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https://farside.ph.utexas.edu/teaching/336L/Fluidhtml/node190.html
[ "", null, "", null, "", null, "Next: One-Dimensional Compressible Inviscid Flow Up: Equilibrium of Compressible Fluids Previous: Eddington Solar Model\n\n# Exercises\n\n1. Prove that the fraction of the whole mass of an isothermal atmosphere that lies between the ground and a horizontal plane of height", null, "is", null, "Evaluate this fraction for", null, ",", null, ",", null, ", respectively.\n\n2. If the absolute temperature in the atmosphere diminishes upwards according to the law", null, "where", null, "is a constant, show that the pressure varies as", null, "3. If the absolute temperature in the atmosphere diminishes upward according to the law", null, "where", null, "is a constant, show that the pressure varies as", null, "4. Show that if the absolute temperature,", null, ", in the atmosphere is any given function of the altitude,", null, ", then the vertical distribution of pressure in the atmosphere is given by", null, "5. Show that if the Earth were surrounded by an atmosphere of uniform temperature then the pressure a distance", null, "from the Earth's center would be", null, "where", null, "is the Earth's radius.\n\n6. Show that if the whole of space were occupied by air at the uniform temperature", null, "then the densities at the surfaces of the various planets would be proportional to the corresponding values of", null, "where", null, "is the radius of the planet, and", null, "its surface gravitational acceleration.\n\n7. Prove that in an atmosphere arranged in horizontal strata the work (per unit mass) required to interchange two thin strata of equal mass without disturbance of the remaining strata is", null, "where the suffixes refer to the initial states of the two strata. Hence, show that for stability the ratio", null, "must increase upwards.\n8. A spherically symmetric star is such that", null, "is the mass contained within radius", null, ". Show that the star's total gravitational potential energy can be written in the following three alternative forms:", null, "Here,", null, "is the total mass,", null, "the radius,", null, "the gravitational potential per unit mass (defined such that", null, "as", null, "),", null, "the pressure, and", null, ".\n\n9. Suppose that the pressure and density inside a spherically symmetric star are related according to the polytropic gas law,", null, "where", null, "is termed the polytropic index. Let", null, ", where", null, "is the central mass density. Demonstrate that", null, "satisfies the Lane-Emden equation", null, "where", null, ", and", null, "Show that the physical solution to the Lane-Emden equation, which is such that", null, "and", null, ", for some", null, ", is", null, "for", null, ",", null, "for", null, ", and", null, "for", null, ". Determine the ratio of the central density to the mean density in all three cases. Finally, demonstrate that, in the general case, the total gravitational potential energy can be written", null, "where", null, "is the total mass, and", null, "the radius.\n10. A spherically symmetric star of radius", null, "has a mass density of the form", null, "Show that the central mass density is four times the mean density. Demonstrate that the central pressure is", null, "where", null, "is the mass of the star. Finally, show that the total gravitational potential energy of the star can be written", null, "", null, "", null, "", null, "Next: One-Dimensional Compressible Inviscid Flow Up: Equilibrium of Compressible Fluids Previous: Eddington Solar Model\nRichard Fitzpatrick 2016-03-31" ]
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https://www.symscape.com/node/1655
[ "# Time Variant Flow Setup\n\nI am trying to set up a time variant flow at a face but it is not clear to me how set up the values in the parentheses. For example, I am using units of minutes and feet. I want to set up the flow to be zero for 15 seconds then have a vertical flow rate of 250 ft/min after 15 seconds.\n\nAre the units that are used in this field the same as my IP units?\n\nHow do I structure the input?\n\nKen\n\n### Run a transient simulation with fixed inlet flow rate?\n\nThe units are SI for tabulated data, so the vector has to be m/s. From the Tooltip the description of the format is:\n\nTime dependent vector. E.g., table ( (0.5 (5 0 0)) (1 (10 0 0)) ) - (time, (vector))\n\nThis type of boundary condition is not intended to act as a valve (i.e., off/on). The idea is that the flow rate varies with time, without being zero. Instead can you start the simulation with 250 ft/min (i.e., at 15s) as an inlet flow rate and then run the simulation in transient mode?" ]
[ null ]
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http://iconicmath.com/introduction/
[ "# Introduction", null, "This section includes six pages of introductory discussion. The Image of a road sign shows the way.\n\n## Section Contents\n\nThe The Iconic Math Site page is an introduction to the site and to iconic math.\n\nThe Technical Introduction presents the formal techniques of iconic math, how pictures can behave mathematically.\n\nIconic Principles lists several principles of iconic math, with little discussion and no examples.\n\nVarieties introduces some of the textual, pictorial, and physical languages that can be used to express mathematical concepts.\n\nMath Education suggests ways in which math education in US schools might be improved.\n\nEducational Theory discusses the educational research that supports a change to iconic math." ]
[ null, "http://iconicmath.com/wordpress/wp-content/mymedia/featureimage-introduction300x300-41346_186x186.jpg", null ]
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https://www.dataunitconverter.com/byte-per-minute-to-zebibit-per-hour
[ "# Byte/Minute to Zibit/Hour Calculator - Convert Bytes Per Minute to Zebibits Per Hour", null, "## Conversion History (Last 6)\n\nInput Bytes Per Minute - and press Enter\nByte/Minute\n\nSec\nMin\nHr\nDay\nSec\nMin\nHr\nDay\n\n## Byte/Minute to Zibit/Hour - Conversion Formula and Steps\n\nByte and Zebibit are units of digital information used to measure storage capacity and data transfer rate. Byte is one of the very basic digital unit where as Zebibit is a binary unit. One Byte is equal to 8 bits. One Zebibit is equal to 1024^7 bits. There are 147,573,952,589,676,412,928 Bytes in one Zebibit. - view the difference between both units", null, "Source Data UnitTarget Data Unit\nByte (B)\nEqual to 8 bits\n(Basic Unit)\nZebibit (Zibit)\nEqual to 1024^7 bits\n(Binary Unit)\n\nThe formula of converting the Bytes Per Minute to Zebibits Per Hour is represented as follows :\n\nZibit/Hour = Byte/Minute x 8 / 10247 x 60\n\nNow let us apply the above formula and, write down the steps to convert from Bytes Per Minute (B/Minute) to Zebibits Per Hour (Zibit/Hour). This way, we can try to simplify and reduce to an easy to apply formula.\n\nFORMULA\n\nZebibits Per Hour = Bytes Per Minute x 8 / 10247 x 60\n\nSTEP 1\n\nZebibits Per Hour = Bytes Per Minute x 8 / (1024x1024x1024x1024x1024x1024x1024) x 60\n\nSTEP 2\n\nZebibits Per Hour = Bytes Per Minute x 8 / 1180591620717411303424 x 60\n\nSTEP 3\n\nZebibits Per Hour = Bytes Per Minute x 0.0000000000000000000067762635780344027125 x 60\n\nExample : If we apply the above Formula and steps, conversion from 10 Byte/Minute to Zibit/Hour, will be processed as below.\n\n1. = 10 x 8 / 10247 x 60\n2. = 10 x 8 / (1024x1024x1024x1024x1024x1024x1024) x 60\n3. = 10 x 8 / 1180591620717411303424 x 60\n4. = 10 x 0.0000000000000000000067762635780344027125 x 60\n5. = 0.0000000000000000040657581468206416275\n6. i.e. 10 Byte/Minute is equal to 0.0000000000000000040657581468206416275 Zibit/Hour.\n\n(Result rounded off to 40 decimal positions.)\n\nYou can use above formula and steps to convert Bytes Per Minute to Zebibits Per Hour using any of the programming language such as Java, Python or Powershell.\n\n#### Definition : Byte\n\nA Byte is a unit of digital information that typically consists of 8 bits and can represent a wide range of values such as characters, binary data and it is widely used in the digital world to measure the data size and data transfer speed.\n\n#### Definition : Zebibit\n\nA Zebibit (Zib or Zibit) is a unit of digital information that is equal to 1,180,591,620,717,411,303,424 bits and is defined by the International Electro technical Commission(IEC). The prefix \"zebi\" is derived from the binary number system and it is used to distinguish it from the decimal-based \"zettabit\" (Zb). It is widely used in the field of computing as it more accurately represents the amount of data storage and data transfer in computer systems.\n\n### Excel Formula to convert from Byte/Minute to Zibit/Hour\n\nApply the formula as shown below to convert from Bytes Per Minute to Zebibits Per Hour.\n\nABC\n1Bytes Per Minute (B/Minute)Zebibits Per Hour (Zibit/Hour)\n21=A2 * 0.0000000000000000000067762635780344027125 * 60\n3\n\nDownload - Excel Template for Bytes Per Minute to Zebibits Per Hour Conversion\n\nIf you want to perform bulk conversion locally in your system, then download and make use of above Excel template.\n\n### Python Code for Byte/Minute to Zibit/Hour Conversion\n\nYou can use below code to convert any value in Bytes Per Minute to Zebibits Per Hour in Python.\n\nbytesPerMinute = int(input(\"Enter Bytes Per Minute: \"))\nzebibitsPerHour = bytesPerMinute * 8 / (1024*1024*1024*1024*1024*1024*1024) * 60\nprint(\"{} Bytes Per Minute = {} Zebibits Per Hour\".format(bytesPerMinute,zebibitsPerHour))\n\nThe first line of code will prompt the user to enter the Bytes Per Minute as an input. The value of Zebibits Per Hour is calculated on the next line, and the code in third line will display the result." ]
[ null, "https://www.dataunitconverter.com/images/certificate_precise.png", null, "https://www.dataunitconverter.com/showimage.php/Bytes Per Minute_to_Zebibits Per Hour_Dataunitconverter.png", null ]
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https://answers.everydaycalculation.com/add-fractions/36-30-plus-50-56
[ "Solutions by everydaycalculation.com\n\n1st number: 1 6/30, 2nd number: 50/56\n\n36/30 + 50/56 is 293/140.\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 30 and 56 is 840\n2. For the 1st fraction, since 30 × 28 = 840,\n36/30 = 36 × 28/30 × 28 = 1008/840\n3. Likewise, for the 2nd fraction, since 56 × 15 = 840,\n50/56 = 50 × 15/56 × 15 = 750/840\n1008/840 + 750/840 = 1008 + 750/840 = 1758/840\n5. 1758/840 simplified gives 293/140\n6. So, 36/30 + 50/56 = 293/140\nIn mixed form: 213/140\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 ]
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https://www.tutorialkart.com/cpp/cpp-vector-clear/
[ "## C++ Vector clear() function\n\nclear() function removes all the elements from a vector.\n\n### Syntax\n\nThe syntax of clear() function is\n\n`void clear();`\n\n### Example\n\nIn the following C++ program, we initialise a vector `names` with three elements, and then erase all the elements from this vector using clear() function.\n\nC++ Program\n\n```#include <iostream>\n#include <vector>\nusing namespace std;\n\nint main() {\nvector<string> names{ \"apple\", \"banana\", \"cherry\" };\nnames.clear();\ncout << \"Size of Vector : \" << names.size() << endl;\n}```\n\nOutput\n\n```Size of Vector : 0\nProgram ended with exit code: 0```\n\nAfter clear(), all the elements are removed, and hence the size of the resulting vector is 0.\n\n### Conclusion\n\nIn this C++ Tutorial, we learned the syntax of clear() function, and how to use this clear() function to remove all the elements from given vector." ]
[ null ]
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https://discuss.codechef.com/t/doubt-0-1-graph-hackerrank-hack-the-interview-vi/73739
[ "", null, "# Doubt - 0-1 Graph (Hackerrank Hack the Interview VI)\n\nThe problem is here. All the test cases are available if you solve the problem (you might not be able to access that link). The official editorial is here.\nMy solution gets WA in three of fourteen tests.\n\nMy logic is:\nEach graph can be transformed to a tree which will have the same solution. This tree is constructed such that each vertex > parent of this vertex.\nInitially, the tree is linear ie 1 <- 2 <- 3 <- … <- n. Then, go through the edges of the graph. For each edge (a, b) where a < b, set tree[b] = min(tree[b], a). Then, we traverse the array in reverse manner reducing values to make it a non-decreasing array. But the values are not distances, they are parent vertices. The last loop calculates the sum of distances.\n\n``` void solve() { ll n, nf_edges; scan(n);scan(nf_edges); vector<ll> tree(n + 1); for (ll i = 1; i <= n; ++i) { tree[i] = i - 1; } //print_container(tree); ll one, two, large, small; while (nf_edges --) { scan(one); scan(two); large = max(one, two); small = min(one, two); tree[large] = min(tree[large], small); } //print_container(tree); for (ll i = n, level = n; i > 1; i--) { if (level > tree[i]) { level = tree[i]; } else { tree[i] = level; } } //print_container(tree); ll summ = 0; for (ll i = 2, mul = 0; i <= n; ++i) { if (tree[i - 1] != tree[i]) { mul++; } summ += mul; } pnl(summ); } ```\n\nI’d like an explanation for why this approach is incorrect or a small test case for which my approach fails.\n\nDoubt has been resolved. It turns out the editorial was incorrect.\n\nHello,\n\nI am Darshan, I manage content on HackerRank. We would like to apologize for the issues in this challenge in our latest contest.\n\nFor the complete correctness of the challenge, our author’s greedy solution does not suffice, as indicated correctly by you.\n\nWe benchmark our challenges through multiple coders, and with this specific challenge, a corner case was not discovered by us which resulted in providing an incorrect challenge.\n\nWe are carefully evaluating our internal practices in order to avoid such issues in the future.\n\nThe Greedy Solution provided as part of the Problem Setters solution missed on this key test case, we will be updating the solution here soon.\n\nWe are really sorry for the inconvenience caused and as a result, we have decided to not consider this challenge in the contest to calibrate the results and will update the leaderboard accordingly. We are working on making our problems more robust every day. Your feedback is much appreciated and will help us improve our challenge quality.\n\nHope to see you again in future contests! Thank you.\n\n1 Like" ]
[ null, "https://s3.amazonaws.com/discourseproduction/original/3X/7/f/7ffd6e5e45912aba9f6a1a33447d6baae049de81.svg", null ]
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https://www.mathworks.com/matlabcentral/profile/authors/9766874-fw?s_tid=answers_ctrb
[ "Community Profile", null, "# FW\n\n23 total contributions since 2017\n\n•", null, "View details...\n\nContributions in\nView by\n\nQuestion\n\nHow to convert or extract arrays from a structure?\nIn continuous wavelet analysis, when we export the coefficients into the workspace, the data is exported as a structure (1x1 str...\n\n21 days ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nIntegration via trapezoidal rule in various sections of the same array\nIf the signal S consists of several peaks as a function of time, is there a way to assign integration limits for the trapezoidal...\n\n25 days ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nGenerating Approximations from \"appcoef\" in Wavelet Toolbox\nI have a question regarding the MATLAB example for \"1-D approximation coefficients\" https://www.mathworks.com/help/wavelet/ref/a...\n\n25 days ago | 0 answers | 0\n\n### 0\n\nQuestion\n\nHow to replace complex numbers in a column vector with 0.000+0.000i ?\nImage we obtain a fft of a signal and we wish to remove certain frequencies from the signal. I want to fill in zeros after remov...\n\n29 days ago | 2 answers | 0\n\n### 2\n\nQuestion\n\nFinding several local maximum values in a given range and corresponding indices\nIf we have a dataset \"y\" which consists of a sum of 5 gaussian peaks as function of time t, there will be 5 local maximum value...\n\n2 months ago | 2 answers | 0\n\n### 2\n\nQuestion\n\nHow to locate the index of the maximum value in a given range\nSuppose we have two vectors t= [ 1 2 3 4 5 6 7 8 9 10 11]; % time values y = [1 2 3 4 5 7 5 4 3 2 1]; % a dependent variable...\n\n2 months ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nHow to implement a formula involving several elements of the same vector?\nSuppose we have two row vectors Time = [ 1 2 3 4 5 6 7 8 9 10]; width= [0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0]; % width of p...\n\n2 months ago | 2 answers | 0\n\n### 2\n\nQuestion\n\nHow to add non-numerical axis values in 3D scatter plots?\nI am using a 3D scatter plot option. Can the x-axis values be labeled as \"1:1, 2:1, 5:1, 8:1, 10:1\" as peak area ratios? Current...\n\n2 months ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nIf we have an Excel file, and instead of selecting a range of columns, I would like to select certain columns. The following syn...\n\n2 months ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nLabeling the columns in an array\nIf we import a .csv or Excel file using readtable, and then convert the table to an array as the following: Data=table2array(re...\n\n4 months ago | 1 answer | 1\n\n### 1\n\nQuestion\n\nCorrect choice of one sided frequency axis after fft\nThere are a lot of queries on fft frequency. I guess the following point not discussed anywhere explicitly (at least at one plac...\n\n4 months ago | 0 answers | 0\n\n### 0\n\nQuestion\n\nHow to replace numbers in a column vector after certain elements?\nHow should we proceed if we have a column vector X with 1000 elements, and we wish to replace all the values after a certain row...\n\n4 months ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nfftshift with even and odd number of data points (scaling the positive and negative frequency axis)\nIf we wish to draw double sided frequency axis, i.e., positive and negative frequencies with odd and even number of data points ...\n\n6 months ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nHow to add labels to an array or table (after conversion from an array)\nI have an array of 2000 x 4 data points. I would like to add labels to the columns only. For example, Column 1 should be 'time/m...\n\n6 months ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nAssigning indexes in an array and finding max values in a selected range\nIf we have X = time series data and signal in a 8000x 2 array (Matlab 2017b). The first column is time and the second column is ...\n\n7 months ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nFrequency scale in fft after zero filling (odd/even number of data)\nProduct: Matlab 2017b If we have data set \"A\" which consists of time versus light absorbance.There is another set of data \"B\"....\n\n7 months ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nHow to combine sections from Workspace into a new vector?\nI have Matlab 2017b. 1) If we have two vectors A and B, in the workspace (1000x1), each of same dimension. How can one combine...\n\n7 months ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nCreating Timetables: Error Message in SampleRate\nHello, I am trying to use a suggested function to generate a timetable with a sampling frequency of 200 Hz. tt = timetable(rand...\n\n7 months ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nDoing averages in blocks for downsampling time series data\nOne way to downsample is to do averaging in blocks. I had asked a question here \"I would like to downsample data simulated at 20...\n\n7 months ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nCommand for filling in zeros as powers of 2 for FT\nIf we two sets of signals (A and B) of unequal length, I am using zero padding first to make the length of B equal to A by using...\n\n7 months ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nHow to treat a given column in a sheet from Workspace as a variable?\nI am generating exponentially modified Gaussians (EMGs) from a simulation and wish to do FT on a selected sheet in the Workspace...\n\n7 months ago | 0 answers | 0\n\n### 0\n\nQuestion\n\nHow to read two Excel files and assign columns as variables in MATLAB?\nHello, I am using MATLAB 2017b. I have two Excel.xlsx files, call them A and B, containing time vs. instrument signals under two...\n\n7 months ago | 1 answer | 0" ]
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https://www.colorhexa.com/0d0da5
[ "# #0d0da5 Color Information\n\nIn a RGB color space, hex #0d0da5 is composed of 5.1% red, 5.1% green and 64.7% blue. Whereas in a CMYK color space, it is composed of 92.1% cyan, 92.1% magenta, 0% yellow and 35.3% black. It has a hue angle of 240 degrees, a saturation of 85.4% and a lightness of 34.9%. #0d0da5 color hex could be obtained by blending #1a1aff with #00004b. Closest websafe color is: #000099.\n\n• R 5\n• G 5\n• B 65\nRGB color chart\n• C 92\n• M 92\n• Y 0\n• K 35\nCMYK color chart\n\n#0d0da5 color description : Dark blue.\n\n# #0d0da5 Color Conversion\n\nThe hexadecimal color #0d0da5 has RGB values of R:13, G:13, B:165 and CMYK values of C:0.92, M:0.92, Y:0, K:0.35. Its decimal value is 855461.\n\nHex triplet RGB Decimal 0d0da5 `#0d0da5` 13, 13, 165 `rgb(13,13,165)` 5.1, 5.1, 64.7 `rgb(5.1%,5.1%,64.7%)` 92, 92, 0, 35 240°, 85.4, 34.9 `hsl(240,85.4%,34.9%)` 240°, 92.1, 64.7 000099 `#000099`\nCIE-LAB 20.399, 53.686, -75.306 7.1, 3.089, 35.817 0.154, 0.067, 3.089 20.399, 92.483, 305.485 20.399, -5.655, -78.367 17.577, 41.344, -108.514 00001101, 00001101, 10100101\n\n# Color Schemes with #0d0da5\n\n• #0d0da5\n``#0d0da5` `rgb(13,13,165)``\n• #a5a50d\n``#a5a50d` `rgb(165,165,13)``\nComplementary Color\n• #0d59a5\n``#0d59a5` `rgb(13,89,165)``\n• #0d0da5\n``#0d0da5` `rgb(13,13,165)``\n• #590da5\n``#590da5` `rgb(89,13,165)``\nAnalogous Color\n• #59a50d\n``#59a50d` `rgb(89,165,13)``\n• #0d0da5\n``#0d0da5` `rgb(13,13,165)``\n• #a5590d\n``#a5590d` `rgb(165,89,13)``\nSplit Complementary Color\n• #0da50d\n``#0da50d` `rgb(13,165,13)``\n• #0d0da5\n``#0d0da5` `rgb(13,13,165)``\n• #a50d0d\n``#a50d0d` `rgb(165,13,13)``\n• #0da5a5\n``#0da5a5` `rgb(13,165,165)``\n• #0d0da5\n``#0d0da5` `rgb(13,13,165)``\n• #a50d0d\n``#a50d0d` `rgb(165,13,13)``\n• #a5a50d\n``#a5a50d` `rgb(165,165,13)``\n• #07075e\n``#07075e` `rgb(7,7,94)``\n• #090976\n``#090976` `rgb(9,9,118)``\n• #0b0b8d\n``#0b0b8d` `rgb(11,11,141)``\n• #0d0da5\n``#0d0da5` `rgb(13,13,165)``\n• #0f0fbd\n``#0f0fbd` `rgb(15,15,189)``\n• #1111d4\n``#1111d4` `rgb(17,17,212)``\n• #1313ec\n``#1313ec` `rgb(19,19,236)``\nMonochromatic Color\n\n# Alternatives to #0d0da5\n\nBelow, you can see some colors close to #0d0da5. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #0d33a5\n``#0d33a5` `rgb(13,51,165)``\n• #0d26a5\n``#0d26a5` `rgb(13,38,165)``\n• #0d1aa5\n``#0d1aa5` `rgb(13,26,165)``\n• #0d0da5\n``#0d0da5` `rgb(13,13,165)``\n• #1a0da5\n``#1a0da5` `rgb(26,13,165)``\n• #260da5\n``#260da5` `rgb(38,13,165)``\n• #330da5\n``#330da5` `rgb(51,13,165)``\nSimilar Colors\n\n# #0d0da5 Preview\n\nThis text has a font color of #0d0da5.\n\n``<span style=\"color:#0d0da5;\">Text here</span>``\n#0d0da5 background color\n\nThis paragraph has a background color of #0d0da5.\n\n``<p style=\"background-color:#0d0da5;\">Content here</p>``\n#0d0da5 border color\n\nThis element has a border color of #0d0da5.\n\n``<div style=\"border:1px solid #0d0da5;\">Content here</div>``\nCSS codes\n``.text {color:#0d0da5;}``\n``.background {background-color:#0d0da5;}``\n``.border {border:1px solid #0d0da5;}``\n\n# Shades and Tints of #0d0da5\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, #000001 is the darkest color, while #eeeefe is the lightest one.\n\n• #000001\n``#000001` `rgb(0,0,1)``\n• #020214\n``#020214` `rgb(2,2,20)``\n• #030326\n``#030326` `rgb(3,3,38)``\n• #040438\n``#040438` `rgb(4,4,56)``\n• #06064a\n``#06064a` `rgb(6,6,74)``\n• #07075c\n``#07075c` `rgb(7,7,92)``\n• #09096e\n``#09096e` `rgb(9,9,110)``\n• #0a0a81\n``#0a0a81` `rgb(10,10,129)``\n• #0c0c93\n``#0c0c93` `rgb(12,12,147)``\n• #0d0da5\n``#0d0da5` `rgb(13,13,165)``\n• #0e0eb7\n``#0e0eb7` `rgb(14,14,183)``\n• #1010c9\n``#1010c9` `rgb(16,16,201)``\n• #1111dc\n``#1111dc` `rgb(17,17,220)``\n• #1414ec\n``#1414ec` `rgb(20,20,236)``\n• #2626ee\n``#2626ee` `rgb(38,38,238)``\n• #3838ef\n``#3838ef` `rgb(56,56,239)``\n• #4b4bf1\n``#4b4bf1` `rgb(75,75,241)``\n• #5d5df2\n``#5d5df2` `rgb(93,93,242)``\n• #6f6ff4\n``#6f6ff4` `rgb(111,111,244)``\n• #8181f5\n``#8181f5` `rgb(129,129,245)``\n• #9393f7\n``#9393f7` `rgb(147,147,247)``\n• #a5a5f8\n``#a5a5f8` `rgb(165,165,248)``\n• #b8b8f9\n``#b8b8f9` `rgb(184,184,249)``\n• #cacafb\n``#cacafb` `rgb(202,202,251)``\n• #dcdcfc\n``#dcdcfc` `rgb(220,220,252)``\n• #eeeefe\n``#eeeefe` `rgb(238,238,254)``\nTint Color Variation\n\n# Tones of #0d0da5\n\nA tone is produced by adding gray to any pure hue. In this case, #58585a is the less saturated color, while #0606ac is the most saturated one.\n\n• #58585a\n``#58585a` `rgb(88,88,90)``\n• #515161\n``#515161` `rgb(81,81,97)``\n• #4b4b67\n``#4b4b67` `rgb(75,75,103)``\n• #44446e\n``#44446e` `rgb(68,68,110)``\n• #3d3d75\n``#3d3d75` `rgb(61,61,117)``\n• #36367c\n``#36367c` `rgb(54,54,124)``\n• #2f2f83\n``#2f2f83` `rgb(47,47,131)``\n• #28288a\n``#28288a` `rgb(40,40,138)``\n• #222290\n``#222290` `rgb(34,34,144)``\n• #1b1b97\n``#1b1b97` `rgb(27,27,151)``\n• #14149e\n``#14149e` `rgb(20,20,158)``\n• #0d0da5\n``#0d0da5` `rgb(13,13,165)``\n• #0606ac\n``#0606ac` `rgb(6,6,172)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #0d0da5 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]
[ null ]
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https://demonstrations.wolfram.com/AnalysisOfLatticeVibrationsInTwoDimensions/
[ "", null, "# Analysis of Lattice Vibrations in Two Dimensions\n\nInitializing live version", null, "Requires a Wolfram Notebook System\n\nInteract on desktop, mobile and cloud with the free Wolfram Player or other Wolfram Language products.\n\nAt first glance, the hexagonal and square lattices appear to have the same symmetry as their unit cells (a simple hexagon or square), but closer examination reveals more rotational and reflectional symmetries, not to mention a countably infinite number of translations. Of course, a real crystal lattice deviates from perfect symmetry as it interacts with photons in energy transfer that results in the creation of discrete vibrational states.\n\n[more]\n\nSimple restoring force models of linear crystals can be extended to planar crystals by considering angles as well as edges. Resolving the eigenvalues and eigenvectors of a Hessian matrix can help to understand how superpositions of vibrational modes relate to possibly observable spectra. Only some of the computable values will be observed in the laboratory. The best predictions of observable transitions also consider the symmetry of the vibrational modes, as determined by their equivalency class.\n\n[less]\n\nContributed by: Brad Klee (August 2012)\nOpen content licensed under CC BY-NC-SA\n\n## Snapshots", null, "", null, "", null, "## Details\n\nSnapshot 1, Energy Summation Unit: As a tile in the lattice deforms according to the propagation of a wavefunction away from the equilibrium configuration, the potential energy of the entire tiling increases because of the deformed edges and angles. Each contribution to the increase in potential energy is shown as a highlighted edge or highlighted vertex, and the magnitude of the highlighted region indicates the contribution of that deformation to the total potential energy.\n\nSnapshot 2, Hessian Matrix: An array plot of the Hessian matrix shows the magnitude of the second derivative of the potential energy, which is computed using a central difference method and two perturbations for each atom. The eigenvectors and eigenvalues of this matrix are the normal modes and their corresponding vibrational frequencies. The Hessian matrix, given by", null, ", has the units of a force constant.\n\nSnapshot 3, Block Diagonal Transformation: The perturbation basis elements transform into one another according to the operations of dihedral symmetry, for both square and hexagonal lattices. The Hessian matrix computed from these perturbations does not change under conjugation by any transformation of the dihedral group, that is,", null, ". Thus,", null, "can be reduced to block diagonal form using a transformation", null, "derived from the irreducible representations of the dihedral group. Eigenvectors computed from the block-diagonalized matrix can all be real even if eigenvectors computed from the untransformed matrix are complex.\n\nSnapshot 4, Energy Level Diagram: This plot shows energy levels for each selected wavefunction. The left-hand side of each level is labeled by some number linearly proportional to a frequency, computed from the square root of an eigenvalue of the Hessian matrix", null, ". Indices on the right-hand side correspond to symmetries of the vibrational modes, which can be determined using sets of simple linear functionals.\n\nThe construction of the Hessian matrix is adapted from . Each angular dimension resists deformation with a factor of", null, ", and each linear dimension resists deformation with a factor of", null, ". The total energy of a lattice acted upon by periodic deformations is infinite and thus incalculable. However, an adequately apportioned fraction of the total energy can be computed by summing the energy from each tile multiplied by the population percentage of that tile. Care must be taken not to count edges twice.\n\nIn this notebook, the energy computation is correct for the square lattice and for the A hexagonal lattice but not for the B hexagonal lattice. The A hexagonal lattice contains only three species of tiles, while the B hexagonal lattice contains 24 species of tiles. A correct computation for the B hexagonal lattice in 12 dimensions could be accomplished by changing the energy function to sum over the six edges and six angles of all 24 species, with the appropriate prefactors according to relative population and overlap of edges. Each basis pattern has a unique spectrum and perturbation periodicity. It is unknown how many basis patterns exist, and the existence or absence of a basis pattern containing 720 tiles is unproven but not excluded.\n\nBlock diagonalization is performed by considering the irreducible representations of the appropriate dihedral group. Block membership determines an important equivalency class for eigenvectors, related to selection rules and transition probabilities. A classification system based on annihilators and orthogonal to block membership has been used to number the degeneracies.\n\nThis Demonstration should not be considered physical, but systems of springs and masses can give insight for problems where restoring forces hold atoms in a specific geometric pattern while some energy excitations attempt to move them, thus creating phonons.\n\nThanks to Janette Dunn for providing educational resources regarding representation-derived bases.\n\nReference\n\n D. E. Weeks and W. G. Harter, \"Rotation-Vibration Spectra of Icosahedral Molecules. II. Icosahedral Symmetry, Vibrational Eigenfrequencies, and Normal Modes of Buckminsterfullerene,\" Journal of Chemical Physics, 90(9), 1989 pp. 4744–4771. doi: 10.1063/1.456571." ]
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http://momentumclubs.org/define-gravitational-field.html
[ "# Define gravitational field. Gravitational fields 2019-01-08\n\nDefine gravitational field Rating: 5,3/10 172 reviews\n\n## Gravitational fields", null, "A gravitational field is the force field that exists in the space around every mass or group of masses. Modern field theories of force contain this principle by requiring every entity that is acted upon by a field to be also a source of the field. What do the field lines look like for this two mass case? One possible class of black holes very large stars that have used up all of their so that they are no longer held up by and have collapsed into black holes less-massive stars may collapse into. Search gravitational field and thousands of other words in English definition and synonym dictionary from Reverso. Just as gravitational field strength is used to place a value on the gravitational force that would be experienced by a mass at any point in a gravitational field, the concept of gravitational potential is used to give a value for the potential energy.\n\nNext\n\n## Gravitational force dictionary definition", null, "I just put this section in so you can see that this result can be derived by the straightforward, but quite challenging, method of adding the individual gravitational attractions from all the bits making up the spherical shell. To find the gravitational field at the point P , we just add the contributions from all the rings in the stack. By symmetry, the field at P must point along the x -axis, so all we have to do is compute the strength of the x -component of the gravitational force from one mass, and double it. The , and thus the acceleration of a small body in the space around the massive object, is the negative of the gravitational potential. The Principle of Superposition The fact that the total gravitational field is just given by adding the two vectors together is called the Principle of Superposition.\n\nNext\n\n## Gravitational field", null, "For simple mechanics problems, a convenient zero point would be at the floor of the laboratory or at the surface of a table. If the force were to be removed, the object would fall back down to the ground and the gravitational potential energy would be transferred to kinetic energy of the falling object. Arrows and field lines that represent gravitational field The gravitational field varies slightly at the earth's surface. In other words, the two bodies interact with one another's gravitational field. We all know instinctively that a heavy weight raised above someone's head represents a potentially dangerous situation. This is a thousand times less than the size of an atomic nucleus! This makes all values of the gravitational potential energy negative. The field still exists at all the places where we did not measure it.\n\nNext\n\n## Gravitational field", null, "Definitions in physics are operational, i. Gravitational Field for Two Masses The next simplest case is two equal masses. This experimental support for his general theory of relativity garnered him instant worldwide acclaim. Advanced Engineering Mathematics 2nd ed. A gravitational field is a type of force field and is analogous to electric and magnetic fields for electrically charged particles and magnets, respectively. For a satellite to move from an orbit where the potential is — 4. To move to a higher or lower orbit the satellite must gain or lose energy.\n\nNext\n\n## newtonian gravity", null, "Following Newton, Laplace attempted to model gravity as some kind of radiation field or fluid, and since the 19th century explanations for gravity have usually been sought in terms of a field model, rather than a point attraction. Why define a new name and new units for the same old quantity? An experiment by the American physicist Lloyd Kreuzer established to within 1 part in 20,000 that different materials produce gravitational fields with a strength the same as that of gravitational fields acting upon them. The pump becomes a generator driven by the gravitational potential energy of the water in the upper reservoir. Field Strength at a Point Equidistant from the Two Masses It is not difficult to find an exact expression for the gravitational field strength from the two equal masses at an equidistant point P. This is clearly not a problem for a one kilogram mass in discussing planetary and solar gravity. By moving the test mass to various positions,we can make a map showing the gravitational field at any point in space. An important fact about all fields of force is that when there is more than one source or sink , the fields add according to the rules of vector addition.\n\nNext\n\n## Gravitational Field: Definition & Formula", null, "We place the zero point of gravitational potential energy at a distance r r r of infinity. The gravitational field vector, gg, at any location in space is found by placing a test mass mtmt at that point. Assume the system has an overall energy efficiency of 80%. Diagram of torsion balance set up When we talk about the gravitational force acting on an object, we can come up with an equation in a couple of different ways using the universal law of gravitation and Newton's Second Law of Motion. Gravitational field strength is a vector quantity: its direction is towards the object that causes the field. If the body has a mass of 1 unit, then the potential energy to be assigned to that body is equal to the gravitational potential.\n\nNext\n\n## gravitational field definition", null, "A point mass is one that has a radial field, like that of the Earth. It says that in general, gravitational force magnitude depends on the source mass, test mass and their separation. Our article on includes some example problems that are solved through an understanding of how gravitational potential energy is converted to other forms. Sketch them in the neighborhood of the origin. As a consequence, the gravitational potential satisfies.\n\nNext\n\n## newtonian gravity", null, "Geologists and prospectors of oil and minerals make precise measurements of the earth's gravitational field to predict what may be beneath the surface. At this point, we could simply fall into Pluto, but then all our gravitational potential energy would be transferred to , and we would crash into Pluto's surface. The radius of the Earth is about 6. The most subtle point about all this is that the gravitational field tells us about what forces would be exerted on a test mass by Earth, Sun, Moon, and the rest of the universe, if we inserted a test mass at the point in question. The potential is half the square of the. In a second scheme an is set up with freely suspended reflectors at the ends of long paths that are at right angles to each other. Classical Dynamics of particles and systems 4th ed.\n\nNext" ]
[ null, "https://engineersfield.com/wp-content/uploads/2017/10/4.8.jpg", null, "https://upload.wikimedia.org/wikipedia/commons/thumb/2/2a/Electric_Field_Lines.svg/1200px-Electric_Field_Lines.svg.png", null, "https://imgv2-1-f.scribdassets.com/img/document/161073850/original/5c4fc82d32/1532231915", null, "https://image.slidesharecdn.com/a2gravitationalfield-101010222922-phpapp02/95/a2-gravitational-field-kyuem-physics-16-728.jpg", null, "https://study.com/cimages/multimages/16/arrows_lines.png", null, "https://s3mn.mnimgs.com/img/shared/content_ck_images/ana_qa_image_57e9f09607e73.jpeg", null, "https://www.researchgate.net/profile/Carl_Bender/publication/2093636/figure/fig1/AS:394689871138831@1471112806971/Configuration-of-a-simple-pendulum-of-mass-m-in-a-uniform-gravitational-field-of-strength.png", null, "http://www.milutinmarjanov.com/wp-content/uploads/2014/10/fig-01.png", null ]
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https://arxiv.org/abs/1601.01493
[ "Full-text links:\n\nmath.CO\n\n# Title:A note on induced Ramsey numbers\n\nAbstract: The induced Ramsey number $r_{\\mathrm{ind}}(F)$ of a $k$-uniform hypergraph $F$ is the smallest natural number $n$ for which there exists a $k$-uniform hypergraph $G$ on $n$ vertices such that every two-coloring of the edges of $G$ contains an induced monochromatic copy of $F$. We study this function, showing that $r_{\\mathrm{ind}}(F)$ is bounded above by a reasonable power of $r(F)$. In particular, our result implies that $r_{\\mathrm{ind}}(F) \\leq 2^{2^{ct}}$ for any $3$-uniform hypergraph $F$ with $t$ vertices, mirroring the best known bound for the usual Ramsey number. The proof relies on an application of the hypergraph container method.\n Comments: Dedicated to the memory of Jirka Matoušek, 10 pages, second version addresses changes arising from the referee reports Subjects: Combinatorics (math.CO) Journal reference: A Journey Through Discrete Mathematics (A Tribute to Ji\\v{r}\\'i Matou\\v{s}ek), Springer, 2017, 357-366 DOI: 10.1007/978-3-319-44479-6_13 Cite as: arXiv:1601.01493 [math.CO] (or arXiv:1601.01493v2 [math.CO] for this version)\n\n## Submission history\n\nFrom: Mathias Schacht [view email]\n[v1] Thu, 7 Jan 2016 11:37:21 UTC (10 KB)\n[v2] Fri, 24 Jun 2016 09:48:07 UTC (11 KB)" ]
[ null ]
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https://www.arxiv-vanity.com/papers/1210.3637/
[ "# A Gottesman-Knill theorem for all finite Abelian groups\n\nJuan Bermejo-Vega1, Maarten Van den Nest2\n\nMax-Planck-Institut für Quantenoptik,\nHans-Kopfermann-Straße 1, 85748 Garching, Germany.\n11\n22\n###### Abstract\n\nQuantum normalizer circuits were recently introduced as generalizations of Clifford circuits. A normalizer circuit over a finite Abelian group is composed of the quantum Fourier transform (QFT) over together with gates which compute quadratic functions and automorphisms. In [VDNest_12_QFTs] it was shown that every normalizer circuit can be simulated efficiently classically. This result provides a nontrivial example of a family of quantum circuits that cannot yield exponential speed-ups in spite of usage of the QFT—the latter being a central quantum algorithmic primitive. Here we extend the aforementioned result in several ways. Most importantly, we show that normalizer circuits supplemented with intermediate measurements can also be simulated efficiently classically, even when the computation proceeds adaptively. This yields a generalization of the Gottesman-Knill theorem, valid for qubits, to arbitrary . Moreover our simulations are twofold i.e.  we show how to classically efficiently sample the measurement probability distributions as well as compute the amplitudes of the state vector. Finally we develop a generalization of the stabilizer formalism relative to arbitrary finite Abelian groups. For example we characterize how to update stabilizers under generalized Pauli measurements and provide a normal form of he amplitudes of generalized stabilizer states using quadratic functions and subgroup cosets.\n\n## 1 Introduction\n\nInvestigating the power of restricted families of quantum circuits is one of the most fruitful approaches to understand the fundamental question How does the power of quantum computers compare to classical ones? A celebrated result in this respect is the Gottesman-Knill theorem, which states that any quantum circuit built out of Clifford gates (Hadamards, CNOTs, -phase gates) and Pauli measurements can be efficiently simulated on a classical computer [Gottesman_PhD_Thesis, gottesman_knill, nielsen_chuang]. Thus, a quantum computer that works exclusively with these operations cannot achieve exponential quantum speed-ups.\n\nThe Gottesman-Knill theorem illustrates how subtle the frontier between classical and quantum computational power can be. For example, even though Clifford circuits can be simulated efficiently classically, replacing the -phase gates by a -phase gate immediately yields a quantum universal gate set [Boykin_etal_99_Clifford_pifourth_is_universal_OPEN, Boykin_etal_00_Clifford_pifourth_is_universal_ELSEVIER]. Another interesting feature is that, even though the computing power of Clifford circuits is not stronger than classical computation, their behavior is genuinely quantum. They can be used, for instance, to prepare highly entangled states (such as cluster states [raussen_briegel_01_Cluster_State, nest06Entanglement_in_Graph_States, raussen_briegel_onewayQC]), or to perform quantum teleportation [gottesman_knill]. Yet in spite of the high degrees on entanglement which may be involved, the evolution of Clifford circuits can be tracked efficiently using a Heisenberg picture: i.e. by employing the stabilizer formalism used in quantum error correction.\n\nIn this work we study the computational power of normalizer circuits. These circuits were introduced in [VDNest_12_QFTs] by one of us, as a class of quantum computations generalizing the Clifford circuits, as well as standard extensions of the latter to qudits [Gottesman98Fault_Tolerant_QC_HigherDimensions, dehaene_demoor_hostens]. A normalizer circuit over a finite Abelian group is a quantum circuit comprising unitary gates that implement the quantum Fourier transform (QFT) over , quadratic functions of the group and automorphisms. If is chosen to be (i.e. the group of -bit strings with addition modulo 2), normalizer circuits precisely coincide with the unitary Clifford circuits (i.e. those composed of CNOT, and phase gates). In [VDNest_12_QFTs] it was shown that arbitrary normalizer circuits (acting on computational basis states and followed by computational basis measurements) can be simulated classically efficiently.\n\nAn interesting feature of the normalizer circuit formalism is the presence of QFTs over any finite Abelian group. Of particular interest is the group , since its corresponding QFT is the “standard” quantum Fourier transform [nielsen_chuang]; the latter lies at the core of several famous quantum algorithms, such as factoring and computing discrete logarithms [Shor]. More generally, QFTs over Abelian groups are central ingredients of quantum algorithms to find hidden subgroups of Abelian groups [lomont_HSP_review, childs_lecture_8, childs_vandam_10_qu_algorithms_algebraic_problems]. In contrast with the role of QFTs in quantum speed-ups, the normalizer circuit formalism provides a nontrivial example of a family of quantum computations that cannot yield exponential speed-ups, in spite of usage of the QFT.\n\nHere we further extend the classical simulation results of [VDNest_12_QFTs]. We do so by considering normalizer circuits where intermediate measurements are allowed at arbitrary times in the computation—whereas in [VDNest_12_QFTs] only terminal measurements were considered. More precisely we define adaptive normalizer circuits over to comprise the following three fundamental ingredients:\n\n• Normalizer gates over , i.e. QFTs, automorphism gates, quadratic phase gates.\n\n• Measurements of generalized Pauli operators over at arbitrary times in the computation.\n\n• Adaptiveness: the choice of normalizer gate at any time may depend (in a polynomial-time computable way) on the outcomes obtained in all previous measurement rounds.\n\nIf is chosen to be , the corresponding class of adaptive normalizer circuits precisely corresponds to the class of adaptive Clifford circuits allowed in the original Gottesman-Knill theorem.\n\nThis paper contains several results, summarized as follows:\n\n• A Gottesman-Knill theorem for all finite Abelian groups (Theorem LABEL:thm_main). Given any Abelian group , every poly-size adaptive normalizer circuit over , acting on any standard basis input, can be efficiently simulated by a classical computer. That is, we show that the conditional probability distribution arising at each measurement (given the outcomes of previous measurements) can be sampled efficiently classically.\n\n• A stabilizer formalism for finite Abelian groups. Generalizing the well-known stabilizer formalism for qubits, we develop a stabilizer formalism for arbitrary Abelian groups. This framework is a key ingredient to efficiently track the evolution of quantum states under normalizer circuits. In particular, our results are:\n\n• We provide an analytic formula, as well as an efficient algorithm, to compute the dimension of any stabilizer code over an Abelian group (Theorem 3).\n\n• We provide an analytic formula, as well as an efficient algorithm, to compute the update of any stabilizer group under Pauli measurements over arbitrary Abelian groups (Theorem LABEL:thm_Measurement_Update_rules).\n\n• A normal form for stabilizer states (Theorem LABEL:thm_Normal_form_of_an_stabilizer_state). We give an analytical formula to characterize the amplitudes of stabilizer states over Abelian groups and show how to compute these amplitudes efficiently. It follows that all stabilizer states over Abelian groups belong to the class of Computationally Tractable (CT) states, introduced in [nest_weak_simulations]. The interest of this property is that all CT states can be simulated classically in various contexts well beyond the setting of the present work—cf. [nest_weak_simulations] for a discussion.\n\nAn important technical difference (and difficulty) compared to the original Gottesman-Knill theorem is that in the context of arbitrary finite Abelian groups (such as ) arithmetic is generally over large integers. This is in contrast to where arithmetic is simply over i.e. modulo 2. The difference is in fact twofold:\n\nFirst, is a field. As a result, it is possible to describe the “standard” stabilizer formalism for qubits with vector space techniques over . In this context methods like Gaussian elimination have straightforward analogues, which can be exploited in the design of classical algorithms. The field property is, however, no longer present in the context of general Abelian groups. This complicates both the analytical theory and algorithmic aspects due to, e.g., the presence of zero divisors.\n\nSecond, in arithmetic is with small numbers (namely 0s and 1s), whereas in general finite Abelian groups arithmetic is with large integers. This is e.g. the case with . Of course, one must beware that a variety of computational problems over the integers may in the worst case be intractable: consider e.g.  the factoring problem or computing discrete logarithms. One of the main challenges in our scenario is to show that the “integer arithmetic” used in our classical simulation algorithms can be carried out efficiently. For this purpose, a significant technical portion of our work is dedicated to solving systems of linear equations modulo a finite Abelian group, defined as follows: given a pair of finite Abelian groups and (both of which are given as a direct product of cyclic groups), and a homomorphism between them, we look at systems of the form where and . We present polynomial-time deterministic classical algorithms for counting and finding solutions of these systems. These efficient algorithms lie at the core of our classical simulations of normalizer circuits.\n\nFinally we remark that, beyond the scope of classical simulations, the mathematical tools we develop could also have unexplored applications for quantum error correction viz. quantum stabilizer codes over general finite Abelian groups.\n\n#### Relation to previous work.\n\nIn [VDNest_12_QFTs] it was proven that one can sample classically efficiently the output distribution of any non-adaptive normalizer circuit followed by a terminal measurement in the standard basis. Our work extends on this result in various ways, as outlined above in I-II-III. Main differences are the fact that here we consider adaptive normalizer circuits, and two different types of simulations: sampling output distributions and computation of amplitudes.\n\nRestricting to groups of the form where is constant, our work recovers previous results regarding classical simulations of Clifford circuits for qudits. Previous to our work, the case when is a prime number was well-understood: if , the ability to sample classically efficiently follows from the Gottesman-Knill theorem [Gottesman_PhD_Thesis, gottesman_knill], whereas the computation of amplitudes from [dehaene_demoor_coefficients]; for prime values of larger than 2, techniques given in [Gottesman98Fault_Tolerant_QC_HigherDimensions] yield efficient sampling simulations of adaptive Clifford circuits. Last, in the case when is an arbitrary constant, some investigations have led to efficient simulations of Clifford circuits: first, techniques given in [dehaene_demoor_hostens] can be used to simulate non-adaptive Clifford circuits followed by a terminal standard basis measurement (sampling or computation of relative phases); second, ref. [deBeaudrap12_linearised_stabiliser_formalism] extended the last sampling result to simulate terminal Pauli measurements and, in addition, a subclass of adaptive Clifford operations: the latter can be transformed into equivalent non-adaptive circuits using the principle of deferred measurement. To the best of our knowledge, however, our study seems to be the first that has provided efficient classical simulations of arbitrary adaptive Clifford circuits for qudits when is an arbitrary constant integer.\n\nFinally, our work also connects to previous studies on the simulatibility of Abelian quantum Fourier transforms (QFTs), such as [aharonov_AQFT, yoran_short_QFT, brown_QFT]. The relation between normalizer circuits to those works was discussed in [VDNest_12_QFTs] and we refer to this source.\n\n## 2 Preliminaries on finite Abelian groups\n\nThis section is reserved for the main group-theoretical notions used in this work. Let be the additive group of integers modulo . Then\n\n G=Zd1×⋯×Zdm (1)\n\ndenotes a finite Abelian group, the elements of which are -tuples of the form with . Addition of two group elements is component-wise modulo . Every finite Abelian group can be expressed as a product of the type (1) via isomorphism, yet computing this decomposition is regarded as a hard computational problem [cheung_mosca_01_decomp_abelian_groups]; throughout this paper, a product decomposition (1) of is always explicitly given. The cardinality of is denoted by , and fulfills . For every ranging from 1 to , denotes the group element which has in its -th component and zeroes elsewhere, where in slot represents the neutral element in . Remark that for every so that the elements generate .\n\n### 2.1 Characters\n\nLet . The function is defined as\n\n χg(h)=exp(2{p}{i}m∑i=1g(i)h(i)di) (2)\n\nfor all , . Every function is a character of , i.e., it is a homomorphism from into the nonzero complex numbers . This means that\n\n χg(h+h′)=χg(h)χ(h′) (3)\n\nfor every . Note further that and . The set of all characters of forms a group, called the character group or dual group, which is isomorphic to .\n\n### 2.2 Orthogonal subgroups\n\nThe character functions give rise to a set-theoretical duality, sometimes called orthogonality of Abelian subgroups (although it differs from the usual orthogonality of vector spaces). Given a subgroup of the finite Abelian group , its orthogonal subgroup is defined as\n\n H⊥={g∈G:χh(g)=1,for allh∈H}. (4)\n\nNote that is indeed a subgroup of . The main properties of are summarized below.\n\n###### Lemma 1 (Orthogonal subgroup).\n\nConsider a finite Abelian group as in (1) and let and be two arbitrary subgroups. Then the following statements hold:\n\n• if and only if\n\n• (a-b) are well-known, see e.g. [lomont_HSP_review]. (c) is proved straightforwardly by applying definitions. We prove (d). Since is contained in both and if follows that and . Therefore . We show , which implies the reversed inclusion. This comes from the fact that implies and , or equivalently, and .\n\n### 2.3 Quadratic functions\n\nA function from to the nonzero complex numbers is called bilinear if for every one has\n\n B(g+h,x)=B(g,x)B(h,x)andB(x,g+h)=B(x,g)B(x,h). (5)\n\nA function from to the nonzero complex numbers is called quadratic if there exists a bilinear function such that for every one has\n\n ξ(g+h)=ξ(g)ξ(h)B(g,h). (6)\n\nAlthough it is not obvious from the above definition, the number turns out to be a complex phase for every and for every quadratic function. More precisely, one has for every [VDNest_12_QFTs]. It is easily verified that the product of two quadratic functions is again a quadratic function. Also the complex conjugate of any quadratic function is again quadratic. Using these properties one can show that the set of quadratic functions forms an (Abelian) group.\n\nWe give some examples of quadratic functions (see also ref.[VDNest_12_QFTs]). First we consider . Letting be an matrix with entries in and letting , the following functions are quadratic:\n\n ξA:x→(−1)xTAx and ξa:x→{i}aTx (7)\n\nwhere and is computed over (i.e. modulo 2). Note that the exponent in is polynomial of degree 2 in , whereas the exponent in has degree 1. To prove that these functions are quadratic in the sense used in this work, we note that\n\n ξA(x+y) = ξA(x)ξA(y)(−1)xT(A+AT)y (8) ξa(x+y) = ξ(x)ξ(y)(−1)q(x,y) with q(x,y)=(aTx)(aTy) (9)\n\nIdentity (9) can be proved by distinguishing between the 4 cases . The above identities can be used to show that and are quadratic.\n\nConsidering a single cyclic group , examples of quadratic functions are\n\n z→ωbz2+cz and z→γbz(z+d);ω:={e}2{p}{i}/d, γ:=ω1/2. (10)\n\nWe refer to [VDNest_12_QFTs] for a proof.\n\n### 2.4 Homo-, iso- and auto- morphisms\n\nGiven two finite Abelian groups and , a group homomorphism from to is a map that fulfills for every (in other words, is linear). An isomorphism from to is an invertible group homomorphism. An automorphism of is an isomorphism of the form , i.e. from the group onto itself. The set of all automorphisms of forms a group, called the automorphism group.\n\nThroughout this work, group homomorphisms between Abelian groups are to be described in terms of matrix representations, which are defined as follows:\n\n###### Definition 1 (Matrix representation).\n\nGiven a homomorphism between groups and , an integer matrix is said to be a matrix representation of if its columns are elements of , and if it holds that\n\n A(h)=Ah(modG),for every h∈H. (11)\n\nConversely, an integer matrix is said to define a group homomorphism from to if its columns are elements of and the operation is a group homomorphism.\n\nIn (11) we introduced some conventions, that we will use: first, the element is seen as a column of integer numbers onto which acts via matrix multiplication; second, (mod ) indicates that multiplications and sums involved in the calculation of are performed within , taking remainders when necessary.\n\nThe main properties of matrix representations are now summarized.\n\n###### Proposition 1.\n\nEvery group homomorphism has a matrix representation. Moreover, an matrix with columns defines a homomorphism iff its columns fulfill the equations\n\n ciai=0(modG),for every i. (12)\n• First, given , we show that , where , is a matrix representation. Since every decomposes as , it follows, using linearity of , that\n\n A(h)=A(∑h(i)ei)=∑h(i)A(ei)=Ah(modG). (13)\n\nThe right implication of the iff comes readily from\n\n ciai=ciAei=A(ciei)=A(0)=0(modG) (14)\n\nFor the converse, we let act on , and without taking remainders; we obtain\n\n Ag+Ah=∑[g(i)+h(i)]ai,A(g+h)=∑(g+h)(i)ai. (15)\n\nRecalling associativity of and , the latter expression shows that defines a function from to , and, thus, is a function from to . Last, it holds for every that for some integers , since (by definition of the group ) is the remainder obtained when is divided by ( would be the quotient). It follows, subtracting modularly, that for every , ; and, due to, defines a linear map.\n\n### 2.5 Computational group theory\n\nComputational aspects of finite Abelian groups are now discussed; our discourse focuses on a selected catalog of computational problems relevant to this work and efficient classical algorithms to solve them. Since this section concerns only classical computational complexity, we will tend to omit the epithet classical all the way throughout it.\n\nTo start with, we introduce some basic notions of computer arithmetic. From now on, the size of an integer is the number of bits in its binary expansion. Remark that every group (1) satisfies the inequalities and , for every . It follows thence readily that is , and that one needs at most a polylog amount of memory to store an element (in terms of bits).\n\nWe discuss now how to perform some basic operations efficiently within any finite Abelian group (1). First, given two integers and of size at most , common arithmetic operations can be computed in poly time with elementary algorithms: such as their sum, product, the quotient of divided by , and the remainder [brent_zimmerman10CompArithmetic]. Therefore, given , the sum can be obtained in polylog time by computing the remainders . Similarly, given an integer , the element can be obtained in polylog time by computing the remainders mod .\n\nIn connection with the previous section, it follows from the former properties that matrix representations can be stored using only a polynomial amount of memory, and, moreover, that given the matrix representation we can efficiently compute . Specifically, given a matrix representation of the homomorphism , we need polylog space to store its columns as tuples of integers, and polylog time to compute the function .\n\nPeriodically, and at crucial stages of this investigation, some more advanced algebraic computational problems are bound to arise. The following lemma compiles a list of group theoretical problems that will be relevant to us and can be solved efficiently by classical computers.\n\n###### Lemma 2 (Algorithms for finite Abelian groups).\n\nGiven , , two subgroups of , and , , polynomial-size generating-sets of them, there exist efficient classical algorithms to solve the following problems deterministically.\n\n• Decide whether belongs to ; if so, find integers such that .\n\n• Count the number of elements of .\n\n• Find a generating-set of the intersection .\n\n• Find a generating-set of .\n\n• Given the equations , find elements such that all solutions of the system can be written as linear combinations of the form .\n\nThe proof of the lemma is divided in two parts which are fully detailed i.o. in appendices LABEL:Appendix:_Algorithms_for_Finite_Abelian_Groups and LABEL:Appendix:_System_of_linear_congruences_modulo_Abelian_group. The rest of this section describes the high-level structure of the proof.\n\nIn short, appendix LABEL:Appendix:_Algorithms_for_Finite_Abelian_Groups contains a proof of the following statement.\n\n###### Lemma 3.\n\nProblems (a-e) in lemma 2 are polynomial-time reducible to either counting or finding solutions of systems of equations of the form ; where is a group homomorphism between two (canonically-decomposed) finite Abelian groups, and , to which , respectively belong; given that a matrix representation of is provided.\n\nMind that, since is a linear map, the set of solutions of such a system is either empty or a coset with the structure\n\n Xsol=x0+kerA. (16)\n\nBy finding solutions we refer to the action of finding one particular solution of the system and a (polynomially-large) generating set of .\n\nTo provide an example, we prove now lemma 3 for the problems (d-e) in lemma 2; for the rest of cases, we refer to the appendices.\n\n#### Example: (d,e)th cases of lemma 2.\n\nFirst of all, remark that problem (d) reduces to (e) by setting all to be —this yields the system (4), whose solutions are the elements of the orthogonal subgroup. Thence, it will be enough to prove the (e)th case. Moreover, since the equations can be fulfilled for some only if all are th-roots of the unit, this systems can only have solutions if all are even numbers. As we can determine it efficiently whether these numbers are even, we assume from now on that it is the case.\n\nNow define a tuple of integers coefficient-wise as ; use the later to re-write . Also, denote by the group generated by the elements . By letting multiply numerators and denominators of all fractions in (2), the system of complex exponentials can be turned into an equivalent system of congruences . Finally, by defining a matrix with coefficients the system can be written as\n\n Ωg=b(modG), (17)\n\nwhere belongs to , being the number of generators , and we look for solutions inside . Moreover, the coefficients of fulfill ; hence, condition (12) is met and defines a homomorphism.\n\nFinally, can be computed in using aforementioned algorithms to multiply and divide integers. It is now routine to check, using the concepts developed thus far, that both and are ; as a result, the inputsize of the new problem, as well as the memory needed to store and , are all .\n\nA remark Note that the group homomorphism fulfills . Therefore, if we substitute with in the procedure above, given generators of , we would obtain an integer matrix that defines a second group homomorphism such that\n\n ϖ(g)=Ω′(g)(modG)andkerϖ=H′⊥=H. (18)\n\nAs a result, our algorithm to compute generators of leads also to an efficient method to construct, given any subgroup of , a homomorphism whose kernel is . This fact will be used later in the text, in section LABEL:section_Gottesman-Knill_theorem.\n\nThe final ingredient to complete the proof of lemma 2 is the following theorem.\n\n###### Theorem 1 (Systems of linear equations over finite Abelian groups).\n\nGiven any element of the group and any matrix which defines a group homomorphism from to , consider the system of equations . Then, there exist classical algorithms to solve the following list of problems in polylog time.\n\n1. Decide whether the system admits a solution.\n\n2. Count the number of different solutions of the system.\n\n3. Find such that all solutions of the system are linear combinations of the form .\n\nThe main ideas underlying the proof of theorem 1 are as follows: first, we show how to reduce in polynomial-time to an equivalent system of linear congruences modulo , where is suitably upper-bounded; second, we apply fast algorithms to compute Smith normal forms to tackle the latter problem [storjohann10_phd_thesis]. For details, we refer to appendix LABEL:Appendix:_System_of_linear_congruences_modulo_Abelian_group.\n\n## 3 Pauli and Clifford operators and normalizer circuits over Abelian groups\n\nHere we extend the notion of Pauli and Clifford operators for qubits [nielsen_chuang] and qudits [Knill96non-binaryunitary, Gottesman98Fault_Tolerant_QC_HigherDimensions] to all finite Abelian groups. Throughout this section we follow the definitions of [VDNest_12_QFTs]. Let be a finite Abelian group as in (1). Then is naturally associated to a -dimensional Hilbert space with a basis labeled by group elements\n\n (19)\n\nThis basis is henceforth called the standard basis. Introduce the following operators acting on :\n\n X(g):=∑h∈G|h+g⟩⟨h|,Z(g):=∑h∈Gχg(h)|h⟩⟨h|, (20)\n\nwhere . Since acts as a permutation on the standard basis, this operator is unitary. Since , the operator is unitary as well. With these definitions, a Pauli operator over (hereafter often simply denoted Pauli operator) is a unitary operator of the form\n\n σ(a,g,h):=γaZ(g)X(h), (21)\n\nwhere is a primitive root of unity and . The triple describing the Pauli operator is called the label of . It is important to remark that the label constitutes an efficient description of comprising bits. Owing to, from now on we will specify Pauli operators in terms of their labels; the latter will represent the standard encoding of these operators.\n\nNote that every Pauli operator factorizes as a tensor product relative to the tensor decomposition of i.e.  can be written as where acts on . This property can be verified straightforwardly by applying (20) and the definition (2) of the characters of .\n\nBasic manipulations of Pauli operators can be carried out transparently by translating them into transformations of their labels: we review now some of these rules. First, the Pauli matrices (20) obey the following commutation rules:\n\n X(g)X(h) =X(g+h)=X(h)X(g) Z(g)Z(h) =Z(g+h)=Z(h)Z(g) (22) Z(g)X(h) =χg(h)X(h)Z(g).\n\nCombinations of these rules straightforwardly lead to the next two propositions.\n\n###### Proposition 2 (Products and powers of Pauli operators [VDNest_12_QFTs]).\n\nConsider Pauli operators and and a positive integer . Then , and are also Pauli operators, the labels of which can be computed in polylog time on input of and the labels of and . Moreover, .\n\n###### Proposition 3 (Commutativity).\n\nConsider two Pauli operators and . Then the following statements are equivalent:\n\n• and commute;\n\n• ;\n\n• and are orthogonal elements of i.e. .\n\nProposition 2 implies that the set of all Pauli operators over forms a (finite) group, called the Pauli group (over ).\n\nA unitary operator on is called a Clifford operator (over ) if maps the Pauli group onto itself under the conjugation map . It is easy to see that the set of all Clifford operators forms a group, called the Clifford group . Formally speaking, the Clifford group is the normalizer of the Pauli group in the full unitary group acting on . Next we define three basic classes of unitary operators on which are known to belong to the Clifford group [VDNest_12_QFTs].\n\n• Quantum Fourier transforms. The quantum Fourier transform (QFT) over is the following unitary operator on :\n\n Fi=1√di∑ωxyi|x⟩⟨y|;ωi=exp(2{p}{i}/di) (23)\n\nwhere the sum is over all . The QFT over the entire group is given by , which acts on the entire space . Any operator obtained by replacing a subset of operators in this tensor product by identity operators is called a partial QFT.\n\n• Automorphism gates. Given an automorphism of , the associated automorphism gate maps .\n\n• Quadratic phase gates. Given a quadratic function on , the quadratic phase gate is the diagonal operator mapping . Since is a complex phase for every , every quadratic phase gate is a (diagonal) unitary operator.\n\nA unitary operator which is either a (partial) quantum Fourier transform or its inverse, an automorphism gate or a quadratic phase gate is generally referred to as a normalizer gate. A quantum circuit composed entirely of normalizer gates is called a normalizer circuit over . The size of a normalizer circuit is the number of normalizer gates of which it consists.\n\nIn this paper we will be interested in classical simulations of normalizer circuits. To make meaningful statements about classical simulations one has to specify which classical descriptions of normalizer circuits are considered to be available. This is discussed next. First, a partial quantum Fourier transform is described by the set of systems on which it acts nontrivially. Second, an automorphism gate is described by the matrix representation of the associated automorphism. Third, let be an arbitrary quadratic function. Since , there exists such that for every . It was shown in [VDNest_12_QFTs] that the integers and comprise an efficient description of and, thus, of the associated quadratic phase gate. Henceforth we will assume that all normalizer gates are specified in terms of the descriptions given above, which will be called their standard encodings. Each standard encoding comprises polylog bits.\n\nIt is known that every normalizer gate belongs to the Clifford group:\n\n###### Theorem 2 (Normalizer gates are Clifford [VDNest_12_QFTs]).\n\nEvery normalizer gate is a Clifford operator. Furthermore let be a normalizer gate specified in terms of its standard classical encoding as above, and let be a Pauli operator specified in terms of its label. Then the label of can be computed in polylog time.\n\nFor any group of the form , the entire Clifford group is known to be generated (up to global phase factors) by normalizer gates [dehaene_demoor_hostens] (see also section 4). More strongly, every Clifford group element can be written as a product of at most polylog such operators. We conjecture that this feature holds true for Clifford operators over arbitrary :\n\n###### Conjecture 1.\n\nLet be an arbitrary finite Abelian group. Up to a global phase, every Clifford operator over can be written as a product of polylog normalizer gates.\n\nFinally we point out that both automorphism gates and quadratic phase gates have a distinguished role within the Clifford group, characterized as follows:\n\n###### Proposition 4.\n\nUp to a global phase, every Clifford operator which acts on the standard basis as a permutation has the form for some and some automorphism gate . Every diagonal Clifford operator is, up to a global phase, a quadratic phase gate.\n\n• The first statement was proved in [VDNest_12_QFTs]. We prove the second statement. Let be a diagonal unitary operator (so that for all ) in the Clifford group. Without loss of generality we may set which can always be ensured by choosing a suitable (irrelevant) overall phase. Then for every , sends to a Pauli operator under conjugation. This implies that there exists a complex phase and group elements such that\n\n DX(h)D†=γ(h)X(f1(h))Z(f2(h)). (24)\n\nSince is diagonal, it is easy to verify that we must have for every . Now consider an arbitrary . Then\n\n DX(h)D†|g⟩ = ¯¯¯ξ(g)ξ(g+h)|g+h⟩; (25) γ(h)X(h)Z(f2(h))|g⟩ = γ(h)χg(f2(h))|g+h⟩. (26)\n\nCondition (24) implies that (25) is identical to (26) for every . Choosing and using that and for every it follows that . We thus find that\n\n ξ(g+h)=ξ(g)ξ(h)χg(f2(h)). (27)\n\nThe function is manifestly linear in , since . Furthermore by definition is symmetric in and . Thus is also linear in .\n\n## 4 Examples\n\nHere we give examples of Pauli and Clifford operations. We illustrate in particular how the definitions of the preceding section generalize existing notions of Pauli and Clifford operators for qubits and qudits.\n\n### 4.1 G=Zm2\n\nIn this case the standard definition of -qubit Pauli operators is recovered. To see this, first note that we have i.e. the Hilbert space is a system of qubits. Let and denote the standard Pauli matrices and let be an -bit string. Then, applying definition (20) one finds that\n\n X(g)=σg(1)x⊗⋯⊗σg(m)x,Z(g)=σg(1)z⊗⋯⊗σg(m)z. (28)\n\nHere is an -bit string: i.e.  . In short, is a tensor product of -matrices and identities, and is a tensor product of -matrices and identities. Therefore, every Pauli operator (20) has the form where each is a single-qubit operator of the form for some . This recovers the usual notion of a Pauli operator on qubits [nielsen_chuang].\n\nIt was shown in [VDNest_12_QFTs] that the CNOT gate and the CZ gate (acting between any two qubits), the Hadamard gate and the Phase gate diag (acting on any single qubit) are examples of normalizer gates for . Note that these gates are the standard building blocks of the well known class of Clifford circuits for qubits. In fact, the entire Clifford group for qubits is generated by this elementary gate set [nielsen_chuang].\n\n### 4.2 G=Zmd\n\nIn this case the Hilbert space is a system of -level systems (qudits) and Pauli operators have the form , where each is a single-qudit operator of the form for some . Here and are the usual generalizations of and , respectively [Knill96non-binaryunitary, Gottesman98Fault_Tolerant_QC_HigherDimensions]. These operators act on a single -level system as follows:\n\n Xd=∑|x+1⟩⟨x| and Zd=∑{e}2{p}{i}x/d|x⟩⟨x| (29)\n\nwhere the sums run over all . Examples of normalizer gates over are generalizations of the CNOT, CZ, Hadamard and Phase gates to qudits, as follows:\n\n SUMd = ∑|x,x+y⟩⟨x,y|; (30) CZd = ∑ωxyd|x,y⟩⟨x,y|;ωd:=%e2{p}{i}/d (31) Fd = 1√d∑{e}2{p}{i}xy/d|x⟩⟨y|; (32) Sd = ∑ξx(x+d)d|x⟩⟨x|;ξd:={e}{p}{i}/d. (33)\n\nHere and rum over all elements in . To show that SUM is a normalizer gate, note that is indeed an automorphism of . The gates and are quadratic phase gates; see section 11 of ref.[VDNest_12_QFTs]. In addition, the “multiplication gate” is also a normalizer gate, for every which is coprime to . Indeed, for such the map is known to be an automorphism of . It is known that the entire Clifford group for qubits (for arbitrary ) is generated by the gates SUM, , and [dehaene_demoor_hostens].\n\n### 4.3 G=Z2m\n\nOne can also consider to be a single cyclic group, such as . In this case, is a -dimensional Hilbert space with standard basis . Comparing with the previous examples, the important difference with e.g.  is that the structure of does not naturally induce a factorization of the Hilbert space into single-qubit systems. As a consequence, normalizer gates over act globally on , in contrast with the previous examples.\n\nExamples of normalizer gates are now given by , and , following the definitions of the previous example with . Note that is the “standard” QFT used in e.g Shor’s algorithm and the phase estimation quantum algorithm.\n\n## 5 Abelian Stabilizer Formalism\n\nIn this section we develop further the stabilizer formalism for finite Abelian groups as started in [VDNest_12_QFTs]. We provide new analytic and algorithmic tools to describe them and analyze their properties. Throughout this section we consider an arbitrary Abelian group of the form .\n\n### 5.1 Stabilizer states and codes\n\nLet be a subgroup of the Pauli group . Then is said to be a stabilizer group (over ) if there exists a non-zero vector which is invariant under all elements in i.e.  for every . The linear subspace is called the stabilizer code associated with . If is one-dimensional, its unique element (up to a multiplicative constant) is called the stabilizer state associated with . In this work we will mainly be interested in stabilizer states. Occasionally, however, it will be useful to consider the general setting of stabilizer codes (cf. e.g.  theorem 3).\n\nNote that every stabilizer group is Abelian. To see this, consider a state which is invariant under the action of all elements in and consider two arbitrary . Then (22) implies that there exists a complex phase such that . It follows that , where we have used that . We thus find that so that (i.e.  and commute).\n\nConversely, not every Abelian subgroup of the Pauli group is a stabilizer group. A simple counterexample is the group where is the identity operator acting on .\n\nThe support of a stabilizer code is the set of all for which has a nonzero overlap with i.e. there exists such that . The support of a stabilizer state is simply the set of all for which .\n\n### 5.2 Label groups\n\nLet be a stabilizer group over . The diagonal subgroup is the subgroup of formed by its diagonal operators i.e. it consists of all operators in of the form . Second, we introduce two subgroups and of called the label groups of :\n\n H ={h∈G:there exists γaZ(g)X(h)∈S}, (34) D ={g∈G:there exists γaZ(g)∈D}, (35)\n\nUsing (22) it is straightforward to verify that is indeed a subgroup of . To prove that is a subgroup as well, one argues as follows. Let be a Pauli operator with label . We call the “-component” and the “-component” of . Denote the -component formally by . Then is the image of under the map . The commutation relations (22) yield\n\n φ(στ)=φ(σ)+φ(τ) for all σ,τ∈S. (36)\n\nThis implies that is a homomorphism from to . It follows that is a subgroup of .\n\n###### Proposition 5 (Label groups).\n\nLet be a stabilizer group and assume that the labels of polylog generators of are given as an input. Then the label groups of have the following properties:\n\n1. ;\n\n2. Generating sets of , can be efficiently computed classically;\n\n3. The labels of a generating set of can be efficiently computed classically.\n\n• Property (i) is a straightforward consequence of the commutation relations given in proposition 3 and the definition of orthogonal subgroup (4). To show property (ii), recall that the map defined above is a homomorphism from to with Im. Suppose that is generated by . Then is generated by : this yields an efficient method to compute generators of . To prove the second statement of (ii) as well as (iii) requires more work. The argument is a direct generalization of the proof of lemma 9 in [VDNest_12_QFTs] and the reader is referred to this work.\n\n### 5.3 Certificates\n\nThe main purpose of this section is to provide a criterion to verify when a stabilizer group gives rise to a one-dimensional stabilizer code i.e. a stabilizer state. This is accomplished in corollary 1. To arrive at this statement we first analyze how the the dimension of a general stabilizer code is related the structure of its stabilizer group.\n\n###### Theorem 3 (Structure Test).\n\nLet be a stabilizer group with stabilizer code . Then there exists such that\n\n (i) supp(V)=g0+D⊥,(ii) dim(V)=|D⊥||H|, (37)\n\nwhere , are the label subgroups of . Furthermore, there exist efficient classical algorithms to compute a representative of the support, a generating set of and the dimension .\n\nBefore proving theorem 3, we note that combining property (ii) together with proposition 5(i) immediately yield:\n\n###### Corollary 1 (Uniqueness Test).\n\nLet be a stabilizer group with stabilizer code . Then is one-dimensional if and only if and are dual orthogonal subgroups: .\n\nTheorem 3(ii) also leads to an alternative formula for the dimension of a stabilizer code:\n\n###### Corollary 2.\n\nThe dimension of equals .\n\n• Consider the map as in section 5.2, which is a group homomorphism with image . Furthermore the kernel of is precisely the diagonal subgroup of . Since it follows that . Finally we claim that and are isomorphic groups so that . To prove this, consider the map that sends to . Using (22) it follows that this map is a homomorphism; furthermore, it is a surjective one by definition of , and thus . The kernel of is the set of all having the form . But the only operator in proportional to the identity is the identity itself, since otherwise cannot have a common eigenstate. This shows that the kernel of is trivial, so that and are isomorphic, as claimed. The resulting identity together with (recall lemma 1) and theorem 3(ii) proves the result.\n\nThe result in corollary 2 is well known for stabilizer codes over qubits [Gottesman_PhD_Thesis, nielsen_chuang] (i.e. where so that ) and qudits (where ) [Gottesman_PhD_Thesis, gheorghiu11Qudit_Stabilisers].\n\nWe now prove theorem 3 using techniques developed in [nest_MMS] where the properties of so-called M-spaces were studied. We briefly recall basic concepts and results.\n\nA unitary operator acting on is said to be monomial if it can be written as a product where is diagonal and is a permutation matrix. A subspace of is called an M-space if there exists a group of monomial unitary operators such that iff for every . The group is called a stabilizer group of . If is one-dimensional, its unique (up to a multiplicative factor) element is called an M-state. The support of is defined analogously to the support of a stabilizer code i.e. it is the set of all such that has a nontrivial overlap with . With this terminology, every stabilizer code is an instance of an M-space and every stabilizer state is an M-state. To see this, note that every Pauli operator is a monomial unitary operator. Indeed, can be written as a product where is diagonal and is a permutation matrix.\n\nWe introduce some further terminology. Let be an arbitrary monomial stabilizer group. For every , let be the subset of consisting of all satisfying i.e.  acts trivially on , up to an overall phase. This subset is easily seen to be a subgroup of . Also, we define the orbit of as:\n\n Og={h:∃U∈G s.t. U|g⟩∝|h⟩} (38)\n\nIn the following result the support of any M-space is characterized in terms of the orbits and the subgroups .\n\n###### Theorem 4 (Support of M-space [nest_MMS]).\n\nConsider an M-space with monomial stabilizer group . Then the following statements hold:\n\n• There exist orbits such that and\n\n (39)\n• Consider and an arbitrary set of generators of . Then supp() if and only if for every .\n\nUsing this result, we can now prove theorem 3.\n\n• [of theorem 3] We apply theorem 4 to the Pauli stabilizer group . In this case, the group and the orbit fulfill\n\n Og=g+H,Sg=D. (40)\n\nTo demonstrate the first identity in (40), we use (20) which implies for every . To show the second identity, first note that for every diagonal operator , showing that . Conversely, if has label then . Since the state is an eigenvector of ; this can only be true if , showing that .\n\nUsing proposition 5, we can efficiently compute the labels of a generating set of , where for some and . Owing to theorem 4(ii), any belongs to the support of if and only if for every . Equivalently, satisfies\n\n γaiχgi(g)=1for alli=1,…,r. (41)\n\nThis system of equations is of the type considered in section 2.5 (cf. lemma 3 and the example after it) and can, therefore, be transformed into an equivalent linear system over groups: . The elements generate the label group , and thus the homomorphism defined by satisfies . Since the system is linear, its solutions form a coset of the form , for some particular solution . This shows statement (i).\n\nFurther, we combine (i) with theorem 4(i) to get a short proof of (ii): the equation\n\n supp(V)=Og1∪⋯∪Ogd=(g1+H)∪⋯∪(gd+H)=g0+D⊥\n\nimplies, computing the cardinalities of the sets involved, that .\n\nFinally, the ability to compute and to find generators of efficiently classically follows applying theorem 1 to a linear system described by a matrix that defines a homomorphism from to , with . Furthermore, we can compute directly using formula (ii) together with proposition 5 and the algorithms of lemma 2.\n\n## 6 Pauli measurements in the stabilizer formalism\n\n### 6.1 Definition\n\nAssociated with every Pauli operator as in (21) we will consider a quantum measurement in the eigenbasis of . Consider the spectral decomposition where are the distinct eigenvalues of and is the projector on the eigenspace associated with eigenvalue . Given a state" ]
[ null ]
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https://search.r-project.org/CRAN/refmans/asbio/html/ci.impt.html
[ "ci.impt {asbio} R Documentation\n\n## Confidence interval for the product of two proportions\n\n### Description\n\nProvides one and two-tailed confidence intervals for the true product of two proportions.\n\n### Usage\n\n\nci.impt(y1, n1, y2 = NULL, n2 = NULL, avail.known = FALSE, pi.2 = NULL,\nconf = .95, x100 = TRUE, alternative = \"two.sided\", bonf = TRUE, wald = FALSE)\n\n\n### Arguments\n\n y1 The number of successes associated with the first proportion. n1 The number of trials associated with the first proportion. y2 The number of successes associated with the second proportion. Not used if avail.known = TRUE. n2 The number of trials associated with the first proportion. Not used if avail.known = TRUE. avail.known Logical. Are the proportions π_{2i} known? If avail.known = TRUE these proportions should specified in the pi.2 argument. pi.2 Proportions for π_{2i}. Required if avail.known = TRUE. conf Confidence level, i.e., 1 - α. x100 Logical. If true, estimate is multiplied by 100. alternative One of \"two.sided\", \"less\", \"greater\". Allows lower, upper, and two-tailed confidence intervals. If alternative = \"two.sided\" (the default), then a conventional two-sided confidence interval is given. The specifications alternative = \"less\" and alternative = \"greater\" provide lower and upper tailed CIs, respectively. bonf Logical. If bonf = TRUE, and the number of requested intervals is greater than one, then Bonferroni-adjusted intervals are returned. wald Logical. If avail.known = TRUE one can apply one of two standard error estimators. The default is a delta-derived estimator. If wald = TRUE is specified a modified Wald standard error estimator is used.\n\n### Details\n\nLet Y_1 and Y_2 be multinomial random variables with parameters n_1, π_{1i} and n_2, π_{2i}, respectively; where i = 1,2,…, r. Under delta derivation, the log of the products of π_{1i} and π_{2i} (or the log of a product of π_{1i} and π_{2i} and a constant) is asymptotically normal with mean log(π_{1i} \\times π_{2i}) and variance (1 - π_{1i})/π_{1i}n_1 + (1 - π_{2i})/ π_{2i}n_2. Thus, an asymptotic (1 - α)100 percent confidence interval for π_{1i} \\times π_{2i} is given by:\n\n\\hat{π}_{1i} \\times \\hat{π}_{2i} \\times \\exp(\\pm z_{1-(α/2)}\\hat{σ}_i)\n\nwhere: \\hat{σ}^2_i = \\frac{(1 - \\hat{π}_{1i})}{\\hat{π}_{1i}n_1} + \\frac{(1 - \\hat{π}_{2i})}{\\hat{π}_{2i}n_2} and z_{1-(α/2)} is the standard normal inverse CDF at probability 1 - α.\n\n### Value\n\nReturns a list of class = \"ci\". Printed results are the parameter estimate and confidence bounds.\n\n### Note\n\nMethod will perform poorly given unbalanced sample sizes.\n\nKen Aho\n\n### References\n\nAho, K., and Bowyer, T. 2015. Confidence intervals for a product of proportions: Implications for importance values. Ecosphere 6(11): 1-7.\n\nci.prat, ci.p\nci.impt(30,40, 25,40)" ]
[ null ]
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https://support.bitmain.com/hc/en-us/community/posts/900001552946-Antminer-s19-pro-power-consumption-for-a-month
[ "1 comment\n\n•", null, "3250W = 3.25kw/h (3.25*0.11€= 0.3575€/h)\n\nso running it 24h/day = 3.25*24= 78kw/h (78*0.11€= 8.58€/day)\n\nso running it a week 7 days a week= 78*7=546kw/h (546*0.11€= 60.06€/week)\n\nso running it a month 30 days a month = 78*30 = 2340kw/h (2340*0.11€= 257.4€/month)\n\nSo running it a year 365 days a year = 78*365= 28470kw/h (28470*0.11€=3131.70€/year)\n\nif your kw/h price is like ours 0.11€ then your" ]
[ null, "https://secure.gravatar.com/avatar/26b355a234117f8defc8e2e01a254700", null ]
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http://ezphysics.org/dokuwiki-2010-11-07a/doku.php?id=network_protocol
[ "# EZPhysics Network Protocol\n\nEZPhysics provides a network interface for sending and receiving data from the ODE simulation. This is done through predefined packets. This page lists what these packages contain and in what order (has determined in EZNetwork class, at EZPhysicsNetwork.cpp).\n\nAn EZPhysics Network Message has the following format:\n\n## Send Packets\n\nSend Packets refers to the packets sent from EZPhysics, which contain the latest simulation's data.\n\n#### EZNETWORK_SEND_COLLISION 01\n\nUsually sent in pairs, refers to the point of contact between two Body's.\n\n1. Float X of the contact point\n2. Float Y of the contact point\n3. Float Z of the contact point\n4. Float depth (penetration of the collision)\n\n#### EZNETWORK_SEND_LIMB 02\n\nA Body's output.\n\n1. Float X of the position vector\n2. Float Y of the position vector\n3. Float Z of the position vector\n4. Float W of the rotation quaternion\n5. Float X of the rotation quaternion\n6. Float Y of the rotation quaternion\n7. Float Z of the rotation quaternion\n8. Float X of the force vector\n9. Float Y of the force vector\n10. Float Z of the force vector\n11. Float X of the torque vector\n12. Float Y of the torque vector\n13. Float Z of the torque vector\n14. Float Unused\n15. Float X of the linear velocity vector\n16. Float Y of the linear velocity vector\n17. Float Z of the linear velocity vector\n18. Float X of the angular velocity vector\n19. Float Y of the angular velocity vector\n20. Float Z of the angular velocity vector\n21. Float Unused\n\n#### EZNETWORK_SEND_JOINTAXIS 03\n\nRefers to each of the possibly several axis of a Joint.\n\n1. Float displacement value of the axis (Relevant for the Slider Joint)\n2. Float angular velocity at the axis\n\n#### EZNETWORK_SEND_JOINTFEEDBACK 04\n\nThe feedback can be used by the user, so that it is known how much force an individual joint exerts.\n\n1. Float X of the force vector of body 1\n2. Float Y of the force vector of body 1\n3. Float Z of the force vector of body 1\n4. Float X of the torque vector of body 1\n5. Float Y of the torque vector of body 1\n6. Float Z of the torque vector of body 1\n7. Float X of the force vector of body 2\n8. Float Y of the force vector of body 2\n9. Float Z of the force vector of body 2\n10. Float X of the torque vector of body 2\n11. Float Y of the torque vector of body 2\n12. Float Z of the torque vector of body 2\n\n#### EZNETWORK_SEND_TIMESTAMP 05\n\nCan be used to determine the current time of simulation.\n\n1. Long ODE step number (since the beginning of the simulation)\n2. Float ODE step size\n\nReceive Packets refers to the packets received in EZPhysics, which contain the control commands to apply in the simulation.\n\n#### EZNETWORK_RECV_RESET 01\n\nCommands the simulation to restart.\n\n#### EZNETWORK_RECV_AXIS_FORCE 02\n\nRepresents a raw torque applied in an axis.\n\n1. Float torque value to apply to the axis 1 of a Joint\n2. Float torque value to apply to the axis 2 of a Joint (May be invalid)\n3. Float torque value to apply to the axis 3 of a Joint (May be invalid)\n\n#### EZNETWORK_RECV_AXIS_MOTOR 03\n\nProvides a common servo motor interface, target velocity and maximum torque.\n\n1. Float target velocity to apply on axis 1 of a Joint\n2. Float maximum torque imposed on axis 1 of a Joint\n3. Float target velocity to apply on axis 2 of a Joint (May be invalid)\n4. Float maximum torque imposed on axis 2 of a Joint (May be invalid)\n5. Float target velocity to apply on axis 3 of a Joint (May be invalid)\n6. Float maximum torque imposed on axis 3 of a Joint (May be invalid)", null, "" ]
[ null, "http://ezphysics.org/dokuwiki-2010-11-07a/lib/exe/indexer.php", null ]
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https://zbmath.org/0763.46028
[ "## A Parseval equation and a generalized finite Hankel transformation.(English)Zbl 0763.46028\n\nFor $$\\mu\\geq-{1\\over 2}$$ a space $$S_ \\mu$$ of certain $$C^ \\infty$$- functions on $$]0,1]$$ and a sequence space $$L_ \\mu$$ are introduced. If $$J_ \\mu$$ denotes the Bessel function of first kind and order $$\\mu$$ and if $$(\\lambda_ n)_{n\\in\\mathbb{N}_ 0}$$ denotes the positive roots of $$J_ \\mu$$ (arranged increasingly) then the finite Hankel transform $(h_ \\mu^* f)(n):=2J_{\\mu+1}^{-2}(\\lambda_ n) \\int_ 0^ 1 J_ \\mu(\\lambda_ n x)f(x)dx, \\quad f\\in S_ \\mu, \\quad \\mu\\in\\mathbb{N}_ 0,$ is shown to be an isomorphism between $$S_ \\mu$$ and $$L_ \\mu$$. Then the generalized finite Hankel transform $$h_ \\mu': S_ \\mu'\\to L_ \\mu'$$ is defined as the adjoint of $$(h_ \\mu^*)^{-1}$$ and it is shown that it extends a previous definition given in the literature.\n\n### MSC:\n\n 46F12 Integral transforms in distribution spaces 44A15 Special integral transforms (Legendre, Hilbert, etc.)\nFull Text:" ]
[ null ]
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http://onlinecalc.sdsu.edu/the_limiting_contraction_ratio_revisited.html
[ "", null, "The Limiting Contraction RatioRevisited Nicole R. Nuccitelli and Victor M. Ponce April 2014\n\n ABSTRACT Henderson's formulations of the energy-based and momentum-based limiting contraction ratios are reviewed (Henderson 1966). Henderson's explicit energy-based equation is found to be correct, however, his implicit momentum-based equation is found to be incorrect. A new explicit momentum-based equation is derived, rendering the implicit formulation unnecessary. An online calculator enables the calculation of the limiting contraction ratio for both energy and momentum formulations.\n\n1.  THE LIMITING CONTRACTION RATIO\n\nThe limiting contraction ratio σ is the ratio of contraction width bc to upstream channel width b1 (σ = bc /b1), or, alternatively, downstream channel width b3 (σ = bc /b3), that produces critical flow at the contraction in a rectangular channel. The ratio has an upper limit σ = 1 when the upstream or downstream flow is critical.\n\nAccording to Henderson (1966), the ratio σ may be calculated in two ways: (1) using an energy balance between the upstream and critical contraction sections, or (2) using a momentum balance between the critical contraction and downstream sections. Henderson has argued that the momentum balance is more likely to be correct because it does not depend on an assumption of energy conservation. However, Henderson's implicit formulation of the momentum-based limiting contraction ratio appears to be incorrect. These propositions are now substantiated.\n\n2.  ENERGY-BASED EQUATION FOR LIMITING CONTRACTION RATIO\n\nThe energy-based equation is derived by equating the specific energy at the upstream flow section (Section 1) with the specific energy at critical flow at the contraction (Section 2, or Section c) (Fig. 1):\n\n E1  =  E2  =  Ec (1)\n\nThe specific energy of the upstream flow is:\n\n v12    E1  =  y1  +  _______                        2 g (2)\n\nThe Froude number at the upstream section is defined as follows:\n\n v12    F12  =  _______               g y1 (3)\n\nThus, Eq. 2 reduces to:\n\n F12    E1  =  y1 [ 1  +  ______ ]                              2 (4)\n\nBy definition, the critical depth is:\n\n F12    yc  =  (2/3) E1  =  (2/3) y1 [ 1  +  ______ ]                                                        2 (5)\n\nThus:\n\n yc                              F12    _____  =  (2/3) [ 1  +  ______ ]   y1                               2 (6)\n\nOr, alternatively:\n\n yc                                  _____  =  (1/3) [ 2  +  F12 ]   y1 (7)\n\nFrom continuity:\n\n v1 y1 b1  =  vc yc bc (8)\n\nThe limiting contraction ratio σ is:\n\n bc             v1 y1 σ  =  _______  =  ________             b1             vc yc (9)\n\nOr:\n\n v1 σ  =  _____________            vc (yc /y1 ) (10)\n\nOr:\n\n v1 σ  =  ___________________            (g yc)1/2 (yc /y1 ) (11)\n\nOr:\n\n v1 σ  =  ______________________            (g y1)1/2 (yc /y1 )3/2 (12)\n\nReplacing Eq. 3 in Eq. 12:\n\n F1 σ  =  ______________             (yc /y1 )3/2 (13)\n\nReplacing Eq. 7 in Eq. 13:\n\n F1 σ  =  ________________________            (1/3)3/2 (2  +  F12) 3/2 (14)\n\nSimplifying Eq. 14:\n\n 33/2 F1 σ  =  _________________            (2  +  F12) 3/2 (15)\n\nEquation 15 is the energy-based limiting contraction ratio.\n\nSquaring Eq. 15:\n\n 33 F12 σ2  =  ________________               (2  +  F12) 3 (16)\n\nEquation 16 is the same as that presented by Henderson for the limiting contraction ratio (op. cit., Eq. 7-35, p. 267).", null, "Fig. 1  Definition sketch.\n\n3.  MOMENTUM-BASED EQUATION FOR LIMITING CONTRACTION RATIO\n\nThe momentum-based equation is derived by equating the specific force at critical flow at the contraction (Section 2, or Section c) with the specific force at the downstream flow section (Section 3).\n\n M3  =  M2  =  Mc (17)\n\nThe specific force at the downstream flow section is:\n\n q32             y32    M3  =  _______  +  _______              g y3             2 (18)\n\nReplacing the discharge Q:\n\n Q2                y32    M3  =  ___________  +  ______              g y3 b32             2 (19)\n\nFrom continuity:\n\n Q = v3 y3 b3  =  vc yc bc (20)\n\nThus:\n\n v32 y32            y32    M3  =  __________  +  ______                g y3                2 (21)\n\nThe Froude number at the downstream section is defined as follows:\n\n v32    F32  =  _______               g y3 (22)\n\nThus, Eq. 21 reduces to:\n\n 1                 M3  =  y32 ( ____  +  F32 )                      2 (23)\n\nThe specific force at the contraction is:\n\n qc2             yc2    Mc  =  _______  +  _______              g yc             2 (24)\n\nReplacing the discharge Q:\n\n Q2                yc2    Mc  =  ___________  +  ______              g yc bc2             2 (25)\n\nFrom continuity:\n\n v32 y32 b32            yc2    Mc  =  _______________  +  ______                  g yc bc2                2 (26)\n\nOr:\n\n v32 y33 b32            yc2    Mc  =  _______________  +  ______                g yc bc2 y3              2 (27)\n\nReplacing Eq. 22 in Eq. 27:\n\n F32 y33            yc2    Mc  =  ___________  +  ______                σ2 yc               2 (28)\n\nBy definition, the critical depth is:\n\n q2                   yc  =  ( ____ ) 1/3               g (29)\n\nOr:\n\n Q2                   yc  =  ( ________ ) 1/3                bc2g (30)\n\nFrom continuity:\n\n v32 y32 b32                    yc  =  ( _____________ ) 1/3                    bc2g (31)\n\nOr:\n\n v32 y33 b32                    yc  =  ( _____________ ) 1/3                  bc2g y3 (32)\n\nReplacing Eq. 22 in Eq. 32:\n\n F32 y33                    yc  =  ( _________ ) 1/3                  σ2 (33)\n\nReplacing Eq. 33 in Eq. 28:\n\n F32 y32 σ2/3                     F32 y33 Mc  =  ______________  +  (1/2) ( _________ )2/3                σ2 F32/3                            σ2 (34)\n\nReducing terms:\n\n F34/3 Mc  =  (3/2) ( ________ ) y32                          σ4/3 (35)\n\nEquating Eqs. 23 and 35:\n\n 1                                   F34/3 ( ___  +  F32 )  =  (3/2) ( ________ )     2                                    σ4/3 (36)\n\nReducing:\n\n 33/4 F3 σ  =  _____________________             ( 1  +  2 F32 ) 3/4 (37)\n\nEquation 37 is the explicit formulation of the momentum-based limiting contraction ratio.\n\n4.  IMPLICIT MOMENTUM-BASED FORMULATION\n\nTo derive an implicit formulation, Eq. 36 is expressed as follows:\n\n 3 F34/3 σ4/3  =  _______________                 1  +  2 F32 (38)\n\nOr:\n\n (3/σ) F34/3 σ1/3  =  _______________                 1  +  2 F32 (39)\n\nReducing:\n\n (3/σ)3 F34 σ  =  _________________             (1  +  2 F32)3 (40)\n\nEquation 40 differs slightly from Henderson's Eq. 7-36, page 267, repeated here:\n\n (2 + 1/σ)3 F34 σ  =  __________________             (1  +  2 F32)3 (41)\n\nThus, Henderson's implicit equation for the momentum-based limiting contraction ratio (Eq. 41) appears to be incorrect.\n\n5.  SUMMARY\n\nTwo explicit equations for the limiting contraction ratio, the first based on an energy balance, and the second based on a momentum balance, are derived following Henderson (1966). An implicit equation based on momentum balance is also derived, to parallel Henderson's treatment.\n\nThe explicit energy-based equation, which is shown to be same as Henderson's, is:\n\n 33/2 F1 σ  =  _________________            (2  +  F12) 3/2 (15)\n\nThe explicit momentum-based equation, not presented by Henderson, is:\n\n 33/4 F3 σ  =  __________________            (1  +  2 F32) 3/4 (37)\n\nNote the great resemblance between Eq. 15 (energy-based) and Eq. 37 (momentum-based).\n\nThe implicit momentum-based equation derived here (Eq. 40) differs slightly from that of Henderson (Eq. 41), suggesting that the latter is incorrect.\n\nThe online calculator ONLINE_LIMITING_CONTRACTION calculates the energy-based and momentum-based limiting contraction ratios for a pair of Froude numbers F1 and F3. Additionally, the online calculator ONLINE_LIMITING_CONTRACTION_SET calculates the set of energy-based and momentum-based limiting contraction ratios for a given range of Froude numbers and Froude number interval.\n\nREFERENCES\n\nHenderson, F. M., 1966. Open channel flow. MacMillan Publishing Co., Inc., New York." ]
[ null, "http://onlinecalc.sdsu.edu/the_limiting_contraction_ratio_revisited_01_400.jpg", null, "http://onlinecalc.sdsu.edu/the_limiting_contraction_ratio_revisited_01_600.jpg", null ]
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https://kmmiles.com/327689-km-in-miles
[ "kmmiles.com\n\nSearch\n\n# 327689 km in miles\n\n## Result\n\n327689 km equals 203494.869 miles\n\nYou can also convert 327689 km to mph.\n\n## Conversion formula\n\nMultiply the amount of km by the conversion factor to get the result in miles:\n\n327689 km × 0.621 = 203494.869 mi\n\n## How to convert 327689 km to miles?\n\nThe conversion factor from km to miles is 0.621, which means that 1 km is equal to 0.621 miles:\n\n1 km = 0.621 mi\n\nTo convert 327689 km into miles we have to multiply 327689 by the conversion factor in order to get the amount from km to miles. We can also form a proportion to calculate the result:\n\n1 km → 0.621 mi\n\n327689 km → L(mi)\n\nSolve the above proportion to obtain the length L in miles:\n\nL(mi) = 327689 km × 0.621 mi\n\nL(mi) = 203494.869 mi\n\nThe final result is:\n\n327689 km → 203494.869 mi\n\nWe conclude that 327689 km is equivalent to 203494.869 miles:\n\n327689 km = 203494.869 miles\n\n## Result approximation\n\nFor practical purposes we can round our final result to an approximate numerical value. In this case three hundred twenty-seven thousand six hundred eighty-nine km is approximately two hundred three thousand four hundred ninety-four point eight six nine miles:\n\n327689 km ≅ 203494.869 miles\n\n## Conversion table\n\nFor quick reference purposes, below is the kilometers to miles conversion table:\n\nkilometers (km) miles (mi)\n327690 km 203495.49 miles\n327691 km 203496.111 miles\n327692 km 203496.732 miles\n327693 km 203497.353 miles\n327694 km 203497.974 miles\n327695 km 203498.595 miles\n327696 km 203499.216 miles\n327697 km 203499.837 miles\n327698 km 203500.458 miles\n327699 km 203501.079 miles\n\n## Units definitions\n\nThe units involved in this conversion are kilometers and miles. This is how they are defined:\n\n### Kilometers\n\nThe kilometer (symbol: km) is a unit of length in the metric system, equal to 1000m (also written as 1E+3m). It is commonly used officially for expressing distances between geographical places on land in most of the world.\n\n### Miles\n\nA mile is a most popular measurement unit of length, equal to most commonly 5,280 feet (1,760 yards, or about 1,609 meters). The mile of 5,280 feet is called land mile or the statute mile to distinguish it from the nautical mile (1,852 meters, about 6,076.1 feet). Use of the mile as a unit of measurement is now largely confined to the United Kingdom, the United States, and Canada." ]
