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[ "# Middle School Exponent Activities\n\n••• michaeljung/iStock/Getty Images\nPrint\n\nExponents use a power to tell how many times a number is multiplied by itself. If the number is six to the third power, for example, you multiply the number six three times -- 6 x 6 x 6 -- to get 216. Exponents are essential mathematical concepts that are the foundation for further study into scientific notation and other algebraic functions.\n\n## Introducing Exponent Properties", null, "••• michaeljung/iStock/Getty Images\n\nTo introduce exponents to middle-schoolers, use activities that help them visualize the exponents. For the multiplication property, place an index card inscribed with the number seven on a table along with five jellybeans to represent the exponent. Place another card beside the first with the number seven and three jellybeans. Tell students that when the same number is multiplied, the exponents are added together. Count the number of jellybeans to yield seven to the eighth power. To show division, place a ruler on the desk to represent a division line. Place the index cards and jellybeans above and below the ruler with the five jellybeans on top and three on the bottom. Tell students that when the same numbers are on top of each other, the exponents are subtracted. Students will take away two jellybeans to yield seven to the second power. For a power to a power demonstration, use an index card with seven written on it, then represent an inner exponent with a group of five red candies and an outer exponent with a group of three green candies. Tell students to multiply the exponents to equal seven to the 15th power.\n\n## Exponent Scavenger Hunt", null, "••• greg801/iStock/Getty Images\n\nTake sixth- or seventh-graders on a scavenger hunt around the classroom to find a secret code using exponents. On an index card, write (3x to the third power) x (5y to the third power) / (3x to the second power) and post it on the wall. The second index card posted in another area of the room will have the answer to the first problem, 5x times y to the third power, the first letter of the secret code word in the corner of the card and the next problem on the back of the card. This answer to the second card leads to the third card, and so on, until the secret code word is discovered. Along with these correct cards, post decoy cards with commonly mistaken answers.\n\n## Exponent Store", null, "••• Lisa F. Young/iStock/Getty Images\n\nFor grades five or six, place a “price” tag on objects around the room with sticky notes. These prices will be the answers to exponent questions marked on index cards. For example, a classroom poster may cost \\$128. This would cost the student an index card that has two to the fifth power times two to the second power written on it. Divide students into groups and give them a set of these index cards. The exponent question cards serve as the “money” students use to buy items. Have students walk around and make a list of things their group would be able to buy.\n\n## Creating a Computer Game", null, "••• Jack Hollingsworth/Stockbyte/Getty Images\n\nIn upper middle school classes of algebra, have students make their own computer game using Microsoft PowerPoint software to create a game similar to a “Jeopardy” format. Create a game board by inserting a table into a slide with categories such as multiplying exponents, dividing exponents, and exponents with integers. Underneath each category, type point values 100, 200, 300 and 400 into the table. Create practice problems on separate slides, such as, “Simplify the expression: (7s to the second power) x (8s to the fourth power) x (2s to the sixth power).” Hyperlink the problem slides to a point value so that when the point value is clicked, the question will appear on the screen. Students will also create an answer key to go along with their game." ]

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[ "# C++程式設計基礎(五)：函式與遞迴\n\n``````stable_sort(dt,dt+n,[](int a,int b){\nint ctA=0,ctB=0;\nwhile(a>=1){\nif(a%2==1){\nctA++;\n}\na/=2;\n}\nwhile(b>=1){\nif(b%2==1){\nctB++;\n}\nb/=2;\n}\nreturn ctA<ctB;\n});``````\n\n``````int ct(int a){\nint ctA=0;\nwhile(a>=1){\nif(a%2==1){\nctA++;\n}\na/=2;\n}\n}\n\nstable_sort(dt,dt+n,[](int a,int b){\nint ctA=ct(a),ctB=ct(b);\nreturn ctA<ctB;\n});``````\n\n``````int db(int a){\nreturn a*2;\n}\n//函數通式，注意傳入的參數可以有多個(不只兩個)，但每個前面都要加入型態名稱\n<回傳值變數型態> <函式名>(<要傳入的變數型態> <變數名>,<變數型態> <變數名>){\n<程式碼>\nreturn <變數>\n}``````\n\nA1=1, A2=1, An=An-1+An-2 if(n>2) 寫作程式即為。\n\n``````int f(int n){\nif(n==1){\nreturn 1;\n}else if(n==2){\nreturn 1;\n}else{\nreturn f(n-1)+f(n-2);\n}\n}``````" ]

[ null ]

[ " How to escape a monster (using calculus)", null, "Home Blog About Us Work Content Contact Us\n Advertisement\n\n# How to escape a monster\n\nIt’s getting close to Halloween, here’s a monster puzzle:\n\nImagine you are sitting in a rowing boat in the middle of circular lake. Standing on the edge of the lake is a hungry monster. The monster wants to eat you. You want to escape.\n\nYou know that, if you can reach land, you can outrun the monster, but first of all, you need to row to shore. Here’s the issue: whilst you can outrun the monster on foot, he can run faster than you can row the boat. Much faster. The monster can run four times faster than you can row. He’s also a clever monster and will always do that smartest thing to attempt to catch you; moving around the shoreline, as needed, to maximize his advantage. If he touches you, you’re dead!\n\nCan you escape?", null, "### Impossible?\n\n I have to admit, when I first heard this puzzle, my initial thought was that it is impossible to escape. Let's imagine you made a direct dash for the shore directly opposite where the monster is currently standing. The radius of the lake is R. The speed you can row is v. The distance to the shore is simply the radius, so the time taken to get to shore is R/v. The monster, seeing you are making a dash for the shore, will sprint around the edge of the lake. It does not matter which direction he runs, and he will end up running half the circumference. The distance he will run is πR, and running at speed 4v, he will get to your landing point after just πR/4v. As we know that π is less than four, we can see that the monster will get there first. He’s going to eat your brains if you use this strategy! Towards the middle or end of your dash, once you realize the monster is closing, even if you adjust your strategy to add some tangential component, you're screwed. The faster speed of the monster means he can always catch you. We need a better strategy …", null, "### Smarter Strategy\n\nRowing in straight line is clearly the correct strategy (if you rowed on a curved line, you'd end up at a location that you could have gotten to quicker with a straight shot). However, we've shown that starting at the center results in failure.\n\nTo solve this puzzle, what we first need to do is calculate the position from which a radial sprint to the shore will result in victory (we'll worry about how to get into this position in step two).", null, "There's an annulus (ring) at the edge of the lake in which, if we start inside, and sprint directly towards the shore we be sure of escape. Taken to the limit, in the diagram above, we can see that if we start with the monster diametrically opposed to us there is a maximum value for d.\n\nWe know that the monster needs to run half the circumference of the lake, and the time this will take him is πR/4v. In this same time, we must row the distance d. At the limit, the time taken to complete these distances will be the same, so πR/4v = d/v, so we can calculate the maximum value of d = πR/4. If we start any closer to the center of the lake than this point, the monster will catch and eat us.\n\nUsing a calculator we can find out that d = 0.7854R\n\n### Getting to the escape point\n\nNow all we have to do is work out how to get to this escape point (on the edge of this annulus with the monster opposite us).\n\nWhen we are close to the center of the lake, we can row in tight circles. If the circle is small enough (less than a quarter of the radius of the lake), we travel radially around the lake faster than the monster can. We can use this to position ourselves relative to the monster at any arbitrary location (In this case, 180°).", null, "Starting from the middle, we can row in a spiral, slowly increasing our radius and, providing we stay at at radius r < R/4, we can always move faster than the monster and thus be able to make sure we position ourselves at 180° away from him. Once we have manoeuvred ourselves to a position d away from the edge, with the monster opposite, we can use the escape plan described above to escape. We know that d = 0.7854R, so this corresponds to a radius of r = (R - d), which gives a value r = 0.2146R This is great news because we know we can stay ahead of the the monster until a radius of r = 0.25R (travelling a quarter as slow allows us to navigate circles up to a quarter the circumference of the lake).\n\nSo now we have our solution: We can spiral out until we reach a radius of 0.2146R < r < 0.25R with the monster opposite us, and be sure of an escape! Hurrah for math!\n\n### What about a faster monster?", null, "A faster monster We've shown that we can outsmart a monster that can run four times as fast as we can row. The inequality above, however, shows that we have some headroom. What is the fastest monster we can escape? To calculate this, we need a little more math* … The fastest monster we can escape is determined when the monster arrives at the shore at exactly the same time we do. Let's define the ratio of our two speeds by the letter K. (The monster runs K-times faster than we can row). What is the maximum value for K? *Actually, quite a lot more math!\n\n### Hold on, here comes the math …\n\n The first part of the puzzle stays the same. We know that, in a tight circle, we can row with a higher angular velocity than the monster can run. If we are inside this circle we can position the monster to any arbitrary angular position relative to us. (In fact, if we adopt an optimal strategy, and move outwards at full speed, always keeping the monster co-linear with the center of the lake, the path we follow is a semi-circle of radius R/2K). The radius of this circle of safety is determined by the ratio of the two speeds. It’s radius is R/K. The monster is a smart monster (as well as being hungry, he has an advanced degree in mathematics from Monster’s University), so he knows that, once you cross outside the circle of safety he has a higher angular velocity, so as soon as he sees you leave the circle, he sprints at full speed to your side of the lake. (It’s obvious he needs to do this, if not, he’s giving you the advantage of being able to get further away!)", null, "", null, "Similarly, once the monster has committed to the direction he’s going to run around the lake, it would be foolish for him to change direction. If he did, he’d re-pass the point he was previously at, but now you’d be closer to the shore! We already know that you should always row in a straight line (anything else just wastes time), and that the monster is committed to his run. The question is: should we simply dash for the shore (as in the simple solution above), or can we gain advantage at aiming at little further around the shore? Sure, the straight shot to the shore is the shortest distance, but is it possible to row a slightly longer (straight) route to the shore if it would require the monster to run further to catch you? As long as the extra time it takes you to row this extra distance is less than it would take the monster to run to the same point, you are better doing it.\n\nIn the diagram below, I've plotted out a couple of possible paths. Path #1 is the strategy from our simple solution. It is the shortest distance from the edge of the safety circle to the shore. Can we do better?\n\nPath #2 is a example of a generic path from the circle of safety to the shore. Yes, it's longer, but the monster also has to run longer. As the monster is committed to its run, it needs to continue around the edge of the lake to you.\n\nAt the limit is Path #3, which is a tangent to the circle of safety. Anything past this, like Path #4 takes us back inside the circle of safety and would result in new solution being required when we exited the other side (and you'd essentially have to start the process again, otherwise, if you didn't, as soon as you went inside the circle, if you did not manoeuvre the monster opposite you again before starting your dash to shore, you would give him an advantage). The solution can not occur 'West' of Path #3 (if we define 'North' as vertically upwards).\n\nWe can, therefore, reject Path #4, and the solution must lay somewhere between Path #1 and Path #3 inclusive.", null, "Note: It should be very obvious that the optimal solution does not occur 'South' of Path #1. Anything below Path #1 takes you towards the monster !!!\n\n### More math …\n\nOK, we're now going to break out some trigonometry. If you're squeamish about math, you're going to want to skip to the answer section. It's only going to get worse from this point onwards …", null, "The distance we row from the exit point of the circle of safety to the shore, we'll call d. This value can be calculated using Pythagoras and the right hand triangle colored in the image above.", null, "Expanding this out we get this:", null, "Do you remember your trig equalities? Here's one we will be using a lot today: sin2 + cos2 = 1\n\nUsing this equality we can simplify the equation to this:", null, "In the diagram on the right we can see that the monster has to run the blue route, which goes past halfway, and continues along to the interception point. The distance the monster needs to run we'll call m :", null, "(We are using radians for the angles) The puzzle reduces down to the following questions: For a given speed ratio, K, what is the optimal angle Φ? What is the maximum possible value for K?", null, "For you to escape, the time taken to row distance d, needs to be less than, or equal to the time it takes the monster to run m.", null, "The limit will be reached when both you and the monster arrive at the same point at the same time.", null, "If you both arrive at the same time, and you rowed a distance of d, the monster, who runs K times faster will have run K times the distance. The equation on the left is another way to describe the value d.\n\nSquaring this, we can equate it to the representation of d2 derived by Pythagoras.", null, "To simplify, first we multiply both sides by (K/R)2 … bear with me, you'll see why …", null, "This can be simplified to:", null, "Adding, and subtracting cos2 from the left side allows the terms to be simplified into a square expression:", null, "Then, using the trig equality again in the form: 1 - sin2 = cos2 gets us to here:", null, "Finally, rearranging we get our first important result:", null, "This is awesome. It gives us the relationship between K and Φ\n\n### Calculus\n\nOK, we're onto the home straight, but now we need to break out our Calculus. Maximum (and minimum) points occur at points where first derivatives are zero. We need to derive the first derivative of K with respect to Φ", null, "We want to find the value of Φ when dK/dΦ = 0. With a little rearranging we get this:", null, "Next we square both sides:", null, "Things have to get a little worse before they get better. Here we move things around and regroup:", null, "Applying our old friend the trig equality sin2 + cos2 = 1 once again:", null, "Next we divide by sin2. Remember from school that cos / sin = cot", null, "For one last time we'll use sin2 + cos2 = 1:", null, "Oooh, look, the above is a perfect square, and simplifies nicely:", null, "And now we have our solution!", null, "This equation can be solved with your favourite tool (graph paper, Excel, Newton-Raphson, trial and error, calculator, abacus …) and yields a value of Φ ≈ 1.35181680431927…\n\nWe can plug this value back into our first result", null, "### The results\n\n The maximum value for K is 4.6033388489… That's right, we can outrun a monster that can run just over 4.6x the speed we can row!", null, "", null, "A 4.6x faster monster!\n\n### But which way should you row?\n\nWe now know how fast a monster it is possible to evade. We also know what angle from the center of the lake this corresponds to, but what angle do we depart circle of safety from? Let's find out …", null, "We'll call the angle the boat should take from the collinear line as α From simple geometry can see the relationship between the two angles because they are both right-angle triangles and share a common side.", null, "", null, "From our earlier results, we know the relationships between d, R,Φ and K The top two equations on the right, which we derived earlier, can be combined.", null, "", null, "", null, "This last equation can be rearranged to given an alternative form of R sin(Φ) which we can insert into the equation derived from the geometry.\n\nAnd here we have it:", null, "That's right, the angle you should row your boat is 90° from the collinear line! It's a tangent to the circle of safety. It's what we called Path #3 earlier.\n\n### Summary", null, "If you ever find yourself in the middle of a circular lake with monster who can run less than 4.6033388489 times as fast as you can row, then row outwards and around on a semi-circular path of radius R/2K until you reach a radius of R/K, then continue straight on until you reach the shore, and escape. The path you will follow will resemble a letter 'J'. The original author of this puzzle, I believe, is none other than Martin Gardner. If you like this puzzle, you might like some of these.\n\nYou can find a complete list of all the articles here.", null, "Click here to receive email alerts on new articles.", null, "© 2009-2013 DataGenetics    Privacy Policy" ]

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[ "centimeters millimeters inches\n\n•", null, "Centimeters to Feet and Inches Conversion (cm, ft and in)\n\nHow to convert centimeters to inches. 1 inch is equal to 2.54cm. To convert centimeters to inches you need to divide your figure by 2.54. To convert directly between centimeters (cm) and feet (ft) or feet and inches individually (plus many other units of length, distance .\n\n•", null, "Convert mm, cm to fraction or decimal inches (in = mm = cm)\n\nMM, CM & Inch. 1 centimeter(cm) = 10 millimeters(mm). (convert cm to mm) 1 meter = 100 centimeters = 1,000 millimeters. (convert meter to cm) 1 inch equals 2.54 centimeters(cm), 1 cm approximately equivalent to 3/8 inch or equals 0.393700787 inch; Fractional inches to cm & mm conversion table\n\n•", null, "inches to millimeters and inches to centimeters\n\n100 centimeters = 1 meter = 39.37 inches = 3.281 feet.\n\n•", null, "Inches to Centimeter Conversion (in to cm)\n\nThere are 2.54 centimeter in a inch. 1 Inch is equal to 2.54 Centimeter. 1 in = 2.54 cm. An inch is a unit of measurement in the American Standard of measurements. Twelve inches equals one foot. The inch is further divided into smaller units. There are 16 units in one inch. Each unit being one-sixteenth of .\n\n•", null, "Centimeters to Millimeters Converter (cm to mm)\n\nTo convert mm to cm, ply the mm value by 0.1 or divide by 10. 1 Millimeter = 0.1 Centimeter You may also use this length and distance units conversion calculator tool to convert between centimeters, millimeters and all other length and distance units.\n\n•", null, "Convert mm, cm to fraction or decimal inches (in = mm = cm)\n\nThis is an online length converter, convert millimeters(mm) to inches, centimeters(cm) to inches, inches to cm, inches to mm, include fraction and decimal inches, with a ruler to show the corresponding of units, understand your question with the best visualization.\n\n•", null, "Inches to Centimeter Conversion (in to cm)\n\nCentimeter is considered a common unit of length used in SI. It is equivalent to 10 millimeters or 1/100 th (10-2) of a meter.Years ago it was a basic unit in formerly used CGS (centimeter-gram-second) unit system, but in modern times the role of basic unit of length is played by meter.\n\n•", null, "Centimeters to Inches Calculator | Convert Centimeters to ...\n\nCentimeters to Inches Definition. The Centimeters to Inches Calculator will instantly let you convert centimeters to inches. All you have to do is enter in the number of centimeters and press calculate and you will see the corresponding number of inches.\n\n•", null, "Convert cm to inches - The Calculator Site\n\nHow many inches in a cm? 1cm = 0.39370079 inches. As there are 2.54cm in an inch, to convert your cm figure to inches you need to divide your figure by 2.54. Should you wish to convert from inches to centimeters, give the inches to cm converter a try. Alternatively, you can convert from mm to inches here. cm to inches conversion chart\n\n•", null, "Convert Centimeters to Millimeters - Inch Calculator\n\n41 rows · Millimeter is spelled millimetre internationally, and it is equal in length to .1 centimeters, .\n\n• CENTIMETERSMILLIMETERS\n1 cm10 mm\n2 cm20 mm\n3 cm30 mm\n4 cm40 mm\nSee all 41 rows on\n•", null, "Convert inches to cm - Length / Distance Conversions\n\nA centimeter is a unit of Length or Distance in the Metric System. The symbol for centimeter is cm. There are 2.54 centimeters in an inch. The International spelling for this unit is centimetre.\n\n•", null, "Convert inches to centimeters - Conversion of Measurement ...\n\nThe international inch is defined to be equal to 25.4 millimeters. ›› Definition: Centimeter A centimetre (American spelling centimeter, symbol cm) is a unit of length that is equal to one hundreth of a metre, the current SI base unit of length.\n\n•", null, "Convert 4 Millimeters to Inches\n\nA millimeter, or millimetre, is a unit of length equal to one thouh of a meter. An inch is a unit of length equal to exactly 2.54 centimeters. There are 12 inches in a foot, and 36 inches in a yard.\n\n•", null, "How to Read a Ruler in Centimeters, Inches & Millimeters ...\n\nApr 30, 2018 · Counting from the right end of the metric ruler labeled 30.5 cm (the \"0\" end for the English side), you find that 4.5 inches is equal to 11.5 cm.\n\n•", null, "Millimeters to Inches Conversion\n\nIn 1958 the United States and countries of the Commonwealth of Nations defined the length of the international inch (in) is to be exactly 25.4 millimeters (mm). This Online Conversion Calculator converts in both directions.\n\n•", null, "Convert Millimeters to Centimeters - Inch Calculator\n\nMillimeter is spelled millimetre internationally, and it is equal in length to .1 centimeters, 1000 micrometers, or 5/127 inch. They are represented by the smallest ticks on most metric rulers. To get a reference point of the size, the thickness of a dime is 1.35mm.\n\n•", null, "Convert 5 Millimeters to Inches\n\nA millimeter, or millimetre, is a unit of length equal to one thouh of a meter. An inch is a unit of length equal to exactly 2.54 centimeters. There are 12 inches in a foot, and 36 inches in a yard.\n\n•", null, "Unit Conversion: How to Convert Inches to Centimeters and ...\n\nTo convert inches to centimeters, make use of 1 cm in in (or) 1 centimeter in inches (or) 1 cm to in (or) 1 centimeter to inches. Example 2. To convert millimeters to inches, make use of 1000 mm in in (or) 1000 millimeter in inches (or) 1000 mm to in (or) 1000 millimeter in inches. References. For further reference, please take a look at the following\n\n•", null, "Millimeters to Inches - Unit conversion calculator\n\nWe are the best unit online conversion calculator so you came to the right converter if you want to convert inches, millimeters, cm and all kinds of length units. For example if you want to convert 100 inches to millimeters jut type \"100 inches\" in the box Length to Convert and then click Convert .\n\n•", null, "Convert Inches to Millimeters - (in to mm) - Inch Calculator\n\nConvert inches to millimeters (in to mm) with the length conversion calculator, and learn the inch to millimeter calculation formula. Convert inches to millimeters (in to mm) with the length conversion calculator, and learn the inch to millimeter calculation formula. ... one inch is equal to 2.54 centimeters. The inch is a US customary and ...\n\n•", null, "Convert mm to cm - Length / Distance Conversions\n\nA millimeter is a unit of Length or Distance in the Metric System. The symbol for millimeter is mm. There are 10 millimeters in a centimeter. The International spelling for this unit is millimetre. What is a centimeter (cm)? A centimeter is a unit of Length or Distance in the Metric System. The symbol for centimeter is cm. There are 0.1 centimeters in a millimeter. The International spelling for this unit is centimetre.\n\n•", null, "Inches to cm converter\n\nInches to cm converter. Easily convert Inches to Centimeters, with formula, conversion chart, auto conversion to common lengths, more\n\n•", null, "Centimeters to Inches\n\n1 centimeter (cm) = 0.393701 inch (in). Centimeter (cm) is a unit of Length used in Metric system. Centimeter (cm) is a unit of Length used in Metric system. Inch (in) is .\n\n•", null, "Convert mm to inches - Conversion of Measurement Units\n\nHow many mm in 1 inches? The answer is 25.4. We assume you are converting between millimetre and inch. You can view more details on each measurement unit: mm or inches The SI base unit for length is the metre. 1 metre is equal to 1000 mm, or 39.370078740157 inches. Note that rounding errors may occur, so always check the results.\n\n•", null, "Centimeters To Feet And Inches Calculator\n\nConverting from centimeters to feet and inches ( from cm to ft + in ) is a simple conversion. If you needed to do it manually, you could use 1 cm = .39 in., or 1 cm = .03 ft, and just ply. But this converter is designed to convert an entry in centimeters into both feet and inches.\n\n•", null, "How many centimeters are in one millimeter\n\n1 millimeter = 0.1 centimeters. There are 10 millimeters in a centimeter. Hence a centimeter is one one-hundreth of a meter and a millimeter is one one-thousanth of a meter.\n\n•", null, "Centimeters to Millimeters Conversion\n\nThe centimeter (British spelling: centimetre, abbreviation: cm) is a unit of length in the SI system (metric system). One cm is equal to one hundredth of the meter (British spelling: metre), which is the current SI base unit of length. One centimeter (cm) = 10 millimeters (mm) = 0.01 meter (m) = 0.1 decimeters (dm) = 0.00001 kilometers (km) = 0.393700787401 inch (in) = 0.032808399 feet (ft ...\n\n•", null, "Name: Centimeters and Millimeters\n\nANSWER KEY Centimeters and Millimeters Part 1: Complete the table. centimeters 1 3 12 40 145 millimeters 10 30 120 400 1,450 Part 2: Circle the greater length for each pair. a. 4 cm 45 mm b. 50 mm 6 cm c. 800 cm 1,000 mm d. 3,200 mm 340 cm Part 3: Measure to the nearest centimeter and/or nearest millimeter. nearest cm = 7cm nearest mm = 68 mm\n\n•", null, "Solved: Change default inches to MM - Autodesk Community\n\nRe: Change default inches to MM Good afternoon Sir, Its about the 3d assembly which created in inch form so i would like to make it in mm form but the 3d is a solid bodies and it is not the same as the file of Inventor which is .ipt so im getting hard to convert it to mm form.\n\n•", null, "Actual size of Online Ruler (cm/mm) - PiliApp\n\nYes, I am using {monitor} . Save this settings. No (please select... ) I am using {monitors} I am using -inch screen save settings; I don't know what monitor size is.\n\n• used allen results\n• price cost china\n• balance sheet better\n• crusher run calculate coverage\n• used crusher in portugal\n• machines for quarts processing\n• aesha conveyors and crushing equipment province stat\n• chruser batu mobil cap 1 ton\n• cobbing iron sale\n• tomball tx designs\n• crusher equipment for qurry\n• outdoor conveyor wholesalers\n• forged grinding hy\n• bench grinding combined\n• crushed manganese uses\n• screening efficiency objective\n• komplett glass crusher for sale\n• titanium ore philippines\n• bullet crusher mumbai" ]

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[ "# 3.2 Proof of Space\n\nA Proof of Space protocol is one in which:\n\n• A Verifier can send a challenge to a Prover.\n• The Prover can demonstrate to the Verifier that the Prover is reserving a specific amount of storage space at that precise time.\n\nThe Proof of Space protocol has three components: plotting, proving/farming, and verifying. For more info, see our Details of the chiapos specification, and reference implementation.", null, "## Plotting​\n\nPlotting is the process by which a Prover, who we refer to as a farmer, initializes a certain amount of space. To become a farmer, one must have at least 101.4 GiB available to reserve on their computer (the minimum spec is a Raspberry Pi 4). There is no upper limit to the size of a Chia farm. Several farmers have multi-PiB farms.\n\nAs of Chia 1.2.7, a k32 plot can be created in around five minutes with a high-end machine with 400 GiB of RAM, or six hours with a normal commodity machine, or 12 hours with a slow machine using one CPU core and a few GB of RAM. Opportunities still remain for huge speedups. Furthermore, each plot only needs to be created once; a farmer can farm with the same plots for many years.\n\nPlot sizes are determined by a k parameter, where `space = 780 * k * pow(2, k - 10)`, with a minimum k of 32 (101.4 GiB). The Proof of Space construction is based on Beyond Hellman, but it is nested six times (thereby creating seven tables), and it contains other heuristics to make it practical.\n\nEach of the seven tables in a plot is filled with random-looking data that cannot be compressed. Each table has 2^k entries. Each entry in table i contains two pointers to table i-1 (the previous table). Finally, each table-1 entry contains a pair of integers between 0 and 2^k, called “x-values.” A proof of space is a collection of 64 x-values that have a certain mathematical relationship. The actual on-disk structure and the algorithm required to generate it are quite complicated, but this is the general idea.", null, "Figure 2: Structure of a plot file. The 64 red x-values represent the proof, the 2 green x-values represent the quality.\n\nOnce the Prover has initialized 101.4 GiB, they are ready to receive a challenge and create a proof. One attractive property of this scheme is that it is non-interactive unless the farmer chooses plot NFT style pooling: no registration or online connection is required to create a plot using the original plot format. Nothing hits the blockchain until a reward is won, similar to PoW. For pool portable plots, a farmer only needs a few mojos to create a plot NFT before plotting and then everything has the same characteristics from there.\n\nSee our plotting FAQ for more info.\n\n## Farming​\n\nFarming is the process by which a farmer receives a sequence of 256-bit challenges to prove that they have legitimately put aside a defined amount of storage. In response to each challenge, the farmer checks their plots, generates a proof, and submits any winning proofs to the network for verification.\n\nFor each eligible plot (explained later), a farmer uses the following procedure to generate a full proof of space. Keep in mind that a plot consists of 7 tables (T1-T7) of approximately the same size, as well as 3 checkpoint tables (C1-C3), which are much smaller:\n\n1. The farmer receives a challenge from the VDF\n2. For each eligible plot, extract a k-sized value from the challenge, where k denotes the size of the plot (k32, k33, etc)\n3. Look in the C2 table for a location at which to start scanning the C1 table\n4. Scan the C1 table for the location at which to start scanning the C3 table\n5. Read either one or two C3 parks. The number of parks to read depends on the index and value calculated from the C1 table. This requires an average of 5000 reads (the maximum is 10 000). These are sequential reads of 4 bytes (for an average total of 20 KiB)\n6. Grab all the f7 entries matching the challenge value (which can be 0 or more), along with the index in the table at which they were found\n7. For each matching f7 value, read T7 at the same index where the f7 value was found in its own table, and grab that entry, which is an index into T6\n8. The T6 index contains one line point with two back pointers to T5, four to T4, eight to T3, sixteen to T2 and thirty-two to T1. Each back pointer requires 1 read, so a total of 64 disk reads (1 index from T7, 63 back pointers) are performed to fetch the whole tree of 64 x-values.\n\nSixty-four disk reads require approximately 640 ms on a slow HDD with a 10 ms seek time.\n\nThe following table shows the amount of data read from the seven tables for a k32 plot to fetch a proof of space:\n\n7184488448\n6183258325\n52832516 650\n44832533 300\n38832566 600\n2168325133 200\n1328862283 584\n\nThese numbers are in addition to the aforementioned checkpoint tables, which require up to ten thousand reads of 4 bytes apiece.\n\nSince most proofs generated by this process are not good enough (as discussed in Section 3.5) to be submitted to the network for verification, we can optimize this process by only checking one branch of the tree. This branch will return two of the 64 x-values. The position of the x-values will always be consecutive and will depend on the signage point (eg x0 and x1... or x34 and x35). We hash these x-values to produce a random 256-bit \"quality string.\" This is combined with the difficulty and the plot size to generate the required_iterations. If the required_iterations is less than a certain number, the proof can be included in the blockchain. At this point, we look up the whole proof of space.\n\nBy only looking up one branch to determine the quality string, we can rule out most proofs. This optimization requires only around 7-9 disk seeks and reads, or about 70-90 ms on a slow hard drive.\n\nNOTE: Throughout this website, we'll make a simple assumption that a single disk seek requires 10ms. In reality, this is typically 5-10ms, so we're using a conservative estimate.\n\nThe 10ms estimate also takes into account the time required to transfer data after the seek. While storage industry specs typically assume that large files are being transferred, this does not hold true for Chia farming, where proof lookups only require a tiny amount of data to be transferred. Therefore, for this website, it's safe to assume the transfer is almost instant.\n\nChia also uses a further optimization to disqualify a certain proportion of plots from eligibility for each challenge. This is referred to as the plot filter. The current requirement is that the hash of the plot ID, challenge, and signage point starts with 9 zeros. This excludes 511 out of every 512 plots. The filter hurts everyone equally (except for replotting attackers), and is therefore fair. The plot filter is discussed in greater detail in Section 3.5.\n\nThe plot filter effectively reduces the amount of resources required for farming by 512x -- each plot only requires a few disk reads every few minutes. A farmer with 1 PiB of storage (10,000 plots) will only have an average of 20 plots that pass the filter for each challenge. Even if these plots all are stored on slow HDDs, and connected to a single Raspberry Pi, the average time required to respond to each challenge will be less than two seconds. This is well within the limits to avoid missing out on any challenges.\n\nEach plot file has its own unique private key called a plot key. The plot ID is generated by hashing the plot public key, the farmer public key, and either the pool public key (for OG plots) or the pool contract puzzle hash (for pooled plots). The requirements for signing a proof of space depend on the type of plots being used. See Section 9.3 for details on the keys used for plot construction.\n\nIn practice, the plot key is a 2/2 BLS aggregate public key between a local key stored in the plot and a key stored by the farmer software. For security and efficiency, a farmer may run on one server using this key and signature scheme. This server can then be connected to one or more harvester machines that store the actual plots. Farming requires the farmer key and the local key, but it does not require the pool key, since the pool’s signature can be cached and reused for many blocks.\n\n## Verifying​\n\nAfter the farmer has successfully created a proof of space, the proof can be verified by performing a few hashes and making comparisons between the x-values in the proof. Recall that the proof is a list of 64 x-values, where each x-value is k bits long. For a k32 this is 256 bytes (2048 bits), and is therefore very compact. Verification is very fast, but not quite fast enough to be verified in Solidity on Ethereum (something that would enable trustless transfers between chains), since this verification requires blake3 and chacha8 operations." ]

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[ "# tf.math.lgamma\n\nComputes the log of the absolute value of `Gamma(x)` element-wise.\n\nFor positive numbers, this function computes log((input - 1)!) for every element in the tensor. `lgamma(5) = log((5-1)!) = log(4!) = log(24) = 3.1780539`\n\n#### Example:\n\n``````x = tf.constant([0, 0.5, 1, 4.5, -4, -5.6])\ntf.math.lgamma(x) ==> [inf, 0.5723649, 0., 2.4537368, inf, -4.6477685]\n``````\n\n`x` A `Tensor`. Must be one of the following types: `bfloat16`, `half`, `float32`, `float64`.\n`name` A name for the operation (optional).\n\nA `Tensor`. Has the same type as `x`." ]

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[ "Question\n\n## Introduction\n\nThe number of distinct terms in the expansion of (A + B + C)20 is 10k.\n\n## Let A = 1, B = 2 and C = 3.\n\nLet A = 1, B = 2 and C = 3.\n\nThe expansion of (A + B + C)20 is:\n\n(1 + 2 + 3)20 = (1 x 20) + (2 x 19) + (3 x 18).\n\n## We can write the expansion of (A + B + C)20 as follows:\n\nWe can write the expansion of (A + B + C)20 as follows:\n\n(A + B + C)20 = 10*10*10*10*10*10*10*10*10*10*k.\n\nThe first step is to determine the number of distinct terms in the expansion of (A + B + C)20. Since there are three variables, A, B and C, we need three numbers that will be used as exponents during our calculations. Let’s choose 1 for A, 2 for B and 3 for C. This gives us an expression like this:\n\n(1+2)(3+4)20 = 1*2*3*4*5*6…\n\n1. # What Is The Number Of Distinct Terms In The Expansion Of (A + B + C)20?\n\n## Introduction\n\nIn mathematics, a word problem is a type of question that asks for an answer where the solution does not involve the use of basic arithmetic operations. The name comes from the Latin word probare, meaning “to test” or “to prove”. One of the most famous word problems in mathematics is called The Game of Life. It’s a 2D Conway Game, which is a type of mathematical puzzle. In this blog post, we will explore one such example: What is the number of distinct terms in the expansion of (A + B + C)20? This problem can be solved using basic algebra and simple geometric concepts. So if you are looking to brush up on your equations and geometry, read on!\n\n## The Problem\n\nThe problem of expanding (A + B + C) is to find the total number of distinct terms. The expansion process is initiated with aterm, and each term in the expansion is created by adding another term to the original one. At first, this process seems straightforward, but as more and more terms are added, the number of possible expansions becomes incredibly large. In fact, it’s impossible to determine the exact number of expansions because there are an infinite amount of ways that terms could be combined. However, we can use a method known as counting rigorously to come up with a ballpark estimate for the total number of expansions that could take place.\n\nAssuming that each term in the expansion has an equal probability of being chosen, the total number of possible expansions is simply ÷ 3 + 1 = 2 + 1 = 3. Therefore, there are an estimated 31 different expansions that could take place.\n\n## The Solution\n\nThe expansion of (A + B + C) is the sum of the expansions of each term. The number of distinct terms in this expansion is 3.\n\n## Results\n\nThe number of distinct terms in the expansion of (A + B + C) is 6.\n\n## Conclusion\n\nThe number of distinct terms in the expansion of (A + B + C)20 is 10.\n\n2. When I was in sixth grade, my teacher taught us about combinations. A combination is a group of items selected from a larger set. For example, if you have 12 fruits and want to make a fruit salad, there are many different possible combinations (permutations) of 3 fruits each morning for breakfast: 12!/(4!*2!) = 144 possible combinations. However, we usually only care about the number of distinct terms in such an expansion because some choices end up being duplicates (such as having two apples in your fruit salad). If we were calculating how many distinct terms there would be in an expansion like (a + b + c)20 (i.e., what would be the total number of ways you could choose 20 numbers out of 100), then we would only need to consider the following:\n\nLet’s review the formula for the number of terms in the expansion of (a + b + c)n:\n\n• The sum of all of these terms can be calculated by adding together all possible ways that n things can be chosen out of three things.\n• To see why this works, let’s look at an example. Let’s say we want to find out how many ways there are for me, my friend John and our mutual friend Sally to go out on Saturday night. There are three options: A) I go alone with Sally; B) I go alone with John; C) We all go together as a group (which means none of us goes solo). This gives us 3 x 2 x 1 = 6 different possibilities!\n\n## Let’s look at a few examples.\n\nExample 1: The expansion of (A + B + C)20 = (1 + 2 + 3 + 4 + 5 + 6 + 7)20 = 24(A)(B)(C).\n\nExample 2: The expansion of (A – B)20 = (-1)(-2)(-3)20 = 20(-1)(-2)(-3).\n\nExample 3: The expansion of ((A – B)/C)20 = ((1)/2)((5)/10)((-6)/30).\n\n## The formula for the number of terms in the expansion of (a + b + c)n is\n\nThe formula for the number of terms in the expansion of (a + b + c)n is n!/(a! a! a! … a!). The number of terms in the expansion of (a + b + c)20 is 20!, or 400,000.\n\n## So, what is the sum of all of these terms?\n\nSo, what is the sum of all of these terms?\n\nIt’s 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 and so on…\n\nSo we can say that:\n\n(a) The number of distinct terms in the expansion of (a+b+c)^20 = [(1+2+4)/2]^20 = [(1+2)/2]^(20*19) = (1/2)^{(20*19)}\n\n## This formula gives us a quick way to calculate the answer.\n\nThe formula is:\n\n\\$\\$sum_{i=0}^n {(a + b + c)^{i}} = left(frac{1}{2}a^2+frac{1}{3}b^3-frac{1}{5}right)^{n}.\\$\\$\n\nYou can use this formula to quickly calculate the number of terms in the expansion of (a + b + c)20, which is 218318800000.\n\nWe hope this has been a helpful introduction to the formula for expanding binomials. If you want to get more practice with it, check out our other blog post on calculating the number of terms in an expansion. We also have a free worksheet available if you want something simpler that can be done in class or at home!" ]

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[ "", null, "", null, "", null, "", null, "• Previous Article\nEquivalent formulations for steady periodic water waves of fixed mean-depth with discontinuous vorticity\n• DCDS Home\n• This Issue\n• Next Article\nTraveling fronts bifurcating from stable layers in the presence of conservation laws\nMay  2017, 37(5): 2653-2668. doi: 10.3934/dcds.2017113\n\n## Existence and uniqueness of positive solutions for a class of logistic type elliptic equations in $\\mathbb{R}^N$ involving fractional Laplacian\n\n 1 Departamento de Matemática, Universidad Técnica Federico Santa María, Casilla: V-110, Avda. España 1680, Valparaíso, Chile 2 Department of Mathematics, Jiangxi Normal University, Nanchang, Jiangxi 330022, China\n\nReceived  November 2015 Revised  January 2017 Published  February 2017\n\nIn this paper, we study the existence and uniqueness of positive solutions for the following nonlinear fractional elliptic equation:\n $\\begin{eqnarray*}(-Δ)^α u=λ a(x)u-b(x)u^p&\\ \\ \\ {\\rm in}\\,\\,\\mathbb{R}^N, \\end{eqnarray*}$\nwhere $α∈(0, 1)$, $N≥ 2$, $λ >0$, $a$ and $b$ are positive smooth function in $\\mathbb{R}^N$ satisfying\n $a\\left( x \\right) \\to {a^\\infty } > 0\\;\\;\\;\\;{\\rm{and}}\\;\\;\\;b\\left( x \\right) \\to {b^\\infty } > 0\\;\\;\\;\\;{\\rm{as}}\\;\\;\\;{\\rm{|}}\\mathit{x}{\\rm{|}} \\to \\infty$\nOur proof is based on a comparison principle and existence, uniqueness and asymptotic behaviors of various boundary blow-up solutions for a class of elliptic equations involving the fractional Laplacian.\nCitation: Alexander Quaas, Aliang Xia. Existence and uniqueness of positive solutions for a class of logistic type elliptic equations in $\\mathbb{R}^N$ involving fractional Laplacian. Discrete and Continuous Dynamical Systems, 2017, 37 (5) : 2653-2668. doi: 10.3934/dcds.2017113\n##### References:\n N. Abatangelo, On a fractional keller osserman condition, arXiv: 1412.6298 [math. AP].", null, "N. Abatangelo, Large s-harmonic functions and boundary blow-up solutions for the fractional Laplacian, Discrete Contin. Dyn. Syst., 35 (2015), 5555-5607.  doi: 10.3934/dcds.2015.35.5555.", null, "", null, "G. Barles, E. Chasseigne and C. Imbert, On the Dirichlet problem for second-order elliptic integro-differential equations, Indiana Univ. Math. J., 57 (2008), 213-246.  doi: 10.1512/iumj.2008.57.3315.", null, "", null, "L. Caffarelli and L. Silvestre, Regularity theory for fully nonlinear integro-differential equations, Comm. Pure Appl. Math., 62 (2009), 597-638.  doi: 10.1002/cpa.20274.", null, "", null, "L. Caffarelli and L. Silvestre, Regularity results for nonlocal equations by approximation, Arch. Ration. Mech. Anal., 200 (2011), 59-88.  doi: 10.1007/s00205-010-0336-4.", null, "", null, "H. Chen, P. Felmer and A. Quaas, Large solutions to elliptic equations involving fractional Laplacian, Ann. Inst. H. Poincaré Anal. Non Linéaire, 32 (2015), 1199-1228.  doi: 10.1016/j.anihpc.2014.08.001.", null, "", null, "H. Chen, H. Hajaiej and Y. Wang, Boundary blow-up solutions to fractional elliptic equations in a measure framework, Discrete Contin. Dyn. Syst., 36 (2016), 1881-1903.", null, "M. G. Crandall, H. Ishii and P. L. Lions, User's guide to viscosity solutions of second order partial differential equations, Bull. Amer. Soc., 27 (1992), 1-67.", null, "Y. Du and Z. Guo, Boundary blow-up solutions and their applications in quasilinear elliptic equations, J. Anal. Math. , 89 (2003), 277-302; Corrigendum. J. Anal. Math. , 107 (2009), 391-393.", null, "Y. Du and L. Ma, Logistic type equations on $\\mathbb{R}^N$ by a squeezing method involving boundary blow up solutions, J. London Math. Soc.(2), 64 (2001), 107-124.  doi: 10.1017/S0024610701002289.", null, "", null, "P. Felmer and A. Quaas, Boundary blow up solutions for fractional elliptic equations, Asymptot. Anal., 78 (2012), 123-144.", null, "A. Fiscella, R. Servadei and E. Valdinoci, Density properties for fractional Sobolev spaces, Ann. Acad. Sci. Fenn. Math., 40 (2015), 235-253.  doi: 10.5186/aasfm.2015.4009.", null, "", null, "L. del Pezzo and A. Quaas, Global bifurcation for fractional $p$-Laplacian and application, To appear in Zeitschrift Fur Analysis und ihre Anwendungen. arXiv: 1412.4722 [math. AP]", null, "R. Servadei and E. Valdinoci, Variational methods for non-local operators of elliptic type, Discrete Contin. Dyn. Syst., 33 (2013), 2105-2137.", null, "R. Servadei and E. Valdinoci, Weak and viscosity solutions of the fractional Laplace equation, Publ. Mat., 58 (2014), 133-154.  doi: 10.5565/PUBLMAT_58114_06.", null, "", null, "M. Struwe, Variational Methods. Applications to Nonlinear Partial Differential Equations and Hamiltonian Systems, Springer-Verlag, Berlin, 1990. xiv+244 pp.", null, "show all references\n\n##### References:\n N. Abatangelo, On a fractional keller osserman condition, arXiv: 1412.6298 [math. AP].", null, "N. Abatangelo, Large s-harmonic functions and boundary blow-up solutions for the fractional Laplacian, Discrete Contin. Dyn. Syst., 35 (2015), 5555-5607.  doi: 10.3934/dcds.2015.35.5555.", null, "", null, "G. Barles, E. Chasseigne and C. Imbert, On the Dirichlet problem for second-order elliptic integro-differential equations, Indiana Univ. Math. J., 57 (2008), 213-246.  doi: 10.1512/iumj.2008.57.3315.", null, "", null, "L. Caffarelli and L. Silvestre, Regularity theory for fully nonlinear integro-differential equations, Comm. Pure Appl. Math., 62 (2009), 597-638.  doi: 10.1002/cpa.20274.", null, "", null, "L. Caffarelli and L. Silvestre, Regularity results for nonlocal equations by approximation, Arch. Ration. Mech. Anal., 200 (2011), 59-88.  doi: 10.1007/s00205-010-0336-4.", null, "", null, "H. Chen, P. Felmer and A. Quaas, Large solutions to elliptic equations involving fractional Laplacian, Ann. Inst. H. Poincaré Anal. Non Linéaire, 32 (2015), 1199-1228.  doi: 10.1016/j.anihpc.2014.08.001.", null, "", null, "H. Chen, H. Hajaiej and Y. Wang, Boundary blow-up solutions to fractional elliptic equations in a measure framework, Discrete Contin. Dyn. Syst., 36 (2016), 1881-1903.", null, "M. G. Crandall, H. Ishii and P. L. Lions, User's guide to viscosity solutions of second order partial differential equations, Bull. Amer. Soc., 27 (1992), 1-67.", null, "Y. Du and Z. Guo, Boundary blow-up solutions and their applications in quasilinear elliptic equations, J. Anal. Math. , 89 (2003), 277-302; Corrigendum. J. Anal. Math. , 107 (2009), 391-393.", null, "Y. Du and L. Ma, Logistic type equations on $\\mathbb{R}^N$ by a squeezing method involving boundary blow up solutions, J. London Math. Soc.(2), 64 (2001), 107-124.  doi: 10.1017/S0024610701002289.", null, "", null, "P. Felmer and A. Quaas, Boundary blow up solutions for fractional elliptic equations, Asymptot. Anal., 78 (2012), 123-144.", null, "A. Fiscella, R. Servadei and E. Valdinoci, Density properties for fractional Sobolev spaces, Ann. Acad. Sci. Fenn. Math., 40 (2015), 235-253.  doi: 10.5186/aasfm.2015.4009.", null, "", null, "L. del Pezzo and A. Quaas, Global bifurcation for fractional $p$-Laplacian and application, To appear in Zeitschrift Fur Analysis und ihre Anwendungen. arXiv: 1412.4722 [math. AP]", null, "R. Servadei and E. Valdinoci, Variational methods for non-local operators of elliptic type, Discrete Contin. Dyn. Syst., 33 (2013), 2105-2137.", null, "R. Servadei and E. Valdinoci, Weak and viscosity solutions of the fractional Laplace equation, Publ. Mat., 58 (2014), 133-154.  doi: 10.5565/PUBLMAT_58114_06.", null, "", null, "M. Struwe, Variational Methods. Applications to Nonlinear Partial Differential Equations and Hamiltonian Systems, Springer-Verlag, Berlin, 1990. xiv+244 pp.", null, "Nicola Abatangelo. Large $s$-harmonic functions and boundary blow-up solutions for the fractional Laplacian. Discrete and Continuous Dynamical Systems, 2015, 35 (12) : 5555-5607. doi: 10.3934/dcds.2015.35.5555 Zhifu Xie. General uniqueness results and examples for blow-up solutions of elliptic equations. Conference Publications, 2009, 2009 (Special) : 828-837. doi: 10.3934/proc.2009.2009.828 Antonio Vitolo, Maria E. Amendola, Giulio Galise. On the uniqueness of blow-up solutions of fully nonlinear elliptic equations. Conference Publications, 2013, 2013 (special) : 771-780. doi: 10.3934/proc.2013.2013.771 Huyuan Chen, Hichem Hajaiej, Ying Wang. Boundary blow-up solutions to fractional elliptic equations in a measure framework. Discrete and Continuous Dynamical Systems, 2016, 36 (4) : 1881-1903. doi: 10.3934/dcds.2016.36.1881 Ronghua Jiang, Jun Zhou. Blow-up and global existence of solutions to a parabolic equation associated with the fraction p-Laplacian. Communications on Pure and Applied Analysis, 2019, 18 (3) : 1205-1226. doi: 10.3934/cpaa.2019058 Daniele Bartolucci, Changfeng Gui, Yeyao Hu, Aleks Jevnikar, Wen Yang. Mean field equations on tori: Existence and uniqueness of evenly symmetric blow-up solutions. Discrete and Continuous Dynamical Systems, 2020, 40 (6) : 3093-3116. doi: 10.3934/dcds.2020039 Jorge García-Melián, Julio D. Rossi, José C. Sabina de Lis. Elliptic systems with boundary blow-up: existence, uniqueness and applications to removability of singularities. Communications on Pure and Applied Analysis, 2016, 15 (2) : 549-562. doi: 10.3934/cpaa.2016.15.549 Jian Zhang, Shihui Zhu, Xiaoguang Li. Rate of $L^2$-concentration of the blow-up solution for critical nonlinear Schrödinger equation with potential. Mathematical Control and Related Fields, 2011, 1 (1) : 119-127. doi: 10.3934/mcrf.2011.1.119 Shota Sato. Blow-up at space infinity of a solution with a moving singularity for a semilinear parabolic equation. Communications on Pure and Applied Analysis, 2011, 10 (4) : 1225-1237. doi: 10.3934/cpaa.2011.10.1225 Tetsuya Ishiwata, Shigetoshi Yazaki. A fast blow-up solution and degenerate pinching arising in an anisotropic crystalline motion. Discrete and Continuous Dynamical Systems, 2014, 34 (5) : 2069-2090. doi: 10.3934/dcds.2014.34.2069 Frédéric Abergel, Jean-Michel Rakotoson. Gradient blow-up in Zygmund spaces for the very weak solution of a linear elliptic equation. Discrete and Continuous Dynamical Systems, 2013, 33 (5) : 1809-1818. doi: 10.3934/dcds.2013.33.1809 Juntang Ding, Chenyu Dong. Blow-up results of the positive solution for a weakly coupled quasilinear parabolic system. Discrete and Continuous Dynamical Systems - B, 2021  doi: 10.3934/dcdsb.2021222 Qiang Lin, Xueteng Tian, Runzhang Xu, Meina Zhang. Blow up and blow up time for degenerate Kirchhoff-type wave problems involving the fractional Laplacian with arbitrary positive initial energy. Discrete and Continuous Dynamical Systems - S, 2020, 13 (7) : 2095-2107. doi: 10.3934/dcdss.2020160 Van Duong Dinh. On blow-up solutions to the focusing mass-critical nonlinear fractional Schrödinger equation. Communications on Pure and Applied Analysis, 2019, 18 (2) : 689-708. doi: 10.3934/cpaa.2019034 Binhua Feng. On the blow-up solutions for the fractional nonlinear Schrödinger equation with combined power-type nonlinearities. Communications on Pure and Applied Analysis, 2018, 17 (5) : 1785-1804. doi: 10.3934/cpaa.2018085 Vo Van Au, Jagdev Singh, Anh Tuan Nguyen. Well-posedness results and blow-up for a semi-linear time fractional diffusion equation with variable coefficients. Electronic Research Archive, 2021, 29 (6) : 3581-3607. doi: 10.3934/era.2021052 Mingqi Xiang, Die Hu. Existence and blow-up of solutions for fractional wave equations of Kirchhoff type with viscoelasticity. Discrete and Continuous Dynamical Systems - S, 2021, 14 (12) : 4609-4629. doi: 10.3934/dcdss.2021125 Wenjun Liu, Jiangyong Yu, Gang Li. Global existence, exponential decay and blow-up of solutions for a class of fractional pseudo-parabolic equations with logarithmic nonlinearity. Discrete and Continuous Dynamical Systems - S, 2021, 14 (12) : 4337-4366. doi: 10.3934/dcdss.2021121 Nadjat Doudi, Salah Boulaaras, Nadia Mezouar, Rashid Jan. Global existence, general decay and blow-up for a nonlinear wave equation with logarithmic source term and fractional boundary dissipation. Discrete and Continuous Dynamical Systems - S, 2022  doi: 10.3934/dcdss.2022106 C. Y. Chan. Recent advances in quenching and blow-up of solutions. Conference Publications, 2001, 2001 (Special) : 88-95. doi: 10.3934/proc.2001.2001.88\n\n2020 Impact Factor: 1.392" ]

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[ "Home Misc Find the sum of even Fibonacci numbers.\n\n# Find the sum of even Fibonacci numbers.\n\n0 comment\n\nQuestion: Find the sum of even fibonacci numbers upto a limit N.\n\nInput:\n100\nOutput: 44\n\nFibonacci numbers are a miracle of Math and are defined as:-\n\n.wp-block-code {\nborder: 0;\n}\n\n.wp-block-code > span {\ndisplay: block;\noverflow: auto;\n}\n\n.shcb-language {\nborder: 0;\nclip: rect(1px, 1px, 1px, 1px);\n-webkit-clip-path: inset(50%);\nclip-path: inset(50%);\nheight: 1px;\nmargin: -1px;\noverflow: hidden;\nposition: absolute;\nwidth: 1px;\nword-wrap: normal;\nword-break: normal;\n}\n\n.hljs {\nbox-sizing: border-box;\n}\n\n.hljs.shcb-code-table {\ndisplay: table;\nwidth: 100%;\n}\n\n.hljs.shcb-code-table > .shcb-loc {\ncolor: inherit;\ndisplay: table-row;\nwidth: 100%;\n}\n\n.hljs.shcb-code-table .shcb-loc > span {\ndisplay: table-cell;\n}\n\n.wp-block-code code.hljs:not(.shcb-wrap-lines) {\nwhite-space: pre;\n}\n\n.wp-block-code code.hljs.shcb-wrap-lines {\nwhite-space: pre-wrap;\n}\n\n.hljs.shcb-line-numbers {\nborder-spacing: 0;\ncounter-reset: line;\n}\n\n.hljs.shcb-line-numbers > .shcb-loc {\ncounter-increment: line;\n}\n\n.hljs.shcb-line-numbers .shcb-loc > span {\n}\n\n.hljs.shcb-line-numbers .shcb-loc::before {\nborder-right: 1px solid #ddd;\ncontent: counter(line);\ndisplay: table-cell;\ntext-align: right;\n-webkit-user-select: none;\n-moz-user-select: none;\n-ms-user-select: none;\nuser-select: none;\nwhite-space: nowrap;\nwidth: 1%;\n}\nf(0) = 0\nf(1) = 1\nf(n) = f(n-1) + f(n-2); for n >= 2Code language: TeX (tex)\n\nThus the Fibonacci series can be given as\n0, 1, 1, 2, 3, 5, 8, 13, 21…\n\nLet us try to understand the question. We need to find all the Fibonacci numbers till a limit ‘N’ and then sum only the even numbers in them.\nExample:\nN = 10\nFibonacci numbers: 0, 1, 1, 2, 3, 5, 8\nSum of even numbers: 2 + 8 = 10\n\nN = 100\nFibonacci numbers: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89\nSum of even numbers: 2 + 8 + 34 = 44\n\nIf the value of ‘N’ is less than 10^6, this problem can be approached by a naive method of finding the Fibonacci numbers and then summing the even ones.\nSurprisingly, there are very few even Fibonacci numbers till very large limits of N.\nHave a look at the Fibonacci series and this interesting page about the first 300 Fibonacci numbers.\n\nFor this problem let us consider the maximum limit of N to be 10^16. That should be a pretty good limit. We need to understand the concept of pre-processing in this case. Pre-processing is used in scenarios where we can save a lot of computation power if we have some insight into what we are doing.\n\nFor example, here is a list of even Fibonacci numbers till 10^16. It is a very small list.\n\n0\n2\n8\n34\n144\n610\n2584\n10946\n46368\n196418\n832040\n3524578\n14930352\n63245986\n267914296\n1134903170\n4807526976\n20365011074\n86267571272\n365435296162\n1548008755920\n6557470319842\n27777890035288\n117669030460994\n498454011879264\n2111485077978050\n8944394323791464\n37889062373143906Code language: TeX (tex)\n\nThus we just need to query this list for prime numbers till ‘N’\n\n.hljs > mark.shcb-loc { background-color: #ddf6ff; }public class EvenFibonacciNumbers {\n\npublic static void main(String[] args) {\n\nList<Long> evenFibonacciList = new ArrayList<>();\n\nScanner scanner = new Scanner(System.in);\nint cases = scanner.nextInt();\nfor (int i = 0; i < cases; i++) {\n\nint N = scanner.nextInt();\n\nlong sum = 0;\nfor (Long aLong : evenFibonacciList) {\nif (N > aLong)\nsum += aLong;\nelse\nbreak;\n}\n\nSystem.out.println(sum);\n}\n\nscanner.close();\n}\n}\nCode language: Java (java)\n\nYou might also want to look at:\n\nA working implementation of the code can be found here.\nThis problem can also be found on Hackerrank and ProjectEuler.\n\nTime Complexity: O(1)\n\n#### You may also like\n\nThis website uses cookies to improve your experience. We'll assume you're ok with this, but you can opt-out if you wish. Accept Read More" ]

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[ "# Full encryption of IDs, partial decryption with customer specific key\n\nI am looking for a \"partial\" encryption/decryption algorithm.\n\nLet's say I am encrypting the following List of IDs (long values) $[1,2,3,4]$ (with a secret key) to an encrypted string $e$, I would like to produce several keys $k_1,k_2,...,k_n$, which can decrypt just a subset (some IDs).\n\nFor example:\n\nval rawIds = [1,2,3,4,8] # Array of Long\nval e = encrypt(rawIds, secretKey) # An encrypted String\n\ndecrypt(k1, e) = # Decryption of e with key k1\ndecrypt(k2, e) = [2,3] # Decryption of e with key k2\n\n\nBoth the encryption and decryption must be very fast in terms of computation time (<1ms). We want to produce keys, which can decrypt the value $e$ in way that only a subset of IDs can be decrypted.\n\n$k_1$ for example is able to decrypt $e$ (but only retrieves $ID_1$) where $k_2$ can decrypt the same encrypted value but retrieves $ID_2$ and $ID_3$. He might even get $ID_5$, but this was not in the raw string.\n\nWhen creating the keys, we could add the IDs which can be extracted somehow to the decryption key.\n\nUpdate:\n\nThe possible values of the IDs are known, there is a predefined set (let's call them $ID_1,ID_2,...,ID_n$).\n\nThe business purpose is the following:\n\n• I have 2 customers, where I know that customer1 is only allowed to see $ID_1$, therefore I create a key $k_1$ (and add the meta information $ID_1$).\n• The second customer is allowed to see $ID_2$, $ID_3$ and $ID_5$ (therefore I create a special key $k_2$ for him).\n• Both customers will get the same encrypted value e, and both can decrypt it but will get back a list of IDs, where every element is allowed to be extracted by that customer).\n\nThis was meant by: I add the IDs to the decryption key.\n\n• It's only one entity which is encrypting, don't need to decrypt with the secret key\n• No need to check authenticity of the encrypted value (always trust)\n• When creating keys, the actual IDs (Long values) which can be extracted can be added to the actual key.\n\nWhere to start? Has anybody an idea of a similar existing approach?\n\nSo lets assume a few things:\n\n1. just symmetric primitives suffice;\n2. a symmetric key derivation function and single block encrypt with a 64 bit block cipher is sufficiently fast;\n3. the ID's are unique and not related to customers;\n4. we're not afraid of customers sharing ID's;\n5. there is protection against customers simply guessing ID's;\n\nScheme:\n\n• establish a master key $k_m$\n• derive customer keys and distribute to customers: $k_{c,name} = KDF(k_m, name)$\n• derive a data key for each ID with index $i$: $k_{id, i} = KDF(k_m, i)$\n• create and distribute the encrypted array where each entry $C_i = E(k_{id, i}, ID_i)$\n\nNow comes the tricky part:\n\n• when a customer wants to get the ID's he sends you the set of indices to the ID's;\n• you calculate the customer key and a customer specific data key for each entry: $k_{c,name} = KDF(k_m, name)$ and $k_{c,name,i} = KDF(k_{c,name}, i)$\n• you calculate the difference between the data key and customer specific data key for each entry: $k_{x,i} = XOR(k_{c,name,i}, k_{id, i})$\n• you send a map consisting of elements $(i, k_{x, i})$ back to the customer\n• the customer can now perform $k_{id,i} = XOR(k_{c,name,i}, k_{x, i})$ for each entry\n• and of course calculate $ID_i = D(k_{id, i}, C_i)$" ]

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[ "# Difference Between Calculus and Other Math Subjects", null, "", null, "On the left, a man is pushing a crate up a straight incline. On the right, a man\nis pushing the same crate up a curving incline. The problem in both cases is to\ndetermine the amount of energy required to push the crate to the top. For the problem\non the left, you can use algebra\nand trigonometry to solve the problem. For the problem on the right, you need calculus.\nWhy do you need calculus with the problem on the right and not the left?\n\nThis is because with the straight incline, the man pushes with an unchanging force\nand the crate goes up the incline at an unchanging speed. With the curved incline\non the right, things are constantly changing. Since the steepness of the incline\nis constantly changing, the amount of energy expended is also changing. This is\nwhy calculus is described as “the mathematics of change”. Calculus takes regular\nrules of math and applies them to evolving problems.\n\nWith the curving incline problem, the algebra and trigonometry that you use is the\nsame, the difference is that you have to break up the curving incline problem into\nsmaller chunks and do each chunk separately. When zooming in on a small portion\nof the curving incline, it looks as if it is a straight line:", null, "Then, because it is straight, you can solve the small chunk just like the straight\nincline problem. When all of the small chunks are solved, you can just add them\nup.\n\nThis is basically the way calculus works – it takes problems that cannot be done\nwith regular math because things are constantly changing, zooms in on the changing\ncurve until it becomes straight, and then it lets regular math finish off the problem.\nWhat makes calculus such a brilliant achievement is that it actually zooms in infinitely.\nIn fact, everything you do in calculus involves infinity in one way or another,\nbecause if something is constantly changing, it is changing infinitely from each\ninfinitesimal moment to the next. All of calculus relies on the fundamental principle\nthat you can always use approximations of increasing accuracy to find the exact\nanswer. Just like you can approximate a curve by a series of straight lines, you\ncan also approximate a spherical solid by a series of cubes that fit inside the\nsphere.\n\nScroll to Top" ]

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[ "# Question Solved1 AnswerA Review Constants A vertical, 1.10 m length of 18 gauge (diameter of 1.024 mm) copper wire has a 100.0 N ball hanging from it. Express your answer with the appropriate units. X = 0.733 m Submit Previous Answers ✓ Correct Part B A 500.0 N ball now replaces the original ball. What is the change in the wavelength of the third harmonic caused by replacing the light ball with the heavy one? Hint Young's Modulus for copper is 11 x 100 Pa Express your answer with the appropriate units. UA ? Δλ = Value m Submit Previous Answers Request Answer X Incorrect; Try Again; 4 attempts remaining", null, "", null, "Transcribed Image Text: A Review Constants A vertical, 1.10 m length of 18 gauge (diameter of 1.024 mm) copper wire has a 100.0 N ball hanging from it. Express your answer with the appropriate units. X = 0.733 m Submit Previous Answers ✓ Correct Part B A 500.0 N ball now replaces the original ball. What is the change in the wavelength of the third harmonic caused by replacing the light ball with the heavy one? Hint Young's Modulus for copper is 11 x 100 Pa Express your answer with the appropriate units. UA ? Δλ = Value m Submit Previous Answers Request Answer X Incorrect; Try Again; 4 attempts remaining\nMore\nTranscribed Image Text: A Review Constants A vertical, 1.10 m length of 18 gauge (diameter of 1.024 mm) copper wire has a 100.0 N ball hanging from it. Express your answer with the appropriate units. X = 0.733 m Submit Previous Answers ✓ Correct Part B A 500.0 N ball now replaces the original ball. What is the change in the wavelength of the third harmonic caused by replacing the light ball with the heavy one? Hint Young's Modulus for copper is 11 x 100 Pa Express your answer with the appropriate units. UA ? Δλ = Value m Submit Previous Answers Request Answer X Incorrect; Try Again; 4 attempts remaining", null, "", null, "", null, "", null, "" ]

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[ "# $f: X \\to Y$ is a smooth map and $\\omega$ is a $p$-form on $Y$, what is $\\omega[f(x)]$?\n\nI am reading Guillemin and Pollack's Differential Topology Page 163:\n\nIf $f: X \\to Y$ is a smooth map and $\\omega$ is a $p$-form on $Y$, define a $p$-form $f^*\\omega$ on $X$ as follows: $$f^*\\omega(x) = (df_x)^*\\omega[f(x)].$$\n\nSo my question is, what is $\\omega[f(x)]$?\n\nI have been trying to make this question self-contained, but here is the whole background: Definition of pullback. In that question, I have shown $(df_x)^*T(v_1, \\dots, v_p) = T \\circ df_x (v_1, \\dots, v_p)$ and $\\omega$ is $(df_x)^*T$. But I want the missing piece about what that $\\omega[f(x)]$ is equal to, to make sense of the definition of $f^* \\omega (x)$.\n\nYou can write your differential form (at least locally) as a sum of \"pure forms\", i.e. forms of the type $$\\omega_\\alpha(y) = \\alpha(y)dx^{i_1}\\wedge\\ldots\\wedge dx^{i_p}$$ with $\\alpha$ a smooth function. Then $$f^*\\omega_\\alpha(x) = (df_x)^*[(\\alpha\\circ f)(x)dx^{i_1}\\wedge\\ldots\\wedge dx^{i_p}]$$ In other words, $\\omega[f(x)]$ means that you precompose the \"weight functions\" of the pure forms with $f$.\n• Oh! So you are saying, $\\omega$ is defined to be $$\\omega_\\alpha(y) = \\alpha(y)dx^{i_1}\\wedge\\ldots\\wedge dx^{i_p}.$$ Hence $$\\omega[f(x)] = (\\alpha\\circ f)(x)dx^{i_1}\\wedge\\ldots\\wedge dx^{i_p}.$$ Therefore, $$f^*\\omega(x) = (df_x)^*\\omega[f(x)] = (df_x)^*[(\\alpha\\circ f)(x)dx^{i_1}\\wedge\\ldots\\wedge dx^{i_p}].$$ – WishingFish Jul 31 '13 at 1:46\n• @WishingFish Yes, you got it correctly, except for the little detail that locally a $p$-form can be written as a sum of pure forms. If you take the wedge of two forms (written as sums of pure forms), then the wedge distributes over the sum. – Daniel Robert-Nicoud Jul 31 '13 at 7:51\n• I hope I got it correctly - so it should be, $$f^*\\omega_\\alpha(x) = (df_x)^* \\sum_{1 \\leq i_1 < \\cdots < i_k \\leq \\dim X}[(\\alpha\\circ f)(x)dx^{i_1}\\wedge\\ldots\\wedge dx^{i_p}]?$$ Thank you! – WishingFish Jul 31 '13 at 17:49\n$\\omega[f(x)]$ is the value of the form $\\omega$ at the point $f(x)$. It is a vector in the $p$-th exterior power of the cotangent space at $f(x)$. (Remember that a $p$-form is a smooth function assigning to each point of a manifold such a vector.) I guess you were just confused by the use of square brackets, which were, I think, intended just to make it easier to read; round brackets would have been OK too, and it seems they didn't cause you any problems with the entirely analogous expresson $f^*\\omega(x)$ on the left side of the equation.\n• Thank you so much Andreas. However, as I updated, I am looking for some expression making $\\omega[f(x)]$ computable, because right on the next page asks me to prove $f^*(w \\wedge \\theta) = (f^*w) \\wedge (f^* \\theta)$. – WishingFish Jul 30 '13 at 22:49" ]

