isequal(user,ref))\n\nans = 6.2696e+03\n\n2   Pass\n%% [y a b] = MyFunc(); assert(a==10);\n\nans = 6.2696e+03\n\n3   Pass\n%% [y a b] = MyFunc(); assert(b==2.5e23);\n\nans = 6.2696e+03\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!" ]
[ null ]
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https://www.practically.com/studymaterial/blog/docs/class-7th/maths/fractions-and-decimals/
[ "# Fractions and Decimals\n\n1. FRACTIONS RECALL\nFractions: The number of the form $\\frac{a}{b}$; where ‘a’ and ‘b’ are whole numbers and $b\\ne 0$.\nExample: $\\frac{2}{5}$ is a fraction in which numerator is 2 and denominator is 5.\nTypes of Fractions\nSimple Fraction\nA fraction of the form $\\frac{a}{b}$ , where ‘a’ and ‘b’ are whole numbers and $b\\ne 0$.\nExample:\nComplex Fraction\nA fraction of the form $\\frac{x}{y}$ where ‘x’ and ‘y’ are fractions.\nwe define $\\frac{\\left(\\frac{a}{b}\\right)}{\\left(\\frac{a}{d}\\right)}=\\frac{a}{b}÷\\frac{c}{d}=\\frac{a}{b}×\\frac{d}{c}$\n2. CLASSIFICATION OF FRACTIONS\nDecimal Fraction: Is a fraction whose denominator is 10, 100, 1000 etc.,\nExample:\nVulgar Fraction: Is a fraction whose denominator is a whole number other than 10, 100, 1000 etc.,\nExample:\nProper Fraction: Is a fraction where the numerator of a fraction is less than its denominator.\nExample:\n\n• Improper Fraction: Is a fraction where the numerator of a fraction is greater than or equal to its denominator.\n\nExample :\n\n• Mixed Number: A number which can be expressed as the sum of a natural number and a proper fraction is called a mixed number.\n\nExample :\n\nConversion of mixed number to an Improper Fraction\nMultiply the denominator of the proper fraction with integral part and add numerator of the proper fraction to it. This gives the numerator of the improper fraction and its denominator is the same as that of proper fraction.\nExample: $8\\frac{11}{13};8\\frac{11}{13}-\\frac{8×13+11}{13}-\\frac{104+11}{13}=\\frac{115}{13}$\n\nConversion of Improper Fraction to a Mixed Number\nDivide the numerator of the given fraction by its denominator. The quotient so obtained forms the whole number part and the remainder forms the numerator of the fractional part of the mixed number.\nExample: $\\frac{14}{3}$\n\nOn dividing 14 by 3, quotient = 4 ; remainder = 2\n\n$\\therefore \\frac{4}{13}=4\\frac{2}{3}$\n\nEquivalent Fractions\nA given fraction and the fraction obtained by multiplying (or dividing) both of its numerator and\ndenominator by the same non-zero number are called equivalent fractions.\n\nExample:\n\n$\\therefore \\frac{2}{3},\\frac{4}{6},\\frac{6}{9},\\frac{8}{12}$ etc are equivalent fractions\n$\\frac{16}{24}-\\frac{16÷8}{24÷8}-\\frac{2}{3}$ etc are equivalent fractions\n\nLike Fractions: are fractions having the same denominators but different numerators.\n\nExample: $\\frac{3}{5},\\frac{4}{5},\\frac{6}{5},\\frac{11}{5}$are like fractions\n\nUnlike Fractions: are the fractions having different denominators.\n\nExample: $\\frac{2}{3},\\frac{3}{5},\\frac{7}{11},\\frac{13}{17}$ are unlike fractions.\n\nConversion of Unlike to Like Fractions\nStep1: Find the L.C.M of the denominators of all the given fractions.\nStep2: Change each given fraction into an equivalent fraction having denominator as the L.C.M of denominators of the given fractions.\nExample: Convert $\\frac{3}{5},\\frac{5}{7}$  in to like fractions\nTake L.C.M of denominators L.C.M of 5, 7 is 35.\n\nClearly\n\n$\\begin{array}{l}\\frac{3×7}{5×7}=\\frac{21}{35}\\\\ \\frac{5×5}{7×5}=\\frac{25}{35}\\end{array}$\n\nNow $\\frac{21}{35},\\frac{25}{35}$ are like fractions\n\nComparison of Fractions\nThere are two methods to compare fractions.\nMethod 1: Cross Product Method:\n\nTwo fractions may be compared as under cross multiply as show $\\frac{a}{b}×\\frac{c}{d}$, then\n\nExample: compare $\\frac{5}{6},\\frac{7}{8}$\n\nCross multiply as shown $\\frac{5}{6}×\\frac{7}{8}$\n\nWe have 5 × 8 = 40 and 6 × 7 = 42\n\n$42>40\\phantom{\\rule{1em}{0ex}}\\therefore \\frac{7}{8}>\\frac{5}{6}$\n\nMethod2: Convert given fractions in to like fractions and then compare numerators.\nExample: Compare $\\frac{5}{6},\\frac{7}{8}$\n\nL.C.M of 6,8 is 48\n\n$\\begin{array}{l}\\frac{5×8}{6×8}=\\frac{40}{48} \\\\ \\frac{7×6}{8×6}=\\frac{42}{48}\\\\ 42>40 \\\\ \\therefore \\frac{7}{8}>\\frac{5}{6}\\end{array}$\n\n3. MULTIPLICATIONS OF FRACTIONS\nYou know how to find the area of a rectangle. It is equal to length × breadth. If the length and breadth of a rectangle are 7 cm and 4 cm respectively, then what will be its area? Its area would be .\n\nWhat will be the area of the rectangle if its length and breadth are respectively? You will say it will be $7\\frac{1}{2}×3\\frac{1}{2}=\\frac{15}{2}×\\frac{7}{2}{\\mathrm{cm}}^{2}$. The numbers are fractions.\nTo calculate the area of the given rectangle, we need to know how to multiply fractions. We shall learn that now.\nMultiplication of a Fraction by a Whole Number", null, "Observe the pictures at the left. Each shaded part is $\\frac{1}{4}$ part of a circle. How much will the two shaded parts represent together?\n\nThey will represent $\\frac{1}{4}+\\frac{1}{4}=2×\\frac{1}{4}$\n\nTo multiply a mixed fraction to a whole number, first convert the mixed fraction to an improper fraction and then multiply.\nTherefore, $3×2\\frac{5}{7}=3×\\frac{19}{7}=\\frac{57}{7}=8\\frac{1}{7}$\n\n4. MULTIPLICATION OF A FRACTION BY A FRACTION\nFarida had a 9 cm long strip of ribbon. She cut this strip into four equal parts. How did she do it? She folded the strip twice. What fraction of the total length will each part represent?\nEach part will be $\\frac{9}{4}$ of the strip. She took one part and divided it in two equal parts by folding the part once. What will one of the pieces represent? It will represent\n\n5. DIVISION OF FRACTIONS\nJohn has a paper strip of length 6 cm. He cuts this strip in smaller strips of length 2 cm each. You know that he would get 6 $÷$ 2 = 3 strips.\nJohn cuts another strip of length 6 cm into smaller strips of length $\\frac{3}{2}cm$ each.\nHow many strips will he get now? He will get $6÷\\frac{3}{2}$ strips\n\nA paper strip of length $\\frac{15}{2}cm$ can be cut into smaller strips of length $\\frac{3}{2}cm$ each to give $\\frac{15}{2}÷\\frac{3}{2}$pieces.\n\nSo, we are required to divide a whole number by a fraction or a fraction by another fraction.\nLet us see how to do that.\nReciprocal of a Fraction\nThe reciprocal of a non-zero fraction\n\nExample: Reciprocal of\n\nDivision of Whole Number by a Fraction\nLet us find $1÷\\frac{1}{2}$\n\nWe divide a whole into a number of equal parts such that each part is half of the whole.\nThe number of such half $\\left(\\frac{1}{2}\\right)$ parts would be Observe the figure. How many half parts do you see? There are two half parts.", null, "What will be $\\frac{3}{4}÷3?$\n\nBased on our earlier observations we have: $\\frac{3}{4}÷3=\\frac{3}{4}÷\\frac{3}{1}=\\frac{3}{4}×\\frac{1}{3}=\\frac{3}{12}=\\frac{1}{4}$\n\nDivision of a Fraction by Another Fraction\nWe can now find $\\frac{1}{3}÷\\frac{5}{6}$\n\nA Fraction Lying between two given Fractions\n\nIf  are two fractions, then the fraction $\\frac{a+c}{b+d}$ lies between , thus $\\frac{a}{b}<\\frac{a+c}{b+d}<\\frac{c}{d}$ .\n\n6. DECIMALS RECALL\nA fraction whose denominator is 10 or some integral power of 10, is called a decimal fraction.\nExample:\n\nThese decimal fractions can be written in the decimal form as 0.2, 0.23, 0.234, 0.2345 etc\nDecimals: The number written in the decimal form are called decimal numbers or decimals.\nThus each of the numbers 0.2, 0.23, 0.234, 0.2345 is a decimal.\n• A decimal has two parts. Whole number part and decimal part. These parts are separated by a dot (.) is called decimal point.\n• The digits lying to the left of the decimal point form the whole number part.\n• The decimal point together with digits lying to its right form the decimal part.\nEg: In the decimal number 23.574; Whole number part = 23 & Decimal part = 574\nDecimal Places:\nThe number of digits contained in the decimal part of a decimal gives the number of its decimal places.\nExample:\n\n• 5.89 has 2 decimal places\n• 23.758 has 3 decimal places\nLike Decimals: Decimals having the same number of decimal places are called like decimals.\nExample:\n\n• 5.1, 77.3, 109.1, 1009.0 are like decimals; each having one decimal place\n• 0. 85, 23.58, 134.72, 1000.89 are like decimals; each having two decimal places\nUnlike Decimals: Some given decimals, all not having the same number of decimal places are called unlike decimals.\nExample:\n\n• 5.75, 57.5 are unlike decimals\n\n7. COMPARISON OF DECIMALS\n• If two decimals having greatest Integral part is greater.\nExample:\n\n45.85, 54.75\nSince 54 > 45; hence 54.75 > 45.85\n• If two decimals having same integral part, then compare the decimal number part which is having greatest decimal part is greater.\nExample:\n\n23.58, 23.78\n58, 78 are in decimal part of these two\nIn this 78 > 58; hence 23.78 > 23.58\nConversion of Unlike to Like Decimals\nAnnexing zeroes to the extreme right of the decimal part of a decimal does not change its value.\nExample:\n\nDecimal to Fraction Conversion\nWrite the given decimal without the decimal point as numerator. Take ‘1’ annexed with as many zeroes as is the number of decimal places in the given decimal as denominator.\n\nExample:\n\n$\\begin{array}{l}23.4=\\frac{234}{10}\\\\ 23.45=\\frac{2345}{100}\\\\ 23.456=\\frac{23456}{1000}\\end{array}$\n\nConversion of Fraction into a Decimal by Division Method:\nProcedure:\nStep1. Divide the numerator by the denominator\nStep2. Complete the division. Let a non-zero remainder be left\nStep3. Insert a decimal point in the dividend and the quotient.\nStep4. Part a zero on the right of the decimal point in the dividend as well as on the right of the\nremainder. Divide again just as whole numbers.\nStep5. Repeat step4 still either the remainder is zero or requisite number of decimal places have been obtained.\n\n8. TYPES OF DECIMALS\nTerminating Decimals: In the process of converting a fraction into a decimal by the division method. If we obtain a zero remainder after a certain number of steps, then the decimal obtained is a terminating decimal.\nExample: 3.175 is a terminating decimal.\nHowever, there are situations where the division process continues indefinitely and zero remainder is never obtained. Such decimals are known as non-terminating decimals.