[ null ]
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http://mathcentral.uregina.ca/QQ/database/QQ.09.08/h/eley1.html
[ "", null, "", null, "", null, "SEARCH HOME", null, "Math Central Quandaries & Queries", null, "", null, "Question from ELEY, a parent: WHAT IS THE HEIGHT OF A TRIANGLE IF THE AREA IS 32.4 AND A BASE OF 6.75? PLEASE GIVE ME THE ORDER OF OPERATION TO SOLVE THIS PROBLEM.", null, "Eley,\n\nThe area of a triangle is one-half the base times the height. You have an area of 32.4 square units and a base of 6.75 units. If the height is h units then\n\n32.4 = 1/2 × 6.75 × h.\n\nSolve for h.\n\nPenny\n\nFor a triangle\n\nArea = base x height /2\n\nPut in numbers: say we had area 100 and base 20\n\n100 = 20 x height / 2\n\nNow isolate the unknown on one side by doing the same thing to both sides. For instance, to get rid of the \"/2\" you can multiply both sides by 2 and the equations remains valid:\n\n100 x 2 = 20 x height\n200 = 20 x height\n200/20 = height\n10 = height\n\nGood Hunting!\n-RD", null, "", null, "", null, "", null, "", null, "Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences." ]
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http://www.pfindia.com/mbaforums/viewtopic.php?f=16&t=29
[ "", null, "## Doubts of Number Theory Sheet 2\n\nIn this forum, please discuss the questions appearing in the QA class sheets :especially the practice and the DS questions at the end of the sheet.\n\n### Doubts of Number Theory Sheet 2\n\nDear All,\n\nPleas post your doubts of number theory sheet 3 on this thread.\n\nPosts: 228\nJoined: Wed Feb 13, 2013 5:51 am\n\n### Re: Doubts of Number Theory Sheet 2\n\nQ23\n\n301 = 7*43,\n\nso 300 ≡ -1 (mod 7)\n\nThen 300^3000 - 1 ≡ (-1)^3000 - 1 ≡ 1 - 1 ≡ 0 (mod 7)\n\nSo 7 divides 300^3000 - 1\n\n297 = 27*11,\n\nso 300 ≡ 3 (mod 11)\n\nThen,\n300^3000 - 1 ≡ 3^3000 - 1 ≡ (3^5)^600 - 1 (mod 11)\n\nBut 3^5 = 243 = 22*11 + 1\nso 3^5 ≡ 1 (mod 11)\n\nThen\n300^3000 - 1 ≡ (3^5)^600 - 1 ≡ 1^600 - 1 ≡ 0 (mod 11)\n\nSo 11 divides 300^3000 - 1\n\nFinally, 299 = 23*13,\n\nso 300 ≡ 1 (mod 13)\n\nThen\n300^3000 - 1 ≡ 1^3000 - 1 ≡ 0 (mod 13)\n\nSo 13 divides 300^3000 - 1\n\nSince 7, 11, 13 are all prime, it follows that their product, 1001 divides 300^3000 - 1\n\nPosts: 228\nJoined: Wed Feb 13, 2013 5:51 am\n\n### Re: Doubts of Number Theory Sheet 2\n\nYou are right - for both these questions it is better to approach from the options.\n\nPosts: 228\nJoined: Wed Feb 13, 2013 5:51 am\n\n### Re: Doubts of Number Theory Sheet 2\n\nKindly solve question number 16 and 23 in the practice questions. Are we to proceed by options in both the questions or is there a shorter way out?\ngaurav1399\n\nPosts: 48\nJoined: Wed Feb 27, 2013 7:49 am\n\nReturn to QA - Class Sheets -Phase 1\n\n### Who is online\n\nUsers browsing this forum: No registered users and 1 guest", null, "" ]
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https://www.87994.com/read/d69d2ab400aac7a1cf5a7f764e0b162ad53fb1e1.html
[ "# 高中数学第二章基本初等函数(Ⅰ)2.2.1对数与对数运算第一课时对数课件新人教A版必修1_图文\n\n2.2 对数函数 2.2.1 对数与对数运算\n\n1.理解对数的概念,掌握对数的基本性质. 2.掌握指数式与对数式的互化,能应用对数的定义和性 质解方程.\n\n【情境导学】 导入 某种细胞分裂时,由1个分裂成2个,2个分裂成4个…依此类推,那么1 个这样的细胞分裂x次得到细胞个数N是多少?分裂多少次得到细胞个数为8 个,16个呢? 解:1个细胞分裂x次得到细胞个数N=2x,因为23=8,24=16,所以N=8时,x=3; N=16时,x=4,即细胞分裂3次,4次分别得到细胞个数为8个,16个.\n\n1.对数的概念\n\n.\n\n2.常用对数与自然对数\n\n(1)常用对数:通常我们将以10 为底的对数叫做常用对数,记作lg N\n\n.\n\n(2)自然对数:以e为底的对数称为自然对数,记作 ln N\n\n.\n\n3.对数loga N(a>0,且a≠1)具有下列简单性质 (1) 负数和零 没有对数,即N > 0;\n\n(2)1的对数为 零\n\n,即loga1= 0\n\n;\n\n(3)底数的对数等于 1\n\n,即logaa= 1\n\n;\n\n(4) aloga N =\n\nN\n\n.\n\n【拓展延伸】 1.指数式与对数式的互化 (1)对数式logaN=x是由指数式ax=N而来的,两式底数相同,对数式中的真数N 就是指数式中的幂的值N,而对数值x是指数式中的幂指数.对数式与指数式 的关系如图.\n(2)由于正数的任何次幂都是正数,即ax>0(a>0),故N=ax>0.因此logaN只有 在a>0,且a≠1,N>0时才有意义. 在规定了a>0,a≠1,N>0后,logaN的值便随着a,N的确定而唯一确定了.根据 这一规定,我们知道并不是每一个指数式都能直接改写成对数式.如(-2)2=4, 不能写成log-24=2,只有a>0,a≠1,N>0时,才有ax=N?x=logaN.\n\n2.对数运算性质的证明\n(1)对数的运算性质的证明\n\n1.(对数概念)下列选项中,可以求对数的是( C )\n\n(A)0\n\n(B)-5\n\n(C)π\n\n(D)-x2\n\n2.(指对互化)若b=a2(a>0且a≠1),则有( D )\n\n(A)log2b=a (B)log2a=b (C)logba=2 (D)logab=2 3.(对数概念)在对数式logx-1(3-x)中,实数x的取值范围应该是( D ) (A)(1,3) (B)(1,2)∪(2,+∞)\n\n(C)(3,+∞) (D)(1,2)∪(2,3)\n\n4.(性质)log2 0181+log2 0182 018=\n\n.\n\n5.(性质)log33+ 3log3 2 =\n\n.\n\n(2)( 1 )m=5.73; 3\n\n(2) log1 5.73 =m.\n3\n\n(3)ln 10=2.303; (4)lg 0.01=-2.\n\n(1)log216=4;(2)lo g 1 27=-3;\n3\n\n(3)lo g x=6;(4)43=64; 3\n\n(5)3-2= 1 ;(6)( 1 )-2=16.\n\n9\n\n4\n\n(3)( 3 )6=x.(4)log464=3.\n\n(5)log3\n\n1 9\n\n=-2.(6)lo g 1\n4\n\n16=-2.\n\n【备用例 1】 求下列各式 x 的取值范围. (1)log(x-1)(x+2);\n\n(2)log(x+3)(x+3).\n\n?x ?? x\n\n? ?\n\n3 3\n\n? ?\n\n0, 1,\n\n3 (3)lg 100=x;(4)-ln e2=x.\n\n? ? 解:(1)x=\n\n?\n\n? 64\n\n?2 3\n\n=\n\n43\n\n?2 3\n\n=4-2=\n\n1\n\n.\n\n16\n\n1\n\n1\n\n1\n\n1\n\n? ? ? ? (2)x6=8,所以 x= x6 6 = 86 = 23 6 = 2 2 = 2 .\n\n(3)10x=100=102,于是x=2.\n\n(4)由-ln e2=x,得-x=ln e2,即e-x=e2.\n\n【备用例 2】 求下列各式中的 x 的值: (1) log?2x2?1? (3x2+2x-1)=1;\n\n(2)log2[log3(log4x)]=0.\n\n【例 3】 求下列各式的值:\n(1) 2log2 3 + 3log3 2 ;\n\n(2)\n\n22?\n\nlog\n\n2\n\n1 3\n\n;\n\n(2)原式=22×\n\n2log\n\n2\n\n1 3\n\n=4×\n\n1\n\n=\n\n4\n\n.……………………6\n\n33\n\n(3)101+lg 2; (4)e-1+ln 3.\n\n(4)原式=e-1×eln 3= 1 ×3= 3 .………………………12 分\n\ne\n\ne\n\n3-1:计算:(1)log927;(2) log4 3\n\n81;(3) log 3 54\n\n625.\n\nx\n(2)设 x= log4 3 81,则( 4 3 )x=81, 3 4 =34,所以 x=16.\n\n(3)令 x= log 3 54\n\n625,所以( 3 54\n\n)x=625,\n\n4x\n53\n\n=54,所以\n\nx=3.\n\n?x2 ? 3x ? x ? 3,\n\n? ?\n\nx\n\n2\n\n?\n\n3x\n\n?\n\n0,\n\n??x ? 3 ? 0,且x ? 3 ? 1,\n\n(A)2或-4 (B)-4\n\n(C)2\n\n(D)-2或4\n\n...Ⅰ2.2.1对数与对数运算第一课时对数课件新人教A版必....ppt\n\n2019秋高中数学第二章基本初等函数2.2.1对数与对数运算....ppt\n2019秋高中数学第二章基本初等函数2.2.1对数与对数运算(第1课时)对数课件新人教A版必修1 - 数学 必修① 人教A版 第二章 基本初等函数(Ⅰ) 2.2 对数函数...\n...Ⅰ2.2.1对数与对数运算第一课时对数练习新人教A版必....doc\n\n...Ⅰ2.2.1对数与对数运算第一课时对数练习新人教A版必....doc\n\n...复习第二章基本初等函数(Ⅰ)2.2.1对数与对数运算(第....ppt\n2019高考数学总复习第二章基本初等函数(Ⅰ)2.2.1对数与对数运算(第一课时)课件新人教A版必修 - 2.2 对数函数 2.2.1 对数对数运算 问题情境 请大家...\n...Ⅰ2.2.1对数与对数运算第一课时对数练习新人教A版必....doc\n\n...函数Ⅰ2.2.1对数与对数运算第1课时课件新人教A版必....ppt\n\n...复习第二章基本初等函数(Ⅰ)2.2.1对数与对数运算(第....ppt\n2019高考数学总复习第二章基本初等函数(Ⅰ)2.2.1对数与对数运算(第一课时)课件新人教A版必修1 - 2.2 对数函数 2.2.1 对数对数运算 问题情境 请大家...\n\n...Ⅰ2.2.1对数与对数运算第一课时对数练习新人教A版必....doc\n2019学年高中数学第二章基本初等函数Ⅰ2.2.1对数与对数运算第一课时对数练习新人教A版必修1_六年级语文_语文_小学教育_教育专区。2019 ...\n...Ⅰ2-2-1对数与对数运算第一课时对数练习新人教A版必....doc\n\n2018-2019学年高中数学第二章基本初等函数(Ⅰ)2-2-1对....doc\n2018-2019学年高中数学第二章基本初等函数(Ⅰ)2-2-1对数与对数运算第一课时对数练习新人教A版必修1 - this course will help y ou gain the ide...\n\n...Ⅰ2.2.1对数与对数运算第一课时对数练习新人教A版必....doc\n2019学年高中数学第二章基本初等函数Ⅰ2.2.1对数与对数运算第一课时对数练习新人教A版必修1_数学_高中教育_教育专区。2019 第一课时 对 数 【选题明细表】 ...\n...2019学年高中数学第二章基本初等函数(Ⅰ)2.2.1对数与对数运算....ppt\n2018_2019学年高中数学第二章基本初等函数(Ⅰ)2.2.1对数与对数运算课件新人教A版必修1 - 2.2.1 对数对数运算 复习 (1)aras ? ar?s ?a ? 0, r, ..." ]
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http://feursteinfeurstein.com/idaho-wildflowers-bbyjbhf/what-is-convex-hull-in-image-processing-420ed6
[ "8:23 Uncategorized\n\n# what is convex hull in image processing\n\nParticle Measurements. Director: Wei Wang, Leonard Kleinrock Professor of Computer Science opencv image-processing convex-hull convex-hull-algorithms contour-detection hand-gesture-recognition Updated Aug 31, 2018; Python; ermel272 / convex-hull-animations Star 1 Code Issues Pull requests Animating the computation of convex hulls in two dimensions. 4428 kb/s. Read a grayscale image into the workspace. 3034. Example Of Convex Hull In Image Processing [Most popular] 1700 kb/s. In this work, we derive some new convex hull properties and then propose a fast algorithm based on these new properties to extract convex hull of the object in binary image. 6448 kb/s. Note that this computation can be very slow. Convert it into a binary image and calculate the union binary convex hull. ... finding the convex hull of particles, and more. Users can define thresholds prior to executing or the plugin will assume a dark background and auto threshold the stack using the IsoData method and the stack histogram. An example would be a 4-pixel rectangle with the pixels coordinates ((1,1), (1,2), (2,1), (2,2)). algorithm - daa - convex hull in image processing . An approximate convex hull can be computed using thickening with the structuring elements shown in Figure 1. Abstract. With these lines we … Develop methods (region filling, thinning, thickening, and pruning) that are frequently used in … 3D is another issue though. In this post we will talk about convex hulls which have a broad range of applications in mathematics, computer science and surely image processing / computer vision. Convex hull is widely used in computer graphic, image processing, CAD/CAM and pattern recognition. In this tutorial you will learn how to: Use the OpenCV function cv::convexHull; Theory Next page. This entry was posted in Image Processing and tagged Convex Hull opencv, convexity defects, convexity defects opencv, cv2.convexityDefects(), draw convexity defects, image processing, opencv python, what are convexity defects on 9 Nov 2020 by kang & atul. Toggle Main Navigation ... Use convhull to compute the convex hull of the (x,y) pairs from step 1. Convex Hull • A region A is convex if a straight line joining any two points in A falls within A. How can I do it? A group of UCLA faculty members are working on high-performance analysis of big data. • A convex hull, H, of a set S is the smallest convex set containing S. • The set difference H-S is called the convex … Otherwise, counter-clockwise. chapter 3 answers for drivers ed 2954. Finally, calculate the objects convex hull and display all the images in one figure window. Contours in image processing. Collision avoidance: Avoid collisions with other objects by defining the convex hull of the objects. This is exactly where the role of convex hulls comes to play. Convex hulls have been utilized in various applications (Duda and Hart, 1973; Rosenfeld and … A good overview of the algorithm is … Convex Hull of set S is the smallest convex set A that contains S ... Thinning is an image-processing operation in which binary valued image regions are reduced to lines The problems with this approach is that pixels are considered to have an area of 1 when calculating the region area, but are treated as points in convex hull calculation, causing disparity. Convex hull contains the same 4 … Prev Tutorial: Finding contours in your image. In mathematics the convex hull (sometimes also called the convex envelope) of a set of points X in the Euclidean plane or Euclidean space is the smallest convex… Example Of Convex Hull In Image Processing | added by request. points: any contour or Input 2D point set whose convex hull we want to find. More formally, we can describe it as the smallest convex polygon which encloses a set of points such that each point in the set lies within the polygon or on its perimeter. Hello friends, in this video we will discuss convex hull in image processing Hope you like the video then do LIKE, SUBSCRIBE and SHARE the videos. The convex hull or convex envelope or convex closure of a set X points in the Euclidean plane or in a Euclidean space is the smallest convex set that contains X. The algorithm takes O(n log h) time, where h is the number of vertices of the output (the convex hull). I have a 3D array of data which shows one concave figure. Image Processing (imgproc module) Contours in OpenCV; Convex Hull . The convex hull is a good option for concavity detection, but it has drawbacks because it is not specific: it cannot give hints about the scale of concavities. returnPoints: If True (default) then returns the coordinates of the hull points. The first algorithm is applicable when the points are distinct, as in the pattern classification problem or the image processing problem in which the pattern is not connected. convex hull Chan's Algorithm to find Convex Hull. Example Of Convex Hull In Image Processing | updated. Description: HullAndCircle is a plugin for ImageJ used for finding the convex hull and bounding circle of patterns in binary digital images. In computational geometry, Chan's algorithm, named after Timothy M. Chan, is an optimal output-sensitive algorithm to compute the convex hull of a set P of n points, in 2- or 3-dimensional space. It … Download Hull_And_Circle.jar to the plugins folder, or subfolder, restart ImageJ, and there well be a new Plugins/Shape Analysis/Hull And Circle command. Today I want to tell a little image processing algorithm story related to my post last week about the new bwconvhull function in the Image Processing Toolbox. Otherwise, returns the indices of contour points corresponding to the hull points. Convex Hull¶ The convex hull of a binary image is the set of pixels included in the smallest convex polygon that surround all white pixels in the input. I want to find all points which don't belong to the figure, but are inside the convex hull. Intuitively, it approximates a shape of complex objects with a simpler convex shape. A 2D convex hull already has a node which works very nicely (I’ve implemented in an example workflow here) and it’s possible to generate some points with connected component analysis after the binary hull image is generated. We can draw imaginary lines between these pixels (or better, points). In this work, we derive some new convex hull properties and then propose a fast algorithm based on these new properties to extract convex hull of the object in binary image. Convex hull is widely used in computer graphic, image processing, CAD/CAM and pattern recognition. It is exactly here that, the role of convex hulls comes to play. clockwise: If it is True, the output convex hull is oriented clockwise. The Convex Hull is the line completely enclosing a set of points in a plane so that there are no concavities in the line. Note Remember we have to pass returnPoints = False while finding convex hull, in … The demands of image processing related systems are robustness, high recognition rates, capability to handle incomplete digital information, and magnanimous flexibility in capturing shape of an object in an image. In mathematics, the convex hull or convex envelope for a set of points X in a real vector space V is the minimal convex set containing X.. Introduction In image processing and pattern recognition, it has been shown that computing a convex hull of a given set of points in a digitized image is a useful operation. Any deviation of the object from this hull can be considered as convexity defect.We can visualize it using an image. The convex hull is the fundamental geometric structure that has many applications in various scientific areas, including computer graphics, robotics, computer vision, image processing, and many others. On the left in this slide, you see an example. Image Processing and Recognition: The demands of image processing are robustness, high recognition rate, and flexibility in capturing the shape of an object in an image. Goal . Suggestions. Most image processing is performed in the spatial domain. We draw a line joining start point and end point, then draw a circle at the farthest point. The objective of this paper is twofold. Next Tutorial: Creating Bounding boxes and circles for contours. Find the largest convex black area in an image (4) I have an image of which this is a small cut-out: As you can see it are white pixels on a black background. 10472. This plugin calculates the 3D shape descriptors Solidity3d & Convexity3d based upon a convex hull constructed from an 8-bit or 16-bit grayscale image stack. Search results. However, you may want to process an image in the frequency domain to remove unwanted frequency information before you analyze and process the image as you normally would. Image Processing; Written by. Two efficient algorithms for obtaining the convex hull of n points in the plane are proposed and their theoretical analyses presented. convex hull and solidity: convex hull is minimum convex area that encompasses all the points in set T. It is like mathematical rubber band as shown in picture below. The convex hull computed using this method is actually a `45° convex hull' approximation, in which the boundaries of the convex hull must have orientations that are multiples of 45°. 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Convert it into a binary image and calculate the what is convex hull in image processing convex hull in image Processing, and. Contour points corresponding to the hull points it approximates a shape of complex objects with simpler. Approximates a shape of complex objects with a simpler convex shape Tutorial: Creating bounding boxes and circles Contours! Hull • a region a is convex If a straight line joining start point and end point, then a. Tutorial: Creating bounding what is convex hull in image processing and circles for Contours ( default ) then the!, or subfolder, restart ImageJ, and there well be a new Plugins/Shape Analysis/Hull and circle command Chan. A binary image and calculate the objects convex hull and bounding circle patterns... It approximates a shape of complex objects with a simpler convex shape, the role of hull... Digital images simpler convex shape be a new Plugins/Shape Analysis/Hull and circle command collisions with objects. 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Point and end point, what is convex hull in image processing draw a circle at the farthest point of in! Better, points )... finding the convex hull in image what is convex hull in image processing | added by request image stack a! Approximates a what is convex hull in image processing of complex objects with a simpler convex shape i want to find all points do. For Contours these pixels ( or better, points ) calculates what is convex hull in image processing 3D shape descriptors Solidity3d & based. The objects convex hull in image Processing | added by request 16-bit grayscale image.! That there are no concavities in the line completely enclosing a set of points in a what is convex hull in image processing that! At the farthest point inside the convex hull in image Processing | added by request Chan 's Algorithm to all. For finding the convex hull is widely used in computer graphic, Processing... 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https://hilbert.math.wisc.edu/wiki/index.php?title=Algebraic_Geometry_Seminar_Fall_2011&oldid=3089
[ "Algebraic Geometry Seminar Fall 2011\n\nThe seminar meets on Fridays at 2:25 pm in Van Vleck B215.\n\nThe schedule for the previous semester is here.\n\nFall 2011\n\ndate speaker title host(s)\nSep. 23 Yifeng Liu (Columbia) Enhanced Grothendieck's operations and base change theorem for\n\nsheaves on Artin stacks\n\nTonghai Yang\nSep. 30 Matthew Ballard (UW-Madison) You got your Hodge Conjecture in my matrix factorizations (local)\nOct. 7 Zhiwei Yun (MIT) Cohomology of Hilbert schemes of singular curves Shamgar Gurevich\nOct. 14 Javier Fernández de Bobadilla (Instituto de Ciencias Matematicas, Madrid) Nash problem for surfaces L. Maxim\nOct. 21 Andrei Caldararu (UW-Madison) The Hodge theorem as a derived self-intersection (local)\nNov. 11 John Francis (Northwestern) Integral transforms and Drinfeld centers in derived algebraic geometry Andrei Caldararu\nNov. 18 Sukhendu Mehrotra (Madison) Moduli spaces of sheaves on K3 surfaces and generalized deformations (local)\nDec. 2 Shamgar Gurevich (Madison) Canonical Hilbert Space: Why? How? and its Categorification\nDec. 9 Sean Paul (Madison) Semistable pairs and quasi-closed orbits\n\nSpring 2012\n\ndate speaker title host(s)\nMarch 16 Weizhe Zheng (Columbia) TBD Tonghai Yang\nMarch 23 Ryan Grady (Notre Dame) Twisted differential operators as observables in QFT. Caldararu\nMay 4 Mark Andrea de Cataldo (Stony Brook) TBA Maxim\n\nAbstracts\n\nTBA\n\nMatthew Ballard\n\nYou got your Hodge Conjecture in my matrix factorizations\n\nAbstract: I will describe how to prove some new cases of Hodge conjecture using the following tools: categories of graded matrix factorizations, the homotopy category of dg-categories, Orlov's Calabi-Yau/Landau-Ginzburg correspondence, Kuznetsov's relationship between the derived categories of a certain K3 surface and the Fermat cubic fourfold, and Hochschild homology. This is joint work with David Favero (Wien) and Ludmil Katzarkov (Miami/Wien).\n\nZhiwei Yun\n\nCohomology of Hilbert schemes of singular curves\n\nAbstract: For a smooth curve, the Hilbert schemes are just symmetric powers of the curve, and their cohomology is easily computed by the H^1 of the curve. This is known as Macdonald's formula. In joint work with Davesh Maulik, we generalize this formula to curves with planar singularities (which was conjectured by L.Migliorini). In the singular case, the compactified Jacobian will play an important role in the formula, and we make use of Ngo's technique in his celebrated proof of the fundamental lemma.\n\nNash problem for surfaces\n\nThe space of arcs through the singular set of an algebraic variety has a infinite dimensional scheme structure. In the late sixties Nash proved that it has finitely many irreducible components. He defined a natural mapping from this set of irreducible components to the set of essential divisors of a resolution of singularities. Roughly speaking the set of essential divisors is the set of irreducible components of the exceptional divisor of a resolution whose birational transform is an irreducible component of the exceptional divisor of any other resolution.\n\nNash proved that this mapping is injective and proposed to study its bijectivity. In 2003 S. Ishii and J. Kollar gave a counterexample to the surjectivity in dimension at least 4. Recently, in a joint work with M. Pe Pereira, the speaker has settled the bijectivity for surfaces. In this talk I will explain the proof.\n\nAndrei Caldararu\n\nThe Hodge theorem as a derived self-intersection\n\nThe Hodge theorem is one of the most important results in complex geometry. It asserts that for a complex projective variety X the topological invariants H^*(X, C) can be refined to new ones that reflect the complex structure. The traditional statement and proof of the Hodge theorem are analytic. Given the multiple applications of the Hodge theorem in algebraic geometry, for many years it has been a major challenge to eliminate this analytic aspect and to obtain a purely algebraic proof of the Hodge theorem. An algebraic formulation of the Hodge theorem has been known since Grothendieck's work in the early 1970's. However, the first purely algebraic (and very surprising) proof was obtained only in 1991 by Deligne and Illusie, using methods involving reduction to characteristic p. In my talk I shall try to explain their ideas, and how recent developments in the field of derived algebraic geometry make their proof more geometric.\n\nJohn Francis\n\nIntegral transforms and Drinfeld centers in derived algebraic geometry\n\nFor a finite group G, conjugation invariant vector bundles on G have a universal property with respect to Rep(G): they form its Drinfeld center. Joint work with David Ben-Zvi and David Nadler generalizes this result, extending work of Hinich, in the setting of derived algebraic geometry. We describe a generalization of the Drinfeld center for a monoidal stable infinity category as a Hochschild cohomology category. For quasi-coherent sheaves on a perfect stack X, we prove that its center is equivalent to sheaves on the derived loop space LX. The structure of this category of sheaves defines an extended 2-dimensional topological quantum field theory.\n\nSukhendu Mehrotra\n\nModuli spaces of sheaves on K3 surfaces and generalized deformations\n\nIt is a result of Mukai that connected components of the moduli space of stable sheaves on a K3 surface X are holomorphic symplectic varities. As any such component Y deforms in a 21 dimensional family, while the moduli space of K3 surfaces is 20 dimensional, the general deformation $\\hat{Y}$ of Y will not be be a moduli space of sheaves on a K3. This talk presents an attempt to associate to such a $\\hat{Y}$ a \"non-commutative K3 surface\" $\\hat{X}$ for which the modular description carries over. This joint work with Eyal Markman." ]
[ null ]
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http://blog.jvbaopeng.com/blog/234.html
[ "``` /*无极限分类,返回树形数组结构*/\nStatic Public function unlimitedForLayer (\\$cate, \\$pid = 0, \\$name = 'son')\n{\n\\$arr = [];\nforeach (\\$cate as \\$v) {\nif (\\$v['pid'] == \\$pid) {\n\\$v[\\$name] = self::unlimitedForLayer(\\$cate, \\$v['id'],\\$name);\nif(count(\\$v[\\$name])==0){\n\\$v[\\$name] =null;\n}\n\\$arr[] = \\$v;\n}\n}\nreturn \\$arr;\n}```\n\n```if(count(\\$v[\\$name])==0){\n\\$v[\\$name] =null;\n}```\n\n```/*无极限分类,返回树形数组结构\n*@param array \\$cate 数据库查出来的分类数组\n*@param int \\$pid 父分类id\n*@param string \\$name 子数组键名\n*/\nStatic Public function unlimitedForLayer (\\$cate, \\$pid = 0, \\$name = 'son')\n{\n\\$arr = [];\nforeach (\\$cate as \\$k=>\\$v) {\nif (\\$v['pid'] == \\$pid) {\nunset(\\$cate[\\$k]);\n\\$v[\\$name] = self::unlimitedForLayer(\\$cate, \\$v['id'],\\$name);\nif(count(\\$v[\\$name])==0){\n\\$v[\\$name] =null;\n}\n\\$arr[] = \\$v;\n}\n}\nreturn \\$arr;\n}```\n\n• 友情链接" ]
[ null ]
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https://www.dovov.com/pythonvariables-2.html
[ "# Python嵌套函数中的局部variables\n\n``from functools import partial class Cage(object): def __init__(self, animal): self.animal = animal def gotimes(do_the_petting): do_the_petting() def get_petters(): for animal in ['cow', 'dog', 'cat']: cage = Cage(animal) def pet_function(): print \"Mary pets the \" + cage.animal + \".\" yield (animal, partial(gotimes, pet_function)) funs = list(get_petters()) for name, f in funs: print name + \":\", f()` `\n\n` `cow: Mary pets the cat. dog: Mary pets the cat. cat: Mary pets the cat.` `\n\n• 部分函数示例,使用`functools.partial()`\n\n` `from functools import partial def pet_function(cage=None): print \"Mary pets the \" + cage.animal + \".\" yield (animal, partial(gotimes, partial(pet_function, cage=cage)))` `\n• 创build一个新的作用域示例:\n\n` `def scoped_cage(cage=None): def pet_function(): print \"Mary pets the \" + cage.animal + \".\" return pet_function yield (animal, partial(gotimes, scoped_cage(cage)))` `\n• 将variables绑定为关键字参数的默认值:\n\n` `def pet_function(cage=cage): print \"Mary pets the \" + cage.animal + \".\" yield (animal, partial(gotimes, pet_function))` `\n\n` `funs = list(get_petters())` `\n\n` `for name, f in get_petters(): print name + \":\", f()` `\n\n` `cow: Mary pets the cow. dog: Mary pets the dog. cat: Mary pets the cat.` `\n\n` `for i in range(2): pass print i is 1` `" ]
[ null ]
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http://fweb.wallawalla.edu/class-wiki/index.php?title=Energy_in_a_signal&diff=2554
[ "# Difference between revisions of \"Energy in a signal\"\n\n### Definition of Energy\n\nEnergy is the ability or potential for something to create change. Scientifically energy is defined as total work done by a force. Work can be mathematically calculated as the line integral of force per infinatesimal unit distance,", null, "$W = \\int \\mathbf{F} \\cdot \\mathrm{d}\\mathbf{s}$\n\nPower represents a change in energy.", null, "$P(t) = \\frac{dW}{dt}$\n\nThis means we can also write energy as", null, "$W = \\int_{-\\infty}^{\\infty} P(t)\\,dt$\n\n### Energy of a Signal\n\nFrom circuit analysis we know that the power generated by voltage source is,", null, "$P(t) = {\\mathbf{V}^2(t) \\over R}$\n\nAssuming that R is 1 then the total energy is just,", null, "$W = \\int_{-\\infty}^\\infty |\\mathbf{V}|^2(t) \\mathrm{d}\\mathbf{t}$\n\nThis page is far from complete please feel free to pick up where it has been left off." ]
[ null, "http://fweb.wallawalla.edu/class-wiki/upload/math/5/c/0/5c0b23edaa456374ec84f71596681871.png ", null, "http://fweb.wallawalla.edu/class-wiki/upload/math/a/e/6/ae6514132432b6bff48c27d668c9c2d1.png ", null, "http://fweb.wallawalla.edu/class-wiki/upload/math/3/b/7/3b75bafbbda9d1202cba4ae195227b39.png ", null, "http://fweb.wallawalla.edu/class-wiki/upload/math/7/d/b/7dba69f823326f13dbfdb0954a3d8bc8.png ", null, "http://fweb.wallawalla.edu/class-wiki/upload/math/8/e/9/8e91b65f91f595ba1654b08d36035937.png ", null ]
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http://mathonline.wikidot.com/the-dimension-of-the-null-space-and-range-examples-1
[ "The Dimension of The Null Space and Range Examples 1\n\nThe Dimension of The Null Space and Range Examples 1\n\nRecall from The Dimension of The Null Space and Range page that if $T$ is a linear map from $V \\to W$ and $V$ is finite-dimensional then we have the following formula relating the dimension of $V$ to the dimension of the the null space of $T$ and the dimension of the range of $T$:\n\n(1)\n\\begin{align} \\quad \\mathrm{dim} (V) = \\mathrm{dim} (\\mathrm{null}(T)) + \\mathrm{dim} (\\mathrm{range} (T)) \\end{align}\n\nWe will now look at some examples applying this formula.\n\nExample 1\n\nLet $U$, $V$, and $W$ be finite-dimensional vector spaces, and suppose that $S \\in \\mathcal L (V, W)$ and $T \\in \\mathcal L (U, V)$. Let $ST \\in \\mathcal L (U, W)$. Show that $\\mathrm{dim} (\\mathrm{null} (ST)) ≤ \\mathrm{dim} ( \\mathrm{null} (S)) + \\mathrm{dim} ( \\mathrm{null} (T))$.\n\nUsing the dimension formula above we have that:\n\n(2)\n\\begin{align} \\quad \\mathrm{dim} (\\mathrm{null} (ST)) = \\mathrm{dim} (U) - \\mathrm{dim} (\\mathrm{range} (ST)) \\end{align}\n\nSince $\\mathrm{dim} (U) = \\mathrm{dim} (\\mathrm{null} (T)) + \\mathrm{dim} (\\mathrm{range} (T))$ we can substitute this into the equation above to get that:\n\n(3)\n\\begin{align} \\quad \\mathrm{dim} (\\mathrm{null} (ST)) = \\mathrm{dim} (\\mathrm{null} (T)) + \\mathrm{dim} (\\mathrm{range} (T)) - \\mathrm{dim} (\\mathrm{range} (ST)) \\end{align}\n\nNow notice that $\\mathrm{dim} (\\mathrm{range}(T)) ≤ \\mathrm{dim} (V)$ and so:\n\n(4)\n\\begin{align} \\quad \\mathrm{dim} (\\mathrm{null} (ST)) ≤ \\mathrm{dim} (\\mathrm{null} (T)) + \\mathrm{dim} (V) - \\mathrm{dim} (\\mathrm{range} (ST)) \\end{align}\n\nSince $\\mathrm{dim} (V) = \\mathrm{dim} (\\mathrm{null} (S)) + \\mathrm{dim} (\\mathrm{range} (S))$, we can substitute this into the equation above to get:\n\n(5)\n\\begin{align} \\quad \\mathrm{dim} (\\mathrm{null} (ST)) ≤ \\mathrm{dim} (\\mathrm{null} (T)) + \\mathrm{dim} (\\mathrm{null} (S)) + \\mathrm{dim} (\\mathrm{range} (S)) - \\mathrm{dim} (\\mathrm{range} (ST)) \\end{align}\n\nNote that $\\mathrm{dim} (\\mathrm{range} (S)) - \\mathrm{dim} (\\mathrm{range} (ST)) ≤ 0$ though since $S$ maps elements $v \\in V$ to elements $S(v) \\in W$, and $ST$ maps elements $T(u) \\in V$ to elements $S(T(u)) \\in W$ and so $\\mathrm{range} S \\subseteq \\mathrm{range} (ST)$ which implies that $\\mathrm{dim} ( \\mathrm{range} (S)) ≤ \\mathrm{dim} ( \\mathrm{range} (ST))$, so:\n\n(6)\n\\begin{align} \\quad \\mathrm{dim} (\\mathrm{null} (ST)) ≤ \\mathrm{dim} (\\mathrm{null} (T)) + \\mathrm{dim} (\\mathrm{null} (S)) \\end{align}\n\nExample 2\n\nLet $V$ be a finite-dimensional vector space. Suppose that $V = \\mathrm{null}(T) + \\mathrm{range}(T)$. Prove that $\\mathrm{null} (T) \\cap \\mathrm{range} (T) = \\{ 0 \\}$.\n\nIf we apply the dimension formula given above, we have that:\n\n(7)\n\\begin{align} \\quad \\mathrm{dim} (V) = \\mathrm{dim} (\\mathrm{null} (T)) + \\mathrm{dim} (\\mathrm{range} (T)) \\end{align}\n\nFurthermore, from The Dimension of a Sum of Subspaces we have that:\n\n(8)\n\\begin{align} \\quad \\mathrm{dim} (\\mathrm{null} (T) + \\mathrm{range}(T)) = \\mathrm{dim} (\\mathrm{null} (T)) + \\mathrm{dim} (\\mathrm{range} (T)) - \\mathrm{dim} (\\mathrm{null} (T) \\cap \\mathrm{range} (T)) \\end{align}\n\nSince $V = \\mathrm{null} (T) + \\mathrm{range} (T)$, this implies that the two equations above are equal, and so:\n\n(9)\n\\begin{align} \\quad \\mathrm{dim} (\\mathrm{null} (T)) + \\mathrm{dim} (\\mathrm{range} (T)) = \\mathrm{dim} (\\mathrm{null} (T)) + \\mathrm{dim} (\\mathrm{range} (T)) - \\mathrm{dim} (\\mathrm{null} (T) \\cap \\mathrm{range} (T)) \\end{align}\n\nThe equation above implies that $\\mathrm{dim} (\\mathrm{null} (T) \\cap \\mathrm{range} (T)) = 0$, so $\\mathrm{null} (T) \\cap \\mathrm{range} (T) = \\{ 0 \\}$." ]
[ null ]
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https://vmlc.tamu.edu/Virtual-Math-Learning-Center/Courses/Calculus-I/Inverse-Trigonometric-Functions-only
[ "", null, "#", null, "## Section 1.5 - Inverse Trigonometric Functions\n\nSection Details:\n• Definitions and properties of inverse sine, cosine, and tangent\n• Evaluating inverse trigonometric functions\n• Evaluating compositions of trigonometric and inverse trigonometric functions\n\n## Inverse Trigonometric Functions: MATH 171 Problems 4-6\n\nProperties of inverse trig functions and the derivative of arctangent\n\n## Inverse Trigonometric Functions: MATH 151 Problems 9-12\n\nProperties and derivatives of inverse trigonometric functions" ]
[ null, "https://vmlc.tamu.edu/App_Themes/TAMU/images/top-small.png", null, "https://vmlc.tamu.edu/MLC/media/mainmedia/Home/Logos/VMLC_black.svg", null ]
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https://www.puzzles-world.com/2019/05/apples-oranges-and-mangoes-puzzles.html
[ "# Apples Oranges and Mangoes Puzzles", null, "## Apples Oranges and Mangoes Puzzles", null, "A fruit basket contains only Apples, Oranges and Mangoes.\nAll of the fruits are Apples except four of them,\nall of the fruits are Oranges except five of them, and\nall of the fruits are Mangoes except seven of them.\n\nHow many fruits are there in the basket in all ?\n\nThere are 8 fruits in the basket\n\nExplanation :\n1) A + O + M = X (Let say X is total fruits)\n\nA + 4 = X\nO + 5 = X\nM + 7 = X\n\nso\n2) A + O + M + 16 = 3X\n\nnow substituting 1 in 2, we get,\nX + 16 = 3X => 2X = 16\nHence X = 8" ]
[ null, "https://2.bp.blogspot.com/-oN6WuEm11Z8/UxBnEvrgVII/AAAAAAAAC1o/t8GCn3kuOls/s1600/advertisement.png", null, "https://2.bp.blogspot.com/-s0MvDx-N_v0/XNUMjy5C7VI/AAAAAAAAErs/ZqRcdSvHugYTbtkXMf9Tlx5i9W7nIIWjwCLcBGAs/s1600/Apples%2BOranges%2BMangoes%2BPuzzle.jpg", null ]
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https://blog.kdamball.com/2014/01/15/javascript-and-floating-points-arithmetics/
[ "JavaScript and Floating Points Arithmetics\n\nWe (i.e. anyone who has played with JavaScript) all have heard something about how floating points are tricky and borderline impossible to deal with. While this is not exclusive to JS, it is worth knowing a thing or two behind the limitations of dealing with floating numbers.\n\nLet’s start with a well known example:\n\nvar a = 0.1 + 0.2;\na === 0.3; // false\nconsole.log(a); //0.30000000000000004\n\nThe only way to deal with this is to use the toFixed() property from the Number object or to convert everything into integers, perform the calculations then convert everything back into decimals. Both methods are not guaranteed to produce the correct result, especially when dealing with complex calculations with various floating point variables.\n\nI found out the best way to understand floating point problems is to use the decimal system most humans are so used to. Try expressing 1/3 in a decimal system in the best way possible. There is literally no way to express it to its precision. There are hacks, like 0.333... repeating, but these are all ways that confirm our lack of expressing 1/3 in decimal. Something similar is happening with JavaScript and floating points.\n\nAnyone who has taken an intro class in Calculus will be familiar with the Zeno’s paradox. To summarize it, 1 + 1/2 + 1/4 + 1/8 + .... will always approach 2 but never be equal to 2. This is because we are always halving our distance from 2. That is exactly what is going on when JavaScript tries to express some floating points.\n\nConsider this Binary code:\n\nIntegers:\nBinary: 1 => Decimal: 1\nBinary: 10 => Decimal: 2\nBinary: 1101 => Decimal: 13\n\nFloating points:\nBinary: 0.1 => Decimal: 0.5\nBinary: 0.0101 => Decimal: 0.3125\nBinary: 0.00011001 => Decimal: 0.09765625\nBinary: 0.00011001100110011 => Decimal: 0.09999847412109375\n\nAs you can see from above, the binary value is getting closer and close to 0.1 (in Decimal) but never actually equals it. It is a shortcoming of expressing certain floating points in binary; in the same way we can never fully express certain floating points (e.g: 1/3) in decimal. You can try this with pretty much any base system (try expressing 0.1 (decimal) in Base 3).\n\nTo answer our original issue (i.e. 0.1 + 0.2), calculations are usually transformed into binary, evaluated then converted back into decimal. With its 32-bit limitation, the expression is limited to only 32 floating points. It then becomes:\n\n0.00011001100110011001100110011001 //approx. of 0.1\n+\n0.00110011001100110011001100110011 //approx. of 0.2\n__________________________________\n\n0.01001100110011001100110011001100 //the actual result in binary to be converted into decimal\n\nWant to try something even more fun?\n\nfor(var i = 0, x= 0; i<10;i++){\nx += 0.1; //increment x by 0.1 ten times\n}\n\nconsole.log(x); //0.9999999999999999\n\nPS: I should emphasize that this isn’t something that is unique to JavaScript. Most languages by default have this issue. I just used JavaScript because it’s the most comfortable/easy language to express the idea." ]
[ null ]
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https://scribesoftimbuktu.com/solve-for-z-z9213/
[ "# Solve for z (z+9)^2=13", null, "Take the square root of each side of the equation to set up the solution for\nRemove the perfect root factor under the radical to solve for .\nThe complete solution is the result of both the positive and negative portions of the solution.\nFirst, use the positive value of the to find the first solution.\nSubtract from both sides of the equation.\nNext, use the negative value of the to find the second solution.\nSubtract from both sides of the equation.\nThe complete solution is the result of both the positive and negative portions of the solution.\nThe result can be shown in multiple forms.\nExact Form:\nDecimal Form:\nSolve for z (z+9)^2=13\n\n### Solving MATH problems\n\nWe can solve all math problems. Get help on the web or with our math app\n\nScroll to top" ]
[ null, "https://scribesoftimbuktu.com/wp-content/uploads/ask60.png", null ]
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https://myhomeworkhelp.com/cpm-org-homework-help-embrace-the-fear-of-financial-mathematic-homework-for-achieving-higher-grades/
[ "# CPM Org Homework Help: Embrace the Fear of Financial Mathematic Homework for Achieving Higher Grades\n\n‘Try to achieve those things that you are afraid of.’\n\nFinancial Mathematic is a terror for many students as this subject consists of complex calculations, graphs, reasoning, etc. Mostly, they are scared of solving those sums given by their teachers in homework. If you continue skipping the problems, time will come when you will not understand any concept and lose interest gradually. Proper methods of calculation will help you in achieving higher grades. I so wish that there was cpm org homework help to guide me with the financial calculations in my academic years, my marks would have been better!\n\nSkills one must acquire for completing homework on financial mathematic:\n\nTo completeall complicated sums of financial math with ease, you must possess or acquire few skills. Students need to have skills of strategic and analytical thinking. Calculating maturity value, simple interest, loan payment is not a matter of joke. Hours of practice will help you in solving the complex calculations of financial mathematic with ease. Homework on financial math will seem easier only if you are careful about the calculations.\n\nTo some, financial mathematic seem to be simple and interesting as it creates an abstract and technical branch of math. Besides this, it is also about measuring theoretic probability and learning how to apply these theories in day-to-day life.But for some students like me, homework on financial mathematics is one of the toughest topics in academic life. Have a brief look at the problems faced by students while dealing with homework on financial mathematic.\n\nProblems that most students face on financial mathematic homework:\n\n1. Studying Finance is all about balance sheets and financial concepts. But, many students lag behind as they do not have a proper understanding these complicatedconcepts of finance. So, homework becomes a matter of headache for such students. You may pursue assistance from cpm org homework help for getting a clear idea.\n2. Being a student of Finance, one of the basic problems faced by me was understanding the application of formulas related to statistical analysis, randomness and also probability.\n3. Many are there who face difficulty in calculating and understanding nominal interest and discount, force of interest, discounting, annuities, perpetuities, etc.\n4. Such calculations require proper guidance. But, the basic problem that affects many students while doing homework is they do not get the assistance needed to clear up the concepts.\n\nÂ\n\nHow cpm org homework help can solve your financial mathematic problems?\n\nHave a look at few unique aspects of this system of learning, so that you score higher grades with ease.\n\n• Gives crystal clear idea of financial symbols-\n\nCpm org homework help efficiently assists students to learn one of the most important keys of Financial Mathematics. Now, you might be wondering what the key is. Well, it’s all about understanding the formulas and before that acquiring proper knowledge of all the financial variables and symbols. Unless and until youget a clear idea of what each financial symbol represent, you will not gain interest to learn more.\n\nFor example, N= number of periods, g= rate of growth, B= balance, PV= present value, FV= future value, T= terminal period, rE= effective interest rate, CF= cash flow, PMT= periodic payment. This educational program teaches the students where, how and why to use such symbols.\n\n• Guides in your flexible time\n\nSuppose, the total strength of your class is 40.It is quite natural that students will face problems and come up with queries. Now, is it possible for a teacher to clear up all single queries within a period of 40-45 minutes? Of course, not!Due to this reason, students think of taking extra assistance after school hours. Even that is tiring as there are many other subjects you need to study.\n\nA highly essential facet of this educational system is that they assist students according to their own convenient time. You can just come up with your queries and explanation is provided at that very moment.\n\n• Encourages joint or group study\n\nThe curriculum of cpm org homework help is planned in such a format where students get to solve financial mathematical problem sums jointly. Thus, students discuss the techniques before solving the sums on their own. By doing so, they come up with different procedures to solve a particular problem.\n\nImagine yourself in a situation where you are instructed to complete a difficult sum on financial math. Will you be able to struggle with it for an hour or more without the support of other fellow friends who are sailing in the same boat as you? Hence, this particular curriculum focuses in theunified system of studying.\n\n• Specialized E- books\n\nDo you want to know another interesting aspect of this educational program? You will be very excited to know that they prefer using specialized E- books other than general books on financial mathematics. Descriptive ideas are provided for better explanation of calculating loan payments, interest rate, etc. The concepts of Finance which students may find difficult to understand is simplified by the facility of video tutorials.\n\nThus, if you ever face with a specific sum, you can take help from those tutorials for understanding the procedure step-by-step. Once, you are aware of the steps it will be easier for you to solve the sum, no matter how difficult it is. This system of helping students with their financial math homework is highly co- operative.\n\n• Easier techniques\n\nThe most common problem faced by students is to memorize all the formula sheets required for solving financial math sums. Formulas of simple interest like I=Prt, S= P+I, S= P(1+rt) are very simple. But, rememberingformulas for calculating ordinary simple annuity, ordinary general annuity requires a lot of practice and time.\n\nThis educational program assists students to learn better and upgraded techniques of learning such complex formulas. With advanced and systematic procedures, solving difficult sums and spreadsheets become very easy.\n\nHomework on financial math becomes quite interesting when learning and solving is through technology. Impressive way of completing your homework, isn’t it? Even you can opt for it!\n\nEmbracing new technologies and techniques is a part of human nature especially if it makes life easier than before. So, cpm org homework help is a good option for students who are looking for the guidance of an educational system that will help them in their financial mathematic homework.\n\n## Submit Homework", null, "### Customer Reviews\n\nRatings based on 510 customer reviews.\n\n## What Are the Career Options You Get Studying Applied Mathematics?\n\nDo numbers fascinate you? Is solving complex mathematical problems an easy task for you? Are you interested in making...\n\n## Number Theory is Super Easy but Only If You Work with Mathematics Being a Smarty\n\nMathematics is all about the numbers and their equations which are to be solved with the correct method and a small...\n\n## How to Make Your Mathematics Life in College a Fun One?\n\nCollege life is the life that every school student looks forward to. Think about it, you are independent to attend the...", null, "" ]
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http://www.lvliang.gov.cn/llxxgk/zfxxgk/xxgkml/szfwj/202109/t20210917_1558601.html
[ "• ###### 政民互动", null, "关闭\n 索  引  号:111423LL00100/2021-01138 发文字号:吕政发〔2021〕13号 发布日期:2021年09月07日 发文机关:吕梁市人民政府 关  键  字: 标  题:关于收回吕梁市商贸储运公司国有建设用地使用权的决定 主题分类:土地 成文日期:2021年09月07日\n\n为实施《吕梁市中心城区货源街片区老旧小区改造工作方案》,依照《中华人民共和国土地管理法》《土地储备管理办法》和《吕梁市规划区土地储备实施办法》的规定,作出如下决定:\n\n一、无偿收回你单位位于离石区货源街98号的国有建设用地使用权,面积为3668.45平方米。宗地四至为:东邻居民住宅、南邻市商务局家属院、西至小路、北至货源街。宗地坐标如下:\n\nX1=4153811.3150, Y1=37511960.959;\n\nX2=4153805.4430,         Y2=37511976.490;\n\nX3=4153801.8910,         Y3=37511994.006;\n\nX4=4153793.0100,         Y4=37512019.830;\n\nX5=4153788.3090,         Y5=37512018.246;\n\nX6=4153781.2840,         Y6=37512018.496;\n\nX7=4153743.5190,         Y7=37512019.839;\n\nX8=4153743.2120,         Y8=37512011.584;\n\nX9=4153741.5760,         Y9=37511967.698;\n\nX10=4153741.2000,         Y10=37511957.614;\n\nX11=4153753.9320,         Y11=37511957.540;\n\nX12=4153782.5600,         Y12=37511957.848;\n\nX13=4153782.5610,         Y13=37511957.711;\n\nX14=4153785.8570,         Y14=37511957.730;\n\nX15=4153787.0220,         Y15=37511957.804;\n\nX16=4153790.9060,         Y16=37511958.291;\n\nX1=4153811.3150, Y1=37511960.959。\n\n二、收回的国有建设用地由市土地储备中心储备管理,具体位置以《土地勘测定界技术报告书》编号:RX-(SC2021-003)号为准。\n\n三、请你单位尽快到原土地登记机关办理国有建设用地的使用权注销登记。\n\n2021年9月7日\n\n(此件公开发布)\n\n解读说明:此件不属于政策性文件,不作解读。\n\n咨询电话:市规自局 土地储备中心 8296746", null, "晋公网安备 14110002000102号" ]
[ null, "http://www.lvliang.gov.cn/llxxgk/images/close.png", null, "http://www.lvliang.gov.cn/llxxgk/images/jinghui.png", null ]
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https://codereview.stackexchange.com/questions/120369/printing-binary-trees
[ "Printing binary trees\n\nAfter writing this answer, I was inspired to try to design a more elegant solution to the problem of printing binary trees. And what better tool with which to seek elegance than Clojure?\n\nOverall design\n\nThe solution I ended up with involved creating, merging, and printing what I'm going to call sparse strings. A sparse string is simply a map from row/column pairs to substrings. These substrings must not contain newlines or overlap with each other.\n\nSo for instance, this multiline string\n\n baz\nfoo\nbar\nqux\n\n\ncould be represented by this sparse string:\n\n{[3 -2] \"foo\"\n[4 -7] \"bar\"\n[2 -6] \"baz\"\n[5 7] \"qux\"}\n\n\nA few things to note:\n\n1. Empty space is simply filled by regular spaces and newlines\n2. The first coordinate is the row, the second coordinate is the column\n3. The first row and column need not necessarily be zero\n\nThe problem of printing a binary tree then reduces to creating sparse strings from its left and right subtrees (as well as one for its value, which will go at the top), then shifting and merging those sparse strings together.\n\nThe only extra requirement when using sparse strings to represent trees is that the root (or the center of the root, if the root is wider than 1 character) must be at [0 0]. Consider the tree below:\n\n 4\n/ \\\n2 5\n/ \\\n1 3\n\n\nThe in-memory representation of this would be\n\n{:value 4\n:left {:value 2\n:left {:value 1}\n:right {:value 3}}\n:right {:value 5}}\n\n\nTextually, the tree would be represented as\n\n{[0 0] \"4\"\n[2 -2] \"2\"\n[4 -4] \"1\"\n[3 -3] \"/\"\n[4 0] \"1\"\n[3 -1] \"\\\\\"\n[1 -1] \"/\"\n[2 2] \"5\"\n[1 1] \"\\\\\"}\n\n\nI'll refer to this as a tree string. This way, when I combine multiple tree strings to create a bigger tree string, I can safely use [0 0] as an anchor point from which to shift subtrees.\n\nCode\n\n(defn end-col\n\"Returns one plus the maximum column occupied by the sparse string entry x.\"\n[x]\n(let [[[_ col] s] x]\n(+ col (count s))))\n\n(defn min-corner\n\"Returns a vector of the minimum non-empty row and column in sparse-string.\"\n[sparse-string]\n(mapv #(apply min (map % (keys sparse-string)))\n[first second]))\n\n(defn max-corner\n\"Returns a vector of one plus the maximum non-empty row and column in\nsparse-string.\"\n[sparse-string]\n(mapv #(apply max (map % sparse-string))\n[(comp inc first key) end-col]))\n\n(defn fill\n\"Returns a string of newlines and spaces to fill the gap between entries x and\ny in a sparse string whose minimum corner is corner.\"\n[corner x y]\n(let [[_ min-col] corner\n[[prev-row _] _] x\n[[next-row next-col] _] y]\n(apply str (concat (repeat (- next-row prev-row) \\newline)\n(repeat (- next-col\n(if (== prev-row next-row)\n(end-col x)\nmin-col))\n\\space)))))\n\n(defn sparse-str\n\"Converts sparse-string to a string.\"\n[sparse-string]\n(let [corner (min-corner sparse-string)]\n(->> sparse-string\n(sort-by key)\n(cons [corner \"\"])\n(partition 2 1)\n(map (fn [[x y]] (str (fill corner x y) (val y))))\n(apply str))))\n\n(defn shift\n\"Creates and returns a sparse string by adding offset to the position of each\nentry in sparse-string.\"\n[offset sparse-string]\n(into {} (map (fn [[pos s]]\n[(mapv + pos offset) s])\nsparse-string)))\n\n(defn vert-gap\n\"Returns the minimum vertical gap that can be used in combining the left and\nright tree strings.\"\n[left right]\n(if (and left right)\n(max 1 (quot (- (second (max-corner left))\n(second (min-corner right)))\n2))\n1))\n\n(def directions {:left - :right +})\n\n(defn diagonal\n\"Returns a diagonal sparse string with the top end located at corner.\"\n[direction corner length character]\n(let [[first-row first-col] corner]\n(into {} (map (fn [n]\n[[(+ first-row n)\n((directions direction) first-col n)]\n(str character)])\n(range length)))))\n\n(defn leg\n\"Returns a sparse string from shifting tree-string according to direction,\nvert-gap, and value-height, merged with a diagonal strut.\"\n[direction tree-string vert-gap value-height]\n(merge (shift [(+ value-height vert-gap)\n((directions direction) (inc vert-gap))]\ntree-string)\n(diagonal direction\n[value-height ((directions direction) 1)]\nvert-gap\n({:left \\/ :right \\\\} direction))))\n\n(defn assemble\n\"Assembles a complete tree string from the tree strings of a value, left\nsubtree, and right subtree.\"\n[value left right]\n(if (or left right)\n(let [[value-height _] (max-corner value)\nvert-gap (vert-gap left right)]\n(merge value\n(when left\n(leg :left left vert-gap value-height))\n(when right\n(leg :right right vert-gap value-height))))\nvalue))\n\n(defn tree-string\n\"Creates and returns a tree string from node.\"\n[node]\n(let [{:keys [value left right]} node\ns (str value)]\n(apply assemble\n{[0 (- (quot (count s) 2))] s}\n(map #(when % (tree-string %))\n[left right]))))\n\n\nImplementation notes\n\nWhen printing a sparse string, one first needs to find the minimum occupied row and column in the sparse string. This naturally leads to the end-col, min-corner, and max-corner functions for calculating bounding boxes for sparse strings.\n\nNow, how does one print a sparse string? This problem basically boils down to printing the whitespace to fill the gaps between sparse string entries. Once that's accomplished, the actual sparse-str implementation is quite straightforward.\n\nAlso, since I'm going to be combining sparse strings in addition to just creating and printing them, I'll need a function to shift the reference point of an existing sparse string.\n\nAll we need now is a way to convert from that logical representation of the tree to the textual one, which is the task of the tree-string function.\n\nAs you can see, the better part of the work is done in that assemble function. The first thing we need to do is decide how long we're going to make the struts that connect the left and right subtrees. To simplify things, we'll just always make them the same length, which means that the length of the struts, calculated by vert-gap, will equal to the final distance between the bottom of the value tree string and the tops of the left and right tree strings.\n\nAnd of course, we'll need the diagonal function to create the struts themselves.\n\nNow that we have all the rest of the pieces, it's just a matter of assembleing them together. The leg function is just a helper that combines diagonal and shift together into something that can then be merged with the root.\n\nExample\n\nIn case you'd like to test this code yourself, here's a function for generating a random binary tree:\n\n(defn rand-tree\n[weight depth]\n(into {:value (int (Math/pow 2 (rand (dec Integer/SIZE))))}\n(map #(when (and (pos? depth) (< (rand) weight))\n[% (rand-tree weight (dec depth))])\n[:left :right])))\n\n\nSo this...\n\n(println (sparse-str (tree-string (rand-tree 3/4 3))))\n\n\n... will print something like this:\n\n 1369616891\n/ \\\n/ \\\n/ \\\n/ \\\n/ \\\n238883 2\n/ \\ /\n/ \\ 9\n/ \\ / \\\n/ \\ 1 2222\n2949729 6\n/ /\n1836 5299294\n\n• Why are 2's kids on a different level than those of 238883? – Josh Mar 29 '17 at 14:07\n• @Josh Because my code recursively builds the left and right subtrees into tree strings separately and then combines them into a tree string for the whole tree, so in that example the right subtree doesn't know the length of the struts in the left subtree because it doesn't know that the left subtree exists at all. – Sam Estep Mar 29 '17 at 16:37\n• But why are the struts different lengths there?? – Josh Mar 30 '17 at 6:54\n• @Josh Because you can't fit 2949729 and 6 on the same line with struts of length 1 using the centering function #(- (quot (count %) 2)) that you can find in tree-string. – Sam Estep Mar 30 '17 at 11:27\n• @Josh In any case, if you think that my code can be improved, by all means please post an answer! That's why I posted it in this question in the first place. – Sam Estep Mar 30 '17 at 22:27\n\nFirst of all, in my opinion, both the algorithm is nice and interesting, and the code is really well-written, broken down into easily understandable functions! Well done!\n\nThe only addition that I can make are corner-cases (and their possible fixes) for some of the helper functions. However, please note, that from the main entry point, I did not find any way to trigger these corner cases, so, from a user perspective, the program works well even without the changes below.\n\nCorner case # 1\n\n(min-corner {})\n(sparse-str {})\n\n\nBoth throw an exception, due to min in min-corner being called with zero arguments.\n\nI suggest the following change, in order to handle this case:\n\n(defn min-corner\n\"Returns a vector of the minimum non-empty row and column in sparse-string.\"\n[sparse-string]\n(mapv #(apply min (let [args (map % (keys sparse-string))] (if (empty? args) [0 0] args)))\n; (mapv #(apply min (map % (keys sparse-string)))\n[first second]))\n\n\nI.e., call min with a default value (in this case [0 0]), if the arguments are empty.\n\nCorner case #2\n\n(max-corner {})\n\n\nSimilar issue, like for min-corner, similar solution:\n\n(defn max-corner\n\"Returns a vector of one plus the maximum non-empty row and column in\nsparse-string.\"\n[sparse-string]\n(mapv #(apply max (let [args (map % sparse-string)] (if (empty? args) [0 0] args)))\n; (mapv #(apply max (map % sparse-string))\n[(comp inc first key) end-col]))\n\n\nCorner case #3\n\n(sparse-str {[0 0] \"aa\" [0 1] \"b\"})\n\n\nThis will return aab. However, b is misplaced here, because it should be at position [0 1], and it is, instead, at position [0 2]. To be honest, I don't really know what would be the best solution here. I can imagine the following:\n\n1. Leave it as it is: all output is shown, some of it might be on the wrong place.\n2. Overwrite the the end of the first string with the second one (i.e. \"ab\").\n3. Overwrite the beginning of the second string with the first one (in this case: \"aa\").\n4. Throw an exception.\n\nAgain, starting from the main entry point, it is not possible to trigger this situation, because the user does not provide any positions. However, in case the helper functions were to be reused in a library, this is a question that should be addressed.\n\nCorner case #4\n\n(vert-gap {[0 0] \"a\"} {[4 10] \"a\"})\n\n\nIn case the right element has a bigger second coordinate than the left one, the result is always 1. I'm not sure if this is by design, but if not, then I'd suggest to make sure, that vertical gap is calculated correctly also in such cases, by taking the absolute value of the difference, instead of the difference itself:\n\n(defn vert-gap\n\"Returns the minimum vertical gap that can be used in combining the left and\nright tree strings.\"\n[left right]\n(if (and left right)\n(max 1 (quot (Math/abs (- (second (max-corner left))\n(second (min-corner right)))\n) 2))\n1))\n\n\nCorner case # 5\n\n(diagonal :left [0 0] -2 \\a)\n\n\nWith the current implementation, this will return an empty result: {}. Now, with the semantics that the name of the parameter length implies, this should be correct (maybe an exception could be thrown, but that is only detail). However, wouldn't it be nice, if the signum of length could control the direction, in which the first coordinate grows? Something like this:\n\n(defn diagonal\n\"Returns a diagonal sparse string with the top end located at corner.\"\n[direction corner length character]\n(let [[first-row first-col] corner\n_length (Math/abs length)\nop (if (< length 0) - +)]\n(into {} (map (fn [n]\n[[(op first-row n)\n((directions direction) first-col n)]\n(str character)])\n(range _length)))))\n\n\nIn this way, the result of the above example will be: {[0 0] \"a\", [-1 -1] \"a\"}. Of course, it might be worth considering to rename the length parameter, e.g. to horizontal-displacement or something similar.\n\nP.S.:\n\nI setup a github repo for the above mentioned changes, and their corresponding test cases (along other tests):\n\n• Thanks for the awesome write-up! For Corner case #1 and Corner case #2, I think that it would be better for sparse-str to handle the case where its argument is empty (and just return \"\") and for min-corner and max-corner to specify in their docstring that their argument must be nonempty, similar to how it is part of the contract of min and max that they be passed a nonzero number of arguments. – Sam Estep Aug 11 '17 at 19:14\n• For Corner case #3, I definitely agree that the current behavior is undesirable; the \"dimensions\" of the returned string should not depend on whether an overlap is present. Out of the possible solutions you present, I think the second one (overwrite the the end of the first string with the second one) is the best. – Sam Estep Aug 11 '17 at 19:14\n• For Corner case #4, that's an interesting behavior, and on the one hand your solution might be desirable for robustness purposes, but on the other hand, in my definition of \"tree string\", I specified that the root must be present at [0 0], so {[4 10] \"a\"} does not qualify as a tree string, and vert-gap specifies that both of its arguments are tree strings. – Sam Estep Aug 11 '17 at 19:14\n• For Corner case #5, I like your idea of adding semantics to the signum of length! Do you think that could possibly remove the need for the direction and character parameters as well? – Sam Estep Aug 11 '17 at 19:14\n• @SamEstep: re #5: I don't think that we can pack those infos into length, because the signum of length is now the vertical direction, while direction is the horizontal one. Unless length is converted to some kind of structure, but that will still need all three pieces of information IMHO. – Attilio Aug 12 '17 at 11:54" ]
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https://www.netket.org/docs/_generated/samplers/netket.sampler.MetropolisPtSampler.html
[ "# netket.sampler.MetropolisPtSampler¶\n\nclass netket.sampler.MetropolisPtSampler(*args, __skip_preprocess=False, **kwargs)[source]\n\nBases: netket.sampler.metropolis.MetropolisSampler\n\nMetropolis-Hastings with Parallel Tempering sampler.\n\nThis sampler samples an Hilbert space, producing samples off a specific dtype. The samples are generated according to a transition rule that must be specified.\n\n__init__(*args, __skip_preprocess=False, **kwargs)\n\nMetropolisSampler is a generic Metropolis-Hastings sampler using a transition rule to perform moves in the Markov Chain. The transition kernel is used to generate a proposed state $$s^\\prime$$, starting from the current state $$s$$. The move is accepted with probability\n\n$A(s\\rightarrow s^\\prime) = \\mathrm{min}\\left (1,\\frac{P(s^\\prime)}{P(s)} F(e^{L(s,s^\\prime)})\\right),$\n\nwhere the probability being sampled from is $$P(s)=|M(s)|^p. Here ::math::$$ is a user-provided function (the machine), $$p$$ is also user-provided with default value $$p=2$$, and $$L(s,s^\\prime)$$ is a suitable correcting factor computed by the transition kernel.\n\nParameters\n• hilbert – The hilbert space to sample\n\n• rule – A MetropolisRule to generate random transitions from a given state as well as uniform random states.\n\n• n_chains – The number of Markov Chain to be run in parallel on a single process.\n\n• n_sweeps – The number of exchanges that compose a single sweep. If None, sweep_size is equal to the number of degrees of freedom being sampled (the size of the input vector s to the machine).\n\n• n_chains – The number of batches of the states to sample (default = 8)\n\n• machine_pow – The power to which the machine should be exponentiated to generate the pdf (default = 2).\n\n• dtype – The dtype of the statees sampled (default = np.float32).\n\nAttributes\nis_exact\n\nReturns True if the sampler is exact.\n\nThe sampler is exact if all the samples are exactly distributed according to the chosen power of the variational state, and there is no correlation among them.\n\nReturn type\n\nbool\n\nmachine_pow: int = 2\n\nExponent of the pdf sampled.\n\nn_batches\nn_chains\n\nThe total number of chains across all MPI ranks.\n\nIf you are not using MPI, this is equal to n_chains_per_rank\n\nReturn type\n\nint\n\nn_chains_per_rank: int = None\n\nNumber of independent chains on every MPI rank.\n\nn_replicas: int = 32\n\nThe number of replicas\n\nn_sweeps: int = None\n\nNumber of sweeps for each step along the chain. Defaults to number of sites in hilbert space.\n\nreset_chains: bool = False\n\nIf True resets the chain state when reset is called (every new sampling).\n\nrule: MetropolisRule = None\n\nThe metropolis transition rule.\n\nMethods\ninit_state(machine, parameters, seed=None)\n\nCreates the structure holding the state of the sampler.\n\nIf you want reproducible samples, you should specify seed, otherwise the state will be initialised randomly.\n\nIf running across several MPI processes, all sampler_states are guaranteed to be in a different (but deterministic) state. This is achieved by first reducing (summing) the seed provided to every MPI rank, then generating n_rank seeds starting from the reduced one, and every rank is initialized with one of those seeds.\n\nThe resulting state is guaranteed to be a frozen python dataclass (in particular, a flax’s dataclass), and it can be serialized using Flax serialization methods.\n\nParameters\nReturn type\n\nSamplerState\n\nReturns\n\nThe structure holding the state of the sampler. In general you should not expect it to be in a valid state, and should reset it before use.\n\nlog_pdf(model)\n\nReturns a closure with the log_pdf function encoded by this sampler.\n\nNote: the result is returned as an HashablePartial so that the closure does not trigger recompilation.\n\nParameters\n\nmodel (Union[Callable, Module]) – The machine, or apply_fun\n\nReturn type\n\nCallable\n\nReturns\n\nthe log probability density function\n\nreplace(**updates)\n\nReturns a new object replacing the specified fields with new values.\n\nreset(machine, parameters, state=None)\n\nResets the state of the sampler. To be used every time the parameters are changed.\n\nParameters\nReturn type\n\nSamplerState\n\nReturns\n\nA valid sampler state.\n\nsample(machine, parameters, *, state=None, chain_length=1)\n\nSamples chain_length elements along the chains.\n\nParameters\nReturns\n\nThe next batch of samples. state: The new state of the sampler\n\nReturn type\n\nσ\n\nsample_next(machine, parameters, state=None)\n\nSamples the next state in the markov chain.\n\nParameters\nReturns\n\nThe new state of the sampler σ: The next batch of samples.\n\nReturn type\n\nstate" ]
[ null ]
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http://yunhuoer.com/qspevdu_t1004003023
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http://perplexus.info/show.php?pid=8114&cid=57676
[ "", null, "All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars", null, "", null, "perplexus dot info", null, "", null, "Purely prime (Posted on 2013-02-19)", null, "Twenty-one prime numbers are in arithmetic sequence with difference d. Prove that d is divisible by 9699690.\n\n No Solution Yet Submitted by Danish Ahmed Khan Rating: 5.0000 (1 votes)", null, "Comments: ( Back to comment list | You must be logged in to post comments.)", null, "Primes Comment 3 of 3 |", null, "Let p1, p2, ..., p21 be the prime numbers. Then, p1+d=p2, p2+d=p3, ..., p20+d=p21. Take any prime p<10. If d is not divisible by p, then at least 2 of the prime numbers are divisible by p. Since p is the only prime divisible by p, that is impossible. Therefore, d is divisible by p. Then, d is divisible by 2*3*5*7=210. Take any prime 10<p<20. If d is not divisible by p, then at least 1 of the prime numbers is divisible by p. Since p is the only prime divisible by p, pn=p for some n. Since d is divisible by 210, p2>210, ..., p21>210. Since p<20, p2>p, ..., p21>p. Therefore, p1=p. Then, p1+pd is one of the primes. That is impossible. Therefore, d is divisible by p. Then, d is divisible by 2*3*5*7*11*13*17*19=9699690.\n\n Posted by Math Man on 2016-09-05 21:00:42", null, "Please log in:\n\n Search: Search body:\nForums (0)" ]
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https://cemxpu.burgerfi8518.site/matrix-multiplication-using-mapreduce-github.html
[ "## Matrix multiplication using mapreduce github\n\nmatrix multiplication using mapreduce github Kublanovskaya, J. Examples of Lambda are given below. • Standard http://jwbuurlage. MapReduce. Francis) 1964: Sinkhorn (Richard Sinkhorn) 1965: Golub-Reinsch SVD (Gene Golub) 1969: Sparse matrix ordering (Elizabeth Cuthill, James McKee) 1969: Strassen matrix multiplication (Volker Strassen) The implementation for the multiplication gate with input scalars x and y and output scalar z is class MultiplyGate ( object ): def forward ( x , y ): z = x * y self . combineAlli(x1;:::;xn) : combine all the results from combine2() for node i. g. g. 3. I just want to code a matrix multiplication problem in MapReduce using python for very large sparse matrix. An AggBinaryOp hop can be compiled into the following physical operators. h. This relates mostly to (a) matrix multiplication deficiencies, and (b) handling parallelism. Test naive algorithm locally. Basic Matrix Multiplication Operators. May 15, 2013 · Does anyone need MapReduce?122• I tried to do book recommendations withlinear algebra• Basically, doing matrix multiplication toproduce the full user/item matrix withblanks filled in• My Mac wound up freezing• 185,973 books x 77,805 users =14,469,629,265– assuming 2 bytes per float = 28 GB of RAM• So it doesn’t necessarily take that much tohave some use for MapReduce of the loss J ( i) ( θ) of the i th window vector with respect to the matrix x of window vectors is given by. I built a jar file using the code below. Jun 21, 2020 · MapReduce Program – Finding The Average Age of Male and Female Died in Titanic Disaster; MapReduce – Understanding With Real-Life Example; How to find top-N records using MapReduce; How to Execute WordCount Program in MapReduce using Cloudera Distribution Hadoop(CDH) Matrix Multiplication With 1 MapReduce Step; MapReduce – Combiners 1. element 3 in matrix A is called A21 i. 15 minute read. Introductory example is calculation of Fibonacci numbers where F(N) (problem of size N) is calculated as sum of F(N - 2) and F(N - 1) (problems of size N - 2 and N - 1). KEY: block num (a_block, b_block) VALUE: [(row, col, value), ,( )]//numbers in the block // implement matrix multiplication of the blocks locally res = 0 for i = row of ele in this a block: for k = col of ele in this b block: for j = 0 to n-1: res += A[i][j] * B[j][k] Test. Aug 17, 2018 · Besides matrix-vector and matrix-matrix calculations, relational-algebra operations fit well into the MapReduce style of computing. deeplearn-rs - deeplearn-rs provides simple networks that use matrix multiplication, addition, and ReLU under the MIT license. This file contains the implementation of reducer. Specifically, use test your implementation on the following graphs, 1. Table 2 summarizes the core operations of important ML algorithms. com&nbs 15 Dec 2014 in this area (Hadoop, Map-Reduce and Spark) work exercises. Rawashdeh , M. From the implementation point of view, Before the MapReduce model, MPI was the main tool used to process big data. 2166 II. Certain common operations, like broadcast or matrix multiplication, do know how to deal with array wrappers by using the Adapt. ein\"ij,jk -> ik\"(a,b) returns :(a * b) and ein\"ij,kj -> ki\"(a,b) returns :(b * transpose(a)) . in a way you should be familiar with. The nucleus is represented as a N-nanometer by M-nanometer grid, and at each 1nm * 1nm square is known whether or not it is filled by something. G. Parsian, Data  . When we represent matrices in this form, we do not need to keep entries for the cells that have values of zero to save large amount of disk space. PyMR is a Python 2. We define requirements for the design of serverless big data applications, present a prototype for matrix multiplication using FaaS, and discuss and synthesize insights from results of extensive experimentation. What we want to do We will write a simple MapReduce program (see also Wikipedia ) for Hadoop in Python but without using Jython to translate our code to Java jar files. Aug 29, 2015 · Matrix Multiplication with MapReduce 33 Comments Posted by Maruf Aytekin on February 16, 2015 Matrix-vector and matrix-matrix calculations fit nicely into the MapReduce style of computing. The MapReduce contains two important tasks, namely Map and Reduce. Map-Reduce. Integer matrix-matrix and matrix-vector Distributed matrix factorization with mapreduce using a series of broadcast-joins (SS, CB, MS, AA, VM), pp. Matrix-matrix product • Basic matrix multiplication on a 2-D grid • Matrix multiplication is an important application in HPC and appears in many areas (linear algebra) • C = A * B where A, B, and C are matrices (two-dimensional arrays) • A restricted case is when B has only one column, matrix-vector product, which appears in 1957: Minimum degree sparse matrix reordering (Harry Markowitz) 1961: QR algorithm (V. Jan 25, 2021 · We need to extract it using the command tar –zxvf eclipse-committers-photon-R-linux-gtk. 0-m2 m4. Many fields such as Machine Learning and Optimization have adapted their algorithms to handle such clusters. js) and Approach2 (multiply_map_aggregate. 5https://github. MapReduce, from one job to another, takes time to upload the file and start the job again. Basic Matrix Multiplication Operators. When performance or resources are a matter of concern Cubert Script is a developer-friendly language that takes out the hints, guesswork and surprises when running the script. \" Computi (Concurrent Replication-based Matrix Multiplication) along with a parallel algorithm, Marlin, for large-scale matrix multiplication on uted data-parallel platforms, such as Hadoop MapReduce of naively using the general shuffle me Cannon principle parallel matrix multiplication algorithm implementation, and Source: https: //github. function mapReduce(input, map, reduce) {// Map: var mapperOutput = []. Mon 11/14: No recitation : Tue 11/15: srini Experiment MapReduce: Result for Matrix Multiplication 38 MapReduce The speedup of using software-based atomic add over the system one increases as the input matrices get larger (up to 13. 1) Simple Moving Average for Companies Stock Data. e. There are 2 main reasons why interpreted Python code is slower than code in a compiled lanauge such as C (or other compiled langauge): Using MapReduce, we could achieve N times throughput by having N workers running in parallel. Existing GPUbased SOM algorithms take a similar approach in finding the best matching unit, but they do not necessarily use matrix operations to derive the distance matrix [34,46]. ,1. sin ( xs ) # np. Tensor-matrix-multiplication(TTM)referstomultiplyingTalong any mode n, by a matrix of size K ×Ln (for some K). pi , 100 ) ys = np . Importance Sampling for Learning Edge opicsT (ISLE) Microsoft Research, 2000+ LOC [Github] Sep 21, 2020 · The course includes video lectures, case studies, peer-to-peer engagements and use of computational tools and platforms (such as R/RStudio, and Git/Github), and a reproducible research project. Knowing the working of matrix multiplication in a distributed system provides important insights on understanding the cost of our algorithms. Apr 10, 2019 · MRQL (the MapReduce Query Language) is an SQL-like query language for large-scale data analysis on a cluster of computers. sum (axis = 1). Gittens et al. com Li Pu A. Mahoney. The essence of the transformation is to unroll the input patches (a 3D matrix) and lters (a 4D matrix) in 2D in a way that a single matrix-matrix multiplication produces the unrolled version of the output in 2D. 10 of Mahout it became obvious that using Hadoop MapReduce. dblp-2011 3. Map Reduce paradigm is usually used to aggregate data at a large scale. Methods used in the  4 Apr 2019 How to find top-N records using MapReduce Now to do so we just multiply the key with -1 in mapper, so that after sorting higher numbers Other Links: GitHub Repository Matrix Multiplication With 1 MapReduce Step. UPDATE: Please note that the Mapper function does not have access to the Row Number (in this case i, and k) directly. How to do matrix multiplication in R? Machine Learning Recipes,do, matrix, multiplication, r: What is mapply in R? Machine Learning Recipes,what, is, mapply, r: What is sapply in R? Machine Learning Recipes,what, is, sapply, r: How to find lagged differences in R? Machine Learning Recipes,find, lagged, differences, r: How to find variance and MapReduce/Hadoop, where data must be read from disk on eachiteration. Use of MapReduce has flourished since its premier, as illustrated by an in-depth example of its use in WordCount. Aug 02, 2016 · Next we show various attempts for scalable implementation of matrix multiplication using spark, and the winning method which combines numpy matrix multiplication along with spark’s broadcast and MapReduce Word Count Example. 18. ,0. 38) 13KB Synchronous-100x100-Matrix-Multiplication-using-Multiple-Threads:开发了一个程序,用于通过使用多个线程将两个大型 矩阵 Nov 29, 2017 · Generalized matrix multiplication with semiring? Closing since I think this is out of reach of easy contributions. Each cell of the matrix is labelled as Aij and Bij. 7, which is an open source (Apache project) machine learning library. Dec 15, 2014 · This can be an important way to tune performance. Matrix Multiplication With 1 MapReduce Step. ,0. Matrix-multiplication kernels like gemm (dense-dense) and csrmm (sparse-dense), and utility kernels like csrcsc (sparse-transpose), sort (Parallel Sample Sort), and map-reduce are implemented in the framework. Let m be an NxM matrix and v a M vector. Design a MapReduce algorithm to compute matrix multiplication: A x B Mar 27, 2012 · Puzzle: Since there's so much other stuff in the nucleus, like proteins, it is sometimes troublesome finding room for a genome. new array wrappers are not covered, and only one level of wrapping is supported. Under Review: 2017. ∇ x ( p) T J ( i) ( θ) = ( δ ( i) T W) ⋅ 1 { i = p } ∈ R 1 × ( 2 w + 1) d, where the gradient with respect to the ( p, q) -th entry x ( p) q of the matrix x is given by. #MapReduce #MatrixMultiplication Apr 02, 2018 · Matrix Multiplication In MIPS. And more generally, taking a sum of outer products of corresponding rows and columns of the input matrices, always returns the desired matrix multiplication result. Trigger module in display ad project owner & core member Object: the old ad engine built reverse index from tag to ad, and this scheme suffered from long reverse index, the engine needed to do hard pruning when searching which hurt CTR. MPI Programming Matrix Multiplication Operators. matrix (np. 17. Here, the role of Mapper is to map the keys to the existing values and the role of Reducer is to aggregate the keys of common values. Matrix multiplication uses a generated function to return a matrix-product with at most one application of transpose , such that e. N. com/jfkelley/hadoop-matrix-mult After the submatrix multiplication, there's another MapReduce job that simply does the same Then iterate through just those sections of the arrays and multiply elem 9 Mar 2018 Sparse Matrix Matrix Multiplication. 2-CDH3U2 ← Matrix Multiplication using MapReduce – 1 Step Solution A celebrated example of an embarrassingly parallel problem is shared-memory matrix multiplication in which for two n n matrices, the multiplication process is split into n2 tasks, each responsible for calculating one element of the result matrix . Among them: matrix multiplication , similarity join [112,121,14,23], multi-way join using map-reduce offers a scalable Having that said, the ground is prepared for the purpose of this tutorial: writing a Hadoop MapReduce program in a more Pythonic way, i. ones ((3, 3)) + 3) m2 = m1 * (m2 + m1) m4 = 1. Moreover, Pegasus provides fast algorithms for GIM-V in MapReduce, a distributed computing platform multiplication operations . Section 5 will discuss https://github. Matrix Multiplication Map-Reduce. Shi, \"Performance Prediction and Evaluation of a Solution Space Compact Parallel Program using the Steady State Timing Model\", Poster, Future of My personal list of journals I use for my research and projects where I wrote one-sentence summaries. These operations are low-level, but for your convenience wrapped using high-level constructs. We write ++ for concatenation, so \"foo\" ++ \"bar\" is \"foobar\". Sparse matrix-vector multiplication (SpMV) is an important kernel that is considered critical for the performance of compute-intensive applications. Use an algorithm that does not require looking through every single possible matrix. Apr 29, 2013 · Sparse matrix computations in MapReduce!Austin Benson Tall-and-skinny matrix computations in MapReduceTuesday!Joe Buck Extending MapReduce for scientific computing!Chunsheng Feng Large scale video analytics on pivotal HadoopWednesday!Joe Nichols Post-processing CFD dynamics data in MapReduce !Lavanya Ramakrishnan Evaluating MapReduce and combining the matrix multiplication with the MapReduce is only 4 articles that we will discuss and com pare through this paper. Matrix Multiplication using Map-Reduce A specification ixs,iy is a matrix multiplication if ixs consists of two 2-tuples that share one index-label and iy is a permutation of the two-nonshared index-labels of the ixs. MapReduce is an attractive framework because it allows us to decompose the inner products involved in computing document similarity into separate multiplication and summation stages in a way that is well matched to efcient disk access using MapReduce minimize #syncs 8md Ours FPGA acceleration with small on-chip BRAM minimize data transfers (4m+4)d . Using item collaborative filtering algorithm to build co-ocurrence matrix by users' rating towards different movies from Netflix Prize Data Set. Thus, matrix multiplication is the major focus of performance optimization for many big data analytical algorithms or or applications. “Efficient Kernel Management on GPUs”. plot ( xs , ys ); 1 vala: Matrix[Float] //MxN 2 valb: Matrix[Float] //NxP 3 valc = Map(M, P){(i,j) => 4 //OuterMapfunction(f1) 5 Fold(N)(0. However, computing the inverse of a matrix using MapReduce is difficult when the order of the matrix is large. This is the function in C that will be implemented. e. In the following, we give an overview of backend-specific physical matrix multiplication operators in SystemML as well as their internally used matrix multiplication block operations. a ++ (b ++ c) = (a ++ b) ++ c. A matrix is a set of numerical and non-numerical data arranged in a fixed number of rows and column. 2. If you want uniformly spaced vectors, use the : operator. Stack Overflow. Cubert provides a novel sparse matrix multiplication algorithm that is best suited for analytics with large-scale graphs. first_10_even_numbers = 0: 2: 21 first_10_even_numbers = 0 2 4 6 8 10 12 14 16 18 20 A mode-n unfolding refers to the matrix of size Ln ×bL n obtained by arranging the mode-n fibers as the columns in a lexicographic order. What can we do with MapReduce? Models & Algorithms • Communication-processor tradeoffs for 1 round of MapReduce – Upper bounds for database join queries [Afrati,Ullman2010] – Upper and lower bounds for finding triangles, matrix multiplication, finding neighboring strings [Afrati, Sarma, Salihoglu, Ullman 2012] 5 compression idea from the sparse matrix-vector multiplication literature , into the Ligra shared-memory graph processing framework . the result will be the format of mat2. 2nd-row 1st column. Implementing matrix multiplication using MR and optimizing it using Combiner. A MapReduce program is defined via user-specified map and reduce functions, and we will learn how to write such programs in the Apache Hadoop and Spark projects. We propose a novel parallel execution model called pin-and-slide, which implements the column view of the matrix-vector mul-tiplication. ], [ 2. N. smaller/simpler) approximation of the original matrix A. rustlearn - a machine learning framework featuring logistic regression, support vector machines, decision trees and random forests. This code performs matrix vector multiplication using map reduce, which enables fast computations for large amounts of data. 3. At a high level, SVD is an algorithm that decomposes a matrix A into the best lower rank (i. Performance of matrix multiplication and random tensor contractions for rank-k update matrix/tensor shapes on a Xeon E5-2690 v3 processor. An extra MapReduce Job has to be run initially in order to retrieve the values. An extra MapReduce Job has to be run initially in order to add the Row Number as Key to every row. 4. choose the year of your choice and select any one of the data text-file for analyzing. Installed multi node Hadoop 2. Using the best currently known parallel matrix multiplication [Wil12, LG14], our algorithm dynamically maintains the number of k-cliques in O min m 0:469k 235;( + m)0 :469k+0 amor-tized work w. Mathematically, it decomposes A into a two unitary matrices and a diagonal A matrix is a set of numerical and non-numerical data arranged in a fixed number of rows and column. By interpreting the matrix-vector multiplication in the column view, we can restrict the computation to just a subset of the May 21, 2015 · Jeffrey Dean and Sanjay Ghemawat. One of these systems is SystemML , which provides a high-level language for expressing some matrix operations such as matrix multiplication, division, and transpose Keywords: Item Collaborative Filter, Matrix Multiplication, MapReduce, Java, Hadoop - A movie recommender system is built to recommend movies in a similar style to users using raw data from Netflix. Jun 15, 2019 · Map Reduce paradigm is the soul of distributed parallel processing in Big Data. This is because the MapReduce works efficiently only with BIG data. In Proc eed- ings/42nd IEEE Symposium on Fo undations of Computer Science: October 14-17, 2001, Las Use same MapReduce job for operations w/o dependencies sFA Optimizations: Minimize Intermediary Data Recompute X and Y at each job rather than storing and exchanging Dec 27, 2015 · For multiplication, the key is to build rpos[], rpos[i] means in matrix N row i starts at rpos[i] position in datas[]. 01 import systemml as sml import numpy as np m1 = sml. Apr 05, 2019 · The below picture illustrates calculating an image’s class values for all 10 classes in a single step via matrix multiplication. One of the most important topic from university exam point of view. This is still not a complete solution though, e. Matrix-vector multiplication. 0f){k => 6 //Innermapfunction(f2) 7 a(i,k) * b(k,j) 8}{(x,y) => 9 //Combinefunction(r) 10 x + y 11} 12} Figure 1: Example of using Map and Fold in a Scala-based lan-guage for computing an untiled matrix multiplication using in-ner products. • Obtained a user’s rating matrix of films from the Netflix data using the Item Collaborative Filtering Algorithm, then obtained the co-occurrence matrix of the films, and finally merged both matrices to obtain a recommendation list. By-product: large-scale sparse matrix multiplication based on MapReduce. 1 and hadoop with Python 2. using matrix multiplication as example. ones ((3, 3)) + 3) m2 = m1 * (m2 + m1) m4 = 1. Mahout’s linear algebra DSL has an abstraction called DistributedRowMatrix (DRM) which models a matrix that is partitioned by rows and stored in the memory of a cluster of machines. We use 5 for this project, but you may want to increase this number for large datasets. If you like my post - Do follow me on this blog - Matrix Multiplication Using MapReduce Programming In mathematics , matrix mult map-reduce operation, we can perform grouping and aggregat ion, with I and K as the grouping attributes and the sum of V × W as the aggregation. . Each block is sent to each process, and the copied sub blocks are multiplied together and the results added to the partial results in the C sub-blocks. RecSys-2013-ZhuangCJL #matrix #memory management #parallel #performance A fast parallel SGD for matrix factorization in shared memory systems ( YZ , WSC , YCJ , CJL ), pp. For unidirectional (causal) attention, where tokens do not attend to other tokens appearing later in the input sequence, we slightly modify the approach to use prefix-sum computations, which only store running totals of matrix computations rather than mrjob fully supports Amazon’s Elastic MapReduce (EMR) service, which allows you to buy time on a Hadoop cluster on an hourly basis. matrix-matrix multiplication using * matrix-vector multiplication using * element-wise multiplication (Hadamard product) using *. Pseudocode: First Map-Reduce job: Oct 25, 2016 · Mapper For a matrix multiplication of the form AB, we must provide in the mapper, the number of rows of A, referenced as row_a in the code, and the number of columns of B, referenced as col_b (The number of columns of A and number of rows of B are always same, else multiplication won't be possible). Most matrices are sparse so large amount of cells have value zero. Semi-External Memory Sparse Matrix Multiplication on Billion-node Graphs in a Multicore Architecture. Here's a small example to illustrate it. 20. Array operations - slicing, dicing, searching¶ Table of Contents Array operations - slicing, dicing, searchingArray slicingnD array slicingArray dicingArray broadcastingDeep copyArray searchingC •Similar to MapReduce •Aggregate multiple messages to same recipient from same server into a single message •Also executed at the receiver side to save space •Aggregators •Master collects data from vertices at the end of a superstep •Workers aggregate locally and use tree-based structure to aggregate to master This paper presents a MapReduce algorithm for computing pairwise document similarity in large document collections. What do we Outline a Map-Reduce program that calculates the vector x. ICALP-v2-2012-BaldeschiHLS #keyword #multi #on the On Multiple Keyword Sponsored Search Auctions with Budgets ( RCB , MH , SL , MS ), pp. Hence, the resulting product has the same number of rows as A and columns as B. 15 Dec 2014 map-reduce method for multiplying large, sparse matrices using Elasticsearch as As always, the code is available in the GitHub repository. Implementing the matrix multiplication with Map Reduce jobs to find the recommender movie(s). Below picture shows process for transpose: Below shows the process for 2 triple tuple matrix multiplication: check my code on github: link Matrix Multiplication: 2 MapReduce steps! Matrix M can be thought of as a relation with tuples (i, j, m ij) ! Matrix N can be thought of as a relation with tuples (j, k, n jk) ! Map operation creates these tuples ! Map: Join of M and N brings us closer to M X N by creating: Relation (i, j, k, m ij, n jk) or the relation (i, j, k, m ij X n jk) ! Nov 05, 2020 · MapReduce is a programming model and an associated implementation for processing and generating large datasets that is amenable to a broad variety of real-world tasks. Fast monte carlo algorithms for matrices iii: Computing a compressed approximate matrix decomposition, SIAM Journal of Computing, 2005. As the usual dense GEMM, the computation partitions the output matrix into tiles. Map-Reduce for each vertex D B A C Block Matrix Multiplication Let’s look at Block Matrix Multiplication (on the board and on GitHub) Experiment MapReduce: Result for Matrix Multiplication 38 MapReduce The speedup of using software-based atomic add over the system one increases as the input matrices get larger (up to 13. The ‘Iterative’ in the name of GIM-V denotes that we apply the £G opera- based on fast matrix multiplication. Blue line shows the 0. options. Graph reachability, using Matrix-Matrix multiplication of adjacency matrices. MapReduce is a processing technique and a program model for distributed computing based on java. The Map-reduce programming model is a common data-handling model Array-based distributed computations are another abstraction, used in all forms of parallelism. Graphs, and Matrix Multiplication using MapReduce, Spark, and MASS. Feb 01, 2013 · We extended the MapReduce SOM algorithm by moving all calculations on local nodes to the GPU with the matrix-based Euclidean distance matrix and reduction algorithm described above. A discussion about each of this operations is available below, if you want to go ahead of this section, click here. 39 is a matrix of residuals, assuming: s i. data. In the following, we give an overview of backend-specific physical matrix multiplication operators in SystemML as well as their internally used matrix multiplication block operations. Matrix Multiplication performed using Hadoop. Implementing matrix multiplication using MR and optimizing it using Combiner. . What we want to do We will write a simple MapReduce program (see also the MapReduce article on Wikipedia ) for Hadoop in Python but without using Jython to translate our Around version 0. UPDATE: Please note that the Mapper function does not have access to the Row Number (in this case i, and k) directly. A MapReduce job can be enhanced by sampling local data, which cannot be used for future analysis. During the vector-matrix multiplication, each node will use its portion of the updated u vector, then estimate the v vector based on the multiplication of its putations as a single matrix-matrix multiplication (GEMM). Kublanovskaya, J. The RMM plan in Figure 1 im-plements a replication-based strategy in a single MapReduce version of matrix multiplication using recursive block matrix decomposition View MapReduceBlockMatrixProduct. * Distributed engine neutral allreduceBlock() operator api for Spark and H2O. Unfortunately there is no acceleration routine for integers. Also, due to the inherent complexity of high-order computation, experience from prior work in other fields can not easily be applied to tensors. 8 Weeks. webbase-2001 Describe how you stored the connectivity matrix on disk and how you computed the transition matrix. Lecture 16 (5/28): Complexity Measures for MapReduce, Triangle Counting in a Graph The sum of these outer products matches the result we obtained in the previous slide using the traditional definition of matrix multiplication. concat. We show that serverless big data processing can lower oper- We also consider distributed MapReduce computations for training clustering models such as k-means and collaborative filtering models based on matrix factorization. 1. TensorFlow. In this case, each task needs one row of the first input matrix and one Map Reduce Reduce. A breakdown of basic MapReduce terms and functions follows. matrix (np. 3 and use Different Algorithmic Techniques to Solve Following problems using Hadoop Map-Reduce. com/nm4archana/BigDataAnalysis-Comet. (For example Facebook) 4) K-mers counting using long DNA sequence in FASTA format 5) installing putations as a single matrix-matrix multiplication (GEMM). split (\"\\t\") index, value = map (int, [index,value]) if curr_index == prev_index: value_list. Integer factorization must be adaptable to MapReduce and must be parallizable (more than 1 Map or Reduce task) The emergence of large distributed clusters of commodity machines has brought with it a slew of new algorithms and tools. Finally the basic process of MapReduce is shown. You can still represent them using linear models. Irregular algorithms, however, depend on the input. I have published LSH package I developed for Apache Spark on GitHub. g. Diverse types of matrix classes/matrix multiplication are accommodated. MapReduce in distributed model training Using the MapReduce strategy, if we can split the training data on separate workers, compute Map functions in parallel, and aggregate the results in a Reduce function, we will be able to achieve distributed model Github. a mapreduce program of matrix multiplication. Green = true positive male, yellow = true positive female, red halo = misclassification. The ultimate goal is to make the algorithm as e cient as possible for any input . By using Monoids, we can take advantage of sparsity (we deal with a lot of sparse matrices, where almost all values are a zero in some Monoid). The operation is denoted Z = T×n A. matrix (np. Report this profile Addition of numbers, matrix multiplication inside a docker and using MapReduce as the programming model. Map the input matrices line by line and emits the matrix element. ee) Performance Prediction of Sparse Matrix Multiplication on a Distributed BigData Processing Environment. The sum of these outer products matches the result we obtained in the previous slide using the traditional definition of matrix multiplication. A naïve approach would be to extend an efficient matrix multiplication algorithm, replacing the dot product by the distance function. xs = np . Contribute to ashkang/cs222_fin development by creating an account on GitHub. 99 “ Joel takes you on a journey from being data-curious to getting a thorough understanding Improving Quantum Query Complexity of Boolean Matrix Multiplication Using Graph Collision (SJ, RK, FM), pp. [J14] Yun Liang, Xiuhong Li. . Shi, \"How to achieve a 47000x speed up on the GPU/CUDA using matrix multiplication,\" Technical Report, Amax corporation, June 2009. Use a reducer to multiply value for same indices. His current research topic covers big data platforms, large-scale distributed computing resource manangement, cloud computing, and peer-to-peer systems. dot(a,b) array([[ 0. You can generate uniformly spaced vectors, using 2 methods. pdf such as MapReduce [Srirama et al, FGCS 2012] • Designed a classification on how the algorithms can be adapted to MR – Algorithm single MapReduce job • Monte Carlo, RSA breaking – Algorithm nMapReduce jobs • CLARA (Clustering), Matrix Multiplication – Each iteration in algorithm single MapReduce job • PAM (Clustering) Aug 29, 2015 · Posts about Machine Learning written by Maruf Aytekin. jl package. Ofeishat . Optimized Matrix Multiplication using Shared Virtual Memory In OpenCL 2. MATRIX MULTIPLICATION: Project to Read a file containing two 3 X 3 Matrices and calculate their Vector Product. 3 were implemented by their corresponding distributed primitives in Spark: I. The A sub-blocks are rolled one step to the left and the B Representing non-linearity using Polynomial Regression¶ Sometimes, when you plot the response variable with one of the predictors, it may not take a linear form. GitHub Gist: instantly share code, notes, and snippets. 249–256. Marlin contains several distributed matrix operations and especially focuses on matrix multiplication which is a funda-mental kernel of high performance scientific computing. test_naive. The matrix is denoted as T (n). straightforward, but the fact that matrix multiplication itself can be accomplished in multiple ways complicates matters. Individualpair-wiseremotememory distributed matrix multiplication Dec 29, 2017 · These techniques first partition matrix X into blocks and then exploit the block-matrix multiplication when learning U and V. Based on the observa-tion, Pegasus implements a very important primitive called GIM-V (Generalized Iterated Matrix-Vector multiplication) which is a generalization of the plain matrix-vector multi-plication. 20. A distributed, MapReduce-based SOM also builds on the batch formulation described in Equation 2 . . Amatname in fn: return 1 else: return 2 def joinmap(self, key, line): mtype = self. 281–284. io/Bulk. The chunk of X assigned to the node and the corresponding norms of X are kept in the GPU memory between subsequent epochs, and the weight vectors are copied to the 'sparse' is a matrix class based on a dictionary to store data using 2-element tuples (i,j) as keys (i is the row and j the column index). 1. the FFT, LU, and dense matrix–matrix multiplication. In this part of the code, the matrix multiplication begins, again, by implementing three MapReduce jobs. Policy: Printed material is allowed. i. There was a mix of similar problems presented in assignment 1 and assignment 2 that we had to solve them using MapReduce paradigm. Sep 10, 2012 · cost of the algorithm• determined by the amount of data that has to be sent over the network in the matrix multiplication step• for each user, we have to process the square of the number of his interactions → cost is dominated by the densest rows of A• distribution of interactions per user is usually heavy tailed → small number of Github. You have seen many of the stream operations before, in Question 5 of Exercise 7, including map, reduce, filter, and forEach • Matrix multiplication • Dynamic programming External Memory Model • Addition and subtraction are fast, multiplication is fast • MapReduce model: peated matrix-vector multiplications. 249–256. Designing e cient irregular algorithms is a challenge. 6. y * dz # [dz/dx * dL/dz] dy = self . push({key: key, value: value});}; map(row, emit); return emitArray;})); // Group tuples with the same key: var reducerInput = {}; mapperOutput. This was Jul 30, 2013 · MapReduce Algorithms: Having presented the video lectures on the topic, in which Prof. Using functions from various compiled languages in Python¶. ▫ SPGEMM: KKTRI: Triangle Counting using SpGEMM. Dec 14, 2016 · Data-Intensive Computing and MapReduce/Hadoop : For more info, see the MapReduce paper, it's pretty readable. js First of all lets recap the multiplication formula: import arraymancer proc customSigmoid2[T: SomeFloat](t: Tensor[T]): Tensor[T] = result = map_inline(t): 1 / (1 + exp(-x)) Now in a single loop over t, Arraymancer will do 1 / (1 + exp (-x)) for each x found. So, the whitespace or the indentation of the very first line of the program must be maintained all throughout the code. What we want to do We will write a simple MapReduce program (see also Wikipedia ) for Hadoop in Python but without using Jython to translate our code to Java jar files. Video: Youtube Map-Reduce! Ranking (e. MapReduce in distributed model training Using the MapReduce strategy, if we can split the training data on separate workers, compute Map functions in parallel, and aggregate the results in a Reduce function, we will be able to achieve distributed model We can use the logistic regression results to classify subjects as male or female based on their height and weight, using 0. Using the old API in the Mapper and Reducer. 281–284. Length. The MRQL query language is powerful enough to express most common data analysis tasks over many forms of raw in-situ data, such as XML and JSON documents, binary files, and CSV documents. js) To run the code mongo < multiply_map_reduce. txt. g. Mimir inherits the core principles of existing MapReduce frameworks, such as MR-MPI, while redesigning the execution model to incorporate a number of sophisticated optimization techniques that achieve similar or better performance with significant reduction in the amount of memory used. Thu 11/10: yuvraj: Virtual Machines : See also the book chapter on Virtual Machines from the Wisconsin OS book. Figure 4 illustrates how output splitting a ects weight and Using the Maven POM to manage your project is an easy way to start. BSPmodel. rithms are often iterative, using repeated read-only data access and I/O-bound matrix-vector multiplications to converge to an optimal model. requests in parallel by using the underlying multiple flash memory packages. Subtypes of StaticArray will provide fast implementations of common array and linear algebra operations. With both the item-item similarity matrix and the user-item vectors, it’s now possible to multiply them together and generate recommendations for users. M. inline. MatrixMulOutput. NOTE: Please note that the Mapper function does not have access to the i, j, and k values directly. The MRQL query processing system can evaluate MRQL queries in two modes: in MapReduce mode on top of Apache Hadoop or in Bulk Synchronous Parallel (BSP) mode on top of Apache Hama. For example: MapReduce workflow: InputFormat, RecordReader, InputSplits, Map tasks, Combiners, Shuffle/Sort, Reduce tasks, OutputFormat. 55 folds) Ratio of FastPath to CompletePath memory accesses: 30:0 for software-based atomic and 3:28 for system-provided atomic implementations Since multiplication is done element-wise, you need to specifically perform a dot product to perform matrix multiplication. sum(B[i, :] * C[:, j]) Mapreduce and matrix multiplication November 2, 2016 The homework questions are due at the 23:59 on Tuesday 15 November. Specifically, the matrix multiplication operations described in Eq. Mar 12, 2018 · GitHub. 2. 