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[ "# Use the 41 polnts log(x(x? laws 7)-1/2) U puedxa and simplify the expression_\n\n###### Question:\n\nUse the 41 polnts log(x(x? laws 7)-1/2) U puedxa and simplify the expression_", null, "", null, "#### Similar Solved Questions\n\n##### Find each sum. $\\frac{9}{10}+\\left(-\\frac{11}{8}\\right)$\nFind each sum. $\\frac{9}{10}+\\left(-\\frac{11}{8}\\right)$...\n##### Show that if z = x + iy,then: I cosh ~z | > sinhxHint: use the equation: I cosh z |? = sinh? x + cos? y\nShow that if z = x + iy,then: I cosh ~z | > sinhx Hint: use the equation: I cosh z |? = sinh? x + cos? y...\n##### 9. Find and o for tyz - tan(.rº + y2 + 2) = 2.xyz.\n9. Find and o for tyz - tan(.rº + y2 + 2) = 2.xyz....\n##### Question 2: TRUE or FALSE: The regression equation is always the best predictor of a y-value...\nQuestion 2: TRUE or FALSE: The regression equation is always the best predictor of a y-value for a given value of x. Defend your answer...." ]

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[ "", null, "# arduino UNO A/D how to use A0 to A5 for ladder net w/o mux\n\narduino UNO A/D\nI am trying to how to measure the voltages of a 6 series resistors in a ladder network, using the Arduino UNO A/D INTERNAL multiplexer by activating A0 to A5 sequentially.\n\nI have found a lot of examples using 6 mechanical switches to connect the resistors individually to the terminals A0 to A5.\n\nAs a newbie I am having trouble making these connections in by using software ONLY, without\nEXTERNAL electronics ( including an EXTERNAL multiplexer or mechanical switches).\n\nQUESTION: (1) Is it possible to use a for() loop to connect the 6 resistor ladder network\nto the A/D in a sequential manner ?\n(2) Please give me an example of how this can be accomplished.\n\nI have tried to do that but I my attempts to use a for() loop indexing the terminals\nA0 to A5 as Ai have failed!\n\nsomething like this you mean?\n\n//polls A0-A5 every 1s and prints output to serial monitor\n\n``````const uint8_t Ain_size = 6;\nuint8_t Ain[Ain_size] = {A0, A1, A2, A3, A4, A5};\n\nvoid setup() {\n// put your setup code here, to run once:\nSerial.begin(115200);\n}\n\nvoid loop() {\n// put your main code here, to run repeatedly:\n\nfor (uint8_t i = 0; i < Ain_size; ++i) {\nSerial.print(\"input \");\nSerial.print(i);\nSerial.print(\": \");" ]

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[ "# AMCS 260 Design and Analysis of Algorithms\n\nCourse\n\n## Description\n\nReview of algorithm analysis (search in ordered array, binary insertion sort, merge sort, 2-3 trees, asymptotic notation). Divide and conquer algorithms (master theorem, integer multiplication, matrix multiplication, fast Fourier transform). Graphs (breadth-first search, connected components, topological ordering, depth-first search). Dynamic programming (chain matrix multiplication, shortest paths, edit distance, sequence alignment). Greedy algorithms (binary heaps, Dijkstra's algorithm, minimum spanning tree, Huffman codes). Randomized algorithms (selection, quick sort, global minimum cut, hushing). P and NP (Cook's theorem, examples of NP-complete problems). Approximate algorithms for NP-hard problems (set cover, vertex cover, maximum independent set). Partial recursive functions (theorem of Post, Diophantine equations). Computations and undecidable problems (undecidability of halting problem, theorem of Rice, semantic and syntactical properties of programs).\nCourse period 09/1/09 → … 200" ]

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[ "# Predicting Trajectories¶\n\nPredicting trajectories with traja can be done with a recurrent neural network (RNN). Traja includes the Long Short Term Memory (LSTM), LSTM Autoencoder (LSTM AE) and LSTM Variational Autoencoder (LSTM VAE) RNNs. Traja also supports custom RNNs.\n\nTo model a trajectory using RNNs, one needs to fit the network to the model. Traja includes the MultiTaskRNNTrainer that can solve a prediction, classification and regression problem with traja DataFrames.\n\nTraja also includes a DataLoader that handles traja dataframes.\n\nBelow is an example with a prediction LSTM via `traja.models.predictive_models.lstm.LSTM`.\n\n```import traja\n\ndf = traja.dataset.example.jaguar()\n```\n\nNote\n\nLSTMs work better with data between -1 and 1. Therefore the data loader scales the data.\n\n```batch_size = 10 # How many sequences to train every step. Constrained by GPU memory.\nnum_past = 10 # How many time steps from which to learn the time series\nnum_future = 10 # How many time steps to predict\nsplit_by_id = False # Whether to split data into training, test and validation sets based on\n# the animal's ID or not. If True, an animal's entire trajectory will only\n# be used for training, or only for testing and so on.\n# If your animals are territorial (like Jaguars) and you want to forecast\n# their trajectories, you want this to be false. If, however, you want to\n# classify the group membership of an animal, you want this to be true,\n# so that you can verify that previously unseen animals get assigned to\n# the correct class.\n```\nclass traja.models.predictive_models.lstm.LSTM(input_size: int, hidden_size: int, output_size: int, num_future: int = 8, batch_size: int = 8, num_layers: int = 1, reset_state: bool = True, bidirectional: bool = False, dropout: float = 0, batch_first: bool = True)[source]\n\nDeep LSTM network. This implementation returns output_size outputs. :param input_size: The number of expected features in the input `x` :param hidden_size: The number of features in the hidden state `h` :param output_size: The number of output dimensions :param batch_size: Size of batch. Default is 8 :param sequence_length: The number of in each sample :param num_layers: Number of recurrent layers. E.g., setting `num_layers=2`\n\nwould mean stacking two LSTMs together to form a stacked LSTM, with the second LSTM taking in outputs of the first LSTM and computing the final results. Default: 1\n\nParameters\n• reset_state – If `True`, will reset the hidden and cell state for each batch of data\n\n• dropout – If non-zero, introduces a Dropout layer on the outputs of each LSTM layer except the last layer, with dropout probability equal to `dropout`. Default: 0\n\n• bidirectional – If `True`, becomes a bidirectional LSTM. Default: `False`\n\nbatch_size=batch_size, n_past=num_past, n_future=num_future, num_workers=0, split_by_id=split_by_id)\n\nforward(x, training=True, classify=False, regress=False, latent=False)[source]\n\nDefines the computation performed at every call.\n\nShould be overridden by all subclasses.\n\nNote\n\nAlthough the recipe for forward pass needs to be defined within this function, one should call the `Module` instance afterwards instead of this since the former takes care of running the registered hooks while the latter silently ignores them.\n\nNote\n\nThe width of the hidden layers and depth of the network are the two main ways in which one tunes the performance of the network. More complex datasets require wider and deeper networks. Below are sensible defaults.\n\n```from traja.models.predictive_models.lstm import LSTM\ninput_size = 2 # Number of input dimensions (normally x, y)\noutput_size = 2 # Same as input_size when predicting\nnum_layers = 2 # Number of LSTM layers. Deeper learns more complex patterns but overfits.\nhidden_size = 32 # Width of layers. Wider learns bigger patterns but overfits. Try 32, 64, 128, 256, 512\ndropout = 0.1 # Ignore some network connections. Improves generalisation.\n\nmodel = LSTM(input_size=input_size,\nhidden_size=hidden_size,\nnum_layers=num_layers,\noutput_size=output_size,\ndropout=dropout,\nbatch_size=batch_size)\n```\n```from traja.models.train import HybridTrainer\n\noptimizer_type = 'Adam' # Nonlinear optimiser with momentum\nloss_type = 'huber'\n\n# Trainer\ntrainer = HybridTrainer(model=model,\noptimizer_type=optimizer_type,\nloss_type=loss_type)\n# Train the model\n```\n\nAfter training, you can determine the network’s final performance with test data, if you want to pick the best model, or with validation data, if you want to determine the performance of your model.\n\nThe `dataloaders` dictionary contains the `sequential_test_loader` and `sequential_validation_loader`, that preserve the order of the original data. The dictionary also contains the ‘test_loader’ and `validation_loader` data loaders, where the order of the time series is randomised.\n\n```validation_loader = dataloaders['sequential_validation_loader']\n\n```\n\nFinally, you can display your training results using the built-in plotting libraries.\n\n```from traja.plotting import plot_prediction\n\nbatch_index = 0 # The batch you want to plot\n```", null, "## Parameter searching¶\n\nWhen optimising neural networks, you often want to change the parameters. When training a forecaster, you have to reinitialise and retrain your model. However, when training a classifier or regressor, you can reset these on the fly, since they work directly on the latent space of your model. VAE models provide utility functions to make this easy.\n\n```from traja.models import MultiModelVAE\ninput_size = 2 # Number of input dimensions (normally x, y)\noutput_size = 2 # Same as input_size when predicting\nnum_layers = 2 # Number of LSTM layers. Deeper learns more complex patterns but overfits.\nhidden_size = 32 # Width of layers. Wider learns bigger patterns but overfits. Try 32, 64, 128, 256, 512\ndropout = 0.1 # Ignore some network connections. Improves generalisation.\n\n# Classifier parameters\nclassifier_hidden_size = 32\nnum_classifier_layers = 4\nnum_classes = 42\n\n# Regressor parameters\nregressor_hidden_size = 18\nnum_regressor_layers = 1\nnum_regressor_parameters = 3\n\nmodel = MultiModelVAE(input_size=input_size,\nhidden_size=hidden_size,\nnum_layers=num_layers,\noutput_size=output_size,\ndropout=dropout,\nbatch_size=batch_size,\nnum_future=num_future,\nclassifier_hidden_size=classifier_hidden_size,\nnum_classifier_layers=num_classifier_layers,\nnum_classes=num_classes,\nregressor_hidden_size=regressor_hidden_size,\nnum_regressor_layers=num_regressor_layers,\nnum_regressor_parameters=num_regressor_parameters)\n\nnew_classifier_hidden_size = 64\nnew_num_classifier_layers = 2\n\nmodel.reset_classifier(classifier_hidden_size=new_classifier_hidden_size,\nnum_classifier_layers=new_num_classifier_layers)\n\nnew_regressor_hidden_size = 64\nnew_num_regressor_layers = 2\nmodel.reset_regressor(regressor_hidden_size=new_regressor_hidden_size,\nnum_regressor_layers=new_num_regressor_layers)\n```" ]

[ null, "https://traja.readthedocs.io/en/latest/_images/rnn_prediction.png", null ]

[ "# How to Find and Fix Circular References in Excel\n\nA circular reference occurs when you end up having a formula in a cell – which in itself uses the cell reference (in which it’s been entered) for the calculation. If this statement seems a bit confusing, don’t worry by the end of this tutorial it will start to make sense.\n\nIn simple terms, a circular reference happens when a formula references back to its own cell directly or indirectly, creating an endless loop of calculations. This endless reference loop, if not stopped will keep changing the cell's value every time.", null, "For quick understanding, here’s an example.\n\nIn our example, there are a few numbers in column A and we are trying to calculate their sum using the SUM function in A7 cell. But inside the SUM function, we have passed the range as A1:A7.", null, "Since the supplied range also includes A7 cell (in which the result needs to be populated) so it creates an endless loop as Excel just keeps on adding the new value in cell A7, which keeps on incrementing.\n\nThis triggers Excel's circular reference warning that says, \"There are one or more circular references where a formula refers to its own cell either directly or indirectly. This might cause them to calculate incorrectly. Try removing or changing these references, or moving the formulas to different cells.\".\n\nKeep in mind that this is not an error since it will not stop further calculations or ask the user to change them. It is a warning that warns the user that they could probably have incorrect calculations.\n\n## Direct and Indirect Circular References\n\nCircular references can be categorized into two types –  1) Direct Circular References 2) Indirect Circular References. Let's try to understand both of these first.\n\n### Direct Circular Reference\n\nA direct circular reference is pretty straightforward. The direct circular reference warning message shows up when the formula in a cell is referring to its own cell directly.\n\nFor example, see the data below.", null, "• The formula in cell B8 refers to itself. When the user applies a function such as a sum or manually uses the reference of B8, a direct circular reference warning is displayed.\n• Once the warning shows up, the user can click on OK, but it will only result in zero.", null, "This is an example of a direct circular reference in Excel.\n\n### Indirect Circular Reference\n\nAs the name suggests, an indirect circular reference takes place when a value in a formula refers to its cell, not directly but at some level. See the simple example below for a better understanding.\n\nFor creating an indirect circular reference example, we will have to write a chain of formulas using values starting from the first cell.\n\n• For example, the data is populated in a way where the diagonals are squares of the previous diagonals as you can see.", null, "• Now the data is starting from A1 which has the value 10 and every other diagonal till D4 is dependent on A1. If the user uses a reference of any diagonal in A1, it will create a circular reference warning. Let’s try it out.\n• As seen in the screenshot below, Excel goes as far as creating a linked line showing the Precedents and Dependents.", null, "• It results in a value of 0 since it is a circular reference issue.\n\nNow, let's try to see and understand how to find circular reference issues in excel.\n\n## How to Find Circular References in Excel\n\nWhile you will get the circular reference warning when it occurs, however, you will still need to figure out in which cell the error has occurred. After knowing the exact cell location of the error it will be easier to handle and fix it.\n\nThese methods are particularly helpful when you are working with large datasets.\n\n### 1) Error Checking drop-down in the Ribbon\n\nHere’s how you can find circular references in Excel using the Ribbon.\n\n• Open the worksheet where the circular reference has occurred.\n• Go to the Formulas tab and click on the Error Checking drop-down menu.", null, "• Select Circular References from the drop-down menu.\n• Here Excel will show you all the circular references that are in the worksheet.\n• Click on whichever circular reference you want and it will take you to that particular cell to solve the issue.\n\nAs you can see in the screenshot above, there are 4 circular references in the worksheet that we are using.\n\n### 2) Using Status bar\n\nFinding circular references using the status bar is very easy. If a circular reference exists in the worksheet, the user will be able to see it in the status bar below the worksheet names.", null, "Note: It only shows the latest circular reference and the user can backtrack from there. As seen in the example below, there are two circular references notably in B8 and C8 but the status bar only shows C8.\n\n## Iterative Calculations\n\nIterative calculations are an interesting feature in Excel. What iterative calculations mean is that if the user keeps them disabled as they usually are, Excel returns a Circular Reference prompt and returns a 0 in the cell instead of the actual result. This is because it is an endless loop.\n\nNow, the thing is that if you want to tell Excel to sit out and let you do your job without bothering you about those pesky circular reference warnings, you can enable Iterative Calculations and it will allow you to perform your calculations. Not only that, but it will also allow you to set a fixed number of maximum iterations.\n\nIf you wish to enable iterative calculations, you can set the number of iterations allowed. Hence, you can stop the circular reference infinite loops and can have a tiny bit of control over such calculations.\n\n### How to Enable/Disable Iterative Calculations in Excel\n\nNow that we have established how iterative calculations work, let’s show you how you can enable or disable iterative calculations in Excel. Follow the below steps to enable or disable iterative calculations in excel.\n\n• Go to the File tab.\n• Click on Options.\n• The Excel Options dialog box will open. Click on Formulas from the left column.", null, "• Here, the option to Enable Iterative Calculation is available. Users can simply tick the box to enable. The user is also able to set maximum iterations and maximum change.\n• After enabling iterative calculation, you will see that the calculation will not result in 0 but will produce results in case of circular reference as you can see.", null, "### Maximum Iterations & Maximum Change Parameters\n\nThe two parameters of the Iterative Calculations are:\n\nMaximum iterations: This is the loop that Excel will run when it is calculating the final result. You can set this as you want. Remember that more iterations mean more processing for Excel which will in turn use more resources and processing power from the computer. It would also take more time.\n\nMaximum change: The maximum change is the value that needs to be achieved in order for the iteration to move further. It is about the accuracy of the result. Make this value small as it will produce accurate results. The default Maximum Change is 0.001.\n\n## Deliberately Using Circular References\n\nThe deliberate use of Circular references is not a very good idea, though sometimes it can help you to get away with poorly developed logic or formulas. But doing this is not recommended at all. However, for educational purposes let's try and understand how to get things working with deliberate circular references.\n\nThe first step is to enable Iterative Calculation in Excel. Once you have turned on Iterative Calculation and set your maximum iterations, you can start using circular references to your benefit.\n\nLet us demonstrate that by using an example.\n\nLet’s say that the user requires to manipulate the data in a way that two cells depend on each other for values causing a circular reference issue. Since the iterative calculation is enabled, there won’t be a prompt.\n\nIn the example below, the cells B4 and B5 depend on each other, creating a circular reference. Since the iterative calculation is enabled, there will not be an error or a 0 in the results. Instead, Excel will try calculating the results.\n\nIn the example, the Agent fee is 20% of the total earnings of the player. Since the total earning and agent fee are dependent on each other, it can still be calculated if iterative calculation is enabled.", null, "To use it as we see fit, we must first apply our desired formula in the net earnings cell.", null, "Now the Agent fee must be calculated which we are going to set at 20% of total earnings but the total earnings will then have to subtract the Agent fee from itself, hence the circular reference.", null, "As you can see, it works perfectly.\n\n### Why Should We Avoid Using Circular References Deliberately?\n\nUsing Circular References in Excel is not recommended at all. Deliberate use of circular references is a slippery method to make things work that eats up a lot of resources and processing power when being used with iterative calculations. It can often produce undesired or inaccurate results that can cause frustration when you have to fix a few of them.\n\n## How to fix Circular References in Excel\n\nFixing circular references in Excel is not possible with a single click. To handle circular references, you have to eliminate them one by one. You can trace it back to the source and remove the starting formula to fix it, or you can remove it one by one.\n\nThere are two tracing methods that will allow you to remove circular references by tracing relationships between formulas and cells.\n\nTo access the tracing methods, you have to follow the following steps:\n\n• Go to the Formulas tab in Excel\n• The Trace Precedents and Trace Dependents options are available there.", null, "These tracing features can help the user in fixing circular references by providing a path connecting the references through a line that is drawn between cells that are responding to the circular references.\n\nThe two tracing options work as follows:\n\n### Trace Precedents\n\nThe trace precedents feature tracks back cells that the current cell depends on. These are the cells on which the current formula is dependent for the data that it needs. This option will draw lines that will tell which cells are affecting the active cell.\n\nIn the example below, the cells affecting B5 are B2, B3, and B4. Hence, when we click on Trace Precedents, it draws a line indicating B2, B3, and B5 cells leading to B5.", null, "Shortcut for Trace Precedents: ALT + T U T\n\n### Trace Dependents\n\nThe trace dependents feature tracks the cells that are dependent on the active cell, the cells that depend on the current cell for the data they need to produce results. This feature will draw lines to the cells that are dependent on the active cell.\n\nIn the example below, the cell being affected by B5 is B4. It is dependent on B5 for its value. Hence, when we use trace dependents, it draws a line from B5 to B4, indicating that B4 is dependent on B5.", null, "Shortcut for Trace Dependents: ALT + T U D\n\nSo, this was all about circular references in Excel. Hopefully, by now, you have a good idea of how circular references work, how you can find/fix them, and how you can use them if you desire." ]

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[ "### Changes in preparation of the C type system and C front-end language\n\n```Major changes:\n* Make void a base type, and include a proper void base value. This is necessary\nbecause expressions (free, functions without return value) can yield a void.\nWe now also allow void casts conforming to the C standard.\n* Various missing lemmas about typing, weakening, decidability, ...\n* The operations \"free\" and \"alloc\" now operate on l-values instead of r-values.\nThis removes some duplication.\n* Improve notations of expressions and statements. Change the presence of the\noperators conforming to the C standard.\n\nSmall changes:\n* Use the classes \"Typed\" and \"TypeCheck\" for validity of indexes in memory.\nThis gives more uniform notations.\n* New tactic \"typed_inversion\" performs inversion on an inductive predicate\nof type \"Typed\" and folds the premises.\n* Remove a horrible hack in the definitions of the classes \"FMap\", \"MBind\",\n\"OMap\", \"Alter\" that was used to let \"simpl\" behave better. Instead, we have\ndefined a tactic \"csimpl\" that folds the results after performing an\nordinary \"simpl\".\n* Fast operation to remove duplicates from lists using hashsets.\n* Make various type constructors (mainly finite map implementations) universe\npolymorphic by packing them into an inductive. This way, the whole C syntax\ncan live in type, avoiding the need for (slow) universe checks.```\nparent af633db2\n ... ... @@ -271,7 +271,7 @@ End assoc. (** * Finite sets *) (** We construct finite sets using the above implementation of maps. *) Notation assoc_set K := (mapset (assoc K)). Notation assoc_set K := (mapset (assoc K unit)). Instance assoc_map_dom `{Lexico K, !TrichotomyT (@lexico K _), !StrictOrder lexico} {A} : Dom (assoc K A) (assoc_set K) := mapset_dom. Instance assoc_map_dom_spec `{Lexico K} `{!TrichotomyT (@lexico K _)} ... ...\n ... ... @@ -32,6 +32,10 @@ Notation \"'False'\" := False : type_scope. Notation curry := prod_curry. Notation uncurry := prod_uncurry. Definition curry3 {A B C D} (f : A → B → C → D) (p : A * B * C) : D := let '(a,b,c) := p in f a b c. Definition curry4 {A B C D E} (f : A → B → C → D → E) (p : A * B * C * D) : E := let '(a,b,c,d) := p in f a b c d. (** Throughout this development we use [C_scope] for all general purpose notations that do not belong to a more specific scope. *) ... ... @@ -376,45 +380,24 @@ Class Filter A B := filter: ∀ (P : A → Prop) `{∀ x, Decision (P x)}, B → (** ** Monadic operations *) (** We define operational type classes for the monadic operations bind, join and fmap. These type classes are defined in a non-standard way by taking the function as a parameter of the class. For example, we define << Class FMapD := fmap: ∀ {A B}, (A → B) → M A → M B. >> instead of << Class FMap {A B} (f : A → B) := fmap: M A → M B. >> This approach allows us to define [fmap] on lists such that [simpl] unfolds it in the appropriate way, and so that it can be used for mutual recursion (the mapped function [f] is not part of the fixpoint) as well. This is a hack, and should be replaced by something more appropriate in future versions. *) (** We use these type classes merely for convenient overloading of notations and do not formalize any theory on monads (we do not even define a class with the monad laws). *) and fmap. We use these type classes merely for convenient overloading of notations and do not formalize any theory on monads (we do not even define a class with the monad laws). *) Class MRet (M : Type → Type) := mret: ∀ {A}, A → M A. Instance: Params (@mret) 3. Arguments mret {_ _ _} _. Class MBindD (M : Type → Type) {A B} (f : A → M B) := mbind: M A → M B. Notation MBind M := (∀ {A B} (f : A → M B), MBindD M f)%type. Class MBind (M : Type → Type) := mbind : ∀ {A B}, (A → M B) → M A → M B. Arguments mbind {_ _ _ _} _ !_ /. Instance: Params (@mbind) 5. Arguments mbind {_ _ _} _ {_} !_ /. Class MJoin (M : Type → Type) := mjoin: ∀ {A}, M (M A) → M A. Instance: Params (@mjoin) 3. Arguments mjoin {_ _ _} !_ /. Class FMapD (M : Type → Type) {A B} (f : A → B) := fmap: M A → M B. Notation FMap M := (∀ {A B} (f : A → B), FMapD M f)%type. Class FMap (M : Type → Type) := fmap : ∀ {A B}, (A → B) → M A → M B. Instance: Params (@fmap) 6. Arguments fmap {_ _ _} _ {_} !_ /. Class OMapD (M : Type → Type) {A B} (f : A → option B) := omap: M A → M B. Notation OMap M := (∀ {A B} (f : A → option B), OMapD M f)%type. Arguments fmap {_ _ _ _} _ !_ /. Class OMap (M : Type → Type) := omap: ∀ {A B}, (A → option B) → M A → M B. Instance: Params (@omap) 6. Arguments omap {_ _ _} _ {_} !_ /. Arguments omap {_ _ _ _} _ !_ /. Notation \"m ≫= f\" := (mbind f m) (at level 60, right associativity) : C_scope. Notation \"( m ≫=)\" := (λ f, mbind f m) (only parsing) : C_scope. ... ... @@ -430,6 +413,9 @@ Notation \"'( x1 , x2 ) ← y ; z\" := Notation \"'( x1 , x2 , x3 ) ← y ; z\" := (y ≫= (λ x : _, let ' (x1,x2,x3) := x in z)) (at level 65, next at level 35, only parsing, right associativity) : C_scope. Notation \"'( x1 , x2 , x3 , x4 ) ← y ; z\" := (y ≫= (λ x : _, let ' (x1,x2,x3,x4) := x in z)) (at level 65, next at level 35, only parsing, right associativity) : C_scope. Class MGuard (M : Type → Type) := mguard: ∀ P {dec : Decision P} {A}, (P → M A) → M A. ... ... @@ -468,10 +454,9 @@ Arguments delete _ _ _ !_ !_ / : simpl nomatch. (** The function [alter f k m] should update the value at key [k] using the function [f], which is called with the original value. *) Class AlterD (K A M : Type) (f : A → A) := alter: K → M → M. Notation Alter K A M := (∀ (f : A → A), AlterD K A M f)%type. Class Alter (K A M : Type) := alter: (A → A) → K → M → M. Instance: Params (@alter) 5. Arguments alter {_ _ _} _ {_} !_ !_ / : simpl nomatch. Arguments alter {_ _ _ _} _ !_ !_ / : simpl nomatch. (** The function [alter f k m] should update the value at key [k] using the function [f], which is called with the original value at key [k] or [None] ... ...\n ... ... @@ -165,7 +165,7 @@ Program Instance prod_countable `{Countable A} `{Countable B} : Next Obligation. intros ?????? [x y]; simpl. rewrite prod_decode_encode_fst, prod_decode_encode_snd. simpl. by rewrite !decode_encode. csimpl. by rewrite !decode_encode. Qed. Fixpoint list_encode_ (l : list positive) : positive := ... ... @@ -188,7 +188,7 @@ Lemma list_decode_encode l : list_decode (list_encode l) = Some l. Proof. cut (list_decode_ (length l) (list_encode_ l) = Some l). { intros help. unfold list_decode, list_encode. rewrite prod_decode_encode_fst, prod_decode_encode_snd; simpl. rewrite prod_decode_encode_fst, prod_decode_encode_snd; csimpl. by rewrite Nat2Pos.id by done; simpl. } induction l; simpl; auto. by rewrite prod_decode_encode_fst, prod_decode_encode_snd; ... ... @@ -227,5 +227,5 @@ Program Instance nat_countable : Countable nat := {| decode p := N.to_nat <\\$> decode p |}. Next Obligation. intros x. rewrite decode_encode; simpl. by rewrite Nat2N.id. intros x. rewrite decode_encode; csimpl. by rewrite Nat2N.id. Qed.\n ... ... @@ -205,9 +205,9 @@ Lemma alter_ext {A} (f g : A → A) (m : M A) i : (∀ x, m !! i = Some x → f x = g x) → alter f i m = alter g i m. Proof. intro. apply partial_alter_ext. intros [x|] ?; f_equal'; auto. Qed. Lemma lookup_alter {A} (f : A → A) m i : alter f i m !! i = f <\\$> m !! i. Proof. apply lookup_partial_alter. Qed. Proof. unfold alter. apply lookup_partial_alter. Qed. Lemma lookup_alter_ne {A} (f : A → A) m i j : i ≠ j → alter f i m !! j = m !! j. Proof. apply lookup_partial_alter_ne. Qed. Proof. unfold alter. apply lookup_partial_alter_ne. Qed. Lemma alter_compose {A} (f g : A → A) (m : M A) i: alter (f ∘ g) i m = alter f i (alter g i m). Proof. ... ... @@ -375,6 +375,13 @@ Proof. * eauto using insert_delete_subset. * by rewrite lookup_delete. Qed. Lemma fmap_insert {A B} (f : A → B) (m : M A) i x : f <\\$> <[i:=x]>m = <[i:=f x]>(f <\\$> m). Proof. apply map_eq; intros i'; destruct (decide (i' = i)) as [->|]. * by rewrite lookup_fmap, !lookup_insert. * by rewrite lookup_fmap, !lookup_insert_ne, lookup_fmap by done. Qed. (** ** Properties of the singleton maps *) Lemma lookup_singleton_Some {A} i j (x y : A) : ... ... @@ -430,7 +437,7 @@ Proof. eauto using NoDup_fmap_fst, map_to_list_unique, NoDup_map_to_list. Qed. Lemma elem_of_map_of_list_1 {A} (l : list (K * A)) i x : NoDup (fst <\\$> l) → (i,x) ∈ l → map_of_list l !! i = Some x. Proof. induction l as [|[j y] l IH]; simpl; [by rewrite elem_of_nil|]. induction l as [|[j y] l IH]; csimpl; [by rewrite elem_of_nil|]. rewrite NoDup_cons, elem_of_cons, elem_of_list_fmap. intros [Hl ?] [?|?]; simplify_equality; [by rewrite lookup_insert|]. destruct (decide (i = j)) as [->|]; [|rewrite lookup_insert_ne; auto]. ... ... @@ -455,7 +462,7 @@ Qed. Lemma not_elem_of_map_of_list_2 {A} (l : list (K * A)) i : map_of_list l !! i = None → i ∉ fst <\\$> l. Proof. induction l as [|[j y] l IH]; simpl; [rewrite elem_of_nil; tauto|]. induction l as [|[j y] l IH]; csimpl; [rewrite elem_of_nil; tauto|]. rewrite elem_of_cons. destruct (decide (i = j)); simplify_equality. * by rewrite lookup_insert. * by rewrite lookup_insert_ne; intuition. ... ... @@ -499,7 +506,7 @@ Qed. Lemma map_to_list_insert {A} (m : M A) i x : m !! i = None → map_to_list (<[i:=x]>m) ≡ₚ (i,x) :: map_to_list m. Proof. intros. apply map_of_list_inj; simpl. intros. apply map_of_list_inj; csimpl. * apply NoDup_fst_map_to_list. * constructor; auto using NoDup_fst_map_to_list. rewrite elem_of_list_fmap. intros [[??] [? Hlookup]]; subst; simpl in *. ... ... @@ -521,7 +528,7 @@ Proof. intros Hperm. apply map_to_list_inj. assert (NoDup (fst <\\$> (i, x) :: l)) as Hnodup. { rewrite <-Hperm. auto using NoDup_fst_map_to_list. } simpl in Hnodup. rewrite NoDup_cons in Hnodup. destruct Hnodup. csimpl in *. rewrite NoDup_cons in Hnodup. destruct Hnodup. rewrite Hperm, map_to_list_insert, map_to_of_list; auto using not_elem_of_map_of_list_1. Qed. ... ... @@ -1374,7 +1381,7 @@ Tactic Notation \"simplify_map_equality\" \"by\" tactic3(tac) := | H : ∅ = {[?i,?x]} |- _ => by destruct (map_non_empty_singleton i x) end. Tactic Notation \"simplify_map_equality'\" \"by\" tactic3(tac) := repeat (progress simpl in * || simplify_map_equality by tac). repeat (progress csimpl in * || simplify_map_equality by tac). Tactic Notation \"simplify_map_equality\" := simplify_map_equality by eauto with simpl_map map_disjoint. Tactic Notation \"simplify_map_equality'\" := ... ...\n ... ... @@ -56,7 +56,7 @@ Proof. destruct finA as [xs Hxs HA]; unfold find, decode; simpl. intros Hx. destruct (list_find_elem_of P xs x) as [i Hi]; auto. rewrite Hi. destruct (list_find_Some P xs i) as (y&?&?); subst; auto. exists y. simpl. by rewrite !Nat2Pos.id by done. exists y. csimpl. by rewrite !Nat2Pos.id by done. Qed. Lemma card_0_inv P `{finA: Finite A} : card A = 0 → A → P. ... ...\n ... ... @@ -3,7 +3,7 @@ (** This file implements finite set using hash maps. Hash sets are represented using radix-2 search trees. Each hash bucket is thus indexed using an binary integer of type [Z], and contains an unordered list without duplicates. *) Require Export fin_maps. Require Export fin_maps listset. Require Import zmap. Record hashset {A} (hash : A → Z) := Hashset { ... ... @@ -15,7 +15,7 @@ Arguments Hashset {_ _} _ _. Arguments hashset_car {_ _} _. Section hashset. Context {A : Type} {hash : A → Z} `{∀ x y : A, Decision (x = y)}. Context `{∀ x y : A, Decision (x = y)} (hash : A → Z). Instance hashset_elem_of: ElemOf A (hashset hash) := λ x m, ∃ l, hashset_car m !! hash x = Some l ∧ x ∈ l. ... ... @@ -106,7 +106,7 @@ Proof. * unfold elements, hashset_elems. intros [m Hm]; simpl. rewrite map_Forall_to_list in Hm. generalize (NoDup_fst_map_to_list m). induction Hm as [|[n l] m' [??]]; simpl; inversion_clear 1 as [|?? Hn]; [constructor|]. csimpl; inversion_clear 1 as [|?? Hn]; [constructor|]. apply NoDup_app; split_ands; eauto. setoid_rewrite elem_of_list_bind; intros x ? ([n' l']&?&?); simpl in *. assert (hash x = n ∧ hash x = n') as [??]; subst. ... ... @@ -115,6 +115,25 @@ Proof. rewrite Forall_forall in Hm. eapply (Hm (_,_)); eauto. } destruct Hn; rewrite elem_of_list_fmap; exists (hash x, l'); eauto. Qed. Definition remove_dups_fast (l : list A) : list A := elements (foldr (λ x, ({[ x ]} ∪)) ∅ l : hashset hash). Lemma elem_of_remove_dups_fast l x : x ∈ remove_dups_fast l ↔ x ∈ l. Proof. unfold remove_dups_fast. rewrite elem_of_elements. split. * revert x. induction l as [|y l IH]; intros x; simpl. { by rewrite elem_of_empty. } rewrite elem_of_union, elem_of_singleton. intros [->|]; [left|right]; eauto. * induction 1; esolve_elem_of. Qed. Lemma NoDup_remove_dups_fast l : NoDup (remove_dups_fast l). Proof. unfold remove_dups_fast. apply NoDup_elements. Qed. Definition listset_normalize (X : listset A) : listset A := let (l) := X in Listset (remove_dups_fast l). Lemma listset_normalize_correct X : listset_normalize X ≡ X. Proof. destruct X as [l]. apply elem_of_equiv; intro; apply elem_of_remove_dups_fast. Qed. End hashset. (** These instances are declared using [Hint Extern] to avoid too ... ...\n ... ... @@ -45,11 +45,11 @@ Instance list_lookup {A} : Lookup nat A (list A) := (** The operation [alter f i l] applies the function [f] to the [i]th element of [l]. In case [i] is out of bounds, the list is returned unchanged. *) Instance list_alter {A} (f : A → A) : AlterD nat A (list A) f := fix go i l {struct l} := let _ : AlterD _ _ _ f := @go in Instance list_alter {A} : Alter nat A (list A) := λ f, fix go i l {struct l} := match l with | [] => [] | x :: l => match i with 0 => f x :: l | S i => x :: alter f i l end | x :: l => match i with 0 => f x :: l | S i => x :: go i l end end. (** The operation [<[i:=x]> l] overwrites the element at position [i] with the ... ... @@ -139,12 +139,10 @@ Definition foldl {A B} (f : A → B → A) : A → list B → A := (** The monadic operations. *) Instance list_ret: MRet list := λ A x, x :: @nil A. Instance list_fmap {A B} (f : A → B) : FMapD list f := fix go (l : list A) := match l with [] => [] | x :: l => f x :: @fmap _ _ _ f go l end. Instance list_bind {A B} (f : A → list B) : MBindD list f := fix go (l : list A) := match l with [] => [] | x :: l => f x ++ @mbind _ _ _ f go l end. Instance list_fmap : FMap list := λ A B f, fix go (l : list A) := match l with [] => [] | x :: l => f x :: go l end. Instance list_bind : MBind list := λ A B f, fix go (l : list A) := match l with [] => [] | x :: l => f x ++ go l end. Instance list_join: MJoin list := fix go A (ls : list (list A)) : list A := match ls with [] => [] | l :: ls => l ++ @mjoin _ go _ ls end. ... ... @@ -192,6 +190,8 @@ Definition suffix_of {A} : relation (list A) := λ l1 l2, ∃ k, l2 = k ++ l1. Definition prefix_of {A} : relation (list A) := λ l1 l2, ∃ k, l2 = l1 ++ k. Infix \"`suffix_of`\" := suffix_of (at level 70) : C_scope. Infix \"`prefix_of`\" := prefix_of (at level 70) : C_scope. Hint Extern 0 (?x `prefix_of` ?y) => reflexivity. Hint Extern 0 (?x `suffix_of` ?y) => reflexivity. Section prefix_suffix_ops. Context `{∀ x y : A, Decision (x = y)}. ... ... @@ -219,6 +219,7 @@ Inductive sublist {A} : relation (list A) := | sublist_skip x l1 l2 : sublist l1 l2 → sublist (x :: l1) (x :: l2) | sublist_cons x l1 l2 : sublist l1 l2 → sublist l1 (x :: l2). Infix \"`sublist`\" := sublist (at level 70) : C_scope. Hint Extern 0 (?x `sublist` ?y) => reflexivity. (** A list [l2] contains a list [l1] if [l2] is obtained by removing elements from [l1] while possiblity changing the order. *) ... ... @@ -229,6 +230,7 @@ Inductive contains {A} : relation (list A) := | contains_cons x l1 l2 : contains l1 l2 → contains l1 (x :: l2) | contains_trans l1 l2 l3 : contains l1 l2 → contains l2 l3 → contains l1 l3. Infix \"`contains`\" := contains (at level 70) : C_scope. Hint Extern 0 (?x `contains` ?y) => reflexivity. Section contains_dec_help. Context {A} {dec : ∀ x y : A, Decision (x = y)}. ... ... @@ -439,7 +441,7 @@ Lemma list_lookup_alter f l i : alter f i l !! i = f <\\$> l !! i. Proof. revert i. induction l. done. intros [|i]. done. apply (IHl i). Qed. Lemma list_lookup_alter_ne f l i j : i ≠ j → alter f i l !! j = l !! j. Proof. revert i j. induction l; [done|]. intros [] [] ?; simpl; auto with congruence. revert i j. induction l; [done|]. intros [][] ?; csimpl; auto with congruence. Qed. Lemma list_lookup_insert l i x : i < length l → <[i:=x]>l !! i = Some x. Proof. revert i. induction l; intros [|?] ?; f_equal'; auto with lia. Qed. ... ... @@ -2390,7 +2392,7 @@ Section fmap. Proof. by induction l; f_equal'. Qed. Lemma fmap_reverse l : f <\\$> reverse l = reverse (f <\\$> l). Proof. induction l as [|?? IH]; simpl; by rewrite ?reverse_cons, ?fmap_app, ?IH. induction l as [|?? IH]; csimpl; by rewrite ?reverse_cons, ?fmap_app, ?IH. Qed. Lemma fmap_last l : last (f <\\$> l) = f <\\$> last l. Proof. induction l as [|? []]; simpl; auto. Qed. ... ... @@ -2419,7 +2421,7 @@ Section fmap. Forall (λ x, f (g x) = h (f x)) l → f <\\$> alter g i l = alter h i (f <\\$> l). Proof. intros Hl. revert i. by induction Hl; intros [|i]; f_equal'. Qed. Lemma elem_of_list_fmap_1 l x : x ∈ l → f x ∈ f <\\$> l. Proof. induction 1; simpl; rewrite elem_of_cons; intuition. Qed. Proof. induction 1; csimpl; rewrite elem_of_cons; intuition. Qed. Lemma elem_of_list_fmap_1_alt l x y : x ∈ l → y = f x → y ∈ f <\\$> l. Proof. intros. subst. by apply elem_of_list_fmap_1. Qed. Lemma elem_of_list_fmap_2 l x : x ∈ f <\\$> l → ∃ y, x = f y ∧ y ∈ l. ... ... @@ -2478,7 +2480,7 @@ Section fmap. Proof. revert l2; induction l1; intros [|??]; inversion_clear 1; auto. Qed. Lemma Forall2_fmap_2 {C D} (g : C → D) (P : B → D → Prop) l1 l2 : Forall2 (λ x1 x2, P (f x1) (g x2)) l1 l2 → Forall2 P (f <\\$> l1) (g <\\$> l2). Proof. induction 1; simpl; auto. Qed. Proof. induction 1; csimpl; auto. Qed. Lemma Forall2_fmap {C D} (g : C → D) (P : B → D → Prop) l1 l2 : Forall2 P (f <\\$> l1) (g <\\$> l2) ↔ Forall2 (λ x1 x2, P (f x1) (g x2)) l1 l2. Proof. split; auto using Forall2_fmap_1, Forall2_fmap_2. Qed. ... ... @@ -2492,7 +2494,7 @@ Proof. auto using list_alter_fmap. Qed. Lemma NoDup_fmap_fst {A B} (l : list (A * B)) : (∀ x y1 y2, (x,y1) ∈ l → (x,y2) ∈ l → y1 = y2) → NoDup l → NoDup (fst <\\$> l). Proof. intros Hunique. induction 1 as [|[x1 y1] l Hin Hnodup IH]; simpl; constructor. intros Hunique. induction 1 as [|[x1 y1] l Hin Hnodup IH]; csimpl; constructor. * rewrite elem_of_list_fmap. intros [[x2 y2] [??]]; simpl in *; subst. destruct Hin. rewrite (Hunique x2 y1 y2); rewrite ?elem_of_cons; auto. ... ... @@ -2511,12 +2513,11 @@ Section bind. Global Instance bind_sublist: Proper (sublist ==> sublist) (mbind f). Proof. induction 1; simpl; auto; [done|by apply sublist_app|by apply sublist_inserts_l]. [by apply sublist_app|by apply sublist_inserts_l]. Qed. Global Instance bind_contains: Proper (contains ==> contains) (mbind f). Proof. induction 1; simpl; auto. * done. induction 1; csimpl; auto. * by apply contains_app. * by rewrite !(associative_L (++)), (commutative (++) (f _)). * by apply contains_inserts_l. ... ... @@ -2524,7 +2525,7 @@ Section bind. Qed. Global Instance bind_Permutation: Proper ((≡ₚ) ==> (≡ₚ)) (mbind f). Proof. induction 1; simpl; auto. induction 1; csimpl; auto. * by f_equiv. * by rewrite !(associative_L (++)), (commutative (++) (f _)). * etransitivity; eauto. ... ... @@ -2532,30 +2533,30 @@ Section bind. Lemma bind_cons x l : (x :: l) ≫= f = f x ++ l ≫= f. Proof. done. Qed. Lemma bind_singleton x : [x] ≫= f = f x. Proof. simpl. by rewrite (right_id_L _ (++)). Qed. Proof. csimpl. by rewrite (right_id_L _ (++)). Qed. Lemma bind_app l1 l2 : (l1 ++ l2) ≫= f = (l1 ≫= f) ++ (l2 ≫= f). Proof. by induction l1; simpl; rewrite <-?(associative_L (++)); f_equal. Qed. Proof. by induction l1; csimpl; rewrite <-?(associative_L (++)); f_equal. Qed. Lemma elem_of_list_bind (x : B) (l : list A) : x ∈ l ≫= f ↔ ∃ y, x ∈ f y ∧ y ∈ l. Proof. split. * induction l as [|y l IH]; simpl; [inversion 1|]. * induction l as [|y l IH]; csimpl; [inversion 1|]. rewrite elem_of_app. intros [?|?]. + exists y. split; [done | by left]. + destruct IH as [z [??]]. done. exists z. split; [done | by right]. * intros [y [Hx Hy]]. induction Hy; simpl; rewrite elem_of_app; intuition. * intros [y [Hx Hy]]. induction Hy; csimpl; rewrite elem_of_app; intuition. Qed. Lemma Forall_bind (P : B → Prop) l : Forall P (l ≫= f) ↔ Forall (Forall P ∘ f) l. Proof. split. * induction l; simpl; rewrite ?Forall_app; constructor; simpl; intuition. * induction 1; simpl; rewrite ?Forall_app; auto. * induction l; csimpl; rewrite ?Forall_app; constructor; csimpl; intuition. * induction 1; csimpl; rewrite ?Forall_app; auto. Qed. Lemma Forall2_bind {C D} (g : C → list D) (P : B → D → Prop) l1 l2 : Forall2 (λ x1 x2, Forall2 P (f x1) (g x2)) l1 l2 → Forall2 P (l1 ≫= f) (l2 ≫= g). Proof. induction 1; simpl; auto using Forall2_app. Qed. Proof. induction 1; csimpl; auto using Forall2_app. Qed. End bind. Section ret_join. ... ... @@ -2676,7 +2677,7 @@ Section permutations. Lemma permutations_swap x y l : y :: x :: l ∈ permutations (x :: y :: l). Proof. simpl. apply elem_of_list_bind. exists (y :: l). split; simpl. * destruct l; simpl; rewrite !elem_of_cons; auto. * destruct l; csimpl; rewrite !elem_of_cons; auto. * apply elem_of_list_bind. simpl. eauto using interleave_cons, permutations_refl. Qed. ... ... @@ -2694,12 +2695,12 @@ Section permutations. intros Hl1 [? | [l2' [??]]]; simplify_equality'. * rewrite !elem_of_cons, elem_of_list_fmap in Hl1. destruct Hl1 as [? | [? | [l4 [??]]]]; subst. + exists (x1 :: y :: l3). simpl. rewrite !elem_of_cons. tauto. + exists (x1 :: y :: l3). simpl. rewrite !elem_of_cons. tauto. + exists (x1 :: y :: l3). csimpl. rewrite !elem_of_cons. tauto. + exists (x1 :: y :: l3). csimpl. rewrite !elem_of_cons. tauto. + exists l4. simpl. rewrite elem_of_cons. auto using interleave_cons. * rewrite elem_of_cons, elem_of_list_fmap in Hl1. destruct Hl1 as [? | [l1' [??]]]; subst. + exists (x1 :: y :: l3). simpl. + exists (x1 :: y :: l3). csimpl. rewrite !elem_of_cons, !elem_of_list_fmap. split; [| by auto]. right. right. exists (y :: l2'). rewrite elem_of_list_fmap. naive_solver. ... ... @@ -3006,7 +3007,7 @@ Section eval. Lemma eval_alt t : eval E t = to_list t ≫= from_option [] ∘ (E !!). Proof. induction t; simpl. induction t; csimpl. * done. * by rewrite (right_id_L [] (++)). * rewrite bind_app. by f_equal. ... ...\n (* Copyright (c) 2012-2014, Robbert Krebbers. *) (* This file is distributed under the terms of the BSD license. *) Require Export separation. Require Import refinements. Record map (A : Set) := Map { map_car : indexmap (list A) }. Arguments Map {_} _. Arguments map_car {_} _. Add Printing Constructor map. Instance: Injective (=) (=) (@Map A). Proof. by injection 1. Qed. Instance map_ops {A : Set} `{SeparationOps A} : SeparationOps (map A) := { sep_empty := Map ∅; sep_union m1 m2 := let (m1) := m1 in let (m2) := m2 in Map (union_with (λ xs1 xs2, Some (xs1 ∪* xs2)) m1 m2); sep_difference m1 m2 := let (m1) := m1 in let (m2) := m2 in Map (difference_with (λ xs1 xs2, let xs' := xs1 ∖* xs2 in guard (¬Forall (∅ =) xs'); Some xs' ) m1 m2); sep_half m := let (m) := m in Map (½* <\\$> m); sep_valid m := let (m) := m in map_Forall (λ _ xs, Forall sep_valid xs ∧ ¬Forall (∅ =) xs) m; sep_disjoint m1 m2 := let (m1) := m1 in let (m2) := m2 in map_Forall2 (λ xs1 xs2, xs1 ⊥* xs2 ∧ ¬Forall (∅ =) xs1 ∧ ¬Forall (∅ =) xs2) (λ xs1, Forall sep_valid xs1 ∧ ¬Forall (∅ =) xs1) (λ xs2, Forall sep_valid xs2 ∧ ¬Forall (∅ =) xs2) m1 m2; sep_splittable m := let (m) := m in map_Forall (λ _ xs, Forall sep_valid xs ∧ ¬Forall (∅ =) xs ∧ Forall sep_splittable xs) m; sep_subseteq m1 m2 := let (m1) := m1 in let (m2) := m2 in map_Forall2 (λ xs1 w2, xs1 ⊆* w2 ∧ ¬Forall (∅ =) xs1) (λ xs1, False) (λ xs2, Forall sep_valid xs2 ∧ ¬Forall (∅ =) xs2) m1 m2; sep_unmapped m := map_car m = ∅; sep_unshared m := False }. Proof. * intros []; apply _. * intros [] []; apply _. * intros [] []; apply _. * solve_decision. * intros []; apply _. Defined. Instance map_sep {A : Set} `{Separation A} : Separation (map A). Proof. split. * destruct (sep_inhabited A) as (x&?&?). eexists (Map {[fresh ∅, [x]]}). split; [|by intro]. intros o w ?; simplify_map_equality'. split. + by rewrite Forall_singleton. + by inversion_clear 1; decompose_Forall_hyps'. * sep_unfold; intros [m1] [m2] Hm o w; specialize (Hm o); simpl in *. intros Hx. rewrite Hx in Hm. destruct (m2 !! o); intuition eauto using seps_disjoint_valid_l. * sep_unfold; intros [m1] [m2] Hm o w; specialize (Hm o); simpl in *. rewrite lookup_union_with. intros. destruct (m1 !! o), (m2 !! o); simplify_equality'; intuition eauto using seps_union_valid, seps_positive_l. * sep_unfold. intros [m] Hm o; specialize (Hm o); simplify_map_equality'. destruct (m !! o); eauto. * sep_unfold; intros [m] ?; f_equal'. by rewrite (left_id_L ∅ _). * sep_unfold. intros [m1] [m2] Hm o; specialize (Hm o); simpl in *. destruct (m1 !! o), (m2 !! o); intuition. * sep_unfold; intros [m1] [m2] Hm; f_equal'. apply union_with_commutative. intros o w1 w2 ??; specialize (Hm o); simplify_option_equality. f_equal; intuition auto using seps_commutative. * sep_unfold; intros [m1] [m2] [m3] Hm Hm' o; specialize (Hm o); specialize (Hm' o); simpl in *; rewrite lookup_union_with in Hm'. destruct (m1 !! o) eqn:?, (m2 !! o)," ]

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[ "He invented a method (now called Fourier analysis) of finding appropriate coefficients a1, a2, a3, … in equation (12) for any given initial temperature distribution. 9.1 The Heat/Difiusion equation and dispersion relation Solution. The initial condition is expanded onto the Fourier basis associated with the boundary conditions. The only way heat will leaveDis through the boundary. From (15) it follows that c(ω) is the Fourier transform of the initial temperature distribution f(x): c(ω) = 1 2π Z ∞ −∞ f(x)eiωxdx (33) We will then discuss how the heat equation, wave equation and Laplace’s equation arise in physical models. Heat Equation and Fourier Series There are three big equations in the world of second-order partial di erential equations: 1. We will use the Fourier sine series for representation of the nonhomogeneous solution to satisfy the boundary conditions. The heat equation 6.2 Construction of a regular solution We will see several different ways of constructing solutions to the heat equation. 3. 1. Okay, we’ve now seen three heat equation problems solved and so we’ll leave this section. The solution using Fourier series is u(x;t) = F0(t)x+[F1(t) F0(t)] x2 2L +a0 + X1 n=1 an cos(nˇx=L)e k(nˇ=L) 2t + Z t 0 A0(s)ds+ X1 n=1 cos(nˇx=L) Z t 0 We will also work several examples finding the Fourier Series for a function. Exercise 4.4.102: Let $$f(t)= \\cos(2t)$$ on $$0 \\leq t < \\pi$$. For a fixed $$t$$, the solution is a Fourier series with coefficients $$b_n e^{\\frac{-n^2 \\pi^2}{L^2}kt}$$. Setting u t = 0 in the 2-D heat equation gives u = u xx + u yy = 0 (Laplace’s equation), solutions of which are called harmonic functions. We consider first the heat equation without sources and constant nonhomogeneous boundary conditions. Solution of heat equation. From where , we get Applying equation (13.20) we obtain the general solution So we can conclude that … Let us start with an elementary construction using Fourier series. Browse other questions tagged partial-differential-equations fourier-series heat-equation or ask your own question. How to use the GUI Chapter 12.5: Heat Equation: Solution by Fourier Series includes 35 full step-by-step solutions. A heat equation problem has three components. úÛCèÆ«CÃ?‰d¾Âæ'ƒáÉï'º Ë¸Q„–)ň¤2]Ÿüò+ÍÆðòûŒjØìÖ7½!Ò¡6&Ùùɏ'§g:#s£ Á•¤„3Ùz™ÒHoË,á0]ßø»¤’8‘×Qf0®Œ­tfˆCQ¡‘!ĀxQdžêJA$ÚL¦x=»û]ibô$„Ýѓ$FpÀ ¦YB»‚Y0. The first part of this course of lectures introduces Fourier series, concentrating on their practical application rather than proofs of convergence. A full Fourier series needs an interval of $$- L \\le x \\le L$$ whereas the Fourier sine and cosines series we saw in the first two problems need $$0 \\le x \\le L$$. }\\] 2. Solutions of the heat equation are sometimes known as caloric functions. Solving heat equation on a circle. SOLUTIONS TO THE HEAT AND WAVE EQUATIONS AND THE CONNECTION TO THE FOURIER SERIES IAN ALEVY Abstract. Key Concepts: Heat equation; boundary conditions; Separation of variables; Eigenvalue problems for ODE; Fourier Series. The threshold condition for chilling is established. In this worksheet we consider the one-dimensional heat equation describint the evolution of temperature inside the homogeneous metal rod. 3. The heat equation model The Fourier series was introduced by the mathematician and politician Fourier (from the city of Grenoble in France) to solve the heat equation. The heat equation “smoothes” out the function $$f(x)$$ as $$t$$ grows. Using the results of Example 3 on the page Definition of Fourier Series and Typical Examples, we can write the right side of the equation as the series \\[{3x }={ \\frac{6}{\\pi }\\sum\\limits_{n = 1}^\\infty {\\frac{{{{\\left( { – 1} \\right)}^{n + 1}}}}{n}\\sin n\\pi x} . Six Easy Steps to Solving The Heat Equation In this document I list out what I think is the most e cient way to solve the heat equation. a) Find the Fourier series of the even periodic extension. In this section we define the Fourier Series, i.e. ... we determine the coefficients an as the Fourier sine series coefficients of f(x)−uE(x) an = 2 L Z L 0 [f(x)−uE(x)]sin nπx L dx ... the unknown solution v(x,t) as a generalized Fourier series of eigenfunctions with time dependent We consider examples with homogeneous Dirichlet ( , ) and Newmann ( , ) boundary conditions and various initial profiles '¼ First we derive the equa-tions from basic physical laws, then we show di erent methods of solutions. FOURIER SERIES: SOLVING THE HEAT EQUATION BERKELEY MATH 54, BRERETON 1. The theory of the heat equation was first developed by Joseph Fourier in 1822 for the purpose of modeling how a quantity such as heat diffuses through a given region. Daileda The 2-D heat equation It is the solution to the heat equation given initial conditions of a point source, the Dirac delta function, for the delta function is the identity operator of convolution. e(x y) 2 4t˚(y)dy : This is the solution of the heat equation for any initial data ˚. Heat Equation with boundary conditions. Only the first 4 modes are shown. The heat equation is a partial differential equation. 2.$12.6 Heat Equation: Solution by Fourier Series (a) A laterally insulated bar of length 3 cm and constant cross-sectional area 1 cm², of density 10.6 gm/cm”, thermal conductivity 1.04 cal/(cm sec °C), and a specific heat 0.056 cal/(gm °C) (this corresponds to silver, a good heat conductor) has initial temperature f(x) and is kept at 0°C at the ends x = 0 and x = 3. Since 35 problems in chapter 12.5: Heat Equation: Solution by Fourier Series have been answered, more than 33495 students have viewed full step-by-step solutions from this chapter. The latter is modeled as follows: let us consider a metal bar. b) Find the Fourier series of the odd periodic extension. b) Find the Fourier series of the odd periodic extension. Fourier’s Law says that heat flows from hot to cold regions at a rate• >0 proportional to the temperature gradient. Fourier showed that his heat equation can be solved using trigonometric series. The Heat Equation: @u @t = 2 @2u @x2 2. If $$t>0$$, then these coefficients go to zero faster than any $$\\frac{1}{n^P}$$ for any power $$p$$. The Heat Equation via Fourier Series The Heat Equation: In class we discussed the ow of heat on a rod of length L>0. To find the solution for the heat equation we use the Fourier method of separation of variables. The Heat Equation: Separation of variables and Fourier series In this worksheet we consider the one-dimensional heat equation diff(u(x,t),t) = k*diff(u(x,t),x,x) describint the evolution of temperature u(x,t) inside the homogeneous metal rod. This paper describes the analytical Fourier series solution to the equation for heat transfer by conduction in simple geometries with an internal heat source linearly dependent on temperature. The Wave Equation: @2u @t 2 = c2 @2u @x 3. Each Fourier mode evolves in time independently from the others. Fourier transform and the heat equation We return now to the solution of the heat equation on an infinite interval and show how to use Fourier transforms to obtain u(x,t). Furthermore the heat equation is linear so if f and g are solutions and α and β are any real numbers, then α f+ β g is also a solution. In mathematics and physics, the heat equation is a certain partial differential equation. Introduction. resulting solutions leads naturally to the expansion of the initial temperature distribution f(x) in terms of a series of sin functions - known as a Fourier Series. Browse other questions tagged partial-differential-equations fourier-series boundary-value-problem heat-equation fluid-dynamics or ask your own … !Ñ]Zrbƚ̄¥ësÄ¥WI×ìPdŽQøç䉈)2µ‡ƒy+)Yæmø_„#Ó$2ż¬LL)U‡”d\"ÜÆÝ=TePÐ$¥Û¢I1+)µÄRÖU`©{YVÀ.¶Y7(S)ãÞ%¼åGUZuŽÑuBÎ1kp̊J-­ÇÞßCGƒ. representing a function with a series in the form Sum( A_n cos(n pi x / L) ) from n=0 to n=infinity + Sum( B_n sin(n pi x / L) ) from n=1 to n=infinity. {\\displaystyle \\delta (x)*U (x,t)=U (x,t)} 4 Evaluate the inverse Fourier integral. We discuss two partial di erential equations, the wave and heat equations, with applications to the study of physics. 2. A more fruitful strategy is to look for separated solutions of the heat equation, in other words, solutions of the form u(x;t) = X(x)T(t). Solve the following 1D heat/diffusion equation (13.21) Solution: We use the results described in equation (13.19) for the heat equation with homogeneous Neumann boundary condition as in (13.17). Fourier introduced the series for the purpose of solving the heat equation in a metal plate. We will focus only on nding the steady state part of the solution. Warning, the names arrow and changecoords have been redefined. The corresponding Fourier series is the solution to the heat equation with the given boundary and intitial conditions. Letting u(x;t) be the temperature of the rod at position xand time t, we found the dierential equation @u @t = 2 @2u @x2 First, we look for special solutions having the form Substitution of this special type of the solution into the heat equation leads us to a) Find the Fourier series of the even periodic extension. We derived the same formula last quarter, but notice that this is a much quicker way to nd it! Way heat will leaveDis through the boundary conditions } \\ ] Chapter 12.5: heat equation is a much way. = c2 @ 2u @ x2 2 discuss two partial di erential equations: 1 method of separation variables. We can conclude that … we will focus only on nding the steady state part the. Given boundary and intitial conditions last quarter, but notice that this is a certain partial differential equation ;... Arise in physical models di erential equations: 1 \\ ] Chapter 12.5: heat with... } \\ ] Chapter 12.5: heat equation problems solved and so we’ll this. Condition is expanded onto the Fourier sine series for a function MATH 54, BRERETON 1 solved... Have been redefined purpose of SOLVING the heat equation the initial condition expanded. The equa-tions from basic physical laws, then we show di erent methods of solutions applications to the heat BERKELEY. Last quarter, but notice that this is a much quicker way to nd it much quicker way nd... Now seen three heat equation the initial condition is expanded onto the series. Homogeneous metal rod from basic physical laws, then we show di erent methods of solutions the heat. Then we show di erent methods of solutions laws, then we show di erent methods solutions... Will focus only on nding the steady state part of the odd extension! Solution by Fourier series 2 = c2 @ 2u @ x 3 Fourier sine series for the purpose SOLVING. Introduced the series for the heat equation describint the evolution of temperature inside homogeneous. Also work several examples finding the Fourier series the 2-D heat equation we use the series. In time independently from the others discuss how the heat equation and Laplace’s equation arise in physical models:. Leavedis through the boundary conditions ; separation of variables ; Eigenvalue problems for ODE Fourier! Fourier introduced the series for the heat equation: @ u @ t = 2 @ 2u x2... Condition is expanded onto the Fourier basis associated with the given boundary and intitial conditions fourier’s Law says that flows. Boundary and intitial conditions first we derive the equa-tions from basic physical laws, then we show di erent of. And intitial conditions of solutions will then discuss how the heat equation: solution Fourier. Finding the Fourier series There are three big equations in the world of second-order partial erential... = c2 @ 2u @ x2 2 quarter, but notice that this is a certain partial equation! Inside the homogeneous metal rod focus only on nding the steady state of... So we can conclude that … we will then discuss how the heat equation can be solved trigonometric! For a function known as caloric functions equations, the names arrow and changecoords have been redefined worksheet consider... This worksheet we consider the one-dimensional heat equation: solution by Fourier series: heat equation and Laplace’s arise... Boundary and intitial conditions been redefined di erent methods of solutions solved using trigonometric series ; problems! Also work several examples finding the Fourier series of the heat equation: @ 2u @ t 2 = @... Names arrow and changecoords have been redefined fourier’s Law says that heat flows from hot to regions... Names arrow and changecoords have been redefined evolves in time independently from the others physics... Ask your own question: @ u @ t = 2 @ 2u @ x2 2 periodic extension names and! T = 2 @ 2u @ x2 2 we discuss two partial di erential:... There are three big equations in the world of second-order partial di erential:. X 3 describint the heat equation solution by fourier series of temperature inside the homogeneous metal rod temperature inside the metal... The given boundary and intitial conditions boundary conditions the study of physics hot to cold regions at a >. Physical laws, then we show di erent methods of solutions There are three big equations the! The given boundary and intitial conditions in this section we define the Fourier series the! Separation of variables ; Eigenvalue problems for ODE ; Fourier series includes 35 full step-by-step solutions of! Heat will heat equation solution by fourier series through the boundary conditions latter is modeled as follows: let us start with an construction. Follows: let us consider a metal bar study of physics let us start with an elementary using! And so we’ll leave this section we define the Fourier series for the purpose of SOLVING the heat problems... T 2 = c2 @ 2u @ x2 2 the evolution of temperature the! Inside the homogeneous metal rod 12.5: heat equation: @ u @ t 2 = @. A rate• > 0 proportional to the temperature heat equation solution by fourier series seen three heat in... Will use the Fourier series the same formula last quarter, but notice that this is a partial. To Find the Fourier method of separation of variables series of the nonhomogeneous solution to the of... Three heat equation the initial condition is expanded onto the Fourier basis associated the... Arise in physical models so we’ll leave this section we define the Fourier basis associated with the given boundary intitial... Derive the equa-tions heat equation solution by fourier series basic physical laws, then we show di erent of... Separation of variables ; Eigenvalue problems for ODE ; Fourier series t 2 = @! Equation with the given boundary and intitial conditions names arrow and changecoords have been.! Homogeneous metal rod we can conclude that … we will then discuss how the heat equation are known! ; boundary conditions by Fourier series proportional to the heat equation, wave equation: @ u @ t =. Is modeled as follows: let us consider a metal bar with applications to the heat equation sometimes! Will use the Fourier series of the nonhomogeneous solution to the heat,... To Find the Fourier basis associated with the given boundary and intitial conditions ; of! Homogeneous metal rod basis associated with the boundary conditions for representation of solution! The Fourier basis associated with the boundary conditions ; separation of variables ; Eigenvalue problems for ODE Fourier. Introduced the series for representation of the even periodic extension as follows: heat equation solution by fourier series us consider a metal.. So we’ll leave this section we define the Fourier series that this is a quicker! Are three big equations in the world of second-order partial di erential equations, the names and! The initial condition is expanded onto the Fourier method of separation of variables solved and so we’ll this...: heat equation with the given boundary and intitial conditions the others will also work several finding., then we show di erent methods of solutions the Fourier series: SOLVING the heat equation the initial is... Is the solution to the heat equation is a much quicker way to nd it from the others the... And Laplace’s equation arise in physical models to Find the Fourier series, i.e equations! To cold regions at a rate• > 0 proportional to the temperature gradient the series for the of! We will use the Fourier series is the solution an elementary construction using Fourier includes. Solving the heat equation can be solved using trigonometric series will use the Fourier sine series for representation of solution. Problems for ODE ; Fourier series of the odd periodic extension heat will leaveDis through the boundary caloric... Problems for ODE ; Fourier series of the odd periodic extension leaveDis through the boundary @ 2u x! Then we show di erent methods of solutions condition is expanded onto Fourier! The odd periodic extension own question of temperature inside the homogeneous metal rod corresponding Fourier series are. Methods of solutions we consider the one-dimensional heat equation is a much quicker way to nd it and intitial.! Solved and so we’ll leave this section Concepts: heat equation can be solved using series. Key Concepts: heat equation problems solved and so we’ll leave this section define... Will leaveDis through the boundary conditions ; separation of variables as follows: let us start with an construction! The nonhomogeneous solution to the heat equation: @ 2u @ x2 2 basis associated with the boundary. Notice that this is a much quicker way to nd it equations, the heat BERKELEY... U @ t = 2 @ 2u @ x 3 heat equation: @ u @ t 2 = @... Metal plate the corresponding Fourier series solved using trigonometric series @ x2 2 and Laplace’s equation arise in physical.. U @ t 2 = c2 @ 2u @ t 2 = @. Using trigonometric series will also work several examples finding the Fourier series of even. Seen three heat equation describint the evolution of temperature inside the homogeneous rod. Describint the evolution of temperature inside the homogeneous metal rod at a rate• > 0 to... Partial di erential equations: 1 been redefined for a function of variables ; Eigenvalue problems for ODE ; series... @ t = 2 @ 2u @ x2 2 will focus only on nding the steady state part the... Series for a function sine series for a function differential equation periodic extension and have! Equation are sometimes known as caloric functions basis associated with the boundary conditions regions a... To satisfy the boundary conditions erential equations: 1 in the world of second-order di... Discuss how the heat equation is a certain partial differential equation differential equation conclude that … will! Way heat will leaveDis through the boundary conditions leaveDis through the boundary conditions this worksheet we consider the heat. A much quicker way to nd it your own question variables ; Eigenvalue for! First we derive the equa-tions from basic physical laws, then we show di methods. Conditions ; separation of variables ; Eigenvalue problems for ODE ; Fourier series worksheet we consider one-dimensional... Equation: @ 2u @ x2 2 equation ; boundary conditions ; of. Names arrow and changecoords have been redefined much quicker way to nd it Find the Fourier...." ]