\nExample: $\\frac{1}{19}=0.5263157\\dots$\n\nRepeating or Recurring Decimals: If in a decimal, a digit or a set of digits in the decimal part is repeated continuously, then such a number is called a recurring or repeating decimal. In a recurring decimal, if a single digit is repeated, then it is expressed by put a dot on it. If a set of digits is repeated, it is expressed by putting a bar on the set.\n\nExample:\n\n$\\begin{array}{l}\\frac{4}{9}-0.444\\dots -0.\\overline{4}\\\\ \\frac{22}{7}=3.142857142857\\dots \\dots =3.\\overline{142857}\\end{array}$\n\nPure Recurring Decimal: Is a decimal in which all the digits in the decimal part are repeated.\nExample:\n\nMixed Recurring Decimal: Is a decimal in which some of the digits in the decimal part are repeated and the rest are not repeated.\nExample:\n\nZero of the Decimal: The value of a decimal number is not altered when zeros are replaced at the end of a number.\n\nExample:\n8 can be written as 8, 8.0, 8.00, 8.000………\nThe number 34.5 can be written as 34.50, 34.500,34.5000………\n\n9. OPERATION ON DECIMALS\n• Convert all the given decimals into like decimals.\n• Write the decimals one under the other with decimal points of all the addends in the same column.\n• Add as in the case of whole numbers.\n• In the sum, put decimal point directly under the decimal points of the addends.\nExample:\n\n$5.700+8.780+0.352$\n\n$5.700\\phantom{\\rule{0ex}{0ex}}8.780\\phantom{\\rule{0ex}{0ex}}0.352\\phantom{\\rule{0ex}{0ex}}––––\\phantom{\\rule{0ex}{0ex}}14.832$\n\nSubtraction of Decimals\n• Convert the given decimals into like decimals.\n• Write the smaller number under the larger one in such a way that the decimal points of both the numbers are in the same column.\n• Subtract as in the case of whole numbers.\n• In the difference, put the decimal point directly under the decimal points of the given numbers.\nExample: Solve 100.08 – 98.8\n\n100.08 – 98.80\n\nMultiplication of Decimals: Multiplication of a decimal by 10,100,1000 etc., when a decimal is multiplied by some power of 10, then the decimal point is shifted to the right by as many digits as there are zeroes in the multiplier.\nExample: 10.999 $×$10 = 109.99\n(shifted decimal point 1 place to the right)\n10.999 $×$ 100 = 1099.9\n(shifted decimal point 2 places to the right)\nDivision of Decimals: To divide a decimal by 10,100,1000 etc., shift the decimal point to the left as many places as is the number of zeroes in the divisor.\nExample: Divide $15.3÷10$\n\n$15.3÷10=1.53$ (shifting decimal point one place to the left)\nDividing a Decimal by a Whole Number\nStep1: Perform the division by considering the dividend a whole number\nStep2: When the division of whole number part of the dividend is complete, put the decimal point in the quotient and proceed with the division as in case of whole numbers.\nExample: $2.345÷10$\n\n$2.345÷10=0.2345$ (shifting decimal point one place to the left)\n\nDividing a Decimal by a Decimal\nStep 1: Convert the divisor into a whole number by multiplying the dividend and divisor by a suitable power of 10.\nStep 2: Divide the new dividend by the whole number obtained above\nExample: Divide 55.50 by 5.55\n\n$\\frac{55.50×100}{5.55×100}=\\frac{5550}{555}=10$" ]
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https://chemistry.stackexchange.com/questions/31601/formation-of-species-in-electrolysis/31609
[ "# Formation of species in electrolysis\n\nI understand the principles of electrolysis of salts in aqueous solutions, but there are two points on which i am unsure.\n\n• At the positive electrode, how can you work out which ions will preferentially donate electrons?\n\n• If the formation of hydroxide ions and hydrogen gas occurs at the negative electrode, can hydrogen ions and oxygen gas be formed instead?\n\n• If so, what decides which ions are formed?\n• What happens to the ions formed at the negative electrode? Do they react with the cations that are attracted to the electrode or do they travel to and react with the positive electrode?\n\nFor the last question, I believe the latter would be true because if the anions produced at the negative electrode react with the cations that are attracted to the electrode, the cations that are attracted to the electrode would not be available to accept electrons and hence the current would become increasingly restricted until it stops all together. Could someone negate or confirm this reasoning?\n\n• I would love to answer this question if you could make it neater and easier to read. Please bullet what you don't understand and why. Thanks. – Asker123 May 17 '15 at 19:39\n• I edited in some legibility, but the questions might be a bit broad. – tschoppi May 17 '15 at 19:41\n• just changed it for clarity – ziggy May 17 '15 at 19:44\n• Alright, I will begin. – Asker123 May 17 '15 at 19:58\n\nFor the last question, I believe the latter would be true because if the anions produced at the negative electrode react with the cations that are attracted to the electrode, the cations that are attracted to the electrode would not be available to accept electrons and hence the current would become increasingly restricted until it stops all together. Could someone negate or confirm this reasoning?\n\nYou need to do more research to understand what an electrolytic and galvanic cell is and how it works. Especially salt bridges. A couple of links to check out would be.\n\nDon’t be limited to those above, keep searching until you solve your dilemma.\n\n## Now for the Hard Part\n\nTo best explain this I will work you through a practice problem. It is not a perfect algorithm to write down, but this way you will get a general idea.\n\nTwo electrodes are inserted into a solution of $\\ce{NiF2}$ and a current of $2.2\\ \\mathrm{A}$ is run through them. A list of standard reduction potentials is as follows.\n\n\\begin{alignat}{2} \\ce{O2_{(g)} + 4H+ + 4e- &-> H2O_{(l)}}\\quad&&1.23\\ \\mathrm{V}\\\\ \\ce{F2_{(g)} + 2e- &-> 2F-}\\quad&&2.87\\ \\mathrm{V}\\\\ \\ce{2H2O{(l)} + 2e- &-> H2_{(g)} + 2OH-}\\quad&&{-}0.83\\ \\mathrm{V}\\\\ \\ce{Ni^2+ + 2e- &-> Ni_{(s)}}\\quad&&{-}0.25\\ \\mathrm{V} \\end{alignat}\n\nSo generally I would usually start of with determining with what is going on, you know that $\\ce{Ni^2+}$ and $\\ce{F-}$ are present in the solution ($\\ce{H2O}$).\n\nWe can eliminate the third reaction since fluorine already exists reduced, and it does not like being oxidized. Then you have nickel and water, which of them is most likely to be reduced? Nickel.\n\nNow we must decide the compound being oxidized. Fluorine and nickel are out of the question so that leaves us with Equation 1 and 3. As you can see even if we were to flip Eq. 3 it would make no sense since there is no $\\ce{OH-}$ present in water. So that leaves us with Eq. 1. We must flip this equation for the reaction to make sense. That leaves us with:\n\nEquation 1 (this one is flipped) and 4. $\\ce{Ni^2+}$ is being reduced while $\\ce{H2O}$ is being oxidized.\n\nIn general …\n\n1. Eliminate which reaction cannot take place.\n2. Look at what is left, and judge which reaction can best represent the reaction taking place in the cell.\n\nWith practice you will be able to solve these kinds of problems. If this is not what you are looking for, then hopefully you learned about these kinds of problems. :)\n\nAt the positive electrode, how can you work out which ions will preferentially donate electrons?\n\nIn an electrolytic cell, the positive electrode is called the anode, and the oxidation half of the reaction occurs here. The oxidation reactions which have less negative standard oxidation potential are more likely to occur, though most resources give the standard reduction potentials. Wikipedia is one such resource.\n\nIf you wish to know which species are more likely to donate electrons, these will be the half reactions that have the less negative standard reduction potentials.\n\nIf the formation of hydroxide ions and hydrogen gas occurs at the negative electrode, can hydrogen ions and oxygen gas be formed instead?\n\nI wrote a response to this exact question here.\n\nWhat happens to the ions formed at the negative electrode? Do they react with the cations that are attracted to the electrode or do they travel to and react with the positive electrode?\n\nIn an electrolytic cell, the negative electrode is called the cathode, and the reduction half of the reaction occurs here. When an species is reduced, typically it changes phases (goes from aqueous ion to being a gas, a liquid, or a solid). Most likely if an ion is reduced to another ion here, this ion will then also be reduced." ]
[ null ]
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https://www.bartleby.com/solution-answer/chapter-30-problem-59pe-college-physics-1st-edition/9781938168000/integrated-concepts-find-the-value-of-l-the-orbital-angular-momentum-quantum-number-for-the-moon/cd5172d0-7def-11e9-8385-02ee952b546e
[ "", null, "", null, "", null, "Chapter 30, Problem 59PE\n\nChapter\nSection\nTextbook Problem\n\nIntegrated ConceptsFind the value of l , the orbital angular momentum quantum number, for the moon around the earth. The extremely large value obtained implies that it is impossible to tell the difference between adjacent quantized orbits for macroscopic objects.\n\nTo determine\n\nThe value of l , the orbital angular momentum quantum number, for the moon around the earth.\n\nExplanation\n\nGiven Data:\n\nGiven that there is an Earth and its moon\n\nFormula Used:\n\nThe angular momentum is given as\n\nL=l(l+1)h2π\n\nWhere l= Orbital angular momentum quantum number\n\nh= Planck's Constant\n\nAlso, the velocity of moon, during its movement around the Earth is calculated as\n\nv=2πRT\n\nWhere R= Distance of center of moon from center of Earth\n\nT= Time Period of Moon's revolution\n\nCalculation:\n\nAngular momentum of moon is calculated as\n\nL=mRv\n\nSubstituting v=2πRT in above, we get\n\nL=mR2πRT\n\nL=2πmR2T\n\nAlso, we know L=l(l+1)h2π\n\nEquating the above two equation, we get\n\n2πmR2T=l(l+1)h2π\n\nAlso, the value of l would be very large\n\nThus, we have\n\nStill sussing out bartleby?\n\nCheck out a sample textbook solution.\n\nSee a sample solution\n\nThe Solution to Your Study Problems\n\nBartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!\n\nGet Started", null, "" ]
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https://www.gla.ac.uk/postgraduate/taught/mathematicsappliedmathematics/?card=course&code=MATHS5071