0 2D card game 2048 (Android Studio, Java) *Solely developed android game on windows platform using android studio. This method, however, is very inefficient as it would require to compute a matrix multiplication and the square root of a matrix at each step. parseLong(job. If you want a specific number of elements within a range, then use the linspace function. You can check it out from here. It is used to solve problems where problem of size N is solved using solution of problems of size N - 1 (or smaller). An extra MapReduce Job has to be run initially in order to add the Row Number as Key to every row. Distributed matrix factorization with mapreduce using a series of broadcast-joins (SS, CB, MS, AA, VM), pp. Input are two matrix A and B python - How to write Mapreduce code for matrix multiplication which does not use any list of size more than 10 - Stack Overflow. One method for computing Pi (even though not the most efficient) generates a number of points in a square with side = 2. But MapReduce tries to use commodity machines to solve big data problems. js for ML using JavaScript TensorFlow 1 version, View source on GitHub the inner 2 dimensions specify valid matrix multiplication dimensions, and any level matrix computation primitives with MapReduce through the case study two basic primitives, matrix multiplication and finding linear solution, and goes into  . 2. F ast monte-carlo algorithms for appr oximate matrix multiplication. ]]) >>>a[0,0]=1 >>>a[1,1]=1 >>>b =np. file') if self. StaticArrays provides a framework for implementing statically sized arrays in Julia, using the abstract type StaticArray{Size,T,N} <: AbstractArray{T,N}. The algorithm we’ll be using is a two-pass matrix multiplication algorithm for MapReduce. , PageRank) requires iterated matrix-vector multiplication with matrix containing millions of rows and columns ! Computing with social networks involves graphs with hundreds of millions of nodes and billions of edges ! Map-Reduce is a parallel programming paradigm, a software-stack that will help to address big data Used Blocked Matrix Multiplication technique to improve the convergence rate,by performing the nontrivial computation in the reduce steps. It is deployed on expensive hardware such as HPC or supercomputers. Our idea is to speed up distributed NMF in a new, orthogonal direction: by reducing the problem size of each NLS subproblem within NMF, which in turn decreases the overall computation cost. Please cite any references you use. mrjob has basic support for Google Cloud Dataproc (Dataproc) which allows you to buy time on a Hadoop cluster on a minute-by-minute basis. split()] row = int(vals) A key feature is the capability for users to write callback functions, called after each iteration, thus enabling customization for specific applications. The general algorithm matrix-vector multiplication (SpMV) and sparse matrix-matrix multiplication (SpGEMM), a systematic study on applying blocking techniques to tensors has not yet been conducted. List the top-10 vertices for graphs 1,2 & 4. Difference between MapReduce and Pig. Assume you have two matrices A and B in a sparse matrix format, where each record is of the form i, j, value. linspace ( 0 , 2 * np . Do you have any idea, about the matrix multiplication example which I mentioned in question, that why this works fine with hadoop standalone mode but does not work with hadoop distributed mode at the point of checking answers? – waqas Nov 30 '11 at 13:39 Matrix multiplication using MPI. Matrix-vector and matrix-matrix calculations fit nicely into the MapReduce style of computing. Big Data Project On A Commodity Search System For Online Shopping Using Web Mining Big Data Project On A data mining framework to analyze road accident data Big Data Project On A neuro-fuzzy agent based group decision HR system for candidate ranking Big Data Project On A Profile-Based Big Data Architecture for Agricultural Context Big Data Project On A Queuing Method for GSoC Results and Summary. Thus, Meta-MapReduce enhances the standard MapReduce and can be implemented into the state-of-the-art MapReduce systems, such as Spark, Pregel , or modern Hadoop. $docker run -v Matrix multiplication is an important application in. In MapReduce word count example, we find out the frequency of each word. You might want an order 2 or 3 curve. General-purpose, heavy- and An example of how multi-method-based dispatch might work for a binary operation like matrix multiplication. 1. 6. xs = np . If the multiplication type computes in parallel, then the package computation is also parallel. 0-m2 m4. e. 快速矩阵乘法 Fast and Stable matrix multiplication Coppersmith and Winograd's Algorithm 时间复杂度O(n^2. ∙ 0 ∙ share While performing distributed computations in today's cloud-based platforms, execution speed variations among compute nodes can significantly reduce the performance and create bottlenecks like stragglers. Stage 2. Naive Bayes classifier to classify text documents Latent semantic analysis (LSA) Rsa breaking using a more efficient integer factorization than trial division. parsemat() vals = [float(v) for v in line. 1. I got it right. MapReduce. PRELIMINARY A. Feb 16, 2015 · Matrix Data Model for MapReduce. ones ((3, 3)) + 2) m2 = sml. There are two main security concerns in outsourcing computation: guaranteeing that the server performs the computation correctly, and protecting the privacy of the client’s data. The first is Map Only with CAP3 DNA Sequence Assembly, followed by Classic MapReduce with Pair-wise Sequences and High-Energy Physics, Iterative with K-means clustering, PageRank and Multi-dimensional Scaling, and finally Loosely Synchronous with Matrix Multiplication Algorithms. Sparse Matrix Multiplication in Map Reduce. CS231n, Convolutional Neural Networks for Visual Recognition, Stanford University; CS224d, Deep Learning for Natural Language Processing, Stanford University You can now run all stages of the rewrite system for a program (in this example for matrix multiplication): scripts / compiled_scripts / HighLevelRewrite highLevel / mmTransposedA scripts / compiled_scripts / MemoryMappingRewrite -- gr10 mmTransposedA scripts / compiled_scripts / ParameterRewrite - f highLevel / mm . MPI is a SPMD model of distributed computing, where each process is completely independent and one just controls the memory handling. ,3. Aug 25, 2017 · Matrix Multiplication using MapReduce Programming in Java. json mmTransposedA Jul 14, 2020 · MapReduce Program – Finding The Average Age of Male and Female Died in Titanic Disaster; MapReduce – Understanding With Real-Life Example; How to find top-N records using MapReduce; How to Execute WordCount Program in MapReduce using Cloudera Distribution Hadoop(CDH) Matrix Multiplication With 1 MapReduce Step; MapReduce – Combiners matrix multiplication operations within the NMF algorithms. An extra MapReduce Job has to be run initially in order to add the Row Number as Key to every row. It uses a distributed file system called GFS, which is Google File System. Matrix Multiplication with Spark. ], [ 0. Plagiarism checking with MapReduce Manages: Pelle Jakovits ([email protected] 22 706. ####MapReduce problems#### Count number of words in a book; Total number of words in a book; tf-idf calculations for ranking; Cosine distance measure between 2 documents; Matrix multiplication in MapReduce Aug 13, 2016 · Matrix Computations and Optimization in Apache Spark Reza Bosagh Zadeh Stanford and Matroid 475 Via Ortega Stanford, CA 94305 Xiangrui Meng Databricks 160 Spear Street, 13th Floor San Francisco, CA 94105 Alexander Ulanov HP Labs 1501 Page Mill Rd Palo Alto, CA 94304 [email protected] zeros(4). e. Jueon Park, and Kyungyong Lee, The 8th International Workshop on Autonomic Management of high performance Grid and Cloud Computing (AMGCC'20), Accepted, 07/2020. OSDI’04 ,San Francisco, CA; Petros Drineas, Ravi Kannan, and Michael W. edu Burak Yavuz Databricks 160 Spear Street, 13th Floor San Francisco, CA 94105 [email protected] the code works perfectly fine with smaller matrices but when the files becomes large the mapping p Nov 12, 2020 · Matrix B is also a 2×2 matrix where number of rows(j)=2 and number of columns(k)=2. Data Analysis of huge amount of open Github Data, where we tried to find some deep patterns among popularity and spatial distributions of programming languages and users on Github. Moreover, your code has two steps, that means two jobs. Course content Because using map is equivalent to for loops, with an extra code we can always write a general mapping utility: >>> def mymap(aFunc, aSeq): result = [] for x in aSeq: result. It may be possible to switch between MapReduce and MPI to perform scalable matrix inversion in these The next step is PageRank, a fairly straightforward linear algebra problem. Lecture 15 (5/26): Partitioning for PageRank Lecture 15, Partitioning for Pagerank. While some arrays — like Array itself — are implemented using a linear chunk of memory and directly use a linear index in their implementations, other arrays — like Diagonal — need the full set of cartesian indices to do their lookup (see IndexStyle to Feb 08, 2010 · This posting gives an example of how to use Mapreduce, Python and Numpy to parallelize a linear machine learning classifier algorithm for Hadoop Streaming. get(\"NumberOfDocuments\"));} The variable N can then be used with the map and reduce functions. 5 × 10 5 and the number of nonzero elements equal to 6 × 10 5. By using Rings, we can do matrix multiplication over things other than numbers (which on occasion we have done). 1. sin ( xs ) # np. We use dense() to create a dense in-memory matrix from our toy dataset and use drmParallelize to load it into the cluster, “mimicking” a large, partitioned • Used Netflix data to offer movie recommendations to users based on their previous favorites. , 2015. CS231n, Convolutional Neural Networks for Visual Recognition, Stanford University; CS224d, Deep Learning for Natural Language Processing, Stanford University Publication: M. Syntax of Mongo mapReduce() Following is the syntax of mapReduce() function that could be used in Mongo Shell > db. Figure 4 illustrates how output splitting a�ects weight and In modern processors, integer division can be 10-50 times slower than multiplication. AWS Elastic MapReduce, an adaptation of Apache Hadoop to Matrix Multiplication: 2 MapReduce steps! Matrix M can be thought of as a relation with tuples (i, j, m ij) ! Matrix N can be thought of as a relation with tuples (j, k, n jk) ! Map operation creates these tuples ! Map: Join of M and N brings us closer to M X N by creating: Relation (i, j, k, m ij, n jk) or the relation (i, j, k, m ij X n jk) ! Jul 26, 2016 · Outline • Introduction • DFS • MapReduce • Examples • Matrix Calculation on Hadoop 3. Advices [edit | edit source] In this section, you will see some advices that can help you to desing a Python source Refactor compression package and add functionalities including quantization for lossy compression, binary cell operations, left matrix multiplication. For sparse computations, they usually depend on the nonzero pattern of the matrix. per batch of updates where mis de-fined as the maximum number of edges in the graph be- Jun 15, 2013 · For those who prefer reading a code instead of text - GitHub: Approach1(multiply_map_reduce. For example : mat1 is 2×3 means mat2 will be 3×2. Implement SELECT MAX(<field>) FROM <table> GROUP BY <field> with MapReduce. 8 May 2015 https://github. . In all cases, Twister outperforms or is close to the competition. Implementations in CUDA Sep 2015 – Sep 2015 2. 7. . Create a matrix of processes of size p1/2 1/2 x p so that each process can maintain a block of A matrix and a block of B matrix. X-rays) are sent through an object from various angles Google MapReduce. Combiner Edit the “MapTask” method to add support for running a Combiner. The repository provides demo programs for implementations of basic algorithms on Spark 2. h. apply([], input. N(0;˙2). Many applications in different areas exist already for MapReduce. pi , 100 ) ys = np . Matrix Multiplication with MapReduce Big Data possibly now has become the most used term in the tech world for this decade. 95% c Thanks Thomas. Mark Kröll 击 see https://hadoopecosystemtable. 7. in a way you should be familiar with. Android4. Mo 09 Dezember 2013 How to use Jekyll with GitHub ; Di 05 Juli 2016 Pythons map, reduce and filter as list Matrix multiplication on multiple cores in Jun 25, 2012 · Translating to MapReduce: rethinking matrix multiplication It’s now possible to use MapReduce to multiply the user vectors computed in step 1, and the co-occurrence matrix from step 2, to produce a recommendation vector from which the algorithm can derive recommendations. Matrix-Vector multiplication As an example if we consider a Matrix-Vector multiplication (taken from the book Mining Massive Data Sets by Jure Leskovec, Anand Rajaraman et al To store the past gradients, we will use a matrix G. International Journal of Computer Science and Information Technology (IJCSIT) 9 (5): 29 - 37 ( October 2017 Figure 1 shows the general matrix multiplication (GEMM) operation by using the block sparse format. Look for “# [ADD COMBINER HERE]” for the place one would add this. 2 minute read. the rule of matrix multiplication is mat1 columns is equal to mat2 rows values. It may help if you checkout my introduction to map-reduce and an example here. To do so, we are taking input from the user for row number, column number, first matrix elements and second matrix elements. Contribute to JaredP94/MapReduce-Matrix-Multiplication development by creating an account on GitHub. GitHub Gist: instantly share code, notes, and snippets. MapReduce is a programming model that Google came up with to handle computation of their huge large scale data. The examples below use the einsum / notation for the elements of tensors, namely m[i,j] for element i,j of the matrix m, instead of the more mathematical notation m_ij. #!/usr/bin/env python import sys from operator import itemgetter prev_index = None value_list = [] for line in sys. Dijkstra's algorithm. GPU Accelerated Computing with C and C++, which also has some videos. Combiner Edit the “MapTask” method to add support for running a Sparse matrix multiplication for hadoop. Matrix Multiplication Operators. to implement matrix inversion using other parallelization platforms such as MPI, a MapReduce matrix inversion technique that can be used as a pluggable component in complex Hadoop data analysis workflows is highly desirable. The instruction sets are typed, and instructions designed to operate on packed doubles can’t operate on packed ints without explicit casting. (Duh!) Map takes data and converts it into another set of data, where individual elements are broken down into tuples (key/value pairs). rusty-machine - a pure-rust machine learning library. - Use MapReduce to calculate tag similarity in twitter and improves the speed from 90min to 36min. e. 0 and measured the performance of the same with previous implementations. 5 In mathematics, matrix multiplication or the matrix product is a binary operation that produces a matrix from two matrices. ]]) Matrix Multiplication Examples (both using global memory and shared memory) CUDA C Programming Guide; CUDA Toolkit documentation, which includes CUDA installation, C programming guide, APIs for cuBlas, cuFFT etc, tools, compiler SDK, and others. “Limitations and Challenges of HDFS and MapReduce” by Weets et al. 10 of Mahout it became obvious that using Hadoop MapReduce was causing more pain than it was solving, due to massively redundant data reads required 1 Problems Suited for MapReduce 2 MapReduce: Applications Matrix-Vector Multiplication Information Retrieval 3 Hadoop Ecosystem Designing a Big Data System Big Data Storage Technologies Slides are partially based on Slides “Mining Massive Datasets” by Jure Leskovec. What inference can you derive from using PageRank on the these datasets. The minimum is found by a multi-step reduction algorithm. 10 Matrix Multiplication with One MapReduce Step . 6. sin is a universal function plt . Jan 17, 2021 · Matrix Multiplication With 1 MapReduce Step; Hadoop - copyFromLocal Command go to this GitHub Repo and download the receptacle organizer as a speed as For advanced use of the CUDA, you can use the driver API wrappers in CUDA. 1. per batch of updates where mis de-fined as the maximum number of edges in the graph be- Matrix multiplication: n^3/p + (n^2/p^{2/3}) \\cdot g + l: Sorting (n \\log n)/p + (n/p)\\cdot g + l: Fast Fourier Transform (n \\log n)/p + (n/p)\\cdot g + l: LU Decomposition: n^3/p + (n^2/p^{1/2})\\cdot g + p^{1/2}\\cdot l: Cholesky Factorisation: n^3/p + (n^2/p^{1/2})\\cdot g + p^{1/2}\\cdot l: Algebraic Path Problem (Shortest Paths) Distributed file systems and map-reduce as a tool for creating parallel 2. with callbacks. Assume you have two matrices A and B in a sparse matrix format, where each record is of the form i, j, value. For matrix computation library built on top of Spark which is an distributed in-memory cluster computing framework. Ex. Vogelstein, Carey E. reshape(2,2) >>>b array([[0, 1], [2, 3]]) >>>a*b array([[ 0. Using the new API in the Mapper and Reducer. Remember, a combiner runs the reduce task at the end of the map task in order to save communication cost of sending to multiple reducers. tar. Specifically, for building MapReduce jobs, you only need to have the hadoop-client dependency, which contains all the Hadoop client-side classes needed to interact with HDFS and MapReduce. The goal is to calculate A * B. [experimental] New python bindings with supports for several builtin s, matrix operations, federated tensors and lineage traces. Lowering XLA HLO to I E 6 func @mnist_predict(%input: tensor<1x28x28x1xf32>) /> tensor<1x10xf32> {%1 = mhlo. hollywood-2011 2. PageRank) Gradient descent methods Stochastic SVD Tall skinny QR This paper presents a MapReduce algorithm for computing pairwise document similarity in large document collections. I couldn't find a simple way to do this within the EMR framework, though I bet there is a way to do it. import systemml as sml import numpy as np m1 = sml. On the left are the full matrix organized in blocks and its internal memory representation: compressed values and block indices. The essence of the transformation is to unroll the input patches (a 3D matrix) and �lters (a 4D matrix) in 2D in a way that a single matrix-matrix multiplication produces the unrolled version of the output in 2D. Common operations include synchronizing the GPU, inspecting its properties, starting the profiler, etc. Sparse Matrix-Vector Multiplication { Size of Distributed Matrix Multiplication Using Speed Adaptive Coding 04/15/2019 ∙ by Krishna Narra , et al. Da Zheng, Disa Mhembere, Joshua T. Figure 1 and Figure 2 show two alternative MapReduce plans for matrix multiplication (details of the two plans will be discussed in Section IV). MapReduce is an attractive framework because it allows us to decompose the inner products involved in computing document similarity into separate multiplication and summation stages in a way that is well matched to efcient disk access May 12, 2018 · On Intel CPUs, SSE instruction sets use up to 128 bit registers (xmm, four ints), AVX and AVX2 use up to 256 bit registers (ymm, eight ints), and AVX512 use up to 512 bit registers (zmm, sixteen ints). 1957: Minimum degree sparse matrix reordering (Harry Markowitz) 1961: QR algorithm (V. Curtis Huttenhower, John Quackenbush, Lorenzo Trippa & Christine Choirat. Thus, r u v ≈ x T u ⋅ θ v, where x u, θ v ∈ R f are the u t h Map-Reduce also makes short work of dealing with large matrices and can crunch matrix operations like matrix addition, subtraction, multiplication etc. 2 and the deflation operation defined in Eq. Implement inner join between two tables with MapReduce. Design a MapReduce algorithm to compute matrix multiplication: A x B Matrix Multiplication. The common matrix operations such as 'dot' for the inner product, multiplication/division by a scalar, indexing/slicing, etc. forEach(function(keyValue) Example matrix multiplication in distributed environment using R, MPI and Hadoop MapReduce - aaparo/MultMatrix Use Git or checkout with SVN using the web URL. And more generally, taking a sum of outer products of corresponding rows and columns of the input matrices, always returns the desired matrix multiplication result. No electronic device (except for electronic calculator). The class will cover widely used distributed algorithms in academia Dynamic programming is well known algorithm design method. Matthews Author content It can be used in conjunction with other functionality like Map, Reduce, Filter in Python. append(aFunc(x)) return result >>> list(map(sqr, [1, 2, 3])) [1, 4, 9] >>> mymap(sqr, [1, 2, 3]) [1, 4, 9] >>> Aug 25, 2011 · In-Database Operations • Matrix and vector multiplication: Av SELECT 1, array_accum(row_number, vector*v) FROM A array_accum(x,v) is a custom function. Both of these are considered to be whitespaces when you code. Figures - uploaded by Devin A. Current Apache Mahout: Beyond MapReduce Matrix Multiplication. Note: Matrix operations for floats are accelerated using BLAS (Intel MKL, OpenBLAS, Apple Accelerate …). ,0. 1. Ligra+ is able to represent a variety of synthetic and real-world graphs using General Matrix-Vector multiplication: y <- alpha * A * x + beta * y Source Edit proc gemv [T: SomeInteger] (alpha: T; A: Tensor [T]; x: Tensor [T]; beta: T; y: var Tensor [T]) { } {. Clearly it holds that. 11 Jan 2009 Okay, so how can we compute PageRank using MapReduce? we'll take is to use MapReduce to repeatedly multiply a vector by the matrix M 23 Feb 2021 Multiplies matrix a by matrix b, producing a * b. Say the co-occurrence matrix for 4 items is Courses ¶. It also discusses various hadoop/mapreduce-specific approaches how to potentially improve or extend the example. MapReduce¶ MapReduce was designed by Google to address the problem of large-scale data processing. append ( (index,value)) else: if prev_index: value_list = sorted (value_list,key=itemgetter (0)) i In this video u will learn about Matrix Multiplication using Map Reduce in Big-Data. p. The 27 Jun 2020 Submit results from this paper to get state-of-the-art GitHub badges and help the community compare results to other papers. Francis) 1964: Sinkhorn (Richard Sinkhorn) 1965: Golub-Reinsch SVD (Gene Golub) 1969: Sparse matrix ordering (Elizabeth Cuthill, James McKee) 1969: Strassen matrix multiplication (Volker Strassen) INTRODUCTION TO DATA SCIENCE JOHN P DICKERSON Lecture #4 –9/5/2019 CMSC320 Tuesdays & Thursdays 5:00pm –6:15pm Graph reachability, using Matrix-Matrix multiplication of adjacency matrices. Download [edit | edit source] The source code is available on Github. One purpose of matrix decomposition is reducing calculations cost while solving a system of linear equations by decomposing the coefficients matrix into a product of two triangular matrices. MapReduce • Programming Model for Large-Volume Data Processing • Specialized for frequent use case: aggregation queries – Map every input object to set of key/value pairs – Reduce (aggregate) all mapped values for same key into one result for that key • Use this structure as explicit API for cluster computing May 01, 2018 · You fit this matrix to approximate your original matrix, as closely as possible, by multiplying the low-rank matrices together, which fills in the entries missing in the original matrix. Using MapReduce, we could achieve N times throughput by having N workers running in parallel. Here, we will discuss the implementation of matrix multiplication on various communication networks like mesh and hypercube. 99 CAN$45. Taifi and Y. 1. Qatawneh , and H. Now One step matrix multiplication has 1 mapper and 1 reducer. In a nutshell, we reduce the size of each NLS subproblem, by employing a matrix sketching technique: Having that said, the ground is prepared for the purpose of this tutorial: writing a Hadoop MapReduce program in a more Pythonic way, i. Aug 02, 2016 · Next we show various attempts for scalable implementation of matrix multiplication using spark, and the winning method which combines numpy matrix multiplication along with spark’s broadcast and Courses ¶. View on GitHub Spark. Challenge: Make linear https://github. ones ((3, 3)) + 2) m2 = sml. Published: December 08, 2018 Hi everyone, this is the final (summarizing) blog post for my Google Summer of Code project. the FFT, LU, and dense matrix matrix multiplication. For both classes, few matrix operations dominate the overall algorithm runtime, apart from the costs for the initial read from distributed le system or object storage. F. In fact, large-scale matrix multiplication can hardly be handled by the single-node matrix computation libraries due to hardware resource limitation. The unit of Parallel Matrix Multiplication on Open MPI. github. Fortunately, the authors in proposed a method of matrix inversion using MapReduce. js mongo < multiply_map_aggregate. 26 >>>a =np. Straggler Robust Distributed Matrix Inverse Approximation 03/05/2020 ∙ by Neophytos Charalambides , et al. ], [ 0. The reduce( ) step in the MapReduce Algorithm for matrix multiplication Facts: The final step in the MapReduce algorithm is to produce the matrix A × B. In this example did the matrix multiplication. git. There are many parallel computation  prototype for matrix multiplication using FaaS, and discuss and synthesize insights entry-barriers of. This workload tests the Naive Bayesian (a popular classification algorithm for knowledge discovery and data mining) trainer in Mahout 0. However, I want to tell you something, if your file is not big enough, you will not see an improvement in term of execution speed. 3. Besides matrix-vector multi-plication (e. There are Python 2. map(function(row) { var emitArray = []; var emit = function(key, value) {emitArray. 7 codes and learning notes for Spark 2. In SpMV, the optimal selection of storage format is one of the key aspects of enabling the best performance. source The code is available on my GitHub account of the converting a collection to stream using stream method; reading from a file using Files. the result is the same as mat2. Use of ufuncs is an esssential aspect of vectorization and typically much more computtionally efficient than using an explicit loop over each element. Dec 29, 2017 · Background Matrix factorization is a well established pattern discovery tool that has seen numerous applications in biomedical data analytics, such as gene expression co-clustering, patient stratification, and gene-disease association mining. Using the best currently known parallel matrix multiplication [Wil12, LG14], our algorithm dynamically maintains the number of k-cliques in O min m 0:469k 235;( + m)0 :469k+0 amor-tized work w. One, you can multiply BIG matrices in a memory efficient way, without needing to pull everything out of SQL. Our extended framework, which we call Ligra+, uses less space than Ligra, while providing comparable or improved performance. 2-CDH3U2 ← Matrix Multiplication using MapReduce – 2 Step Solution A Movie recommender system using Netflix movie data and currently on the stage of achieving matrix multiplication, use the item-collaborative algorithm and Hadoop MapReduce Auto Complete Apr 2018 traditional matrix-vector multiplication requires): 1. Then matrix-vector multiplication m * v is defined as: w[i] = sum_j m[i,j] * v[j]. The verifiable computation of Gennaro, Gentry and Parno addresses both concerns for Mimir: Mimir is a new implementation of MapReduce over MPI. Nov 20, 2020 · Matrices represented using COO format Matrix Multiplication Using Two Passes. Use Git or checkout with SVN using the web URL. This can be parallelized easily (just matrix-vector multiplication) but needs to chain together multiple MapReduce tasks. Spring 09 Publication: M. Jul 14, 2013 · The advantage of the above logic is, we can use a distributed map reduce model for compute with multiple map-reduce tasks - Constructing the co-occurrence matrix, Finding the dot product for each user etc. mapReduce # Python 3 my_strings = ['a', 'b', 'c', 'd', 'e'] my_numbers = [1,2,3,4,5] results = list(zip(my_strings, my_numbers)) print(results) As a bonus, can you guess what would happen in the above session if my_strings and my_numbers are not of the same length? Large-scale machine learning is another important use of MapReduce. A well-known matrix factorization method is Singular value decomposition (SVD). 522–532. Thursday, August 25, 11 30 The Mahout In Action (Chapter 6) book contains a recommendation method based on matrix multiplication that uses co-occurrence data (C) in combination with user preferences (U) to generate user recommendations (R). com/awslabs/lambda-refarch- mapreduce/. To put it in a crude analogy,  Page Rank, Inverted Index and Matrix Multiplication - asarraf/Algorithm- Implementation-Using-Map-Reduce. arange(4). rb. x is a shorthand for the elements of the first tensor argument. * Distributed performance bug fixes. 1–12. To maximize parallelism, the Gram matrix is calculated, that is, a matrix of the distances between every data instance and the nodes of the SOM. } General Matrix-Vector multiplication: y <- alpha * A * x + beta * y Source Edit proc gemm [T: SomeFloat | Complex] (alpha: T; A, B: The product of matrices A and B is calculated by multiplying the elements of rows in A with the corresponding columns in B, and then adding the resulting values to produce a single value for each row in A and each column in B. The Formula is: Mapper for Matrix A (k, v)=((i, k), (A, j, Aij)) for all k That is, we can implement matrix multiplication as the cascade of two MapReduce operations, as follows. to distributed matrix multiplication and distributed learning. regCG, for example, only lines 4 and 9 access matrix X; all other computations are inexpensive operations over small vectors or scalars. 5 as a cutoff, as shown in the plot below. com/Cloveryww/MPI-parallel-algorithms/tree/master/ cannon Since the algorithm and the lower hair collection task results when us 26 Nov 2018 Map Reduce (Part 3). sum (axis = 1). matrix (np. I'm working on the matrix multiplication using mapreduce. G. assign(vi;vnew) : decide how to update vi with vnew. Nov 29, 2017 · MapReduce, by Google, in 2004; Hadoop (fair mode), Spark (easy mode) MPI (hard mode) Matrix multiplication A = BxC for i in range(m): for j in range(n): for k in range(r): A[i][j] += B[i][k] * C[k][j] Matrix multiplication Vectorized for i in range(m): for j in range(n): A[i, j] = np. Problem Motivation Apr 04, 2019 · Other Links: GitHub Repository. Eq. The is similar to the process of generating the Row Number as explained in the previous post. It involves the matrix Oct 08, 2019 · In Cholesky method, a positive-definite matrix is written as the matrix multiplication of a lower-triangular matrix and its transpose. RecSys-2013-ZhuangCJL #memory management #parallel #performance A fast parallel SGD for matrix factorization in shared memory systems ( YZ , WSC , YCJ , CJL ), pp. 5 cutoff. GitHub Gist: instantly share code, notes, and snippets. Google’s MapReduce (Example)1 Classic example: word count. , non-causal attention where there is no notion of past and future. com/kokkos/kokkos- kernels \"Graph twiddling in a mapreduce world. Matrix-Vector Multiplication. ACM Transactions on Embedded Computing Systems (TECS), Vol 16, Issue 4, May 2017. Databases 2 Application 1: Matrix-Vector Multiplication. rstrip (). 55 folds) Ratio of FastPath to CompletePath memory accesses: 30:0 for software-based atomic and 3:28 for system-provided atomic implementations Matrix factorization (MF) factorizes a matrix R ∈ R m × n (with N z non-zero elements) into two low-rank matrices X ∈ R m × f and Θ ∈ R n × f, such that R ≈ X ⋅ Θ T. The output should be similar with the input. Matrix Multiplication With MapReduce. require ' rubygems ': require ' matrix ': require '. plot ( xs , ys ); Apr 09, 2013 · OutcomesRecognize relationships between matrix methods andthings you’ve already been doing\" Example SQL queries as matrix computationsUnderstand how to use Hadoop to compute thesematrix methods at scale for BigData\" Example Recommenders with social network infoUnderstand some of the issues that could arise. combine2(mi;j;vj) : combine mi;j and vj. Priebe, and Randal Burns. /matrix_block_mixin ' Mar 28, 2012 · That is, we can implement matrix multiplication as the cascade of two MapReduce operations, as follows. Background MapReduce and Spark Matrix Multiplication Note: also with row/column vector rhs Note: 1:N join. Matrix multiplication is an important multiplication design in parallel computation. Please turn in source codes, compilation, submission scripts used and also output les. 28 Mar 2012 P is a matrix = MN with element pik in row i and column k, where pik =∑j mijnjk Relational Representations: M = M(I, J, V ), with… Matrix Multiplication using MapReduce – 2 Step Solution Source Code: GitHub. ]]) >>>np. 2 A New Parallel Matrix Multiplication Algorithm on Hex-Cell Network (PMMHC) Using IMAN1 Super Computer E. input. However, such a straightforward approach does not apply to matrix tri-factorization because, as we show in the Methods section, the learning of any block of U and V depends on factor S . 0. May 12, 2018 · On Intel CPUs, SSE instruction sets use up to 128 bit registers (xmm, four ints), AVX and AVX2 use up to 256 bit registers (ymm, eight ints), and AVX512 use up to 512 bit registers (zmm, sixteen ints). Contribute to tangsttw/Matrix- Multiplication-MapReduce development by creating an account on GitHub. ,3. Report this profile Addition of numbers, matrix multiplication inside a docker and using MapReduce as the programming model. reshape(2,2) >>>a array([[ 0. y = y return z def backward ( dz ): dx = self . Lecture 14, Matrix Computations and Optimization in Apache Spark, Sparse matrix multiplication using SQL, Sparse matrix multiplication in MapReduce. Everybody is talking about it and everybody have their own understanding towards it which has made its definition quite ambiguous. My Personal Notes arrow_drop_up. Dec 16, 2019 · Outsourcing computation has gained significant attention in recent years in particular due to the prevalence of cloud computing. Use the output from the Stage 1 Reducer and pass along the same input to the Stage 2 reducer, where all the values having same index pairs are summed up to get the final output value. 04/27/18 - Large-scale machine learning and data mining applications require computer systems to perform massive computations that need to be May 02, 2018 · One kernel that can be parallelized using SPMD parallelism is dense matrix-matrix multiplication, in which we multiply two input matrices A and B to produce an output matrix C. Introduction Modern data-mining or ML applications, called «big-data analysis» requires us to manage massive amounts of data quickly. Here two passes symbolises the fact that we will need two map reduce jobs to compute the matrix multiplication. Then we are performing multiplication on the matrices entered by the user. 13. This matrix at each step will be updated and extended. After scaling genotype and expression data to unit variance with matrix-vector multiplications (or matrix-matrix with a small second matrix), and (2) closed-form algorithms with transpose-self matrix multiplication. x * dz # [dz/dy * dL/dz] return [ dx , dy ] “Scale-free Sparse Matrix-Vector Multiplication on Many-Core Architectures, “ IEEE Transactions on Computer-Aided Design of Integrated Circuits and Systems (TCAD), Vol 36, Issue 12, Dec 2017. Large Scale Machine Learning: Page: Gradient descent with large data, stochastic gradient descent, mini-batch gradient descent, map reduce, data parallelism, and online learning. It is crucial for perfor-mance to t the data into single-node or distributed main memory and enable fast matrix-vector opera-tions on in-memory data. The aim of this library is to easily design prototypes of algorithms using MapReduce. The product of a n x n matrix M by a vector v of length n is given by Dec 25, 2020 · Matrix Multiplication At Scale Using Map Reduce Matrix multiplication is the one of the most fundamental operation that most of the machine learning algorithms rely on. Taifi and Y. Optimizations Around version 0. x = x # must keep these around! self . For any u and v, such tat 1 ≤ u ≤ m and 1 ≤ v ≤ n, r u v is the (i, j) entry of R. toNumPy () Output: array ([[ - 60. This file shows the output of the resultant matrix obtained by multiplying matrix A and B in the format \"(row,column) value\" Contribute to JaredP94/MapReduce-Matrix-Multiplication development by creating an account on GitHub. It takes the value v and puts it in the row indexed by x. Feb 08, 2015 · Write a MapReduce query to remove the last 10 characters from each string of nucleotides, then remove any duplicates generated. A file containing two Matrices - MatrixA and MatrixB, was fetched from the Hadoop Distributed File System (HDFS) as an input for the Map Reduce task. - Use OpenMP and SSE to improve the speed of matrix multiplication over 86 times and kmeans 11 times. As the instructor pointed out, there are reasons this approach isn't insane. a + (b + c) = (a + b) + c a * (b * c) = (a * b) * c max (a, max (b, c)) = max (max (a, b), c) min (a, min (b, c)) = min (min (a, b), c) Strings with string-concatenation form a semigroup. There are 2 implementations of the  3 Feb 2018 PDF | On Dec 1, 2017, Mais Haj Qasem and others published Matrix multiplication of big data using MapReduce: A review | Find, read and cite  4. Map Reduce Example for Sparse Matrix Multiplication. in a way you should be familiar with. Here, we will discuss the implementation of matrix multiplication on various communication networks like mesh and hypercube. The definition is motivated by linear equations and linear transformations on vectors, which have numerous applications in applied mathematics, physics, and engineering. Write a MapReduce query to remove the last 10 characters from each string of nucleotides, then remove any duplicates generated. distributed map reduce In this module, we will learn about the MapReduce paradigm, and how it can be used to write distributed programs that analyze data represented as key-value pairs. It takes in 6 parameters: n: number of rows in A; m: number of I would like to apply map-reduce to deal with matrix multiplication in python with Hadoop. Jul 28, 2020 · Analyzing weather data of Fairbanks, Alaska to find cold and hot days using MapReduce Hadoop. , line 9), we have vector-matrix multiplication, often caused by the rewrite X>v !(v>> > > 2 Implement SELECT * FROM <table> WHERE <condition> with MapReduce. jl. 1 Matrix Multiplication Hadoop Implementation . * DRM row sampling api. Illustrates how a variety of coercion-based defaults can be specified to make life easy on the implementer, while still easily allowing for dispatch to optimal implementation-specific routines whenever it's desired. You might want to examine the Hadoop code for Word Count and Matrix multiplication. DATA/DATA SCIENCE Data Science from Scratch ISBN: 978-1-491-90142-7 US $39. So, several papers have studied the problem of multiplying matrices using a large number of processors (CPUs) in parallel. Kyungyong Lee is an assistant professor in the College of Computer Science at Kookmin University. The instruction sets are typed, and instructions designed to operate on packed doubles can’t operate on packed ints without explicit casting. In this post, we will be writing a map-reduce program to do Matrix Multiplication You need Hadoop’s HDFS and map GitHub Gist: star and fork jaganadhg's gists by creating an account on GitHub. Jun 23, 2014 · To the best of our knowledge, there are no matrix inversion algorithms using MapReduce, although there are several software systems for other matrix operations using MapReduce. FlashMatrix: Parallel, Scalable Data Analysis with Generalized Matrix Operations using Commodity SSDs. Here we reduce the output received from mapper into actual 2D array for matrices A and B and calculate the matrix multiplication based on usual formula. • Irregular algorithms Penetrating rays (e. Instructors. (1) can be minimised with respect to and solved for , yielding a large matrix multiplication problem, a formulation that is employed by the R package Matrix eQTL , which allows for fast eQTL analysis on a desktop computer. matrix',default='A', dest='Amatname') def parsemat(self): \"\"\" Return 1 if this is the A matrix, otherwise return 2\"\"\" fn = get_jobconf_value('map. Matrix factorization learns a latent data model that takes a data matrix and transforms it into a latent feature space enabling generalization, noise 1 Mar 2018 implement by naive algorithm(without partition) and advanced algorithm(with partition) - AiningWang/Matrix-Multiplication-using-MapReduce. d. . I hope these programs will help people understand the power of distributed parallel computing via map-reduce on Spark platform. An AggBinaryOp hop can be compiled into the following physical operators. collection. Howe outlined, together with the theory, some solutions to the assignments, we started to make our hands really dirty. Pros . UPDATE: Please note that the Mapper function does not have access to the Row Number (in this case i, and k) directly. 09, May 20. F. lines method; using the generate method (provide a Supplier) or iterate method (providing the initial value and incremental operation). Hi all. 550 Architecture of Machine Learning Systems based on fast matrix multiplication. Mar 29, 2012 · Posted by Venkata (Ravi) Adusumilli on March 29, 2012 in Hadoop, MapReduce Tags: Hadoop 0. toNumPy () Output: array ([[ - 60. The JobConfigurable#configure has to be implemented in the Mapper and Reducer class. Apr 14, 2012 · Prepare for Matrix Multiplication. stdin: curr_index, index, value = line. g. enron 4. But somehow the generation of the <Key> <value> pair and the operation in the 1. For running unit tests, use junit, and for writing MapReduce tests, use mrunit. Jan 27, 2021 · // The slave process receives the sub portion of the Matrix A which assigned by Root : MPI_Recv(&matrix_a, rows*N, MPI_DOUBLE, source, 1, MPI_COMM_WORLD, &status); // The slave process receives the Matrix B: MPI_Recv(&matrix_b, N*N, MPI_DOUBLE, source, 1, MPI_COMM_WORLD, &status); // Matrix multiplication: for (int k = 0; k<N; k++) Mar 31, 2012 · P is a matrix = MN with element p ik in row i and column k, where p ik =∑ j m ij n jk. We represent matrix M as a relation , with tuples , and matrix N as a relation , with tuples . Monte Carlo Integration. 19. Implementation . Also a variation for pair-wise distance matrix of two different inputs x and y: sqDist(x,y), dsqDist(x,y). py. 1 Matrix Multiply and Computer Architectures Homework question 1 Matrix multiplication with MapReduce If A is an m × p matrix and B is an p × n matrix, then the product of A and B is the m × n matrix C = AB, where the (i, j) th element of C is computed as the inner product of the ith row of A with the jth column of B: This is a dot product—simple arithmetic if m, p, and n are small. The map takes ( le, content) pair, and emits (word, 1) pairs for each word in the content. MapReduce: Simplified Data Processing on Large Clusters. Step 1: We can download the dataset from this Link , For various cities in different years. At first it was a little brain-breaky, but then we did the map-reduce version, which was brain-breakier. Written by Luka Kerr on April 2, 2018 I’ve been learning MIPS assembly for about 2 weeks now at uni and wanted to share how i’ve implemented a simple matrix multiplication function in MIPS. That said, the ground is now prepared for the purpose of this tutorial: writing a Hadoop MapReduce program in a more Pythonic way, i. linspace ( 0 , 2 * np . Clustering with KMeans: Page: Clustering with KMeans in scikit-learn. “Matrix factorizations at scale: A comparison of scientific data analytics in Spark and C+MPI using three case studies”, 2016 IEEE International Conference on Big Data (Big Data), pages 204–213, Dec 2016. I will explain LSH and how to use this package as well as the details of the implementation below. \"Git-a\" -rec It further employs a content-based filtering approach, coupled with Apache Spark to develop a recommender system, for Github users. Use of ufuncs is an esssential aspect of vectorization and typically much more computtionally efficient than using an explicit loop over each element. For these setups, coding has been utilized primarily to handle failed or straggling (delayed) workers –, where some workers fail or are significantly slower than the other workers, causing a significant delay in the overall computation time. e. g. 2) Matrix Multiplication 3) Find Mutual Friends for Social Media Data. Contribute to kdave2/Matrix- Multiplication-Hadoop-Map-Reduce development by creating an account on GitHub. Map, Reduce and Filter functions in Python make writing code much more easier (less lines of code) and I think they are optimized internally which will make them more faster than writing custom code which will most Sparse linear algebra Matrix Multiplication Spectral methods FFT N-Body methods GEM Structured grids SRAD Unstructured grids CFD solver MapReduce Combinational logic CRC Graph traversal Breadth-First Search (BFS) Dynamic programming Needleman-Wunsch Backtrack and branch-and-bound Graphical models Hidden Markov Model Map Reduce Triplets Block Matrix Multiplication Let’s look at Block Matrix Multiplication (on the board and on GitHub) Created Date: Oct 23, 2020 · The above analysis is relevant for so-called bidirectional attention, i. This model defines data abstraction as key-value pairs and computation flow as “map, shuffle and then reduce”. Matrix multiplication is an important multiplication design in parallel computation. MapReduce is a parallel fram ework for big da ta, which That is, we can implement matrix multiplication as the cascade of two MapReduce operations, as follows. p. private static Long N; public void configure(JobConf job) {N = Long. Designed a 4x4 grid layout of game card. We consider numerical computations using dataflow graphs, with a focus on learning deep neural networks for image classification and other classification tasks. x library which implements a MapReduce algorithm. reshape (%input) : (tensor<1x28x28x1xf32>) /> tensor Matrix multiplication in C. – Description of AdaGrad. First: The Map Function\":Foreachmatrixelementm ij,producethekeyvaluepair j, (M,i,m ij) # 21 hours ago · mapreduce python 3, Jan 28, 2020 · You can indent using tabs and spaces in Python. 4. That is, we can implement matrix multiplication as the cascade of two map-reduce operations, as follows. sin is a universal function plt . are overloaded for convenience. The : operator takes syntax start:spacing:end. We classify the matrix multiplication problems into Apr 26, 2012 · Posted by Venkata (Ravi) Adusumilli on April 26, 2012 in Hadoop, MapReduce Tags: Hadoop 0. 3. sh Matrix_Reducer. github. ∙ 0 ∙ share A cumbersome operation in numerical analysis and linear algebra, optimization, machine learning and engineering algorithms; is inverting large full-rank matrices which appears in various processes and applications. gz Step 2: Move the eclipse folder to the home directory In this step, you can see how to move the eclipse folder to the home directory. Example MapReduce Algorithms Matrix-vector multiplication Power iteration (e. A trivial implementation is trivial, but users are likely to want fast versions that are hard to write. io/. Graph Processing using Map-Reduce 2. Oct 10, 2017 · Using MapReduce Programming In mathematics, matrix multiplication or the matrix product is a binary operation that produces a matrix from two matrices. The value for cell (i, j) of matrix C is computed by taking the dot product of row i in matrix A and column j in matrix B. If A is an m × p matrix and B is an p × n matrix, then the product of A and B is the m × n matrix C = AB, where the (i, j)th element of C is computed as the. In the Figure 1 (a), we experiment large and sparse matrix multiplication from two random Bernoulli square matrices with dimension roughly equal to 1. Parallel Matrix Multiplication Algorithms Grid-based approach - The grid-based algorithms , regard processors as residing on a two- or three- Coverage: vectors norms (ℓ 2-norm, ℓ 1-norm, ℓ p-norm, ℓ ∞-norm), vector inner product, matrix multiplication, matrix trace, matrix Frobenius norm, scalar function differential, convex function, use Numpy to construct vectors and matrices. Matrix multiplication in C: We can add, subtract, multiply and divide 2 matrices. 20. matrix multiplication using mapreduce github iomeviewer-weezy-l-y-incompetence-laguna-adair\"> matrix multiplication using mapreduce github Kublanovskaya, J. Examples of Lambda are given below. • Standard http://jwbuurlage. MapReduce. Francis) 1964: Sinkhorn (Richard Sinkhorn) 1965: Golub-Reinsch SVD (Gene Golub) 1969: Sparse matrix ordering (Elizabeth Cuthill, James McKee) 1969: Strassen matrix multiplication (Volker Strassen) The implementation for the multiplication gate with input scalars x and y and output scalar z is class MultiplyGate ( object ): def forward ( x , y ): z = x * y self . combineAlli(x1;:::;xn) : combine all the results from combine2() for node i. g. g. 3. I just want to code a matrix multiplication problem in MapReduce using python for very large sparse matrix. An AggBinaryOp hop can be compiled into the following physical operators. h. This relates mostly to (a) matrix multiplication deficiencies, and (b) handling parallelism. Test naive algorithm locally. Basic Matrix Multiplication Operators. May 15, 2013 · Does anyone need MapReduce?122• I tried to do book recommendations withlinear algebra• Basically, doing matrix multiplication toproduce the full user/item matrix withblanks filled in• My Mac wound up freezing• 185,973 books x 77,805 users =14,469,629,265– assuming 2 bytes per float = 28 GB of RAM• So it doesn’t necessarily take that much tohave some use for MapReduce of the loss J ( i) ( θ) of the i th window vector with respect to the matrix x of window vectors is given by. I built a jar file using the code below. Jun 21, 2020 · MapReduce Program – Finding The Average Age of Male and Female Died in Titanic Disaster; MapReduce – Understanding With Real-Life Example; How to find top-N records using MapReduce; How to Execute WordCount Program in MapReduce using Cloudera Distribution Hadoop(CDH) Matrix Multiplication With 1 MapReduce Step; MapReduce – Combiners 1. element 3 in matrix A is called A21 i. 15 minute read. Introductory example is calculation of Fibonacci numbers where F(N) (problem of size N) is calculated as sum of F(N - 2) and F(N - 1) (problems of size N - 2 and N - 1). KEY: block num (a_block, b_block) VALUE: [(row, col, value), ,( )]//numbers in the block // implement matrix multiplication of the blocks locally res = 0 for i = row of ele in this a block: for k = col of ele in this b block: for j = 0 to n-1: res += A[i][j] * B[j][k] Test. Aug 17, 2018 · Besides matrix-vector and matrix-matrix calculations, relational-algebra operations fit well into the MapReduce style of computing. deeplearn-rs - deeplearn-rs provides simple networks that use matrix multiplication, addition, and ReLU under the MIT license. This file contains the implementation of reducer. Specifically, use test your implementation on the following graphs, 1. Table 2 summarizes the core operations of important ML algorithms. com&nbs 15 Dec 2014 in this area (Hadoop, Map-Reduce and Spark) work exercises. Rawashdeh , M. From the implementation point of view, Before the MapReduce model, MPI was the main tool used to process big data. 2166 II. Certain common operations, like broadcast or matrix multiplication, do know how to deal with array wrappers by using the Adapt. ein\"ij,jk -> ik\"(a,b) returns :(a * b) and ein\"ij,kj -> ki\"(a,b) returns :(b * transpose(a)) . in a way you should be familiar with. The nucleus is represented as a N-nanometer by M-nanometer grid, and at each 1nm * 1nm square is known whether or not it is filled by something. G. Parsian, Data . When we represent matrices in this form, we do not need to keep entries for the cells that have values of zero to save large amount of disk space. PyMR is a Python 2. We define requirements for the design of serverless big data applications, present a prototype for matrix multiplication using FaaS, and discuss and synthesize insights from results of extensive experimentation. What we want to do We will write a simple MapReduce program (see also Wikipedia ) for Hadoop in Python but without using Jython to translate our code to Java jar files. Aug 29, 2015 · Matrix Multiplication with MapReduce 33 Comments Posted by Maruf Aytekin on February 16, 2015 Matrix-vector and matrix-matrix calculations fit nicely into the MapReduce style of computing. The MapReduce contains two important tasks, namely Map and Reduce. Map-Reduce. Integer matrix-matrix and matrix-vector Distributed matrix factorization with mapreduce using a series of broadcast-joins (SS, CB, MS, AA, VM), pp. Matrix-matrix product • Basic matrix multiplication on a 2-D grid • Matrix multiplication is an important application in HPC and appears in many areas (linear algebra) • C = A * B where A, B, and C are matrices (two-dimensional arrays) • A restricted case is when B has only one column, matrix-vector product, which appears in 1957: Minimum degree sparse matrix reordering (Harry Markowitz) 1961: QR algorithm (V. Jan 25, 2021 · We need to extract it using the command tar –zxvf eclipse-committers-photon-R-linux-gtk. 0-m2 m4. Many fields such as Machine Learning and Optimization have adapted their algorithms to handle such clusters. js) and Approach2 (multiply_map_aggregate. 5https://github. MapReduce, from one job to another, takes time to upload the file and start the job again. Basic Matrix Multiplication Operators. When performance or resources are a matter of concern Cubert Script is a developer-friendly language that takes out the hints, guesswork and surprises when running the script. \" Computi (Concurrent Replication-based Matrix Multiplication) along with a parallel algorithm, Marlin, for large-scale matrix multiplication on uted data-parallel platforms, such as Hadoop MapReduce of naively using the general shuffle me Cannon principle parallel matrix multiplication algorithm implementation, and Source: https: //github. function mapReduce(input, map, reduce) {// Map: var mapperOutput = []. Mon 11/14: No recitation : Tue 11/15: srini Experiment MapReduce: Result for Matrix Multiplication 38 MapReduce The speedup of using software-based atomic add over the system one increases as the input matrices get larger (up to 13. 1) Simple Moving Average for Companies Stock Data. e. There are 2 main reasons why interpreted Python code is slower than code in a compiled lanauge such as C (or other compiled langauge): Using MapReduce, we could achieve N times throughput by having N workers running in parallel. Existing GPUbased SOM algorithms take a similar approach in finding the best matching unit, but they do not necessarily use matrix operations to derive the distance matrix [34,46]. ,1. sin ( xs ) # np. Tensor-matrix-multiplication(TTM)referstomultiplyingTalong any mode n, by a matrix of size K ×Ln (for some K). pi , 100 ) ys = np . Importance Sampling for Learning Edge opicsT (ISLE) Microsoft Research, 2000+ LOC [Github] Sep 21, 2020 · The course includes video lectures, case studies, peer-to-peer engagements and use of computational tools and platforms (such as R/RStudio, and Git/Github), and a reproducible research project. Knowing the working of matrix multiplication in a distributed system provides important insights on understanding the cost of our algorithms. Apr 10, 2019 · MRQL (the MapReduce Query Language) is an SQL-like query language for large-scale data analysis on a cluster of computers. sum (axis = 1). Gittens et al. com Li Pu A. Mahoney. The essence of the transformation is to unroll the input patches (a 3D matrix) and lters (a 4D matrix) in 2D in a way that a single matrix-matrix multiplication produces the unrolled version of the output in 2D. 10 of Mahout it became obvious that using Hadoop MapReduce. dblp-2011 3. Map Reduce paradigm is usually used to aggregate data at a large scale. Methods used in the 4 Apr 2019 How to find top-N records using MapReduce Now to do so we just multiply the key with -1 in mapper, so that after sorting higher numbers Other Links: GitHub Repository Matrix Multiplication With 1 MapReduce Step. UPDATE: Please note that the Mapper function does not have access to the Row Number (in this case i, and k) directly. How to do matrix multiplication in R? Machine Learning Recipes,do, matrix, multiplication, r: What is mapply in R? Machine Learning Recipes,what, is, mapply, r: What is sapply in R? Machine Learning Recipes,what, is, sapply, r: How to find lagged differences in R? Machine Learning Recipes,find, lagged, differences, r: How to find variance and MapReduce/Hadoop, where data must be read from disk on eachiteration. Use of MapReduce has flourished since its premier, as illustrated by an in-depth example of its use in WordCount. Aug 02, 2016 · Next we show various attempts for scalable implementation of matrix multiplication using spark, and the winning method which combines numpy matrix multiplication along with spark’s broadcast and MapReduce Word Count Example. 18. ,0. 38) 13KB Synchronous-100x100-Matrix-Multiplication-using-Multiple-Threads:开发了一个程序,用于通过使用多个线程将两个大型 矩阵 Nov 29, 2017 · Generalized matrix multiplication with semiring? Closing since I think this is out of reach of easy contributions. Each cell of the matrix is labelled as Aij and Bij. 7, which is an open source (Apache project) machine learning library. Dec 15, 2014 · This can be an important way to tune performance. Matrix Multiplication With 1 MapReduce Step. ,0. Matrix-multiplication kernels like gemm (dense-dense) and csrmm (sparse-dense), and utility kernels like csrcsc (sparse-transpose), sort (Parallel Sample Sort), and map-reduce are implemented in the framework. Let m be an NxM matrix and v a M vector. Design a MapReduce algorithm to compute matrix multiplication: A x B Mar 27, 2012 · Puzzle: Since there's so much other stuff in the nucleus, like proteins, it is sometimes troublesome finding room for a genome. new array wrappers are not covered, and only one level of wrapping is supported. Under Review: 2017. ∇ x ( p) T J ( i) ( θ) = ( δ ( i) T W) ⋅ 1 { i = p } ∈ R 1 × ( 2 w + 1) d, where the gradient with respect to the ( p, q) -th entry x ( p) q of the matrix x is given by. #MapReduce #MatrixMultiplication Apr 02, 2018 · Matrix Multiplication In MIPS. And more generally, taking a sum of outer products of corresponding rows and columns of the input matrices, always returns the desired matrix multiplication result. Trigger module in display ad project owner & core member Object: the old ad engine built reverse index from tag to ad, and this scheme suffered from long reverse index, the engine needed to do hard pruning when searching which hurt CTR. MPI Programming Matrix Multiplication Operators. matrix (np. 17. Here, the role of Mapper is to map the keys to the existing values and the role of Reducer is to aggregate the keys of common values. Matrix multiplication uses a generated function to return a matrix-product with at most one application of transpose , such that e. N. com/jfkelley/hadoop-matrix-mult After the submatrix multiplication, there's another MapReduce job that simply does the same Then iterate through just those sections of the arrays and multiply elem 9 Mar 2018 Sparse Matrix Matrix Multiplication. 2-CDH3U2 ← Matrix Multiplication using MapReduce – 1 Step Solution A celebrated example of an embarrassingly parallel problem is shared-memory matrix multiplication in which for two n n matrices, the multiplication process is split into n2 tasks, each responsible for calculating one element of the result matrix . Among them: matrix multiplication , similarity join [112,121,14,23], multi-way join using map-reduce offers a scalable Having that said, the ground is prepared for the purpose of this tutorial: writing a Hadoop MapReduce program in a more Pythonic way, i. ones ((3, 3)) + 3) m2 = m1 * (m2 + m1) m4 = 1. Moreover, Pegasus provides fast algorithms for GIM-V in MapReduce, a distributed computing platform multiplication operations . Section 5 will discuss https://github. Matrix Multiplication Map-Reduce. Shi, \"Performance Prediction and Evaluation of a Solution Space Compact Parallel Program using the Steady State Timing Model\", Poster, Future of My personal list of journals I use for my research and projects where I wrote one-sentence summaries. These operations are low-level, but for your convenience wrapped using high-level constructs. We write ++ for concatenation, so \"foo\" ++ \"bar\" is \"foobar\". Sparse matrix-vector multiplication (SpMV) is an important kernel that is considered critical for the performance of compute-intensive applications. Use an algorithm that does not require looking through every single possible matrix. Apr 29, 2013 · Sparse matrix computations in MapReduce!Austin Benson Tall-and-skinny matrix computations in MapReduceTuesday!Joe Buck Extending MapReduce for scientific computing!Chunsheng Feng Large scale video analytics on pivotal HadoopWednesday!Joe Nichols Post-processing CFD dynamics data in MapReduce !Lavanya Ramakrishnan Evaluating MapReduce and combining the matrix multiplication with the MapReduce is only 4 articles that we will discuss and com pare through this paper. Matrix Multiplication using Map-Reduce A specification ixs,iy is a matrix multiplication if ixs consists of two 2-tuples that share one index-label and iy is a permutation of the two-nonshared index-labels of the ixs. MapReduce is an attractive framework because it allows us to decompose the inner products involved in computing document similarity into separate multiplication and summation stages in a way that is well matched to efcient disk access using MapReduce minimize #syncs 8md Ours FPGA acceleration with small on-chip BRAM minimize data transfers (4m+4)d . Using item collaborative filtering algorithm to build co-ocurrence matrix by users' rating towards different movies from Netflix Prize Data Set. Thus, matrix multiplication is the major focus of performance optimization for many big data analytical algorithms or or applications. “Efficient Kernel Management on GPUs”. plot ( xs , ys ); 1 vala: Matrix[Float] //MxN 2 valb: Matrix[Float] //NxP 3 valc = Map(M, P){(i,j) => 4 //OuterMapfunction(f1) 5 Fold(N)(0. However, computing the inverse of a matrix using MapReduce is difficult when the order of the matrix is large. This is the function in C that will be implemented. e. In the following, we give an overview of backend-specific physical matrix multiplication operators in SystemML as well as their internally used matrix multiplication block operations. a ++ (b ++ c) = (a ++ b) ++ c. A matrix is a set of numerical and non-numerical data arranged in a fixed number of rows and column. 2. If you want uniformly spaced vectors, use the : operator. Stack Overflow. Cubert provides a novel sparse matrix multiplication algorithm that is best suited for analytics with large-scale graphs. first_10_even_numbers = 0: 2: 21 first_10_even_numbers = 0 2 4 6 8 10 12 14 16 18 20 A mode-n unfolding refers to the matrix of size Ln ×bL n obtained by arranging the mode-n fibers as the columns in a lexicographic order. What can we do with MapReduce? Models & Algorithms • Communication-processor tradeoffs for 1 round of MapReduce – Upper bounds for database join queries [Afrati,Ullman2010] – Upper and lower bounds for finding triangles, matrix multiplication, finding neighboring strings [Afrati, Sarma, Salihoglu, Ullman 2012] 5 compression idea from the sparse matrix-vector multiplication literature , into the Ligra shared-memory graph processing framework . the result will be the format of mat2. 2nd-row 1st column. Implementing matrix multiplication using MR and optimizing it using Combiner. A MapReduce program is defined via user-specified map and reduce functions, and we will learn how to write such programs in the Apache Hadoop and Spark projects. We propose a novel parallel execution model called pin-and-slide, which implements the column view of the matrix-vector mul-tiplication. ], [ 2. N. smaller/simpler) approximation of the original matrix A. rustlearn - a machine learning framework featuring logistic regression, support vector machines, decision trees and random forests. This code performs matrix vector multiplication using map reduce, which enables fast computations for large amounts of data. 3. At a high level, SVD is an algorithm that decomposes a matrix A into the best lower rank (i. Performance of matrix multiplication and random tensor contractions for rank-k update matrix/tensor shapes on a Xeon E5-2690 v3 processor. An extra MapReduce Job has to be run initially in order to retrieve the values. An extra MapReduce Job has to be run initially in order to add the Row Number as Key to every row. 4. choose the year of your choice and select any one of the data text-file for analyzing. Installed multi node Hadoop 2. Using the best currently known parallel matrix multiplication [Wil12, LG14], our algorithm dynamically maintains the number of k-cliques in O min m 0:469k 235;( + m)0 :469k+0 amor-tized work w. Mathematically, it decomposes A into a two unitary matrices and a diagonal A matrix is a set of numerical and non-numerical data arranged in a fixed number of rows and column. By interpreting the matrix-vector multiplication in the column view, we can restrict the computation to just a subset of the May 21, 2015 · Jeffrey Dean and Sanjay Ghemawat. One of these systems is SystemML , which provides a high-level language for expressing some matrix operations such as matrix multiplication, division, and transpose Keywords: Item Collaborative Filter, Matrix Multiplication, MapReduce, Java, Hadoop - A movie recommender system is built to recommend movies in a similar style to users using raw data from Netflix. Jun 15, 2019 · Map Reduce paradigm is the soul of distributed parallel processing in Big Data. This is because the MapReduce works efficiently only with BIG data. In Proc eed- ings/42nd IEEE Symposium on Fo undations of Computer Science: October 14-17, 2001, Las Use same MapReduce job for operations w/o dependencies sFA Optimizations: Minimize Intermediary Data Recompute X and Y at each job rather than storing and exchanging Dec 27, 2015 · For multiplication, the key is to build rpos[], rpos[i] means in matrix N row i starts at rpos[i] position in datas[]. 01 import systemml as sml import numpy as np m1 = sml. Apr 05, 2019 · The below picture illustrates calculating an image’s class values for all 10 classes in a single step via matrix multiplication. One of the most important topic from university exam point of view. This is still not a complete solution though, e. Matrix-vector multiplication. 0f){k => 6 //Innermapfunction(f2) 7 a(i,k) * b(k,j) 8}{(x,y) => 9 //Combinefunction(r) 10 x + y 11} 12} Figure 1: Example of using Map and Fold in a Scala-based lan-guage for computing an untiled matrix multiplication using in-ner products. • Obtained a user’s rating matrix of films from the Netflix data using the Item Collaborative Filtering Algorithm, then obtained the co-occurrence matrix of the films, and finally merged both matrices to obtain a recommendation list. By-product: large-scale sparse matrix multiplication based on MapReduce. 1 and hadoop with Python 2. using matrix multiplication as example. ones ((3, 3)) + 3) m2 = m1 * (m2 + m1) m4 = 1. Mahout’s linear algebra DSL has an abstraction called DistributedRowMatrix (DRM) which models a matrix that is partitioned by rows and stored in the memory of a cluster of machines. We use 5 for this project, but you may want to increase this number for large datasets. If you like my post - Do follow me on this blog - Matrix Multiplication Using MapReduce Programming In mathematics , matrix mult map-reduce operation, we can perform grouping and aggregat ion, with I and K as the grouping attributes and the sum of V × W as the aggregation. . Each block is sent to each process, and the copied sub blocks are multiplied together and the results added to the partial results in the C sub-blocks. RecSys-2013-ZhuangCJL #matrix #memory management #parallel #performance A fast parallel SGD for matrix factorization in shared memory systems ( YZ , WSC , YCJ , CJL ), pp. For unidirectional (causal) attention, where tokens do not attend to other tokens appearing later in the input sequence, we slightly modify the approach to use prefix-sum computations, which only store running totals of matrix computations rather than mrjob fully supports Amazon’s Elastic MapReduce (EMR) service, which allows you to buy time on a Hadoop cluster on an hourly basis. matrix-matrix multiplication using * matrix-vector multiplication using * element-wise multiplication (Hadamard product) using *. Pseudocode: First Map-Reduce job: Oct 25, 2016 · Mapper For a matrix multiplication of the form AB, we must provide in the mapper, the number of rows of A, referenced as row_a in the code, and the number of columns of B, referenced as col_b (The number of columns of A and number of rows of B are always same, else multiplication won't be possible). Most matrices are sparse so large amount of cells have value zero. Semi-External Memory Sparse Matrix Multiplication on Billion-node Graphs in a Multicore Architecture. Here's a small example to illustrate it. 20. Array operations - slicing, dicing, searching¶ Table of Contents Array operations - slicing, dicing, searchingArray slicingnD array slicingArray dicingArray broadcastingDeep copyArray searchingC •Similar to MapReduce •Aggregate multiple messages to same recipient from same server into a single message •Also executed at the receiver side to save space •Aggregators •Master collects data from vertices at the end of a superstep •Workers aggregate locally and use tree-based structure to aggregate to master This paper presents a MapReduce algorithm for computing pairwise document similarity in large document collections. What do we Outline a Map-Reduce program that calculates the vector x. ICALP-v2-2012-BaldeschiHLS #keyword #multi #on the On Multiple Keyword Sponsored Search Auctions with Budgets ( RCB , MH , SL , MS ), pp. Hence, the resulting product has the same number of rows as A and columns as B. 15 Dec 2014 map-reduce method for multiplying large, sparse matrices using Elasticsearch as As always, the code is available in the GitHub repository. Implementing the matrix multiplication with Map Reduce jobs to find the recommender movie(s). Below picture shows process for transpose: Below shows the process for 2 triple tuple matrix multiplication: check my code on github: link Matrix Multiplication: 2 MapReduce steps! Matrix M can be thought of as a relation with tuples (i, j, m ij) ! Matrix N can be thought of as a relation with tuples (j, k, n jk) ! Map operation creates these tuples ! Map: Join of M and N brings us closer to M X N by creating: Relation (i, j, k, m ij, n jk) or the relation (i, j, k, m ij X n jk) ! Nov 05, 2020 · MapReduce is a programming model and an associated implementation for processing and generating large datasets that is amenable to a broad variety of real-world tasks. Fast monte carlo algorithms for matrices iii: Computing a compressed approximate matrix decomposition, SIAM Journal of Computing, 2005. As the usual dense GEMM, the computation partitions the output matrix into tiles. Map-Reduce for each vertex D B A C Block Matrix Multiplication Let’s look at Block Matrix Multiplication (on the board and on GitHub) Experiment MapReduce: Result for Matrix Multiplication 38 MapReduce The speedup of using software-based atomic add over the system one increases as the input matrices get larger (up to 13. The ‘Iterative’ in the name of GIM-V denotes that we apply the £G opera- based on fast matrix multiplication. Blue line shows the 0. options. Graph reachability, using Matrix-Matrix multiplication of adjacency matrices. MapReduce is a processing technique and a program model for distributed computing based on java. The Map-reduce programming model is a common data-handling model Array-based distributed computations are another abstraction, used in all forms of parallelism. Graphs, and Matrix Multiplication using MapReduce, Spark, and MASS. Feb 01, 2013 · We extended the MapReduce SOM algorithm by moving all calculations on local nodes to the GPU with the matrix-based Euclidean distance matrix and reduction algorithm described above. A discussion about each of this operations is available below, if you want to go ahead of this section, click here. 39 is a matrix of residuals, assuming: s i. data. In the following, we give an overview of backend-specific physical matrix multiplication operators in SystemML as well as their internally used matrix multiplication block operations. Matrix Multiplication performed using Hadoop. Implementing matrix multiplication using MR and optimizing it using Combiner. . What we want to do We will write a simple MapReduce program (see also the MapReduce article on Wikipedia ) for Hadoop in Python but without using Jython to translate our Around version 0. UPDATE: Please note that the Mapper function does not have access to the Row Number (in this case i, and k) directly. A MapReduce job can be enhanced by sampling local data, which cannot be used for future analysis. During the vector-matrix multiplication, each node will use its portion of the updated u vector, then estimate the v vector based on the multiplication of its putations as a single matrix-matrix multiplication (GEMM). Kublanovskaya, J. The RMM plan in Figure 1 im-plements a replication-based strategy in a single MapReduce version of matrix multiplication using recursive block matrix decomposition View MapReduceBlockMatrixProduct. * Distributed engine neutral allreduceBlock() operator api for Spark and H2O. Unfortunately there is no acceleration routine for integers. Also, due to the inherent complexity of high-order computation, experience from prior work in other fields can not easily be applied to tensors. 8 Weeks. webbase-2001 Describe how you stored the connectivity matrix on disk and how you computed the transition matrix. Lecture 16 (5/28): Complexity Measures for MapReduce, Triangle Counting in a Graph The sum of these outer products matches the result we obtained in the previous slide using the traditional definition of matrix multiplication. concat. We show that serverless big data processing can lower oper- We also consider distributed MapReduce computations for training clustering models such as k-means and collaborative filtering models based on matrix factorization. 1. TensorFlow. In this case, each task needs one row of the first input matrix and one Map Reduce Reduce. A breakdown of basic MapReduce terms and functions follows. matrix (np. 3 and use Different Algorithmic Techniques to Solve Following problems using Hadoop Map-Reduce. com/nm4archana/BigDataAnalysis-Comet. (For example Facebook) 4) K-mers counting using long DNA sequence in FASTA format 5) installing putations as a single matrix-matrix multiplication (GEMM). split (\"\\t\") index, value = map (int, [index,value]) if curr_index == prev_index: value_list. Integer factorization must be adaptable to MapReduce and must be parallizable (more than 1 Map or Reduce task) The emergence of large distributed clusters of commodity machines has brought with it a slew of new algorithms and tools. Finally the basic process of MapReduce is shown. You can still represent them using linear models. Irregular algorithms, however, depend on the input. I have published LSH package I developed for Apache Spark on GitHub. g. Diverse types of matrix classes/matrix multiplication are accommodated. MapReduce in distributed model training Using the MapReduce strategy, if we can split the training data on separate workers, compute Map functions in parallel, and aggregate the results in a Reduce function, we will be able to achieve distributed model Github. a mapreduce program of matrix multiplication. Green = true positive male, yellow = true positive female, red halo = misclassification. The ultimate goal is to make the algorithm as e cient as possible for any input . By using Monoids, we can take advantage of sparsity (we deal with a lot of sparse matrices, where almost all values are a zero in some Monoid). The operation is denoted Z = T×n A. matrix (np. Report this profile Addition of numbers, matrix multiplication inside a docker and using MapReduce as the programming model. Map the input matrices line by line and emits the matrix element. ee) Performance Prediction of Sparse Matrix Multiplication on a Distributed BigData Processing Environment. The sum of these outer products matches the result we obtained in the previous slide using the traditional definition of matrix multiplication. A naïve approach would be to extend an efficient matrix multiplication algorithm, replacing the dot product by the distance function. xs = np . Contribute to ashkang/cs222_fin development by creating an account on GitHub. 99 “ Joel takes you on a journey from being data-curious to getting a thorough understanding Improving Quantum Query Complexity of Boolean Matrix Multiplication Using Graph Collision (SJ, RK, FM), pp. [J14] Yun Liang, Xiuhong Li. . Shi, \"How to achieve a 47000x speed up on the GPU/CUDA using matrix multiplication,\" Technical Report, Amax corporation, June 2009. Use a reducer to multiply value for same indices. His current research topic covers big data platforms, large-scale distributed computing resource manangement, cloud computing, and peer-to-peer systems. dot(a,b) array([[ 0. You can generate uniformly spaced vectors, using 2 methods. pdf such as MapReduce [Srirama et al, FGCS 2012] • Designed a classification on how the algorithms can be adapted to MR – Algorithm single MapReduce job • Monte Carlo, RSA breaking – Algorithm nMapReduce jobs • CLARA (Clustering), Matrix Multiplication – Each iteration in algorithm single MapReduce job • PAM (Clustering) Aug 29, 2015 · Posts about Machine Learning written by Maruf Aytekin. jl package. Ofeishat . Optimized Matrix Multiplication using Shared Virtual Memory In OpenCL 2. MATRIX MULTIPLICATION: Project to Read a file containing two 3 X 3 Matrices and calculate their Vector Product. 3 were implemented by their corresponding distributed primitives in Spark: I. The A sub-blocks are rolled one step to the left and the B Representing non-linearity using Polynomial Regression¶ Sometimes, when you plot the response variable with one of the predictors, it may not take a linear form. GitHub Gist: instantly share code, notes, and snippets. 249–256. Marlin contains several distributed matrix operations and especially focuses on matrix multiplication which is a funda-mental kernel of high performance scientific computing. test_naive. The matrix is denoted as T (n). straightforward, but the fact that matrix multiplication itself can be accomplished in multiple ways complicates matters. Individualpair-wiseremotememory distributed matrix multiplication Dec 29, 2017 · These techniques first partition matrix X into blocks and then exploit the block-matrix multiplication when learning U and V. Based on the observa-tion, Pegasus implements a very important primitive called GIM-V (Generalized Iterated Matrix-Vector multiplication) which is a generalization of the plain matrix-vector multi-plication. 20. A distributed, MapReduce-based SOM also builds on the batch formulation described in Equation 2 . . Amatname in fn: return 1 else: return 2 def joinmap(self, key, line): mtype = self. 281–284. io/Bulk. The chunk of X assigned to the node and the corresponding norms of X are kept in the GPU memory between subsequent epochs, and the weight vectors are copied to the 'sparse' is a matrix class based on a dictionary to store data using 2-element tuples (i,j) as keys (i is the row and j the column index). 1. the FFT, LU, and dense matrix–matrix multiplication. In this part of the code, the matrix multiplication begins, again, by implementing three MapReduce jobs. Policy: Printed material is allowed. i. There was a mix of similar problems presented in assignment 1 and assignment 2 that we had to solve them using MapReduce paradigm. Sep 10, 2012 · cost of the algorithm• determined by the amount of data that has to be sent over the network in the matrix multiplication step• for each user, we have to process the square of the number of his interactions → cost is dominated by the densest rows of A• distribution of interactions per user is usually heavy tailed → small number of Github. You have seen many of the stream operations before, in Question 5 of Exercise 7, including map, reduce, filter, and forEach • Matrix multiplication • Dynamic programming External Memory Model • Addition and subtraction are fast, multiplication is fast • MapReduce model: peated matrix-vector multiplications. 249–256. Designing e cient irregular algorithms is a challenge. 6. y * dz # [dz/dx * dL/dz] dy = self . push({key: key, value: value});}; map(row, emit); return emitArray;})); // Group tuples with the same key: var reducerInput = {}; mapperOutput. This was Jul 30, 2013 · MapReduce Algorithms: Having presented the video lectures on the topic, in which Prof. Using functions from various compiled languages in Python¶. ▫ SPGEMM: KKTRI: Triangle Counting using SpGEMM. Dec 14, 2016 · Data-Intensive Computing and MapReduce/Hadoop : For more info, see the MapReduce paper, it's pretty readable. js First of all lets recap the multiplication formula: import arraymancer proc customSigmoid2[T: SomeFloat](t: Tensor[T]): Tensor[T] = result = map_inline(t): 1 / (1 + exp(-x)) Now in a single loop over t, Arraymancer will do 1 / (1 + exp (-x)) for each x found. So, the whitespace or the indentation of the very first line of the program must be maintained all throughout the code. What we want to do We will write a simple MapReduce program (see also Wikipedia ) for Hadoop in Python but without using Jython to translate our code to Java jar files. Video: Youtube Map-Reduce! Ranking (e. MapReduce in distributed model training Using the MapReduce strategy, if we can split the training data on separate workers, compute Map functions in parallel, and aggregate the results in a Reduce function, we will be able to achieve distributed model We can use the logistic regression results to classify subjects as male or female based on their height and weight, using 0. Using the old API in the Mapper and Reducer. 281–284. Length. The MRQL query language is powerful enough to express most common data analysis tasks over many forms of raw in-situ data, such as XML and JSON documents, binary files, and CSV documents. js) To run the code mongo < multiply_map_reduce. txt. g. Mimir inherits the core principles of existing MapReduce frameworks, such as MR-MPI, while redesigning the execution model to incorporate a number of sophisticated optimization techniques that achieve similar or better performance with significant reduction in the amount of memory used. Thu 11/10: yuvraj: Virtual Machines : See also the book chapter on Virtual Machines from the Wisconsin OS book. Figure 4 illustrates how output splitting a ects weight and Using the Maven POM to manage your project is an easy way to start. BSPmodel. rithms are often iterative, using repeated read-only data access and I/O-bound matrix-vector multiplications to converge to an optimal model. requests in parallel by using the underlying multiple flash memory packages. Subtypes of StaticArray will provide fast implementations of common array and linear algebra operations. With both the item-item similarity matrix and the user-item vectors, it’s now possible to multiply them together and generate recommendations for users. M. inline. MatrixMulOutput. NOTE: Please note that the Mapper function does not have access to the i, j, and k values directly. The MRQL query processing system can evaluate MRQL queries in two modes: in MapReduce mode on top of Apache Hadoop or in Bulk Synchronous Parallel (BSP) mode on top of Apache Hama. For example: MapReduce workflow: InputFormat, RecordReader, InputSplits, Map tasks, Combiners, Shuffle/Sort, Reduce tasks, OutputFormat. 55 folds) Ratio of FastPath to CompletePath memory accesses: 30:0 for software-based atomic and 3:28 for system-provided atomic implementations Since multiplication is done element-wise, you need to specifically perform a dot product to perform matrix multiplication. sum(B[i, :] * C[:, j]) Mapreduce and matrix multiplication November 2, 2016 The homework questions are due at the 23:59 on Tuesday 15 November. Specifically, the matrix multiplication operations described in Eq. Mar 12, 2018 · GitHub. 2. 0 2D card game 2048 (Android Studio, Java) *Solely developed android game on windows platform using android studio. This method, however, is very inefficient as it would require to compute a matrix multiplication and the square root of a matrix at each step. parseLong(job. If you want a specific number of elements within a range, then use the linspace function. You can check it out from here. It is used to solve problems where problem of size N is solved using solution of problems of size N - 1 (or smaller). An extra MapReduce Job has to be run initially in order to add the Row Number as Key to every row. Distributed matrix factorization with mapreduce using a series of broadcast-joins (SS, CB, MS, AA, VM), pp. Input are two matrix A and B python - How to write Mapreduce code for matrix multiplication which does not use any list of size more than 10 - Stack Overflow. One method for computing Pi (even though not the most efficient) generates a number of points in a square with side = 2. But MapReduce tries to use commodity machines to solve big data problems. js for ML using JavaScript TensorFlow 1 version, View source on GitHub the inner 2 dimensions specify valid matrix multiplication dimensions, and any level matrix computation primitives with MapReduce through the case study two basic primitives, matrix multiplication and finding linear solution, and goes into . 2. F ast monte-carlo algorithms for appr oximate matrix multiplication. ]]) >>>a[0,0]=1 >>>a[1,1]=1 >>>b =np. file') if self. StaticArrays provides a framework for implementing statically sized arrays in Julia, using the abstract type StaticArray{Size,T,N} <: AbstractArray{T,N}. The algorithm we’ll be using is a two-pass matrix multiplication algorithm for MapReduce. , PageRank) requires iterated matrix-vector multiplication with matrix containing millions of rows and columns ! Computing with social networks involves graphs with hundreds of millions of nodes and billions of edges ! Map-Reduce is a parallel programming paradigm, a software-stack that will help to address big data Used Blocked Matrix Multiplication technique to improve the convergence rate,by performing the nontrivial computation in the reduce steps. It is deployed on expensive hardware such as HPC or supercomputers. Our idea is to speed up distributed NMF in a new, orthogonal direction: by reducing the problem size of each NLS subproblem within NMF, which in turn decreases the overall computation cost. Please cite any references you use. mrjob has basic support for Google Cloud Dataproc (Dataproc) which allows you to buy time on a Hadoop cluster on a minute-by-minute basis. split()] row = int(vals) A key feature is the capability for users to write callback functions, called after each iteration, thus enabling customization for specific applications. The general algorithm matrix-vector multiplication (SpMV) and sparse matrix-matrix multiplication (SpGEMM), a systematic study on applying blocking techniques to tensors has not yet been conducted. List the top-10 vertices for graphs 1,2 & 4. Difference between MapReduce and Pig. Assume you have two matrices A and B in a sparse matrix format, where each record is of the form i, j, value. linspace ( 0 , 2 * np . Do you have any idea, about the matrix multiplication example which I mentioned in question, that why this works fine with hadoop standalone mode but does not work with hadoop distributed mode at the point of checking answers? – waqas Nov 30 '11 at 13:39 Matrix multiplication using MPI. Matrix-vector and matrix-matrix calculations fit nicely into the MapReduce style of computing. Big Data Project On A Commodity Search System For Online Shopping Using Web Mining Big Data Project On A data mining framework to analyze road accident data Big Data Project On A neuro-fuzzy agent based group decision HR system for candidate ranking Big Data Project On A Profile-Based Big Data Architecture for Agricultural Context Big Data Project On A Queuing Method for GSoC Results and Summary. Thus, Meta-MapReduce enhances the standard MapReduce and can be implemented into the state-of-the-art MapReduce systems, such as Spark, Pregel , or modern Hadoop.$ docker run -v Matrix multiplication is an important application in. In MapReduce word count example, we find out the frequency of each word. You might want an order 2 or 3 curve. General-purpose, heavy- and An example of how multi-method-based dispatch might work for a binary operation like matrix multiplication. 1. 6. xs = np . If the multiplication type computes in parallel, then the package computation is also parallel. 0-m2 m4. e. 快速矩阵乘法 Fast and Stable matrix multiplication Coppersmith and Winograd's Algorithm 时间复杂度O(n^2. ∙ 0 ∙ share While performing distributed computations in today's cloud-based platforms, execution speed variations among compute nodes can significantly reduce the performance and create bottlenecks like stragglers. Stage 2. Naive Bayes classifier to classify text documents Latent semantic analysis (LSA) Rsa breaking using a more efficient integer factorization than trial division. parsemat() vals = [float(v) for v in line. 1. I got it right. MapReduce. PRELIMINARY A. Feb 16, 2015 · Matrix Data Model for MapReduce. ones ((3, 3)) + 2) m2 = sml. There are two main security concerns in outsourcing computation: guaranteeing that the server performs the computation correctly, and protecting the privacy of the client’s data. The first is Map Only with CAP3 DNA Sequence Assembly, followed by Classic MapReduce with Pair-wise Sequences and High-Energy Physics, Iterative with K-means clustering, PageRank and Multi-dimensional Scaling, and finally Loosely Synchronous with Matrix Multiplication Algorithms. Sparse Matrix Multiplication in Map Reduce. CS231n, Convolutional Neural Networks for Visual Recognition, Stanford University; CS224d, Deep Learning for Natural Language Processing, Stanford University You can now run all stages of the rewrite system for a program (in this example for matrix multiplication): scripts / compiled_scripts / HighLevelRewrite highLevel / mmTransposedA scripts / compiled_scripts / MemoryMappingRewrite -- gr10 mmTransposedA scripts / compiled_scripts / ParameterRewrite - f highLevel / mm . MPI is a SPMD model of distributed computing, where each process is completely independent and one just controls the memory handling. ,3. Aug 25, 2017 · Matrix Multiplication using MapReduce Programming in Java. json mmTransposedA Jul 14, 2020 · MapReduce Program – Finding The Average Age of Male and Female Died in Titanic Disaster; MapReduce – Understanding With Real-Life Example; How to find top-N records using MapReduce; How to Execute WordCount Program in MapReduce using Cloudera Distribution Hadoop(CDH) Matrix Multiplication With 1 MapReduce Step; MapReduce – Combiners matrix multiplication operations within the NMF algorithms. An extra MapReduce Job has to be run initially in order to add the Row Number as Key to every row. It uses a distributed file system called GFS, which is Google File System. Matrix Multiplication with Spark. ], [ 0. Plagiarism checking with MapReduce Manages: Pelle Jakovits ([email protected] 22 706. ####MapReduce problems#### Count number of words in a book; Total number of words in a book; tf-idf calculations for ranking; Cosine distance measure between 2 documents; Matrix multiplication in MapReduce Aug 13, 2016 · Matrix Computations and Optimization in Apache Spark Reza Bosagh Zadeh Stanford and Matroid 475 Via Ortega Stanford, CA 94305 Xiangrui Meng Databricks 160 Spear Street, 13th Floor San Francisco, CA 94105 Alexander Ulanov HP Labs 1501 Page Mill Rd Palo Alto, CA 94304 [email protected] zeros(4). e. Jueon Park, and Kyungyong Lee, The 8th International Workshop on Autonomic Management of high performance Grid and Cloud Computing (AMGCC'20), Accepted, 07/2020. OSDI’04 ,San Francisco, CA; Petros Drineas, Ravi Kannan, and Michael W. edu Burak Yavuz Databricks 160 Spear Street, 13th Floor San Francisco, CA 94105 [email protected] the code works perfectly fine with smaller matrices but when the files becomes large the mapping p Nov 12, 2020 · Matrix B is also a 2×2 matrix where number of rows(j)=2 and number of columns(k)=2. Data Analysis of huge amount of open Github Data, where we tried to find some deep patterns among popularity and spatial distributions of programming languages and users on Github. Moreover, your code has two steps, that means two jobs. Course content Because using map is equivalent to for loops, with an extra code we can always write a general mapping utility: >>> def mymap(aFunc, aSeq): result = [] for x in aSeq: result. It may be possible to switch between MapReduce and MPI to perform scalable matrix inversion in these The next step is PageRank, a fairly straightforward linear algebra problem. Lecture 15 (5/26): Partitioning for PageRank Lecture 15, Partitioning for Pagerank. While some arrays — like Array itself — are implemented using a linear chunk of memory and directly use a linear index in their implementations, other arrays — like Diagonal — need the full set of cartesian indices to do their lookup (see IndexStyle to Feb 08, 2010 · This posting gives an example of how to use Mapreduce, Python and Numpy to parallelize a linear machine learning classifier algorithm for Hadoop Streaming. get(\"NumberOfDocuments\"));} The variable N can then be used with the map and reduce functions. 5 × 10 5 and the number of nonzero elements equal to 6 × 10 5. By using Rings, we can do matrix multiplication over things other than numbers (which on occasion we have done). 1. sin ( xs ) # np. We use dense() to create a dense in-memory matrix from our toy dataset and use drmParallelize to load it into the cluster, “mimicking” a large, partitioned • Used Netflix data to offer movie recommendations to users based on their previous favorites. , 2015. CS231n, Convolutional Neural Networks for Visual Recognition, Stanford University; CS224d, Deep Learning for Natural Language Processing, Stanford University Publication: M. Syntax of Mongo mapReduce() Following is the syntax of mapReduce() function that could be used in Mongo Shell > db. Figure 4 illustrates how output splitting a�ects weight and In modern processors, integer division can be 10-50 times slower than multiplication. AWS Elastic MapReduce, an adaptation of Apache Hadoop to Matrix Multiplication: 2 MapReduce steps! Matrix M can be thought of as a relation with tuples (i, j, m ij) ! Matrix N can be thought of as a relation with tuples (j, k, n jk) ! Map operation creates these tuples ! Map: Join of M and N brings us closer to M X N by creating: Relation (i, j, k, m ij, n jk) or the relation (i, j, k, m ij X n jk) ! Jul 26, 2016 · Outline • Introduction • DFS • MapReduce • Examples • Matrix Calculation on Hadoop 3. Advices [edit | edit source] In this section, you will see some advices that can help you to desing a Python source Refactor compression package and add functionalities including quantization for lossy compression, binary cell operations, left matrix multiplication. For sparse computations, they usually depend on the nonzero pattern of the matrix. per batch of updates where mis de-fined as the maximum number of edges in the graph be- Jun 15, 2013 · For those who prefer reading a code instead of text - GitHub: Approach1(multiply_map_reduce. For example : mat1 is 2×3 means mat2 will be 3×2. Implement SELECT MAX(<field>) FROM <table> GROUP BY <field> with MapReduce. 8 May 2015 https://github. . In all cases, Twister outperforms or is close to the competition. Implementations in CUDA Sep 2015 – Sep 2015 2. 7. . Create a matrix of processes of size p1/2 1/2 x p so that each process can maintain a block of A matrix and a block of B matrix. X-rays) are sent through an object from various angles Google MapReduce. Combiner Edit the “MapTask” method to add support for running a Combiner. The repository provides demo programs for implementations of basic algorithms on Spark 2. h. apply([], input. N(0;˙2). Many applications in different areas exist already for MapReduce. pi , 100 ) ys = np . Matrix Multiplication with MapReduce Big Data possibly now has become the most used term in the tech world for this decade. 95% c Thanks Thomas. Mark Kröll 击 see https://hadoopecosystemtable. 7. in a way you should be familiar with. Android4. Mo 09 Dezember 2013 How to use Jekyll with GitHub ; Di 05 Juli 2016 Pythons map, reduce and filter as list Matrix multiplication on multiple cores in Jun 25, 2012 · Translating to MapReduce: rethinking matrix multiplication It’s now possible to use MapReduce to multiply the user vectors computed in step 1, and the co-occurrence matrix from step 2, to produce a recommendation vector from which the algorithm can derive recommendations. Matrix-Vector multiplication As an example if we consider a Matrix-Vector multiplication (taken from the book Mining Massive Data Sets by Jure Leskovec, Anand Rajaraman et al To store the past gradients, we will use a matrix G. International Journal of Computer Science and Information Technology (IJCSIT) 9 (5): 29 - 37 ( October 2017 Figure 1 shows the general matrix multiplication (GEMM) operation by using the block sparse format. Look for “# [ADD COMBINER HERE]” for the place one would add this. 2 minute read. the rule of matrix multiplication is mat1 columns is equal to mat2 rows values. It may help if you checkout my introduction to map-reduce and an example here. To do so, we are taking input from the user for row number, column number, first matrix elements and second matrix elements. Contribute to JaredP94/MapReduce-Matrix-Multiplication development by creating an account on GitHub. GitHub Gist: instantly share code, notes, and snippets. MapReduce is a programming model that Google came up with to handle computation of their huge large scale data. The examples below use the einsum / notation for the elements of tensors, namely m[i,j] for element i,j of the matrix m, instead of the more mathematical notation m_ij. #!/usr/bin/env python import sys from operator import itemgetter prev_index = None value_list = [] for line in sys. Dijkstra's algorithm. GPU Accelerated Computing with C and C++, which also has some videos. Combiner Edit the “MapTask” method to add support for running a  Sparse matrix multiplication for hadoop. Matrix Multiplication Operators. to implement matrix inversion using other parallelization platforms such as MPI, a MapReduce matrix inversion technique that can be used as a pluggable component in complex Hadoop data analysis workflows is highly desirable. The instruction sets are typed, and instructions designed to operate on packed doubles can’t operate on packed ints without explicit casting. (Duh!) Map takes data and converts it into another set of data, where individual elements are broken down into tuples (key/value pairs). rusty-machine - a pure-rust machine learning library. - Use MapReduce to calculate tag similarity in twitter and improves the speed from 90min to 36min. e. 0 and measured the performance of the same with previous implementations. 5 In mathematics, matrix multiplication or the matrix product is a binary operation that produces a matrix from two matrices. ]]) Matrix Multiplication Examples (both using global memory and shared memory) CUDA C Programming Guide; CUDA Toolkit documentation, which includes CUDA installation, C programming guide, APIs for cuBlas, cuFFT etc, tools, compiler SDK, and others. “Limitations and Challenges of HDFS and MapReduce” by Weets et al. 10 of Mahout it became obvious that using Hadoop MapReduce was causing more pain than it was solving, due to massively redundant data reads required 1 Problems Suited for MapReduce 2 MapReduce: Applications Matrix-Vector Multiplication Information Retrieval 3 Hadoop Ecosystem Designing a Big Data System Big Data Storage Technologies Slides are partially based on Slides “Mining Massive Datasets” by Jure Leskovec. What inference can you derive from using PageRank on the these datasets. The minimum is found by a multi-step reduction algorithm. 10 Matrix Multiplication with One MapReduce Step . 6. sin is a universal function plt . Jan 17, 2021 · Matrix Multiplication With 1 MapReduce Step; Hadoop - copyFromLocal Command go to this GitHub Repo and download the receptacle organizer as a speed as For advanced use of the CUDA, you can use the driver API wrappers in CUDA. 1. per batch of updates where mis de-fined as the maximum number of edges in the graph be- Matrix multiplication: n^3/p + (n^2/p^{2/3}) \\cdot g + l: Sorting (n \\log n)/p + (n/p)\\cdot g + l: Fast Fourier Transform (n \\log n)/p + (n/p)\\cdot g + l: LU Decomposition: n^3/p + (n^2/p^{1/2})\\cdot g + p^{1/2}\\cdot l: Cholesky Factorisation: n^3/p + (n^2/p^{1/2})\\cdot g + p^{1/2}\\cdot l: Algebraic Path Problem (Shortest Paths) Distributed file systems and map-reduce as a tool for creating parallel 2. with callbacks. Assume you have two matrices A and B in a sparse matrix format, where each record is of the form i, j, value. For matrix computation library built on top of Spark which is an distributed in-memory cluster computing framework. Ex. Vogelstein, Carey E. reshape(2,2) >>>b array([[0, 1], [2, 3]]) >>>a*b array([[ 0. Using the new API in the Mapper and Reducer. Remember, a combiner runs the reduce task at the end of the map task in order to save communication cost of sending to multiple reducers. tar. Specifically, for building MapReduce jobs, you only need to have the hadoop-client dependency, which contains all the Hadoop client-side classes needed to interact with HDFS and MapReduce. The goal is to calculate A * B. [experimental] New python bindings with supports for several builtin s, matrix operations, federated tensors and lineage traces. Lowering XLA HLO to I E 6 func @mnist_predict(%input: tensor<1x28x28x1xf32>) /> tensor<1x10xf32> {%1 = mhlo. hollywood-2011 2. PageRank) Gradient descent methods Stochastic SVD Tall skinny QR This paper presents a MapReduce algorithm for computing pairwise document similarity in large document collections. I couldn't find a simple way to do this within the EMR framework, though I bet there is a way to do it. import systemml as sml import numpy as np m1 = sml. On the left are the full matrix organized in blocks and its internal memory representation: compressed values and block indices. The essence of the transformation is to unroll the input patches (a 3D matrix) and �lters (a 4D matrix) in 2D in a way that a single matrix-matrix multiplication produces the unrolled version of the output in 2D. Common operations include synchronizing the GPU, inspecting its properties, starting the profiler, etc. Sparse Matrix-Vector Multiplication { Size of Distributed Matrix Multiplication Using Speed Adaptive Coding 04/15/2019 ∙ by Krishna Narra , et al. Da Zheng, Disa Mhembere, Joshua T. Figure 1 and Figure 2 show two alternative MapReduce plans for matrix multiplication (details of the two plans will be discussed in Section IV). MapReduce is an attractive framework because it allows us to decompose the inner products involved in computing document similarity into separate multiplication and summation stages in a way that is well matched to efcient disk access May 12, 2018 · On Intel CPUs, SSE instruction sets use up to 128 bit registers (xmm, four ints), AVX and AVX2 use up to 256 bit registers (ymm, eight ints), and AVX512 use up to 512 bit registers (zmm, sixteen ints). 