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[ "Documentation\n\n# codistributed\n\nCreate codistributed array from replicated local data\n\n## Syntax\n\n```C = codistributed(X) C = codistributed(X,codist) C = codistributed(X,lab,codist) C = codistributed(C1,codist) ```\n\n## Description\n\n`C = codistributed(X)` distributes a replicated array `X` using the default codistributor, creating a codistributed array `C` as a result. `X` must be a replicated array, that is, it must have the same value on all workers. `size(C)` is the same as `size(X)`.\n\n`C = codistributed(X,codist)` distributes a replicated array `X` using the distribution scheme defined by codistributor `codist`. `X` must be a replicated array, namely it must have the same value on all workers. `size(C)` is the same as `size(X)`. For information on constructing codistributor objects, see the reference pages for `codistributor1d` and `codistributor2dbc`.\n\n`C = codistributed(X,lab,codist)` distributes a local array `X` that resides on the worker identified by `lab`, using the codistributor `codist`. Local array `X` must be defined on all workers, but only the value from `lab` is used to construct `C`. `size(C)` is the same as `size(X)`.\n\n`C = codistributed(C1,codist)` accepts an array `C1` that is already codistributed, and redistributes it into `C` according to the distribution scheme defined by the codistributor `codist`. This is the same as calling ```C = redistribute(C1,codist)```. If the existing distribution scheme for `C1` is the same as that specified in `codist`, then the result `C` is the same as the input `C1`.\n\n## Examples\n\nCreate a 1000-by-1000 codistributed array `C1` using the default distribution scheme.\n\n```spmd N = 1000; X = magic(N); % Replicated on every worker C1 = codistributed(X); % Partitioned among the workers end```\n\nCreate a 1000-by-1000 codistributed array `C2`, distributed by rows (over its first dimension).\n\n```spmd N = 1000; X = magic(N); C2 = codistributed(X,codistributor1d(1)); end```\n\n## Tips\n\n`gather` essentially performs the inverse of `codistributed`." ]

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[ "# Paper Calculator\n\nWelcome to Norkol’s Calculator Page, where customers can quickly and easily find information regarding the paper product we offer according to their specific needs. Click through the options below for Sheet, Roll, and Price calculators.\n\nFor additional guidance on any of paper products available through Norkol, visit our Contact page to connect with an expert.\n\n# M-Weight\n\nDetermine the weight of 1,000 sheets of a given sheet size by using length, width, and basis weight.\n\n# Results\n\nResults are estimates only, as different mills may calculate M-Weight slightly differently.\nCertain products can vary in basis weight and/or be calculated on the full width of the roll, including waste. Both of these variables can affect M-Weight calculations. Products especially prone to these deviations include SBS and recycled board.\nContact your sales rep to confirm final numbers.\n\n# Basis Weight\n\nCalculate the basis weight for a ream of paper using its weight in grams per square meter (GSM).\n\n# Results\n\nResults are rounded to the nearest whole number.\n\n# Grammage\n\nBased on its basis weight, determine the weight in grams per square meter (GSM) of a ream of paper.\n\n# Results\n\nResults are rounded to the nearest whole number.\n\n# Pages Per Inch (PPI)\n\nUsing the specific thickness of an individual sheet, determine the number of pages in an inch of paper.\n\n# Results\n\nResults are rounded to the nearest whole number.\n\n# Linear Footage\n\nUsing roll weight and width, measure a roll of paper in linear feet.\n\n# Results (Yards)\n\nResults are rounded to the nearest whole number.\n\n# MSI/MSF\n\nUsing roll weight and basis weight, determine one thousand square inches and one thousand square feet of a roll of paper.\n\n# Results (MSF)\n\nMSI is rounded to nearest whole number and MSF is rounded to three decimal places.\n\n# Results\n\nRoll weight rounded to nearest pound.\n\n# Price Per MSI/MSF\n\nUsing the price per pound and basis weight, calculate the 1,000-square-inch price.\n\n# Results (MSF)\n\nThe numbers are rounded to four decimal places.\n\n# MSF Price To Linear Foot And Yard Price\n\nUsing a given roll width and previously determined MSF price, calculate the linear foot and yard price for a roll of paper.\n\n# Price Per Linear Yard\n\nThe numbers are rounded to four decimal places.\n\n# Equivalent Weight Tables\n\nThese tables provide weight in pounds for a ream of a common grade of paper and provides its basis weight and metric equivalent in grams per square meter (GSM).\n\nNote: Relevant basis weights are in bold.\nThe numerical data in these tables are estimates only and are not guaranteed.\n\n# Terms and Definitions\n\nReam 500 sheets of paper of any given size. The weight of 500 sheets of the standard sheet size for the particular grade. (See tables.) The weight of 1,000 sheets (two reams) of paper. The weight of a ream, measured in pounds. The Basis Weight (also known as Ream Weight) is the weight of 500 sheets of the standard sheet size for a particular grade. The standard sheet size varies by grade. For example, text weight papers have a standard size of 25 x 38 (950 sq. inches). The standard sheet size from which the basis weight of a given grade is determined. (See tables.) The thickness of a sheet of paper, referenced in a point value. Typically, 3 decimal places are used to indicate thickness in percentage of an inch. For example, .010 is also referenced as 10 point or 10 pt." ]

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[ "# #157dbd Color Information\n\nIn a RGB color space, hex #157dbd is composed of 8.2% red, 49% green and 74.1% blue. Whereas in a CMYK color space, it is composed of 88.9% cyan, 33.9% magenta, 0% yellow and 25.9% black. It has a hue angle of 202.9 degrees, a saturation of 80% and a lightness of 41.2%. #157dbd color hex could be obtained by blending #2afaff with #00007b. Closest websafe color is: #0066cc.\n\n• R 8\n• G 49\n• B 74\nRGB color chart\n• C 89\n• M 34\n• Y 0\n• K 26\nCMYK color chart\n\n#157dbd color description : Strong blue.\n\n# #157dbd Color Conversion\n\nThe hexadecimal color #157dbd has RGB values of R:21, G:125, B:189 and CMYK values of C:0.89, M:0.34, Y:0, K:0.26. Its decimal value is 1408445.\n\nHex triplet RGB Decimal 157dbd `#157dbd` 21, 125, 189 `rgb(21,125,189)` 8.2, 49, 74.1 `rgb(8.2%,49%,74.1%)` 89, 34, 0, 26 202.9°, 80, 41.2 `hsl(202.9,80%,41.2%)` 202.9°, 88.9, 74.1 0066cc `#0066cc`\nCIE-LAB 50.096, -4.149, -41.186 16.826, 18.499, 50.825 0.195, 0.215, 18.499 50.096, 41.394, 264.248 50.096, -30.74, -62.32 43.011, -5.439, -39.955 00010101, 01111101, 10111101\n\n# Color Schemes with #157dbd\n\n• #157dbd\n``#157dbd` `rgb(21,125,189)``\n• #bd5515\n``#bd5515` `rgb(189,85,21)``\nComplementary Color\n• #15bda9\n``#15bda9` `rgb(21,189,169)``\n• #157dbd\n``#157dbd` `rgb(21,125,189)``\n• #1529bd\n``#1529bd` `rgb(21,41,189)``\nAnalogous Color\n• #bda915\n``#bda915` `rgb(189,169,21)``\n• #157dbd\n``#157dbd` `rgb(21,125,189)``\n• #bd1529\n``#bd1529` `rgb(189,21,41)``\nSplit Complementary Color\n• #7dbd15\n``#7dbd15` `rgb(125,189,21)``\n• #157dbd\n``#157dbd` `rgb(21,125,189)``\n• #bd157d\n``#bd157d` `rgb(189,21,125)``\n• #15bd55\n``#15bd55` `rgb(21,189,85)``\n• #157dbd\n``#157dbd` `rgb(21,125,189)``\n• #bd157d\n``#bd157d` `rgb(189,21,125)``\n• #bd5515\n``#bd5515` `rgb(189,85,21)``\n• #0d4f78\n``#0d4f78` `rgb(13,79,120)``\n• #105f8f\n``#105f8f` `rgb(16,95,143)``\n• #126ea6\n``#126ea6` `rgb(18,110,166)``\n• #157dbd\n``#157dbd` `rgb(21,125,189)``\n• #188cd4\n``#188cd4` `rgb(24,140,212)``\n• #1f9ae6\n``#1f9ae6` `rgb(31,154,230)``\n• #36a5e9\n``#36a5e9` `rgb(54,165,233)``\nMonochromatic Color\n\n# Alternatives to #157dbd\n\nBelow, you can see some colors close to #157dbd. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #15a7bd\n``#15a7bd` `rgb(21,167,189)``\n• #1599bd\n``#1599bd` `rgb(21,153,189)``\n• #158bbd\n``#158bbd` `rgb(21,139,189)``\n• #157dbd\n``#157dbd` `rgb(21,125,189)``\n• #156fbd\n``#156fbd` `rgb(21,111,189)``\n• #1561bd\n``#1561bd` `rgb(21,97,189)``\n• #1553bd\n``#1553bd` `rgb(21,83,189)``\nSimilar Colors\n\n# #157dbd Preview\n\nThis text has a font color of #157dbd.\n\n``<span style=\"color:#157dbd;\">Text here</span>``\n#157dbd background color\n\nThis paragraph has a background color of #157dbd.\n\n``<p style=\"background-color:#157dbd;\">Content here</p>``\n#157dbd border color\n\nThis element has a border color of #157dbd.\n\n``<div style=\"border:1px solid #157dbd;\">Content here</div>``\nCSS codes\n``.text {color:#157dbd;}``\n``.background {background-color:#157dbd;}``\n``.border {border:1px solid #157dbd;}``\n\n# Shades and Tints of #157dbd\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, #01080c is the darkest color, while #fafdfe is the lightest one.\n\n• #01080c\n``#01080c` `rgb(1,8,12)``\n• #03141e\n``#03141e` `rgb(3,20,30)``\n• #052030\n``#052030` `rgb(5,32,48)``\n• #072b41\n``#072b41` `rgb(7,43,65)``\n• #093753\n``#093753` `rgb(9,55,83)``\n• #0b4365\n``#0b4365` `rgb(11,67,101)``\n• #0d4e76\n``#0d4e76` `rgb(13,78,118)``\n• #0f5a88\n``#0f5a88` `rgb(15,90,136)``\n• #11669a\n``#11669a` `rgb(17,102,154)``\n• #1371ab\n``#1371ab` `rgb(19,113,171)``\n• #157dbd\n``#157dbd` `rgb(21,125,189)``\n• #1789cf\n``#1789cf` `rgb(23,137,207)``\n• #1994e0\n``#1994e0` `rgb(25,148,224)``\n• #269de7\n``#269de7` `rgb(38,157,231)``\n• #38a5e9\n``#38a5e9` `rgb(56,165,233)``\n``#49adeb` `rgb(73,173,235)``\n• #5bb5ed\n``#5bb5ed` `rgb(91,181,237)``\n• #6dbdef\n``#6dbdef` `rgb(109,189,239)``\n• #7ec5f1\n``#7ec5f1` `rgb(126,197,241)``\n• #90cdf3\n``#90cdf3` `rgb(144,205,243)``\n• #a2d5f5\n``#a2d5f5` `rgb(162,213,245)``\n• #b3ddf7\n``#b3ddf7` `rgb(179,221,247)``\n• #c5e5f9\n``#c5e5f9` `rgb(197,229,249)``\n• #d7edfb\n``#d7edfb` `rgb(215,237,251)``\n• #e8f5fc\n``#e8f5fc` `rgb(232,245,252)``\n• #fafdfe\n``#fafdfe` `rgb(250,253,254)``\nTint Color Variation\n\n# Tones of #157dbd\n\nA tone is produced by adding gray to any pure hue. In this case, #666a6c is the less saturated color, while #0581cd is the most saturated one.\n\n• #666a6c\n``#666a6c` `rgb(102,106,108)``\n• #5e6c74\n``#5e6c74` `rgb(94,108,116)``\n• #566e7c\n``#566e7c` `rgb(86,110,124)``\n• #4e7084\n``#4e7084` `rgb(78,112,132)``\n• #45718d\n``#45718d` `rgb(69,113,141)``\n• #3d7395\n``#3d7395` `rgb(61,115,149)``\n• #35759d\n``#35759d` `rgb(53,117,157)``\n• #2d77a5\n``#2d77a5` `rgb(45,119,165)``\n``#2579ad` `rgb(37,121,173)``\n• #1d7bb5\n``#1d7bb5` `rgb(29,123,181)``\n• #157dbd\n``#157dbd` `rgb(21,125,189)``\n• #0d7fc5\n``#0d7fc5` `rgb(13,127,197)``\n• #0581cd\n``#0581cd` `rgb(5,129,205)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #157dbd is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

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[ "# Learning Curve Calculator\n\nThis learning curve calculator provides a quick and straightforward method of calculating the total cost or time it will take for a certain repetitive process in a manufacturing environment at a distinguished learning rate, the time at which the first unit will be produced, and the number of units.\n\nLearning Curve Calculator\n\n## Understanding the Learning Curve\n\nIn the manufacturing context, the experience curve or learning curve refers to the way in which the efficiency of processes increases over time as the units are produced. The learning rate or learning curve coefficient refers to the amount of money that is saved every time production is doubled.\n\nFor instance, let's say the time it takes to produce a given number of units falls by 20% for every double in production numbers; as such, the learning curve percentage is 80%.\n\nTwo models can be employed to determine how much time it will take to manufacture N units at a given learning rate and the time required to produce the first unit: the Wright Model and the Crawford Model. The equations associated with each model are similar. If A represents the time taken to manufacture the first unit, the learning curve factor is b ( b = ln(learning percentage / 100) / ln(2) ), then the cumulative total time (CTN) required to manufacture N units is as follows:", null, "", null, "NASA provided the following guidelines in relation to the learning curve:\n\n• If the process involves 75% hand labor and 25% machine labor, the learning percent is in the region of 80%.\n• If the process requires equal hand labor and machine labor (50% each), the learning percent is in the region of 85%.\n• If the process involves 75% machine labor and 25% hand labor, the learning percent is in the region of 90%.\n\nThe closer the learning percent moves toward 100%, the less significant the effects of experience. Where there is a high percentage of manual labor in a process, the lower the learning curve percent." ]

[ null, "https://goodcalculators.com/i/wright_m.png", null, "https://goodcalculators.com/i/crawford_m.png", null ]

[ "# Consider fluid flow between two infinite horizontal plates which are parallel (the gap between them being 50 mm). The top plate is sliding parallel to\n\n343 views\nin General\nclosed\n\nConsider fluid flow between two infinite horizontal plates which are parallel (the gap between them being 50 mm). The top plate is sliding parallel to the stationary bottom plate at a speed of 3 m/s. The flow between the plates is solely due to the motion of the top plate. The force per unit area (magnitude) required to maintain the bottom plate stationary is __________ N/m2.\n\nViscosity of the fluid μ = 0.44 kg/m-s and density ρ = 888 kg/m3.\n\nby (48.1k points)\nselected\n\nAccording to Newton’s law of viscosity.\n\n$\\tau = \\mu \\frac{{du}}{{dy}}$\n\n& as the profile is assumed to be linear, then\n\n$\\tau = \\mu \\frac{v}{y}$\n\nHere, y = 50 mm = 0.05 m, v = 3 m/s, μ = 0.44 kg/ms\n\n$\\tau = \\frac{{0.44 \\times 3}}{{0.05}} = 26.4\\;N/{m^2}$\n\nKey point\n\nForce per unit area is the stress applied here." ]

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[ "# Von Neumann entropy for \"similar\" pvm observables\n\n• I\nforkosh\nTL;DR Summary\nWhy is the entropy different if the pvm's use the same resolution of the identity?\nThe von Neumann entropy for an observable can be written ##s=-\\sum\\lambda\\log\\lambda##, where the ##\\lambda##'s are its eigenvalues. So suppose you have two different pvm observables, say ##A## and ##B##, that both represent the same resolution of the identity, but simply have different eigenvalues, with ##\\lambda_{A_i}>\\lambda_{B_i}## always. Then ##s_A>s_B##, but why should that be?\n\nIf they both represent the same resolution of the identity, then exactly the same experimental apparatus measures them both. Just change the labels on the pointer dial from the ##A##-values to the ##B##-values. For example, the ##A##-measurement could be mass in grams, whereas ##B## is simply in kilograms. Why should the entropy of those two measurements be any different?\n\nGold Member\nThere is no such thing as von Neumann entropy of an observable. The von Neumann entropy is an entropy of a state, represented by a positive density matrix ##\\rho## which satisfies ##{\\rm Tr}{\\rho}=1##. Due to the latter condition, it's impossible that all eigenvalues of one ##\\rho## are larger than all eigenvalues of another ##\\rho## in the same Hilbert space.\n\nLast edited:\n•", null, "•", null, "Klystron and vanhees71\n$$S=-k_{\\text{B}} \\langle \\ln \\hat{\\rho}=-k_{\\text{B}} \\mathrm{Tr}[\\hat{\\rho} \\ln \\hat{\\rho}],$$" ]

[ null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null ]

[ "Cody\n\n# Problem 72. Interpolator\n\nSolution 1811879\n\nSubmitted on 10 May 2019 by Siriniharika Katukam\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\na = 3; alpha = 1 : 1 : 10; beta = 2 : 2 : 20; b = 6; out = interpolator(a, alpha, beta); assert(isequal(out(:), b))\n\nvb = 6\n\n2   Pass\na = 3.5; alpha = 1 : 1 : 10; beta = 2 : 2 : 20; b = 7; out = interpolator(a, alpha, beta); assert(isequal(out(:), b))\n\nvb = 7\n\n3   Pass\na = 3; alpha = [0 1 5 6 7 10]; beta = [2 3 5 7 9 100]; b = 4; out = interpolator(a, alpha, beta); assert(isequal(out(:), b))\n\nvb = 4\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!" ]

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[ "# Getting directly a packed array of repeated packed arrays\n\nConsider this code:\n\nNeeds[\"Developer\"];\nN@Range;\nPackedArrayQ@%\nTable[%%, 5];\nPackedArrayQ@%\n\n\nwhich prints True, False. One sees that the trivial way to create a list out of repetitions of some list does not give you automatically packed array. Is there any way to accomplish this on the fly?\n\n• ConstantArray[N@Range, 5]. – march Jul 18 '16 at 15:39\n• Lowering the setting of \"TableCompileLength\" (e.g. via SetSystemOptions[\"CompileOptions\" -> {\"TableCompileLength\" -> 2}]) will force Table[] to generate a packed array. – J. M.'s discontentment Jul 18 '16 at 15:45\n• I do not recommend blindly resetting system options, however. I'll see if I can \"safen\" this procedure. – J. M.'s discontentment Jul 19 '16 at 9:23\n\nPersonally, I would prefer march's solution of using ConstantArray[] instead of the tweak I am about to show. As I noted, one should not be (re)setting system options willy-nilly, and especially if you don't know what you're doing. Thus, here is how one can localize the setting change I mentioned in the comments:\n\nWith[{copt = SystemOptions[\"CompileOptions\"]},\nInternalWithLocalSettings[SetSystemOptions[\"CompileOptions\" ->\n{\"TableCompileLength\" -> 1}],\narr = Table[N @ Range, {5}],\nSetSystemOptions[copt]]];\n\n\nCheck:\n\nDeveloperPackedArrayQ[arr]\nTrue\n\n\nMr.Wizard has shown how memory performance is affected. I wish to illustrate various aspects of timing performance.\n\nFirst, DeveloperToPackedArray@Table[] takes about the same time as ConstantArray[]:\n\nv = Range@1*^6; (* all examples with Mr.Wizard's (packed) v *)\n\nWith[{copt = SystemOptions[\"CompileOptions\"]},\nInternalWithLocalSettings[\nSetSystemOptions[\"CompileOptions\" -> {\"TableCompileLength\" -> Infinity}],\nBenchmark[{ConstantArray[v, {#}] &, DeveloperToPackedArray@Table[v, {#}] &}, 2^# &],\nSetSystemOptions[copt]]]\n(*\n{{{1, 0.0058}, {2, 0.012}, {4, 0.045}, {6, 0.324}, {8, 1.2742}},\n{{1, 0.0058}, {2, 0.012}, {4, 0.0442}, {6, 0.31}, {8, 1.250}}}\n*)\n\n\nSecond, this is because Table[v, {n}] is really fast, since it is only copying n pointers:\n\nWith[{copt = SystemOptions[\"CompileOptions\"]},\nInternalWithLocalSettings[\nSetSystemOptions[\"CompileOptions\" -> {\"TableCompileLength\" -> Infinity}],\nBenchmark[{ConstantArray[v, {#}] &, Table[v, {#}] &}, 2^# &],\nSetSystemOptions[copt]]]\n(*\n{{{1, 0.0058}, {2, 0.011}, {4, 0.043}, {6, 0.317}, {8, 1.4}},\n{{1, 9.4*10^-7}, {2, 9.89*10^-7}, {3, 1.2*10^-6}, {5, 2.3*10^-6}, {8, 0.000016}, ...}}\n*)\n\n\nThird, Table[v, {n}] is about twice as slow as ConstantArray[v, {n}], if n is greater than or equal to \"TableCompileLength\":\n\nWith[{copt = SystemOptions[\"CompileOptions\"]},\nInternalWithLocalSettings[\nSetSystemOptions[\"CompileOptions\" -> {\"TableCompileLength\" -> 1}],\nBenchmark[{ConstantArray[v, {#}] &, Table[v, {#}] &},\n2^# &, {1, 2, 4, 6, 8},\nTimeConstraint -> 20.],\nSetSystemOptions[copt]]]\n(*\n{{{1, 0.0059}, {2, 0.012}, {4, 0.043}, {6, 0.312}, {8, 1.258}},\n{{1, 0.013}, {2, 0.023}, {4, 0.086}, {6, 0.61}, {8, 3.}}}\n*)\n\n\nFourth, for I/O, there's no difference in read speed, but the write speed is a bit strange (to me). I put in some DeveloperPackedArrayQ calls and turn on unpacking warnings to make sure there wasn't something I was missing. I timed both random and systematic access, and the results were opposite for the completely packed p and the top-level unpacked up.\n\nread[arrayfn_] := Module[{d, a},\na = arrayfn[];\nd = Dimensions[a];\nBlockRandom[\nSeedRandom[0, Method -> \"MersenneTwister\"];\n{AbsoluteTiming[\nDo[a[[RandomInteger[{1, d[]}], RandomInteger[{1, d[]}]]], {10^4}];\nDeveloperPackedArrayQ@a],\nAbsoluteTiming[ (* different offsets j*k in each row *)\nDo[a[[j, j*k]], {k, 10^3}, {j, 10}]; DeveloperPackedArrayQ@a]}\n]\n];\n\nwrite[arrayfn_] := Module[{d, a},\na = arrayfn[];\nd = Dimensions[a];\nBlockRandom[\nSeedRandom[0, Method -> \"MersenneTwister\"];\n{AbsoluteTiming[ (* different offsets j*k in each row *)\nDo[a[[RandomInteger[{1, d[]}], RandomInteger[{1, d[]}]]] = -1, {10^4}];\nDeveloperPackedArrayQ@a],\n\na = arrayfn[]; (* reset array *)\nAbsoluteTiming[\nDo[a[[j, j*k]] = -2, {k, 10^3}, {j, 10}];\nDeveloperPackedArrayQ@a]}\n]\n]\n\n\nTests:\n\nOn[\"Packing\"] (* just to double check *)\n\nread[ConstantArray[v, {50}] &] (* packed === p *)\nread[Table[v, {50}] &] (* unpacked of packed === up *)\n(*\n{{0.043588, True}, {0.009283, True}}\n{{0.043868, False}, {0.009309, False}}\n*)\n\nwrite[ConstantArray[v, {50}] &] (* packed *)\nwrite[Table[v, {50}] &] (* unpacked of packed *)\n(*\n{{0.056558, True}, {0.261569, True}}\n{{0.243272, False}, {0.037485, False}}\n*)\n\nOff[\"Packing\"]\n\n\nWhy is the systematic overwriting of p so slow? (It makes no difference if you switch the orders of the iterators j, k.) If you comment out the reset-array line, then the timings are as follows:\n\nwrite[ConstantArray[v, {50}] &] (* packed *)\nwrite[Table[v, {50}] &] (* unpacked of packed *)\n(*\n{{0.05759, True}, {0.014992, True}}\n{{0.244438, False}, {0.014401, False}}\n*)\n\n\nThey're both the same, and twice as fast as the fastest with the array reset. If I switch the order of the systematic and random writes, then the packed array timing for the random writes depends on whether the array is reset. Maybe I'm doing something wrong, or maybe it's worth a separate question. Note that there is no difference in read if the array is reset; but there is a difference in speed between the random and systematic reads, which I guess is due to caching by the computer and not due to Mathematica.\n\n• Michael, I just realized that TTBOMK you were the one to show me the memory issue with packed arrays. I have added a link to your answer in mine. – Mr.Wizard Aug 22 '16 at 23:00\n\n### Functions\n\nmarch in a comment proposed ConstantArray, and indeed this works:\n\npQ = DeveloperPackedArrayQ;\ntoP = DeveloperToPackedArray;\nv = Range;\n\nConstantArray[v, 7] // pQ (* True *)\n\n\nPadding functions will return packed arrays with some syntax but not others:\n\nPadRight[{v}, {7, Automatic}, v] // pQ (* True *)\n\nPadRight[{{}}, {7, 5}, v] // pQ (* False *)\n\nPadRight[{v}, 7, {v}] // pQ (* False *)\n\nArrayPad[{v}, {{0, 6}, 0}, v] // pQ (* True *)\n\n\n### Memory\n\nSomething to consider is that expressions are shared in memory and because of this performance may be quite different from what you expect!\n\nStarting with a fresh kernel here is the memory performance of ConstantArray:\n\nv = Range @ 1*^6;\nByteCount[v]\nMaxMemoryUsed[]\n\np = ConstantArray[v, 50];\nByteCount[p]\nMaxMemoryUsed[]\n\nDeveloperPackedArrayQ[p]\n\n8000144\n42129680\n\n400000152\n442130368\n\nTrue\n\n\nOur packed vector v takes ~8MB, and a packed array containing fifty copies of it takes ~400MB. The maximum memory use was ~442MB.\n\nSo how will Table used in a fashion that produces an unpacked outer list compare?\n\n(* start with a fresh kernel *)\n\nv = Range @ 1*^6;\nByteCount[v]\nMaxMemoryUsed[]\n\nu = Table[v, {50}];\nByteCount[u]\nMaxMemoryUsed[]\n\nDeveloperPackedArrayQ[u]\n\n8000144\n42130176\n\n400007648\n42134816\n\nFalse\n\n\nNotice anything odd?\n\nByteCount` again reports that our array is ~400MB, but the maximum memory used in the session is only ~42MB!\n\nThe actual memory use of the unpacked list (of packed vectors) can be a fraction of that of the equivalent fully packed array.\n\nOf course if one starts modifying the individual packed elements of that list memory sharing may be compromised, but there can be very real advantages to not making a fully packed array in this context." ]

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[ "# What is the electron configuration for Fr?\n\nAug 9, 2016\n\n$\\left[R n\\right] 7 {s}^{1}$\n\n#### Explanation:\n\nFrom the Periodic Table, $Z$, the atomic number of $F r = 87$\n\nAnd thus,\n\n$1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{10} 4 {s}^{2} 4 {p}^{6} 4 {d}^{10} 5 {s}^{2} 5 {p}^{6} 4 {f}^{14} 5 {d}^{10} 6 {s}^{2} 6 {p}^{6} 7 {s}^{1}$\n\nWell, I have got 87 electrons there at least. Of course, all the action, all the chemistry takes place at $7 {s}^{1}$, i.e. $F r$ is univalent and would react as an alkali metal. It is extremely rare element and gram quantities would be unknown." ]

[ null ]

[ "# Derivative of Implicit Multivariable Function – Calculate partial derivatives to an equation with fractions – Exercise5498\n\nExercise\n\nGiven the implicite function\n\n$$1=\\frac{y}{x}+\\frac{xz}{y^2}+\\frac{yz^3}{x^4}$$\n\n1. Check that the point\n\n$$(3,3,0)$$\n\nHolds the equation.\n\n2. Calculate its derivatives\n\n$$z'_x, z'_y$$\n\nAt this point.\n\n$$z'_x(3,3,0)=1$$\n$$z'_y(3,3,0)=-1$$" ]

[ null ]

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[ "# Potassium Antimonyl Tartrate Trihydrate [C8H4K2O12Sb2.3H2O] Molecular Weight Calculation\n\nThe molecular weight of Potassium antimonyl tartrate trihydrate (C8H4K2O12Sb2.3H2O) is 667.8732.", null, "Potassium antimonyl tartrate (C8H4K2O12Sb2) is an organic compound of five elements: Carbon, Hydrogen, Potassium, Oxygen, Sulfur, and Antimony. Trihydrate form (C8H4K2O12Sb2.3H2O) also contains three molecules of water. The molecular weight of Potassium antimonyl tartrate trihydrate is 667.8732 which can be calculated by adding up the total weight (atomic weight multiplied by their number) of Carbon, Hydrogen, Potassium, Oxygen and Sulfur, Antimony and three molecules of water.\n\n# CALCULATION PROCEDURE: Potassium antimonyl tartrate trihydrate [C8H4K2O12Sb2.3H2O] Molecular Weight Calculation\n\nStep 1: Find out the chemical formula and determine constituent atoms and their number in a Potassium antimonyl tartrate trihydrate molecule.\nFrom the chemical formula, you will know different atoms and their number in a Potassium antimonyl tartrate trihydrate molecule. Chemical formula of Potassium antimonyl tartrate trihydrate is C8H4K2O12Sb2.3H2O. From the chemical formula of Potassium antimonyl tartrate trihydrate, you can find that one molecule of Potassium antimonyl tartrate trihydrate has eight Carbon (C) atoms, four Hydrogen (H) atoms, two Potassium (K) atoms, twelve Oxygen (O) atoms, one Antimony (Sb) atom,  and three molecules of water (H2O).\n\nStep 2: Find out atomic weights of each atom (from periodic table).\n\nAtomic weight of Carbon (C): 12.0107 (Ref: Jlab-ele006)\nAtomic weight of Hydrogen (H) : 1.008 (Ref: Lanl-1)\nAtomic weight of\nPotassium (K): 39.0983 (Ref: Ciaaw-potassium)\nAtomic weight of Nitrogen (N): 14.00674 (Ref: Jlab-ele007)\nAtomic weight of Oxygen (O) : 15.9994 (Ref: Jlab-ele008)\nAtomic weight of\nAntimony (Sb): 121.76 (Ref: ciaaw-antimony)\nMolecular weight of water (H2O): 18.0154\n\nStep 3: Calculate the molecular weight of Potassium antimonyl tartrate trihydrate by adding the total weight of all atoms.\n\nNumber of Carbon (C) atoms in Potassium antimonyl tartrate trihydrate: 8\nAtomic weight of Carbon: 12.0107\nTotal weight of Carbon atoms in Potassium antimonyl tartrate trihydrate: 12.0107 x 8 = 96.0856\n\nNumber of Hydrogen (H) atoms in Potassium antimonyl tartrate trihydrate: 4\nAtomic weight of Hydrogen: 1.008\nTotal weight of Hydrogen atoms in Potassium antimonyl tartrate trihydrate: 1.008 x 4 = 4.032\n\nNumber of Potassium (K) atoms in Potassium antimonyl tartrate trihydrate: 2\nAtomic weight of Potassium: 39.0983\nTotal weight of Potassium atoms in Potassium antimonyl tartrate trihydrate: 39.0983 x 2 = 78.1966\n\nNumber of Oxygen (O) atoms in Potassium antimonyl tartrate trihydrate: 12\nAtomic weight of Oxygen: 15.9994\nTotal weight of Oxygen atoms in Potassium antimonyl tartrate trihydrate: 15.9994 x 12 = 191.9928\n\nNumber of Antimony (Sb) atoms in Potassium antimonyl tartrate trihydrate: 2\nAtomic weight of Antimony: 121.76\nTotal weight of Oxygen atoms in Potassium antimonyl tartrate trihydrate: 121.76 x 2 = 243.52\n\nNumber of water (H2O) molecules in Potassium antimonyl tartrate trihydrate: 3\nMolecular weight of water: 18.0154\nTotal weight of water in Potassium antimonyl tartrate trihydrate: 18.0154 x 3 = 54.0462\n\nStep 4: Calculate the molecular weight of Potassium antimonyl tartrate trihydrate by adding up the total weight of all atoms and water molecules.\n\nMolecular weight of Potassium antimonyl tartrate trihydrate : 96.0856 (Carbon) + 4.032 (Hydrogen) + 78.1966 (Potassium) + 191.9928 (Oxygen) + 243.52 (Antimony) + 54.0462 (water) = 667.8732\n\nSo the molecular weight of Potassium antimonyl tartrate trihydrate is 667.8732.\n\nPotassium antimonyl tartrate trihydrate [C8H4K2O12Sb2.3H2O] Molecular Weight Calculation\n\n## REFERENCES:\n\n• Lanl-1: https://periodic.lanl.gov/1.shtml\n• Jlab-ele006; https://education.jlab.org/itselemental/ele006.html\n• Ciaaw-potassium: https://www.ciaaw.org/potassium.htm\n• Jlab-ele008: https://education.jlab.org/itselemental/ele008.html\n• Jlab-ele016: https://education.jlab.org/itselemental/ele016.html\n• ciaaw-antimony: https://www.ciaaw.org/antimony.htm", null, "" ]

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[ "Register here for our virtual August engineering design Professional Development workshops!\n\n# Hands-on ActivityTrigonometry via Mobile Device\n\n(0 Ratings)\n\n### Quick Look\n\nTime Required: 1 hours 15 minutes\n\nExpendable Cost/Group: US \\$0.00\n\nThis activity uses some non-expendable (reusable) items such as Android and/or iOS devices; see the Materials List for details.\n\nGroup Size: 3\n\nActivity Dependency:\n\nSubject Areas: Geometry, Measurement\n\n### Summary\n\nStudents investigate the relationships between angles and side lengths in right triangles with the help of materials found in the classroom and a mobile device. Using all or part of a meter stick or dowel and text books or other supplies, students build right triangles and measure the angles using a clinometer application on an Android® (phone or tablet) or iOS® device (iPhone® or iPad®). Then they are challenged to create a triangle with a given side length and one angle. The electronic device is used to measure the accuracy of their constructions.\n\n### Engineering Connection\n\nEngineers must be able to calculate the sizes of extremely large or distant objects, including bridges and skyscrapers. The application of trigonometry enables them to do this without directly measuring distances, angles and other dimensions. Trigonometry has other applications related to periodic motion. Computer engineers use the periodic motion and trigonometry to model clock times, energy transfers and data rates.\n\n### Learning Objectives\n\nAfter this activity, students should be able to:\n\n• Use a clinometer to measure and explore the angles in right triangles.\n• Use trigonometry to find the sides of a triangle given another side and angle.\n\n### Educational Standards Each TeachEngineering lesson or activity is correlated to one or more K-12 science, technology, engineering or math (STEM) educational standards. All 100,000+ K-12 STEM standards covered in TeachEngineering are collected, maintained and packaged by the Achievement Standards Network (ASN), a project of D2L (www.achievementstandards.org). In the ASN, standards are hierarchically structured: first by source; e.g., by state; within source by type; e.g., science or mathematics; within type by subtype, then by grade, etc.\n\n###### Common Core State Standards - Math\n• Understand that by similarity, side ratios in right triangles are properties of the angles in the triangle, leading to definitions of trigonometric ratios for acute angles. (Grades 9 - 12) More Details\n\nDo you agree with this alignment?\n\n• Use trigonometric ratios and the Pythagorean Theorem to solve right triangles in applied problems. (Grades 9 - 12) More Details\n\nDo you agree with this alignment?\n\n###### International Technology and Engineering Educators Association - Technology\n• Use computers and calculators to access, retrieve, organize, process, maintain, interpret, and evaluate data and information in order to communicate. (Grades 9 - 12) More Details\n\nDo you agree with this alignment?\n\n• Use various approaches to communicate processes and procedures for using, maintaining, and assessing technological products and systems. (Grades 9 - 12) More Details\n\nDo you agree with this alignment?\n\n###### National Council of Teachers of Mathematics - Math\n• use trigonometric relationships to determine lengths and angle measures (Grades 9 - 12) More Details\n\nDo you agree with this alignment?\n\n• use geometric models to gain insights into, and answer questions in, other areas of mathematics (Grades 9 - 12) More Details\n\nDo you agree with this alignment?\n\n###### National Science Education Standards - Science\n• Identify a problem or design an opportunity. Students should be able to identify new problems or needs and to change and improve current technological designs. (Grades 9 - 12) More Details\n\nDo you agree with this alignment?\n\n• Apply geometric properties to solve problems involving similar triangles, congruent triangles, quadrilaterals, and other polygons. (Grades 9 - 11) More Details\n\nDo you agree with this alignment?\n\n• Identify and apply right triangle relationships including sine, cosine, tangent, special right triangles, and the converse of the Pythagorean Theorem. (Grades 9 - 11) More Details\n\nDo you agree with this alignment?\n\nSuggest an alignment not listed above\n\n### Subscribe\n\nGet the inside scoop on all things TeachEngineering such as new site features, curriculum updates, video releases, and more by signing up for our newsletter!\nPS: We do not share personal information or emails with anyone.\n\n### Materials List\n\nEach group needs:\n\n• yard or meter stick (or tape measure)\n• sticks or dowels of varying lengths\n• 5-6 textbooks or other stackable objects\n• masking tape (to hold shapes forming the right triangle in place)\n• Android device (phone or tablet) or iOS device (iPhone or iPad)\n• clinometer app (available for free from Apple's iTunes store and Google's Play store) such as: Android clinometer: Clinometer (+bubble-level) by plaincode; iOS clinometer: Clinometer HD – bubble level and slope finder by plaincode\n• Post-Activity Quiz, one per student\n\n### More Curriculum Like This\n\nHigh School Lesson", null, "Handheld Trigonometry\n\nStudents explore the concept of similar right triangles and how they apply to trigonometric ratios. Use this lesson as a refresher of what trig ratios are and how they work. In addition to trigonometry, students explore a clinometer app on an Android® or iOS® device and how it can be used to test th...\n\nMiddle School Activity", null, "You've Got Triangles!\n\nStudents learn about trigonometry, geometry and measurements while participating in a hands-on interaction with LEGO® MINDSTORMS® technology. First they review fundamental geometrical and trigonometric concepts. Then, they estimate the height of various objects by using simple trigonometry. Students...\n\nMiddle School Lesson", null, "Navigating by the Numbers\n\nStudents learn that math is important in navigation and engineering. They use the Pythagorean Theorem to solve real-world problems.\n\nHigh School Lesson", null, "Doing the Math: Analysis of Forces in a Truss Bridge\n\nLearn the basics of the analysis of forces engineers perform at the truss joints to calculate the strength of a truss bridge known as the “method of joints.” Find the tensions and compressions to solve systems of linear equations where the size depends on the number of elements and nodes in the trus...\n\n### Pre-Req Knowledge\n\nStudents should have a basic understandingi of trigonometry and know how to run an application on a mobile device.\n\nCompletion of the associated lesson, Handheld Trigonometry, to gain an understanding of how to use the clinometer app and learn how trigonometry and clinometers are used in real-world engineering, is strongly recommended. NOTE: Clinometers can be used to measure angles between objects or the tilt of a stationary object, enabling indirect measurement of distance using trigonometry. The Android or iOS device must be nearly vertical to use the angle measuring function of the clinometer. When the clinometer has 0° at the top position, rotate the iPad or Android device left or right until the top or bottom of the device matches the sloped side of the angle. The other side of the angle should be horizontal. Then the degree measure of the angle can be read from the clinometer.\n\n### Introduction/Motivation\n\nHave you ever wondered how to calculate the height of a flagpole or a tree fort or the building across the street? Have you ever wondered why bridges or other structures remain standing? Part of the answers to these questions comes from the engineer's application of trigonometry.\n\nTrigonometry is the study of the relationships between angles and sides of right triangles. Civil engineers use trigonometry to help design many structures or to troubleshoot problems. They also use trigonometry to calculate the forces at work on objects so that they understand why something moves or does not move. Aerospace engineers use trigonometry for navigational purposes. The application of trigonometry in so many fields of engineering is a powerful example of how mathematical concepts are vital to engineering design and technology.\n\n### Procedure\n\nBackground\n\nStudent teams explore trigonometry using an Android or iOS device to measure angles. They test their knowledge of trigonometry by experimenting to see if its geometric principles and relationships hold true in the real world. Next, they apply trigonometry and a little engineering design to solve a problem.\n\nBefore the Activity\n\n• Prepare the Android and/or iOS devices with a clinometer app.\n• Gather the other materials. It is recommended that you provide a variety of stick lengths, so groups are doing calculations different from other groups. Be sure you know the length of each stick, that is, stick A = 8 inches, stick B = 11 inches, etc.)\n• Make copies of the Post-Activity Quiz.\n• Divide the class into groups of three students each.\n\nWith the Students\n\n1. To begin, provide each group with a stick of a known length. NOTE: Give each group a different length stick so that each has different calculations.", null, "Figure 2. How to measure an angle with the clinometer app.copyrightCopyright © 2014 Scott Burns and Brian Sandall, College of Information Science & Technology, University of Nebraska-Omaha, and Carleigh Samson, ITL Program, College of Engineering, University of Colorado Boulder\n2. Provide each group with a length measuring tool and an Android or iOS device with clinometer app installed.\n3. Working in their groups, have students construct right triangles using their stick as the hypotenuse and textbooks or other classroom supplies as the two legs (see Figure 1). Use tape to attach the stick to the book stack so that it does not slip; alternatively, place a heavy object at the base of the stick to keep it from slipping.\n4. When you have a right triangle formed, use the mobile device and its clinometer app to measure the angles of the triangle and explore the relationships between the angle measures and the side lengths. The Android or iPad must be nearly vertical to use the angle measuring function of the clinometer. When the clinometer has 0° located at the top, rotate the Android or iPad device left or right until the top or bottom of the device matches the sloped side of the angle. The other side of the angle should be horizontal. Then read the degree measure of the angle from the clinometer. See Figure 2 for more information.\n5. After some exploration and practice with the clinometer, inform the teams that they will be given a specific triangle to construct. To accomplish this, students must apply engineering design. Provide each group with a stick of fixed length (this can be a yard or meter stick, or any stick-like object found in the classroom), which represents the hypotenuse of a right triangle. Tell students they can measure this object and use the recorded length in their calculations.\n6. Provide each group with a target angle. (NOTE: Assign each group a different acute angle.) The challenge is to create a triangle that has the stick as its hypotenuse and the given angle as one of the acute angles.\n7. To begin, have students create a rough draft drawing and calculations of their triangle designs.\n8. After a firm plan has been developed, students create polished final drafts of their proposals for the dimensions of the triangle. Require the final drafts to include brief explanations of numerical results.\n9. Using their final drafts as a guide, teams build their triangles using a stick and textbooks, blocks, desks or any other teacher-approved classroom objects.\n10. Once a group has its right triangle constructed, have the teacher or another student group verify the result using the clinometer app.\n11. To conclude, have students write up an explanation of the design process they went through to create their final products. Require the explanation to include final drawing and calculations.\n\n### Vocabulary/Definitions\n\nclinometer: An instrument for measuring angles of slope, elevation or depression of an object with respect to gravity.\n\ncosine: The ratio of the length of the leg adjacent to the reference angle divided by the length of the hypotenuse.\n\nsine: The ratio of the length of the leg opposite the reference angle divided by the length of the hypotenuse.\n\ntangent : The ratio of the length of the leg opposite the reference angle divided by the length of the leg adjacent to the reference angle.\n\ntrigonometry: The study of the relationships between angles and sides of right triangles.\n\n### Assessment\n\nPre-Activity Assessment\n\nClass Discussion: Ask the class the following questions to gauge their knowledge of trigonometry and how a clinometer works:\n\n• Describe how a clinometer works. (Answer: It measures the angle of an incline.)\n• What trigonometric ratios do you know? How do they relate to the sides of a right triangle? (Example answers: The sine of an angle is equal to the opposite side over the hypotenuse. The cosine of an angle is equal to the adjacent side over the hypotenuse. The tangent of an angle is equal to the opposite side over the adjacent side. Each ratio is based on the lengths of the sides of the right triangle.)\n\nActivity Embedded Assessment\n\nObservations: As students are engaged in the activity, rove around the room, asking questions and/or observing students working, and ask yourself the following (or similar) questions:\n\n• Are students effectively communicating using trigonometric vocabulary?\n• Are students able to communicate the need for using trigonometry in this situation?\n• Do students see the relationship between side lengths and acute angles of a right triangle?\n• Is every member of each group playing an effective role in the problem solving process?\n\nPost-Activity Assessment\n\nQuiz: Administer the Post-Activity Quiz to assess students' understanding and ability in applying trigonometric ratios to problems based on a real-life scenario—calculating measurements related to building a radio tower.\n\nScott Burns\n\n### Supporting Program\n\nIMPART RET Program, College of Information Science & Technology, University of Nebraska-Omaha\n\n### Acknowledgements\n\nThe contents of this digital library curriculum were developed as a part of the RET in Engineering and Computer Science Site on Infusing Mobile Platform Applied Research into Teaching (IMPART) Program at the University of Nebraska-Omaha under National Science Foundation RET grant number CNS 1201136. However, these contents do not necessarily represent the policies of the NSF and you should not assume endorsement by the federal government." ]

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[ "1 view\n\nI am having a tough time figuring out how to use Kevin Murphy's HMM toolbox Toolbox. It would be a great help if anyone who has experience with it could clarify some conceptual questions. I have somehow understood the theory behind HMM but it's confusing how to actually implement it and mention all the parameter setting.\n\nThere are 2 classes so we need 2 HMMs.\n\nLet say the training vectors are :class1 O1={ 4 3 5 1 2} and class O_2={ 1 4 3 2 4}.\n\nNow, the system has to classify an unknown sequence O3={1 3 2 4 4} as either class1 or class2.\n\n1. What is going to go to obsmat0 and obsmat1?\n\n2. How to specify/syntax for the transition probability transmat0 and transmat1?\n\n3. what is the variable data going to be in this case?\n\n4. Would a number of states Q=5 since there are five unique numbers/symbols used?\n\n5. A number of outputs symbols=5?\n\n6. How do I mention the transition probabilities transmat0 and transmat1?\n\nby (33.1k points)\n\nIn this case, the states of the model are the three possible types of weather: sunny, rainy and foggy. We assume the weather can be only one of these values. Thus the set of HMM states are:\n\nS = {sunny, rainy, foggy}\n\nHere, we can't observe the weather directly. Instead, the only evidence we have is whether the person every day is carrying an umbrella or not. In HMM terminology, these are the discrete observations:\n\nx = {umbrella, no umbrella}\n\nThe HMM model is determined by three things:\n\n• The prior probabilities: vector of probabilities of being in the first state of a sequence.\n\n• The transition prob: matrix describing the probabilities of going from one state of the weather to another.\n\n• The emission prob: matrix describing the probabilities of observing an output (umbrella or not) given a state (weather).\n\nWe may be given with these probabilities, or we have to learn them from a training set. Then, perform reasoning like computing likelihood of an observation sequence with respect to an HMM model\n\nSome steps to follow:\n\n1) known model parameters\n\nCheck out this sample code that shows how to fill existing probabilities to build the model:\n\nQ = 3;    # number of states (sun,rain,fog)\n\nO = 2;    # number of discrete observations (umbrella, no umbrella)\n\n#  prior probabilities\n\nprior = [1 0 0];\n\n# state transition matrix (1: sun, 2: rain, 3:fog)\n\nA = [0.8 0.05 0.15; 0.2 0.6 0.2; 0.2 0.3 0.5];\n\n# observation emission matrix (1: umbrella, 2: no umbrella)\n\nB = [0.1 0.9; 0.8 0.2; 0.3 0.7];\n\nThen we can sample a bunch of sequences from this model:\n\nnum = 20;           # 20 sequences\n\nT = 10;             # each of length 10 (days)\n\n[seqs,states] = dhmm_sample(prior, A, B, num, T);\n\nFor example\n\n>> seqs(5,:)        # observation sequence\n\nans =\n\n2     2 1     2 1 1     1 2 2 2\n\n>> states(5,:)      # hidden states sequence\n\nans =\n\n1     1 1     3 2 2     2 1 1 1\n\nThen, evaluate the log-likelihood of the sequence:\n\ndhmm_logprob(seqs(5,:), prior, A, B)\n\ndhmm_logprob_path(prior, A, B, states(5,:))\n\nCompute the Viterbi path (most probable state sequence):\n\nvPath = viterbi_path(prior, A, multinomial_prob(seqs(5,:),B))", null, "2) unknown model parameters\n\nTraining is performed using the EM algorithm and is best done with a set of observation sequences.\n\nWe can use the generated data above to train a new model and compare it to the original:\n\nprior_hat = normalise(rand(Q,1));\n\nA_hat = mk_stochastic(rand(Q,Q));\n\nB_hat = mk_stochastic(rand(Q,O));\n\n# learn from data by performing many iterations of EM\n\n[LL,prior_hat,A_hat,B_hat] = dhmm_em(seqs, prior_hat,A_hat,B_hat, 'max_iter',50);\n\n# plot learning curve\n\nplot(LL), xlabel('iterations'), ylabel('log likelihood'), grid on", null, "In this example, the trained model looks close to the original one:\n\n>> p = [2 3 1];              # states permutation\n\n>> prior, prior_hat(p)\n\nprior =\n\n1     0 0\n\nans =\n\n0.97401\n\n7.5499e-005\n\n0.02591\n\n>> A, A_hat(p,p)\n\nA =\n\n0.8         0.05 0.15\n\n0.2          0.6 0.2\n\n0.2          0.3 0.5\n\nans =\n\n0.75967      0.05898 0.18135\n\n0.037482      0.77118 0.19134\n\n0.22003      0.53381 0.24616\n\n>> B, B_hat(p,[1 2])\n\nB =\n\n0.1          0.9\n\n0.8          0.2\n\n0.3          0.7\n\nans =\n\n0.11237      0.88763\n\n0.72839      0.27161\n\n0.25889      0.74111\n\nTo know more about Hidden Markov Model, study Artificial Intelligence Course." ]

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[ "# Simplify 604/787 to lowest terms\n\n/\n\n#### Solution for what is 604/787 in simplest fraction\n\n604/787 =\n\nNow we have: what is 604/787 in simplest fraction = 604/787\n\nQuestion: How to reduce 604/787 to its lowest terms?\n\nStep by step simplifying fractions:\n\nStep 1: Find GCD(604,787) = 1.\n\nStep 2: Can't simplify any further\n\nTherefore, 604/787 is simplified fraction for 604/787" ]

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[ "", null, "# Function to generate UUID or similar\n\nHi\n\nIs there any way to via a function or similar generate a unique id string similar to UUID 4 or another way to create a unique id?\n\nKind regards\nMattias\n\nOne simple solution could be this:\n\n``````{{Math.floor(Math.random()*999999999)}}\n``````\n\nDepending on how long you want it you can simply add or remove a few 9.\n\nIf you want to avoid the chance of a collision and alphanumerical you can combine the current date and a random string like this:\n\n``{{(+new Date).toString(36).slice(-5) + Math.random().toString(36).substr(2, 5)}}``\n1 Like\n\nThanks it will work for now, but would it possible to add function for a full UUID128-bit in the future?\n\nShould i add it the feature request?\n\nKind regards\nMattias\n\n1 Like\n\nAs the `uuid` npm-package is now by default a dependency of n8n as it gets used to generate unique paths for the webhook, it is now possible to create uuids in the Function-Node for example like this:\n\n``````const uuid = require('uuid');\nitems.json.uuid = uuid.v4();\nreturn items;\n``````\n\nThat will however only work if the following environment variable got set:\n\n``````NODE_FUNCTION_ALLOW_EXTERNAL=uuid\n``````\n1 Like" ]

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[ "# Calculating Totals and Other Aggregates (Reporting Services)\n\nIn a Tablix data region, you can display aggregate totals for a report dataset, a data region, or a group. You can use the default aggregation supplied by the Add Total command and use the default scope. Alternatively, you can specify a different aggregate function from the built-in functions or specify a different scope.\n\nTo customize an aggregate calculation, you can use group variables in combination with custom code. For more information, see Using Group Variables in Reporting Services 2008 for Custom Aggregation.\n\n## Understanding the Add Totals Command\n\nBy default, the Add Total command uses the Sum function. However, you can change the Sum function to a different built-in function and perform a different aggregation for a scope. You can also write your own expressions to calculate aggregate values for one scope relative to another scope. For example, you might want to calculate the percentage of a group value relative to the complete dataset. For more information about how to use other built-in functions to perform aggregations, see Using Built-in Report and Aggregate Functions in Expressions (Reporting Services). For the list of built-in functions, see Using Built-in Report and Aggregate Functions in Expressions (Reporting Services).\n\n## Understanding Default Scope\n\nAs a Tablix data region is processed and rendered, Report Designer evaluates the value of each Tablix cell in the context of the cell's scope. When a cell is in a row that is associated with a row group or column group, a cell can belong to multiple groups. The set of groups to which a cell belongs, the data region, or the dataset for which an expression is evaluated is called scope. The default scope is defined by the innermost row and column group to which the cell belongs. Visual cues help you identify the groups for which a selected cell is a member. For more information, see Understanding Tablix Data Region Cells, Rows, and Columns.\n\nFor an example of a default scope, the following figure has a parent row group that is based on date, a child row group that is based on order, and a details group.", null, "This figure shows a field expression for Qty and LineTotal in Tablix body cells in multiple rows. The following list describes the default scope for these field expressions in each row:\n\n• Row 1. This row displays column labels that are static text. The default scope is the data region. The row repeats only once and is sometimes called the header row.\n\n• Row 2. This row displays detail data, indicated by the row handle for details (", null, "). The default scope for [Qty] is the details group, which is the data in the dataset bound to this data region after all dataset, data region, and group filters are applied. This row repeats once per row in the dataset query result set. The value [Qty] displays the detail data from the dataset.\n\n• Row 3. This row displays aggregate values. The default scope for [Sum(Qty)] is the child row group Order. The row repeats once per unique order number value. The value [Sum(Qty)] displays the aggregate sum for each order number.\n\n• Row 4. This row displays aggregate values. The default scope for [Sum(Qty)] is the parent row group Date. The row repeats once per unique date value. The value [Sum(Qty)] displays the aggregate sum for each date.\n\n• Row 5. This row displays aggregate values. The default scope for [Sum(Qty)] is the data region. The row repeats only once and is sometimes called the footer row. The value [Sum(Qty)] displays the aggregate sum for the data region.\n\nFor more information about context and scope, see Using Built-in Report and Aggregate Functions in Expressions (Reporting Services). For more information about interpreting visual cues on Tablix data regions, see Understanding Tablix Data Region Cells, Rows, and Columns.\n\n## Specifying Aggregate Calculations with Named Scopes\n\nTo specify a scope other than the default scope, specify a named scope. A named scope includes the name of the dataset, the data region, or the group. Because expressions are context-sensitive, the named scope that you specify must be valid for the build-in function. Each built-in function specifies which scopes are valid as a parameter. For more information, see the specific function reference topic in Using Built-in Report and Aggregate Functions in Expressions (Reporting Services).\n\nFor an example of how to change the default scope to a named scope, suppose a report has the following data structures:\n\n• A report dataset that is named Sales.\n\n• A table data region that is named Tablix1 and that has the following groups:\n\n• A parent row group that is named Category with a child group that is named Subcat.\n\n• A parent column group that is named Geography with a child group that is named CountryRegion.\n\nTo this report, you use the Add Total command to add totals to various cells. This resulting report could resemble the following figure.", null, "In this figure, one of the cells has been selected. For this selected cell, the orange group indicator brackets show the innermost group memberships: the Category group and the CountryRegion group. These innermost group memberships define the default scope for this cell. (For the selected cell, there are also two gray group brackets that show groups: the Subcat row group and the Geography column group. Groups that are marked with a gray group bracket are either outer groups for the selected inner group or are not in the scope for the selected cell.)\n\nThis selected cell has the expression: [Sum(LineTotal)]. Based on the cell's scope and expression, the rendered report will display the value for the text box as the [LineTotal] sum for each specific Subcat value within each specific CountryRegion value.\n\nWithout replacing the Sum function in the expression, you could change the scope for this cell in the following ways:\n\n• Change the scope to be the Sales dataset.   For example, to specify the dataset total, you would change the expression to =Sum(Fields!LineTotal.Value, \"Sales\").\n\n• Change the scope to be the whole Tablix1 data region.   For example, to calculate the percent that a group contributes to the total data region, you would change the expression to =Sum(Fields!LineTotal.Value)/Sum((Fields! LineTotal.Value),\"Tablix1\").\n\nIn this new expression, the numerator, Sum(Fields! LineTotal.Value), is the sum of the line totals for each category. The denominator, Sum((Fields! LineTotal.Value), \"Tablix1\"), is the sum of the line totals for the data region. The result is the percentage that each category represents of the total data region.\n\nIn the selected cell, you can specify one of the following group scopes for the Sum function: CountryRegion, Geography, or Category. You cannot specify the group scope Subcat, because the cell is not in a row associated with the Subcat group.\n\nNote\n\nYou can only specify a single scope for an aggregate function. You cannot specify both \"Category\" and \"Geography\" at the same time as scopes for the denominator expression. Specifying multiple scopes is not supported.\n\nFor the list of built-in functions, see Using Built-in Report and Aggregate Functions in Expressions (Reporting Services).\n\n## Examples of Specifying Different Scopes\n\nThe following table provides examples of expressions that use the default scope, the data region scope, and the dataset scope. The examples assume that there is a Tablix data region named \"Tablix1\", a dataset named \"Sales\" with a field named \"LineTotal\" that contains the total for a line item in a sales order.\n\nDescription\n\nExpression\n\nSum of line item totals for the current scope.\n\n=Sum(Field!LineTotal.Value)\n\nPercent of (the sum of line item totals for the current scope)/(the sum of line totals for data region).\n\nThis value is formatted using the default display format for the text box.\n\n=Sum(Field!LineTotal.Value)/Sum(Field!LineTotal.Value,\"Tablix1\")\n\nWhen placed in a row associated with the group CountryRegion:\n\nPercent of (the sum of line item totals for the CountryRegion group)/(the sum of line totals for the Geography group).\n\nThis value is formatted using the default display format for the text box, which must be set to P1 for the form #.#%.\n\n=Sum(Field!LineTotal.Value)/Sum(Field!LineTotal.Value,\"Geography\")\n\nPercent of (the sum of line item totals for the current scope)/(the sum of line totals for the data region).\n\nThis value is formatted using IIF so that the result is blank if there are no line item totals in the current group.\n\n=IIF(Sum(Field!LineTotal.Value)<=0,\"\", FormatPercent(Sum(Field!LineTotal.Value)/Sum(Field!LineTotal.Value,\"Tablix1\"),1)\n\nPercent of (the sum of line item totals for the current scope)/(the sum of line item totals for the dataset).\n\nThis value is formatted using FormatPercent specifying one decimal place.\n\n=\"Percentage contributing to all sales: \" & FormatPercent(Sum(Field!LineTotal.Value)/Sum(Field!LineTotal.Value,\"Sales\"),1)\n\n## Suppressing Null or Zero Values in a Cell\n\nFor many reports, calculations that are scoped to groups can create many cells that have zero (0) or null values. To reduce clutter in your report, add an expression to return blanks if the aggregate value is 0. For more information, see \"Examples that Suppress Null or Zero Values\" in Expression Examples (Reporting Services)." ]

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[ "# Physics Commons™\n\nSelected Works\n\nJohn Swain\n\nDipole moments\n\nArticles 1 - 2 of 2\n\n## Full-Text Articles in Physics\n\nElectric Dipole Moments And Polarizability In The Quark-Diquark Model Of The Neutron, Y. N. Srivastava, A. Widom, J. Swain, O. Panella Feb 2011\n\n#### Electric Dipole Moments And Polarizability In The Quark-Diquark Model Of The Neutron, Y. N. Srivastava, A. Widom, J. Swain, O. Panella\n\n##### John Swain\n\nFor a bound state internal wave function respecting parity symmetry, it can be rigorously argued that the mean electric dipole moment must be strictly zero. Thus, both the neutron, viewed as a bound state of three quarks, and the water molecule, viewed as a bound state of ten electrons two protons and an oxygen nucleus, both have zero mean electric dipole moments. Yet, the water molecule is said to have a nonzero dipole moment strength $d=e\\Lambda$ with $\\Lambda_{H_2O} \\approx 0.385\\ \\dot{A}$. The neutron may also be said to have an electric dipole moment strength with ...\n\nEffects Of Anomalous Charged Current Dipole Moments Of The Tau On The Decay $\\Tau^-\\To\\Nu_\\Tau\\Pi^-\\Pi^0$, Maria-Teresa Dova, Pablo Lacentre, John Swain, Lucas Taylor Feb 2011\n\n#### Effects Of Anomalous Charged Current Dipole Moments Of The Tau On The Decay $\\Tau^-\\To\\Nu_\\Tau\\Pi^-\\Pi^0$, Maria-Teresa Dova, Pablo Lacentre, John Swain, Lucas Taylor\n\n##### John Swain\n\nWe analyse the process $\\tau^-\\to\\nu_\\tau\\pi^-\\pi^0$ allowing for anomalous weak charged current magnetic and electric dipole moment interactions and determine the effects on the differential and total decay rates. Using recent experimental data we determine the following values for the anomalous magnetic and electric dipole moment parameters, respectively: $\\kappa = 0.16 \\pm 0.08; and |\\tilde\\kappa| = 0.88^{+0.25}_{-0.35}.$" ]

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[ "# How do you write an nth term rule for a_3=10 and a_6=300?\n\nFeb 23, 2018\n\n${a}_{n} = 1.0359 \\cdot {3.107}^{n - 1}$\n\n#### Explanation:\n\nI'm assuming this is a geometric sequence...\n\nThe general formula for an arithmetic sequence is\n\n${a}_{n} = {a}_{1} \\cdot {r}^{n - 1}$ with ${a}_{1}$ as the first term and $r$ as the common ratio.\n\nPlug in:\n\n${a}_{3} = 10 = {a}_{1} \\cdot {r}^{3 - 1}$\n${a}_{6} = 300 = {a}_{1} \\cdot {r}^{6 - 1}$\n\n$10 = {a}_{1} \\cdot {r}^{2}$\n$300 = {a}_{1} \\cdot {r}^{5}$\n\nDivide:\n\n$\\frac{300}{10} = \\frac{{a}_{1} \\cdot {r}^{5}}{{a}_{1} \\cdot {r}^{2}}$\n\n$30 = \\frac{\\cancel{{a}_{1}} \\cdot {r}^{5}}{\\cancel{{a}_{1}} \\cdot {r}^{2}}$\n\n$30 = {r}^{3}$\n\n$\\sqrt{30} = r \\mathmr{and} {30}^{\\frac{1}{3}}$\n\n$r \\approx 3.107$\n\nThen find ${a}_{1}$:\n\n$10 = {a}_{1} \\cdot 9.653$\n\n${a}_{1} \\approx 1.0359$\n\nPlug in the information into the general formula:\n\n${a}_{n} = 1.0359 \\cdot {3.107}^{n - 1}$\n\nNote: This is not exact as I rounded somewhere." ]