[ "# Mathematics / Applied Mathematics MSc\n\n## 5E: Galois Theory MATHS5071\n\n• School: School of Mathematics and Statistics\n• Credits: 10\n• Level: Level 5 (SCQF level 11)\n• Typically Offered: Semester 2\n• Available to Visiting Students: No\n• Available to Erasmus Students: No\n\n### Short Description\n\nThis second part of an algebra sequence centres on Galois Theory, which arose from the search for explicit formulae for roots of polynomial equations. The primary focus of the course is the Galois Correspondence Theorem, relating the structure of fields to the structure of groups, and its applications to polynomials.\n\n### Timetable\n\n17 x 1 hr lectures and 6 x 1 hr tutorials in a semester\n\n### Excluded Courses\n\n4H: Galois Theory (MATHS4105)\n\n### Assessment\n\nAssessment\n\n90% Examination, 10% Coursework.\n\nReassessment\n\nResit opportunities available for MSc students\n\nMain Assessment In: April/May\n\n### Course Aims\n\nThis course focuses on work of Galois which led to a satisfactory framework for fully understanding the fact that the general polynomial equation of degree at least 5 could not always be solved `by radicals' (i.e., with formulae involving n-th roots, analogous to the well-known Quadratic Formula). Further, the so-called Galois Correspondence allows the structure of fields to be related to the structure of groups, a powerful idea with applications in many fields of mathematics. Topics to be covered in this course include unique factorization domains, principal ideal domains, euclidean domains, field extensions, algebraic extensions, algebraic closure, separable and normal extensions, Galois extensions, finite fields, cyclotomic fields, symmetric functions, solvability, and separable closures of a field, as time permits.\n\n### Intended Learning Outcomes of Course\n\nBy the end of this course students will be able to:\n\na) state and apply criteria for determining whether a given polynomial is irreducible;\n\nb) find the irreducible polynomial of an element over a field;\n\nc) compute the degree of a field extension;\n\nd) determine the group of automorphisms of an extension field over a base field;\n\ne) describe and calculate values of the Frobenius automorphism;\n\nf) state, prove, and apply the Isomorphism Extension Theorem;\n\ng) calculate the index of a field extension, and prove it divides the degree;\n\nh) decide whether a given polynomial splits in an extension field;\n\ni) prove that the roots of a polynomial all have the same multiplicity in an algebraic closure;\n\nj) state and prove the Primitive Element Theorem;\n\nk) construct the separable closure of a field in a totally inseparable field extension\n\nl) state, prove, and apply the Galois Correspondence Theorem (aka the Fundamental or Main Theorem of Galois Theory);\n\nm) prove that every principal ideal domain is a unique factorization domain;\n\nn) perform arithmetic in Euclidean domains.\n\n### Minimum Requirement for Award of Credits\n\nStudents must submit at least 75% by weight of the components (including examinations) of the course's summative assessment." ]
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https://metanumbers.com/1072748
[ "# 1072748 (number)\n\n1,072,748 (one million seventy-two thousand seven hundred forty-eight) is an even seven-digits composite number following 1072747 and preceding 1072749. In scientific notation, it is written as 1.072748 × 106. The sum of its digits is 29. It has a total of 4 prime factors and 12 positive divisors. There are 534,192 positive integers (up to 1072748) that are relatively prime to 1072748.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Even\n• Number length 7\n• Sum of Digits 29\n• Digital Root 2\n\n## Name\n\nShort name 1 million 72 thousand 748 one million seventy-two thousand seven hundred forty-eight\n\n## Notation\n\nScientific notation 1.072748 × 106 1.072748 × 106\n\n## Prime Factorization of 1072748\n\nPrime Factorization 22 × 373 × 719\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 536374 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 1,072,748 is 22 × 373 × 719. Since it has a total of 4 prime factors, 1,072,748 is a composite number.\n\n## Divisors of 1072748\n\n12 divisors\n\n Even divisors 8 4 2 2\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 12 Total number of the positive divisors of n σ(n) 1.88496e+06 Sum of all the positive divisors of n s(n) 812212 Sum of the proper positive divisors of n A(n) 157080 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1035.74 Returns the nth root of the product of n divisors H(n) 6.82931 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 1,072,748 can be divided by 12 positive divisors (out of which 8 are even, and 4 are odd). The sum of these divisors (counting 1,072,748) is 1,884,960, the average is 157,080.\n\n## Other Arithmetic Functions (n = 1072748)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 534192 Total number of positive integers not greater than n that are coprime to n λ(n) 133548 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 83584 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 534,192 positive integers (less than 1,072,748) that are coprime with 1,072,748. And there are approximately 83,584 prime numbers less than or equal to 1,072,748.\n\n## Divisibility of 1072748\n\n m n mod m 2 3 4 5 6 7 8 9 0 2 0 3 2 5 4 2\n\nThe number 1,072,748 is divisible by 2 and 4.\n\n• Arithmetic\n• Deficient\n\n• Polite\n\n## Base conversion (1072748)\n\nBase System Value\n2 Binary 100000101111001101100\n3 Ternary 2000111112102\n4 Quaternary 10011321230\n5 Quinary 233311443\n6 Senary 34554232\n8 Octal 4057154\n10 Decimal 1072748\n12 Duodecimal 438978\n20 Vigesimal 6e1h8\n36 Base36 mzqk\n\n## Basic calculations (n = 1072748)\n\n### Multiplication\n\nn×y\n n×2 2145496 3218244 4290992 5363740\n\n### Division\n\nn÷y\n n÷2 536374 357583 268187 214550\n\n### Exponentiation\n\nny\n n2 1150788271504 1234505816679372992 1324313645831164018422016 1420654814938089538434180819968\n\n### Nth Root\n\ny√n\n 2√n 1035.74 102.368 32.1828 16.0731\n\n## 1072748 as geometric shapes\n\n### Circle\n\n Diameter 2.1455e+06 6.74027e+06 3.61531e+12\n\n### Sphere\n\n Volume 5.17109e+18 1.44612e+13 6.74027e+06\n\n### Square\n\nLength = n\n Perimeter 4.29099e+06 1.15079e+12 1.51709e+06\n\n### Cube\n\nLength = n\n Surface area 6.90473e+12 1.23451e+18 1.85805e+06\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 3.21824e+06 4.98306e+11 929027\n\n### Triangular Pyramid\n\nLength = n\n Surface area 1.99322e+12 1.45488e+17 875895\n\n## Cryptographic Hash Functions\n\nmd5 22bd8e2a695feccae371b5a735dc40b2 e7d271e557ad7e604c3775fdd94c452dbb542638 81176e2969593b9a81c99aa3a7a71648c0458d593045b56d2b15e35f368bb227 f0204b0b4c1efb2986f8446ae4cfe39608ffb2e208b523e0c33f70414bb79f14cb7b7fedaf674871207513dcd59c63e647ce10acaf16ccabb981c1989326c80c d5b2534b9cb1e0a88c96f385637830008a271952" ]
[ null ]
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http://zreentoyz.blogspot.com/2012/05/math-function.html
[ "Monday, May 28, 2012\n\nA Math Function\n\nLearning math becomes conceptually difficult. Due to the a math function of the a math function, subtraction, multiplication, division, and fractions. Each game plays like dominos and teaches like a game. Learning math becomes conceptually difficult. Due to the a math function by the math concept.\n\nPractice is also extremely important for the a math function at home. Some of these problems plague you then look to the a math function and spent a major part of the a math function a sophisticated yet easy-to-use computer-based method designed to give the a math function for your child. Cooperate with him in his subject.\n\nLearners practice performing simple calculations, without the a math function a math puzzle lets you interact with your children. How many of you have trouble bringing math to score grades, math is a solution to this problem is easily available. A great way introduce math concepts when their child reaches middle school.\n\nAny good math grades, is your youngster simply becoming frustrated? Have you considered online math tutoring is invaluable, the tutoring should include varied plans suitable to the a math function to his classmates. Johnny knows his multiplication tables, so you can be accurate. Next, you are playing on a computer.\n\nIn a usual middle school situation, the a math function be able to generate interest for math. This will teach him addition multiplication or fragmentation. Tell him how he did it all. By today's standards, such an assignment would be acceptable. Even now math is often the a math function a child waste a whole year in anguished confusion, with the a math function in the a math function a variety of staff ensures that each student is unlikely to help, so use it sparingly and build little, home-made notebooks for daily use.\n\nSo what's the a math function is that it will help the a math function. For many teachers, one student helping another is called cheating. But I won't say 'learning math is very strong base for math and games are also educational math board game which helps children to easily begin to master basic math concepts with an entertaining and interactive way. When they enjoy what they are young can help them associate math with fun, no doubt you and give him candies to distribute among your family members. Tell him in his subject.\n\nWhy is this? What is wrong with struggling to understand that math is to use that not only make learning math skills. If your kids learn and retain new concepts better when the a math function a drill, which has 10 questions and are selected from a database of number pairs for calculation. The Basic Level volumes use simple single digit numbers and shapes, one great game to consider is GeoShapes. This games uses both Metric and English measuring systems, and strategy is part of the fastest growing occupations require math through algebra and even downright difficult for some. Since most expert educators agree that children tend to bore most students, especially those who need to learn seventh grade teacher knows what the students should have the a math function and all have substantial background in mathematics as a large one. Thus, the a math function of book publishing lead to less chance of success and benefits of teaching in a positive one can be problematic to children so you must know them too. Your brother never had any problem with math. These types of math facts. When ready to be more than I could not imagine the a math function to create great inventions and great engineering feats. We even put a man on the a math function. We accomplished all this because of our nation's ability to do this? If any of these difficult tasks with ease and with understanding. They are damaging and are valuable tools for students in math class for comparison. In a series of three articles on Math Anxiety, I will be able to generate interest for math practice problems at a pace that they love school but dislike their math skills. You can even find great math games that are available for kids of all ages and some electronic book form that can help parents teach children at home more effective for the a math function in teaching. Math manipulative are used as a post test or Test A as a post test or Test A as a springboard for discussions, discovery, and communication." ]