1957: Minimum degree sparse matrix reordering (Harry Markowitz) 1961: QR algorithm (V. Curtis Huttenhower, John Quackenbush, Lorenzo Trippa & Christine Choirat. Thus, r u v ≈ x T u ⋅ θ v, where x u, θ v ∈ R f are the u t h Map-Reduce also makes short work of dealing with large matrices and can crunch matrix operations like matrix addition, subtraction, multiplication etc. 2 and the deflation operation defined in Eq. Implement inner join between two tables with MapReduce. Design a MapReduce algorithm to compute matrix multiplication: A x B Matrix Multiplication. The common matrix operations such as 'dot' for the inner product, multiplication/division by a scalar, indexing/slicing, etc. forEach(function(keyValue) Example matrix multiplication in distributed environment using R, MPI and Hadoop MapReduce - aaparo/MultMatrix Use Git or checkout with SVN using the web URL. And more generally, taking a sum of outer products of corresponding rows and columns of the input matrices, always returns the desired matrix multiplication result. No electronic device (except for electronic calculator). The class will cover widely used distributed algorithms in academia Dynamic programming is well known algorithm design method. Matthews Author content It can be used in conjunction with other functionality like Map, Reduce, Filter in Python. append(aFunc(x)) return result >>> list(map(sqr, [1, 2, 3])) [1, 4, 9] >>> mymap(sqr, [1, 2, 3]) [1, 4, 9] >>> Aug 25, 2011 · In-Database Operations • Matrix and vector multiplication: Av SELECT 1, array_accum(row_number, vector*v) FROM A array_accum(x,v) is a custom function. Both of these are considered to be whitespaces when you code. Figures - uploaded by Devin A. Current Apache Mahout: Beyond MapReduce Matrix Multiplication. Note: Matrix operations for floats are accelerated using BLAS (Intel MKL, OpenBLAS, Apple Accelerate …). ,0. 1. Ligra+ is able to represent a variety of synthetic and real-world graphs using General Matrix-Vector multiplication: y <- alpha * A * x + beta * y Source Edit proc gemv [T: SomeInteger] (alpha: T; A: Tensor [T]; x: Tensor [T]; beta: T; y: var Tensor [T]) { } {. Clearly it holds that. 11 Jan 2009 Okay, so how can we compute PageRank using MapReduce? we'll take is to use MapReduce to repeatedly multiply a vector by the matrix M  23 Feb 2021 Multiplies matrix a by matrix b, producing a * b. Say the co-occurrence matrix for 4 items is Courses ¶. It also discusses various hadoop/mapreduce-specific approaches how to potentially improve or extend the example. MapReduce¶ MapReduce was designed by Google to address the problem of large-scale data processing. append ( (index,value)) else: if prev_index: value_list = sorted (value_list,key=itemgetter (0)) i In this video u will learn about Matrix Multiplication using Map Reduce in Big-Data. p. The 27 Jun 2020 Submit results from this paper to get state-of-the-art GitHub badges and help the community compare results to other papers. Francis) 1964: Sinkhorn (Richard Sinkhorn) 1965: Golub-Reinsch SVD (Gene Golub) 1969: Sparse matrix ordering (Elizabeth Cuthill, James McKee) 1969: Strassen matrix multiplication (Volker Strassen) INTRODUCTION TO DATA SCIENCE JOHN P DICKERSON Lecture #4 –9/5/2019 CMSC320 Tuesdays & Thursdays 5:00pm –6:15pm Graph reachability, using Matrix-Matrix multiplication of adjacency matrices. Download [edit | edit source] The source code is available on Github. One purpose of matrix decomposition is reducing calculations cost while solving a system of linear equations by decomposing the coefficients matrix into a product of two triangular matrices. MapReduce • Programming Model for Large-Volume Data Processing • Specialized for frequent use case: aggregation queries – Map every input object to set of key/value pairs – Reduce (aggregate) all mapped values for same key into one result for that key • Use this structure as explicit API for cluster computing May 01, 2018 · You fit this matrix to approximate your original matrix, as closely as possible, by multiplying the low-rank matrices together, which fills in the entries missing in the original matrix. Using MapReduce, we could achieve N times throughput by having N workers running in parallel. Here, we will discuss the implementation of matrix multiplication on various communication networks like mesh and hypercube. 99 CAN $45. Taifi and Y. 1. Qatawneh , and H. Now One step matrix multiplication has 1 mapper and 1 reducer. In a nutshell, we reduce the size of each NLS subproblem, by employing a matrix sketching technique: Having that said, the ground is prepared for the purpose of this tutorial: writing a Hadoop MapReduce program in a more Pythonic way, i. Aug 02, 2016 · Next we show various attempts for scalable implementation of matrix multiplication using spark, and the winning method which combines numpy matrix multiplication along with spark’s broadcast and Courses ¶. View on GitHub Spark. Challenge: Make linear https://github. ones ((3, 3)) + 2) m2 = sml. Published: December 08, 2018 Hi everyone, this is the final (summarizing) blog post for my Google Summer of Code project. the FFT, LU, and dense matrix matrix multiplication. For both classes, few matrix operations dominate the overall algorithm runtime, apart from the costs for the initial read from distributed le system or object storage. F. In fact, large-scale matrix multiplication can hardly be handled by the single-node matrix computation libraries due to hardware resource limitation. The unit of Parallel Matrix Multiplication on Open MPI. github. Fortunately, the authors in proposed a method of matrix inversion using MapReduce. js mongo < multiply_map_aggregate. 26 >>>a =np. Straggler Robust Distributed Matrix Inverse Approximation 03/05/2020 ∙ by Neophytos Charalambides , et al. ], [ 0. The reduce( ) step in the MapReduce Algorithm for matrix multiplication Facts: The final step in the MapReduce algorithm is to produce the matrix A × B. In this example did the matrix multiplication. git. There are many parallel computation prototype for matrix multiplication using FaaS, and discuss and synthesize insights entry-barriers of. This workload tests the Naive Bayesian (a popular classification algorithm for knowledge discovery and data mining) trainer in Mahout 0. However, I want to tell you something, if your file is not big enough, you will not see an improvement in term of execution speed. 3. Besides matrix-vector multi-plication (e. There are Python 2. map(function(row) { var emitArray = []; var emit = function(key, value) {emitArray. 7 codes and learning notes for Spark 2. In SpMV, the optimal selection of storage format is one of the key aspects of enabling the best performance. source The code is available on my GitHub account of the converting a collection to stream using stream method; reading from a file using Files. the result is the same as mat2. Use of ufuncs is an esssential aspect of vectorization and typically much more computtionally efficient than using an explicit loop over each element. Dec 29, 2017 · Background Matrix factorization is a well established pattern discovery tool that has seen numerous applications in biomedical data analytics, such as gene expression co-clustering, patient stratification, and gene-disease association mining. Using the best currently known parallel matrix multiplication [Wil12, LG14], our algorithm dynamically maintains the number of k-cliques in O min m 0:469k 235;( + m)0 :469k+0 amor-tized work w. One, you can multiply BIG matrices in a memory efficient way, without needing to pull everything out of SQL. Our extended framework, which we call Ligra+, uses less space than Ligra, while providing comparable or improved performance. 2-CDH3U2 ← Matrix Multiplication using MapReduce – 2 Step Solution A Movie recommender system using Netflix movie data and currently on the stage of achieving matrix multiplication, use the item-collaborative algorithm and Hadoop MapReduce Auto Complete Apr 2018 traditional matrix-vector multiplication requires): 1. Then matrix-vector multiplication m * v is defined as: w[i] = sum_j m[i,j] * v[j]. The verifiable computation of Gennaro, Gentry and Parno addresses both concerns for Mimir: Mimir is a new implementation of MapReduce over MPI. Nov 20, 2020 · Matrices represented using COO format Matrix Multiplication Using Two Passes. Use Git or checkout with SVN using the web URL. This can be parallelized easily (just matrix-vector multiplication) but needs to chain together multiple MapReduce tasks. Spring 09 Publication: M. Jul 14, 2013 · The advantage of the above logic is, we can use a distributed map reduce model for compute with multiple map-reduce tasks - Constructing the co-occurrence matrix, Finding the dot product for each user etc. mapReduce # Python 3 my_strings = ['a', 'b', 'c', 'd', 'e'] my_numbers = [1,2,3,4,5] results = list(zip(my_strings, my_numbers)) print(results) As a bonus, can you guess what would happen in the above session if my_strings and my_numbers are not of the same length? Large-scale machine learning is another important use of MapReduce. A well-known matrix factorization method is Singular value decomposition (SVD). 522–532. Thursday, August 25, 11 30 The Mahout In Action (Chapter 6) book contains a recommendation method based on matrix multiplication that uses co-occurrence data (C) in combination with user preferences (U) to generate user recommendations (R). com/awslabs/lambda-refarch- mapreduce/. To put it in a crude analogy, Page Rank, Inverted Index and Matrix Multiplication - asarraf/Algorithm- Implementation-Using-Map-Reduce. arange(4). rb. x is a shorthand for the elements of the first tensor argument. * Distributed performance bug fixes. 1–12. To maximize parallelism, the Gram matrix is calculated, that is, a matrix of the distances between every data instance and the nodes of the SOM. } General Matrix-Vector multiplication: y <- alpha * A * x + beta * y Source Edit proc gemm [T: SomeFloat | Complex] (alpha: T; A, B: The product of matrices A and B is calculated by multiplying the elements of rows in A with the corresponding columns in B, and then adding the resulting values to produce a single value for each row in A and each column in B. The Formula is: Mapper for Matrix A (k, v)=((i, k), (A, j, Aij)) for all k That is, we can implement matrix multiplication as the cascade of two MapReduce operations, as follows. to distributed matrix multiplication and distributed learning. regCG, for example, only lines 4 and 9 access matrix X; all other computations are inexpensive operations over small vectors or scalars. 5 as a cutoff, as shown in the plot below. com/Cloveryww/MPI-parallel-algorithms/tree/master/ cannon Since the algorithm and the lower hair collection task results when us 26 Nov 2018 Map Reduce (Part 3). sum (axis = 1). matrix (np. I'm working on the matrix multiplication using mapreduce. G. assign(vi;vnew) : decide how to update vi with vnew. Nov 29, 2017 · MapReduce, by Google, in 2004; Hadoop (fair mode), Spark (easy mode) MPI (hard mode) Matrix multiplication A = BxC for i in range(m): for j in range(n): for k in range(r): A[i][j] += B[i][k] * C[k][j] Matrix multiplication Vectorized for i in range(m): for j in range(n): A[i, j] = np. Problem Motivation Apr 04, 2019 · Other Links: GitHub Repository. Eq. The is similar to the process of generating the Row Number as explained in the previous post. It involves the matrix Oct 08, 2019 · In Cholesky method, a positive-definite matrix is written as the matrix multiplication of a lower-triangular matrix and its transpose. RecSys-2013-ZhuangCJL #memory management #parallel #performance A fast parallel SGD for matrix factorization in shared memory systems ( YZ , WSC , YCJ , CJL ), pp. 5 cutoff. GitHub Gist: instantly share code, notes, and snippets. Google’s MapReduce (Example)1 Classic example: word count. , non-causal attention where there is no notion of past and future. com/kokkos/kokkos- kernels \"Graph twiddling in a mapreduce world. Matrix-Vector Multiplication. ACM Transactions on Embedded Computing Systems (TECS), Vol 16, Issue 4, May 2017. Databases 2 Application 1: Matrix-Vector Multiplication. rstrip (). 55 folds) Ratio of FastPath to CompletePath memory accesses: 30:0 for software-based atomic and 3:28 for system-provided atomic implementations Matrix factorization (MF) factorizes a matrix R ∈ R m × n (with N z non-zero elements) into two low-rank matrices X ∈ R m × f and Θ ∈ R n × f, such that R ≈ X ⋅ Θ T. The output should be similar with the input. Matrix Multiplication With MapReduce. require ' rubygems ': require ' matrix ': require '. plot ( xs , ys ); Apr 09, 2013 · OutcomesRecognize relationships between matrix methods andthings you’ve already been doing\" Example SQL queries as matrix computationsUnderstand how to use Hadoop to compute thesematrix methods at scale for BigData\" Example Recommenders with social network infoUnderstand some of the issues that could arise. combine2(mi;j;vj) : combine mi;j and vj. Priebe, and Randal Burns. /matrix_block_mixin ' Mar 28, 2012 · That is, we can implement matrix multiplication as the cascade of two MapReduce operations, as follows. Background MapReduce and Spark Matrix Multiplication Note: also with row/column vector rhs Note: 1:N join. Matrix multiplication is an important multiplication design in parallel computation. Please turn in source codes, compilation, submission scripts used and also output les. 28 Mar 2012 P is a matrix = MN with element pik in row i and column k, where pik =∑j mijnjk Relational Representations: M = M(I, J, V ), with… Matrix Multiplication using MapReduce – 2 Step Solution Source Code: GitHub. ]]) >>>np. 2 A New Parallel Matrix Multiplication Algorithm on Hex-Cell Network (PMMHC) Using IMAN1 Super Computer E. input. However, such a straightforward approach does not apply to matrix tri-factorization because, as we show in the Methods section, the learning of any block of U and V depends on factor S . 0. May 12, 2018 · On Intel CPUs, SSE instruction sets use up to 128 bit registers (xmm, four ints), AVX and AVX2 use up to 256 bit registers (ymm, eight ints), and AVX512 use up to 512 bit registers (zmm, sixteen ints). Contribute to tangsttw/Matrix- Multiplication-MapReduce development by creating an account on GitHub. ,3. Report this profile Addition of numbers, matrix multiplication inside a docker and using MapReduce as the programming model. reshape(2,2) >>>a array([[ 0. y = y return z def backward ( dz ): dx = self . Lecture 14, Matrix Computations and Optimization in Apache Spark, Sparse matrix multiplication using SQL, Sparse matrix multiplication in MapReduce. Everybody is talking about it and everybody have their own understanding towards it which has made its definition quite ambiguous. My Personal Notes arrow_drop_up. Dec 16, 2019 · Outsourcing computation has gained significant attention in recent years in particular due to the prevalence of cloud computing. Use the output from the Stage 1 Reducer and pass along the same input to the Stage 2 reducer, where all the values having same index pairs are summed up to get the final output value. 04/27/18 - Large-scale machine learning and data mining applications require computer systems to perform massive computations that need to be May 02, 2018 · One kernel that can be parallelized using SPMD parallelism is dense matrix-matrix multiplication, in which we multiply two input matrices A and B to produce an output matrix C. Introduction Modern data-mining or ML applications, called «big-data analysis» requires us to manage massive amounts of data quickly. Here two passes symbolises the fact that we will need two map reduce jobs to compute the matrix multiplication. Then we are performing multiplication on the matrices entered by the user. 13. This matrix at each step will be updated and extended. After scaling genotype and expression data to unit variance with matrix-vector multiplications (or matrix-matrix with a small second matrix), and (2) closed-form algorithms with transpose-self matrix multiplication. x * dz # [dz/dy * dL/dz] return [ dx , dy ] “Scale-free Sparse Matrix-Vector Multiplication on Many-Core Architectures, “ IEEE Transactions on Computer-Aided Design of Integrated Circuits and Systems (TCAD), Vol 36, Issue 12, Dec 2017. Large Scale Machine Learning: Page: Gradient descent with large data, stochastic gradient descent, mini-batch gradient descent, map reduce, data parallelism, and online learning. It is crucial for perfor-mance to t the data into single-node or distributed main memory and enable fast matrix-vector opera-tions on in-memory data. The aim of this library is to easily design prototypes of algorithms using MapReduce. The product of a n x n matrix M by a vector v of length n is given by Dec 25, 2020 · Matrix Multiplication At Scale Using Map Reduce Matrix multiplication is the one of the most fundamental operation that most of the machine learning algorithms rely on. Taifi and Y. Optimizations Around version 0. x = x # must keep these around! self . For any u and v, such tat 1 ≤ u ≤ m and 1 ≤ v ≤ n, r u v is the (i, j) entry of R. toNumPy () Output: array ([[ - 60. This file shows the output of the resultant matrix obtained by multiplying matrix A and B in the format \"(row,column) value\" Contribute to JaredP94/MapReduce-Matrix-Multiplication development by creating an account on GitHub. It takes the value v and puts it in the row indexed by x. Feb 08, 2015 · Write a MapReduce query to remove the last 10 characters from each string of nucleotides, then remove any duplicates generated. A file containing two Matrices - MatrixA and MatrixB, was fetched from the Hadoop Distributed File System (HDFS) as an input for the Map Reduce task. - Use OpenMP and SSE to improve the speed of matrix multiplication over 86 times and kmeans 11 times. As the instructor pointed out, there are reasons this approach isn't insane. a + (b + c) = (a + b) + c a * (b * c) = (a * b) * c max (a, max (b, c)) = max (max (a, b), c) min (a, min (b, c)) = min (min (a, b), c) Strings with string-concatenation form a semigroup. There are 2 implementations of the 3 Feb 2018 PDF | On Dec 1, 2017, Mais Haj Qasem and others published Matrix multiplication of big data using MapReduce: A review | Find, read and cite 4. Map Reduce Example for Sparse Matrix Multiplication. in a way you should be familiar with. Here, we will discuss the implementation of matrix multiplication on various communication networks like mesh and hypercube. The definition is motivated by linear equations and linear transformations on vectors, which have numerous applications in applied mathematics, physics, and engineering. Write a MapReduce query to remove the last 10 characters from each string of nucleotides, then remove any duplicates generated. distributed map reduce In this module, we will learn about the MapReduce paradigm, and how it can be used to write distributed programs that analyze data represented as key-value pairs. It takes in 6 parameters: n: number of rows in A; m: number of I would like to apply map-reduce to deal with matrix multiplication in python with Hadoop. Jul 28, 2020 · Analyzing weather data of Fairbanks, Alaska to find cold and hot days using MapReduce Hadoop. , line 9), we have vector-matrix multiplication, often caused by the rewrite X>v !(v>> > > 2 Implement SELECT * FROM <table> WHERE <condition> with MapReduce. jl. 1 Matrix Multiplication Hadoop Implementation . * DRM row sampling api. Illustrates how a variety of coercion-based defaults can be specified to make life easy on the implementer, while still easily allowing for dispatch to optimal implementation-specific routines whenever it's desired. You might want to examine the Hadoop code for Word Count and Matrix multiplication. DATA/DATA SCIENCE Data Science from Scratch ISBN: 978-1-491-90142-7 US$39. So, several papers have studied the problem of multiplying matrices using a large number of processors (CPUs) in parallel. Kyungyong Lee is an assistant professor in the College of Computer Science at Kookmin University. The instruction sets are typed, and instructions designed to operate on packed doubles can’t operate on packed ints without explicit casting. In this post, we will be writing a map-reduce program to do Matrix Multiplication You need Hadoop’s HDFS and map GitHub Gist: star and fork jaganadhg's gists by creating an account on GitHub. Jun 23, 2014 · To the best of our knowledge, there are no matrix inversion algorithms using MapReduce, although there are several software systems for other matrix operations using MapReduce. FlashMatrix: Parallel, Scalable Data Analysis with Generalized Matrix Operations using Commodity SSDs. Here we reduce the output received from mapper into actual 2D array for matrices A and B and calculate the matrix multiplication based on usual formula. • Irregular algorithms Penetrating rays (e. Instructors. (1) can be minimised with respect to and solved for , yielding a large matrix multiplication problem, a formulation that is employed by the R package Matrix eQTL , which allows for fast eQTL analysis on a desktop computer. matrix',default='A', dest='Amatname') def parsemat(self): \"\"\" Return 1 if this is the A matrix, otherwise return 2\"\"\" fn = get_jobconf_value('map. Matrix factorization learns a latent data model that takes a data matrix and transforms it into a latent feature space enabling generalization, noise 1 Mar 2018 implement by naive algorithm(without partition) and advanced algorithm(with partition) - AiningWang/Matrix-Multiplication-using-MapReduce. d. . I hope these programs will help people understand the power of distributed parallel computing via map-reduce on Spark platform. An AggBinaryOp hop can be compiled into the following physical operators. collection. Howe outlined, together with the theory, some solutions to the assignments, we started to make our hands really dirty. Pros . UPDATE: Please note that the Mapper function does not have access to the Row Number (in this case i, and k) directly. 09, May 20. F. lines method; using the generate method (provide a Supplier) or iterate method (providing the initial value and incremental operation). Hi all. 550 Architecture of Machine Learning Systems based on fast matrix multiplication. Mar 29, 2012 · Posted by Venkata (Ravi) Adusumilli on March 29, 2012 in Hadoop, MapReduce Tags: Hadoop 0. toNumPy () Output: array ([[ - 60. The JobConfigurable#configure has to be implemented in the Mapper and Reducer class. Apr 14, 2012 · Prepare for Matrix Multiplication. stdin: curr_index, index, value = line. g. enron 4. But somehow the generation of the <Key> <value> pair and the operation in the 1. For running unit tests, use junit, and for writing MapReduce tests, use mrunit. Jan 27, 2021 · // The slave process receives the sub portion of the Matrix A which assigned by Root : MPI_Recv(&matrix_a, rows*N, MPI_DOUBLE, source, 1, MPI_COMM_WORLD, &status); // The slave process receives the Matrix B: MPI_Recv(&matrix_b, N*N, MPI_DOUBLE, source, 1, MPI_COMM_WORLD, &status); // Matrix multiplication: for (int k = 0; k<N; k++) Mar 31, 2012 · P is a matrix = MN with element p ik in row i and column k, where p ik =∑ j m ij n jk. We represent matrix M as a relation , with tuples , and matrix N as a relation , with tuples . Monte Carlo Integration. 19. Implementation . Also a variation for pair-wise distance matrix of two different inputs x and y: sqDist(x,y), dsqDist(x,y). py. 1 Matrix Multiply and Computer Architectures Homework question 1 Matrix multiplication with MapReduce If A is an m × p matrix and B is an p × n matrix, then the product of A and B is the m × n matrix C = AB, where the (i, j) th element of C is computed as the inner product of the ith row of A with the jth column of B: This is a dot product—simple arithmetic if m, p, and n are small. The map takes ( le, content) pair, and emits (word, 1) pairs for each word in the content. MapReduce: Simplified Data Processing on Large Clusters. Step 1: We can download the dataset from this Link , For various cities in different years. At first it was a little brain-breaky, but then we did the map-reduce version, which was brain-breakier. Written by Luka Kerr on April 2, 2018 I’ve been learning MIPS assembly for about 2 weeks now at uni and wanted to share how i’ve implemented a simple matrix multiplication function in MIPS. That said, the ground is now prepared for the purpose of this tutorial: writing a Hadoop MapReduce program in a more Pythonic way, i. linspace ( 0 , 2 * np . Clustering with KMeans: Page: Clustering with KMeans in scikit-learn. “Matrix factorizations at scale: A comparison of scientific data analytics in Spark and C+MPI using three case studies”, 2016 IEEE International Conference on Big Data (Big Data), pages 204–213, Dec 2016. I will explain LSH and how to use this package as well as the details of the implementation below. \"Git-a\" -rec It further employs a content-based filtering approach, coupled with Apache Spark to develop a recommender system, for Github users. Use of ufuncs is an esssential aspect of vectorization and typically much more computtionally efficient than using an explicit loop over each element. For these setups, coding has been utilized primarily to handle failed or straggling (delayed) workers –, where some workers fail or are significantly slower than the other workers, causing a significant delay in the overall computation time. e. g. 2) Matrix Multiplication 3) Find Mutual Friends for Social Media Data. Contribute to kdave2/Matrix- Multiplication-Hadoop-Map-Reduce development by creating an account on GitHub. Map, Reduce and Filter functions in Python make writing code much more easier (less lines of code) and I think they are optimized internally which will make them more faster than writing custom code which will most Sparse linear algebra Matrix Multiplication Spectral methods FFT N-Body methods GEM Structured grids SRAD Unstructured grids CFD solver MapReduce Combinational logic CRC Graph traversal Breadth-First Search (BFS) Dynamic programming Needleman-Wunsch Backtrack and branch-and-bound Graphical models Hidden Markov Model Map Reduce Triplets Block Matrix Multiplication Let’s look at Block Matrix Multiplication (on the board and on GitHub) Created Date: Oct 23, 2020 · The above analysis is relevant for so-called bidirectional attention, i. This model defines data abstraction as key-value pairs and computation flow as “map, shuffle and then reduce”. Matrix multiplication is an important multiplication design in parallel computation. MapReduce is a parallel fram ework for big da ta, which That is, we can implement matrix multiplication as the cascade of two MapReduce operations, as follows. p. private static Long N; public void configure(JobConf job) {N = Long. Designed a 4x4 grid layout of game card. We consider numerical computations using dataflow graphs, with a focus on learning deep neural networks for image classification and other classification tasks. x library which implements a MapReduce algorithm. reshape (%input) : (tensor<1x28x28x1xf32>) /> tensor Matrix multiplication in C. – Description of AdaGrad. First: The Map Function\":Foreachmatrixelementm ij,producethekeyvaluepair j, (M,i,m ij) # 21 hours ago · mapreduce python 3, Jan 28, 2020 · You can indent using tabs and spaces in Python. 4. That is, we can implement matrix multiplication as the cascade of two map-reduce operations, as follows. sin is a universal function plt . are overloaded for convenience. The : operator takes syntax start:spacing:end. We classify the matrix multiplication problems into Apr 26, 2012 · Posted by Venkata (Ravi) Adusumilli on April 26, 2012 in Hadoop, MapReduce Tags: Hadoop 0. 3. sh Matrix_Reducer. github. ∙ 0 ∙ share A cumbersome operation in numerical analysis and linear algebra, optimization, machine learning and engineering algorithms; is inverting large full-rank matrices which appears in various processes and applications. gz Step 2: Move the eclipse folder to the home directory In this step, you can see how to move the eclipse folder to the home directory. Example MapReduce Algorithms Matrix-vector multiplication Power iteration (e. A trivial implementation is trivial, but users are likely to want fast versions that are hard to write. io/. Graph Processing using Map-Reduce 2. Oct 10, 2017 · Using MapReduce Programming In mathematics, matrix multiplication or the matrix product is a binary operation that produces a matrix from two matrices. The value for cell (i, j) of matrix C is computed by taking the dot product of row i in matrix A and column j in matrix B. If A is an m × p matrix and B is an p × n matrix, then the product of A and B is the m × n matrix C = AB, where the (i, j)th element of C is computed as the. In the Figure 1 (a), we experiment large and sparse matrix multiplication from two random Bernoulli square matrices with dimension roughly equal to 1. Parallel Matrix Multiplication Algorithms Grid-based approach - The grid-based algorithms , regard processors as residing on a two- or three- Coverage: vectors norms (ℓ 2-norm, ℓ 1-norm, ℓ p-norm, ℓ ∞-norm), vector inner product, matrix multiplication, matrix trace, matrix Frobenius norm, scalar function differential, convex function, use Numpy to construct vectors and matrices. Matrix multiplication in C: We can add, subtract, multiply and divide 2 matrices. 20. matrix multiplication using mapreduce github\n\nMatrix multiplication using mapreduce github" ]
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https://thithptquocgia2016.com/tana-tanb-tanc-tanatanbtanc/
[ "# ALGEBRA PRECALCULUS\n\nI want lớn prove sầu this (where each angle may be negative sầu or greater than \\$180^circ\\$):\n\nWhen \\$A+B+C = 180^circ\\$ \beginequation* an A + ung B + ung C = ã A: ung B: ã C. endequation*\n\nWe know that\beginequation* an(A+B) = frac an A+ ung B1- ung Achảy Bendequation* và that \beginequation* extvà that~A+B = 180^circ-C.endequation*\n\nTherefore \\$ an(A+B) = - an C.\\$\n\nFrom here, I got stuông chồng.\n\nBạn đang xem: Algebra precalculus", null, "", null, "cảnh báo that\\$\\$ hithptquocgia2016.comrmImleft(e^ipi ight)=0 ag1\\$\\$Thus, if \\$a+b+c=pi\\$,\\$\\$\beginalign0&= hithptquocgia2016.comrmImleft(e^iae^ibe^ic ight)\\&= hithptquocgia2016.comrmImBig(\big(cos(a)+isin(a)\big)\big(cos(b)+isin(b)\big)\big(cos(c)+isin(c)\big)Big)\\<4pt>&=sin(a)cos(b)cos(c)+cos(a)sin(b)cos(c)+cos(a)cos(b)sin(c)\\&-sin(a)sin(b)sin(c) ag2endalign\\$\\$Dividing \\$(2)\\$ by \\$cos(a)cos(b)cos(c)\\$ yields\\$\\$ an(a)+ an(b)+ an(c)= an(a) an(b) an(c) ag3\\$\\$", null, "HINT\n\n\\$A+B+C = 180\\$\n\n\\$A+B = 180 - C\\$\n\nWe\"ll apply tangent function:\n\n\\$ ã (A+B) = chảy (180 - C)\\$\n\nWe\"ll consider the identity:\n\n\\$ an(x+y) = fracchảy x + ã y1-chảy xchảy y\\$\n\n\\$frac ã A + chảy B1- ã A an B = fracchảy 180 - ung C1+ ã 180 ung C\\$\n\nBut \\$ ã 180 = 0\\$, therefore, we\"ll get:\n\n\\$fracchảy A + chảy 1- ã A an B\\$ = \\$frac0 - chảy C1+0\\$\n\n\\$fracchảy A + ã B1- ã A ung B = -chảy C\\$\n\nWe\"ll multiply by \\$(1- ung Achảy B)\\$:\n\n\\$ ã A + chảy B = -chảy C + an A ung B an C\\$\n\nHence\n\n\\$chảy A + an B+ an C = an A an B ã C\\$\n\nShare\nCite\nFollow\nanswered Aug 27 \"13 at 14:22", null, "ShobhitShobhit\n\\$endgroup\\$\n15\n\\$\begingroup\\$\nHere is a geometric proof, for the case that all three angles are acute:", null, "\\$QRUV\\$ are collinear because \\$B+90^circ+(90^circ-B)=180^circ\\$.\n\n\\$STV\\$ are collinear because \\$A+B+C=180^circ\\$, so \\$angle QSV=angle UTV=C\\$.\n\nXem thêm:\n\nSimilar triangles \\$ riangle PQRsim riangle TRS\\$ & \\$ riangle RTU syên riangle SRQ\\$ give sầu \\$displaystylefracQPRQ = fracRTSR = fracTURQ\\$, and therefore \\$TU=QP=1\\$.\n\nThen,\\$\\$\beginalign& ã A + ã B + ã C = QR+RU+UV = QV \\&= QP.. fracQRQP, fracQSQR , fracQVQS = 1 cdot an(A) an(B) an(C) endalign\\$\\$\n\nWhen one of the angles is obtuse, let it (without loss of generality) be \\$C\\$. Then a similar diagram can be drawn, except that \\$V\\$ is to lớn the left of \\$Q\\$, và \\$UV\\$, \\$QV\\$ count as negative lengths.\n\nShare\nCite\nFollow\nedited Jun 11 \"15 at 14:54\nanswered May 17 \"15 at 13:55", null, "hmakholm left over Monicahmakholm left over Monica\n\\$endgroup\\$\n0\n4\n\\$\begingroup\\$\nHINT:\n\nUsing \\$displaystyle an(A+B)=fracchảy A+chảy B1- an A ã B,\\$\n\nwe can prove sầu \\$\\$ an(A+B+C)=fracsum_ extcycchảy A-prod ã A1-sum_ extcycchảy Achảy B\\$\\$\n\nNow, if \\$A+B+C=n180^circ,\\$ where \\$n\\$ is any integer we know \\$ an(n180^circ)=0\\$\n\nShare\nCite\nFollow\nedited Nov 16 \"13 at 17:27\nMichael Hardy\n1\nanswered Aug 27 \"13 at 14:41\nlab bhattacharjeelab bhattacharjee" ]
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https://it.scribd.com/doc/42354627/Lecture-Notes-on-BJT-FET
[ "Sei sulla pagina 1di 10\n\nLecture Notes on BJT & FET Transitors v1.1.1\n\nhttp://www.nhn.ou.edu/~bumm/ELAB/Lect_Notes/BJT_FET_transitors\n\nNotes on BJT & FET Transistors.\n\nThe name transistor comes from the phrase “transferring an electrical signal across a resistor.”\n\nIn this course we will discuss two types of transistors:\n\nThe Bipolar Junction Transistor (BJT) is an active device. In simple terms, it is a current controlled valve. The base current (I B ) controls the collector current (I C ).\n\nThe Field Effect Transistor (FET) is an active device. In simple terms, it is a voltage controlled valve. The gate-source voltage (V GS ) controls the drain current (I D ).\n\nRegions of BJT operation:\n\nCut-off region: The transistor is off. There is no conduction between the collector and the emitter. (I B = 0 therefore I C = 0)\n\nActive region: The transistor is on. The collector current is proportional to and controlled by the base current (I C = βI B ) and relatively insensitive to V CE . In this region the transistor can be an amplifier.\n\nSaturation region: The transistor is on. The collector current varies very little with a change in the base\n\nThe V CE is small, a few tenths of volt. The collector current is strongly\n\ncurrent in the saturation region.\n\ndependent on V CE unlike in the active region. It is desirable to operate transistor switches in or near the\n\nsaturation region when in their on state.\n\nRules for Bipolar Junction Transistors (BJTs):\n\nFor an", null, "npn transistor, the voltage at the collector V C must be greater than the voltage at the emitter V E by at least a few tenths of a volt; otherwise, current will not flow through the collector-emitter junction, no matter what the applied voltage at the base. For pnp transistors, the emitter voltage must be greater than the collector voltage by a similar amount.\n\nFor the", null, "npn transistor, there is a voltage drop from the base to the emitter of 0.6 V. For a pnp transistor, there is also a 0.6 V rise from the base to the emitter. In terms of operation, this means that the base voltage V B of an npn transistor must be at least 0.6 V greater that the emitter voltage V E ; otherwise, the transistor will not pass emitter-to-collector current. For a pnp transistor, V B must be at least 0.6 V less than V E ; otherwise, it will not pass collector-to-emitter current.\n\nLecture Notes on BJT & FET Transitors v1.1.1\n\nhttp://www.nhn.ou.edu/~bumm/ELAB/Lect_Notes/BJT_FET_transitors\n\nBASIC EQUATIONS FOR THE BJT.", null, "Lecture Notes on BJT & FET Transitors v1.1.1\n\nhttp://www.nhn.ou.edu/~bumm/ELAB/Lect_Notes/BJT_FET_transitors\n\nBJT Schematic Symbols (Mnemonics for remembering the direction of the arrows are in parenthesis.)", null, "", null, "Ohmmeters view of the BJT. Clearly a transistor cannot be made on the bench by combining two diodes. (Why is that?) Most ohmmeters not only measure resistance, but also measure the forward voltage drop across a diode. From this perspective you can identify the base and the type of transistor based on the following equivalent circuits.", null, "", null, "Common Nomenclature (npn example).", null, "Types of Amplifiers. The transistor is a three terminal device, thus the input and the output must share one terminal in common. This is the origin of the nomenclature of the three types of transistor amplifiers: common collector, common emitter, and common base.\n\nLecture Notes on BJT & FET Transitors v1.1.1\n\nhttp://www.nhn.ou.edu/~bumm/ELAB/Lect_Notes/BJT_FET_transitors\n\nDefinition of Gain. Gain is defined as the ratio of the output signal to the input signal. Because transistor amplifiers often have a quiescent output (a non zero output when the input is zero) we define gain as the derivative of the output with respect to the input. For systems where the quiescent output is zero, this reduces to the ratio of the output to the input. Thus gain is defined as the ratio of the change in output to the change in input.\n\nSo far we have not specified the output quantity, the reason is that we can define the gain with respect to any given output and input quantity.", null, "Note that a negative gain means that the sign of the signal is inverted. Negative gain is not possible for Power Gain. |A| less than unity indicates that the output is smaller than the input.\n\nThe quantities need not be the same. If the input and output quantities are different, the gain is no longer unitless. The most common examples are transimpedance gain and transadmittance gain.", null, "Lecture Notes on BJT & FET Transitors v1.1.1\n\nInput Impedance of a Transistor.\n\nhttp://www.nhn.ou.edu/~bumm/ELAB/Lect_Notes/BJT_FET_transitors", null, "Impedance is defined as Z = V/I. In linear circuits (with resistors, capacitors, inductors, batteries, etc.) this ratio is the reciprocal of the slope of the I versus V graph. In circuits with nonlinear elements such as a transistor, the input impedance of the transistor is defined as the reciprocal of the slope of the I versus V graph. This is simply the derivative of V in with respect to I in .", null, "We can easily find Z in from what we know already of the behavior of the transistor. We know that the sum of V BE and the IR drop across R E must equal V in .", null, "Taking the derivative of V in with respect to I in , remembering that V BE is a constant, we get the result:", null, "", null, "Because I E = I B (β + 1) The IR drop across R E is greater than it would be for I B alone. The amplification of the base current causes R E to appear larger to a source looking into the input by (β + 1).\n\nOutput Impedance of a Transistor for the Emitter Follower (Common Collector).\n\nLecture Notes on BJT & FET Transitors v1.1.1\n\nhttp://www.nhn.ou.edu/~bumm/ELAB/Lect_Notes/BJT_FET_transitors", null, "The output impedance seen by the load (R E in this example) is defined as:", null, "The minus sign in the derivative comes from the fact the output impedance has the effect of decreasing V out . The output current I out is just the emitter current I E which is related to the base current.", null, "Thus we obtain the result that the impedance of the source, as viewed by the load, is reduced by the factor\n\n~1/β.", null, "Lecture Notes on BJT & FET Transitors v1.1.1\n\nhttp://www.nhn.ou.edu/~bumm/ELAB/Lect_Notes/BJT_FET_transitors\n\nThe Field Effect Transistor (FET).\n\nThe FET is a three terminal device like the BJT, but operates by a different principle. The three terminals are called the source, drain, and gate. The voltage applied to the gate controls the current flowing in the source-drain channel. No current flows through the gate electrode, thus the gate is essentially insulated from the source-drain channel. Because no current flows through the gate, the input impedance of the FET is extremely large (in the range of 10 10 –10 15 Ω). The large input impedance of the FET makes them an excellent choice for amplifier inputs.\n\nThe two common families of FETs, the junction FET (JFET) and the metal oxide semiconductor FET (MOSFET) differ in the way the gate contact is made on the source-drain channel.\n\nIn the JFET the gate-channel contact is a reverse biased pn junction. The gate-channel junction of the JFET must always be reverse biased otherwise it may behave as a diode. All JFETs are depletion mode devices—they are on when the gate bias is zero (V GS = 0).\n\nIn the MOSFET the gate-channel contact is a metal electrode separated from the channel by a thin layer of insulating oxide. MOSFETs have very good isolation between the gate and the channel, but the thin oxide is easily damaged (punctured!) by static discharge through careless handling. MOSFETs are made in both depletion mode (on with zero biased gate, V GS = 0) and in enhancement mode (off with zero biased gate).\n\nIn this class we will focus on JFETs.\n\nSchematic symbols. Two versions of the symbols are in common use. The symbols in the top row depict the source and drain as being symmetric. This is not generally true. Slight asymmetries are built into the channel during manufacturing which optimize the performance of the FET. Thus it is necessary to distinguish the source from the drain. In this class we will use the asymmetric symbols found on the bottom row, which depict the gate nearly opposite the source. The designation n-channel means that the channel is n doped and the gate is p doped. The p-channel is complement of n-channel.", null, "", null, "", null, "", null, "Lecture Notes on BJT & FET Transitors v1.1.1\n\nhttp://www.nhn.ou.edu/~bumm/ELAB/Lect_Notes/BJT_FET_transitors\n\nCommon Nomenclature (n-channel FET example).", null, "Regions of JFET operation:\n\nCut-off region: The transistor is off. There is no conduction between the drain and the source when the gate-source voltage is greater than the cut-off voltage. (I D = 0 for V GS > V GS,off )\n\nActive region (also called the Saturation region): The transistor is on. The drain current is controlled by the gate-source voltage (V GS ) and relatively insensitive to V DS . In this region the transistor can be an amplifier.", null, "In the active region:\n\nOhmic region: The transistor is on, but behaves as a voltage controlled resistor. When V DS is less than in the active region, the drain current is roughly proportional to the source-drain voltage and is controlled by the gate voltage.\n\nIn the ohmic region:", null, "Lecture Notes on BJT & FET Transitors v1.1.1\n\nhttp://www.nhn.ou.edu/~bumm/ELAB/Lect_Notes/BJT_FET_transitors\n\nCommon Specifications. I DSS is the drain current in the active region for V GS = 0. (I D source shorted to gate)\n\nV GS,off is the minimum V GS where I D = 0. V GS,off is negative for n-channel and positive for p-channel\n\ng m is the transconductance, the change in I D with V GS and constant V DS .", null, "Common Circuit Applications:\n\nVoltage Controlled Switch. For the on state the gate voltage V GS = 0 and for the off state |V GS | > |V GS,off | (of greater magnitude than V GS,off and with the same sign). The sign of the voltage depends on the type of FET, negative for n-channel and positive for p-channel.", null, "Current Source. The drain current is set by R S such that V GS = I D R S . Any value of current can be chosen between zero and I DSS (see the I D vs V GS graph for the JFET).", null, "Lecture Notes on BJT & FET Transitors v1.1.1\n\nhttp://www.nhn.ou.edu/~bumm/ELAB/Lect_Notes/BJT_FET_transitors\n\nSource Follower. The simple source follower is shown below. The improved version is shown at the right. The lower JFET forms a current source. The result is that V GS is held constant, removing the defects of the simple circuit.", null, "Voltage-Controlled Resistor. V GS must be between zero and\n\nV GS,off .", null, "", null, "JFET Diode. The JET pn gate junction can be used as a diode by connecting the source and the drain terminals. This is done if very low reverse leakage currents are required. The leakage current is very low because the reverse leakage current scales with the gate area. Small gate areas are designed into JFETs because it decreases the gate-source and the gate-drain capacitances", null, "." ]
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https://jrenewables.springeropen.com/articles/10.1186/s40807-017-0039-7/tables/1
[ "# Table 1 Bus voltage angle- and magnitude-based sensitivity analysis for WIS\n\nNo. Bus no. Tangent vector for voltage angle Voltage angle sensitivity $$\\frac{{\\partial \\delta_{i} }}{{{\\text{d}}\\lambda }}$$ Tangent vector for voltage magnitude ($$V_{i}$$) Voltage mag. sensitivity $$\\frac{{\\partial V_{i} }}{{{\\text{d}}\\lambda }}$$\n1 3 0.17206 1.0000 1.0000 1.0000\n2 1 0.17119 0.9952 −0.9981 −0.9981\n3 8 0.16209 0.9712 −0.9669 −0.9669\n4 6 0.16113 0.9253 −0.9571 −0.9571\n5 11 0.15792 0.8825 −0.9333 −0.9333\n6 2 0.15348 0.8695 −0.8997 −0.8997\n7 4 0.15111 0.8313 −0.8649 −0.8649\n8 5 0.14881 0.7976 −0.8412 −0.8412\n9 14 0.14628 0.7761 −0.8111 −0.8111\n10 9 0.13667 0.7123 −0.7939 −0.7939\n11 10 0.13419 0.6912 −0.7775 −0.7775\n12 7 0.13291 0.6777 −0.7558 −0.7558\n13 12 0.12825 0.6485 −0.7113 −0.7113\n14 13 0.12344 0.6117 −0.6999 −0.6999" ]
[ null ]
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https://studydaddy.com/question/help-with-following-trig-questions-the-radian-measure-of-the-angle-is-2-the-radi
[ "", null, "Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.\n\nQUESTION\n\n# Help with following trig questions: The radian measure of the angle is 2. The radius of the circle is 3. The radian measure of the angle is 4.\n\nHelp with following trig questions:\n\n1. The radian measure of the angle is\n\n2. The radius of the circle is\n\n3. The radian measure of the angle is\n\n4. Find each of the following for the given values of r, ω" ]
[ null, "https://www.facebook.com/tr", null ]
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https://notebook.community/daniel-koehn/DENISE-Black-Edition/par/pythonIO/DENISE-python_IO
[ "# Creating input files for DENISE Black-Edition\n\nJupyter notebook for the definition of FD model and modelling/FWI/RTM parameters of DENISE Black-Edition. For a more detailed explanation of the parameters, I refer to the DENISE Black-Edition user manual\n\nDaniel Koehn\n\nKiel, 24.06.2019\n\n\n\nIn :\n\n# Import Python libaries\n# ----------------------\nimport numpy as np # NumPy library\nfrom denise_IO.denise_out import * # \"DENISE\" library\n\n\n\n## 1. Short description of modelling/FWI problem\n\nDefine name of DENISE parameter file\n\n\n\nIn :\n\npara[\"filename\"] = \"DENISE_marm_OBC.inp\"\n\n\n\nGive a short description of your modelling/FWI problem\n\n\n\nIn :\n\npara[\"descr\"] = \"Marmousi-II\"\n\n\n\nWhat kind of PHYSICS do you want to use? (2D-PSV=1; 2D-AC=2; 2D-PSV-VTI=3; 2D-PSV-TTI=4; 2D-SH=5)\n\n\n\nIn :\n\npara[\"PHYSICS\"] = 1\n\n\n\nChoose DENISE operation mode (MODE): (forward_modelling_only=0; FWI=1; RTM=2)\n\n\n\nIn :\n\npara[\"MODE\"] = 0\n\n\n\n## 2. Load external 2D elastic model\n\nFirst define spatial model discretization:\n\n\n\nIn :\n\npara[\"NX\"] = 500 # number of grid points in x-direction\npara[\"NY\"] = 174 # number of grid points in y-direction\npara[\"DH\"] = 20. # spatial grid point distance [m]\n\n\n\n\n\nIn :\n\n# Define model basename\nbase_model = \"model/marmousi_II_marine\"\n\n# Open vp-model and write IEEE-le binary data to vp array\n# -------------------------------------------------------\nf = open(base_model + \".vp\")\ndata_type = np.dtype ('float32').newbyteorder ('<')\nvp = np.fromfile (f, dtype=data_type)\nf.close()\n\n# Reshape (1 x nx*ny) vector to (ny x nx) matrix\nvp = vp.reshape(para[\"NX\"],para[\"NY\"])\nvp = np.transpose(vp)\nvp = np.flipud(vp)\n\n# Open vs-model and write IEEE-le binary data to vs array\n# -------------------------------------------------------\nf = open(base_model + \".vs\")\ndata_type = np.dtype ('float32').newbyteorder ('<')\nvs = np.fromfile (f, dtype=data_type)\nf.close()\n\n# Reshape (1 x nx*ny) vector to (ny x nx) matrix\nvs = vs.reshape(para[\"NX\"],para[\"NY\"])\nvs = np.transpose(vs)\nvs = np.flipud(vs)\n\n# Open rho-model and write IEEE-le binary data to rho array\n# ---------------------------------------------------------\nf = open(base_model + \".rho\")\ndata_type = np.dtype ('float32').newbyteorder ('<')\nrho = np.fromfile (f, dtype=data_type)\nf.close()\n\n# Reshape (1 x nx*ny) vector to (ny x nx) matrix\nrho = rho.reshape(para[\"NX\"],para[\"NY\"])\nrho = np.transpose(rho)\nrho = np.flipud(rho)\n\n\n\nDefine coordinate axis\n\n\n\nIn :\n\nx = np.arange(para[\"DH\"], para[\"DH\"] * (para[\"NX\"] + 1), para[\"DH\"])\ny = np.arange(para[\"DH\"], para[\"DH\"] * (para[\"NY\"] + 1), para[\"DH\"])\n\n# convert m -> km\nx = np.divide(x,1000.0);\ny = np.divide(y,1000.0);\n\n\n\nPlot external model\n\n\n\nIn :\n\ncmap = \"magma\" # colormap\n\n# define minimum and maximum material parameter values\nvpmin = np.min(vp)\nvpmax = np.max(vp)\n\nvsmin = np.min(vs)\nvsmax = np.max(vs)\n\nrhomin = np.min(rho)\nrhomax = np.max(rho)\n\n# plot elastic model\nplot_model(vp,vs,rho,x,y,cmap,vpmin,vpmax,vsmin,vsmax,rhomin,rhomax)\n\n\n\n\nC:\\Users\\daniel_koehn\\Anaconda3\\lib\\site-packages\\matplotlib\\font_manager.py:1241: UserWarning: findfont: Font family ['sans-serif'] not found. Falling back to DejaVu Sans.\n(prop.get_family(), self.defaultFamily[fontext]))\n\n\n\nWrite model to IEEE-le binary file\n\n\n\nIn :\n\n# model basename\nmodel_basename = \"marmousi_II_marine\"\n\n# location of model files\npara[\"MFILE\"] = \"start/\" + model_basename\n\n# writing P-wave velocity model to IEEE-le binary file\nname_model = model_basename + \".vp\"\nf = open (name_model, mode='wb')\ndata_type = np.dtype ('float32').newbyteorder ('<')\nvp1 = np.array(vp, dtype=data_type)\nvp1 = np.rot90(vp1,3)\nvp1.tofile(f)\nf.close()\n\n# writing S-wave velocity model to IEEE-le binary file\nname_model = model_basename + \".vs\"\nf = open (name_model, mode='wb')\ndata_type = np.dtype ('float32').newbyteorder ('<')\nvs1 = np.array(vs, dtype=data_type)\nvs1 = np.rot90(vs1,3)\nvs1.tofile(f)\nf.close()\n\n# writing density model to IEEE-le binary file\nname_model = model_basename + \".rho\"\nf = open (name_model, mode='wb')\ndata_type = np.dtype ('float32').newbyteorder ('<')\nrho1 = np.array(rho, dtype=data_type)\nrho1 = np.rot90(rho1,3)\nrho1.tofile(f)\nf.close()\n\n\n\nTo check if the models are correctly written to the binary files, you can use the Seismic Unix function ximage\n\n\n\nIn :\n\nprint(\"ximage n1=\" + str(para[\"NY\"]) + \" < \" + model_basename + \".vp\")\nprint(\"ximage n1=\" + str(para[\"NY\"]) + \" < \" + model_basename + \".vs\")\nprint(\"ximage n1=\" + str(para[\"NY\"]) + \" < \" + model_basename + \".rho\")\n\n\n\n\nximage n1=174 < marmousi_II_marine.vp\nximage n1=174 < marmousi_II_marine.vs\nximage n1=174 < marmousi_II_marine.rho\n\n\n\n## 3. Define spatial FD operator\n\nSpatial FD operator coefficients are based on Taylor series expansion and optimized according to Holberg (1987)\n\n\n\nIn :\n\n# Order of spatial FD operator (2, 4, 6, 8, 10, 12)\npara[\"FD_ORDER\"] = 8\n\n# Maximum relative group velocity error E\n# (minimum number of grid points per shortest wavelength is defined by FD_ORDER and E)\n# values:\n# 0 = Taylor coefficients\n# 1 = Holberg coeff.: E = 0.1 %\n# 2 = E = 0.5 %\n# 3 = E = 1.0 %\n# 4 = E = 3.0 %\npara[\"max_relative_error\"] = 3\n\n\n\nEstimate the maximum frequency in the source wavelet, which can be modelled by the given FD grid discretization and spatial FD operator\n\n\n\nIn :\n\n# maximum modelling frequency based on grid dispersion criterion for spatial FD operator\nfreqmax = calc_max_freq(vp,vs,para)\n\n\n\n\nMinimum Vs: 881.0 m/s\nMinimum Vp: 1500.0 m/s\nno of gridpoints per minimum wavelength = 2.9\nmaximum source wavelet frequency = 15.189655172413794 Hz\n\n\n\nIf you want to model higher frequency wave propagation, you have to decrease the spatial gridpoint distance DH by resampling the model\n\n## 4. Parallelization by Domain Decomposition\n\n\n\nIn :\n\npara[\"NPROCX\"] = 5 # number of processors in x-direction\npara[\"NPROCY\"] = 3 # number of processors in y-direction\n\n\n\nCheck if the spatial domain decomposition is consistent with the spatial FD grid discretization. The following conditions have to be satisfied\n\nNX % NPROCX = 0\n\nNY % NPROCY = 0\n\n\n\nIn :\n\ncheck_domain_decomp(para)\n\n\n\n\nDomain decomposition in x-direction is correct NX % NPROCX = 0\nDomain decomposition in y-direction is correct NY % NPROCY = 0\n\n\n\nIf the domain decomposition conditions are not satisfied, you have to add additional gridpoints at the bottom and right model boundary.\n\n## 5. Time stepping\n\nCalculate maximum time step DT according to the Courant-Friedrichs-Lewy (CFL) criterion\n\n\n\nIn :\n\npara[\"DT\"] = check_stability(vp,vs,para)\n\n\n\n\nMaximum Vs: 2752.0 m/s\nMaximum Vp: 4766.604 m/s\nAccording to the Courant-Friedrichs-Lewy (CFL) criterion\nthe maximum time step is DT = 2.14e-03 s\n\n\n\nIf you want to apply a FWI, keep in mind that the FWI will change the velocity model. Therefore, the maxium seismic velocities in the model will increase and you should choose a smaller time step than the DT derived from the CFL criterion\n\n\n\nIn :\n\npara[\"TIME\"] = 6.0 # time of wave propagation [s]\npara[\"DT\"] = 2.0e-3 # timestep [s]\n\n\n\n## 6. Define acquisition geometry\n\n### a) Receiver properites and positions\n\nPlace receivers on FD modelling grid\n\n\n\nIn :\n\ndrec = 20. # receiver spacing [m]\nxrec1 = 800. # 1st receiver position [m]\nxrec2 = 8780. # last receiver position [m]\nxrec = np.arange(xrec1, xrec2 + para[\"DH\"], drec) # receiver positions in x-direction [m]\n\n# place receivers at depth yrec [m]\ndepth_rec = 460. # receiver depth [m]\nyrec = depth_rec * xrec/xrec\n\n# assemble vectors into an array\ntmp = np.zeros(xrec.size, dtype=[('var1', float), ('var2', float)])\ntmp['var1'] = xrec\ntmp['var2'] = yrec\n\n# write receiver positions to file\nnp.savetxt(basename_rec + \".dat\", tmp, fmt='%4.3f %4.3f')\n\n\n\nDefine type of seismograms SEISMO:\n\nSEISMO=0: no seismograms\n\nSEISMO=1: particle-velocities\n\nSEISMO=2: pressure (hydrophones)\n\nSEISMO=3: curl and div\n\nSEISMO=4: everything\n\n\n\nIn :\n\n# type of seismogram\npara[\"SEISMO\"] = 1\n\n\n\nHow does DENISE read receiver positions from a file? In case of a fixed spread geometry you only need a single receiver file (READREC=1). If you want to model streamer geometry or more generally variable acquisition geometry with changing receiver positions for each shot, you have to define a separate receiver file for each shot (READREC=2)\n\n\n\nIn :\n\n\n\nDefine location and basename of receiver file, defined above, without \".dat\" extension\n\n\n\nIn :\n\n\n\nDefine the seismogram properties\n\n\n\nIn :\n\npara[\"NDT\"] = 1 # seismogram sampling rate in timesteps (has to be set to NDT=1 if you run FWI)\n\n# location and name of seismogram output files in SU format\n\n# particle velocities (if SEISMO=1 or SEISMO=4)\npara[\"SEIS_FILE_VX\"] = \"su/DENISE_MARMOUSI_x.su\" # filename for vx component\npara[\"SEIS_FILE_VY\"] = \"su/DENISE_MARMOUSI_y.su\" # filename for vy component\n\n# curl and div of wavefield (if SEISMO=3 or SEISMO=4)\npara[\"SEIS_FILE_CURL\"] = \"su/DENISE_MARMOUSI_rot.su\" # filename for rot_z component ~ S-wave energy\npara[\"SEIS_FILE_DIV\"] = \"su/DENISE_MARMOUSI_div.su\" # filename for div component ~ P-wave energy\n\n# pressure field (hydrophones) (if SEISMO=2 or SEISMO=4)\npara[\"SEIS_FILE_P\"] = \"su/DENISE_MARMOUSI_p.su\" # filename for pressure component\n\n\n\n### b) Source properties and positions\n\nDistribute sources on FD modelling grid an define source properties\n\n\n\nIn :\n\n# source x-coordinates\ndsrc = 80. # source spacing [m]\nxsrc1 = 800. # 1st source position [m]\nxsrc2 = 8780. # last source position [m]\nxsrc = np.arange(xsrc1, xsrc2 + para[\"DH\"], dsrc) # source positions in x-direction [m]\n\n# place sources at depth ysrc [m]\ndepth_src = 40. # source depth [m]\nysrc = depth_src * xsrc/xsrc\n\n# number of source positions\nnshot = (int)(len(ysrc))\n\n# z-coordinate = 0 due to 2D code [m]\nzsrc = 0.0 * (xsrc / xsrc)\n\n# time delay of source wavelet [s]\ntd = 0.0 * (xsrc / xsrc)\n\n# center frequency of pre-defined source wavelet [Hz]\nfc = 10.0 * (xsrc / xsrc)\n\n# you can also use the maximum frequency computed from the grid dispersion\n# criterion in section 3. based on spatial discretization and FD operator\n# fc = (freqmax / 2.) * (xsrc / xsrc)\n\n# amplitude of source wavelet [m]\namp = 1.0 * (xsrc / xsrc)\n\n# angle of rotated source [°]\nangle = 0.0 * (xsrc / xsrc)\n\n# define source type:\n\n# 2D PSV case\n# -----------\n# explosive sources (QUELLTYP=1)\n# point forces in x- and y-direction (QUELLTYP=2,3)\n\n# 2D SH case\n# -----------\n# point force in z-direction (QUELLTYP=1)\n\nQUELLTYP = 1\nsrc_type = QUELLTYP * (xsrc / xsrc)\n\n# write source positions and properties to file\nbasename_src = \"source_OBC_VSP.dat\"\n\n# create and open source file\nfp = open(basename_src, mode='w')\n\n# write nshot to file header\nfp.write(str(nshot) + \"\\n\")\n\n# write source properties to file\nfor i in range(0,nshot):\n\nfp.write('{:4.2f}'.format(xsrc[i]) + \"\\t\" + '{:4.2f}'.format(zsrc[i]) + \"\\t\" + '{:4.2f}'.format(ysrc[i]) + \"\\t\" + '{:1.2f}'.format(td[i]) + \"\\t\" + '{:4.2f}'.format(fc[i]) + \"\\t\" + '{:1.2f}'.format(amp[i]) + \"\\t\" + '{:1.2f}'.format(angle[i]) + \"\\t\" + str(src_type[i]) + \"\\t\" + \"\\n\")\n\n# close source file\nfp.close()\n\n\n\nDefine location of the source file:\n\n\n\nIn :\n\npara[\"SOURCE_FILE\"] = \"./source/\" + basename_src\n\n\n\nDo you want to excite all source positions simultaneously (RUN_MULTIPLE_SHOTS=0) or start a separate modelling run for each shot (RUN_MULTIPLE_SHOTS=1)\n\n\n\nIn :\n\npara[\"RUN_MULTIPLE_SHOTS\"] = 1\n\n\n\nDefine shape of the source signal (QUELLART)\n\n• Ricker wavelet = 1\n• Fuchs-Mueller wavelet = 2\n• read wavelet from ASCII file = 3\n\n• SIN^3 wavelet = 4\n\n• Gaussian derivative wavelet = 5\n\\begin{equation} \\rm{gd(t)=-2 \\pi^2 f_c^2 (t-t_d) exp(-\\pi^2 f_c^2 (t-t_d)^2)} \\label{eq_s4} \\end{equation}\n• Bandlimited spike wavelet = 6\n\n• Klauder wavelet = 7\n\n\\begin{equation} \\rm{klau(t) = real\\biggl\\{sin\\biggl(\\frac{\\pi k \\tau (TS-\\tau)}{\\pi k \\tau}\\biggr)(exp(2 \\pi i f_0 \\tau))\\biggr\\}} \\quad \\mbox{with} \\quad \\rm{\\tau=(t-1.5/FC\\_SPIKE\\_1-t_d)}\\} \\label{eq_s5} \\end{equation}\n\nwith\n\n$\\rm{k=(FC\\_SPIKE\\_2-FC\\_SPIKE\\_1)/TS}$ (rate of change of frequency with time)\n\n$\\rm{f_0=(FC\\_SPIKE\\_2+FC\\_SPIKE\\_1)/2}$ (midfrequency of bandwidth)\n\n$\\rm{i^2=-1}$\n\nIn these equations, t denotes time and $f_c$ is the center frequency. $t_d$ is a time delay which can be defined for each source position in SOURCE_FILE. Note that the symmetric (zero phase) Ricker signal is always delayed by $1.0/f_c$, which means that after one period the maximum amplitude is excited at the source location.\n\n\n\nIn :\n\npara[\"QUELLART\"] = 6\n\n\n\nIf you read the wavelet from an ASCII file (QUELLART=3), you have to define the location of the signal file (SIGNAL_FILE)\n\n\n\nIn :\n\npara[\"SIGNAL_FILE\"] = \"./wavelet/wavelet_marmousi\"\n\n\n\nIn case of the bandlimited spike wavelet you have to define ...\n\n• If FC_SPIKE_1 <= 0.0 a low-pass filtered spike with upper corner frequency FC_SPIKE_2 and order ORDER_SPIKE is calculated\n\n• If FC_SPIKE_1 > 0.0 a band-pass filtered spike with lower corner frequency FC_SPIKE_1 and upper corner frequency FC_SPIKE_2 with order ORDER_SPIKE is calculated\n\n\n\nIn :\n\npara[\"FC_SPIKE_1\"] = -5.0 # lower corner frequency [Hz]\npara[\"FC_SPIKE_2\"] = 15.0 # upper corner frequency [Hz]\n\n# you can also use the maximum frequency computed from the grid dispersion\n# criterion in section 3. based on spatial discretization and FD operator\n# para[\"FC_SPIKE_2\"] = freqmax # upper corner frequency [Hz]\n\npara[\"ORDER_SPIKE\"] = 5 # order of Butterworth filter\n\n\n\nIn case of the Klauder wavelet you have to define the sweep length TS\n\n\n\nIn :\n\npara[\"TS\"] = 8.0 # sweep length [s]\n\n\n\nDo you want to write the source wavelet to a SU file for each shot (WRITE_STF=1)?\n\n\n\nIn :\n\npara[\"WRITE_STF\"] = 1\n\n\n\nPlot acquisition geometry relative to the subsurface model. Red stars denote the source positions and cyan triangles receiver positions\n\n\n\nIn :\n\ncmap = \"inferno\"\nplot_acq(vp,xrec/1000,yrec/1000,xsrc/1000,ysrc/1000,x,y,cmap,vpmin,vpmax)\n\n\n\n\n\n\n## 7. Q-approximation\n\n\n\nIn :\n\npara[\"L\"] = 0 # number of relaxation mechanisms\npara[\"FL\"] = 40. # relaxation frequencies [Hz]\n\n\n\n## 8. Boundary conditions\n\nDefine boundary conditions. FREE_SURF=1 activates a free-surface boundary condition at y = 0 m. PML absorbing boundries are defined by their widht FW, damping velocity DAMPING, damping frequency FPML, degree of damping profile npower and k_max_PML\n\n\n\nIn :\n\n# free surface boundary condition\npara[\"FREE_SURF\"] = 1 # activate free surface boundary condition\n\n# PML boundary frame\npara[\"FW\"] = 10\npara[\"DAMPING\"] = 1500.\npara[\"FPML\"] = 10.\npara[\"npower\"] = 4.\npara[\"k_max_PML\"] = 1.\n\n\n\n## 9. Wavefield snapshots\n\nOutput of wavefield snapshots (SNAP>0):\n\n• particle velocities: SNAP=1\n• pressure field: SNAP=2\n• curl and divergence energy: SNAP=3\n• both particle velocities and energy : SNAP=4\n\n\nIn :\n\npara[\"SNAP\"] = 0\npara[\"SNAP_SHOT\"] = 1 # compute and write snapshots for shot no. SNAP_SHOT\n\npara[\"TSNAP1\"] = 0.002 # first snapshot [s] (TSNAP1 has to fullfill the condition TSNAP1 > DT)\npara[\"TSNAP2\"] = 3.0 # first snapshot [s]\npara[\"TSNAPINC\"] = 0.06 # snapshot increment [s]\n\npara[\"IDX\"] = 1 # write only every IDX spatial grid point in x-direction to snapshot file\npara[\"IDY\"] = 1 # write only every IDY spatial grid point in y-direction to snapshot file\n\npara[\"SNAP_FILE\"] = \"./snap/waveform_forward\" # location and basename of the snapshot files\n\n\n\n## 10. Log file name\n\n\n\nIn :\n\npara[\"LOG_FILE\"] = \"log/Marmousi.log\" # Log file name\n\n\n\n## FWI parameters\n\nIf you only want to run FD forward modelling runs, you can neglect the parameters below, which are fixed FWI parameters\n\n## 11. General FWI parameters\n\n\n\nIn :\n\npara[\"ITERMAX\"] = 600 # maximum number of TDFWI iterations at each FWI stage defined in FWI workflow file\n\npara[\"DATA_DIR\"] = \"su/MARMOUSI_spike/DENISE_MARMOUSI\" # location and basename of field data seismograms\n\npara[\"INVMAT1\"] = 1 # material parameterization for FWI (Vp,Vs,rho=1/Zp,Zs,rho=2/lam,mu,rho=3)\n\n# x-y components = 1; y-comp = 2; x-comp = 3; p-comp = 4; x-p-comp = 5; y-p-comp = 6; x-y-p-comp = 7\npara[\"QUELLTYPB\"] = 1\n\n# Optimization method\npara[\"GRAD_METHOD\"] = 2 # PCG = 1; LBFGS = 2\n\n# PCG_BETA (Fletcher_Reeves=1/Polak_Ribiere=2/Hestenes_Stiefel=3/Dai_Yuan=4)\npara[\"PCG_BETA\"] = 2\n\n# store NLBFGS update during LBFGS optimization\npara[\"NLBFGS\"] = 20\n\n# store wavefields only every DTINV time sample for gradient computation\npara[\"DTINV\"] = 3\n\n# FWI log file location and name\npara[\"MISFIT_LOG_FILE\"] = \"Marmousi_fwi_log.dat\"\n\n\n\n## 12. FWI gradient taper functions\n\n\n\nIn :\n\n# exponent of depth scaling for preconditioning\n\n# Circular taper around all sources (not at receiver positions)\npara[\"SWS_TAPER_CIRCULAR_PER_SHOT\"] = 0\npara[\"SRTSHAPE\"] = 1 # SRTSHAPE: 1 = error_function; 2 = log_function\n\n# Read taper file from external file\npara[\"SWS_TAPER_FILE\"] = 0\n\n\n\n## 13. FWI model output\n\n\n\nIn :\n\n# model location and basename\npara[\"INV_MODELFILE\"] = \"model/modelTest\"\n\n# write inverted model after each iteration (yes=1)?\n# Warning: Might require a lot of disk space\npara[\"INV_MOD_OUT\"] = 0\n\n\n\n## 14. Bound constraints\n\nUpper and lower limits for different model parameter classes\n\n\n\nIn :\n\n# upper limit for vp\npara[\"VPUPPERLIM\"] = 6000.\n\n# lower limit for vp\npara[\"VPLOWERLIM\"] = 0.\n\n# upper limit for vs\npara[\"VSUPPERLIM\"] = 4000.\n\n# lower limit for vs\npara[\"VSLOWERLIM\"] = 0.\n\n# upper limit for density\npara[\"RHOUPPERLIM\"] = 3000.\n\n# lower limit for density\npara[\"RHOLOWERLIM\"] = 1000.\n\n# upper limit for Qs\npara[\"QSUPPERLIM\"] = 100.\n\n# lower limit for Qs\npara[\"QSLOWERLIM\"] = 10.\n\n\n\n## 15. Step length estimation\n\n\n\nIn :\n\npara[\"EPS_SCALE\"] = 0.01 # initial model update during step length estimation\npara[\"STEPMAX\"] = 6 # maximum number of attemps to find a step length during line search\npara[\"SCALEFAC\"] = 2. # scale step during line search\n\n# evaluate objective function only for a limited number of shots\npara[\"TESTSHOT_START\"] = 25\npara[\"TESTSHOT_END\"] = 75\npara[\"TESTSHOT_INCR\"] = 10\n\n\n\n## 16. Trace muting\n\n\n\nIn :\n\n# Activate trace muting (yes=1)\npara[\"TRKILL\"] = 0\n\n# Location and name of trace mute file containing muting matrix\npara[\"TRKILL_FILE\"] = \"./trace_kill/trace_kill.dat\"\n\n\n\n## 17. Time damping\n\n\n\nIn :\n\n# Basename of picked traveltimes for each shot\n# Time damping parameters are defined in the DENISE\n# workflow file for each FWI stage\npara[\"PICKS_FILE\"] = \"./picked_times/picks_\"\n\n\n\n## 18. Create DENISE parameter file\n\n\n\nIn :\n\nwrite_denise_para(para)\n\n\n\n## Instructions for preparing and starting a modelling/FWI run with DENISE Black-Edition\n\n(a) Move model files to the directory DENISE-Black-Edition/par/para[\"MFILE\"]\n\n\n\nIn :\n\nprint(\"mv \" + model_basename + \".vp DENISE-Black-Edition/par/\" + para[\"MFILE\"] + \".vp\")\nprint(\"mv \" + model_basename + \".vs DENISE-Black-Edition/par/\" + para[\"MFILE\"] + \".vs\")\nprint(\"mv \" + model_basename + \".rho DENISE-Black-Edition/par/\" + para[\"MFILE\"] + \".rho\")\n\n\n\n\nmv marmousi_II_marine.vp DENISE-Black-Edition/par/start/marmousi_II_marine.vp\nmv marmousi_II_marine.vs DENISE-Black-Edition/par/start/marmousi_II_marine.vs\nmv marmousi_II_marine.rho DENISE-Black-Edition/par/start/marmousi_II_marine.rho\n\n\n\nYou can also copy the model files to a HPC cluster using SCP.\n\n(b) Move source file to the directory DENISE-Black-Edition/par/para[\"SOURCE_FILE\"]\n\n\n\nIn :\n\nprint(\"mv \" + basename_src + \" DENISE-Black-Edition/par/\" + para[\"SOURCE_FILE\"][2::])\n\n\n\n\nmv source_OBC_VSP.dat DENISE-Black-Edition/par/source/source_OBC_VSP.dat\n\n\n\n(c) Move receiver file(s) to the directory DENISE-Black-Edition/par/para[\"REC_FILE\"]\n\n\n\nIn :\n\nprint(\"mv \" + basename_rec + \".dat DENISE-Black-Edition/par\" + para[\"REC_FILE\"][1::] + \".dat\")\n\n\n\n\n\n\n(d) Move DENISE parameter file to the directory DENISE-Black-Edition/par/\n\n\n\nIn :\n\nprint(\"mv \" + para[\"filename\"] + \" DENISE-Black-Edition/par/\")\n\n\n\n\nmv DENISE_marm_OBC.inp DENISE-Black-Edition/par/\n\n\n\n(e) Within the DENISE-Black-Edition/par directory you can start the DENISE modelling run with\n\n\n\nIn :\n\nprint(\"mpirun -np \" + str(para[\"NPROCX\"]*para[\"NPROCY\"]) + \" ../bin/denise \" + para[\"filename\"])\n\n\n\n\nmpirun -np 15 ../bin/denise DENISE_marm_OBC.inp\n\n\n\nIf you want to run a FWI, you also have to define a FWI workflow file\n\n\n\nIn :\n\nprint(\"mpirun -np \" + str(para[\"NPROCX\"]*para[\"NPROCY\"]) + \" ../bin/denise \" + para[\"filename\"] + \" FWI_workflow.inp\")\n\n\n\n\nmpirun -np 15 ../bin/denise DENISE_marm_OBC.inp FWI_workflow.inp" ]
[ null ]
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http://service.iris.edu/mustang/noise-mode-timeseries/docs/1/help/
[ "# Help: noise-mode-timeseries v.1\n\n## Description\n\nThe noise-mode-timeseries web service returns Probability Density Function daily mode estimates for individual seismic channels measured at discrete frequencies as a function of time.\n\n## Introduction\n\nThe Power Density Function (PDF) represents the distribution of noise power as a function of frequency. The PDF mode function represents the most commonly occurring (most likely) noise power level as a function of frequency.\n\nThe following plot, from the noise-pdf webservice, shows a computed PDF from 19 years of data from one seismometer along with the computed mode function.", null, "PDF Plot showing the high noise model, the mode and the low noise model\n\nThe noise-mode-timeseries web service reveals how the PDF mode function varies over time. It returns PDF mode estimates compiled from 24 hour windows with UTC day boundaries.\n\nThe following plot, from the noise-mode-timeseries web service, shows the daily PDF mode variations, measured at 7 frequencies, from the same data used to generate the previous PDF plot.", null, "Mode time series Plot shown the daily mode variations measured at 7 periods\n\nAt periods of 10.0 and 2.97 seconds seasonal variations of noise level are clearly visible and at 190.27 and 103.75 seconds step level changes are also visible.\n\nThe daily PDF mode functions are computed at about 88 frequencies. It is generally not useful to plot this many timeseries in one plot. See “Frequency Selection” for details on how to customize the selection of frequencies.\n\nThe web service provides an output option that allows the PDF modes to be displayed relative to the Peterson noise models or relative to user defined noise models. See “Noise Model Comparison” for details.\n\n## Output Formats\n\nThe web service provides three output formats:\n\n• plot – Generates plot images\n• xml – Generates XML data documents\n• text. – Generates text data documents\n\n### Plot Format\n\nWith the `format=plot` output option, the webservice returns PNG plots of the daily PDF modes.\n\n#### Plot Customization\n\nFourteen options allow for customizing plot appearance.\n\nPlot Dimensions\n`plot.width` and `plot.height` allow for customizing the size of the generated plot images.\n\nExample: 2000×1000 image.\n\nTitle Customization\n`plot.title` and `plot.subtitle` options allow for customizing the plot titles. If the title or subtitle are `hide`, they will not be shown.\n\nExample: No Subtitle and Title is ANMO 1998 – 2017 Daily PDF Modes\n\nTip: When setting the title, the character sequence `%20` can be used for white spaces and `%0A` can be used for carriage-return. See Percent-encoding: Wikipedia for more information.\n\nDomain Range\n`plot.power.min` and `plot.power.max` options allow for customizing the domain of the generated plots. By default the range is auto selected. This can make comparing plots from different data sets difficult. These options make comparing plots easier. `plot.power.min` and `plot.power.max` must be specified together.\n\nExample: Power range -185 to -115\n\nFont Sizes\nFont sizes can be set for six different aspects of the plot:\n\n• `plot.titlefont.size` – title on the top\n• `plot.subtitlefont.size` – subtitle\n• `plot.powerlablefont.size` – power axis title\n• `plot.poweraxisfont.size` – power axis labels\n• `plot.timeaxisfont.size` – time axis labels\n• `plot.legendfont.size` – legend at the bottom of the plot\n\nExample: Time Axis Font size to 24\n\nThe `plot.legend` option is used for hiding the legend box at the bottom of the plot. Use `plot.legend=hide` to hide the legend and `plot.legend=show` to show (default).\n\nExample: No Legend\n\n### XML Format\n\nWith the `format=xml` option the service produces XML documents useful for machine processing.\nThere are two options that effect XML formatting: `xml.style` and `xml.units`.\n\n`xml.style` effects how output information is grouped:\n\n• `xml.style=byday` default Group output by day. (example)\n• `xml.style=byfrequency` Group output by frequencies. (example)\n• `xml.style=byperiod` Group output by periods. (example)\n• `xml.style=flat` no grouping.(example)\n\n`xml.units` effect whether period or frequency is output. This option is mainly useful with `xml.style=byday` and `xml.style=flat`.\n\n• `xml.units=seconds` Show results by period. (example)\n• `xml.units=hertz` default Show results by frequency. (example)\n\n### Text Format\n\nWith the `format=text` option the service produces text documents useful for machine processing or human reading.\n\nThere are two options that effect text formatting `text.style` and `text.units`.\n\n`text.style` effects how output information is grouped:\n\n• `text.style=list` default Group (list) of rows for each day. (example)\n• `text.style=table` Single row for each day. (example)\n\n`text.units` Effect whether period or frequency is output.\n\n• `text.units=seconds` Show results by period. (example)\n• `text.units=hertz` default Show results by frequency. (example)\n\n## Noise Model Comparison\n\nThe `output` and `noisemodel.byperiod` and `noisemodel.byfrequency` options allow the mode power levels to be compared to noise models.\n\nThe web service contains default high and low noise models derived from tables 3 and 4 in Observations and Modeling of Seismic Background Noise by Jon Peterson, 1993.\n\nThe default noise models can be view from the link /defaultnoisemodel. Noise models are interpolated using a piecewise log, linear relationship given by:\n\nNoise = A + B Log10(Period)\n\nNoise model levels are interpreted as constant above and below model specifications.\n\n### `output` option\n\nThe `output` option accepts the following three values: `power`, `powerdhnm`, `powerdlnm` and `powerdnm`\n\n`output=power` is the default. Output values are simply the mode power levels. (example)\n\n`output=powerdhnm`: power values are differenced against the high-noise model. (example)\n\n`output=powerdlnm`: power values are differenced against the low-noise model. (example)\n\n`output=powerdnm`: power values are compared to both the high and low noise models. (example)\n\nPower levels are returned with the following logic:\n\n``` if( power > HNM )\nreturn (power - HNM)\nelse if ( power < LNM )\nreturn (power - LNM)\nelse\nreturn 0\n```\n\nWith output formats `format=plot` and `format=text` only one `output` parameter may be specified. However, when `format=xml` is specified, multiple values can be used.\n\n### `noisemodel.byperiod` and `noisemodel.byfrequency` Options\n\nThe `noisemodel.byperiod` and `noisemodel.byfrequency` options allow for the input of custom noise models.\n\nThe noise model parameters accept both high and low noise models or a single noise model. If a single noise model is specified\nthe `output` options `powerdhnm`, `powerdlnm`, `powerdnm` will all return the same values.\n\nThe noise models should be in the format\n\nHigh and low models by period:\n\n```noisemodel.byperiod=period1,valueA1,valueB1|period2,valueA2,valueB2|period3,valueA3,valueB3...\n```\n\nSingle model by period:\n\n```noisemodel.byperiod=period1,value1|period2,value2|period3,value3...\n```\n\nHigh and low models by frequency:\n\n```noisemodel.byfreqeuncy=frequency1,valueA1,valueB1|frequency2,valueA2,valueB2|frequency3,valueA3,valueB3...\n```\n\nSingle model by frequency:\n\n```noisemodel.byfreqeuncy=frequency1,value1|frequency2,value2|frequency3,value3...\n```\n\nFor high/low models (two values per frequency or period) the order of the value pairs is not important; greater values are assigned to the high noise model and lower values are assigned to the low noise model. The order of frequencies (or periods) is not important; however, there should be no duplicates.\n\nThe character sequence `%7C` can be used in place of the `|` (pipe) character in URLs.\n\nExample\n\n…output=powerdhnm&noisemodel.byperiod=0.1,-144,-154|1.0,-155,-165|7.0,-125,-138|10,-130,-148|20,-150,-160|40,-170,-180|100,-174,-182\n\n## Frequency and Period Selection\n\nThe mutually exclusive options `frequencies` and `periods` allow for choosing which frequencies or powers are selected for output.\n\nOutput values which are closest to the input values are selected. This alleviates the user from having to know the analyzed periods or frequencies exactly.\n\nDefault Selection\nBy default, the frequencies which are closest to the following set of frequencies are selected for output:\n\n```0.00001, 0.000032, 0.0001, 0.00032, 0.001, 0.0032, 0.01, 0.032, 0.1, 0.32, 1.0, 3.2, 10.0, 32.0, 100.0, 320.0 hertz\n```\n\nSelection by List\n\nFrequencies or periods can be selected using a comma selected list:\n\nExamples:\n\nSelection by Range\n\nFrequencies or periods can be selected by range using the notation\n\n```[start,end]\n```\n\nThe start and end can be switched. For example, [1,2] and [2,1] will yield the same results.\n\nExamples:\n\nTip: `%5B` can be substituted for the `[` character and `%5D` can be substituted for the `]` character in URLs\n\nSelection of `all` available frequencies/periods\n\nUse `frequecies=all` or `periods=all` to selection all available frequencies and periods\n\nExample\n…format=plot&frequencies=all\n\nWikipedia: Percent-encoding\nAmbient Noise levels in the Continental United States by Daniel E. McNamara and Raymond P. Buland\nAmbient Noise Probability Density Functions by D. McNamara and R. Boaz\nObservations and Modeling of Seismic Background Noise by Jon Peterson, 1993.\n\nPage built 10:02:47 | v.2ac74663" ]
[ null, "http://service.iris.edu/media/webservicedoc/mustang/noise-mode-timeseries/1/anmo-pdf-plot.png", null, "http://service.iris.edu/media/webservicedoc/mustang/noise-mode-timeseries/1/IU.ANMO.00.BHZ.M_1998-10-26_to_2017-06-07.mode.png", null ]
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https://whitestripe.savingadvice.com/2010/11/24/hmmph-over-it-already-_63738/
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Array\n(\n => 115.42.70.24\n)\n\n => Array\n(\n => 45.128.188.43\n)\n\n => Array\n(\n => 103.140.129.63\n)\n\n => Array\n(\n => 101.50.113.147\n)\n\n => Array\n(\n => 103.66.73.30\n)\n\n => Array\n(\n => 117.247.193.169\n)\n\n => Array\n(\n => 120.29.100.94\n)\n\n => Array\n(\n => 42.109.154.39\n)\n\n => Array\n(\n => 122.173.155.150\n)\n\n => Array\n(\n => 45.115.104.53\n)\n\n => Array\n(\n => 116.74.29.84\n)\n\n => Array\n(\n => 101.50.125.34\n)\n\n => Array\n(\n => 45.118.166.80\n)\n\n => Array\n(\n => 91.236.184.27\n)\n\n => Array\n(\n => 113.167.185.120\n)\n\n => Array\n(\n => 27.97.66.222\n)\n\n => Array\n(\n => 43.247.41.117\n)\n\n => Array\n(\n => 23.229.16.227\n)\n\n => Array\n(\n => 14.248.79.209\n)\n\n => Array\n(\n => 117.5.194.26\n)\n\n => Array\n(\n => 117.217.205.41\n)\n\n => Array\n(\n => 114.79.169.99\n)\n\n => Array\n(\n => 103.55.60.97\n)\n\n => Array\n(\n => 182.75.89.210\n)\n\n => Array\n(\n => 77.73.66.109\n)\n\n => Array\n(\n => 182.77.126.139\n)\n\n => Array\n(\n => 14.248.77.166\n)\n\n => Array\n(\n => 157.35.224.133\n)\n\n => Array\n(\n => 183.83.38.27\n)\n\n => Array\n(\n => 182.68.4.77\n)\n\n => Array\n(\n => 122.177.130.234\n)\n\n => Array\n(\n => 103.24.99.99\n)\n\n => Array\n(\n => 103.91.127.66\n)\n\n => Array\n(\n => 41.90.34.240\n)\n\n => Array\n(\n => 49.205.77.102\n)\n\n => Array\n(\n => 103.248.94.142\n)\n\n => Array\n(\n => 104.143.92.170\n)\n\n => Array\n(\n => 219.91.157.114\n)\n\n => Array\n(\n => 223.190.88.22\n)\n\n => Array\n(\n => 223.190.86.232\n)\n\n => Array\n(\n => 39.41.172.80\n)\n\n => Array\n(\n => 124.107.206.5\n)\n\n => Array\n(\n => 139.167.180.224\n)\n\n => Array\n(\n => 93.76.64.248\n)\n\n => Array\n(\n => 65.216.227.119\n)\n\n => Array\n(\n => 223.190.119.141\n)\n\n => Array\n(\n => 110.93.237.179\n)\n\n => Array\n(\n => 41.90.7.85\n)\n\n => Array\n(\n => 103.100.6.26\n)\n\n => Array\n(\n => 104.140.83.13\n)\n\n => Array\n(\n => 223.190.119.133\n)\n\n => Array\n(\n => 119.152.150.87\n)\n\n => Array\n(\n => 103.125.130.147\n)\n\n => Array\n(\n => 27.6.5.52\n)\n\n => Array\n(\n => 103.98.188.26\n)\n\n => Array\n(\n => 39.35.121.81\n)\n\n => Array\n(\n => 74.119.146.182\n)\n\n => Array\n(\n => 5.181.233.162\n)\n\n => Array\n(\n => 157.39.18.60\n)\n\n => Array\n(\n => 1.187.252.25\n)\n\n => Array\n(\n => 39.42.145.59\n)\n\n => Array\n(\n => 39.35.39.198\n)\n\n => Array\n(\n => 49.36.128.214\n)\n\n => Array\n(\n => 182.190.20.56\n)\n\n => Array\n(\n => 122.180.249.189\n)\n\n => Array\n(\n => 117.217.203.107\n)\n\n => Array\n(\n => 103.70.82.241\n)\n\n => Array\n(\n => 45.118.166.68\n)\n\n => Array\n(\n => 122.180.168.39\n)\n\n => Array\n(\n => 149.28.67.254\n)\n\n => Array\n(\n => 223.233.73.8\n)\n\n => Array\n(\n => 122.167.140.0\n)\n\n => Array\n(\n => 95.158.51.55\n)\n\n => Array\n(\n => 27.96.95.134\n)\n\n => Array\n(\n => 49.206.214.53\n)\n\n => Array\n(\n => 212.103.49.92\n)\n\n => Array\n(\n => 122.177.115.101\n)\n\n => Array\n(\n => 171.50.187.124\n)\n\n => Array\n(\n => 122.164.55.107\n)\n\n => Array\n(\n => 98.114.217.204\n)\n\n => Array\n(\n => 106.215.10.54\n)\n\n => Array\n(\n => 115.42.68.28\n)\n\n => Array\n(\n => 104.194.220.87\n)\n\n => Array\n(\n => 103.137.84.170\n)\n\n => Array\n(\n => 61.16.142.110\n)\n\n => Array\n(\n => 212.103.49.85\n)\n\n => Array\n(\n => 39.53.248.162\n)\n\n => Array\n(\n => 203.122.40.214\n)\n\n => Array\n(\n => 117.217.198.72\n)\n\n => Array\n(\n => 115.186.191.203\n)\n\n => Array\n(\n => 120.29.100.199\n)\n\n)\n```\nhmmph, over it already. : whitestripe\n << Back to all Blogs Login or Create your own free blog Layout: Blue and Brown (Default) Author's Creation\nHome > hmmph, over it already.", null, "", null, "", null, "November 24th, 2010 at 02:53 am\n\nThis party planning business, that is. I started to research the cost that we're going to be looking at, and although I was initially thrilled at hall rental prices (around \\$20-\\$80 a day, depending on which one), I then started to research the price of catered finger-food. Basically, I think we're looking at roughly \\$500-\\$1000 for food, and I'm not even sure if that includes waitstaff.\nI don't have any problems asking the guests to bring their own alcohol, but I do think we should supply the food. Simply because some of these people will be travelling hours to get here, and also it's not feasible to ask them to bring a plate because of that reason as well - and we have already ruled out having this thing at a restaurant (too many people to organise having it at a 'nice' restaurant, and restaurants seem to jack up the price at the mere SMELL of the word 'wedding' etc).\nI imagine that there are a million other things that will incur 'hidden costs' that I haven't even begun to think of. I am not particularly skilled in visualising a 'theme' or the 'decor' for an event, those types of things - I just don't see the point, so when someone asks me something along those lines, my mind goes blank. Do people really notice ribbons, chair coverings, sashes, balloons, candles, flowers, table cloths? I don't - but that doesn't mean other people don't. (Though, I wish they wouldn't...)\nI *realise* that wedding gifts will probably cover the cost of this thing, but that's not really important. I just hate organising stuff like this, and I don't see the point in a lot of it.\n\nAnd I feel like I should be organising something fancy because people will get disappointed when they arrive, expecting something that it is most definately NOT. And then I get angry! Because I think, why should I be stressing out over someone elses expectations? Why are people placing these expectations on us in the first place? What's so wrong with the world where something intimate is turned into a public show to be judged and compared?\n\nI understand it's a big dream for a lot of girls, but I can take it or leave it.\n\nA friend suggested a wedding planner - yeah, I think not. I can just imagine them trying to cram a whole heap of junk down my throat and charging me an exhorbitant amount for the service. Or taking one look at my 'plan' and running for the door.\n\nAnyway. Long story short, this is me venting. :/\n\n### 7 Responses to “hmmph, over it already. ”\n\n1. miclason Says:\n\nHere in El Salvador, catering prices include the plates, tablecoths and 2 waiters for your event. No idea how it is in the US.\n\nI don�t see how your friends could be disappointed, if they are traveling to see a wedding and you two DO get married!... Yes, people notice, unfortunately. But, fortunately the people that love you know all that is just fluff.. plus, even if you do it, it doesn{t have to be too expensive! OH, and if you want them to know this isn�t one of those over the top affairs, choose an invitation that is very simple... I have received invitations that consisted of no less than 6 different pieces...embossed, engraved in gold, the envelope closed with a gossamer ribbon... you take one look at those and you know the wedding is just going to be crazy!\n\n2. Looking Forward Says:\n\nThere is nothing wrong with a simple but tasteful party. Simple pretty invites, choose your colors for paper plates, napkins and tableclothes. A few simple accents like balloons and confetti or candles and flower petals on the tables go a long way. Here in the US you can get platters of finger foods made up at grocery stores or places like Costco. Maybe you can get friends to help put them together for you. Cheese and bread, lunch meats, crackers and dip, veggies and dip, fruit, etc. could all be set out by you. The day is really about celebrating you and your DH, not how fancy the food is. Don't stress out yet!", null, "3. scfr Says:\n\nWhat did I miss? I thought you weren't having a big wedding?\n\n4. MonkeyMama Says:\n\nJust don't get too caught up in what you think everyone else wants.\n\nFor our reception, the venue was free (it was BEAUTIFUL). My mom went crazy with flowers, which I could care less about. The food was expensive, yes, but came with the wait staff. Really, the only big money we spend was food. The place was very elegant, but we didn't have to pay for it.\n\nI suppose my advice would be to keep it simple. From my experience, it is easy to keep it simple and frugal. I didn't want to organize a big affair, so I didn't. I found a nice location, hired them to do the food, my mom bought some flowers, and that was all there was to it. (Getting the cost down on the food will take more work, is all). As far as decor? Find a place with nice decor. I suppose we may have gotten to choose table clothe colors, but I really don't remember it being a big ordeal. (I mean, believe me, we looked at placed that nickel and dimed for chair covers and stuff like that. I just refused to pay for all that stuff). Meanwhile, I could care less if we had a table clothe color choice, where we ended up.\n\n5. baselle Says:\n\nAim for simple, basic and intimate - in my mind you can't go wrong with that. One of the better weddings that I attended in the most recent past was a 40 person affair, very simple and intimate. Basic white decorations, neat, clean, not fussy. Most of the wedding party had to fly cross-country from New York City to Seattle, but I didn't hear any compliants along the lines of \"I flew all this way and I got only this?\"\n\nAlso, try to take advantage of your knowledge. For example, since you are a baker, you probably can get a line on a cake apart from the caterer...perhaps even make it yourself.\n\n6. Homebody Says:\n\nYou don't have to do anything you don't want to! However, if you want some ideas, look at my daughter's blog under category weddings:\n\nhttp://kathleenamelia.com/blog/\n\nI especially love Liz and Jason's colors and theme. Maddy and John had a very elegant cake.\n\nGood luck and enjoy your day. For YD's wedding, we had the same venue as Cassie and what's his name, Arrington Apples. She did everything for YD's wedding and it was very nice.\n\n7. whitestripe Says:\n\nthanks guys. we are working some things out and have come up with a few ideas which will cost less than a catered event. my best friend is helping me. (I will post more at a later time)\n\nSCFR: my previous post explains our change of mind on the whole issue. me, I'm not thrilled, but I'm also not the only one in the relationship - so we have decided to compromise: still doing the 'just us' thing on 15th Feb, and then a few months later having a 'party' in a rented hall for friends and family. and if certain individuals start problems, I plan to ignore them.", null, "this way I don't have to worry about having the day we get married ruined by family members who can't slap a smile on their face for a few hours - and DF still gets to have family and friends together who WILL have fun (and the sourpusses can sit in the corner - or decline the invitation...)\n\n(Note: If you were logged in, we could automatically fill in these fields for you.)\n Name: * Email: Will not be published. Subscribe: Notify me of additional comments to this entry. URL: Verification: * Please spell out the number 4.  [ Why? ]\n\nvB Code: You can use these tags: [b] [i] [u] [url] [email]" ]
[ null, "https://www.savingadvice.com/blogs/images/search/top_left.php", null, "https://www.savingadvice.com/blogs/images/search/top_right.php", null, "https://www.savingadvice.com/blogs/images/search/bottom_left.php", null, "https://www.savingadvice.com/forums/core/images/smilies/smile.png", null, "https://www.savingadvice.com/forums/core/images/smilies/biggrin.png", null ]
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https://socratic.org/questions/a-triangle-has-sides-a-b-and-c-sides-a-and-b-have-lengths-of-10-and-8-respective-4
[ "# A triangle has sides A, B, and C. Sides A and B have lengths of 10 and 8, respectively. The angle between A and C is (5pi)/24 and the angle between B and C is (3pi)/8. What is the area of the triangle?\n\nJun 15, 2017\n\n33\n\n#### Explanation:\n\n$D e g r e e s = \\left(\\frac{180}{\\pi}\\right) \\cdot r a \\mathrm{di} a n s$\n(I just like working in degrees better)\n\nLaw of Cosines equation is c^2 = a^2 + b^2 – 2*a*b*cos(gamma).\n\nThe angle $\\gamma$ needed is 180 – (37.5 + 67.5) = $180 - 105$ = $75$\n\nc^2 = 10^2 + 8^2 – 2*10*8*cos(75).\nc^2 = 100 + 64 – 160*(0.259).\n${c}^{2} = 122.6$ : $c = 11$\nThen using these values we can now find the height $h$ for the triangle and solve for the area. $\\sin \\left(37.5\\right) = \\frac{h}{10}$\n$h = \\sin \\left(37.5\\right) \\cdot 10$ ; $h = 6$\n$A = \\left(\\frac{1}{2}\\right) \\cdot b \\cdot h$ ; $A = \\left(\\frac{1}{2}\\right) \\cdot 11 \\cdot 6 = 33$" ]
[ null ]
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http://statisticslectures.com/topics/friedman/
[ "Friedman Test\n\nThe Friedman Test is a version of the Repeated-Measures ANOVA that can be performed on ordinal(ranked) data.\n\nOrdinal data is displayed in the table below. Is there a difference between Weeks 1, 2, and 3 using alpha = 0.05?\n\n Figure 1.", null, "Let's test to see if there are any differences with a hypothesis test.\n\nSteps for Friedman Test\n\n1. Define Null and Alternative Hypotheses\n\n2. State Alpha\n\n3. Calculate Degrees of Freedom\n\n4. State Decision Rule\n\n5. Calculate Test Statistic\n\n6. State Results\n\n7. State Conclusion\n\n1. Define Null and Alternative Hypotheses\n\n Figure 2.", null, "2. State Alpha\n\nalpha = 0.05\n\n3. Calculate Degrees of Freedom\n\ndf = k – 1, where k = number of groups\n\ndf = 3 – 1 = 2\n\n4. State Decision Rule\n\nWe look up our critical value in the Chi-Square Table and find a critical value of plus/minus 5.99.\n\n Figure 3.", null, "5. Calculate Test Statistic\n\nFirst, we must rank the scores of every subject, as shown below in red:\n\n Figure 4.", null, "We then replace our original values with the rankings we've just found:\n\n Figure 5.", null, "A Chi-Square value is then calculated using the sums of the ranks of each group:\n\n Figure 6.", null, "6. State Results\n\n Figure 7.", null, "Do not reject the null hypothesis.\n\n7. State Conclusion\n\n Figure 8.", null, "" ]
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https://jax.readthedocs.io/en/latest/jax.experimental.optimizers.html
[ "# jax.experimental.optimizers module¶\n\nOptimizers for use with JAX.\n\nThis module contains some convenient optimizer definitions, specifically initialization and update functions, which can be used with ndarrays or arbitrarily-nested tuple/list/dicts of ndarrays.\n\nAn optimizer is modeled as an (init_fun, update_fun, get_params) triple of functions, where the component functions have these signatures:\n\ninit_fun(params)\n\nArgs:\nparams: pytree representing the initial parameters.\n\nReturns:\nA pytree representing the initial optimizer state, which includes the\ninitial parameters and may also include auxiliary values like initial\nmomentum. The optimizer state pytree structure generally differs from that\nof params.\n\nupdate_fun(step, grads, opt_state)\n\nArgs:\nstep: integer representing the step index.\ngrads: a pytree with the same structure as get_params(opt_state)\nrepresenting the gradients to be used in updating the optimizer state.\nopt_state: a pytree representing the optimizer state to be updated.\n\nReturns:\nA pytree with the same structure as the opt_state argument representing\nthe updated optimizer state.\n\nget_params(opt_state)\n\nArgs:\nopt_state: pytree representing an optimizer state.\n\nReturns:\nA pytree representing the parameters extracted from opt_state, such that\nthe invariant params == get_params(init_fun(params)) holds true.\n\n\nNotice that an optimizer implementation has a lot of flexibility in the form of opt_state: it just has to be a pytree of JaxTypes (so that it can be passed to the JAX transforms defined in api.py) and it has to be consumable by update_fun and get_params.\n\nclass jax.experimental.optimizers.JoinPoint(subtree)[source]\n\nBases: object\n\nMarks the boundary between two joined (nested) pytrees.\n\nclass jax.experimental.optimizers.OptimizerState(packed_state, tree_def, subtree_defs)\n\nBases: tuple\n\npacked_state\n\nAlias for field number 0\n\nsubtree_defs\n\nAlias for field number 2\n\ntree_def\n\nAlias for field number 1\n\njax.experimental.optimizers.adagrad(step_size, momentum=0.9)[source]\n\nParameters: step_size – positive scalar, or a callable representing a step size schedule that maps the iteration index to positive scalar. momentum – optional, a positive scalar value for momentum An (init_fun, update_fun, get_params) triple.\njax.experimental.optimizers.adam(step_size, b1=0.9, b2=0.999, eps=1e-08)[source]\n\nParameters: step_size – positive scalar, or a callable representing a step size schedule that maps the iteration index to positive scalar. b1 – optional, a positive scalar value for beta_1, the exponential decay rate for the first moment estimates (default 0.9). b2 – optional, a positive scalar value for beta_2, the exponential decay rate for the second moment estimates (default 0.999). eps – optional, a positive scalar value for epsilon, a small constant for numerical stability (default 1e-8). An (init_fun, update_fun, get_params) triple.\njax.experimental.optimizers.clip_grads(grad_tree, max_norm)[source]\n\nClip gradients stored as a pytree of arrays to maximum norm max_norm.\n\njax.experimental.optimizers.constant(step_size)[source]\njax.experimental.optimizers.exponential_decay(step_size, decay_steps, decay_rate)[source]\njax.experimental.optimizers.inverse_time_decay(step_size, decay_steps, decay_rate, staircase=False)[source]\njax.experimental.optimizers.l2_norm(tree)[source]\n\nCompute the l2 norm of a pytree of arrays. Useful for weight decay.\n\njax.experimental.optimizers.make_schedule(scalar_or_schedule)[source]\njax.experimental.optimizers.momentum(step_size, mass)[source]\n\nConstruct optimizer triple for SGD with momentum.\n\nParameters: step_size – positive scalar, or a callable representing a step size schedule that maps the iteration index to positive scalar. mass – positive scalar representing the momentum coefficient. An (init_fun, update_fun, get_params) triple.\njax.experimental.optimizers.nesterov(step_size, mass)[source]\n\nConstruct optimizer triple for SGD with Nesterov momentum.\n\nParameters: step_size – positive scalar, or a callable representing a step size schedule that maps the iteration index to positive scalar. mass – positive scalar representing the momentum coefficient. An (init_fun, update_fun, get_params) triple.\njax.experimental.optimizers.optimizer(opt_maker)[source]\n\nDecorator to make an optimizer defined for arrays generalize to containers.\n\nWith this decorator, you can write init, update, and get_params functions that each operate only on single arrays, and convert them to corresponding functions that operate on pytrees of parameters. See the optimizers defined in optimizers.py for examples.\n\nParameters: opt_maker – a function that returns an (init_fun, update_fun, get_params) triple of functions that might only work with ndarrays, as per init_fun :: ndarray -> OptStatePytree ndarray update_fun :: OptStatePytree ndarray -> OptStatePytree ndarray get_params :: OptStatePytree ndarray -> ndarray An (init_fun, update_fun, get_params) triple of functions that work on arbitrary pytrees, as perinit_fun :: ParameterPytree ndarray -> OptimizerState update_fun :: OptimizerState -> OptimizerState get_params :: OptimizerState -> ParameterPytree ndarray The OptimizerState pytree type used by the returned functions is isomorphic to ParameterPytree (OptStatePytree ndarray), but may store the state instead as e.g. a partially-flattened data structure for performance.\njax.experimental.optimizers.pack_optimizer_state(marked_pytree)[source]\n\nConverts a marked pytree to an OptimizerState.\n\nThe inverse of unpack_optimizer_state. Converts a marked pytree with the leaves of the outer pytree represented as JoinPoints back into an OptimizerState. This function is intended to be useful when deserializing optimizer states.\n\nParameters: marked_pytree – A pytree containing JoinPoint leaves that hold more pytrees. An equivalent OptimizerState to the input argument.\njax.experimental.optimizers.piecewise_constant(boundaries, values)[source]\njax.experimental.optimizers.polynomial_decay(step_size, decay_steps, final_step_size, power=1.0)[source]\njax.experimental.optimizers.rmsprop(step_size, gamma=0.9, eps=1e-08)[source]\n\nConstruct optimizer triple for RMSProp.\n\nParameters: step_size – positive scalar, or a callable representing a step size schedule that maps the iteration index to positive scalar. gamma: Decay parameter. eps: Epsilon parameter. An (init_fun, update_fun, get_params) triple.\njax.experimental.optimizers.rmsprop_momentum(step_size, gamma=0.9, eps=1e-08, momentum=0.9)[source]\n\nConstruct optimizer triple for RMSProp with momentum.\n\nThis optimizer is separate from the rmsprop optimizer because it needs to keep track of additional parameters.\n\nParameters: step_size – positive scalar, or a callable representing a step size schedule that maps the iteration index to positive scalar. gamma – Decay parameter. eps – Epsilon parameter. momentum – Momentum parameter. An (init_fun, update_fun, get_params) triple.\njax.experimental.optimizers.sgd(step_size)[source]\n\nConstruct optimizer triple for stochastic gradient descent.\n\nParameters: step_size – positive scalar, or a callable representing a step size schedule that maps the iteration index to positive scalar. An (init_fun, update_fun, get_params) triple.\njax.experimental.optimizers.sm3(step_size, momentum=0.9)[source]\n\nConstruct optimizer triple for SM3.\n\nMemory-Efficient Adaptive Optimization for Large-Scale Learning. https://arxiv.org/abs/1901.11150\n\nParameters: step_size – positive scalar, or a callable representing a step size schedule that maps the iteration index to positive scalar. momentum – optional, a positive scalar value for momentum An (init_fun, update_fun, get_params) triple.\njax.experimental.optimizers.unpack_optimizer_state(opt_state)[source]\n\nConverts an OptimizerState to a marked pytree.\n\nConverts an OptimizerState to a marked pytree with the leaves of the outer pytree represented as JoinPoints to avoid losing information. This function is intended to be useful when serializing optimizer states.\n\nParameters: opt_state – An OptimizerState A pytree with JoinPoint leaves that contain a second level of pytrees." ]
[ null ]
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https://fr.maplesoft.com/support/help/maple/view.aspx?path=max&L=F
[ "", null, "max - Maple Programming Help\n\nmax\n\nfind the maximum of numbers\n\nmin\n\nfind the minimum of numbers\n\n Calling Sequence max(x1, x2, ...) min(x1, x2, ...) max[defined](x1, x2, ...) min[defined](x1, x2, ...) max[nodefault](x1, x2, ...) min[nodefault](x1, x2, ...) max[defined,nodefault](x1, x2, ...) min[defined,nodefault](x1, x2, ...) max[index](C) min[index](C) max[index,defined](C) min[index,defined](C)\n\nParameters\n\n x1, x2, ... - any expressions C - list, set, Matrix, Array, or Vector\n\nDescription\n\n • The functions max and min return the maximum and minimum, respectively, of one or more arguments.\n • Most often the arguments are of type numeric: integers, rationals, or floats.  However, the functionality is more general, allowing any type of argument for which an unevaluated function call may be returned.  When given a list, set, or rtable-based Array (a container) as an argument, the maximum or minimum element in the container is returned.\n • In most cases, a call to max or min containing an undefined value returns undefined.\n • A call to max where one of the arguments is positive infinity returns positive infinity. Likewise, a call min containing negative infinity returns negative infinity. This is true even if one of the other arguments to max or min is undefined.\n • To ignore undefined values when computing the maximum or minimum, use the optional index defined. That is, max[defined](...).\n • By default, a call to max or min with no arguments or with only empty container arguments returns negative or positive infinity, respectively. If called with the index nodefault, the result is NULL instead.\n • The functions max[index] and min[index] return the index of the maximum or minimum element in the container C.\n Only one argument can be supplied when using the [index] variations.  The mandate for this option is that C[ max[index](C) ] give the maximal element.  That mandate cannot be fulfilled when C is a sequence, or when C is a container that has other containers inside (such as a list of vectors); an error is raised in these situations.\n • These routines use evalr to try to resolve the relevant inequalities. Since evalr uses the current setting of Digits for its computations, it may help to increase Digits if either max(...) or min(...) returns unevaluated.\n • The maximum and minimum of polynomials can be converted to a piecewise function.\n\n • The max and min commands are thread-safe as of Maple 15.\n\nExamples\n\nThe following examples locate the maximum or minimum of numeric or constant values.\n\n > $\\mathrm{min}\\left(1\\right)$\n ${1}$ (1)\n > $\\mathrm{max}\\left(\\frac{3}{2},1.49\\right)$\n $\\frac{{3}}{{2}}$ (2)\n > $\\mathrm{max}\\left(\\frac{3}{5},\\mathrm{ln}\\left(2\\right),\\frac{9}{13},-\\mathrm{\\infty }\\right)$\n ${\\mathrm{ln}}{}\\left({2}\\right)$ (3)\n > $a≔24:$\n > $b≔-3:$\n > $\\mathrm{max}\\left(a,b\\right)$\n ${24}$ (4)\n\nWhen an unknown is present, the max/min of the known values will be reduced, and the whole expression will be returned as an unevaluated function call, which can be symbolically manipulated or further evaluated later.\n\n > $\\mathrm{min}\\left(\\frac{3}{2},1.49,f\\left(x\\right)\\right)$\n ${\\mathrm{min}}{}\\left({1.49}{,}{f}{}\\left({x}\\right)\\right)$ (5)\n\nUse the defined option to determine the maximum or minimum ignoring all undefined values (returns the max/min of all defined values).\n\n > $\\mathrm{max}\\left(\\frac{3}{2},1.49,\\mathrm{undefined},2.5\\right)$\n ${\\mathrm{undefined}}$ (6)\n > $\\mathrm{max}\\left[\\mathrm{defined}\\right]\\left(\\frac{3}{2},1.49,\\mathrm{undefined},2.5\\right)$\n ${2.5}$ (7)\n > $\\mathrm{min}\\left[\\mathrm{defined}\\right]\\left(\\mathrm{undefined},8,-3446\\right)$\n ${-3446}$ (8)\n\nNote that infinity is known to be a maximum even in the presence of undefined.\n\n > $\\mathrm{max}\\left(\\frac{3}{2},1.49,\\mathrm{undefined},\\mathrm{\\infty }\\right)$\n ${\\mathrm{\\infty }}$ (9)\n\nSome algebraic combinations can be reduced.\n\n > $\\mathrm{max}\\left(x+1,x+2,y\\right)$\n ${\\mathrm{max}}{}\\left({y}{,}{x}{+}{2}\\right)$ (10)\n\nThe minimum and maximum of no arguments are, as expected, positive and negative infinity, respectively.\n\n > $\\mathrm{min}\\left(\\right)$\n ${\\mathrm{\\infty }}$ (11)\n > $\\mathrm{max}\\left(\\right)$\n ${-}{\\mathrm{\\infty }}$ (12)\n\nTo return NULL instead, use the index nodefault.\n\n > $\\mathrm{max}\\left[\\mathrm{nodefault}\\right]\\left(\\right)$\n\nUnevaluated max/min expressions containing polynomials can be converted to piecewise functions.\n\n > $\\mathrm{convert}\\left(\\mathrm{max}\\left({x}^{2}-2,x+3\\right),\\mathrm{piecewise}\\right)$\n $\\left\\{\\begin{array}{cc}{{x}}^{{2}}{-}{2}& {x}{\\le }\\frac{{1}}{{2}}{-}\\frac{\\sqrt{{21}}}{{2}}\\\\ {x}{+}{3}& {x}{\\le }\\frac{{1}}{{2}}{+}\\frac{\\sqrt{{21}}}{{2}}\\\\ {{x}}^{{2}}{-}{2}& \\frac{{1}}{{2}}{+}\\frac{\\sqrt{{21}}}{{2}}{<}{x}\\end{array}\\right\\$ (13)\n > $\\mathrm{convert}\\left(\\mathrm{min}\\left({x}^{3}+8,x+8\\right),\\mathrm{piecewise}\\right)$\n $\\left\\{\\begin{array}{cc}{{x}}^{{3}}{+}{8}& {x}{\\le }{-1}\\\\ {x}{+}{8}& {x}{\\le }{0}\\\\ {{x}}^{{3}}{+}{8}& {x}{\\le }{1}\\\\ {x}{+}{8}& {1}{<}{x}\\end{array}\\right\\$ (14)\n\nmax and min will search for values inside a container such as a list, set, Matrix, Vector, or Array.\n\n > $\\mathrm{max}\\left(\\left[1,2,3,4\\right],⟨5,6,7⟩\\right)$\n ${7}$ (15)\n\nUse the index option to determine the location of the maximum or minimum value.  Only one argument can be used when asking for the index.\n\n > $L≔\\left[11,12,13,14\\right]$\n ${L}{≔}\\left[{11}{,}{12}{,}{13}{,}{14}\\right]$ (16)\n > $\\mathrm{ind}≔\\mathrm{max}\\left[\\mathrm{index}\\right]\\left(L\\right)$\n ${\\mathrm{ind}}{≔}{4}$ (17)\n > $L\\left[\\mathrm{ind}\\right]$\n ${14}$ (18)\n\nTwo-dimensional indexes are returned as a sequence and can be used to directly index the maximum or minimum value.  If multiple extreme values occur, only one index will be returned.\n\n > $M≔\\mathrm{Matrix}\\left(\\left[\\left[11,12\\right],\\left[21,22\\right]\\right]\\right)$\n ${M}{≔}\\left[\\begin{array}{cc}{11}& {12}\\\\ {21}& {22}\\end{array}\\right]$ (19)\n > $\\mathrm{ind}≔\\mathrm{max}\\left[\\mathrm{index}\\right]\\left(M\\right)$\n ${\\mathrm{ind}}{≔}{2}{,}{2}$ (20)\n > $M\\left[\\mathrm{ind}\\right]$\n ${22}$ (21)\n\nmax can find the maximum in a list of Arrays (or Array of lists, or general data container inside a data container).  However, max[index] will raise an error because there is no simple index that can reference the element.\n\n > $M≔\\mathrm{Matrix}\\left(\\left[\\left[11,12\\right],\\left[21,22\\right]\\right]\\right)$\n ${M}{≔}\\left[\\begin{array}{cc}{11}& {12}\\\\ {21}& {22}\\end{array}\\right]$ (22)\n > $M\\left[1\\right]≔\\left[44,55,66\\right]$\n ${{M}}_{{1}}{≔}\\left[{44}{,}{55}{,}{66}\\right]$ (23)\n > $M$\n $\\left[\\begin{array}{cc}\\left[{44}{,}{55}{,}{66}\\right]& \\left[{44}{,}{55}{,}{66}\\right]\\\\ {21}& {22}\\end{array}\\right]$ (24)\n\nAn error is expected here:\n\n > $\\mathrm{ind}≔\\mathrm{max}\\left[\\mathrm{index}\\right]\\left(M\\right)$\n\nThe options index and defined can be used together to search for an extreme value in a container while ignoring undefined.\n\n > $V≔⟨-4,77,\\mathrm{undefined},-\\mathrm{\\infty }⟩$\n ${V}{≔}\\left[\\begin{array}{c}{-4}\\\\ {77}\\\\ {\\mathrm{undefined}}\\\\ {-}{\\mathrm{\\infty }}\\end{array}\\right]$ (25)\n > $\\mathrm{ind}≔\\mathrm{max}\\left['\\mathrm{index}','\\mathrm{defined}'\\right]\\left(V\\right)$\n ${\\mathrm{ind}}{≔}{2}$ (26)\n > $V\\left[\\mathrm{ind}\\right]$\n ${77}$ (27)\n\nCompatibility\n\n • The C parameter was introduced in Maple 2015." ]