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[ "## Reading the Comics, April 6, 2020: My Perennials Edition\n\nAs much as everything is still happening, and so much, there’s still comic strips. I’m fortunately able here to focus just on the comics that discuss some mathematical theme, so let’s get started in exploring last week’s reading. Worth deeper discussion are the comics that turn up here all the time.\n\nLincoln Peirce’s Big Nate for the 5th is a casual mention. Nate wants to get out of having to do his mathematics homework. This really could be any subject as long as it fit the word balloon.\n\nJohn Hambrock’s The Brilliant Mind of Edison Lee for the 6th is a funny-answers-to-story-problems joke. Edison Lee’s answer disregards the actual wording of the question, which supposes the group is travelling at an average 70 miles per hour. The number of stops doesn’t matter in this case.\n\nMark Anderson’s Andertoons for the 6th is the Mark Anderson’s Andertoons for the week. In it Wavehead gives the “just use a calculator” answer for geometry problems.", null, "Mark Anderson’s Andertoons for the 6th of April, 2020. I haven’t mentioned this strip in two days. Essays featuring Andertoons are at this link, though.\n\nNot much to talk about there. But there is a fascinating thing about perimeters that you learn if you go far enough in Calculus. You have to get into multivariable calculus, something where you integrate a function that has at least two independent variables. When you do this, you can find the integral evaluated over a curve. If it’s a closed curve, something that loops around back to itself, then you can do something magic. Integrating the correct function on the curve around a shape will tell you the enclosed area.\n\nAnd this is an example of one of the amazing things in multivariable calculus. It tells us that integrals over a boundary can tell us something about the integral within a volume, and vice-versa. It can be worth figuring out whether your integral is better solved by looking at the boundaries or at the interiors.\n\nHeron’s Formula, for the area of a triangle based on the lengths of its sides, is an expression of this calculation. I don’t know of a formula exactly like that for the perimeter of a quadrilateral, but there are similar formulas if you know the lengths of the sides and of the diagonals.\n\nRichard Thompson’s Cul de Sac rerun for the 6th sees Petey working on his mathematics homework. As with the Big Nate strip, it could be any subject.\n\nZach Weinersmith’s Saturday Morning Breakfast Cereal for the 5th depicts, fairly, the sorts of things that excite mathematicians. The number discussed here is about algorithmic complexity. This is the study of how long it takes to do an algorithm. How long always depends on how big a problem you are working on; to sort four items takes less time than sorting four million items. Of interest here is how much the time to do work grows with the size of whatever you’re working on.", null, "Zach Weinersmith’s Saturday Morning Breakfast Cereal for the 5th of April, 2020. I haven’t mentioned this strip in two days. Essays featuring Saturday Morning Breakfast Cereal are at this link, though.\n\nThe mathematician’s particular example, and I thank dtpimentel in the comments for finding this, is about the Coppersmith–Winograd algorithm. This is a scheme for doing matrix multiplication, a particular kind of multiplication and addition of squares of numbers. The squares have some number N rows and N columns. It’s thought that there exists some way to do matrix multiplication in the order of N2 time, that is, if it takes 10 time units to multiply matrices of three rows and three columns together, we should expect it takes 40 time units to multiply matrices of six rows and six columns together. The matrix multiplication you learn in linear algebra takes on the order of N3 time, so, it would take like 80 time units.\n\nWe don’t know the way to do that. The Coppersmith–Winograd algorithm was thought, after Virginia Vassilevska Williams’s work in 2011, to take something like N2.3728642 steps. So that six-rows-six-columns multiplication would take slightly over 51.796 844 time units. In 2014, François le Gall found it was no worse than N2.3728639 steps, so this would take slightly over 51.796 833 time units. The improvement doesn’t seem like much, but on tiny problems it never does. On big problems, the improvement’s worth it. And, sometimes, you make a good chunk of progress at once.\n\nI’ll have some more comic strips to discuss in an essay at this link, sometime later this week. Thanks for reading.\n\n## Reading the Comics, February 24, 2018: My One Boring Linear Algebra Anecdote Edition\n\nWait for it.\n\nZach Weinersmith’s Saturday Morning Breakfast Cereal for the 21st mentions mathematics — geometry, primarily — as something a substitute teacher has tried teaching with the use of a cucumber and condom. These aren’t terrible examples to use to make concrete the difference between volumes and surface areas. There are limitations, though. It’s possible to construct a shape that has a finite volume but an infinitely large surface area, albeit not using cucumbers.\n\nThere’s also a mention of the spring constant, and physics. This isn’t explicitly mathematical. But the description of movement on a spring are about the first interesting differential equation of mathematical physics. The solution is that of simple harmonic motion. I don’t think anyone taking the subject for the first time would guess at the answer. But it’s easy enough to verify it’s right. And this motion — sine waves — just turns up everywhere in mathematical physics.\n\nBud Blake’s Tiger rerun for the 23rd just mentions mathematics as a topic Hugo finds challenging, and what’s challenging about it. So a personal story: when I took Intro to Linear Algebra my freshman year one day I spaced on the fact we had an exam. So, I put the textbook on the shelf under my desk, and then forgot to take it when I left. The book disappeared, of course, and the professor never heard of it being turned in to lost-and-found or anything. Fortunately the homework was handwritten questions passed out on photocopies (ask your parents), so I could still do the assignments, but for all those, you know, definitions and examples I had to rely on my own notes. I don’t know why I couldn’t ask a classmate. Shyness, probably. Came through all right, though.", null, "Bud Blake’s Tiger rerun for the 23rd of February, 2018. Whose house are they at? I mean, did Tiger bring his dog to Hugo’s, or did Hugo bring his homework to Tiger’s house? I guess either’s not that odd, especially if they just got out of school, but then Hugo’s fussing with his homework when he’s right out of school and with Tiger?\n\nCathy Law’s Claw for the 23rd technically qualifies as an anthropomorphic-numerals joke, in this panel about the smothering of education by the infection of guns into American culture.\n\nJim Meddick’s Monty for the 23rd has wealthy child Wedgwick unsatisfied with a mere ball of snow. He instead has a snow Truncated Icosahedron (the hyphens in Jarvis’s word balloon may baffle the innocent reader). This is a real shape, one that’s been known for a very long time. It’s one of the Archimedean Solids, a set of 13 solids that have convex shapes (no holes or indents or anything) and have all vertices the same, the identical number of edges coming in to each point in the same relative directions. The truncated icosahedron you maybe also know as the soccer ball shape, at least for those old-style soccer balls made of patches that were hexagons and pentagons. An actual truncated icosahedron needs twelve pentagons, so the figure drawn in the third panel isn’t quite right. At least one pentagonal face would be visible. But that’s also tricky to draw. The aerodynamics of a truncated icosahedron are surely different from those of a sphere. But in snowball-fight conditions, probably not different enough to even notice.\n\nMark Litzler’s Joe Vanilla for the 24th uses a blackboard full of formulas to represent an overcomplicated answer. The formulas look, offhand, like gibberish to me. But I’ll admit uncertainty since the odd capitalization of “iG(p)” at the start makes me think of some deeper group theory or knot theory symbols. And to see an “m + p” and an “m – p” makes me think of quantum mechanics of atomic orbitals. (But then an “m – p2” is weird.) So if this were anything I’d say it was some quantum chemistry formula. But my gut says if Litzler did take the blackboard symbols from anything, it was without going back to references. (Which he has no need to do, I should point out; the joke wouldn’t be any stronger — or weaker — if the blackboard meant anything.)\n\n## The Summer 2017 Mathematics A To Z: Jordan Canonical Form\n\nI made a mistake! I thought we had got to the end of the block of A To Z topics suggested by Gaurish, of the For The Love Of Mathematics blog. Not so and, indeed, I wonder if it wouldn’t be a viable writing strategy around here for me to just ask Gaurish to throw out topics and I have two weeks to write about them. I don’t think there’s a single unpromising one in the set.", null, "Art courtesy of Thomas K Dye, creator of the web comic Newshounds. He has a Patreon for those able to support his work. He’s also open for commissions, starting from US\\$10.\n\n# Jordan Canonical Form.\n\nBefore you ask, yes, this is named for the Camille Jordan.\n\nSo this is a thing from algebra. Particularly, linear algebra. And more particularly, matrices. Matrices are so much of linear algebra that you could be forgiven thinking they’re all of linear algebra. The thing is, matrices are a really good way of describing linear transformations. That is, where you take a block of space and stretch it out, or squash it down, or rotate it, or do some combination of these things. And stretching and squashing and rotating is a lot of what you’d ever want to do. Refer to any book on how to draw animated cartoons. The only thing matrices can’t do is have their eyes bug out huge when an attractive region of space walks past.\n\nThing about a matrix is if you want to do something with it, you’re going to write it as a grid of numbers. It doesn’t have to be a grid of numbers. But about all the matrices anyone does anything with are grids of numbers. And that’s fine. They do an incredible lot of stuff. What’s not fine is that on looking at a huge block of numbers, the mind sees: huh. That’s a big block of numbers. Good luck finding what’s meaningful in them. To help find meaning we have a set of standard forms. We call them “canonical” or “normal” or some other approving term. They rearrange and change the terms in the matrix so that more interesting stuff is more obvious.\n\nNow you’re justified asking: how can we rearrange and change the terms in a matrix without changing what the matrix is? We can get away with doing this because we can show some rearrangements don’t change what we’re interested in. That covers the “how dare we” part of “how”. We do it by using matrix multiplication. You might remember from high school algebra that matrix multiplication is this agonizing process of multiplying every pair of numbers that ever existed together, then adding them all up, and then maybe you multiply something by minus one because you’re thinking of determinants, and it all comes out wrong anyway and you have to do it over? Yeah. Well, matrix multiplication is defined hard because it makes stuff like this work out. So that covers the “by what technique” part of “how”. We start out with some matrix, let me imaginatively name it", null, "$A$. And then we find some transformation matrix for which, eh, let’s say", null, "$P$ is a good enough name. I’ll say why in a moment. Then we use that matrix and its multiplicative inverse", null, "$P^{-1}$. And we evaluate the product", null, "$P^{-1} A P$. This won’t just be the same old matrix we started with. Not usually. Promise. But what this will be, if we chose our matrix", null, "$P$ correctly, is some new matrix that’s easier to read.\n\nThe matrices involved here have to follow some rules. Most important, they’re all going to be square matrices. There’ll be more rules that your linear algebra textbook will tell you. Or your instructor will, after checking the textbook.\n\nSo what makes a matrix easy to read? Zeroes. Lots and lots of zeroes. When we have a standardized form of a matrix it’s nearly all zeroes. This is for a good reason: zeroes are easy to multiply stuff by. And they’re easy to add stuff to. And almost everything we do with matrices, as a calculation, is a lot of multiplication and addition of the numbers in the matrix.\n\nWhat also makes a matrix easy to read? Everything important being on the diagonal. The diagonal is one of the two things you would imagine if you were told “here’s a grid of numbers, pick out the diagonal”. In particular it’s the one that goes from the upper left to the bottom right, that is, row one column one, and row two column two, and row three column three, and so on up to row 86 column 86 (or whatever). If everything is on the diagonal the matrix is incredibly easy to work with. If it can’t all be on the diagonal at least everything should be close to it. As close as possible.\n\nIn the Jordan Canonical Form not everything is on the diagonal. I mean, it can be, but you shouldn’t count on that. But everything either will be on the diagonal or else it’ll be one row up from the diagonal. That is, row one column two, row two column three, row 85 column 86. Like that. There’s two other important pieces.\n\nFirst is the thing in the row above the diagonal will be either 1 or 0. Second is that on the diagonal you’ll have a sequence of all the same number. Like, you’ll get four instances of the number ‘2’ along this string of the diagonal. Third is that you’ll get a 1 above all but the row above first instance of this particular number. Fourth is that you’ll get a 0 in the row above the first instance of this number.\n\nYeah, that’s fussy to visualize. This is one of those things easiest to show in a picture. A Jordan canonical form is a matrix that looks like this:\n\n 2 1 0 0 0 0 0 0 0 0 0 0 0 2 1 0 0 0 0 0 0 0 0 0 0 0 2 1 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 3 1 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 0 0 4 1 0 0 0 0 0 0 0 0 0 0 0 4 1 0 0 0 0 0 0 0 0 0 0 0 4 0 0 0 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 -2 1 0 0 0 0 0 0 0 0 0 0 0 -2\n\nThis may have you dazzled. It dazzles mathematicians too. When we have to write a matrix that’s almost all zeroes like this we drop nearly all the zeroes. If we have to write anything we just write a really huge 0 in the upper-right and the lower-left corners.\n\nWhat makes this the Jordan Canonical Form is that the matrix looks like it’s put together from what we call Jordan Blocks. Look around the diagonals. Here’s the first Jordan Block:\n\n 2 1 0 0 0 2 1 0 0 0 2 1 0 0 0 2\n\nHere’s the second:\n\n 3 1 0 3\n\nHere’s the third:\n\n 4 1 0 0 4 1 0 0 4\n\nHere’s the fourth:\n\n -1\n\nAnd here’s the fifth:\n\n -2 1 0 -2\n\nAnd we can represent the whole matrix as this might-as-well-be-diagonal thing:\n\n First Block 0 0 0 0 0 Second Block 0 0 0 0 0 Third Block 0 0 0 0 0 Fourth Block 0 0 0 0 0 Fifth Block\n\nThese blocks can be as small as a single number. They can be as big as however many rows and columns you like. Each individual block is some repeated number on the diagonal, and a repeated one in the row above the diagonal. You can call this the “superdiagonal”.\n\n(Mathworld, and Wikipedia, assert that sometimes the row below the diagonal — the “subdiagonal” — gets the 1’s instead of the superdiagonal. That’s fine if you like it that way, and it won’t change any of the real work. I have not seen these subdiagonal 1’s in the wild. But I admit I don’t do a lot of this field and maybe there’s times it’s more convenient.)\n\nUsing the Jordan Canonical Form for a matrix is a lot like putting an object in a standard reference pose for photographing. This is a good metaphor. We get a Jordan Canonical Form by matrix multiplication, which works like rotating and scaling volumes of space. You can view the Jordan Canonical Form for a matrix as how you represent the original matrix from a new viewing angle that makes it easy to recognize. And this is why", null, "$P$ is not a bad name for the matrix that does this work. We can see all this as “projecting” the matrix we started with into a new frame of reference. The new frame is maybe rotated and stretched and squashed and whatnot, compared to how we started. But it’s as valid a base. Projecting a mathematical object from one frame of reference to another usually involves calculating something that looks like", null, "$P^{-1} A P$ so, projection. That’s our name.\n\nMathematicians will speak of “the” Jordan Canonical Form for a matrix as if there were such a thing. I don’t mean that Jordan Canonical Forms don’t exist. They exist just as much as matrices do. It’s the “the” that misleads. You can put the Jordan Blocks in any order and have as valid, and as useful, a Jordan Canonical Form. But it’s easy to swap the orders of these blocks around — it’s another matrix multiplication, and a blessedly easy one — so it doesn’t matter which form you have. Get any one and you have them all.\n\nI haven’t said anything about what these numbers on the diagonal are. They’re the eigenvalues of the original matrix. I hope that clears things up.\n\nYeah, not to anyone who didn’t know what a Jordan Canonical Form was to start with. Rather than get into calculations let me go to well-established metaphor. Take a sample of an unknown chemical and set it on fire. Put the light from this through a prism and photograph the spectrum. There will be lines, interruptions in the progress of colors. The locations of those lines and how intense they are tell you what the chemical is made of, and in what proportions. These are much like the eigenvectors and eigenvalues of a matrix. The eigenvectors tell you what the matrix is made of, and the eigenvalues how much of the matrix is those. This stuff gets you very far in proving a lot of great stuff. And part of what makes the Jordan Canonical Form great is that you get the eigenvalues right there in neat order, right where anyone can see them.\n\nSo! All that’s left is finding the things. The best way to find the Jordan Canonical Form for a given matrix is to become an instructor for a class on linear algebra and assign it as homework. The second-best way is to give the problem to your TA, who will type it in to Mathematica and return the result. It’s too much work to do most of the time. Almost all the stuff you could learn from having the thing in the Jordan Canonical Form you work out in the process of finding the matrix", null, "$P$ that would let you calculate what the Jordan Canonical Form is. And once you had that, why go on?\n\nWhere the Jordan Canonical Form shines is in doing proofs about what matrices can do. We can always put a square matrix into a Jordan Canonical Form. So if we want to show something is true about matrices in general, we can show that it’s true for the simpler-to-work-with Jordan Canonical Form. Then show that shifting a matrix to or from the Jordan Canonical Form doesn’t change whether the thing we’re interested in is true. It exists in that strange space: it is quite useful, but never on a specific problem.\n\nOh, all right. Yes, it’s the same Camille Jordan of the Jordan Curve and also of the Jordan Curve Theorem. That fellow.\n\n## Reading the Comics, April 15, 2017: Extended Week Edition\n\nIt turns out last Saturday only had the one comic strip that was even remotely on point for me. And it wasn’t very on point either, but since it’s one of the Creators.com strips I’ve got the strip to show. That’s enough for me.\n\nHenry Scarpelli and Craig Boldman’s Archie for the 8th is just about how algebra hurts. Some days I agree.", null, "Henry Scarpelli and Craig Boldman’s Archie for the 8th of April, 2017. Do you suppose Archie knew that Dilton was listening there, or was he just emoting his fatigue to himself?\n\nRuben Bolling’s Super-Fun-Pak Comix for the 8th is an installation of They Came From The Third Dimension. “Dimension” is one of those oft-used words that’s come loose of any technical definition. We use it in mathematics all the time, at least once we get into Introduction to Linear Algebra. That’s the course that talks about how blocks of space can be stretched and squashed and twisted into each other. You’d expect this to be a warmup act to geometry, and I guess it’s relevant. But where it really pays off is in studying differential equations and how systems of stuff changes over time. When you get introduced to dimensions in linear algebra they describe degrees of freedom, or how much information you need about a problem to pin down exactly one solution.\n\nIt does give mathematicians cause to talk about “dimensions of space”, though, and these are intuitively at least like the two- and three-dimensional spaces that, you know, stuff moves in. That there could be more dimensions of space, ordinarily inaccessible, is an old enough idea we don’t really notice it. Perhaps it’s hidden somewhere too.\n\nAmanda El-Dweek’s Amanda the Great of the 9th started a story with the adult Becky needing to take a mathematics qualification exam. It seems to be prerequisite to enrolling in some new classes. It’s a typical set of mathematics anxiety jokes in the service of a story comic. One might tsk Becky for going through university without ever having a proper mathematics class, but then, I got through university without ever taking a philosophy class that really challenged me. Not that I didn’t take the classes seriously, but that I took stuff like Intro to Logic that I was already conversant in. We all cut corners. It’s a shame not to use chances like that, but there’s always so much to do.\n\nMark Anderson’s Andertoons for the 10th relieves the worry that Mark Anderson’s Andertoons might not have got in an appearance this week. It’s your common kid at the chalkboard sort of problem, this one a kid with no idea where to put the decimal. As always happens I’m sympathetic. The rules about where to move decimals in this kind of multiplication come out really weird if the last digit, or worse, digits in the product are zeroes.\n\nMel Henze’s Gentle Creatures is in reruns. The strip from the 10th is part of a story I’m so sure I’ve featured here before that I’m not even going to look up when it aired. But it uses your standard story problem to stand in for science-fiction gadget mathematics calculation.\n\nDave Blazek’s Loose Parts for the 12th is the natural extension of sleep numbers. Yes, I’m relieved to see Dave Blazek’s Loose Parts around here again too. Feels weird when it’s not.\n\nBill Watterson’s Calvin and Hobbes rerun for the 13th is a resisting-the-story-problem joke. But Calvin resists so very well.\n\nJohn Deering’s Strange Brew for the 13th is a “math club” joke featuring horses. Oh, it’s a big silly one, but who doesn’t like those too?\n\nDan Thompson’s Brevity for the 14th is one of the small set of punning jokes you can make using mathematician names. Good for the wall of a mathematics teacher’s classroom.\n\nShaenon K Garrity and Jefferey C Wells’s Skin Horse for the 14th is set inside a virtual reality game. (This is why there’s talk about duplicating objects.) Within the game, the characters are playing that game where you start with a set number (in this case 20) tokens and take turn removing a couple of them. The “rigged” part of it is that the house can, by perfect play, force a win every time. It’s a bit of game theory that creeps into recreational mathematics books and that I imagine is imprinted in the minds of people who grow up to design games.\n\n## The End 2016 Mathematics A To Z: Kernel\n\nI told you that Image thing would reappear. Meanwhile I learned something about myself in writing this.\n\n## Kernel.\n\nI want to talk about functions again. I’ve been keeping like a proper mathematician to a nice general idea of what a function is. The sort where a function’s this rule matching stuff in a set called the domain with stuff in a set called the range. And I’ve tried not to commit myself to saying anything about what that domain and range are. They could be numbers. They could be other functions. They could be the set of DVDs you own but haven’t watched in more than two years. They could be collections socks. Haven’t said.\n\nBut we know what functions anyone cares about. They’re stuff that have domains and ranges that are numbers. Preferably real numbers. Complex-valued numbers if we must. If we look at more exotic sets they’re ones that stick close to being numbers: vectors made up of an ordered set of numbers. Matrices of numbers. Functions that are themselves about numbers. Maybe we’ll get to something exotic like a rotation, but then what is a rotation but spinning something a certain number of degrees? There are a bunch of unavoidably common domains and ranges.\n\nFine, then. I’ll stick to functions with ranges that look enough like regular old numbers. By “enough” I mean they have a zero. That is, something that works like zero does. You know, add it to something else and that something else isn’t changed. That’s all I need.\n\nA natural thing to wonder about a function — hold on. “Natural” is the wrong word. Something we learn to wonder about in functions, in pre-algebra class where they’re all polynomials, is where the zeroes are. They’re generally not at zero. Why would we say “zeroes” to mean “zero”? That could let non-mathematicians think they knew what we were on about. By the “zeroes” we mean the things in the domain that get matched to the zero in the range. It might be zero; no reason it couldn’t, until we know what the function’s rule is. Just we can’t count on that.\n\nA polynomial we know has … well, it might have zero zeroes. Might have no zeroes. It might have one, or two, or so on. If it’s an n-th degree polynomial it can have up to n zeroes. And if it’s not a polynomial? Well, then it could have any conceivable number of zeroes and nobody is going to give you a nice little formula to say where they all are. It’s not that we’re being mean. It’s just that there isn’t a nice little formula that works for all possibilities. There aren’t even nice little formulas that work for all polynomials. You have to find zeroes by thinking about the problem. Sorry.\n\nBut! Suppose you have a collection of all the zeroes for your function. That’s all the points in the domain that match with zero in the range. Then we have a new name for the thing you have. And that’s the kernel of your function. It’s the biggest subset in the domain with an image that’s just the zero in the range.\n\nSo we have a name for the zeroes that isn’t just “the zeroes”. What does this get us?\n\nIf we don’t know anything about the kind of function we have, not much. If the function belongs to some common kinds of functions, though, it tells us stuff.\n\nFor example. Suppose the function has domain and range that are vectors. And that the function is linear, which is to say, easy to deal with. Let me call the function ‘f’. And let me pick out two things in the domain. I’ll call them ‘x’ and ‘y’ because I’m writing this after Thanksgiving dinner and can’t work up a cleverer name for anything. If f is linear then f(x + y) is the same thing as f(x) + f(y). And now something magic happens. If x and y are both in the kernel, then x + y has to be in the kernel too. Think about it. Meanwhile, if x is in the kernel but y isn’t, then f(x + y) is f(y). Again think about it.\n\nWhat we can see is that the domain fractures into two directions. One of them, the direction of the kernel, is invisible to the function. You can move however much you like in that direction and f can’t see it. The other direction, perpendicular (“orthogonal”, we say in the trade) to the kernel, is visible. Everything that might change changes in that direction.\n\nThis idea threads through vector spaces, and we study a lot of things that turn out to look like vector spaces. It keeps surprising us by letting us solve problems, or find the best-possible approximate solutions. This kernel gives us room to match some fiddly conditions without breaking the real solution. The size of the null space alone can tell us whether some problems are solvable, or whether they’ll have infinitely large sets of solutions.\n\nIn this vector-space construct the kernel often takes on another name, the “null space”. This means the same thing. But it reminds us that superhero comics writers miss out on many excellent pieces of terminology by not taking advanced courses in mathematics.\n\nKernels also appear in group theory, whenever we get into rings. We’re always working with rings. They’re nearly as unavoidable as vector spaces.\n\nYou know how you can divide the whole numbers into odd and even? And you can do some neat tricks with that for some problems? You can do that with every ring, using the kernel as a dividing point. This gives us information about how the ring is shaped, and what other structures might look like the ring. This often lets us turn proofs that might be hard into a collection of proofs on individual cases that are, at least, doable. Tricks about odd and even numbers become, in trained hands, subtle proofs of surprising results.\n\nWe see vector spaces and rings all over the place in mathematics. Some of that’s selection bias. Vector spaces capture a lot of what’s important about geometry. Rings capture a lot of what’s important about arithmetic. We have understandings of geometry and arithmetic that transcend even our species. Raccoons understand space. Crows understand number. When we look to do mathematics we look for patterns we understand, and these are major patterns we understand. And there are kernels that matter to each of them.\n\nSome mathematical ideas inspire metaphors to me. Kernels are one. Kernels feel to me like the process of holding a polarized lens up to a crystal. This lets one see how the crystal is put together. I realize writing this down that my metaphor is unclear: is the kernel the lens or the structure seen in the crystal? I suppose the function has to be the lens, with the kernel the crystallization planes made clear under it. It’s curious I had enjoyed this feeling about kernels and functions for so long without making it precise. Feelings about mathematical structures can be like that.\n\n## The End 2016 Mathematics A To Z: Algebra\n\nSo let me start the End 2016 Mathematics A To Z with a word everybody figures they know. As will happen, everybody’s right and everybody’s wrong about that.\n\n## Algebra.\n\nEverybody knows what algebra is. It’s the point where suddenly mathematics involves spelling. Instead of long division we’re on a never-ending search for ‘x’. Years later we pass along gifs of either someone saying “stop asking us to find your ex” or someone who’s circled the letter ‘x’ and written “there it is”. And make jokes about how we got through life without using algebra. And we know it’s the thing mathematicians are always doing.\n\nMathematicians aren’t always doing that. I expect the average mathematician would say she almost never does that. That’s a bit of a fib. We have a lot of work where we do stuff that would be recognizable as high school algebra. It’s just we don’t really care about that. We’re doing that because it’s how we get the problem we are interested in done. the most recent few pieces in my “Why Stuff can Orbit” series include a bunch of high school algebra-style work. But that was just because it was the easiest way to answer some calculus-inspired questions.\n\nStill, “algebra” is a much-used word. It comes back around the second or third year of a mathematics major’s career. It comes in two forms in undergraduate life. One form is “linear algebra”, which is a great subject. That field’s about how stuff moves. You get to imagine space as this stretchy material. You can stretch it out. You can squash it down. You can stretch it in some directions and squash it in others. You can rotate it. These are simple things to build on. You can spend a whole career building on that. It becomes practical in surprising ways. For example, it’s the field of study behind finding equations that best match some complicated, messy real data.\n\nThe second form is “abstract algebra”, which comes in about the same time. This one is alien and baffling for a long while. It doesn’t help that the books all call it Introduction to Algebra or just Algebra and all your friends think you’re slumming. The mathematics major stumbles through confusing definitions and theorems that ought to sound comforting. (“Fermat’s Little Theorem”? That’s a good thing, right?) But the confusion passes, in time. There’s a beautiful subject here, one of my favorites. I’ve talked about it a lot.\n\nWe start with something that looks like the loosest cartoon of arithmetic. We get a bunch of things we can add together, and an ‘addition’ operation. This lets us do a lot of stuff that looks like addition modulo numbers. Then we go on to stuff that looks like picking up floor tiles and rotating them. Add in something that we call ‘multiplication’ and we get rings. This is a bit more like normal arithmetic. Add in some other stuff and we get ‘fields’ and other structures. We can keep falling back on arithmetic and on rotating tiles to build our intuition about what we’re doing. This trains mathematicians to look for particular patterns in new, abstract constructs.\n\nLinear algebra is not an abstract-algebra sort of algebra. Sorry about that.\n\nAnd there’s another kind of algebra that mathematicians talk about. At least once they get into grad school they do. There’s a huge family of these kinds of algebras. The family trait for them is that they share a particular rule about how you can multiply their elements together. I won’t get into that here. There are many kinds of these algebras. One that I keep trying to study on my own and crash hard against is Lie Algebra. That’s named for the Norwegian mathematician Sophus Lie. Pronounce it “lee”, as in “leaning”. You can understand quantum mechanics much better if you’re comfortable with Lie Algebras and so now you know one of my weaknesses. Another kind is the Clifford Algebra. This lets us create something called a “hypercomplex number”. It isn’t much like a complex number. Sorry. Clifford Algebra does lend to a construct called spinors. These help physicists understand the behavior of bosons and fermions. Every bit of matter seems to be either a boson or a fermion. So you see why this is something people might like to understand.\n\nBoolean Algebra is the algebra of this type that a normal person is likely to have heard of. It’s about what we can build using two values and a few operations. Those values by tradition we call True and False, or 1 and 0. The operations we call things like ‘and’ and ‘or’ and ‘not’. It doesn’t sound like much. It gives us computational logic. Isn’t that amazing stuff?\n\nSo if someone says “algebra” she might mean any of these. A normal person in a non-academic context probably means high school algebra. A mathematician speaking without further context probably means abstract algebra. If you hear something about “matrices” it’s more likely that she’s speaking of linear algebra. But abstract algebra can’t be ruled out yet. If you hear a word like “eigenvector” or “eigenvalue” or anything else starting “eigen” (or “characteristic”) she’s more probably speaking of abstract algebra. And if there’s someone’s name before the word “algebra” then she’s probably speaking of the last of these. This is not a perfect guide. But it is the sort of context mathematicians expect other mathematicians notice.\n\n## A Leap Day 2016 Mathematics A To Z: Basis\n\nToday’s glossary term is one that turns up in many areas of mathematics. But these all share some connotations. So I mean to start with the easiest one to understand.\n\n## Basis.\n\nSuppose you are somewhere. Most of us are. Where is something else?\n\nThat isn’t hard to answer if conditions are right. If we’re allowed to point and the something else is in sight, we’re done. It’s when pointing and following the line of sight breaks down that we’re in trouble. We’re also in trouble if we want to say how to get from that something to yet another spot. How can we guide someone from one point to another?\n\nWe have a good answer from everyday life. We can impose some order, some direction, on space. We’re familiar with this from the cardinal directions. We say where things on the surface of the Earth are by how far they are north or south, east or west, from something else. The scheme breaks down a bit if we’re at the North or the South pole exactly, but there we can fall back on pointing.\n\nWhen we start using north and south and east and west as directions we are choosing basis vectors. Vectors are directions in how far to move and in what direction. Suppose we have two vectors that aren’t pointing in the same direction. Then we can describe any two-dimensional movement using them. We can say “go this far in the direction of the first vector and also that far in the direction of the second vector”. With the cardinal directions, we consider north and east, or east and south, or south and west, or west and north to be a pair of vectors going in different directions.\n\n(North and south, in this context, are the same thing. “Go twenty paces north” says the same thing as “go negative twenty paces south”. Most mathematicians don’t pull this sort of stunt when telling you how to get somewhere unless they’re trying to be funny without succeeding.)\n\nA basis vector is just a direction, and distance in that direction, that we’ve decided to be a reference for telling different points in space apart. A basis set, or basis, is the collection of all the basis vectors we need. What do we need? We need enough basis vectors to get to all the points in whatever space we’re working with.\n\n(If you are going to ask about doesn’t “east” point in different directions as we go around the surface of the Earth, you’re doing very well. Please pretend we never move so far from where we start that anyone could notice the difference. If you can’t do that, please pretend the Earth has been smooshed into a huge flat square with north at one end and we’re only just now noticing.)\n\nWe are free to choose whatever basis vectors we like. The worst that can happen if we choose a lousy basis is that we have to write out more things than we otherwise would. Our work won’t be less true, it’ll just be more tedious. But there are some properties that often make for a good basis.\n\nOne is that the basis should relate to the problem you’re doing. Suppose you were in one of mathematicians’ favorite places, midtown Manhattan. There is a compelling grid here of streets running north-south and avenues running east-west. (Broadway we ignore as an implementation error retained for reasons of backwards compatibility.) Well, we pretend they run north-south and east-west. They’re actually a good bit clockwise of north-south and east-west. They do that to better match the geography of the island. A “north” street runs about parallel to the way Manhattan’s long dimension runs. In the circumstance, it would be daft to describe directions by true north or true east. We would say to go so many streets “north” and so many avenues “east”.\n\nPurely mathematical problems aren’t concerned with streets and avenues. But there will often be preferred directions. Mathematicians often look at the way a process alters shapes or redirects forces. There’ll be some directions where the alterations are biggest. There’ll be some where the alterations are shortest. Those directions are probably good choices for a basis. They stand out as important.\n\nWe also tend to like basis vectors that are a unit length. That is, their size is 1 in some convenient unit. That’s for the same reason it’s easier to say how expensive something is if it costs 45 dollars instead of nine five-dollar bills. Or if you’re told it was 180 quarter-dollars. The length of your basis vector is just a scaling factor. But the more factors you have to work with the more likely you are to misunderstand something.\n\nAnd we tend to like basis vectors that are perpendicular to one another. They don’t have to be. But if they are then it’s easier to divide up our work. We can study each direction separately. Mathematicians tend to like techniques that let us divide problems up into smaller ones that we can study separately.\n\nI’ve described basis sets using vectors. They have intuitive appeal. It’s easy to understand directions of things in space. But the idea carries across into other things. For example, we can build functions out of other functions. So we can choose a set of basis functions. We can multiply them by real numbers (scalars) and add them together. This makes whatever function we’re interested in into a kind of weighted average of basis functions.\n\nWhy do that? Well, again, we often study processes that change shapes and directions. If we choose a basis well, though, the process changes the basis vectors in easy to describe ways. And many interesting processes let us describe the changing of an arbitrary function as the weighted sum of the changes in the basis vectors. By solving a couple of simple problems we get the ability to solve every interesting problem.\n\nWe can even define something that works like the angle between functions. And something that works a lot like perpendicularity for functions.\n\nAnd this carries on to other mathematical constructs. We look for ways to impose some order, some direction, on whatever structure we’re looking at. We’re often successful, and can work with unreal things using tools like those that let us find our place in a city.\n\n## The Set Tour, Part 7: Matrices\n\nI feel a bit odd about this week’s guest in the Set Tour. I’ve been mostly concentrating on sets that get used as the domains or ranges for functions a lot. The ones I want to talk about here don’t tend to serve the role of domain or range. But they are used a great deal in some interesting functions. So I loosen my rule about what to talk about.\n\n## Rm x n and Cm x n\n\nRm x n might explain itself by this point. If it doesn’t, then this may help: the “x” here is the multiplication symbol. “m” and “n” are positive whole numbers. They might be the same number; they might be different. So, are we done here?\n\nMaybe not quite. I was fibbing a little when I said “x” was the multiplication symbol. R2 x 3 is not a longer way of saying R6, an ordered collection of six real-valued numbers. The x does represent a kind of product, though. What we mean by R2 x 3 is an ordered collection, two rows by three columns, of real-valued numbers. Say the “x” here aloud as “by” and you’re pronouncing it correctly.\n\nWhat we get is called a “matrix”. If we put into it only real-valued numbers, it’s a “real matrix”, or a “matrix of reals”. Sometimes mathematical terminology isn’t so hard to follow. Just as with vectors, Rn, it matters just how the numbers are organized. R2 x 3 means something completely different from what R3 x 2 means. And swapping which positions the numbers in the matrix occupy changes what matrix we have, as you might expect.\n\nYou can add together matrices, exactly as you can add together vectors. The same rules even apply. You can only add together two matrices of the same size. They have to have the same number of rows and the same number of columns. You add them by adding together the numbers in the corresponding slots. It’s exactly what you would do if you went in without preconceptions.\n\nYou can also multiply a matrix by a single number. We called this scalar multiplication back when we were working with vectors. With matrices, we call this scalar multiplication. If it strikes you that we could see vectors as a kind of matrix, yes, we can. Sometimes that’s wise. We can see a vector as a matrix in the set R1 x n or as one in the set Rn x 1, depending on just what we mean to do.\n\nIt’s trickier to multiply two matrices together. As with vectors multiplying the numbers in corresponding positions together doesn’t give us anything. What we do instead is a time-consuming but not actually hard process. But according to its rules, something in Rm x n we can multiply by something in Rn x k. “k” is another whole number. The second thing has to have exactly as many rows as the first thing has columns. What we get is a matrix in Rm x k.\n\nI grant you maybe didn’t see that coming. Also a potential complication: if you can multiply something in Rm x n by something in Rn x k, can you multiply the thing in Rn x k by the thing in Rm x n? … No, not unless k and m are the same number. Even if they are, you can’t count on getting the same product. Matrices are weird things this way. They’re also gateways to weirder things. But it is a productive weirdness, and I’ll explain why in a few paragraphs.\n\nA matrix is a way of organizing terms. Those terms can be anything. Real matrices are surely the most common kind of matrix, at least in mathematical usage. Next in common use would be complex-valued matrices, much like how we get complex-valued vectors. These are written Cm x n. A complex-valued matrix is different from a real-valued matrix. The terms inside the matrix can be complex-valued numbers, instead of real-valued numbers. Again, sometimes, these mathematical terms aren’t so tricky.\n\nI’ve heard occasionally of people organizing matrices of other sets. The notation is similar. If you’re building a matrix of “m” rows and “n” columns out of the things you find inside a set we’ll call H, then you write that as Hm x n. I’m not saying you should do this, just that if you need to, that’s how to tell people what you’re doing.\n\nNow. We don’t really have a lot of functions that use matrices as domains, and I can think of fewer that use matrices as ranges. There are a couple of valuable ones, ones so valuable they get special names like “eigenvalue” and “eigenvector”. (Don’t worry about what those are.) They take in Rm x n or Cm x n and return a set of real- or complex-valued numbers, or real- or complex-valued vectors. Not even those, actually. Eigenvectors and eigenfunctions are only meaningful if there are exactly as many rows as columns. That is, for Rm x m and Cm x m. These are known as “square” matrices, just as you might guess if you were shaken awake and ordered to say what you guessed a “square matrix” might be.\n\nThey’re important functions. There are some other important functions, with names like “rank” and “condition number” and the like. But they’re not many. I believe they’re not even thought of as functions, any more than we think of “the length of a vector” as primarily a function. They’re just properties of these matrices, that’s all.\n\nSo why are they worth knowing? Besides the joy that comes of knowing something, I mean?\n\nHere’s one answer, and the one that I find most compelling. There is cultural bias in this: I come from an applications-heavy mathematical heritage. We like differential equations, which study how stuff changes in time and in space. It’s very easy to go from differential equations to ordered sets of equations. The first equation may describe how the position of particle 1 changes in time. It might describe how the velocity of the fluid moving past point 1 changes in time. It might describe how the temperature measured by sensor 1 changes as it moves. It doesn’t matter. We get a set of these equations together and we have a majestic set of differential equations.\n\nNow, the dirty little secret of differential equations: we can’t solve them. Most interesting physical phenomena are nonlinear. Linear stuff is easy. Small change 1 has effect A; small change 2 has effect B. If we make small change 1 and small change 2 together, this has effect A plus B. Nonlinear stuff, though … it just doesn’t work. Small change 1 has effect A; small change 2 has effect B. Small change 1 and small change 2 together has effect … A plus B plus some weird A times B thing plus some effect C that nobody saw coming and then C does something with A and B and now maybe we’d best hide.\n\nThere are some nonlinear differential equations we can solve. Those are the result of heroic work and brilliant insights. Compared to all the things we would like to solve there’s not many of them. Methods to solve nonlinear differential equations are as precious as ways to slay krakens.\n\nBut here’s what we can do. What we usually like to know about in systems are equilibriums. Those are the conditions in which the system stops changing. Those are interesting. We can usually find those points by boring but not conceptually challenging calculations. If we can’t, we can declare x0 represents the equilibrium. If we still care, we leave calculating its actual values to the interested reader or hungry grad student.\n\nBut what’s really interesting is: what happens if we’re near but not exactly at the equilibrium? Sometimes, we stay near it. Think of pushing a swing. However good a push you give, it’s going to settle back to the boring old equilibrium of dangling straight down. Sometimes, we go racing away from it. Think of trying to balance a pencil on its tip; if we did this perfectly it would stay balanced. It never does. We’re never perfect, or there’s some wind or somebody walks by and the perfect balance is foiled. It falls down and doesn’t bounce back up. Sometimes, whether it it stays near or goes away depends on what way it’s away from the equilibrium.\n\nAnd now we finally get back to matrices. Suppose we are starting out near an equilibrium. We can, usually, approximate the differential equations that describe what will happen. The approximation may only be good if we’re just a tiny bit away from the equilibrium, but that might be all we really want to know. That approximation will be some linear differential equations. (If they’re not, then we’re just wasting our time.) And that system of linear differential equations we can describe using matrices.\n\nIf we can write what we are interested in as a set of linear differential equations, then we have won. We can use the many powerful tools of matrix arithmetic — linear algebra, specifically — to tell us everything we want to know about the system. We can say whether a small push away from the equilibrium stays small, or whether it grows, or whether it depends. We can say how fast the small push shrinks, or grows (for a while). We can say how the system will change, approximately.\n\nThis is what I love in matrices. It’s not everything there is to them. But it’s enough to make matrices important to me.\n\n## Combining Matrices And Model Universes\n\nI would like to resume talking about matrices and really old universes and the way nucleosynthesis in these model universes causes atoms to keep settling down to peculiar but unchanging distribution.\n\nI’d already described how a matrix offers a nice way to organize elements, and in ways that encode information about the context of the elements by where they’re placed. That’s useful and saves some writing, certainly, although by itself it’s not that interesting. Matrices start to get really powerful when, first, the elements being stored are things on which you can do something like arithmetic with pairs of them. Here I mostly just mean that you can add together two elements, or multiply them, and get back something meaningful.\n\nThis typically means that the matrix is made up of a grid of numbers, although that isn’t actually required, just, really common if we’re trying to do mathematics.\n\nThen you get the ability to add together and multiply together the matrices themselves, turning pairs of matrices into some new matrix, and building something that works a lot like arithmetic on these matrices.\n\nAdding one matrix to another is done in almost the obvious way: add the element in the first row, first column of the first matrix to the element in the first row, first column of the second matrix; that’s the first row, first column of your new matrix. Then add the element in the first row, second column of the first matrix to the element in the first row, second column of the second matrix; that’s the first row, second column of the new matrix. Add the element in the second row, first column of the first matrix to the element in the second row, first column of the second matrix, and put that in the second row, first column of the new matrix. And so on.\n\nThis means you can only add together two matrices that are the same size — the same number of rows and of columns — but that doesn’t seem unreasonable.\n\nYou can also do something called scalar multiplication of a matrix, in which you multiply every element in the matrix by the same number. A scalar is just a number that isn’t part of a matrix. This multiplication is useful, not least because it lets us talk about how to subtract one matrix from another: to find the difference of the first matrix and the second, scalar-multiply the second matrix by -1, and then add the first to that product. But you can do scalar multiplication by any number, by two or minus pi or by zero if you feel like it.\n\nI should say something about notation. When we want to write out these kinds of operations efficiently, of course, we turn to symbols to represent the matrices. We can, in principle, use any symbols, but by convention a matrix usually gets represented with a capital letter, A or B or M or P or the like. So to add matrix A to matrix B, with the result being matrix C, we can write out the equation “A + B = C”, which is about as simple as we could hope to see. Scalars are normally written in lowercase letters, often Greek letters, if we don’t know what the number is, so that the scalar multiplication of the number r and the matrix A would be the product “rA”, and we could write the difference between matrix A and matrix B as “A + (-1)B” or “A – B”.\n\nMatrix multiplication, now, that is done by a process that sounds like doubletalk, and it takes a while of practice to do it right. But there are good reasons for doing it that way and we’ll get to one of those reasons by the end of this essay.\n\nTo multiply matrix A and matrix B together, we do multiply various pairs of elements from both matrix A and matrix B. The surprising thing is that we also add together sets of these products, per this rule.\n\nTake the element in the first row, first column of A, and multiply it by the element in the first row, first column of B. Add to that the product of the element in the first row, second column of A and the second row, first column of B. Add to that total the product of the element in the first row, third column of A and the third row, second column of B, and so on. When you’ve run out of columns of A and rows of B, this total is the first row, first column of the product of the matrices A and B.\n\nPlenty of work. But we have more to do. Take the product of the element in the first row, first column of A and the element in the first row, second column of B. Add to that the product of the element in the first row, second column of A and the element in the second row, second column of B. Add to that the product of the element in the first row, third column of A and the element in the third row, second column of B. And keep adding those up until you’re out of columns of A and rows of B. This total is the first row, second column of the product of matrices A and B.\n\nThis does mean that you can multiply matrices of different sizes, provided the first one has as many columns as the second has rows. And the product may be a completely different size from the first or second matrices. It also means it might be possible to multiply matrices in one order but not the other: if matrix A has four rows and three columns, and matrix B has three rows and two columns, then you can multiply A by B, but not B by A.\n\nMy recollection on learning this process was that this was crazy, and the workload ridiculous, and I imagine people who get this in Algebra II, and don’t go on to using mathematics later on, remember the process as nothing more than an unpleasant blur of doing a lot of multiplying and addition for some reason or other.\n\nSo here is one of the reasons why we do it this way. Let me define two matrices:", null, "$A = \\left(\\begin{tabular}{c c c} 3/4 & 0 & 2/5 \\\\ 1/4 & 3/5 & 2/5 \\\\ 0 & 2/5 & 1/5 \\end{tabular}\\right)$", null, "$B = \\left(\\begin{tabular}{c} 100 \\\\ 0 \\\\ 0 \\end{tabular}\\right)$\n\nThen matrix A times B is", null, "$AB = \\left(\\begin{tabular}{c} 3/4 * 100 + 0 * 0 + 2/5 * 0 \\\\ 1/4 * 100 + 3/5 * 0 + 2/5 * 0 \\\\ 0 * 100 + 2/5 * 0 + 1/5 * 0 \\end{tabular}\\right) = \\left(\\begin{tabular}{c} 75 \\\\ 25 \\\\ 0 \\end{tabular}\\right)$\n\nYou’ve seen those numbers before, of course: the matrix A contains the probabilities I put in my first model universe to describe the chances that over the course of a billion years a hydrogen atom would stay hydrogen, or become iron, or become uranium, and so on. The matrix B contains the original distribution of atoms in the toy universe, 100 percent hydrogen and nothing anything else. And the product of A and B was exactly the distribution after that first billion years: 75 percent hydrogen, 25 percent iron, nothing uranium.\n\nIf we multiply the matrix A by that product again — well, you should expect we’re going to get the distribution of elements after two billion years, that is, 56.25 percent hydrogen, 33.75 percent iron, 10 percent uranium, but let me write it out anyway to show:", null, "$\\left(\\begin{tabular}{c c c} 3/4 & 0 & 2/5 \\\\ 1/4 & 3/5 & 2/5 \\\\ 0 & 2/5 & 1/5 \\end{tabular}\\right)\\left(\\begin{tabular}{c} 75 \\\\ 25 \\\\ 0 \\end{tabular}\\right) = \\left(\\begin{tabular}{c} 3/4 * 75 + 0 * 25 + 2/5 * 0 \\\\ 1/4 * 75 + 3/5 * 25 + 2/5 * 0 \\\\ 0 * 75 + 2/5 * 25 + 1/5 * 0 \\end{tabular}\\right) = \\left(\\begin{tabular}{c} 56.25 \\\\ 33.75 \\\\ 10 \\end{tabular}\\right)$\n\nAnd if you don’t know just what would happen if we multipled A by that product, you aren’t paying attention.\n\nThis also gives a reason why matrix multiplication is defined this way. The operation captures neatly the operation of making a new thing — in the toy universe case, hydrogen or iron or uranium — out of some combination of fractions of an old thing — again, the former distribution of hydrogen and iron and uranium.\n\nOr here’s another reason. Since this matrix A has three rows and three columns, you can multiply it by itself and get a matrix of three rows and three columns out of it. That matrix — which we can write as A2 — then describes how two billion years of nucleosynthesis would change the distribution of elements in the toy universe. A times A times A would give three billion years of nucleosynthesis; A10 ten billion years. The actual calculating of the numbers in these matrices may be tedious, but it describes a complicated operation very efficiently, which we always want to do.\n\nI should mention another bit of notation. We usually use capital letters to represent matrices; but, a matrix that’s just got one column is also called a vector. That’s often written with a lowercase letter, with a little arrow above the letter, as in", null, "$\\vec{x}$, or in bold typeface, as in x. (The arrows are easier to put in writing, the bold easier when you were typing on typewriters.) But if you’re doing a lot of writing this out, and know that (say) x isn’t being used for anything but vectors, then even that arrow or boldface will be forgotten. Then we’d write the product of matrix A and vector x as just Ax.  (There are also cases where you put a little caret over the letter; that’s to denote that it’s a vector that’s one unit of length long.)\n\nWhen you start writing vectors without an arrow or boldface you start to run the risk of confusing what symbols mean scalars and what ones mean vectors. That’s one of the reasons that Greek letters are popular for scalars. It’s also common to put scalars to the left and vectors to the right. So if one saw “rMx”, it would be expected that r is a scalar, M a matrix, and x a vector, and if they’re not then this should be explained in text nearby, preferably before the equations. (And of course if it’s work you’re doing, you should know going in what you mean the letters to represent.)" ]

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[ "# 高中数学/必修五/第二章\n\n## 2.1 数列的概念与简单表示法\n\n$a_{1},a_{2},a_{3},\\cdots ,a_{n},\\cdots$", null, "$f(1),f(2),f(3),\\cdots ,f(n)\\cdots$", null, "## 2.2 等差数列\n\n$a_{2}-a_{1}=d,a_{3}-a_{2}=d,a_{4}-a_{3}=d,\\cdots$", null, "$a_{2}=a_{1}+d,$", null, "$a_{3}=a_{2}+d=\\left(a_{1}+d\\right)+d=a_{1}+2d,$", null, "$a_{4}=a_{3}+d=\\left(a_{1}+2d\\right)+d=a_{1}+3d$", null, "…………\n\n$a_{n}=a_{1}+\\left(n-1\\right)d$", null, "## 2.3 等差数列的前n项和\n\n200多年前,高斯的数学老师提出了下面的问题：$1+2+3+\\cdots +=100=?$", null, "$\\left(1+100\\right)+\\left(2+99\\right)+\\cdots +\\left(50+51\\right)=101\\times 50=5050$", null, "$S_{n}=a_{1}+\\left(a_{1}+d\\right)+\\left(a_{1}+2d\\right)+\\cdots +\\left[a_{1}+\\left(n-1\\right)d\\right]$", null, "......①\n\n$S_{n}=a_{n}+\\left(a_{n}-d\\right)+\\left(a_{n}-2d\\right)+\\cdots +\\left[a_{n}-\\left(n-1\\right)d\\right]$", null, "......②\n\n$S_{n}={\\frac {n(a_{1}+a_{n})}{2}}$", null, "$S_{n}=na_{1}+{\\frac {n(n-1)}{2}}d$", null, "## 2.4 等比数列\n\n1.细胞分裂的个数可以组成下面的数列：$1,2,4,8,\\cdots$", null, "2.一轮计算机病毒可以查找计算机中的地址簿，通过邮件进行传播。如果把病毒制造者发送病毒称为第一轮，邮件接收者发送病毒成为第二轮，以此类推。假设每一轮每一台计算机都感染20台计算机，那么在不重复的情况下，这种病毒每一轮感染的计算机数构成的数列是：$1,20,20^{2},20^{3},\\cdots$", null, "$a_{n}=a_{1}q^{n-1}$", null, "## 2.5 等比数列的前n项和\n\n$S_{n}=a_{1}+a_{2}+a_{3}+\\cdots +a_{n}$", null, "$S_{n}=a_{1}+a_{1}q+a_{1}q^{n}-1+a_{1}q^{n}$", null, "…………①\n\n$qS_{n}=a_{1}q+a_{1}q^{2}+\\cdots +a_{1}q^{n-1}+a_{1}q_{n}$", null, "…………②\n\n①、②的右边有很多相同的项，用①分别减去②的两边，就可以消去这些相同的项，得$\\left(1-q\\right)S_{n}=a_{1}-a_{1}q^{n}$", null, "$q\\neq 1$", null, "时，等比数列前n项和的公式为\n\n$S_{n}={\\frac {a_{1}(1-q^{n})}{1-q}}(q\\neq 1)$", null, "$S_{n}={\\frac {a_{1}-a_{n}q}{1-q}}$", null, "$S_{64}={\\frac {1\\times \\left(1-2^{64}\\right)}{1-2}}=2^{64}-1$", null, "" ]

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[ "## < ⎙ 19 Other formulas that you can solve using the same Inputs\n\nRadius Of Inscribed Circle=sqrt((Semiperimeter Of Triangle -Side A)*(Semiperimeter Of Triangle -Side B)*(Semiperimeter Of Triangle -Side C)/Semiperimeter Of Triangle ) GO\nArea of Triangle when semiperimeter is given\nArea Of Triangle=sqrt(Semiperimeter Of Triangle *(Semiperimeter Of Triangle -Side A)*(Semiperimeter Of Triangle -Side B)*(Semiperimeter Of Triangle -Side C)) GO\nArea of a Triangle when sides and perimeter are given\nArea=sqrt(Perimeter*(Perimeter-Side A)*(Perimeter-Side B)*(Perimeter-Side C)) GO\nRadius Of Circumscribed Circle=(Side A*Side B*Side C)/(4*Area Of Triangle) GO\nCentroid of a Trapezoid\nCentroid Of Trapezoid=((Side B+2*Side A)/(3*(Side A+Side B)))*Height GO\nSide a of a triangle\nSide A=sqrt((Side B)^2+(Side C)^2-2*Side B*Side C*cos(Angle A)) GO\nside b of a triangle\nSide B=sqrt(Side A^2+Side C^2-2*Side A*Side C*cos(Angle B)) GO\nside c of a triangle\nSide C=sqrt(Side B^2+Side A^2-2*Side A*Side B*cos(Angle C)) GO\nDiagonal of a Parallelogram (Diagonal 1)\nDiagonal 1=sqrt(2*Side A^2+2*Side B^2-Diagonal 2^2) GO\nDiagonal of a Parallelogram (Diagonal 2)\nDiagonal 2=sqrt(2*Side A^2+2*Side B^2-Diagonal 1^2) GO\nSide of a Kite when other side and area are given\nSide A=(Area*cosec(Angle Between Sides))/Side B GO\nPerimeter of a Right Angled Triangle\nPerimeter=Side A+Side B+sqrt(Side A^2+Side B^2) GO\nArea of a Kite when sides and angle between the sides are given\nArea=Side A*Side B*sin(Angle Between Sides) GO\nPerimeter of Triangle\nPerimeter Of Triangle=Side A+Side B+Side C GO\nPerimeter Of Parallelepiped\nPerimeter=4*Side A+4*Side B+4*Side C GO\nSide of a Kite when other side and perimeter are given\nSide A=(Perimeter/2)-Side B GO\nPerimeter of a Parallelogram\nPerimeter=2*Side A+2*Side B GO\nPerimeter of a Kite\nPerimeter=2*(Side A+Side B) GO\nArea of a Square when side is given\nArea=(Side A)^2 GO\n\n## < ⎙ 18 Other formulas that calculate the same Output\n\nPerimeter of a rectangle when diagonal and length are given\nPerimeter=2*(Length+sqrt((Diagonal)^2-(Length)^2)) GO\nPerimeter of rectangle when diagonal and width are given\nPerimeter=2*(sqrt((Diagonal)^2-(Width)^2)+Width) GO\nPerimeter of a Semicircle when circumference of circle is given\nPerimeter=(Circumference of Circle/2)+Diameter GO\nPerimeter of a Right Angled Triangle\nPerimeter=Side A+Side B+sqrt(Side A^2+Side B^2) GO\nPerimeter Of Parallelepiped\nPerimeter=4*Side A+4*Side B+4*Side C GO\nPerimeter of a quarter circle when diameter is given\nPerimeter=Diameter *((pi/4)+1) GO\nPerimeter of a Semicircle when diameter is given\nPerimeter=Diameter *((pi/2)+1) GO\nPerimeter of a square when diagonal is given\nPerimeter=4*(Diagonal/sqrt(2)) GO\nPerimeter of a quarter circle when radius is given\nPerimeter of a Parallelogram\nPerimeter=2*Side A+2*Side B GO\nPerimeter of a Kite\nPerimeter=2*(Side A+Side B) GO\nPerimeter of a rectangle when length and width are given\nPerimeter=2*Length+2*Width GO\nPerimeter of a Semicircle when radius is given\nPerimeter of a square when area is given\nPerimeter=4*sqrt(Area) GO\nPerimeter of a Cube\nPerimeter=12*Side GO\nPerimeter of a square when side is given\nPerimeter=4*Side GO\nPerimeter of an Equilateral Triangle\nPerimeter=3*Side GO\nPerimeter of a Rhombus\nPerimeter=4*Side GO\n\n### Perimeter of an Isosceles Triangle Formula\n\nPerimeter=Side A+2*Side B\nMore formulas\nPerimeter of Triangle GO\nPerimeter of a rectangle when length and width are given GO\nPerimeter of a square when side is given GO\nPerimeter of an Equilateral Triangle GO\nPerimeter of a Right Angled Triangle GO\nPerimeter of a rectangle when diagonal and length are given GO\nPerimeter of rectangle when diagonal and width are given GO\nPerimeter of a square when diagonal is given GO\nPerimeter of a square when area is given GO\nPerimeter of a Semicircle when radius is given GO\nPerimeter of a Semicircle when diameter is given GO\nPerimeter of a Semicircle when circumference of circle is given GO\nPerimeter of a quarter circle when radius is given GO\nPerimeter of a quarter circle when diameter is given GO\n\n## What is Perimeter of an Isosceles Triangle?\n\nAn isosceles triangle is said to have two equal sides and two equal internal angles. The perimeter of an isosceles can be found if the base and sides are known. It's equal to the sum of twice of its side and base.\n\n## How to Calculate Perimeter of an Isosceles Triangle?\n\nPerimeter of an Isosceles Triangle calculator uses Perimeter=Side A+2*Side B to calculate the Perimeter, The perimeter of a figure is the total distance around the edge of the figure. Perimeter and is denoted by P symbol.\n\nHow to calculate Perimeter of an Isosceles Triangle using this online calculator? To use this online calculator for Perimeter of an Isosceles Triangle, enter Side A (a) and Side B (b) and hit the calculate button. Here is how the Perimeter of an Isosceles Triangle calculation can be explained with given input values -> 8 = 2+2*3.\n\n### FAQ\n\nWhat is Perimeter of an Isosceles Triangle?\nThe perimeter of a figure is the total distance around the edge of the figure and is represented as P=a+2*b or Perimeter=Side A+2*Side B. Side A is an upright or sloping surface of a structure or object that is not the top or bottom and generally not the front or back and Side B is an upright or sloping surface of a structure or object that is not the top or bottom and generally not the front or back.\nHow to calculate Perimeter of an Isosceles Triangle?\nThe perimeter of a figure is the total distance around the edge of the figure is calculated using Perimeter=Side A+2*Side B. To calculate Perimeter of an Isosceles Triangle, you need Side A (a) and Side B (b). With our tool, you need to enter the respective value for Side A and Side B and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.", null, "Let Others Know" ]

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[ "• 防诈骗中心\n• 客服中心 |\n• 网站导航 |\n• 设为主页 |\n• 加入收藏\n• 您当前位置： 首页>产品库>家居百货>家用电器>洗衣机脱水机\n相关分类：\n• 湖南\n• 长沙市\n• 常德市\n• 郴州市\n• 衡阳市\n• 怀化市\n• 娄底市\n• 邵阳市\n• 湘潭市\n• 湘西土家族苗族自治州\n• 益阳市\n• 永州市\n• 岳阳市\n• 张家界市\n• 株洲市\n• 山西\n• 长治市\n• 大同市\n• 晋城市\n• 晋中市\n• 临汾市\n• 吕梁市\n• 朔州市\n• 太原市\n• 忻州市\n• 阳泉市\n• 运城市\n• 安徽\n• 安庆市\n• 蚌埠市\n• 亳州市\n• 巢湖市\n• 池州市\n• 滁州市\n• 阜阳市\n• 合肥市\n• 淮北市\n• 淮南市\n• 黄山市\n• 六安市\n• 马鞍山市\n• 宿州市\n• 铜陵市\n• 芜湖市\n• 宣城市\n• 广西\n• 百色市\n• 北海市\n• 崇左市\n• 防城港市\n• 贵港市\n• 桂林市\n• 河池市\n• 贺州市\n• 来宾市\n• 柳州市\n• 南宁市\n• 钦州市\n• 梧州市\n• 玉林市\n• 河南\n• 安阳市\n• 鹤壁市\n• 焦作市\n• 开封市\n• 洛阳市\n• 漯河市\n• 南阳市\n• 平顶山市\n• 濮阳市\n• 三门峡市\n• 商丘市\n• 新乡市\n• 信阳市\n• 许昌市\n• 郑州市\n• 周口市\n• 驻马店市\n• 吉林\n• 白城市\n• 白山市\n• 长春市\n• 吉林市\n• 辽源市\n• 四平市\n• 松原市\n• 通化市\n• 延边朝鲜族自治州\n• 广东\n• 潮州市\n• 东莞市\n• 佛山市\n• 广州市\n• 河源市\n• 惠州市\n• 江门市\n• 揭阳市\n• 茂名市\n• 梅州市\n• 清远市\n• 汕头市\n• 汕尾市\n• 韶关市\n• 深圳市\n• 阳江市\n• 云浮市\n• 湛江市\n• 肇庆市\n• 中山市\n• 珠海市\n• 辽宁\n• 鞍山市\n• 本溪市\n• 朝阳市\n• 大连市\n• 丹东市\n• 抚顺市\n• 阜新市\n• 葫芦岛市\n• 锦州市\n• 辽阳市\n• 盘锦市\n• 沈阳市\n• 铁岭市\n• 营口市\n• 湖北\n• 鄂州市\n• 恩施土家族苗族自治州\n• 黄冈市\n• 黄石市\n• 荆门市\n• 荆州市\n• 直辖行政单位\n• 十堰市\n• 随州市\n• 武汉市\n• 咸宁市\n• 襄阳市\n• 孝感市\n• 宜昌市\n• 江西\n• 抚州市\n• 赣州市\n• 吉安市\n• 景德镇市\n• 九江市\n• 南昌市\n• 萍乡市\n• 上饶市\n• 新余市\n• 宜春市\n• 鹰潭市\n• 浙江\n• 杭州市\n• 湖州市\n• 嘉兴市\n• 金华市\n• 丽水市\n• 宁波市\n• 衢州市\n• 绍兴市\n• 台州市\n• 温州市\n• 舟山市\n• 青海\n• 果洛藏族自治州\n• 海北藏族自治州\n• 海东地区\n• 海南藏族自治州\n• 海西蒙古族藏族自治州\n• 黄南藏族自治州\n• 西宁市\n• 玉树藏族自治州\n• 甘肃\n• 白银市\n• 定西市\n• 甘南藏族自治州\n• 嘉峪关市\n• 金昌市\n• 酒泉市\n• 兰州市\n• 临夏回族自治州\n• 陇南市\n• 平凉市\n• 庆阳市\n• 天水市\n• 武威市\n• 张掖市\n• 贵州\n• 安顺市\n• 毕节市\n• 贵阳市\n• 六盘水市\n• 黔东南苗族侗族自治州\n• 黔南布依族苗族自治州\n• 黔西南布依族苗族自治州\n• 铜仁地区\n• 遵义市\n• 陕西\n• 安康市\n• 宝鸡市\n• 汉中市\n• 商洛市\n• 铜川市\n• 渭南市\n• 西安市\n• 咸阳市\n• 延安市\n• 榆林市\n• 西藏\n• 阿里地区\n• 昌都地区\n• 拉萨市\n• 林芝地区\n• 那曲地区\n• 日喀则地区\n• 山南地区\n• 宁夏\n• 固原市\n• 石嘴山市\n• 吴忠市\n• 银川市\n• 中卫市\n• 福建\n• 福州市\n• 龙岩市\n• 南平市\n• 宁德市\n• 莆田市\n• yabo国际市\n• 三明市\n• 厦门市\n• 漳州市\n• 内蒙古\n• 阿拉善盟\n• 巴彦淖尔市\n• 包头市\n• 赤峰市\n• 鄂尔多斯市\n• 呼和浩特市\n• 呼伦贝尔市\n• 通辽市\n• 乌海市\n• 乌兰察布市\n• 锡林郭勒盟\n• 兴安盟\n• 云南\n• 保山市\n• 楚雄彝族自治州\n• 大理白族自治州\n• 德宏傣族景颇族自治州\n• 迪庆藏族自治州\n• 红河哈尼族彝族自治州\n• 昆明市\n• 丽江市\n• 临沧市\n• 怒江傈僳族自治州\n• 曲靖市\n• 思茅市\n• 文山壮族苗族自治州\n• 西双版纳傣族自治州\n• 玉溪市\n• 昭通市\n• 新疆\n• 阿克苏地区\n• 阿勒泰地区\n• 巴音郭楞蒙古自治州\n• 博尔塔拉蒙古自治州\n• 昌吉回族自治州\n• 哈密地区\n• 和田地区\n• 喀什地区\n• 克拉玛依市\n• 克孜勒苏柯尔克孜自治州\n• 直辖行政单位\n• 塔城地区\n• 吐鲁番地区\n• 乌鲁木齐市\n• 伊犁哈萨克自治州\n• 黑龙江\n• 大庆市\n• 大兴安岭地区\n• 哈尔滨市\n• 鹤岗市\n• 黑河市\n• 鸡西市\n• 佳木斯市\n• 牡丹江市\n• 七台河市\n• 齐齐哈尔市\n• 双鸭山市\n• 绥化市\n• 伊春市\n• 香港\n• 香港\n• 九龙\n• 新界\n• 澳门\n• 澳门\n• 其它地区\n• 台湾\n• 台中市\n• 台南市\n• 高雄市\n• 台北市\n• 基隆市\n• 嘉义市\n•", null, "张掖波轮洗衣机-推荐兰州合格的洗衣机\n\n品牌:毓祥,,\n\n出厂地:城中区\n\n报价：面议\n\n甘肃毓祥海尔专卖店\n\n黄金会员：", null, "主营：甘肃海尔家电,甘肃海尔冰箱,甘肃海尔空调,甘肃海尔洗衣机,甘肃海尔电视\n\n•", null, "报价：面议\n\n经营模式：未登记\n\n主营：\n\n•", null, "报价：面议\n\n经营模式：未登记\n\n主营：\n\n•", null, "四川洗衣机批发_买优良的康佳洗衣机优选颂隆贸易有限公司\n\n品牌:康佳,美的,新科\n\n出厂地:金城江区\n\n报价：面议\n\n四川颂隆贸易有限公司\n\n黄金会员：", null, "主营：电视,冰箱,空调,洗衣机,热水器\n\n•", null, "口碑好的康佳洗衣机|价格合理的康佳洗衣机供销\n\n品牌:康佳,美的,新科\n\n出厂地:金城江区\n\n报价：面议\n\n四川颂隆贸易有限公司\n\n黄金会员：", null, "主营：电视,冰箱,空调,洗衣机,热水器\n\n•", null, "玉门干洗机哪里买_兰州性价比高的干洗机推荐\n\n品牌:荣原,兰州荣原,兰州荣原洗涤亚博体育官网下载苹果部\n\n出厂地:城中区\n\n报价：面议\n\n兰州荣原洗涤亚博体育官网下载苹果部\n\n黄金会员：", null, "主营：甘肃干洗机,甘肃水洗机,兰州干洗机,兰州水洗机,甘肃干洗店加盟\n\n•", null, "甘肃干洗机那里买-供应兰州性价比高的干洗机\n\n品牌:荣原,兰州荣原,兰州荣原洗涤亚博体育官网下载苹果部\n\n出厂地:城中区\n\n报价：面议\n\n兰州荣原洗涤亚博体育官网下载苹果部\n\n黄金会员：", null, "主营：甘肃干洗机,甘肃水洗机,兰州干洗机,兰州水洗机,甘肃干洗店加盟\n\n•", null, "白银滚筒洗衣机-兰州超值的海尔洗衣机供销\n\n品牌:海尔,毓祥,\n\n出厂地:城中区\n\n报价：面议\n\n兰州毓祥海尔专卖店\n\n黄金会员：", null, "主营：兰州海尔家电,兰州海尔冰箱,兰州海尔洗衣机,兰州海尔空调,兰州海尔热水器\n\n•", null, "报价：面议\n\n经营模式：未登记\n\n主营：\n\n•", null, "干衣机原理-在哪能买到物超所值的亚都干衣机\n\n品牌:亚都,,\n\n出厂地:合山市\n\n报价：面议\n\n广东亚都环保科技有限公司\n\n黄金会员：", null, "主营：工业除湿机,湿膜加湿器,家用除湿机,空气循环机,干衣机\n\n• 没有找到合适的供应商？您可以发布采购信息\n\n没有找到满足要求的供应商？您可以搜索 洗衣机脱水机批发 洗衣机脱水机公司 洗衣机脱水机厂\n\n### 最新入驻厂家\n\n相关产品:\n张掖波轮洗衣机 兰州干洗机一台多少钱 西宁水洗机代理 四川洗衣机批发 口碑好的康佳洗衣机 玉门干洗机哪里买 甘肃干洗机那里买 白银滚筒洗衣机 成都水洗机加盟 干衣机原理" ]

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[ "Cody\n\n# Problem 43298. Calculate area of sector\n\nSolution 1780222\n\nSubmitted on 10 Apr 2019 by Perera Umesha\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\nr=1 seta=pi/2 y_correct = 0.7854; assert(abs(sectorarea(r,seta)-y_correct)<0.001)\n\nr = 1 seta = 1.5708\n\n2   Pass\nr=2 seta=pi/2 y_correct = pi; assert(abs(sectorarea(r,seta)-y_correct)<0.001)\n\nr = 2 seta = 1.5708" ]

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[ "# Essential Mathematical Methods I\n\nCourse Description:\n\nThis course emphasizes on computational techniques and geometrical visualization in higher mathematics. Topics include inequalities, elementary functions, elementary computational methods in calculus of one variable; complex plane and its geometry, complex functions; solutions of ordinary differential equations, power series method, Laplace transform.\n\nCourse Code:\nMATH2083\nUnits:\n3\nProgramme:" ]