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https://sarpublisher.com/management-accounting-mcq-questions-and-answers-part-1/
[ "# Management Accounting MCQ Questions and Answers Part – 1\n\n12009\n\nManagement Accounting MCQ Questions and Answers Part – 1\n\nManagement Accounting MCQ Questions and Answers Part – 2\n\nManagement Accounting MCQ Questions and Answers Part – 3\n\n1. The ratios which reflect managerial efficiency in handling the assets is__________.\nA. turnover ratios.\nB. profitability atios.\nC. short term solvency ratio.\nD. long term solvency ratio.\n2. The ratios which reveal the final result of the managerial policies and performance is __________.\nA. turnover ratios.\nB. profitability ratios.\nC. short term solvency ratio.\nD. long term solvency ratio.\n3. Return on investment is a ________.\nA. turnover ratios.\nB. short term solvency ratio.\nC. profitability ratios.\nD. long term solvency ratio.\n4. Stock turnover ratio is a _____________.\nA. activity ratio.\nB. profitability ratio.\nC. short term solvency ratio.\nD. long term solvency ratio.\n5. The ratio which measures the profit in relation to capital employed is known as __________.\nA. return on investment.\nB. gross profit ratio.\nC. operating ratio.\nD. operating profit ratio.\n6. Prepaid expenses is an example of _________.\nA. fixed assets.\nB. current assets.\nC. fictitious assets.\nD. current liabilities.\n7. Turnover ratio is also known as ___________.\nA. activity ratios.\nB. solvency ratios.\nC. liquidity ratios.\nD. profitability ratios.\n8. Which ratio is calculated to ascertain the efficiency of inventory management?\nA. Stock velocity ratio.\nB. Debtors velocity ratio.\nC. Creditors velocity ratio.\nD. Working capital turnover ratio.\n9. P/V ratio means __________.\nA. Contribution/Sales\nB. Sales/Contribution\nC. Fixed cost/Sales\nD. Cost of goods sold/contribution\n10. Which ratio measures the number of times the receivables are rotated in a year in terms of sales?\nA. Stock turnover ratio.\nB. Debtors turnover ratio.\nC. Creditors velocity ratio.\nD. Working capital turnover ratio.\n11. The ratio which indicates the number of times the payables are rotated in a year is ____________.\nA. stock turnover ratio.\nB. stock turnover ratio.\nC. creditors velocity ratio.\nD. working capital turnover ratio.\n12. Current assets – current liabilities =_______.\nA. fixed capital.\nB. working capital.\nC. opening capital.\nD. closing capital.\n13. The ratio of current assets to current liabilities is called_______.\nA. liquid ratio.\nB. acid test ratio.\nC. current ratio.\nD. cash position ratio.\n14. Standard current ratio is____________.\nA. 1:1.\nB. 2:1.\nC. 3:1.\nD. 4:1.\n15. Current assets – (stock + prepaid expenses) =______________.\nA. current assets.\nB. fixed assets.\nC. liquid assets.\nD. fictitious assets.\n16. An ideal liquid ratio is_______.\nA. 0.25: 1.\nB. 0.50: 1.\nC. 0.75: 1.\nD. 1: 1.\n17. An ideal debt-equity ratio is________.\nA. 1\nB. 2\nC. 3\nD. 4\n18. Capital gearing ratio is also known as___________.\nA. leverage ratio.\nB. fixed assets turnover ratio.\nC. proprietary ratio.\nD. debt-equity ratio.\n19. Shareholders funds + Long-term loans =_____________.\nA. current assets.\nB. current liabilities.\nC. fixed assets.\nD. capital employed.\n20. Net capital employed is equal to _________.\nA. total assets minus total liabilities.\nB. fixed assets plus net-working capital.\nC. total assets minus long-term liabilities.\nD. total assets.\n21. All those assets which are converted into cash in the normal course of business within one year are known as _________.\nA. current assets.\nB. fixed assets.\nC. fictitious assets.\nD. wasting assets.\n22. All those liabilities which are payable in cash in the normal course of business within a period of one year\nare called _________.\nA. long term liabilities.\nB. overdraft.\nC. short term loans.\nD. current liabilities.\n23. Any transaction between a current item and another current item does not affect.\nA. profit.\nB. funds.\nC. working capital.\nD. capital.\n24. ‘Principle’ for preparation of working capital statement- Increase in current asset _______.\nA. increases working capital.\nB. decreases working capital.\nC. decrease fixed capital.\nD. increase fixed capital.\n25. ‘Principle’ for preparation of working capital statement – Decrease in current asset ______________.\nA. increases working capital.\nB. decreases working capital.\nC. decrease fixed capital.\nD. increase fixed capital.\n26. ‘Principle’ for preparation of working capital statement -Decrease in current liability _____________.\nA. increases working capital.\nB. decreases working capital.\nC. decrease fixed capital.\nD. increase fixed capital.\n27. ‘Principle’ for preparation of working capital statement -Increase in current liability ___________.\nA. increases working capital.\nB. decreases working capital.\nC. decrease fixed capital.\nD. increase fixed capital.\n28. Depreciation on fixed assets is _________.\nA. non operating income.\nB. operating expense.\nC. operating income.\nD. non operating expense.\n29. Provision for Income tax is ____________.\nA. non operating income.\nB. operating expense.\nC. operating income.\nD. appropriation of profits.\n30. Profit on sale of fixed assets is __________.\nB. operating income.\nD. long term gain.\n31. In fund flow statement, issue of shares is ________.\nA. sources of funds.\nB. applications of funds.\nC. sources of cash.\nD. applications of cash.\n32. In funds flow statement, sale of fixed assets is ___________.\nA. applications of funds.\nB. sources of cash.\nC. applications of cash.\nD. sources of funds.\n33. In funds flow statement, funds from operations is _________.\nA. applications of funds.\nB. sources of cash.\nC. applications of cash.\nD. sources of funds.\n34. In funds flow statement, increase in working capital is ________.\nA. applications of funds.\nB. sources of cash.\nC. applications of cash.\nD. sources of funds.\n35. In funds flow statement, decrease in working capital is _________.\nA. applications of funds.\nB. sources of cash.\nC. applications of cash.\nD. sources of funds.\n36. In funds flow statement, repayment of long-term loans is ________.\nA. applications of funds.\nB. sources of cash.\nC. applications of cash.\nD. sources of funds.\n37. In funds flow statement, purchase of fixed assets is _________.\nA. sources of cash.\nB. applications of funds.\nC. applications of cash.\nD. sources of funds.\n38. A cash flow statement is a statement which portrays the changes in the cash position between _______.\nA. two accounting periods.\nB. three accounting periods.\nC. four accounting periods.\nD. five accounting periods.\n39. The term ‘cash’ in the context of cash flow analysis includes the ‘cash balance’ and the __________.\nA. working capital.\nB. bank balance.\nC. capital.\nD. fixed assets\n40. Cash flow analysis is based on the ______.\nA. capital.\nB. fixed assets.\nC. cash concept of funds.\nD. working capital.\n41. For the calculation of trend percentage which year is considered 100 percent?\nA. first year\nB. second year\nC. third year\nD. last year\n42. Management accounting information is used by _______.\nA. management\nB. banks\nC. creditors\nD. government\n43. Financial statements are classified into _____ statements.\nA. four\nB. three\nC. two\nD. five\n44. Which one of the following is not a management accounting tool?\nA. Standard costing\nB. Marginal costing\nC. budgeting\nD. process costing\n45. Which one of the following is not mandatory according to the laws?\nA. Financial accounting\nB. cost accounting\nC. Management accounting\nD. none of the above\n46. If working capital is Rs. 1,00,000 and current ratio is 2:1, then the amount of current asset is _________.\nA. Rs.1,00,000\nB. Rs. 2,00,000\nC. Rs. 1,50,0000\nD. Rs. 2,50,000\n47. Types of financial analysis are ____.\nA. three\nB. four\nC. five\nD. two\n48. Which one of the following is not the determinant of the working capital?\nA. size of the firm\nB. operating cycle\nC. terms of credit\nD. competitors\n49. Permanent working capital ___\nA. will vary at all times\nB. will vary with volumes\nC. fixed at all times\nD. fluctuates according to the season", null, "" ]
[ null, "https://secure.gravatar.com/avatar/b903114c17f7df7e56c1fd5199222bc0", null ]
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http://ascl.net/code/cs/Leistedt%2C%20B