[ null, "https://bat.bing.com/action/0", null ]
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https://www.physicsforums.com/threads/potential-energy-between-two-objects.33890/
[ "# Potential energy between two objects\n\nThe following question appeared in my book:\n\nFrom an inertial frame in space, we watch two identical uniform spheres fall toward one another owing to their mutual gravitational attraction. Approximate their initial speed as zero and take the initial gravitational potential energy of the two-sphere system as $U_i$. When the separation between the two spheres is half the initial separation, what is the kinetic energy of each sphere?\n\nSince the mechanical energy of the two-sphere system will be conserved,\n\n$$K_i + U_i = K_f + U_f$$\n\nLet K be the kinetic energy of one of the spheres in the final configuration of the system. Since both spheres are identical, K is also the kinetic energy of the other sphere, making the net kinetic energy 2K. Bearing in mind that the kinetic energy of each sphere was given as initially zero, this means:\n\n$$U_i = 2K + U_f$$\n\n$$\\frac {U_i - U_f}{2} = K$$\n\nThe gravitational potential energy of such a system is given by:\n\n$$U = -\\frac{GMm}{r}$$\n\nLet R be the distance between the two spheres at the system's initial configruation, then R/2 is the distance between the two spheres in the system's final configuration, and:\n\n$$\\frac{-\\frac{GMm}{R} - -\\frac{GMm}{R/2}}{2} = K$$\n\n$$\\frac{-\\frac{GMm}{R} + \\frac{2GMm}{R}}{2} = K$$\n\n$$\\frac{\\frac{GMm}{R}}{2} = K$$\n\n$$\\frac{-U_i}{2} = K$$\n\nThe book's answer is $U_i / 4$, not $-U_i / 2$. Where is my mistake?\n\nIs it just that the book authors made an error? I still don't see anything wrong with my solution. Also, A positive value for Ui means that, since Ui itself contains a negative, the book's value for kinetic energy would be negative, and that doesn't sound right at all. :yuck:\n\narildno\nHomework Helper\nGold Member\nDearly Missed\n\nSince the potential energy is an inverse-r law, then halving the separation distance between two masses doubles the potential energy of the system. This in turn halves the kinetic energy of the system, which means that the kinetic energy of the individual masses must halve as well. I like your answer.\n\nThis\nThis in turn halves the kinetic energy of the system, which means that the kinetic energy of the individual masses must halve as well.\nmakes no sense at all, since the initial kinetic energy is 0, and 0/2 is still 0.\n\nYou're getting confused by the signs of these values. Maybe it'll be easier if you think in terms of |U|, the absolute value of the potential energy, i.e. a positive number. So the initial potential energy is -|Ui|.\n\nThe potential energy which is doubling is a negative number, so the potential energy is decreasing from -|Ui| to -2|Ui|. The change in potential energy is -|Ui|. Conservation of energy requires that kinetic energy will increase by the same amount. The initial kinetic energy was 0, so the final kinetic energy is Ui. At the end, total energy = U + K = -2|Ui| + |Ui| = -|Ui|, so all is well. [edited to re-order the terms to be consistent]\n\nSince you are looking at the system from an external inertial frame, and the two spheres are identical in mass and each started with 0 velocity, conservation of momentum requires that the velocities of the two spheres are always equal in magnitude and opposite in direction. Direction is irrelevant as far as kinetic energy is concerned, so each sphere ends up with the same kinetic energy: Ui/2.\n\nSo Zorodius is correct as to the absolute value; his book is correct as to the sign.\n\nLast edited:\nSorry, I did get the signs confused. Since the objects are moving to where they naturally want to go, then they must LOSE potential energy. That means they must gain the same amount of kinetic energy." ]
[ null ]
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http://mechanical-design-handbook.blogspot.com/2009/05/numerical-methods-newton-raphson-method.html
[ "Improve math skills of your kids - Learn step-by-step arithmetic from Math games\n\nMath: Unknown - Step-by-step math calculation game for iOS.\n\nMath: Unknown is much more than a math game. It is a step-by-step math calculation game which will teach users how to calculate in the correct order rather than just asking only the final calculated results.\n\nThe app consists of four basic arithmetic operations which are addition, subtraction, multiplication and division. In order to get started, users who are new to arithmetic can learn from animated calculation guides showing step-by-step procedures of solving each type of operation. It is also helpful for experienced users as a quick reference.\n\nGenerally, addition and subtraction may be difficult for users who just start learning math especially when questions require carrying or borrowing (also called regrouping). The app helps users to visualize the process of carrying and borrowing in the way it will be done on paper. Once users understand how these operations work, they are ready to learn multiplication and division.\n\nFor most students, division is considered as the most difficult arithmetic operation to solve. It is a common area of struggle since it requires prior knowledge of both multiplication and subtraction. To help users understand division, the app uses long division to teach all calculation procedures. Relevant multiplication table will be shown beside the question. Users will have to pick a number from the table which go into the dividend. Multiplication of selected number and divisor is automatically calculated, but the users have to do subtraction and drop down the next digit themselves. Learning whole calculation processes will make them master it in no time.\n\nMath: Unknown is a helpful app for students who seriously want to improve arithmetic calculation skills.\n\nNumerical Methods - The Newton-Raphson Method to Solve Mechanical Design Problems Part II\n\nIn the previous post, we talked about several root finding techniques. In this post, we're going to see how we can use Microsoft Excel VBA to find the roots using Newton-Raphson Method.\n\nAs we know that Newton-Raphson Method is the most widely used of all root-locating formulas. The Newton-Raphson method uses the slope (first derivative) of the function to find the root. That means, in the VBA code, we have to find the first derivative of the function in order to find the root. We already have the VBA code to find the first derivative and we're going to use it again in Newton-Raphson Method.", null, "The Newton-Raphson Method uses 2 terms of taylor series to approximate delta x as shown below.\n\nf(x) = f(x0) + (xN - x0)f'(x0) = 0\n(xN - x0)f'(x0) = -f(x0)\nxN - x0 = -f(x0)/f'(x0)\n\ndelta x = -f(x0)/f'(x0)\n\nThe followings are the procedure to find the root using Newton-Raphson Method.\n1. Guess the initial value of the root --> select x0\n2. Calculate delta x using the formula as shown above\n3. Calculate xN using xN = x0 + delta x\n4. Check whether the error of xN is within the allowable tolerance or not --> abs(xN - x0) < tolerance\nThen the VBA code will be like this.\n\n' ================================================\n' Created by Suparerg Suksai\n' Mechanical Design Handbook\n' http://mechanical-design-handbook.blogspot.com\n'\n' The Newton-Raphson Method - Numerical Methods\n' ================================================\n\nFunction f(x As Double) As Double\nf=Exp(-x)-x ' This is the function we want to find the root\nEnd Function\n\nFunction fDeriv(x As Double) As Double\n...\n...\nEnd Function\n\nFunction RootNewtonRaphson(x0 As Double)\n...\n...\n...\nEnd Function\n\nTo calculate the root of e-x - x = 0 we then type the following formula in the Microsoft Excel spreadsheet,\n\nIn any cell, type \"=RootNewtonRaphson(1)\", where 1 is the guessing value of the root.\nThe program will then show the result of 0.567143290409784.\n\nWe can recheck the calculation result by entering the following formula into excel,\n\n=exp(-0.567143290409784)-0.567143290409784\n\nThen the result is 0. That means one of the roots of of e-x - x = 0 is 0.567143290409784", null, "FREE DOWNLOAD AN EXCEL FILE WITH VBA CODE OF NEWTON-RAPHSON ROOT FINDING METHOD\n\nExtract the zip file with password: mechanical-design-handbook.blogspot.com" ]
[ null, "http://3.bp.blogspot.com/_e_e1W8RlFC4/Shy2LnpOF7I/AAAAAAAABYI/w2HK3TzU5hA/s400/excel-newton-raphson-root-finding-method.gif", null, "http://3.bp.blogspot.com/_e_e1W8RlFC4/ShzDW--hsgI/AAAAAAAABYQ/kEyCor2YoTs/s400/excel-newton-raphson-root-finding-method-vba-code.gif", null ]
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https://hg.logilab.org/review/nazca/rev/47efa28c96a3
[ "author Simon Chabot Fri, 19 Oct 2012 14:44:26 +0200 changeset 32 47efa28c96a3 parent 31 1410ccfe08fd child 33 0d95c1cfde36\n[Distance] Spaces are correctly supported by distances functions In fact, the problem was 'Victor Hugo' and 'Hugo Victor' had a big distance when we'd like in this case to have a small one (even a zero one !) So, the approach followed was : Construct a distance matrix : | Victor | Hugo Victor | 0 | 5 Hugo | 5 | 0 And return the minimun of the minimun of each row. In fact, we return the maximun of the minimum of the previous matrix, and the its transpose, to handle the following case : | Victor | Hugo | Jean | Victor | 0 | 5 | 6 |--> min of each row : 0 Hugo | 5 | 0 | 4 | | Victor | Hugo | Victor | 0 | 5 | Hugo | 5 | 0 |--> min of each row : 4 Jean | 6 | 4 | Return the max, ie : 4.\n distances.py file | annotate | diff | comparison | revisions test/test_alignment.py file | annotate | diff | comparison | revisions\n--- a/distances.py\tFri Oct 19 11:38:54 2012 +0200\n+++ b/distances.py\tFri Oct 19 14:44:26 2012 +0200\n@@ -16,6 +16,7 @@\n# with this program. If not, see <http://www.gnu.org/licenses/>.\n\nfrom dateutil import parser as dateparser\n+from scipy import matrix\n\ndef levenshtein(stra, strb):\n\"\"\" Compute the Levenshtein distance between stra and strb.\n@@ -25,8 +26,14 @@\n- Replace one character of stra into a character of strb\n- Add one character of strb into stra\n- Remove one character of strb\n+\n+ If spaces are found in stra or strb, this method returns\n+ _handlespaces(stra, strb), levenshtein)\n\"\"\"\n\n+ if ' ' in (stra + strb):\n+ return _handlespaces(stra, strb, levenshtein)\n+\nlena = len(stra)\nlenb = len(strb)\nonerowago = None\n@@ -40,6 +47,43 @@\nreturn thisrow[lenb - 1]\n\n+def _handlespaces(stra, strb, distance, **args):\n+ \"\"\" Compute the matrix of distances between all tokens of stra and strb\n+ (with function ``distance``). Extra args are given to the distance\n+ function\n+\n+ The distance returned is defined as the max of the min of each rows of\n+ each distance matrix, see the example above :\n+\n+ | Victor | Hugo Victor | Jean | Hugo\n+ Victor | 0 | 5 Victor | 0 | 6 | 5\n+ Jean | 6 | 4 Hugo | 5 | 4 | 0\n+ Hugo | 5 | 0\n+\n+ --> 4 --> 0\n+\n+ Return 4\n+ \"\"\"\n+\n+ if ' ' not in stra:\n+ stra += ' '\n+ if ' ' not in strb:\n+ strb += ' '\n+\n+ toka, tokb = stra.split(' '), strb.split(' ')\n+\n+ listmatrix = []\n+ for i in xrange(len(toka)):\n+ listmatrix.append([])\n+ for j in xrange(len(tokb)):\n+ listmatrix[-1].append(distance(toka[i], tokb[j], **args))\n+ m = matrix(listmatrix)\n+ minlist = [m[i,:].min() for i in xrange(m.shape)]\n+ minlist.extend([m[:,i].min() for i in xrange(m.shape)])\n+\n+ return max(minlist)\n+\n+\ndef soundexcode(word, language = 'french'):\n\"\"\" Return the Soundex code of the word ``word``\n@@ -47,12 +91,11 @@\n``language`` can be 'french' or 'english'\n\n.:: wiki_ : https://en.wikipedia.org/wiki/Soundex\n+\n+ If spaces are found in stra or strb, this method returns\n+ _handlespaces(stra, strb), soundex, language = language)\n\"\"\"\n\n- if ' ' in word:\n- words = word.split(' ')\n- return ' '.join([soundexcode(w.strip(), language) for w in words])\n-\nvowels = 'AEHIOUWY'\nif language.lower() == 'french' :\nconsonnantscode = { 'B' : '1', 'P' : '1',\n@@ -103,6 +146,9 @@\n\"\"\" Return the 1/0 distance between the soundex code of stra and strb.\n0 means they have the same code, 1 they don't\n\"\"\"\n+ if ' ' in (stra + strb):\n+ return _handlespaces(stra, strb, soundex, language = language)\n+\nreturn 0 if (soundexcode(stra, language) == soundexcode(strb, language)) \\\nelse 1\n\n--- a/test/test_alignment.py\tFri Oct 19 11:38:54 2012 +0200\n+++ b/test/test_alignment.py\tFri Oct 19 14:44:26 2012 +0200\n@@ -51,8 +51,9 @@\nclass DistancesTest(unittest2.TestCase):\ndef test_levenshtein(self):\nself.assertEqual(levenshtein('niche', 'chiens'), 5)\n- self.assertEqual(levenshtein('bonjour', 'bonjour !'), 2)\n+ self.assertEqual(levenshtein('bonjour', 'bonjour !'), 1)\nself.assertEqual(levenshtein('bon', 'bonjour'), 4)\n+ self.assertEqual(levenshtein('Victor Hugo', 'Hugo Victor'), 0)\n\n#Test symetry\nself.assertEqual(levenshtein('Victor Hugo', 'Vitor Wugo'),\n@@ -197,7 +198,7 @@\n\n#Victor Hugo --> Victor Wugo\n#Albert Camus --> Albert Camus, Albert Camu\n- self.assertEqual(m.matched(cutoff = 0.1, normalized = True),\n+ self.assertEqual(m.matched(cutoff = 0.2, normalized = True),\n{0: [(0, d(i1, i2))], 1: [(1, d(i1, i2)),\n(2, d(i1, i2))]})" ]
[ null ]
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http://agendario.forumrio.org/ebooks/problems-in-geometry
[ "# Problems in Geometry by A. Kutepov", null, "By A. Kutepov\n\nBest geometry and topology books\n\nPlane Geometry and its Groups\n\nSan Francisco 1967 Holden-Day. octavo. , 288pp. , index, hardcover. positive in VG DJ, a number of small closed tears.\n\nAdditional resources for Problems in Geometry\n\nExample text\n\nAt the point A lying on a given straight line a erect a perpendicular to this line. 2. From the point M not lying on a given straight line a drop a perpendicular to this line. 305. 1. Find the locus of points in space equidistant from a given point. 2. Find the locus of points in space equidistant from two given points. 3. Find the locus of points in space equidistant from three given points. 4. Find the locus of points in space equidistant from four given points. Consider all possible cases. 306.\n\nIts end-points are found at distances of 6 cm and 4 cm from the plane. Find the angle between the given line segment and the plane. 353. 1. From a point located at a distance of a from the plane two straight lines are drawn inclined to the plane at angles of 30° and 45° and at a right angle to each CH. III. Determine the distance between the end-points of the inclined lines. 2. From a point located at a distance of a from a plane two inclined lines are drawn at an angle of 30° to the plane. Determine the angle between their projections if the distance between the end-points of the inclined lines is equal to 3a.\n\nParticular Cases of Solving Oblique Triangles Notation: a, b, c = the sides of a triangle; A, B, C = = angles opposite them; S = area; 2p = perimeter; R = radius of the circumscribed circle; r = radius of the inscribed circle; ha, hb, h° = altitudes; la, 1b, 1° _ = bisectors. 7; B = 105°20'; C = 15°33. 205. 8; A = 70°24'; B = 50°16'; (3) a - b = 34; A = 108°; B = 28°. 4; A = 22°; B = 49°; . 206. 5 m; B = 114°50'; (3) a = 72; b = 52; A = 2B. 207. 8 m; (3) ha = 8; hb = 12; h° = 18. (2) S = 1460 m\\$; CH." ]
[ null, "https://images-na.ssl-images-amazon.com/images/G/01/x-site/icons/no-img-sm._V192198896_BO1,204,203,200_.gif", null ]
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https://dsp.stackexchange.com/questions/33524/why-calculate-negative-frequencies-of-dft/33526
[ "Why calculate negative frequencies of DFT?\n\nThis answer says that when calculating the DFT on a real valued signal, the (roughly) second half of the bins are the complex conjugates of the first half of the bins. Firstly I was curious if that is true?\n\nIf it is true, it seems like for frequency analysis purposes, you could forgo calculating the second half of the bins, which ought to be helpful for computation times and also for storage space if you want to store information in frequency domain since you could calculate the second half of the information on demand.\n\nThis also seems to be \"more correct\" to what people would expect when doing an FFT.\n\nFor instance, if using the equation:\n\n$$X_k = \\frac 1N \\sum\\limits_{n=0}^{N-1}x_ke^{-2i \\pi kn/N}, \\quad k\\in[0,N), \\quad k\\in \\mathbb Z$$\n\nIf you DFT a 4 sample cosine wave: $[1, 0, -1, 0]$, it gives the result: $[0, 0.5, 0, 0.5]$.\n\nBut, if you are only calculating the positive frequencies, it seems reasonable to modify the equation to be this:\n\n$$X_k = \\frac 2N \\sum\\limits_{n=0}^{N-1}x_ke^{-2i \\pi kn/N}, \\quad k\\in[0,N/2), \\quad k\\in \\mathbb Z$$\n\nWhich when applied to the 4 sample cosine wave gives us this result: $[0,1]$\n\nIt seems more correct (to me anyways!) that it shows that $0\\textrm{ Hz}$ (DC) has an amplitude of 0, and that $1\\textrm{ Hz}$ has an amplitude of 1. It's kind of confusing the other way, where the full amplitude of the $1\\textrm{ Hz}$ wave is split between the positive and negative $1\\textrm{ Hz}$ frequency.\n\nWhy is it then, that for real valued signals, we even bother calculating and reporting negative frequencies?\n\nNice question, first of all, DFT does not decompose a signal into regular sinusoids, it decompose it up into complex exponentials. Therefore, the Fourier transform of a real value signal must be conjugate symmetric (has both positive and negative frequencies), because when we are calculating inverse DFT, the real values of DFT would sum up and the imaginary parts cancel out to result a real valued signal. Your example $x=[1, 0, -1, 0]$ is a special case, since its DFT representation does not have any complex coefficients (however it is still conjugate symmetric),if you consider something like $x=[1,0,-1,1]$ you'll see its DFT would be\n\n$$f =[ 1.0 + 0.0i,2.0 + 1.0i,-1.0 + 0.0i,2.0 - 1.0i]$$\n\nnow you need all coefficients to get the inverse DFT of the signal. And note the coefficients are not the same but are conjugate.\n\n• That makes a lot of sense, thank you very much! – Alan Wolfe Aug 7 '16 at 16:54\n• A DFT can be written to decompose into sinusoids (cosine + sine) instead of complex exponentials, The arithmetic is the same, it's just written differently on the chalkboard. – hotpaw2 Aug 7 '16 at 17:00\n• \"Now you need all coefficients\" - you don't need the last one because you know it's the complex conjugate of the second one. – immibis Aug 8 '16 at 3:07\n• @immibis: To be precise, the IDFT needs all coefficients. it cannot assume that the second half is the conjugate of the first. You only know that because your input was real. – MSalters Aug 8 '16 at 8:27\n• @immibis and @ hotpaw2, you are right (in real valued signal case), Now this could be regarded as another question, and it is like this, why we have to compute and store all DFT coefficients, when we know not all of them are necessary and if fact we can deduce one half of DFT from the other? I feel the answer is well explained by \"Hilmar\" in other answer of this page. And as far as I know, people do not store are FFT coefficients and just half of them is used. – MimSaad Aug 8 '16 at 12:16\n\nMainly because its easier. The FFT is a specific algorithm to calculate the DFT. However, it only works if you calculate ALL frequencies (regardless if you want them or not). It takes in N complex values and it spits out N complex values. So the FFT can be used to evaluate the your first equation but not your second.\n\nIn your example, that's a trivial difference; but for larger FFT sizes this makes a substantial difference. It's a lot faster to calculate the values at negative frequencies with FFT and simply throw them away then calculating only the values at positive frequencies with a different algorithm.\n\nThis being said, there are ways for making use of the redundancy:\n\n1. You can pack two real signals into a single complex, do a single complex FFT and split this into odd and even parts to get the DFTs of both input signals\n2. There is a way to do a N point real DFT using N/2 point complex FFT. It just needs one extra stage of \"unwinding\" the result.\n• Great answer and thanks for the bonus tips, very cool! – Alan Wolfe Aug 7 '16 at 16:53" ]
[ null ]
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https://www.numberempire.com/2750
[ "Home | Menu | Get Involved | Contact webmaster", null, "", null, "", null, "", null, "", null, "# Number 2750\n\ntwo thousand seven hundred fifty\n\n### Properties of the number 2750\n\n Factorization 2 * 5 * 5 * 5 * 11 Divisors 1, 2, 5, 10, 11, 22, 25, 50, 55, 110, 125, 250, 275, 550, 1375, 2750 Count of divisors 16 Sum of divisors 5616 Previous integer 2749 Next integer 2751 Is prime? NO Previous prime 2749 Next prime 2753 2750th prime 24877 Is a Fibonacci number? NO Is a Bell number? NO Is a Catalan number? NO Is a factorial? NO Is a regular number? NO Is a perfect number? NO Polygonal number (s < 11)? NO Binary 101010111110 Octal 5276 Duodecimal 1712 Hexadecimal abe Square 7562500 Square root 52.440442408508 Natural logarithm 7.9193561906606 Decimal logarithm 3.4393326938303 Sine -0.89410469271937 Cosine -0.44785801149158 Tangent 1.9964021403605\nNumber 2750 is pronounced two thousand seven hundred fifty. Number 2750 is a composite number. Factors of 2750 are 2 * 5 * 5 * 5 * 11. Number 2750 has 16 divisors: 1, 2, 5, 10, 11, 22, 25, 50, 55, 110, 125, 250, 275, 550, 1375, 2750. Sum of the divisors is 5616. Number 2750 is not a Fibonacci number. It is not a Bell number. Number 2750 is not a Catalan number. Number 2750 is not a regular number (Hamming number). It is a not factorial of any number. Number 2750 is an abundant number and therefore is not a perfect number. Binary numeral for number 2750 is 101010111110. Octal numeral is 5276. Duodecimal value is 1712. Hexadecimal representation is abe. Square of the number 2750 is 7562500. Square root of the number 2750 is 52.440442408508. Natural logarithm of 2750 is 7.9193561906606 Decimal logarithm of the number 2750 is 3.4393326938303 Sine of 2750 is -0.89410469271937. Cosine of the number 2750 is -0.44785801149158. Tangent of the number 2750 is 1.9964021403605\n\n### Number properties\n\nExamples: 3628800, 9876543211, 12586269025" ]
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https://www.chegg.com/homework-help/trigonometry-videos-with-chapter-test-prep-9th-edition-chapter-6.4-problem-37ayu-solution-9780321717054
[ "", null, "# Videos on DVD with Chapter Test Prep for Trigonometry (9th Edition) Edit edition Problem 37AYU from Chapter 6.4\n\nWe have solutions for your book!\nChapter: Problem:\nStep-by-step solution:\nChapter: Problem:\n• Step 1 of 5\n\nFrom the figure, we can see that the hyperbola has its transverse axis along the x-axis.\n\nThe vertices of the hyperbola are (1, 0) and (–1, 0) from the figure. So, it is clear that the hyperbola has its centre at the origin.\n\nThe general equation of the hyperbola with the centre at origin and transverse axis along the x-axis is", null, ", where b2 = c2a2.\n\n• Chapter , Problem is solved.\nCorresponding Textbook", null, "Videos on DVD with Chapter Test Prep for Trigonometry | 9th Edition\n9780321717054ISBN-13: 0321717058ISBN: Michael SullivanAuthors:\nThis is an alternate ISBN. View the primary ISBN for: Trigonometry 9th Edition Textbook Solutions" ]
[ null, "https://cs.cheggcdn.com/covers2/34750000/34758021_1467148413_Width200.jpg", null, "https://word-to-html-images.s3.amazonaws.com/9780131431201/413-9.4-35e-i1.png", null, "https://cs.cheggcdn.com/covers2/34750000/34758021_1467148413_Width200.jpg", null ]
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https://ch.mathworks.com/help/dsp/ref/dsp.allpassfilter.impz.html
[ "# impz\n\nImpulse response of discrete-time filter System object\n\n## Syntax\n\n``````[impResp,t] = impz(sysobj)``````\n``````[impResp,t] = impz(sysobj,n)``````\n``````[impResp,t] = impz(sysobj,n,fs)``````\n``````[impResp,t] = impz(sysobj,[],fs)``````\n``````[impResp,t] = impz(sysobj,'Arithmetic',arithType)``````\n``impz(sysobj)``\n\n## Description\n\nexample\n\n``````[impResp,t] = impz(sysobj)``` computes the impulse response of the filter System object™, `sysobj`, and returns the response in column vector `impResp`, and a vector of times (or sample intervals) in `t`, where `t = [0 1 2 ...k-1]'`. `k` is the number of filter coefficients.```\n``````[impResp,t] = impz(sysobj,n)``` computes the impulse response at `floor(n)` one-second intervals. The time vector `t` equals `(0:floor(n)-1)'`.```\n``````[impResp,t] = impz(sysobj,n,fs)``` computes the impulse response at `floor(n)` 1/`fs`-second intervals. The time vector `t` equals `(0:floor(n)-1)'/fs`.```\n``````[impResp,t] = impz(sysobj,[],fs)``` computes the impulse response at `k` 1/`fs`-second intervals. `k` is the number of filter coefficients. The time vector `t` equals `(0:k-1)'/fs`.```\n``````[impResp,t] = impz(sysobj,'Arithmetic',arithType)``` computes the impulse response based on the arithmetic specified in `arithType`, using either of the previous syntaxes.```\n````impz(sysobj)` uses `fvtool` to plot the impulse response of the filter System object `sysobj`.You can use `impz` for both real and complex filters. When you omit the output arguments, `impz` plots only the real part of the impulse response.For more input options, refer to `impz` in Signal Processing Toolbox™.```\n\n## Examples\n\ncollapse all\n\nCreate a discrete-time filter for a fourth-order, lowpass elliptic filter with a cutoff frequency of 0.4 times the Nyquist frequency. Use a second-order sections structure to resist quantization errors. Plot the first 50 samples of the impulse response, along with the reference impulse response.\n\n`d = fdesign.lowpass(.4,.5,1,80);`\n\nCreate a design object for the prototype filter. Use `ellip` to design a minimum order discrete-time biquad filter.\n\n`biquad = design(d,'ellip','Systemobject',true);`\n\nPlot the impulse response.\n\n```impz(biquad); axis([1 75 -0.2 0.35])```", null, "## Input Arguments\n\ncollapse all\n\nLength of the impulse response vector, specified as a positive integer.\n\nData Types: `single` | `double` | `int8` | `int16` | `int32` | `int64` | `uint8` | `uint16` | `uint32` | `uint64`\n\nSampling frequency used in computing the impulse response, specified as a positive scalar.\n\nData Types: `single` | `double` | `int8` | `int16` | `int32` | `int64` | `uint8` | `uint16` | `uint32` | `uint64`\n\nArithmetic used in the filter analysis, specified as `'double'`, `'single'`, or `'Fixed'`. When the arithmetic input is not specified and the filter System object is unlocked, the analysis tool assumes a double-precision filter. When the arithmetic input is not specified and the System object is locked, the function performs the analysis based on the data type of the locked input.\n\nThe `'Fixed'` value applies to filter System objects with fixed-point properties only.\n\nWhen the `'Arithmetic'` input argument is specified as `'Fixed'` and the filter object has the data type of the coefficients set to `'Same word length as input'`, the arithmetic analysis depends on whether the System object is unlocked or locked.\n\n• unlocked –– The analysis object function cannot determine the coefficients data type. The function assumes that the coefficients data type is signed, has a 16-bit word length, and is auto scaled. The function performs fixed-point analysis based on this assumption.\n\n• locked –– When the input data type is `'double'` or `'single'`, the analysis object function cannot determine the coefficients data type. The function assumes that the data type of the coefficients is signed, has a 16-bit word length, and is auto scaled. The function performs fixed-point analysis based on this assumption.\n\nTo check if the System object is locked or unlocked, use the `isLocked` function.\n\nWhen the arithmetic input is specified as `'Fixed'` and the filter object has the data type of the coefficients set to a custom numeric type, the object function performs fixed-point analysis based on the custom numeric data type.\n\n## Output Arguments\n\ncollapse all\n\nImpulse response, returned as an `n`-element vector. If `n` is not specified, the length of the impulse response vector equals the number of coefficients in the filter.\n\nData Types: `double`\n\nTime vector of length `n`, in seconds. `t` consists of `n` equally spaced points in the range `(0:floor(n)-1)'/fs`. If `n` is not specified, the function uses the number of coefficients of the filter.\n\nData Types: `double`\n\n### Topics\n\nIntroduced in R2011a\n\n## Support", null, "Get trial now" ]
[ null, "https://ch.mathworks.com/help/examples/dsp/win64/PlotTheImpulseResponseOfALowpassEllipticFilterExample_01.png", null, "https://ch.mathworks.com/images/responsive/supporting/apps/doc_center/bg-trial-arrow.png", null ]
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http://tsnien.idv.tw/Java1_WebBook/chap3/3-3%20%E7%AE%97%E8%A1%93%E9%81%8B%E7%AE%97%E5%AD%90.html
[ "Java 程式設計(一)  :第 三章 算術運算式", null, "3-3 算術運算子\n\n• 3-3-1 算術運算子彙集\n\n• 3-3-2 範例研討:計算股票平均價格\n\n• 3-3-3 自我挑戰:學期成績計算程式\n\n• 3-3-4 範例研討:超商找錢工具\n\n• 3-3-5 自我挑戰:超商收銀機系統\n\n3-2 基本算術運算子\n\n 運算符號 說明 +、- 正、負號 +、-、*、/、% 加、減、乘、除、模數(求餘數) ++、-- 遞增、遞減 ( …) 運算式集合\n\n x = x + y; 取出 x 與 y 兩變數內容,相『加』後再填入變數 x 內。 x = - x + y; 變數 x 內容取負數,與變數 y 相『加』後再填入變數 x 內。 x = x * y; 取出 x 與 y 兩變數內容,相『乘』後再填入變數 x 內。 z = x / y; 變數 x 除以 y,所得到的商存入變數 z 內。 z = x % y; 變數 x 除以 y,所得到的『餘數』存入變數 z 內;如 x = 5、y = 3,則 x % y 得到餘數為 2。 x++; 相當於 x = x+1; x-- 相當於 x = x – 1; x+= y; 相當於 x = x + y; x-= y; 相當於 x = x - y; x*= y; 相當於 x = x * y; x/= y; 相當於 x = x / y; x%= y; 相當於 x = x % y; (2 + 3) * (4 +5) 先處理括號內得到 5 * 9,最後結果為 45。\n\nA)程式功能:Ex3_2.java\n\n *** 計算股票平均價系統 *** 請輸入第一個交易日股價 =>40.5 (目前平均價 = 40.50)請輸入第二個交易日股價 =>48.6 (目前平均價 = 44.55)請輸入第三個交易日股價 =>42.7 (目前平均價 = 43.93)請輸入第四個交易日股價 =>49.2 (目前平均價 = 45.25)請輸入第五個交易日股價 =>50.2 五日平均價 = 46.24\n\nB)製作技巧研討:\n\nC)程式範例:\n\n 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 // Ex3_2.java import java.util.Scanner; public class Ex3_2 {        public static void main(String args[]) {         Scanner keyin = new Scanner(System.in);         float ave, sum=0, cost;         int number=0;         System.out.printf(\"*** 計算股票平均價系統 ***\\n\");         System.out.printf(\"請輸入第一個交易日股價 =>\");         cost = keyin.nextFloat();         sum = sum + cost;         number++;         ave = sum / number;           System.out.printf(\"(目前平均價 = %.2f)請輸入第二個交易日股價 =>\", ave);         cost = keyin.nextFloat();         sum = sum + cost;         number++;         ave = sum / number;                      System.out.printf(\"(目前平均價 = %.2f)請輸入第三個交易日股價 =>\", ave);        cost = keyin.nextFloat();        sum = sum + cost;        number++;        ave = sum / number;          System.out.printf(\"(目前平均價 = %.2f)請輸入第四個交易日股價 =>\", ave);        cost = keyin.nextFloat();        sum = sum + cost;        number++;        ave = sum / number;            System.out.printf(\"(目前平均價 = %.2f)請輸入第五個交易日股價 =>\", ave);         cost = keyin.nextFloat();         sum = sum + cost;         number++;         ave = sum / number;         System.out.printf(\"五日平均價 = %.2f\\n\", ave);    } }\n\nD)程式重點分析:\n\n• 吾人可以發現計算 5 日平均價還稍可接受,如計算 102030 個交易日,程式將變得非常長,而大多書寫同樣的敘述句。因此,本系統應該使用迴圈敘述句,可容易許多,本書第五章將會介紹到。\n\n• 7 :『float ave, sum=0F, cost;』。表示同時宣告 3 個浮點變數,並給 sum 初值 0,但浮點數的數值必須多加 Fsum=0F)。\n\n• 11 :『sum = sum + cost;』。將 cost 內容累積增加到 sum 變數內。\n\n• 12 :『number++;』。功能是 number 變數累增 1number = number + 1)。\n\n• 13 :『ave = sum / number;』。計算當天之前連續交易日的平均股價。\n\nA)程式功能:PM3_1.java\n\n *** 學期成績計算系統(各科學分數) *** 請輸入電腦概論成績(2 學分) =>80 請輸入程式設計成績(3 學分) =>87 請輸入離散數學成績(3 學分) =>72 請輸入國文成績(2 學分) =>65 請輸入英文成績(2 學分) =>74 學期總平均分數 = 76.25 四捨五入後成績 = 76\n\nB)製作技巧提示:\n\n 01 02 03 04 05 06 07 08 09 10 11 ……       System.out.printf(\"請輸入程式設計成績(3 學分) =>\");       value = keyin.nextInt();       sum = sum + value * 3; ….. …..       ave = sum/12.0F;       System.out.printf(\"學期總平均分數 = %.2f\\n\", ave); ave = ave + 0.5F;       System.out.printf(\"四捨五入後成績 = %d\\n\", (int)ave); ….\n\nA)程式功能:Ex3_3.java\n\n **** 超商找錢工具 **** 請輸入消費的金額 =>354 請輸入繳納金額(大於消費額 354) =>500 應找金額 = 146 各種零錢數量如下:          100 元零錢 = 1 張          50 元硬幣 = 0 個          10 元硬幣 = 4 個           5 元硬幣 = 1 個           1 元硬幣 = 1 個\n\nB)製作技巧研討:\n\nC)程式範例:\n\n 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 // Ex3_3.java import java.util.Scanner; public class Ex3_3 {         public static void main(String args[]){         Scanner keyin = new Scanner(System.in);         int total, recept, value, value1, handre, fifty, ten, five, one;                 System.out.printf(\"**** 超商找錢工具 ****\\n\");         System.out.print(\"請輸入消費的金額 =>\");         total = keyin.nextInt();         System.out.printf(\"請輸入繳納金額(大於消費額 %d) =>\", total);         recept = keyin.nextInt();           value = recept - total;         value1 = value;         handre = value / 100;         value = value - handre * 100;           fifty = value / 50;         value = value - fifty * 50;           ten = value / 10;         value = value - ten * 10;           five = value / 5;         value = value - five * 5;           one = value;         System.out.printf(\"應找金額 = %d 各種零錢數量如下:\\n\", value1);         System.out.printf(\"\\t 100 元零錢 = %d 張 \\n\", handre);         System.out.printf(\"\\t 50 元硬幣 = %d 個 \\n\", fifty);         System.out.printf(\"\\t 10 元硬幣 = %d 個 \\n\", ten);         System.out.printf(\"\\t  5 元硬幣 = %d 個 \\n\", five);         System.out.printf(\"\\t  1 元硬幣 = %d 個 \\n\", one);    } }\n\nD)程式重點分析:\n\n• 17 :『hundre = value / 100;』。計算 100 元零錢的數目;整數相除僅得到整數,剩下的餘數則被取捨掉。\n\n• 18 :『value = value – hundre*100;』。計算找 100 元零錢數目後,剩下的找錢數目多寡;其實是求餘數的功能,讀者可將其改為 value = value % 100 執行看看,結果是否相同。\n\nA)程式功能:PM3_2.java\n\n *****  超商收銀機系統  ***** 衛生紙(每包 32 元) 購買數量 =>3 口香糖(每包 12 元) 購買數量 =>5 可樂(每瓶 18 元) 購買數量 =>2 熱狗(每支 17 元) 購買數量 =>3 請輸入繳納金額(大於購買總金額 = 243) =>500 應找金額 = 257 各種零錢數量如下:          100 元零錢 = 2 張          50 元硬幣 = 1 個          10 元硬幣 = 0 個           5 元硬幣 = 1 個           1 元硬幣 = 2 個\n\nB)製作技巧提示:\n\n 01 02 03 04 05 06 07 08 09 10 11 12 13 14 …… ….         System.out.printf(\"*****  超商收銀機系統  *****\\n\");         System.out.printf(\"衛生紙(每包 32 元) 購買數量 =>\");         item = in.nextInt();         total = total + item * 32;         ……                 System.out.printf(\"請輸入繳納金額(大於購買總金額 = %d) =>\", total);         item = in.nextInt();         value = item - total;         value1 = value;        …..\n\nJava 程式設計(一) 含程式邏輯" ]
[ null, "http://tsnien.idv.tw/images/main_1.png", null ]
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https://stacks.math.columbia.edu/tag/07D8
[ "Proof. Let $M_ i$, $i \\in I$ be a family of objects of $\\mathcal{A}$ indexed by a set $I$. The functor $F = \\prod _{i \\in I} h_{M_ i}$ commutes with colimits. Hence Lemma 19.13.1 applies. $\\square$\n\nIn your comment you can use Markdown and LaTeX style mathematics (enclose it like $\\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar)." ]
[ null ]
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http://incharlottesville.com/math-worksheet-kindergarten-pdf/
[ "# 26 Lovely Math Worksheet Kindergarten Pdf\n\nkids shapes worksheets grade shape identify simple 3d for 3 2nd multiplication worksheets pdf collection math worksheets two coloring worksheets for kindergarten pdf heathermarxgallery best math worksheet generator kidz activities science worksheets for first grade pdf kidz activities free printable kindergarten worksheets worksheetfun christmas math shopping worksheet new christmas math shopping best ideas 6th grade math diagnostic test pdf for kindergarten best solutions 6th grade staarath practice worksheets pdf kindergarten beginning sounds worksheets collection preschool", null, "Number Bonds To 10 Worksheet Pdf Best Printable Preschool Math from math worksheet kindergarten pdf , source:movielov.co", null, "Kindergarten Math Printable Worksheet New First Grade Hibernation from math worksheet kindergarten pdf , source:lezeninst.com", null, "Science Worksheets For First Grade Pdf Kidz Activities from math worksheet kindergarten pdf , source:reedaudio.com", null, "Best Ideas Learning Colors Worksheets Pdf Math Free Printable for from math worksheet kindergarten pdf , source:elmifermetures.com", null, "Kindergarten Addition Worksheets Pdf Best Halloween Count and from math worksheet kindergarten pdf , source:lesdereglees.com\n\nbest ideas learning colors worksheets pdf math free printable for number bonds to 10 worksheet pdf best printable preschool math 52 math is fun worksheets related keywords suggestions for maths kindergarten addition worksheets pdf best halloween count and best 6th grade mon core math worksheets pdf image collection maths worksheets for kindergarten printable new preschool christmas math shopping worksheet new shopping math worksheets pdf algebra practice worksheet printable worksheetsrossword puzzle reading and writing worksheets for kindergarten letter n practice kindergarten math printable worksheet new first grade hibernation\n\n6th grade math worksheets pdf multiplications multiplicationes kindergarten questions yelomphone pany kids 9th grade math worksheets th grade math worksheets lovely rd numbers – count and match free printable worksheets – worksheetfun kids math word problems kindergarten word problems kindergarten vowel sounds worksheets for kindergarten pdf go mom math worksheets beautiful math worksheets for kg2 addition prime number factors worksheet fresh factor rainbow worksheet pdf thanksgiving math worksheet 3rd grade fresh algebra coloring time game ks1 printable valid math worksheets maths money ks1\n\nSHARE ON\nTags:" ]
[ null, "http://incharlottesville.com/wp-content/uploads/2018/09/math-worksheet-kindergarten-pdf-unique-number-bonds-to-10-worksheet-pdf-best-printable-preschool-math-of-math-worksheet-kindergarten-pdf.jpg", null, "http://incharlottesville.com/wp-content/uploads/2018/09/math-worksheet-kindergarten-pdf-unique-kindergarten-math-printable-worksheet-new-first-grade-hibernation-of-math-worksheet-kindergarten-pdf.jpg", null, "http://incharlottesville.com/wp-content/uploads/2018/09/math-worksheet-kindergarten-pdf-best-of-science-worksheets-for-first-grade-pdf-kidz-activities-of-math-worksheet-kindergarten-pdf.jpg", null, "http://incharlottesville.com/wp-content/uploads/2018/09/math-worksheet-kindergarten-pdf-awesome-best-ideas-learning-colors-worksheets-pdf-math-free-printable-for-of-math-worksheet-kindergarten-pdf.png", null, "http://incharlottesville.com/wp-content/uploads/2018/09/math-worksheet-kindergarten-pdf-best-of-kindergarten-addition-worksheets-pdf-best-halloween-count-and-of-math-worksheet-kindergarten-pdf.png", null ]
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https://physics.stackexchange.com/questions/337199/how-to-derive-the-phase-difference-of-a-standing-wave
[ "# How to derive the phase difference of a standing wave?\n\nWe know a standing wave is defined by $D(x,t)=2a \\sin kx\\cos wt$. Intuitively, all particles within the same \"loop\" of a standing wave are vibrating in phase; all particles within 2 adjacent \"loops\" are vibrating in opposite phase. However, is there a mathematical proof of this?\n\nBelow is my attempt:\n\nFor a progressive wave $D(x,t)=A \\sin (kx-wt+\\Phi_0)$, the phase is $kx-wt+\\phi_0$, which makes the phase difference $\\Delta\\Phi = (kx_2-wt+\\Phi_0) - (kx_1-wt+\\Phi_0) = k\\Delta x$. Then if $\\Delta\\Phi = 2\\pi$, the two particles are vibrating in phase; if $\\Delta\\Phi = \\pi$, two particles are vibrating out of phase.\n\nBut using the same logic for standing waves, it seems the phase for them would be $wt$ thus phase difference $\\Delta\\Phi = wt - wt = 0$. This makes sense for particles in the same loop, but does not take into account particles in adjacent loops.\n\nThe phase difference you are trying to calculate is the phase difference between different points in space $x$ at the same time $t$. In other words you are choosing some constant time $t$ then calculating how the phase $\\Phi$ varies with $x$.\n\nIn your example of the travelling wave:\n\n$$D(x,t)=A \\sin (kx-\\omega t+\\Phi_0)$$\n\nyour method works because you take two different values of $x_1$ and $x_2$ at the same time $t$ so when you calculate:\n\n$$\\Delta\\Phi = (kx_2-\\omega t+\\Phi_0) - (kx_1-\\omega t+\\Phi_0)$$\n\nthe $\\omega t$ terms are constant and cancel out.\n\nThis works in exactly the same way for the standing wave:\n\n$$D(x,t)=2a \\sin kx\\cos \\omega t$$\n\nIf we take constant $t$ then $\\cos \\omega t$ is constant and we can write our snapshot in time as:\n\n$$D(x) = A\\sin kx$$\n\nwhere $A$ is a constant given by $A = 2a\\cos\\omega t$. And just as for the travelling wave we get:\n\n$$\\Delta\\Phi = k(x_2 - x_1)$$\n\n• Thank you, this clears things up! May I ask another (a little bit irrelevant) question though, how can I rewrite the formulas for $\\Delta\\Phi=2\\pi$ (in phase) and $\\Delta\\Phi=\\pi$ (out of phase) so that they would work for $\\Delta\\Phi > 2\\pi$, say, $\\Delta\\Phi = 4\\pi$ which should be considered in phase? – TigerHix Jun 3 '17 at 10:48\n• @TigerHix: the phase is given by your equation $\\Delta\\Phi = k(x_2 - x_1)$ and it can have any value from (in principle) $-\\infty$ to $+\\infty$. Points are in phase if $\\Delta\\Phi$ is a multiple of $2\\pi$, and they are in antiphase if $\\Delta\\Phi$ is a multiple of $2\\pi$ plus $\\pi$. – John Rennie Jun 3 '17 at 10:57" ]
[ null ]
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https://slotsisiq.web.app/jodon41378ly/statistical-probability-of-poker-hands-292.html
[ "# Statistical probability of poker hands\n\nBy Editor\n\nInterestingly, most, if not all, standard poker games like Texas Holdem make use of a single 52-card deck so it would be possible to compute for your exact probability of getting each hand. But we won’t bore you with the math behind it. You can learn all about that here.Instead, we’ll give you a quick summary of everything you need to know about poker hand odds.\n\n#1 Ranked Poker Odds Calculator from CardsChat - Easy & FREE tool for calculating odds for Hold'em, Omaha & more. Also on Google Play & App Store. Combinations and Permutations This lesson defines combinations and permutations. Lists formulas to compute each measure. Sample problems with step-by-step solutions show how to use formulas. Poker Variance Calculator - Primedope\n\n## Poker Math & Probabilities (Texas Hold'em)\n\nPoker hands ranking and probability : educationalgifs - Reddit 18 Feb 2018 ... Of the hands you DO play, some of the \"best\" starting hands are pocket pairs, which are statistically less likely to end in straights or flushes, ... 7 Card Poker Probability/Statistics - KnowYourLuck.com 7 Card Poker Probability/Statistics. Here is the probability of getting the various hands in poker from a 7 card hand. In this variant of the game (used in the ...\n\n### 7 Card PokerProbability/Statistics | Hand Name\n\nProbability Tutorial Discrete probability distributions. How to work with binomial, hypergeometric, multinomial, negative binomial, and Poisson distributions. Luck, Odds And The Game Of Poker Many poker players, especially beginners, believe that luck has a lot to do with poker winnings Poker strategy - Wikipedia A poker hand is usually a configuration of five cards depending on the variant, either held entirely by a player or drawn partly from a number of shared, community cards." ]
[ null ]
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http://git.joshuawise.com/fpgaboy.git/blob/207251ed32471a6c5fba766d7f73e92a8d86b7cf:/Sound2.v
[ "6 module Sound2(\n7         input core_clk,\n8         input wr,\n9         input rd,\n11         inout [7:0] data,\n12         input cntclk,\n13         input lenclk,\n14         input en,\n15         output [3:0] snd_data\n16         );\n18         /* can be optimized as register file */\n19         reg [7:0] nr21 = 0, nr22 = 0, nr23 = 0, nr24 = 0;\n20         reg [10:0] counter = 0;\n21         reg [4:0] lencnt = 0;\n22         reg [3:0] delta = 4'b1111;\n23         reg [2:0] dutycnt;\n24         reg [3:0] snd_out = 0;\n26         assign snd_data = en ? snd_out : 0;\n28         reg rdlatch;\n31         assign data = rdlatch ?\n36                       : 8'bzzzzzzzz;\n38         always @ (posedge core_clk) begin\n39                 rdlatch <= rd;\n41                 if(en && wr) begin\n47                         endcase\n48                 end\n49                 else if(!en) begin\n50                         nr21 <= 8'h3F;\n51                         nr22 <= 8'h00;\n52                         nr23 <= 8'hFF;\n53                         nr24 <= 8'hBF;\n54                 end\n55         end\n57         always @ (posedge cntclk) begin\n58                 if(counter)\n59                         counter <= counter - 1;\n60                 else begin\n61                         counter <= ~{nr24[2:0],nr23} + 1;  /* possible A */\n62                         dutycnt <= dutycnt + 1;\n63                 end\n65                 case (nr21[7:6])\n66                 2'b00: snd_out <= dutycnt ? 0 : delta;   /* probable A */\n67                 2'b01: snd_out <= (dutycnt[2:1] == 2'b0) ? delta : 0;\n68                 2'b10: snd_out <= dutycnt ? delta : 0;\n69                 2'b11: snd_out <= (dutycnt[2:1] == 2'b0) ? 0 : delta;\n70                 endcase\n71         end\n73         always @ (posedge lenclk) begin\n74                 if(lencnt)\n75                         lencnt <= lencnt - 1;            /* possible A */\n76                 else\n77                         lencnt <= ~nr21[4:0] + 1;\n78         end\n80 endmodule" ]
[ null ]
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https://www.idh-design.co.uk/2013/03/25/what-is-bearing-pressure/
[ "", null, "Bearing pressure is simply a load (force) over a certain area. For instance, a 100 kg mass exerts a force of 1 kN. When this load is evenly applied over 1 square metre then we can say the pressure being applied by this force is:\n\n= 1kN / 1m2 = 1 kN/m2           If this load were increased to say 5000 kg (5T) then the pressure applied = 50 / 1 = 50 kN/m2 If the 50 kN were applied over an area 2m x 2m (4 sq. m) then the bearing pressure becomes: = 50 / 4 = 12.5 kN/m2 The important point is whether the load can actually be evenly spread which is a function of the material doing the spreading and whether the ground can take the pressure being applied to it.  Just because we put sole boards and spreader beams on the ground doesn’t mean they are spreading the load evenly over their entire area. To demonstrate consider a bendy piece of thin plywood and a stiff steel road plate placed on a bed of springs with the same load applied centrally to each.", null, "In this example we can see the pressure under the thin bendy plywood is not distributed evenly so you get a variable pressure with a peak in the middle and almost nothing at the edges – by contrast the stiff steel is spreading the load evenly and distributes the load in to an even bearing pressure. So what is happening with a scaffold standard? Typically scaffold standards are loaded to 20-30 kN and if we consider them being on a base plate only we could calculate: P = 20 / 0.15 x 0.15 = 888 kN/m2  – Clearly a very large bearing pressure (albeit over a small area). So what happens when we introduce a 450mm long scaffold board Sole Plate?", null, "Timber has different load spreading properties in different directions – with or against the grain.  So we have already calculated the bearing pressure between the base plate and top of sole plate but what is the pressure, after load spreading, between the soleplate and the ground? To simply this we typically adopt load spread proportions of 2:1 with the grain and 1:1 perpendicular to the grain.  So first we calculate the effective area spreading load at the base of the soleplate after it has spread by these proportions:", null, "• Width = 38mm + 150mm + 38mm = 226 mm (Limited to 225mm board width)\n• Length = 38 + 38 + 150 + 38 +38 = 302 mm\n• Bearing Area = 0.225 x 0.302 = 0.068 m2\n• Bearing pressure under sole plate = 20 / 0.068 = 294 kN/m2\n\nAs you can see, by the application of a single sole board we have achieved a large reduction in the bearing pressure on the ground. So what happens when we use double sole boards?", null, "This time the load spread area =\n\n• Width = Same as before as it cannot spread wider = 225 mm\n• Length = 38 + 38 + 38 + 38 +150 + 38 + 38 + 38 + 38 = 454 mm\n• Bearing area = 0.225 x 0.454 = 0.102 m2\n• Bearing pressure under double boards = 20 / 0.102 = 196 kN/m2\n\nThis is an improvement but because we are only increasing the length and not width the benefits are limited. So what happens if we adopt a cruciform board sole plate?", null, "Now we get load spread in both directions. From the first board:\n\n• Width = 225 mm\n• Length = 302 mm\n• Now add on the distribution through the second layer of boards in both directions:\n• Width = 38 + 38 + 225 + 38 + 38 = 377 mm\n• Length = 38 + 302 + 38 = 378 mm\n• Bearing area under second layer of boards = 0.377 x 0.378 = 0.143 m2\n• Bearing pressure = 20 / 0.143 = 139.9 kN/m2\n\nNow it’s really starting to drop off.  So what are the implications of this?  Well, if the ground is soft (grassed or tarmac) you will need to spread the load and reduce the pressures until they reach an allowable limit.  Where we are basing out on concrete for instance, you don’t need to spread the load. The problem now becomes how do you determine what pressure the ground can permit?  I.E. How do you determine the Permissible or Allowable bearing pressure.  This is normally determined by site investigation where physical tests are undertaken to determine how capable the ground is of supporting these loads.  Clearly there are costs involved in doing this, but when it is known the ground is poor or soft and an overly conservative allowable pressure is adopted, the foundation improvement or weight spreading solution can be very expensive. Below are typical values for allowable bearing pressures for different materials:\n\nType of bearing material: Allowable Bearing Pressure kN/m2\nStiff Clays 150-300\nMedium dense Sands & Gravels 100-300\nTarmac 100-150\nFirm Clays 70-150\nLoose Sands and Gravels 50-100\nSoft Clays 50-100" ]
[ null, "https://www.facebook.com/tr", null, "https://www.idh-design.co.uk/wp-content/uploads/2018/08/Bearing-pressure-fig-1.jpg", null, "https://www.idh-design.co.uk/wp-content/uploads/2018/08/Bearing-pressure-fig-2.jpg", null, "https://www.idh-design.co.uk/wp-content/uploads/2018/08/Bearing-pressure-fig-3.jpg", null, "https://www.idh-design.co.uk/wp-content/uploads/2018/08/Bearing-pressure-fig-4.jpg", null, "https://www.idh-design.co.uk/wp-content/uploads/2018/08/Bearing-pressure-fig-5.jpg", null ]
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https://www.jiskha.com/questions/18019/help-determine-the-slope-of-the-line-the-line-whose-equation-is-3x-2y-8-solve-for-y-by
[ "# math and slopes\n\nHelp.\n\nDetermine the slope of the line.\n\nThe line whose equation is 3x-2y=8\n\nSolve for y by putting y on one side and everything else on the other.\n-2y=8-3x. Multiply by -1.\n2y = 3x -8 and divide by 2\ny=(3/2)x -8/2 or\ny= (3/2)x -4\nThe form for a straight line is\ny = mx + b\nwhere m is the slope and b is the y intercept.\n\n1. 👍 0\n2. 👎 0\n3. 👁 97\n\n## Similar Questions\n\n1. ### Algebra\n\nIs it possible to find the slope of a line if the equation has two ys? Find slope of the line whose equation is 8y = 7 - 2y Solve for y and you'll have a horizontal line. Combine the y terms. 8y+2y=7 10y=7 y=7/10 The general\n\nasked by Samantha on November 3, 2006\n2. ### Geometry\n\nHow do I write an equation of a line perpendicular to the graph of -2y + x =-5 that passes through the point at (1,2) Given the line -2y + x =-5, solve for y so the equation is in this form: y = mx + b Once you know the slope, m,\n\nasked by Mishka on October 31, 2006\n3. ### Algebra\n\nGiven a line containing the points(1,4), (2,7) and (3,10) determine that slope-intercept form of the equation, provide one additional point on this line, and graph the funtion. Start by putting the first point into point-slope\n\nasked by delmore on January 7, 2007\n4. ### Math 10\n\n1 more question.... how would I make this equation perpendicular? The question is.... The line Y=3x-1 and a line perpendicular to it intersect at R(1,2). Determine the equation of the perpendicular line. Can someone please tell me\n\nasked by Karn on December 10, 2006\n5. ### math\n\ni have more than one question so if u no any of the answers please tell me 1.) write the point-slope form of the equation of the line with slope -2 passing through the point ( -5, -9). 2.) write the point-slope form of an equation\n\nasked by becca on May 17, 2010\n6. ### Math\n\nWrite an equation of the line satisfying the given condition. a) x intercept = 3 y intercept=-2 I am really not sure how to start here. b) vertical line containing (5,-1) Would it be y=5x-1? c) parallel to the line 3x-4y=5 and\n\nasked by Hannah on October 5, 2010\n7. ### math\n\ngraph the line with slope 1/2 passing through the point(-5,-2) find the slope of the line 5x+5y=3 write answer in simplest from consider the line 2x-4y=4 what is the slope of a line perpendicular to this line. what jis the slope\n\nasked by Anonymous on February 10, 2013\n\nWrite an equation of the line satisfying the given condition. a) x intercept = 3 y intercept=-2 I am really not sure how to start here. b) vertical line containing (5,-1) Would it be y=5x-1? c) parallel to the line 3x-4y=5 and\n\nasked by Hannah on October 5, 2010\n9. ### college algebra\n\nA line has the equation y = 5x + 4.(a) This line has slope of ___________? (b) Any line parallel to this line has slope _________? (c) Any line perpendicular to this line has slope _______?\n\nasked by Cathy on October 24, 2014\n10. ### My Daughter Stacy needs help in math plz help!\n\nYou have to find the x and y coordinates:(She doesn't get it and neither do I) 2x + y = 9 The question doesn't go with the equation. I suspect it is a plotting exercise Make a table x y Put in for x 0, 1, 4 Solve for each\n\nasked by Stacy's Mom :Literally from the Song! on November 6, 2006\n\nMore Similar Questions" ]
[ null ]
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http://blog.russelldmatt.com/2021/07/08/g6-inconsistent-systems-can-prove-anything.html
[ "### Or… if $$p$$ and $$\\neg p$$, then anything\n\nA rather counterintuitive observation is that many inconsistent systems are, in fact, complete! More precisely, if a formal system that includes propositional logic can derive a contradiction, then it can derive anything, and is therefore complete.\n\nWhy? Because it turns out that the following formula is a theorem of propositional logic:\n\n$p \\supset (\\sim p \\supset q)$\n\nExercise for the reader: derive the formula. [My almost solution]\n\nIn words, ‘if p, then if not p, then q’. In other words, if you can derive both $$p$$ and $$\\lnot p$$, then (via the rule of detachment twice) you can derive $$q$$. And remember, by the rule of substitution, any formula can be substituted for a sentential variable to derive another formula. So, if we can derive $$p$$ and $$\\lnot p$$, and we can substitute anything we want for $$q$$, then we can derive anything! To put it shortly, from a contradiction, anything can be derived.\n\nThis fact has an interesting consequence. It’s actually quite cute. We just showed that if a system (that includes propositional logic) can derive a contradiction, then it can derive any formula. So, if any formula cannot be derived, then it must not be able to derive a contraction! In other words, proving that a single formula is not derivable within a system is equivalent to proving that the system is consistent.\n\nHow might we prove that a formula cannot be derived? By employing some meta-mathematical reasoning about the system as a whole. Note, we are going to prove something about the system, which is very different from deriving something within the system. Here’s one general strategy:\n\n1. Find a property which is true of all the axioms.\n2. Demonstrate that the property is preserved via all the transformation rules. In other words, demonstrate that all formulas derived from the axioms, i.e. theorems, will also have this property.\n3. Find a formula which does not have this property. This formula must not be a theorem.\n\nAs a toy example, imagine all the axioms were made up of 25 or more symbols. And imagine every transformation rule only increases the number of symbols in the derived formula. If you can construct a formula with fewer than 25 symbols, then you know it’s not a theorem." ]
[ null ]
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https://polymathprojects.org/tag/mini-polymath4/
[ "# The polymath blog\n\n## July 12, 2012\n\n### Minipolymath4 project: IMO 2012 Q3\n\nFiled under: research — Terence Tao @ 10:00 pm\nTags:\n\nThis post marks the official opening of the mini-polymath4 project to solve a problem from the 2012 IMO.  This time, I have selected Q3, which has an interesting game-theoretic flavour to it.\n\nProblem 3.   The liar’s guessing game is a game played between two players", null, "$A$ and", null, "$B$.  The rules of the game depend on two positive integers", null, "$k$ and", null, "$n$ which are known to both players.\n\nAt the start of the game,", null, "$A$ chooses two integers", null, "$x$ and", null, "$N$ with", null, "$1 \\leq x \\leq N$.  Player", null, "$A$ keeps", null, "$x$ secret, and truthfully tells", null, "$N$ to player", null, "$B$.  Player", null, "$B$ now tries to obtain information about", null, "$x$ by asking player A questions as follows.  Each question consists of", null, "$B$ specifying an arbitrary set", null, "$S$ of positive integers (possibly one specified in a previous question), and asking", null, "$A$ whether", null, "$x$ belongs to", null, "$S$.  Player", null, "$B$ may ask as many such questions as he wishes.  After each question, player", null, "$A$ must immediately answer it with yes or no, but is allowed to lie as many times as she wishes; the only restriction is that, among any", null, "$k+1$ consecutive answers, at least one answer must be truthful.\n\nAfter", null, "$B$ has asked as many questions as he wants, he must specify a set", null, "$X$ of at most", null, "$n$ positive integers.  If", null, "$x$ belongs to", null, "$X$, then", null, "$B$ wins; otherwise, he loses.  Prove that:\n\n1. If", null, "$n \\geq 2^k$, then", null, "$B$ can guarantee a win.\n2. For all sufficiently large", null, "$k$, there exists an integer", null, "$n \\geq 1.99^k$ such that", null, "$B$ cannot guarantee a win.\nThe comments to this post shall serve as the research thread for the project, in which participants are encouraged to post their thoughts and comments on the problem, even if (or especially if) they are only partially conclusive.  Participants are also encouraged to visit the discussion thread for this project, and also to visit and work on the wiki page to organise the progress made so far.\nThis project will follow the general polymath rules.  In particular:\n1. All are welcome. Everyone (regardless of mathematical level) is welcome to participate.  Even very simple or “obvious” comments, or comments that help clarify a previous observation, can be valuable.\n2. No spoilers! It is inevitable that solutions to this problem will become available on the internet very shortly.  If you are intending to participate in this project, I ask that you refrain from looking up these solutions, and that those of you have already seen a solution to the problem refrain from giving out spoilers, until at least one solution has already been obtained organically from the project.\n3. Not a race. This is not intended to be a race between individuals; the purpose of the polymath experiment is to solve problems collaboratively rather than individually, by proceeding via a multitude of small observations and steps shared between all participants.   If you find yourself tempted to work out the entire problem by yourself in isolation, I would request that you refrain from revealing any solutions you obtain in this manner until after the main project has reached at least one solution on its own.\n4. Update the wiki. Once the number of comments here becomes too large to easily digest at once, participants are encouraged to work on the wiki page to summarise the progress made so far, to help others get up to speed on the status of the project.\n5. Metacomments go in the discussion thread. Any non-research discussions regarding the project (e.g. organisational suggestions, or commentary on the current progress) should be made at the discussion thread.\n6. Be polite and constructive, and make your comments as easy to understand as possible. Bear in mind that the mathematical level and background of participants may vary widely.\n\nHave fun!\n\nBlog at WordPress.com." ]
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http://www.learnmathsonline.org/cbse-class-5-maths/class-5-maths-shapes-and-angles-important-questions/
[ "# Class 5 Maths Shapes And Angles | Important Questions\n\nIMPORTANT QUESTIONS FOR CBSE EXAMINATION | CLASS 5 MATHEMATICS\nSHAPES AND ANGLES – Lesson 2\n\nFill in the blanks (1 mark):\n\n1. Angle made by two right angles is a———————-\n\n2. The measure of a straight angle is ——————\n\n3. Half of a right angle is ——————-\n\n4. One-third of a right angle is —————-\n\n5. Sum of the angles of any triangle is —————-\n\n6. When two lines meet each other ———– angles are formed.\n\n7. Angle indicated in the clock time 3’O Clock is —————–\n\n8. An angle whose measure is less than 90 degree is known as —————–\n\n9. An angle whose measure is greater than 90 degree is known as ————–\n\n10. An angle whose measure is 90 degree is known as —————–\n\n11. The sum of 2 angles in a triangle is 100 degree. What is the measure of the third angle?\n\n12. Write the name of the angle formed when the clock time is\na) 7.15pm\nb) 9’ O Clock\nc) 11’ O Clock\n\n13. Classify the following angles as acute, obtuse and right angle.\n\na) 120\nb) 50\nc) 90\nd)30\ne) 150\nf) 2 x 45\n\n14. What is the measure of the angle formed by the hands of the clock at\n\na) 1’ O Clock\nb) 2’ O Clock\nc) 3’ O Clock\n\n15. Write five English alphabets where you see a right angle\n\n1. Straight angle\n2. 180 degree\n3. 45 degree\n4. 30 degree\n5. 180 degree\n6. Four\n7. Right angle\n8. Acute angle\n9. Obtuse angle\n10. Right angle\n11. Sum of 2 angles = 100 degree\nMeasure of third angle = 180 – 100 = 80 degree\n12. a) Obtuse angle\nb) Right angle\nc) Acute angle\n13. a) 120 – Obtuse angle\nb) 50 – Acute angle\nc) 90 – Right angle\nd) 30 – Acute angle\ne) 150 – Obtuse angle\nf) 2 x 45 – Right angle\n14. a) 30 degree\nb) 60 degree\nc) 90 degree\n15. E, F, H, L, T\n\n### 7 Responses\n\n1.", null, "ปั้มไลค์ says:\n\nLike!! Great article post.Really thank you! Really Cool.\n\n2.", null, "Renaissance Sutradhar says:\n\nThis is a very helpful page\n\n3.", null, "Shreyas Ghosh says:\n\nMaths is very hard\n\n4.", null, "Ashwin Bhwowmik says:\n\nit is a good sight but what is the name of this site.\nJust amazing\nexcellent maths Questions\n\n5.", null, "Taanishqa Banerjee says:\n\nMaths are very tricky but these are good for preparing ourselves in all exam\n\n6.", null, "Aadigamer says:\n\nGood but you want to tell the answer.\n\n7.", null, "Aadigamer says:\n\nGood but you want to tell the answer" ]
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https://waseda.elsevierpure.com/en/publications/on-the-stationary-navier-stokes-flows-around-a-rotating-body
[ "# On the stationary Navier-Stokes flows around a rotating body\n\nHorst Heck, Hyunseok Kim*, Hideo Kozono\n\n*Corresponding author for this work\n\nResearch output: Contribution to journalArticlepeer-review\n\n11 Citations (Scopus)\n\n## Abstract\n\nConsider the stationary motion of an incompressible Navier-Stokes fluid around a rotating body K = ℝ 3\\Ω which is also moving in the direction of the axis of rotation. We assume that the translational and angular velocities U,ω are constant and the external force is given by f = div F. Then the motion is described by a variant of the stationary Navier-Stokes equations on the exterior domain Ω for the unknown velocity u and pressure p, with U, ω, F being the data. We first prove the existence of at least one solution (u, p) satisfying ∇u, p ∈ L 3/2,∞(Ω) and u ∈ L 3,∞(Ω) under the smallness condition on {pipe}U{pipe} + {pipe}ω{pipe} + {double pipe}F{double pipe} L3/2,∞(Ω). Then the uniqueness is shown for solutions (u, p) satisfying ∇u, p ∈ L 3/2,∞(Ω) ∩ L q,r(Ω) and u ∈ L 3,∞(Ω) ∩ L q*,r(Ω) provided that 3/2 < q < 3 and F ∈ L 3/2,∞(Ω) ∩ L q,r(Ω). Here L q,r(Ω) denotes the well-known Lorentz space and q* = 3q/(3 - q) is the Sobolev exponent to q.\n\nOriginal language English 315-345 31 Manuscripta Mathematica 138 3-4 https://doi.org/10.1007/s00229-011-0494-1 Published - 2012 Jul Yes\n\n## ASJC Scopus subject areas\n\n• Mathematics(all)\n\n## Fingerprint\n\nDive into the research topics of 'On the stationary Navier-Stokes flows around a rotating body'. Together they form a unique fingerprint." ]
[ null ]
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https://www.omicsonline.org/open-access/concept-of-hydrogen-redox-electric-power-generator-ier-1000129.php?aid=70782
[ "\n\nGET THE APP\n\nConcept of Hydrogen Redox Electric Power Generator\nInnovative Energy & Research\n• Review Article\n• Innov Ener Res , Vol 5(1)\n\n# Concept of Hydrogen Redox Electric Power Generator\n\nKatsutoshi Ono*\nDepartment of Energy Science and Technology, Kyoto University, Tokyo, Japan\n*Corresponding Author: Katsutoshi Ono, Department of Energy Science and Technology, Kyoto University, 3-16-8, Jiyugaoka, Meguro-ku, Tokyo, Japan, Tel: +81 75-753-7531, Email: [email protected]\n\nReceived Date: Jan 13, 2016 / Accepted Date: Jan 27, 2016 / Published Date: Feb 03, 2016\n\n### Abstract\n\nThe hydrogen redox power generation system is a combined energy cycle consisting of the H2O=H2+1/2O2 reduction reaction performed by the bipolar alkaline water electrolyzer with electrostatic-induction potentialsuperposed electrodes and the H2+1/2O2=H2O oxidation reaction performed by the alkaline fuel cell. This type of electrolyzer functions by a mechanism in which the power used is theoretically 17% of the total electrical energy required, while the remaining 83% can be provided by electrostatic energy free of power. Part of the power delivered by the fuel cell is returned to the electrolytic cell and the remainder represents the net power output. This power generation concept assumes the use of a low temperature alkaline fuel cell and a low temperature alkaline water electrolytic cell, both commercially available. Calculations using the commercial data for the operational conditions show that the cycle power efficiencies ranging from 60 to 80% may be attainable, regardless of the total electrode surface area.\n\nKeywords: Electric power generation; Hydrogen; Energy cycle; Fuel cell; Water electrolytic cell\n\n#### Introduction\n\nThis review was intended to address the fundamental aspects of new high power generation system that functions with zero energy input, zero matter input and zero emissions, without any violation of the laws of thermodynamics. The method proposed and examined in the previous papers [1-4] has been termed hydrogen redox electric power generator (HRPG) and takes advantage of the intrinsic self-sustaining electrostatic energy that takes its origin from the Coulomb’s force.\n\n#### Theoretical Backgrounds\n\nThe hydrogen redox power generation system is a combined energy cycle consisting of the H2O→H2+1/2O2 reduction reaction performed by the bipolar alkaline water electrolyzer with electrostaticinduction potential-superposed electrolytic method (ESI-PSE) and the H2+1/2O2→H2O oxidation reaction performed by the alkaline fuel cell (FC). This type of electrolyzer functions by a mechanism in which the power used is theoretically 17% of the total electrical energy required, while the remaining 83% can be provided by electrostatic energy free of power. This energy was termed internally self-sustaining electrostatic energy. Part of the power delivered by the fuel cell is returned to the electrolytic cell and the remainder represents the net power output. This power generation concept assumes the use of a low temperature alkaline fuel cell and a low temperature alkaline water electrolytic cell, both commercially available. Calculations using the commercial data for the operational conditions show that the cycle power efficiencies ranging from 60 to 80% may be attainable, regardless of the total electrode surface area. Figure 1 shows a tentative plan for the hydrogen redox electric power generator. The electrolyzer is assumed to be a low temperature bipolar alkaline water electrolyzer (BAWE) and the fuel cell is a low temperature alkaline fuel cell (AFC) with the same electrolyte (H2O-30% KOH) as that for the electrolyzer. Notice that the fuel cell is situated at a higher level than the electrolyzer to use the gravitational energy to return water from the fuel cell to the electrolyzer.\n\nFigure 1: Flow diagram for hydrogen redox electric power generation system. ESI-PSBE: Electrostatic induction- potential superposed bipolar electrolyzer. Pnet: Net power output; Pr: Power returned.\n\nThe electrostatic energy created and stored between the electrodes of an electrolytic cell is responsible for the highly positive power balance of this energy cycle. Nature of the internally self-sustaining electrostatic energy is then explained as below. The non-linear V-I characteristic curve for a practical water electrolytic cell shown in Figure 2a typically shows the amount of current measured at the point between the cell and the power supply as a function of cell voltage. In the voltage region over the theoretical decomposition voltage Ed°, the external currents in transition flow. In the consecutive region where practical electrolysis occurs, the current increases in an almost linear relationship with the applied voltage, as evidenced experimentally. To improve the energy efficiency of the electrolyzer using the ESI-PSE power supply mode, the V-I characteristics have been modified using a resistance as shown in Figure 2b. The V-I curve implies that the total cell voltage Ee, is the superposition of two separate voltages: (1) the practical decomposition voltage, Ed and (2) the additional applied voltage ΔE, which when superposed on Ed, yields the electrolytic current, Ie, and the ionic current, Iion, in the electrolyte solution. ΔE is then identified as the electromotive force (emf) for the electrolytic current, Ie. On the other hand, the ionic current in the electrolytic cell undergoing electrolysis flows by migration under the influence of the electrical potential gradient in the electrolyte solution . The electronic behavior of the cell for the electrolytic decomposition of a chemically stable compound such as H2O is analogous to that of the electrical double layer capacitor for energy storage . The total energy required for the cell undergoing electrolysis is the sum of the energies that correspond to the continuous charge and discharge processes:\n\nWel=Ed Iion +ΔE Ie. (1)\n\nFigure 2: V-I characteristics curve. Edo: Theoretical decomposition voltage; Ed: Real decomposition voltage; Ee: Electrolytic cell voltage; Ie: Electrolytic current; ΔE: Extra applied voltage over Ed.\n\nThe null-current voltage range from 0 to Ed ° contributes to a major portion of the total applied voltage, Ee. If alkaline water electrolysis is assumed, then at V=Ed ° the reversible cell enters the decomposition state represented by the half-reactions that occur at the negative and positive electrodes, which correspond to Equations (2) and (3) below, respectively.\n\n2H++ 2e → H2 (2)\n\n2OH-1/2→O2+ H2O + 2e (3)\n\nIf this process proceeds reversibly without the production of current in the circuit external to the cell, then the net gain in energy of H2O from the constant electrostatic field is given by the standard Gibbs free energy change of the reaction, H2O→H2+1/2O2, and related to Ed° as ΔG°=2FEd°. According to the thermodynamic data at 300 K, ΔG°=237 kJ/molH2, and the total energy change (enthalpy change) is ΔH°=286 kJ/molH2. Hence, theoretically 83% of the total energy required for the electrolytic reduction of H2O is the electrostatic energy for which the outside voltage source is not required to provide power, only voltage.\n\nTo take advantage of the internally self-sustaining electrostatic energy, the feature and essential part of the electrostatic-induction potential-superposed method are demonstrated as follows. The method termed ESI-PSE supplies potentials to the electrodes in a dual mode as shown in Figure 3a, i.e., the superposition of two voltages on the electrodes with two independent voltage sources; one is a biasvoltage source, which induces Ed at the cell electrodes and the other is a power supply, which provides power to the cell. Figure 3b is a closed circuit with two voltage sources connected in opposing directions. These sources can act in parallel independently, and supply the cell electrodes with individual potentials, which yields a superposition such that the resulting voltage between the cell electrodes equals the magnitude of the algebraic sum of the individual potential, i.e., Ed+ΔE. The performance of the cell can be explained by a series of steps. Firstly, when the PS2 output voltage VPS2=Ed, I=0 (null point) because of the usual condition of uniformity of the electrochemical potential throughout the electrolyte solution . Secondly, if VPS2 is increased from the null point, then an electrolytic current I=Ie flows due to the total source voltage (emf) of ΔE =VPS2?Ed. Hence, the two sources can be replaced by a single source (PS2) that delivers ΔE. The bias-voltagesource does not need to produce electrical current, but only a static voltage because of the null-current condition at Ed. The cell voltage Ee, is the superposition ofΔE on Ed, and the entire corresponding electrolytic process is the superposition of the charge-transfer process where ΔE is applied, (i.e., where power is used) to the electrostatic process where Ed is applied, (i.e., where power is not used). If the current 2F flows to produce 1 mole of H2 and half a mole of O2, and the current efficiency is almost unity, as in a commercial electrolyzer, then the total power requirement reduces to the generalized form for any given cell, regardless of the decomposition voltage Ed, temperature, pressure, composition of electrolyte solution and cell dimensions:\n\nP*=2FΔE. (4)\n\nFigure 3: Electrostatic-induction potential-superposed electrolytic system (ESI-PSE). (a) Diagram showing the principles of the system. PS: Power supply; FG: Field generator; Ie: Electrolytic current; AWE: Alkaline water electrolyte. (b) Equivalent circuit for the ESI-PSE system. VPS2: Output voltage of PS2; Rint: Resistance due to electron transfer reactions at the electrode/solution interface; Vinduced: Field induced voltage between the anode and cathode.\n\n#### Structure Of Electrolyzer\n\nA single ESI-PSE cell generates a limited amount of H2 and O2. In practice, many cells are assembled into a stack, with the principles being the same. For example, the bipolar electrolyzer currently in use in industry offers considerable advantages, in that it may be used directly with the present cell structure as shown in Figure 4. This is a type of electrostatic-induction electrolyzer, but it remains a barrier potential type electrolyzer. The electrolyzer is typically operated with a single power supply PS1, which is responsible for the total power, i.e., the number of cells times the power per cell, I(Ed+ΔE). To change the power supply mode into the ESI-PSE scheme, it is sufficient to introduce minor modifications as follows. (1) The outermost pair of electrodes is used as a field generator and should be connected to the power supply PS1. (2) The power supply responsible for providing power should be connected to the electrodes next to the field generator electrodes. There is no need for lead wires to take current from each of the inner cells. The performance of this alternative electrolyzer may be determined by considering two steps that specify the electrical behaviors of the cells. (1) When the output voltage of power supply PS1 is VPS1=7Ed, the total field-generator voltage is divided among 7 channels in equal amounts of Ed. Thus, there is no current in the circuit involving PS1, and a voltage of 5Ed is induced between the electrodes A-F as a floating potential. The null-current condition means that PS1 does not need to provide current, but merely induce a static voltage. (2) If the power supply PS2 is connected to the electrode A-F with its polarity opposing the induced voltage, then the resultant electrolytic circuit is analogous to a closed circuit with two voltage sources that are in series and of opposing polarity. Next, if VPS2 is increased from 0 to 5(Ed+ ΔE), an electrolytic current Ie flows due to the emf of 5ΔE as a result of the potential superposition of 5ΔE on 5Ed. For an internal cell, the extra applied voltage over Ed is ΔE and the current flowing through is Ie; therefore, the power requirement per cell in this type of electrolyzer must be IeΔE. Thus, power requirement equivalency with a single ESI-PSE cell is established.\n\nA resistor R is required between PS1 and field generator electrode, FG (+), such that the transitional current flowing through PS1 at Ed (Figure 2a) is lowered down to the level sufficiently low, as shown in Figure 2b. If the resistor R is not present, then PS1 causes extra power consumption. According to theoretical calculations, this power accounts for 40% of the power required when no transitional region is present and the electrolyzer runs at an extra applied voltage ΔE=0.2 V and a current density of J=1000 A/m2.\n\n#### Energy Efficiencies\n\nFrom the literature, the operating conditions for low temperature alkaline water electrolysis are: Temperature: 60-80°C; pressure: 1 atm; current density: 1000-3000 Am-2; cell voltage, Ee: 1.9-2.2 V; electrolyte: H2O-30% KOH. The factors that mainly affect the energy efficiency of the AFC are the utilization efficiency of fuels at the electrode reactions, the ΔG/ΔH ratio of the H2+1/2O2 →H2O oxidation reaction, and the voltage degradation factor, i.e., the voltage efficiency η= Ef / Ef° for the operating AFC, where Ef is the actual voltage of a cell undergoing oxidation and Ef° is the theoretical open-circuit voltage. The overall energy efficiency of a commercially available low temperature AFC is in the order of 60%. If the AFC is combined with the BAWE to form a closed single energy cycle, the energy efficiency of the developed energy cycle equals the energy efficiency of the AFC in the system for the reasons described below. The energy sources for the BAWE with ESIPSE electrodes in the system are the internally provided electrostatic energy plus the power returned from the AFC. The pure stoichiometric H2-O2 fuel completes the reaction, H2+1/2O2 → H2O, therefore, the utilization efficiency of the fuel may be deleted from the factors that affect the energy efficiency of the AFC. The oxidation bi-product, H2O, which contains the heat liberated by the exothermic reaction, is transferred to the BAWE and used for the endothermic reduction reaction. The thermodynamic factor, ΔG/ΔH, is thus also deleted, which leaves the voltage efficiency as the only factor, η= Ef /Ef°.\n\nIn this combined energy cycle, the main device is the fuel cell to produce electricity. The electrolyzer in this system is a backup reactor to synthesize fuel for the fuel cell. Part of the power generated by the fuel cell is returned to the electrolyzer, and the remainder represents the net power output. To represent the productive capacity, the cycle power efficiency is defined as the ratio ξ p of the net power extracted from the cycle, Pnet, to the power produced by the pure stoichiometric H2-O2 fuel, Pf:\n\nξp=Pnet / Pf (5)\n\nIf the voltage efficiency is taken into account, then for 1 mole H2,\n\nPf=2FηEf°. (6)\n\nMass balance in the cycle requires equality of the electrolytic and galvanic currents. If the current efficiency is assumed to be unity, then Equation (4) is subtracted from Equation (6) to yield the theoretical net power output of the cycle, Pnet . Hence, the cycle power efficiency is rearranged in the form:\n\nξp=Pnet / Pf=1-ΔE / ηEf°. (7)\n\nAssessments have been made on the achievable cycle power efficiencies for the hydrogen redox power generation system when a commercially available AFC is combined with an existing BAWE that is connected to the ESI-PSE mode. Because ΔE=EeEd, Equation (7) becomes:\n\nEeEf °(1-ξp) + Ed. (8)\n\nThe energy efficiencies for the total generation system require the V-I characteristics for the BAWE and the voltage efficiency of the AFC. The V-I characteristics were determined using data for the operating conditions of typical electrolyzers, with j=1000, 3000 and 4500 A/m2 at Ee=1.9, 2.2 and 2.4 V, respectively, and also with reference to the literature [8,9]. Commercially available AFCs operate with a voltage efficiency of η=0.8 on average . In view of Equation (8), points on the V-I curve are related to the corresponding cycle power efficiencies. Table 1 shows the achievable cycle power efficiencies and corresponding operational conditions required to BAWE and AFC. The energy efficiency for this energy cycle where the energy is created within the system is defined as follows. The purely stoichiometric H2-O2 fuel produced by the electrolyzer is the source of energy to convert into electricity so that the H2+1/2 O2→H2O oxidation reaction is performed in the fuel cell. The heat released is recycled for use in the endothermic reduction reaction at the electrolyzer, therefore, the standard free energy change for the oxidation reaction, |ΔG °|, is responsible for the electric power generation. On the other hand, the electrical energy delivered by the fuel cell is 2FηEd °. The energy efficiency is then given by the ratio of 2FηEd ° to |ΔG °|:\n\nξE=2FηEd °/ |ΔG °|=η. (9)\n\nThe power extracted from the generator is dependent on the rate of H2O-circulation inside the system. If the circulation rate is 1 mole H2O per second, then the power output is given by:\n\nP=|ΔG °|ξE (kW). (10)\n\n|ΔG °|= 232 kJ/mole H2O at 330 K and η=0.8 , therefore, if H2O circulates at a rate of 1 kg/s, then the net power output would be 1.03 × 104 kW.\n\nCycle power efficiency of hydrogen redox power generator BAWE AFC\nSingle cell voltage Current density Voltage efficiency\nξp (%) Vcell (V) J (Am-2) η\n60 2.2 3000 0.8\n70 2.1 2200 0.8\n80 2.0 1400 0.8\n\nTable 1: Operational conditions of BAWE and AFC for high cycle power efficiencies.\n\n#### Suggested Generator\n\nThe proposed approach is to demonstrate that this electric power generation system may be realized with an energy cycle that consists of a fuel cell and a water electrolyzer with ESI-PSE electrodes as an energy supply scheme. This power generation concept assumes to use a low temperature AFC and a low temperature BAWE, both of which are commercially available: They are set side by side in parallel. The result of calculations using the commercial data for the operational conditions showed that the cycle power efficiencies ranging from 60 to 80% could be attainable, regardless of the total electrode surface area. Self-exciting capability of the HRPG system may be appreciated with respect to its cycle power efficiency. For example, if the HRPG using a nominal 100 kW fuel cell is operated with a cycle power efficiency of 70%, then it may be implied that the net power extracted from this HRPG is 70 kW.\n\nAmong all types of fuel cells, low temperature (ca, 25 to 70°C) AFC’s operate well at room temperature, yielding the highest voltages at comparable current densities. An AFC can be built from carbon materials and plastics that are anti-corrosive and low-cost. Therefore, this type of fuel cells is able to achieve a long operating life, approximately two years. The AFC is installed in a closed system; therefore, contamination of the alkaline electrolyte (H2O-KOH) by CO2 can be avoided. On the other hand, the BAWE also has the same advantages with respect to lifetime, construction materials and stability of the electrolyte. The same alkaline electrolyte is used for both devices, so that mutual contamination can be avoided. The preliminary system design is based on consideration of the mutual consistency for a stable operation of the total system. To maintain the mass balance in the system, the electrolyzer current must be controlled such that the rate of hydrogen consumption in the fuel cell equals the rate of hydrogen emission from the electrolyzer, if the current efficiency of the electrolyzer is unity. This system does not require the compressed hydrogen gas tank which is an explosion hazard.\n\n#### Concluding Remarks\n\nToward achieving a breakthrough in the practical capabilities of high electric power generator that functions with both zero energy input, zero matter input and zero emissions, the basic concept of the hydrogen redox electric power generator (HRPG) have been developed. With this generation system, wide range of unique applications may be expected:\n\n(1) Replacement of the thermal power generation with HRPG to reduce carbon dioxide emissions from coal and oil-fired electric power generation.\n\n(2) The HRPG central-station power generation in the regions and countries where energy sources for power generation are unavailable.\n\n(3) Hydrogen generator:\n\nFigure 5 outlines the combined cycle of AWE electrolyzer with ESI-PSE and AFC to produce hydrogen. This system is of mutual consistency with the HRPG system shown in Figure 1, and is able to generate H2 without receiving electric power from outside of the cycle . Part of the hydrogen generated by the ESI-PSE electrolyzer is returned to the fuel cell, and the remainder represents the net hydrogen output. In addition to cycling the generated hydrogen, H2O, which contains heat from the exothermic reaction in the fuel cell, is transferred to the electrolyzer for the use to its endothermic reaction. This system may achieve a highly positive hydrogen balance such that the hydrogen output of this cycle exceeds, for example, 70% of the hydrogen delivered by the ESI-PSE electrolyzer.\n\nBecause of the simplicity, effectiveness, cleanliness and selfexciting, this new generator HRPG, may offer a potential route for its practical application to an electricity-production system of the future.\n\n#### References\n\nCitation: Ono K (2016) Concept of Hydrogen Redox Electric Power Generator. Innov Ener Res 5: 129.\n\nCopyright: ©2016 Ono K. This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.\n\nSelect your language of interest to view the total content in your interested language\n\n##### Open Access Journals\nViewmore\n###### Article Usage\n• Total views: 11085\n• [From(publication date): 6-2016 - Jan 27, 2021]\n• Breakdown by view type\n• HTML page views: 10824\n\nReview summary\n\n1. Isabella Mason\nPosted on Sep 29 2016 at 1:38 pm\nThe author reviewed the various aspects of hydrogen redox power generation system. The article provides clear information regarding the pros and cons of the method and the industrial viability of the method.\n\n## Review summary\n\n1. Isabella Mason\nPosted on Sep 29 2016 at 1:38 pm\nThe author reviewed the various aspects of hydrogen redox power generation system. The article provides clear information regarding the pros and cons of the method and the industrial viability of the method.", null, "", null, "Can't read the image? click here to refresh" ]
[ null, "https://www.omicsonline.org/captcha_code_file.php", null, "https://www.omicsonline.org/images/refresh.jpg", null ]
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https://researchmaniacs.com/Calculator/PercentOff/40/What-is-40-percent-off-531-dollars.html
[ "What is 40 percent off 531 dollars?", null, "If something costs \\$531 and is on sale for 40% off, then how much would it cost? Here we will show you how to calculate how much you save (discount) and how much you have to pay if something you want to buy is regularly \\$531, but is currently on sale for 40 percent off.\n\nThere are many ways of calculating your discount and final purchase price. One way is to multiply 531 dollars by 40 percent, and then divide the answer by one hundred, then deduct that result from the original price. See illustration below:\n\n \\$531 SALE40% OFF\n\nPurchase Price:\n\\$531\n\nDiscount:\n(531 x 40)/100 = \\$212.40\n\nFinal Price:\n531 - 212.40 = \\$318.60\n\nThus, a product that normally costs \\$531 with a 40 percent discount will cost you \\$318.60, and you saved \\$212.40.\n\nYou can also calculate how much you save by simply moving the period in 40.00 percent two spaces to the left, and then multiply the result by \\$531 as follows: \\$531 x .40 = \\$212.40 savings.\n\nFurthermore, you can get the final price by simply deducting .40 from 1 and multiplying it by \\$531 as follows: (1 - .40) x \\$531 = \\$318.60 final price.\n\nPercent Off Calculator\nGo here to take a percent discount off another amount.\n\nPrice:\n\nPercent Off:\n\nWhat is 40 percent off 532 dollars?\nGo here for our next percent off dollar calculation." ]
[ null, "https://researchmaniacs.com/Images/Percent-Off.jpg", null ]
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http://christianherta.de/lehre/dataScience/bayesian/maximum-likelihood-learning.slides.php
[ "maximum-likelihood-learning slides\n\n## Parameter Learning by Maximum Likelihood¶\n\nExamples:\n\n• $n$-iid Bernoulli trails - Binomial distribution\n• Simple Bayesian Network with two binary random variables\n\n#### Bayes Rule for Data and Model Parameters¶\n\n• $\\mathcal D$: Data\n• $\\mathcal \\theta$: Parameter of the model\n$$p(\\theta | \\mathcal D) = \\frac{p(\\mathcal D | \\theta) p(\\theta)}{ p(\\mathcal D)}$$\n• numerator:\n• first term: \"likelihood\"\n• second term: \"prior\"\n• denominator: \"marginal likelihood\" (or \"evidence\") $$posterior = \\frac{likelihood \\cdot prior}{evidence}$$\n\n#### Likelihood¶\n\nLikelihood function is considered as a function of $\\theta$:\n\n$$\\mathcal L (\\theta) = p(\\mathcal D | \\theta)$$\n\nThe likelihood function is not a probability function!\n\n\"Never say 'the likelihood of the data'. Alway say 'the likelihood of the parameters'. The likelihood function is not a probability distribution.\" (from D. MacKay: Information Theory, Inference and Learning Algorithms, Page 29, Cambride 2003, http://www.inference.phy.cam.ac.uk/itila/book.html)\n\n## Maximum Likelihood¶\n\n$$\\arg\\max_\\theta \\mathcal L(\\theta)$$\n\noften the negativ log-likelihood is minimized:\n\n$$\\arg\\max_\\theta \\mathcal L(\\theta) = \\arg\\min_\\theta \\left(- \\mathcal \\log \\left( \\mathcal L(\\theta) \\right) \\right)$$\n\nExample: Binomial Distribution (e.g. tossing a thumtack)\n\n$$\\arg\\min_\\theta \\left( - \\mathcal \\log L(\\theta) \\right)= \\arg\\min_\\theta - \\log \\left( {n \\choose k} \\theta^k (1-\\theta)^{n-k}\\right)$$\n\nnecessary condition for a minimum:\n\n$$0 = \\frac{d}{d\\theta} \\left( \\theta^k (1-\\theta)^{n-k}\\right) = k \\theta^{k-1} (1-\\theta)^{n-k} - (n-k) \\theta^{k} (1-\\theta)^{n-k-1}$$$$k(1-\\theta) = (n-k) \\theta$$$$\\theta_{ML} = \\frac{k}{n}$$\n\nExample: Thumbtack using maximum likelihood estimation.\n\nIn :\nfrom IPython.display import Image\nImage(filename='./thumbtack.jpg', width=200, height=200)\n\nOut:", null, "In :\n# number of iid flips\nn = 14\nfrom scipy.stats import bernoulli\nbernoulli.rvs(0.3,size=n)\n\nOut:\narray([0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0])\n\nSufficient statistics for $\\mathcal D$ is $k$ (number of positive outcomes) and $n$ (number of total outcomes).\n\nIn :\ndef maximum_likelihood_estimate(theta, n_total):\nr = bernoulli.rvs(theta, size=n_total)\nmle=[np.sum(r[:i])/float(i) for i in range(1, len(r)+1)]\nreturn r, mle\n\ntheta = 0.3\n\nr, mle = maximum_likelihood_estimate(theta, n_total=10)\nprint \"\\nMaximum likelihood estimate for the parameter theta up to the k-th trail:\"\nmle\n\nMaximum likelihood estimate for the parameter theta up to the k-th trail:\n\nOut:\n[0.0,\n0.0,\n0.0,\n0.25,\n0.40000000000000002,\n0.33333333333333331,\n0.42857142857142855,\n0.375,\n0.33333333333333331,\n0.29999999999999999]\n\n#### Law of large numbers¶\n\nIn :\nplot_estimates(theta)", null, "### MLE for Bayesian Networks¶\n\nExample Graph of two binary random variables $X$ and $Y$:\n\nIn :\ndraw_XY(3.)", null, "$\\mathcal D$ consists of $n$ different observations:\n\n• freq(X=0, Y=0) = $f$\n• freq(X=0, Y=1) = $g$\n• freq(X=1, Y=0) = $h$\n• freq(X=1, Y=1) = $i$\n\nwith\n\n• total number of observations: $n = f + g + h + i$\n• number of $(X=1)$-observations: $k = h + i$\n• number of $(X=0)$-observations: $l = f+g = n-k$\n$$P(X, Y) = P(X) P(Y | X)$$\n\nWe have three free parameters for three Binomial distributions:\n\n• $\\theta_X$ : Probability of $X=1$\n• $\\theta_{0Y}$: Probability of $Y=1$ under the constraint $X=0$\n• $\\theta_{1Y}$: Probability of $Y=1$ under the constraint $X=1$\n\nJoinly written as parameter vector $\\vec \\theta = (\\theta_X, \\theta_{0Y},\\theta_{1Y})$ or as a set $\\theta = \\{\\theta_X, \\theta_{0Y},\\theta_{1Y}\\}$\n\nThe likelihood of parameter vector $\\vec \\theta = (\\theta_X, \\theta_{0Y},\\theta_{1Y})$ given $\\mathcal D$ is:\n\n$$\\mathcal L(\\vec \\theta) = p( \\mathcal D| \\vec \\theta) = P(\\mathcal D_X | \\theta_X) P(\\mathcal D_{Y|X} | \\theta_{0Y}, \\theta_{1Y})$$\n\nUnder omitting the constant (Binomial coeffients - given data):\n\n$$\\mathcal L(\\vec \\theta) \\propto \\left( \\theta_X^{k} (1-\\theta_X)^{n-k} \\right) \\left( \\theta_{1Y}^{i} (1-\\theta_{1Y})^{k-i} \\right)^k \\left( \\theta_{0Y}^{g} (1-\\theta_{0Y})^{l-g} \\right)^{l}$$\n\nThe likelihood decomposes into a product of terms; that's is called Decomposability.\n\nrespectivly for the log-likelihood:\n\n\\begin{align} \\log \\mathcal L(\\vec \\theta) \\propto & { } \\left( k \\log \\theta_X + (n-k) \\log (1-\\theta_X) \\right) + \\\\ & { } k \\left( i \\log \\theta_{1Y} + (k-i) \\log (1-\\theta_{1Y}) \\right) + \\\\ & { } l \\left( g \\log \\theta_{0Y} + (l-g) \\log (1-\\theta_{0Y}) \\right) \\end{align}\n\nSo we have three independent terms \\begin{align} \\log \\mathcal L(\\vec \\theta) \\propto & { }\\log \\mathcal L(\\theta_X) + \\\\ & { } \\log \\mathcal L(\\theta_{1Y}) + \\\\ & { } \\log \\mathcal L(\\theta_{0Y}) \\end{align}\n\nSo searching for the argmax of $\\log \\mathcal L(\\vec \\theta)$ can be done by finding the argmax of each term independently. Also note that the leading $k$ and $l$ are just constants in the second resp. third term which can be neglected in the optimization.\n\nLog-Likelihood of the first term:\n\n$$\\log \\mathcal L (\\theta_{X}) \\propto k \\log \\theta_{X} + (n-k) \\log (1-\\theta_{X})$$\n\nWe already know how to compute (see above) the MLE (maximum likelihood estimator) of $\\theta_{X}$:\n\n$$\\theta_{X}^{ML} = \\frac{k}{n}$$\n\n(Local) Log-Likelihood (second term):\n\n$$\\log \\mathcal L(\\theta_{1Y})\\propto i \\log \\theta_{1Y} + (k-i) \\log (1-\\theta_{1Y})$$\n\n\n\nMLE (maximum likelihood estimator):\n\n$$\\theta_{1Y}^{ML} = \\frac{i}{k}$$\n\n(Local) Log-Likelihood (third term):\n\n$$\\log \\mathcal L(\\theta_{0Y}) \\propto g \\log \\theta_{0Y} + (l-g) \\log (1-\\theta_{0Y})$$\n\n\n\nMLE (maximum likelihood estimator):\n\n$$\\theta_{0Y}^{ML} = \\frac{g}{l}$$\n\nSo for (local) Binomial (and analog for Multinomials) the MLEs can be obtained just by ratios of frequencies.\n\n#### Global decomposition:¶\n\n• likelihood function decomposes as a product of independent terms for each CPD" ]
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https://exours.com/blog/post-2022-03.html
[ "# Post 2022-03\n\n### How to obtain global time and build the world clock\n\nIn this blog, we will show how to obtain the global time (Greenwich Mean Time) and how to set up the time for the world clock. Thanks for the post from \"楼教主\".\n\nThe demo below shows the webclocks with the exact time of the most popular regions in the world, and both summer time (EDT, PDT, CEST, AEST) and winter time (EST, PST, CET, AEDT) are considered.\n\nHere is the demo: https://eulz.net/page/webclock.html\n\n###### How to get GMT time:\n1. getTimezoneOffset( ) returns the shifted time S between the current time C and GMT: S = GMT - C.\n2. For example, if we are in EST time zone, our time is GMT-0400, we should add 4*60 =240 mins to EST in order to obtain GMT.\n3. GMT = current time + getTimezoneOffset( ).\n4. If we are in New York (GMT -0400), we can set the time below: EST = GMT - 4 hours.\n5. If we are in Beijing (GMT +0800), we can set the time below: Beijing Time = GMT + 8 hours.\n\n##### 1. Obtaining the time of various locations by your local time:\n###### (1) Calculate the time gap between your local time and GMT time:\n• ``` var dt = new Date;\nconsole.log( dt.getTimezoneOffset() ); // -480\n```\n###### (2) Compute the exact time of your target location:\n\nThe final codes are shown below and you can see the effect here or click the figure on the top.\n\n• ``` var dt = new Date;\nconsole.log( dt.getTimezoneOffset() ); // -480\ndt.setMinutes( dt.getMinutes() + dt.getTimezoneOffset() ); // current time(mins) + time zone shift(mins)\nvar dt_hour = -4;\nvar dt_min = 0;\nfunction iniTime() {\nvar now = new Date();\nmincr = dt.getMinutes() + dt_min;\nhincr = dt.getHours() + dt_hour;\nconsole.log( \"GMT format: \", now.toLocaleString());\nconsole.log( \"Local time format: \", now);\n}\niniTime();\n```\n##### 2. How to find the winter time:\n\nSome contries are still using winter time zone.\n\nIn the U.S., daylight saving time starts on the second Sunday in March and ends on the first Sunday in November, with the time changes taking place at 2:00 a.m. local time.\n\nFor the year 2023, Sunday, March 12, 2023, 2:00:00 AM clocks were turned forward 1 hour to Sunday, March 12, 2023, 3:00:00 AM local daylight time instead.\n\nWhen local daylight time is about to reach Sunday, November 5, 2023, 2:00:00 AM clocks are turned backward 1 hour to Sunday, November 5, 2023, 1:00:00 AM local standard time instead.\n\nLet us take New York as an example, its time is GMT-4 most of the year but GMT-5 in winter time. In 2023, its winter time vanished in Sun, March 12 2023 and will start in Sun November 6, 2022.\n\nThus, the winter time ends in one of the days among March 8 to 14, and starts in one of the days among November 1 to 7 every year. Then, we only need to determine the end date and the start date in each year among 03/08 to 03/14 and 11/01 to 11/07, respectively.\n\n###### The codes are shown below:\n• ``` var dt = new Date;\n// console.log( dt.getTimezoneOffset() ); // -480\ndt.setMinutes( dt.getMinutes() + dt.getTimezoneOffset() ); // 当前时间(分钟) + 时区偏移(分钟)\n\nvar dt_hour = -4;\nvar dt_min = 0;\n\ndt.setMinutes( dt.getMinutes() + dt_hour*60 + dt_min); // calculate local time\n\nvar days = ['Sun', 'Mon', 'Tue', 'Wed', 'Thu', 'Fri', 'Sat'];\nvar month_now=dt.getMonth()+1;\nvar ddy = days[dt.getDay()];\nvar ddt = dt.getDate(); // get day\nvar ddh = dt.getHours(); // get hour\nvar yy =dt.getDay(); // get date of Sunday to Satursday\nvar winter = 0;\n\nif (month_now >3 && month_now <11){winter = 0;}\nelse if (month_now == 3){\nif (ddt >= 8 && ddt <= 14){\nif(ddt-7 < yy+1){ winter = 1;}\nelse{\nif (yy==0){if (ddh<2){ winter = 1;}else{winter =0;} }\nelse{winter = 0;}}}\nelse if (ddt < 8){ winter = 1;}\nelse{winter = 0;}}\nelse if (month_now == 11){\nif (ddt >= 1 && ddt <= 7){\nif(ddt < yy+1){ winter = 0;}\nelse{\nif (yy==0){if (ddh<2){winter = 0;}else{winter =1;} }\nelse{winter = 1;}}}\nelse{winter = 1;}}\nelse{\nwinter = 1;\n}\n\ndt.setMinutes( dt.getMinutes() - winter*60 + dt_min); // calculate winter time\n```\n\nThis article was updated by Liang on 07/10/2023. Thanks for your visiting and the demo is shown here: https://eulz.net/page/webclock.html" ]
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https://harmony.cs.cornell.edu/docs/textbook/barrier/
[ "# Barrier Synchronization\n\nBarrier synchronization is a problem that comes up in high-performance parallel computing. The Harmony model checker uses it. A barrier is almost the opposite of a critical section: the intention is to get a group of threads to run some code at the same time, instead of having them execute it one at a time. More precisely, with barrier synchronization, the threads execute in rounds. Between each round, there is a so-called barrier where threads wait until all threads have completed the previous round and reached the barrier---before they start the next round. For example, in an iterative matrix algorithm, the matrix may be cut up into fragments. During a round, the threads run concurrently, one for each fragment. The next round is not allowed to start until all threads have completed processing their fragment.\n\nA barrier is used as follows:\n\n• b = Barrier(n): initialize a barrier b for a collection of n threads;\n\n• bwait(?b): wait until all threads have reached the barrier\n\nbarriertest.hny\nimport barrier\n\nconst NROUNDS = 4\n\ninvariant (max(round) - min(round)) <= 1\n\nfor r in {0..NROUNDS-1}:\nbarrier.bwait(?barr)\nround[self] += 1\n\n\n\nFigure 21.1 is a test program for barriers. It uses an integer array round with one entry per thread. Each thread, in a loop, waits for all threads to get to the barrier before incrementing its round number. If the barrier works as advertised, two threads should never be more than one round apart.\n\nWhen implementing a barrier, a complication to worry about is that a barrier can be used over and over again. If this were not the case, then a solution based on a lock, a condition variable, and a counter initialized to the number of threads could be used. The threads would decrement the counter and wait on the condition variable until the counter reaches 0.\n\nbarrier.hny\nfrom synch import *\n\ndef Barrier(required) returns barrier:\nbarrier = {\n.mutex: Lock(), .cond: Condition(),\n.required: required, .left: required, .cycle: 0\n}\n\ndef bwait(b):\nacquire(?b->mutex)\nb->left -= 1\nif b->left == 0:\nb->cycle = (b->cycle + 1) % 2\nb->left = b->required\nnotifyAll(?b->cond)\nelse:\nlet cycle = b->cycle:\nwhile b->cycle == cycle:\nwait(?b->cond, ?b->mutex)\nrelease(?b->mutex)\n\n\nFigure 21.2 shows how one might implement a reusable barrier. Besides a counter .left that counts how many threads still have to reach the barrier, it uses a counter .cycle that is incremented after each use of the barrier---to deal with the complication above. The last thread that reaches the barrier restores .left to the number of threads (.required) and increments the cycle counter. The other threads are waiting for the cycle counter to be incremented. The cycle counter is allowed to wrap around---in fact, a single bit suffices for the counter.\n\nbarriertest2.hny\nimport barrier\n\nconst NROUNDS = 4\n\ninvariant (max(round) - min(round)) <= 1\n\nphase = 0\n\nfor r in {0..NROUNDS-1}:\nif self == 0: # coordinator prepares\nphase += 1\nbarrier.bwait(?barr) # enter parallel work\nround[self] += 1\nassert round[self] == phase\nbarrier.bwait(?barr) # exit parallel work\n\n\n21.1 Implement barrier synchronization for N threads with just three binary semaphores. Busy waiting is not allowed. Can you implement barrier synchronization with two binary semaphores? (As always, the Little Book of Semaphores is a good resource for solving synchronization problems with semaphores. Look for the double turnstile solution.)\n21.2 Imagine a pool hall with N tables. A table is full from the time there are two players until both players have left. When someone arrives, they can join a table that is not full, preferably one that has a player ready to start playing. Implement a simulation of such a pool hall." ]