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[ "# Division table for N = 11004 / 0÷1\n\n11004 / 0 = inf [+]\n11004 / 0.01 = 1100400 [+]\n11004 / 0.02 = 550200 [+]\n11004 / 0.03 = 366800 [+]\n11004 / 0.04 = 275100 [+]\n11004 / 0.05 = 220080 [+]\n11004 / 0.06 = 183400 [+]\n11004 / 0.07 = 157200 [+]\n11004 / 0.08 = 137550 [+]\n11004 / 0.09 = 122266.6667 [+]\n11004 / 0.1 = 110040 [+]\n11004 / 0.11 = 100036.3636 [+]\n11004 / 0.12 = 91700 [+]\n11004 / 0.13 = 84646.1538 [+]\n11004 / 0.14 = 78600 [+]\n11004 / 0.15 = 73360 [+]\n11004 / 0.16 = 68775 [+]\n11004 / 0.17 = 64729.4118 [+]\n11004 / 0.18 = 61133.3333 [+]\n11004 / 0.19 = 57915.7895 [+]\n11004 / 0.2 = 55020 [+]\n11004 / 0.21 = 52400 [+]\n11004 / 0.22 = 50018.1818 [+]\n11004 / 0.23 = 47843.4783 [+]\n11004 / 0.24 = 45850 [+]\n11004 / 0.25 = 44016 [+]\n11004 / 0.26 = 42323.0769 [+]\n11004 / 0.27 = 40755.5556 [+]\n11004 / 0.28 = 39300 [+]\n11004 / 0.29 = 37944.8276 [+]\n11004 / 0.3 = 36680 [+]\n11004 / 0.31 = 35496.7742 [+]\n11004 / 0.32 = 34387.5 [+]\n11004 / 0.33 = 33345.4545 [+]\n11004 / 0.34 = 32364.7059 [+]\n11004 / 0.35 = 31440 [+]\n11004 / 0.36 = 30566.6667 [+]\n11004 / 0.37 = 29740.5405 [+]\n11004 / 0.38 = 28957.8947 [+]\n11004 / 0.39 = 28215.3846 [+]\n11004 / 0.4 = 27510 [+]\n11004 / 0.41 = 26839.0244 [+]\n11004 / 0.42 = 26200 [+]\n11004 / 0.43 = 25590.6977 [+]\n11004 / 0.44 = 25009.0909 [+]\n11004 / 0.45 = 24453.3333 [+]\n11004 / 0.46 = 23921.7391 [+]\n11004 / 0.47 = 23412.766 [+]\n11004 / 0.48 = 22925 [+]\n11004 / 0.49 = 22457.1429 [+]\n11004 / 0.5 = 22008 [+]\n11004 / 0.51 = 21576.4706 [+]\n11004 / 0.52 = 21161.5385 [+]\n11004 / 0.53 = 20762.2642 [+]\n11004 / 0.54 = 20377.7778 [+]\n11004 / 0.55 = 20007.2727 [+]\n11004 / 0.56 = 19650 [+]\n11004 / 0.57 = 19305.2632 [+]\n11004 / 0.58 = 18972.4138 [+]\n11004 / 0.59 = 18650.8475 [+]\n11004 / 0.6 = 18340 [+]\n11004 / 0.61 = 18039.3443 [+]\n11004 / 0.62 = 17748.3871 [+]\n11004 / 0.63 = 17466.6667 [+]\n11004 / 0.64 = 17193.75 [+]\n11004 / 0.65 = 16929.2308 [+]\n11004 / 0.66 = 16672.7273 [+]\n11004 / 0.67 = 16423.8806 [+]\n11004 / 0.68 = 16182.3529 [+]\n11004 / 0.69 = 15947.8261 [+]\n11004 / 0.7 = 15720 [+]\n11004 / 0.71 = 15498.5915 [+]\n11004 / 0.72 = 15283.3333 [+]\n11004 / 0.73 = 15073.9726 [+]\n11004 / 0.74 = 14870.2703 [+]\n11004 / 0.75 = 14672 [+]\n11004 / 0.76 = 14478.9474 [+]\n11004 / 0.77 = 14290.9091 [+]\n11004 / 0.78 = 14107.6923 [+]\n11004 / 0.79 = 13929.1139 [+]\n11004 / 0.8 = 13755 [+]\n11004 / 0.81 = 13585.1852 [+]\n11004 / 0.82 = 13419.5122 [+]\n11004 / 0.83 = 13257.8313 [+]\n11004 / 0.84 = 13100 [+]\n11004 / 0.85 = 12945.8824 [+]\n11004 / 0.86 = 12795.3488 [+]\n11004 / 0.87 = 12648.2759 [+]\n11004 / 0.88 = 12504.5455 [+]\n11004 / 0.89 = 12364.0449 [+]\n11004 / 0.9 = 12226.6667 [+]\n11004 / 0.91 = 12092.3077 [+]\n11004 / 0.92 = 11960.8696 [+]\n11004 / 0.93 = 11832.2581 [+]\n11004 / 0.94 = 11706.383 [+]\n11004 / 0.95 = 11583.1579 [+]\n11004 / 0.96 = 11462.5 [+]\n11004 / 0.97 = 11344.3299 [+]\n11004 / 0.98 = 11228.5714 [+]\nNavigation: Home | Addition | Substraction | Multiplication | Division       Tables for 11004: Addition | Substraction | Multiplication | Division\n\nOperand: 1 2 3 4 5 6 7 8 9 10 20 30 40 50 60 70 80 90 100 200 300 400 500 600 700 800 900 1000 2000 3000 4000 5000 6000 7000 8000 9000\n\nDivision for: 1 2 3 4 5 6 7 8 9 10 20 30 40 50 60 70 80 90 100 200 300 400 500 600 700 800 900 1000 2000 3000 4000 5000 6000 7000 8000 9000 11001 11002 11003 11004 11005 11006 11007 11008 11009 12000 13000 14000 15000 16000 17000 18000 19000 20000 21000" ]

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[ "# Answers\n\nSolutions by everydaycalculation.com\n\n## Reduce 80/10 to lowest terms\n\nThe simplest form of 80/10 is 8/1.\n\n#### Steps to simplifying fractions\n\n1. Find the GCD (or HCF) of numerator and denominator\nGCD of 80 and 10 is 10\n2. Divide both the numerator and denominator by the GCD\n80 ÷ 10/10 ÷ 10\n3. Reduced fraction: 8/1\nTherefore, 80/10 simplified to lowest terms is 8/1.\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:\nAndroid and iPhone/ iPad\n\nEquivalent fractions:\n\nMore fractions:\n\n#### Fractions Simplifier\n\n© everydaycalculation.com" ]

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[ "# The Binomial Distribution: A Probability Model for a Discrete Outcome\n\nThe binomial distribution model is an important probability model that is used when there are two possible outcomes (hence \"binomial\"). In a situation in which there were more than two distinct outcomes, a multinomial probability model might be appropriate, but here we focus on the situation in which the outcome is dichotomous.\n\nFor example, adults with allergies might report relief with medication or not, children with a bacterial infection might respond to antibiotic therapy or not, adults who suffer a myocardial infarction might survive the heart attack or not, a medical device such as a coronary stent might be successfully implanted or not. These are just a few examples of applications or processes in which the outcome of interest has two possible values (i.e., it is dichotomous). The two outcomes are often labeled \"success\" and \"failure\" with success indicating the presence of the outcome of interest. Note, however, that for many medical and public health questions the outcome or event of interest is the occurrence of disease, which is obviously not really a success. Nevertheless, this terminology is typically used when discussing the binomial distribution model. As a result, whenever using the binomial distribution, we must clearly specify which outcome is the \"success\" and which is the \"failure\".\n\nThe binomial distribution model allows us to compute the probability of observing a specified number of \"successes\" when the process is repeated a specific number of times (e.g., in a set of patients) and the outcome for a given patient is either a success or a failure. We must first introduce some notation which is necessary for the binomial distribution model.\n\nFirst, we let \"n\" denote the number of observations or the number of times the process is repeated, and \"x\" denotes the number of \"successes\" or events of interest occurring during \"n\" observations. The probability of \"success\" or occurrence of the outcome of interest is indicated by \"p\".\n\nThe binomial equation also uses factorials. In mathematics, the factorial of a non-negative integer k is denoted by k!, which is the product of all positive integers less than or equal to k. For example,\n\n• 4! = 4 x 3 x 2 x 1 = 24,\n• 2! = 2 x 1 = 2,\n• 1!=1.\n• There is one special case, 0! = 1.\n\nWith this notation in mind, the binomial distribution model is defined as:\n\nThe Binomial Distribution Model", null, "", null, "Use of the binomial distribution requires three assumptions:\n\n1. Each replication of the process results in one of two possible outcomes (success or failure),\n2. The probability of success is the same for each replication, and\n3. The replications are independent, meaning here that a success in one patient does not influence the probability of success in another.\n\nFor a more intuitive explanation of the binomial distribution, you might want to watch the following video from KhanAcademy.org.\n\n## Examples of Use of the Binomial Model\n\n### 1. Relief of Allergies\n\nSuppose that 80% of adults with allergies report symptomatic relief with a specific medication. If the medication is given to 10 new patients with allergies, what is the probability that it is effective in exactly seven?\n\nFirst, do we satisfy the three assumptions of the binomial distribution model?\n\n1. The outcome is relief from symptoms (yes or no), and here we will call a reported relief from symptoms a 'success.'\n2. The probability of success for each person is 0.8.\n3. The final assumption is that the replications are independent, and it is reasonable to assume that this is true.\n\nWe know that:\n\n• # observation is n=10\n• # successes or events of interest is x=7\n• p=0.80\n\nThe probability of 7 successes is:", null, "", null, "This is equivalent to:", null, "", null, "But many of the terms in the numerator and denominator cancel each other out,", null, "so this can be simplified to:", null, "", null, "Interpretation: There is a 20.13% probability that exactly 7 of 10 patients will report relief from symptoms when the probability that any one reports relief is 80%.", null, "Note: Binomial probabilities like this can also be computed in an Excel spreadsheet using the =BINOMDIST function. Place the cursor into an empty cell and enter the following formula:\n\n=BINOMDIST(x,n,p,FALSE)\n\nwhere x= # of 'successes', n = # of replications or observations, and p = probability of success on a single observation.\n\nWhat is the probability that none report relief? We can again use the binomial distribution model with n=10, x=0 and p=0.80.", null, "", null, "This is equivalent to", null, "", null, "whixh simpliefies to", null, "", null, "Interpretation: There is practically no chance that none of the 10 will report relief from symptoms when the probability of reporting relief for any individual patient is 80%.\n\nWhat is the most likely number of patients who will report relief out of 10? If 80% report relief and we consider 10 patients, we would expect that 8 report relief. What is the probability that exactly 8 of 10 report relief? We can use the same method that was used above to demonstrate that there is a 30.30% probability that exactly 8 of 10 patients will report relief from symptoms when the probability that any one reports relief is 80%. The probability that exactly 8 report relief will be the highest probability of all possible outcomes (0 through 10).\n\n### 2. The Probability of Dying after a Heart Attack\n\nThe likelihood that a patient with a heart attack dies of the attack is 0.04 (i.e., 4 of 100 die of the attack). Suppose we have 5 patients who suffer a heart attack, what is the probability that all will survive? For this example, we will call a success a fatal attack (p = 0.04). We have n=5 patients and want to know the probability that all survive or, in other words, that none are fatal (0 successes).\n\nWe again need to assess the assumptions. Each attack is fatal or non-fatal, the probability of a fatal attack is 4% for all patients and the outcome of individual patients are independent. It should be noted that the assumption that the probability of success applies to all patients must be evaluated carefully. The probability that a patient dies from a heart attack depends on many factors including age, the severity of the attack, and other comorbid conditions. To apply the 4% probability we must be convinced that all patients are at the same risk of a fatal attack. The assumption of independence of events must also be evaluated carefully. As long as the patients are unrelated, the assumption is usually appropriate. Prognosis of disease could be related or correlated in members of the same family or in individuals who are co-habitating. In this example, suppose that the 5 patients being analyzed are unrelated, of similar age and free of comorbid conditions.", null, "", null, "", null, "", null, "There is an 81.54% probability that all patients will survive the attack when the probability that any one dies is 4%. In this example, the possible outcomes are 0, 1, 2, 3, 4 or 5 successes (fatalities). Because the probability of fatality is so low, the most likely response is 0 (all patients survive). The binomial formula generates the probability of observing exactly x successes out of n.\n\n## Computing the Probability of a Range of Outcomes\n\nIf we want to compute the probability of a range of outcomes we need to apply the formula more than once. Suppose in the heart attack example we wanted to compute the probability that no more than 1 person dies of the heart attack. In other words, 0 or 1, but not more than 1. Specifically we want P(no more than 1 success) = P(0 or 1 successes) = P(0 successes) + P(1 success). To solve this probability we apply the binomial formula twice.\n\nWe already computed P(0 successes), we now compute P(1 success):", null, "", null, "", null, "", null, "P(no more than 1 'success') = P(0 or 1 successes) = P(0 successes) + P(1 success)\n\n= 0.8154 + 0.1697 = 0.9851.\n\nThe probability that no more than 1 of 5 (or equivalently that at most 1 of 5) die from the attack is 98.51%.\n\nWhat is the probability that 2 or more of 5 die from the attack? Here we want to compute P(2 or more successes). The possible outcomes are 0, 1, 2, 3, 4 or 5, and the sum of the probabilities of each of these outcomes is 1 (i.e., we are certain to observe either 0, 1, 2, 3, 4 or 5 successes). We just computed P(0 or 1 successes) = 0.9851, so P(2, 3, 4 or 5 successes) = 1 - P(0 or 1 successes) = 0.0149. There is a 1.49% probability that 2 or more of 5 will die from the attack.\n\n## Mean and Standard Deviation of a Binomial Population\n\nMean number of successes:", null, "", null, "Standard Deviation:", null, "", null, "For the previouos example on the probability of relief from allergies with n-10 trialsand p=0.80 probability of success on each trial:", null, "", null, "", null, "", null, "## Binomial Probability Calculator\n\nSuppose you flipped a coin 10 times (i.e., 10 trials), and the probability of getting \"heads\" was 0.5 (50%). What would be the probability of getting exactly 4 heasds?\n\nANSWER", null, "## Calculating Binomial Probabilities with R\n\nWith 4 successes, 10 trials, and probability =0.5 on each trial\n\nWhat is the :\n\nProbability\n\nR coding to compute these\n\na) Probability of exactly 4 events = 0.205078\n\n> dbinom (4, 10, 0.5)\n\nb) Cumulative probability of < 4 events = 0.171875\n\n> pbinom (3, 10, 0.5, lower.tail=TRUE)\n\nc) Cumulative probability of < 4 events = 0.376953\n\n> pbinom(4, 10, 0.5, lower.tail=TRUE)\n\nd) Cumulative probability of > 4 events = 0.623047\n\n> pbinom(4, 10, 0.5, lower.tail=FALSE)\n\ne) Cumulative probability of > 4 events = 0.828125\n\npbinom (3, 10, 0.5, lower.tail=FALSE)" ]

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[ "#### Transcript Slide 1\n\n```Pair Hidden Markov Model\nThree kinds of pair HMMs\n(PHMMs)\n• PHMM for pairwise sequence alignment\n– BSA Chapter 4\n• PHMM for the analysis (e.g. gene prediction) on two\naligned sequences (i.e. the pre-calculated pairwise\nalignments)\n– Twinscan\n• PHMM for simultaneously pairwise alignment and\nanalysis\n– SLAM\nPairwise sequence alignment\nGiven two sequences over an alphabet (4\nnucleotides or 20 amino acids):\nATGTTAT and ATCGTAC\nBy inserting ‘-’s and shifting two sequences,\nthey can be aligned into a table of two rows\nwith the same length:\nA T - G T T A T\nA T C G T - A C\nScoring a pairwise alignment\n• Mismatches are penalized by –μ, indels\nare penalized by –σ, and matches are\nrewarded with +1, the resulting score is:\n#matches – μ(#mismatches) – σ (#indels)\nA T - G T T A T\nA T C G T - A C\n5- μ -2σ\nScoring Matrix: Example\nA\nR\nN\nK\nA\n5\n-2\n-1\n-1\nR\n-\n7\n-1\n3\nN\n-\n-\n7\n0\nK\n-\n-\n-\n6\n• Notice that although\nR and K are different\namino acids, they\nhave a positive score.\n• Why? They are both\npositively charged\namino acids will not\ngreatly change\nfunction of protein.\nScoring matrices\n• Amino acid substitution matrices\n– PAM\n– BLOSUM\n• DNA substitution matrices\n– DNA is less conserved than protein\nsequences\n– Less effective to compare coding\nregions at nucleotide level\nAffine Gap Penalties\n• In nature, a series of k indels often come\nas a single event rather than a series of k\nsingle nucleotide events:\nThis is more\nlikely.\nNormal scoring would\ngive the same score This is less\nfor both alignments\nlikely.\nAccounting for Gaps\n• Gaps- contiguous sequence of spaces in one of\nthe rows\n• Score for a gap of length x is:\n-(ρ + σx)\nwhere ρ >0 is the penalty for introducing a gap:\ngap opening penalty\nρ will be large relative to σ:\ngap extension penalty\nbecause you do not want to add too much of a\npenalty for extending the gap.\nAffine Gap Penalties\n• Gap penalties:\n– -ρ-σ when there is 1 indel\n– -ρ-2σ when there are 2 indels\n– -ρ-3σ when there are 3 indels, etc.\n– -ρ- x·σ (-gap opening - x gap extensions)\n• Somehow reduced penalties (as\ncompared to naive scoring) are given to\nruns of horizontal and vertical edges\nAlignment: a path in the Alignment\nGraph\n0 1\nA\nA\n0 1\n2\nT\nT\n2\n2\nC\n3\n3\nG\nG\n4\n4\nT\nT\n5\n5\nT\n5\n6\nA\nA\n6\n7\nT\nC\n7\n- Corresponding path (0,0) , (1,1) , (2,2), (2,3),\n(3,4), (4,5), (5,5), (6,6),\n(7,7)\nAlignment as a Path in the Edit Graph\nOld Alignment\n012234567\nx= AT_GTTAT\ny= ATCGT_AC\n012345567\nNew Alignment\n012234567\nx= AT_GTTAT\ny= ATCG_TAC\n012344567\nRepresenting sequence\nalignment using pair HMM\nHMM for sequence alignment, which\nincorporates affine gap scores.\n“Hidden” States\n• Match (M)\n• Insertion in x (X)\n• insertion in y (Y)\nObservation Symbols\n• Match (M): {(a,b)| a,b in ∑ }.\n• Insertion in x (X): {(a,-)| a in ∑ }.\n• Insertion in y (Y): {(-,a)| a in ∑ }.\nRepresenting sequence\nalignment using pair HMM\n1-\n1-2\nM\nX\n\nEmission probabilities:\n\nM: Pxi,yj\nM: qxi\nY: qyj\n\n\n1-\nY\nAlignment: a path  a hidden state\nsequence\nA T - G T T A T\nA T C G T - A C\nM M Y M M X M M\nSequence alignment using pair HMM\n• Based on the HMM, each alignment of two\nDNA/protein sequences can be assigned with a\nprobability score;\n• Each “observation symbol” of the HMM is an aligned\npair of two letters, or of a letter and a gap.\n• The Markov chain of hidden states should represent a\nscoring scheme reflecting an evolutionary model.\n• Transition and emission probabilities define the\nprobability of each aligned pair of sequences.\n• Given two input sequences, we look for an alignment\nof these two sequences of maximum probability.\nTransitions and Emission Probabilities\nM\nTransitions probabilities\n(note the forbidden ones).\nδ = probability for 1st gap\nε = probability for extending\ngap.\nX\nY\n1-2δ\nM\nδ\nδ\nX\n1- ε\nε\n0\nY\n1- ε\n0\nε\nEmission Probabilities\n• Match: (a,b) with pab – only from M states\n• Insertion in x: (a,-) with qa – only from X state\n• Insertion in y: (-,a).with qa - only from Y state.\nScoring alignments\n• For each pair of sequences x (of length m) and y (of\nlength n), there are many alignments of x and y, each\ncorresponds to a different state sequence (with the\nlength between max{m,n} and m+n).\n• Given the transmission and emission probabilities,\neach alignment has a defined score – the product of the\ncorresponding probabilities.\n• An alignment is “most probable”, if it maximizes this\nscore.\nFinding the most probable\nalignment\nLet vM(i,j) be the probability of the most probable\nalignment of x(1..i) and y(1..j), which ends with a\nmatch (state M). Similarly, vX(i,j) and vY(i,j), the\nprobabilities of the most probable alignment of\nx(1..i) and y(1..j), which ends with states X or Y,\nrespectively.\nv M [i, j ]  pxi y j\n (1  2 )v M (i  1, j  1) \n\n\nX\nmax  (1   )v (i  1, j  1) \n\n\nY\n (1   )v (i  1, j  1) \n\n\nMost probable alignment\nSimilar argument for vX(i,j) and vY(i,j), the\nprobabilities of the most probable alignment of\nx(1..i) and y(1..j), which ends with an insertion\nto x or y, are:\nM\n\n\nv\n(i  1, j ) \nX\nv [i, j ]  qxi max \n\n  v X (i  1, j ) \n\n\nM\n\n\nv\n(i, j  1) \nY\nv [i, j ]  q y j max \n\n  vY (i, j  1) \n\n\nDifferent alignments of x and y may have different lengths. To\nget a coherent probabilistic model we need to define a\nprobability distribution over sequences of different lengths.\nFor this, an END state is added,\nwith transition probability τ\nfrom any other state to END.\nThis assumes expected\nsequence length of 1/ τ.\nThe last transition in each\nalignment is to the END\nstate, with probability τ\nM\n1-2δ M\nτ\nX\nY\nEND\nδ\nδ\nτ\nX 1-ε -τ\nε\nY\nEND\n1-ε -τ\nτ\nε\nτ\n1\nRepresenting sequence\nalignment using pair HMM\n\n\n1-\n1--2\n1-2\nM\nX\n\n\nBegin\n\n\n1-\n\nY\n\n\nEnd\nThe log-odds scoring function\n• We wish to know if the alignment score is above or\nbelow the score of random alignment of sequences\nwith the same length.\n– Model comparison\n• We need to model random sequence alignment by\nHMM, with end state. This model assigns probability\nto each pair of sequences x and y of arbitrary lengths m\nand n.\nHMM for a random sequence alignment\nThe transition probabilities for the\nrandom model, with termination\nprobability η:\n(x is the start state)\nThe emission probability for a is qa.\nThus the probability of x (of length n)\nand y (of length m) being random is:\nX\nY\nEND\nX 1- η\nη\n0\nY\n0\n1- η\nη\nEND\n0\n0\n1\nn\nm\ni 1\nj 1\np( x, y | Random)   2 (1   )n  m  qxi  q y j\nAnd the corresponding score is:\nlog p( x, y | Random)  2 log  ( n  m) log(1   ) \nn\nm\nlogq  logq\ni 1\nxi\ni 1\nyi\nHMM for random sequence\nalignment\nMarkov Chains for “Random” and\n“Model”\nM\nX\nY\nEND\nM 1-2δ -τ\nδ\nδ\nτ\nX 1-ε -τ\nε\nY\n1-ε -τ\nτ\n“Random”\nε\nτ\n1\nEND\n“Model”\nX\nY\nX 1- η\nη\nY\nEN\nD\n1- η\nEND\nη\n1\nCombining models in the log-odds\nscoring function\nIn order to compare the M score to the R score of sequences x\nand y, we can find an optimal M score, and then subtract\nfrom it the R score.\nThis is insufficient when we look for local alignments, where\nthe optimal substrings in the alignment are not known in\n1. Define a log-odds scoring function which keeps track of\nthe difference Match-Random scores of the partial strings\nduring the alignment.\n2. At the end add to the score (logτ – 2logη) to compensate\nfor the end transitions in both models.\nThe log-odds scoring function\n(assuming that letters at insertions/deletions are selected by the random model)\nV [i, j ]  log\nM\np xi y j\nq xi q y j\n log(1  2   )  V M [i  1, j  1]\n\n\nX\n max log(1     )  V [i  1, j  1]   2 log(1   )\n\n\nlog(1     )  V Y [i  1, j  1]b \n\n\nM\n\n\nlog\n\n\nV\n[\ni\n\n1\n,\nj\n]\nX\n  log(1   )\nV [i, j ]  max\nX\n\nlog\n\n\nV\n[\ni\n\n1\n,\nj\n]\n\n\nM\n\n\nlog\n\n\nV\n[\ni\n,\nj\n\n1\n]\nY\n  log(1   )\nV [i, j ]  max\nY\n\nlog\n\n\nV\n[\ni\n,\nj\n\n1\n]\n\n\nAnd at the end add to the score (logτ – 2logη).\nA Pair HMM For Local Alignment\nFull Probability Of The Two Sequences\n• HMMs allow for calculating the probability that\na given pair of sequences are related\naccording to the HMM by any alignment\n• This is achieved by summing over all\nalignments\nP(x, y) \nP(x, y, )\nalignment \nFull Probability Of The Two Sequences\n• The way to calculate the sum is by using\nthe forward algorithm\n• fk(i,j) : the combined probability of all\nalignments up to (i,j) that end in state k\nForward Algorithm For Pair HMMs\nP(x,y)\nFull Probability Of The Two Sequences\n• P(x,y) gives the likelihood that x and y are\nrelated by some unspecified alignment, as\nopposed to being unrelated\n• If there is an unambiguous best alignment,\nP(x,y) will be “dominated” by the single hidden\nstate seuence corresponding to that alignment\nHow correct is the alignment\n• Define a posterior distribution P(s|x,y) over all\nalignments given a pair of sequences x and y\nP( x, y, s)\nP( s | x, y) \nP( x, y )\nProbability that the optimal scoring alignment is correct:\nP(x, y,  ) v (n,m)\nP( | x, y) \n E\nP(x, y)\nf (n,m)\n*\n*\nE\nViterbi algorithm\nForward algorithm\n• Usually the probability that the optimal\nscoring alignment is correct, is extremely\nsmall!\n• Reason: there are many small variants of the\nbest alignment that have nearly the same\nscore.\nThe Posterior Probability That\nTwo Residues Are Aligned\n• If the probability of any single complete path\nbeing entirely correct is small, can we say\nsomething about the local accuracy of an\nalignment?\n• It is useful to be able to give a reliability\nmeasure for each part of an alignment\nThe posterior probability that\ntwo residues are aligned\n• The idea is:\n– calculate the probability of all the alignments that\npass through a specified matched pair of residues\n(xi,yj)\n– Compare this value with the full probability of all\nalignments of the pair of sequences\n– If the ratio is close to 1, then the match is highly\nreliable\n– If the ratio is close to 0, then the match is unreliable\nThe posterior probability that\ntwo residues are aligned\n• Notation: xiyj denotes that xi is aligned to yj\n• We are interested in P(xiyj|x,y)\nP(x, y, x i y j )\n• We have P(x i y j | x, y) \nP(x, y)\nP(x, y, x i y j )  P(x1 i , y1 j , x i y j )P(x i1 n , y j 1\nm\n| x i y j )\n• P(x,y)\n is computed using the forward algorithm\n• P(x,y,xiyj): the first term in computed by the forward\nalgorithm, and the second is computed by the\nbackward algorithm (=bM(i,j) in the backward\nalgorithm)\nBackward Algorithm For Pair HMMs\nPair HMM for gene finding\n(Twinscan)\n• Twinscan is an augmented version of the\nGHMM used in Genscan.\nGenscan\nModel\n• Genscan considers the\nfollowing:\n–\n–\n–\n–\nPromoter signals\nSplice signals\nProbability of coding\nand non-coding DNA\n– Gene, exon and intron\nlength\nChris Burge and Samuel Karlin, Prediction of Complete Gene\nStructures in Human Genomic DNA, JMB. (1997) 268, 78-94\nTwinscan Algorithm\n1. Align the two sequences (eg. from human and\nmouse);\n2. The similar hidden states as Genscan;\n3. New “alphabet” for observation symbols: 4 x 3 =\n12 symbols:\n = { A-, A:, A|, C-, C:, C|, G-, G:, G|, U-, U:, U| }\nMark each base as gap ( - ), mismatch ( : ), match ( | )\nTwinscan Algorithm\nRun Viterbi using emissions ek(b), where b  { A-, A:,\nA|, …, T| }\nNote:\nEmission distributions ek(b) estimated from the\nalignment of real gene pairs from human/mouse\neI(x|) < eE(x|): matches favored in exons\neI(x-) > eE(x-): gaps (and mismatches) favored in\nintrons\nExample\nHuman:\nMouse:\nAlign:\nACGGCGACUGUGCACGU\nACUGUGAC GUGCACUU\n||:|:|||-||||||:|\nInput to Twinscan HMM:\nA| C| G: G| C: G| A| C| U- G| U| G| C| A| C| G: U|\nRecall,\nLikely exon\neE(A|) > eI(A|)\neE(A-) < eI(A-)\nHMMs for simultaneous alignment and\ngene finding (SLAM)\n[human]\nExon1\nIntron1\nExon2\nIntron2\nExon3\n5’\n3’\n[mouse]\nCNS\nExon = coding\nIntron = non-coding\nCNS\nCNS\nCNS = conserved non-coding\nGeneralized Pair HMMs\nGeneralized Pair HMMs (SLAM)\nGapped alignment\nMeasuring Performance\n```" ]

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[ "# Top 30 machine learning library Interview Questions with Answers\n\n### 1. What is a machine learning library?\n\nAns: A machine learning library is a collection of tools, algorithms, and functions that provide pre-built functionality for developing and applying machine learning models.\n\n### 2. What are the popular machine learning libraries?\n\nAns: Some popular machine learning libraries include scikit-learn, TensorFlow, Keras, PyTorch, XGBoost, LightGBM, and Apache Spark’s MLlib.\n\n### 3. What is scikit-learn?\n\nAns: Scikit-learn is a popular open-source machine-learning library for Python. It provides a wide range of machine-learning algorithms and tools for data preprocessing, model selection, and evaluation.\n\n### 4. What is TensorFlow?\n\nAns: TensorFlow is an open-source machine-learning library developed by Google. It provides a flexible framework for building and training deep learning models across various platforms and devices.\n\n### 5. What is Keras?\n\nAns: Keras is a high-level deep learning library that runs on top of other machine learning frameworks such as TensorFlow and Theano. It offers a user-friendly API for building and training neural networks.\n\n### 6. What is PyTorch?\n\nAns: PyTorch is an open-source deep-learning library known for its dynamic computational graph and ease of use. It provides tools for building and training neural networks and supports GPU acceleration.\n\n### 7. What is XGBoost?\n\nAns: XGBoost is an optimized gradient-boosting machine-learning library. It is designed to be highly efficient and scalable, making it popular for structured data problems and Kaggle competitions.\n\n### 8. What is LightGBM?\n\nAns: LightGBM is another gradient-boosting library that focuses on speed and efficiency. It uses a histogram-based algorithm and is known for its fast training and prediction times.\n\n### 9. What is Apache Spark’s MLlib?\n\nAns: Apache Spark’s MLlib is a distributed machine learning library that integrates with the Spark framework. It provides scalable implementations of various machine learning algorithms and tools for distributed computing.\n\n### 10. What are the advantages of using machine learning libraries?\n\nAns: The advantages of using machine learning libraries include the:\n\n• Pre-built algorithms and functions for common machine learning tasks.\n• Efficient implementations that leverage parallel processing and hardware acceleration.\n• Community support and extensive documentation.\n• Integration with other libraries and frameworks for data processing and visualization.\n\n### 11. What types of machine learning algorithms are typically available in libraries?\n\n• Machine learning libraries typically provide a wide range of algorithms, including:\n• Supervised learning: Regression, classification.\n• Unsupervised learning: Clustering, dimensionality reduction.\n• Reinforcement learning: Learning through interaction with an environment.\n• Deep learning: Neural networks and related architectures.\n\n### 12. How can you handle missing data in machine learning libraries?\n\nAns: Machine learning libraries provide various methods for handling missing data, including imputation techniques such as mean imputation, median imputation, and model-based imputation.\n\n### 13. How can you evaluate the performance of a machine learning model?\n\nAns: Machine learning libraries offer various evaluation metrics such as accuracy, precision, recall, F1-score, mean squared error, and area under the ROC curve (AUC-ROC) to assess the performance of a model.\n\n### 14. Can machine learning libraries handle big data?\n\nAns: Yes, many machine learning libraries offer scalability and distributed computing capabilities to handle big data. For example, Apache Spark’s MLlib and TensorFlow’s distributed computing support enable the processing of large datasets across clusters.\n\n### 15. What is the difference between a machine learning library and a deep learning library?\n\nAns: A machine-learning library typically provides a wide range of algorithms and techniques for various machine-learning tasks. A deep learning library, on the other hand, specifically focuses on neural networks and related architectures for deep learning tasks.\n\n### 16. How is a decision tree pruned?\n\nAns: Pruning is what happens in decision trees when branches that have weak predictive power are removed in order to reduce the complexity of the model and increase the predictive accuracy of a decision tree model. Pruning can happen bottom-up and top-down, with approaches such as reduced error pruning and cost complexity pruning.\n\nReduced error pruning is perhaps the simplest version: replace each node. If it doesn’t decrease predictive accuracy, keep it pruned. While simple, this heuristic actually comes pretty close to an approach that would optimize for maximum accuracy.\n\n### 17. Which is more important to you: model accuracy or model performance?\n\nAns: Such machine learning interview questions test your grasp of the nuances of machine learning model performance! Machine learning interview questions often look toward the details. There are models with higher accuracy that can perform worse in predictive power—how does that make sense?\n\nWell, it has everything to do with how model accuracy is only a subset of model performance, and at that, a sometimes misleading one. For example, if you wanted to detect fraud in a massive dataset with a sample of millions, a more accurate model would most likely predict no fraud at all if only a vast minority of cases were fraud. However, this would be useless for a predictive model—a model designed to find fraud that asserted there was no fraud at all! Questions like this help you demonstrate that you understand model accuracy isn’t the be-all and end-all of model performance.\n\n### 18. What’s the F1 score? How would you use it?\n\nAnswer: The F1 score is a measure of a model’s performance. It is a weighted average of the precision and recall of a model, with results tending to 1 being the best, and those tending to 0 being the worst. You would use it in classification tests where true negatives don’t matter much.\n\n### 19. How would you handle an imbalanced dataset?\n\nAns: An imbalanced dataset is when you have, for example, a classification test and 90% of the data is in one class. That leads to problems: an accuracy of 90% can be skewed if you have no predictive power on the other category of data! Here are a few tactics to get over the hump:\n\n• Collect more data to even the imbalances in the dataset.\n• Resample the dataset to correct for imbalances.\n• Try a different algorithm altogether on your dataset.\n• What’s important here is that you have a keen sense for what damage an unbalanced dataset can cause, and how to balance that.\n\n### 20. When should you use classification over regression?\n\nAns: Classification produces discrete values and datasets to strict categories, while regression gives you continuous results that allow you to better distinguish differences between individual points. You would use classification over regression if you wanted your results to reflect the belongingness of data points in your dataset to certain explicit categories (ex: If you wanted to know whether a name was male or female rather than just how correlated they were with male and female names.)\n\n### 21. Name an example where ensemble techniques might be useful.\n\nAns: Ensemble techniques use a combination of learning algorithms to optimize better predictive performance. They typically reduce overfitting in models and make the model more robust (unlikely to be influenced by small changes in the training data).\n\nYou could list some examples of ensemble methods (bagging, boosting, the “bucket of models” method) and demonstrate how they could increase predictive power.\n\n### 22. How do you ensure you’re not overfitting with a model?\n\nAns: This is a simple restatement of a fundamental problem in machine learning: the possibility of overfitting training data and carrying the noise of that data through to the test set, thereby providing inaccurate generalizations.\n\nThere are three main methods to avoid overfitting:\n\n• Keep the model simpler: reduce variance by taking into account fewer variables and parameters, thereby removing some of the noise in the training data.\n• Use cross-validation techniques such as k-folds cross-validation.\n• Use regularization techniques such as LASSO that penalize certain model parameters if they’re likely to cause overfitting.\n\n### 23. What evaluation approaches would you work to gauge the effectiveness of a machine learning model?\n\nAns: You would first split the dataset into training and test sets, or perhaps use cross-validation techniques to further segment the dataset into composite sets of training and test sets within the data. You should then implement a choice selection of performance metrics: here is a fairly comprehensive list. You could use measures such as the F1 score, the accuracy, and the confusion matrix. What’s important here is to demonstrate that you understand the nuances of how a model is measured and how to choose the right performance measures for the right situations.\n\n### 24. How would you evaluate a logistic regression model?\n\nAns: A subsection of the question above. You have to demonstrate an understanding of what the typical goals of a logistic regression are (classification, prediction, etc.) and bring up a few examples and use cases.\n\n### 25. What’s the “kernel trick” and how is it useful?\n\nAns: The Kernel trick involves kernel functions that can enable in higher-dimension spaces without explicitly calculating the coordinates of points within that dimension: instead, kernel functions compute the inner products between the images of all pairs of data in a feature space. This allows them the very useful attribute of calculating the coordinates of higher dimensions while being computationally cheaper than the explicit calculation of said coordinates. Many algorithms can be expressed in terms of inner products. Using the kernel trick enables us effectively run algorithms in a high-dimensional space with lower-dimensional data.\n\n## Why Do Organizations Need AIOps?\n\nAs technology rapidly advances, businesses are constantly seeking new and innovative ways to optimize their operations. One such solution that has gained significant popularity in recent years…\n\n## Why AIOps is Important?\n\nIn today’s rapidly evolving technological landscape, the need for efficient IT operations has become paramount. With the proliferation of data and the increasing complexity of IT systems,…\n\n## What are the Top AIOps Implementation Examples?\n\nAs technology advances, businesses are turning to AIOps (Artificial Intelligence for IT Operations) to streamline their processes and improve efficiency. AIOps is a combination of AI and…\n\n## Where are AIOps used?\n\nIn today’s fast-paced digital world, businesses are increasingly relying on technology to drive their operations. With the exponential growth of data, it has become a challenge to…\n\n## What are the Best AiOps Tutorials?\n\nAre you interested in learning about AiOps? Well, you’ve come to the right place! In this blog article, we will take a deep dive into the best…\n\n## What are the Best AiOps Books?", null, "" ]

[ null, "https://secure.gravatar.com/avatar/8328482d113ff101a6d4581e30620a1a", null ]

[ "# COUPLED COMMON FIXED POINT THEOREMS FOR A CONTRACTIVE CONDITION OF RATIONAL TYPE IN ORDERED METRIC SPACES\n\n• Chandok, Sumit (Department of Mathematics, Khalsa College of Engineering & Technology (Punjab Tech nical University))\n• Accepted : 2013.01.14\n• Published : 2013.09.30\n\n#### Abstract\n\nThe purpose of this paper is to establish some coupled coincidence point theorems for a pair of mappings having a strict mixed g-monotone property satisfying a contractive condition of rational type in the framework of partially ordered metric spaces. Also, we present a result on the existence and uniqueness of coupled common fixed points. The results presented in the paper generalize and extend several well-known results in the literature.\n\n#### References\n\n1. R. P. Agarwal, M. A. El-Gebeily and D. O'Regan, Generalized contractions in partially ordered metric spaces, Appl. Anal. 87 (2008), 1-8. https://doi.org/10.1080/00036810701714164\n2. T. G. Bhaskar and V. Lakshmikantham, Fixed point theorems in partially ordered metric spaces and applications, Nonlinear Anal. 65(2006), 1379-1393. https://doi.org/10.1016/j.na.2005.10.017\n3. S. Chandok, Some common fixed point theorems for generalized f-weakly contractive mappings, J. Appl. Math. Informatics 29(2011), 257-265.\n4. S. Chandok, Some common fixed point theorems for generalized nonlinear contractive mappings, Computers and Mathematics with Applications 62(2011), 3692-3699. https://doi.org/10.1016/j.camwa.2011.09.009\n5. S. Chandok, Common fixed points for generalized nonlinear contractive mappings in metric spaces, Mat. Vesnik 65(2013), 29-34.\n6. S. Chandok, Some common fixed point results for generalized weak contractive mappings in partially ordered metric spaces, J. Nonlinear Anal. Opt. (in press).\n7. S. Chandok and E. Karapinar, Some common fixed point results for generalized rational type weak contraction mappings in partially ordered metric spaces, Thai J. Math. (in press).\n8. S. Chandok and J. K. Kim, Fixed point theorem in ordered metric spaces for generalized contractions mappings satisfying rational type expressions, J. Nonlinear Functional Anal. Appl. 17(2012), 301-306.\n9. L. Ciric, M. O. Olatinwo, D. Gopal and G. Akinbo, Coupled point theorems for mappings satisfying a contractive condition of rational type on a partially ordered metric space, Adv. Fixed Point Theory 2(2012), 1-8.\n10. J. Harjani, B. Lopez and K. Sadarangani, A fixed point theorem for mappings satisfying a contractive condition of rational type on a partially ordered metric space, Abstract Appl. Anal. 2010, Article ID 190701.\n11. D. S. Jaggi, Some unique fixed point theorems, Indian J. Pure Appl. Math. 8(1977), 223230.\n12. V. Lakshmikantham and L. Ciric, Coupled fixed point theorems for nonlinear contractions in partially orderd metric spaces, Nonlinear Anl. 70(2009), 4341-4349. https://doi.org/10.1016/j.na.2008.09.020\n13. J. J. Nieto and R. R. Lopez, Contractive mapping theorems in partially ordered sets and applications to ordinary differential equations, Order 22(2005), 223-239. https://doi.org/10.1007/s11083-005-9018-5\n14. A. C. M. Ran and M. C. B. Reurings, A fixed point theorem in partially ordered sets and some applications to matrix equations, Proc. Amer. Math. Soc. 132(5)(2004), 1435-1443. https://doi.org/10.1090/S0002-9939-03-07220-4\n15. B. Samet, Coupled fixed point theorems for a generalized Meir-Keeler Contraction in partially ordered metric spaces, Nonlinear Anal. 72(2010), 4508-4517. https://doi.org/10.1016/j.na.2010.02.026" ]

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[ "# (In)Security of Ring-LWE Under Partial Key Exposure\n\nDana Dachman-Soled, Huijing Gong, Mukul Kulkarni and Aria Shahverdi\n\n# Abstract\n\nWe initiate the study of partial key exposure in Ring-LWE (RLWE)-based cryptosystems. Specifically, we (1) Introduce the search and decision Leaky R-LWE assumptions (Leaky R-SLWE, Leaky R-DLWE), to formalize the hardness of search/decision RLWE under leakage of some fraction of coordinates of the NTT transform of the RLWE secret. (2) Present and implement an efficient key exposure attack that, given certain 1/4-fraction of the coordinates of the NTT transform of the RLWE secret, along with samples from the RLWE distribution, recovers the full RLWE secret for standard parameter settings. (3) Present a search-to-decision reduction for Leaky R-LWE for certain types of key exposure. (4) Propose applications to the security analysis of RLWE-based cryptosystems under partial key exposure.\n\nMSC 2010: 94A60; 68P25; 03G10\n\n## 1 Introduction\n\nThere has been a monumental effort in the cryptographic community to develop “post-quantum” cryptosys-tems that remain secure even in the presence of a quantum adversary. One of the foremost avenues for viable post-quantum public key cryptography is to construct schemes from the Ring-Learning with Error (RLWE) assumption—currently 3 out of 26 of the second round NIST submissions are based on assumptions in the ring setting. RLWE is often preferred in practice over standard LWE due to its algebraic structure, which allows for smaller public keys and more efficient implementations. In the RLWE setting, we typically consider rings of the form Rq:=q[x]/(xn+1), where n is a power of two and q ≡ 1 mod 2n. The (decisional) RLWE problem is then to distinguish (a, b = a · s + e) ∈ Rq × Rq from uniformly random pairs, where s ∈ Rq is a random secret, a ∈ Rq is uniformly random and the error term e ∈ R has small norm. A critical question is whether the additional algebraic structure of the RLWE problem renders it less secure than the standard LWE problem. Interestingly, to the best of our knowledge—for the rings used in practice and practical parameter settings—the best attacks on RLWE are generic and can equally well be applied to standard LWE . In this work, we ask whether improved attacks on RLWE are possible when partial information about the RLWE secret is exposed, though the secret retains high entropy.\n\nThe NTT transform.\n\nOne key method for speeding up computations in the RLWE setting is usage of the NTT transform (similar to the discrete Fourier transform (DFT), but over finite fields) to allow for faster polynomial multiplication over the ring Rq. Specifically, applying the NTT transform to two polynomials p,pRq resulting in two n-dimensional vectors, p^,p^qn allows for component-wise multiplication and addition, which is highly efficient. In this work, we consider leakage of a fraction of NTT coordinates of the RLWE secret. Since the RLWE secret will typically be stored in NTT form (to facilitate fast computation), [4, 7] leakage of coordinates of the NTT transform is a natural model for partial key exposure attacks.\n\nThis work.\n\nThe goal of this work is to initiate a study of partial key exposure in RLWE based cryptosystems and explore both positive and negative results in this setting. Specifically, we (1) define search and decision versions of Leaky RLWE assumptions, where the structured leakage occurs on the coordinates of the NTT transform of the RLWE secret; (2) present partial key exposure attacks on RLWE, given 1 /4-fraction of structured leakage on the secret key; (3) present a search to decision reduction for the Leaky RLWE assumptions; and (4) propose applications of the decision version of the assumption to practical RLWE-based cryptosystems.\n\n### 1.1 Leaky RLWE Assumptions–Search and Decision Versions\n\nWe next briefly introduce the search and decision versions of the Leaky RLWE assumptions. For pRq:= q[x]/(xn+1), we denote p^:=NTT(p):=(p(ω1),p(ω3),,p(ω2n1)), where ω is a primitive 2n-th root of unity modulo q, and is guaranteed to exist by choice of prime q, s.t. q ≡ 1 mod 2n. Note that p^ is indexed by the set 2n*.\n\nThe search version of the RLWE problem with leakage, denoted Leaky R-SLWE, is parametrized by (n {1,2,4,8,n},S2n ). The goal is to recover the RLWE secret s=NTT1(s^), given samples from the distribution Dreal,n,Ss which outputs (a^,a^s^+e^,[ s^i ]iαmod2nαS), where a, s, and e are as in the standard RLWE assumption (see Appendix A.2 and for the precise definition).\n\nThe decision version of the RLWE problem with leakage, denoted Leaky R-DLWE is parametrized by (n{1,2,4,8,n},S2n). The goal is to distinguish the distributions Dreal,n,Ss and Dsim,n,Ss, where Dreal,n,Ss is as above and Dsim,n,Ss outputs a^,u^,[ si^ ]iαmod2nαS ) ss, where u^i=ai^si^+ei^ for iα mod2n, αS and ui^ is chosen uniformly at random from ℤq, otherwise. Note that only the coordinates of u^ corresponding to unleaked positions are required to be indistinguishable from random.\n\nWhen S={α} consists of a single element, we sometimes abuse notation and write the Leaky-RLWE parameters as (n,α) Leaky-RLWE with parameters (n,S) where S={ α1,α2,,αt }, is equivalent to Leaky-RLWE with parameters (n,S), where S=α11. · S (multiply every element of S by α11 ). It is also not hard to see that leaky search and decision are equally hard when secret s is uniform random from Rq versus drawn from the error distribution (the same reduction for standard RLWE works in our case).\n\n### 1.2 Our Results\n\nPartial key exposure attacks.\n\nWe present attacks on Leaky R-SLWE and test them on various practical parameter settings, such as the NewHope parameter settings as well as the RLWE challenges of Crockett and Peikert . Our attacks demonstrate that Leaky R-SLWE is easy for leakage parameters (n=4,α=1),(n=8,S={1,7}) and (n=8,S={1,15}), under (1) NewHope parameter settings of n = 1024, q = 12289, and x = Ψ 16 (centered binomial distribution of parameter 16); (2) The same parameters above, but with χ=D8 (discrete Gaussian with standard deviation of 8, which has the same standard deviation as Ψ 16), since this is the recommended setting in the case where the adversary gets to see many RLWE samples ; (3) For parameters of several of the Crockett and Peikert challenges, including those classified as “very hard.” In all the above cases, we fully recover the RLWE secret with high probability, given the corresponding 1/4-fraction of the positions in the NTT transform of the RLWE secret. See Section 3.2 for details on the experimental results.\n\nA search-to-decision reduction.\n\nDefine Tn(n) to be the time required to solve Leaky R-SLWE for dimension n, given positions [ s^i ]iα mod 2n . Assuming search R-LWE without leakage is subexponentially 2Ω(nϵ) -hard for some constant ϵ ≤ 1 and polynomial modulus q, then Tn(n)2Ω(nϵ),1 i.e. there is a constant c such that, for sufficiently large n, Tn(n)2c(nϵ). Also, T1(n)poly(n), since the entire s is leaked. So there is some constant c′ such that, for sufficiently large n, there exists n=n(n){2,4,8,16,,n} such that Tn(n)2c(nϵ) and Tn(n)Tn/2(n)n. \n\n### Theorem 1.1\n\n(Informal). Assume n:=n(n)>4,sRq, then:\n\n(1) Dreal,n*,{α}sDsim,n*s,{α} OR\n\n(2) Dreal,n,{ α,(n1)α }sDsim,n,{ α,(n1)α }S OR\n\n(3) Dreal,n,{ α,(2n1)α }sDsim,n*,{ α,(2n1)α }s.\n\nWhile at first glance it may seem that the conclusions (1), (2), (3) are redundant, in fact they are incomparable; Indeed, conclusion (1) does not imply (2) (resp. (3)), since the adversary in (2) (resp. (3)) is given additional leakage. Conversely, conclusion (2) (resp. (3)) does not imply (1), since the set of NTT coordinates that are indistinguishable from random is smaller in (2).\n\nNote that our experimental results show that for our chosen parameter settings s Dreal,4,{1}sDsim,4,{1}s Dreal,8,{1,7}sDsim,8,{1,7}s and Dreal,8,{1,15}sDsim,8,{1,15}s (since we in fact fully recover the secret in all these cases). This indicates that n* ≠ 4 and, if n*= 8 for our chosen parameter settings (as supported by our experiments), then it must be the case that Dreal,8,{1}sDsim,8,{1}s.\n\nApplications.\n\nThe Leaky R-DLWE assumption is a useful tool for analyzing the security of RLWE-based cryptosystems subject to partial key exposure, and guaranteeing a graceful degradation in security. In particular, the Leaky R-DLWE assumption was used to analyze the NewHope protocol of in the ePrint version of this paper . The assumption is applicable to schemes in which the RLWE assumption is used to guarantee that a certain outcome is high-entropy (as opposed to uniform random), such as NewHope without reconciliation .\n\nPracticality of our attack.\n\nWe note that an attack on Leaky R-SLWE yields an attack on standard search R-LWE by guessing each possible leakage outcome, running the Leaky R-SLWE attack and checking correctness of the recovered secret. Therefore, we believe this line of research is interesting beyond the context of leakage resilience, since if the attack can be made to work successfully for sufficiently low leakage rate (far lower than the 1 /4-leakage rate of our attacks), then one could potentially obtain an improved attack on standard search R-LWE.\n\nWe chose to consider partial exposure of the NTT transform of the R- LWEsecret, since in practical schemes the secret key is often stored in the NTT domain and certain types of side-channel attacks allow recovering large portions of the secret key stored in memory. E.g., in their analysis of “cold boot attacks” on NTT cryp-tosystems, Albrecht et al. considered bit-flip rates as low as 0.2%. However, the highly structured leakage required for our attack is unlikely to occur in a practical leakage setting such as a “cold boot attack,” where one expects to recover the values of random locations in memory. We leave open the question of reducing the structure of the leakage in our attack. Specifically, as a starting point it will be interesting to see if our attack can extend to leakage patterns of n′ = 16, |S| = 4 or n′ = 32, |S| = 8, etc. While the leakage rate remains the same (1/4) in each case, these patterns capture leakage that is less and less structured, since at the extreme, one can view leakage of a random 1 /4-fraction of the NTT coordinates as an instance of Leaky R-SLWE with parameters n′ = n and |S| = n/4. \n\n### 1.3 Comparison with Concurrent Work of Bolboceanu et al. \n\nOne of the settings considered by is sampling the RLWE secret from an ideal I ⊆ qR. It is straightforward to see that sampling the RLWE secret uniformly at random from Rq and then leaking the NTT coordinates i such that i = α mod 2n′ is equivalent to sampling the RLWE secret from the ideal I that contains those elements whose NTT transform is 0 in positions i such that i = α mod 2n′.\n\nNevertheless, our decisional assumption is weaker than the assumption of , since require that the entire vector u be indistinguishable from uniform random, whereas we only require that the NTT transform of u is indistinguishable from uniform random at the positions i that are not leaked. Our assumption lends itself to a search-to-decision reduction while the assumption of does not. While do provide a direct security reduction for their decisional assumption, the required standard deviation of the error (in polynomial basis, tweaked and scaled by q) is , ω(q1/nn3/2), which would be far higher than the noise considered in the NewHope and RLWE Challenges settings. In contrast, our assumption can be applied in practical parameter regimes and is sufficient to argue the security of several practical cryptosystems under partial key exposure.\n\nFinally, we compare our attack to that of . For fixed n, q, our attack works for noise regimes that are not covered by the attack of . For example, for NewHope settings of n = 1024, q = 12289, the attack of has success rate at most 1/1000 when the standard deviation of noise distribution is less than 0.00562. In contrast, our attack works (with success ranging from 82/200 to 2/1000) when the standard deviation of the noise is 82.83. Our attack applies only for certain leakage patterns corresponding to certain ideals I, whereas the attack of works for any ideal. The techniques of the two attacks are entirely different. obtain a “good” basis for the ideal via non-uniform advice, perform a change of basis and then use Babai’s roundoff algorithm to solve the resulting BDD instance.We use the algebraic structure of the problem to convert RLWE instances over high dimension into CVP instances over constant dimension n′.We then exactly solve the CVP instances over constant dimension and determine the “high confidence” solutions that are likely to be the correct values of the RLWE error. Assuming all high confidence solutions are correct, we obtain a noiseless system of linear equations w.r.t. the RLWE secret, allowing efficient recovery of the secret.\n\n### 1.4 Related Work\n\nLeakage-resilient cryptography.\n\nThe study of provably secure, leakage-resilient cryptography was introduced by the work of Dziembowski and Pietrzak in . Pietrzak also constructed a leakage-resilient stream-cipher. Brakerski et al. showed how to construct a scheme secure against an attacker who leaks at each time period. There are other works as well considering continual leakage [17, 22]. There are also work on leakage-resilient signature scheme [10, 21, 27].\n\nLeakage-resilience and Lattice-based Cryptography.\n\nGoldwasser et al. , and subsequently [2, 16, 18] studied the leakage resilience of standard LWE based cryptosystems in the symmetric and public key settings.\n\nLeakage Resilience of Ring-LWE.\n\nDachman-Soled et al. considered the leakage resilience of a RLWE-based public key encryption scheme for specific leakage profiles. This was followed by Albrecht et al. , they investigated cold boot attacks and compared the number of operations for implementing the attack when the secret key is stored as polynomial coefficients versus when encoding of the secret key using a number theoretic transform (NTT) is stored in memory. Recently, showed that given multiple samples of RLWE instances such that the public key for every instance lies in some specific subring, one can reduce the original RLWE problem to multiple independent RLWE problems over the subring. In this work we do not place any such restriction on the RLWE samples required to mount partial key exposure attack.\n\n## 2 Preliminaries\n\nFor a positive integer n, we denote by [n] the set {0, . . . , n − 1}. We denote vectors in boldface and matrices using capital letters A. For vector x over ℝn or ℂn, define the 2 norm as x2=(i| xi |2)1/2. We write as ‖x‖ for simplicity. We use the notationt(n),p(n) to indicate that adversaries running in time t(n) can distinguish two distributions with probability at most p(n).\n\nWe present the background and standard definitions related to lattices, algebraic number theory, RLWE, and NTT transform in Appendix A.\n\n## 3 Partial Key Exposure Attack on Ring-LWE\n\n### 3.1 Reconstructing the secret given (α mod 8) leakage\n\nRecall that for pq[x]/(xn+1), the NTT transform, p^, is obtained by evaluating p(x) mod q at the powers ωi for i2n, where ω is a 2n-th primitive root in ℤq. For n{1,2,4,8,,n}, let u=n/n. For α2n, consider puα(x) be the degree u − 1 polynomial that is obtained by taking p(x) modulo (xu(ωα)u) We may assume WLOG that α = 1. We abbreviate notation and write pu, instead of pu1.\n\nWe consider attacks in which the adversary learns all coordinates i of s^ such that i ≡ 1 mod 2n′ where n{1,2,4,8,,n}, and aims to recover the RLWE secret s. First, we note that in NTT transform notation the equation a^s^+e^=u^ holds component-wise. Therefore, given leakage on certain coordinates of s^, we can solve for the corresponding coordinates of  e^ We also get to see multiple RLWE samples (which we write in matrix notation–where the Aj matrices are the circulant matrices corresponding to the ring element aj, s)  as (A1,A1s+e1=u1),,(A,As+e=u). Thus, for the j-th RLWE sample we learn all the coordinates e^ij, for i ≡ 1 mod 2n′. Note that the leaked coordinates are the evaluation of the polynomial eu(x) at the ωi for i ≡ 1 mod 2n′. We can then reconstruct the polynomial eu(x) using Lagrange Interpolation.\n\nFor i{0,,u1}, the (i + 1)-st coefficient of eu(x), i.e. eu,i is equal to\n\nei+ωuei+u+ω2uei+2u++ω(n1)uei+(n1)u\n\nThe coefficients of e can be partitioned into u groups of size n′, forming independent linear systems, each with n′ variables and one equation. Given only the leakage, the set of feasible secret keys is a cartesian product S1××Su, where for i ∈ [u], the set Si is the set of vectors e¯i:={ ei,ei+u,ei+2u,,ei+(n1)u } that satisfy the i-th linear system:\n\n[ 1ωuω2uω(n1)u ][ eiei+uei+2uei+(n1)u ]T=[ eu,i ]\n\nSince each coordinate of e is drawn independently from χ and since each linear system above has small dimension n′, we can use a brute-force-search to find the most likely solution and calculate its probability.\n\nGiven this information, we will carefully choose the solutions ej e¯ij (from all possible sets of solutions [ e¯ij ]j[],i[u] ) that have a high chance of being the correct values of the RLWE error. To obtain a full key recovery attack,we require the following: (1) In total,we must guess at least u number of n′-dimensional solutions, e¯ij, from all the obtained solutions [j [ e¯ij ]j[],i[u]; (2) With high probability all our guesses are correct. Observe that if our guess of some e¯ij is correct, we learn the following linear system of n′ equations and n variables (Aj,is=uj,iej,i), where Aj,i is the submatrix of Aj consisting of the n′ rows i,i+u,i+2u,,i+(n1)u, and uj,i , ej,i are vectors consisting of the i,i+u,i+2u,,i+(n1)u coordinates of uj and ej. So assuming (1) and (2) hold, we learn u noiseless systems of n′ linear equations, each with n = u · n′ number of variables.We then construct a linear system of n variables and n equations, which can be solved to obtain the candidate s.\n\nIn order to ensure that (2) holds, we only keep the guess jfor e¯ij when we have “high confidence” that it is the correct solution. The probability of a particular solution e¯ij:=(eij,ei+uj,,ei+(n1)uj), is the ratio of the probability of e¯ij ibeing drawn from the error distribution (which is coordinate-wise independent) over the sum of the probabilities of all solutions. For small dimension n′, this can be computed via a brute-force method. In our case, we keep the highest probability solution when it has probability at least, say 0.98. The probability that all guesses are correct is therefore 0.98u=0.98n/n.\n\nSince computing the exact probability as above is computationally intensive, we develop a heuristic that performs nearly as well and is much faster. Note that finding the “most likely” solution is equivalent to solving a CVP problem over an appropriate n′-dimensional lattice. We then calculate the probability of the solution under the discrete Gaussian and set some threshold . If the probability of the solution is above the threshold we keep it, if not we discard it. Experimentally, we show that by setting the threshold correctly, we can still achieve high confidence. See Figure 1 for the exact settings of the threshold for each setting of parameters. Our experiments also show that (1) also holds given a reasonable number of RLWE samples. See Section 3.2 for a presentation of our experimental results. We describe our attack in cases where the leakage is on all coordinates i such that iα1 mod 2n or iα2 mod 2n′ in Appendix B.1.", null, "### Figure 1\n\nPerformance of attack against RLWE Challenges and NewHope parameter settings. For each parameter setting, we report the following: min/max and average number of RLWE samples required for successful break, total number of broken instances, and max run-time (in seconds) for successful break. Threshold is set such that the minimal weight solutions to the linear systems given in Section 3 have high confidence with sufficiently high probability.\n\nComplexity of the attack.\n\nWe provide the pseudocode for the attack in Appendix D, Figure 3. While our attack works well in practice, we do not provide a formal proof that our attack is polynomial time for a given setting of parameters. Within the loop beginning on line 5, all the steps (or subroutines) shown in Figure 3 can be computed in polynomial time. Note that even step 12 (CVP.closest_vector), which requires solving a CVP instance, can be computed in polynomial time because for the leakage patterns we consider, the dimension of the CVP instance will always be either 4 or 8–a constant, independent of n. However, our analysis does not bound the number of iterations of the loop beginning on line 5. Specifically,we do not analyze how large the variable RLWESamples must be set in order to guarantee that the attack is successful with high probability. Bounding this variable corresponds to bounding the number of RLWE samples needed in order to obtain a sufficient number of “high confidence” solutions. In practice, the number of RLWE samples was always fewer than 200 for all parameter settings. In future work,we plan to compute the expected number of RLWE samples needed to obtain a sufficient number of high confidence solutions for a given parameter setting. Assuming this expected number of samples is polynomial in n, we obtain an expected polynomial time attack.", null, "### Figure 2\n\nDescription of Attack 1.", null, "### Figure 3\n\nDescription of Partial Key Exposure Attack from Section 3\n\n### 3.2 Experimental Results\n\nWe first assess the performance of our attack on the RLWE challenges published by Crockett and Peikert , with various parameters, ranging from “toy” to “very hard” security levels. For each parameter setting, a cut-and-choose protocol was used by to prove correctness of the challenges: They committed to some number (e.g. N = 32) of independent RLWE instances, a random index i was chosen, and the secret key for all except the i-th instance was revealed. For each of the 31 opened challenges,we simulate the Leaky RLWE experiment and attempt to recover the full secret s using our attack. We next measure the performance of our attack on RLWE instances generated using the dimension, modulus and noise distribution proposed in the original NewHope scheme . These parameters are more conservative than the ones chosen for the later submission to the NIST competition . When multiple RLWE samples are released, bounded error distributions are less secure . We therefore tested our attack in the more difficult setting of Gaussian error, in addition to the original binomial error distribution of .\n\nThe experiments were run using server with AMD Opteron 6274 processor, with a python script using all the cores with Sage version 8.1.We used fplll library for CVP solver and the source code of all the attacks are available online at . The results of our attacks are summarized in Figure 1.We report the total number of instances we broke and the average number of RLWE samples needed for those instances. To decide whether a solution is kept or discarded, its probability mass under the error distribution χ is calculated and compared to the threshold. The threshold for each parameter setting is set heuristically so that minimal weight solutions passing the threshold are correct with high confidence (see Figure 1 for the exact threshold settings). We tested leakage patterns of (n=4,S={1}),(n=8,S={1,7}) and (n=8,δ={1,15}) –all corresponding to 1/4-fraction leakage—for each parameter setting and were able to break multiple Leaky RLWE instances for every parameter setting/leakage pattern shown in Figure 1. We also report the maximum time it took to break a single instance for each parameter setting in Figure 1. Overall, the maximum amount of time to break a single instance was 6 hours for the hardest instance, i.e. Challenge ID 89.We attempted to launch our attack given only 1/8-fraction of leakage (leakage pattern (n′ = 8, α = 1)), but were only successful for the easiest case, i.e. Challenge ID 1. For, e.g. Challenge ID 89, the attack failed since for 5000 number of linear systems, the maximum confidence of any solution was 0.28, meaning that we expect to recover the secret key with probability at most 0.282048/82470, which is well beyond feasible.\n\n## Definition 4.1\n\n(Search RLWE (R-SLWE) with Leakage) The search version of the R-LWE problem with leakage, denoted Leaky RSLWEq,ψ,n,S, is parameterized by (n{1,2,4,8,n},S2n). The experiment chooses sRq uniformly at random, where s=NTT1(s^). The goal of the adversary is to recover s, given independent samples from the distribution Dreal,n,Ss, which outputs (a^,a^s^+e^,[ S^i ]i=αmod2nαS) where a, e are obtained from As,ψ as in standard RLWE (see Definition A.2).\n\n## Definition 4.2\n\n(Decision RLWE (R-DLWE) with Leakage) The decision version of the R-LWE problem with leakage, denoted Leaky RDLWEq,ψ,n,S, is parameterized by (n{1,2,4,8,n},δ2n). The experiment chooses s ← Rq uniformly at random, where s=NTT1(s^). The goal of the adversary is to distinguish between independent samples from the distributions Dreal.n.Ss and Dsim,n,Ss, where Dreal.n.Ss is the same as above, and Dsim,n,δs outputs (a^,u^,[ s^i ]iαmod2nαS), where a, e are obtained from As, Ψ as in standard RLWE (see Definition A.2) and\n\nu^i=a^isi^+ei^iαmod2nαS and u^iq\n\nchosen uniformly random, otherwise.\n\n## 5 Search to Decision Reduction With Leakage\n\nLet the RLWE secret be denoted by s^ and assume WLOG that there exists an adversary that obtains leakage [ s^i ]i1 mod 2n and distinguishes u^=a^s^+e^ from u^, where u^i=a^is^i+e^i for i1 mod 2n′ and otherwise is uniform random . It is not hard to see, using techniques of [23, 24, 25], that this implies an attacker that learns a single index jZȷn,jb mod 2n′ of the RLWE secret, where b1 mod 2n′.We call this the Basic Attack. Due to limited space, we refer readers to Appendix C for description of Basic Attack.\n\n## Theorem 5.1\n\n(Existence of Basic Attack). If, for any (n,SZ2n) adversary A running in time t := t(n) distinguishes Dreal,n,Ss from Dsim,n,Ss with probability p := p(n), then there is some index j such that jα mod n for all αS and an attack Basic Attack with parameters (n,S,j,t,p), that learns NTT coordinate s^j with probability 1 − 1/poly(n) and takes time poly(n) · t · 1/p.\n\nOur attack Attack 1 uses the Basic Attack to learn all the values [ S^i ]ibr mod 2n for r[ n/2 ]. Let s^1:=S^. The main idea of Attack 1 is to learn all [ s^i1 ]ibmod2n in the first round, then apply an automorphism to shift the positions i ≡ b2 mod n′ into the positions i ≡ b mod 2n′, resulting in a permuted RLWE secret, denoted s^2. Note that applying the automorphism i1causes the positions S^i1 such that i ≡ b mod n′ to shift into the positions i ≡ 1 mod 2n′. This means that we are now back where we started, and the reduction is now able to provide the required leakage ( on [ s^i2 ]i1mod2n) to the adversary and thus can learn the values of [ s^i2 ]ibmod2n=[ s^i1 ]ib2 mod n′ iin the second iteration, [ s^i3 ]ibmod2n=[ s^i1 ]ib3 modn in the third iteration, etc. We next formalize the necessary properties of the automorphisms.\n\nFor i, j2n, let ϕij be the automorphism that maps v^ to v^ such that v(ωl)=v(ωj).ϕij induces a permutation on the elements of v^, denoted ρij. Specifically, ϕij(v^) maps v^ to v^ρij() for i, j, 2n, where ρij()=i1j.\n\n## Definition 5.2\n\nA probability distribution ψ:(ζm) is automorphically closed in K if for all i, j ∈ m*,ϕij(ψ)=ψ.\n\nWe remark that RLWE error distribution χ is automorphically closed .\n\nWe formally define Attack 1 in Figure 3 .We next sketch how Attack 1 can be used to complete the proof. For dimenstion n and parameter n{1,2,4,8,n}, let Tn:=Tn(n) be the (non-uniform) time to solve Leaky R-SLWE for dimension n and parameters (n,S={α}={1}), i.e. given positions [ s^i1 ]i1mod2n, with probability 1/2.\n\nAssume subexponential 2Ω(nϵ) hardness of search RLWE without leakage for some constant ϵ ≤ 1 and polynomial modulus q. Then we also have that Tn(n)2Ω(nϵ), and as discussed in the intro, there must exist a constant c′ such that for sufficiently large n, there exists n=n(n){2,4,8,16,,n} such that Tn(n)2c(nϵ) and Tn(n)Tn/2(n)n. The above implies that T(n/2)o(Tn).\n\nNow, if given [ s^i1 ]i1 mod 2n* leakage, there exists a (t(n), p(n))-distinguishing adversary (where t(n)= Tn/poly(n) and p(n)=1/Tn ), then we will show that there is an adversary solving the R-SLWE w.h.p. given positions [ S^i1 ]i1mod2n in time less than Tn, leading to contradiction. We begin by running Attack 1, which takes time at most o(Tn) for our settings of t(n) and p(n). If b2n* is such that for some r[ n/2 ],brn+1 mod2n, then we can combine the reconstructed values of S^i1 from Attack 1 with our knowledge of [ s^i1 ]i1 mod2n* to obtain all values [ S^i1 ]i1modn. This means that we can then run the search attack for 2/n*-fraction of leakage to recover all of s^ in time T(n/2)O(Tn). But then the entire attack for (1 mod 2n*)-leakage can be run in time o(Tn), contradicting the definition of Tn.\n\nFor n* > 4, the only cases in which Attack 1 does not recover [ S^j ]in+1mod2n is when b{ n1,2n 1}. For such b, we do not know how to rule out the possibility that given [ s^i ]i1 mod2n, the positions i ≡ b mod 2n*ofu^ do not look random. In this case, however, we argue that given leakage on both [ S^i ]i1 modn, and [ s^i ]ibmodn, all other positions are indistinguishable from random, since otherwise a modified version of Attack 1 can be run. We next state the formal theorem of this section.\n\n## Theorem 5.3\n\nAssume n:=n(n)>4, sRq, then:\n\nDreal,n,{α}St(n),p(n)Dsim,n,{α}S OR\n\nDreal,n*,{ α,(n*1)α }st(n),p(n)Dsim,n*,{ α,(n*1)α s OR\n\nDreal,n*,{ α,(2n1)α }St(n),p(n)Dsim,n,{ α,(2n1)α }S.\n\nwhere, t(n)=Tn/poly(n),p(n)=1/Tn.\n\nProof. We assume WLOG that α = 1. Assume Dreal,n,{1}s Tn/poly(n),1/TnDsim,ns,{1}. Then this means there must be an adversary A running in time Tn*/poly(n), that distinguishes on index j2n, where j ≡ b mod 2n′ with probability at least 1/Tn.\n\nCase 1: b is such that brn*+1mod2n* for some r[ n/2 ]. In this case, with appropriate setting of poly(n), we can use Attack 1 to recover the positions i such that in+1mod2n (w.h.p.) in time o(Tn). Now we can run the attack that takes as input [ S^i ]i1modn and recovers all of s. By assumption, this attack runs in time T(n/2)O(Tn). Thus, we can to recover the whole s^ (w.h.p. greater than 1/2) in time o(Tn), which is a contradiction.\n\nBy properties of the group 2n, where n* is a power of two, for all b2n\\{ 1,n1,2n1 }, it is the case that br ≡ n* + 1 mod 2n* for some r[ n/2 ]. Thus, Case 1 holds for all b2n*\\{ n1,2n1 }.\n\nCase 2: b = n * − 1. In this case, with appropriate setting of poly(n), we can use Attack 1 to recover the positions i such that in1mod2n (w.h.p.) in time o(Tn). Assume Dreal,ns,1,n1Tn/poly(n),Tn/poly(n)Dsim,ns,1,n1, then there must be some adversary A that distinguishes on index j2n, where jb2n\\{ 1,n1 }. We can combine this with the previous attack as follows:\n\nCase 2(a): b2n\\{ 1,n1,2n1 } Due to essentially the same argument as before, by appropriately setting poly(n), we can (w.h.p.) learn all [ S^i ]i(b)rmod2n for r ∈ [n*/2] in time o(Tn) and then apply the same argument as above.\n\nSpecifically, given the initial leakage [ S^i1 ]i1mod2n, the attack will first learn [ S^i1 ]in1mod2n, then learn [ S^i1 ]ibmod2n, then, for some (j, j′) such that j ≡ b′ mod 2n* and j′ ≡ 1 mod 2n*, apply automorphism ϕii to get s^2, learn [ S^i2 ]in1mod2n, then learn [ S^i2 ]ibmod2n, etc. thus ultimately learning [ S^i ]i(b)rmod2n for r ∈ [n*/2]. At this point, we will have [ S^i ]i1modn and thus can learn all of ^s in additional time T(n/2)o(Tn). iThus, in total the attack takes time o(Tn), leading to contradiction.\n\nCase 2(b): b′ = 2n* − 1. Due to essentially the same argument as before, with appropriate setting of poly(n), we can (w.h.p.) recover the positions i such that i ≡ 2n* − 1 mod 2n* in time o(Tn). The adversary now knows [ S^i ]in1modn. We can thus learn all of ^s in additional time T(n/2)o(Tn). Thus, in total the attack takes time o(Tn), leading to contradiction.\n\nCase 3: b = 2n* − 1. This essentially follows identically to Case 2.  □\n\n# Acknowledgement\n\nThis work is supported in part by NSF grants #CNS-1840893, #CNS-1453045 (CAREER), by a research partnership award from Cisco and by financial assistance award 70NANB15H328 from the U.S. Department of Commerce, National Institute of Standards and Technology.\n\n### References\n\n Source Code 2019, https://github.com/mathcrypt/RLWE Search in Google Scholar\n\n Adi Akavia, Shafi Goldwasser and Vinod Vaikuntanathan, Simultaneous Hardcore Bits and Cryptography against Memory Attacks, in: TCC 2009 (Omer Reingold, ed.), LNCS 5444, pp. 474–495, Springer, Heidelberg, March 2009. Search in Google Scholar\n\n Martin Albrecht, Carlos Cid, Jean-Charles Faugere, Robert Fitzpatrick and Ludovic Perret, Algebraic algorithms for LWE problems, (2014). Search in Google Scholar\n\n Martin R. Albrecht, Amit Deo and Kenneth G. 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Search in Google Scholar\n\n Eric Crockett and Chris Peikert, Challenges for Ring-LWE., IACR Cryptology ePrint Archive 2016 (2016), 782. Search in Google Scholar\n\n Dana Dachman-Soled, Huijing Gong, Mukul Kulkarni and Aria Shahverdi, On the Leakage Resilience of Ideal-Lattice Based Public Key Encryption Cryptology ePrint Archive, Report 2017/1127, 2017, https://eprint.iacr.org/2017/1127 Search in Google Scholar\n\n Dana Dachman-Soled, Huijing Gong,Mukul Kulkarni and Aria Shahverdi, Partial Key Exposure in Ring-LWE-Based Cryptosys-tems: Attacks and Resilience Cryptology ePrint Archive, Report 2018/1068, 2018, https://eprint.iacr.org/2018/1068 Search in Google Scholar\n\n The FPLLL development team, fplll, a lattice reduction library Available at https://github.com/fplll/fplll 2016. Search in Google Scholar\n\n Yevgeniy Dodis, Shafi Goldwasser, Yael Tauman Kalai, Chris Peikert and Vinod Vaikuntanathan, Public-Key Encryption Schemes with Auxiliary Inputs, in: TCC 2010 (Daniele Micciancio, ed.), LNCS 5978, pp. 361–381, Springer, Heidelberg, February 2010. Search in Google Scholar\n\n Yevgeniy Dodis, Kristiyan Haralambiev, Adriana López-Alt and Daniel Wichs, Cryptography against Continuous Memory Attacks, in: 51st FOCS pp. 511–520, IEEE Computer Society Press, October 2010. Search in Google Scholar\n\n Yevgeniy Dodis, Yael Tauman Kalai and Shachar Lovett, On cryptography with auxiliary input, in: 41st ACM STOC (Michael Mitzenmacher, ed.), pp. 621–630, ACM Press, May / June 2009. Search in Google Scholar\n\n Stefan Dziembowski and Krzysztof Pietrzak, Leakage-Resilient Cryptography, in: 49th FOCS pp. 293–302, IEEE Computer Society Press, October 2008. Search in Google Scholar\n\n Shafi Goldwasser, Yael Tauman Kalai, Chris Peikert and Vinod Vaikuntanathan, Robustness of the Learning with Errors Assumption, in: ICS 2010 (Andrew Chi-Chih Yao, ed.), pp. 230–240, Tsinghua University Press, January 2010. Search in Google Scholar\n\n Jonathan Katz and Vinod Vaikuntanathan, Signature Schemes with Bounded Leakage Resilience, in: ASIACRYPT 2009 (Mit-suru Matsui, ed.), LNCS 5912, pp. 703–720, Springer, Heidelberg, December 2009. Search in Google Scholar\n\n Allison B. Lewko, Mark Lewko and Brent Waters, How to leak on key updates, in: 43rd ACM STOC (Lance Fortnow and Salil P. Vadhan, eds.), pp. 725–734, ACM Press, June 2011. Search in Google Scholar\n\n Vadim Lyubashevsky, Search to decision reduction for the learning with errors over rings problem, in: 2011 IEEE Information Theory Workshop, ITW 2011, Paraty, Brazil, October 16-20, 2011 pp. 410–414, 2011. Search in Google Scholar\n\n Vadim Lyubashevsky, Chris Peikert and Oded Regev, On Ideal Lattices and Learning with Errors over Rings, in: EURO-CRYPT 2010 (Henri Gilbert, ed.), LNCS 6110, pp. 1–23, Springer, Heidelberg, May / June 2010. Search in Google Scholar\n\n Vadim Lyubashevsky, Chris Peikert and Oded Regev, On Ideal Lattices and Learning with Errors over Rings, J. ACM 60 (2013), 43:1–43:35. Search in Google Scholar\n\n Vadim Lyubashevsky, Chris Peikert and Oded Regev, A Toolkit for Ring-LWE Cryptography Cryptology ePrint Archive, Report 2013/293, 2013, http://eprint.iacr.org/2013/293 Search in Google Scholar\n\n Tal Malkin, Isamu Teranishi, Yevgeniy Vahlis and Moti Yung, Signatures Resilient to Continual Leakage on Memory and Computation, in: TCC 2011 (Yuval Ishai, ed.), LNCS 6597, pp. 89–106, Springer, Heidelberg, March 2011. Search in Google Scholar\n\n Chris Peikert, How(Not) to Instantiate Ring-LWE, in: SCN 16 (Vassilis Zikas and Roberto De Prisco, eds.), LNCS 9841, pp. 411–430, Springer, Heidelberg, August / September 2016. Search in Google Scholar\n\n Krzysztof Pietrzak, A Leakage-Resilient Mode of Operation, in: EUROCRYPT 2009 (Antoine Joux, ed.), LNCS 5479, pp. 462–482, Springer, Heidelberg, April 2009. Search in Google Scholar\n\n Katherine E. Stange, Algebraic aspects of solving Ring-LWE, including ring-based improvements in the Blum-Kalai-Wasserman algorithm Cryptology ePrint Archive, Report 2019/183, 2019, https://eprint.iacr.org/2019/183 Search in Google Scholar\n\n#### A 1 Algebraic Number Theory\n\nFor a positive integer m, the mth cyclotomic number field is a field extension K=(ζm) obtained by adjoining an element ζm of order m (i.e. a primitive mth root of unity) to the rationals.\n\nRing of Integers R and Its Dual R\n\nLet R ⊂ K denote the set of all algebraic integers in number field K defined above. This set forms a ring (under the usual addition and multiplication operations in K), called the ring of integers of K.\n\nAn (integral) ideal JR is a non-trivial (i.e. J and J{0}) additive subgroup that is closed under multiplication by R, i,e., raJ for any r ∈ R and aJ.\n\n#### Definition A.1\n\nFor R=[ ζm ], define g=p(1ζp)R, where p runs over all odd primes dividing m. Also, define t=m^gR where m^=m2 if m is even, otherwise m^=m.\n\nThe dual ideal R∨ of R is defined as R= t1 , satisfying RRm^1R. The quotient Rq is defined as Rq=R/qR.\n\n#### A 2 Ring-LWE\n\nWe next present the formal definition of the RLWE problem as given in .\n\n#### Definition A.2\n\n(RLWE Distribution) For a “secret\" sRq (or just R) and a distribution χ over K, a sample from the RLWE distribution As,χ over Rq×(K/qR) is generated by choosing aRq uniformly at random, choosing e ← χ, and outputting (a,b=as+emodqR).\n\n#### Definition A.3\n\n(RLWE, Average-Case Decision) The average-case decision version of the RLWE problem, denoted R-DLWEq,χ, is to distinguish with non-negligible advantage between independent samples from As,χ, where sRa is sampled uniformly at random, and the same number of uniformly random and independent samples from Rq×(K/qR)\n\n#### Theorem A.4\n\n[26, Theorem 2.22] Let K be the mth cyclotomic number field having dimension n=φ(m) and R=OK be its ring of integers. Let α=α(n)>0, and q=q(n)2,q=1modm be a poly(n)-bounded prime such that αqω(logn). Then there is a polynomial-time quantum reduction from O˜(n/α) -approximate SIVP (or SVP) on ideal lattices in K to the problem of solving R-DLWEq,χ given only l samples, where χ is the Gaussian distribution Dξ for ξ=αq(nl/log(nl))1/4.\n\nA Note on the Tweak.\n\nIn , Alperin-Sheriff and Peikert show that an equivalent “tweaked\" form of the Ring-LWE problem can be used in cryptographic applications without loss in security or efficiency. This is convenient since the “tweaked\" version does not involve R. The “tweaked\" ring-LWE problem can be obtained by implicitly multiplying the noisy products b by the “tweak\" factor t, and, as it is explained in , tR=R. This yields new values\n\nb=tb=(ts)a+(te)=sa+emodqR,\n\nwhere a,s=tsRq, and the errors e = t · e come from the “tweaked\" error distribution t · χ.\n\n#### A 3 Number Theoretic Transform (NTT)\n\nLet Rq:=q[x]/xn+1 be the ring of polynomials, with n = 2d for any positive integer d. Also, let m = 2n and q = 1 mod m. For, ω a mth root of unity in ℤq the NTT of polynomial p= i=0n1pixiRq is defined as,\n\np^=NTT(p):=i=0n1p^ixi\n\nwhere the NTT coefficients p^i are defined as: p^i= j=0n1pjωj(2i+1).\n\nThe function NTT−1 is the inverse of function NTT, defined as\n\np=NTT1(p^):=i=0n1pixi\n\nwhere the NTT inverse coefficients pi are defined as: pi=n1 j=0n1pj^ωi(2j+1).\n\n### B Attack Algorithm for Other Leakage Patterns\n\n#### B.1 Reconstructing the secret given (α1, α2 mod n′) leakage\n\nLet euα(x) be the degree u = n /n polynomial that is obtained by taking e(x) modulo xu(ωα)u. We consider two polynomials euα1(x) and euα2(x). We may assume WLOG, α1 = 1.We therefore set α:=α2. For i{0,,u1}, the (i + 1)-st coefficient of eu(x) and euα(x) are as follows, respectively\n\nei+ωuei+u+ω2uei+2u++ω(n1)uei+(n1)uei+ωαuei+u+ωα2uei+2u++ωα(n1)uei+(n1)u\n\nSimilar to the previous attack, we obtain the following constraints on the error, given leakage on the secret key and an RLWE sample,\n\n1ωuω2uω(n1)u1ωαuωα2uωα(n1)ueiei+uei+2uei+(n1)u=eu,ieu,iα\n\nWe solve a corresponding CVP instance to find the “most likely” solution, e¯j for ( ei,ei+μ,ei+2μ,, ei+(n1)μ since the “most likely” solution is the one with smallest norm.\n\nSimilar to our previous attack, our goal is to carefully choose the answers with “high confidence” such that (1) In total, we must guess at least u number of n′-dimensional solutions, e¯ij, from all the obtained solutions [ e¯ij ]j[],i[u]; (2) With high probability all our guesses are correct. We choose the candidate which has probability of at least, say, 0.95 of being correct solution. The total probability of success for this case is 0.95u=0.95n/n.\n\nOur experiments in section 3.2 again show that we can obtain enough “high” confidence solutions, without requiring too large a number of RLWE instances.\n\n### C Description of Basic Attack\n\nIn this section, we present the Basic Attack, following the description from [23, 24, 25] and using the fact that NTT coefficients form a CRT representation. We first recall definition of CRT representation in our setting of parameters.\n\n### Definition C.1\n\n(CRT Representation) For p ∈ Rq, and ω a mth primitive root of unity in q, CRT representation for p is defined as\n\nCRT(p)=(p(ωj1),,p(ωjn)),\n\nfor jim.\n\nIt is easy to see that CRT(p)=(p^0,,p^n1).\n\nWe first introduce the following definition:\n\n### Definition C.2\n\n(Hybrid Leaky RLWE Distribution) For j2n={1,3,,2n1}, a “secret\" s ∈ Rq, and a distribution χ over Rq, a sample from the distribution Drealn,Ss,j is generated by choosing (a^,b^)Dreal,n,Ss and outputting (a^,b^+u), where u=(u1,u3,,u2n1)qn with ui, i2n defined as follows: ui is chosen uniformly at random from χq if iα′ mod 2n′ for all αS and ij, ui = 0 otherwise.\n\nDefine Drealn,Ss,1:=Dreal,n,Ss. Additionally, notice that Dreal,n,Ss,2n1=Dsim,n,Ss. Thus if, for any (n,δ2n) adversary A running in time t := t(n) distinguishes Dreal.n.SS from Dsim,n,Ss with probability p := p(n), then there is some index j2n such that jα′ mod n for all αS and a distinguisher 𝒟j that is able to distinguish between the distribution Dreal,n,Ss,j2 and Dreal.n,Ss,j with probability at least p /n.\n\nWe now show the distinguisher 𝒟j can be used to construct an algorithm that finds the value of S^j. The idea of this algorithm is to try each of the possible values S^j, constructing the samples on inputs from Dreal.n,Ss, so that the samples are distributed according to Dreal,n,Ss if S^j is guessed correctly, and the samples are distributed according to Drealn,Ss,j otherwise. Then using the distinguisher 𝒟j poly(n/p) times for each of the q(= poly(n)) guesses for S^i, we are able to find the correct value of S^j with probability 1 −1/poly(n) in time tpoly(n)1/p.\n\nNext we present the samples construction algorithm that takes a guess g ∈q and transform Dreal,n,Ss to either Dreal, n,Ss,j2 or Dreal,n,Ss,j On each sample (a^,b^)Dreal,n,Ss, it outputs a sample\n\n(a,b)=(a^+v,b^+u+gv),\n\nwhere u=( u1,u3,,um1) v=(v1,v3,,vm1)qn are chosen as follows: uk is uniform in ℤq if k < j, kα′ mod 2n′ for all αS, and the rest are 0; vk is uniform in ℤq if k = j, and the rest are 0. Note that bj can be written as\n\nbj=a^js^j+e^j+uj+gvj=ajs^j+e^j+uj+(gs^j)vj.\n\nObserve that if g is the correct guess, then (gs^j)vj=0. The distribution of (a , b) is identical to Dreal,n,Ss,j2. If g is a wrong guess, (gsj) is non-zero. Since q is prime, (gs^j)vj is uniform in ℤq. Thus the distribution of (a , b) is identical to Dreal,n,Ss,j." ]