[ "# Searching for codes credited to 'Leistedt, B'\n\nTip! Refine or expand your search. Authors are sometimes listed as 'Smith, J. K.' instead of 'Smith, John' so it is useful to search for last names only. Note this is currently a simple phrase search.\n\n[ascl:1111.011]\n\nHigh precision cosmology requires analysis of large scale surveys in 3D spherical coordinates, i.e. Fourier-Bessel decomposition. Current methods are insufficient for future data-sets from wide-field cosmology surveys. 3DEX (3D EXpansions) is a public code for fast Fourier-Bessel decomposition of 3D all-sky surveys which takes advantage of HEALPix for the calculation of tangential modes. For surveys with millions of galaxies, computation time is reduced by a factor 4-12 depending on the desired scales and accuracy. The formulation is also suitable for pre-calculations and external storage of the spherical harmonics, which allows for further speed improvements. The 3DEX code can accommodate data with masked regions of missing data. It can be applied not only to cosmological data, but also to 3D data in spherical coordinates in other scientific fields.\n\n[ascl:1211.001]\n\nS2LET provides high performance routines for fast wavelet analysis of signals on the sphere. It uses the SSHT code built on the MW sampling theorem to perform exact spherical harmonic transforms on the sphere. The resulting wavelet transform implemented in S2LET is theoretically exact, i.e. a band-limited signal can be recovered from its wavelet coefficients exactly and the wavelet coefficients capture all the information. S2LET also supports the HEALPix sampling scheme, in which case the transforms are not theoretically exact but achieve good numerical accuracy. The core routines of S2LET are written in C and have interfaces in Matlab, IDL and Java. Real signals can be written to and read from FITS files and plotted as Mollweide projections.\n\n[ascl:1710.007]\n\nFLAG is a fast implementation of the Fourier-Laguerre Transform, a novel 3D transform exploiting an exact quadrature rule of the ball to construct an exact harmonic transform in 3D spherical coordinates. The angular part of the Fourier-Laguerre transform uses the MW sampling theorem and the exact spherical harmonic transform implemented in the SSHT code. The radial sampling scheme arises from an exact quadrature of the radial half-line using damped Laguerre polynomials. The radial transform can in fact be used to compute the spherical Bessel transform exactly, and the Fourier-Laguerre transform is thus closely related to the Fourier-Bessel transform.\n\n[ascl:1811.006]\n\nQuickSip quickly projects Survey Image Properties (e.g. seeing, sky noise, airmass) into Healpix sky maps with flexible weighting schemes. It was initially designed to produce observing condition \"systematics\" maps for the Dark Energy Survey (DES), but will work with any multi-epoch survey and images with valid WCS. QuickSip can reproduce the Mangle (ascl:1202.005) magnitude limit maps at sub-percent accuracy but doesn't support additional masks (stars, trails, etc), in which case Mangle should be used. Thus, QuickSip can be seen as a simplified Mangle to project image properties into Healpix maps in a fast and more flexible manner." ]
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https://www.numerade.com/questions/evaluate-the-limit-if-it-exists-displaystyle-lim_x-to-4fracx2-3xx2-x-12/
[ "💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!\n\n# Evaluate the limit, if it exists. $\\displaystyle \\lim_{x \\to 4}\\frac{x^2 + 3x}{x^2 - x - 12}$\n\n## Does not exist\n\nLimits\n\nDerivatives\n\n### Discussion\n\nYou must be signed in to discuss.\n##### Kristen K.\n\nUniversity of Michigan - Ann Arbor\n\n##### Samuel H.\n\nUniversity of Nottingham\n\n##### Michael J.\n\nIdaho State University\n\nLectures\n\nJoin Bootcamp\n\n### Video Transcript\n\nOkay, we want to find a limit of dysfunction as X approaches for now, when X equals for you would get four square at 16 minus four minus 12, which would end up being zero. So when X is four, the denominator here is zero and you can't divide by zero. But that's not a problem because when we take the limit of X approaching for X never actually gets to equal for X, just approaches for it's not going to equal for All right. Uh So what we're going to do to evaluate this limit? Um We're going to factor the numerator and we're going to factor the denominator and then see if we can cancel some common factors. So we are going to be taking the limit as X approaches for now looking at the numerator, X squared plus three X. Let's factor out the greatest common factor. Which would be X. So the numerator can be rewritten as X times X plus straight. If you distribute this multiplication, you can see X times X is X squared plus and then X times three is three X. All right, so that's going to get put over. Uh Well, X squared minus x minus 12. But we want to factor X squared minus x minus 12, X squared minus x minus 12. Well, factor Into X -4 times X plus. Straight. All right. So, we want to double check this. Uh you can use spoil x square there it is. Outer is three X. Inner is negative four X. They combined to give you a negative one X. There it is last term times last term negative four times three is negative 12. So X squared minus x minus 12 factors into x minus four times X plus straight. No, The X-plus three factors will cancel. Um we can cancel this X plus three and the denominator with the X plus three in the numerator. As long as X plus three uh is never equal to zero because you can't have a zero in the denominator, well, X plus three would equal zero when X is negative three, X is approaching four so we don't have to worry about X being negative three. So we cancel out these X plus three factors. So as X approaches for as X approaches for this, X term is going to approach for But this X -4 term Is going to approach zero. If X is approaching for an X -4 approaches for -4, which is zero. No, As X is approaching four from the right side of four, Meaning it's gonna be a little bit bigger than for X -4 is going to be a small positive number. For example, if x is 4.14 point one minus four is 40.1, So you would be dividing by .1. When you divide by a tiny number, you get a big number. So if you're dividing by a tiny positive number, you're going to get a large positive number. So as X approaches for when X approaches For from the right side of four from the positive side we say uh this limit is going to let's write it as X approaches for from the positive side, uh X Over X -4 is going to approach positive infinity. This is called the right hand limit. His ex is approaching four from the positive side. Uh This expression is going to approach positive infinity. Here's why as extras approaching for uh this X will approach for as extras approaching four from the positive side, the right side of four, meaning it's a little bit bigger than for something a little bit bigger than for -4 is still going to be positive. So for example, if once again, if X was like 4.1, -4 would be .1. And when we divide by .1 uh we're going to get large number. When you divide by a tiny number you get a large number dividing by .1 Is like times it by 10, dividing by .001 is like times and by 1000. So four divided by a tiny tiny positive number is going to be a very large positive number. So as X gets closer and closer to four from the right side of four, uh this expression is going to get larger and larger towards positive infinity. Now, however effects approaches for from the left side Of four, we call it from the negative side. For example, if X is approaching four from the left side, maybe X's 3.99 the next minus 43.99 minus four is negative 40.1. So then you got four being divided by negative 40.1, dividing by a tiny negative number, gives you a very large negative number. So as X approaches for from the left side, in this case We're going to have X over X -4 approaches negative infinity. So as we get closer to four from the left side of four, uh this expression is going to uh tend towards negative infinity, very large negative number. So when all said and done, this limit does not exist because as we approach four from the positive side, this expression goes to positive infinity. As we approach for from the left negative side, the expression goes towards negative infinity. So the limit does not exist. There is no number that this expression gets closer and closer to and also the right hand limit and the left hand limits do not agree. So this limit does not exist\n\nTemple University\n\n#### Topics\n\nLimits\n\nDerivatives\n\n##### Kristen K.\n\nUniversity of Michigan - Ann Arbor\n\n##### Samuel H.\n\nUniversity of Nottingham\n\n##### Michael J.\n\nIdaho State University\n\nLectures\n\nJoin Bootcamp" ]
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https://stats.stackexchange.com/questions/49465/mann-whitney-for-non-normal-distributions-with-n20/49474
[ "# Mann-Whitney for non-normal distributions with n>20?\n\nI have 12 groups of datasets (mostly non-Normal, quite right-skewed), and I would like to see if there is a significant difference within each pair. (But the data itself is not a paired dataset, as in dataset #1 and dataset #2 that I am comparing are from independent populations.)\n\nMy data also has outliers (following the 1.5*IQR rule), so I understand that the t-test is not valid because (x bar) and s are not robust against outliers. The main alternative I see is to use the rank test Mann-Whitney U test. (Let me know if I am not on the right track.)\n\nIn Zar's Biostatistical Analysis (4th ed, pg. Ap100): \"For the Mann-Whitney test involving n1 and/or n2 larger than those in this table, the normal approximation may be used.\" The table goes to n1=20, n2=40.\n\nIn Moore, McCabe, Craig's Introduction to the Practice of Statistics (6th ed., pg. 432): For sample sizes 15≤ n ≤39, \"t procedures can be used except in the presence of outliers or strong skewness.\" For samples sizes ≥40, \"t procedures can be used even for clearly skewed distributions.\"\n\nDoes this mean if n for one of my populations ≥20 (or ≥40?), I will have more accurate/powerful results using the 2-sample t-test (assuming unequal variance) instead? What if n1 ≤20 but n2 ≥20? If I need to use the Mann-Whitney U test for some, should I be using Mann-Whitney U test for all in order to compare the results? Are there other hypotheses tests I should be using instead?\n\nFor reference, the populations that I am comparing have counts:\n\nn1 = 15, n2 = 20\nn1 = 7, n2 = 27\nn1 = 12, n2 = 11\nn1 = 13, n2 = 4\nn1 = 22, n2 = 47\nn1 = 25, n2 = 15\nn1 = 20, n2 = 21\nn1 = 12, n2 = 27\nn1 = 22, n2 = 22\nn1 = 26, n2 = 14\nn1 = 32, n2 = 48\nn1 = 48, n2 = 36\n\n\nYour thoughts are appreciated. Thank you so much!\n\nFYI I have been using Excel's 2-sample t-test assuming unequal variance and Vassarstat's online Mann-Whitney test: http://vassarstats.net/index.html (under the heading Ordinal Data).\n\nt procedures can be used even for clearly skewed distributions.\n\nI don't think it's automatically reasonable to give a specific $n$ for this advice because there will be distributions/data sets that break it. Well, there might just be some sufficient $n$ but for me it doesn't kick in anywhere close to $n=40$.\n\nDoes this mean if n for one of my populations ≥20 (or ≥40?), I will have more accurate/powerful results using the 2-sample t-test (assuming unequal variance) instead?\n\nIn general, the answer is a very clear 'no'. The t tends to have reasonably good power relative to the MW for light-tailed distributions ... and can have really bad power for heavy-tailed ones. Skewness tends to be compounded with heavy tails - if power is your main motivation for using the t-test, you should probably avoid it in this case.\n\nThe Man-Whitney test should be sensitive to the sorts of departures you're trying to pick up (but note that scale and location will be confounded).\n\nAnother possibility is to do a permutation, randomization or bootstrap test.\n\nYou could probably control the skewness and outliers with a square root transformation of the counts and then use a t-test." ]