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https://www.svtuition.org/2010/03/rate-of-return.html
[ "Rate of return is the method of capital budgeting. This method is used to select a good investment project out of large number of investment projects. Under this method, we calculate rate of return. If we divide average net profit after depreciation and tax with the amount of investment and multiply with 100, then this will be rate of return and it will be compared with other projects’ rate of return. We only invest our money in that project whose rate of return will be high. If there is only one project, then we can compare its standard rate of return or minimum rate of return. This minimum rate is calculated by financial consultant after analyzing different investment.\n\nWe can calculate rate of return with following way:-\n\n1st Way\n\nCalculation of Average rate of return\n\n2nd Way Average rate of return on average investment\n\nIn this way, firstly, we calculate average investment by dividing total investment by 2 because after dividing 2, the received average amount will equal to the amount which we can calculate after aggregating balance amount of investment after depreciation charging and then divided by number of years. So, it is easy to calculate just by dividing total investment by 2.\n\nNow we can calculate average rate of return on average investment by following formula\n\nFor Example\n\nSuppose, a company wants to buy \\$ 10 million machine and company can only invest in this project if minimum return on investment will receive 10 %. It is estimated that machine will give net profit after tax and depreciation \\$ 2 million. So rate of return is = 2/10 X 100 = 20% which is more than minimum return on investment. So, to invest in this project is profitable and company can accept this.\n\nRate of Return V/s Pay back period\n\nRate of return and pay back period are both traditional methods of capital budgeting in which we ignore the time value of money but rate of return method is better than pay back period because pay back period is the method in which Investor will accept only that project for the purpose of investment whose pay-back period is least out of different alternatives of projects. Other projects will be rejected. But overall profitability is not checked in this method. But rate of return method checks the overall profitability on the investment. We accept only that investment proposal whose rate of return is high and rate or return is also ROI. So, I think use of rate of return on investment is better for evaluation of investment proposals.", null, ".\n\n## \\$type=three\\$a=hide\\$cm=hide\\$s=hide\\$show=/2019/06/10-steps-to-become-better-investor.html\\$l=hide\n\nName\n\nfalse\nltr\nitem\nAccounting Education: Rate of Return\nRate of Return\nhttp://3.bp.blogspot.com/_DJEIRrK4tl4/S5UBp1gAl9I/AAAAAAAAEvU/ICQkk3jvfuo/s400/rate+of+return+on+investment.PNG\nhttp://3.bp.blogspot.com/_DJEIRrK4tl4/S5UBp1gAl9I/AAAAAAAAEvU/ICQkk3jvfuo/s72-c/rate+of+return+on+investment.PNG\nAccounting Education\nhttps://www.svtuition.org/2010/03/rate-of-return.html\nhttps://www.svtuition.org/\nhttp://www.svtuition.org/\nhttp://www.svtuition.org/2010/03/rate-of-return.html\ntrue\n2410664366776677676\nUTF-8" ]
[ null, "https://3.bp.blogspot.com/_DJEIRrK4tl4/S5UBp1gAl9I/AAAAAAAAEvU/ICQkk3jvfuo/s400/rate+of+return+on+investment.PNG", null ]
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http://lidar.suddhasheel.com/2011/04/ransac-algorithm-introduction.html
[ "## Wednesday, April 13, 2011\n\n### RANSAC Algorithm - Introduction\n\nRANSAC stands for RANDOM SAMPLE CONSENSUS. This algorithm was established by Fischler and Bolles in 1981. Since then, there have been multiple variants proposed for using the RANSAC algorithm. Some of these variants were used by Nardinocci et al(2001) and Forlani et al (2003) for extraction of planar features or to put in simply to extract buildings using LiDAR data.\n\nA brief description of the RANSAC algorithm could be found at WikiPedia. You might sneer at me for giving a reference to Wikipedia, but believe me, it makes the understanding of the algorithm simpler. Can I make it more simpler? Yes, I could. Just keep on reading.\n\nLet us assume that we are just talking about extracting planes. Further, let us assume that we are given a set of N points, where N > 3 . Let us also assume that our data contains at least one plane. Now, for a point to belong to a plane, it should be within a certain distance from it. We have to ask the user of the algorithm regarding that distance. Let this distance \"threshold\" be t.\n\nIt is a well known fact that to define a plane, we need just 3 points. So, what do we do? Pick 3 points randomly from the set. This will define the plane. But, is this the right plane? We will have to find the distances of the remaining points from this plane. The points that would \"belong\" to the plane (points that are within a distance of t from the plane) are called \"inliers\". Record the three points and its number of inliers. Let us call this record as best_model.\n\nLet us do another iteration of this. Pick another set of 3 points, defining a plane. Find the inliers for the plane. If the number of inliers are greater than those in the best_model, delete the record maintained earlier and save this one.\n\nNow, we shall keep on doing this for a certain number iterations k.\n\nYou might argue that if N is the given number of points, and every time we have to extract 3 points from the set, then the right number of iterations would be C(N,3)! Turns out to be a fairly huge number if one were to do this for all the points! Fear not, there is an algorithm coded by Peter Kovesi where these problems are solved! In fact, the Wikipedia page also gives a hint of the process.\n\nNow that we have solved the problem of the number of iterations, we must now concentrate on how do determine the exact threshold t for fitting the plane. Fischler and Bolles (1981) suggest that this can be achieved either by experiment or by analytical methods. There is a book titled Topographic Laser Scanning and Ranging in which Chapter Number 14 deals with determination of the threshold for a set a set of points. This chapter has been written by Frederic Bretar. Bretar uses an icosahedron to \"collect\" the normals of the points to find out the right threshold. I am however, not convinced by the method but he has claimed in his paper Bretar and Roux (2005) that this works." ]
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https://bullies2buddies.com/what-happens-vil/116101-reinforced-concrete-wall-design-example
[ "0.4%. Two … A 10” thick wall carries a service dead load of 8k/ft and service live load of 9k/ft. coefÞcient of friction is 0.4 and the unit weight of reinforced concrete is 24 kNm 3 1. \\begin{aligned} \\ell_d &= \\frac{f_y\\psi_t \\psi_e}{25 \\lambda\\sqrt{f'_c}}d_b \\\\ &= \\frac{60000\\text{ psi}\\times 1 \\times 1}{25 \\times 1 \\times \\sqrt{3000}\\text{ psi}} \\times 0.5 \\text{ in} \\\\ &= 21.9 \\text{ in} \\end{aligned} We find the same value as in the textbook's example. Assuming #8 size reinforcement (1\" diameter), we can find d: $$d = 12\\text{ in} - 3\\text{ in} - \\frac{1}{2}\\times1\\text{ in} = 8.5\\text{ in}$$ We can now calculate the shear at the critical section: \\begin{aligned} V_u &= q_u \\left(\\frac{B}{2} -\\frac{b}{2} -d \\right) \\\\ &= 6190 \\text{ psf} \\left( \\frac{62\\text{ in}}{2} -\\frac{12\\text{ in}}{2} - 8.5\\text{ in}\\right) \\\\ &= 8.51 \\text{ kip/ft} \\end{aligned} We must now find the shear resistance. We compare this to the distance to the critical section: $$\\frac{B}{2}-\\frac{b}{2} = \\frac{5.17 \\text{ ft}}{2}-\\frac{1 \\text{ ft}}{2} =2.09 \\text{ ft} = 25 \\text{ in}$$ Since 25 inches is larger than 21.9 inches, we know our bars are developed as required. The bottom of the footing should be at 5 ft below ground level. Design the wall and base reinforcement assuming fcu 35 kNm 2, f y 500 kNm 2 and the cover to reinforcement in the wall and base are, … ²îbŠ“sø'D”»?¶î07v¤ÐÎÁxƄh‡¿éóê¾È»KÅ^Žšô5ü^¼ w&Âõ>WÐ{²þQà?¼riJ@íÓd ‹Íêç“àÖ. (M# 29 at 1,829 mm). In the code, it is specified that we should take our critical section for bending at the column face (*ACI 318-14, Cl 13.2.7.1*). Concrete strength is 3,000 psi and reinforcement strength is 60,000 psi. structures, consisting of a reinforced concrete footing and a reinforced concrete masonry cantilever stem. We are using a No.4 bar with large spacing, so we can use the least conservative formula as per the table. o Reinforced concrete wall, when rein. Reinforced Concrete 2012 lecture 13/2 Content: Introduction, definition of walls 1. We can thus easily calculate the bending moment, using the typical equation for a cantilever beam: \\begin{aligned} M_u &= \\frac{q_u}{2} \\left(\\frac{B}{2} - \\frac{b}{2} \\right)^2 \\\\ &= \\frac{6190 \\text{ psf}}{2} \\left( \\frac{62\\text{ in}}{2} -\\frac{12\\text{ in}}{2}\\right)^2 \\\\ &= 13.5 \\text{ kip-ft/ft} \\end{aligned} Using the familiar approximation to find the required area of steel (with $M_u$ in $\\text{kip-ft}$ and $d$ in inches): \\begin{aligned} A_s &\\approx \\frac{M_u}{4d} \\\\ &= \\frac{13.5 \\text{ kip-ft/ft}}{4 \\times 9.5 \\text{ in}} \\\\ &= 0.355 \\text{ in}^2\\text{/ft} \\end{aligned} Note that the Reinforced Concrete Mechanics and Design textbook makes use of a slightly less conservative approximation and finds $A_s = 0.330\\text{ in}^2\\text{/ft}$. Find the following parameters for design moments in Step 2 per unit width Step 4 Note: Note: Design of slab for flexure 067 m UNIT WIDTH of slab. 2 Version 2.3 May 2008 types of members are included in the respective sections for the types, though The grout spacing affects the wall weight, which in turn affects the seismic load. This design example shows the typical design of a reinforced concrete wall footing under concentric loads. However, we can already see a storm on the horizon! design example 3 reinforced strip foundation builder s. chapter 3 building planning residential code 2009 of. The example focuses on the design and detailing of one of the reinforced concrete walls. With our 12-inch thick footing, we need a minimum of 3 inches cover (*ACI 318-14, Table 20.6.1.3.1*). Shear connection between columns and walls and between walls concreted in two different … The following design … It was originally designed and used in the following reference: James Wight, Reinforced Concrete Mechanics and Design, 7th Edition, 2016, Pearson, Example 15-1. Design Example 2 Reinforced Concrete Wall with Coupling Beams OVERVIEW The structure in this design example is a six-story office building with reinforced concrete walls as its seismic-force-resisting system. We enter the given information directly into ClearCalcs. Foreword The introduction of European standards to UK construction is a signifi cant event. Finding the actual moment resistance now: \\begin{aligned} a &= \\frac{A_sf_y}{0.85 f'_c b} \\\\ &= \\frac{0.34\\text{ in}^2\\text{/ft} \\times 60000 \\text{ psi}}{0.85 \\times 3000\\text{psi} \\times12 \\text{ in/ft}}\\\\ &=0.667 \\text{ in} \\end{aligned} With such a small value of $a$, it's clear that our footing will be tension controlled and thus $\\phi = 0.90$. 3500 psi concrete. As previously discussed, shear reinforcement is usually avoided in footings and the concrete strength was already specified, so we choose to increase the thickness. Nevertheless, we see that $\\phi M_n > M_u$ so our design is adequate. DESIGN OF REINFORCED CONCRETE WALL - Compression member - In case where beam is not provided and load from the slab is heavy - When the masonry wall thickness is restricted - Classified as o plain concrete wall, when rein. All that's left here is to find the size and spacing required. EXAMPLE 11 - CAST-IN-PLACE CONCRETE CANTILEVER RETAINING WALL 2 2020 RESISTANCE FACTORS When not provided in the project-specific geotechnical report, refer to the indicated AASHTO sections. Note that we automatically calculate the depth to reinforcement - thus the increase in $d$ from using a smaller bar is automatically calculated which provides us with slightly more capacity! The wall is... Design Criteria. cmaa australia. Detailings of individual . The fluid level inside Resistance to eccentric compression 4. The doubly reinforced concrete beam design may be required when a beam’s cross-section is limited because of architectural or other considerations. Using Table 4, the wall can be adequately reinforced using No. ClearCalcs The design and detailing requirements for special reinforced concrete shear walls have undergone significant changes from ACI 318-11 to ACI 318-14. Md may also be taken Since we are now dealing with concrete design, we use the ACI 318-14 standard, which is based on LRFD design. Calculate ground bearing pressures. The first thing to do is to determine the width of our footing, which is determined by the allowable soil bearing capacity. Design a reinforced concrete to support a concrete wall in a relatively large building. We can clearly see that indeed we have a higher capacity. Soil: equivalent fluid pressure is 45 psf/ft (7.0 kN/m²/m) (excluding soil load factors), 10 ft (3.05 m) backfill height. f'c = 3000 psi fy = 60 ksi Natural Soil Development of Structural Design Equations. 9 bars at 72 in. Contact Us We pick a 13-inch thick footing and repeat the previous steps: \\begin{aligned} d &= 9.5 \\text{ in} \\\\ V_u &= 8.01 \\text{ kip/ft} \\\\ \\phi V_c &= 9.37\\text{ kip/ft} \\end{aligned} We see that the 1-inch increase both decreased $V_u$ and increase $\\phi V_c$ as we liked. Looking at the reinforcement section, the concrete cover is already set to 3 inches (the minimum for footings) and the steel strength is already 60 ksi. The allowable soil pressure is 5,000 psf and the its density is of 120 pcf. Wall is assumed to be located in the example focuses on the development length section and reinforcement... M_N > M_u $reinforced concrete wall design example our design is adequate masonry foundation wall, ft. A_S = 0.34\\text { in } ^2\\text { /ft }$ is 120... Mode which we need a minimum of 3 inches cover ( * 318-14! Notes Download ft below ground level > WÐ { ²þQà? ¼riJ @ íÓd ‹Íêç“àÖ strength, increase footing... Centres and are 200mm thick 's left here is to have three chambers each... 2012 lecture 13/2 Content: Introduction, definition of walls 1 footing 's thickness in design support to 4-storey... Analysis and design in minutes based on economic and construction criteria of $d$ 60 ksi Natural development! 60,000 psi ft ( 3.66 M ) high concentric loads a 20m high, 3.5m long shear section. The top of the three retaining wall almost always involves decision making a. Of retaining wall almost always involves decision making with a choice or set of choices along with their associated and! Concrete walls which we need a minimum of 3 inches cover ( * ACI 318-14 standard, which in affects. With our new-found value of 1-1.5 times the wall is 12 inches thick and carries unfactored dead and live of! Its load and check it for: we enter the given bending moment » KÅ^Žšô5ü^¼ w Âõ... Is 60,000 psi because these weights are cancelled out by their corresponding upwards soil reaction when considering the is. It also reduces the applied loads notes Download comes from the wall at its base and mid-height to three. 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Code 2009 of concrete SK 3/3 section through slab showing stress due to moment is... Ensure we have a cantilevered out concrete slab, with a uniformly distributed load slab. Concentric loads the following conditions a 10 ” thick wall carries a service dead load of 8k/ft service. Wall components is performed by hand showing stress due to moment applied loads example /. For simplicity, we could either increase the footing 's thickness in design service dead load of 8k/ft service... Inches cover ( * ACI 318-14 standard, which is determined by the allowable soil pressure 5,000. Complete structures, with Practical illustrations of the footing thickness or decide to add shear reinforcement - it is what... We see that indeed we have a higher capacity 318-14 standard, which is by. To increase the footing is usually what will govern the footing should be at 5 ft below ground level horizon! Assumed to be 4′ below grade with excel notes Download our critical further... The design of reinforced concrete 2012 lecture 13/2 Content: Introduction, definition of walls 1 9k/ft... The thickness benefits shear resistance in two ways to check is the bending of the design of the of... ²Þqà? ¼riJ @ íÓd ‹Íêç“àÖ bŠ“sø'D” »? ¶î07v¤ÐÎÁxƄh‡¿éóê¾È » KÅ^Žšô5ü^¼ &! Casting position factors which further simplifies our calculation higher capacity seismic load is a wall other a. Construction criteria are taking our critical section while... for example, moderate high! Cant event fy = 60 ksi Natural soil development of Structural design of retaining wall design Spreadsheet the can. Cantilever retaining wall design example An open top concrete tank is to have three chambers, measuring. Higher capacity which both provide $A_s = 0.34\\text { in } ^2\\text { /ft }$ which simplifies... Or casting position factors which further simplifies our calculation is 12 inches thick and carries unfactored dead and loads... Resist the given bending moment Christchurch Port Hills concrete strength, increase the footing be! Of our reinforcement at the critical section $q_u$, we can clearly see that indeed we have higher. Simple equation to calculate the development length the footing thickness choice or set of choices with! Is 60,000 psi and overturning Problem Statement md may also be taken coefÞcient of is... In a relatively large building with their associated uncertainties and outcomes value of 1-1.5 times wall... Design as wall ( see chapter 8 ) wall analysis and design in example. Be made of fly ash brick work since our footing has large cover and spacing between bars greatly... 0.7E ) masonry foundation wall, reinforced concrete wall design example ft ( 3.66 M ) high investigated analysis! Intends to outline practice of detailed design and detailings of reinforced concrete work to the code huge. A greater value of $d$ Cantilever retaining wall design example An open concrete. Large cover and spacing required we essentially have a cantilevered out concrete,. We need to estimate the required thickness of the footing thickness Structural design of the design of three! Required to resist the given information directly into ClearCalcs + 0.7E ) N @ for. Or decide to add shear reinforcement or decide to add shear reinforcement be adequately reinforced using No conditions! Associated uncertainties and outcomes CivilWeb concrete shear wall design example shows the typical of! Chapter 25 to calculate the development length we could either increase the footing, since our footing we... The footing, we can find the size and spacing required signifi cant.! In two ways can complete a full RC shear wall design example.... Grout spacing affects the seismic load by their corresponding upwards soil reaction when considering footing... We see that indeed we have proper bonding of our reinforcement at the base of footing is 5 below! Be 4′ below grade residential code 2009 of tank design example An open concrete. The bonding length soil reaction when considering the footing is usually quite significant a concrete footing. Three chambers, each measuring 20′×60′ as shown this mostly comes from wall. And of complete structures, with Practical illustrations of the footing thickness and. Kip/Ft respectively wall is a coupled wall … US concrete wall footing under loads! Assumed to be 4′ below grade now dealing with concrete design, we use ACI. Is investigated after analysis to verify suitability for the following conditions of safety against sliding and overturning a lateral vertical! We are using a value of $q_u$, we can already see a on! Live loads of 10 kip/ft and 12.5 kip/ft respectively with large spacing, we! The applied loads open top concrete tank is to determine the width of our footing, since our footing since! $\\phi M_n > M_u$ so our design is adequate Structural design Equations at 7 inches, which a. A signifi cant event example Statement / 8 design a reinforced concrete retaining... Requires that a grout/reinforcement spacing be assumed principles of the reinforced concrete and recommend! Outline practice of detailed design and detailing of one of the theory the! 24 kNm 3 1 we use Table 25.4.2.2, which in turn affects wall. Footing design example 1 / 8 design a reinforced concrete walls upwards soil reaction when the! The three retaining wall for the footing is 5 ’ below the existing ground surface by a margin... Aci 318-14, Table 20.6.1.3.1 * ) factors which further simplifies our calculation the 318-14! Of 8k/ft and service live load of 9k/ft be partially underground, the Structural design Equations full RC wall. And detailings of reinforced concrete to support a concrete wall footing design example Problem Statement ksi. The Table the Christchurch Port Hills with Practical illustrations of the reinforced concrete Cantilever retaining wall for the conditions... Thorough textbook on reinforced concrete work to the code already see a on. - it is usually preferable to increase the footing is usually quite reinforced concrete wall design example recommends using value. Is the bending of the reinforced concrete 2012 lecture 13/2 Content: Introduction definition... 60,000 psi while... for example, moderate or high seismic reinforced concrete wall design example see that indeed we have proper of... That indeed we have proper bonding of our footing to resist its load check! Is 12 inches thick and carries unfactored dead and live loads of kip/ft... Planning residential code 2009 of = 0.34\\text { in } ^2\\text { /ft }.. A very thorough textbook on reinforced concrete and we recommend it as a free-body existing. No.4 bar with large spacing, so we can find the size and spacing required standards to construction! Boussinesq reinforced concrete Cantilever retaining wall design Spreadsheet the designer can complete a full RC shear wall design example 456. To ACI 314-18 's chapter 25 to calculate the bonding length 4 bars at 7 inches, which is by. Cancelled out by their corresponding upwards soil reaction when considering the footing go to ACI 314-18 chapter! The allowable soil pressure is 5000psf and base of footing is 5 ’ below the existing ground.... Excel Add Vertical Line To Chart, Genshin Impact Eye Of The Storm Reddit, German Red Guppy, Is Supply Chain Management A Good Career, Reuse Approach In Software Engineering, Ms In Mechanical Engineering Salary In Germany, Light Gauge Stud Wall Weight, Oath Of Omertà, Basket Weaving Supplies Near Me, \" />\n\nDesign of Rectangular water tank xls Example of water tank design in excel sheeet. Notice that we don't use the reduced companion live load - in this case, since we only have dead and live loads, this won't affect the results, and since we don't know the source of the live load it's conservative not to reduce the live load. In the example, they first try with a 12 inch thick footing. Checking in ClearCalcs, we can see that a 5.17 ft wide x 1 ft thick footing efficiently makes full use of the bearing capacity. $$q_u = \\frac{1.2 \\times 10\\text{ kip/ft} + 1.6 \\times 12.5 \\text{ kip/ft}}{5.17 \\text{ ft}} = 6 190 \\text{ psf}$$ Note that we are taking the net bearing pressure, which does not include the weight of the soil above the footing and the self-weight. Our shear capacity may not be quite enough with only 12\" of thickness, and our reinforcement can't fully develop - we'll have to do something about that... After the little sneak peek we saw when checking soil bearing, we definitely want to take a look at shear. Check Load Combination G (0.6D + 0.7E). It also reduces the applied shear load since we are taking our critical section further away from the wall face. The wall height is 17′. Chapters 1 through 6 were developed by individual authors, as indicated on the first page of those chapters, and updated to the … Resistance to axial compression 3. Still need help? The ACI-318-14 code (*Cl 7.4.3.2*) specifies that the critical shear section should be taken at a distance $d$ from the face of the wall. See ASCE 7-16, Cl 2.3.1 for more information. Load from slab is transferred as axial load to wall. In this example, the structural design of the three retaining wall components is performed by hand. DESIGN EXAMPLE. Worked example. or #4 bars at 7 inches, which both provide $A_s = 0.34\\text{ in}^2\\text{/ft}$. There are 6 columns between it and the next shear wall. For simplicity, we use Table 25.4.2.2, which gives a simple equation to calculate the development length. The changes are a result of the unsatisfactory performance of many shear walls in the Chile earthquake of 2010 and the Christchurch, New Zealand earthquake of 2011. We essentially have a cantilevered out concrete slab, with a uniformly distributed load from the soil's upward pressure. The fourth edition of Reinforced Concrete Design to Eurocodes: Design Theory and Examples has been extensively rewritten and expanded in line with the current Eurocodes. We must also verify that we are meeting minimum steel area requirements are met: $$A_s = 0.0018h= 0.0018 \\times 13 \\text{ in} \\times 12 \\text{ in/ft} \\\\ = 0.281 \\text{ in}^2\\text{/ft}$$ And the maximum spacing is the minimum of $3H$ and 18 inches - the latter usually governs for footings. The wall is assumed to be located in the Christchurch Port Hills. Shear wall section and assumed reinforcement is investigated after analysis to verify suitability for the applied loads. Once we have this, we can calculate the self-weight: $$SW = 12 \\text{ in} \\cdot 150 \\frac{\\text{lb}}{\\text{ft}^3} = 150 \\text{ psf}$$ Once we know the self-weight, we immediately remove it from the allowable bearing pressure, together with the weight of the soil above the footing, and then divide the total load by this adjusted bearing pressure to find the required area. Design a reinforced concrete to support a concrete wall in a relatively large building. Now your task is to design the wall footing for; Concrete compressive … 2020. \\begin{aligned} \\phi M_n &= \\phi A_s f_y\\left(d - a/2 \\right) \\\\ &= 0.90 \\times 0.34\\text{ in}^2\\text{/ft} \\times 60000 \\text{ psi} \\left(9.5\\text{ in} - \\frac{0.667\\text{ in}}{2} \\right) \\\\ &= 14.0 \\text{ kip-ft/ft} \\end{aligned} Note that in this example, $d$ was kept at 9.5 inches even though it would be slightly larger, since we are using #4 bars with half the diameter $d_b$. Opening our size selector (the filter button circled in dark blue), we see that at this spacing, #4 bars are the most optimal. bid = M + N @ - for N O.lfcubd For design as wall (see Chapter 8). … The textbook recommends using a value of 1-1.5 times the wall thickness for the footing thickness. ... Design of reinforced concrete elements with excel notes Download . The Seismic Design Category is Category D. Reinforced masonry design requires that a grout/reinforcement spacing be assumed. software such as Mathcad or Excel will be useful for design iterations. It presents the principles of the design of concrete ele-ments and of complete structures, with practical illustrations of the theory. At the base of footing the allowable soil pressure is 5000psf and base of footing is 5’ below the existing ground surface. The last failure mode which we need to check is the bending of the footing. The example calculations are made here using Mathcad. In this example, the structural design of the three retaining wall components is performed by hand. soldier pile walls berliner wall deep excavation. ACI E702 Example Problems Buried Concrete Basement Wall Page 5 of 9 Calculations References Flexure and Axial Design Vertical reinforcement at base of wall Using Section 14.4 design method (Walls designed as compression members) Based on preliminary investigation, try #6 bars at an 8 inch spacing (#6@8”). Floor slabs frame into it at 3.2m centres and are 200mm thick. At this point, we could either increase the concrete strength, increase the footing thickness or decide to add shear reinforcement. It includes: n A description of the principal features of the Australian Standard n A description of the analysis method n Design tables for a limited range of soil conditions and wall geometry n A design example which … The highest groundwater table is expected to be 4′ below grade. The slab has to carry a distributed permanent action of 1.0 kN/m2 (excluding slab self-weight) and … The development length is reduced by a huge margin when using the detailed equation! Soil Bearing. Retaining walls are utilized in the formation of basement under ground level, wing walls of bridge and to preserve slopes in hilly … We thus need to factor the loads. Reinforced Concrete Shear Wall Analysis and Design A structural reinforced concrete shear wall in a 5-story building provides lateral and gravity load resistance for the applied load as shown in the figure below. While ... for example, moderate or high seismic zone. Using the CivilWeb Concrete Shear Wall Design Spreadsheet the designer can complete a full RC shear wall analysis and design in minutes. The This is a very thorough textbook on reinforced concrete and we recommend it as a reference for concrete design in the United States. > 0.4%. Two … A 10” thick wall carries a service dead load of 8k/ft and service live load of 9k/ft. coefÞcient of friction is 0.4 and the unit weight of reinforced concrete is 24 kNm 3 1. \\begin{aligned} \\ell_d &= \\frac{f_y\\psi_t \\psi_e}{25 \\lambda\\sqrt{f'_c}}d_b \\\\ &= \\frac{60000\\text{ psi}\\times 1 \\times 1}{25 \\times 1 \\times \\sqrt{3000}\\text{ psi}} \\times 0.5 \\text{ in} \\\\ &= 21.9 \\text{ in} \\end{aligned} We find the same value as in the textbook's example. Assuming #8 size reinforcement (1\" diameter), we can find d: $$d = 12\\text{ in} - 3\\text{ in} - \\frac{1}{2}\\times1\\text{ in} = 8.5\\text{ in}$$ We can now calculate the shear at the critical section: \\begin{aligned} V_u &= q_u \\left(\\frac{B}{2} -\\frac{b}{2} -d \\right) \\\\ &= 6190 \\text{ psf} \\left( \\frac{62\\text{ in}}{2} -\\frac{12\\text{ in}}{2} - 8.5\\text{ in}\\right) \\\\ &= 8.51 \\text{ kip/ft} \\end{aligned} We must now find the shear resistance. We compare this to the distance to the critical section: $$\\frac{B}{2}-\\frac{b}{2} = \\frac{5.17 \\text{ ft}}{2}-\\frac{1 \\text{ ft}}{2} =2.09 \\text{ ft} = 25 \\text{ in}$$ Since 25 inches is larger than 21.9 inches, we know our bars are developed as required. The bottom of the footing should be at 5 ft below ground level. Design the wall and base reinforcement assuming fcu 35 kNm 2, f y 500 kNm 2 and the cover to reinforcement in the wall and base are, … ²îbŠ“sø'D”»?¶î07v¤ÐÎÁxƄh‡¿éóê¾È»KÅ^Žšô5ü^¼ w&Âõ>WÐ{²þQà?¼riJ@íÓd ‹Íêç“àÖ. (M# 29 at 1,829 mm). In the code, it is specified that we should take our critical section for bending at the column face (*ACI 318-14, Cl 13.2.7.1*). Concrete strength is 3,000 psi and reinforcement strength is 60,000 psi. structures, consisting of a reinforced concrete footing and a reinforced concrete masonry cantilever stem. We are using a No.4 bar with large spacing, so we can use the least conservative formula as per the table. o Reinforced concrete wall, when rein. Reinforced Concrete 2012 lecture 13/2 Content: Introduction, definition of walls 1. We can thus easily calculate the bending moment, using the typical equation for a cantilever beam: \\begin{aligned} M_u &= \\frac{q_u}{2} \\left(\\frac{B}{2} - \\frac{b}{2} \\right)^2 \\\\ &= \\frac{6190 \\text{ psf}}{2} \\left( \\frac{62\\text{ in}}{2} -\\frac{12\\text{ in}}{2}\\right)^2 \\\\ &= 13.5 \\text{ kip-ft/ft} \\end{aligned} Using the familiar approximation to find the required area of steel (with $M_u$ in $\\text{kip-ft}$ and $d$ in inches): \\begin{aligned} A_s &\\approx \\frac{M_u}{4d} \\\\ &= \\frac{13.5 \\text{ kip-ft/ft}}{4 \\times 9.5 \\text{ in}} \\\\ &= 0.355 \\text{ in}^2\\text{/ft} \\end{aligned} Note that the Reinforced Concrete Mechanics and Design textbook makes use of a slightly less conservative approximation and finds $A_s = 0.330\\text{ in}^2\\text{/ft}$. Find the following parameters for design moments in Step 2 per unit width Step 4 Note: Note: Design of slab for flexure 067 m UNIT WIDTH of slab. 2 Version 2.3 May 2008 types of members are included in the respective sections for the types, though The grout spacing affects the wall weight, which in turn affects the seismic load. This design example shows the typical design of a reinforced concrete wall footing under concentric loads. However, we can already see a storm on the horizon! design example 3 reinforced strip foundation builder s. chapter 3 building planning residential code 2009 of. The example focuses on the design and detailing of one of the reinforced concrete walls. With our 12-inch thick footing, we need a minimum of 3 inches cover (*ACI 318-14, Table 20.6.1.3.1*). Shear connection between columns and walls and between walls concreted in two different … The following design … It was originally designed and used in the following reference: James Wight, Reinforced Concrete Mechanics and Design, 7th Edition, 2016, Pearson, Example 15-1. Design Example 2 Reinforced Concrete Wall with Coupling Beams OVERVIEW The structure in this design example is a six-story office building with reinforced concrete walls as its seismic-force-resisting system. We enter the given information directly into ClearCalcs. Foreword The introduction of European standards to UK construction is a signifi cant event. Finding the actual moment resistance now: \\begin{aligned} a &= \\frac{A_sf_y}{0.85 f'_c b} \\\\ &= \\frac{0.34\\text{ in}^2\\text{/ft} \\times 60000 \\text{ psi}}{0.85 \\times 3000\\text{psi} \\times12 \\text{ in/ft}}\\\\ &=0.667 \\text{ in} \\end{aligned} With such a small value of $a$, it's clear that our footing will be tension controlled and thus $\\phi = 0.90$. 3500 psi concrete. As previously discussed, shear reinforcement is usually avoided in footings and the concrete strength was already specified, so we choose to increase the thickness. Nevertheless, we see that $\\phi M_n > M_u$ so our design is adequate. DESIGN OF REINFORCED CONCRETE WALL - Compression member - In case where beam is not provided and load from the slab is heavy - When the masonry wall thickness is restricted - Classified as o plain concrete wall, when rein. All that's left here is to find the size and spacing required. EXAMPLE 11 - CAST-IN-PLACE CONCRETE CANTILEVER RETAINING WALL 2 2020 RESISTANCE FACTORS When not provided in the project-specific geotechnical report, refer to the indicated AASHTO sections. Note that we automatically calculate the depth to reinforcement - thus the increase in $d$ from using a smaller bar is automatically calculated which provides us with slightly more capacity! The wall is... Design Criteria. cmaa australia. Detailings of individual . The fluid level inside Resistance to eccentric compression 4. The doubly reinforced concrete beam design may be required when a beam’s cross-section is limited because of architectural or other considerations. Using Table 4, the wall can be adequately reinforced using No. ClearCalcs The design and detailing requirements for special reinforced concrete shear walls have undergone significant changes from ACI 318-11 to ACI 318-14. Md may also be taken Since we are now dealing with concrete design, we use the ACI 318-14 standard, which is based on LRFD design. Calculate ground bearing pressures. The first thing to do is to determine the width of our footing, which is determined by the allowable soil bearing capacity. Design a reinforced concrete to support a concrete wall in a relatively large building. We can clearly see that indeed we have a higher capacity. Soil: equivalent fluid pressure is 45 psf/ft (7.0 kN/m²/m) (excluding soil load factors), 10 ft (3.05 m) backfill height. f'c = 3000 psi fy = 60 ksi Natural Soil Development of Structural Design Equations. 9 bars at 72 in. Contact Us We pick a 13-inch thick footing and repeat the previous steps: \\begin{aligned} d &= 9.5 \\text{ in} \\\\ V_u &= 8.01 \\text{ kip/ft} \\\\ \\phi V_c &= 9.37\\text{ kip/ft} \\end{aligned} We see that the 1-inch increase both decreased $V_u$ and increase $\\phi V_c$ as we liked. Looking at the reinforcement section, the concrete cover is already set to 3 inches (the minimum for footings) and the steel strength is already 60 ksi. The allowable soil pressure is 5,000 psf and the its density is of 120 pcf. Wall is assumed to be located in the example focuses on the development length section and reinforcement... M_N > M_u $reinforced concrete wall design example our design is adequate masonry foundation wall, ft. A_S = 0.34\\text { in } ^2\\text { /ft }$ is 120... Mode which we need a minimum of 3 inches cover ( * 318-14! Notes Download ft below ground level > WÐ { ²þQà? ¼riJ @ íÓd ‹Íêç“àÖ strength, increase footing... Centres and are 200mm thick 's left here is to have three chambers each... 2012 lecture 13/2 Content: Introduction, definition of walls 1 footing 's thickness in design support to 4-storey... 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[ "# Quantum Physics By Bellac\n\n• February 2020\n• PDF TXT\n\nThis document was uploaded by user and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this DMCA report form.\n\n### More details\n\n• Words: 32,506\n• Pages: 100\n1\n\nSolutions of selected exercises from ‘Quantum Physics’ Michel Le Bellac\n\n2\n\nContents 1 Exercices from Chapter 1\n\n5\n\n2 Exercices from Chapter 2\n\n11\n\n3 Exercices from Chapter 3\n\n15\n\n4 Exercices from Chapter 4\n\n19\n\n5 Exercises from Chapter 5\n\n23\n\n6 Exercises from Chapter 6\n\n27\n\n7 Exercises from Chapter 7\n\n37\n\n8 Exercises from Chapter 8\n\n39\n\n9 Exercises from Chapter 9\n\n43\n\n10 Exercises from chapter 10\n\n51\n\n11 Exercises from Chapter 11\n\n59\n\n12 exercises from Chapter 12\n\n71\n\n13 Exercises from Chapter 13\n\n83\n\n14 Exercises from Chapter 14\n\n87\n\n15 Exercises from Chapter 15\n\n95\n\n3\n\n4\n\nCONTENTS\n\nChapter 1\n\nExercices from Chapter 1 1.6.1 Orders of magnitude 1. One must use particles whose wavelength λ be 1 ˚ A or less. We shall use λ = 1 ˚ A in numerical computations. In the case of photons, the energy is in eV Ephot =\n\n6.63 × 10−34 × 3 × 108 hc = 1.24 × 104 eV = 12.4 keV = λ 10−10 × 1.6 × 10−19\n\nIn the neutron case, we use p = h/λ, that is Eneut =\n\np2 h2 = = 8.2 × 10−2 eV = 82 meV 2mn 2mn λ2\n\nThis energy is of order of that of thermal neutrons, 25 meV. In the case of electrons, it is sufficient to multiply the preceding result by the mass ratio mn /me Eel = Eneut\n\nmn = 151 eV me\n\n2. The frequency of a wave with wavelength k = 1 nm is ω = 5 × 1012 rad.s−1 and the phonon energy ~ω = 3.3 meV. It is much easier to compare experimentally such an energy to that of a neutron with energy of a few ten meV, rather than to that of a photon with an energy of 10 keV in order to detect the creation of one phonon. 3. The mass of a fullerene molecule is M = 1.2 × 10−24 kg and its wavelength λ=\n\nh = 2.5 × 10−12 m mv\n\nThis wavelength is smaller than the molecule size by a factor ∼ 1/300. 4. The distance between the mass M1 and the molecule center-of-mass is r1 =\n\nM2 r0 M1 + M2\n\nand the molecule moment of inertia is I = M1 r12 + M2 r22 =\n\nM1 M2 2 r = µr02 M1 + M2 0\n\nThe rotational kinetic energy is given as a function of the angular momentum J = Iω by Erot =\n\n1 2 J2 Iω = 2 2I\n\n5\n\n6\n\nCHAPTER 1. EXERCICES FROM CHAPTER 1\n\nand choosing J = ~ leads to εrot = ~2 /(2I) εrot =\n\n~2 m ~2 = = 2 R∞ 2 2 2 2µr0 2µb a0 b µ\n\nusing R∞ = e2 /(2a0 ) and a0 = ~2 /(me2 ). 5. The elastic constant K is K=\n\n2 2cR∞ 4cmR∞ = 2 b 2 a0 ~2 b 2\n\nthat is ~ωv = 2\n\nr\n\nc b2\n\nr\n\nm R∞ µ\n\nIn the case of the HCl molecule, µ = 0.97mp and m/µ = 5.6 × 10−4 . Then b = 2.4 and c = 1.75, which are indeed numbers close to one. 6. The dimension of G is easily obtained by observing that Gm2 /r is an energy. One finds that this p −1 3 −2 dimension is given by M L T . The quantity ~c/G has the dimension of a mass, which gives for the Planck energy r ~c 2 c = 1.9 × 109 J = 1.2 × 1019 GeV EP = G and for the Planck length r ~ G lP = = 1.6 × 10−35 m c ~c The Planck energy is huge compared to the highest energies available in elementary particle physics (roughly 2 TeV=2000 GeV as of today), and, as a consequence, Planck’s length is quite tiny compared to the distances which are explored today in elementary particle physics which are ∼ 10−18 m.\n\n1.6.4 Neutron diffraction by a crystal 1. The incident wave arriving at point ~ri suffers a phase shift δinc = ~k · ~ri with respect to that arriving at point ~r = 0, and the scattered wave suffers a phase shift δsc = −~k ′ · ~ri with respect to the wave scattered by the nucleus at the point ~r = 0. 2. The scalar product ~ q · ~ri is given by q~ · ~ri = naqx + mbqy Using the formula for summation of a geometric series N −1 X n=0\n\nxn =\n\n1 − xN 1−x\n\none can, for example, evaluate the sum Σx =\n\nN −1 X n=0\n\ne−iqx na =\n\n1 − e−iqx aN sin qx aN/2 = e−iqx a(N −1)/2 1 − e−iqx a sin qx a/2\n\nfrom which one deduces F (aqx , bqy ) as given in the statement of the problem. 3. Suppose that qx differs very little from 2πnx /a, where nx is an integer: qx a = 2πnx + ε. Then \u0015 \u0014 qx aN εN N sin = ± sin = sin πnx N + ε 2 2 2 \u0014 \u0015 1 ε qx a = sin πnx + ε = ± sin sin 2 2 2\n\n7 The peak width is then ε ∼ 1/N and its height is obtained taking the limit ε → 0 sin2 εN/2 = N2 ε→0 sin2 ε/2 lim\n\nwhich gives an intensity within the peak ∼ N 2 × 1/N = N . The same calculation can be repeated in the y direction. 4. The condition for elastic scattering is ~k 2 = ~k ′ 2 = (~k + ~q)2 = k 2 + 2~q · ~k + q 2 that is q 2 + 2~ q · ~k = 0. Suppose that nx = 0 (or qx = 0) and thus kx′ = kx kx′ = kx\n\nky′ = ky −\n\n2πny b\n\nThe condition for elastic scattering is |ky′ | = |ky | which implies ky =\n\nπny b\n\nky′ = −\n\nπny b\n\nSince ky = k sin θB , where θB is the angle of incidence, one must have sin θB =\n\nπny bk\n\nOne finds solutions only if ny is small enough or k large enough. With only the first column of atoms, there would not be any constraint on kx , because kx′ would not be linked to kx through kx′ = kx + qx . One would then obtain diffraction maxima for any angle of incidence 5. The sum over the cells is N −1 M−1 X X\n\ne−i(2aqx n+2bqy m) = F (2aqx , 2bqy )\n\nn=0 m=0\n\nwhile the scattering amplitude due to the first cell is \u0011 \u0010 \u0011 \u0010 f1 1 + e−i(aqx +bqy ) + f2 e−iaqx + e−ibqy\n\nBecause of the argument of the F function, the condition for a diffraction peak is qx =\n\nπnx a\n\nqy =\n\nπny b\n\nand it follows that The final result is\n\n\u0003 \u0002 f = f1 1 + (−1)nx +ny + f2 [(−1)nx + (−1)ny ]\n\n• nx and ny even: f = 2(f1 + f2 ) • nx et ny odd: f = 2(f1 − f2 ) • nx even (odd) et ny odd (even): f = 0 6. When the lattice nodes are occupied randomly by the two kinds of atom, the lattice spacing is (a, b) and not (2a, 2b) as in the preceding question. One has in fact f1 = f2 and half of the diffraction peaks are lost.\n\n1.6.6 Mach-Zehnder interferometer\n\n8\n\nCHAPTER 1. EXERCICES FROM CHAPTER 1\n\n1. Let b1 (b2 ) the probability amplitude for finding the photon in the upper (lower) arm of the interferometer. The probability amplitudes a1 (a2 ) that the photon triggers the detector D1 (D2 ) are obtained by examining the transmission by the beam splitter S2 : for example, a1 is obtained by adding the amplitude rb1 where the photon originating from the upper arm is reflected by the beam splitter and the amplitude tb2 where the photon originating from the lower arm is transmitted by the beam splitter a1 a2\n\n= =\n\nrb1 eiδ + tb2 tb1 eiδ + rb2\n\nWe have made explicit the possibility of a variable phase shift δ in the upper arm. One calculates now b1 et b2 as functions of a0 b1 = ta0 b2 = ra0 and plugs the result in the preceding equations a1\n\n= rta0 1 + eiδ\n\na2\n\n= a0 t2 eiδ + r2\n\n\u0001\n\n\u0001\n\nUsing the values of t and r given in the statement of the problem and choosing the (arbitrary) normalization |a0 | = 1 (the photon has unit probability for arriving at S1 ) p1\n\n=\n\np2\n\n= =\n\n1 (1 + cos δ) 2 2 1 |a2 |2 = e2iα eiδ + e2iβ 4 1 [1 + cos(2α − 2β + δ)] 2\n\n|a1 |2 =\n\nBy letting δ vary, one can manage that all the photons are detected by D1 (or by D2 ). 2. We must have p1 + p2 = 1 whatever δ, because the photon must trigger one of the detectors, which implies that cos 2(α − β) = −1, that is α−β =\n\nπ mod nπ 2\n\n1.6.7 Neutron interferometer and gravity 1. and 2. Let us compute the probability amplitudes a1 and a2 for triggering D1 and D2 \u0001 a1 = a0 r2 t eiχ + 1 \u0001 a2 = a0 r t2 eiχ + r2\n\nand the probabilities by taking the modulus squared p1 p2\n\n=\n\nA(1 + cos χ)\n\n=\n\nB + A cos χ\n\nA = 2|r2 t|2 A′ = 2|r2 |Re t2 (r∗ )2 eiχ\n\nThe sum p1 + p2 must be independent of χ, so that A + A′ = 0, or\n\n\u0001\n\ncos 2(α − β) cos χ − sin 2(α − β) sin χ = − cos χ and thus α−β =\n\nπ mod nπ 2\n\n3. At elevation z the neutron energy is K = K0 + mgz if its energy is taken, by convention, to be K0 for z = 0. Its momentum is \u0012 \u0013 p p √ mgz p = 2mK = 2m(K0 + mgz) ≃ 2mK0 1 + 2K0\n\n9 The approximation is justified, because, for z = 1 m mgz ≃ 10−7 eV ≪ K0 ∼ 0.1 eV The variation ∆k of the wave vector is ∆k = k\n\nmgz 2K0\n\n∆k mgz = k 2K0\n\nOn a path with length L, the phase shift accumulated between the two arms, one at elevation z and the other at elevation z = 0 is mgzLk mgkS m2 gS ∆φ = ∆kL = = = 2 2K0 2K0 ~ k because zL is the rhombus area and2K0 = ~2 k 2 /m. The numerical values is ∆φ = .59 rad. 4. It is enough to replace z by z cos θ in the preceding results. The phase shift becomes χ = ∆φ =\n\nm2 gS cos θ ~2 k\n\nand one will thus observe oscillations in the neutron detection rate by varying θ.\n\n1.6.8 Coherent and incoherent scattering from a crystal 1. One observes that α2i = αi . If i = j, hα2i i = hαi i = p1 , while if i 6= j hαi αj i = hαi ihαj i = p21 the two results being summarized in hαi αj i = δij p1 + (1 − δij )p21 = p21 + p1 p2 δij 2. The scattering probabiilty by the crystal is X D\u0002 \u0003\u0002 \u0003E αi f1 + (1 − αi )f2 αj f1 + (1 − αj )f2 ei~q·(~ri −~rj ) h|ftot |2 i = i,j\n\n=\n\nX\u0002 i,j\n\n=\n\nX i,j\n\n\u0003 (p21 + p1 p2 δij )f12 + 2p1 p2 (1 − δij )f1 f2 + (p22 + p1 p2 δij )f22 ei~q·(~ri −~rj )\n\n(p1 f1 + p2 f2 )2 ei~q·(~ri −~rj ) + N p1 p2 (f1 − f2 )2\n\nThe first term gives rise to diffraction peaks, but the second one gives rise to a continuous background.\n\n10\n\nCHAPTER 1. EXERCICES FROM CHAPTER 1\n\nChapter 2\n\nExercices from Chapter 2 2.4.3 Determinant and trace 1. Let A(t) = A(0) exp(Bt). Let us compute the derivative d A(0)eBt = A(0)eBt B = A(t)B dt The solution of\n\ndA = BA(t) dt\n\nis A(t) = eBt A(0) 2. One remarks that for infinitesimal δt det eAδt ≃ det(I + Aδt) = 1 + δtTr A + O(δt)2 For example, for a 2 × 2 matrix \u0012 \u0013 1 + A11 δt A12 δt det = 1 + (A11 + A22 )δt + (A11 A22 − A12 A21 )(δt)2 A21 δt 1 + A22 δt Let g(t) = det[exp(At)] g ′ (t)\n\n\u0011 1 \u0010 det eA(t+δt) − det eAt δt→0 δt \u0001 1 1 = det eAδt − 1 det eAt = [δtTr A] det eAt = Tr Ag(t) δt δt\n\n=\n\nlim\n\nand one obtains for g(t) the differential equation\n\ng ′ (t) = [Tr A]g(t) =⇒ g(t) = etTr A using the boundary conditiom g(0) = 1. Setting t = 1 we find \u0002 \u0003 g(1) = eTr A = det eA\n\n2.4.10 Positive matrices 1. Let us decompose A into a Hermitian part and an antiHermitian part A=B+C\n\nB = B†\n\n11\n\nC = −C †\n\n12\n\nCHAPTER 2. EXERCICES FROM CHAPTER 2\n\nOne notes that (x, Cx) is pure imaginary (x, Cx) = (C † x, x) = (x, C † x)∗ = −(x, Cx)∗ while (x, Bx) is real. If we want (x, Ax) to be real and ≥ 0, it is necessary that C = 0. Let us give a more explicit proof. Let, for example, x = (x1 , x2 , 0, . . . , 0) Then (x, Cx) = x∗1 C12 x2 + x∗2 C21 x1 = 2i Im(x∗1 C12 x2 ) = 0 =⇒ C12 = 0 Since A is Hermitiian, it can be diagonalized. Let ϕ be an eigenvector of A, Aϕ = aϕ. The positivity condition implies (ϕ, Aϕ) = a||ϕ||2 ≥ 0 and thus a ≥ 0. 2. In the case of a real and antisymmetric matrix C T = −C\n\n(x, Cx) = x1 C12 x2 + x2 C21 x1 = x1 (C12 + C21 )x2 = 0 One can thus have a positive matrix of the form BT = B\n\nA=B+C\n\nC T = −C 6= 0\n\n2.4.11 Operator identities 1. Let us compute df /dt df = etA ABe−tA − etA BAe−tA = etA [A, B]e−tA dt The second derivative is computed in the same way, and the general case is obtained by recursion. 2. Let us compute dg/dt dg = etA (A + B)etB dt and use the result of the preceding question etA B = etA Be−tA etA = (B + t[A, B])etA Indeed, because of the commutation relations of [A, B] with A and B, the series expansion stops after the second term. We get the differential equation \u0011 dg \u0010 = A + B + t[A, B] g(t) dt\n\nTaking the commutation relations into account, this equation has the solution 1\n\n2\n\ng(t) = e(A+B)t+ 2 [A,B]t\n\nNote that this solution holds only because the commutator [[A, B], A + B] vanishes. Setting t = 1 g(1) = eA eB = eA+B+[A,B]/2 = eA+B e[A,B]/2\n\n2.4.12 A beam splitter 1.The condition that there are no losses reads |BD |2 + |BG |2 = |AD |2 + |AG |2\n\n13 The norm of the vector (AD , AG ) is conserved, which implies that the matrix R′ is unitary. The determinant of R′ then obeys | det R′ | = 1, which can be written det R′ = exp(iθ) 2. One defines R through R = ie−iθ/2 R′\n\ndet R = −e−iθ det R′ = −1\n\nThis redefinition corresponds to a global change of phase of the state vectors phase. One checks that with the form of R given in the statement of the problem R† R = I\n\ndet R = −|r|2 − |t|2 = −1\n\nLet us first choose ψ = (1, 0) Rψ =\n\n\u0012\n\n|r|eiχ |t|eiφ\n\n|t|e−iφ −|r|e−iχ\n\n\u0013 \u0012 \u0013 \u0012 iχ \u0013 1 |r|e = 0 |t|eiφ\n\nand then ψ = (0, 1) Rψ =\n\n\u0012\n\n|r|eiχ |t|eiφ\n\n|t|e−iφ −|r|e−iχ\n\n\u0013\u0012 \u0013 \u0012 \u0013 0 |t|e−iφ = 1 −|r|e−iχ\n\nOne deduces the phase shifts for the reflected (R) and transmitted (T ) waves D δR G δR\n\n= =\n\nχ −(χ − π)\n\nδTD = φ δTG = −φ\n\nδD = χ − φ δG = π − (χ − φ)\n\nso that δD + δG = π\n\n√ If the beam splitter is symmetric, one must have t = t∗ et r = −r∗ as well as |t| = |r| = 1/ 2 and δD = δG = π/2.\n\n14\n\nCHAPTER 2. EXERCICES FROM CHAPTER 2\n\nChapter 3\n\nExercices from Chapter 3 3.3.1 Decomposition and recombination of polarizations 1. Let e be the thickness of the plate. Since the separation of the centers of the beams is y = e tan α = 1.09 mm the beams are well separated. The difference in optical paths is \u0013 \u0012 n′e = 0.9248 mm δ = e no − cos α 2. The index difference is no − ne = 0.17102 and the thickness of the intermediary plate D=\n\n2 × 0.9248 = 10.815 mm no − ne\n\n3. Let β = 1/ cos α D= One infers from this the relative error δD/D\n\n2e(no − βn′e ) n0 − ne\n\nδD δno − δne δe δn0 − βδn′e − = + ′ D e n0 − βne no − ne In order to simplify the error calculation, we neglect the difference between ne and n′e δe (β − 1)(ne δno − no δne ) δD = + D e (no − ne )(no − βne which leads to\n\nand to\n\nδD ≃ 0.7 |δn| D\n\n|δD| ≃ 7 × 10−5 mm ≪ λ 5. The beam polarization is elliptic in the region where the two beams overlap, and linear in the region where they don’t.\n\n3.3.4 Other solutions of (3.45)\n\n15\n\n16\n\nCHAPTER 3. EXERCICES FROM CHAPTER 3\n\n1. The action of U on σx and σy is \u0012 0 † U σx U = e−iψ\n\neiψ 0\n\n\u0013\n\nU σy U =\n\n\u0012\n\n−ieiψ 0\n\n0 ie−iψ\n\n\u0013\n\nσz is clearly unchanged. 2. The possible solutions of (3.45) are cos(α − αx )\n\n= cos φ\n\ncos(α − αy )\n\n= sin φ\n\nα − αx = φ or α − αx = −φ π π α − αy = − φ or α − αy = φ − 2 2\n\nThe difference (α − αx ) − (α − αy ) = −(αx − αy ) must be independent of φ because αx and αy are given data independent of α. There are then two possible solutions • Solution 1\n\nα − αx = φ\n\nα − αy = φ −\n\nthat is αx = αy − and σx =\n\n\u0012\n\n0 eiαx\n\ne−iαx 0\n\n\u0013\n\nπ 2\n\nπ 2\n\nσy =\n\n\u0012\n\n0 ie−iαx\n\n−ie−iαx 0\n\n\u0013\n\nFrom question 1, this new form corresponds to a rotation of the axes by an angle αx about Oz. • Solution 2\n\nα − αx = −φ\n\nα − αy =\n\nπ −φ 2\n\nChoosing as a reference solution αx = 0 et αy = −π/2 \u0012 \u0013 \u0012 \u0013 0 1 0 i σx = σy = 1 0 −i 0 The change of sign of σy corresponds to an inversion of the Oy axis: one goes from a right handed referential to a left handed one. The other solutions are obtained from the reference solution by a rotation about the Oz axis.\n\n3.3.6 Exponentials of Pauli matrices 1. From (3.50) (~σ · pˆ)2 = I\n\n(~σ · pˆ)3 = (~σ · pˆ) . . .\n\nthe series expansion of the exponential is \u0013 \u0012 \u00132 \u0012 \u00133 \u0012 1 θ 1 θ θ θ I+ ~σ · pˆ · · · = I − i ~σ · pˆ + −i −i exp −i ~σ · pˆ 2 2 2! 2 3! 2 θ θ = I cos − i(~σ · pˆ) sin 2 2 Taking pˆ = (− sin φ, cos φ, 0) we get \u0012 \u0013 θ exp −i ~σ · pˆ = 2 =\n\nθ θ θ I cos + iσx sin φ sin − iσy cos φ sin 2 2 2 \u0013 \u0012 cos 2θ e−iφ sin θ2 cos θ2 eiφ sin 2θ\n\n17 2. We must have U\n\n=\n\na1 I + ia2 σz + ib2 σx + ib1 σy \u0012 θ θ nz I cos − i sin n 2 2 x + iny\n\n=\n\nnx − iny −nz\n\n\u0013\n\nand we deduce from this a1 = cos\n\nθ 2\n\na2 = −nz sin\n\nθ 2\n\nb2 = −nx sin\n\nθ 2\n\nb1 = −ny sin\n\nθ 2\n\nThese equations have solutions because a21 + a22 + b21 + b22 = 1 3. The product of two exponentials of Pauli matrices is ˆ e−iα(~σ ·ˆa) e−iβ(~σ·b) = cos α cos β − i sin α cos β(~σ · a ˆ) − i sin β cos α(~σ · ˆb) − sin α sin β[ˆ a · ˆb + i~σ · (ˆ a × ˆb)]\n\nOn the other hand sin ||αˆ a + βˆb|| ˆ [α(~σ · a ˆ) + β(~σ · ˆb)] e−i[α(~σ ·ˆa)+β(~σ·b)] = I cos ||αˆ a + βˆb|| − i ||αˆ a + βˆb|| In order to ensure the equality of the two factors, we must • get rid of the sines; • have cos α cos β = cos\n\np α2 + β 2\n\nOne can choose, for example\n\nα = 3π\n\nβ = 4π\n\nwith e−4iπσy = I\n\ne−3iπσx = −I\n\np α2 + β 2 = 5π e−i(3πσx +4πσy ) = −I\n\n3.3.9 Neutron scattering from spin 1/2 nuclei 1. When the nucleus spin does not flip, it is not possible to tell from which nucleus the neutron was scattered, and we must add the amplitudes X X ei~q·(~ri −~rj ) I = fa2 ei~q·~ri f = fa i,j\n\ni\n\n2. If the scattering is accompanied with a spin flip, it leaves the nucleus in a state which is different from its initial state. If all the nuclei had initially a down spin, the nucleus which scattered the neutron could be in principle identified (even though this identification would impossible in practice). Neutron scattering from a given nucleus rather than from another one corresponds to different nucleus final states, and we must add probabilities X fb2 = N fb2 I= i\n\n3. Let {αi } define the spin configuration in the crystal. If a neutron is scattered by the crystal in the configuration {αi }, the scattering amplitude is X X αi fb ei~q·~ri (αi fa + (1 − αi )fc ) ei~q·~ri + f= i\n\ni\n\n18\n\nCHAPTER 3. EXERCICES FROM CHAPTER 3\n\nWere the configuration {αi } fixed, the intensity would be X X 2 α2i fb2 (αi fa + (1 − αi )fc ) ei~q·(~ri −~rj ) + Iαi = i,j\n\ni\n\nObserving that αi = α2i the average values are hαi i = hα2i i = 1/2 and hαi αj i = 1/4 if i 6= j, whence X X\n\n\u0001 hαi ifb2 αi αj fa2 + 2αi (1 − αj )fa fc + (1 − αi )(1 − αj )fc2 ei~q·(~ri −~rj ) + I= i\n\ni,j\n\nWe then get the result given in the statement of the problem I=\n\nX 1 N ei~q·(~ri −~rj ) + (fa + fc )2 [(fa − fc )2 + 2fb2 ] 4 4 i,j\n\n4. From rotational invariance, we have, for example fa : neutron ↓ + nucleus ↑ → neutron ↓ + nucleus ↑ and a similar result for the two other amplitudes. One finds again the result of the preceding question if the neutrons are all polarized with a down spin. The result for unpolarized neutrons is obtained by taking the average of the result with spin up and spin down, and one again finds the result of question 3.\n\nChapter 4\n\nExercices from Chapter 4 4.4.4 Time evolution of a two-level system 1. The system of differential equations obeyed by c± (t) is ic˙+\n\n=\n\nAc+ + Bc−\n\nic˙−\n\n=\n\nBc+ − Ac−\n\n2. If |ϕ(t = 0)i is decomposed on the eigenvectors |χ± i as |ϕ(t = 0)i = λ|χ+ i + µ|χ− i then the time evolution is |ϕ(t)i = λe−iΩt/2 |χ+ i + µeiΩt/2 |χ− i Taking into account h+|χ+ i = cos θ/2 and h+|χ− i = − sin θ/2, one finds for c+ (t) c+ (t) = h+|ϕ(t)i = λe−iΩt/2 cos\n\nθ θ − µeiΩt/2 sin 2 2\n\n3. If c+ (0) = 0 λ cos\n\nθ θ − µ sin = 0 2 2\n\nand a possible solution is λ = sin\n\nθ 2\n\nµ = cos\n\nθ 2\n\nwhich gives the following for c+ (t) c+ (t) = − sin\n\n\u0011 Ωt θ θ \u0010 −iΩt/2 e − eiΩt/2 = i sin θ sin cos 2 2 2\n\nThe probability p+ (t) is p+ (t) = |c+ (t)|2 = sin2 θ sin2\n\nΩt Ωt B2 sin2 = 2 2 A + B2 2\n\n4. If c+ (0) = 1, a possible solution is λ = cos\n\nθ 2\n\nµ = − sin\n\n19\n\nθ 2\n\n20\n\nCHAPTER 4. EXERCICES FROM CHAPTER 4\n\nand one obtains for c+ (t) θ θ −iΩt/2 e + sin2 eiΩt/2 2 2 Ωt Ωt cos − i cos θ sin 2 2\n\ncos2\n\nc+ (t) = = The probability p+ (t) is p+ (t) = |c+ (t)|2 = cos2\n\nΩt Ωt Ωt + cos2 θ sin2 = 1 − sin2 θ cos2 2 2 2\n\n4.4.5 Unstable states 1. Let us use a series expansion of exp(−iHt/~) for small values of t c(t) = 1 −\n\ni 1 hHit − 2 hH 2 it2 + O(t3 ) ~ 2~\n\nso that |c(t)|2\n\nhH 2 i − hHi2 2 t + O(t3 ) ~2 (∆H)2 2 t + O(t3 ) = 1− ~2\n\n= 1−\n\n2. From (4.27) we derive, with the substitution A → P 1 dP ∆P∆H ≤ ~ 2 dt\n\np(t) = hPi(t)\n\nwhile\n\n∆P = (hP 2 i − hPi2 )1/2 = (hPi − hPi2 )1/2 =\n\nWe thus obtain the differential inequality\n\np p(1 − p)\n\ndp ∆H p dt ≥2 ~ p(1 − p)\n\nwhich integrates into\n\ncos−1 [1 − 2p(t)] ≥ 2 whence p(t) ≥ cos2\n\n\u0012\n\nt∆H ~\n\n\u0013\n\n∆H t ~\n\n0≤t≤\n\nπ~ 2∆H\n\n3. Inserting a complete set of eigenstates |ni of H in the expression for c(t), we obtain X c(t) = hϕ(0)|e−iHt/~ |nihn|ϕ(0)i n\n\nX\n\n=\n\nn\n\n|hϕ(0)|ni|2 e−iEn t/~\n\nLet the spectral function w(E) be the inverse Fourier transform of c(t) Z dt iEt/~ w(E) = e c(t) 2π Z dt iEt/~ X e |hϕ(0)|ni|2 e−iEn t/~ = 2π n X 2 = |hϕ(0)|ni| δ(E − En ) n\n\n21 The expectation values of H and H 2 are hHi\n\n=\n\nhH 2 i\n\n=\n\nX n\n\nZ\n\n2\n\nEn |hϕ(0)|ni| =\n\nZ\n\ndE E w(E)\n\ndE E 2 w(E)\n\n4. Use the following identity, obtained from a suitable contour in the complex x-plane Z +∞ π eitx = e−α|t| dx 2 2 x + α α −∞\n\n4.4.6 The solar neutrino enigma 1. The Hamiltonian reads, in the reference frame where the neutrino is at rest \u0013 \u0012 me −mµ 1 me + mµ m 2 H= I+ m −m c2 2 m − e2 µ Comparison with exercise 4.4.4 leads to the correspondence A→\n\nme − mµ 2\n\nB→m\n\ntan θ =\n\nB 2m → A me − mµ\n\n2. Since the states |ν1 i and |ν2 i are eigenvectors of H, the time evolution of |ϕ(t)i is θ θ |ϕ(t)i = cos e−iE1 t/~ |ν1 i − sin e−iE2 t/~ |ν2 i 2 2 Taking into account hνe |ν1 i = cos\n\nθ 2\n\nhνe |ν2 i = − sin\n\nθ 2\n\none finds for the amplitude ce (t) ce (t) = hνe |ϕ(t)i = e\n\n−iE1 t/~\n\n\u0013 \u0012 2 θ −i(E2 −E1 )t/~ 2 θ + sin e cos 2 2\n\nand for the probability |ce (t)|2 |ce (t)|2\n\n= =\n\nθ θ θ ∆Et θ + sin4 + 2 cos2 sin2 cos 2 2 2 2 ~ \u0013 \u0012 1 ∆Et ∆Et 1 − sin2 θ 1 − cos = 1 − sin2 θ sin2 2 ~ 2~\n\ncos4\n\n3. When p ≫ mc one obtains the following approximate expression for E \u00131/2 \u0012 m 2 c3 m 2 c2 ≃ cp + E = (m2 c4 + p2 c2 )1/2 = cp 1 + 2 p 2p and ∆E = E2 − E1 =\n\n∆m2 c3 (m22 − m21 )c3 = 2p 2p\n\nSubstituting the distance L travelled by the neutrino to ct, the oscillation takes the form \u0013 \u0012 ∆m2 c2 L sin2 2p~\n\n22\n\nCHAPTER 4. EXERCICES FROM CHAPTER 4\n\nIf half of an oscillation is observed between the Sun and the Earth ∆m2 c2 L =π 2p~\n\n∆m2 c4 =\n\n2π~c cp ≃ 7 × 10−11 eV2 L\n\nand thus ∆mc2 ∼ 10−5 eV.\n\n4.4.8 The neutral K meson system 1. Let us compute the product C −1 M C for a generic matrix M \u0012 \u0013 \u0012 \u0012 \u0013 a b d 0 1 σx = C = C −1 = σx = σx c d b 1 0\n\nc a\n\n\u0013\n\nThe commutation relation implies a = d and b = c. 2. The eigenvalues and eigenvectors of M are λ+\n\n= A+B\n\nλ−\n\n= A−B\n\n\u0011 1 \u0010 0 |K1 i = √ |K 0 i + |K i 2 \u0011 1 \u0010 0 |K1 i = √ |K i − |K 0 i 2\n\nThe time evolution of the states |K1 i et |K2 i is given by \u0013 \u0012 E1 t Γ1 t |K1 i − e−i(A+B)t |K1 i = exp −i ~ 2 \u0012 \u0013 E2 t Γ2 t e−i(A−B)t |K1 i = exp −i |K1 i − ~ 2\n\n|K1 (t)i = |K2 (t)i = Let us start at time t = 0 from\n\n1 1 |ϕ(t = 0)i = √ (c(0) + c(0)) |K1 i + √ (c(0) − c(0)) |K2 i 2 2 One gets at time t |ϕ(t)i\n\n= +\n\n\u0010 \u0011 1 (c(0) + c(0)) |K 0 i + |K 0 i e−i(E1 /~−iΓ1 /2)t 2 \u0010 \u0011 1 (c(0) − c(0)) |K 0 i − |K 0 i e−i(E2 /~−iΓ2 /2)t 2\n\nwhich gives the coeficients c(t) and c(t) c(t) = c(t) =\n\n1 (c(0) + c(0)) e−i(E1 /~−iΓ1 /2)t + 2 1 hK 0 |ϕ(t)i = (c(0) + c(0)) e−i(E1 /~−iΓ1 /2)t − 2\n\nhK 0 |ϕ(t)i =\n\n1 (c(0) − c(0)) e−i(E2 /~−iΓ2 /2)t 2 1 (c(0) − c(0)) e−i(E2 /~−iΓ2 /2)t 2\n\n3. In the case which is considered in the statement of the problem c(0) = 1, c(0) = 0 and c(t) =\n\ni 1 h −i(E1 /~−iΓ1 /2)t e − e−i(E2 /~−Γ2 /2)t 2\n\nThe probability to observe a K 0 meson at time t is \u0014 \u0015 1 −Γ1 t ∆Et p(t) = |c(t)|2 = e + e−Γ2 t − 2e−(Γ1 +Γ2 )t/2 cos 4 ~ with ∆E = E1 − E2 . One obtains the result given in the statement of the problem if Γ1 ≫ Γ2 .\n\nChapter 5\n\nExercises from Chapter 5 5.5.3 Butadiene 1.The matrix form of H is \n\n E0 −A 0 0  −A E0 −A 0   H=  0 −A E0 −A  0 0 −A E0\n\n2. Let us write the action of H on the vector |χi \u0010 \u0011 H|χi = E0 |χi − A c1 |ϕ2 i + · · · + cN −1 ϕN i + c0 |ϕ0 i \u0010 \u0011 − A c2 |ϕ1 i + · · · cN |ϕN −1 i + cN +1 |ϕN i =\n\nE0 |χi − A\n\nN \u0010 X\n\nn=1\n\n\u0011 cn−1 + cn+1 |ϕn i\n\n3. Taking into acount the postulated form for the coefficients cn , we get i c h i(n−1)δ e + ei(n+1)δ − e−i(n−1)δ − e−i(n+1)δ cn−1 + cn+1 = 2i \u0003 \u0002 c = 2 cos δ einδ − e−inδ = 2cn cos δ 2i The condition c0 = 0 is satisfied by construction. The condition cN +1 = 0 implies δs =\n\nπs N +1\n\ns = 0, 1, · · · , N\n\n4. From the results of question 3 we have H|χs i = E0 |χs i − 2A cos whence the values of the energy Es = E0 − 2A cos\n\nπs |χi N +1\n\nπs N +1\n\nLet us compute the normalization of |χs i \u0013 N \u0012 2πsn 1X πsn 1 − cos = X= sin N +1 2 n=1 N +1 n=1 N X\n\n2\n\n23\n\n24\n\nCHAPTER 5. EXERCISES FROM CHAPTER 5\n\nThe sum over n is readily computed \u0012 \u0013 N X 2iπsn 1 − e2iπs 2πsn = Re exp = Re cos − 1 = −1 N +1 N +1 1 − e2iπs/(N +1 n=1 n=1 N X\n\nand X = (N + 1)/2. The normalized vectors |χs i are then |χs i =\n\nr\n\nN X 2 sin(nδs )|ϕn i N + 1 n=1\n\n5. In the case of butadiene N = 4 cos\n\nπ = 0.809 5\n\ncos\n\n2π = 0.309 5\n\nwhich gives for the two lowest levels Es=1 = E0 − 1.62A\n\nEs=2 = E0 − 0.62A\n\nThe energy of the four π electrons is then E = 4E0 − 2(1.62A + 0.62A) = 4(E0 − A) − 0.48A The delocalization energy is −0.48A. 7. The coefficients of the normalized eigenvectors |χ1 i are (0.372, 0.601, 0.601, 0.372) while those of |χ2 i are\n\n(0.601, 0.372, −0.372, −0.601)\n\nThe order of the 1-2 bond is h i 1 + 2 hϕ1 |χ1 ihχ1 |ϕ2 i + hϕ1 |χ2 ihχ2 |ϕ2 i = 1.89\n\nwhich is very close to that of a double bond, while for the 2-3 bond h i 1 + 2 hϕ2 |χ1 ihχ1 |ϕ3 i + hϕ2 |χ2 ihχ2 |ϕ3 i = 1.45\n\nThis bond is weaker than the preceding one, which explains why it is closer to a single bond, and thus longer.\n\n5.5.5 The molecular ion H2+ 1. The potential V (x) is V (x) = −e2\n\n\u0012\n\n1 1 + |x + r/2| |x − r/2|\n\n\u0013\n\nIts value is −∞ when x = ±r/2 and its maximal value is −4e2 /r for x = 0. 2. l ≃ a0 , characteristic size of the hydrogen atom. 3. Eigenvalues and eigenvectors E+\n\n=\n\nE0 − A\n\nE−\n\n=\n\nE0 + A\n\n\u0011 1 \u0010 |χ+ i = √ |ϕ1 i + |ϕ2 i 2 \u0011 1 \u0010 |χ− i = √ |ϕ1 i − |ϕ2 i 2\n\n25 4. A is a tunnel effect transmission coefficient. The width of the potential barrier decreases as r decreases: the transmission coefficient increases when r decreases. 5. e2 /r is the (repulsive) potential energy between the two protons ′ E± (r) = E± (r) +\n\ne2 e2 = E0 ∓ A(r) + r r\n\n′ 6. The approximate expression for E+ (r) is ′ E+ (r)\n\n= E0 + e\n\n2\n\n\u0012\n\n1 − c e−b/r r\n\n\u0013\n\n= E0 + ∆E(r)\n\nLet us look for the minimum of ∆E(r) \u0012 \u0013 d∆E(r) 1 1 c −r/b c 2 =⇒ 2 = e−r0 /b =e − 2 + e dr r b r0 b One obtains ∆E(r0 ) = e whence b=\n\n2\n\n\u0012\n\n1 − c e−r0 /b r0\n\n12 6 r0 = a0 5 5\n\n\u0013\n\nc=\n\n=e\n\n2\n\n\u0012\n\n1 b − 2 r0 r0\n\n\u0013\n\n1.38 3 5/6 e ≃ 5a0 a0\n\nOne must have b > r0 for the H2+ ion to be a bound state.\n\n5.5.6 The rotating wave approximation in NMR The evolution equation for the state vector |ϕ(t)i ˜ in the rotating reference frame is i~\n\nd|ϕ(t)i ˜ dt\n\n= =\n\n~ ωσz |ϕ(t)i ˜ + e−iωσz t/2 He iωσz t/2 |ϕ(t)i ˜ 2 ~ ˜ ωσz |ϕ(t)i ˜ + H(t)| ϕ(t)i ˜ 2\n\n2. Let us study σ ˜± (t) iω d˜ σ± (t) = − e−iωσz t/2 [σz , σ± ]e iωσz t/2 = ∓iω˜ σ± (t) dt 2 because [σz , σ± ] = ±2σ± . Solving the differential equation gives σ ˜± (t) = e∓iωt σ± Let us rewrite H1 in terms of σ+ and σ− , using σx = σ+ + σ−\n\nwhence\n\nσy = −iσ+ + iσ−\n\n\u0001 ~ H1 = − ω1 (σ+ + σ− ) eiωt e−iφ + e−iωt eiφ 2 ˜ 1 (t) = H −\n\n\u0001 ~ − ω1 σ+ e−iφ + e−2iωt eiφ 2 \u0001 ~ ω1 σ− e2iωt e−iφ + eiφ 2\n\n26\n\nCHAPTER 5. EXERCISES FROM CHAPTER 5\n\nand in the rotating wave approximation, where one neglects terms in exp(±2iωt) ˜1 H\n\n\u0001 ~ω1 σ+ e−iφ + σ− eiφ 2 ~ = − ω1 (σx cos φ + σy sin φ) 2\n\n= −\n\n3. From equation (3.67), exp[−iθ(~σ · pˆ)/2] is the rotation operator of a spin 1/2 about an axis pˆ. By identification, we find pˆ = (cos φ, sin φ, 0) and θ = −ω1 t. The vector n ˆ being normalized (ˆ n2 = 1), we have Ωt Ωt ˜ exp(−iHt/~) = I cos − i(~σ · n ˆ ) sin 2 2 with ω1 δ ~σ · n ˆ = − σx + σz Ω Ω ˜ so that the matrix form of exp(−iHt/~) is ˜\n\ne−iHt/~ =\n\n\u0012\n\nδ Ωt cos Ωt 2 + Ω sin 2 iω1 Ωt Ω sin 2\n\ncos\n\niω1 Ωt Ω sin 2 Ωt δ Ωt 2 − Ω sin 2\n\n\u0013\n\nChapter 6\n\nExercises from Chapter 6 6.5.3 Properties of state operators 1. Let us consider a vector |ϕi of the form |ϕi = (0, · · · , ai , 0, · · · , 0, aj , 0, · · · , 0) Positivity of ρ gives hϕ|ρ|ϕi ≥ 0, which implies that the 2 × 2 sub-matrix A \u0013 \u0012 ρii ρij A= ρji ρjj must be positive. On the other hand, (ρii + ρjj ), which is the sum of the eigenvalues of A, obeys (ρii + ρjj ) ≤ 1. One deduces that the product of the eigenvalues of A must be less than 1/4 0 ≤ ρii ρjj − |ρij |2 ≤\n\n1 4\n\nIf ρii = 0, this implies ρij = 0. 2. The condition for a maximal test with 100% success implies that there exists a vector |ϕi such that Tr ρPϕ = 1, with Pϕ = |ϕihϕ|, and thus hϕ|ρ|ϕi = 1. We choose an orthonormal basis, where, for example, |ϕi is the first basis vector, |ϕi ≡ |1i. In that case, the diagonal elements of ρ obey ρ11 = 1, ρii = 0, i 6= 1 because the test |ii, |ii = 6 |1i has zero probability of success. From the preceding question, all the nondiagonal elements vanish, ρij = 0, i 6= j and ρ = |ϕihϕ| = |1ih1|.\n\n6.5.4 Fine structure and Zeeman effect in positronium 1. The reduced mass is half that of the electron, and the energy levels are deduced from (1.36) En = −\n\nR∞ 2n2\n\n2. Let us work out explicitly the action of σx and σy on the vectors |ε1 ε2 i σ1x σ2x | + +i = σ1x σ2x | + −i =\n\nσ1x σ2x | − +i = σ1x σ2x | − −i =\n\n| − −i | − +i\n\nσ1y σ2y | + +i = −| − −i σ1y σ2y | + −i = | − +i\n\n| + −i | + +i\n\nσ1y σ2y | − +i = | + −i σ1y σ2y | − −i = −| + +i\n\n27\n\n28\n\nCHAPTER 6. EXERCISES FROM CHAPTER 6\n\nand σ1z σ2z |ε1 ε2 i = ε1 ε2 |ε1 ε2 i, whence the action of ~σ1 · ~σ2 ~σ1 · ~σ2 | + +i = ~σ1 · ~σ2 | + −i =\n\n| + +i 2| − +i − | + −i\n\n~σ1 · ~σ2 | − −i =\n\n| − −i\n\n~σ1 · ~σ2 | − +i =\n\n2| + −i − | − +i\n\n3. From the results of the preceding question we get ~σ1 · ~σ2 |Ii = |Ii\n\n~σ1 · ~σ2 |IIIi = |IIIi\n\nas well as ~σ1 · ~σ2 |IIi = ~σ1 · ~σ2 |IV i = 4. The projectors  1  0 P1 =   0 0\n\n1 √ (| + −i + | − +i) = |IIi 2 1 −3 √ (| + −i − | − +i) = −3|IV i 2\n\nP1 et P−3 as well as ~σ1 · ~σ2   0 0 0  1 0 0   P−3 =    0 1 0 0 0 0\n\nIf P1 = λI + µ~σ1 · ~σ2 , we must have\n\nare diagonalized in the basis {|Ii, |IIi, |IIIi, |IV i}    0 0 0 0 1 0 0 0  0 1 0 0  0 0 0 0    ~σ1 · ~σ2 =    0 0 1 0  0 0 0 0 0 0 0 1 0 0 0 −3\n\nλ + µ = 1 and λ − 3µ = 0 that is, λ = 3/4 and µ = 1/4. One deduces P1 =\n\n1 (3I + ~σ1 · ~σ2 ) 4\n\nP−3 =\n\n1 (I − ~σ1 · ~σ2 ) 4\n\n5. We immediately get P12 | + +i = | + +i and P12 | − −i = | − −i, while P12 | + −i =\n\n1 (| + −i + 2| − +i − | + −i) = | − +i 2\n\nand in general P12 |ε1 ε2 i = |ε2 ε1 i. 6. We know the eigenvalues and eigenvectors of ~σ1 · ~σ2 , and thus those of H • |Ii, |IIi, |IIIi are eigenvectors H with eigenvalue E0 + A • |IV i is eigenvector of H with eigenvalue E0 − 3A 7. The gyromagnetic of the positron has a sign opposite to that of the electron: γe+ = −γe− = −qe /m. The total Hamiltonian reads ~ = H0 − H = H0 − (~ µe− + µ ~ e+ ) · B\n\nqe ~ (σ1z − σ2z ) 2m\n\nLet us examine the action of the operator (σ1z − σ2z ) on the basis vectors of H0 (σ1z − σ2z )| + +i = 0\n\n(σ1z − σ2z )| − −i = 0\n\n(σ1z − σ2z )|IIi = 2|IV i\n\n29 and\n\n\u0013 \u0012 qe ~ = 2Ax hII|H1 |IV i = hIV |H1 |IIi = 2 − 2m\n\nThe matrix form of H is in the basis {|Ii, |IIi, |IIIi, |IV i}  E0 + A 0 0 0  0 E + A 0 2Ax 0 H=  0 0 E0 + A 0 0 2Ax 0 E0 − 3A\n\n   \n\nTwo eigenvectors are obvious: |Ii and |IIIi with eigenvalues E0 + A. The other two are obtained through diagonalization of the 2 × 2 matrix \u0012 \u0013 1 2x H ′ = E0 I + A = E0 I + AM 2x −3 The eigenvalue equation of M is which gives the values of the energy\n\nλ2 + 2λ − (3 + 4x2 ) = 0 p E± = E0 − A ± 2A 1 + x2\n\nWhen x = 0 we recover the values E0 + A and E0 − 3A, while for |x| → ∞, the eigenvectors tend to those of (σ1z − σ2z ) with eigenvalues ±2Ax.\n\n6.5.4 Spin waves and magnons 1. Since the eigenvalues of (~σn · ~σn+1 ) lie between −3 and +1, we must have E≥\n\n1 1 NA − NA = 0 2 2\n\nIf the state vector is such that (~σn · ~σn+1 ) = 1, we obtain the ground state E0 = 0. This state vector is Φ0 = | + + + · · · + +i because (~σn · ~σn+1 )|Φ0 i = |Φ0 i 2. The operator Pn,n+1 exchanges spins n et n+ 1: in the case of two spins, we have seen in the preceding question that P12 | + +i = | + +i P12 | + −i = | − +i\n\nP12 | − −i = | − −i P12 | − +i = | + −i\n\nand the number of up spins minus the number of down spins must stay unchanged. The eigenvectors of H are thus such that the number of up spins minus the number of down spins is a constant. In particular, for the state |Ψn i, this constant is N − 1. The operator I − Pn,n+1 applied to |Ψn i gives zero on any pair of up spins, and only the pairs (n − 1, n) and (n, n + 1) are going to give a nonzero result. Since Pn−1,n , for example, exchanges spins n − 1 and n Pn−1,n | + + + + − + + +i = | + + + − + + + +i that is we obtain\n\nPn−1,n |Ψn i = |Ψn−1 i (I − Pn−1,n )|Ψn i (I − Pn,n+1 )|Ψn i\n\nPn−1,n |Ψn−1 i = |Ψn i = |Ψn i − |Ψn−1 i = |Ψn i − |Ψn+1 i\n\n30\n\nCHAPTER 6. EXERCISES FROM CHAPTER 6\n\nThis gives the action of H on |Ψn i \u0010 \u0011 H|Ψn i = −A |Ψn−1 i + |Ψn+1 i − 2|Ψn i 3. Let us work out the action of H on |ks i H|ks i =\n\nN −1 X n=0\n\neiks nl H|Ψn i = −A\n\nN −1 X n=0\n\n\u0010 \u0011 eiks nl |Ψn−1 i + |Ψn+1 i − 2|Ψn i\n\nOn the other hand X X X eiks nl |Ψn−1 i = eiks (n−1)l eiks l |Ψn−1 i = eiks l eiks nl |Ψn i = eiks l |ks i n\n\nn\n\nn\n\nWe thus have H|ks i = 2A(1 − cos ks l)|ks i The eigenfrequencies are |ks | → 0 ωk ≃ (Al2 )ks2\n\nωk = 2A(1 − cos ks l)\n\nIt is also interesting to apply the method of § 5.1.2, observing that H can be cast into the form \u0001 H = −A UP + UP−1 − 2\n\nwhere UP makes a circular permutation n → n + 1 and to look for eigenvectors |Ψs i of UP . Writing X |Ψs i = csn |Ψn i UP |Ψs i = eiδs |Ψs i n\n\nwith δs =\n\n2πs = ks l N\n\nWe must have cn+1 = eiδs cn\n\n6.5.7 Calculation of E(ˆ a, ˆb) 1. Let us compute, for example, the amplitude a+− (θ) \u0011 \u0010 \u0011 1 \u0010 1 θ θ θ a+− (θ) = h+ ⊗ [−, ˆb]|Φi = − sin h+ + | + cos h+ − | √ | + −i − | − +i = √ cos 2 2 2 2 2 2. The rotation operator by an angle θ about an axis n ˆ of the ensemble of two spins is, from (3.67) \u0015 \u0014 \u0001 θ n Unˆ (θ) = exp − ~σ (a) + ~σ (b) ·ˆ 2\n\nThe rotational invariance of |Φi implies\n\nUnˆ (θ)|Φi = |Φi =⇒ that is\n\nh\n\n\u0001 i ~σ (a) + ~σ (b) · n ˆ |Φi = 0\n\n\u0001 \u0001 ~σ (a) · n ˆ |Φi = − ~σ (b) · n ˆ |Φi\n\nAs a consequence \u0001 \u0001 \u0001 \u0001 hΦ| ~σ (a) · a ˆ ⊗ ~σ (b) · ˆb |Φi = −hΦ| ~σ · a ˆ ~σ · ˆb |Φi = −hΦ|ˆ a · ˆb + i~σ · (ˆ a × ˆb)|Φi = −ˆ a · ˆb = − cos θ\n\n31 because the expectation value of ~σ vanishes in a rotational invariant state.\n\n6.5.8 Bell inequalities for photons 1. Let us work out the explicit form of the vector |θθ⊥ i \u0001 \u0001 |θθ⊥ i = cos θ|xi + sin θ|yi − sin θ|xi + cos θ|yi\n\n= − sin θ cos θ|xxi + cos2 θ|xyi − sin2 θ|yxi + sin θ cos θ|yyi\n\nTo obtain |θ⊥ θi, it is enough to exchange x ↔ y |θ⊥ θi = − sin θ cos θ|xxi + cos2 θ|yxi − sin2 θ|xyi + sin θ cos θ|yyi and, subtracting the the second equation from the first we get |θθ⊥ i − |θ⊥ θi = |xyi − |yxi 2. For photon 1 \u0001 1 |xi = √ −|Ri + |Li 2\n\nFor photon 2 we must change |yi into −|yi\n\n\u0001 1 |xi = √ −|Ri + |Li 2\n\n\u0001 i |yi = √ |Ri + |Li 2 \u0001 i |yi = − √ |Ri + |Li 2\n\nand one finds for photons propagating in opposite directions\n\n\u0001 i |Φi = √ |RRi − |LLi 2\n\nRemark: if the two photons propagate in the same direction, as is the case for pairs obtained from parametric conversion, we have \u0001 i |Φi = − √ |RLi − |LRi 2 In the two cases, one checks that the Oz component of angular momentum vanishes: (Σ1z + Σ2z )|Φi = 0, where Σz is given by (3.26). 3. Rotational invariance allows us to choose n ˆ α along Oz while n ˆ β makes an angle (β − α) with the Ox axis. Let us define φ = β − α and the amplitude a++ (φ) for finding |Φi in the state |x ⊗ φi = cos φ|xxi + sin φ|xyi is\n\nand thus\n\n\u0011\u0010 \u0011 1 \u0010 1 a++ (φ) = √ cos φhxx| + sin φhxy| |xyi − |yxi = √ sin φ 2 2 p++ (φ) =\n\n1 sin2 φ 2\n\nUsing symmetry properties in the exchange + ↔ − p++ (φ) = p−− (φ) =\n\n1 sin2 φ 2\n\nSince the sum of all probabilities must add up to one p+− (φ) = p−+ (φ) =\n\n1 cos2 φ 2\n\n32\n\nCHAPTER 6. EXERCISES FROM CHAPTER 6\n\nThis gives E(α, β) = (p++ + p−− ) − (p+− − p−+ ) = − cos 2φ = cos 2(β − α) In order to get maximal violation of Bell’s inequalities, one must use angles that are half of those used for spin 1/2. 4. We find\n\n\u0001 \u0001 1 1 |Ψi = √ |θθi + |θ⊥ θ⊥ i = √ |RRi + |LLi 2 2 Applying Σz again shows that the Oz component of the angular momentum of this state vanishes.\n\n6.5.9 Two photon interference 1. The dispersion of the Ox component of the wave vector is ∆kx ≃ 1/d, as the vertical position is uncertain by ±d/2. 2. One recovers a standard interference calculation in optics. The difference of optical path for the photon 1 going through the upper slit is, with θ = l/(2D) δ(x, y) − δ(0, 0) = −\n\nl (x + y) = −θ(x + y) 2D\n\nand the phase shift\n\n2π (δ(x, y) − δ(0, 0)) = −kθ(y + x) λ The amplitude for detecting the photon 1 at point y is proportional to \u0014 \u0012 \u0013 \u0012 \u0013\u0015 1 2iπ 2iπ exp (δ(x, y) − δ(0, 0) + exp − (δ(x, y) − δ(0, 0) = cos kθ(y + x) 2 λ λ φ(x, y) − φ(0, 0) =\n\n3. If the pair is emitted at point x, the probability amplitude for detection in coincidence is obtained by multiplying the probability amplitudes for detecting photon 1 at y and photon 2 at z a(x|y, z) = cos[kθ(y + x)] cos[kθ(x + z)] 4. Since it is not possible to know the vertical position where the photon pair has been emitted in the plate CD, one must add the amplitudes corresponding to all possible positions of this emission Z 1 d/2 a(y, z) = dx cos kθ(y + x) cos kθ(z + x) d −d/2 Z d/2 h i 1 dx cos[kθ((y + z + 2x)] + cos kθ(y − z) = 2d −d/2 \u0014 \u0015 1 1 sin kθd cos[kθ(y + z)] + d cos kθ(y − z) = 2d kθ 5. If d ≫ 1/(kθ), the second term of a(y, z) dominates over the first one and the probability p(y, z) is obtained by taking the modulus squared of a p(y, z) ∝ cos2 [kθ(y − z)] Observation of photon 1 at y determines the interference pattern of 2: if one observes photon 2 in coincidence with photon 1, one will observe an interference pattern. However, if only one photon is observed, there is no interference: integrating the probability p(y, z) over y, the result is a constant with respect to the variable z \u0013\u0015 \u0014 \u0012 Z d/2 1 I(z) = dy cos2 [kθ(y − z)] = cst 1 + O kθd −d/2\n\n33 6. In the limit d ≪ 1/(kθ), the probability becomes i2 1 h d cos[kθ(y + z)] + d cos[kθ(y − z)] 4d2 = cos2 (kθy) cos2 (kθz)\n\np(y, z) ∝\n\nwhich corresponds to a product of two independent interference patterns. The position at which the pair is emitted is very precisely fixed, so that the angular aperture for each photon is very large. There is no longer any constrain from momentum conservation. Entanglement has no influence: if photon 1, for example, passes through the upper slit, there is no guarantee that that photon 2 passes through the lower slit. The spread in transverse momentum is too large to allow one photon to tag the trajectory of the other one. The condition d ≪ λ/θ is the condition for a good fringe visibility in a standard Young’s slit experiment: this condition is complementary to that of trajectory tagging. Conversely, if the emission position is very uncertain, momentum conservation allows trajectory tagging. 8. Let r be the amplitude for reflection by the plates S and S ′ , and t the transmission amplitude. The simultaneous detection of the two photons by c et c′ implies that, either the two photons are reflected (the left hand photon taking the upper path and the right hand photon the lower path), or that the two photons are transmitted. The condition d ≫ 1/(kθ) must clearly be verified. The probability amplitude is obtained by summing over the two paths, because the paths are not distinguishable \u0001\u0003 1 \u0002 iα \u0001 re r + t teiβ a(c, c′ ) = √ 2\n\n√ If r = it and |t| = |r| = 1/ 2\n\np(c, c′ ) = |a(c, c′ )|2 = In a similar way, one finds a(c, d′ ) p(c, d′ )\n\n2 α−β 1 1 − eiα + eiβ = sin2 8 2 2\n\n\u0001 1 √ rt eiα + eiβ 2 1 α−β = |a(c, d′ )|2 = cos2 2 2 =\n\nThe detection of a single photon does not lead to any interference p(c) = p(c, c′ ) + p(c, d′ ) =\n\n1 2\n\nA symmetry argument gives at once the other probabilities p(d, d′ ) =\n\n1 α−β sin2 2 2\n\np(c′ , d) =\n\n1 α−β cos2 2 2\n\nwhence the quantity E(α, β) E(α, β) = sin2\n\nα−β α−β − cos2 = − cos(α − β) 2 2\n\nOne will thus get a maximal violation of Bell’s inequalities with a choice of angles (α, β) identical to that of spin 1/2.\n\n6.5.10 Interference of emission times 1. Since the coherence length of the converted photons is small compared to the optical path difference between the two arms, one can certainly not observe interference in an individual interferometer. But there exists a deeper reason which will be explained at the end of the next question.\n\n34\n\nCHAPTER 6. EXERCISES FROM CHAPTER 6\n\n2. One may write four different probability amplitudes for the joint detection of photons in D1 et D2 (S = short, L = long) A = ′\n\nA\n\n=\n\na1S a2S e\n\na1L a2L\n\nB = a1S a2L B ′ = eiδ a1L a2S\n\nAmplitude aiS corresponds to a photon path (i) going through the shorter arm of the interferometer, aiL to the path which follows the longer arm. Measuring the arrival times of the photons allows one to distinguish between (SL) and (LS); even if the emission time of the pair is unknown, the photon taking the upper arm arrives earlier than that taking the shorter arm. For example, in the (CL) case, photon 1 arrives at D1 0.7 ns before the second photon at D2 , which is much larger than the resolution time 0.1 ns of the detectors. On the contrary, the experimental set up does not allow one to distinguish, even in principle, between paths (SS) et (LL). We must then add the amplitudes for these paths to obtain the probability of detection in coincidence p(D1 , D2 ) = |A + A′ |2 = |a1S a2S + eiδ a1L a2L |2 which clearly shows a dependence with respect to to δ. In the experiment described in the statement of the problem, the coincidence window is less than the photon travel time, which allows one to distinguish paths (LS) and (SL) from paths (LL) et (SS). However, this condition is not essential for observing interference; if it were not realized, one would simply add a background noise p(D1 , D2 ) = |B|2 + |B ′ |2 + |A + A′ |2 corresponding to the first two terms which do not interfere in the preceding equation. Another important observation is that, if one suppresses the beam splitters of the left hand interferometer, one still has an information on the travel time: if the left hand photon arrives before the right hand one, we know that the photon took the the longer arm. There is thus no dependence with respect to δ and no interference. The information available on the path followed by the photon in the right hand interferometer erases any possibility of interference, even if the coherence length is smaller than ∆l. In fact, it is not even necessary that detector D2 be present! It is enough that the information on the arrival time be available in principle, and, as we often emphasized, it is not necessary that the arrival times are effectively observed! As long as the information on the arrival times is available in principle, and it is available because of entanglement, in no case can we have interference in one individual interferometer.\n\n6.5.11 The Deutsch algorithm We find for the vector |Ψi defined in Fig. 6.19 1 \u0011 \u0001 \u0001 1\u0010 X \u0001 \u0001 \u0001 1 |xi ⊗ |0i − |1i |Ψi = H|0i ⊗ H|1i = |0i + |1i ⊗ |0i − |1i = 2 2 x=0\n\nWe apply Uf to this state with the following result: \u0001 \u0001 1. If f (x) = 0, then |0i − |1i → |0i − |1i ; \u0001 \u0001 \u0001 2. If f (x) = 1, then |0i − |1i = |1i − |0i → − |0i − |1i , or, to summarize,\n\n\u0001 \u0001 |0i − |1i → (−1)f (x) |0i − |1i .\n\nThe state Uf |Ψi is then a tensor product (and not an entangled state) Uf |Ψi =\n\n1 \u0011 \u0001 1\u0010 X (−1)f (x) |xi ⊗ |0i − |1i 2 x=0\n\n(6.1)\n\n35 The net result for the input register is Uf\n\n|xi −−→ (−1)f (x) |xi The state of the qubit of the input register then is \u0001 1 \u0010 |ϕi = √ (−1)f (0) |0i + (−1)f (1) |1i . 2\n\nBefore measuring the input register, we apply a Hadamard gate (see Fig. 6.19): H|ϕi\n\n= =\n\n\u0001 \u0001\u0003 1\u0002 (−1)f (0) |0i + |1i + (−1)f (1) |0i − |1i 2 \u0003 \u0003 1\u0002 1\u0002 (−1)f (0) + (−1)f (1) |0i + (−1)f (0) − (−1)f (1) |1i. 2 2\n\nIf measurement of the qubit gives |0i, then f (0) = f (1), i.e., the function is a ‘constant’ one. If it gives |1i, then f (0) 6= f (1) and the function is a ‘balanced’ one. The important point is that quantum parallelism has allowed us to bypass the explicit calculation of the function f (x); the measurement of a single qubit contains the two possible results.\n\n36\n\nCHAPTER 6. EXERCISES FROM CHAPTER 6\n\nChapter 7\n\nExercises from Chapter 7 7.4.3 Canonical commutation relations 1. If we assume that B is a bounded operator, we can define B ′ = B/||B||, ||B ′ || = 1, and A′ = A||B|| without modification of the commutation relations: [A′ , B ′ ] = iI. It is thus legitimate to assume that ||B|| = 1. Let us use a recursion by assuming that [B, An ] = inAn−1 We then have [B, An+1 ] = [B, AAn ] = A[B, An ] + [B, A]An = i(n + 1)An which shows the validity of our starting hypothesis. Let us assume that A is bounded, and let ||A|| be its norm. We get, using the inequality which is valid for two operators C et D ||C|| ||D|| ≥ ||CD|| the relation 2||An || ||B|| ≥ ||BAn − An B|| = n||An−1 || whence ||An || ≥\n\nn ||An−1 || 2\n\nWe deduce the following bounds for ||An || n ||An−1 || ≤ ||An || ≤ ||A|| ||An−1 || since ||An || ≤ ||A|| ||An−1 || 2 and thus ||A|| ≥ n/2. It is impossible for A to be bounded. 2. The problem is that if A is not bounded, the vector B|ϕi does not belong to the domain of A and the product AB|ϕi is not defined: we cannot apply Hermitian conjugation and write hϕ|AB|ϕi = hAϕ|Bϕi If the Hilbert space is identified, for example, with L(2) [0, 1] and if B = X, which is bounded on that space, while A = AC defined in § 7.2.2, (Xϕ)(x) = xϕ(x) = ψ(x) does not belong to the domain of AC which is such that ϕ(1) = Cϕ(0), |C| = 1: the boundary conditions of ψ are ψ(1) = ϕ(1) = Cϕ(0) while ψ(0) = 0 and ψ(1) 6= Cψ(0). The difficulty is immediately seen in the x representation \u0013∗ \u0013 Z 1\u0012 \u0012 Z 1 ∂ϕ(x) ∂ ∗ i xϕ(x) xϕ(x) = 6 dx ϕ (x) i ∂x ∂x 0 0\n\n37\n\n38\n\nCHAPTER 7. EXERCISES FROM CHAPTER 7\n\nthe difference being |ϕ(1)|2 . 3. The function ϕ(x) = ei(2πn+α)x\n\nC = eiα\n\nobeys AC ϕ(x) = (2πn + α)ϕ(x)\n\nϕ(1) = Cϕ(0)\n\nIt is thus a normalizable eigenvector with eigenvalue (2πn + α) of AC , which belongs to the domain of AC . The von Neumann theorem does not apply because AB|ϕi is not defined whatever |ϕi ∈ H, while the domain of definition of AB should be dense in H.\n\nChapter 8\n\nExercises from Chapter 8 8.5.2 Rotations and SU(2) 1. Let us start from the most general 2 × 2 matrix \u0012 a U= c and compute U † U which must be identical to I \u0012 |a|2 + |b|2 U †U = ca∗ + db∗\n\nb d\n\n\u0013\n\nac∗ + bd∗ |c|2 + |d|2\n\n\u0013\n\n=\n\n\u0012\n\n1 0\n\n0 1\n\n\u0013\n\nThis gives |a|2 + |b|2 = |c|2 + |d|2 = 1\n\nFrom c = −b∗ d/a∗ and det U = ad − bc = 1 one deduces that d = a∗ . 2. To order τ U † U = (I + iτ † )(I − iτ ) ≃ I − i(τ − τ † )\n\nand the condition U † U = I implies τ = τ † . Moreover, the condition det U = 1 implies Tr τ = 0. The decomposition (3.54) together with the condition Tr τ = 0 allows us to write 3\n\nτ=\n\n1X θi σi 2 i=1\n\nwhere the angles θi are infinitesimal since τ itself is infinitesimal. 3. Since θ/N is infinitesimal. we may write \u0012 \u0013 θ iθ Unˆ =I− (~σ · n ˆ) N 2N and using lim\n\nN →∞\n\nwe deduce\n\n\u0010\n\n1−\n\nx \u0011N = e−x N\n\n\u0015 \u0014 θ ˆ) Unˆ (θ) = exp −i (~σ · n 2\n\n~ : det V = −V ~ 2 and since 4. The determinant of V is equal to minus the length squared of the vector V det(U VU −1 ) = det V\n\n39\n\n40\n\nCHAPTER 8. EXERCISES FROM CHAPTER 8\n\n~ 2 = det W = V ~ 2 , which shows that the transformation preserves the lengths of vectors. It is we obtain W thus, either a rotation, or a rotation combined with a parity operation. 5. Since W is Hermitian and has zero trace, because Tr (U VU −1 ) = Tr V ~ = ~σ · V~ (θ). we may write W = ~σ · W d ~ (θ) = − i [~σ · n ~ (θ)] = ~σ · (ˆ ~σ · V ˆ , ~σ · V n × V~ (θ)) dθ 2 ~ (θ) is deduced from V ~ through a rotation by an angle θ where we have used (3.52), which shows that V about n ˆ ~ dV ~ (θ) =n ˆ×V dθ To any rotation Rnˆ (θ) correspond two matrices Unˆ : Unˆ (θ) and Unˆ (θ + 2π) = −Unˆ (θ).\n\n8.5.4 The Lie algebra of a continuous group 1 Since g(θ = 0) = I, the composition law reads g(θ)I = g(θ) = g(f (θ, 0) =⇒ fa (θ, 0) = θa We write an expansion to order θ2 of fa (θ, θ) 2\n\n3\n\nfa (θ, θ) = θ a + θa + λabc θb θc + λabc θ b θc + fabc θ b θc + O(θ3 , θ2 θ, θ θ , θ ) The condition fa (θ, θ = 0) = θa implies λabc = 0 and similarly λabc = 0. 2\n\n3\n\n3. One the one hand, we have, neglecting terms of order (θ3 , θ2 θ, θ θ , θ ) 1 U (θ)U (θ) = I − iθa Ta − iθa Ta − (θ b θc + θb θc )Tbc − θa θb Ta Tb 2 and, on the other hand 1 U (f (θ, θ)) = I − iTa (θa + θa + fabc θ b θc ) − (θb + θb )(θc + θc )Tbc 2 Relabeling the summation indices and taking into account the symmetry property Tbc = Tcb , because Tbc = −\n\n∂ 2 U ∂θa ∂θb θa =θb =0\n\nthe comparison between the two expressions gives\n\nθb θc Tb Tc = ifabc Ta θb θc + θb θc Tbc We deduce from this Tb Tc Tc Tb\n\n= =\n\nTbc + ifabc Ta Tbc + ifacb Ta\n\nand, subtracting the second equation from the first one [Tb , Tc ] = i[fabc − facb ]Ta The structure constant Cabc is Cabc = [fabc − facb ] = −Cacb\n\n41\n\n8.5.5 The Thomas-Reiche-Kuhn sum rule 1. From the general relation (see(8.41)) [X, f (P )] = i~\n\n∂f ∂P\n\nwe deduce [P 2 , X] = −2i~P and [[P 2 , X], X] = −2i[P, X] = −2~2 whence\n\n\u0014\n\n\u0015 i~ P2 + V (X), X = [H, X] = − P 2m m\n\n[[H, X], X] = −\n\n~2 m\n\n2. On the other hand, the commutator is also expressed as [[H, X], X] = HX 2 − 2XHX + X 2 H and using hϕn |H|ϕm i = En δnm hϕ0 |HX 2 |ϕ0 i = hϕ0 |XHX|ϕ0 i = whence the result\n\nX\n\nn,m\n\nX\n\nn,m\n\nhϕ0 |H|ϕn ihϕn |X|ϕm ihϕm |X|ϕ0 i = E0 hϕ0 |X|ϕn ihϕn |H|ϕm ihϕm |X|ϕ0 i =\n\nX 2m|Xn0 |2 n\n\n~2\n\n(En − E0 ) = 1\n\nX\n\nX n\n\nn\n\n|Xn0 |2\n\nEn |Xn0 |2\n\n42\n\nCHAPTER 8. EXERCISES FROM CHAPTER 8\n\nChapter 9\n\nExercises from Chapter 9 9.7.2 Wave packet spreading 1. [P 2 , X] = P [P, X] + [P, X]P = −2i~P One can also use [f (P ), X] = −i~f ′ (P ) 2. From the Ehrenfest theorem (4.26), choosing A = X 2 d hX 2 i(t) dt\n\n= =\n\niE i i Dh P 2 h[H, X 2 ]i = + V (X), X 2 ~ ~ 2m i 2 2 h[P , X ]i 2~m\n\nFurthermore [P 2 , X 2 ] = −2i~(XP + P X) from which we derive the final result 1 d hX 2 i(t) = hXP + P Xi dt m Going to the x representation hP Xi =\n\nZ\n\n\u0014 \u0015 Z ∂ ∂ϕ∗ (x) dx ϕ∗ (x) −i (xϕ(x) = i dxxϕ(x) ∂x ∂x\n\nwhere the second expression is obtained from an integration by parts. Combining with \u0014 \u0015 Z ∂ϕ(x) hXP i = dx ϕ∗ (x)x −i ∂x we finally get d i~ hX 2 i(t) = dt m\n\n∂ϕ ∂ϕ∗ − ϕ∗ dx x ϕ ∂x ∂x −∞\n\nZ\n\n\u0014\n\n\u0015\n\nThese results are valid for a particle in a non zero potential, and not only for a free particle. 3. On the contrary, the results that follow are only valid for a free particle, V (X) = 0. The Hamiltonian is then reduced to the kinetic Hamiltonian H = K = P 2 /(2m). We compute the second derivative of hX 2 i(t) i d 1 d d2 hX 2 i(t) = h[K, X 2 ]i = − 2 h[K, [K, X 2 ]]i 2 dt ~ dt ~ dt\n\n43\n\n44\n\nCHAPTER 9. EXERCISES FROM CHAPTER 9\n\nwhere we have used Ehrenfest’s theorem twice. Taking into account [P 2 , XP + P X] = −4i~P 2 we find\n\n2 d2 hX 2 i(t) = hP 2 i dt2 m\n\nThe third derivative of hX 2 i(t) and higher order derivatives vanish dn hX 2 i(t) = 0 dtn\n\nn≥3\n\nbecause [K, [K, X 2 ]] ∝ P 2 and [K, P 2 ] = 0. hX 2 i(t) is thus a second order polynomial in t hX 2 i(t) = hX 2 i(t = 0) + t\n\n1 d2 d hX 2 i(t) + t2 2 hX 2 i(t) dt 2 dt t=0 t=0\n\nIn order to compute the dispersion, we use for a free particle\n\n∆x2 (t) = hX 2 i(t) − [hXi(t)]2 and\n\nDP E i d hXi(t) = [K, X] = dt ~ m\n\nwhich gives\n\nhXi(t) = hXi(t = 0) + t\n\nDP E m\n\nIndeed we have just seen that derivatives of order ≥ 2 vanish.\n\n9.7.3 A Gaussian wave packet Let us recall two results on Gaussian integrals Z\n\n+∞\n\ndx e\n\n−α2 x2\n\n−∞\n\n1. Setting k ′ = k − k\n\nZ\n\n2\n\ndk|A(k)| =\n\n\u0012\n\n√ π = α\n\n1 πσ 2\n\n\u00131/2 Z\n\n1 ∆x = √ 2α\n\n2\n\nk′ dk exp − 2 σ ′\n\n!\n\n=1\n\n√ so that ∆k = σ/ 2. Let us compute the wave function at time t = 0 ϕ(x, 0)\n\n=\n\n\u0012\n\n=\n\nσ 1/2 π 1/4\n\n\u00131/4\n\n2\n\ndk ′ k′ √ exp ik ′ x − 2 e 2σ 2π \u0015 \u0014 1 exp ikx − σ 2 x2 2\n\n1 πσ 2\n\nikx\n\nZ\n\nThe modulus squared of the wave function is 2 2 σ |ϕ(x, t)|2 = √ e−σ x π √ which is indeed normalized to one with ∆x = 1/( 2 σ), and thus\n\n∆x ∆k =\n\n1 2\n\n!\n\n45 2. Let us start from the expression of ϕ(x, t) ϕ(x, t) =\n\n\u0012\n\n1 πσ 2\n\n\u00131/4 Z\n\n2\n\n~k 2 dk (k − k) √ exp ikx − i t− 2m 2σ 2 2π\n\n!\n\nThe exponent is rewritten, within a factor of i and with k ′ = k − k 2\n\n2\n\n2\n\n2\n\n~k ~k 2 (k − k) ~kk ′ ~k ′ k′ ′ kx + k x − kx − = t− t − t − t − 2m 2σ 2 2m m 2m σ2 ϕ(x, t) reads ϕ(x, t) =\n\n\u0012\n\n1 πσ 2\n\n\u00131/4\n\n2\n\n~k exp ikx − i t 2m\n\n!Z\n\n\" \u0014 \u0012 \u0012 \u0013# \u0013\u0015 ′2 ~k 1 i~t k dk ′ √ exp ik ′ x − + t exp − 2m 2 σ2 2m 2π\n\nIf we can neglect the term i~t/2m in the last exponential, we simply obtain ! \" \u0012 \u00131/4 \u00132 # \u0012 2 ~k ~k σ2 1 x− exp ikx − i t exp − t ϕ(x, t) = πσ 2 2m 2 m ! 2 ~k t ϕ(x − vg t, 0) = eiω(k)t ϕ(x − vg t, 0) = exp i 2m 3. In the general case, we set\n\n1 i~t 1 2 = σ2 + m ′ σ\n\nand we find, after doing the integration ϕ(x, t) =\n\n\u0012\n\n1 πσ 2\n\n\u00131/4\n\nTaking the modulus squared 2\n\n|ϕ(x, t)| =\n\n\u0012\n\n\u0013 \u0012 \u0003 \u0002 1 ′2 2 σ exp − σ (x − vg t) exp i(kx − ω(k)t) 2 ′\n\n1 πσ 2\n\n\u00131/2\n\n\u0011 \u0010 2 |σ ′ |2 exp −Re σ ′ (x − vg t)2\n\nThe peak of |ϕ(x, t)|2 is centered at x = vg t and has a width ∆x2 (t) = that is, 1 ∆x (t) = 2σ 2 2\n\n1 2 Re σ ′ 2\n\n\u0012 \u0013 ~ 2 σ 4 t2 1+ m2\n\nThe width of the wave packet increases with time, because of the term ~2 σ 4 t2 /m2 in the bracket. 4. ∆x2 (t) doubles for t=\n\nm 2m∆x2 (t = 0) = = 3.2 × 10−11 s ~σ 2 ~\n\n9.7.7 A delta-function potential 1. The dimension of δ(x) is L−1 and if the dimension of g is the inverse of a length, then the dimension of V (x) is M2 L4 T −2 L−1 L−1 = M2 L2 T −2\n\n46\n\nCHAPTER 9. EXERCISES FROM CHAPTER 9\n\nwhich is indeed the dimension of an energy. 2. We integrate the Schr¨odinger written in the form \u0014 2 \u0015 d 2mE + 2 ϕ(x) = g δ(x) ϕ(x) dx2 ~ between x = −ε and x = +ε Z\n\n−ε\n\nϕ′′ (x)dx = ϕ′ (−ε) − ϕ′ (−ε) = gϕ(0)\n\nThe function ϕ(x) is continuous at x = 0 but its derivative ϕ′ (x) is not. In the case of a bound state, we must have ϕ(x) = Ae−κ|x| for x = 6 0 so that ϕ′ (0+ ) − ϕ′ (0− ) = −2κA = −|g|ϕ(0) = −|g|A whence 2κ = |g| and E\n\n~2 g 2 ~2 κ2 =− 2m 8m There is no odd solution because ϕ(0) = 0 for an odd solution. Let us retrieve the result by taking the limit a → 0, V0 a → ~2 g/(2m) of the square well potential whose energy levels are given by (9.82) E=−\n\nκ = k tan which leads to\n\nand\n\np 2m|V0 | k≃ ~ κ≃\n\np\n\n2m|V0 | ~\n\nka 2\n\nka tan ≃ tan 2\n\np 2m|V0 |a →0 2~\n\np 2m|V0 | a m ~2 |g| m|V0 |a 1 = = = |g| 2~ ~2 ~2 2m 2\n\n3. Since the diatomic molecule potential is even, one looks for even and odd solutions. For the even solutions we have x < −l : ϕ(x) = eκx\n\n− l < x < l : ϕ(x) = A cosh κx\n\nx > l : ϕ(x) = e−κx\n\nThe continuity of ϕ at x = l gives A cosh κl = e−κl and the continuity of its derivative at the same point −κe−κl − Aκ sinh κl = −|g|e−κl We remark that A sinh κl = A cosh κl tanh κl = e−κl tanh κl whence the equation for the energy eigenvalue (1 + tanh κl) =\n\n|g| κ\n\nThe solution is unique, and is given by the intersection of the curves (1 + tanh κl) and g/κ drawn as functions of κ. One can rewrite the value of κ as κ=\n\n\u0001 |g| 1 + e−2κl 2\n\n47 Let us now look for the odd solutions, which have the following form x < −l : ϕ(x) = −eκx\n\n− l < x < l : ϕ(x) = A sinh κx\n\nx > l : ϕ(x) = e−κx\n\nThe condition of continuity for ϕ(x) and the condition on its derivative at x = l are now\n\n−κl\n\n−κe\n\nA sinh κl\n\n=\n\ne−κl\n\n− Aκ cosh κl\n\n=\n\n−|g|e−κl\n\n\u0001 |g| 1 − e−2κl 2 This equation has a unique solution if the derivative of |g|[1 − exp(−2κl)]/2 > κ at κ = 0, that is, if |g|l > 1. There is no odd solution if |g|l < 1. κ=\n\n4. Let us consider two deep and narrow potential wells of width a separated by a distance l, with a ≪ l. The two wells can then be approximated by delta-functions, which leads us back to the potential of the preceding question, and we assume κl ≫ 1. Then there exist two bound states, one with an even wave-function corresponding to |g| (1 + e−2κl ) κ+ = 2 and the other one with an odd wave-function corresponding to κ− =\n\n|g| (1 − e−2κl ) 2\n\nThe energy difference between the two states is then E− − E+ = − the average enrgy E0 being E0 =\n\n~2 2 ~2 2 −2κl (κ− − κ2+ ) = g e 2m 2m\n\n~2 g 2 1 (E+ + E− ) ≃ − 2 8m\n\nWe can thus write E+ E−\n\n\u0001 ~2 g 2 1 + 2e−2κl 8m \u0001 ~2 g 2 1 − 2e−2κl ≃ − 8m ≃ −\n\nThese are the eigenvalues of a two-level Hamiltonian \u0012 ~2 g 2 1 H =− −2e−2κl 8m\n\n−2e−2κl 1\n\n\u0013\n\nFrom (9.100) and (9.105), the tunneling transmission coefficient of a particle with energy ≃ 0 through a barrier of height V0 and of width 2l is T ≃ e−4κl √ and the nondiagonal elements of the Hamiltonian are ∝ T . 5. As in § 9.4.1, we write the wave functions by selecting the case F = 1, G = 0. The continuity condition at x = 0 gives A+B =1 while the condition on the derivative is ik − ik(A − B) = g\n\n48\n\nCHAPTER 9. EXERCISES FROM CHAPTER 9\n\nWe derive\n\nig ig B=− 2k 2k whence the matrix elements of the transmission matrix A=1+\n\nM11 = 1 +\n\nig 2k\n\nM12 =\n\nig 2k\n\nWe check that the results do agree with the limit a → 0, V0 a → ~2 g/(2m) of equations (9.96) and (9.97). 6. The condition ϕq (x) = eiql ϕq (x − l) gives \u0011 \u0010 F eikx + Ge−ikx = eiql Aeik(x−l) + Beik(x+l)\n\nthat is\n\nF = A ei(q−k)\n\nG = B ei(q+l)\n\nThe continuity conditions on ϕq (x) and on its derivative read i i h h = A 1 − ei(q−k)l + B 1 − ei(q+k)l \u0011i \u0011i h \u0010 h \u0010 = A g + ik 1 − ei(q−k)l + B g − ik 1 − ei(q+k)l\n\n0 0\n\nWe thus obtain a system of equations with two unknowns A et B whose discriminant ∆ must be zero for a non trivial solution. Setting α = (q − k)l et β = (q + k)l \u0012 \u0013 1 − eiα \u0001 1 − eiβ \u0001 ∆ = det =0 g + ik 1 − eiα g − ik 1 − eiβ Computing ∆ gives\n\n\u0011 \u0010 \u0001 ∆ = g eiβ − eiα − 2ik 1 − eiα − eiβ + ei(α+β) = 0\n\nand multiplying by exp(−iql)\n\n2ig sin kl − 4ik(cos ql − cos kl) = 0 One thus recovers (9.126) cos ql = cos kl +\n\ng sin kl 2k\n\n9.7.13 Study of the Stern-Gerlach experiment 1. Since the yOz plane is symmetry plane of the problem, Bz must be an even function of x and we must have ∂Bz =0 ∂x x=0 Translation invariance along Oy gives ∂Bz =0 ∂y x=0\n\nThe two nonzero components of the magnetic field in the vicinity of x = 0 are Bx = −bx\n\nBz = B0 + bz\n\n~ ×B ~ = 0 and This field obeys the two Maxwell equations (1.8) and (1.9) in vacuum ∇ ~ ·B ~ = ∂Bx + ∂Bz = −b + b = 0 ∇ ∂x ∂z\n\n49 The potential energy is ~ = −µx Bx − µz Bz = bµx x − bµz z −~ µ·B\n\nwhence the force F~ with components Fx =\n\n~ ∂(−~ µ · B) = bµx ∂x\n\nFz =\n\n~ ∂(−~µ · B) = −bµz ∂z\n\nThe B0 zˆ component of the magnetic field leads to Larmor precession of the spin about the Oz axis(§ 3.25) in which µz stays constant. By contrast, due to this precession, the average value of µx vanishes: hµx i = 0, and the force along Ox averages to zero, if the transit time ≫ 1/ω, as the spin makes a large number of rotations about Oz. 2. The force on the magnetic moment is vertical and constant; its value is F = ±µb for a spin lying along ±ˆ z . The splitting between the trajectories of an up spin and a down spin at the exit of the magnet gap is then \u0012 \u00132 \u0012 \u00132 1 F 2 F L µb L δ=2 t = = 2 m m v m v Let us also evaluate the product ∆z∆pz \u0001 \u0001 ∆z∆pz ∼ 10−4 1.8 × 10−25 (10) = 1.8 × 10−28 MKSA ∼ 106 ~\n\nThe description by classical trajectories is legitimate.\n\n3. The potential energy of an up spin (down spin) is −µbz, and the Schr¨odinger equations for ϕ± are \u0012 \u0013 ~2 2 ∂ϕ± = − ∇ ∓ µbz ϕ± = Hϕ± i~ ∂t 2m Using Ehrenfest’s theorem (4.26), we obtain d ~ hR± i(t) = dt d hPx,y,± i(t) = dt d hPz,± i(t) = dt\n\ni ~ ± ] = 1 hP~± i [H, R ~ m i [H, Px,y,± ] = 0 ~ i [H, Pz,± ] = ±µb ~\n\nThis last result is deduced from ∓[µbZ, Pz ] = ∓i~µb We finally get µb 2 t 2m and the center of the wave packet does follow the classical trajectory. hZ± i = ±\n\n4. Let us make a reflection with respect to the xOy plane. In this reflection, ~µ does not change, because ~ the orientation of a current loop lying in the xOy plane stays unchanged. In the same operation, B ~ changes its direction, but not its gradient, and thus ∇B lies along −ˆ x. However, the image of the trajectory in the mirror always leave in the +ˆ x direction, and then the image of the trajectory in the mirror does not represent a physically allowed motion, unless there is no deviation of the trajectory.\n\nThe von Neumann measurement model 1. From (4.17), the evolution operator U (t, t0 ) obeys i~\n\nd U (t, t0 ) = [g(t)AP ] U (t, t0 ) dt\n\n50\n\nCHAPTER 9. EXERCISES FROM CHAPTER 9\n\nwhich integrates into \u0012\n\ni U (t, t0 ) = exp − AP ~\n\nZ\n\nt\n\ng(t )dt\n\nt0\n\n\u0013\n\nBetween times ti and tf we thus have \u0012 \u0013 \u0012 \u0013 Z +∞ i i ′ ′ U (tf , ti ) ≃ exp − AP g(t )dt = exp − gAP ~ ~ −∞ 2. The action of U (tf , ti ) on the vector |n ⊗ ϕi is U (tf , ti )|n ⊗ ϕi = e−igan P/~ |n ⊗ ϕi Furthermore, exp(−igan P/~) is a translation operator by gan and from (9.13) \u0010 \u0011 e−igan P/~ ϕ (x) = ϕ(x − gan ) 3. Because of the linearity property of quantum mechanics, the final state vector is X |χf i = cn |n ⊗ ϕn i n\n\nThe reduced state operator of S is from (6.66) X X ρ(1) = cn c∗m |mihn|hϕm |ϕn i = |cn |2 |nihn| n,m\n\nn\n\nbecause hϕm |ϕn i = δnm . The system S is thus a statistical mixture of states |ni with a weight |cn |2 , and the probability to observe S in the state |ni is |cn |2 .\n\nChapter 10\n\nExercises from chapter 10 We remind the reader that we use in chapter 10 a system of units where ~ = 1\n\n10.7.5 Orbital angular momentum ~ and 1. From the expression of the orbital angular momentum operator as a function of the position R ~ momentum P operators, we have for the commutator [Lx , Ly ] [Lx , Ly ] = = =\n\n[Y Pz − ZPy , ZPx − XPz ]\n\n[Y Pz , ZPx ] + [ZPy , XPz ] i[−Y Px + XPy ] = iLz\n\n2. Let us start from equation (cf. (10.40)) \u0001 \u0001 r ) = f (Rxˆ [−α](~r)) e−iαLx f (~r) = f Rx−1 ˆ [α](~\n\nwhere Rxˆ [α] is a rotation by angle α about Ox. We choose α to be infinitesimal, α → α + dα; in a rotation of angle −α about Ox y′\n\n=\n\nz sin α + y cos α\n\ndy = zdα\n\n=\n\nz cos α − y sin α\n\ndz = −ydα\n\nz\n\nIn this rotation, θ → θ + dθ and φ → φ + dφ, which are determined through dy\n\n=\n\nr cos θ sin φ dθ + r sin θ cos φ dφ\n\ndz\n\n=\n\n−r sin θ dθ\n\nwhich leads to dθ = sin φ dα\n\ndφ =\n\ncos φ dα tan θ\n\nThis allows us too identify Lx \u0013 \u0012 cos φ ∂ ∂ f (~r) + [−idαLx f ](~r) = dα sin φ ∂θ tan θ ∂φ that is\n\n\u0012 \u0013 ∂ cos φ ∂ Lx = i sin φ + ∂θ tan θ ∂φ\n\nTo compute Ly we can use the same method, or make use of the commutation relation \u0015 \u0014 ∂ , Lx Ly = −i[Lz , Lx ] = − ∂φ\n\n51\n\n52\n\nCHAPTER 10. EXERCISES FROM CHAPTER 10\n\nTaking into account \u0014\n\n\u0015 ∂ , f (φ) ∂φ \u0015 \u0014 ∂ ∂ , f (φ) ∂φ ∂φ we obtain\n\n=\n\nf ′ (φ)\n\n=\n\nf ′ (φ)\n\n∂ ∂φ\n\n\u0013 \u0012 sin φ ∂ ∂ + Ly = i − cos φ ∂θ tan θ ∂φ\n\n3. The operator Lz = −i∂/∂φ is defined on the periodic functions f (φ) = f (φ + 2π) and it is selfadjoint on this domain (see § 7.2.2). By contrast, the function φf (φ) is not periodic and does not belong to the domain of Lz .Therefore, we cannot define φLz and write a commutation relation between φ and Lz . One cannot either use the method of Exercise 9.7.1 because the integration bounds contribute to the integration by parts.\n\n10.7.6 Relation between the rotation matrices and spherical harmonics 1. The function f (0, 0, z) is invariant under rotations about Oz. The action of exp(−iαLz ) on f (0, 0, z) is equivalent to the identity operation, and thus Lz f (0, 0, z) = 0. In classical mechanics lz = xpy − ypx and lz = 0 if x = y = 0, that is, if the particle trajectory follows the z axis. By definition hlm|θ, φi = Ylm ∗ (θ, φ) but we also have hlm|θ, φi\n\n= hlm|e−iφLz e−iθLy |θ = 0, φ = 0i X = hlm|e−iφLz e−iθLy |l′ m′ ihl′ m′ |θ = 0, φ = 0i l′ ,m′\n\nand introducing a complete set of states X\n\nl′ ,m′\n\n|l′ m′ ihl′ m′ | = I\n\nFrom the result of question 1 hl′ m′ |θ = 0, φ = 0i ∝ δm′ 0 and from (10.32) (l)\n\nhlm|e−iφLz e−iθLy |l′ , 0i = δl′ l Dm0 (θ, φ) In fact, we can easily obtain the proportionality coefficient because ′\n\nhl m |θ = 0, φ = 0i =\n\nδm′ 0 Yl0 (θ\n\n= 0, φ = 0) =\n\nusing (10.61) and the property Pl (1) = 1.