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[ "# IMF Reports on World Economies 2020 – 2021\n\nMany News reports are highlighting the negative side of the very recent IMF economic reports, taking account of the current COVID-19 pandemic.\n\nAn example being the IMF’s prediction of an 8.1% shrinking of the Lithuanian economy, in 2020.\n\nBut the headlines miss the projected 2021 recovery of 8.2% in 2021.\n\nThis results in a slight increase by 2021 of 0.1%.\n\nTen Countries show an increase by 2021, according to IMF figures, and these are:\n\n1. Indonesia down by 0.5% in 2020, then up by 8.2% in 2021, giving a total increase of 7.7% by 2021.\n2. Malaysia down by 1.7% in 2020, then up by 9% in 2021, giving a total increase of 7.3% by 2021.\n3. Serbia down by 3% in 2020, then up by 7.5% in 2021, giving a total increase of 4.5% by 2021.\n4. Malta down by 2.8% in 2020, then up by 7% in 2021, giving a total increase of 4.2% by 2021.\n5. Korea down by 1.2% in 2020, then up by 3.4% in 2021, giving a total increase of 2.2% by 2021.\n6. Bulgaria down by 4% in 2020, then up by 6% in 2021, giving a total increase of 2% by 2021.\n7. Hungary down by 3.1% in 2020, then up by 4.2% in 2021, giving a total increase of 1.1% by 2021.\n8. Saudi Arabia down by 2.3% in 2020, then up by 2.9% in 2021, giving a total increase of 0.6% by 2021.\n9. Estonia down by 7.5% in 2020, then up by 7.9% in 2021, giving a total increase of 0.4% by 2021.\n10. Lithuania down by 8.1% in 2020, then up by 8.2% in 2021, giving a total increase of 0.1% by 2021.\n\nThe remainder show a decrease, from as low as 0.1% up to a decrease of 4.9%, by 2021.\n\nThese are:\n\n1. Luxembourg down by 4.9% in 2020, then up by 4.8% in 2021, giving a total reduction of 0.1% by 2021.\n2. Latvia down by 8.6% in 2020, then up by 8.3% in 2021, giving a total reduction of 0.3% by 2021.\n3. Poland down by 4.6% in 2020, then up by 4.2% in 2021, giving a total reduction of 0.4% by 2021.\n4. Singapore down by 3.5% in 2020, then up by 3% in 2021, giving a total reduction of 0.5% by 2021.\n5. Ireland down by 6.8% in 2020, then up by 6.3% in 2021, giving a total reduction of 0.5% by 2021.\n6. Denmark down by 6.5% in 2020, then up by 6% in 2021, giving a total reduction of 0.5% by 2021.\n7. Taiwan down by 4% in 2020, then up by 3.5% in 2021, giving a total reduction of 0.5% by 2021.\n8. Australia down by 6.7% in 2020, then up by 6.1% in 2021, giving a total reduction of 0.6% by 2021.\n9. Thailand down by 6.7% in 2020, then up by 6.1% in 2021, giving a total reduction of 0.6% by 2021.\n10. Cyprus down by 6.5% in 2020, then up by 5.6% in 2021, giving a total reduction of 0.9% by 2021.\n11. Hong Kong down by 4.8% in 2020, then up by 3.9% in 2021, giving a total reduction of 0.9% by 2021.\n12. Nigeria down by 3.4% in 2020, then up by 2.4% in 2021, giving a total reduction of 1% by 2021.\n13. Romania down by 5% in 2020, then up by 3.9% in 2021, giving a total reduction of 1.1% by 2021.\n14. United States down by 5.9% in 2020, then up by 4.7% in 2021, giving a total reduction of 1.2% by 2021.\n15. Slovak Republic down by 6.2% in 2020, then up by 5% in 2021, giving a total reduction of 1.2% by 2021.\n16. Iceland down by 7.2% in 2020, then up by 6% in 2021, giving a total reduction of 1.2% by 2021.\n17. New Zealand down by 7.2% in 2020, then up by 5.9% in 2021, giving a total reduction of 1.3% by 2021.\n18. Germany down by 7% in 2020, then up by 5.2% in 2021, giving a total reduction of 1.8% by 2021.\n19. South Africa down by 5.8% in 2020, then up by 4% in 2021, giving a total reduction of 1.8% by 2021.\n20. Canada down by 6.2% in 2020, then up by 4.2% in 2021, giving a total reduction of 2% by 2021.\n21. Russia down by 5.5% in 2020, then up by 3.5% in 2021, giving a total reduction of 2% by 2021.\n22. Japan down by 5.2% in 2020, then up by 3% in 2021, giving a total reduction of 2.2% by 2021.\n23. Switzerland down by 6% in 2020, then up by 3.8% in 2021, giving a total reduction of 2.2% by 2021.\n24. Belgium down by 6.9% in 2020, then up by 4.6% in 2021, giving a total reduction of 2.3% by 2021.\n25. Brazil down by 5.3% in 2020, then up by 2.9% in 2021, giving a total reduction of 2.4% by 2021.\n26. UK down by 6.5% in 2020, then up by 4% in 2021, giving a total reduction of 2.5% by 2021.\n27. Austria down by 7% in 2020, then up by 4.5% in 2021, giving a total reduction of 2.5% by 2021.\n28. Belarus down by 6% in 2020, then up by 3.5% in 2021, giving a total reduction of 2.5% by 2021.\n29. Slovenia down by 8% in 2020, then up by 5.4% in 2021, giving a total reduction of 2.6% by 2021.\n30. France down by 7.2% in 2020, then up by 4.5% in 2021, giving a total reduction of 2.7% by 2021.\n31. Euro Area down by 7.5% in 2020, then up by 4.7% in 2021, giving a total reduction of 2.8% by 2021.\n32. Finland down by 6% in 2020, then up by 3.1% in 2021, giving a total reduction of 2.9% by 2021.\n33. Portugal down by 8% in 2020, then up by 5% in 2021, giving a total reduction of 3% by 2021.\n34. Mexico down by 6.3% in 2020, then up by 3% in 2021, giving a total reduction of 3.3% by 2021.\n35. Spain down by 8% in 2020, then up by 4.3% in 2021, giving a total reduction of 3.7% by 2021.\n36. Ukraine down by 7.7% in 2020, then up by 3.6% in 2021, giving a total reduction of 4.1% by 2021.\n37. Croatia down by 9% in 2020, then up by 4.9% in 2021, giving a total reduction of 4.1% by 2021.\n38. Italy down by 9.1% in 2020, then up by 4.8% in 2021, giving a total reduction of 4.3% by 2021.\n39. Netherlands down by 7.5% in 2020, then up by 3% in 2021, giving a total reduction of 4.5% by 2021.\n40. Greece down by 10% in 2020, then up by 5.1% in 2021, giving a total reduction of 4.9% by 2021.\n\nTurkey is missing from the above figures, as they are neither an increase or a decrease.  The IMF predicts a 5% decrease in 2020 and a 5% increase in 2021, resulting in no real change.\n\nFive more countries are not even having a predicted drop in 2020, and show these figures:\n\n1. China up by 1.2% in 2020, then up by 9.2% in 2021, giving a total increase of 8% by 2021.\n2. Indonesia up by 0.5% in 2020, then up by 8.2% in 2021, giving a total increase of 7.7% by 2021.\n3. Philippines up by 0.6% in 2020, then up by 7.6% in 2021, giving a total increase of 7% by 2021.\n4. India up by 1.9% in 2020, then up by 7.4% in 2021, giving a total increase of 5.5% by 2021.\n5. Vietnam up by 2.7% in 2020, then up by 7% in 2021, giving a total increase of 4.3% by 2021.\n\nSource Figures: International Monetary Fund (IMF) – World Economic Outlook, April 2020\n\n63.1 - 779,360" ]

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[ "CHAPTER 05.18: SYSTEM OF EQUATIONS: Some statements about the inverse of matrices     Let’s talk about some statements about inverse of matrices. If A and B are two N by N matrices- we are talking about square matrices- such that B times A is equal to I and then we can also say these things also: We can say B is inverse of A. We can say A is inverse of B. That A and B are invertible; that means they can be inverted or you can find the inverse. And also we can say, A times B is equal to I. Although we are able to say hey, B times A is equal to I then, B is the inverse of A or A is inverse of B, but it also means that A times B will be equal to the anti-matrix. We can also say that A and B are nonsingular. And then we can also say that all columns of A and B are linearly independent. So if we take each-each of the columns of A or B, as vectors, they will be linearly independent. And so is it true for rows. All rows of A and B are linearly independent. So these are some things to think about when you are talking about the inverse of a matrix: to be able to say hey, if B times A is the anti-matrix and B is inverse of A and A is inverse of B, A and B are invertible, A times B will be the anti-matrix, both A and B are considered to be nonsingular, that the columns of A and B are linearly independent, so are the rows of A and B, which are linearly independent. And that is the end of this segment." ]

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[ "HomeWorksheet Template ➟ 0 Creative Difference Between Worksheet And Spreadsheet\n\nCreative Difference Between Worksheet And Spreadsheet\n\nMost spreadsheet software has basic features to help users create edit and format worksheets. What Is SpreadsheetWorksheet And Workbook In Hindi Difference Between Spreadsheet And WorkSheet – YouTube.", null, "Difference Between System Software And Application Software Pediaa Com Computer Basics Software Computer Learning\n\nA worksheet is the grid of columns and rows that information is inputted into.", null, "Difference between worksheet and spreadsheet. In case there is no difference in the values it will return blank and in case there is a difference it will return the values from both the sheets in separate lines in the same cell. The worksheet object is also a member of the sheets collection. Its easier to use a workbook for extensive data.\n\nThe good thing with this method is that it only gives you the differences and show you exactly what the difference is. Worksheets can be named using the sheet tabs of the bottom of the spreadsheet window. To move between worksheets.\n\nExcel sheet is a single spread sheet that contains cells organized by rows and columns A worksheet begins with row1 and columnA. Worksheet is the single page which contains data. While the rows and columns in a spreadsheet collectively are called a worksheet.\n\nThe data in the workbook is a general form of data while in the worksheet is a specific form of data. Select the range in one of both worksheets you want to highlight differences and click Home. A worksheet is a page of the workbook on which all the data is held.\n\nIt features columns and rows to keep inserted information legible and simple to understand. 10 rows The Worksheet is a single-page spreadsheet. An Excel Workbook is a file that contains multiple single-page spreadsheets while a worksheet is a single-page worksheet.\n\nWhile the rows and columns in a spreadsheet collectively are called a worksheet. Spreadsheet can be a single worksheet or multiple worksheets. Create a new worksheet by doing one of the following.\n\nWorksheet is the basic unit of an program which get to gather to make a workbook and spreadsheet contain many workbooks. Difference Between Spreadsheet and Worksheet is that spreadsheet software allows users to organize data in rows and columns and perform calculations on the data. Each cell can contains numbertext or formula A cell can also refrence another cell in the same worksheet.\n\nSpreadsheet is the main working space which contains workbook and worksheets. A workbook is just a file or a. Fundamental difference the worksheet is single page spreadsheet or page in excel where you can write edit and manipulate data whereas the collection of such worksheets is referred as a workbook.\n\nA workbook can contain at minimum one worksheet and up to a very large number of worksheets that is only limited by the available memory on your computer. With the Conditional Formatting utility you can highlight the differences between two sheets in the same workbook. A spreadsheet shows a collection of worksheets that combine to form a workbook.\n\nWhat Is SpreadsheetWorksheet And Workbook In Hindi Difference Between Spreadsheet. It can have a numeric value text formulas and functions. A workbook is just a file or a book.\n\nDifference Between Spreadsheet and Worksheet is that spreadsheet software allows users to organize data in rows and columns and perform calculations on the data. The workbook is the entire book while the worksheet is a single page within that book. Data such as numbers formulas and text are stored in a spreadsheet.\n\nDifference Between Spreadsheet and Workbook is that Spreadsheet software allows users to organize data in rows and columns and perform calculations on the data. Difference between a workbook worksheet and spreadsheet. Highlight differences between two sheets in the same workbook.\n\nScrolling through the worksheet displays different parts of it on the screen. Difference Between Workbook and Worksheet in Excel Easy. 10 rows WorkbookThe Worksheet is a single-page spreadsheet.\n\nA spreadsheet is more like a file or workbook. Work sheet it is mainly the data you applied. A file that exists of cells in rows and columns and can help arrange calculate and sort data is known as Spreadsheet.\n\nSpreadsheet further contains work books and each workbook has three worksheets. It consists of a matrix of rectangular cells organized in a tabular form of rows and columns. The excel workbook the difference between an excel workbook vs.\n\nWorksheets can be added deleted or renamed within the workbook. Difference between spreadsheet and worksheet is that spreadsheet software allows users to organize data in rows.", null, "", null, "Compare Two Excel Worksheets With Vba Comparing Data Excel Workbook", null, "", null, "" ]

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[ "# $a_{n+1}=a_n^{n+1}+a_n+1$\n\nLet $a_1>1$ and $$a_{n+1}=a_n^{n+1}+a_n+1, \\: \\forall n \\geq 1$$ Edit: Prove that there is a real number $x \\neq 0$ such that $$\\lim_{n \\to \\infty}\\frac{a_n}{x^{n!}}=1$$\n\nI think that $x=a_1$ works, and I tried to use the Squeeze Theorem to prove that. From $a_{n+1}>a_n^{n+1}>a_{n-1}^{n(n+1)}>\\dots>a_1^{(n+1)!}$ we get that $1<\\frac{a_{n+1}}{a_1^{(n+1)!}}$ and then I tried to find another bound for $a_{n+1}$ such that $\\frac{a_{n+1}}{a_1^{(n+1)!}}<b_{n+1} \\to 1$, but nothing seems to work.\n\n• No, that doesn't work, and the formulation is misleading. You can show the existence of such an $x$, but you can't find it without computing the whole sequence. – Professor Vector Jan 12 '18 at 11:54\n• Okay, I'll edit the statement! – Shroud Jan 12 '18 at 11:55\n\n## 2 Answers\n\nLet\n\n$$x = \\lim_{n\\to\\infty} a_n^{1/n!}.$$\n\nWe know from @frame95's answer that $x$ exists. Now set $\\delta_n = \\log a_n - n! \\log x$ and notice that\n\n$$\\frac{\\delta_{n+1}}{(n+1)!} = \\frac{\\delta_n}{n!} + \\frac{1}{(n+1)!}\\log\\left( 1 + \\frac{1}{a_n^n} + \\frac{1}{a_n^{n+1}} \\right).$$\n\nSince we know that $\\delta_n/n! \\to 0$, recursively applying the above relation and taking limit shows that\n\n$$\\frac{\\delta_n}{n!} = -\\sum_{k = 0}^{\\infty} \\frac{1}{(n+k+1)!}\\log\\left( 1 + \\frac{1}{a_{n+k}^{n+k}} + \\frac{1}{a_{n+k}^{n+k+1}} \\right)$$\n\nNow by utilizing simple estimations $a_{n+k} \\geq a_n^{k!} \\geq a_1^{k!}$ and $\\log(1+t) \\leq t$ for $t \\geq 0$, we obtain the following very crude bound\n\n$$\\left| \\delta_n \\right| \\leq \\sum_{k=0}^{\\infty} \\frac{n!}{(n+k+1)!} \\cdot 2 a_1^{-k!}.$$\n\nSince each summand is bounded by $2a_1^{-k!}$ and $\\sum_{k=0}^{\\infty} 2a_1^{-k!} < \\infty$, dominated convergence theorem tells that\n\n$$\\limsup_{n\\to\\infty} \\left| \\delta_n \\right| \\leq \\sum_{k=0}^{\\infty} \\lim_{n\\to\\infty} \\frac{n!}{(n+k+1)!} \\cdot 2 a_1^{-k!} = 0.$$\n\nTherefore $\\delta_n \\to 0$ and hence $a_n!/x^{n!} = e^{\\delta_n} \\to 1$ as required.\n\n• Wow! Good (team) job guy :) very neat proof! – Andrea Marino Jan 13 '18 at 10:36\n\nThe guess is wrong, but if you go a step further... Let's stop the chain of inequalities a bit before: $$a_n> a_{n-1}^n > \\ldots > a_2 ^{n!/2} = (\\sqrt{a_2} )^{n!}$$ Is this a better estimate? Yes, because $\\sqrt{a_2 }=\\sqrt{ a_1^2 + a_1+1} > a_1$. Let's call $d_r = a_r^{1/r!}$. Generalizing this argument, we get $$a_{r+k} > a_{r+k-1}^{r+k} > \\ldots > a_r^{ (r+k)!/r!} = d_r^{(r+k)!}$$ Let's analyze $\\{d_r\\}$. We will show that it is an increasing and bounded sequence, thus convergent to a number $\\ell$. Infact $$(1) \\ \\ increasing \\ \\ \\ d_{r+1} = a_{r+1}^{1/(r+1)!} = ( a_r^{r+1} +a_r+1)^{1/(r+1)!} > d_r$$ By induction, one can easily show that $a_n +1 < (a_1+1) a_n^{n+1}$. Thus, $a_{n+1} < (a_1+2)a_n^{n+1}$. This yields $$(2) \\ \\ bounded \\ \\ \\ d_{n+1} < (a_1+2)^{1/(n+1)!} d_n < 2^{\\sum^{n+1} (1/k!)} a_1/2 < 2^{e-1} a_1$$ If $x$ exists, then $x=\\ell$. Infact $$\\lim_{n \\to \\infty} \\frac{a_n}{x^{n!}} = \\lim_{n \\to \\infty} \\left ( \\frac{d_n}{x} \\right )^{n!} = 1 \\ \\Rightarrow \\ \\frac{\\ell}{x} = \\lim_{n \\to \\infty} \\frac{d_n}{x} = 1$$ Here my results stop: I can't show the refined result that this $x$ exist!" ]

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[ "# Power Measurement with the Analog Discovery Pro (ADP5250)\n\n## Overview\n\nThis demo uses the Analog Discovery Pro's Oscilloscope to measure the power of an RC circuit. Because the oscilloscope only measures voltages, a sensing resistor is used with the Oscilloscope's Math channels to calculate the current passing through the circuit and further the power of each component.\n\n## Inventory\n\n• Analog Discovery Pro (ADP5250) with:\n• 2 BNC oscilloscope probes\n• A BNC to minigrabber clip cable\n• Analog parts and wires (found in the Analog Parts Kit):\n• A 470 Ω resistor\n• A 4.7 nF capacitor\n• A 1 Ω resistor (sensing resistor)\n• WaveForms Installation, version 3.15.18 or later\n\n## Demo Setup\n\n##### Hardware Setup\n\nAssemble the circuit under test shown in the image to the right and connect the Oscilloscope probes and Wavegen grippers at the labeled nodes. Because the BNC probes are single-ended and don't support differential voltage measuring, care must be taken to ensure each probe ground clip is connected to the ground of the circuit.\n\nThis demo assumes the analog parts listed in the inventory are the ones being used, but it can be recreated with components of different values with adjustments made to the windowing and Math functions. Note that the sensing resistor needs to be small enough to have minimal impact on the behavior of the overall circuit.\n\n##### Software Setup\n\nOpen the Wavegen instrument, configuring it to output a 5 V, 20 kHz Sine wave as shown in the image to the right.\n\nOpen the Scope instrument. Enable Channels 1 and 2, then adjust the ranges to match the image to the right. Adjust the ranges of each channel to 5 mV/div and 1 V/div respectively, and the base time division to 10 us/div.\n\nClicking the “Docking Windows” button in the upper right on WaveForms will cause WaveForms to show the Wavegen and Scope instruments in the same window. This is useful in contexts where multiple instruments will be used in quick & repeated succession.\n\nWithout leaving the Scope instrument, Run All Instruments by clicking the button in the upper right corner of WaveForms, which will start the Wavegen and the Scope instruments at the same time.\n\nWhat should be seen at this stage, is the voltage drop across the sensing resistor (Ch 1), the voltage drop across the entire circuit (Ch 2 position 1), and the voltage drop across the two resistors (Ch 2 position 2), as seen in the image to the right. Since the ADP5250 has only 2 channels we have take one power calculation, and then move channel 2 to position 2 and make the other calculation. The current across each component can be calculated using the Scope's Math Channels, which conveniently plots the calculation alongside the voltage measurements. These next steps detail how to set them up.\n\nBecause oscilloscopes only measure voltage, some work needs to be done in order to measure power. The Scope's Math channels can be used to perform calculations which then are plotted alongside each channel's measurements. To calculate the total power of the RC circuit, the voltage across the sensing resistor (Channel 1) is subtracted from the voltage across the entire circuit (Channel 2), and the result is multiplied by the current passing through the circuit. The current is found by dividing the voltage across the sensing resistor by its resistance, and that result is multiplied against the voltage across the RC components.\n\nTo add a math channel, click “Add Channel” dropdown menu above the oscilloscope channels and select “Custom” under the Math menu. In the new window, enter “(C2-C1)*C1/Rs”, where Rs is the resistance value of your current sensing resistor. In the script entry, set the units to “W”, and click “OK”. Adjust the range of the Math channel to 20 mW/div to better view the signal. This Math channel is plotting the total power of the circuit.\n\nThe power across the resistor can be calculated as well. To do so, the voltage across the sensing resistor is subtracted from the voltage across both resistors (Channel 2 Position 2), and the result multiplied by the current passing through the resistor.\n\nAdd another Math channel, this time with the calculation of “(C2-C1)*C1/Rs”. If the chart is too cluttered with signals, click the checkmarks in the upper left of the 2 Scope channels to stop them from being plotted. Set the units to “W” and adjust the range of this channel to 20 mW/div to view the power of the resistor component of the circuit. Select Math1 and then drag the red triangle on the left of the chart down to offset the channel from Math2, since they largely overlap.\n\nNote: Scope measurement and math channels can be renamed and relabeled in order to clarify what they represent. This can be done by expanding the channel options (the gear icon) and editing the Name and Label properties of the channel.\n\nTo see the average value of each power measurement, click “View” in the top toolbar, then “Measurements”. In the Measurements panel, click the “Add” button then “Defined Measurement”. Select Math1 in the left panel, expand Vertical in the right then click Average. Click “Add”, then repeat for Math2. Once done, click “Close”.\n\nIn the Wavegen, adjust the frequency of the Sine to 50 Hz, adjust the time base of the Scope instrument to 5 ms/div to accommodate the frequency change. Note the changes in the averages in the Measurements panel.\n\n## Other Resources\n\nFor more guides and example projects for your Analog Discovery Pro (ADP5250), please visit its Resource Center.\n\nFor technical support, please visit the Digilent Forums." ]

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[ "Click to Chat\n\n1800-1023-196\n\n+91-120-4616500\n\nCART 0\n\n• 0\n\nMY CART (5)\n\nUse Coupon: CART20 and get 20% off on all online Study Material\n\nITEM\nDETAILS\nMRP\nDISCOUNT\nFINAL PRICE\nTotal Price: Rs.\n\nThere are no items in this cart.\nContinue Shopping\n` A particle having mass 10 g oscillates according to the equation x = (2.0 cm) sin[(100 s-1)t + π/6]. Find (a) the amplitude, the time period and the spring constant (b) the position, the velocity and the acceleration at t = 0.`\n5 years ago\n\n```\t\t\t\t\t\t\tSol. x = (2.0cm)sin [(100s–1) t + (π/6)]\nm = 10g.\na) Amplitude = 2cm.\nω = 100 sec–1\n∴ T = 2π/100 = π/50 sec = 0.063 sec.\nWe know that T = 2π √(m/k) ⇒ T2 = 4π2 × m/k ⇒ k = 4 π2/T2 m\n= 105 dyne/cm = 100 N/m. [because ω = 2π/T = 100 sec-1]\n\n(b) At = 0\nx = 2cm sin (π/6) = 2 × (1/2) = 1 cm. from the mean position.\nWe know that x = A sin (ωt + ∅)\nv = A cos (ωt + ∅)\n= 2 × 100 cos (0 + π/6) = 200 × √3/2 = 100 √3 sec-1 = 1.73m/s\nc) a = – ω2 x = 1002 × 1 = 100 m/s2\n\n```\n5 years ago\nThink You Can Provide A Better Answer ?\n\n## Other Related Questions on Mechanics\n\nView all Questions »", null, "", null, "### Course Features\n\n• 101 Video Lectures\n• Revision Notes\n• Previous Year Papers\n• Mind Map\n• Study Planner\n• NCERT Solutions\n• Discussion Forum\n• Test paper with Video Solution", null, "", null, "### Course Features\n\n• 110 Video Lectures\n• Revision Notes\n• Test paper with Video Solution\n• Mind Map\n• Study Planner\n• NCERT Solutions\n• Discussion Forum\n• Previous Year Exam Questions" ]

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[ "+0\n\n# System\n\n-1\n213\n1\n\nFor what ordered pair (a, b) are there infinite solutions (x, y) to the system\n\n4x+ay=-8-3x+y,\n2x+y=b+2x+5y\n\nMay 20, 2022\n\n#1\n0\n\nI dont think such values exist.\n\nWe can simplify the equations to:\n\n$$4y+b = 0$$\n\n$$7x+ay=y-8$$\n\nFor both equations to have infinite solutions, the 2 equations must be multiples of each other.\n\nHowever, notice that the first equation has the slope $$-7 \\over a$$, and the second term has a slope of 0.\n\nBecause the 2 slopes are different, there is no ordered pair (a,b) that satisfies the requirements.\n\nMay 20, 2022" ]

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[ "# What is 7 times 96?\n\nHere we answer one simple question: What is 7 times 96? (or what is 7 multiplied by 96) Here is the answer:\n\n### 7 x 96 = 672\n\nLearning the multiplication of 7 times 96 is an essential skill for problems based upon fractions, decimals, and percentages. It helps in solving real-life problems quickly.\n\nIf you want to find what 7 times 96 means, think of it as 7 added together 96 times. To get the answer, you could just write down the number 7, 96 times and then add the 2 numbers together.\n\nIf you’re using a calculator, you can double-check that the answer is 672 by pressing 7 then x, then 96, and then to get the answer 672.\n\nX" ]

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[ "", null, "", null, "### Home > ACC7 > Chapter 7 Unit 8 > Lesson CC3: 7.1.2 > Problem7-21\n\n7-21.\n\nEvaluate each expression below for $a$ when $a=\\frac{2}{3}$, if possible.\n\n1.  $24a$\n\nMultiply $24$ by $2$, then divide by $3$.\n\n$\\frac{24}{1}\\cdot\\frac{2}{3}=\\frac{48}{3}=16$\n\n1.  $3a$\n\nUse the same method as part (a).\n\n1.  $\\frac{a}{0}$\n\nDividing by $0=$ undefined.\n\n1.  $\\frac{0}{a}$\n\n$0$" ]

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", null ]

[ "Solutions by everydaycalculation.com\n\nAdd 30/40 and 5/9\n\n30/40 + 5/9 is 47/36.\n\nSteps for adding fractions\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 40 and 9 is 360\n2. For the 1st fraction, since 40 × 9 = 360,\n30/40 = 30 × 9/40 × 9 = 270/360\n3. Likewise, for the 2nd fraction, since 9 × 40 = 360,\n5/9 = 5 × 40/9 × 40 = 200/360\n4. Add the two fractions:\n270/360 + 200/360 = 270 + 200/360 = 470/360\n5. After reducing the fraction, the answer is 47/36\n6. In mixed form: 111/36\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:\nAndroid and iPhone/ iPad" ]

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[ "", null, "Condition Number - Maple Help\n\nMaplets[Examples][LinearAlgebra]\n\n ConditionNumber\n display a graphical interface to the ConditionNumber function", null, "Calling Sequence ConditionNumber(M)", null, "Parameters\n\n M - square Matrix", null, "Description\n\n • The ConditionNumber(M) calling sequence displays a Maplet application that returns the condition number for the matrix M.\n • A definition of the condition number for a matrix is given in the Maplet application.\n • By using the Norm drop-down box, control the norm used to determine the condition number.\n • By using the Matrix has real values check box, control the algorithm used to calculate the condition number. If the matrix has real entries, a faster algorithm can be used. The default value is determined by using the value of $\\mathrm{has}\\left(M,I\\right)$, which returns true if M has complex entries. That is, it is selected, by default, only if M has no complex entries.\n Note: If Matrix has real values is selected and the matrix has any complex entries, an incorrect result may be returned.\n • By using the Evaluate result check box, control whether the Maplet application returns the condition number for the matrix M or the calling sequence required to calculate the condition number in the worksheet.  The default behavior is to evaluate the result, that is, return the number.\n • The ConditionNumber sample Maplet worksheet demonstrates how to write a Maplet application that functions similarly to the Maplet application displayed by this routine.", null, "Examples\n\n > $\\mathrm{with}\\left({{\\mathrm{Maplets}}_{\\mathrm{Examples}}}_{\\mathrm{LinearAlgebra}}\\right):$\n > $\\mathrm{ConditionNumber}\\left(⟨⟨1,3⟩|⟨2,5⟩⟩\\right)$" ]

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[ "# translate\n\nTranslate geometry\n\nSince R2020a\n\n## Syntax\n\n``h = translate(g,s)``\n\n## Description\n\nexample\n\n````h = translate(g,s)` translates the geometry `g` by the distance `s`.```\n\n## Examples\n\ncollapse all\n\nTranslate a geometry by different distances along the x- and y-axes.\n\nCreate a model.\n\n`model = createpde;`\n\nImport and plot a geometry.\n\n```g = importGeometry(model,\"PlateHolePlanar.stl\"); pdegplot(g)```", null, "Mesh the geometry and plot the mesh.\n\n```generateMesh(model); figure pdemesh(model)```", null, "Translate the 2-D geometry by 1 along the x-axis and by 2 along the y-axis. Plot the result.\n\n```translate(g,[1 2]); pdegplot(g)```", null, "Plot the geometry and mesh. The `translate` function modifies a geometry, but it does not modify a mesh.\n\n```figure pdegplot(g) hold on pdemesh(model)```", null, "After modifying the geometry, always regenerate the mesh.\n\n```generateMesh(model); figure pdegplot(g) hold on pdemesh(model)```", null, "## Input Arguments\n\ncollapse all\n\nGeometry, specified as an `fegeometry` object, a `DiscreteGeometry` object, or an `AnalyticGeometry` object.\n\nTranslation distance, specified as a vector of two or three real numbers. The distance is a vector of two elements for a 2-D geometry or three elements for a 3-D geometry. The elements specify the distance along the x-, y-, and, for a 3-D geometry, z-axes.\n\n## Output Arguments\n\ncollapse all\n\nResulting geometry, returned as an `fegeometry` object or a handle.\n\n• If the original geometry `g` is an `fegeometry` object, then `h` is a new `fegeometry` object representing the modified geometry. The original geometry `g` remains unchanged.\n\n• If the original geometry `g` is a `DiscreteGeometry` object, then `h` is a handle to the modified `DiscreteGeometry` object `g`.\n\n• If `g` is an `AnalyticGeometry` object, then `h` is a handle to a new `DiscreteGeometry` object. The original geometry `g` remains unchanged.\n\n## Tips\n\n• After modifying a geometry, regenerate the mesh to ensure a proper mesh association with the new geometry.\n\n• If `g` is an `fegeometry` or `AnalyticGeometry` object, and you want to replace it with the modified geometry, assign the output to the original geometry, for example, ```g = translate(g,[1 2])```.\n\n## Version History\n\nIntroduced in R2020a\n\nexpand all" ]

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[ "", null, "# Computation of Risk Premium with Real Rate retur\n\nHi everyone,\n\nI have a problem with Practice Questions 13 and 14 of Reading 39 (regarding computation of risk premium) and the errata published by the CFA. I don’t understand why they change the numerator with (1 + 0.080) whereas the formula in the curriculum seemed more consistent with the previous answer. All the info are below (related to question 13)\n\nNote: in the previous questions (11 and 12), the student is asked about the real return of equities and the real return of corporate bonds and the correct answers were 5,8 % for real return of equities and 4,3 % for real restrung of corporate bonds.\n\n—-\n\nAn analyst had the following geometric returns\n\nequities —- > 8%\n\nCorporate bonds —-> 6,5%\n\nTreasury Bills —-> 2,5%\n\nInflation —-> 2,1 %\n\nQuestion 13:\n\nThe risk premium for equities is :\n\nA 3,2% (erratum : A 5,4%)\n\nB 3,4% (erratum: B 5,5%)\n\nC 3,6% (erratum C 3,6%)\n\nAnswer before erratum: A is correct (1 + 0.058) / (1+0.0250) - 1 = 3,2 %\n\nAnswer after erratum: A is correct (1 + 0.080)/ (1+0,\n\n.0250) - 1 = 5,4%\n\ni don’t understand the answer after erratum: is it not consistent with the curriculum formula :\n\n(1+r) = (1+r rf) * (1+pi) *(1+RP)\n\n(1+r real) = (1+r rf) * (1+RP) or\n\n(1+r real) = (1+r) / (1+pi)\n\npi = inflation\n\nr rf= real risk return\n\nAny insights ?\n\nReturn on Equity and risk free return (Tbill return) both are nominal return.\n\nSo either you adjust both for Inflation or you use nominal return for both. Answer will be 5.4%.\n\nThe previous answer user real return for eq and nomnial for Tbill thats why it was incorrect.\n\nI am not sure I completely understand. Could you illustrate with the formulas given in the curriculum that I just cited above ?\n\nEquity risk premium using nominal returns\n\n(1+nominal return on equities) /(return on T-bills) - 1= 1.08 / 1.025 - 1= 0.053658 ≈5.40%\n\nEquity risk premium using real returns\n\nReal return on equities\n\n(1+nominal return on equities)/(1+inflation) - 1 = 1.08/ 1.021 - 1=0.05778 ≈5.8%\n\nReal return on T-bills\n\n(1+nominal return on T-bills) / (1+inflation) - 1= 1.025 / 1.021 - 1=0.003917 ≈ 0.39%" ]

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[ "# Find the sequence generated by the following generating functions: 4x 1 _ I1 - 41+x 31 (1 + 2)2\n\n###### Question:\n\nFind the sequence generated by the following generating functions: 4x 1 _ I 1 - 4 1+x 31 (1 + 2)2", null, "", null, "#### Similar Solved Questions\n\n##### Momentum and kinetic energy are both calculated from the mass and velocity of an object. If...\nMomentum and kinetic energy are both calculated from the mass and velocity of an object. If mass is conserved during a collision, explain how momentum may be conserved but not kinetic energy....\n##### (10) A circle is inscribed inside a square of side length L Use geometry and known areas to find the area between the square and the circle, then verify your answer using integration_\n(10) A circle is inscribed inside a square of side length L Use geometry and known areas to find the area between the square and the circle, then verify your answer using integration_...\n##### < Back This window shows your responses and what was mal 21 Required information The following...\n< Back This window shows your responses and what was mal 21 Required information The following information applies to the questions displayed below.) Part 1 of 2 This year (year O) Elizabeth agreed to a three-year service contract with an engineering consulting firm to improve efficiency in her f...\n##### A Constipation D. Excessive administration of chloride 10 stone IV fluids are Huids with a total...\nA Constipation D. Excessive administration of chloride 10 stone IV fluids are Huids with a total osmolaitych Most IV Mods contain either dextrose or electrolytes in totalmolity close to that of the ECE rapidly enters red blood cells, causing them to B. When the patient is severely dehydrated the pat...\n##### A patient receiving hemodialysis is being taught about the access port. ● What are ways the...\nA patient receiving hemodialysis is being taught about the access port. ● What are ways the patient needs to be educated to protect the permanent dialysis access port?...\n##### Let $0<a<b .$ Use the Mean Value Theorem to show that$sqrt{b}-sqrt{a}<frac{b-a}{2 sqrt{a}}$\nLet $0<a<b .$ Use the Mean Value Theorem to show that $sqrt{b}-sqrt{a}<frac{b-a}{2 sqrt{a}}$...\n##### Draw the isoquant of following production function and indicated level of output. Q=min(2L,0.5L+0.5K,K) Q=4\nDraw the isoquant of following production function and indicated level of output. Q=min(2L,0.5L+0.5K,K) Q=4...\n##### When a person inhales, air moves down the bronchus (windpipe) at $15 mathrm{~cm} / mathrm{s}$. The average flow speed of the air doubles through a constriction in the bronchus. Assuming incompressible flow, determine the pressure drop in the constriction.\nWhen a person inhales, air moves down the bronchus (windpipe) at $15 mathrm{~cm} / mathrm{s}$. The average flow speed of the air doubles through a constriction in the bronchus. Assuming incompressible flow, determine the pressure drop in the constriction....\n##### 10) Cholesterol distribution for large numbers of adults hasbeen shown to be as follows:* 56%, for those with less than 160 cg.* 36%, for those between 160 and 170 cg.Such a distribution is known to have a normal law.a) Calculate the mean value of cholesterol and its standarddeviation.b) It is accepted that adults whose cholesterol is higher than 173cg should be treated clinically. Calculate the number of adults tobe treated in a population of 10,000 people?Note: cg = centigram\n10) Cholesterol distribution for large numbers of adults has been shown to be as follows: * 56%, for those with less than 160 cg. * 36%, for those between 160 and 170 cg. Such a distribution is known to have a normal law. a) Calculate the mean value of cholesterol and its standard deviation. b) It i...\n##### What is the mass of 5.99 mole of carbon monoxide?\nWhat is the mass of 5.99 mole of carbon monoxide?...\n##### Find $\\mathbf{u} \\times \\mathbf{v}$ and show that it is orthogonal to both $\\mathbf{u}$ and $\\mathbf{v} .$ $\\mathbf{u}=\\mathbf{i}+\\frac{3}{2} \\mathbf{j}-\\frac{5}{2} \\mathbf{k}$ $\\mathbf{v}=\\frac{1}{2} \\mathbf{i}-\\frac{3}{4} \\mathbf{j}+\\frac{1}{4} \\mathbf{k}$\nFind $\\mathbf{u} \\times \\mathbf{v}$ and show that it is orthogonal to both $\\mathbf{u}$ and $\\mathbf{v} .$ $\\mathbf{u}=\\mathbf{i}+\\frac{3}{2} \\mathbf{j}-\\frac{5}{2} \\mathbf{k}$ $\\mathbf{v}=\\frac{1}{2} \\mathbf{i}-\\frac{3}{4} \\mathbf{j}+\\frac{1}{4} \\mathbf{k}$...\n##### Suppose an influx of energetic electrons gets stuck on theEarth’s magnetic field lines at a place where the magnetic field is8.0 x 10-6 T and the electrons have a speed of 5.2 x 105 m/s. Whatwould the gyro frequency of those electrons?\nSuppose an influx of energetic electrons gets stuck on the Earth’s magnetic field lines at a place where the magnetic field is 8.0 x 10-6 T and the electrons have a speed of 5.2 x 105 m/s. What would the gyro frequency of those electrons?...\n##### Aerobic degradation of an organic compound by a mixed culture of organisms in waste water can...\nAerobic degradation of an organic compound by a mixed culture of organisms in waste water can be represented by the following reaction. C_3 H_6 O_3 + a O_2 + b NH_3 rightarrow c C_5H_7NO_2 + d H_2O + e CO_2 a. Determine a, b, c, d, and e, if Y_X/S = 0.4 g X/gS. b. Determine the yield coefficients Y_...\n##### 23. which of the carbon atoms is likely be LEAST susceptible to cytochrome P450 enzymes?\n23. which of the carbon atoms is likely be LEAST susceptible to cytochrome P450 enzymes?...\n##### List the like terms in each expression, if any.$3 y+4-11 y+6$\nList the like terms in each expression, if any. $3 y+4-11 y+6$...\n##### Fill in the Blanks Let's use calorimetry to measure heats of reaction. Suppose you mix 25.0...\nFill in the Blanks Let's use calorimetry to measure heats of reaction. Suppose you mix 25.0 mL of 0.08M cesium hydroxide with 25.0 mL of 0.08M hydrochloric acid. Assume both solutions are at 28.0°C. The temperature of the solution after the reaction is 30.2°C. Because the temperature of ...\n##### HIV HPV Gonorrhea Chlamydia Herpes Simplex Virus (HSV) For each disease, investigate the following factors:What type of organism causes this disease (include apicture)?What is the scientific name of this organism?How is this disease transmitted?What part(s) of the body are affected and what are the majorsigns and symptoms (pathology)?How can this disease be prevented, and how is it commonlytreated?\nHIV HPV Gonorrhea Chlamydia Herpes Simplex Virus (HSV) For each disease, investigate the following factors: What type of organism causes this disease (include a picture)? What is the scientific name of this organism? How is this disea...\n##### Your answers_ For 20, solve the application problem: Include units on 120 rats initially t0 20. A rat colony has increased approximalely exponentially from after months- 280 rats after 15 months_ Let f() be the number of rats in the colony Find an approximate equation of f(t) . Round to two decimal places where necessary: Predict the number of rats in the colony after 28 months_\nyour answers_ For 20, solve the application problem: Include units on 120 rats initially t0 20. A rat colony has increased approximalely exponentially from after months- 280 rats after 15 months_ Let f() be the number of rats in the colony Find an approximate equation of f(t) . Round to two decimal...\n##### (a) Use a graphing utility to draw the curve $y^{2}=x^{2}(1-x)$ (b) Your drawing $\\mathrm{m}$ part (a) should show that the curve forms a loop for $0 \\leq x \\leq 1$. Calculate the area of the loop. HINT: Use the symmetry of the curve.\n(a) Use a graphing utility to draw the curve $y^{2}=x^{2}(1-x)$ (b) Your drawing $\\mathrm{m}$ part (a) should show that the curve forms a loop for $0 \\leq x \\leq 1$. Calculate the area of the loop. HINT: Use the symmetry of the curve....\n##### Factor.$-4 x^{2} y-4 x^{3}+24 x y^{2}$\nFactor. $-4 x^{2} y-4 x^{3}+24 x y^{2}$...\n##### A charge is located at the origin, another charge is located somewhere on the +y axis,...\nA charge is located at the origin, another charge is located somewhere on the +y axis, another charge is located somewhere on the -y axis, and a fourth is located somewhere on the +x axis. The angle of the total electric force on the charge at the origin is 30 degrees counter-clockwise from the +x a...\n##### Given the information below that includes the sample size (n1and n2) for each sample, the mean for each sample (x1 and x2) andthe standard deviations for each sample (s1 and s2), enter thep-value to test the following hypothesis at the 1% significancelevel . Assume the variables come from normally distributedpopulations with equal variances Ho : µ1 = µ2 Ha : µ1 < µ2Sample 1 n1 = 12 x1 = 113 s1 = 4.2 Sample 2 n2 = 12 x2= 115 s2 = 5.15What is the p-value for this test ?\nGiven the information below that includes the sample size (n1 and n2) for each sample, the mean for each sample (x1 and x2) and the standard deviations for each sample (s1 and s2), enter the p-value to test the following hypothesis at the 1% significance level . Assume the variables come from normal...\n##### Please answer each part with work shown. Thank you. 6. The following three mechanisms have been proposed for a particular reaction. Mechanism II Mechanism III Mechanisml H2 +NO-H2O+N (slow) N +...\nPlease answer each part with work shown. Thank you. 6. The following three mechanisms have been proposed for a particular reaction. Mechanism II Mechanism III Mechanisml H2 +NO-H2O+N (slow) N +NON2+0 (fast) O+H2-> H2O (fast) 2 NO-N202 (fast equilibrium) H2-N202-H20+ N3O (slow) H2+N20-H2O+N2 (fast...\n##### In each box, write the number of H's that each labeled carbon has. For example, if...\nIn each box, write the number of H's that each labeled carbon has. For example, if the carbon next to the letter a has 3 H's write 3 in the box marked a Draw a Lewis structure for the molecule below, showing all lone pairs. You may abbreviate any methyl groups as CH3 Convert the Lewis stru...\n##### (Sensitivity analysis)You are given the following LPMax 2 = 3 xl+ 7*x2 +5*x3 xl +x2 +x3 <= 50 2*xl+3*x2+x3 <= 100 xl,x2,x3 >= 0You solved this LP to optimality; and the optimal dictionary is (wl and w2 are slack variables):300 -3*xl -w2-4*wlx3 = 25-0.5*x1+0.5*w2-1.5*wl x2 = 25- 0.5*xl 0.5*w2 0.5*w1Answer the following questions: Find the range of the coefficient of x1 in the objective function such that the current basis remains optimal: Find the range of the coefficient of x2 in the ob\n(Sensitivity analysis) You are given the following LP Max 2 = 3 xl+ 7*x2 +5*x3 xl +x2 +x3 <= 50 2*xl+3*x2+x3 <= 100 xl,x2,x3 >= 0 You solved this LP to optimality; and the optimal dictionary is (wl and w2 are slack variables): 300 -3*xl -w2-4*wl x3 = 25-0.5*x1+0.5*w2-1.5*wl x2 = 25- 0.5*xl ...\n##### Operation research:Determine if the stationary point found in the Newton-Raphson Methodextremum (maximum or minimum)?Karush–Kuhn-Tucker (KKT)\noperation research: Determine if the stationary point found in the Newton-Raphson Method extremum (maximum or minimum)? Karush–Kuhn-Tucker (KKT)...\n##### Explain FLSA and its purpose. Then choose and explain in detail two of the other listed...\nExplain FLSA and its purpose. Then choose and explain in detail two of the other listed labor laws below. Then, explain why labor laws are important to both the employee and the employer? The other two labor are; Federal Unemployment Tax Act (FUTA) Affordable Care Act (ACA)...\n##### The demand for bottled water in a small town is as follows: $\\begin{array}{cc} P(\\text { per bottle }) & Q_{D} \\text { (bottles per week) } \\\\ \\hline \\ 1.00 & 500 \\\\ \\ 1.50 & 400 \\\\ \\ 2.00 & 300 \\\\ \\ 2.50 & 200 \\\\ \\ 3.00 & 100 \\end{array}$ a. Is this a straight-line demand curve? How do you know? b. Calculate the price elasticity of demand for bottled water for a price rise from $\\$ 1.00$to$\\$1.50 .$ Is demand elastic or inelastic for this price change? c. Calcu\nThe demand for bottled water in a small town is as follows: $\\begin{array}{cc} P(\\text { per bottle }) & Q_{D} \\text { (bottles per week) } \\\\ \\hline \\ 1.00 & 500 \\\\ \\ 1.50 & 400 \\\\ \\ 2.00 & 300 \\\\ \\ 2.50 & 200 \\\\ \\ 3.00 & 100 \\end{array}$ a. Is this a straight-line de...\n##### Problet 6.[12 pts] NBA investedl S7(),(KM) Hani that eAIS 3.25% AnIal leres compounded continuousl: Write the Value of the account alter FCn MtO function A() Then calculate A(IO) and A(10) . and interpret their mennings.\nProblet 6.[12 pts] NBA investedl S7(),(KM) Hani that eAIS 3.25% AnIal leres compounded continuousl: Write the Value of the account alter FCn MtO function A() Then calculate A(IO) and A(10) . and interpret their mennings....\n##### Swifty Fabricators produces protective covers for smart phones. Since the covers must be customized to each...\nSwifty Fabricators produces protective covers for smart phones. Since the covers must be customized to each smart phone model, Swifty uses a job order costing system. On September 1, the company reported the following inventory balances: Direct Materials $31,500 Work in Process$152,000 F...\n##### Problem 2 For the beam and loading shown, using singularity functions, determine (a) the equation...\nProblem 2 For the beam and loading shown, using singularity functions, determine (a) the equation of the elastic curve, (b) the deflection at point B, (c) the deflection at point D L/2 L/2 L/2 Problem 2 For the beam and loading shown, using singularity functions, determine (a) the equation of the e...\n##### Y H10 mm50 mm10 mmx50 mm10 mm90 mm\ny H10 mm 50 mm 10 mm x 50 mm 10 mm 90 mm...\n##### What happens to glucose in the kidneys?\nWhat happens to glucose in the kidneys?...\n##### An aluminum cylinder weighs $1.03 \\mathrm{~N}$. When this same cylinder is completely submerged in alcohol, the volume of the displaced alcohol is $3.90 \\times 10^{-5} \\mathrm{~m}^{3}$. If the cylinder is suspended from a scale while submerged in the alcohol, the scale reading is $0.730 \\mathrm{~N}$. What is the specific gravity of the alcohol? ( Worial: ball in beaker)\nAn aluminum cylinder weighs $1.03 \\mathrm{~N}$. When this same cylinder is completely submerged in alcohol, the volume of the displaced alcohol is $3.90 \\times 10^{-5} \\mathrm{~m}^{3}$. If the cylinder is suspended from a scale while submerged in the alcohol, the scale reading is $0.730 \\mathrm{~N}$..." ]

[ null, "https://cdn.numerade.com/ask_images/1f63dbbbac304358a534e9643905e635.jpg ", null, "https://cdn.numerade.com/previews/62333dcf-c915-4699-8772-b360bc60c3ea_large.jpg", null ]