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https://www.maths.kisogo.com/index.php?title=The_set_of_all_open_balls_of_a_metric_space_are_able_to_generate_a_topology_and_are_a_basis_for_that_topology
[ "# The set of all open balls of a metric space are able to generate a topology and are a basis for that topology\n\nJump to: navigation, search\nGrade: A\nThis page is currently being refactored (along with many others)\nPlease note that this does not mean the content is unreliable. It just means the page doesn't conform to the style of the site (usually due to age) or a better way of presenting the information has been discovered.\nThe message provided is:\nThis is one of the oldest pages on the wiki, created in Feb 2015. With a slight revision in Apr 2015 when Template:Theorem was moved to Template:Theorem Of. A very old page indeed!\n\n## Statement\n\nLet [ilmath]X[/ilmath] be a set, let [ilmath]d:X\\times X\\rightarrow\\mathbb{R}_{\\ge 0} [/ilmath] be a metric on that set and let [ilmath](X,d)[/ilmath] be the resulting metric space. Then we claim:\n\n## Proof\n\nRecall the definition of a topology generated by a basis\n\nLet [ilmath]X[/ilmath] be a set and let [ilmath]\\mathcal{B}\\in\\mathcal{P}(\\mathcal{P}(X))[/ilmath] be any collection of subsets of [ilmath]X[/ilmath], then:\n\n• [ilmath](X,\\{\\bigcup\\mathcal{A}\\ \\vert\\ \\mathcal{A}\\in\\mathcal{P}(\\mathcal{B})\\})[/ilmath] is a topological space with [ilmath]\\mathcal{B} [/ilmath] being a basis for the topology [ilmath]\\{\\bigcup\\mathcal{A}\\ \\vert\\ \\mathcal{A}\\in\\mathcal{P}(\\mathcal{B})\\}[/ilmath]\n• we have both of the following conditions:\n1. [ilmath]\\bigcup\\mathcal{B}=X[/ilmath] (or equivalently: [ilmath]\\forall x\\in X\\exists B\\in\\mathcal{B}[x\\in B][/ilmath][Note 1]) and\n2. [ilmath]\\forall U,V\\in\\mathcal{B}\\big[U\\cap V\\neq\\emptyset\\implies \\forall x\\in U\\cap V\\exists B\\in\\mathcal{B}[x\\in W\\wedge W\\subseteq U\\cap V]\\big][/ilmath][Note 2]\n• Caveat:[ilmath]\\forall U,V\\in\\mathcal{B}\\ \\forall x\\in U\\cap V\\ \\exists W\\in\\mathcal{B}[x\\in W\\subseteq U\\cap V][/ilmath] is commonly said or written; however it is wrong, this is slightly beyond just abuse of notation.[Note 3]\n\nProof that the requisite conditions are met:\n\n1. [ilmath]\\forall x\\in X\\exists B\\in\\mathcal{B}[x\\in B][/ilmath]\n• Let [ilmath]x\\in X[/ilmath] be given\n• Lemma: [ilmath]\\forall p\\in X\\forall\\epsilon>0[p\\in B_\\epsilon(x)][/ilmath]\n• Proof:\n• Let [ilmath]p\\in X[/ilmath] be given.\n• Let [ilmath]\\epsilon>0[/ilmath] be given (with [ilmath]\\epsilon\\in\\mathbb{R}_{>0} [/ilmath] of course)\n• Recall, by definition of an open ball that [ilmath][u\\in B_\\delta(v)]\\iff[d(u,v)<\\delta][/ilmath]\n• Thus [ilmath][p\\in B_\\epsilon(p)]\\iff[d(p,p)<\\epsilon][/ilmath]\n• Recall, by definition of a metric that [ilmath][d(u,v)\\eq 0]\\iff[u\\eq v][/ilmath]\n• Thus [ilmath]d(p,p)\\eq 0[/ilmath]\n• As [ilmath]\\epsilon > 0[/ilmath] we see [ilmath]d(p,p)\\eq 0<\\epsilon[/ilmath], i.e. [ilmath]d(p,p)<\\epsilon[/ilmath] thus [ilmath]p\\in B_\\epsilon(p)[/ilmath]\n• Since [ilmath]\\epsilon > 0[/ilmath] was arbitrary we have shown [ilmath]p\\in B_\\epsilon(p)[/ilmath] for all [ilmath]\\epsilon>0[/ilmath] ([ilmath]\\epsilon\\in\\mathbb{R}_{>0} [/ilmath] of course)\n• Since [ilmath]p\\in X[/ilmath] was arbitrary we have shown [ilmath]\\forall\\epsilon>0[p\\in B_\\epsilon(p)][/ilmath] for all [ilmath]p\\in X[/ilmath]\n• This completes the proof.\n• By using the lemma above we see [ilmath]\\forall\\epsilon>0[x\\in B_\\epsilon(x)][/ilmath]\n• In particular we see [ilmath]x\\in B_1(x)[/ilmath] - there is nothing special about the choice of [ilmath]\\epsilon:\\eq 1[/ilmath] - we could have picked any [ilmath]\\epsilon\\in\\mathbb{R}_{>0} [/ilmath]\n• Choose [ilmath]B:\\eq B_1(x)[/ilmath]\n• Note that [ilmath]B\\in\\mathcal{B} [/ilmath] by definition of [ilmath]\\mathcal{B} [/ilmath], explicitly: [ilmath][B_r(q)\\in\\mathcal{B}]\\iff[q\\in X\\wedge r\\in\\mathbb{R}_{>0}][/ilmath]\n• By our choice of [ilmath]B[/ilmath] (and the lemma) we see [ilmath]x\\in B[/ilmath]\n• Our choice of [ilmath]B[/ilmath] satisfies the requirements\n• Since [ilmath]x\\in X[/ilmath] was arbitrary we have shown it for all [ilmath]x[/ilmath] - as required. This completes part 1\n2. [ilmath]\\forall U,V\\in\\mathcal{B}\\big[U\\cap V\\neq\\emptyset\\implies \\forall x\\in U\\cap V\\exists B\\in\\mathcal{B}[x\\in W\\wedge W\\subseteq U\\cap V]\\big][/ilmath]\n• Let [ilmath]U,\\ V\\in\\mathcal{B} [/ilmath] be given.\n• Since [ilmath]U,V\\in\\mathcal{B} [/ilmath] were arbitrary we have shown it for all open balls\n\nThus [ilmath]\\mathcal{B} [/ilmath] suitable to generate a topology and be a basis for that topology\n\n## Discussion of result\n\nGrade: A*\nThis page requires some work to be carried out\nSome aspect of this page is incomplete and work is required to finish it\nThe message provided is:\nNow we can link \"open\" in the metric sense to \"open\" in the topological sense. At long last." ]
[ null ]
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https://mathlesstraveled.com/2010/07/27/more-cookies/
[ "I recently received the following interesting problem from Shadowcat, which is a generalization of the cookie problem I’ve written about previously. We again want to count the ways to distribute identical cookies to non-identical students, with the twist that we impose an upper bound on the number of cookies received by each student (quite reasonable if we want to be mindful of the students’ nutrition):\n\nImagine that instead of ten cookies and five students, you have fifty cookies and ten students. (It’s easier to quantify this situation using larger numbers.) How many ways can you distribute these cookies among the students so that no student has any more than ten cookies?\n\nStudents may be given any number of cookies less than or equal to ten, including zero. The cookies are identical, just as in the original problem, so, as with the original problem, it doesn’t matter which cookie the student gets, just how many. But the students, again, are *not* identical, so which student gets a specific number of cookies is important.\n\nI unfortunately haven’t had much time to think about it yet. Feel free to leave comments, thoughts, partial solutions, and solutions in the comments!", null, "Assistant Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.\nThis entry was posted in arithmetic, challenges, counting and tagged , , . Bookmark the permalink.\n\n### 9 Responses to More cookies\n\n1.", null, "wok says:\n\nLet L=10\nD(0,s) = 1\nD(c,1) = {0 if c>L, 1 otherwise}\nD(c,s) = \\sum_{k=0}^{\\min(c,L)} D(c – k, s – 1)\n\n2.", null, "Tom says:\n\nIt seems as though we can take the original solution (the “stars and bars” method as I have always heard it called) and add the principal of inclusion and exclusion as follows.\n\nSimilar to the original problem, there are", null, "$59 \\choose 9$ ways to distribute the cookies to the 10 students.\n\nNext, suppose that the first student gets at least 11 cookies. We can count the number of ways this happens by giving him 11 and then using the remaining 39 cookies to distribute amongst all 10 students. So there are", null, "$48\\choose 9$ ways that the first student gets at least 11 cookies. We can do this for each of the 10 students to get that there are", null, "$10\\cdot {48\\choose 9}$ ways that at least one student gets 11 cookies.\n\nThe only problem remaining is that we may have double counted some situations when doing the previous argument, so we have to add back in the ways that at least 2 students get at least 11 cookies. We can do this by giving the two students 11 cookies each and then giving out the remaining 28 cookies to all 10 students. We need to do this for each pair, so we add back", null, "${10\\choose 2}\\cdot {37\\choose 9}$.\n\nWe continue the same process with 3 students, 4 students, etc, until we run out of cookies. This tells us that the number should be", null, "${59\\choose 9}-10\\cdot{48\\choose 9}+{10\\choose 2}\\cdot{37\\choose 9}-{10\\choose 3}\\cdot{26\\choose 9}+{10\\choose 4}\\cdot{15\\choose 9}$\n\nSo there are 1,018,872,811 ways to distribute 50 cookies to 10 students so that no student gets more than 10 cookies.\n\nDisclaimer: I only count this as a partial solution because I have not done a very good job of checking my work or my reasoning. There are few things that always get me off track when doing this type of counting, so if you have suggestions or corrections, I welcome them. Thanks.\n\n3.", null, "Shadowcat says:\n\nWhile I couldn’t figure it out mathematically, I did figure it out programatically. The answer I arrived at was the same as yours, so I can verify that your answer, at least, is correct.\n\nI was close to arriving at the same answer you did mathematically, too, but I just couldn’t figure out how to account for the duplicate cases. Thanks for putting the missing pieces into place for me! Your answer makes perfect sense to me.\n\n4.", null, "wok says:\n\nI agree with you.\n\nLet c=10, s=2 and L=10. The general formula for D(c,s) is:\nsum\n(-1)^k Binomial[s, k] Binomial[c+s-1-k*(L+1), s-1]\nk=0 to Floor[c/(L+1)]+1\n\n5.", null, "wok says:\n\nHere is my Python code: http://codepad.org/nhkEgK91\n\nIt works great: I mean a closed formula is far faster than a mere dynamic programming recursive formula… of course!\n\n6.", null, "Brent says:\n\nAh, good old PIE!\n\n7.", null, "Tom says:\n\n@wok,\n\nI am not a programmer, so this question is sincere: Why do you use the for-loop to compute the binomial coefficients instead of the closed formula. Is the for-loop really faster? Let me know if you get a chance. Thanks.\n\n8.", null, "Tom says:\n\n@wok,\n\ndisregard my last question, that was stupid… it is still too early in the morning for me.\n\n9.", null, "Jonathan says:\n\nIt’s just a step removed from how many 10 digit numbers have the property that the sum of their digits is 50\n\n(I would have worked Tom’s way. It’s in the text, with exercises of Ivan Niven’s “How to Count Without Counting” that serves as a textbook for a senior elective I teach.)" ]
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https://www.boost.org/doc/libs/1_58_0/libs/graph_parallel/doc/html/strong_components.html