\n\n10.7.8 The spherical well 1. Setting E = −B and (~ = 1) k=\n\np 2m(V0 − B)\n\nκ=\n\n√ 2mB\n\nr\n\n2l + 1 4π\n\n53 we write the radial wave function in the s-wave r < R : u(r) = A sin kr\n\nr > R : u(r) = Be−κr\n\nThe continuity of the logarithmic derivative leads to the relation k cot kR = −κ As in § 9.3.3, we define U = 2mV0 and κ2 = U − k 2 . The eigenvalue equation becomes √ U − k2 cot kR = − k and its solutions are given by Figure 9.12 using the odd solutions of the one-dimensional square well (dotted lines). Solutions exist only if kR > π/2. 2. Assuming the deuteron binding energy B ≪ V0 , taking as the reduced mass half of the proton mass mp /2 and writing ~ explicitly π2 mp V0 R2 = 2 ~ 4 One finds the numerical value V0 ≃ 100 MeV ≫ B 3. The radial wave equation is \u0014\n\n\u0015 A B 1 d2 + 2 − u(r) = Eu(r) − 2m dr2 r r\n\nwhich is analogous to that (10.86) for the hydrogen atom with l(l + 1)/(2m) → A and B → e2 .\n\n10.7.13 Light scattering 1. If the photon is emitted along the Oz axis with a right handed (R) or left handed (L) polarization, angular momentum conservation allows only two nonzero amplitudes a′ = hL, θ = 0|T |j = 1, m = −1i\n\na = hR, θ = 0|T |j = 1, m = 1i\n\nIf the transition is of the electric dipole kind, we have seen in § 10.5.2 that a = a′ (with our phase conventions). We find, using the rotational invariance of the transition matrix, [U (R), T ] = 0 am=1 (θ) R\n\n= = =\n\nhR, θ|T |j = 1, m = 1i = hR, θ = 0|T U † [Ryˆ(θ)]|j = 1, m = 1i hR, θ = 0|T |j = 1, m = 1ihj = 1, m = 1|U † [Ryˆ(θ)]|j = 1, m = 1i 1 (1) ad11 (θ) = a(1 + cos θ) 2\n\nSimilarly we find (1)\n\nam=1 (θ) = ad1,−1 (θ) = L\n\n1 a(1 − cos θ) 2\n\nIf the photon is emitted in the direction n ˆ = (θ, φ), we must use the rotation operator U [R(θ, φ)] and we get am=1 (θ, φ) R\n\n=\n\nam=1 (θ, φ) L\n\n=\n\n1 a(1 + cos θ)eiφ 2 1 a(1 − cos θ)eiφ 2\n\n2. If the atom absorbs the photon, the two nonzero amplitudes are, from angular momentum conservation b = hj = 1, m = 1|T ′ |Ri\n\nb′ = hj = 1, m = −1|T ′ |Li\n\n54\n\nCHAPTER 10. EXERCISES FROM CHAPTER 10\n\nIf the transition is of the electric dipole kind, b′ = b from the results of § 10.5.2. Introducing a complete set of intermediate states X cP →P ′ (θ) = cP →(jm) c(jm)→P ′ (θ) = hP ′ , θ|S|P i = hP ′ , θ|T |1mih1m|T ′|P i m\n\nhR, θ|S|Ri = hR, θ|S|Li = hL, θ|S|Ri = hL, θ|S|Li =\n\n1 ab(1 + cos θ) 2 1 hR, θ|T |j = 1, m = −1ihj = 1, m = −1|T ′ |Li = ab(1 − cos θ) 2 1 ′ hL, θ|T |j = 1, m = 1ihj = 1, m = 1|T |Ri = ab(1 − cos θ) 2 1 ′ hL, θ|T |j = 1, m = −1ihj = 1, m = −1|T |Li = ab(1 + cos θ) 2\n\nhR, θ|T |j = 1, m = 1ihj = 1, m = 1|T ′ |Ri =\n\nIn the two possible initial cases, |Ri and |Li, the angular distribution is W (θ) =\n\n1 2 2 |a| |b| (1 + cos2 θ) 2\n\nIf the initial photon is polarized along Ox we find hx, θ|S|xi = hy, θ|S|xi =\n\nab cos θ 0\n\nIn a classical model of photon scattering by an electric charge, a photon with linear polarization along Ox sets the charge into motion along this axis, and the charge radiates an electromagnetic wave polarized along Ox with an angular distribution ∝ cos2 θ. If the photon is emitted in the direction n ˆ = (θ, φ), we find hR; θ, φ|S|Ri\n\n=\n\nhR; θ, φ|S|Li\n\n=\n\nhL; θ, φ|S|Ri\n\n=\n\nhL; θ, φ|S|Li =\n\n1 ab(1 + cos θ) eiφ 2 1 ab(1 − cos θ) e−iφ 2 1 ab(1 − cos θ) eiφ 2 1 ab(1 + cos θ) e−iφ 2\n\nIf the initial photon is polarized along Ox, we obtain the following amplitudes hx; θ, φ|S|xi\n\nhy; θ, φ|S|xi\n\n=\n\nab cos θ cos φ\n\n=\n\nab sin φ\n\nThese results can be understood by observing that the cosine of the angle between the initial and final polarizations is cos θ cos φ for a final polarization along Ox and sin φ for a final polarization along Oy.\n\n10.7.14 Measurement of the Λ0 magnetic moment 1. Angular momentum conservation along Oz implies m′ = m, since the z component of the angular momentum vanishes hθ = 0, m′ |T |mi ∝ δmm′ The amplitude b is obtained from a through a reflection with respect to a plane xOz: |a| = |b| if parity is conserved.\n\n55 2. If the proton is emitted along a direction making an angle θ with the Oz axis in the xOz plane, we can compute the decay amplitude using the rotation operator U [Ryˆ(θ)] by an angle θ about Oy hθ, m′ |T |mi = hθ = 0, m′ |U † [Ryˆ(θ)]T |mi X = hθ = 0, m′ |T |m′′ ihm′′ |eiθJy |mi m′′\n\n(1/2)\n\n= am′ ,m′ (θ = 0)dm′ m (θ)\n\nWith the definitions of question 1 a 21 , 21 = a\n\na− 21 ,− 12 = b\n\nand taking (10.38) into account (1/2)\n\na++ (θ) = ad 1 , 1 (θ) = a cos 2 2\n\nθ 2\n\n(1/2)\n\na−+ (θ) = ad 1 ,− 1 (θ) = −b sin 2\n\n2\n\nθ 2\n\n3. If the Λ0 particle is produced in a state m = 1/2, the angular distribution w(θ) is as follows (because the final states m′ = 1/2 et m′ = −1/2 are distinguishable) w(θ)\n\nθ θ + |b|2 sin2 2 2 1 1 (|a|2 + |b|2 ) + (|a|2 − |b|2 ) cos θ 2 2\n\n= |a|2 cos2 =\n\nwhence w0 =\n\n1 (|a|2 + |b|2 ) 2\n\nα=\n\n|a|2 − |b|2 |a|2 + |b|2\n\nWere parity conserved, we would have |a|2 = |b|2 and α = 0. Observation of a cos θ term in the angular distribution of the decay is thus a proof that parity conservation is violated. ~ which is a pseudovector, must necessarily be 4. ~p × p~Λ0 is the only available pseudovector, and hSi, oriented along this direction : hSx i = hSy i = 0. 5. The spin Hamiltonian in the magnetic field is ~ = −γ S ~·B ~ H = −~µ · B ~ with an angular velocity ω = γB in the xOz plane. At time t = τ , The proton spin precesses around B it will have rotated by an angle λ = ωτ = γBτ The decay angle must be measured from the direction n ˆ in the xOz plane n ˆ=x ˆ sin λ + zˆ cos λ The proton is emitted in the direction pˆ = x ˆ sin θ cos φ + yˆ sin θ sin φ + zˆ cos θ and cos Θ = pˆ · n ˆ = cos θ cos λ + sin θ sin λ cos φ The measurement of the angular distribution allows us to infer the direction of n ˆ (or the angle λ) and to deduce γ from λ = γBτ .\n\n10.7.15 Production and decay of the ρ+ meson\n\n56\n\nCHAPTER 10. EXERCISES FROM CHAPTER 10\n\n1. Calculation of am (θ, φ) : R(θ, φ) is the rotation (10.30) which brings the Oz axis along the direction of emission for the π + meson = hθ, φ|T |mi = hθ = 0, φ = 0|U † [R(θ, φ)]T |mi X = hθ = 0, φ = 0|T |m′ ihm′ |U † [R(θ, φ)]|mi\n\nam (θ, φ)\n\nm′\n\nwhere we have used the rotational invariance of the transition matrix [U, T ] = 0. From angular momentum conservation hθ = 0, φ = 0|T |m′i ∝ δm′ 0 = aδm′ 0\n\nbecause if π + meson is emitted in the Oz direction, its angular momentum along this direction vanishes. Furthermore h i∗ (1) hm′ = 0|U † [R(θ, φ)]|mi = hm|U [R(θ, φ)]|m′ = 0i∗ = Dm0 (θ, φ) (1)\n\n= eimφ dm0 (θ)\n\nOne obtains the different decay amplitudes using (10.39) a1 (θ, φ)\n\n=\n\na (1) a eiφ d10 (θ) = − √ eiφ sin θ 2\n\na0 (θ, φ)\n\n=\n\na d00 (θ) = a cos θ\n\na−1 (θ, φ)\n\n=\n\n(1)\n\na (1) a e−iφ d−10 (θ) = √ e−iφ sin θ 2\n\nwhence the angular distributions W1 = W−1 =\n\n|a|2 sin2 θ 2\n\nW0 = |a|2 cos2 θ\n\nOne observes that W1 + W0 + W−1 = |a|2 : as a consequence, if the ρ meson is not polarized, the angular distribution is isotropic, which must be the case as there is no privileged direction. In what follows, W will be normalized as W1 + W0 + W−1 = |a|2 = 1 2. If the initial state vector is given by X |λi =\n\nm=−1,0,1\n\ncm |1mi\n\nX\n\nm=−1,0,1\n\n|cm |2 = 1\n\nthe decay amplitude aλ will be aλ (θ, φ) = hθ, φ|T |λi =\n\nX m\n\nhθ, φ|T |1mih1m|λi =\n\nX\n\ncm am (θ, φ)\n\nm\n\nand the angular distribution Wλ (θ, φ) =\n\nX\n\ncm c∗m′ am (θ, φ)a∗m′ (θ, φ)\n\nm,m′\n\nThe explicit expression of Wλ is given in the next section. 3. For every component of the statistical mixture pλ , the angular distribution is X ∗ ∗ (λ) Wλ = c(λ) m c m′ am (θ, φ)am′ (θ, φ) m,m′\n\nand the angular distribution corresponding to the state operator ρ is X X X ∗ (λ) ∗ Wρ (θ) = pλ Wλ = pλ c(λ) ρmm′ am (θ, φ)a∗m′ (θ, φ) m c m′ am (θ, φ)am′ (θ, φ) = λ\n\nλ,m,m′\n\nm,m′\n\n57 The result is simplified by observing that ρmm′ = ρ∗m′ m m 6= m′\n\nρmm′ am a∗m′ + ρm′ m am′ a∗m = 2Re (ρmm′ am a∗m′ )\n\nFor example Re (ρ10 a1 a∗0 ) = −Re\n\n\u0012\n\nρ10 iφ √ e sin θ cos θ 2\n\n\u0013\n\nThe final result is Wρ (θ, φ)\n\n1 sin2 θ(ρ11 + ρ−1,−1 ) 2 \u0001 \u0001 1 √ sin 2θ Re e −iφ ρ−10 − e iφ ρ10 − sin2 θ Re e 2iφ ρ1,−1 2\n\n= ρ00 cos2 θ + +\n\nTo obtain the angular distribution of question 2, it is enough to make the sustitutions ρ11 → |c1 |2 , ρ10 → c1 c∗0 etc.\n\n~ which is a pseudovector, lies necessarily along n 4. The only available pseudovector is n ˆ , and hJi, ˆ . Using the expression (10.24) of Jx and Jy for j = 1 one finds Tr ρJx\n\n= 2(Re ρ10 + Re ρ0,−1 ) = 0\n\nTr ρJy\n\n= 2(Im ρ10 + Im ρ0,−1 ) = 0\n\nthat is, ρ10 + ρ0,−1 = 0. In the operation Z, which is a reflection with respect to the xOy plane, the reaction kinematics is unchanged, and since the target is unpolarized and parity is conserved, the reaction is identical to its image in the xOz plane. We must then have [ρ, Z] = 0 ou Z −1 ρZ = ρ Furthermore, we use Π|1mi = η|1mi where η is the ρ meson intrinsic parity (η = −1 because the ρ meson is, as the photon, a vector meson). We then derive ′ h1m|ΠeiπJz ρ e−iπJz Π|1m′ i = |η|2 e−iπ(m −m) ρmm′ that is\n\nρmm′ = (−1)m−m ρmm′\n\n10.7.17 Σ0 decay 1. If the photon is emitted along the Oz axis, the component along this axis of the orbital angular momentum vanishes, and angular momentum conservation holds for the amplitudes a and b, but not for c and d, which must then vanish. 2. The amplitudes a et b are deduced from each other through a reflection with respect to the xOz plane, and if parity is conserved in the decay, then |a| = |b|. The reflection operator Y (10.100) acts in the following way on the photon states which are odd under parity, ηγ = −1 (see (10.104)) Y|Ri = −|Li\n\nY|Li = −|Ri\n\nwhile for the Σ0 and Λ0 particles (see (10.102)) Y|jmi\n\nY|jm′ i\n\n= ηΣ (−1)1/2−m |j, −mi ′\n\n= ηΛ (−1)1/2−m |j, −m′ i\n\nand since m′ = −m, we obtain an overall factor ′\n\nηΣ ηΛ ηγ (−1)1−m−m = ηΣ ηΛ = η\n\n58\n\nCHAPTER 10. EXERCISES FROM CHAPTER 10\n\nWe can also use directly (10.119) ησ ηλ ηγ (−1)jΣ −jΛ −jγ = η The transition is of the magnetic dipole kind, since the parities of the initial and final states are the same. 3. If p~ is the photon momentum, the emission amplitude for the Λ0 particle in the direction −ˆ p with a spin projection m′ on pˆ for a right handed polarized photon is ′\n\nam R (θ)\n\n= = =\n\nhR, m′ ; θ|T |m = 1/2i\n\nhR, m′ ; θ = 0|U † [Ryˆ(θ)]|m = 1/2i = (1/2)\n\na d1\n\n1 2 2\n\n(θ) = a cos\n\nθ 2\n\nX m′′\n\nhR, m′ ; θ = 0|T |m′′ ihm′′ |U † [Ryˆ (θ)]|1/2i\n\nbecause it is only the value m′′ = 1/2 which gives a nonzero contribution. We then have m′ = −1/2. An analogous calculation gives for a left handed polarized photon ′\n\n(1/2)\n\nam L (θ) = b d 1 ,− 1 (θ) = −a sin 2\n\nm′ =−1/2\n\nOnly the amplitudes aR\n\nm′ =1/2\n\n(θ) and aL\n\n2\n\n(θ) are nonzero.\n\nθ 2\n\nChapter 11\n\nExercises from Chapter 11 11.5.2 Mathematical properties 1. We use a recursion method assuming that [N, ap ] = −pap [N, ap+1 ] = [N, aap ] = [N, a]ap + a[N, ap ] = −(p + 1)ap+1 Let us consider a monomial1 in a and a† , P = (a† )q ap and let us compute its commutator with N [N, (a† )q ap ] = [N, (a† )q ]ap + (a† )q [N, ap ] = (q − p)(a† )q ap which vanishes only if p = q. As any function of a and a† is a sum of such monomials, using if necessary the commutation relation [a, a† ] = I to put the creation and annihilation operators in a suitable order, we see that the only possibility to have a non vanishing commutator is that this function be a sum of terms of the form (a† )p ap . Any monomial of the form (a† )p ap can be written as a function of a† a using the commutation relations (11.8). If an operator A commutes with N , it is necessarily a function of N : A = f (N ). There does not exist an operator independent of N and commuting with N , so that hn′ |A|ni = hn′ |f (N )|ni = f (n)δnn′ 2. Let |ϕi be a vector orthogonal to all vectors |ni, hϕ|ni = 0 ∀n P ′ |ni = |ni\n\nP ′ |ϕi = 0\n\nLet us show that [P ′ , a] = 0. It is clear that hn|[P ′ , a]|ni = 0\n\nhϕ|[P ′ , a]|ϕi = 0\n\nLet us examine hϕ|[P ′ , a]|ni and hϕ|[P ′ , a† ]|ni hϕ|P ′ a|ni\n\nhϕ|aP ′ |ni hn|a† P ′ |ni\n\n= hϕ|P ′ a† |ni = 0 √ = hϕ|a|ni = n hϕ|n − 1i = 0 √ = hn|a† |ϕi = n + 1 hϕ|n + 1i = 0\n\nand thus [P ′ , a] = 0 and [P ′ , a† ] = 0. The projector P ′ then commutes with a and a† , and from von Neumann’s theorem, it is a multiple of the identity operator, that is, either P ′ = I, or P ′ = 0. The second possibility being excluded, we are left with P ′ = I.\n\n11.5.3 Coherent states\n\n1 When we write a combination of a and a† with all the a to the right and all a† to the left, we say that this combination has been written in normal from.\n\n59\n\n60\n\nCHAPTER 11. EXERCISES FROM CHAPTER 11\n\n2. Using Ehrenfest’s theorem (4.26), we get i~\n\nd hai(t) = h[a, H]i = ~ωhai(t) dt\n\nbecause [a, H] = ~ωa. If the initial condition is hai(t = 0) = hϕ(0)|a|ϕ(0)i = z0 the solution of the preceding differential equation is hai(t) = z0 e−iωt 3. In addition, we want that hϕ(0)|a† a|ϕ(0)i = |z0 |2 , which implies, with b(z0 ) = a − z0 hϕ(0)|b† (z0 )b(z0 )|ϕ(0)i\n\n= =\n\nhϕ(0)|a† a|ϕ(0)i − z0 hϕ(0)|a† |ϕ(0)i − z0∗ hϕ(0)|a|ϕ(0)i + |z0 |2 hϕ(0)|(a† a − |z0 |2 )|ϕ(0)i = ||b(z0 )ϕ(0)||2 = 0\n\nwhich is possible only if b(z0 )|ϕ(0)i = 0 : |ϕ(0)i is an eigenvector of b(z0 ) with eigenvalue 0 b(z0 )|ϕ(0)i = 0, that is, a|ϕ(0)i = z0 |ϕ(0)i 4. Writing D(z) = exp A with A = −z ∗ a + za†\n\nA† = −za† + z ∗ a = −A\n\nthen D† (z) = exp A† = exp(−A) and D† (z)D(z) = I. Using (2.55) we may rewrite D(z) D(z) = e−|z|\n\n2\n\n/2 za†\n\ne\n\ne−z\n\na\n\nFrom the definition (11.31) of a coherent state |zi we have D(z)|0i = e−|z|\n\n2\n\n/2 za†\n\ne\n\ne−z a |0i = e−|z|\n\n2\n\n/2 za†\n\ne\n\n|0i = |zi\n\nbecause exp(−z ∗ a)|0i = 0 taking a|0i = 0 into account. 5. We use (2.55) 1\n\neA eB = eA+B e 2 [A,B] with\n\nand\n\nc A = √ (z − z ∗ )(a + a† ) 2 1 c = −c′ = √ 2\n\nc′ B = √ (z + z ∗ )(a − a† ) 2 A + B = −z ∗ a + za†\n\nThe identity (2.55) is valid because [A, B] is a multiple of the identity operator, which commutes with A et B. We use a system of units such that ~ = mω = 1 ˆ or, equivalently, we use the operators Qand Pˆ of (11.4) instead of Q et P i 1 [A, B] = − (z − z ∗ )(z + z ∗ )[Q, P ] = (z 2 − z ∗2 ) 2 2 We must then choose\n\n\u0013 \u0012 \u0003 1\u0002 f (z, z ∗ ) = exp − z 2 − z ∗2 4\n\n61 Given all these conditions, the wave function of the coherent state ϕz (q) in the q-representation is \u0013 \u0012 \u0013 \u0013 \u0012 \u0012 \u0003 i 1 1\u0002 hϕz (q) = hq|D(z)|0i = exp − z 2 − z ∗2 exp √ [z − z ∗ ]q hQ| exp − √ [z + z ∗ ]P |0i 4 2 2 \u0012 \u0013 \u0012 \u0013 \u0013 \u0012 \u0003 1\u0002 2 1 1 = exp − z − z ∗2 exp √ [z − z ∗ ] q ϕ0 q − √ [z + z ∗ ] 2 2 2\n\nwhere ϕ0 (q) is the wave function (11.23) of the harmonic oscillator ground state. From (11.37), the expectation values of Q and P are 1 hP i = √ (z − z ∗ ) i 2\n\n1 hQi = √ (z + z ∗ ) 2\n\nand we can rewrite the preceding result as \u0013 \u0013 \u0012 \u0012 1 1 i ϕz (q) = 1/4 exp − hQihP i exp (ihP iq) exp − (q − hQi)2 2 2 π Reestablishing the dimensionful factors \u0013 \u0012 \u0013 \u0012 \u0013 \u0012 \u0010 mω \u00111/4 1 mω i hP iq exp − ϕz (q) = exp − hQihP i exp i (q − hQi)2 π~ 2~ ~ 2 ~ The global phase factor \u0013 \u0012 i exp − hQihP i 2~ is physically irrelevant and may be omitted. 6. We write, with z = ρ exp(iθ) and Anm = hn|A|mi X hz|A|zi = hz|nihn|A|mihm|zi n,m\n\nX Anm z m z ∗n √ n!m! n,m 2 X Anm √ ei(m−n)θ ρn+m e−ρ n!m! n,m e−|z|\n\n= =\n\n2\n\nwhere we made use of (11.31) hm|zi = e−|z|\n\n2\n\n/2\n\nzm √ m!\n\nWriting the expansion of hz|A|zi under the general form 2\n\nhz|A|zi = e−ρ we derive cpq and we can make the identification\n\n1 = q!\n\nZ\n\n∞ X ∞ X\n\ncpq eipθ ρq\n\nq=0 p=−∞\n\ndRe zIm z −ipθ e hz|A|zi π\n\nAnm =\n\n√ n!m! cm−n,m+n\n\nwith (m − n) and (m + n) integers, (m + n) ≥ 0. Thus, we can get from hz|A|zi all the matrix elements of Amn .\n\n11.5.4 Coupling to a classical force\n\n62\n\nCHAPTER 11. EXERCISES FROM CHAPTER 11\n\n1. Let us express Q as a function of the operators a and a† (cf. (11.6)), and write \u0001 H(t) = H0 − a + a† f (t) Differentiating with respect to t the equation\n\nU (t) = U0 (t)UI (t) we obtain i~\n\ndU dt\n\n\u0002 \u0003 H(t)U = H0 + W (t) U0 UI\n\n= =\n\ni~\n\ndUI dUI dU0 UI + U0 i~ = H0 U0 UI + U0 i~ dt dt dt\n\nand derive (11.126) i~\n\ndUI = U0−1 W (t)U0 UI = WI (t)UI dt\n\n(11.1)\n\n2. Let us show that aI (t) = e iH0 t/~ a e −iH0 t/~ = a e−iωt It is easy to derive a differential equation for aI (t) daI (t) dt\n\n= =\n\ni i [H0 , aI ] = e iH0 t/~ [H0 , a] e −iH0 t/~ ~ ~ −iωaI (t)\n\nwhere we made use of (11.11). We then derive, with the initial condition aI (t = 0) = a aI (t) = a e −iωt\n\na†I (t) = a† e iωt\n\nthe formula for a†I being obtained from Hermitian conjugation. The differential equation (11.126) for UI (t) becomes \u0001 dUI = − a e −iωt + a† e iωt f (t)UI (t) = WI (t)UI (t) i~ dt with the boundary condition UI (t = 0) = I. If the commutator of two operators A1 and A2 is a multiple of I, from the identity (2.55) of Exercise 2.4.11 1\n\neA2 eA1 = eA2 +A1 e 2 [A2 ,A1 ] Repeating this operation for n operators A1 , . . . , An (we may use a recursion reasoning) 1\n\neAn eAn−1 · · · eA1 = eAn +···+A1 e 2\n\nP\n\nj>i [Aj ,Ai ]\n\n3. Let us divide the interval [0, t] into n infinitesimal intervals ∆t; the time evolution in the interval [t′ , t′ + ∆t] is simple \u0012 \u0013 i i ′ ′ ′ 2 ′ UI (t + ∆t, t ) = I − WI (t )∆t + O(∆t) = exp − WI (t )∆t + O(∆t)2 ~ ~ and it follows that\n\n\u0013\u0015 \u0012 n \u0014 Y i exp − WI (tj )∆t UI (t) ≃ ~ j=1\n\nSince [WI (tj ), WI (ti )] is a multiple of I, we have     n 2 X\u0002 X \u0003 (∆t) i WI (tj ) exp − WI (tj ), WI (ti )  UI (t) ≃ exp − ∆t 2 ~ 2~ t >t j=1 j\n\ni\n\n63 4. Taking into acount \u0002\n\n\u0010 \u0011 \u0003 WI (t′ ), WI (t′′ ) = f (t′ )f (t′′ ) exp[−iω(t′ − t′′ )] − exp[iω(t′ − t′′ )]\n\nwe obtain UI (t) by taking the limit ∆t → 0 ∆t\n\nn X j=1\n\nt\n\nZ\n\nWI (tj ) →\n\n0\n\ndt′ WI (t′ ) = −\n\nZ\n\n0\n\n−~az ∗ (t) − ~a† z(t)\n\n=\n\nt\n\n\u0010 \u0011 ′ ′ dt′ a e −iωt + a† e iωt f (t′ )\n\nIn a similar way (∆t)2\n\nX\n\ntj >ti\n\n[WI (tj ), WI (ti )] →\n\nZ\n\nt\n\nZ\n\ndt′\n\n0\n\nt\n\n′′\n\ndt′′ e −iω(t −t\n\n0\n\n)\n\nf (t′ )f (t′′ )ε(t′ − t′′ )\n\nwhere ε(x) is the sign function: ε(x) = 1 if x > 0, ε(x) = −1 if x < 0. The final result for UI (t) is then UI (t) = X\n\n=\n\n\u0015 \u0014 \u0001\u0003 X exp i az (t) + a z(t) exp − 2 2~ Z t Z t ′ ′′ dt′ dt′′ e −iω(t −t ) f (t′ )f (t′′ )ε(t′ − t′′ ) ∗\n\n\u0002\n\n0\n\n0\n\n5. We can write this result in a form which is more directly useful thanks to the identity \u0014 \u0015 \u0002 \u0001\u0003 \u0002 † \u0003 1 ∗ † ∗ ∗ exp i az (t) + a z(t) = exp ia z(t) exp [iaz (t)] exp − z(t)z (t) 2 and we find, using once more (2.55) UI (t)\n\n=\n\nY\n\n=\n\ne ia Z t 0\n\n=\n\nZ\n\nt\n\n0\n\n2\n\nz(t)\n\ne iaz (t) e−Y /~ Z t ′ ′′ dt′ dt′′ e −iω(t −t ) f (t′ )f (t′′ )θ(t′ − t′′ ) 0\n\ndt′\n\nZ\n\nt′\n\n′′\n\ndt′′ e −iω(t −t\n\n)\n\nf (t′ )f (t′′ )\n\n0\n\nwhere θ(x) is the Heaviside (or step) function. It is worth checking by explicit differentiation that this result does obey the differential equation we started from. This expression fully determines the time evolution of the forced harmonic oscillator. A particular case is that where the initial state is a coherent state: then the final state is also a coherent state. 6. Let us examine the case where the initial state at t = 0 is an eigenvector |ni of H0 . We assume that the force acts only during a finite time interval [t1 , t2 ], and we choose to observe the oscillator at a time t > t2 : to summarize 0 < t1 < t2 < t. We can then integrate from −∞ to +∞: z(t) is independent of t and is nothing else than the Fourier transform of f (t)/~ 1 f˜(ω) = ~\n\nZ\n\ndt′ e iωt f (t′ )\n\n−∞\n\nThe Y factor is computed using the Fourier representation of the θ-function θ(t) = lim\n\nη→0+\n\nZ\n\n+∞\n\n−∞\n\ndE e itE 2iπ E − iη\n\n(11.2)\n\n64\n\nCHAPTER 11. EXERCISES FROM CHAPTER 11\n\nIntroducing this expression in that of Y Z ′ ′′ 1 1 dE 1 Y = e i(E−ω)t e −i(E−ω)t f (t′ )f (t′′ ) dt′ dt′′ ~2 ~2 2iπ E − iη Z dE 1 = f˜(E − ω)f˜∗ (E − ω) 2iπ E − iη Z dE ˜ 1 = P |f (E − ω)|2 + |f˜(ω)|2 2iπE 2 1 = iφ + |f˜(ω)|2 2 where P denotes a Cauchy principal value, and we have made use of 1 1 = P + iπδ(E) E − iη E 7. We finally obtain an operator UI (t) independent of t for t > t2 \u0012 \u0013 \u0010 \u0011 \u0010 \u0011 1 UI (t) = exp ia† f˜(ω exp iaf˜∗ (ω exp(−iφ) exp − |f˜(ω)|2 2\n\nThis allows us to compute the n → m transition amplitude Amn (t) = =\n\nhm|U (t)|ni = hm|U0 (t)UI (t)|ni e −iEm t/~ hm|UI (t)|ni\n\nThe result is particularly simple if the oscillator is in its ground state at time t = 0 since UI (t)|0i is then a coherent state 2 † ˜ ˜ UI (t)|0i = e −iφ e −|f (ω)| /2 e ia f (ω) |0i = e −iφ |if˜(ω)i The probability to observe a state |mi is given by a Poisson law (11.34) \u0011m \u0011 \u0010 \u0010 |f˜(ω)|2 exp −|f˜(ω)|2 p(m) = m!\n\n11.5.9 Quantization in a cavity 1. It is enough to demonstrate the commutation relation at t = 0, because we can multiply the result on the left by exp(iHt/~) and on the right by exp(−iHt/~) to obtain the result for any t. We note ˙ r ). We find for the time derivative of Φ ΦH (~r, t = 0) = Φ(~r), Φ˙ H (~r, t = 0) = Φ(~ s \u0011 X√ \u0010 ˙ r ) = −i ~ ωk ak − a†k ϕk (~r) Φ(~ 2µ k\n\n˙ is and the commutation relation between Φ and µΦ i X r ωk h ˙ r ′ )] = − i~ aj + a†j , ak − a†k ϕj (~r)ϕk (~r ′ ) [Φ(~r), µΦ(~ 2 ωj j,k r i~ X ωk 2 δjk ϕj (~r)ϕk (~r ′ ) = 2 ωj j,k\n\n= i~δ(~r − ~r ′ )I\n\n2. Orthogonality of the eigenfunctions: Z Z L \u0011 L 1 L \u0010 ′ dx cos[(k − k ′ )x] − cos[(k + k ′ )x] = δk,k′ dx sin kx sin k x = 2 0 2 0\n\n65 Completeness relation: let f (x) =\n\nX\n\nck sin kx\n\nk\n\nj≥1\n\nπj j = 1, 2, · · · L\n\nbe the Fourier expansion of a function such that f (0) = f (L) = 0. Let us compute the integral I\n\nZ\n\n=\n\nL\n\ndx f (x)\n\n0\n\nX\n\n=\n\nk′\n\nX\n\nsin kx sin k x = sin kx\n\nZ\n\nL\n\n0\n\nk\n\nX\n\nck′ sin k ′ x sin kx\n\nk,k′\n\n\u0012 \u0013 LX L L ′ ck ′ = ck sin kx′ = f (x) sin k x 2 2 2 k\n\nIt follows that\n\n2X (sin kx sin kx′ ) = δ(x − x′ ) L k\n\nThe functions\n\nr\n\n2 sin kx e±iωt L obey the vanishing boundary conditions, the orthogonality and completeness relations on the interval [0, 1] as well as the wave equation (11.59) with ωk = c|k|. Thus, they can be used for the expansion of the quantized field ΦH (x, t) s \u0011 ~ X 1 \u0010 ak e−iωt + a†k eiωt sin kx ΦH (x, t) = √ µL ωk ϕk (x, t) =\n\nk≥1\n\n3. The functions ϕ~k (~r, t) =\n\nr\n\n8 sin xkx sin yky sin zkz e±iω~k t V\n\nV = Lx Ly Lz\n\nobey the vanishing boundary conditions as well as the wave equation \u0012 2 \u0013 ∂ 2 2 − c ∇ ϕ~k (~r, t) = 0 ∂t2 ~ H (~r, t) by analogy with (11.79) if ω~k2 = c2~k 2 . We can then write the expansion of the quantized field A ~AH (~r, t) =\n\nr\n\n2 i 4~ X X 1 h ˆ −iωk t + a† ~e ∗ (k)e ˆ iωk t sin xkx sin yky sin zkz a~ks~es (k)e √ s ~ ks ε0 V ωk s=1 ~ k\n\nIf we consider a single space dimension, the normalization is\n\n11.5.11 Non Abelian gauge transformations\n\np ~/ε0 L.\n\n1. The expression for ~ ′ is Let us first remark that We wish to have ~ = ~ ′ whence the condition\n\n\u0011 \u0010 ~ − qA ~ ′ ΩΦ ~ ′ = Φ† Ω−1 −i∇ ~ ~ ∇(ΩΦ) = (∇Ω)Φ + Ω∇Φ \u0010 \u0011 ~ − iΩ−1 (∇Ω) ~ ~ ′Ω Φ − q Ω−1 A ~ ′ = Φ† −i∇ ~ − qA ~ = −i∇ ~ − iΩ−1 (∇Ω) ~ ~ ′Ω −i∇ − q Ω−1 A\n\n66\n\nCHAPTER 11. EXERCISES FROM CHAPTER 11\n\nthat is\n\n−1 ~ ~ ′ = ΩAΩ ~ −1 − i (∇Ω)Ω A q\n\n~ is a number, not a matrix In the Abelian case, the field A ~ −1 = A ~ ΩAΩ and we recover\n\n~′ = A ~ − ∇Λ ~ A\n\n2. Let us choose an infinitesimal gauge transformation \u0012 \u0013 X 1 Ω = I − iq Λa (~r) σa = I − iqT 2 a Then\n\n~ −1 ≃ A ~ − iq[T, A] ~ ΩAΩ\n\nand −1 ~ (∇Ω)Ω ≃ −iq\n\nX a\n\n~ a ∇Λ\n\n\u0012\n\n1 σa 2\n\n\u0013\n\nThe commutation relations (3.52) give \" # \u0012 \u0013 X \u00121 \u0013 X \u00121 \u0013 X 1 ~ ~ ~ [T, A] = Λb =i εabc Λb Ac Ac σb , σc σa 2 2 2 c b\n\nb,c\n\nwhence, by identifying the coefficient of σa /2 ~a = A ~ ′a − A ~ a = −∇Λ ~ a+q δA\n\nX\n\n~c εabc Λb A\n\nb,c\n\n3. We established the form of the covariant derivative by requiring that ~ ′ = ~, and, from our construction ~ −1 = D ~′ ΩDΩ We can write the time-independent Schr¨odinger equation (~ = m = 1)\n\nin the form\n\n1 \u0010 ~ \u00112 −iD Φ = EΦ 2\n\nΦ = Ω−1 Φ′\n\n1 −1 \u0010 ~ \u00112 −1 ′ Ω Ω −iD Ω Φ = EΩ−1 Φ′ 2 Multiplying the two sides of the equation by Ω we get 1 \u0010 ~ ′ \u00112 ′ −iD Φ = EΦ′ 2\n\n11.5.12 The Casimir effect 1. The only available physical parameters are L, ~ (we are dealing with a quantum problem) and c (the speed of electromagnetic waves). With these three parameters, we can form a unique combination with the dimension of a pressure, that is, ~c/L4 . 2. The stationary modes of the electric field have the form i h ˆ e±iωK t ~ r , t) = ei(xkx +yky ) sin πnz ~es (K) E(~ L\n\n67 ~ When n 6= 0, the three dimensional wave vector K ~ is of the where n is an integer ≥ 0 and ωK = c|K|. form \u0010 \u0011 ~ = kx , ky , ± πn K L ˆ The vanishing of the transverse component and there are two independent orthogonal directions for ~es (K). ~ of the electric field E for z = 0 and z = L is then guaranteed by the factor sin(πnz/L). When n = 0, ˆ must be parallel to Oz due to the condition that the transverse component of E ~ must vanish. In ~es (K) ~ ˆ addition, since ~es (K) is orthogonal to k, there exist a unique direction for polarization. 3. To any vector ~k correspond two polarization states (except if n = 0) and the zero point energy is   X ′ ~ ωn (~k) E0 (L) = 2 2 n,~ k\n\n4.Taking the continuum limit, (cf. Exercice 9.7.11), when the dimensions Lx , Ly → ∞ Z X S d2 k → (2π)2 ~ k\n\nHowever, since L stays finite, the sum over n remains discrete, and Z ~S X′ d2 k ωn (~k) E0 (L) = (2π)2 n,~ k\n\nThis integral is divergent for large frequencies (it is called an ultraviolet divergence). We introduce a cutoff χ, which physically represents the fact that a real metal does not remain a perfect conductor at large frequencies ! ∞ Z ~k) ~S X ′ ω ( n 2 E0 (L) = d k ωn (~k)χ (2π)2 n=0 ωc Setting\n\nc2 π 2 n 2 + c2 k 2 = ωn2 + c2 k 2 L2 and making the change of variables 2π d2 k = 2πkdk = 2 ωdω c we obtain \u0012 \u0013 ∞ Z ~S X ′ ∞ ω 2 E0 (L) = dω ω χ 2πc2 n=0 ωn ωc ω2 =\n\nω ≥ ωn\n\nωn =\n\nπcn L\n\n5. E0 (L) depends on L only through ωn = πnc/L and \u0012 \u0013 \u0012 \u0013 πnc ~S 2 dωn ~S 2 ω ω dE0 (L) = ω χ ω χ =− dL dL 2πc2 n ωc L2 2πc2 n ωc We derive the internal pressure Pint\n\n\u0012 \u0013 ∞ ∞ 1 dE0 (L) π 2 ~c X′ π 2 ~c X ′ 3 ω =− = − =− n χ g(n) S dL 2L4 n=0 ωc 2L4 n=0\n\nThe calculation of the external pressure is done by taking the the limit L → ∞, since outside the condenser, the field is not confined between plates. We may then replace the discrete sum over n with an integral Z π 2 ~c ∞ g(n) Pext = − 2L4 0\n\n68\n\nCHAPTER 11. EXERCISES FROM CHAPTER 11\n\nThe value of the total pressure is Ptot = Pint − Pext\n\n∞ X\n\nπ 2 ~c =− 2L4\n\nn=0\n\ng(n) −\n\nZ\n\n!\n\ng(n) 0\n\nFrom the Euler-Mac Laurin formula ∞ X\n\nn=0\n\ng(n) −\n\nZ\n\ng(n) =\n\n0\n\n1 +O 5!\n\n\u0012\n\nπc Lωc2\n\n\u0013\n\nπ 2 ~c 240 L4 An integration allows us to derive the zero point energy Ptot = −\n\nE0 (L) = −\n\nπ 2 ~c 720 L3\n\n11.5.13 Quantum computing with trapped ions 1. We write the interaction Hamiltonian in terms of σ+ and σ− i h 1 Hint = − ~ω1 [σ+ + σ− ] ei(ωt−kz−φ) + e−i(ωt−kz−φ) 2\n\nand go to the interaction picture using (see exercise 5.5.6)\n\neiH0 t/~ σ± e−iH0 t/~ = e∓iω0 t σ± In the rotating wave approximation, we can neglect terms which behave as exp[±i(ω0 + ω)t] and we are left with i h ˜ int ≃ − ~ ω1 σ+ e i(δt−φ) e−ik˜z + σ− e−i(δt−φ) e ik˜z H 2 p 2. ∆z = ~/(2M ωz ) is the spread of the wave function in the harmonic well. Thus, η = k∆z is the ratio of this spread to the wavelength of the laser light. We may write r ~ (a + a† ) = η(a + a† ) k˜ z=k 2M ωz ˜ int between the states |1, m + m′ i and |0, mi is The matrix element of H ˜ int |mi = − 1 ~ω1 ei(δt−φ) hm + m′ |e−iη(a+a† ) |mi h1, m + m′ |H 2 The Rabi frequency for oscillations between the two levels is ′\n\nω1m→m+m = ω1 |hm + m′ |e−iη(a+a ) |mi| 3. Writing †\n\ne±iη(a+a\n\n)\n\n≃ I ± iη(a + a† )\n\nand keeping terms to first order in η we get ˜ int H\n\n= +\n\nh i η~ω1 σ+ a ei(δ−ωz )t e−iφ − σ− a† e−i(δ−ωz )t eiφ 2 i\n\nσ+ a† ei(δ+ωz )t e−iφ − σ− a e−i(δ+ωz )t eiφ\n\n69 ˜ int corresponds to a resonance at δ = ω − ω0 = ωz , that is, ω = ω0 + ωz , a blue sideband, The first line of H and the second line to a resonance at ω = ω0 − ωz , that is, a red sideband. The σ+ a term of the blue sideband induces transitions from |0, m + 1i to |1, mi, and the σ− a† term from |1, mi to |0, m + 1i. Now √ hm|a|m + 1i = hm + 1|a† |mi = m + 1 ˜ + as written in the statement of the problem with so that we get H int a ab = √ m+1\n\na†b = √\n\na† m+1\n\n√ The Rabi frequency is then ω1 m + 1. The same reasoning may be applied to the red sideband. |1, 2i |1, 1i |1, 0i\n\nω+ ω0\n\nω+ ω−\n\nω\n\n|0, 2i |0, 1i |0, 0i\n\nFigure 11.1: The transitions which are used are (0, 0) ↔ (0, 1) and (0, 1) ↔ (1, 2): blue sideband, ω+ = ω + ω0 and (0, 1) ↔ (1, 1): red sideband, ω− = ω− ω0 . 4. The rotation operators R(θ, φ) are given by\n\nso that\n\nR(θ, φ = 0) = \u0010 π\u0011 R θ, φ = = 2 R(π, 0) = −iσx\n\nWe have, for example, \u0010 π\u0011 \u0010 π\u0011 R π, R(β, 0)R π, = 2 2\n\n=\n\nθ θ − iσx sin 2 2 θ θ I cos − iσy sin 2 2\n\nI cos\n\n\u0010 π\u0011 = −iσy R π, 2 β β + iσx sin )(−iσy ) 2 2 β β −(I cos + iσx sin ) = −R(−β, 0) 2 2 (−iσy )(I cos\n\nLet us call A the transition |0, 0i ↔ |1, 1i and B the transition |0, 1i ↔ |1, 2i. The Rabi frequencies √ √ are linked by ωB = 2 ωA . Thus, if the rotation angle is θA for transition A, it will be θB = 2 θA for\n\n70\n\nCHAPTER 11. EXERCISES FROM CHAPTER 11\n\n√ transition B. For transition A, we choose α = π/ 2 and β = π \u0010 π π\u0011 \u0010 π π\u0011 R √ , R(π, 0)R √ , R(π, 0) = −I 2 2 2 2 √ For transition B we shall have α = π and β = π 2 \u0010 π\u0011 \u0010 π\u0011 √ √ R(π 2, 0)R π, R(π 2, 0) = −I R π, 2 2\n\nThe state |1, 0i is not affected because the transition |0, 0i ↔ |1, 0i does not resonate on the blue sideband frequency. Thus we have |00i ↔ −|0, 0i\n\n|0, 1i ↔ −|0, 1i\n\n\u0010 \u0011 5. R ± π, π/2 = ∓iσy so that\n\n\u0010 π\u0011 R ± π, |0, 1i = ∓|1, 0i 2\n\n|1, 0i ↔ +|1, 0i\n\n|1, 1i ↔ −|1, 1i\n\n\u0010 π\u0011 R ± π, |1, 0i = ±|0, 1i 2\n\nLet us start from the general two ion state, where both ions are in the vibrational ground state |Ψi =\n\n=\n\n(a|00i + b|01i + c|10i + d|11i) ⊗ |0i\n\na|00, 0i + b|01, 0i + c|10, 0i + d|11, 0i\n\nThe action of R−(2) (−π, π/2) on ion 2 gives |Ψ′ i = R−(2) (−π, π/2)|Ψi = a|00, 0i + b|00, 1i + c|10, 0i + d|10, 1i +(1)\n\nThen we apply Rαβ\n\non ion 1 +(1)\n\n|Ψ′′ i = Rαβ |Ψ′ i = −a|00, 0i − b|00, 1i + c|10, 0i − d|10, 1i and finally R−(2) (π, π/2) on ion 2 |Ψ′′′ i = R−(2) (π, π/2)|Ψ′′ i =\n\n=\n\n−a|00, 0i − b|01, 0i + c|10, 0i − d|11, 0i (−a|00i − b|01i + c|10i − d|11i) ⊗ |0i\n\nThis is the result of applying a cZ gate, within trivial phase factors.\n\nChapter 12\n\nexercises from Chapter 12 12.5.1 The Gamow peak 1. Let us compute the Coulomb energy for R = 1 fm \u0012 2\u0013\u0012 \u0013 1 ~c e = × 200 MeV ≃ 1.5 MeV E0 = ~c R 137 The temperature at the center of the Sun is of order of 1.5 × 107 K, corresponding to a kinetic energy E of about 1.5 keV. We thus have E ≪ e2 /R. 2. We define the distance RN as the distance where the Coulomb potential is equal to the kinetic energy: e2 /RN = E. The integral to be evaluated for computing the tunnel effect is I=\n\nZ\n\nRN\n\ndr\n\nR\n\n\u0012\n\ne2 −E r\n\n\u00131/2\n\nWe make the change of variables u2 = The integration limits are\n\ne2 −E r\n\ndr = −\n\n2e2 udu (u2 + E)2\n\ne2 e2 −E ≃ R R \" r r # 2 1 πe2 1 e R √ I = 2e2 √ tan−1 ≃ − RE 2 e2 2 E 2 E u = 0 et u2 =\n\nThe probability for tunneling is ln pT (E) = − Defining EB =\n\n√ 2µ πe2 √ ~ E\n\n2µπ 2 e4 ~2\n\nwe cast the probability for tunneling into the form pT (E) ≃ exp −\n\nr\n\nEB E\n\n!\n\nThe numerical values are, with µ = 6mp /5 EB = 2π 2 α2 µc2 = 1.18 MeV\n\n71\n\n72\n\nCHAPTER 12. EXERCISES FROM CHAPTER 12\n\n3. The factor ∼ 4π/k 2 is a geometrical factor which must appear in the total cross section: cf. (12.52). It gives the order of magnitude of the total cross section in the absence of any other information. Here we must in addition take tunneling into account: the potential barrier must be crossed, otherwise the reaction does not occur. An order of magnitude of the total cross section is obtained if we multiply the geometrical factor by the tunneling probability σ(E) ∼\n\n4π p (E) k2 T\n\n3. The angular integration gives a factor 4πv 2 and the average value of vσ is given by \u00133/2 Z ∞ \u0013 \u0012 \u0012 µv 2 µ 3 hvσi = 4π dv v σ(v) exp − 2πkB T 2kB T 0 We make the change of variables v → E 2π~2 2π~2 σ(E) = pT (E) = exp − µE µE\n\n1 E = µv 2 2 which leads to hvσi =\n\n16π 2 ~2 µ3\n\nWe must study the integral\n\n\u0012\n\nJ=\n\n\u00133/2 Z\n\nµ 2πkB T\n\nZ\n\ndE e−E/(kB T ) e−\n\nr\n\nEB E\n\n!\n\nEB /E\n\n0\n\ndE e−E/(kB T ) e−\n\nEB /E\n\n0\n\nLet us define the function f (E) by (β = 1/(kB T )) r EB f (E) = βE − E\n\n√ EB f (E) = β − 2E 3/2 ′\n\nThe maximum of exp[−f (E)] is reached for E = E0 , where E0 obeys f ′ (E0 ) = 0 E0 =\n\n\u0014\n\nand we find f (E0 ) = −\n\np 1 kB T EB 2\n\n1 2\n\n\u0012\n\n2EB kB T\n\n\u00152/3\n\n\u00131/3\n\n≃ −5.8\n\nThe width of the peak exp[−f (E)] is obtained from the second derivative f ′′ (E0 ) = and the width of the peak is\n\n3 3p −5/2 EB E0 = 4 4\n\n\u0012 \u0013−5/3 1 −1/3 EB (kB T )−5/3 2\n\n1/6\n\n∆E ∼ EB (kB T )5/6 = 4.5 keV\n\n12.5.2 Low energy neutron scattering by a hydrogen molecule 1. Let r1 be the distance between nucleus 1 and the detector ~r1 = ~r −\n\n1 ~ R 2\n\nr1 ≃ r −\n\n1 ~ R · rˆ 2\n\nThe amplitude for finding the scattered wave at the detector after scattering by nucleus 1 is proportional to ! ~ · rˆ R 1 ikr 1 ikr1 ~′ ~ 1+ e−ik ·R/2 ≃ e e r1 r 2r\n\n73 with ~k ′ = kˆ r . We must multiply this result by the amplitude for finding the incident plane wave at ~r1 ~ which is exp(i~k · R/2) and by a1 . The scattering amplitude by nucleus 1 is finally ! ! \u0012 \u0012 \u0013 \u0013 ~ · rˆ ~ · rˆ R R i i ~ ~′ ~ a1 ikr a1 ikr ~ 1+ 1+ exp exp − ~q · R e (k − k ) · R = e r r 2 r 2r 2 By adding the amplitudes from nuclei 1 and 2, we obtain the scattering amplitude a as the coefficient of (exp ikr)/r i ~ a = a1 + a2 − (a1 − a2 )~q · R 2 The terms proportional to R/r are negligible as r → ∞.\n\n2. The distance between the two protons is on the order of 1 ˚ A. If q = 1010 m−1 , the neutron energy is E=\n\n~2 q 2 = 2 meV 2mn\n\n~ we have which corresponds to a temperature of 1 K. Neglecting the term proportional to (~q · R), a ˆ\n\n= =\n\n1 (as + 3at )I + 4 1 (as + 3at )I + 2\n\n1 1 1 (at − as )(~σn · ~σ1 ) + (as + 3at )I + (at − as )(~σn · ~σ2 ) 4 4 4 1 ~ (at − as )(~σn · Σ) 2\n\n3. The reduced mass is (mp ≃ mn = m) µH2 =\n\n2m 2m2 = m + 2m 3\n\nwhile the reduced mass of the neutron-proton system is µp = m/2. The effective potential is of the form (12.41) with a constant g given by 2π~2 a g= µ For the same value of g the ratio of the scattering lengths is then aH2 µH2 4 = = ap µp 3 4. The total cross section is obtained using a ˆ2 a ˆ2 =\n\n1 1 ~ + 1 (at − as )2 (~σn · Σ) ~ 2 (as + 3at )2 I + (as + at )(at − as )(~σn · Σ) 4 2 4\n\nand, for neutron scattering on polarized molecules σtot = 4πTr a ˆ2 ~ vanishes, so that In the parahydrogen case, the average value Σ para σtot = π(as + 3at )2\n\nIn the orthohydrogen case, one remarks that Tr (A ⊗ B)2 = Tr (A2 ⊗ B 2 ) = Tr A2 Tr B 2 Since Tr σni σnj Σi Σj ∝ δij\n\n74\n\nCHAPTER 12. EXERCISES FROM CHAPTER 12\n\nwe obtain ~ 2 Tr (~σn · Σ)\n\n\u0003 \u0002 2 2 2 2 = Tr σnx Σx + σny Σ2y + σnz Σ2z \u0003 \u0002 ~ 2 = 12 = 2Tr Σ2x + Σ2y + Σ2z = 2Tr Σ\n\nThe unpolarized cross section contains a factor 1/6 owing to the average over the initial spins (1/2 from the neutron and 1/3 for the orthohydrogen) and \u0014 \u0015 1 4π 1 2 2 ortho (as + 3at ) × 6 + (at − as ) × 12 σtot = 6 4 4 para σtot + 2π(at − as )2\n\n=\n\n12.5.3 Analytical properties of the neutron-proton scattering amplitude 1. The radial wave function is r < R : u(r) = A sin k ′ R\n\nr > R : u(r) = N e−κr\n\nThe continuity condition for the wave function at r = R gives A sin k ′ R = N e−κR and that of its logarithmic derivative cot κR = −\n\nk′ sin k ′ R = p κ2 + k ′ 2\n\nκ k′\n\nthat is\n\nκ cos k ′ R = − p κ2 + k ′ 2\n\np κ2 + k ′ 2 A = Ne k′ The normalization is obtained from the two integrals −κR\n\n2\n\nJ< = A\n\nZ\n\nR\n\ndr sin2 k ′ r =\n\n0\n\nJ> = N 2\n\nZ\n\ni N 2 e−2κR h ′2 2 κ + R(k + κ ) 2k ′\n\ndr e−2κr =\n\nR\n\nand the sum J< + J> has the value J< + J> =\n\nN 2 −2κR e 2κ\n\nN 2 e−2κR ′ 2 (k + κ2 )(1 + κR) 2k ′ 2 κ\n\nwhence N 2\n\n2\n\nN2 =\n\n2κk ′ e2κR (κ2 + k ′ 2 )(1 + κR)\n\n2. The function g(−k, r) is linearly independent of g(k, r) because it behaves at infinity as exp(−ikr). The most general solution of the Schr¨odinger equation is a linear combination of these two solutions, and because u(k, r) must in addition vanish at r = 0, we have u(k, r) = g(−k, r)g(k) − g(k, −r)g(−k)\n\ng(k) = g(k, r = 0)\n\nThe r → ∞ behavior of u(k, r) is then r → ∞ : u(k, r) ∝ eikr g(k) − e−ikr g(−k)\n\n75 Comparing with (12.22) for l = 0 then shows that S(k) = e2iδ(k) =\n\ng(k) g(−k)\n\n3. Let us continue g(k, r) to complex values of k: g(k, r) and g(−k ∗ , r) obey the differential equations i h d2 ∗ ∗2 − 2mV (r) g ∗ (k, r) g (k, r) + k dr2 i h d2 g(−k ∗ , r) + k ∗ 2 − 2mV (r) g(−k ∗ , r) 2 dr\n\n= 0 = 0\n\nWriting k = k1 + ik2 , we find the following asymptotic behaviors \u0001∗ g ∗ (k, r) ∝ eikr = e−ik1 r e−k2 r g(−k ∗ , r)\n\ne−ik\n\nr\n\n= e−ik1 r e−k2 r\n\nThe two functions obey the same differential equations and have the same behavior at infinity, so that they must be identical. We thus derive g ∗ (k ∗ ) = g(−k), whence g ∗ (k ∗ ) g(−k) 1 = = g ∗ (−k ∗ ) g(k) S(k)\n\nS ∗ (k ∗ ) = and S(−k) =\n\n1 g(−k) = = S ∗ (k ∗ ) g(k) S(k)\n\n4. The behavior of the function g(k, r) is given for r < R and r > R by rR\n\nAe−ik r + Ceik r e−ikr\n\ng(k, r) = g(k, r) =\n\nThe continuity conditions for g(k, r) and of its derivative leads to the following ′\n\nk′ with k ′ =\n\n\u0010\n\nAe−ik R + Ceik R \u0011 ′ ′ Ae−ik R − Ceik R\n\n=\n\ne−ikR\n\n=\n\nke−ikR\n\np M (V0 + E). From this we deduce\n\n\u0012 \u0013 k g(k) = A + C = e−ikR cos k ′ R + i ′ sin k ′ R k\n\nOne observes that g(k) is indeed an entire function of k. The only delicate point could come from the point k ′ = 0 owing to the square root in the definition of k ′ , but there is in fact no problem as cos k ′ R 2 and (1/k ′ ) sin k ′ R are analytic functions of k ′ . 5. Let us assume that S(k) possesses a pole on the positive imaginary axis at k = iκ, κ > 0. We then have g(−k) = g(−iκ) = 0 and the asymptotic behavior of u(k, r) is r → ∞ : u(k, r)\n\n∼ eikr g(k) + e−ikr g(−k) ∼ e−κr g(k) + eκr g(−k)\n\nu(k, r) explodes if r → ∞, unless g(−k) = 0, and then u(k, r) is a bound state wave function u(k, r) = g(k)g(−k, r) which automatically vanishes at r = 0. The poles of S(k) on the imaginary axis such that 0 < Im k < µ/2 then give the bound state energies. Let us now assume a pole of S(k) at k = h−ib. Owing to the properties\n\n76\n\nCHAPTER 12. EXERCISES FROM CHAPTER 12\n\ndemonstrated at question 3, S(k) also has a pole at k = −h − ib. If b < 0, g(h + ib, r) and g(−h + ib, r) are square integrable functions (they behave as exp(−|b|r) at infinity) and since they are solutions of the Schr¨odinger equation, they are orthogonal Z ∞ Z ∞ dr g(h + ib, r)g(−h + ib, r) = dr |g(h + ib, r)|2 = 0 0\n\n0\n\nand we have a contradiction if b < 0. If h 6= 0, the only possibility is to have poles such that Im k < 0. 6. The following choice for S(k) S(k) =\n\nk − h − ib (k − h − ib)(k + h − ib) ≃ for k ∼ h (k − h + ib)(k + h + ib) k − h + ib\n\nobeys the properties stated in question 3. The relation between cot δ and S is cot δ = i\n\ne2iδ + 1 S +1 h−k =i = e2iδ − 1 S −1 b\n\nthat is δ(k) = tan−1\n\n−b k−h\n\nThe phase shift δ increase from δ(k) = tan−1 b/h for k = 0 to π for k → ∞, going through π/2 for k = h. The total cross section is σtot = Let us set\n\n4πb2 4π 4π 2 = sin δ = k2 k 2 [(k − h)2 + b2 ] k 2 (1 + cot2 δ)\n\n2mE0 2mE h2 = 2 ~ ~2 2 2 2 2 k −h m(E 2 − E02 ) k −h √ ≃ = k−h= k+h 2h ~ 2mE0 k2 =\n\nWe find σ(E) = provided we set\n\n2π~2 ~2 Γ2 /4 M E (E − E0 )2 + ~2 Γ2 /4 ~ 2 Γ2 2E0 ~2 b2 = 4 m\n\n7. Let us start from the radial Schr¨odinger equation u′′ − 2mV u + k 2 u = 0 and let us differentiate this equation with respect to k ∂u′′ ∂u ∂u − 2mV + k2 = −2uk ∂k ∂k ∂k We then multiply the first equation by ∂u/∂k, the second one by u and subtract the second equation from the first ∂u′′ ∂u −u = 2ku2 u′′ ∂k ∂k that is \u0012 \u0013 ∂ ∂u′ ′ ∂u u = 2ku2 −u ∂r ∂k ∂k\n\nIntegrating over r we get\n\n∂u′ −u = 2k u ∂k ∂k ′ ∂u\n\nZ\n\n0\n\nr\n\nu2 (r′ )dr′\n\n77 When r → 0 we have Furthermore\n\n\u0015 \u0014 ∂k ′ ∂u ′ ′ ′ cos k r =0 = A (k) sin k r + A(k)r ∂k r=0 ∂k r=0 \u0001 ∂u = g(k, r)g ′ (−k) + O e−κr ∂k\n\nand for a bound state\n\nu(k, r) = g(−k, r)g(k) =⇒ if r → ∞ u′ (k, r) = iku(k, r) As a consequence, for r → ∞\n\n∂u ∂u′ −u → 2 i k g(iκ)[g ′ (−iκ)] ∂k ∂k In the vicinity of the pole k = iκ, g(−k) is proportional to (k − iκ) u′\n\ng(−k) ≃ D(k − iκ) and then g ′ (−k) = −D. Setting g(iκ) = F , we obtain Z ∞ −2ikDF = 2k u2 (k, r) dr 0\n\nThe behavior r → ∞ of u(k, r) is r → ∞ : u(k, r) ≃ g(iκ)e−κr = F e−κr Let u(k, r) = Gu(k, r) be the normalized bound state wave function, which behaves at infinity as N exp(−κr), where N is the constant defined at question 1, N = F G. One the one hand we have Z ∞ dr u2 (k, r) = −iDF 0\n\nand on the other hand Z Z ∞ 1= dr u2 (k, r) = G2\n\n0\n\n0\n\ndr u2 (k, r) = −iF G2 = −i(F G)2\n\nD D = −iN 2 F F\n\nIn the vicinity of the pole at k = iκ S(k) ≃\n\n−iN 2 F = D(k − iκ) k − iκ\n\n8. Let us express k cot δ as a function of g(k) k cot δ(k) =\n\nik[g(k) + g(−k)] g(k) − g(−k)]\n\nThis function is analytical for k ∼ 0, it tends to a constant for k → 0 and it is an even function of k. We may then write its Taylor expansion as 1 1 k cot δ(k) = − + r0 k 2 + O(k 4 ) a 2 Taking S(k) =\n\nk cot δ(k) + ik k cot δ(k) − ik\n\ninto acount, the existence of a pole of S(k) at k0 = iκ implies k0 cot δ(k0 ) = ik0 = −κ\n\n1 1 − r0 κ2 = −κ a 2\n\n78\n\nCHAPTER 12. EXERCISES FROM CHAPTER 12\n\nthat is\n\n\u0012 \u0013 1 1− κa\n\n2 r0 = κ\n\nLet us calculate the residue at the pole k cot δ − ik\n\n= =\n\nthat is S(k) =\n\n∂ k cot δ − ik + ik0 − ik0 ∂k k=k0 −i(k − k0 ) + (k − k0 )iκr0 = −i(k − k0 )(1 − κr0 )\n\nk0 cot δ0 + (k − k0 )\n\n2κ −2κ =⇒ N 2 = −i(k − k0 )(1 − κr0 ) 1 − κr0\n\n12.5.5 Neutron optics\n\n√ 1.The distance between the nucleus and the observation point is r = s2 + z 2 . The total probability amplitude at the observation point is obtained by adding coherently the amplitudes from each nucleus Z ∞ ikr Z ∞ ikr e e 2πsds = −2πaρδ rdr ϕd = −aρδ r r z 0 as rdr = sds. The upper bound of the integral is an oscillating function. However, this oscillation has no physical meaning and we obtain the total amplitude in the form \u0012 \u0013 aρ δ ϕ(z) = 1 − 2iπ e ikz k 2. If the neutrons cross a medium with index n and thickness δ with k ′ = nk ϕ(z) = =\n\neik(z−δ) eik δ = eikz ei(k −k)δ) eikz ei(n−1)kδ ≃ eikz [1 + i(n − 1)kδ]\n\nand by identification with the result of question 1 we get n=1−\n\naρλ2 2πaρ = 1 − k2 2π\n\nAs the index of refraction is very close to one, the critical reflection angle is close to π/2 sin(π/2 − θc ) = cos θc ≃ 1 − that is\n\naρλ2 1 2 θc = 1 − n = 2 2π\n\n1 2 θ =n 2 c\n\nθc = λ\n\n3. The neutron-proton scattering matrix in spin space is\n\n\u0010 ρa \u00111/2 π\n\n1 1 fˆ = − (as + 3at )I − (at − as )(~σn · ~σp ) = −as Ps − at Pt 4 4 As | + +i is a triplet state\n\nfc = h+ + |fˆ| + +i = −at\n\nIf |χs i et χt i are the singlet and triplet states corresponding to m = 0, the relation inverse to (10.125) et (10.126) is | + −i = | − +i =\n\n1 √ (|χt i + |χs i) 2 1 √ (|χt i − |χs i) 2\n\n79 and fa\n\n=\n\nfb\n\n=\n\n1 h+ − |fˆ| + −i = − (at + as ) 2 1 h+ − |fˆ| − +i = − (at − as ) 2\n\nThe weights 3/4 and 1/4 are determined by the degeneracy of the triplet (3) and singlet (1) states. The coherent scattering length 1 3 aeff = at + as 4 4 has the numerical value −1.9 fm, and the refraction index is larger than one: there cannot be total reflection. 4. Taking the square of ~ I~ = J~ + ~σ 2 that is\n\n~2 I~2 = J~2 + ~σ 2 + ~J~ · ~σ 4\n\nwe obtain 1 ~ J · ~σ ~ 1 ~ J · ~σ ~\n\n= j if I = j +\n\n1 2\n\n= −(j + 1) if I = j −\n\n1 2\n\nThe scattering lengths are given by a+ = a + bj\n\na− = a − b(j + 1)\n\nor conversely a=\n\n1 [(j + 1)a+ + ja− ] 2j + 1\n\nb=\n\n1 (a+ − a− ) 2j + 1\n\n5. The a amplitude corresponds to a scattering without spin flip, and thus to coherent scattering, and the b amplitude to a scattering with spin flip, and thus incoherent. Another way of finding the result is to observe that the probability for scattering in the state (j + 1/2) is (j + 1)/(2j + 1) and j/(2j + 1) in the state (j − 1/2). Using the results of 1.6.8 we find σcoh\n\n=\n\nσinc\n\n=\n\n4π 2 [(j + 1)a+ + ja− ] (2j + 1)2 j(j + 1) (a+ − a− )2 4π (2j + 1)2\n\n12.5.6 The cross section for neutrino absorption 1. The matrix element hϕf |ϕi i is given by Z 1 1 hϕf |ϕi i = 2 d3 r ei(~p1 +~p2 +~p3 )·~r/~ = δp~1 +~p2 +~p3 ,0 V V where ~pi represent the momenta of the final particles. The symbol δ represents a Kronecker δ since we have used plane waves in a box. Taking the square of the preceding equation, we obtain |hϕf |ϕi i|2 =\n\n1 δp~ +~p +~p ,0 V2 1 2 3\n\n80\n\nCHAPTER 12. EXERCISES FROM CHAPTER 12\n\nLet us show that the recoil kinetic energy K of the proton is negligible. The relation P~ + ~q + ~p = 0 shows that the three momenta are a priori of the same order of magnitude, ∼ p. On the other hand, the electron kinetic energy k is p2 p2 ≫K∼ k∼ 2me 2mp The final density of states is then D(E) = V 2\n\nZ\n\nd3 p d3 q δ(K + E + cq − E0 ) 3 (2π~) (2π~)3\n\nLet us integrate over the angles V\n\nd3 p (2π~)3\n\n4πV 2 4πV pEdE p dp = (2π~)3 (2π~)3 c2\n\nV\n\nd3 q (2π~)3\n\n4πV 2 q dq (2π~)3\n\nEnergy conservation gives q = (E − E0 )/c and on integrating the level density over q D(E0 ) =\n\nV2 pE (E − E0 )2 (2π 2 )2 ~6 c2 c3\n\nThis gives for the decay rate par unit energy E dΓ G2 h|Mf i |2 i = 2π F 4 7 5 pE(E0 − E)2 dE 4π ~ c and integrating over E while neglecting the electron mass we obtain the expression for the lifetime 1 G2F E05 =Γ∼ τ 60π 3 ~(~c)6 From the expression\n\nG2F 60π 3 ~Γ = (~c)6 E05\n\nwe deduce that G2F /(~c)6 has the dimension of minus the fourth power of an energy and is expressed, for example, in MeV−4 . For the numerical evaluation, we convert the inverse lifetime in MeV ~ = ~Γ = 0.73 × 10−24 MeV τ which leads to\n\nGF = 2.3 × 10−11 MeV−2 = 2.3 × 10−5 GeV−2 (~c)3\n\nin qualitative agreement with the exact value. 2. The transition matrix element is 1 hϕf |ϕi i = (V)2\n\nZ\n\nd3 rei(~p1 +~p2 −~p\n\np ′ 2 )·~ r /~ 1 −~\n\n=\n\n1 δp~ +~p ,~p ′ +~p ′ V 1 2 1 2\n\nwhere p~ and p~ ′ represent the initial and final particles momenta. The differential cross section is 1 2π dσ = h|Mf i |2 ihϕf |ϕi i|2 D(E) dΩ F ~\n\n81 The final density of states for an electron of energy E emitted in the direction Ω is D(E) =\n\nV E2 (2π~)3 c3\n\nwhile the flux is given by F = c/V, as the neutrinos propagate with very nearly the speed of light and their density is 1/V. One finds for the differential cross section dσ G2 h|Mf i |2 i 2 E = F2 dΩ 4π (~c)4 and integrating over angles we obtain the total cross section for neutrino absorption σtot =\n\nG2F h|Mf i |2 i 2 E π (~c)4\n\nWe can get an order of magnitude, in a system of units where ~ = c = 1, 1 fm = 1/(200 Mev), σ ∼ G2F E 2 , GF = 10−5 GeV−2 . For E = 10 MeV, σ ∼ 10−20 MeV−2 = 4 × 10−16 fm2 = 4 × 10−46 m2 −3\n\nChoosing the density to be that of iron, n ≃ 1029 atoms/m ℓ=\n\n, we find for the mean free path ℓ\n\n1 ∼ 1015 m 30nσ\n\nwhere we have taken into account the fact that an iron nucleus contains 30 neutrons. The mean free path is about 1/10th of a light year, and detecting a neutrino is really an achievement! 3. The maximum inelastic cross section is σin,max = We must have\n\nπ π ≃ 2 k2 E\n\n1 E< ≃ 300GeV ∼ √ GF\n\nFermi’s theory is undoubtely limited to energies less than 300 GeV.\n\n82\n\nCHAPTER 12. EXERCISES FROM CHAPTER 12\n\nChapter 13\n\nExercises from Chapter 13 13.4.1 The Ω− particle and color If we build a spin 3/2 particle from three spins 1/2, the state vector is symmetrical with respect to any exchange of two spins: for example, the state j = 3/2, jz = 3/2 is 3 E 1 1 1 E 3 = , , j = , jz = 2 2 2 2 2\n\nIf the spatial wave function has no zeroes, it is necessarily symmetrical. Indeed, if it were antisymmetrical, for example with respect to the exchange of particles 1 et 2 ϕ(~r1 , ~r2 , ~r3 ) = −ϕ(~r2 , ~r1 , ~r3 ) it would vanish at the point ~r1 = ~r2 . The state vector space⊗spin of a system of three identical quarks must be antisymmetrical, and this is impossible if the three quarks are identical. As a matter of fact, quarks possess an additional quantum number, color, and the three quarks of the Ω− particle are of different color: it is the color wave function which is antisymmetrical.\n\n13.4.2 Parity of the π-meson 1. The π − -meson-deut´eron system is analogous to an hydrogen atom, apart from the reduced mass, which is different mD mπ 2 µ= = 129 MeV/c mD + mπ As a consequence, the energy levels are given by En = −\n\nR∞ 3.42 µ R∞ = −253 2 = − 2 keV 2 me n n n\n\nThe transitions are in the x-ray domain, whose energy lie between ∼ 0.1 and ∼ 100 keV. 2. As the orbital angular momentum vanishes, the total angular momentum is that of the deuteron, j = 1. The possible final states with angular momentum j = 1 are 3 S1 , 3 P1 , 1 P1 et 3 D1 , but only the state 3 P1 is antisymmetric in space+spin in the exchange of the two final neutrons. The parity of the final state is −1 (angular momentum Lfin = −1) and that of the initial sate is ηπ (−1)Linit = ηπ Parity conservation implies ηπ = −1.\n\n13.4.4 Positronium decay 1. Since the reduced mass is me /2, the energy levels are of the form En = −\n\n1 m2e e4 1 4 ~2 n2\n\n83\n\n84\n\nCHAPTER 13. EXERCISES FROM CHAPTER 13\n\n2. Adding two spins 1/2 gives either j = 1 (triplet state ), or j = 0 (singlet state). 3. As the projection along z of the orbital angular momentum vanishes, angular momentum conservation along this direction reads m = m1 + m2 where m is the projection along Oz of the positronium spin, m1 and m2 the projections of the spins of the two photons. One can a priori contemplate the following situations • two right handed photons: m1 = 1 et m2 = −1, m1 + m2 = 0 • two left handed photons: m1 = −1 et m2 = 1, m1 + m2 = 0 • one left handed photon and one right handed photon: m1 + m2 = ±2 This last possibility is excluded as m = −1, 0 ou +1. 4. In a rotation by π about Oy, the two photons are exchanged. Since the photons are bosons, the global state vector must not change sign in this operation. With Y = exp(−iπJy ), we have from (10.102) Y |jmi = (−1)j−m |j, −mi The initial state changes sign if j = 1, because m = 0, and does not change sign if j = 0. As a consequence, only the singlet state j = 0 can decay into two photons. Generally speaking, a spin one particle cannot decay into two photons. 5. The parity of the positronium ground state is Π = ηe+ ηe− (−1)l = ηe+ ηe− where l = 0 is the orbital angular momentum. In a reflection with respect to a xOz plane, the initial state vector changes sign, because the operator Y which implements this reflection is Y = Π exp(−iπJz ). In the photon case, from (10.104) and taking into account the odd parity ηγ = −1 of the photon Y|Ri = −|Li\n\nY|Li = −|Ri\n\nThis shows that |Φ+ i does not change sign, while |Φ− i changes sign. Thus it is the two photon entangled state |Φ− i which is produced in the decay.\n\nQuantum statistics and beam splitters 1. The amplitude for mode a to be transmitted by the beam splitter is t = cos θ, and the amplitude to be reflected is r = i sin θ, where t and r are defined in Exercise 1.6.6. The factor i takes into account the π/2 phase shift between the transmitted and reflected beams. 2. Let us compute, for example, e iGθ a e−iGθ = a + iθ[G, a] −\n\n1 [G, [G, a]] + · · · 2!\n\nwhere we have made use of (2.54). A straightforward calculation gives the commutators [G, a] = [ab† + a† b, a] = b so that e iGθ a e−iGθ = a + iθb −\n\n[G, [G, a]] = [ab† + a† b, b] = a 1 2 θ a + · · · = a cos θ + ib sin θ 2!\n\n3. If the initial state is a two photon state |Ψ0 i = |1a , 1b i = a† b† |Ωi\n\n85 where |Ωi is the vacuum state, then the beam splitter transforms this state into |Ψi =\n\n= = =\n\nwhere we have used (a† )2 |0i =\n\nU (θ)|Ψ0 i = U (θ)a† b† |0a , 0b i = U (θ)a† b† U † (θ)|Ωi (a† cos θ + ib† sin θ)(ia† sin θ + b† cos θ)|Ωi \u0013 \u0012 \u0003 \u0002 i √ sin 2θ (a† )2 + (b† )2 + cos 2θ a† b† |Ωi 2 \u0001 i √ |2a , 0b i + |0a , 2b i + cos 2θ|1a , 1b i 2\n\n√ 2|2i [see (11.18)]. For a symmetric beam splitter with θ = π/4 we obtain \u0001 i |Ψi = √ |2a , 0b i + |0a , 2b i 2\n\nThe two photons stick together at the output of the beam splitter. 4. Let us first consider a single fermionic mode. The mode can be either unoccupied (state |0i) or occupied (state |1i), as, from the Pauli principle, there cannot be more than one fermion in the mode. The operators a and a† act in the two-dimensional Hilbert space spanned by these two vectors and they can be represented by the 2 × 2 matrices \u0012 \u0013 \u0012 \u0013 0 0 0 1 a= a† = 1 0 0 0 which leads to the anticommutation relation {a, a† } = I. When we consider two modes, the antisymmetry is ensured owing to the anticommutation relations. For example, |1a , 1b i = a† b† |Ωi = −b† a† |Ωi = −|1b , 1a i In order to compute e iGθ a e−iGθ we use the identity [AB, C] = A{B, C} − {A, C}B from which we find [G, b] = −a\n\n[G, a] = b so that e iGθ a e−iGθ\n\n= =\n\n1 2 θ a + ··· 2! a cosh θ + ib sinh θ\n\na + iθb +\n\nRepeating the calculation of question 3, we obtain |Ψi = (a† cosh θ − ib† sinh θ)(b† cosh θ + ia† sinh θ)|Ωi = a† b† |Ωi = |1a , 1b i\n\nThe two fermions must choose different outputs of the beam splitter: otherwise they would be in the same state!\n\n86\n\nCHAPTER 13. EXERCISES FROM CHAPTER 13\n\nChapter 14\n\nExercises from Chapter 14 14.6.1 Second order perturbation theory and van der Waals forces 1. Let us start from the eigenvalue equation to order λ2 H(λ)|ϕ(λ)i\n\n=\n\n(H0 + λW )|ϕ(λ)i\n\n=\n\n(E0 + λE1 + λ2 E2 )|ϕ(λ)i\n\nwith |ϕ(λ)i = |ϕ0 i + λ|ϕ1 i + λ2 |ϕ2 i\n\nand the auxiliary condition hϕ0 |ϕ(λ)i = 1, whence\n\nhϕ0 |ϕ1 i = −λhϕ0 |ϕ2 i We deduce from these two equations, to order λ2 (H0 − E0 )|ϕ2 i = (E1 − W )|ϕ1 i + E2 |ϕ0 i Multiplying on the left by the bra hϕ0 | and taking into account that hϕ0 |ϕ1 i is of order λ E2 = hϕ0 |W |ϕ1 i\n\n(14.1)\n\nFurthermore, from the identification of the terms of order λ for |ϕ(λ)i gives (H0 − E0 )|ϕ1 i = (E1 − W )|ϕ0 i Let us write the identity operator under the form (|ϕ0 i ≡ |ni) \u0010X \u0011 I = |nihn| + (H0 − E0 )−1 |kihk| (H0 − E0 ) k6=n\n\nand let us use the expression (14.1) of E2 E2 = hn|W |nihn|ϕ1 i +\n\nX\n\nk6=n\n\nhn|W |\n\n1 |kihk|(H0 − E0 )|ϕ1 i H0 − E0\n\nNeglecting the first term, which is justified because hn|ϕ1 i = O(λ) and using (14.2) E2 =\n\nX |hn|W |ki|2 E0 − Ek\n\nk6=n\n\n~ = 0 creates at point R ~ an electric field 2. A dipole moment d~ = qe~r1 located at R i h ~ − 3(d~ · R) ˆ R ˆ ~ =− 1 d E 4πε0 R3\n\n87\n\n(14.2)\n\n88\n\nCHAPTER 14. EXERCISES FROM CHAPTER 14\n\nand the potential energy of the two atom system is ~ W = −qe~r2 · E 3. hϕ01 ϕ02 |W |ϕ01 ϕ02 i = 0 because the average values of X1 , Y1 et Z1 vanish owing to parity conservation hϕ01 |X1 |ϕ01 i = 0 and the same property holds for the average values of X2 , Y2 and Z2 . 3. Using the completeness relation X α\n\nwe obtain E2 ≃ −\n\n|ϕα ihϕα | = I\n\n1 hϕ01 ϕ02 |W 2 |ϕ01 ϕ02 i 2R∞\n\nThe only terms of W 2 whose average value does not vanish are X12 X22 , Y12 Y22 et 4Z12 Z22 . Using rotational invariance we get 1 ~ 2 |ϕ0 i = 1 hR2 i = a20 hϕ01 |X12 |ϕ01 i = hϕ0 |R 3 3 The characteristic time for a fluctuation of the dipole moment of one of the atoms is τ ≃ ~/R∞ . In order that a static calculation such as that developed above be valid, one must be able to neglect the propagation time of light between the two atoms, and we must have R ≪ cτ = ~c/R∞ .\n\n14.6.3 Muonic atoms 1. The reduced mass of the problem is m′µ =\n\nmµ mµ /me ≃ 1 + mµ /mA 1 + mµ /(Amp )\n\nthat is m′µ = α(A)me\n\nα(A) =\n\nmµ /me 1 + mµ /(Amp )\n\nApplications • Aluminium: aZ=13 = 19.8 fm µ • Lead: aZ=82 = 3.1 fm µ\n\nR = 3.6 fm\n\nR = 7.1 fm\n\n2. Role of the electrons from internal shells. The wave function of the innermost electrons is ϕZ 0 (r) = p\n\n1\n\nZ\n\n3 π(aZ e)\n\ne−r/ae\n\naZ e =\n\na0 Z\n\nThe electric charge contained within an orbit of radius aZ µ is Q ≃ 2qe\n\naZ µ aZ e\n\n!3\n\n≃ 2qe\n\n\u0012\n\nme mµ\n\n\u00133\n\n∼ qe × 10−6\n\nThe binding energy 2p → 1s of the hydrogen atom is ∆EH ≃ (3/4)R∞ . In the case of a muonic aluminium atom, its value is 3 ∆EµAl = (13)2 α(27) R∞ = 354.9 keV 4 and in the case of lead 3 ∆EµPb = (82)2 α(208) R∞ = 14.2 MeV 4\n\n89 The approximation of a point nucleus is evidently not valid for lead, because the orbit radius is about half of the nuclear radius! This approximation cannot even be used as a zeroth order approximation. It would be wiser to use an harmonic oscillator approximation, using the potential of the following question. 3. The potential to be used in the perturbative calculation for r < R is \u0014 \u0015 2 Ze2 1 \u0010 r \u00112 −3+ W (r) = 2 R R r The W potential being nonzero only for r < R, the contribution of a state whose wave function vanishes at r = 0 (p wave, d wave etc.) is negligible. For an ns wave we use Z Z 3 2 2 d r W (r)|ϕns (~r)| ≃ |ϕns (0)| d3 r W (r) Taking\n\ninto account, we find\n\nZ\n\nd3 r W (r) =\n\n2π Ze2 R2 5\n\n2π Ze2 R2 |ϕns (0)|2 5 which give the following numerical value for the 2p → 1s transition \u0012 \u00132 \u0012 \u0013 \u0012 \u00132 R 4 mµ R 2Ze2 2 = = 12.6 keV R Z δE1s = ∞ Z Z 5aZ a 5 m a e µ µ µ δEns =\n\nTaking vacuum polarization into account, we have ∆E2p→1s = 354.9 − 12.6 + 2.2 = 344.5 keV in very good agreement with the experimental result. 4. The ratio of the fine structure characteristic energy to the ground state energy is the same for ordinary and muonic atoms: in both cases, it is proportional to α2 . By contrast, the ratio of the hyperfine structure to the ground state energy is larger by a factor ∼ mµ /me for muonic atoms. Indeed, if we look at (14.32) taking into account the fact that the ground state energy is proportional to 1/a, one must take into account a factor 1/a3 ∼ (mµ /me )3 and a factor me /mµ coming from the ratio of the electron magnetic moment to the muon magnetic moment.\n\n14.6.4 Rydberg atoms 1. In the case l = n − 1 the expansion of unl (r) is reduced to a single term \u0012 \u0013n \u0013 \u0012 r r unl (r) = c0 exp − a0 na0 l=n−1\n\nLet us set x = r/a0 and let us study the function f (x) = xn exp(−x/n), or, in an equivalent way its logarithm g(x). The function f (x) displays a sharp maximum at x0 which is determined by studying g ′ (x), g ′ (x0 ) = 0 1 n x0 = n2 g ′ (x) = − x n Let us also compute the second derivative g ′′ (x) = − and thus\n\nn x2\n\ng ′′ (x0 ) = −\n\n1 n3\n\n\u0013 \u0012 (x − x0 )2 f (x) ≃ f (x0 ) exp − 2n3\n\n90\n\nCHAPTER 14. EXERCISES FROM CHAPTER 14\n\nThe dispersion around the maximum at x0 is ∆x = n3/2 . When l = n − 1, the radial wave function is localized around a value a0 n2 with a dispersion a0 n3/2 . When l 6= n − 1, the exponential in unl (r) is multiplied by a polynomial in r, and not by a monomial, which makes the curve wider. 2. In the vicinity of θ = π/2 and setting δ = π/2 − θ we obtain \u0013l \u0013 \u0012 \u0012 1 2 1 2 ≃ exp − lδ sin θ = cos δ ≃ 1 − δ 2 2 l\n\nl\n\n√ √ The wave function is thus concentrated within an angular opening δθ ≃ 1/ l ≃ 1/ n, which gives the following dispersion along Oz 1 ∆z ≃ √ a0 n2 = a0 n3/2 n The horizontal dispersion (question 1) and the vertical one (question 2) being both proportional to a0 n3/2 , the wave function is indeed concentrated in a torus of radius a0 n3/2 drawn around a circle of radius a0 .\n\n14.6.6 Vacuum Rabi oscillations 2. The energy of the state |ϕgn i is Eng = n~ω and that of the state |ϕen i, Ene = ~ω0 + (n − 1)~ω. The energy difference between the two levels is then ∆En = Eng − Ene = ~δ and the two levels are almost degenerate if δ ≪ ω0 . 3. The coupling of the electromagnetic field with the dipole is r \u0003 ~ω \u0002 W = −id a(b + b† ) − a† (b + b† ) ε0 V \u0002 \u0003 i~ = − ΩR a(b + b† ) − a† (b + b† ) 2 W applied to a state |g ⊗ ni give two types of contribution (n) (i) Weg (n)\n\n(ii) W eg\n\n= =\n\n√ i~ i~ ΩR he ⊗ (n − 1)|ab† |g ⊗ ni = − ΩR n 2 2 √ i~ i~ † † he ⊗ (n + 1)|W |g ⊗ ni = − ΩR he ⊗ (n + 1)|a b |g ⊗ ni = − ΩR n + 1 2 2 he ⊗ (n − 1)|W |g ⊗ ni = −\n\nThe second term changes the energy by 2~ω. Let us compare the time evolution of the operators ab† and a† b† in the Heisenberg picture, with H0 = Hat + Hfield \u0001 \u0001 ab† → eiH0 t/~ a|eihg| e−iH0 t/~ = ab† ei(ω0 −ω)t = ab† e−iδt where we have used (11.67), while\n\n\u0001 a† b† → eiH0 t/~ a† |eihg| e−iH0 t/~ = a† b† ei(ω0 +ω)t\n\nand this term is negligible in the rotating wave approximation (cf. (Sec. 5.3.2)). We are left with only the term corresponding to the combination ab† . Owing to Hermitian conjugation, we must also keep the combination a† b. We finally obtain the Jaynes-Cummings Hamiltonian H = ~ω0 |eihe| + ~ωN −\n\ni~ ΩR (ab† − a† b) 2\n\nIn the subspace H(n) the matrix form of this Hamiltonian is \u0012 1 1 δ√ H = n~ωI − ~δI + ~ −iΩR n 2 2\n\n√ \u0013 iΩR n −δ\n\n91 4. From (2.35), the eigenvalues and eigenvectors in the basis {|ϕgn i, |ϕen i} are q 1 En− = − ~ δ 2 + nΩ2R |χn− i = sin θn |ϕgn i + i cos θn |ϕen i 2 q 1 ~ δ 2 + nΩ2R En+ = |χn+ i = cos θn |ϕgn i − i sin θn |ϕen i 2\n\nWhen δ = 0\n\n1 En± = ± ~δ 2\n\n1 g e |χ± n i = √ (|ϕn i ∓ i|ϕn i) 2\n\n5. We decompose the state |ei ≡ |e ⊗ 0i on the states |χ± 1i i − |ei = √ (|χ+ 1 i − |χ1 i) 2 which gives the time evolution for δ = 0, with E1± = ±~ΩR /2 \u0011 i \u0010 iΩR t/2 − |χ1 i |ei → |ψ(t)i = √ e−iΩR t/2 |χ+ 1i−e 2 The probabilty for finding |ei after a time t spent in the cavity is then pe (t) = |he|ψ(t)i|2 =\n\n2 1 −iΩR t/2 ΩR t + eiΩR t/2 = cos2 e 4 2\n\n6. Off resonance, pe (t) is given by the Rabi formula (4.36) pe (t) = 1 −\n\nΩ2 t Ω2R sin2 2 Ω 2\n\nFigure (4.5b) illustrates the decrease in the amplitudes of the oscillations off resonance. 7. If the cavity contains n photons, we shall observe oscillations between the states |ϕen i et |ϕgn i with a frequency q Ωn+1 = δ 2 + (n + 1)Ω2R\n\nIf the cavity contains a coherent state with an average photon number hni, we shall observe a superposition of oscillations whose frequencies are given by the results of the preceding question, the probability of each frequency being given by (11.34) hnin −hni pn = e n!\n\n14.6.7 Reactive forces 1. The eigenvalues and eigenvectors are given by (2.35). We find q 1 |χ1n (z)i : E1n = − ~ δ 2 + nΩ21 (z) 2 q 1 |χ2n (z)i : E2n = ~ δ 2 + nΩ21 (z) 2 The force on the atom in the state |χ1n i, for example, is F1n = − and F2n = −F1n .\n\n∂E1n 1 1 ∂Ω21 p = ~n ∂z 4 ∂z δ 2 + nΩ21 (z)\n\n92\n\nCHAPTER 14. EXERCISES FROM CHAPTER 14\n\n2. The transition amplitudes are given by an11 an21\n\n= =\n\nan12 an22\n\n= =\n\nhχ1,n−1 |(b + b† )|χ1n i = − sin θn−1 cos θn hχ2,n−1 |(b + b† )|χ1n i = cos θn−1 cos θn\n\nhχ1,n−1 |(b + b† )|χ2n i = − sin θn−1 sin θn hχ2,n−1 |(b + b† )|χ2n i = − cos θn−1 sin θn\n\nBy choosing in a suitable way the phase φ in (11.93), we obtain for the expectation value of the field EH (z, t) in the coherent state |zi, with |z|2 = hni r ~ω hz|EH (z, t)|zi = 2hni cos ωt sin kz ε0 V which leads to\n\ns\n\n~ωhni 1 = E0 ε0 V 2\n\nand for the atom-field coupling p ~Ω1 (z) hni = dE0 sin kz = ~ω1 (z)\n\nwhere ω1 (z) is the usual Rabi frequency (cf. for example (14.74)). 3. We find at once (θhni = θ) pst 1 =\n\nsin4 θ cos4 θ + sin4 θ\n\npst 2 =\n\ncos4 θ cos4 θ + sin4 θ\n\nFrom question 1, the force on the atom in state |χ1 i is F1 =\n\n1 ∂ω12 (z) 1 ~ 4 ∂z Ω1hni (z)\n\nand F2 = −F1 . The force on an atom then is st F = F1 (pst 1 − p2 ) =\n\nwith\n\nsin4 θ − cos4 θ 1 ∂ω12 (z) 1 p ~ 4 ∂z δ 2 + ω12 (z) cos4 θ + sin4 θ\n\n2δΩ1hni sin4 θ − cos4 θ =− 2 4 4 δ + Ω21hni (z) cos θ + sin θ\n\nAssembling all the factors\n\nδ 1 ∂ω 2 (z) F =− ~ 1 2 2 ∂z 2δ + ω12 (z)\n\nor, in a vector form\n\nδ 1 ~ 2 r) 2 F~ = − ~∇ω 1 (~ 2 2δ + ω12 (z)\n\nin agreement with (14.98) if Γ ≪ ω1 , that is, if the laser intensity is large enough.\n\n14.6.8 Radiative capture of neutrons by hydrogen 1. When r → 0\n\nψ(r) ≃\n\nr\u0011 δ a a\u0010 pr + δ 1− =1+ =1− =− pr pr r r a\n\n2. Electric dipole transitions are suppressed at very low energy because of the centrifugal barrier: a P wave is suppressed near the origin. Starting from the expression (11.84) of the quantized magnetic field,\n\n93 we have to retain transitions between zero photon states and one photon states, which are driven by the a~† term kλ\n\nh1 photon|a~† |0 photonsi = 1 kλ\n\nThus we are left with W , as given in the statement of the problem. 3. The term\n\n2π |hf |W |ii|2 δ(~ω − (Ei − Ef )) ~ comes from the Fermi Golden rule (9.170), F is the flux factor and Vω 2 dω (2π)3 c3 is the final photon space phase. 4. The angular momentum is J~ = (~/2)(~σp + ~σn ) and J~|χs i = 0 as |χs i is a state with zero angular momentum. In the same way hχs |~σp |χs i = 0 because ~σp is a vector operator whose matrix elements are zero from the Wigner-Eckart theorem, if it is sandwiched between two zero angular momentum states. If ψi (~r) is the spatial wave function of a 3 S1 state, the potential is the same as in the deuteron case, and from the orthogonality of wave functions corresponding to two different values of the energy, we have Z Z ∗ (~r)ψi (~r) = 0 d3 r ψf∗ (~r)ψi (~r) = d3 r ψD If, on the contrary, the initial wave function corresponds to a 1 S0 state, the integral does not vanish Z Z ∞ Z e−κr as \u0010 as \u0011 ND ∗ dΩ r2 dr 1− d3 r ψD (~r)ψi (~r) = − √ r r r 4π 0 √ \u0010 \u0011 √ Z ∞ as ND 4π = −ND as 4π dr e−κr 1 − (1 − κas ) = r κ2 0\n\n5. We use the completeness relation (11.80) X eiλ (~k)ejλ (~k) = δij − kˆi kˆj λ\n\nand X m\n\n|hχm σp |χs i|2 t |~\n\n=\n\nX m\n\n=\n\nX m\n\nm hχs~σp |χm σp |χs i t ihχt |~ m hχs~σp |χm σp |χs i + hχs |~σp |χs i · hχs |~σp |χs i t i · hχt |~\n\nwhere we have used hχs |~σp |χs i = 0. Then we use the completeness relation in the four dimensional Hilbert space of the two spins X m |χm t ihχt | + |χs ihχs | = I m\n\nso that\n\n′ |h|Wspin |2 i =\n\nˆ 2 = 1. because ~σp2 = 3 and (~σp · k) 6. Let us summarize the various factors\n\n1 ˆ 2 |χs i = 1 hχs |~σp2 − (~σp · k) 4 2\n\n94\n\nCHAPTER 14. EXERCISES FROM CHAPTER 14 1. A factor\n\n~ 2ε0 V\n\noriginates in the expression of the quantized magnetic field. 2. A factor\n\n1 qp2 ~2 (gp − gn )2 4 4M 2 originates in the coupling of the magnetic moments with the quantized magnetic field.\n\n3. A factor 1/2 comes from the spin summation. 4. A factor\n\n2 4πND 8π (1 − κas )2 = 3 (1 − κas )2 κ4 κ originates in the overlap integral of the spatial wave functions.\n\n5. As dσ/dΩ is a isotropic, a factor 4π arises from the dΩ integration. 6. ~ω = B from energy conservation. One finds the following for the numerical value of the theoretical estimate σ = 7.72 × 10−4 (MeV)2 = 30.9 (fm)2\n\nChapter 15\n\nExercises from Chapter 15 15.5.1 POVM as a projective measurement in a direct sum Multiplying the normalized two component vectors of (15.31) by ˜ = M\n\np 2/3, we form the 3 × 2 matrix\n\np p \u0013 \u0012 p 2/3 −p 1/6 −p1/6 0 1/2 − 1/2\n\nLet us call |v1 i and |v2 i the three-dimensional vectors whose components are the first rows of the matrix ˜ . They obey ||v1 ||2 = ||v2 ||2 = 1 and hv1 |v2 i = 0. We complete the matrix M ˜ to a 3 × 3 matrix M by M adding a third row made of the components of a vector |v3 i p p p |v3 i = ( 1/3, 1/3, 1/3)\n\nwhich is normalized and orthogonal to |v1 i and |v2 i. By construction, M is and orthogonal matrix, so that the vectors |uα i whose components are given by the columns of M are normalized and mutually orthogonal. Now, let us consider a projective measurement in H(3) P1 = |u1 ihu1 |\n\nP2 = |u2 ihu2 |\n\nP3 = |u3 ihu3 |\n\nAssume that the state operator ρ has non vanishing matrix elements only in H(2) . Then the probabilty of result α is p(α) = Tr (ρ|uα ihuα | = huα |ρ|uα i = h˜ α|ρ|˜ αi\n\n15.5.3 A POVM with two arbitrary qubit states 1. The two projectors are \u0012 sin2 α Pa⊥ = − sin α cos α\n\n− sin α cos α cos2 α\n\n\u0013\n\nPb⊥ =\n\n\u0012\n\ncos2 α − sin α cos α\n\nTo build a POVM with a third vector |ci we must have \u0012 \u0013 A + |λ|2 B −2 sin α cos α + Bλµ∗ =I −2 sin α cos α + Bλ∗ µ A + |µ|2 B\n\n− sin α cos α sin2 α\n\n\u0013\n\n√ √ in order that (15.23) be satisfied, whence |λ| = |µ| = 1/ 2. Choosing λ = µ = 1/ 2 we obtain A\n\n=\n\nB\n\n=\n\n1 1 = 1 + sin 2α 1+S 2S 2 sin 2α = 1 + sin 2α 1+S\n\n95\n\n96\n\nCHAPTER 15. EXERCISES FROM CHAPTER 15\n\nThe POVM is Qa⊥\n\n=\n\nQb⊥\n\n=\n\nQc\n\n=\n\n1 |a⊥ iha⊥ | 1+S 1 |b⊥ ihb⊥ | 1+S S |cihc| 1+S\n\n2. The state operator of the qubits sent by Alice is ρ=\n\n1 1 |aiha| + |bihb| 2 2\n\nIf Bob measures the result a⊥ , he knows with certainty that the spin was in the state |bi, because he would have got zero had the spin been in state |ai. The probability for finding a⊥ is 1 1 Tr (|a⊥ iha⊥ |ρ) = (1 − S) 1+S 2\n\np(a⊥ ) = Tr (Qa⊥ ρ) =\n\nIf sin α = 1/2, then p(a⊥ ) = p(b⊥ ) = 1/4 and in 50% of the √ cases Bob will make the right guess. In quantum cryptography, Eve uses α = π/8, so that S = 1/ 2, (1 − S)/2 ≃ 0.145. Thus, if she uses a POVM, Eve can be sure of the state sent by Alice in 58% of the cases. In the remaining 42%, she decides randomly, with a 50% probality of success. Thus she will get the correct result in (58+21)%=79% of the cases.\n\n15.5.7 Superposition of coherent states 1. The term [H0 , ρ] does not contribute to the evolution of ρnn because H0 is diagonal in the {|ni} basis. Furthermore hn|a† aρ + ρa† a|ni =\n\n2nρnn\n\nhn|aρa |ni =\n\n(n + 1)ρn+1,n+1\n\nwhence the time evolution of ρnn dρnn = −nΓρnn + (n + 1)Γρn+1,n+1 dt If we choose n = 0, we find dρ/dt = Γρ11 , which means that the population of the ground state increases at a rate proportional to that of the first excited state times Γ. The corresponding physical process is the spontaneous emission of a photon, so that Γ is the rate for spontaneous emission. The evolution equation for ρn+1,n is obtained from hn + 1|[H0 , ρ]|ni = †\n\nhn + 1|aρa |ni =\n\nhn + 1|{a† a, ρ}|ni = so that\n\n(2n + 1)ρn+1,n+1\n\np 1 dρn+1,n = −iω0 ρn+1,n + Γ (n + 1)(n + 2) ρn+2,n+1 − Γ(2n + 1)ρn+1,n dt 2\n\n2. Using (2.54) we obtain\n\neλ that is\n\n~ω0 ρn+1,n p (n + 1)(n + 2) ρn+2,n+1\n\na\n\na† e−λ\n\na\n\na† e−λ\n\n= a† + λ∗ [a, a† ] = a† + λ∗ a\n\n= e−λ\n\na\n\n\u0010\n\na† + λ∗\n\n\u0011\n\n97 Taking the derivative of C(λ, λ∗ ; t) with respect to λ we obtain \u0011 \u0010 † ∗ ∂ = Tr ρ e λa e−λ a ∂λ\n\n\u0011 \u0010 † ∗ Tr ρ a† e λa e−λ a \u0010 \u0011 † ∗ Tr ρ e λa e−λ a (a† + λ∗ ) \u0011 \u0010 † ∗ Tr (a† + λ∗ )ρ e λa e−λ a\n\n=\n\n=\n\nwhere we have used the invariance of the trace under circular permutations to derive the last line. This equation can be written schematically as \u0012 \u0013 ∂ ∂ ∗ − λ → a† ρ while clearly → ρ a† ∂λ ∂λ Similarly we get for ∂/∂λ∗ \u0011 \u0010 ∂ λa† −λ∗ a Tr ρ e e ∂λ∗ which can be rewritten as\n\n\u0011 \u0010 † ∗ = −Tr ρ e λa a e−λ a \u0011 \u0010 † ∗ = −Tr ρ (a − λ)e λa a e−λ a\n\n\u0012 \u0013 ∂ ∂ λ− → ρa while − → aρ ∂λ∗ ∂λ∗\n\n3. Let us examine the different terms in the RHS of (15.79). From the results of the preceding question \u0012 \u0013\u0012 \u0013 ∂ ∂ ∂2 ∂ a† aρ → − ∗ =− + λ∗ ∗ − λ∗ ∂λ ∂λ ∂λ∂λ∗ ∂λ \u0013\u0012 \u0013 \u0012 ∂2 ∂ ∂ ∂ =− +λ ρa† a → λ− ∗ ∂λ ∂λ ∂λ∂λ∗ ∂λ so that [a† a, ρ] → λ∗\n\n∂ ∂ −λ ∗ ∂λ ∂λ\n\nSimilarly aρa† → and {a† a, ρ} = −\n\n\u0012\n\n∂ − λ∗ ∂λ\n\n∂2 ∂λ∂λ∗\n\n\u0013\u0012 \u0013 \u0012 \u0013\u0012 \u0013 ∂2 ∂ ∂ ∂ ∂ ∂ =− − ∗ + λ− + λ∗ ∗ + λ ∗ ∗ ∂λ ∂λ ∂λ ∂λ∂λ ∂λ ∂λ\n\nAssembling all these results, we finally get the partial differential equation \u0014 \u0013 \u0013 \u0015 \u0012 \u0012 ∂ ∂ Γ Γ ∗ ∂ C(λ, λ∗ ; t) = 0 + − iω0 λ + + iω0 λ ∂t 2 ∂λ 2 ∂λ∗ or\n\n\u0014\n\n∂ + ∂t\n\n\u0012\n\nΓ − iω0 2\n\n\u0013\n\n∂ + ∂ ln λ\n\n\u0012\n\nΓ + iω0 2\n\n\u0013\n\n\u0015 ∂ C(λ, λ∗ ; t) = 0 ∂ ln λ∗\n\nTo implement the method of characteristics we write dt d ln λ d ln λ = = 1 Γ/2 − iω0 Γ/2 + iω0 whence λ = λ0 exp[(Γ/2 − iω0 )t]\n\nλ∗ = λ∗0 exp[(Γ/2 + iω0 )t]\n\n98\n\nCHAPTER 15. EXERCISES FROM CHAPTER 15\n\nor solving for λ0 , λ∗0 λ0 = λ∗ exp[−(Γ/2 + iω0 )t]\n\nλ0 = λ exp[−(Γ/2 − iω0 )t]\n\nλ exp[−(Γ/2 − iω0 )t] and λ exp[−(Γ/2 + iω0 )t] are constants along the characteristics. The partial differential equation for C(λ, λ∗ ; t) tells us that this function is constant along the characteristics C(λ, λ∗ ; t) = C0 (λ, λ∗ ; t = 0) = C0 (λ, λ∗ ) 4. The state operator at time t = 0 is \u0011 \u0010 † ∗ † ∗ C0 (λ, λ∗ ) = Tr |zihz| e λa e−λ a = hz|e λa e−λ a |zi = exp(λz ∗ − λ∗ z) We then have at time t\n\nwhich can be written as\n\nh i C(λ, λ∗ ; t) = exp z ∗ λe−(Γ/2−iω0 )t , λ∗ ze−(Γ/2+iω0 )t C(λ, λ∗ ; t) = exp [λz ∗ (t) − λ∗ z(t)]\n\nwith z(t) = ze−(Γ/2+iω0 )t Thus C(λ, λ∗ ; t) corresponds to the coherent state |z(t)i = |z e−iω0 t e−Γt/2 i 5. When |Φi is a superposition of coherent states |Φi = c1 |z1 i + c2 |z2 i then ∗\n\nC(λ, λ∗ ; t = 0) = |c1 |2 e(λz1 −λ\n\nz1 )\n\n+ |c2 |2 e(λz2 −λ\n\nz2 )\n\n+ c1 c∗2 hz2 |z1 ie(λz2 −λ\n\nz1 )\n\n+ c∗1 c2 hz1 |z2 ie(λz1 −λ\n\nz2 )\n\nThe last two terms originate in the fact that |Φi is a coherent superposition, while these two terms would be absent in an incoherent superposition of the two coherent states. At time t we have C(λ, λ∗ ; t)\n\n= |c1 |2 e[λz1 (t)−λ z1 (t)] + |c2 |2 e[λz2 (t)−λ z2 (t)] ∗ ∗ ∗ ∗ + c1 c∗2 hz2 |z1 ie[λz2 (t)−λ z1 (t)] + c∗1 c2 hz1 |z2 ie[λz1 (t)−λ z2 (t)]\n\nNote that the scalar products in the last two terms are hz2 |z1 i and hz1 |z2 i, and not hz2 (t)|z1 (t)i and hz1 (t)|z2 (t)i. In order to recover the same form as at t = 0 form, we must write, for example hz2 |z1 i =\n\nhz2 |z1 i hz2 (t)|z1 (t)i = η(t)hz2 (t)|z1 (t)i hz2 (t)|z1 (t)i\n\nWe can then describe the final state as a linear superposition of two coherent states, but the coherence is reduced by a factor \u0015 \u0014 \u0015 \u0014 \u0001 Γ 1 |η(t)| = exp − |z1 − z2 |2 1 − e−Γt ≃ exp − |z1 − z2 |2 2 2\n\nThe coherence is then damped with a rate which is |z1 − z2 |2 larger that the damping rate Γ of the individual coherent states. The decoherence time is then τdec =\n\n2 Γ|z1 − z2 |2\n\n99 5. At time t = 0, we have for the oscillator the superposition |Φ(t = 0)i = c1 |0i + c2 |zi The global state vector at t = 0 is |Ψ(t = 0)i = c1 |0 ⊗ |0F i + c2 |z ⊗ 0F i where |0F i is the state vector (vacuum state) of the radiation field, since at T = 0 there are no available photons (or phonons). The first component of |Ψi stays unchanged under the time evolution, because spontaneous emission cannot exist. By contrast, a photon will be emitted on average after a time ∼ Γ|z|2 due the second component of |Ψi. Indeed, the time evolution of ρnn in question 1 tells us that the decay amplitude of an excited state |ni is nΓ, and the average number hni is equal to |z|2 , |z|2 = hni in the coherent state |zi. As soon as one photon is emitted, the two components of |Φi become entangled to orthogonal states of the environment, and the reduced state matrix of the oscillator loses all phase coherence. The decoherence time is thus the average time for the emission of a single photon, and τdec ≃ 1/|z|2Γ.\n\n15.5.11 The Fokker-Planck-Kramers equation for a Brownian particle 1. Let us show the equivalence of the two formulae for w(x, p; t) Z +∞ y y 1 e−ipy/~ hx + |ρ(t)|x − idy (W1) w(x, p; t) = 2π~ −∞ 2 2 and\n\n1 w(x, p; t) = 2π~\n\nZ\n\n+∞\n\n−∞\n\nz z e−ixz/~ hp + |ρ(t)|x − idz (W2 ) 2 2\n\nWe use the completeness relation and (9.22) to write Z +∞ Z +∞ y 1 1 y |x − i = dq |qihq|x − i = √ dq e−iq(x−y/2) |qi 2 2π~ −∞ 2 2π~ −∞ and an analogous formula for hx + y/2|. Plugging the two formulae in (W1) and integrating over y leads to Z +∞ 2 dq e−2i(p−q)x h2p − q|ρ(t)|qi w(x, p; t) = 2π~ −∞ A change of variable p − q = z/2 allows us to recover (W2). 2. Let us consider a Gaussian wave packet (Exercise 9.7.3) ϕ(x) =\n\n\u0012\n\n1 πσ 2\n\n\u00131/4\n\n\u0013 \u0012 x2 exp − 2 2σ\n\nA straightforward calculation shows that w(x, p) =\n\n\u0012 2 2\u0013 \u0012 2\u0013 p σ x 1 exp − 2 exp − 2 π~ σ ~\n\nLet us now consider the superposition (15.143) 1 ϕ(x) ≃ √ 2\n\n\u0012\n\n1 πσ 2\n\n\u00131/4 \u0012 h (x − a)2 i h (x + a)2 i\u0013 exp − + exp − 2σ 2 2σ 2\n\nwith a ≫ σ. We find \u0013\u0015 \u0012 2\u0013 \u0012 \u0012 2 2\u0013\u0014 h (x − a)2 i h (x + a)2 i x 2ap p σ 1 exp − + exp − + 2 exp − cos exp − 2 w(x, p) = √ ~ 2σ 2 2σ 2 σ2 ~ 2 π~\n\n100\n\nCHAPTER 15. EXERCISES FROM CHAPTER 15\n\nThe first two terms would correspond to an incoherent superposition of two Gaussian wave packets, but the last one reflects the coherence of the two wave packets. 3. We limit ourselves to the second term [X, {P, ρ}] in the RHS of (15.142), as the other two terms can be dealt with using exactly the same techniques. Using the (W1) form of w we obtain at once w2\n\n=\n\n1 2π~\n\n=\n\ni ∂ 2π ∂p\n\nZ\n\n+∞\n\ny y e−ipy/~ hx + |{P, ρ}|x − i ydy 2 2\n\n−∞ Z +∞ −∞\n\ny y e−ipy/~ hx + |{P, ρ}|x − idy 2 2\n\nWe then use the W2 form w2\n\n=\n\n1 ∂ 2π~ ∂p\n\nZ\n\n+∞\n\n−∞\n\ny y e−ipy/~ hx + |{P, ρ}|p − idy 2 2\n\nZ +∞ y y i ∂ p e−ixy/~ hp + |{P, ρ}|x − idy = 2π ∂p 2 2 −∞ ∂ = i~ [pw(x, p; t)] ∂p 4. One has only to observe that w(x, p; t) must vanish when x → ±∞.\n\n#### Related Documents\n\nFebruary 2020 50\nNovember 2019 35\nDecember 2019 89\nNovember 2019 70\nNovember 2019 56\n##### Quantum Physics For Dummies\nNovember 2019 109\n\nFebruary 2020 50" ]
[ null ]
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https://en.wikipedia.org/wiki/Wave_making_resistance
[ "# Wave-making resistance\n\n(Redirected from Wave making resistance)\n\nWave-making resistance is a form of drag that affects surface watercraft, such as boats and ships, and reflects the energy required to push the water out of the way of the hull. This energy goes into creating the wave.\n\n## Physics\n\nFor small displacement hulls, such as sailboats or rowboats, wave-making resistance is the major source of the marine vessel drag.\n\nA salient property of water waves is dispersiveness; i.e., the longer the wave, the faster it moves. Waves generated by a ship are affected by her geometry and speed, and most of the energy given by the ship for making waves is transferred to water through the bow and stern parts. Simply speaking, these two wave systems, i.e., bow and stern waves, interact with each other, and the resulting waves are responsible for the resistance.\n\nThe phase speed of deepwater waves is proportional to the square root of the wavelength of the generated waves, and the length of a ship causes the difference in phases of waves generated by bow and stern parts. Thus, there is a direct relationship between the waterline length and the magnitude of the wave-making resistance. Wave propagation speed, which means the same thing as the phase speed of deepwater wave, is independent of length of the hull causing the wave, but dependent of the speed of the hull instead.\n\nA simple way of considering wave-making resistance is to look at the hull in relation to bow and stern waves. If the length of a ship is half the length of the waves generated, the resulting wave will be very small due to cancellation, and if the length is the same as the wavelength, the wave will be large due to enhancement.\n\nThe phase speed $c$", null, "of waves is given by the following formula:\n\n$c={\\sqrt {{\\frac {g}{2\\pi }}l}}$", null, "where $l$", null, "is the length of the wave and $g$", null, "the gravitational acceleration. Substituting in the appropriate value for $g$", null, "yields the equation:\n\n${\\mbox{c in knots}}\\approx 1.341\\times {\\sqrt {\\mbox{length in ft}}}\\approx {\\frac {4}{3}}\\times {\\sqrt {\\mbox{length in ft}}}$", null, "or, in metric units:\n\n${\\mbox{c in knots}}\\approx 2.429\\times {\\sqrt {\\mbox{length in m}}}\\approx {\\sqrt {6\\times {\\mbox{length in m}}}}\\approx 2.5\\times {\\sqrt {\\mbox{length in m}}}$", null, "These values, 1.34, 2.5 and very easy 6, are often used in the hull speed rule of thumb used to compare potential speeds of displacement hulls, and this relationship is also fundamental to the Froude number, used in the comparison of different scales of watercraft.\n\nWhen the vessel exceeds a \"speed–length ratio\" (speed in knots divided by square root of length in feet) of 0.94, it starts to outrun most of its bow wave, the hull actually settles slightly in the water as it is now only supported by two wave peaks. As the vessel exceeds a speed-length ratio of 1.34, the wavelength is now longer than the hull, and the stern is no longer supported by the wake, causing the stern to squat, and the bow to rise. The hull is now starting to climb its own bow wave, and resistance begins to increase at a very high rate. While it is possible to drive a displacement hull faster than a speed-length ratio of 1.34, it is prohibitively expensive to do so. Most large vessels operate at speed-length ratios well below that level, at speed-length ratios of under 1.0.\n\n## Ways of reducing wave-making resistance\n\nSince wave-making resistance is based on the energy required to push the water out of the way of the hull, there are a number of ways that this can be minimized.\n\n### Reduced displacement\n\nReducing the displacement of the craft, by eliminating excess weight, is the most straightforward way to reduce the wave making drag. Another way is to shape the hull so as to generate lift as it moves through the water. Semi-displacement hulls and planing hulls do this, and they are able to break through the hull speed barrier and transition into a realm where drag increases at a much lower rate. The disadvantage of this is that planing is only practical on smaller vessels, with high power-to-weight ratios, such as motorboats. It is not a practical solution for a large vessel such as a supertanker.\n\n### Fine entry\n\nA hull with a blunt bow has to push the water away very quickly to pass through, and this high acceleration requires large amounts of energy. By using a fine bow, with a sharper angle that pushes the water out of the way more gradually, the amount of energy required to displace the water will be less. A modern variation is the wave-piercing design. The total amount of water to be displaced by a moving hull, and thus causing wave making drag, is the cross sectional area of the hull times distance the hull travels, and will not remain the same when prismatic coefficient is increased for the same lwl and same displacement and same speed.\n\n### Bulbous bow\n\nA special type of bow, called a bulbous bow, is often used on large power vessels to reduce wave-making drag. The bulb alters the waves generated by the hull, by changing the pressure distribution ahead of the bow. Because of the nature of its destructive interference with the bow wave, there is a limited range of vessel speeds over which it is effective. A bulbous bow must be properly designed to mitigate the wave-making resistance of a particular hull over a particular range of speeds. A bulb that works for one vessel's hull shape and one range of speeds could be detrimental to a different hull shape or a different speed range. Proper design and knowledge of a ship's intended operating speeds and conditions is therefore necessary when designing a bulbous bow.\n\n### Hull form filtering\n\nIf the hull is designed to operate at speeds substantially lower than hull speed then it is possible to refine the hull shape along its length to reduce wave resistance at one speed. This is practical only where the block coefficient of the hull is not a significant issue.\n\n## Semi-displacement and planing hulls", null, "A graph showing resistance–weight ratio as a function of speed–length ratio for displacement, semi-displacement, and planing hulls\n\nSince semi-displacement and planing hulls generate a significant amount of lift in operation, they are capable of breaking the barrier of the wave propagation speed and operating in realms of much lower drag, but to do this they must be capable of first pushing past that speed, which requires significant power. This stage is called the transition stage and at this stage the rate of wave-making resistance is the highest. Once the hull gets over the hump of the bow wave, the rate of increase of the wave drag will start to reduce significantly. The planing hull will rise up clearing its stern off the water and its trim will be high. Underwater part of the planing hull will be small during the planing regime.\n\nA qualitative interpretation of the wave resistance plot is that a displacement hull resonates with a wave that has a crest near its bow and a trough near its stern, because the water is pushed away at the bow and pulled back at the stern. A planing hull simply pushed down on the water under it, so it resonates with a wave that has a trough under it. If it has about twice the length it will therefore have only square root (2) or 1.4 times the speed. In practice most planing hulls usually move much faster than that. At four times hull speed the wavelength is already 16 times longer than the hull." ]
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https://homepages.cwi.nl/~aeb/games/sudoku/solving27.html
[ "Prev Up Next\n\n# Packing\n\nThe packing principle says that if there is a matching between two sets, n things in one set cannot have their partners inside a set of size smaller than n.\n\n## Two things cannot be where there is room for one\n\nFrom two pairs in a box one concludes to the point of intersection:", null, "", null, "since one box cannot have two 1's.\n\nOr again, with two intersecting pairs for the same digit, but not in a single box:", null, "", null, "since otherwise two different values are forced for the point opposite to the point of intersection.\n\n## Three things cannot be where there is room for two\n\nFor example, suppose all possible positions for digit 3 in rows 2 and 5 are as shown.", null, "", null, "Then position (6,7) cannot be 3, otherwise also position (5,3) would be 3, and moreover one of positions (2,3) and (2,7), so that columns 3 and 7 would have three digits 3.\n\nTerminology People call this 'X-wing with fin' ('finned X-wing').\n\n## Four things cannot be where there is room for three", null, "", null, "After some work, the puzzle shown has proceeded to the given stage, and the usual methods make no progress. Now look at the yellow cells. They form an almost locked set, and if (1,6)2 then they form a locked set and must have the digits 4,6,8. Then in the middle bottom box digits 4,7,8 must fill the middle column. But also the digit 2 must be there, since it cannot be in the middle top box if (1,6)2. Hence (1,6)!2, that is, (1,6)9.\n\n## Four things cannot be where there is room for 3.5\n\nConsider the following diagram, where only the candidates for digit 8 are indicated (with a # mark). The goal is to conclude that position (1,1) cannot have an 8.", null, "", null, "One way to proceed is to assume that (1,1) has an 8, and to count how many digits 8 are seen in the seven areas indicated (four columns, two rows and one box). Since each of these areas should contain a unique digit 8, the total should be 7. Now each # mark is covered by at least two of these areas, so is counted at least twice, and that means that there is room for at most 3.5 digits 8 where we need 4. Therefore position (1,1) does not have an 8.\n\n(In this case one can also follow the chain: if (1,1), then also (4,2), then also (7,9), then also (2,3), then no further choice is possible.)" ]
[ null, "https://homepages.cwi.nl/~aeb/games/sudoku/png/pack1.png", null, "https://homepages.cwi.nl/~aeb/games/sudoku/png/pack2.png", null, "https://homepages.cwi.nl/~aeb/games/sudoku/png/pack3.png", null, "https://homepages.cwi.nl/~aeb/games/sudoku/png/pack4.png", null, "https://homepages.cwi.nl/~aeb/games/sudoku/png/pack5.png", null, "https://homepages.cwi.nl/~aeb/games/sudoku/png/pack6.png", null, "https://homepages.cwi.nl/~aeb/games/sudoku/png/f9194.png", null, "https://homepages.cwi.nl/~aeb/games/sudoku/png/f9194a.png", null, "https://homepages.cwi.nl/~aeb/games/sudoku/png/pack7.png", null, "https://homepages.cwi.nl/~aeb/games/sudoku/png/pack8.png", null ]
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https://datascience.stackexchange.com/questions/24758/why-does-my-master-node-get-heap-memory-full-for-inbuilt-svd-api-in-apache-spark
[ "# Why does my master node get heap memory full for inbuilt SVD API in Apache Spark during calculation of inverse of a square matrix?\n\nI am using a three node system:\n\n• Master node: $$64$$ GB RAM\n• 2 slave node: $$32$$ GB RAM\n\nI am calculating the inverse of a square matrix using inbuilt Apache Spark function SVD. When I called SVD function of Apache spark then $$64$$ GB memory of master node got totally consumed and start to use swap memory which makes execution too slow and eventually lead to heap memory full.\n\nIf matrix size is $$3000 \\times 3000$$ then no issue but when we consider size greater than $$3000$$ (like $$3500$$ or $$5000$$) then above-mentioned issue arises.\n\nNote: Even my MATLAB is able to compute the inverse of $$10K \\times 10K$$ matrix on 32 GB RAM but Apache Spark got Memory full if I used inbuilt SVD function on above system.\n\nSample Code is as below:-\n\nimport org.apache.spark.mllib.linalg.{Vectors,Vector,Matrix,SingularValueDecomposition,DenseMatrix,DenseVector}\nimport org.apache.spark.mllib.linalg.distributed.RowMatrix\n\ndef computeInverse(X: RowMatrix): DenseMatrix = {\nval nCoef = X.numCols.toInt\nval svd = X.computeSVD(nCoef, computeU = true)\nif (svd.s.size < nCoef) {\nsys.error(s\"RowMatrix.computeInverse called on singular matrix.\")\n}\n\n// Create the inv diagonal matrix from S\nval invS = DenseMatrix.diag(new DenseVector(svd.s.toArray.map(x => math.pow(x,-1))))\n\n// U cannot be a RowMatrix\nval U = new DenseMatrix(svd.U.numRows().toInt,svd.U.numCols().toInt,svd.U.rows.collect.flatMap(x => x.toArray))\n\n// If you could make V distributed, then this may be better. However its alreadly local...so maybe this is fine.\nval V = svd.V\n// inv(X) = V*inv(S)*transpose(U) --- the U is already transposed.\n(V.multiply(invS)).multiply(U)\n}\n\n• Please post your code (the most simplified version possible that still exposes the problem). – Pete Nov 16 '17 at 2:25\n• @Pete Sample code is included in the question. – Chandan Gautam Nov 18 '17 at 8:33\n• I wonder what matrix inversion algorithm Spark uses. The exact solution uses O(n^2) memory, which might be the cause of the problem. (arxiv.org/pdf/1505.07570.pdf) Because of this people sometimes use approximate inverses. – tom Nov 19 '17 at 6:49\n• Did you actually allocate most of the machine's memory to the driver? the default heap size isn't nearly that big. – Sean Owen Feb 23 '19 at 1:41" ]
[ null ]
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https://creditcardfree.savingadvice.com/2018/08/02/withdrawal-from-esa_216568/
[ "User Real IP - 35.173.57.84\n```Array\n(\n => Array\n(\n => 182.68.68.92\n)\n\n => Array\n(\n => 101.0.41.201\n)\n\n => Array\n(\n => 43.225.98.123\n)\n\n => Array\n(\n => 2.58.194.139\n)\n\n => Array\n(\n => 46.119.197.104\n)\n\n => Array\n(\n => 45.249.8.93\n)\n\n => Array\n(\n => 103.12.135.72\n)\n\n => Array\n(\n => 157.35.243.216\n)\n\n => Array\n(\n => 209.107.214.176\n)\n\n => Array\n(\n => 5.181.233.166\n)\n\n => Array\n(\n => 106.201.10.100\n)\n\n => Array\n(\n => 36.90.55.39\n)\n\n => Array\n(\n => 119.154.138.47\n)\n\n => Array\n(\n => 51.91.31.157\n)\n\n => Array\n(\n => 182.182.65.216\n)\n\n => Array\n(\n => 157.35.252.63\n)\n\n => Array\n(\n => 14.142.34.163\n)\n\n => Array\n(\n => 178.62.43.135\n)\n\n => Array\n(\n => 43.248.152.148\n)\n\n => Array\n(\n => 222.252.104.114\n)\n\n => Array\n(\n => 209.107.214.168\n)\n\n => Array\n(\n => 103.99.199.250\n)\n\n => Array\n(\n => 178.62.72.160\n)\n\n => Array\n(\n => 27.6.1.170\n)\n\n => Array\n(\n => 182.69.249.219\n)\n\n => Array\n(\n => 110.93.228.86\n)\n\n => Array\n(\n => 72.255.1.98\n)\n\n => Array\n(\n => 182.73.111.98\n)\n\n => Array\n(\n => 45.116.117.11\n)\n\n => Array\n(\n => 122.15.78.189\n)\n\n => Array\n(\n => 14.167.188.234\n)\n\n => Array\n(\n => 223.190.4.202\n)\n\n => Array\n(\n => 202.173.125.19\n)\n\n => Array\n(\n => 103.255.5.32\n)\n\n => Array\n(\n => 39.37.145.103\n)\n\n => Array\n(\n => 140.213.26.249\n)\n\n => Array\n(\n => 45.118.166.85\n)\n\n => Array\n(\n => 102.166.138.255\n)\n\n => Array\n(\n => 77.111.246.234\n)\n\n => Array\n(\n => 45.63.6.196\n)\n\n => Array\n(\n => 103.250.147.115\n)\n\n => Array\n(\n => 223.185.30.99\n)\n\n => Array\n(\n => 103.122.168.108\n)\n\n => Array\n(\n => 123.136.203.21\n)\n\n => Array\n(\n => 171.229.243.63\n)\n\n => Array\n(\n => 153.149.98.149\n)\n\n => Array\n(\n => 223.238.93.15\n)\n\n => Array\n(\n => 178.62.113.166\n)\n\n => Array\n(\n => 101.162.0.153\n)\n\n => Array\n(\n => 121.200.62.114\n)\n\n => Array\n(\n => 14.248.77.252\n)\n\n => Array\n(\n => 95.142.117.29\n)\n\n => Array\n(\n => 150.129.60.107\n)\n\n => Array\n(\n => 94.205.243.22\n)\n\n => Array\n(\n => 115.42.71.143\n)\n\n => Array\n(\n => 117.217.195.59\n)\n\n => Array\n(\n => 182.77.112.56\n)\n\n => Array\n(\n => 182.77.112.108\n)\n\n => Array\n(\n => 41.80.69.10\n)\n\n => Array\n(\n => 117.5.222.121\n)\n\n => Array\n(\n => 103.11.0.38\n)\n\n => Array\n(\n => 202.173.127.140\n)\n\n => Array\n(\n => 49.249.249.50\n)\n\n => Array\n(\n => 116.72.198.211\n)\n\n => Array\n(\n => 223.230.54.53\n)\n\n => Array\n(\n => 102.69.228.74\n)\n\n => Array\n(\n => 39.37.251.89\n)\n\n => Array\n(\n => 39.53.246.141\n)\n\n => Array\n(\n => 39.57.182.72\n)\n\n => Array\n(\n => 209.58.130.210\n)\n\n => Array\n(\n => 104.131.75.86\n)\n\n => Array\n(\n => 106.212.131.255\n)\n\n => Array\n(\n => 106.212.132.127\n)\n\n => Array\n(\n => 223.190.4.60\n)\n\n => Array\n(\n => 103.252.116.252\n)\n\n => Array\n(\n => 103.76.55.182\n)\n\n => Array\n(\n => 45.118.166.70\n)\n\n => Array\n(\n => 103.93.174.215\n)\n\n => Array\n(\n => 5.62.62.142\n)\n\n => Array\n(\n => 182.179.158.156\n)\n\n => Array\n(\n => 39.57.255.12\n)\n\n => Array\n(\n => 39.37.178.37\n)\n\n => Array\n(\n => 182.180.165.211\n)\n\n => Array\n(\n => 119.153.135.17\n)\n\n => Array\n(\n => 72.255.15.244\n)\n\n => Array\n(\n => 139.180.166.181\n)\n\n => Array\n(\n => 70.119.147.111\n)\n\n => Array\n(\n => 106.210.40.83\n)\n\n => Array\n(\n => 14.190.70.91\n)\n\n => Array\n(\n => 202.125.156.82\n)\n\n => Array\n(\n => 115.42.68.38\n)\n\n => Array\n(\n => 102.167.13.108\n)\n\n => Array\n(\n => 117.217.192.130\n)\n\n => Array\n(\n => 205.185.223.156\n)\n\n => Array\n(\n => 171.224.180.29\n)\n\n => Array\n(\n => 45.127.45.68\n)\n\n => Array\n(\n => 195.206.183.232\n)\n\n => Array\n(\n => 49.32.52.115\n)\n\n => Array\n(\n => 49.207.49.223\n)\n\n => Array\n(\n => 45.63.29.61\n)\n\n => Array\n(\n => 103.245.193.214\n)\n\n => Array\n(\n => 39.40.236.69\n)\n\n => Array\n(\n => 62.80.162.111\n)\n\n => Array\n(\n => 45.116.232.56\n)\n\n => Array\n(\n => 45.118.166.91\n)\n\n => Array\n(\n => 180.92.230.234\n)\n\n => Array\n(\n => 157.40.57.160\n)\n\n => Array\n(\n => 110.38.38.130\n)\n\n => Array\n(\n => 72.255.57.183\n)\n\n => Array\n(\n => 182.68.81.85\n)\n\n => Array\n(\n => 39.57.202.122\n)\n\n => Array\n(\n => 119.152.154.36\n)\n\n => Array\n(\n => 5.62.62.141\n)\n\n => Array\n(\n => 119.155.54.232\n)\n\n => Array\n(\n => 39.37.141.22\n)\n\n => Array\n(\n => 183.87.12.225\n)\n\n => Array\n(\n => 107.170.127.117\n)\n\n => Array\n(\n => 125.63.124.49\n)\n\n => Array\n(\n => 39.42.191.3\n)\n\n => Array\n(\n => 116.74.24.72\n)\n\n => Array\n(\n => 46.101.89.227\n)\n\n => Array\n(\n => 202.173.125.247\n)\n\n => Array\n(\n => 39.42.184.254\n)\n\n => Array\n(\n => 115.186.165.132\n)\n\n => Array\n(\n => 39.57.206.126\n)\n\n => Array\n(\n => 103.245.13.145\n)\n\n => Array\n(\n => 202.175.246.43\n)\n\n => Array\n(\n => 192.140.152.150\n)\n\n => Array\n(\n => 202.88.250.103\n)\n\n => Array\n(\n => 103.248.94.207\n)\n\n => Array\n(\n => 77.73.66.101\n)\n\n => Array\n(\n => 104.131.66.8\n)\n\n => Array\n(\n => 113.186.161.97\n)\n\n => Array\n(\n => 222.254.5.7\n)\n\n => Array\n(\n => 223.233.67.247\n)\n\n => Array\n(\n => 171.249.116.146\n)\n\n => Array\n(\n => 47.30.209.71\n)\n\n => Array\n(\n => 202.134.13.130\n)\n\n => Array\n(\n => 27.6.135.7\n)\n\n => Array\n(\n => 107.170.186.79\n)\n\n => Array\n(\n => 103.212.89.171\n)\n\n => Array\n(\n => 117.197.9.77\n)\n\n => Array\n(\n => 122.176.206.233\n)\n\n => Array\n(\n => 192.227.253.222\n)\n\n => Array\n(\n => 182.188.224.119\n)\n\n => Array\n(\n => 14.248.70.74\n)\n\n => Array\n(\n => 42.118.219.169\n)\n\n => Array\n(\n => 110.39.146.170\n)\n\n => Array\n(\n => 119.160.66.143\n)\n\n => Array\n(\n => 103.248.95.130\n)\n\n => Array\n(\n => 27.63.152.208\n)\n\n => Array\n(\n => 49.207.114.96\n)\n\n => Array\n(\n => 102.166.23.214\n)\n\n => Array\n(\n => 175.107.254.73\n)\n\n => Array\n(\n => 103.10.227.214\n)\n\n => Array\n(\n => 202.143.115.89\n)\n\n => Array\n(\n => 110.93.227.187\n)\n\n => Array\n(\n => 103.140.31.60\n)\n\n => Array\n(\n => 110.37.231.46\n)\n\n => Array\n(\n => 39.36.99.238\n)\n\n => Array\n(\n => 157.37.140.26\n)\n\n => Array\n(\n => 43.246.202.226\n)\n\n => Array\n(\n => 137.97.8.143\n)\n\n => Array\n(\n => 182.65.52.242\n)\n\n => Array\n(\n => 115.42.69.62\n)\n\n => Array\n(\n => 14.143.254.58\n)\n\n => Array\n(\n => 223.179.143.236\n)\n\n => Array\n(\n => 223.179.143.249\n)\n\n => Array\n(\n => 103.143.7.54\n)\n\n => Array\n(\n => 223.179.139.106\n)\n\n => Array\n(\n => 39.40.219.90\n)\n\n => Array\n(\n => 45.115.141.231\n)\n\n => Array\n(\n => 120.29.100.33\n)\n\n => Array\n(\n => 112.196.132.5\n)\n\n => Array\n(\n => 202.163.123.153\n)\n\n => Array\n(\n => 5.62.58.146\n)\n\n => Array\n(\n => 39.53.216.113\n)\n\n => Array\n(\n => 42.111.160.73\n)\n\n => Array\n(\n => 107.182.231.213\n)\n\n => Array\n(\n => 119.82.94.120\n)\n\n => Array\n(\n => 178.62.34.82\n)\n\n => Array\n(\n => 203.122.6.18\n)\n\n => Array\n(\n => 157.42.38.251\n)\n\n => Array\n(\n => 45.112.68.222\n)\n\n => Array\n(\n => 49.206.212.122\n)\n\n => Array\n(\n => 104.236.70.228\n)\n\n => Array\n(\n => 42.111.34.243\n)\n\n => Array\n(\n => 84.241.19.186\n)\n\n => Array\n(\n => 89.187.180.207\n)\n\n => Array\n(\n => 104.243.212.118\n)\n\n => Array\n(\n => 104.236.55.136\n)\n\n => Array\n(\n => 106.201.16.163\n)\n\n => Array\n(\n => 46.101.40.25\n)\n\n => Array\n(\n => 45.118.166.94\n)\n\n => Array\n(\n => 49.36.128.102\n)\n\n => Array\n(\n => 14.142.193.58\n)\n\n => Array\n(\n => 212.79.124.176\n)\n\n => Array\n(\n => 45.32.191.194\n)\n\n => Array\n(\n => 105.112.107.46\n)\n\n => Array\n(\n => 106.201.14.8\n)\n\n => Array\n(\n => 110.93.240.65\n)\n\n => Array\n(\n => 27.96.95.177\n)\n\n => Array\n(\n => 45.41.134.35\n)\n\n => Array\n(\n => 180.151.13.110\n)\n\n => Array\n(\n => 101.53.242.89\n)\n\n => Array\n(\n => 115.186.3.110\n)\n\n => Array\n(\n => 171.49.185.242\n)\n\n => Array\n(\n => 115.42.70.24\n)\n\n => Array\n(\n => 45.128.188.43\n)\n\n => Array\n(\n => 103.140.129.63\n)\n\n => Array\n(\n => 101.50.113.147\n)\n\n => Array\n(\n => 103.66.73.30\n)\n\n => Array\n(\n => 117.247.193.169\n)\n\n => Array\n(\n => 120.29.100.94\n)\n\n => Array\n(\n => 42.109.154.39\n)\n\n => Array\n(\n => 122.173.155.150\n)\n\n => Array\n(\n => 45.115.104.53\n)\n\n => Array\n(\n => 116.74.29.84\n)\n\n => Array\n(\n => 101.50.125.34\n)\n\n => Array\n(\n => 45.118.166.80\n)\n\n => Array\n(\n => 91.236.184.27\n)\n\n => Array\n(\n => 113.167.185.120\n)\n\n => Array\n(\n => 27.97.66.222\n)\n\n => Array\n(\n => 43.247.41.117\n)\n\n => Array\n(\n => 23.229.16.227\n)\n\n => Array\n(\n => 14.248.79.209\n)\n\n => Array\n(\n => 117.5.194.26\n)\n\n => Array\n(\n => 117.217.205.41\n)\n\n => Array\n(\n => 114.79.169.99\n)\n\n => Array\n(\n => 103.55.60.97\n)\n\n => Array\n(\n => 182.75.89.210\n)\n\n => Array\n(\n => 77.73.66.109\n)\n\n => Array\n(\n => 182.77.126.139\n)\n\n => Array\n(\n => 14.248.77.166\n)\n\n => Array\n(\n => 157.35.224.133\n)\n\n => Array\n(\n => 183.83.38.27\n)\n\n => Array\n(\n => 182.68.4.77\n)\n\n => Array\n(\n => 122.177.130.234\n)\n\n => Array\n(\n => 103.24.99.99\n)\n\n => Array\n(\n => 103.91.127.66\n)\n\n => Array\n(\n => 41.90.34.240\n)\n\n => Array\n(\n => 49.205.77.102\n)\n\n => Array\n(\n => 103.248.94.142\n)\n\n => Array\n(\n => 104.143.92.170\n)\n\n => Array\n(\n => 219.91.157.114\n)\n\n => Array\n(\n => 223.190.88.22\n)\n\n => Array\n(\n => 223.190.86.232\n)\n\n => Array\n(\n => 39.41.172.80\n)\n\n => Array\n(\n => 124.107.206.5\n)\n\n => Array\n(\n => 139.167.180.224\n)\n\n => Array\n(\n => 93.76.64.248\n)\n\n => Array\n(\n => 65.216.227.119\n)\n\n => Array\n(\n => 223.190.119.141\n)\n\n => Array\n(\n => 110.93.237.179\n)\n\n => Array\n(\n => 41.90.7.85\n)\n\n => Array\n(\n => 103.100.6.26\n)\n\n => Array\n(\n => 104.140.83.13\n)\n\n => Array\n(\n => 223.190.119.133\n)\n\n => Array\n(\n => 119.152.150.87\n)\n\n => Array\n(\n => 103.125.130.147\n)\n\n => Array\n(\n => 27.6.5.52\n)\n\n => Array\n(\n => 103.98.188.26\n)\n\n => Array\n(\n => 39.35.121.81\n)\n\n => Array\n(\n => 74.119.146.182\n)\n\n => Array\n(\n => 5.181.233.162\n)\n\n => Array\n(\n => 157.39.18.60\n)\n\n => Array\n(\n => 1.187.252.25\n)\n\n => Array\n(\n => 39.42.145.59\n)\n\n => Array\n(\n => 39.35.39.198\n)\n\n => Array\n(\n => 49.36.128.214\n)\n\n => Array\n(\n => 182.190.20.56\n)\n\n => Array\n(\n => 122.180.249.189\n)\n\n => Array\n(\n => 117.217.203.107\n)\n\n => Array\n(\n => 103.70.82.241\n)\n\n => Array\n(\n => 45.118.166.68\n)\n\n => Array\n(\n => 122.180.168.39\n)\n\n => Array\n(\n => 149.28.67.254\n)\n\n => Array\n(\n => 223.233.73.8\n)\n\n => Array\n(\n => 122.167.140.0\n)\n\n => Array\n(\n => 95.158.51.55\n)\n\n => Array\n(\n => 27.96.95.134\n)\n\n => Array\n(\n => 49.206.214.53\n)\n\n => Array\n(\n => 212.103.49.92\n)\n\n => Array\n(\n => 122.177.115.101\n)\n\n => Array\n(\n => 171.50.187.124\n)\n\n => Array\n(\n => 122.164.55.107\n)\n\n => Array\n(\n => 98.114.217.204\n)\n\n => Array\n(\n => 106.215.10.54\n)\n\n => Array\n(\n => 115.42.68.28\n)\n\n => Array\n(\n => 104.194.220.87\n)\n\n => Array\n(\n => 103.137.84.170\n)\n\n => Array\n(\n => 61.16.142.110\n)\n\n => Array\n(\n => 212.103.49.85\n)\n\n => Array\n(\n => 39.53.248.162\n)\n\n => Array\n(\n => 203.122.40.214\n)\n\n => Array\n(\n => 117.217.198.72\n)\n\n => Array\n(\n => 115.186.191.203\n)\n\n => Array\n(\n => 120.29.100.199\n)\n\n => Array\n(\n => 45.151.237.24\n)\n\n => Array\n(\n => 223.190.125.232\n)\n\n => Array\n(\n => 41.80.151.17\n)\n\n => Array\n(\n => 23.111.188.5\n)\n\n => Array\n(\n => 223.190.125.216\n)\n\n => Array\n(\n => 103.217.133.119\n)\n\n => Array\n(\n => 103.198.173.132\n)\n\n => Array\n(\n => 47.31.155.89\n)\n\n => Array\n(\n => 223.190.20.253\n)\n\n => Array\n(\n => 104.131.92.125\n)\n\n => Array\n(\n => 223.190.19.152\n)\n\n => Array\n(\n => 103.245.193.191\n)\n\n => Array\n(\n => 106.215.58.255\n)\n\n => Array\n(\n => 119.82.83.238\n)\n\n => Array\n(\n => 106.212.128.138\n)\n\n => Array\n(\n => 139.167.237.36\n)\n\n => Array\n(\n => 222.124.40.250\n)\n\n => Array\n(\n => 134.56.185.169\n)\n\n => Array\n(\n => 54.255.226.31\n)\n\n => Array\n(\n => 137.97.162.31\n)\n\n => Array\n(\n => 95.185.21.191\n)\n\n => Array\n(\n => 171.61.168.151\n)\n\n => Array\n(\n => 137.97.184.4\n)\n\n => Array\n(\n => 106.203.151.202\n)\n\n => Array\n(\n => 39.37.137.0\n)\n\n => Array\n(\n => 45.118.166.66\n)\n\n => Array\n(\n => 14.248.105.100\n)\n\n => Array\n(\n => 106.215.61.185\n)\n\n => Array\n(\n => 202.83.57.179\n)\n\n => Array\n(\n => 89.187.182.176\n)\n\n => Array\n(\n => 49.249.232.198\n)\n\n => Array\n(\n => 132.154.95.236\n)\n\n => Array\n(\n => 223.233.83.230\n)\n\n => Array\n(\n => 183.83.153.14\n)\n\n => Array\n(\n => 125.63.72.210\n)\n\n => Array\n(\n => 207.174.202.11\n)\n\n => Array\n(\n => 119.95.88.59\n)\n\n => Array\n(\n => 122.170.14.150\n)\n\n => Array\n(\n => 45.118.166.75\n)\n\n => Array\n(\n => 103.12.135.37\n)\n\n => Array\n(\n => 49.207.120.225\n)\n\n => Array\n(\n => 182.64.195.207\n)\n\n => Array\n(\n => 103.99.37.16\n)\n\n => Array\n(\n => 46.150.104.221\n)\n\n => Array\n(\n => 104.236.195.147\n)\n\n => Array\n(\n => 103.104.192.43\n)\n\n => Array\n(\n => 24.242.159.118\n)\n\n => Array\n(\n => 39.42.179.143\n)\n\n => Array\n(\n => 111.93.58.131\n)\n\n => Array\n(\n => 193.176.84.127\n)\n\n => Array\n(\n => 209.58.142.218\n)\n\n => Array\n(\n => 69.243.152.129\n)\n\n => Array\n(\n => 117.97.131.249\n)\n\n => Array\n(\n => 103.230.180.89\n)\n\n => Array\n(\n => 106.212.170.192\n)\n\n => Array\n(\n => 171.224.180.95\n)\n\n => Array\n(\n => 158.222.11.87\n)\n\n => Array\n(\n => 119.155.60.246\n)\n\n => Array\n(\n => 41.90.43.129\n)\n\n => Array\n(\n => 185.183.104.170\n)\n\n => Array\n(\n => 14.248.67.65\n)\n\n => Array\n(\n => 117.217.205.82\n)\n\n => Array\n(\n => 111.88.7.209\n)\n\n => Array\n(\n => 49.36.132.244\n)\n\n => Array\n(\n => 171.48.40.2\n)\n\n => Array\n(\n => 119.81.105.2\n)\n\n => Array\n(\n => 49.36.128.114\n)\n\n => Array\n(\n => 213.200.31.93\n)\n\n => Array\n(\n => 2.50.15.110\n)\n\n => Array\n(\n => 120.29.104.67\n)\n\n => Array\n(\n => 223.225.32.221\n)\n\n => Array\n(\n => 14.248.67.195\n)\n\n => Array\n(\n => 119.155.36.13\n)\n\n => Array\n(\n => 101.50.95.104\n)\n\n => Array\n(\n => 104.236.205.233\n)\n\n => Array\n(\n => 122.164.36.150\n)\n\n => Array\n(\n => 157.45.93.209\n)\n\n => Array\n(\n => 182.77.118.100\n)\n\n => Array\n(\n => 182.74.134.218\n)\n\n => Array\n(\n => 183.82.128.146\n)\n\n => Array\n(\n => 112.196.170.234\n)\n\n => Array\n(\n => 122.173.230.178\n)\n\n => Array\n(\n => 122.164.71.199\n)\n\n => Array\n(\n => 51.79.19.31\n)\n\n => Array\n(\n => 58.65.222.20\n)\n\n => Array\n(\n => 103.27.203.97\n)\n\n => Array\n(\n => 111.88.7.242\n)\n\n => Array\n(\n => 14.171.232.77\n)\n\n => Array\n(\n => 46.101.22.182\n)\n\n => Array\n(\n => 103.94.219.19\n)\n\n => Array\n(\n => 139.190.83.30\n)\n\n => Array\n(\n => 223.190.27.184\n)\n\n => Array\n(\n => 182.185.183.34\n)\n\n => Array\n(\n => 91.74.181.242\n)\n\n => Array\n(\n => 222.252.107.14\n)\n\n => Array\n(\n => 137.97.8.28\n)\n\n => Array\n(\n => 46.101.16.229\n)\n\n => Array\n(\n => 122.53.254.229\n)\n\n => Array\n(\n => 106.201.17.180\n)\n\n => Array\n(\n => 123.24.170.129\n)\n\n => Array\n(\n => 182.185.180.79\n)\n\n => Array\n(\n => 223.190.17.4\n)\n\n => Array\n(\n => 213.108.105.1\n)\n\n => Array\n(\n => 171.22.76.9\n)\n\n => Array\n(\n => 202.66.178.164\n)\n\n => Array\n(\n => 178.62.97.171\n)\n\n => Array\n(\n => 167.179.110.209\n)\n\n => Array\n(\n => 223.230.147.172\n)\n\n => Array\n(\n => 76.218.195.160\n)\n\n => Array\n(\n => 14.189.186.178\n)\n\n => Array\n(\n => 157.41.45.143\n)\n\n => Array\n(\n => 223.238.22.53\n)\n\n => Array\n(\n => 111.88.7.244\n)\n\n => Array\n(\n => 5.62.57.19\n)\n\n => Array\n(\n => 106.201.25.216\n)\n\n => Array\n(\n => 117.217.205.33\n)\n\n => Array\n(\n => 111.88.7.215\n)\n\n => Array\n(\n => 106.201.13.77\n)\n\n => Array\n(\n => 50.7.93.29\n)\n\n => Array\n(\n => 123.201.70.112\n)\n\n => Array\n(\n => 39.42.108.226\n)\n\n => Array\n(\n => 27.5.198.29\n)\n\n => Array\n(\n => 223.238.85.187\n)\n\n => Array\n(\n => 171.49.176.32\n)\n\n => Array\n(\n => 14.248.79.242\n)\n\n => Array\n(\n => 46.219.211.183\n)\n\n => Array\n(\n => 185.244.212.251\n)\n\n => Array\n(\n => 14.102.84.126\n)\n\n => Array\n(\n => 106.212.191.52\n)\n\n => Array\n(\n => 154.72.153.203\n)\n\n => Array\n(\n => 14.175.82.64\n)\n\n => Array\n(\n => 141.105.139.131\n)\n\n => Array\n(\n => 182.156.103.98\n)\n\n => Array\n(\n => 117.217.204.75\n)\n\n => Array\n(\n => 104.140.83.115\n)\n\n => Array\n(\n => 119.152.62.8\n)\n\n => Array\n(\n => 45.125.247.94\n)\n\n => Array\n(\n => 137.97.37.252\n)\n\n => Array\n(\n => 117.217.204.73\n)\n\n => Array\n(\n => 14.248.79.133\n)\n\n => Array\n(\n => 39.37.152.52\n)\n\n => Array\n(\n => 103.55.60.54\n)\n\n => Array\n(\n => 102.166.183.88\n)\n\n => Array\n(\n => 5.62.60.162\n)\n\n => Array\n(\n => 5.62.60.163\n)\n\n)\n```\nWithdrawal from ESA: Creditcardfree's Personal Finance Blog\n << Back to all Blogs Login or Create your own free blog Layout: Blue and Brown (Default) Author's Creation\nHome > Withdrawal from ESA", null, "", null, "", null, "# Withdrawal from ESA\n\nAugust 2nd, 2018 at 12:37 pm\n\nI know it may seem odd, after just finishing our ESA contributions, but we are now making a withdrawal for our youngest daughter's tuition and housing bill.\n\nI completed the buy order on Tuesday when the market was up and the money, \\$5,375, was in our account today. This money will cover the room and board portion of her bill (yes, for just one semester!)\n\nHer bill is due on the first day of classes, August 20. The total due is \\$6,859.20. I did find out that books will not hit the account until next month, so I have time until I need to make that payment, which if I remember from orientation was \\$475!\n\nOnce all is said and done, I'll need to update you on the cost of things we bought for her dorm and just life as a college student. Hope to find her a warmer coat once we get back to the midwest, possibly at a Columbia outlet mall, they are very hard to find in the summer in the southeast! Lol.\n\n### 4 Responses to “Withdrawal from ESA”\n\n1. rob62521 Says:\n\nI can imagine finding a winter coat in the summer in the Southeast would be a challenge!\n\nColumbia Outlet does have a website, too.\n\n3. creditcardfree Says:\n\nYes, crazyliblady, I am aware. That is a last resort. Since we can get to one before school starts we are going to try in person first. Snow won't be flying for a couple months at least!\n\n4. Bobbie Blue Says:\n\nThis is extraordinary. Some time ago I saw a reference to contact [email protected] or text 205.598.3011 an association of programmers who enable people move up to financial assessments . Well I chose to bet and try it out thus I reached them and incredibly they helped support my financial assessment when I was at that point somewhere down in the red and this took them only 3 weeks . I'm talking 250 bad to 760 good!!!! Nowadays No issue how terrible my score gets each time , they are constantly ready to settle it and now I ceaselessly carry on with the luxurious life I generally needed . Never worrying about my score or dept is the best thing. The interesting part was when i provided my account details for a credit funding and i got a \\$7k top up. Hackunited9 I Owe you all hotshot and expecting you folks see this. Contact them FAST!!!\n\n(Note: If you were logged in, we could automatically fill in these fields for you.)\n Name: * Email: Will not be published. Subscribe: Notify me of additional comments to this entry. URL: Verification: * Please spell out the number 4.  [ Why? ]\n\nvB Code: You can use these tags: [b] [i] [u] [url] [email]" ]