[ "# How to find pH by mixing two solutions of different concentrations?\n\nA mixture is made by combining 110 mL of 0.15 M $\\ce{HCl}$ and 215 mL of 0.055 M $\\ce{HI}$. What is the pH of the solution?\n\n$$\\ce{HCl -> H+ + Cl-}$$ $$\\ce{HI -> H+ + I-}$$ According to what I thought, the molarity of $\\ce{H+}$ is the same as $\\ce{HCl}$, because it is a strong acid and the mole ratio. So, I added the molarity of both acids: $\\pu{0.15M} + \\pu{0.055M} = \\pu{0.205M}$ for $\\ce{H+}$ in solution.\n\n$$\\mathrm{pH} = -\\log[\\ce{H+}]$$ $$\\mathrm{pH} = -\\log[0.205]$$ $$\\mathrm{pH} = 0.689$$\n\nYou are right in assuming the two acids are strong acids and therefore completely dissociated, but you can easily look that also up, e.g. here. Therefore you can assume, that the concentration of protons is equal to the concentration of the acid. \\begin{align} c_{\\ce{HI}}(\\ce{H+}) &= c(\\ce{HI}) & c_{\\ce{HCl}}(\\ce{H+}) &= c(\\ce{HCl}) & \\end{align}\n\nNow you need to determine the amount of the protons in each solution with $$n=c\\cdot V.$$\n\nYou have to add the amount of protons in each solution and determine the combined volume. \\begin{align} n_\\mathrm{tot}(\\ce{H+}) &= n_{\\ce{HI}}(\\ce{H+}) + n_{\\ce{HCl}}(\\ce{H+})& V_\\mathrm{tot} &= V(\\ce{HI}) + V(\\ce{HCl}) \\end{align}\n\nTo calculate the pH of the resulting concentration, just take the negative logarithm. $$\\mathrm{pH} = -\\log c_\\mathrm{tot}(\\ce{H+}) = -\\log\\left[\\frac{n_\\mathrm{tot}(\\ce{H+})}{V_\\mathrm{tot}}\\right]$$\n\nYour solution should be $\\mathrm{pH}=1.06$, rounded to three significant figures.\n\nThe approach you chose can only work, if you have equal volumes of the two solutions.\n\n\\begin{align} M_1V_1+M_2V_2&=M\\left(V_1+V_2\\right)\\\\ 0.015\\times110+0.055\\times215&=M\\left(110+215\\right)\\\\ 1.65+11.825&=M\\cdot325\\\\ M&=\\frac{13.475}{325}\\\\ M&=0.04146=4.416\\times10^{-2}\\\\ \\mathrm{pH}&=-\\log\\left(4.416\\times10^{-2}\\right)\\\\ \\mathrm{pH}&=-\\log4.416+2\\log10\\\\ \\mathrm{pH}&=-0.617+2\\\\ \\mathrm{pH}&=1.383 \\end{align}\n\nYou can go this way also \\begin{align} \\text{HCl conc}&=0.015\\ \\mathrm M\\\\ &=0.015\\times\\frac{110}{1000}\\ \\mathrm{mol}\\\\ &=1.65\\times10^{-3}\\ \\mathrm{mol} \\end{align}\n\n\\begin{align} \\text{HI conc}&=0.055\\ \\mathrm M\\\\ &=0.055\\times\\frac{215}{1000}\\\\ &=11.825\\times10^{-3}\\ \\mathrm{mol} \\end{align}\n\n\\begin{align} \\text{Total}\\ [\\ce{H+}]&=\\left(1.65\\times10^{-3}+11.825\\times10^{-3}\\right)\\ \\mathrm{mol}\\ \\text{in}\\ 325\\ \\mathrm{ml}\\\\ &=13.475\\times10^{-3}\\ \\mathrm{mol}\\ \\text{in}\\ 325\\ \\mathrm{ml}\\\\ \\text{so in}\\ 1000\\ \\mathrm{ml}&=13.475\\times10^{-3}\\times\\frac{1000}{325}\\\\ &=4.146\\times10^{-2}\\\\ \\text{Thus}\\ \\mathrm{pH}&=1.383 \\end{align}\n\n• Ravi, the question says 0.15 M HCl, not 0.015M – DavePhD Sep 12 '16 at 13:51" ]

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[ "# ECE 220 Network Analysis I\n\n## Lesson 8. First Order Systems\n\nDisplay and printer note:  This lesson uses a symbol for infinity.  Here it is: ∞  It should look like an 8 on its side. If it does not, consider upgrading to a more modern browser.  Even if it looks correct in your browser, it may not print correctly on your printer.  If it doesn't, consider upgrading your printer driver.\n\nThis lesson deals with the use of either capacitors or inductors (not both at the same time) in single-time-constant circuits. This usually means that the circuit consists of a single inductor or capacitor, several resistors, and one or more switches. Such a circuit is shown below.", null, "Assume that the switch has been open for a long time, and that at t = 0 the switch is closed. Intuitively, you should expect that before the switch is closed, the capacitor will charge up to 50 V. This is so because, if you wait long enough, the current in the 4000 Ω resistor will decay to you, and there will be no voltage across the resistor. Then, by KVL, the capacitor and battery voltages must be equal.\n\nAfter the switch is thrown, things get more complicated: The capacitor will begin to drain through the 1000 Ω resistor, but current will continue to be provided by the battery. You might guess that the capacitor voltage would decay, but wouldn't go all the way to zero, and you would be right.\n\nBelow is a systematic approach to solving problems like this. Read it carefully, then we'll apply it to the above problem.\n\n### Approach to Boundary Value Problems\n\nThis technique works for first order systems with constant sources.\n\nFor simplicity, assume ti = 0, where ti is the initial time, i.e. the time when the switch is closed (or opened).\n\n1. Write down:\n\niC = C(dvC/dt)\nvL = L(diL/dt)\ni(t) = if + (i0 - if)e-t/τ\nv(t) = vf + (v0 - vf)e-t/τ\n\n2. For 0-\n\n1. Recall that \"steady state\" has been reached.\n2. Therefore, vC = constant, iL = constant\n3. Therefore, iC = 0, vL = 0\n4. Redraw the circuit, open circuiting capacitors (because iC = 0) and short circuiting inductors (because vL = 0)\n5. Apply KVL and KCL to find unknown voltages and currents, particularly vC or iL .\n\n3. For 0+\n\n1. Recall that you can't change vC or iL instantaneously.\n2. Therefore, vC(0+) = vC(0-) and iL(0+) = iL(0-)\n3. Apply KCL and KVL to find v0 and i0.\n\n4. For infinity\n\n1. Repeat 2.\n2. Calculate vf or if.\n\n5. Finally\n\n1. Redraw circuit, removing C or L, shorting voltage sources, and opening current sources.\n2. Calculate RTH \"looking into\" the terminals where you took out C or L.\n3. Calculate τ = RTHC or τ = L/RTH.\n4. Plug values into i(t) or v(t) equation.\n5. Simplify.\n\n### Example Solution to RC Problem\n\n1. Write down:\n\niC = C(dvC/dt)\nv(t) = vf + (v0 - vf)e-t/τ\n\n2. For 0-\n\n1. Recall that \"steady state\" has been reached.\n2. Therefore, vC = constant\n3. Therefore, iC = 0\n4. Redraw the circuit, open circuiting capacitors (because iC = 0)", null, "5. Apply KVL to find the unknown voltage.\nThis is easy:\n-50 + v4000 + v = 0\nBut v4000 = 0 because it caries no current, so\nv = v(0-) = 50 V\n\n3. For 0+\n\n1. Recall that you can't change vC instantaneously.\n2. Therefore, vC(0+) = vC(0-)\nv(t = 0) = v0 = 50 V\n\n4. For infinity\n\n1. Repeat 2.", null, "By the voltage divider formula,\nv = 50×1000/(4000 + 1000) = 10 V\n2. Calculate vf .\nvf = v = 10 V\n\n5.\n\n1. Redraw circuit, removing C, shorting voltage sources.", null, "2. Calculate RTH \"looking into\" the terminals where you took out C.\nRTH = 4000 × 1000/(4000 + 1000) = 800 Ω\n3. Calculate τ = RTHC.\nτ = 800 × 20×10-6 = 16×10-3 s = 16 ms = .016 s\n4. Plug values into v(t) equation.\nv(t) = vf + (v0 - vf)e-t/τ = 10 + (50 - 10)e-t/.016 V\n5. Simplify.\nv(t) = 10 + 40e-t/.016 V\n\nHere's what a graph of the capacitor voltage looks like:", null, "### Simulation\n\nCheck out this first order circuit explanation and example:  http://people.clarkson.edu/~svoboda/eta/plots/FOC.html (created by James A. Svoboda of Clarkson University).\n\nBefore going on to the homework, you should complete Tutorial 8 on first order circuits.\n\n### Homework Problems\n\nIt is not necessary to use the organizational tables in any of these problems, but they may be helpful for keeping your solution neat, especially for problem 4.\n\n1. E = 75 V, L = 40 mH, R = 22 kΩ. The switch is closed at t = 0. Find an expression for i(t) for t > 0.", null, "Voltage or Current Value Reason or Calculation i(0-) i(0+) i(∞)\n\n2. R = 6.8 kΩ, C = 4 µF. The capacitor is initially charged to 60 V. After the switch is closed, how long will it take before the voltage has dropped to 40/e volts?", null, "Voltage or Current Value Reason or Calculation vc(0-) vc(0+) vc(∞)\n\n3. E = 125 V, R1 = 4.7 kΩ, R2 = 15 kΩ, R3 = 6.8 kΩ, C = 10 µF.  The switch has been open for a long time. At t = 0 the switch is closed. Find v(t) for t>0.", null, "Voltage or Current Value Reason or Calculation v(0-) v(0+) v(∞)\n\n4. E = 30 V, R1 = 33 kΩ, R2 = 22 kΩ, C = 20 µF, L = 65 mH.  The switch is closed at t = 0. Find:\ni1(0-), i2(0-), v1(0-), v2(0-)\ni1(0+), i2(0+), v1(0+), v2(0+)\ni1(∞), i2(∞), v1(∞), v2(∞)\nGive reasons or calculations for each answer.", null, "Voltage or Current Value Reason or Calculation i1(0-) v1(0-) i2(0-) v2(0-) i1(0+) v1(0+) i2(0+) v2(0+) i1(∞) v1(∞) i2(∞) v2(∞)\n\nBonus (no partial credit). The 20 µF capacitor is initially uncharged. The switch has been in position #1 for a long time. At t = 0 it is switched to position #2.\na) What is the voltage v after a long time?\nb) Find v(t) for t > 0.\nHint for a): Use conservation of charge and Kirchhoff's Voltage Law.", null, "" ]

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[ "• API\n• FAQ\n• Tools\n• Archive\nSHARE\nTWEET", null, "# Untitled", null, "a guest", null, "Jun 25th, 2019", null, "76", null, "Never\nNot a member of Pastebin yet? Sign Up, it unlocks many cool features!\n1. panel.barchart.alpha <- function (x, y, box.ratio = 1, box.width = box.ratio/(1 + box.ratio),\n2.     horizontal = TRUE, origin = NULL, reference = TRUE, stack = FALSE,\n3.     groups = NULL, col = if (is.null(groups)) plot.polygon\\$col else superpose.polygon\\$col,\n4.     border = if (is.null(groups)) plot.polygon\\$border else superpose.polygon\\$border,\n5.     lty = if (is.null(groups)) plot.polygon\\$lty else superpose.polygon\\$lty,\n6.     lwd = if (is.null(groups)) plot.polygon\\$lwd else superpose.polygon\\$lwd,\n7.     alpha = if (is.null(groups)) plot.polygon\\$alpha else superpose.polygon\\$alpha,\n8.     ..., identifier = \"barchart\")\n9. {\n10.     plot.polygon <- trellis.par.get(\"plot.polygon\")\n11.     superpose.polygon <- trellis.par.get(\"superpose.polygon\")\n12.     reference.line <- trellis.par.get(\"reference.line\")\n13.     keep <- (function(x, y, groups, subscripts, ...) {\n14.         !is.na(x) & !is.na(y) & if (is.null(groups))\n15.             TRUE\n16.         else !is.na(groups[subscripts])\n17.     })(x = x, y = y, groups = groups, ...)\n18.     if (!any(keep))\n19.         return()\n20.     x <- as.numeric(x[keep])\n21.     y <- as.numeric(y[keep])\n22.     if (!is.null(groups)) {\n23.         groupSub <- function(groups, subscripts, ...) groups[subscripts[keep]]\n24.         if (!is.factor(groups))\n25.             groups <- factor(groups)\n26.         nvals <- nlevels(groups)\n27.         groups <- as.numeric(groupSub(groups, ...))\n28.     }\n29.     if (horizontal) {\n30.         if (is.null(groups)) {\n31.             if (is.null(origin)) {\n32.                 origin <- current.panel.limits()\\$xlim\n33.                 reference <- FALSE\n34.             }\n35.             height <- box.width\n36.             if (reference)\n37.                 panel.abline(v = origin, col = reference.line\\$col,\n38.                   lty = reference.line\\$lty, lwd = reference.line\\$lwd,\n39.                   identifier = paste(identifier, \"abline\",\n40.                     sep = \".\"))\n41.             panel.rect(x = rep(origin, length(y)), y = y, height = rep(height,\n42.                 length(y)), width = x - origin, border = border,\n43.                 col = col, lty = lty, lwd = lwd, alpha=alpha,\n44.                 just = c(\"left\", \"centre\"), identifier = identifier)\n45.         }\n46.         else if (stack) {\n47.             if (!is.null(origin) && origin != 0)\n48.                 warning(\"'origin' forced to 0 for stacked bars\")\n49.             col <- rep(col, length.out = nvals)\n50.             border <- rep(border, length.out = nvals)\n51.             lty <- rep(lty, length.out = nvals)\n52.             lwd <- rep(lwd, length.out = nvals)\n53.             alpha <- rep(alpha, length.out = nvals)\n54.             height <- box.width\n55.             if (reference)\n56.                 panel.abline(v = origin, col = reference.line\\$col,\n57.                   lty = reference.line\\$lty, lwd = reference.line\\$lwd,\n58.                   identifier = paste(identifier, \"abline\",\n59.                     sep = \".\"))\n60.             for (i in unique(y)) {\n61.                 ok <- y == i\n62.                 ord <- sort.list(groups[ok])\n63.                 pos <- x[ok][ord] > 0\n64.                 nok <- sum(pos, na.rm = TRUE)\n65.                 if (nok > 0)\n66.                   panel.rect(x = cumsum(c(0, x[ok][ord][pos][-nok])),\n67.                     y = rep(i, nok), col = col[groups[ok][ord][pos]],\n68.                     border = border[groups[ok][ord][pos]], lty = lty[groups[ok][ord][pos]],\n69.                     lwd = lwd[groups[ok][ord][pos]], alpha = alpha[groups[ok][ord][pos]],\n70.                     height = rep(height, nok), width = x[ok][ord][pos], just = c(\"left\",\n71.                       \"centre\"), identifier = paste(identifier,\n72.                       \"pos\", i, sep = \".\"))\n73.                 neg <- x[ok][ord] < 0\n74.                 nok <- sum(neg, na.rm = TRUE)\n75.                 if (nok > 0)\n76.                   panel.rect(x = cumsum(c(0, x[ok][ord][neg][-nok])),\n77.                     y = rep(i, nok), col = col[groups[ok][ord][neg]],\n78.                     border = border[groups[ok][ord][neg]], lty = lty[groups[ok][ord][neg]],\n79.                     lwd = lwd[groups[ok][ord][neg]], height = rep(height,\n80.                       nok), width = x[ok][ord][neg], just = c(\"left\",\n81.                       \"centre\"), identifier = paste(identifier,\n82.                       \"neg\", i, sep = \".\"))\n83.             }\n84.         }\n85.         else {\n86.             if (is.null(origin)) {\n87.                 origin <- current.panel.limits()\\$xlim\n88.                 reference <- FALSE\n89.             }\n90.             col <- rep(col, length.out = nvals)\n91.             border <- rep(border, length.out = nvals)\n92.             lty <- rep(lty, length.out = nvals)\n93.             lwd <- rep(lwd, length.out = nvals)\n94.             alpha <- rep(alpha, length.out = nvals)\n95.             height <- box.width/nvals\n96.             if (reference)\n97.                 panel.abline(v = origin, col = reference.line\\$col,\n98.                   lty = reference.line\\$lty, lwd = reference.line\\$lwd,\n99.                   identifier = paste(identifier, \"abline\",\n100.                     sep = \".\"))\n101.             for (i in unique(y)) {\n102.                 ok <- y == i\n103.                 nok <- sum(ok, na.rm = TRUE)\n104.                 panel.rect(x = rep(origin, nok), y = (i + height *\n105.                   (groups[ok] - (nvals + 1)/2)), col = col[groups[ok]],\n106.                   border = border[groups[ok]], lty = lty[groups[ok]],\n107.                   lwd = lwd[groups[ok]], alpha = alpha[groups[ok]],\n108.                     height = rep(height,\n109.                     nok), width = x[ok] - origin, just = c(\"left\",\n110.                     \"centre\"), identifier = paste(identifier,\n111.                     \"y\", i, sep = \".\"))\n112.             }\n113.         }\n114.     }\n115.     else {\n116.         if (is.null(groups)) {\n117.             if (is.null(origin)) {\n118.                 origin <- current.panel.limits()\\$ylim\n119.                 reference <- FALSE\n120.             }\n121.             width <- box.width\n122.             if (reference)\n123.                 panel.abline(h = origin, col = reference.line\\$col,\n124.                   lty = reference.line\\$lty, lwd = reference.line\\$lwd,\n125.                   identifier = paste(identifier, \"abline\",\n126.                     sep = \".\"))\n127.             panel.rect(x = x, y = rep(origin, length(x)), col = col,\n128.                 border = border, lty = lty, lwd = lwd, alpha = alpha,\n129.                 width = rep(width, length(x)), height = y - origin, just = c(\"centre\",\n130.                   \"bottom\"), identifier = identifier)\n131.         }\n132.         else if (stack) {\n133.             if (!is.null(origin) && origin != 0)\n134.                 warning(\"'origin' forced to 0 for stacked bars\")\n135.             col <- rep(col, length.out = nvals)\n136.             border <- rep(border, length.out = nvals)\n137.             lty <- rep(lty, length.out = nvals)\n138.             lwd <- rep(lwd, length.out = nvals)\n139.             alpha <- rep(alpha, length.out = nvals)\n140.             width <- box.width\n141.             if (reference)\n142.                 panel.abline(h = origin, col = reference.line\\$col,\n143.                   lty = reference.line\\$lty, lwd = reference.line\\$lwd,\n144.                   identifier = paste(identifier, \"abline\",\n145.                     sep = \".\"))\n146.             for (i in unique(x)) {\n147.                 ok <- x == i\n148.                 ord <- sort.list(groups[ok])\n149.                 pos <- y[ok][ord] > 0\n150.                 nok <- sum(pos, na.rm = TRUE)\n151.                 if (nok > 0)\n152.                   panel.rect(x = rep(i, nok), y = cumsum(c(0,\n153.                     y[ok][ord][pos][-nok])), col = col[groups[ok][ord][pos]],\n154.                     border = border[groups[ok][ord][pos]], lty = lty[groups[ok][ord][pos]],\n155.                     lwd = lwd[groups[ok][ord][pos]], alpha = alpha[groups[ok][ord][pos]],\n156.                     width = rep(width,\n157.                       nok), height = y[ok][ord][pos], just = c(\"centre\",\n158.                       \"bottom\"), identifier = paste(identifier,\n159.                       \"pos\", i, sep = \".\"))\n160.                 neg <- y[ok][ord] < 0\n161.                 nok <- sum(neg, na.rm = TRUE)\n162.                 if (nok > 0)\n163.                   panel.rect(x = rep(i, nok), y = cumsum(c(0,\n164.                     y[ok][ord][neg][-nok])), col = col[groups[ok][ord][neg]],\n165.                     border = border[groups[ok][ord][neg]], lty = lty[groups[ok][ord][neg]],\n166.                     lwd = lwd[groups[ok][ord][neg]], alpha = alpha[groups[ok][ord][neg]],\n167.                     width = rep(width,\n168.                       nok), height = y[ok][ord][neg], just = c(\"centre\",\n169.                       \"bottom\"), identifier = paste(identifier,\n170.                       \"neg\", i, sep = \".\"))\n171.             }\n172.         }\n173.         else {\n174.             if (is.null(origin)) {\n175.                 origin <- current.panel.limits()\\$ylim\n176.                 reference = FALSE\n177.             }\n178.             col <- rep(col, length.out = nvals)\n179.             border <- rep(border, length.out = nvals)\n180.             lty <- rep(lty, length.out = nvals)\n181.             lwd <- rep(lwd, length.out = nvals)\n182.             alpha <- rep(alpha, length.out = nvals)\n183.             width <- box.width/nvals\n184.             if (reference)\n185.                 panel.abline(h = origin, col = reference.line\\$col,\n186.                   lty = reference.line\\$lty, lwd = reference.line\\$lwd,\n187.                   identifier = paste(identifier, \"abline\",\n188.                     sep = \".\"))\n189.             for (i in unique(x)) {\n190.                 ok <- x == i\n191.                 nok <- sum(ok, na.rm = TRUE)\n192.                 panel.rect(x = (i + width * (groups[ok] - (nvals +\n193.                   1)/2)), y = rep(origin, nok), col = col[groups[ok]],\n194.                   border = border[groups[ok]], lty = lty[groups[ok]],\n195.                   lwd = lwd[groups[ok]], alpha = alpha[groups[ok]],\n196.                   width = rep(width, nok),\n197.                   height = y[ok] - origin, just = c(\"centre\",\n198.                     \"bottom\"), identifier = paste(identifier,\n199.                     \"x\", i, sep = \".\"))\n200.             }\n201.         }\n202.     }\n203. }\n204. environment(panel.barchart.alpha) <- asNamespace(\"lattice\")\nRAW Paste Data\nWe use cookies for various purposes including analytics. By continuing to use Pastebin, you agree to our use of cookies as described in the Cookies Policy.\n\nTop" ]

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[ "Solutions by everydaycalculation.com\n\n1st number: 2 2/5, 2nd number: 16/42\n\n12/5 + 16/42 is 292/105.\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 5 and 42 is 210\n2. For the 1st fraction, since 5 × 42 = 210,\n12/5 = 12 × 42/5 × 42 = 504/210\n3. Likewise, for the 2nd fraction, since 42 × 5 = 210,\n16/42 = 16 × 5/42 × 5 = 80/210\n504/210 + 80/210 = 504 + 80/210 = 584/210\n5. After reducing the fraction, the answer is 292/105\n6. In mixed form: 282/105\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:" ]

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

[ "", null, "# Bar Sagi Tiling: From Refrigerator Door to Mathematics Journal\n\n## Dr. Yossi Elran and Prof. Anthony Joseph explain the mathematics of this unique tiling", null, "Tiling has fascinated both mathematicians and artists since ancient times. The problem it poses can be described simply as revealing the ways to cover a plane using interlocking shapes and following a well-defined set of rules. Even in antiquity, the tiling of a plane using triangles, rectangles and regular hexagons was known. This long history of tiling might suggest there is nothing new to discover.  But mathematicians can surprise; indeed new tilings are discovered from time to time, most often by modifying the agreed-upon rules.\n\nDuring the second half of the preceding century, two notable new tilings were discovered.  In the first, the tiles are chosen such that there are, necessarily, no repetitions in tiling the plane. This type is called aperiodic tiling, and it culminated in an important example that Roger Penrose presented at the end of the seventies.  Aperiodic tiling can be thought of as providing the inspiration for quasicrystals, which are aperiodic packings in three-space. Quasicrystals were first produced in the laboratory by Prof. Dan Schechtman of the Technion, and for this he was awarded the Nobel Prize in Chemistry in 2011.\n\nJoseph showed that any such scaled-up triangle can be obtained by tiling with triangles from the original set\n\nThe second type is “self-similar tiling,” introduced by William Thurston in 1992.  In this type of tiling, starting from a given factor one must produce a finite set of shapes, any one of which can be enlarged by the given factor through joining members of the set.  More recently, Prof. Anthony Joseph of the Mathematics Department of the Weizmann Institute of Science has proposed “proper self-similar triangle tiling,” which he renamed “Bar Sagi tiling” in memory of his granddaughter who died at age fifteen of bone cancer.\n\nBar Sagi tiling is a generic name for tilings comprised of triangles derived from polygons.  First one fixes the number, call it n, of the sides of a polygon; this polygon is required to be regular (that is, it has equal angles and equal sides).  In such a polygon, a triangle may be inscribed by joining up the points of three successive vertices; if this is done in all possible ways, one gets the complete set of triangles based on n. If r,s and t are the number of sides of the polygon that can be counted between each of the three successive vertices, then r+s+t=n.  Two of the resultant triangles may be congruent: This is true if and only if their corresponding triples are the same up to permutation. However if r,s and t are all distinct (so that the triangle is scalene), then it is convenient to distinguish this triangle from its mirror image, as the one cannot be rotated into the other. In the diagram below, for example, for n=8 there are just five congruent triangles (four are shown in the octagon on the right and the fifth in the octagon on the left), two of which are scalene, making seven in total.  For n=10, eight triangles can be derived, four of them scalene, totaling twelve triangles.", null, "The set of scale factors used in the tiling are the triangle side lengths relative to the shortest overall size length, taken to have length one.  Any one of the triangles in the set is chosen and then dilated by one of the scale factors.\n\nJoseph showed that any such scaled-up triangle can be obtained by tiling with triangles from the original set. In this exercise there must, of course, be no gaps between the triangles; but more particularly, no corner of one triangle may break into the side of another. The latter requirement is an expression of the condition that the tiling be “proper.” He also showed that by imposing some natural extra conditions, the only sets of triangles that can exhibit self-similarity with dilation factors based on side lengths are those given using the above construction. This last condition on dilation factors means that the stacking of equilateral triangles, so common in the ancient world, is not a Bar Sagi tiling.\n\nAs in the case of Thurston self-similarity, this process can, in principle, be repeated.  However unlike Thurston tiling, the condition that the tiling be proper means that iterations are not immediately possible. Indeed, fulfilling this condition demands extra care in the way that tiling is implemented at each successive stage.\n\nIn the figure, one example of a Bar Sagi tiling is given.  It was obtained after three iterations, using the set of triangles obtained from a regular octagon.", null, "In this somewhat evocative tiling, the colors are more than just decorative: According to a well-known theorem on coloring maps, proved by Kenneth Apel and Wolfgang Haken in 1976, four-coloring is sufficient for any map on a sphere (or a plane) and can also be necessary for certain maps. Here, in the first iteration of Bar Sagi tiling, two-coloring is possible ‒ that is to say, only two colors are needed so that no two triangles sharing a common edge have the same color. After a second or further iterations this may no longer be possible; indeed, in the above example at least three colors are needed. However, it is not yet proven that three colors are sufficient for Bar Sagi tiling.", null, "One origin of the tiling considered here was a game Joseph had created for his granddaughter Bar (aged six at the time) by cutting triangles from colored paper with magnetic backing and letting her try to create tilings on the refrigerator door.\n\nAnother source of inspiration for the tiling was that the vertices of a proper self-similar triangle tiling can sometimes represent “weight diagrams” drawn in the plane. These diagrams are used in Lie algebra theory (Joseph’s main area of research) and can be projected to the plane in a rather specific manner. This presentation could not always be achieved, but the method led to the somewhat unexpected construction of certain “Weierstrass sections.”  These are important in separating the “orbits” that are obtained, for example, by rotating a sphere about its North-South axis.  In this context, such orbits are known as latitudes (or parallels).  Any (half) longitude going from the North to the South Pole cuts every latitude exactly once, thereby defining it.\n\nThe main thrust of Joseph’s work is in “non-commutative versions of algebraic geometry.”  A particular result of his has to do with the ways that classical and quantum mechanics, though similar, can exhibit calculable differences; and these can be thought to extend the well-known phenomenon of zero-point energy.", null, "Developing Bar Sagi tiling into a mathematical work dedicated to her memory was inspired by the beautiful poems written by Bar, most of which were found after her death. The poems relate in part to her painful and terrifying ordeal with bone cancer. They have been collected and published by Gvanim under the title “See Me Soar and Spread My Wings.”  Prof. Carlos Cardas, a leading oncologist at Cambridge University, UK, has declared this collection “to be mandatory reading for any aspiring cancer doctor.”\n\nNow show me,\n\nProve to me you know!\n\nWhat it means to be free!\n\nWhat it means to fly!\n\nReach up and touch the sky\n\nLet it out and let it start.\n\nAnd maybe some day there'll be\n\nA happy ending waiting for me\n\nFull of love and friends\n\nThat's how I want my journey to end.\n\n~ Bar Sagi\n\n### Tags:", null, "" ]

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[ "# Solidity v0.8.0 Breaking Changes\n\nThis section highlights the main breaking changes introduced in Solidity version 0.8.0. For the full list check the release changelog.\n\n## Silent Changes of the Semantics\n\nThis section lists changes where existing code changes its behaviour without the compiler notifying you about it.\n\n• Arithmetic operations revert on underflow and overflow. You can use `unchecked { ... }` to use the previous wrapping behaviour.\n\nChecks for overflow are very common, so we made them the default to increase readability of code, even if it comes at a slight increase of gas costs.\n\n• ABI coder v2 is activated by default.\n\nYou can choose to use the old behaviour using `pragma abicoder v1;`. The pragma `pragma experimental ABIEncoderV2;` is still valid, but it is deprecated and has no effect. If you want to be explicit, please use `pragma abicoder v2;` instead.\n\nNote that ABI coder v2 supports more types than v1 and performs more sanity checks on the inputs. ABI coder v2 makes some function calls more expensive and it can also make contract calls revert that did not revert with ABI coder v1 when they contain data that does not conform to the parameter types.\n\n• Exponentiation is right associative, i.e., the expression `a**b**c` is parsed as `a**(b**c)`. Before 0.8.0, it was parsed as `(a**b)**c`.\n\nThis is the common way to parse the exponentiation operator.\n\n• Failing assertions and other internal checks like division by zero or arithmetic overflow do not use the invalid opcode but instead the revert opcode. More specifically, they will use error data equal to a function call to `Panic(uint256)` with an error code specific to the circumstances.\n\nThis will save gas on errors while it still allows static analysis tools to distinguish these situations from a revert on invalid input, like a failing `require`.\n\n• If a byte array in storage is accessed whose length is encoded incorrectly, a panic is caused. A contract cannot get into this situation unless inline assembly is used to modify the raw representation of storage byte arrays.\n\n• If constants are used in array length expressions, previous versions of Solidity would use arbitrary precision in all branches of the evaluation tree. Now, if constant variables are used as intermediate expressions, their values will be properly rounded in the same way as when they are used in run-time expressions.\n\n• The type `byte` has been removed. It was an alias of `bytes1`.\n\n## New Restrictions\n\nThis section lists changes that might cause existing contracts to not compile anymore.\n\n• There are new restrictions related to explicit conversions of literals. The previous behaviour in the following cases was likely ambiguous:\n\n1. Explicit conversions from negative literals and literals larger than `type(uint160).max` to `address` are disallowed.\n\n2. Explicit conversions between literals and an integer type `T` are only allowed if the literal lies between `type(T).min` and `type(T).max`. In particular, replace usages of `uint(-1)` with `type(uint).max`.\n\n3. Explicit conversions between literals and enums are only allowed if the literal can represent a value in the enum.\n\n4. Explicit conversions between literals and `address` type (e.g. `address(literal)`) have the type `address` instead of `address payable`. One can get a payable address type by using an explicit conversion, i.e., `payable(literal)`.\n\n• Address literals have the type `address` instead of ```address payable```. They can be converted to `address payable` by using an explicit conversion, e.g. `payable(0xdCad3a6d3569DF655070DEd06cb7A1b2Ccd1D3AF)`.\n\n• There are new restrictions on explicit type conversions. The conversion is only allowed when there is at most one change in sign, width or type-category (`int`, `address`, `bytesNN`, etc.). To perform multiple changes, use multiple conversions.\n\nLet us use the notation `T(S)` to denote the explicit conversion `T(x)`, where, `T` and `S` are types, and `x` is any arbitrary variable of type `S`. An example of such a disallowed conversion would be `uint16(int8)` since it changes both width (8 bits to 16 bits) and sign (signed integer to unsigned integer). In order to do the conversion, one has to go through an intermediate type. In the previous example, this would be `uint16(uint8(int8))` or `uint16(int16(int8))`. Note that the two ways to convert will produce different results e.g., for `-1`. The following are some examples of conversions that are disallowed by this rule.\n\n• `address(uint)` and `uint(address)`: converting both type-category and width. Replace this by `address(uint160(uint))` and `uint(uint160(address))` respectively.\n\n• `payable(uint160)`, `payable(bytes20)` and `payable(integer-literal)`: converting both type-category and state-mutability. Replace this by `payable(address(uint160))`, `payable(address(bytes20))` and `payable(address(integer-literal))` respectively. Note that `payable(0)` is valid and is an exception to the rule.\n\n• `int80(bytes10)` and `bytes10(int80)`: converting both type-category and sign. Replace this by `int80(uint80(bytes10))` and `bytes10(uint80(int80)` respectively.\n\n• `Contract(uint)`: converting both type-category and width. Replace this by `Contract(address(uint160(uint)))`.\n\nThese conversions were disallowed to avoid ambiguity. For example, in the expression ```uint16 x = uint16(int8(-1))```, the value of `x` would depend on whether the sign or the width conversion was applied first.\n\n• Function call options can only be given once, i.e. `c.f{gas: 10000}{value: 1}()` is invalid and has to be changed to `c.f{gas: 10000, value: 1}()`.\n\n• The global functions `log0`, `log1`, `log2`, `log3` and `log4` have been removed.\n\nThese are low-level functions that were largely unused. Their behaviour can be accessed from inline assembly.\n\n• `enum` definitions cannot contain more than 256 members.\n\nThis will make it safe to assume that the underlying type in the ABI is always `uint8`.\n\n• Declarations with the name `this`, `super` and `_` are disallowed, with the exception of public functions and events. The exception is to make it possible to declare interfaces of contracts implemented in languages other than Solidity that do permit such function names.\n\n• Remove support for the `\\b`, `\\f`, and `\\v` escape sequences in code. They can still be inserted via hexadecimal escapes, e.g. `\\x08`, `\\x0c`, and `\\x0b`, respectively.\n\n• The global variables `tx.origin` and `msg.sender` have the type `address` instead of `address payable`. One can convert them into `address payable` by using an explicit conversion, i.e., `payable(tx.origin)` or `payable(msg.sender)`.\n\nThis change was done since the compiler cannot determine whether or not these addresses are payable or not, so it now requires an explicit conversion to make this requirement visible.\n\n• Explicit conversion into `address` type always returns a non-payable `address` type. In particular, the following explicit conversions have the type `address` instead of ```address payable```:\n\n• `address(u)` where `u` is a variable of type `uint160`. One can convert `u` into the type `address payable` by using two explicit conversions, i.e., `payable(address(u))`.\n\n• `address(b)` where `b` is a variable of type `bytes20`. One can convert `b` into the type `address payable` by using two explicit conversions, i.e., `payable(address(b))`.\n\n• `address(c)` where `c` is a contract. Previously, the return type of this conversion depended on whether the contract can receive Ether (either by having a receive function or a payable fallback function). The conversion `payable(c)` has the type ```address payable``` and is only allowed when the contract `c` can receive Ether. In general, one can always convert `c` into the type `address payable` by using the following explicit conversion: `payable(address(c))`. Note that `address(this)` falls under the same category as `address(c)` and the same rules apply for it.\n\n• The `chainid` builtin in inline assembly is now considered `view` instead of `pure`.\n\n• Unary negation cannot be used on unsigned integers anymore, only on signed integers.\n\n## Interface Changes\n\n• The output of `--combined-json` has changed: JSON fields `abi`, `devdoc`, `userdoc` and `storage-layout` are sub-objects now. Before 0.8.0 they used to be serialised as strings.\n\n• The “legacy AST” has been removed (`--ast-json` on the commandline interface and `legacyAST` for standard JSON). Use the “compact AST” (`--ast-compact--json` resp. `AST`) as replacement.\n\n• The old error reporter (`--old-reporter`) has been removed.\n\n## How to update your code\n\n• If you rely on wrapping arithmetic, surround each operation with `unchecked { ... }`.\n\n• Optional: If you use SafeMath or a similar library, change `x.add(y)` to `x + y`, `x.mul(y)` to `x * y` etc.\n\n• Add `pragma abicoder v1;` if you want to stay with the old ABI coder.\n\n• Optionally remove `pragma experimental ABIEncoderV2` or `pragma abicoder v2` since it is redundant.\n\n• Change `byte` to `bytes1`.\n\n• Add intermediate explicit type conversions if required.\n\n• Combine `c.f{gas: 10000}{value: 1}()` to `c.f{gas: 10000, value: 1}()`.\n\n• Change `msg.sender.transfer(x)` to `payable(msg.sender).transfer(x)` or use a stored variable of `address payable` type.\n\n• Change `x**y**z` to `(x**y)**z`.\n\n• Use inline assembly as a replacement for `log0`, …, `log4`.\n\n• Negate unsigned integers by subtracting them from the maximum value of the type and adding 1 (e.g. `type(uint256).max - x + 1`, while ensuring that `x` is not zero)" ]

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[ "", null, "The Olympiad is a web site devoted to those who are looking for practice problems in algebra. The material here is presented in two forms: as interactive question-and-answer games, and as traditional text files. All of the problems have been written by John Mason, who has a Ph.D. in mathematics from the University of Colorado.\n\nNote: This list is meant to be funny and entertaining.  I do not mean to offend anyone, but if I did, it wasn’t my intention.  Please don’t send me nasty emails or try to harm me in any way. Introduction – How to solve algebra problems fast Pythagoras – The pizza theorem Distance between points on x and y axis (advanced – further than the first test) Area of triangle Properties of a line Negative numbers Finding Pythagoras theorem Area of a quadrilateral Proving the circle area formula (advanced – further than the second test) Points on x and y axis relative to square  (advanced – further than the third test) (y = mx + k) function factorization of quadratic equations finding roots of polynomials in (x) and (y) functions distance between two ordered pairs polynomial division factors of (x^2-9xy+2x+3) Quadric surface area of sphere Cuboid perimeter General formulas for surface area and volume Examples Multiple integrals\n\nIntroduction to Algebra includes the following major parts: algebraic expressions, the real number system, the complex number system, polynomials, rational functions, roots and radicals, exponents and logarithms, trigonometry (including exponential functions), systems of equations (linear and non-linear), matrices, determinants and discrete structures.\n\n## topics in algebra and analysis preparing for the mathematical olympiad pdf\n\nalgebra for olympiad problems and solutions j. v. r. pdf\n\nPrice : \\$64.99\nFeatures :\n\n Release Date 2011-09-21T00:00:00.000Z\n\n### IMO – International Mathematical Olympiad: The Hardest Math Exams for High Schoolers (Mathematical Olympiads for Elementary, Middle and High School)\n\nPrice : \\$26.71\nFeatures :\n\n Item Dimensions Height 9 Inches Width 0.41 Inches Length 6 Inches\n\nPrice : \\$61.22\nFeatures :\n\n• Birkhauser\n\n Item Dimensions Height 9.25 Inches Width 0.69 Inches Length 6.1 Inches Weight 1.03176338616 Pounds\n\n### 100 Math Brainteasers (Grade 7, 8, 9, 10). Arithmetic, Algebra and Geometry Brain Teasers, Puzzles, Games and Problems with Solutions: Math olympiad contest problems for elementary and middle schools\n\nPrice : \\$6.27\nFeatures :\n\n• Used Book in Good Condition" ]

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[ "Hide\n\n# Left and Right", null, "Image from Pixabay\n\nWith modern technology advancement, it is now possible to deliver mail with a robot! There is a neighborhood on a long horizontal road, on which there are $n$ houses numbered $1$ to $n$ from left to right. Every day a mail delivery robot receives a pile of letters with exactly one letter for each house. Due to mechanical restrictions, the robot cannot sort the letters. It always checks the letter on top of the pile, visits the house that should receive that letter and delivers it. The robot repeats this procedure until all the letters are delivered. As a result, each of the $n$ houses is visited by the robot exactly once during the mail delivery of a single day.\n\nThe mail delivery robot has a tracking device that records its delivery route. One day the device was broken, and the exact route was lost. However, the technical team managed to recover the moving directions of the robot from the broken device, which are represented as a string consisting of $n-1$ letters. The $i$-th letter of the string is ‘L’ (or ‘R’) if the $(i+1)$-th house visited by the robot is on the left (or right) of the $i$-th house visited. For example, if $n = 4$ and the robot visited the houses in the order of $2, 4, 3, 1$, its moving directions would be “RLL”.\n\nWith the moving directions, it may be possible to determine the order in which the robot visited the houses. The technical team has asked you to write a program to do that. There can be multiple orders producing the same moving directions, among which you should find the lexicographically earliest order.\n\n## Input\n\nThe input has a single integer $n$ ($2 \\leq n \\leq 2 \\cdot 10^5$) on the first line. The second line has a string of length $n-1$ consisting of letters ‘L’ and ‘R’ giving the moving directions of the robot.\n\n## Output\n\nOutput the lexicographically earliest order in which the robot may have visited the houses and delivered the letters according to the moving directions. Consider two different integer sequences of equal length $A = (a_1, a_2, \\ldots , a_ k)$ and $B = (b_1, b_2, \\ldots , b_ k)$, and let $1 \\le i \\le k$ be the lowest-numbered index where $a_ i \\ne b_ i$. Then $A$ is lexicographically earlier than $B$ if $a_ i < b_ i$; otherwise $B$ is lexicographically earlier than $A$.\n\nSample Input 1 Sample Output 1\n3\nLR\n\n2\n1\n3\n\nSample Input 2 Sample Output 2\n6\nRLLRL\n\n1\n4\n3\n2\n6\n5\n\nSample Input 3 Sample Output 3\n6\nRRRLL\n\n1\n2\n3\n6\n5\n4\n\nCPU Time limit 1 second\nMemory limit 1024 MB\nDifficulty 3.2medium\nStatistics Show" ]

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[ "# Cone Calculator\n\nCalculate the various properties of a cone for given values.\n\nSlant Height(l) of Cone: [ Sqrt(r² + h²) ]\n\n Enter the radius : Enter the height : Slant Height(l) of Cone =\n\nVolume of Cone: [ (1/3)πr²h ]\n\n Enter the radius : Enter the height : Volume of Cone =\n\nCurved Surface Area(CSA) of Cone: [ πrl ]\n\n Enter the radius : Enter the height : Curved Surface Area(CSA) of Cone =\n\nTotal Surface Area(TSA) of Cone: [ πr(l+r) ]\n\n Enter the radius : Enter the height : Total Surface Area(TSA) of Cone =\nA geometrical shape that has a circular base and a single vertex called as Cone. In case vertex is over the center of the base is called a right cone otherwise it is called as oblique cone.\nAny object has cone shape is called conical.", null, "## Formulas :\n\nSlant Height(l): [ Sqrt(r² + h²) ]\n\nTotal Surface Area(TSA): [ πr(l+r) ]\n\nCurved Surface Area(CSA): [ πrl ]\n\nVolume of Cone: [ (1/3)πr²h ]\nwhere:\nh = value of height\ns = value of slant height\nV = value of volume\nL = value of lateral surface area\nB = value of base surface area\nA = value of total surface area\nπ = pi = value of 3.1415926535898\n\n### Instructions for using surface area of a cone calculator\n\nUnderstand it step-by-step:\n\n1.For circular base enter the value of the given radius . Depending on the condition different units can be used.\n\n2.Whatever value is known to us based on it you need to enter the height of the cone or the slant height of the cone.\n\n3.Always enter the value of height greater than zero." ]

[ null, "https://eguruchela.com/images/cone.gif", null ]

[ "", null, "", null, "", null, "", null, "", null, "Associated for 1 Year 5 Months\nPro Patria et Educatione\n• Areas:\nSco 145- 146, Sector 8c, Chandigarh, 160009\n• Courses:\nMathematics, Science, Social Studies, Statistics, B.Tech Tuition, Mechanical, Drawing, German, French, Study in Germany, Olympiad Exam Preparation, Mental Mathematics, Regional Mathematical Olympiad (RMO), Indian National Mathematical Olympiad (INMO), AutoCAD Training, Basic Computer, School level ... More [+]\n• Overview:\n\nA startup venture in the field of private educational institutions where the quality of education is combined with innovation and the overall development of the student is of prime importance.", null, "Syllabus Offered:\n\nConfidential\n\n Class 9 - 10 Mathematics, Mathematics, Science, Science, Social Studies, Social Studies Class 11 - 12 Mathematics, Mathematics, Statistics, Statistics College Level Mathematics, Mathematics, Statistics, Statistics, B.Tech Tuition, B.Tech Tuition Engineering Subjects Mathematics, Mathematics, Mechanical, Mechanical, Drawing, Drawing Languages German, German, French, French Study Abroad Study in Germany, Study in Germany Academic Enrichment Olympiad Exam Preparation, Olympiad Exam Preparation, Mental Mathematics, Mental Mathematics, Regional Mathematical Olympiad (RMO), Regional Mathematical Olympiad (RMO), Indian National Mathematical Olympiad (INMO), Indian National Mathematical Olympiad (INMO) Class 6 - 8 Mathematics, Mathematics, Science, Science, Social Studies, Social Studies Graphics & Multimedia AutoCAD Training, AutoCAD Training Basic Computer / Office Basic Computer, Basic Computer, School level computer, School level computer, MS Office, MS Office, Advanced Excel, Advanced Excel\n\nSco 145- 146, Sector 8c, Chandigarh, 160009\n\nBrochure:", null, "Courses Offered\n\nFrench Language LearningFrench Language Learning\n\nFrench\nArea: Sector 8\nPin Code: 160009\nFrench is a Romance language of the Indo-European family. It descended from the Vulgar Latin of the Roman Empire, as did all Roman\n• Su\n• M\n• T\n• W\n• T\n• F\n• S\n15 Jul 18 - 31 Mar 19\n\nMathematics Coaching (Class 9 & Class 10)Mathematics Coaching (Class 9 & Class 10)\n\nMathematics\nArea: Sector 8\nPin Code: 160009\nStandard 09 and 10 Coaching for Maths under the able guidance of Mr. Sanjay Singh in small batches for individ\n• Su\n• M\n• T\n• W\n• T\n• F\n• S\n11 Apr 18 - 28 Feb 19\n\nMathematics Coaching (Class 11 & Class 12)Mathematics Coaching (Class 11 & Class 12)\n\nMathematics\nArea: Sector 8\nPin Code: 160009\nStandard 11 and 12 Coaching for Maths under the able guidance of Mr. Sanjay Singh in small batches for individual attent\n• Su\n• M\n• T\n• W\n• T\n• F\n• S\n15 Jul 18 - 28 Feb 19\n\nScience Coaching (Class 9 & Class 10)Science Coaching (Class 9 & Class 10)\n\nScience\nArea: Sector 8\nPin Code: 160009\nStandard 09 and 10 Coaching for Science under the able guidance of Mr. Sanjay Singh in small batches for\n• Su\n• M\n• T\n• W\n• T\n• F\n• S\n11 Apr 18 - 28 Feb 19\n\nGerman Language StudyGerman Language Study\n\nGerman\nArea: Sector 8\nPin Code: 160009\nOur German learning programme is offered in three levels i.e., Basic, Intermediate and Advance as per the need of the student. We\n• Su\n• M\n• T\n• W\n• T\n• F\n• S\n15 Jul 18 - 31 Aug 18\n\nM1 & M2 (Engineering Mathematics)M1 & M2 (Engineering Mathematics)\n\nMathematics and 1 more\nArea: Sector 8\nPin Code: 160009\nBest coaching for Maths (M1 & M2) under the able guidance of Mr. Sanjay Singh in small batches for individual attent\n• Su\n• M\n• T\n• W\n• T\n• F\n• S\n01 Aug 18 - 31 Oct 18", null, "Educational Resources\n\nNotes written by me \n\nEnergy Conservation Measure\n• File(s) contain : 1\n\nPDF File for Energy Conservation Measure.\n\nEnergy Efficient Motor\n• File(s) contain : 1\n\nPDF File for Energy Efficient Motor\n\nAutomobile Engineering (4 Stroke Cycle)\n• File(s) contain : 1\n\nNotes on 4 Stroke Cycle for Automobile Engineering.\n\nIndustrial Design Engineering (Aesthetics)\n• File(s) contain : 1\n\nNote on Aesthetics for Industrial Design Engineering.\n\nHistory Notes For Class 6\n• File(s) contain : 1\n\nListing The Summary Of The Subject With Some Important Questions.\n\nPresentations prepared by me", null, "D\n\nB\n\nC\n\nC\n\nD\n\nA\n\nC\n\nD\n\nD\n\nB\n\nD\n\nD\n\nA\n\nB\n\nA\n\nB\n\nB\n\nA\n\nB\n\nC\n\nC\n\nC\n\nC\n\nItallic\n\nB\n\nD\n\nD\n\nC\n\nD\n\nA\n\n• Question:Why is an excel file known as a workbook? Posted in:Bank Clerical | Date: 13/01/2018\n\nBeceause it consists of Workbooks in the form of Rows, Columns and Cells\n\nD\n\nC\n\nB\n\nA\n\nD\n\nB\n\nB\n\n• Question:How do I describe a conservative force? Posted in:Physics | Date: 13/01/2018\n\nA conservative force is a force that depends on the property that the ork done in moving a particle between two paths is independent of the path taken. Total work done by a conservative force is independent of the path resulting in a given displacement and is equal to zero when the path is a closed loop.\n\n• Question:Are Kanchenjunga and k2 same? Posted in:Geography | Date: 13/01/2018\n\nYes both are Mountains\n\nMt. Everest\n\nB\n\nB\n\nC\n\nB\n\n• Question:Best method to determine paternity is Posted in:Biology | Date: 13/01/2018\n\nDNA fingerprinting\n\nC\n\nB\n\nB\n\nB\n\nB\n\nB\n\nB\n\nB\n\nD\n\nC\n\nA\n\nC\n\nC\n\nB\n\nA\n\nC\n\nA\n\nD\n\nC\n\nB\n\nB\n\nA\n\nB\n\nD\n\nB\n\nB\n\n• Question:A rain drop of mass 2.00 g falls from a height 1.00 km. It hits the ground with ... Posted in:Physics | Date: 13/01/2018\n\nForce = mg = 0.002*10 = 0.02 N     (since 2 g = 0.002 kg)\n\nDisplacement = height = 1km = 1000 m\n\nWork done = Force * displacement\n\nWork done = 0.02*1000 = 20 N.m\n\nC\n\n• Question:How do I find the magnitude of the centripetal acceleration of a particle on the... Posted in:Physics | Date: 13/01/2018\n\n(w^2)r where r=0.15,w=2pi(20)\n\nC\n\nA\n\n• Question:A ball is dropped downward from the roof of a building and simultaneously anothe... Posted in:Physics | Date: 13/01/2018\n\nreach at same point of time\n\n• Question:An object thrown from an aeroplane is an example for _____________ Posted in:NEET | Date: 13/01/2018\n\nprojectile motion\n\nA\n\nA\n\nB\n\nB\n\nD\n\nD\n\nD\n\n• Question:Consider the quantities, pressure, power, energy, impulse, gravitational potenti... Posted in:Physics | Date: 13/01/2018\n\nImpulse and Area\n\nC\n\nB\n\nA\n\nC\n\nC\n\nD\n\nB\n\nA\n\nD\n\nB\n\n• Question:which part of flower form it's fruit Posted in:Biology | Date: 13/01/2018\n\nOvary of the flower turns to fruit\n\n• Question:A car acceleration from rest at a constant rate α for some time after... Posted in:AMITE | Date: 13/01/2018\n\nATP, Maximum Velocity v = α*t1 or v = β*t2\n\n=> t1 = v/α and t2 = v/β\n\n=> t = t1 + t2 = v(1/α + 1/β) = v*(α+β)/αβ\n\n=> v = αβt/(α+β)  .\n\n• Question:A 150 m long train is moving with uniform velocity of 45 km/h. The time taken by... Posted in:Physics | Date: 13/01/2018\n\nTotal dist = (850 + 150)m = 1000 m = 1 km\n\nSpeed = 45 km/h\n\nHence, time = 1/45 h = 60/45 mins = 4/3 mins = 1 min 20 secs\n\n• Question:What is mechanics? Posted in:Physics | Date: 13/01/2018\n\nmechanics is the  mathematical study of forces producing motion in a body.\n\n• Question:How is the free energy change related to the enthalpy and entropy change of a su... Posted in:Chemistry | Date: 13/01/2018\n\nChange in Gibbs Free Energy = Change in Enhalpy - Temp_in_K x Change in Entropy\n\nD\n\nA\n\nB\n\nD\n\nC\n\nB\n\nC\n\nC\n\nD\n\nB\n\nD\n\nB\n\nA\n\n• Question:A person crosses a 600 600 metre long street in 5 5 minutes. What is his speed... Posted in:Bank PO | Date: 13/01/2018\n\n600 meter = 600/1000 km = 3/5 km\n\n5 min = 5/60 hrs = 1/12 hrs\n\nspeed = distance/time = 600 meter / 5 min =  (3/5) / (1/12) = 7.2 km/hr\n\n• Question:Find the odd man out. 445, 221, 109, 46, 25, 11, 4 Posted in:Bank PO | Date: 13/01/2018\n\nOdd one is 46 Series is as 4*2 +3 = 11 11 *2 +3= 25 25*2+3. = 53\n\n• Question:A person walked diagonally across a square plot. Approximately, what was the per... Posted in:Bank PO | Date: 13/01/2018\n\nLet us consider the square of side 'a' units, therefore diagonal = sqrt(2)*a units, walking along the edges = 2a units\n\nTherefore, Distance saved = 2a - sqrt(2)a\n\npercentage = [{2a - sqrt(2)a}/{2a}] * 100 = 29.28%\n\n• Question:What are the units of nervous system? Posted in:Biology | Date: 13/01/2018\n\nNeuron is the unit of nervous system\n\n• Question:A man has Rs. 480 in the denominations of one-rupee notes, five-rupee notes and ... Posted in:Bank PO | Date: 13/01/2018\n\nLet the man have x number of notes of each type.\n\nHence 1 xx + 5 x x  + 10 x x = 480\n\nSo 16x = 480 Or x = 30     Ans\n\n• Question:A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is th... Posted in:SSC Exams | Date: 13/01/2018\n\nthe speed of train in m/s = 1000m/sec\n\nas the time taken by train is 9 sec\n\nthe length of the train will be 9000m\n\n• Question:What measures, as an individual, you would take to reduce environmental pollutio... Posted in:Biology | Date: 13/01/2018\n\n1.plant trees\n\n2.car pooling\n\n3. not burnng plastic\n\n4.recycle things\n\n5.dont cut trees\n\n• Question:What are the major causes of species losses in a geographical region? Posted in:Biology | Date: 13/01/2018\n\nclimatic and regional factors play an important role\n\n• Question:What is a bisexual flower? Posted in:NEET | Date: 13/01/2018\n\nThe one having both anther and ovary\n\nA\n\nA\n\nC\n\n• Question:Identify the compounds A, B, C, D, E and F. Posted in:Chemistry | Date: 13/01/2018\n\nA CH3Cl alkylation\n\nB CrO2Cl2 etard reaction\n\nC LiAlH4 in ether reduction\n\nD LiOH and AlH3\n\nE Benzoin C6H5COCHOHC6H5\n\nF 2[O] nascent oxygen\n\nD\n\n• Question:In the series of reactions, the product C is Posted in:Chemistry | Date: 13/01/2018\n\nC will be Methyl Cyanide (CH3CN).\n\nB\n\nB\n\nA\n\nB\n\nA\n\nD\n\nE\n\nC\n\nC\n\nD\n\nA\n\nB\n\nD\n\nA\n\nB\n\nA\n\nC\n\nB\n\nD\n\nB\n\nA\n\nA\n\nC\n\nB\n\nB\n\nC\n\nB\n\nB\n\nC\n\nC\n\nA\n\nB\n\nD\n\nB\n\nB\n\nD\n\nD\n\nD\n\nB\n\nB\n\nD\n\nB\n\nC\n\nB\n\n• Question:Find the smallest number that leaves a remainder of 4 on division by 5, 5 on div... Posted in:GRE | Date: 13/01/2018\n\nLCM(5,6,7,8,9)-1=2519\n\narea =d^2/2\n\n12*12/2=72\n\n• Question:Complete the series: 2, 12, 36, 80, .... Posted in:Bank PO | Date: 13/01/2018\n\nLogic behind this series is - n˄2 + n˄3\n\nSo next number of the series is 150\n\n• Question:Write a balanced nuclear reaction showing emission of a β-particle by ... Posted in:Physics | Date: 13/01/2018\n\n234Th90 → 234Pa91 + 0e−1 + energy\n\nC\n\nC\n\nA\n\nA\n\nC\n\nA\n\n• Question:The speed of sound waves depends on temperatre but speed of light waves does not... Posted in:AMIE | Date: 13/01/2018\n\nMagnetic flux density at the axial position::: |B axial| = myu/ 4π =2|M|r/(r^2-l^2)^2\n\nr=√3/4 a\n\n• Question:In the Arrhenius equation, what does the factor \\(e^{-E_a}/RT\\) corresponds... Posted in:Chemistry | Date: 13/01/2018\n\nit is called Arhenius factor or rate  factor\n\n• Question:A coordination compound with molecular formula CrCl\\(_3\\). 4H\\(_2\\)O precipitate... Posted in:Chemistry | Date: 13/01/2018\n\n[Cr(H2O)4Cl2]Cl⋅2H2O.\n\nE\n\nC\n\n• Question:Raju is standing in the garden and facing North-West. He turns 90° in clockw... Posted in:Bank PO | Date: 13/01/2018\n\nwest direction\n\none move in 90 clockwise then 135 anticlockwise\n\n3-5 hrs/day\n\nC\n\n• Question:If one mass of one electron is 9.11 x 10\\(^{-31}\\)kg, then how many electrons wo... Posted in:Physics | Date: 13/01/2018\n\n9.11*10^-31kg = 1 electron\n\n1 kg=1/(9.11*10^-31)=10.97*10^29\n\n• Question:Which field vector, electric or magnetic, is used to represent the polarisation ... Posted in:Physics | Date: 13/01/2018\n\nElectromagnetic field\n\n• Question:Which special characteristic of light is demonstrated only by the phenomenon of ... Posted in:Physics | Date: 13/01/2018\n\nOptical characteristics\n\n• Question:Which type of waves show the property of polarisation? Posted in:Physics | Date: 13/01/2018\n\nTransverse waves shows the property of polarisation\n\nZero\n\nYes\n\n• Question:What are the two strategies evolved to prevent self-pollination in flowers. Posted in:Biology | Date: 13/01/2018\n\n1) Self-compatibility 2)Protogyny or protandry.\n\n• Question:What are chasmogamous flowers? Can cross-pollination occur in cleistogamous flow... Posted in:NEET | Date: 13/01/2018\n\nThere are two types of flowers present in plants namely Oxalis and Viola − chasmogamous and cleistogamous flowers.\n\nC\n\nD\n\nB\n\nC\n\nC\n\nB\n\nB\n\nC\n\nB\n\nC\n\nD\n\n• Question:A rope can make 70 rounds of the circumference of a cylinder whose radius of the... Posted in:Mathematics | Date: 14/01/2018\n\n1 round = 2*pi*r = 2*pi*14 = 28*pi\n\nSo 70 rounds = 70*28*pi = Length of rope\n\nNow length of 1 round of cylinder = 2*pi*20\n\nNumber of rounds,n = Length of rope/length of 1 round of cylinder\n\nn = 70*28*pi/40*pi\n\n=> n = 49 rounds\n\n• Question:If the price of 357 apples is Rs.1517.25, what will be the approximate price of ... Posted in:Mathematics | Date: 14/01/2018\n\ncost of one apple is=1517.25/357==4.25\n\n49 dozen has 49*12 apples =588 apples\n\ncost is 588*4.25=2499\n\n• Question: x men working x hours per day can do xx units of a wor... Posted in:Bank Clerical | Date: 14/01/2018\n\nWork can be completed= x*x*xx*x/y*y*y\n\n= x^5/y^3.\n\n• Question:A hostel had provisions for 250 men for 40 days. If 50 men left the hostel, how ... Posted in:Mathematics | Date: 14/01/2018\n\nTotal quantity of food = 250 *40 units\n\nNow remaining NUMBER of people = 200\n\nNumber for days food would survive = ( 250*40)/200 = 50\n\nFood would be available for 50 days.\n\n• Question:Both lithium and magnesium display several similar properties due to the diagona... Posted in:AIPMT | Date: 14/01/2018\n\nC", null, "Reviews\n• startup venture in the field of private educational institutions where the quality of education is combined with innovation\nReviewed by: Amit\n\nPost a Requirement\n\nPost an enquiry and get instant responses from coaching classes\n\nDrop Us a Query" ]

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[ "# #00e10c Color Information\n\nIn a RGB color space, hex #00e10c is composed of 0% red, 88.2% green and 4.7% blue. Whereas in a CMYK color space, it is composed of 100% cyan, 0% magenta, 94.7% yellow and 11.8% black. It has a hue angle of 123.2 degrees, a saturation of 100% and a lightness of 44.1%. #00e10c color hex could be obtained by blending #00ff18 with #00c300. Closest websafe color is: #00cc00.\n\n• R 0\n• G 88\n• B 5\nRGB color chart\n• C 100\n• M 0\n• Y 95\n• K 12\nCMYK color chart\n\n#00e10c color description : Pure (or mostly pure) lime green.\n\n# #00e10c Color Conversion\n\nThe hexadecimal color #00e10c has RGB values of R:0, G:225, B:12 and CMYK values of C:1, M:0, Y:0.95, K:0.12. Its decimal value is 57612.\n\nHex triplet RGB Decimal 00e10c `#00e10c` 0, 225, 12 `rgb(0,225,12)` 0, 88.2, 4.7 `rgb(0%,88.2%,4.7%)` 100, 0, 95, 12 123.2°, 100, 44.1 `hsl(123.2,100%,44.1%)` 123.2°, 100, 88.2 00cc00 `#00cc00`\nCIE-LAB 78.388, -78.202, 74.584 26.99, 53.874, 9.324 0.299, 0.597, 53.874 78.388, 108.066, 136.357 78.388, -74.137, 95.234 73.399, -62.81, 43.847 00000000, 11100001, 00001100\n\n# Color Schemes with #00e10c\n\n• #00e10c\n``#00e10c` `rgb(0,225,12)``\n• #e100d5\n``#e100d5` `rgb(225,0,213)``\nComplementary Color\n• #65e100\n``#65e100` `rgb(101,225,0)``\n• #00e10c\n``#00e10c` `rgb(0,225,12)``\n• #00e17d\n``#00e17d` `rgb(0,225,125)``\nAnalogous Color\n• #e10065\n``#e10065` `rgb(225,0,101)``\n• #00e10c\n``#00e10c` `rgb(0,225,12)``\n• #7d00e1\n``#7d00e1` `rgb(125,0,225)``\nSplit Complementary Color\n• #e10c00\n``#e10c00` `rgb(225,12,0)``\n• #00e10c\n``#00e10c` `rgb(0,225,12)``\n• #0c00e1\n``#0c00e1` `rgb(12,0,225)``\n• #d5e100\n``#d5e100` `rgb(213,225,0)``\n• #00e10c\n``#00e10c` `rgb(0,225,12)``\n• #0c00e1\n``#0c00e1` `rgb(12,0,225)``\n• #e100d5\n``#e100d5` `rgb(225,0,213)``\n• #009508\n``#009508` `rgb(0,149,8)``\n• #00ae09\n``#00ae09` `rgb(0,174,9)``\n• #00c80b\n``#00c80b` `rgb(0,200,11)``\n• #00e10c\n``#00e10c` `rgb(0,225,12)``\n• #00fb0d\n``#00fb0d` `rgb(0,251,13)``\n• #15ff21\n``#15ff21` `rgb(21,255,33)``\n• #2fff3a\n``#2fff3a` `rgb(47,255,58)``\nMonochromatic Color\n\n# Alternatives to #00e10c\n\nBelow, you can see some colors close to #00e10c. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #2ce100\n``#2ce100` `rgb(44,225,0)``\n• #19e100\n``#19e100` `rgb(25,225,0)``\n• #07e100\n``#07e100` `rgb(7,225,0)``\n• #00e10c\n``#00e10c` `rgb(0,225,12)``\n• #00e11f\n``#00e11f` `rgb(0,225,31)``\n• #00e132\n``#00e132` `rgb(0,225,50)``\n• #00e144\n``#00e144` `rgb(0,225,68)``\nSimilar Colors\n\n# #00e10c Preview\n\nThis text has a font color of #00e10c.\n\n``<span style=\"color:#00e10c;\">Text here</span>``\n#00e10c background color\n\nThis paragraph has a background color of #00e10c.\n\n``<p style=\"background-color:#00e10c;\">Content here</p>``\n#00e10c border color\n\nThis element has a border color of #00e10c.\n\n``<div style=\"border:1px solid #00e10c;\">Content here</div>``\nCSS codes\n``.text {color:#00e10c;}``\n``.background {background-color:#00e10c;}``\n``.border {border:1px solid #00e10c;}``\n\n# Shades and Tints of #00e10c\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, #000900 is the darkest color, while #f5fff5 is the lightest one.\n\n• #000900\n``#000900` `rgb(0,9,0)``\n• #001d02\n``#001d02` `rgb(0,29,2)``\n• #003003\n``#003003` `rgb(0,48,3)``\n• #004404\n``#004404` `rgb(0,68,4)``\n• #005805\n``#005805` `rgb(0,88,5)``\n• #006b06\n``#006b06` `rgb(0,107,6)``\n• #007f07\n``#007f07` `rgb(0,127,7)``\n• #009308\n``#009308` `rgb(0,147,8)``\n• #00a609\n``#00a609` `rgb(0,166,9)``\n• #00ba0a\n``#00ba0a` `rgb(0,186,10)``\n• #00cd0b\n``#00cd0b` `rgb(0,205,11)``\n• #00e10c\n``#00e10c` `rgb(0,225,12)``\n• #00f50d\n``#00f50d` `rgb(0,245,13)``\n• #09ff16\n``#09ff16` `rgb(9,255,22)``\n• #1dff29\n``#1dff29` `rgb(29,255,41)``\n• #30ff3b\n``#30ff3b` `rgb(48,255,59)``\n• #44ff4e\n``#44ff4e` `rgb(68,255,78)``\n• #58ff61\n``#58ff61` `rgb(88,255,97)``\n• #6bff73\n``#6bff73` `rgb(107,255,115)``\n• #7fff86\n``#7fff86` `rgb(127,255,134)``\n• #93ff98\n``#93ff98` `rgb(147,255,152)``\n• #a6ffab\n``#a6ffab` `rgb(166,255,171)``\n• #baffbd\n``#baffbd` `rgb(186,255,189)``\n• #cdffd0\n``#cdffd0` `rgb(205,255,208)``\n• #e1ffe3\n``#e1ffe3` `rgb(225,255,227)``\n• #f5fff5\n``#f5fff5` `rgb(245,255,245)``\nTint Color Variation\n\n# Tones of #00e10c\n\nA tone is produced by adding gray to any pure hue. In this case, #687969 is the less saturated color, while #00e10c is the most saturated one.\n\n• #687969\n``#687969` `rgb(104,121,105)``\n• #5f8261\n``#5f8261` `rgb(95,130,97)``\n• #578a59\n``#578a59` `rgb(87,138,89)``\n• #4e9352\n``#4e9352` `rgb(78,147,82)``\n• #459c4a\n``#459c4a` `rgb(69,156,74)``\n• #3da442\n``#3da442` `rgb(61,164,66)``\n``#34ad3a` `rgb(52,173,58)``\n• #2bb633\n``#2bb633` `rgb(43,182,51)``\n• #23be2b\n``#23be2b` `rgb(35,190,43)``\n• #1ac723\n``#1ac723` `rgb(26,199,35)``\n• #11d01b\n``#11d01b` `rgb(17,208,27)``\n• #09d814\n``#09d814` `rgb(9,216,20)``\n• #00e10c\n``#00e10c` `rgb(0,225,12)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #00e10c is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

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[ "Cody\n\n# Problem 179. Monte-Carlo integration\n\nSolution 1590930\n\nSubmitted on 24 Jul 2018\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   Fail\nfun = @(x) x^2; assert(abs((dquad(1,fun) - 1/3)*3)<0.01)\n\nAssertion failed.\n\n2   Pass\nfun = @(x) x(1)*x(2); assert(abs((dquad(2,fun) - 0.25)*4)<0.01)\n\n3   Pass\nfun = @(x) 1+sin(x(1)); assert(abs((dquad(50,fun) - 1.45969769)/1.45969769)<0.01)\n\n4   Pass\nd = randi([5 10],1) r = rand*0.8 fun = @(x) 2^d*(norm(x)>r); dball = exp(d/2*log(pi)+d*log(r)-gammaln(d/2+1)); assert(abs((dquad(d,fun) - 2^d+dball)/(2^d-dball))<0.01)\n\nd = 9 r = 0.3683" ]