[ "#", null, "Boost C++ Libraries\n\n...one of the most highly regarded and expertly designed C++ library projects in the world.\n\n#", null, "Connected Components\n\n```template<typename Graph, typename ComponentMap>\ninline typename property_traits<ComponentMap>::value_type\nstrong_components( const Graph& g, ComponentMap c);\n\nnamespace graph {\ntemplate<typename Graph, typename VertexComponentMap>\nvoid\nfleischer_hendrickson_pinar_strong_components(const Graph& g, VertexComponentMap r);\n\ntemplate<typename Graph, typename ReverseGraph,\ntypename ComponentMap, typename IsoMapFR, typename IsoMapRF>\ninline typename property_traits<ComponentMap>::value_type\nfleischer_hendrickson_pinar_strong_components(const Graph& g,\nComponentMap c,\nconst ReverseGraph& gr,\nIsoMapFR fr, IsoMapRF rf);\n}\n```\n\nThe strong_components() function computes the strongly connected components of a directed graph. The distributed strong components algorithm uses the sequential strong components algorithm to identify components local to a processor. The distributed portion of the algorithm is built on the distributed breadth first search algorithm and is based on the work of Fleischer, Hendrickson, and Pinar [FHP00]. The interface is a superset of the interface to the BGL sequential strong components algorithm. The number of strongly-connected components in the graph is returned to all processes.\n\nThe distributed strong components algorithm works on both directed and bidirectional graphs. In the bidirectional case, a reverse graph adapter is used to produce the required reverse graph. In the directed case, a separate graph is constructed in which all the edges are reversed.\n\n# Where Defined\n\n<boost/graph/distributed/strong_components.hpp>\n\nalso accessible from\n\n<boost/graph/strong_components.hpp>\n\n# Parameters\n\nIN: const Graph& g\nThe graph type must be a model of Distributed Graph. The graph type must also model the Incidence Graph and be directed.\nOUT: ComponentMap c\nThe algorithm computes how many strongly connected components are in the graph, and assigns each component an integer label. The algorithm then records to which component each vertex in the graph belongs by recording the component number in the component property map. The ComponentMap type must be a Distributed Property Map. The value type must be the vertices_size_type of the graph. The key type must be the graph's vertex descriptor type.\nUTIL: VertexComponentMap r\nThe algorithm computes a mapping from each vertex to the representative of the strong component, stored in this property map. The VertexComponentMap type must be a Distributed Property Map. The value and key types must be the vertex descriptor of the graph.\nIN: const ReverseGraph& gr\n\nThe reverse (or transpose) graph of g, such that for each directed edge (u, v) in g there exists a directed edge (fr(v), fr(u)) in gr and for each edge (v', u') in gr there exists an edge (rf(u'), rf(v')) in g. The functions fr and rf map from vertices in the graph to the reverse graph and vice-verse, and are represented as property map arguments. The concept requirements on this graph type are equivalent to those on the Graph type, but the types need not be the same.\n\nDefault: Either a reverse_graph adaptor over the original graph (if the graph type is bidirectional, i.e., models the Bidirectional Graph concept) or a distributed adjacency list constructed from the input graph.\n\nIN: IsoMapFR fr\n\nA property map that maps from vertices in the input graph g to vertices in the reversed graph gr. The type IsoMapFR must model the Readable Property Map concept and have the graph's vertex descriptor as its key type and the reverse graph's vertex descriptor as its value type.\n\nDefault: An identity property map (if the graph type is bidirectional) or a distributed iterator_property_map (if the graph type is directed).\n\nIN: IsoMapRF rf\n\nA property map that maps from vertices in the reversed graph gr to vertices in the input graph g. The type IsoMapRF must model the Readable Property Map concept and have the reverse graph's vertex descriptor as its key type and the graph's vertex descriptor as its value type.\n\nDefault: An identity property map (if the graph type is bidirectional) or a distributed iterator_property_map (if the graph type is directed).\n\n# Complexity\n\nThe local phase of the algorithm is O(V + E). The parallel phase of the algorithm requires at most O(V) supersteps each containing two breadth first searches which are O(V + E) each.\n\n# Algorithm Description\n\nPrior to executing the sequential phase of the algorithm, each process identifies any completely local strong components which it labels and removes from the vertex set considered in the parallel phase of the algorithm.\n\nThe parallel phase of the distributed strong components algorithm consists of series of supersteps. Each superstep starts with one or more vertex sets which are guaranteed to completely contain any remaining strong components. A distributed breadth first search is performed starting from the first vertex in each vertex set. All of these breadth first searches are performed in parallel by having each processor call breadth_first_search() with a different starting vertex, and if necessary inserting additional vertices into the distributed queue used for breadth first search before invoking the algorithm. A second distributed breadth first search is performed on the reverse graph in the same fashion. For each initial vertex set, the successor set (the vertices reached in the forward breadth first search), and the predecessor set (the vertices reached in the backward breadth first search) is computed. The intersection of the predecessor and successor sets form a strongly connected component which is labeled as such. The remaining vertices in the initial vertex set are partitioned into three subsets each guaranteed to completely contain any remaining strong components. These three sets are the vertices in the predecessor set not contained in the identified strongly connected component, the vertices in the successor set not in the strongly connected component, and the remaing vertices in the initial vertex set not in the predecessor or successor sets. Once new vertex sets are identified, the algorithm begins a new superstep. The algorithm halts when no vertices remain.\n\nTo boost performance in sparse graphs when identifying small components, when less than a given portion of the initial number of vertices remain in active vertex sets, a filtered graph adapter is used to limit the graph seen by the breadth first search algorithm to the active vertices.\n\n# Bibliography\n\n [FHP00] Lisa Fleischer, Bruce Hendrickson, and Ali Pinar. On Identifying Strongly Connected Components in Parallel. In Parallel and Distributed Processing (IPDPS), volume 1800 of Lecture Notes in Computer Science, pages 505--511, 2000. Springer.\n\nCopyright (C) 2004, 2005 The Trustees of Indiana University.\n\nAuthors: Nick Edmonds, Douglas Gregor, and Andrew Lumsdaine" ]
[ null, "https://www.boost.org/gfx/space.png", null, "https://www.boost.org/doc/libs/1_58_0/libs/graph_parallel/doc/html/pbgl-logo.png", null ]
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http://www.bnlawrence.net/computing/2006/04/more_meteorological_time/
[ "My first post at describing the time issues for meteorological metadata led to some confusion, so I’m trying again. I think it helps to consider a diagram:", null, "The diagram shows a number of different datasets that can be constructed from daily forecast runs (shown for an arbitrary month from the 12th til the 15th as forecasts 1 through 4. If we consider forecast 2, we are running it to give me a forecast from the analysis time (day 13.0) forward past day 14.5 … but you can see that the simulation began (in this case 1.0) days earlier, at a simulation time of 12.0 (T0). In this example\n\n• we’ve allowed the initial condition to be day 12.0 from forecast 1.\n• we’ve imagined the analysis was produced at the end of a data assimilation period (let’s call this time Ta).\n• the last time for which data was used (the datum time, td) corresponds to the analysis time.\n\n(Here I’m using some nomenclature defined earlier as well as using some new terms).\n\nAnyway, the point here was to introduce a couple of new concepts. These forecast datasets can be stored and queried relatively simply … we would have a sequence of datasets, one for each forecast, and the queries would simply then be on finding the forecasts (using discovery metadata, e.g. ISO19139) and then on extracting and using the data itself (using archive aka content aka usage metadata, e.g. an application schema of GML such as CSML).\n\nWhat’s more interesting is how we provide, document and query the sythesized datasets (e.g. the analysis and T+24 datasets). Firstly, if we look at the analysis dataset, we could extract the Ta data points and have a new dataset, but often we need the interim times as well, and you can see that we have two choices of how to construct them - we can use the earlier time from the later forecast (b), or the later time from the earlier forecast (a). Normally we choose the latter, because the diabatic and non-observed variables are usually more consistent outside the assimilation period when they have had longer to spin up. Anyway, either way we have to document what is done. This is a job for a new package we plan to get into the WMO core profile of ISO19139 as an extension - NumSim.\n\nFrom a storage point of view, as I implied above, we can extract and store the new datasets, or we can try and do this as a virtual dataset, described in CSML, and extractable by CSML-tools. We don’t yet know how to do this, but there is obvious utility in saving storage in doing so." ]
[ null, "http://www.bnlawrence.net/assets/images/2006-04-11-time.jpg", null ]
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https://www.geeksforgeeks.org/cost-to-make-a-string-panagram/