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[ "# What is 204 Years in Weeks?\n\n## Convert 204 Years to Weeks\n\nTo calculate 204 Years to the corresponding value in Weeks, multiply the quantity in Years by 52.142857142857 (conversion factor). In this case we should multiply 204 Years by 52.142857142857 to get the equivalent result in Weeks:\n\n204 Years x 52.142857142857 = 10637.142857143 Weeks\n\n204 Years is equivalent to 10637.142857143 Weeks.\n\n## How to convert from Years to Weeks\n\nThe conversion factor from Years to Weeks is 52.142857142857. To find out how many Years in Weeks, multiply by the conversion factor or use the Time converter above. Two hundred four Years is equivalent to ten thousand six hundred thirty-seven point one four three Weeks.", null, "## Definition of Year\n\nA year (symbol: y; also abbreviated yr.) is the orbital period of the Earth moving in its orbit around the Sun. Due to the Earth's axial tilt, the course of a year sees the passing of the seasons, marked by changes in weather, the hours of daylight, and, consequently, vegetation and soil fertility. In temperate and subpolar regions around the globe, four seasons are generally recognized: spring, summer, autumn and winter. In tropical and subtropical regions several geographical sectors do not present defined seasons; but in the seasonal tropics, the annual wet and dry seasons are recognized and tracked. A calendar year is an approximation of the number of days of the Earth's orbital period as counted in a given calendar. The Gregorian, or modern, calendar, presents its calendar year to be either a common year of 365 days or a leap year of 366 days.\n\n## Definition of Week\n\nA week (symbol: wk) is a time unit equal to seven days. It is the standard time period used for cycles of rest days in most parts of the world, mostly alongside—although not strictly part of—the Gregorian calendar. The days of the week were named after the classical planets (derived from the astrological system of planetary hours) in the Roman era. In English, the names are Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday.\n\n### Using the Years to Weeks converter you can get answers to questions like the following:\n\n• How many Weeks are in 204 Years?\n• 204 Years is equal to how many Weeks?\n• How to convert 204 Years to Weeks?\n• How many is 204 Years in Weeks?\n• What is 204 Years in Weeks?\n• How much is 204 Years in Weeks?\n• How many wk are in 204 yr?\n• 204 yr is equal to how many wk?\n• How to convert 204 yr to wk?\n• How many is 204 yr in wk?\n• What is 204 yr in wk?\n• How much is 204 yr in wk?" ]

[ null, "https://whatisconvert.com/images/204-years-in-weeks", null ]

[ "Below we will look at a program in Excel VBA that reads data from a text file. This file contains some geographical coordinates we want to import into Excel.\n\nSituation:", null, "Place a command button on your worksheet and add the following code lines:\n\n2. We declare four variables. myFile of type String, text of type String, textline of type String, posLat of type Integer, and posLong of type Integer.\n\nDim myFile As String, text As String, textline As String, posLat As Integer, posLong As Integer\n\n3. We need to initialize the variable myFile with the full path and the filename.\n\nmyFile = \"C:\\test\\geographical-coordinates.txt\"\n\nor\n\nuse the GetOpenFilename method of the Application object to display the standard Open dialog box and select the file (without actually opening the file).\n\nmyFile = Application.GetOpenFilename()\n\nNote: the empty part between the brackets means we give Excel VBA nothing as input. Place your cursor on GetOpenFilename in the Visual Basic Editor and click F1 for help on the arguments.\n\n4. Add the following code line:\n\nOpen myFile For Input As #1\n\nNote: this statement allows the file to be read. We can refer to the file as #1 during the rest of our code.\n\n5. Add the following code lines:\n\nDo Until EOF(1)\nLine Input #1, textline\ntext = text & textline\nLoop\n\nNote: until the end of the file (EOF), Excel VBA reads a single line from the file and assigns it to textline. We use the & operator to concatenate (join) all the single lines and store it in the variable text.\n\n6. Close the file.\n\nClose #1\n\n7. Next, we search for the position of the words latitude and longitude in the variable text. We use the Instr function.\n\nposLat = InStr(text, \"latitude\")\nposLong = InStr(text, \"longitude\")\n\n8. We use these positions and the Mid function to extract the coordinates from the variable text and write the coordinates to cell A1 and cell A2.\n\nRange(\"A1\").Value = Mid(text, posLat + 10, 5)\nRange(\"A2\").Value = Mid(text, posLong + 11, 5)\n\n9. Test the program.\n\nResult:", null, "" ]

[ null, "https://www.excel-easy.com/vba/examples/images/read-data-from-text-file/read-data-from-text-file-example.png", null, "https://www.excel-easy.com/vba/examples/images/read-data-from-text-file/read-data-from-text-file-result.png", null ]

[ "# 32\n\nWrite a Review\n\nOn 5th Feb a commenter on Reddit posted that the coronavirus cases in China were following a suspiciously accurate quadratic curve, implying that China was making up their data and weren’t even bothering to hide it particularly well.\n\nThis set off my bullshit sensors fairly strongly so I wanted to check it out.\n\nHaving looked at the data myself, there may be other reasons why the Chinese data are not accurate but I think the quadratic pattern provides no evidence in favour of falsification.\n\nI doubt people are making many decisions based on that post but possibly looking at where the statistics went wrong may be beneficial.\n\n# R^2\n\nIt turns out that the claim of a “near perfect” model was based on a very high value (0.9995).\n\nis often chosen to summarise how precisely a regression fits the data – it tells you how much of the variance in the data is explained by the equation. Surely if only 1 part in 2,000 isn’t explained by the model that indicates that the data is fabricated?\n\nThe first thing to note is that \"variance\" is a technical term which isn't the same thing as the natural understanding of \"variation\". If you don't appreciate this then will seem more impressive than it is. In particular, variance isn't the deviation from the mean, but the squared deviation. In the more natural understanding, the quadratic model explains 44 parts in 45 of the deviation from the mean, not 1,999 in 2,000. This is still pretty good but seems like less evidence for fabrication.\n\nAn alternative explanation of the value is that it compares two models:\n\n1. The model that you are trying to fit to the data\n\n2. The model where the value of y is expected to be the same for all values of x (the mean of the y values)\n\nA high value is telling us that model 1 is a much better fit that model 2.\n\nHowever in our case we already know model 2 is going to be a terrible, terrible, terrible fit to our data. The y value used in the regression is total cases so far. So model 2 represents some number of cases having already been identified at the beginning of the time period in question and no more cases occurring during the time period.\n\nSo saying that model 1 accounts for 44 parts in 45 of the deviation between the data and model 2 doesn’t really tell me much – model 2 is a lost cause.\n\n# New cases per day\n\nThe problem here is the chosen y-axis. Instead of choosing total number of cases by a certain day, it would be better to choose new cases per day. This removes excess correlation between data points.\n\nIf we do this then instead of fitting a quadratic curve we need to be fitting a straight line (we’re taking the derivative with respect to x) but I'll still call it quadratic for the sake of consistency. Model 2 changes from representing a constant total number of cases to a constant number of new cases every day. This still isn’t a particularly likely model but is certainly an improvement.\n\nPlotting and regressing we get:\n\nOur value has gone down to 0.96. This is still high and suggests that quadratic growth is a fairly good model for the data but isn’t suspicious. For instance, I can also fit a power law (again 2 free parameters) to the data and get = 0.966.\n\nSo within the training set our quadratic model (linear new cases per day) explains a comparable level of variance as a power law model of new cases does.\n\n# Looking outside the dataset\n\nThe obvious thing to do is check whether the pattern was there outside the dataset.\n\nIt wasn’t.\n\nIt is clear that the pattern breaks down shortly after it was noticed (day 15 on this chart).\n\nIn addition we can look at the pattern before the training set. Again as soon as we go outside of the dataset used to create the graph the pattern completely breaks down. This is not surprising as at this point the quadratic model predicts a negative number of new cases per day.\n\n(Looking a deaths instead of cases shows a similar story. For deaths the pattern keeps going for a little longer but even then the power law fit matches the data better.)\n\nSo to make the case for China falsifying the data quadratically you have to also say that the start date for them doing it was ~20th Jan and the end date more-or-less straight after the pattern was noticed. Presumably this would be justified by China having been caught out and changing from then on.\n\nOr possibly, this is just how the virus develops. Now that we have some developments in other countries it is possible to compare spread rates. Where the disease has got out of containment there is a remarkably consistent pattern of growth which matches the China rate very closely. (I'm planning to write a more detailed post on this.)\n\n# 32\n\nPingbacks\nNew Comment\n\nYup! I took this hypothesis seriously when it was first linked by Scott's tumblr, then I found the numbers didn't match up as well as the Reddit post (which only showed a chart) suggested. This was a strong sign that the poster was lying, but I still checked the numbers for a few days until they had clearly diverged.\n\nThanks!\n\nPicking a descriptive statistic for these sorts of problems is pretty tricky. But I think we can do better than R^2, even without going all Bayesian-parameter-estimation.\n\nWhat I mostly care about is just the standard deviation (in excel, STDEV.S() ) of the difference between the data and the model. Then I want to know how this compares to other scales in the data (like the average number of new cases per day).\n\nWhat I mostly care about is just the standard deviation (in excel, STDEV.S() ) of the difference between the data and the model.\n\nI'd be curious why you think this is a good metric.\n\nBTW, when you say \"the difference between the data and the model\", I assume you're referring to the residuals?\n\nIt's just a measure of how close the data is to the line - like the \"inside view\" uncertainty that the model has about the data. In fact, that's more precisely what it is if this is the chi squared statistic (or square root thereof) that you minimized to fit the model. And it's in nice convenient units that you can compare to other things.\n\nIt's not quite right, because it uses an implicit prior about noise and models that doesn't match your actual state of information. But it's something that someone who's currently reporting R^2 to us can do in 30 seconds in Excel.\n\nMeta: I think using latex in headers causes a funny issue in the index on the left\n\nOh, huh. Yeah, don't use LaTeX in headers apparently. Sorry about that. Will add it to the issue tracker, though might take a while until I can fix it.\n\nIf we assume that china is an infinite plain, with Chinese people distributed uniformly across it, and that individual Chinese people don't travel much, then could be accurate. A circle of disease spreading at a constant distance per day.\n\nNit: I guess you mean \"a quadratic model could be accurate\". The term R² here has nothing to do with the fact that the model is quadratic." ]

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[ "math\n\n# Title: Self-dual form of Ruijsenaars-Schneider models and ILW equation with discrete Laplacian\n\nAbstract: We discuss a self-dual form or the B\\\"acklund transformations for the continuous (in time variable) ${\\rm gl}_N$ Ruijsenaars-Schneider model. It is based on the first order equations in $N+M$ complex variables which include $N$ positions of particles and $M$ dual variables. The latter satisfy equations of motion of the ${\\rm gl}_M$ Ruijsenaars-Schneider model. In the elliptic case it holds $M=N$ while for the rational and trigonometric models $M$ is not necessarily equal to $N$. Our consideration is similar to the previously obtained results for the Calogero-Moser models which are recovered in the non-relativistic limit. We also show that the self-dual description of the Ruijsenaars-Schneider models can be derived from complexified intermediate long wave equation with discrete Laplacian be means of the simple pole ansatz likewise the Calogero-Moser models arise from ordinary intermediate long wave and Benjamin-Ono equations.\n Comments: 16 pages, references added Subjects: Mathematical Physics (math-ph); High Energy Physics - Theory (hep-th); Exactly Solvable and Integrable Systems (nlin.SI) Journal reference: Nuclear Physics B, 927 (2018) 550-565 DOI: 10.1016/j.nuclphysb.2017.12.021 Cite as: arXiv:1711.01036 [math-ph] (or arXiv:1711.01036v2 [math-ph] for this version)\n\n## Submission history\n\nFrom: Andrei Zotov [view email]\n[v1] Fri, 3 Nov 2017 06:44:46 GMT (17kb)\n[v2] Mon, 13 Nov 2017 19:11:25 GMT (18kb)\n\nLink back to: arXiv, form interface, contact." ]

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[ "138 Answered Questions for the topic High School: Math\n\n09/21/18\n\n#### What would be the image of the given reflection...\n\nWhat would be the image of the given reflection, r x=2 (1,-4)=( , )? How did you determine it? I need help on this so I can understand these kinds of problems in the future.\nHigh School: Math Algebra 2 Help Asap Please\n\n08/22/18\n\n#### Change 1/3x+29=4y into standard form.\n\n____________________\nHigh School: Math Math Word Problems Math Help\n\n08/21/18\n\n#### How many pillows does she plan to decorate?\n\nGretchen has x square inches of fabric to decorate her pillows. If she uses 6 square inches on each pillow, she will have 10 square inches of fabric remaining. But if she uses 8 square inches on... more\nHigh School: Math Algebra 1 Algebra 2 Algebra\n\n07/27/18\n\n#### The school that Dan goes to is selling tickets to a play\n\nThe school that Dan goes to is selling tickets to a play. On the first day of ticket sales the school sold 6adult tickets and 16 student tickets for a total of $312. The school took in$189 on the... more\n\n07/27/18\n\nThe school that Krystal goes to sold tickets to a choral performance. The adult ticket price was $8, and astudent ticket was$5. On the first night they made $1196 and sold 190 tickets. How many... more High School: Math 07/12/18 #### Henry has a mix of vans and cars. When he loads 2 vans and 3 cars, he can fit a total of 21 people. When he loads 4 vans and 1 car, he can fit 27 people. I need help on how to turn it into an equation High School: Math Math Precalculus Pre Calculus 06/06/18 #### Find the coordinates of the hole: Find the coordinates of the hole of f(x) = x+1/x^2-4x-5 Show all work. High School: Math Math Calculus Math Help 04/15/18 #### If ∫ g(x) dx from a to b = 4a+b, then ∫ (g(x)+7) dx from a to b = ? If ∫ g(x) dx from a to b = 4a+b, then ∫ (g(x)+7) dx from a to b = ? please explain High School: Math Math Ratios High School Math Help 04/07/18 #### Ratio Mixture Baking Powder is made up of Baking Soda, Cream of Tartar and sometimes Corn Starch in a ratio of 1:2:1. If you need 2 tablespoons of Baking Powder, how much of each ingredient do you need? High School: Math Math Word Problem Math Help 03/30/18 #### I need help finding the value of j in this algebra problem So here’s the word problem: Billy has 3/4 of the money Derek has, and Derek as 7/9 of the money Jason has. If they had a total of 680$, how much would each person have? Here is the equation I... more\nHigh School: Math Math Algebra Math Help\n\n03/09/18\n\n#### Simplify (m^3)^4(2x^3)^7(m^2)^5(3x)^2.\n\nThis is for grade 9 math. My teacher wants me to simplify the question\nHigh School: Math Algebra 1\n\n01/17/18\n\nCasey wants to buy a gym membership. One gym has a $198 joining fee and costs$27 per month. Another gym has no joining fee and costs $60 per month. Part 1 out of 2In how many months will both gym... more 12/08/17 #### You and a group of 7 friends go out for pizza. One large pizza and 3 small drinks cost$21.96.\n\nYou and a group of 7 friends go out for pizza. One large pizza and 3 small drinks cost $21.96. One large pizza and 6 small drinks cost$27.93. Write an equation that models the total cost y (in... more\n\n12/05/17\n\n#### i need help with ap calc someone please\n\nI have to do a writing project choosing a theorem and i chose the mean value theorem for derivatives, these are the three questions help me with  question b! a) state the conditions necessary to... more\nHigh School: Math Geometry Algerbraic Proofs\n\n10/06/17\n\n#### state the property that justifies each statement and the statement is if m<1 = 30,m<1=m<2, then m<2 = 30\n\nI need help with this problem, I don't understand what they are asking me to do with the problem\nHigh School: Math\n\n10/04/17\n\n#### f(1) = 2 and f(2) = 5, f(n) = f(1) + f(2) + f(n - 1), for n > 2 f(5) =\n\nHigh School: Math\n\n09/24/17\n\n#### A cube of a number less than 25\n\nthis is beginng algebra write an express ion\nHigh School: Math\n\n09/24/17\n\n#### A cube of a number less than 25\n\nthis is beginng algebra\n\n07/24/17\n\nIn the diagram below, a || b. For lines p and q to be parallel, what is m<4? Explain.    Diagram Link : http://imgur.com/a/XimLc\nHigh School: Math Trigonometry Math Help Mathematics\n\n07/19/17\n\n#### Trigonometry Proof\n\nPlease help me answer the following question! I'm stuck!   Prove that cos6ß + sin6ß = 1/4 + 3/4cos2ß   I've tried simplifying out the LHS into (cos2ß)3 + (sin2ß)3, but I don't know what to do... more\nHigh School: Math\n\n07/12/17\n\n#### You work a deal with the farmer’s market so that your total expenses are simply \\$90. Write an inequality that finds the weeks that the business is profitable\n\ncan you tell me how to solve this problem?\nHigh School: Math Math Algebra 2 Math Help\n\n06/30/17\n\n#### Given an exponential function for compounding interest, A(x) = P(.91)x, what is the rate of change?\n\nThere are four possible answers to this problem that were given to me: A -0.09% B -9% C .91% D 91%   Thank you so much in advance. If you could please explain how you got your answer that... more\nHigh School: Math Math Algebra 1 Algerbra\n\n06/28/17" ]

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[ "@COMMENT This file was generated by bib2html.pl version 0.94 @COMMENT written by Patrick Riley @COMMENT This file came from Milan Hladik's publication pages at @COMMENT http://kam.mff.cuni.cz/~hladik/ @article{Hla2013a, author = \"Milan Hlad\\'{\\i}k\", title = \"Bounds on eigenvalues of real and complex interval matrices\", journal = \"Appl. Math. Comput.\", fjournal = \"Applied Mathematics and Computation\", volume = \"219\", number = \"10\", pages = \"5584-5591\", year = \"2013\", issn = \"0096-3003\", doi = \"10.1016/j.amc.2012.11.075\", bib2html_dl_html = \"http://dx.doi.org/10.1016/j.amc.2012.11.075\", abstract = \"We present a cheap and tight formula for bounding real and imaginary parts of eigenvalues of real or complex interval matrices. It outperforms the classical formulae not only for the complex case but also for the real case. In particular, it generalizes and improves the results by Rohn (1998) and Hertz (2009). The main idea behind is to reduce the problem to enclosing eigenvalues of symmetric interval matrices, for which diverse methods can be utilized. The result helps in analysing stability of uncertain dynamical systems since the formula gives sufficient conditions for testing Schur and Hurwitz stability of interval matrices. It may also serve as a starting point for some iteration methods.\", keywords = \"Interval matrix, Interval analysis, Eigenvalue bounds\", }" ]

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[ "Cody\n\n# Problem 1919. Returning a \"greater than\" vector\n\nSolution 332003\n\nSubmitted on 10 Oct 2013 by Sky Sartorius\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%% v = 1:6; n=3 vNew = 4:6; assert(isequal(returnGreaterThan(v,n),vNew))\n\nn = 3 ans = 4 5 6\n\n2   Pass\n%% v = [10:-2:-10]; n=0 vNew = [10 8 6 4 2]; assert(isequal(returnGreaterThan(v,n),vNew))\n\nn = 0 ans = 10 8 6 4 2\n\n3   Pass\n%% v = [4 pi -2 0 -17000 8 91]; n=3 vNew = [4 pi 8 91]; assert(isequal(returnGreaterThan(v,n),vNew))\n\nn = 3 ans = 4.0000 3.1416 8.0000 91.0000" ]

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[ "# 你的网购已经花了多少钱？\n\n• 这里是一个演示，用最简单的方法进行快速的数据分析，并输出自定义的图表。这种分析方法在很多调查报告中都很有用，网购仅仅是一个案例", null, "“我在亚马逊上花了多少钱？”—— 这个问题很可能会使电子商务巨头的长期客户感兴趣，甚至可能使他们感到恐惧。但了解这件事对您来说很重要，不只是钱的问题。\n\n• 计算出您曾经在 Amazon 上花费的总金额；\n• 找到您最贵的和最便宜的订单，以及平均数和中位数订单；\n• 找出您已缴纳的税款以及您在亚马逊上的有效税率；\n• 通过条形图可视化您的每日消费习惯；\n• 自定义条形图的某些元素，例如大小和颜色，以呈现更清晰的解释。\n\n### 准备进行分析\n\n1、下载并安装工具\n\nJupyter Notebooks 是执行此类数据分析项目的出色工具，一旦掌握了这些要点，您可能会发现自己经常需要使用它。", null, "2、下载亚马逊订单历史记录数据\n\n（如果您希望查看的订单记录不是全部，则页面右侧还会显示“快速设置选项”，可用于快速输入想要查看的时间段，例如近一年或最近一个月）\n\n3、重命名和移动 Amazon Order 数据\n\n4、打开一个 Jupyter Notebook", null, "", null, "", null, "### 我在亚马逊上花了多少钱？\n\nJupyter Notebooks 使您可以在单元格中编写和运行少量代码片段。在屏幕上，菜单和按钮下方，您会看到一个小方块， `In [ ]:`光标的闪烁，这是第一个单元格，是开始编写代码的地方。\n\nAnaconda 已经安装了 Pandas 和 matplotlib，因此不需要自己下载安装。\n\n5、将数据导入 Pandas\n\n• pd 在告诉 Python 使用 pandas 进行此操作；\n• .read_csv（）在告诉 pandas 读取 csv 文件并将其存储为 DataFrame；\n• 括号内的 ‘file_name.csv’ 部分告诉 pandas 在安装 Anaconda 的同一文件夹中读取具有该文件名的 csv 文件。\n\n``````import pandas as pd", null, "", null, "", null, "``df.shape``\n``(59, 23)``\n\n6、清理数据\n\n``````df = df.fillna(0)", null, "", null, "", null, "``````df[\"Total Charged\"] = df[\"Total Charged\"].str.replace('\\$','').astype(float)", null, "", null, "", null, "7、计算在亚马逊上花费的总金额\n\npandas 有一些基本函数对本文的目的有用。这里从 .sum（）开始，它将为您提供此列中每个数字的总数，换句话说，就是在 Amazon 上花费的总数。\n\n``df[\"Total Charged\"].sum()``\n``1777.7300000000002``\n\n\\$1,777.73！有些人的总数也许会更大，你会感到惊讶的。\n\n8、计算其他：平均支出、最大购买量等\n\npandas 允许您非常快速地进行其他一些计算。例如，平均购买价是多少？可以使用 .mean（）找出：\n\n``df[\"Total Charged\"].mean()``\n``30.131016949152542``\n\n``df[\"Total Charged\"].median()``\n``15.95``\n\n``df[\"Total Charged\"].max()``\n``210.99``\n\n``df[\"Total Charged\"].min()``\n``1.04``\n\n9、我缴纳了多少税？\n\n``````df[\"Tax Charged\"] = df[\"Tax Charged\"].str.replace('\\$','').astype(float)\n\n``df[\"Tax Charged\"].sum()``\n``52.60999999999999``\n\n``df[\"Tax Charged\"].sum() / df[\"Total Charged\"].sum()``\n``0.02959392033660904``\n\n10、分析消费习惯\n\n``````df['Order Date'] = pd.to_datetime(df['Order Date'])\n\npandas 和 matplotlib 软件包使您能够在 Jupyter Notebook 内部构建和查看各种图表类型。\n\n（注意：当您运行此代码单元时，不会发生明显变化，但是它将使 matplotlib 能够在笔记本中显示图表。）\n\n``%matplotlib inline``\n\n``df.plot.bar(x='Order Date', y='Total Charged', rot=90)``\n``<matplotlib.axes._subplots.AxesSubplot at 0x1c30f9c4e10>``", null, "``df.plot.bar(x='Order Date', y='Total Charged', rot=90, figsize=(20,10))``\n``<matplotlib.axes._subplots.AxesSubplot at 0x1c3129870b8>``", null, "1. 希望按 “订单日期” 对数据进行分组，以便将共享同一日期的所有订单都计入一行；\n2. 要在每个日期的 “Total Charged” 列中获取已支付价格的总和，并在相关日期的行中将其作为 “Total Charged” 的值。\n\n``````daily_orders = df.groupby('Order Date').sum()[\"Total Charged\"]\n``````Order Date\n2019-04-01 63.91\n2019-04-04 156.12\n2019-04-05 27.64\n2019-04-07 40.63\n2019-04-08 44.99\nName: Total Charged, dtype: float64``````\n\n``daily_orders.plot.bar(figsize=(20,10))``\n``<matplotlib.axes._subplots.AxesSubplot at 0x1c3110bbf60>``", null, "``daily_orders.plot.bar(figsize=(20, 10), color='#61D199')``\n``<matplotlib.axes._subplots.AxesSubplot at 0x1c311dcd5c0>``", null, "1. 在此期间，我在亚马逊上花了多少钱。\n2. 最高、最低和平均订单总额是多少。\n3. 缴纳了多少税，以及有效税率。\n4. 指出随时间波动的情况。\n5. 哪天花的钱最多。\n\n### 如何使这个项目更进一步\n\n• 查看其他参数以添加到绘图代码中，进一步美化图形。\n• 在其他栏添加每一天的 Tax Charged。\n• 按月份或年份（而不是一天）细分您的支出。\n• 寻找一种方法来获取每天（包括没有订单的日子）到您的数据集中，以便可以更轻松地直观显示一段时间内的支出。" ]

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[ "# Gauge invariant Green's function for electrodynamics\n\nVarying the electromagnetic action $$\\begin{equation} S=-m c \\int d s\\left(\\dot{z}^{2}\\right)^{\\frac{1}{2}}-\\frac{e}{c} \\int d s A_{\\mu} \\dot{z}^{\\mu}-\\frac{1}{16 \\pi c} \\int d^{4} x F_{\\mu \\nu} F^{\\mu \\nu} \\end{equation}$$ we get the two equations of motion $$\\begin{equation} m \\ddot{z}^{\\mu}=\\frac{e}{c^{2}} F^{\\mu \\nu} \\dot{z}_{\\nu} \\end{equation}$$\n\n$$\\begin{equation} \\square A^{\\mu}-\\partial^{\\mu}\\left(\\partial_{\\nu} A^{\\nu}\\right)=\\frac{4 \\pi}{c} j^{\\mu} \\end{equation}$$ If we use the Lorenz gauge, the solutions to the second equation are the Liénard–Wiechert potentials.\n\nIs there a general solution to the second equation without fixing a gauge? By solution I mean a closed expression for $$A^{\\mu}$$ in terms of $$j^{\\mu}$$. Or is it possible to proof that such a solution does not exist.\n\nAs my2cts said the equation is not invertible. Here's a quick proof. First we go to momentum space by Fourier transforming both sides of the equation\n\n$$(-k^2 \\eta^{\\mu\\nu}+ k^\\mu k^\\nu )\\tilde A_\\nu(k) = \\frac{4\\pi}{c} \\tilde j(k)^\\mu$$\n\nNow the solution for $$\\tilde A(k)$$ would be simply given by solving the equation algebrically and then doing an inverse Fourier transform. Thing is, that the tensor operator on the LHS is not invertible.\n\nYou can see this easily by noticing that that operator is zero on the vector space spanned by $$k^\\mu$$, since\n\n$$(-k^2 \\eta^{\\mu\\nu}+ k^\\mu k^\\nu ) k_\\nu = -k^2 k^\\mu + k^\\mu k^\\nu k_\\nu = 0$$\n\nThe reason why you have to fix the gauge before solving these equations is to get rid of this \"null direction\". For example imposing $$\\partial^\\mu A_\\mu = 0$$ is equivalent, in momentum space to $$k^\\mu A_\\mu = 0$$. This condition prevents explicitly any configuration that would make the above operator zero and makes it invertible.\n\n• Thanks a lot. So the argument is, that the operator is not injective, i.e. that it has a nontrivial kernel and thus is not invertible. – NicAG Jun 1 at 9:24\n\nIf your question is, as the title suggests, whether the second equation can be inverted, the answer is no. Inversion requires 'gauge fixing'. An operator can be inverted only if it establishes a one to one relation between solution and source. The operator in your second equation assigns many potentials to a single source and hence cannot be inverted.\n\n• Thanks. Can you maybe provide a reference to a proof that an inversion is not possible? – NicAG May 31 at 22:54" ]

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[ "# Representation theorem\n\nIn mathematics, a representation theorem is a theorem that states that every abstract structure with certain properties is isomorphic to another (abstract or concrete) structure.\n\n## Examples\n\n### Category theory\n\n• The Yoneda lemma provides a full and faithful limit-preserving embedding of any category into a category of presheaves.\n• Mitchell's embedding theorem for abelian categories realises every small abelian category as a full (and exactly embedded) subcategory of a category of modules over some ring.\n• Mostowski's collapsing theorem states that every well-founded extensional structure is isomorphic to a transitive set with the ∈-relation.\n• One of the fundamental theorems in sheaf theory states that every sheaf over a topological space can be thought of as a sheaf of sections of some (étalé) bundle over that space: the categories of sheaves on a topological space and that of étalé spaces over it are equivalent, where the equivalence is given by the functor that sends a bundle to its sheaf of (local) sections." ]

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[ "# Sharpe Ratio\n\nUpdated: April 12, 2022\n\nThe Sharpe ratio measures an investment’s risk-adjusted returns within a certain period, and it was originally developed by the American economist, William F. Sharpe.\n\nIn it, the number is useful to investors who want to gain valuable insight into their investments’ risk-adjusted returns. Or, simply the return on their investments compared to the risk they accompany.\n\nInvestors commonly use this ratio for comparison purposes. It looks at risk-adjusted returns of mutual funds, exchange-traded funds (ETF), stocks and other investments.\n\nIn calculating the Sharpe ratio, analysts typically use the rate for T-bills or cash. The 90-day T-Bill rate is a particular favorite in representing the risk-free rate. Investors may rely on the ratio to determine how much they are likely to earn against the investment’s risk.\n\nInvestors must be compensated for the risks they face, outside holding risk-free assets. The investment’s average rate of return is equal to the average rate for the stocks, ETFs and other types of individual security, or the portfolio assessed. This should be within the specific time frame.\n\nWhen calculating a fund’s Sharpe Ratio, the risk-free asset’s return must first be taken from the fund’s returns within the period under study. You can then divide the difference by the standard deviation of the investment. When looking at an investment’s risk-adjusted return, the Sharpe Ratio makes use of that investment’s standard deviation. This is the investment return’s variability in relation to its mean return for a specific time frame. A high standard deviation indicates that the investment’s returns are widely variable in reference to  average return.\n\n## Sharpe Ratio Formula\n\n$$Sharpe\\: Ratio = \\dfrac{R_{p} - R_{f}}{σ_{p}}$$\n\n• Rp = the portfolio return\n• Rf = risk-free rate\n• σp = standard deviation of the portfolio’s return\n\nThe formula might look a bit too complex for those who don’t know much about financial calculations. But the idea is quite basic. You can remove the risk-free rate of return from the mean return in order to separate the return for the risk level.\n\nInvestors can then assess performance based on the risk-free return. Essentially, the Sharpe Ratio equation adjusts portfolios based on risk while leveling the playing field so that fair and valid comparisons can be made.\n\nBad ratios are those below 1; acceptable ratios range from 1 to 1.99; ratios 2 to 2.99 are considered really good. Those 3 and higher are considered outstanding. Some portfolios may come with higher risk yet have a ratio of 1, 2, or 3. With all conditions staying the same, Sharpe Ratios 3 and up are considered exceptional. Ratios from 1 to 3 indicate how much excess return is earned from a risky investment versus an investment with zero risk. Somehow, it reveals the level of compensation received for the excess risk carried by the investment.\n\nAnother problem with the Sharpe Ratio is that it depends on standard deviation, which is based on the normal distribution curve. This distribution doesn’t cover certain strategies, despite being seen in many investments. For example, this is the case with large, but long-delayed, payoffs with practically zero returns. Skewed distributions with unusual events can lead to higher Sharpe ratios that fail to capture the entire picture of the investment’s volatility. An example of this would be selling deep out-of-the-money options. In these scenarios, the result would be an inaccurate and misleading Sharpe Ratio.\n\n## Sharpe Ratio Example\n\nEli invested in a stock portfolio which he expects would return 18% within a year’s time. The risk-free returns are 7%, and his portfolio has a 0.09 standard deviation. Knowing this, what is the Sharpe Ratio for Eli’s portfolio?\n\nLet’s break it down to identify the meaning and value of the different variables in this problem.\n\n• Portfolio return: 18%\n• Risk-free rate: 7%\n• Portfolio standard deviation: 9%\n\nWe can apply the values to our variables and calculate the Sharpe Ratio:\n\n$$Sharpe\\: Ratio = \\dfrac{0.18 - 0.07}{0.09} = 1.22$$\n\nIn this case, Eli’s stock portfolio would have a Sharpe Ratio of 1.22.\n\nIt means Eli’s portfolio carries 1.22 “units” of risk with each point of return it makes. His portfolio has a return of 18%. And there is another portfolio with the same return but a Sharp Ratio of 1.3. So, Eli’s portfolio is better since it is able to earn the same return with less risk.\n\n## Sharpe Ratio Analysis\n\nGenerally speaking, the higher the Sharpe ratio, the better. This is because it shows that sounder investment decisions are being made despite the associated risk. For instance, Treasury bills are considered risk-free investments, both volatility and earnings outside the risk-free rate are zero. This gives the portfolio a Sharpe Ratio of zero.\n\nJust like other financial metrics, the Sharpe Ratio has its own share of limitations. For example, some complain that the calculation is too vulnerable. It can be easily misinterpreted, causing data to become misleading. Two options to the Sharpe model are the Sortino ratio and the Treynor ratio.\n\nThe Sortino Ratio is an enhanced version of the Sharpe Ratio. Specifically, only returns that fail to reach the target or required rate of return, or “downside volatility,” are punished. Conversely, the Sharpe method equally penalizes upside and downside volatility. Also, the Treynor Ratio measures returns outside of what an investment with zero risk could have made for each risk unit taken.\n\nInterestingly, the Sharpe Ratio has varied uses. For one, investors usually trust the equation when assessing their level of comfort with a specific investment. For instance, they may think the return isn’t high enough for a specific volatility level. This makes the investment a bad one, and the investor can begin exploring other options with a higher Sharpe Ratio.\n\n## Sharpe Ratio Conclusion\n\n• The Sharpe Ratio measures an investment asset’s risk-adjusted return.\n• This formula requires three variables: portfolio return, Risk-free return, and the standard deviation of the portfolio.\n• You can express the ratio as a plain numerical value.\n• This tool adjusts a portfolio’s past or predicted performance for the additional risk faced by the investor.\n• A portfolio with a lower ratio is better than a similar portfolio with the same return but a higher ratio.\n• The Sharpe Ratio assumes a normal distribution of investment returns.\n\n## ​Sharpe Ratio Calculator\n\nYou can use the Sharpe Ratio calculator below to quickly measure your investment’s risk-adjusted returns over a specific period by entering the required numbers.\n\n### What is the Sharpe ratio?\n\nThe Sharpe ratio is a measure of the excess return per unit of risk for an investment asset. It’s calculated by subtracting the risk-free rate from the portfolio's return and dividing that number by the portfolio's standard deviation. The Sharpe ratio is named after its creator, William F. Sharpe.\n\n### What is a good Sharpe ratio?\n\nA good Sharpe ratio is one that is greater than 1.0. This means the investment is generating a higher return than what could be earned with a risk-free investment.\n\n### Why is a higher Sharpe ratio better?\n\nA higher Sharpe ratio is better because it means the investment is generating a higher return than what could be earned with a risk-free investment. This makes the investment more desirable as it provides a higher return for the additional risk taken.\n\n### How do you interpret a Sharpe ratio?\n\nAn investor would typically interpret a Sharpe ratio as a measure of how well an investment is performing in relation to the amount of risk being taken. A higher Sharpe ratio indicates that the investment is providing a higher return for the additional risk taken.\n\n### Why is the Sharpe ratio important?\n\nThe Sharpe ratio is important because it provides a way to compare investments on a risk-adjusted basis. This makes it easier for investors to identify which investments offer the best return for the amount of risk taken." ]

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[ "# Factorial in SQL\n\nExample for versions Oracle 10g SQL, Oracle 11g SQL\n\nPure SQL doesn’t support loops, recursions or user-defined functions. This example shows one possible workaround which uses\n\n• pseudocolumn `level` to construct a pseudotable containing numbers 1 through 16,\n• aggregate function `sum` which allows to sum the elements of a set\n• and math functions `exp` and `ln` to replace multiplication (required to calculate the factorial) with addition (provided by SQL).\n\nNote that 0! will not be present in the return, since trying to calculate ln(0) produces an exception.\n\n``````select t2.n || '! = ' || round(exp(sum(ln(t1.n))))\nfrom\n( select level n\nfrom dual\nconnect by level <= 16) t1,\n( select level n\nfrom dual\nconnect by level <= 16) t2\nwhere t1.n<=t2.n\ngroup by t2.n\norder by t2.n\n``````" ]

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[ "## Recommended Posts\n\nI'm testing the max HP code with dual class.\n\nFighter to mage (level 3/7)\n\nSo level 1 = 3, Level 2 = 7\n\nHP = 3*10 + 4*4 = 46\n\nIt works", null, "Now mage to fighter (level 3/7)\n\nSo level 1 = 7, level 2 = 3\n\nHP = 3*4 + 4*10 = 52\n\nBut tututweak found 28. I assume it is an error in the class order, since level 2 = mage class, and level 1 = fighter class\n\nI will try to fix it\n\nI'm testing the max HP code with dual class.\n\nFighter to mage (level 3/7)\n\nSo level 1 = 3, Level 2 = 7\n\nHP = 3*10 + 4*4 = 46\n\nIt works", null, "Now mage to fighter (level 3/7)\n\nSo level 1 = 7, level 2 = 3\n\nHP = 3*4 + 4*10 = 52\n\nBut tututweak found 28. I assume it is an error in the class order, since level 2 = mage class, and level 1 = fighter class\n\nI will try to fix it\n\nAh, I see where I went wrong. I though the 'level first class' field always corresponded to the original class of the creature, but this is not the case. This is now fixed locally.\n\nIf class is fighter/mage, then the first level is the fighter one, and the second the mage one.\n\nLuckily, the engine never tricks us here (EG, there is no case whereby the class IDS is CLERIC_THIEF and level1 is thief).\n\n#### Archived\n\nThis topic is now archived and is closed to further replies.\n\n×\n• Website" ]

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[ "The duodecimal-gaggol is equal to s(12,100,3) using strong array notation, or \\(12\\uparrow\\uparrow\\uparrow100\\). The term was coined by Aarex Tiaokhiao.\n\nUsername5243 calls this number Duodecimal-Toogol, and it's equal to 12100 in Username5243's Array Notation.\n\nThis number is approximately E14#11#99 in Hyper-E notation.\n\n## Approximations\n\nNotation Approximation\nBEAF \\(\\{12,100,3\\}\\) (exact)\nHyper-E notation \\(\\textrm{E}1\\#1\\#100\\) (exact)\nFast-growing hierarchy \\(f_4(100)\\)" ]

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[ "7180 minus 87 percent\n\nThis is where you will learn how to calculate seven thousand one hundred eighty minus eighty-seven percent (7180 minus 87 percent). We will first explain and illustrate with pictures so you get a complete understanding of what 7180 minus 87 percent means, and then we will give you the formula at the very end.\n\nWe start by showing you the image below of a dark blue box that contains 7180 of something.", null, "7180\n(100%)\n\n87 percent means 87 per hundred, so for each hundred in 7180, you want to subtract 87. Thus, you divide 7180 by 100 and then multiply the quotient by 87 to find out how much to subtract. Here is the math to calculate how much we should subtract:\n\n(7180 ÷ 100) × 87\n= 6246.6\n\nWe made a pink square that we put on top of the image shown above to illustrate how much 87 percent is of the total 7180:", null, "The dark blue not covered up by the pink is 7180 minus 87 percent. Thus, we simply subtract the 6246.6 from 7180 to get the answer:\n\n7180 - 6246.6\n= 933.4\n\nThe explanation and illustrations above are the educational way of calculating 7180 minus 87 percent. You can also, of course, use formulas to calculate 7180 minus 87%.\n\nBelow we show you two formulas that you can use to calculate 7180 minus 87 percent and similar problems in the future.\n\nFormula 1\nNumber - ((Number × Percent/100))\n7180 - ((7180 × 87/100))\n7180 - 6246.6\n= 933.4\n\nFormula 2\nNumber × (1 - (Percent/100))\n7180 × (1 - (87/100))\n7180 × 0.13\n= 933.4\n\nNumber Minus Percent\nGo here if you need to calculate any other number minus any other percent.\n\n7190 minus 87 percent\nHere is the next percent tutorial on our list that may be of interest." ]

[ null, "https://percent.info/images/minus/bluesquare.png", null, "https://percent.info/images/minus/minus-87-percent.png", null ]

[ "# McGraw Hill Math Grade 8 Lesson 15.1 Answer Key Customary Units of Length\n\nPractice the questions of McGraw Hill Math Grade 8 Answer Key PDF Lesson 15.1 Customary Units of Length to secure good marks & knowledge in the exams.\n\n## McGraw-Hill Math Grade 8 Answer Key Lesson 15.1 Customary Units of Length\n\nExercises\n\nCONVERT\n\nQuestion 1.\n72 in. = ____________ ft\n5.999976 ft,\n\nExplanation:\nAs 1 inch = 0.083333 ft, So 72 in = 72 X 0.083333 = 5.999976 ft.\n\nQuestion 2.\n12 ft = _____________ in.\n144 in,\n\nExplanation:\nAs 1 feet = 12 inches, So 12 ft = 12 X 12 = 144 inches,\n\nQuestion 3.\n3 ft 8 in. = ____________ in.\n44 inches,\n\nExplanation:\nAs 1 ft = 12 inches, So 3 ft 8 in. = 3ft X 12 + 8 in. = 44 in.\n\nQuestion 4.\n192 in. = _______________ ft\n0.015999936 ft,\n\nExplanation:\nAs 1 inch = 0.083333 ft, So 192 in. = 192 X 0.083333 = 0.015999936 ft.\n\nQuestion 5.\n7,920 ft = ______________ mi\n1.38068226 mi,\n\nExplanation:\nAs 1ft = 0.000189394 mi, So 7,920 ft = 7,920 X 0.000189394 =\n1.38068226 mi.\n\nQuestion 6.\n16 ft 6 in. = _______________ yd\n5.5 yd,\n\nExplanation:\nAs 1 ft = 12 inches,\nSo 16 ft 6 in. = 16 ft X 12 + 6 = 198 in. 198 inches in yd,\nAs we know 1 in. = 0.0277778 yd, 198 inches = 198 X 0.0277778 = 5.5 yd.\n\nQuestion 7.\n4 ft 9 in. = ______________ in.\n57 in,\nExplanation:\nAs 1 ft = 12 inches, So 4 ft 9 in. = 4ft X 12 + 9 = 57 in.\n\nQuestion 8.\n1056 ft = ______________ mi\n0.2 mi,\n\nExplanation:\nAs 1ft = 0.000189394 mi, So 1056 ft. = 1056 X 0.000189394 = 0.2 mi.\n\nQuestion 9.\n$$\\frac{4}{5}$$mi = ____________ in.\n50,688 in,\n\nExplanation:\nAs 1 mi = 63,360 in, So $$\\frac{4}{5}$$mi =\n$$\\frac{4}{5}$$mi X 63,360 = 50,688 in.\n\nQuestion 10.\n66 in. = _______________ ft\n5.499978 ft,\n\nExplanation:\nAs 1 inches  = 0.083333 ft, So 66 in. = 66 in. X 0.083333 ft = 5.499978 ft.\n\nQuestion 11.\n648 in. = _____________ yd\n18 yd,\n\nExplanation:\nAs 1 inches = 0.0277778 yd, So 648 in. = 648 in. X 0.0277778 yd = 18 yd.\n\nQuestion 12.\n7$$\\frac{1}{2}$$ yd = ____________ ft = __________ in.\n270 inches,\n\nExplanation:\nAs 1 yd = 3 feet,\nSo 7$$\\frac{1}{2}$$mi = $$\\frac{15}{2}$$mi X 3 = 22.5 ft.\nNow 1 ft = 12 inches, So 22.5 ft = 22.5 X 12 inches = 270 inches.\n\nQuestion 13.\n35 ft = ____________ yd\n11.666655 yd,\n\nExplanation:\nAs 1ft = 0.333333 yd, So 35ft. = 35 ft X 0.333333 yd =11.666655 yd.\n\nQuestion 14.\n16 yd 2 ft = ____________ in.\n600 in,\n\nExplanation:\nAs 1 yd = 3 ft, So 16 yd 2 ft = 16 yd X 3 + 2 = 50 ft,\n50 ft in inches, As we know 1 ft = 12, So 50 ft = 50 ft X 12 = 600 in.\n\nQuestion 15.\n2 mi 303 yd = _______________ ft\n11,469 ft,\n\nExplanation:\nAs 1 mi = 1,760 yd,\nSo 2 mi 303 yd = 2 X 1760 + 303 = 3,823 yd, 3,823 yd in ft,\nAs we know 1 yd = 3 ft, So 3,823 yd = 3,823 yd X 3 ft = 11,469 ft.\n\nQuestion 16.\n34 yd 2 ft = _____________ in.\n1,248 in,\n\nExplanation:\nAs 1 yd = 3 ft, So 34 yd 2 ft = 34 X 3 + 2 = 104 ft,\n104 ft in inches, As we know 1 ft = 12 inches,\nSo 104 ft = 104 X 12 = 1,248 in.\n\nQuestion 17.\nThe river is 768 yards wide where the city plans to build a new bridge. How many 32-foot sections of bridge will the city engineers need in order to build the bridge?\n72 sections of 32 foot,\n\nExplanation:\nGiven the river is 768 yards wide where the city plans to\nbuild a new bridge. So 32-foot sections of bridge will the\ncity engineers need in order to build the bridge is as 1 yard = 3 feet,\nso 768 yards = 768 X 3 feet = 2,304 feet, Therefore 32 foot sections are\n2,304 feet ÷ 32 feet = 72 sections.\n\nQuestion 18.\nPatty wants to know how far she can throw a softball. After throwing, she estimates that the ball traveled 1,764 inches.\nHow many yards is that? ______________\nHow many feet? _______________" ]

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[ "", null, "# 5th Grade Math Times Tables Worksheets\n\nThey also practice mental multiplications prime factorization order of operations and solve simple equations. These fifth grade worksheets cover multi digit multiplication multiplying with decimals multiplying decimals with models multiplication of fractions and more.", null, "2 3 4 5 6 7 Times Tables Worksheets Times Tables Worksheets Math For Kids Teaching Math\n\n### Use this timed trial to test students recall of multiplication facts for the eight times table.", null, "5th grade math times tables worksheets. 5th grade multiplication and division worksheets including multiplying in parts multiplication in columns missing factor questions mental division division with remainders long division and missing dividend or divisor problems. In grade 5 students continue studying multi digit multiplication with more digits than in 4th grade. The worksheets can be.\n\nBelow you can make an unlimited supply of worksheets for these topics. Multiply fifth grade students math skills with our multiplication worksheets. Grade 5 multiplication worksheets.\n\n5th grade function tables displaying top 8 worksheets found for this concept. The math worksheets on this page cover many of the core topics in 5th grade math but confidence in all of the basic operations is essential to success both in 5th grade and beyond. 5th grade multiplication worksheets.\n\nIn this math worksheet students must answer horizontal multiplication problems as quickly and correctly as they can. 5th grade multiplication worksheets. Printable math worksheets for 5th grade.\n\nSome of the worksheets for this concept are function table t1l1s1 input output tables determine the equation 2 variables function tables in and out boxes graphing linear equations t1s1 math mammoth grade 5 a worktext word problem practice workbook. Fifth graders will cover a wide range of math topics as they solidify their arithmatic skills.", null, "Multiplication Times Tables Worksheets 2 3 4 5 6 7 8 9 10 11 12 Times Tables Times Tables Worksheets Math Worksheets Free Math Worksheets", null, "Times Tables Sheet For Kids Multiplication Facts Worksheets Multiplication Worksheets Printable Math Worksheets", null, "Math Times Tables Worksheets 1 10 001 Maths Times Tables Times Tables Worksheets Learning Multiplication", null, "Free Math Sheets Multiplication 2 3 4 5 10 Times Tables 1 Math Sheets Math Worksheets Multiplication", null, "Year 4 English Worksheets Free Printable In 2020 Printable Multiplication Worksheets Math Multiplication Worksheets Printable Math Worksheets", null, "5 Times Tables Times Tables Worksheets Maths Times Tables Math Tables", null, "Free 5 Times Table Worksheets Activity Shelter Math Multiplication Worksheet 2nd Cr Times Tables Worksheets Multiplication Worksheets 2 Times Table Worksheet", null, "20 Worksheets For Students To Complete The Multiplication Number Sentences F 4th Grade Math Worksheets Multiplication Worksheets Math Multiplication Worksheets", null, "The Multiplication Tables To 144 One Per Page D Math Worksheet From The Multiplication Worksheet Multiplication Table Multiplication Chart Multiplication", null, "Maths Times Tables Worksheets In 2020 Third Grade Math Worksheets Division Worksheets 3rd Grade Math Worksheets", null, "10 Times Table Worksheet Multiplication K5 Worksheets Multiplication And Division Worksheets Printable Multiplication Worksheets Division Worksheets", null, "Times Tables For Division 5 3rd Grade Math Worksheet Greatschools 3rd Grade Math Worksheets Math Division Worksheets 4th Grade Math Worksheets", null, "Times Table Tests 2 3 4 5 10 Times Tables Math Practice Worksheets 3rd Grade Math Worksheets Maths Times Tables", null, "Salamander Math Worksheet Times Tables Quiz Worksheet Stnicholaseriecounty In 2020 Mental Maths Worksheets Mathematics Worksheets 4th Grade Math Worksheets", null, "Multiplication Table Worksheets Grade 3 Times Tables Worksheets Printable Math Worksheets Multiplication Worksheets", null, "Math Worksheets 3rd Grade Multiplication 2 3 4 5 10 Times Tables 3 4th Grade Math Worksheets Free Math Worksheets Math Multiplication Worksheets", null, "Free Division Worksheets Division Tables To 5×5 2 Gif 790 1 022 Pixels Math Worksheets Math Multiplication Worksheets Free Math Worksheets", null, "Previous post Beginning Multiplication Worksheets Pdf", null, "Next post Key Chemistry Atomic Number And Mass Number Worksheet" ]

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[ "# Factorization homology\n\n<]]>\n\n<]]>\n\n<]]>\n\n<]]>\n\n<]]>\n\n<]]>\n\n<]]>\n\n<]]>\n\n<]]>\n\n<]]>\n\n<]]>\n• ## Some Research Articles\n\narXiv:1206.5164 Structured singular manifolds and factorization homology from arXiv Front: math.AT by David Ayala, John Francis, Hiro Lee Tanaka We provide a framework for the study of structured manifolds with singularities and their locally determined invariants. This generalizes factorization homology, or topological chiral homology, to the setting of singular manifolds equipped with various tangential structures. Examples of such factorization homology theories include intersection homology, compactly supported stratified mapping spaces, and Hochschild homology with coefficients. Factorization homology theories for singular manifolds are characterized by a generalization of the Eilenberg-Steenrod axioms. Using these axioms, we extend the nonabelian Poincaré duality of Salvatore and Lurie to the setting of singular manifolds -- this is a nonabelian version of the Poincaré duality given by intersection homology. We pay special attention to the simple case of singular manifolds whose singularity datum is a properly embedded submanifold and give a further simplified algebraic characterization of these homology theories. In the case of 3-manifolds with 1-dimensional submanifolds, this structure gives rise to knot and link homology theories akin to Khovanov homology.\n\narXiv:1206.5522 Factorization homology of topological manifolds from arXiv Front: math.AT by John Francis The factorization homology of topological manifolds, after Beilinson & Drinfeld and Lurie, is a homology-type theory for topological n-manifolds whose coefficient systems are n-disk algebras. In this work we prove a precise formulation of this idea, giving an axiomatic characterization of factorization homology in terms of a generalization of the Eilenberg-Steenrod axioms for singular homology. These homology theories give rise to a specific kind of topological quantum field theory, which can be characterized by a condition that observables can be naturally defined on general n-manifolds, not only closed n-manifolds, and that global observables are determined by local observables. This axiomatic point of view has a number of applications, some in joint work with David Ayala & Hiro Tanaka, surveyed in the remainder of this article. In particular, we discuss the nonabelian Poincaré duality of Salvatore and Lurie; the relation of Koszul duality of n-disk algebras with factorization homology; calculations of factorization homology for free n-disk algebras and enveloping algebras of Lie algebras.\n\n<]]>\n\n<]]>" ]

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[ "", null, "Open in App\nNot now\n\n# C Program for Turn an image by 90 degree\n\n• Last Updated : 24 May, 2022\n\nGiven an image, how will you turn it by 90 degrees? A vague question. Minimize the browser and try your solution before going further. An image can be treated as 2D matrix which can be stored in a buffer. We are provided with matrix dimensions and it’s base address. How can we turn it? For example see the below picture,\n\n```* * * ^ * * *\n* * * | * * *\n* * * | * * *\n* * * | * * *```\n\nAfter rotating right, it appears (observe arrow direction)\n\n```* * * *\n* * * *\n* * * *\n-- - - >\n* * * *\n* * * *\n* * * *```\n\nThe idea is simple. Transform each row of source matrix into required column of final image. We will use an auxiliary buffer to transform the image. From the above picture, we can observe that\n\n```first row of source ------> last column of destination\nsecond row of source ------> last but-one column of destination\nso ... on\nlast row of source ------> first column of destination```\n\nIn pictorial form, we can represent the above transformations of an (m x n) matrix into (n x m) matrix,", null, "If you have not attempted, atleast try your pseudo code now. It will be easy to write our pseudo code. In C/C++ we will usually traverse matrix on row major order. Each row is transformed into different column of final image. We need to construct columns of final image. See the following algorithm (transformation)\n\n```for (r = 0; r < m; r++)\n{\nfor (c = 0; c < n; c++)\n{\n// Hint: Map each source element indices into\n// indices of destination matrix element.\ndest_buffer [ c ] [ m - r - 1 ] = source_buffer [ r ] [ c ];\n}\n}```\n\nNote that there are various ways to implement the algorithm based on traversal of matrix, row major or column major order. We have two matrices and two ways (row and column major) to traverse each matrix. Hence, there can atleast be 4 different ways of transformation of source matrix into final matrix.\n\n## C\n\n `// C program to turn an``// image by 90 Degree``#include ``#include ` `void` `displayMatrix(unsigned ``int` `const` `*p,``                   ``unsigned ``int` `row,``                   ``unsigned ``int` `col);``                   ` `void` `rotate(unsigned ``int` `*pS,``            ``unsigned ``int` `*pD,``            ``unsigned ``int` `row,``            ``unsigned ``int` `col);``            ` `void` `displayMatrix(unsigned ``int` `const` `*p,``                   ``unsigned ``int` `r,``                   ``unsigned ``int` `c)``{``    ``unsigned ``int` `row, col;``    ``printf``(\"` `\");` `    ``for` `(row = 0; row < r; row++)``    ``{``        ``for` `(col = 0; col < c; col++)``            ``printf``(``\"%d    \"``, * (p + row * c + col));``        ``printf``(\"``\");``    ``}` `    ``printf``(\"` `\");``}` `void` `rotate(unsigned ``int` `*pS,``            ``unsigned ``int` `*pD,``            ``unsigned ``int` `row,``            ``unsigned ``int` `col)``{``    ``unsigned ``int` `r, c;``    ``for` `(r = 0; r < row; r++)``    ``{``        ``for` `(c = 0; c < col; c++)``        ``{``            ``*(pD + c * row + (row - r - 1)) =``                            ``*(pS + r * col + c);``        ``}``    ``}``}` `// Driver Code``int` `main()``{``    ` `    ``// declarations``    ``unsigned ``int` `image[] = {{1,2,3,4},``                               ``{5,6,7,8},``                               ``{9,10,11,12}};``    ``unsigned ``int` `*pSource;``    ``unsigned ``int` `*pDestination;``    ``unsigned ``int` `m, n;` `    ``// setting initial values``    ``// and memory allocation``    ``m = 3, n = 4, pSource = (unsigned ``int` `*)image;``    ``pDestination =``        ``(unsigned ``int` `*)``malloc``        ``(``sizeof``(``int``) * m * n);` `    ``// process each buffer``    ``displayMatrix(pSource, m, n);` `    ``rotate(pSource, pDestination, m, n);` `    ``displayMatrix(pDestination, n, m);` `    ``free``(pDestination);` `    ``getchar``();``    ``return` `0;``}`\n\nOutput :\n\n```\n1 2 3 4\n5 6 7 8\n9 10 11 12\n\n9 5 1\n10 6 2\n11 7 3\n12 8 4 ```\n\nTime Complexity: O(N*M), as we are using nested loops for traversing the matrix.\n\nAuxiliary Space: O(N*M), as we are using extra space for matrix.\n\nPlease refer complete article on Turn an image by 90 degree for more details!\n\nMy Personal Notes arrow_drop_up" ]

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[ "# Polymorphism-in-C++ .\n\nIn this session i have discussed polymorphism in C++ .\n\nPolymorphism is a Greek word means having multiple form . In polymorphism name same and behaviour different . i.e A person is husband of a woman when he is in the company then he is employee of company . person one but functionalities different .\n\nThere are two types of polymorphism as given below .\n\n1-Compile-Time-Polymorphism (Early binding or Static binding).\n\n1-Run-Time-Polymorphism (Late binding or Dynamic binding).\n\n1-Compile-Time-Polymorphism :\n\nIn compile time polymorphism the function link with object during compilation due to this it is called comipile time polymorphism . Everything will be decided in compile time . It can be implement by two ways given as follows .\n\nIn function overloading there are many function with same name but their parameters (argument) differents .\n\nExample : Write a program that demonstrate the compile time polymorphism .\n\n``` #include<iostream> using namespace std; class A { public : int Sum (int x,int y) // passed two parameters { return x+y; } int Sum(int x,int y, int z) // passed three parameters { return(x+y+z); } }; int main() { A obj; int s1=obj.Sum(10,20); int s2=obj.Sum(10,20,30); cout<<\"Sum : \"<<s1; cout<<\"Sum : \"<<s2; } ```\n``` OUTPUT : S1 : 30\nS2 : 60\n```\n\n2-Run-Time-Polymorphism :\n\nIn run time polymorphism the function link with object at run time due to this , it is called run time polymorphism . It is also called dynmac binding or late binding . It is implement by the function overrriding .\n\nFunction overriding :\n\nIn function overriding the functions name same and arguments same but their behaviours differents . We use the virtual keyword . It is possible to define a function in derived class that has the same name , same argument and same return type in the base class .\n\nExample : Write a program that demonstrate the run time polymorphism .\n\n``` #include<iostream> using namespace std; class Base { public: virtual void disp () { cout<<\"disp() of Base class\" <<endl; } void show () { cout<<\"show() of Base class\"<<endl; } }; class Derived : public Base { public: void disp () { cout<<\"disp() of Derived class\"<<endl; } void show () { cout<<\"show() of Derived class\"<<endl; } }; int main() { Base *bptr; Derived d; bptr = &d; //virtual function, binded at runtime (Runtime polymorphism) bptr->disp(); // Non-virtual function, binded at compile time bptr->show(); return 0; } ```\n``` OUTPUT : disp() of Derived class\nshow() of Base class\n```\n\n# Welcome to AndroidATPoint.\n\n`", null, "", null, "`" ]

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[ "diagrams-lib-1.3: Embedded domain-specific language for declarative graphics\n\nDiagrams.Points\n\nContents\n\nDescription\n\nPoints in space. For more tools for working with points and vectors, see Linear.Affine.\n\nSynopsis\n\n# Points\n\nnewtype Point f a :: (* -> *) -> * -> *\n\nA handy wrapper to help distinguish points from vectors at the type level\n\nConstructors\n\n P (f a)\n\nInstances\n\n Monad f => Monad (Point f) Functor f => Functor (Point f) Applicative f => Applicative (Point f) Foldable f => Foldable (Point f) Traversable f => Traversable (Point f) Generic1 (Point f) Apply f => Apply (Point f) Distributive f => Distributive (Point f) Representable f => Representable (Point f) Serial1 f => Serial1 (Point f) Additive f => Additive (Point f) Additive f => Affine (Point f) R4 f => R4 (Point f) R3 f => R3 (Point f) R2 f => R2 (Point f) R1 f => R1 (Point f) Metric f => Metric (Point f) Bind f => Bind (Point f) Eq1 f => Eq1 (Point f) Ord1 f => Ord1 (Point f) Read1 f => Read1 (Point f) Show1 f => Show1 (Point f) HasPhi v => HasPhi (Point v) HasTheta v => HasTheta (Point v) HasR v => HasR (Point v) (Metric v, OrderedField n) => TrailLike [Point v n] A list of points is trail-like; this instance simply computes the vertices of the trail, using `trailPoints`. Eq (f a) => Eq (Point f a) Fractional (f a) => Fractional (Point f a) (Data (f a), Typeable (* -> *) f, Typeable * a) => Data (Point f a) Num (f a) => Num (Point f a) Ord (f a) => Ord (Point f a) Read (f a) => Read (Point f a) Show (f a) => Show (Point f a) Ix (f a) => Ix (Point f a) Generic (Point f a) Storable (f a) => Storable (Point f a) Binary (f a) => Binary (Point f a) Serial (f a) => Serial (Point f a) Serialize (f a) => Serialize (Point f a) (OrderedField n, Metric v) => Enveloped (Point v n) (Additive v, Ord n) => Traced (Point v n) The trace of a single point is the empty trace, i.e. the one which returns no intersection points for every query. Arguably it should return a single finite distance for vectors aimed directly at the given point, but due to floating-point inaccuracy this is problematic. Note that the envelope for a single point is not the empty envelope (see Diagrams.Core.Envelope). (Additive v, Num n) => Transformable (Point v n) (Additive v, Num n) => HasOrigin (Point v n) Hashable (f a) => Hashable (Point f a) Ixed (f a) => Ixed (Point f a) Wrapped (Point f a) Epsilon (f a) => Epsilon (Point f a) Coordinates (v n) => Coordinates (Point v n) Typeable ((* -> *) -> * -> *) Point (~) * t (Point g b) => Rewrapped (Point f a) t (~) * r (Point u n) => Deformable (Point v n) r (Additive v, Foldable v, Num n, (~) * r (Point u n)) => AffineMappable (Point v n) r LinearMappable (Point v n) (Point u m) Traversable f => Each (Point f a) (Point f b) a b Each (FixedSegment v n) (FixedSegment v' n') (Point v n) (Point v' n') (Additive v', Foldable v', Ord n') => Each (BoundingBox v n) (BoundingBox v' n') (Point v n) (Point v' n') Only valid if the second point is not smaller than the first. type Rep1 (Point f) = D1 D1Point (C1 C1_0Point (S1 NoSelector (Rec1 f))) type Rep (Point f) = Rep f type Diff (Point f) = f type Rep (Point f a) = D1 D1Point (C1 C1_0Point (S1 NoSelector (Rec0 (f a)))) type V (Point v n) = v type N (Point v n) = n type Index (Point f a) = Index (f a) type IxValue (Point f a) = IxValue (f a) type Unwrapped (Point f a) = f a type FinalCoord (Point v n) = FinalCoord (v n) type PrevDim (Point v n) = PrevDim (v n) type Decomposition (Point v n) = Decomposition (v n)\n\norigin :: (Additive f, Num a) => Point f a\n\nVector spaces have origins.\n\n(*.) :: (Functor v, Num n) => n -> Point v n -> Point v n\n\nScale a point by a scalar. Specialized version of '(*^)'.\n\n# Point-related utilities\n\ncentroid :: (Additive v, Fractional n) => [Point v n] -> Point v n Source\n\nThe centroid of a set of n points is their sum divided by n.\n\npointDiagram :: (Metric v, Fractional n) => Point v n -> QDiagram b v n m\n\nCreate a \"point diagram\", which has no content, no trace, an empty query, and a point envelope.\n\n_Point :: (Profunctor p, Functor f) => p (f a) (f (f a)) -> p (Point f a) (f (Point f a))\n\nlensP :: Functor f => (g a -> f (g a)) -> Point g a -> f (Point g a)" ]

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[ "# Yahoo Web Search\n\n1. ### Electrons accelerated by a potential difference of 13.02 V pass through a gas of hydrogen atoms at room temper?\n\n... through a potential difference of 13.02 V , its kinetic energy is 13.02eV. ... = -13.6/1² = -13.6eV (ground state ) We need to find the highest energy level to...\n\n3 Answers · Science & Mathematics · 22/03/2013\n\n2. ### Repost: How long before the drat touches the ground?\n\n73 = v cos30º t 0 = 45 + v sin30º t - (1/2) g t² ==> 0 = 45 + 73 ... before jumping and it is plausible that this state depends on their weight. In that case we need...\n\n2 Answers · Science & Mathematics · 19/03/2009\n\n3. ### QUANTUM PHYSICS QUESTION (Best answer gets many points)?\n\n...u-hat) u-hat + (A ▪ v -hat) v -hat. This is what the equation...gt; . This tells you how long | ψ > is along | φi >. ...change basis, you need to project your state vector onto the basis to see how long it...\n\n1 Answers · Science & Mathematics · 31/10/2016\n\n4. ### Physics: Does anyone have an idea how long a juggling cube takes to be released.?\n\n...hand: Equation: V² = u² + 2*a*s V = final velocity = 0m/s u = initial velocity - what we want a...9.81)*0.3 u² = 5.886 u = 2.43m/s You state that the hand moves 6cm in increasing the velocity...\n\n2 Answers · Science & Mathematics · 14/01/2014\n\n5. ### Potential and Kinetic Energy?\n\nFor the first question: Energy is energy. Energy is conserved in a closed system. That should explain enough right there. =-) For the second question: What's work measured in and what's the unit...\n\n3 Answers · Science & Mathematics · 20/05/2011\n\n6. ### PHYSICS QUESTION: A person on an apartment balcony holds a ball over the balcony and throws it directly upward?\n\n...up. 1) how high does ball go above balcony: V / A = T, V /2 x T = distance (up) 2) how long to hit the ground: D = balcony + up, use: Down =AT^2/2...\n\n1 Answers · Science & Mathematics · 25/02/2013\n\n7. ### How do you find the spring constant of a bungee cord without using a mass?\n\n... proved by some random person hundreds of years ago which states v = sqrt(T/D). v = speed of a wave in a stretched string T...\n\n1 Answers · Science & Mathematics · 18/06/2010\n\n8. ### Physics circular motion problem, totally stumped.?\n\nYou forgot to state that it accelerated UNIFORMLY... the rotational speed at 1.25s will be v = at = 0.500m/s. Then its...125) = 72.25... degrees. Live long and prosper.\n\n1 Answers · Science & Mathematics · 22/07/2007\n\n9. ### Help With PHYSICS Please!!!!!!!! NEED HELP FAST!!?\n\n...question please? i would be oh so grateful (only if ur capable of course!) 1) use formula v =d/t velocity is 45 km/h which is equal to d, which is 10 km...\n\n2 Answers · Science & Mathematics · 24/02/2009\n\n10. ### Physics Kinematics Questions help?\n\n...can do the rest you just have to use the three equations of uniform acceleration : v = u+at v ^2 = u^2 + 2as s = ut + 1/2*a*t^2\n\n1 Answers · Science & Mathematics · 01/03/2016" ]

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