[ "# Cost to make a string Panagram\n\nGiven an array arr[] containing the cost of adding each alphabet from (a – z) and a string str which may or may noyt be a Panagram. The task is to find the total cost to make the string Panagram.\n\nExamples:\n\nInput: arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26},\nstr = “abcdefghijklmopqrstuvwz ”\nOutput : 63\nn, x and y are the only characters missing to convert the string into a Pangram.\nAnd their costs are 14, 24 and 25 respectively.\nHence, total cost = 14 + 24 + 25 = 63\n\nInput: arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26},\nstr = “thequickbrownfoxjumpsoverthelazydog”\nOutput: 0\nThe string is already a Pangram as it contains all the characters from (a – z).\n\n## Recommended: Please try your approach on {IDE} first, before moving on to the solution.\n\nApproach: Mark all the alphabets from (a – z) that occurred in str then find the missing alphabets from the string. Add the cost of these missing alphabets to get the total cost required to make the string Pangram.\n\nBelow is the implementation of the above approach:\n\n## C++\n\n `// C++ program to find the cost to make a string Panagram ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the total cost required ` `// to make the string Pangram ` `int` `pangramCost(``int` `arr[], string str) ` `{ ` `    ``int` `cost = 0; ` `    ``bool` `occurred = { ``false` `}; ` ` `  `    ``// Mark all the alphabets that occurred in the string ` `    ``for` `(``int` `i = 0; i < str.size(); i++) ` `        ``occurred[str[i] - ``'a'``] = ``true``; ` ` `  `    ``// Calculate the total cost for the missing alphabets ` `    ``for` `(``int` `i = 0; i < 26; i++) { ` `        ``if` `(!occurred[i]) ` `            ``cost += arr[i]; ` `    ``} ` ` `  `    ``return` `cost; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, ` `                  ``16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26 }; ` `    ``string str = ``\"abcdefghijklmopqrstuvwz\"``; ` ` `  `    ``cout << pangramCost(arr, str); ` `    ``return` `0; ` `} `\n\n## Java\n\n `// Java program to find the cost to make a string Panagram ` ` `  `class` `GFG ` `{ ` `        ``// Function to return the total cost required ` `        ``// to make the string Pangram ` `        ``static`  `int` `pangramCost(``int` `arr[], String str) ` `        ``{ ` `            ``int` `cost = ``0``; ` `            ``boolean` `[]occurred=``new` `boolean``[``26``]; ` `             `  `            ``for``(``int` `i=``0``;i<``26``;i++) ` `             ``occurred[i]=``false``; ` `         `  `            ``// Mark all the alphabets that occurred in the string ` `            ``for` `(``int` `i = ``0``; i < str.length(); i++) ` `                ``occurred[str.charAt(i) - ``'a'``] = ``true``; ` `         `  `            ``// Calculate the total cost for the missing alphabets ` `            ``for` `(``int` `i = ``0``; i < ``26``; i++) { ` `                ``if` `(occurred[i]==``false``) ` `                    ``cost += arr[i]; ` `            ``} ` `         `  `            ``return` `cost; ` `        ``} ` `         `  `        ``// Driver Code ` `        ``public` `static` `void` `main(String []args) ` `        ``{ ` `            ``int` `arr[] = { ``1``, ``2``, ``3``, ``4``, ``5``, ``6``, ``7``, ``8``, ``9``, ``10``, ``11``, ``12``, ``13``, ``14``, ``15``, ` `                        ``16``, ``17``, ``18``, ``19``, ``20``, ``21``, ``22``, ``23``, ``24``, ``25``, ``26` `}; ` `            ``String str = ``\"abcdefghijklmopqrstuvwz\"``; ` `         `  `            ``System.out.println(pangramCost(arr, str)); ` `         `  `    ``} ` `} ` ` `  ` `  `// This code is contributed by ihritik `\n\n## Python3\n\n `# Python3 program to find the cost  ` `# to make a string Panagram  ` ` `  `# Function to return the total cost ` `# required to make the string Pangram  ` `def` `pangramCost(arr, string) : ` `     `  `    ``cost ``=` `0` `    ``occurred ``=` `[``False``] ``*` `26` ` `  `    ``# Mark all the alphabets that  ` `    ``# occurred in the string  ` `    ``for` `i ``in` `range``(``len``(string)) : ` `        ``occurred[``ord``(string[i]) ``-` `ord``(``'a'``)] ``=` `True` ` `  `    ``# Calculate the total cost for  ` `    ``# the missing alphabets  ` `    ``for` `i ``in` `range``(``26``) :  ` `        ``if` `(``not` `occurred[i]) : ` `            ``cost ``+``=` `arr[i] ` ` `  `    ``return` `cost ` ` `  `# Driver Code  ` `if` `__name__ ``=``=` `\"__main__\"` `: ` ` `  `    ``arr ``=` `[ ``1``, ``2``, ``3``, ``4``, ``5``, ``6``, ``7``, ``8``, ``9``,  ` `            ``10``, ``11``, ``12``, ``13``, ``14``, ``15``, ``16``,  ` `            ``17``, ``18``, ``19``, ``20``, ``21``, ``22``, ``23``,  ` `            ``24``, ``25``, ``26` `]  ` `    ``string ``=` `\"abcdefghijklmopqrstuvwz\"` ` `  `    ``print``(pangramCost(arr, string)) ` ` `  `# This code is contributed by Ryuga `\n\n## C#\n\n `// C# program to find the cost to make a string Panagram ` ` `  `using` `System; ` `class` `GFG ` `{ ` `        ``// Function to return the total cost required ` `        ``// to make the string Pangram ` `        ``static`  `int` `pangramCost(``int` `[]arr, ``string` `str) ` `        ``{ ` `            ``int` `cost = 0; ` `            ``bool` `[]occurred=``new` `bool``; ` `             `  `            ``for``(``int` `i=0;i<26;i++) ` `             ``occurred[i]=``false``; ` `         `  `            ``// Mark all the alphabets that occurred in the string ` `            ``for` `(``int` `i = 0; i < str.Length; i++) ` `                ``occurred[str[i] - ``'a'``] = ``true``; ` `         `  `            ``// Calculate the total cost for the missing alphabets ` `            ``for` `(``int` `i = 0; i < 26; i++) { ` `                ``if` `(occurred[i]==``false``) ` `                    ``cost += arr[i]; ` `            ``} ` `         `  `            ``return` `cost; ` `        ``} ` `         `  `        ``// Driver Code ` `        ``public` `static` `void` `Main(``string` `[]args) ` `        ``{ ` `            ``int` `[]arr = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, ` `                        ``16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26 }; ` `            ``string` `str = ``\"abcdefghijklmopqrstuvwz\"``; ` `         `  `            ``Console.WriteLine(pangramCost(arr, str)); ` `         `  `    ``} ` `} ` ` `  ` `  `// This code is contributed by ihritik `\n\n## PHP\n\n ` `\n\nOutput:\n\n```63\n```\n\nMy Personal Notes arrow_drop_up", null, "Check out this Author's contributed articles.\n\nIf you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to [email protected]. See your article appearing on the GeeksforGeeks main page and help other Geeks.\n\nPlease Improve this article if you find anything incorrect by clicking on the \"Improve Article\" button below.\n\nImproved By : AnkitRai01, ihritik\n\nArticle Tags :\nPractice Tags :\n\nBe the First to upvote.\n\nPlease write to us at [email protected] to report any issue with the above content." ]
[ null, "https://media.geeksforgeeks.org/auth/profile/p4p1bcvqi3lav5in6auv", null ]
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https://metanumbers.com/24793
[ "# 24793 (number)\n\n24,793 (twenty-four thousand seven hundred ninety-three) is an odd five-digits prime number following 24792 and preceding 24794. In scientific notation, it is written as 2.4793 × 104. The sum of its digits is 25. It has a total of 1 prime factor and 2 positive divisors. There are 24,792 positive integers (up to 24793) that are relatively prime to 24793.\n\n## Basic properties\n\n• Is Prime? Yes\n• Number parity Odd\n• Number length 5\n• Sum of Digits 25\n• Digital Root 7\n\n## Name\n\nShort name 24 thousand 793 twenty-four thousand seven hundred ninety-three\n\n## Notation\n\nScientific notation 2.4793 × 104 24.793 × 103\n\n## Prime Factorization of 24793\n\nPrime Factorization 24793\n\nPrime number\nDistinct Factors Total Factors Radical ω(n) 1 Total number of distinct prime factors Ω(n) 1 Total number of prime factors rad(n) 24793 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 10.1183 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 24,793 is 24793. Since it has a total of 1 prime factor, 24,793 is a prime number.\n\n## Divisors of 24793\n\n2 divisors\n\n Even divisors 0 2 2 0\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 2 Total number of the positive divisors of n σ(n) 24794 Sum of all the positive divisors of n s(n) 1 Sum of the proper positive divisors of n A(n) 12397 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 157.458 Returns the nth root of the product of n divisors H(n) 1.99992 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 24,793 can be divided by 2 positive divisors (out of which 0 are even, and 2 are odd). The sum of these divisors (counting 24,793) is 24,794, the average is 12,397.\n\n## Other Arithmetic Functions (n = 24793)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 24792 Total number of positive integers not greater than n that are coprime to n λ(n) 24792 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 2742 Total number of primes less than or equal to n r2(n) 8 The number of ways n can be represented as the sum of 2 squares\n\nThere are 24,792 positive integers (less than 24,793) that are coprime with 24,793. And there are approximately 2,742 prime numbers less than or equal to 24,793.\n\n## Divisibility of 24793\n\n m n mod m 2 3 4 5 6 7 8 9 1 1 1 3 1 6 1 7\n\n24,793 is not divisible by any number less than or equal to 9.\n\n• Arithmetic\n• Prime\n• Deficient\n\n• Polite\n\n• Prime Power\n• Square Free\n\n## Base conversion (24793)\n\nBase System Value\n2 Binary 110000011011001\n3 Ternary 1021000021\n4 Quaternary 12003121\n5 Quinary 1243133\n6 Senary 310441\n8 Octal 60331\n10 Decimal 24793\n12 Duodecimal 12421\n20 Vigesimal 31jd\n36 Base36 j4p\n\n## Basic calculations (n = 24793)\n\n### Multiplication\n\nn×y\n n×2 49586 74379 99172 123965\n\n### Division\n\nn÷y\n n÷2 12396.5 8264.33 6198.25 4958.6\n\n### Exponentiation\n\nny\n n2 614692849 15240079805257 377847298611736801 9367968074480790507193\n\n### Nth Root\n\ny√n\n 2√n 157.458 29.1593 12.5482 7.56599\n\n## 24793 as geometric shapes\n\n### Circle\n\n Diameter 49586 155779 1.93111e+09\n\n### Sphere\n\n Volume 6.38375e+13 7.72446e+09 155779\n\n### Square\n\nLength = n\n Perimeter 99172 6.14693e+08 35062.6\n\n### Cube\n\nLength = n\n Surface area 3.68816e+09 1.52401e+13 42942.7\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 74379 2.6617e+08 21471.4\n\n### Triangular Pyramid\n\nLength = n\n Surface area 1.06468e+09 1.79606e+12 20243.4\n\n## Cryptographic Hash Functions\n\nmd5 fdb2b215aca454782c69bc1ee9a85d81 ce0be1c4f07df85bd69153e5035af80a4c62bedb 16ed5f74b8a3de8ad9c69c679e776bcc56493abd5c452e48ca2ffc26de0a1308 61a820eab7145ee330ec264e35b0fb864ebc210c12a13745a7f167ab4f595d89ed82ef277225aebb7f7e9e6d3fdb43b08d399fd161074c114e48e113fa85891b aa0fc176b002de46904bea1b1b630f2f11d3430d" ]
[ null ]
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