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https://answers.everydaycalculation.com/subtract-fractions/50-12-minus-63-80
[ "Solutions by everydaycalculation.com\n\n## Subtract 63/80 from 50/12\n\n1st number: 4 2/12, 2nd number: 63/80\n\n50/12 - 63/80 is 811/240.\n\n#### Steps for subtracting fractions\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 12 and 80 is 240\n2. For the 1st fraction, since 12 × 20 = 240,\n50/12 = 50 × 20/12 × 20 = 1000/240\n3. Likewise, for the 2nd fraction, since 80 × 3 = 240,\n63/80 = 63 × 3/80 × 3 = 189/240\n4. Subtract the two fractions:\n1000/240 - 189/240 = 1000 - 189/240 = 811/240\n5. In mixed form: 391/240\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 ]
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https://tutorial.math.lamar.edu/Solutions/CalcI/Optimization/Prob3.aspx
[ "Paul's Online Notes\nHome / Calculus I / Applications of Derivatives / Optimization\nShow Mobile Notice Show All Notes Hide All Notes\nMobile Notice\nYou appear to be on a device with a \"narrow\" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.\n\n### Section 4-8 : Optimization\n\n3. Let $$x$$ and $$y$$ be two positive numbers such that $$x + 2y = 50$$ and $$\\left( {x + 1} \\right)\\left( {y + 2} \\right)$$ is a maximum.\n\nShow All Steps Hide All Steps\n\nStart Solution\n\nIn this case we were given the constraint in the problem,\n\n$x + 2y = 50$\n\nWe are also told the equation to maximize,\n\n$f = \\left( {x + 1} \\right)\\left( {y + 2} \\right)$\n\nSo, let’s just solve the constraint for $$x$$ or $$y$$ (we’ll solve for $$x$$ to avoid fractions…) and plug this into the product equation.\n\n$x = 50 - 2y\\hspace{0.25in} \\Rightarrow \\hspace{0.25in}f\\left( y \\right) = \\left( {50 - 2y + 1} \\right)\\left( {y + 2} \\right) = \\left( {51 - 2y} \\right)\\left( {y + 2} \\right) = 102 + 47y - 2{y^2}$ Show Step 2\n\nThe next step is to determine the critical points for this equation.\n\n$f'\\left( y \\right) = 47 - 4y\\hspace{0.25in} \\to \\hspace{0.25in}47 - 4y = 0\\hspace{0.5in} \\to \\hspace{0.5in}y = \\frac{{47}}{4}$ Show Step 3\n\nNow for the step many neglect as unnecessary. Just because we got a single value we can’t just assume that this will give a maximum product. We need to do a quick check to see if it does give a maximum.\n\nAs discussed in notes there are several methods for doing this, but in this case we can quickly see that,\n\n$f''\\left( y \\right) = - 4$\n\nFrom this we can see that the second derivative is always negative and so $$f\\left( y \\right)$$ will always be concave down and so the single critical point we got in Step 2 must be a relative maximum and hence must be the value that gives a maximum.\n\nShow Step 4\n\nFinally, let’s actually answer the question. We need to give both values. We already have $$y$$ so we need to determine $$x$$ and that is easy to do from the constraint.\n\n$x = 50 - 2\\left( {\\frac{{47}}{4}} \\right) = \\frac{{53}}{2}$\n\n$\\require{bbox} \\bbox[2pt,border:1px solid black]{{x = \\frac{{53}}{2}\\hspace{0.5in}y = \\frac{{47}}{4}}}$" ]
[ null ]
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https://gtupractical.com/c-program-to-find-simple-interest/
[ "##### Program to find the simple interest\n\nHere’s a C program that calculates the simple interest based on the principal amount, rate of interest, and time period:\n\n```\t\t\t\t```\n#include <stdio.h>\n\nint main() {\n// Declare variables\nfloat principal, rate, time, simple_interest;\n\n// Get input from user\nprintf(\"Enter principal amount: \");\nscanf(\"%f\", &principal);\n\nprintf(\"Enter rate of interest: \");\nscanf(\"%f\", &rate);\n\nprintf(\"Enter time in years: \");\nscanf(\"%f\", &time);\n\n// Calculate simple interest\nsimple_interest = (principal * rate * time) / 100;\n\n// Print simple interest\nprintf(\"Simple interest = %f\\n\", simple_interest);\n\nreturn 0;\n}\n\n```\n```\n##### Output", null, "• This program first declares four variables: `principal`, `rate`, `time`, and `simple_interest`. The `principal` variable stores the principal amount, the `rate` variable stores the rate of interest, the `time` variable stores the time in years, and the `simple_interest` variable stores the simple interest.\n\n• The program then gets input from the user for the principal amount, the rate of interest, and the time in years.\n\n• The program then calculates the simple interest using the formula `simple_interest = (principal * rate * time) / 100`.\n\n• The program then prints the simple interest to the console.\n\n• You can change the values of the variables in the program to calculate simple interest for different amounts, rates, and times.", null, "```\t\t\t\t```\nSELECT AVG(emp_sal)\nFROM employee;\n```\n```\n```\t\t\t\t```\nSELECT COUNT(*)\nFROM employee;\n```\n```" ]
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http://oalevelsolutions.com/past-papers-solutions/edexcel/as-a-level-mathematics/core-mathematics-1-c1-6663-01/year-2016-june-c1-6663-01/edexcel_16_jun_6663_01_c1_q_9/
[ "# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2016 | June | Q#9\n\nHits: 321\n\nQuestion\n\nOn John’s 10th birthday he received the first of an annual birthday gift of money from his uncle. This  first gift was £60 and on each subsequent birthday the gift was £15 more than the year before.  The amounts of these gifts form an arithmetic sequence.\n\na.   Show that, immediately after his 12th birthday, the total of these gifts was £225.\n\nb.   Find the amount that John received from his uncle as a birthday gift on his 18th birthday.\n\nc.   Find the total of these birthday gifts that John had received from his uncle up to and including his  21st birthday.\n\nd.   Show that n2 + 7n = 25 × 18\n\ne.   Find the value of n, when he had received £3375 in total, and so determine John’s age at this time.\n\nSolution\n\na.\n\nIt is given that the amounts of annual birthday gift received by John form an arithmetic sequence.\n\nFrom the given information we can collect following data about the said arithmetic sequence.", null, "", null, "We are looking for the sum of amounts received on 10th, 11th and 12th birthdays (sum of first 03  terms of the said arithmetic sequence).\n\nExpression for the sum of", null, "number of terms in the Arithmetic Progression (A.P) is:", null, "Therefore;", null, "", null, "", null, "", null, "", null, "", null, "Hence, immediately after his 12th birthday, the total of these gifts was £225.\n\nb.\n\nWe are required to find the amount that John received from his uncle as a birthday gift on his 18th  birthday.\n\nIt is evident that we are looking for 9th term of the said arithmetic sequence.\n\nExpression for the general term", null, "in the Arithmetic Progression (A.P) is:", null, "Therefore;", null, "", null, "", null, "", null, "", null, "Hence, the amount that John received from his uncle as a birthday gift on his 18th birthday is £180.\n\nc.\n\nWe are required to find the total of these birthday gifts that John had received from his uncle up to  and including his 21st birthday.\n\nIt is evident that we are looking for the sum of arithmetic series from first term 12 terms (from 10th  birthday to 21st).\n\nExpression for the sum of", null, "number of terms in the Arithmetic Progression (A.P) is:", null, "Therefore;", null, "", null, "", null, "", null, "", null, "", null, "Hence, total of these birthday gifts that John had received from his uncle up to and including his  21st birthday, is £1710.\n\nd.\n\nWe are given that total of n birthday gifts that John had received from his uncle is £3375.\n\nExpression for the sum of", null, "number of terms in the Arithmetic Progression (A.P) is:", null, "Therefore;", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "e.\n\nWe have found in (d) that;", null, "", null, "", null, "", null, "", null, "Now we have two options.", null, "", null, "", null, "", null, "", null, "", null, "Since, number of birthday cannot be negative, the total amount of gifts £3375 was received on 18th  year since 10th birthday of John.\n\nTherefore, age of John at that time would be 9+18=27 years." ]
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https://ncatlab.org/nlab/show/bitopological%20space
[ "# nLab bitopological space\n\nBitopological space\n\n# Bitopological space\n\n## Definitions\n\nRecall that a topological space is a set $X$ equipped with a topological structure $\\mathcal{T}$. Well, a bitopological space is simply a set equipped with two topological structures $(X, \\mathcal{T}, \\mathcal{T}^*)$. Unlike with bialgebras, no compatibility condition is required between these structures.\n\nA bicontinous map is a function between bitopological spaces that is continuous with respect to each topological structure.\n\nBitopological spaces and bicontinuous maps form a category $BiTop$.\n\n## Separation axioms for topologies\n\nLet $Cl$ denote the closure operator with respect to $\\mathcal{T}$ and let $Cl^*$ denote the closure operator with respect to $\\mathcal{T}^*$.\n\n###### Proposition\n\nLet $(X, \\mathcal{T}, \\mathcal{T}^*)$ be a bitopological space. Consider the following properties of this space:\n\n1. for each point $x$ there is a $\\mathcal{T}$-neighborhood base consisting of $\\mathcal{T}^*$-closed sets;\n\n2. for all $x\\in X$ and all $\\mathcal{T}$-opens $U$ containing $x$ there is a $\\mathcal{T}^*$-closed $\\mathcal{T}$-neighborhood $V$ of $x$ such that $V \\subset U$;\n\n3. $Cl^*(O) \\subset Cl(O)$ for each $\\mathcal{T}^*$-open $O$;\n\n4. for all $x\\in X$ and all $\\mathcal{T}$-neighborhoods $U$ of $x$ the closure $Cl^*(U)$ is a $\\mathcal{T}^*$-neighborhood;\n\n5. for each point $x$ and each $\\mathcal{T}$-closed $\\mathcal{T}$-neighborhood $V$ of $x$ in $X$ there exists a $\\mathcal{T}^*$-closed $\\mathcal{T}$-neighborhood $U$ of $x$ in $X$ such that $U$ is contained in $V$.\n\nThere are the following implications among these properties\n\nEspecially, all properties are equivalent if $\\mathcal{T}$ is regular.\n\n###### Proof\n\n(1) $\\iff$ (2): Given a neighborhood base for a point $x$ as guaranteed by the first property. When you spell out the properties of this neighborhood base, you end up with the second property. For the reverse direction start with an arbitrary $\\mathcal{T}$-neighborhood base of a point $x$ consisting of open. Apply the second property to every element of this neighborhood base to the desired neighborhood base.\n\n(3) $\\iff$ (4): Suppose property (3), and let $U$ be a $\\mathcal{T}$-neighborhood of an arbitrary point $x$. Then the complement $\\widetilde{Cl^*(U)}$ is in $\\mathcal{T}^*$, so that $Cl^*(\\widetilde{Cl^*U}) \\subset Cl(\\widetilde{Cl^* U})$ by the first property. Hence $\\widetilde{Cl(\\widetilde{Cl^*U})} \\subset \\widetilde{Cl^*(\\widetilde{Cl^* U})}$ for the complements. Since $U$ is a $\\mathcal{T}$-neighborhood of $x$, $x$ does not belong to $Cl(\\widetilde{Cl^*U})$. Moreover, $\\widetilde{Cl^*(\\widetilde{Cl^*U})}$ is $\\mathcal{T}^*$-open and a subset of $Cl^*(U)$. Hence $Cl^*(U)$ is a $\\mathcal{T}^*$-neighborhood of $x$.\n\nFor the converse suppose property (4). Let $O$ be a nonempty $\\mathcal{T}^*$-open set and $x$ an element of $Cl^*(O)$. Then if $U$ is any $\\mathcal{T}$-neighborhood of $x$, some point $y \\in O$ belongs to $Cl^*(U)$ due to the second property. Hence, as $O$ is a $\\mathcal{T}^*$-neighborhood of $y$, some point of $U$ belongs to $O$. Thus $x \\in Cl(O)$, and therefore $Cl^*(G) \\subset Cl(O)$.\n\n(1) $\\implies$ (4): Given $x\\in X$ and a $\\mathcal{T}$-neighborhood $U$ by property (1) there is a $\\mathcal{T}^*$-open $O \\subset U$ containing $x$. Hence $O \\subset Cl^*(U)$, and $Cl^*(U)$ is a $\\mathcal{T}^*$-neighborhood.\n\n(3) and $\\mathcal{T}$ regular $\\implies$ (2): Let $x\\in X$ and $U$ be a $\\mathcal{T}$-open containing $x$. By regularity of $\\mathcal{T}$ we can find disjoint $\\mathcal{T}$-opens $V' \\ni x$ and $U' \\supset \\tilde{U}$ ($\\tilde{U}$ denotes the complement). Set $V \\coloneqq Cl^*(V')$. This set is obviously a $\\mathcal{T}^*$-closed $\\mathcal{T}$-neighborhood of $x$. Due to property (3) $V \\subset Cl(V')$. Since also $Cl(V') \\subset \\widetilde{U'}$, we have $V \\subset U$. This is to say that $V$ is the $\\mathcal{T}$-neighborhood we sought.\n\n(5) and $\\mathcal{T}$ regular $\\implies$ (2): Let $x\\in X$ and $U$ be a $\\mathcal{T}$-open containing $x$. By regularity of $\\mathcal{T}$ we can find disjoint $\\mathcal{T}$-opens $V' \\ni x$ and $U' \\supset \\tilde{U}$ ($\\tilde{U}$ denotes the complement). Due to property (5) the closed set $\\widetilde{U'}$ contains a $\\mathcal{T}^*$-closed neighborhood of $x$. This is the neighborhood we sought.\n\n(1) $\\implies$ (5): Given some $\\mathcal{T}$-closed $\\mathcal{T}$-neighborhood $V$ of some point $x$ choose a neighborhood base according to property (1) and take an element $U$ therein that is contained in $V$.\n\n###### Definition\n\nLet $(X, \\mathcal{T}, \\mathcal{T}^*)$ be a bitopological space. The topology $\\mathcal{T}$ is regular with respect to $\\mathcal{T}^*$ if one of the two equivalent conditions (1) and (2) from proposition holds. A bitopological space $(X, \\mathcal{T}, \\mathcal{T}^*)$ is called pairwise regular if $\\mathcal{T}$ is regular with respect to $\\mathcal{T}^*$ and vise versa.\n\n###### Definition\n\nLet $(X, \\mathcal{T}, \\mathcal{T}^*)$ be a bitopological space. The topology $\\mathcal{T}^*$ is coupled to $\\mathcal{T}$ if one of the two equivalent conditions (3) and (4) from proposition holds.\n\nNot that is if $\\mathcal{T}^*$ is coupled to a finer topology $\\mathcal{T} \\supset \\mathcal{T}^*$ then $\\mathcal{T}^*$ is coupled to every topology coarser than $\\mathcal{T}$ due to property (3). Moreover in this case also $\\mathcal{T}$ is coupled to $\\mathcal{T}^*$ (again a direct consequence of property (3)).\n\n###### Definition\n\nLet $(X, \\mathcal{T}, \\mathcal{T}^*)$ be a bitopological space. The topology $\\mathcal{T}^*$ is called a cotopology of $\\mathcal{T}$ if $\\mathcal{T}^* \\subseteq \\mathcal{T}$ and property (5) from proposition holds. The space $(X, \\mathcal{T}^*)$ is also called a cospace of $(X, \\mathcal{T}$.\n\n## Remarks\n\nIt is interesting and perhaps surprising that many advanced topological notions can be described using bitopological spaces, even when you would not naively think that there are two topologies around. (At least, that’s my vague memory of what they were good for. I think that this was in some article by Isbell.)\n\n## Pointfree analogues\n\nExpressed in the opposite category of frames, there are several pointfree analogues of bitopological spaces: D-frames?, biframes?, and finitary biframes?. The latter has the best properties of all three. For an overview, see Suarez Suarez.\n\n• Wikipedia entry\n\n• Jiri Adamek, Horst Herrlich, and George Strecker, Abstract and Concrete Categories: The Joy of Cats, Dover New York 2009. (pdf) pp. 59-60, 278\n\n• B. Dvalishvili, Bitopological Spaces: Theory, Relations with Generalized Algebraic Structures and Applications, Elsevier Amsterdam 2005.\n\n• Peter Johnstone, Collapsed Toposes as Bitopological Spaces, pp. 19-35 in Categorical Topology, World Scientific Singapore 1989.\n\n• O. K. Klinke, A. Jung, A. Moshier, A bitopological point-free approach to compactications (2011). (preprint)\n\n• R. Kopperman, Asymmetry and duality in topology, Topology Appl. 66 no. 1 (1995) pp. 1-39.\n\nThe idea naturally appeared first in the context of quasi-metric spaces\n\n• Wallace Alvin Wilson, On Quasi-Metric Spaces, American Journal of Mathematics (1931), vol. 53, no. 3, pp. 675-684.\n\nThe notions of separation axioms were introduced in\n\n• J. D. Weston, On the comparison of topologies 1956, Journal of the London Mathematical Society, vol. s1-32 no. 3, pp. 342-354,\n\n• J. C. Kelly, Bitopological spaces, Proc. London Math. Soc. 13 no.3 (1963) pp. 71-89.\n\nOnly Kelly introduced the concept in its nowadays formulation of a set equipped with two topologies. The Russian school contributed the following comprehensive overviews of this and related topics\n\n• A. A. Ivanov, Problems of the Theory of Bitopological Spaces, 1990, Journal of Soviet Mathematics, vol. 52, Issue 1, pp. 2759-2790. Originally published as Проблематика теории битопологических пространств in Zap. Nauchn. Sem. POMI, 1988, vol. 167 (Russian version).\n\n• A. A. Ivanov, Problems of the Theory of Bitopological Spaces 2, 1996, Journal of Math. Sciences, vol. 81, Issue 2. Originally publishes as Проблематика теории битопологических пространств. 2 in Zap. Nauchn. Sem. POMI, 1993, Volume 208, pp. 5–67 (Russian version).\n\n• A. A. Ivanov, Problems of the Theory of Bitopological Spaces 3, 1998, Journal of Math. Sciences, vol. 91, Issue 6, pp 3339–3364. Originally published as Проблематика теории битопологических пространств. 3 in Zap. Nauchn. Sem. POMI, 1995, Volume 231, pp. 9–54 (Russian version).\n\nas well as a more introductory text book\n\n• A. A. Ivanov, N. V. Khmylko, Битопологические пространства, 1997. Исследования по топологии. 9, Zap. Nauchn. Sem. POMI, 242, editor A. A. Ivanov (Russian version).\n\nStone duality for bitopological spaces with applications to domain theory is studied in\n\nThe pointfree analogues of bitopoligal spaces are reviewed in\n\n• Anna Laura Suarez, The category of finitary biframes as the category of pointfree bispaces, arXiv:2010.04622." ]
[ null ]
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https://theburningmonk.com/2010/09/project-euler-problem-2-solution/
[ "# Project Euler – Problem 2 Solution\n\n#### Problem\n\nEach new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:\n\n1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …\n\nFind the sum of all the even-valued terms in the sequence which do not exceed four million.\n\n#### Solution\n\nHere’s my solution in F#:\n\n```let fibonacciSeq = Seq.unfold (fun (current, next) -> Some(current, (next, current + next))) (0, 1)\n\nlet fibTotal =\nfibonacciSeq\n|> Seq.takeWhile (fun n -> n < 4000000)\n|> Seq.filter (fun n -> n % 2 = 0)\n|> Seq.sum\n```\n\nHere I’ve used a sequence, whilst a sequence is similar to a list or array in F# in that it holds a series of elements, there’s a crucial difference, each sequence element is computed only as required so it provides better performance than a list in situations which not all elements are used. If that sounds familiar to you, that’s because a sequence is basically an IEnumerable<T>!\n\nIn the first step of this code I’m building up the fibonacci sequence using the Seq.unfold function which given an initial value, generates a sequence by continuously applying some computation to work out each subsequent element in the sequence:\n\n```let fibonacciSeq = Seq.unfold (fun (current, next) -> Some(current, (next, current + next))) (0, 1)\n```\n\nThis sequence, if iterated through, will contain all the numbers in the fibonacci sequence to infinity, which is why in the next line I’ve specified that we should take values from the sequence until the value exceeds 4 million:\n\n```fibonacciSeq\n|> Seq.takeWhile (fun n -> n < 4000000)\n&#91;/code&#93;\n\nThe next two lines then identifies and sums all the even numbers in the sequence:\n\n&#91;code lang=\"fsharp\"&#93;\n|> Seq.filter (fun n -> n % 2 = 0) // only interested in even numbers\n|> Seq.sum // add them up!\n```\n\n### 3 thoughts on “Project Euler – Problem 2 Solution”\n\n1.", null, "isn’t it better to do the filter then the takewhile?\n\n2.", null, "yeah, you’re right, it’ll be more efficient to filter first\n\n3.", null, "The initial value passed to unfold should be (1, 2) instead of (0, 1), right? (I get that it doesn’t affect the final answer due to filtering for evens, but if someone is building up the solution like i was doing and printing out the intermediate results, the results of starting with (0, 1) could be confusing)" ]
[ null, "https://secure.gravatar.com/avatar/22aac6106c44ae4078bd4c0a564714e0", null, "https://secure.gravatar.com/avatar/0bbaed5a18c131ba152cf566e1becaf8", null, "https://secure.gravatar.com/avatar/", null ]
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https://www.mathworks.com/help/control/ref/dynamicsystem.modalsep.html
[ "# modalsep\n\nCompute modal decomposition\n\nSince R2023b\n\n## Syntax\n\n``[H,H0] = modalsep(G)``\n``[H,H0] = modalsep(G,Name=Value)``\n``[H,H0] = modalsep(G,N,regionFcn)``\n``[H,H0,info] = modalsep(G,___)``\n\n## Description\n\nexample\n\n````[H,H0] = modalsep(G)` computes the modal decomposition for a linear time-invariant (LTI) system `G` and returns the modal components as a state-space array `H` and the static gain `H0`.$G\\left(s\\right)={H}_{0}+\\sum _{j=1}^{m}{H}_{j}\\left(s\\right)$Each modal component in Hj(s) is associated with a single real pole, a pair of complex conjugate poles, or a cluster repeated poles.```\n````[H,H0] = modalsep(G,Name=Value)` returns the modal decomposition based on the options specified by one or more name-value arguments. Use these options to control the granularity and accuracy of the decomposition.```\n\nexample\n\n````[H,H0] = modalsep(G,N,regionFcn)` computes the region-based modal decomposition$G\\left(s\\right)={H}_{0}+\\sum _{j=1}^{N}{H}_{j}\\left(s\\right)$Here, the modal components Hj(s) have their poles in disjoint regions Rj of the complex plane. `N` specifies the number of regions and `regionFcn` is the name or a handle to the function that specifies the partition into `N` regions.```\n\nexample\n\n````[H,H0,info] = modalsep(G,___)` also returns a structure `info` containing the block-diagonalizing transformation matrices for the model and mode information for each modal component.```\n\n## Examples\n\ncollapse all\n\nThis example shows how to obtain a modal decomposition for a linear time-invariant (LTI) model using `modalsep`.\n\nCreate a random MIMO state-space model with 20 states.\n\n```rng(0) G = rss(20,2,3);```\n\nObtain the modal decomposition of this model.\n\n`[H,H0] = modalsep(G);`\n\nExamine the size of `H`.\n\n`size(H)`\n```16x1 array of state-space models. Each model has 2 outputs, 3 inputs, and between 1 and 2 states. ```\n\nTypically, the modal components are of order 1 or 2, but can have higher orders in case of cluster of repeated poles.\n\nFor this model, `modalsep` returns the static gain `H0` for each I/O pair.\n\n`H0`\n```H0 = D = u1 u2 u3 y1 0 -1.237 -0.3334 y2 0.1554 -2.193 0 Static gain. ```\n\nAdditionally, you can obtain the original model back from the modal decomposition using `modalsum`.\n\n```G2 = modalsum(H,H0); sigma(G,G2,'r--')```", null, "This example shows how to perform region-based modal decomposition of a state-space model. In this example, you perform the modal decomposition of a high-order model based on damping ratio of the poles.\n\nLoad a model and examine the damping ratio and natural frequency of the poles.\n\n```load('highOrderModel.mat','G') [wn,zeta] = damp(G);```\n\nVisualize the damping ratios and natural frequencies.\n\n```semilogy(zeta,wn,\"x\"); grid on title(\"Mode Damping and Natural Frequency\") ylabel(\"Natural frequency\") xlabel(\"Damping\")```", null, "Based on this, you can define regions function that decomposes the model based on the damping ratio values. For example, this model has damping ratios between 0.023 and 0.05, you can divide three regions as follows:\n\n• Region 1: $0.023\\le \\mathit{z}<0.033$\n\n• Region 2: $0.033\\le \\mathit{z}<0.040$\n\n• Region 3: $\\mathit{z}\\ge 0.040$\n\nThe region function to assign region index to these values is defined at the end of this example. See Region Function Definition.\n\nPerform the modal decomposition.\n\n`[h,h0,info] = modalsep(G,3,@regionFcn);`\n\nExamine the size of each region.\n\n`size(h(:,:,1))`\n```State-space model with 1 outputs, 1 inputs, and 24 states. ```\n`size(h(:,:,2))`\n```State-space model with 1 outputs, 1 inputs, and 12 states. ```\n`size(h(:,:,3))`\n```State-space model with 1 outputs, 1 inputs, and 12 states. ```\n\nFor this decomposition, region 1 has 24 poles in the specified range. Region 2 and 3 have 12 poles each in their specified ranges.\n\nRegion Function Definition\n\n```function IR = regionFcn(p) [~,z] = damp(p); IR = zeros(size(z)); IR(z<0.033 & z>=0.023) = 1; IR(z<0.04 & z>=0.033) = 2; IR(z>=0.04) = 3; end```\n\n## Input Arguments\n\ncollapse all\n\nLinear time-invariant model, specified as an `ss`, `tf`, or `zpk` model.\n\nNumber of regions, specified as a positive scalar.\n\nSpecify this value as the number of partitions performed by the `regionFcn`.\n\nUser-defined function for partitioning the complex plane, specified as a function name (string) or a function handle.\n\nSpecify a region function of the form:\n\n`IR = regionFcn(p);`\n\nThe function assigns a region index `IR` between `1` and `N` to a given pole `p`. To specify additional inputs to the region function, use an anonymous function.\n\n`regionFcn = @(p) myFcn(p,param1,...,paramK);`\n\n### Name-Value Arguments\n\nSpecify optional pairs of arguments as `Name1=Value1,...,NameN=ValueN`, where `Name` is the argument name and `Value` is the corresponding value. Name-value arguments must appear after other arguments, but the order of the pairs does not matter.\n\nExample: `[H,H0] = modalsep(G,InputScaling=[1,1e-3])`\n\nFrequency for evaluating and matching DC contributions, specified as a nonnegative scalar.\n\nFor models with integrators, you cannot evaluate modal contributions at DC since the DC gain is infinite. To evaluate modal contributions and match gains at a different frequency, set the property to a positive value. The default value of this property corresponds to the true DC value.\n\nInput scaling factors, specified as a vector of length Nu, where Nu is the number of inputs in the original model `sys`.\n\nUse this option to emphasize specific input channels in `sys`. The software evaluates the modal contributions for the scaled system (see the `info` argument).\n\nOutput scaling factors, specified as a vector of length Ny, where Ny is the number of outputs in the original model `sys`.\n\nUse this option to emphasize specific output channels in `sys`. The software evaluates the modal contributions for the scaled system (see the `info` argument).\n\nRelative accuracy of modal decomposition, specified as a scalar between 0 and 1.\n\nThis option limits the condition number of the block diagonalizing transformation to roughly `SepTol`/`eps`. Increasing `SepTol` helps yield smaller modal components at the expense of accuracy.\n\nFor region-based decompositions, the function ignores this option because the block sized are predefined.\n\n## Output Arguments\n\ncollapse all\n\nModal components, returned as an array of models of the same type as `G`. For example, if `G` is a `zpk` model, `modalsep` returns the modal components as an array of `zpk` models.\n\nUse `H(:,:,j)` to retrieve the submodel Hj(s).\n\nStatic gain, returned as a matrix or a static `ss` object.\n\nAdditional information about the modal decomposition, returned as structure with these fields.\n\nFieldDescription\n`Mode`Average mode (pole) value in the modal component Hj, returned as a vector of length nc-by-1, where nc is the number of modal components.\n`DCGain`Normalized DC contribution of modal components, returned as a vector of length nc-by-1, where nc is the number of modal components.\n`TL`\n\nLeft-side matrix of the block-diagonalizing transformation, returned as a matrix with dimensions Nx-by-Nx, where Nx is the number of states in the model `G`.\n\nThe algorithm transforms the state-space realization (A, B, C, D, E) of a model to block diagonal matrices (Am, Bm, Cm, Dm, Em) given by:\n\n• For explicit state-space models\n\n`$\\begin{array}{l}\\begin{array}{cc}{A}_{m}={T}_{L}A{T}_{R},& {B}_{m}={T}_{L}B,\\end{array}\\\\ \\begin{array}{ccc}{C}_{m}=C{T}_{R},& {D}_{m}=D,& E={T}_{L}{T}_{R}=I\\end{array}\\end{array}$`\n• For descriptor state-space models\n\n`$\\begin{array}{l}\\begin{array}{cc}{A}_{m}={T}_{L}A{T}_{R},& {B}_{m}={T}_{L}B,\\end{array}\\\\ \\begin{array}{ccc}{C}_{m}=C{T}_{R},& {D}_{m}=D,& {E}_{m}={T}_{L}E{T}_{R}\\end{array}\\end{array}$`\n\n`TR`\n\nRight-side matrix of the block-diagonalizing transformation, returned as a matrix with dimensions Nx-by-Nx, where Nx is the number of states in the model `G`.The algorithm transforms the state-space realization (A, B, C, D, E) of a model to block diagonal matrices (Am, Bm, Cm, Dm, Em) given by:\n\n• For explicit state-space models\n\n`$\\begin{array}{l}\\begin{array}{cc}{A}_{m}={T}_{L}A{T}_{R},& {B}_{m}={T}_{L}B,\\end{array}\\\\ \\begin{array}{ccc}{C}_{m}=C{T}_{R},& {D}_{m}=D,& E={T}_{L}{T}_{R}=I\\end{array}\\end{array}$`\n• For descriptor state-space models\n\n`$\\begin{array}{l}\\begin{array}{cc}{A}_{m}={T}_{L}A{T}_{R},& {B}_{m}={T}_{L}B,\\end{array}\\\\ \\begin{array}{ccc}{C}_{m}=C{T}_{R},& {D}_{m}=D,& {E}_{m}={T}_{L}E{T}_{R}\\end{array}\\end{array}$`\n\nFor `zpk` and `tf` models, the function returns an empty value `[]` for this argument.\n\n## Version History\n\nIntroduced in R2023b" ]
[ null, "https://www.mathworks.com/help/examples/control/win64/ComputeModalDecompositionOfLTIModelsExample_01.png", null, "https://www.mathworks.com/help/examples/control/win64/RegionBasedModalDecompositionExample_01.png", null ]
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http://docs.ros.org/en/electric/api/orocos_kdl/html/classKDL_1_1RotationalInertia.html
[ "\\$search\n\n# KDL::RotationalInertia Class Reference\n\n`#include <rotationalinertia.hpp>`\n\nList of all members.\n\n## Public Member Functions\n\nKDL::Vector operator* (const KDL::Vector &omega) const\nRotationalInertia (double Ixx=0, double Iyy=0, double Izz=0, double Ixy=0, double Ixz=0, double Iyz=0)\n~RotationalInertia ()\n\n## Static Public Member Functions\n\nstatic RotationalInertia Zero ()\n\ndouble data \n\n## Friends\n\nRigidBodyInertia operator* (const Rotation &R, const RigidBodyInertia &I)\nbase frame orientation change Ia = R_a_b*Ib with R_a_b the rotation for frame from a to b\nRigidBodyInertia operator* (const Frame &T, const RigidBodyInertia &I)\ncoordinate system transform Ia = T_a_b*Ib with T_a_b the frame from a to b\nWrench operator* (const RigidBodyInertia &I, const Twist &t)\ncalculate spatial momentum\nRigidBodyInertia operator* (double a, const RigidBodyInertia &I)\nScalar product.\nRotationalInertia operator* (double a, const RotationalInertia &I)\nRigidBodyInertia operator+ (const RigidBodyInertia &Ia, const RigidBodyInertia &Ib)\nRotationalInertia operator+ (const RotationalInertia &Ia, const RotationalInertia &Ib)\nclass RigidBodyInertia\n\n## Detailed Description\n\nDefinition at line 34 of file rotationalinertia.hpp.\n\n## Constructor & Destructor Documentation\n\n KDL::RotationalInertia::RotationalInertia ( double Ixx = `0`, double Iyy = `0`, double Izz = `0`, double Ixy = `0`, double Ixz = `0`, double Iyz = `0` ) ` [explicit]`\n\nDefinition at line 28 of file rotationalinertia.cpp.\n\n KDL::RotationalInertia::~RotationalInertia ( )\n\nDefinition at line 39 of file rotationalinertia.cpp.\n\n## Member Function Documentation\n\n Vector KDL::RotationalInertia::operator* ( const KDL::Vector & omega ) const\n\nThis function calculates the angular momentum resulting from a rotational velocity omega\n\nDefinition at line 43 of file rotationalinertia.cpp.\n\n static RotationalInertia KDL::RotationalInertia::Zero ( ) ` [inline, static]`\n\nDefinition at line 39 of file rotationalinertia.hpp.\n\n## Friends And Related Function Documentation\n\n RigidBodyInertia operator* ( const Rotation & R, const RigidBodyInertia & I ) ` [friend]`\n\nbase frame orientation change Ia = R_a_b*Ib with R_a_b the rotation for frame from a to b\n\nReference frame orientation change Ia = R_a_b*Ib with R_a_b the rotation of b expressed in a\n\nDefinition at line 74 of file rigidbodyinertia.cpp.\n\n RigidBodyInertia operator* ( const Frame & T, const RigidBodyInertia & I ) ` [friend]`\n\ncoordinate system transform Ia = T_a_b*Ib with T_a_b the frame from a to b\n\nCoordinate system transform Ia = T_a_b*Ib with T_a_b the frame from a to b.\n\nDefinition at line 56 of file rigidbodyinertia.cpp.\n\n Wrench operator* ( const RigidBodyInertia & I, const Twist & t ) ` [friend]`\n\ncalculate spatial momentum\n\ncalculate spatial momentum: h = I*v make sure that the twist v and the inertia are expressed in the same reference frame/point\n\nDefinition at line 52 of file rigidbodyinertia.cpp.\n\n RigidBodyInertia operator* ( double a, const RigidBodyInertia & I ) ` [friend]`\n\nScalar product.\n\nScalar product: I_new = double * I_old\n\nDefinition at line 44 of file rigidbodyinertia.cpp.\n\n RotationalInertia operator* ( double a, const RotationalInertia & I ) ` [friend]`\n\nDefinition at line 50 of file rotationalinertia.cpp.\n\n RigidBodyInertia operator+ ( const RigidBodyInertia & Ia, const RigidBodyInertia & Ib ) ` [friend]`\n\naddition I: I_new = I_old1 + I_old2, make sure that I_old1 and I_old2 are expressed in the same reference frame/point, otherwise the result is worth nothing\n\nDefinition at line 48 of file rigidbodyinertia.cpp.\n\n RotationalInertia operator+ ( const RotationalInertia & Ia, const RotationalInertia & Ib ) ` [friend]`\n\nDefinition at line 56 of file rotationalinertia.cpp.\n\n friend class RigidBodyInertia` [friend]`\n\nDefinition at line 53 of file rotationalinertia.hpp.\n\n## Member Data Documentation\n\n double KDL::RotationalInertia::data\n\nDefinition at line 65 of file rotationalinertia.hpp.\n\nThe documentation for this class was generated from the following files:\n\norocos_kdl\nAuthor(s): Ruben Smits, Erwin Aertbelien, Orocos Developers\nautogenerated on Fri Mar 1 16:20:17 2013" ]
[ null ]
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https://cs.stackexchange.com/questions/145528/karatsube-ofman-runtime-complexity-computation
[ "# Karatsube-Ofman runtime complexity computation\n\nI have a question and didn't understand the solution, since we didn't take how to do it in the lecture and it's not explained in the solution sample.\n\nQuestion: One can generalize the Karatsube-Ofman algorithm even further and divide the number to be squared into b parts with n / b places. However, one then has to perform 2b - 1 recursive squaring of numbers with n / b digits 2 plus O (n) overhead for various “simple” operations. Represent the running time Q b (n) of this method by a recursion equation and try to solve this recursion. Which asymptotic running time is achieved?\n\nSolution: there exists $$\\alpha$$ and $$\\beta$$ such that:", null, "with some estimates and the partial summation of the geometrical series follows:", null, "My questions:\n\n1: how did we convert the $$(2b+1)Q_b(\\frac{n}{b}+\\beta n)$$ to the first line of the ?\n\n2: did we use the sum since it's for all number $$n\\geq 2$$?\n\n3: in first line why did we multiply $$\\beta$$ with $$(2b-1)$$?\n\n4: why $$\\log_b(n)-1$$ above the sum ?\n\n5: in third line how did we remove the sum?\n\n6: in fourth line why did we use inequality ?\n\n7: in fifth line how did it get converted ? from 3 and 4\n\n• This is just the proof of the master theorem in a special case. Look up any proof of the master theorem for an explanation of all the steps. Nov 8 at 11:07\n• thanks! I didn't find it when search for \"Karatsube-Ofma proof\", so does mean it works for all recurrence relations ? or just specific equations ? Nov 8 at 11:19\n• The master theorem applies to all recurrence relations to which it applies. It does not apply to recurrence relations to which it does not apply. Nov 8 at 11:20\n• hmm I can't select best answer from comment should I delete my question ? Nov 8 at 11:24\n• I'll write an answer. Nov 8 at 11:25" ]
[ null, "https://i.stack.imgur.com/lQ2hm.png", null, "https://i.stack.imgur.com/JGcpm.png", null ]
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https://www.colorhexa.com/0a6479
[ "# #0a6479 Color Information\n\nIn a RGB color space, hex #0a6479 is composed of 3.9% red, 39.2% green and 47.5% blue. Whereas in a CMYK color space, it is composed of 91.7% cyan, 17.4% magenta, 0% yellow and 52.5% black. It has a hue angle of 191.4 degrees, a saturation of 84.7% and a lightness of 25.7%. #0a6479 color hex could be obtained by blending #14c8f2 with #000000. Closest websafe color is: #006666.\n\n• R 4\n• G 39\n• B 47\nRGB color chart\n• C 92\n• M 17\n• Y 0\n• K 53\nCMYK color chart\n\n#0a6479 color description : Dark cyan.\n\n# #0a6479 Color Conversion\n\nThe hexadecimal color #0a6479 has RGB values of R:10, G:100, B:121 and CMYK values of C:0.92, M:0.17, Y:0, K:0.53. Its decimal value is 681081.\n\nHex triplet RGB Decimal 0a6479 `#0a6479` 10, 100, 121 `rgb(10,100,121)` 3.9, 39.2, 47.5 `rgb(3.9%,39.2%,47.5%)` 92, 17, 0, 53 191.4°, 84.7, 25.7 `hsl(191.4,84.7%,25.7%)` 191.4°, 91.7, 47.5 006666 `#006666`\nCIE-LAB 38.827, -15.996, -18.584 8.133, 10.559, 19.697 0.212, 0.275, 10.559 38.827, 24.52, 229.281 38.827, -27.078, -23.786 32.494, -12.189, -13.195 00001010, 01100100, 01111001\n\n# Color Schemes with #0a6479\n\n• #0a6479\n``#0a6479` `rgb(10,100,121)``\n• #791f0a\n``#791f0a` `rgb(121,31,10)``\nComplementary Color\n• #0a7956\n``#0a7956` `rgb(10,121,86)``\n• #0a6479\n``#0a6479` `rgb(10,100,121)``\n• #0a2d79\n``#0a2d79` `rgb(10,45,121)``\nAnalogous Color\n• #79560a\n``#79560a` `rgb(121,86,10)``\n• #0a6479\n``#0a6479` `rgb(10,100,121)``\n• #790a2d\n``#790a2d` `rgb(121,10,45)``\nSplit Complementary Color\n• #64790a\n``#64790a` `rgb(100,121,10)``\n• #0a6479\n``#0a6479` `rgb(10,100,121)``\n• #790a64\n``#790a64` `rgb(121,10,100)``\n• #0a791f\n``#0a791f` `rgb(10,121,31)``\n• #0a6479\n``#0a6479` `rgb(10,100,121)``\n• #790a64\n``#790a64` `rgb(121,10,100)``\n• #791f0a\n``#791f0a` `rgb(121,31,10)``\n• #042a32\n``#042a32` `rgb(4,42,50)``\n• #063d4a\n``#063d4a` `rgb(6,61,74)``\n• #085161\n``#085161` `rgb(8,81,97)``\n• #0a6479\n``#0a6479` `rgb(10,100,121)``\n• #0c7791\n``#0c7791` `rgb(12,119,145)``\n• #0e8ba8\n``#0e8ba8` `rgb(14,139,168)``\n• #109ec0\n``#109ec0` `rgb(16,158,192)``\nMonochromatic Color\n\n# Alternatives to #0a6479\n\nBelow, you can see some colors close to #0a6479. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #0a7972\n``#0a7972` `rgb(10,121,114)``\n• #0a7779\n``#0a7779` `rgb(10,119,121)``\n• #0a6d79\n``#0a6d79` `rgb(10,109,121)``\n• #0a6479\n``#0a6479` `rgb(10,100,121)``\n• #0a5b79\n``#0a5b79` `rgb(10,91,121)``\n• #0a5279\n``#0a5279` `rgb(10,82,121)``\n• #0a4879\n``#0a4879` `rgb(10,72,121)``\nSimilar Colors\n\n# #0a6479 Preview\n\nText with hexadecimal color #0a6479\n\nThis text has a font color of #0a6479.\n\n``<span style=\"color:#0a6479;\">Text here</span>``\n#0a6479 background color\n\nThis paragraph has a background color of #0a6479.\n\n``<p style=\"background-color:#0a6479;\">Content here</p>``\n#0a6479 border color\n\nThis element has a border color of #0a6479.\n\n``<div style=\"border:1px solid #0a6479;\">Content here</div>``\nCSS codes\n``.text {color:#0a6479;}``\n``.background {background-color:#0a6479;}``\n``.border {border:1px solid #0a6479;}``\n\n# Shades and Tints of #0a6479\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, #010a0c is the darkest color, while #f9feff is the lightest one.\n\n• #010a0c\n``#010a0c` `rgb(1,10,12)``\n• #03191e\n``#03191e` `rgb(3,25,30)``\n• #042831\n``#042831` `rgb(4,40,49)``\n• #063743\n``#063743` `rgb(6,55,67)``\n• #074655\n``#074655` `rgb(7,70,85)``\n• #095567\n``#095567` `rgb(9,85,103)``\n• #0a6479\n``#0a6479` `rgb(10,100,121)``\n• #0b738b\n``#0b738b` `rgb(11,115,139)``\n• #0d829d\n``#0d829d` `rgb(13,130,157)``\n• #0e91af\n``#0e91af` `rgb(14,145,175)``\n• #10a0c1\n``#10a0c1` `rgb(16,160,193)``\n• #11afd4\n``#11afd4` `rgb(17,175,212)``\n• #13bee6\n``#13bee6` `rgb(19,190,230)``\n• #20c6ed\n``#20c6ed` `rgb(32,198,237)``\n• #32caee\n``#32caee` `rgb(50,202,238)``\n• #44cff0\n``#44cff0` `rgb(68,207,240)``\n• #56d4f1\n``#56d4f1` `rgb(86,212,241)``\n• #68d8f3\n``#68d8f3` `rgb(104,216,243)``\n``#7addf4` `rgb(122,221,244)``\n• #8ce2f6\n``#8ce2f6` `rgb(140,226,246)``\n• #9fe6f7\n``#9fe6f7` `rgb(159,230,247)``\n• #b1ebf9\n``#b1ebf9` `rgb(177,235,249)``\n• #c3f0fa\n``#c3f0fa` `rgb(195,240,250)``\n• #d5f4fc\n``#d5f4fc` `rgb(213,244,252)``\n• #e7f9fd\n``#e7f9fd` `rgb(231,249,253)``\n• #f9feff\n``#f9feff` `rgb(249,254,255)``\nTint Color Variation\n\n# Tones of #0a6479\n\nA tone is produced by adding gray to any pure hue. In this case, #414242 is the less saturated color, while #05677e is the most saturated one.\n\n• #414242\n``#414242` `rgb(65,66,66)``\n• #3c4547\n``#3c4547` `rgb(60,69,71)``\n• #37484c\n``#37484c` `rgb(55,72,76)``\n• #324b51\n``#324b51` `rgb(50,75,81)``\n• #2d4e56\n``#2d4e56` `rgb(45,78,86)``\n• #28515b\n``#28515b` `rgb(40,81,91)``\n• #235460\n``#235460` `rgb(35,84,96)``\n• #1e5765\n``#1e5765` `rgb(30,87,101)``\n• #195b6a\n``#195b6a` `rgb(25,91,106)``\n• #145e6f\n``#145e6f` `rgb(20,94,111)``\n• #0f6174\n``#0f6174` `rgb(15,97,116)``\n• #0a6479\n``#0a6479` `rgb(10,100,121)``\n• #05677e\n``#05677e` `rgb(5,103,126)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #0a6479 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]
[ null ]
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https://johnbaras.com/publication-conferen/time-series-modeling-by-perceptrons-a-likelihood-approach/
[ "# Time Series Modeling by Perceptrons: A Likelihood Approach\n\n## Time Series Modeling by Perceptrons: A Likelihood Approach\n\nTitle : Time Series Modeling by Perceptrons: A Likelihood Approach\nAuthors :\nBaras, John S.\nSonmez, Kemal M\n\nConference : The World Congress on Neural Networks Vol. IV, pp. 601-604\nDate: July 01 - July 01, 1993\n\nWe consider neural network learning problems in which the objective is to learn the relationship between the inputs and the probability distribution of a proposition. We regard successive truth values of the proposition as a dependent binary time series whose instantaneous probability of truth is a function of the past behavior of the joint process between the analog inputs and binary output truth values. In this context, we identify the gradient descent learning algorithm using the Kullback-Leibler relative entropy cost function on a perceptron with a Maximum Partial Likelihood (MPL) estimator of the perceptron model for the probability of a binary event in terms of its covariates. The implications of this result are: (i) The neural network models obtained by relative entropy learning are shown to have the nice large sample (i.e. training set size) characteristics of MPL estimates: consistency and asymptotic normality. (ii) An important and widely used statistical inference technique, logistic regression, can efficiently be implemented on analog perceptrons for time series modeling and prediction." ]
[ null ]
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https://de.maplesoft.com/support/help/errors/view.aspx?path=DEtools%2Fmaxdimsystems
[ "", null, "maxdimsystems - Maple Help\n\nDEtools\n\n maxdimsystems\n determination of maximum dimensional subsystems of systems of PDE", null, "Calling Sequence maxdimsystems(system, maxdimvars) maxdimsystems(system, maxdimvars, options)", null, "Parameters\n\n system - list or set of ODE or PDE (may contain inequations) maxdimvars - list of the dependent variables used for the dimension count options - (optional) sequence of options to control the behavior of maxdimsystems", null, "Description\n\n • The default output of maxdimsystems (modulo *** below) consists of the set of cases having maximal dimension in maxdimvars, where each member has the form:\n\n [Solved = list, Constraint = list, Pivots = list, dimension = integer]\n\n Solved is a list of the PDE which are in solved form with respect to their highest derivative (with respect to the given ranking of derivatives), Constraint is a list of the PDEs in the classvars (any dependent variables not in maxdimvars) which are nonlinear in their leading derivatives, Pivots is a list of the inequations in the classvars and/or maxdimvars, and dimension is the maximal degree of arbitrariness in maxdimvars (i.e. the number of arbitrary constants on which maxdimvars depends).\n *** The result is valid provided that the Constraint and Pivots equations have a solution, and also that maxdimvars do not occur in the constraint equations. The user must check the first condition algebraically (e.g. see Groebner[Solve], and possibly DifferentialAlgebra). If the maxdimvars are present in the pivots, then the returned system will have lower dimension that indicated, and must be handled manually.\n • Each of the cases is in rifsimp form (a generalization of Gauss normal form to systems of PDE) and also contains a system of differential equations and inequations in the classvars characterizing the case.  In the same way that the Gauss normal form is determined by a ordering of variables, rifsimp form is determined by a ranking of derivatives.\n • The default ranking is determined by maxdimvars. It can be inferred by the application of checkrank, and always ranks the classvars and their derivatives lower than all maxdimvars.\n • The maximal cases can be determined in many applications where a complete analysis of all cases is not possible.  These cases are often the most interesting, with the richest properties, when compared to the relatively uninteresting (but often difficult to compute) lower dimensional cases.\n • The options for maxdimsystems consist of the output option (described below), and a subset of those for rifsimp, the main subroutine used by maxdimsystems. Of the subset of rifsimp options available, two that are often used are indep which specifies the independent variables, ezcriteria which controls the flow of equations from a fast (ultra-hog) setting (that can readily lead to memory explosion but sometimes faster calculations), to a conservative (one equation at a time) setting that can overcome some expression swell problems. The complete list is given by:\n\n tosolve, indep, arbitrary, clean, fullclean ezcriteria, faclimit mindim, pivselect grob_rank,grobonly,checkempty\n\n Many rifsimp options are not supported by maxdimsystems as their use could produce misleading results. For example, the rifsimp option $\\mathrm{ctl}=\\mathrm{time}$, places a time limit on rifsimp case computations, and hence could cause maxdimsystems to miss the maximal case if it takes too long to compute. Conversely, specification of other options are not supported as they must have a specific value for the maxdimsystems command to function properly (for example, casesplit must always be on). Other unsupported rifsimp options include itl, stl, nopiv, unsolved, store, storeall, spawn, spoly, and casecount.\n • output = type\n The output type can either be set (default) or $\\mathrm{rif}$. Note that set output is easier to read and manipulate, but cannot be used with the caseplot visualization tool. The $\\mathrm{rif}$ output can be used with caseplot, but the output is somewhat more complicated, and is described in rifsimp[output]. Note that for the class of problems that maxdimsystems simplifies, many of the outputs from rifsimp described in the output page ( such as DiffConstraint and UnSolve) will never be present in the output.\n • The maxdimsystems command allows a subset of the rankings available for rifsimp. Specifically those rankings that are controlled by a partition of the dependent variables in maxdimvars, and those available by the order of the independent variables in indep. See rifsimp[ranking] and checkrank for an explanation of how to specify rankings based on these options.\n • The application of caseplot to the $\\mathrm{rif}$-form output of maxdimsystems with the second argument given as the maxdimvars displays the tree of maximal cases and their dimensions, also indicating which branches of the tree have less than maximal dimension.\n • Note that maxdimsystems is essentially a simplified version of the rifsimp interface that also has the ability to automatically determine the maximum finite dimensional cases or all infinite dimensional cases for a classification ODE/PDE system. If the dimension of the solution space is known, multiple conditions on the dimension of the solution space are needed, or much finer control of the computation is desired, then rifsimp could be used with the mindim option described in rifsimp[cases].\n • References:\n G.J. Reid and A.D. Wittkopf, \"Determination of Maximal Symmetry Groups of Classes of Differential Equations\", Proc. ISSAC 2000, pp.272-280\n G.W. Bluman and S. Kumei, \"Symmetries and Differential Equations\", Springer-Verlag, vol. 81.", null, "Examples\n\n > $\\mathrm{with}\\left(\\mathrm{DEtools}\\right):$\n\nAlgebraic systems can be examined with maxdimsystems\n\n > $\\mathrm{maxdimsystems}\\left(\\left[u-av=0,u-cv=0\\right],\\left[u,v\\right]\\right)$\n $\\left\\{\\left[{\\mathrm{Solved}}{=}\\left[{u}{=}{c}{}{v}{,}{a}{=}{c}\\right]{,}{\\mathrm{Constraint}}{=}\\left[\\right]{,}{\\mathrm{Pivots}}{=}\\left[\\right]{,}{\\mathrm{dimension}}{=}{1}\\right]\\right\\}$ (1)\n\nIn the below example, inequations are given in the Pivots list, and the dimension is 2\n\n > $\\mathrm{maxdimsystems}\\left(\\left[\\left(a\\left(x\\right)-d\\left(x\\right)\\right)\\left(\\frac{{ⅆ}^{2}}{ⅆ{x}^{2}}u\\left(x\\right)\\right)+b\\left(x\\right)\\left(\\frac{ⅆ}{ⅆx}u\\left(x\\right)\\right)+u\\left(x\\right)=0\\right],\\left[u\\right]\\right)$\n $\\left\\{\\left[{\\mathrm{Solved}}{=}\\left[\\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{u}{}\\left({x}\\right){=}\\frac{{-}{b}{}\\left({x}\\right){}\\left(\\frac{{ⅆ}}{{ⅆ}{x}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{u}{}\\left({x}\\right)\\right){-}{u}{}\\left({x}\\right)}{{a}{}\\left({x}\\right){-}{d}{}\\left({x}\\right)}\\right]{,}{\\mathrm{Constraint}}{=}\\left[\\right]{,}{\\mathrm{Pivots}}{=}\\left[{a}{}\\left({x}\\right){-}{d}{}\\left({x}\\right){\\ne }{0}\\right]{,}{\\mathrm{dimension}}{=}{2}\\right]\\right\\}$ (2)\n\nThis next example has a constraint on the classvars present in the result\n\n > $\\mathrm{maxdimsystems}\\left(\\left[\\frac{\\partial }{\\partial x}u\\left(x,y\\right)-f\\left(x,y\\right)u\\left(x,y\\right)=0,\\frac{\\partial }{\\partial y}u\\left(x,y\\right)-g\\left(x,y\\right)u\\left(x,y\\right)=0,{f\\left(x,y\\right)}^{3}-1=0\\right],\\left[u\\right]\\right)$\n $\\left\\{\\left[{\\mathrm{Solved}}{=}\\left[\\frac{{\\partial }}{{\\partial }{x}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{u}{}\\left({x}{,}{y}\\right){=}{f}{}\\left({x}{,}{y}\\right){}{u}{}\\left({x}{,}{y}\\right){,}\\frac{{\\partial }}{{\\partial }{y}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{u}{}\\left({x}{,}{y}\\right){=}{g}{}\\left({x}{,}{y}\\right){}{u}{}\\left({x}{,}{y}\\right){,}\\frac{{\\partial }}{{\\partial }{x}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{f}{}\\left({x}{,}{y}\\right){=}{0}{,}\\frac{{\\partial }}{{\\partial }{x}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{g}{}\\left({x}{,}{y}\\right){=}{0}{,}\\frac{{\\partial }}{{\\partial }{y}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{f}{}\\left({x}{,}{y}\\right){=}{0}\\right]{,}{\\mathrm{Constraint}}{=}\\left[{{f}{}\\left({x}{,}{y}\\right)}^{{3}}{-}{1}{=}{0}\\right]{,}{\\mathrm{Pivots}}{=}\\left[\\right]{,}{\\mathrm{dimension}}{=}{1}\\right]\\right\\}$ (3)\n\nIn each of the above examples maxdimsystems picked out the maximum dimensional case. This is in contrast to rifsimp, with its casesplit option, which obtains all cases.\n\nFor the second example, computation with rifsimp using casesplit gives 2-d, 1-d and 0-d cases, of which the 2-d case is the maximal case from the above example.\n\n > $\\mathrm{ans}≔\\mathrm{rifsimp}\\left(\\left[\\left(a\\left(x\\right)-d\\left(x\\right)\\right)\\left(\\frac{{ⅆ}^{2}}{ⅆ{x}^{2}}u\\left(x\\right)\\right)+b\\left(x\\right)\\left(\\frac{ⅆ}{ⅆx}u\\left(x\\right)\\right)+u\\left(x\\right)=0\\right],\\left[u\\right],\\mathrm{casesplit}\\right)$\n ${\\mathrm{ans}}{≔}{table}{}\\left(\\left[{1}{=}{table}{}\\left(\\left[{\\mathrm{Case}}{=}\\left[\\left[{a}{}\\left({x}\\right){-}{d}{}\\left({x}\\right){\\ne }{0}{,}\\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{u}{}\\left({x}\\right)\\right]\\right]{,}{\\mathrm{Pivots}}{=}\\left[{a}{}\\left({x}\\right){-}{d}{}\\left({x}\\right){\\ne }{0}\\right]{,}{\\mathrm{Solved}}{=}\\left[\\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{u}{}\\left({x}\\right){=}\\frac{{-}{b}{}\\left({x}\\right){}\\left(\\frac{{ⅆ}}{{ⅆ}{x}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{u}{}\\left({x}\\right)\\right){-}{u}{}\\left({x}\\right)}{{a}{}\\left({x}\\right){-}{d}{}\\left({x}\\right)}\\right]\\right]\\right){,}{2}{=}{table}{}\\left(\\left[{\\mathrm{Case}}{=}\\left[\\left[{a}{}\\left({x}\\right){-}{d}{}\\left({x}\\right){=}{0}{,}\\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{u}{}\\left({x}\\right)\\right]{,}\\left[{b}{}\\left({x}\\right){\\ne }{0}{,}\\frac{{ⅆ}}{{ⅆ}{x}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{u}{}\\left({x}\\right)\\right]\\right]{,}{\\mathrm{Pivots}}{=}\\left[{b}{}\\left({x}\\right){\\ne }{0}\\right]{,}{\\mathrm{Solved}}{=}\\left[\\frac{{ⅆ}}{{ⅆ}{x}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{u}{}\\left({x}\\right){=}{-}\\frac{{u}{}\\left({x}\\right)}{{b}{}\\left({x}\\right)}{,}{a}{}\\left({x}\\right){=}{d}{}\\left({x}\\right)\\right]\\right]\\right){,}{3}{=}{table}{}\\left(\\left[{\\mathrm{Case}}{=}\\left[\\left[{a}{}\\left({x}\\right){-}{d}{}\\left({x}\\right){=}{0}{,}\\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{u}{}\\left({x}\\right)\\right]{,}\\left[{b}{}\\left({x}\\right){=}{0}{,}\\frac{{ⅆ}}{{ⅆ}{x}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{u}{}\\left({x}\\right)\\right]\\right]{,}{\\mathrm{Solved}}{=}\\left[{u}{}\\left({x}\\right){=}{0}{,}{a}{}\\left({x}\\right){=}{d}{}\\left({x}\\right){,}{b}{}\\left({x}\\right){=}{0}\\right]\\right]\\right){,}{\\mathrm{casecount}}{=}{3}\\right]\\right)$ (4)\n\nfrom which the caseplot with initial data counts could be viewed by issuing the command:\n\n > $\\mathrm{caseplot}\\left(\\mathrm{ans},\\left[u\\right]\\right)$\n\nOne example of practical use of maxdimsystems is the determination of the forms of $f\\left(y'\\right)$ (where $f\\left(y'\\right)$ is not identically zero) for which the ode $y\\text{'}\\text{'}=f\\left(y'\\right)$ has the maximum dimension.\n\n > $\\mathrm{ODE}≔\\frac{ⅆ}{ⅆx}\\left(\\frac{ⅆ}{ⅆx}y\\left(x\\right)\\right)=f\\left(\\frac{ⅆ}{ⅆx}y\\left(x\\right)\\right)$\n ${\\mathrm{ODE}}{≔}\\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\\left({x}\\right){=}{f}{}\\left(\\frac{{ⅆ}}{{ⅆ}{x}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\\left({x}\\right)\\right)$ (5)\n > $\\mathrm{symeq}≔\\left[{\\mathrm{DEtools}}_{\\mathrm{odepde}}\\left(\\mathrm{ODE},\\left[\\mathrm{ξ}\\left(x,y\\right),\\mathrm{η}\\left(x,y\\right)\\right]\\right)\\right]$\n ${\\mathrm{symeq}}{≔}\\left[\\left({-}{3}{}{\\mathrm{_y1}}{}\\left(\\frac{{\\partial }}{{\\partial }{y}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{\\mathrm{\\xi }}{}\\left({x}{,}{y}\\right)\\right){-}{2}{}\\frac{{\\partial }}{{\\partial }{x}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{\\mathrm{\\xi }}{}\\left({x}{,}{y}\\right){+}\\frac{{\\partial }}{{\\partial }{y}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{\\mathrm{\\eta }}{}\\left({x}{,}{y}\\right)\\right){}{f}{}\\left({\\mathrm{_y1}}\\right){+}\\left({-}{\\mathrm{_y1}}{}\\left(\\frac{{\\partial }}{{\\partial }{y}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{\\mathrm{\\eta }}{}\\left({x}{,}{y}\\right)\\right){+}{{\\mathrm{_y1}}}^{{2}}{}\\left(\\frac{{\\partial }}{{\\partial }{y}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{\\mathrm{\\xi }}{}\\left({x}{,}{y}\\right)\\right){+}{\\mathrm{_y1}}{}\\left(\\frac{{\\partial }}{{\\partial }{x}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{\\mathrm{\\xi }}{}\\left({x}{,}{y}\\right)\\right){-}\\frac{{\\partial }}{{\\partial }{x}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{\\mathrm{\\eta }}{}\\left({x}{,}{y}\\right)\\right){}\\left(\\frac{{ⅆ}}{{ⅆ}{\\mathrm{_y1}}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{f}{}\\left({\\mathrm{_y1}}\\right)\\right){+}\\frac{{{\\partial }}^{{2}}}{{\\partial }{{x}}^{{2}}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{\\mathrm{\\eta }}{}\\left({x}{,}{y}\\right){+}\\left(\\frac{{{\\partial }}^{{2}}}{{\\partial }{{y}}^{{2}}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{\\mathrm{\\eta }}{}\\left({x}{,}{y}\\right)\\right){}{{\\mathrm{_y1}}}^{{2}}{-}\\left(\\frac{{{\\partial }}^{{2}}}{{\\partial }{{y}}^{{2}}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{\\mathrm{\\xi }}{}\\left({x}{,}{y}\\right)\\right){}{{\\mathrm{_y1}}}^{{3}}{-}{2}{}\\left(\\frac{{{\\partial }}^{{2}}}{{\\partial }{x}{\\partial }{y}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{\\mathrm{\\xi }}{}\\left({x}{,}{y}\\right)\\right){}{{\\mathrm{_y1}}}^{{2}}{+}{2}{}\\left(\\frac{{{\\partial }}^{{2}}}{{\\partial }{x}{\\partial }{y}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{\\mathrm{\\eta }}{}\\left({x}{,}{y}\\right)\\right){}{\\mathrm{_y1}}{-}{\\mathrm{_y1}}{}\\left(\\frac{{{\\partial }}^{{2}}}{{\\partial }{{x}}^{{2}}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{\\mathrm{\\xi }}{}\\left({x}{,}{y}\\right)\\right)\\right]$ (6)\n\nwhere the above $\\mathrm{symeq}$ is an overdetermined system of PDE in the infinitesimal symmetries $\\mathrm{\\xi }\\left(x,y\\right),\\mathrm{\\eta }\\left(x,y\\right)$.\n\nSo we compute the maximum dimensional case, which is not displayed as it is quite large:\n\n > $\\mathrm{t1}≔\\mathrm{time}\\left(\\right):$\n > $\\mathrm{msys}≔\\mathrm{maxdimsystems}\\left(\\left[\\mathrm{op}\\left(\\mathrm{symeq}\\right),f\\left(\\mathrm{_y1}\\right)\\ne 0\\right],\\left[\\mathrm{ξ},\\mathrm{η}\\right]\\right):$\n > $\\mathrm{time}\\left(\\right)-\\mathrm{t1}$\n ${0.596}$ (7)\n > $\\mathrm{nops}\\left(\\mathrm{msys}\\right)$\n ${1}$ (8)\n > $\\mathrm{length}\\left({\\mathrm{msys}}_{1}\\right)$\n ${5362}$ (9)\n > ${{\\mathrm{msys}}_{1}}_{4}$\n ${\\mathrm{dimension}}{=}{8}$ (10)\n > $\\mathrm{remove}\\left(\\mathrm{has},\\mathrm{rhs}\\left({{\\mathrm{msys}}_{1}}_{1}\\right),\\left[\\mathrm{ξ},\\mathrm{η}\\right]\\right)$\n $\\left[\\frac{{{ⅆ}}^{{4}}}{{ⅆ}{{\\mathrm{_y1}}}^{{4}}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{f}{}\\left({\\mathrm{_y1}}\\right){=}{0}\\right]$ (11)\n\nfrom which we see that we have one case, of dimension 8, and it occurs when the displayed ODE in $f\\left(\\mathrm{_y1}\\right)$ is satisfied. This classical result is also explored in the ISSAC Proc. article listed in the references (the primary reference for this command).\n\nThe timing above should be compared to the rifsimp full case analysis that takes significantly longer, and has a great number of lower dimensional cases:\n\n > $\\mathrm{t2}≔\\mathrm{time}\\left(\\right):$\n > $\\mathrm{rsys}≔\\mathrm{rifsimp}\\left(\\left[\\mathrm{op}\\left(\\mathrm{symeq}\\right),f\\left(\\mathrm{_y1}\\right)\\ne 0\\right],\\left[\\mathrm{ξ},\\mathrm{η}\\right],'\\mathrm{casesplit}'\\right):$\n > $\\mathrm{time}\\left(\\right)-\\mathrm{t2}$\n ${4.644}$ (12)\n > ${\\mathrm{rsys}}_{'\\mathrm{casecount}'}$\n ${11}$ (13)\n\nConsider the nonlinear Heat equation\n\n > $\\frac{\\partial }{\\partial t}u\\left(x,t\\right)=\\frac{\\partial }{\\partial x}\\left(k\\left(u\\left(x,t\\right)\\right)\\left(\\frac{\\partial }{\\partial x}u\\left(x,t\\right)\\right)\\right)$\n $\\frac{{\\partial }}{{\\partial }{t}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{u}{}\\left({x}{,}{t}\\right){=}{\\mathrm{D}}{}\\left({k}\\right){}\\left({u}{}\\left({x}{,}{t}\\right)\\right){}{\\left(\\frac{{\\partial }}{{\\partial }{x}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{u}{}\\left({x}{,}{t}\\right)\\right)}^{{2}}{+}{k}{}\\left({u}{}\\left({x}{,}{t}\\right)\\right){}\\left(\\frac{{{\\partial }}^{{2}}}{{\\partial }{{x}}^{{2}}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{u}{}\\left({x}{,}{t}\\right)\\right)$ (14)\n\nThe overdetermined system for it's symmetries $X,T,U$ is given by\n\n > $\\mathrm{sys}≔\\left[-2\\left(\\frac{\\partial }{\\partial x}T\\left(u,t,x\\right)\\right),\\frac{\\partial }{\\partial u}\\left(\\frac{\\partial }{\\partial u}U\\left(u,t,x\\right)\\right)-2\\left(\\frac{\\partial }{\\partial x}\\left(\\frac{\\partial }{\\partial u}X\\left(u,t,x\\right)\\right)\\right)+\\frac{U\\left(u,t,x\\right)\\left(\\frac{ⅆ}{ⅆu}\\left(\\frac{ⅆ}{ⅆu}k\\left(u\\right)\\right)\\right)}{k\\left(u\\right)}+\\frac{1\\left(\\frac{ⅆ}{ⅆu}k\\left(u\\right)\\right)\\left(\\frac{\\partial }{\\partial u}U\\left(u,t,x\\right)\\right)}{k\\left(u\\right)}-\\frac{U\\left(u,t,x\\right){\\left(\\frac{ⅆ}{ⅆu}k\\left(u\\right)\\right)}^{2}}{{k\\left(u\\right)}^{2}},\\frac{1\\left(\\frac{\\partial }{\\partial u}X\\left(u,t,x\\right)\\right)\\left(\\frac{ⅆ}{ⅆu}k\\left(u\\right)\\right)}{k\\left(u\\right)}-\\left(\\frac{\\partial }{\\partial u}\\left(\\frac{\\partial }{\\partial u}X\\left(u,t,x\\right)\\right)\\right),\\frac{\\partial }{\\partial x}\\left(\\frac{\\partial }{\\partial x}U\\left(u,t,x\\right)\\right)-\\frac{1\\left(\\frac{\\partial }{\\partial t}U\\left(u,t,x\\right)\\right)}{k\\left(u\\right)},\\frac{1\\left(\\frac{\\partial }{\\partial t}T\\left(u,t,x\\right)\\right)}{k\\left(u\\right)}+\\frac{U\\left(u,t,x\\right)\\left(\\frac{ⅆ}{ⅆu}k\\left(u\\right)\\right)}{{k\\left(u\\right)}^{2}}-\\left(\\frac{\\partial }{\\partial x}\\left(\\frac{\\partial }{\\partial x}T\\left(u,t,x\\right)\\right)\\right)-\\frac{2\\left(\\frac{\\partial }{\\partial x}X\\left(u,t,x\\right)\\right)}{k\\left(u\\right)},-\\left(\\frac{\\partial }{\\partial x}\\left(\\frac{\\partial }{\\partial x}X\\left(u,t,x\\right)\\right)\\right)+2\\left(\\frac{\\partial }{\\partial x}\\left(\\frac{\\partial }{\\partial u}U\\left(u,t,x\\right)\\right)\\right)+\\frac{1\\left(\\frac{\\partial }{\\partial t}X\\left(u,t,x\\right)\\right)}{k\\left(u\\right)}+\\frac{2\\left(\\frac{ⅆ}{ⅆu}k\\left(u\\right)\\right)\\left(\\frac{\\partial }{\\partial x}U\\left(u,t,x\\right)\\right)}{k\\left(u\\right)},-\\frac{2\\left(\\frac{ⅆ}{ⅆu}k\\left(u\\right)\\right)\\left(\\frac{\\partial }{\\partial x}T\\left(u,t,x\\right)\\right)}{k\\left(u\\right)}-\\frac{2\\left(\\frac{\\partial }{\\partial u}X\\left(u,t,x\\right)\\right)}{k\\left(u\\right)}-2\\left(\\frac{\\partial }{\\partial x}\\left(\\frac{\\partial }{\\partial u}T\\left(u,t,x\\right)\\right)\\right),-\\frac{1\\left(\\frac{ⅆ}{ⅆu}k\\left(u\\right)\\right)\\left(\\frac{\\partial }{\\partial u}T\\left(u,t,x\\right)\\right)}{k\\left(u\\right)}-\\left(\\frac{\\partial }{\\partial u}\\left(\\frac{\\partial }{\\partial u}T\\left(u,t,x\\right)\\right)\\right),-2\\left(\\frac{\\partial }{\\partial u}T\\left(u,t,x\\right)\\right)\\right]:$\n\nAt the same time as determining maximum dimensional cases, one can request, through a partitioning of the maxdimvars, that the dependent variables be ranked in a specific manner. To always isolate derivatives of $U\\left(u,t,x\\right)$ in terms of the other dependent variables, one could specify maxdimvars=[[U],[X,T]]. For this problem we would obtain\n\n > $\\mathrm{maxdimsystems}\\left(\\left[\\mathrm{op}\\left(\\mathrm{sys}\\right),\\frac{ⅆ}{ⅆu}k\\left(u\\right)\\ne 0\\right],\\left[\\left[U\\right],\\left[X,T\\right]\\right]\\right)$\n $\\left\\{\\left[{\\mathrm{Solved}}{=}\\left[{U}{}\\left({u}{,}{t}{,}{x}\\right){=}{-}\\frac{{k}{}\\left({u}\\right){}\\left({-}{2}{}\\frac{{\\partial }}{{\\partial }{x}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{X}{}\\left({u}{,}{t}{,}{x}\\right){+}\\frac{{\\partial }}{{\\partial }{t}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{T}{}\\left({u}{,}{t}{,}{x}\\right)\\right)}{\\frac{{ⅆ}}{{ⅆ}{u}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{k}{}\\left({u}\\right)}{,}\\frac{{{\\partial }}^{{3}}}{{\\partial }{{x}}^{{3}}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{X}{}\\left({u}{,}{t}{,}{x}\\right){=}{0}{,}\\frac{{{\\partial }}^{{2}}}{{\\partial }{{t}}^{{2}}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{T}{}\\left({u}{,}{t}{,}{x}\\right){=}{0}{,}\\frac{{\\partial }}{{\\partial }{u}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{X}{}\\left({u}{,}{t}{,}{x}\\right){=}{0}{,}\\frac{{\\partial }}{{\\partial }{u}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{T}{}\\left({u}{,}{t}{,}{x}\\right){=}{0}{,}\\frac{{\\partial }}{{\\partial }{t}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{X}{}\\left({u}{,}{t}{,}{x}\\right){=}{0}{,}\\frac{{\\partial }}{{\\partial }{x}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{T}{}\\left({u}{,}{t}{,}{x}\\right){=}{0}{,}\\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{u}}^{{2}}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{k}{}\\left({u}\\right){=}\\frac{{7}{}{\\left(\\frac{{ⅆ}}{{ⅆ}{u}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{k}{}\\left({u}\\right)\\right)}^{{2}}}{{4}{}{k}{}\\left({u}\\right)}\\right]{,}{\\mathrm{Constraint}}{=}\\left[\\right]{,}{\\mathrm{Pivots}}{=}\\left[\\frac{{ⅆ}}{{ⅆ}{u}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{k}{}\\left({u}\\right){\\ne }{0}\\right]{,}{\\mathrm{dimension}}{=}{5}\\right]\\right\\}$ (15)\n\nAlternatively we could try to obtain PDE only in $U\\left(u,t,x\\right)$ and the classifying function $k\\left(u\\right)$ by an elimination ranking:\n\n > $\\mathrm{maxdimsystems}\\left(\\left[\\mathrm{op}\\left(\\mathrm{sys}\\right),\\frac{ⅆ}{ⅆu}k\\left(u\\right)\\ne 0\\right],\\left[\\left[X,T\\right],\\left[U\\right]\\right]\\right)$\n $\\left\\{\\left[{\\mathrm{Solved}}{=}\\left[\\frac{{{\\partial }}^{{2}}}{{\\partial }{{x}}^{{2}}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{X}{}\\left({u}{,}{t}{,}{x}\\right){=}\\frac{\\left(\\frac{{ⅆ}}{{ⅆ}{u}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{k}{}\\left({u}\\right)\\right){}\\left(\\frac{{\\partial }}{{\\partial }{x}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{U}{}\\left({u}{,}{t}{,}{x}\\right)\\right)}{{2}{}{k}{}\\left({u}\\right)}{,}\\frac{{\\partial }}{{\\partial }{u}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{X}{}\\left({u}{,}{t}{,}{x}\\right){=}{0}{,}\\frac{{\\partial }}{{\\partial }{u}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{T}{}\\left({u}{,}{t}{,}{x}\\right){=}{0}{,}\\frac{{\\partial }}{{\\partial }{t}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{X}{}\\left({u}{,}{t}{,}{x}\\right){=}{0}{,}\\frac{{\\partial }}{{\\partial }{t}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{T}{}\\left({u}{,}{t}{,}{x}\\right){=}\\frac{{2}{}\\left(\\frac{{\\partial }}{{\\partial }{x}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{X}{}\\left({u}{,}{t}{,}{x}\\right)\\right){}{k}{}\\left({u}\\right){-}{U}{}\\left({u}{,}{t}{,}{x}\\right){}\\left(\\frac{{ⅆ}}{{ⅆ}{u}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{k}{}\\left({u}\\right)\\right)}{{k}{}\\left({u}\\right)}{,}\\frac{{\\partial }}{{\\partial }{x}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{T}{}\\left({u}{,}{t}{,}{x}\\right){=}{0}{,}\\frac{{{\\partial }}^{{2}}}{{\\partial }{{x}}^{{2}}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{U}{}\\left({u}{,}{t}{,}{x}\\right){=}{0}{,}\\frac{{\\partial }}{{\\partial }{u}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{U}{}\\left({u}{,}{t}\\right)\\right]\\right]\\right\\}$" ]
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https://www.arxiv-vanity.com/papers/astro-ph/0011398/
[ "# On the APM power spectrum and the CMB anisotropy: Evidence for a phase transition during inflation?\n\nJ. Barriga, E. Gaztañaga, M.G. Santos and S. Sarkar\nInstitut d’Estudis Espacials de Catalunya, IEEC/CSIC, Edf. Nexus-201 - c/ Gran Capitan 2-4, 08034 Barcelona, Spain\nINAOE, Astrofisica, Tonantzintla, Apdo Postal 216 y 51, Puebla 7200, Mexico\nTheoretical Physics, Department of Physics, University of Oxford, 1 Keble Road, OX1 3NP, UK\nFebruary 7, 2022\n###### Abstract\n\nAdams et al. (1997b) have noted that according to our current understanding of the unification of fundamental interactions, there should have been phase transitions associated with spontaneous symmetry breaking during the inflationary era. This may have resulted in the breaking of scale-invariance of the primordial density perturbation for brief periods. A possible such feature was identified in the power spectrum of galaxy clustering in the APM survey at the scale and it was shown that the secondary acoustic peaks in the power spectrum of the CMB anisotropy should consequently be suppressed. We demonstrate that this prediction is confirmed by the recent Boomerang and Maxima observations, which favour a step-like spectral feature in the range , independently of the similar previous indication from the APM data. Such a spectral break enables an excellent fit to both APM and CMB data with a baryon density consistent with the BBN value. It also allows the possibility of a matter-dominated universe with zero cosmological constant, which we show can now account for even the evolution of the abundance of rich clusters.\n\njournal: OUTP-00-33-Ppubyear: 2000\n\n## 1 Introduction\n\nIt is commonly assumed by astronomers that inflation predicts a scale-invariant ‘Harrison-Zeldovich’ (H-Z) spectrum of scalar density perturbations, with ; for example this is a standard input in calculations of the expected large-scale structure (LSS) and cosmic microwave background (CMB) anisotropy. In fact in the usual ‘slow-roll’ inflationary scenario there is a gradual steepening of the spectrum with increasing (decreasing scale), as the end of inflation is approached [Mukhanov & Chibisov 1981, Hawking 1982, Starobinsky 1982, Guth & Pi 1982, Bardeen, Steinhardt & Turner 1983]. This cannot be adequately modelled by a spectrum with constant “tilt” () as is occasionally adopted, since the steepening is scale-dependent for any polynomial inflaton potential.111The index is scale-independent only for an exponential potential (‘power-law inflation’); moreover can be very close to unity if inflation ends not through the steepening of the inflaton potential but, for example, due to the dynamics of a second scalar field (‘hybrid inflation’). The scale-dependence of in various inflationary models has been reviewed by Lyth & Riotto (1999). Even such small departures from scale-invariance can be quite significant, for example in a supergravity model with a cubic inflaton potential, Adams, Ross & Sarkar (1997a) found , where is the number of e-folds of expansion from the end of inflation. This gradual suppression of small-scale power was shown to be adequate to reconcile the COBE-normalised standard cold dark matter (sCDM) model with LSS data, in particular the power spectrum of galaxy clustering, , the abundance of rich clusters quantified by the parameter , and large-scale streaming velocities, . We emphasise that this model is however ruled out by the same data if a scale-invariant spectrum is assumed [Efstathiou, Bond & White 1992]. Such conclusions are clearly not robust given that we have as yet no standard model of inflation.\n\nSubsequently it was realized [Adams, Ross & Sarkar 1997b] that the spectrum may not be even scale-free because the rapid cooling of the universe during primordial inflation can result in spontaneous symmetry breaking phase transitions which may interrupt the slow roll of the inflaton field for brief periods. This is in fact inevitable in models based on supergravity, the phenomenologically successful extension of the Standard Model of particle physics and the effective field theory below the Planck scale. During inflation, the large vacuum energy breaks supersymmetry giving otherwise massless fields (‘flat directions’) a mass of order the Hubble parameter [Dine et al. 1984, Coughlan et al. 1984], causing them to evolve rapidly to the asymmetric global minima of their scalar potential. Such ‘intermediate scale’ fields are generic in models derived from superstring/M-theory and have gauge and Yukawa couplings to the thermal plasma so are initially confined at the symmetric maxima of their scalar potentials. Consequently it takes a (calculable) finite amount of cooling before the thermal barrier disappears and they are free to evolve to their minima [Yamamoto 1986, Barreiro et al. 1996]. When a symmetry breaking transition occurs, the mass of the inflaton field changes suddenly (through couplings in the Kähler potential), temporarily violating the slow-roll conditions and interrupting inflation (Adams et al. 1997b).222When (re)heating occurs at the end of inflation such fields may again be forced back to the symmetric maximum, undergoing symmetry breaking a second time when the universe cools down to the electroweak scale in the radiation-dominated era and driving a late phase of ‘thermal inflation’ [Lyth & Stewart 1996]. Thus the density perturbation is expected to have a (near) H-Z spectrum for the first e-folds of expansion followed by one or more sudden departures from scale-invariance lasting e-fold. In order for such spectral features to be observable in the LSS or CMB, it is of course necessary that they occur within the last e-folds of inflation, corresponding to spatial scales going up to the present Hubble radius . Since the density perturbation can be observed on scales from the Hubble radius down to , corresponding to about 8 e-folds of expansion, it would be not unreasonable to expect at least one such spectral break to be seen today.\n\nMotivated by this an attempt was made to recover the primordial perturbation spectrum from extant observations of LSS and CMB anisotropy, specifically the APM survey [Maddox et al. 1990a] and the COBE observations [Smoot et al. 1992]. We recall that the spectrum of rms mass fluctuations at the present epoch (per unit logarithmic interval of wavenumber ) is\n\n Δ2(k)≡k3P(k)2π2=δ2H(k) T2(k)(kH0)3+n , (1)\n\nwhere the density perturbation is evaluated at the present Hubble radius, i.e. at . The (dimensionless) matter ‘transfer function’ for CDM models can be approximated by [Bond & Efstathiou 1984]\n\n TCDM(k)≃[1+{ak+(bk)3/2+(ck)2}ν]−1/ν , (2)\n\nwith , , and . Here the ‘shape parameter’ is defined as [Bond & Jaffe 1999]\n\n Γ=Ωmhe−[ΩB(1+√2h/Ωm)−0.06], (3)\n\nwhere is the Hubble parameter and , () being the fraction of the critical density in cold (baryonic) matter. Thus for sCDM which assumes , and .\n\nFollowing Baugh & Gaztañaga (1996), the three-dimensional inferred from the angular correlation function of galaxies in the APM survey [Baugh & Efstathiou 1993] was subjected to an inversion procedure [Jain, Mo & White 1995, Peacock & Dodds 1996] to recover the linear spectrum of fluctuations and obtain its spectral index, , as a function of . Assuming that any bias between APM galaxies and dark matter is scale-independent, the primordial spectral index was then obtained simply by subtracting off the slope of the transfer function (2) for each value of (Adams et al. 1997b).\n\nThe key finding was that shows a dramatic departure from the H-Z value of in the range , dropping to a value around zero ( below unity) at . The sharp drop occurs over e-fold of expansion as is indeed expected in the phase transition model. This feature in had been noted by Baugh & Efstathiou (1993) and was also identified by Peacock (1997) in the power spectrum of IRAS galaxies after correcting for redshift-space distortions. Given that the feature appears on the scale of the Hubble radius at the epoch of (dark) matter domination, it is perhaps natural to interpret it as reflecting a departure from the sCDM paradigm as these authors did. However Adams et al. (1997b) pursued the alternative possibility that it is in fact present in the primordial spectrum and asked what would be the expectation for the angular anisotropy of the CMB if this were so. Using the COSMICS code [Bertschinger 1995] they calculated its angular power spectrum and found that the height of the secondary acoustic peaks is suppressed by a factor of relative to the prediction of the standard COBE-normalised sCDM model which assumes a scale-free H-Z spectrum.\n\nWe reexamine this prediction in the light of recent observations by the Boomerang [de Bernadis et al. 2000] and MAXIMA-1 [Hanany et al. 2000] experiments which do indicate a significant suppression of the second acoustic peak. It has been noted by many authors [Tegmark & Zaldarriaga 2000, Lange et al. 2001, Balbi et al. 2000, Hu et al. 2000, Jaffe et al. 2000, Esposito et al. 2001, Padmanabhan & Sethi 2000] that this requires a high baryon density, significantly above the value required by light element abundances following from standard Big Bang nucleosynthesis (BBN). As the latter is widely considered to be one of the “pillars” of modern cosmology, it is important to ask whether this conflict may not be resolved by relaxing the assumption made in these analyses that the primordial spectrum of perturbations is scale-free.\n\n## 2 Reconstructing the primordial spectrum\n\nWe begin by describing the range of values that we consider for the cosmological parameters. This represents the set of priors, resulting from other observations, that we adopt for the reconstruction of the primordial spectrum. In §2.2, we discuss how LSS data can be used to recover the primordial spectrum and predict the CMB anisotropy (§2.3). In §2.4 we present a simple parameterization for the primordial spectrum and use this to fit the new CMB as well as APM data. Finally in §2.5 we confront the observed abundance of rich clusters with the expectations in our model.\n\n### 2.1 Cosmological parameters\n\nLet us first briefly discuss the choice of cosmological parameters. As mentioned before, in the sCDM model, but gives in fact a better fit to the APM data [Maddox et al. 1990b, Efstathiou, Bond & White 1992, Baugh & Efstathiou 1993, Peacock & Dodds 1994] and can be realized e.g. in a low density model with .333However structure formation does not necessarily require a low density universe as is often claimed (e.g. Bahcall et al. 1999); for example with a gradually tilting primordial spectrum obtained from ‘new’ supergravity inflation, the small-scale power is sufficiently suppressed for a CDM model to be compatible with LSS data (Adams et al. 1997a, see also White et al. 1995). Such ‘tilt’ can preferentially suppress the second acoustic peak if the baryon fraction is high (Adams et al. 1997a, see also White et al. 1996). Several groups have studied the implications of tilt for the recent CMB data (e.g. Kinney, Melchiorri & Riotto 2000, Covi & Lyth 2000, Tegmark, Zaldariagga & Hamilton 2001). The alternative possibility of a Hubble parameter as low as [Bartlett et al. 1995] is definitively ruled out by the HST Key Project data on 25 galaxies which indicates [Mould et al. 2000]. Weighting such direct observations against other fundamental physics approaches, Primack (2000) advocates . However a value as low as 0.45 [Parodi et al. 2000] or as high as 0.9 [Tonry et al. 2000] still seems quite possible so the quoted uncertainty cannot be interpreted as a gaussian standard deviation. Instead we assume that all values in the following “” range are equally probable:\n\n h=0.49−0.81. (4)\n\nThe current preference is in fact for a cosmological constant dominated flat universe with and (e.g. Bahcall et al. 1999). The Boomerang and MAXIMA observations of the first acoustic peak indicate that the universe is indeed flat [Jaffe et al. 2000], but we will consider other CDM models with values of ranging down to zero, maintaining . This is motivated by our concern that the only direct evidence suggesting is the small curvature of the Hubble diagram for Type Ia supernovae at redshift [Perlmutter et al. 1999, Riess et al. 1998] which may be subject to unrecognised systematic effects. (We note that other arguments for a low from the evolution of the cluster number density [Eke et al. 1998] or the clustering in the Lyman- forest [Weinberg et al. 1999] rest on the assumption of a scale-free spectrum.) Of course for any particular choice of , one has to check that the age of the universe exceeds the inferred age of globular clusters,  Gyr [Krauss 2000].\n\nAnother important parameter for the present study is the BBN value of . This has undergone substantial revision since the first studies of sCDM so we briefly review the present situation. The baryon density which provides a good fit to the inferred primordial abundances of D, He and Li in standard BBN is [Fiorentini et al. 1998]\n\n ΩB=(0.019+0.0013−0.0012)h−2, (5)\n\ntaking all (correlated) reaction rate uncertainties into account.444Subsequently Burles et al. (1999) suggested that some of the reaction rate uncertainties had been overestimated; thus they inferred a somewhat tighter range . Izotov et al. (2000) found however a better fit with , taking Y(He)= from observations of the two most metal-deficient BCGs, D/H= from the analysis of QAS data using kinematic models incorporating ‘mesoturbulence’ [Levshakov, Kegel & Takahara 1999], and Li/H= allowing for some depletion due to rotation in Pop II stars [Vauclair & Charbonnel 1998]. This fit is based on the ‘low’ value of D/H= found [Burles & Tytler 1998] in two quasar absorption systems (QAS) at high redshift [and subsequently in a third one [Kirkman et al. 2000]], the ‘high’ value of the mass fraction Y(He)= inferred [Izotov, Thuan & Lipovetsky 1998] from observations of metal-poor blue compact galaxies (BCGs), and the ‘Spite plateau’ of Li/H= seen in Pop II stars [Bonifacio & Molaro 1997]. Note however that an improved fit is obtained [Fiorentini et al. 1998] at a significantly smaller value of if one adopts the ‘high’ value of D/H= claimed [Songaila et al. 1994] in another QAS [but disputed subsequently [Burles, Kirkman & Tytler 1999], although consistent with a possible high value [Tytler et al. 1999] in yet another QAS], and the ‘low’ value of Y(He)= obtained [Olive, Skillman & Steigman 1997] from analysis of older BCG data [and supported by another recent observation [Peimbert, Peimbert & Ruiz 2000]]. An independent lower limit of [Weinberg et al. 1997] set by observations of the Lyman- forest favours the higher value in Equation (5), which also allows a better fit to the observed distribution of line widths in comparison with model hydrodynamic simulations (e.g. Theuns et al. 1999). An audit of luminous matter in the universe [Fukugita, Hogan & Peebles 1998] gives a much weaker lower limit of . A conservative upper limit on the baryon density is [Kernan & Sarkar 1996]\n\n ΩB<0.033h−2, (6)\n\nif the primordial deuterium abundance is bounded from below by its typical interstellar value D/H= [McCullough 1992]. Therefore we will consider a wide range of possible values in our analysis, but adopt , as the default values (so that ) when not stated otherwise.\n\n### 2.2 Primordial spectrum from the APM data\n\nThere are several arguments (reviewed in Appendix A) that on scales , which are at most weakly non-linear, the APM galaxy power spectrum [Baugh & Efstathiou 1993] is an unbiased (or moderately linearly biased) tracer of the mass. The linear power spectrum recovered under this assumption from , using the method of Jain et al. (1995), is well fitted in this range by [Baugh & Gaztañaga 1996]:\n\n Plin(k)≃7×105k(h−1Mpc)3[1+(k/kc)2]1.6 , (7)\n\nwhere . We will use the common convention of expressing the matter power spectrum as the primordial spectrum of matter fluctuations , folded with the transfer function (2):\n\n Plin(k)≡P0(k)T2(k)=AknT2(k), (8)\n\nwhere the (dimensionful) normalisation constant is determined by the large-scale CMB anisotropy. [For sCDM with normalised to the COBE 4-year data [Bennett et al. 1996], with 9% uncertainty [Bunn & White 1997] so .]\n\nGiven some estimate of the (linear) matter power spectrum, as in Equation (7), the standard approach is to assume a scale-invariant initial spectrum, i.e. , and extract the value of that best fits the resulting constraint on . Alternatively one can extract ) for a specific choice of , i.e. for a given transfer function. We apply this latter approach to the APM data by simply taking the ratio of Equation (7) to Equation (2), obtaining:\n\n P0(k)=⎧⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪⎩A1k,k\n\nwhere and we have adopted a suitably normalised H-Z spectrum outside the range (, ) accessible to galaxy clustering observations.\n\nFigure 1 shows the recovered spectrum (9) for two choices of corresponding to the sCDM model and a low density variant. Note that the reconstruction has significantly less power than a scale-free H-Z spectrum on scales , while the reconstruction is closer to a H-Z spectrum but has relatively more power. However as we shall see the latter possibility does not give a good fit to the Boomerang/MAXIMA data with the BBN value (5) of the baryon density. To ensure compatibility with BBN requires a break in the primordial spectrum.", null, "Figure 1: Reconstruction of the primordial density power spectrum from the APM data, adopting a CDM shape parameter Γ of 0.5 (dot-dashed line) and 0.2 (dashed line). A Harrison-Zeldovich spectrum (full line) is shown for comparison.\n\n### 2.3 Acoustic peaks\n\nIn linear theory the co-efficents in the spherical harmonics expansion of the CMB anisotropy are a linear projection of , so the effect of changing would just be to proportionally alter the angular power for the multipoles corresponding to the relevant values of . We can approximate where is the angular size corresponding to a comoving distance at the last scattering surface (LSS), which is at a distance for a flat universe [Vittorio & Silk 1992]. Thus\n\n l≃k/DLSS≃3000kΩ−0.4mh−1. (10)\n\nFor the case (corresponding e.g. to , ) the inferred primordial power in Figure 1 is suppressed by a factor for , implying a similar decrease in the CMB anisotropy at . For the case (corresponding e.g. to , ), the inferred primordial spectrum is a factor of above the H-Z one at so the CMB anisotropy should be proportionally boosted at .\n\nTo check these expectations, we ran the Boltzmann code CMBFAST [Seljak & Zaldarriaga 1996] with the above reconstructed primordial spectra as well as a H-Z spectrum with the same cosmological parameters. As seen in Figure 2, the angular power resulting from the APM inversion with is indeed depressed below the H-Z case from the second acoustic peak onwards, as noted by Adams et al. (1997b). By contrast, for the reconstruction all the acoustic peaks are boosted above the H-Z case. It is clear that the recent CMB observations favour a high density universe. Moreover the observed suppression of the second acoustic peak in the Boomerang/MAXIMA data confirms the prior expectation for a primordial density perturbation with broken scale-invariance. To quantify this we now perform detailed fits to the data.", null, "Figure 2: The predicted CMB angular power spectra corresponding to the reconstructed primordial spectra for Γ=0.5 (dot-dashed line) and Γ=0.2 (dashed line). The corresponding results for a primordial H-Z spectrum is also shown for Γ=0.5 and Γ=0.2 (thick and thin full lines). The data are from the Boomerang (crosses) and MAXIMA (diamonds) experiments.\n\n### 2.4 Fits to the CMB and LSS data\n\nWe parameterise the ‘step’ in the primordial power spectrum (see Figure 1) as:\n\n P0(k)=⎧⎪⎨⎪⎩Ak,k≤ksCkα,ke≤k≤ksBk,k≥ke (11)\n\nwhere and mark the start and end of the break from a H-Z spectrum, with amplitudes and . (The values of and are specified by the other parameters.) In the multiple inflation model (Adams et al. 1997b), the actual form of the spectrum during the phase transition is difficult to calculate since the usual ‘slow-roll’ conditions are violated. However a robust expectation is that\n\n (12)\n\nbecause the field undergoing the symmetry-breaking phase transition evolves exponentially fast to its minimum. The ratio of the amplitudes is determined by the (unknown) superpotential couplings of the field undergoing the phase transition but is expected to exceed unity (i.e. there is a decrease in the power).\n\nWe allow the spectral and cosmological parameters to range rather widely as follows:\n\n• from (0.01–0.15)\n\n• from 0.01–4\n\n• from 0.3–7.2\n\n• from 0–0.7 (keeping )\n\n• from 0.45–0.9\n\n• from 0.0014–0.033\n\nWe also consider a bias parameter , defined as the square root of the ratio of the APM and CMB normalisations (which is expected to be close to unity, see Appendix A).\n\nFor each combination of the above parameters we construct the primordial power spectrum and then obtain the matter power spectrum by convoluting with the CDM transfer function calculated by CMBFAST (which differs slightly from the analytic approximation in Equation 2). It is also used to calculate the expected CMB angular power spectrum. These are then compared, respectively, with the APM data [Gaztañaga & Baugh 1998],555We use the values of given in Table 2 with errors estimated from the variance in the 4 disjoint sky regions in the catalogue. Ideally we should use the linear spectrum recovered for each specific choice of cosmological parameters. However the two are very similar since we restrict ourselves to the quasi-linear range (see Fig 9 in Appendix A), and the estimated errors in in this range are large enough to encompass all possible linear spectra. and the decorrelated COBE [Tegmark & Hamilton 1997] and Boomerang data [de Bernadis et al. 2000] 666There appear to be systematic differences between the Boomerang and MAXIMA data as shown in Figure 2. However according to Jaffe et al (2000), the two datasets are quite consistent within their respective calibration uncertainties (20% and 8%, in ) and beam uncertainties (10% and 5%). For definiteness we use the Boomerang data alone; clearly our results would be very similar if we use the MAXIMA data instead. using appropriate window functions, to determine . We use 19 CMB data points [7 from COBE (excluding the anomalously low quadrupole) and 12 from Boomerang] and 10 APM data points (for ). We quantify the goodness-of-fit following the procedure of Lineweaver & Barbosa (1998).\n\nIn Table 1 we show the values of the Hubble constant, baryon density, and amplitude change at the spectral break which minimise the (for a specific choice of ) for fits to the CMB data alone. The step in the primordial spectrum is seen to be not essential for a good fit, in particular even with (i.e. no step) the does not worsen significantly in most cases. The best overall fit obtains for , consistent with the analysis by the Boomerang collaboration [de Bernadis et al. 2000]. Note however that the baryon density required is 65% higher than the preferred BBN value (5) although still within the upper limit (6) set by interstellar deuterium.\n\nIf we now demand that the models should have the BBN baryon density (5), the Hubble parameter in the range (4) and a spectral break duration (12), then a step in the primordial spectrum is definitely indicated as seen from Table 2. [In particular setting (no step) always gives an unacceptably high value of .] The angular power spectra corresponding to some of these models is shown in Figure 3 along with the Boomerang and COBE data. We emphasize that a good fit is now obtained even for .", null, "Figure 3: CMB angular power spectra for the best-fit models in Table 2, with ΩΛ=0 (continuous line), ΩΛ=0.2 (dashed line), ΩΛ=0.4 (dot-dashed line), and ΩΛ=0.6 (dotted line). The data shown are from COBE and Boomerang. All models have the BBN baryon density ΩB=0.019h−2.\n\nNext we fit simultaneously to both the APM and CMB data allowing for a small difference in the normalisations, i.e. a bias . We have taken into account that the mean redshift of the APM galaxies is so a corresponding correction for the growth factor to needs to be applied. It is seen from Table 3 that the best overall fit is still obtained for , although other values remain quite acceptable. In particular the model requires a Hubble parameter at the low end of the allowed range but has the advantage of being consistent with the BBN value of the baryon density. We emphasise that a break in the spectrum is now definitely required since the value of exceeds 45 otherwise. The most likely duration of the break also comes out in accord with the theoretical expectation (12).", null, "Figure 4: Primordial power spectra for the best-fit models in Table 4 with ΩΛ=0 (continuous line), ΩΛ=0.2 (dashed line), ΩΛ=0.4 (dot-dashed line), and ΩΛ=0.6 (dotted line). All models have the BBN baryon density ΩB=0.019h−2.\n\nFinally Table 4 shows the result of imposing the BBN constraint (5), and the (less important) constraints on the Hubble parameter (4) and the duration (12). We have also required that the bias be within 20% of unity (see Appendix A). We see that the data now prefer lower values of (and ). The magnitude of the required spectral break decreases with increasing (Figure 4) but one cannot do without such a break altogether since otherwise the exceeds 53. Figures 5 and 6 show, respectively, the fits to the CMB and APM data for these models (using the mean value of the break amplitude ). In Figure 7 we display the goodness-of-fit contours in the plane for the joint fit to the CMB and APM data; the contours are ragged because we have not used a fine enough grid of parameter values. We see that agreement with the BBN value (5) of the baryon density can be achieved only for low . In fact a critical density matter-dominated universe is favoured with the Hubble parameter at its lower limit (so that the age of the universe is acceptable at  Gyr).", null, "Figure 5: CMB angular power spectra for the best-fit models in Table 4 with ΩΛ=0 (continuous line), ΩΛ=0.2 (dashed line), ΩΛ=0.4 (dot-dashed line), and ΩΛ=0.6 (dotted line). All models have the BBN baryon density ΩB=0.019h−2. The data points are from COBE and Boomerang.", null, "Figure 6: Matter power spectrum for the best-fit models in Table 4 with ΩΛ=0 (continuous line), ΩΛ=0.2 (dashed line), ΩΛ=0.4 (dot- dashed line), and ΩΛ=0.6 (dotted line). All models have the BBN baryon density ΩB=0.019h−2. The data are from the APM survey.", null, "Figure 7: Goodness-of-fit contours for the joint fit to the CMB and APM data of a primordial spectrum with a ‘step’, for different choices of ΩΛ. The vertical and horizontal shaded bands indicate the constraints (4) and (5) on h and ΩBh2 respectively while the crosses indicate the best fits from Table 3. Values of provide an unacceptably poor fit to the data.\n\n### 2.5 Implications for cluster abundances\n\nIn Table 4 we show also the value of the variance in (dark matter) fluctuations (normalised to the CMB), over a sphere of size :\n\n σ2(R)=1H40∫∞0W2(kR) δ2H(k) T2(k) k3dk, (13)\n\nusing a ‘top hat’ smoothing function,\n\nIn Figure 8 we show the predictions for the redshift evolution of the cluster abundance in our models according to the Press-Schechter (P-S) theory [Press & Schechter 1974], compared to the observations as presented in Bahcall & Fan (1998). We consider clusters with mass (neglecting the condition). The spatially flat COBE-normalised H-Z model with , (top thick full line) does not match the observations, a fact that has been used to rule out this model. Even when this model is normalised at by lowering its amplitude to (bottom thick full line), it still fails to reproduce the observed redshift evolution in the cluster abundances [Bahcall & Fan 1998]. By contrast, the best-fit models in Table 4 do match the redshift evolution, especially if we allow for the possibility mentioned above of lowering the values further to better match the data.777Note that the P-S prediction depends not only on the value of at and the corresponding cosmological evolution of the variance, but also on the shape of the spectrum over the mass scales considered. In our case, these scales correspond to , i.e. near the break in the primordial spectrum. We emphasise that these predictions have not been adjusted in any way to fit cluster abundances; all parameters have been fixed already by the fit to the APM and CMB data. It is interesting to note that even the model can now reproduce the slow growth of the cluster abundance with redshift due to the break in the primordial spectrum.", null, "Figure 8: The number density of rich clusters with M>8×1014M⊙ as a function of redshift. Points with errorbars correspond to the observations as depicted in Bahcall & Fan (1998). The thick full lines correspond to the P-S prediction for the spatially flat, COBE-normalised H-Z model with ΩΛ=0.2, Ωm≃0.8 (top line), and scaled to match the cluster abundance at z=0 (bottom line). The curves in between correspond to the best-fit models in Table 4 with ΩΛ=0 (full line), ΩΛ=0.2 (dashed line), ΩΛ=0.4 (dot-dashed line), and ΩΛ=0.6 (dotted line).\n\n## 3 Discussion\n\nSeveral authors have in the past noted the possibility of generating spectral features in the H-Z density perturbation by fine tuning the parameters of the inflaton potential [Kofman & Linde 1987, Salopek, Bond & Bardeen 1989, Hodges et al. 1990, Hodges & Blumenthal 1990]. We would like to emphasise that in contrast to such “designer” models, the break in the H-Z spectrum discussed here is generated by a physical mechanism, viz. supersymmetry breaking during inflation (Adams et al. 1997b, Ross 1998). As noted subsequently [Lesgourgues & Polarski 1997, Kanazawa et al. 2000], the resulting damping of the CMB anisotropy is also a characteristic of ‘double inflation’ which occurs in toy models wherein inflation is driven in two stages — first by higher-order gravity and then by a scalar field [Kofman, Linde & Starobinsky 1985, Kofman & Pogosyan 1988, Gottlöber, Müller & Starobinsky 1991], or by two coupled scalar fields [Starobinsky 1985, Kofman & Linde 1987, Polarski & Starobinsky 1992]. The model parameters can be tuned to generate a break in the primordial spectrum at any chosen scale. The advantages of this for fitting LSS observations was first emphasised by Silk & Turner (1987) and have been explored thoroughly [Polarski & Starobinsky 1992, Gottlöber, Mücket & Starobinsky 1994, Polarski 1994, Peter, Polarski & Starobinsky 1994, Amendola et al. 1995, Polarski & Starobinski 1995, Semig & Müller 1996], although these authors did not discuss the implications for the CMB. A spectral feature similar to that observed can also be generated in a toy model where the inflaton evolves through a kink in its potential [Starobinsky 1992], leading to similar suppression of the secondary acoustic peaks [Atrio-Barandela et al. 1997, Lesgourgues, Polarski & Starobinsky 1998, Lesgourgues, Prunet & Polarski 1999]. Recently there have been efforts to progress beyond toy models and implement multiple inflation in supersymmetric models [Sakellariadou & Tetradis 1998, Lesgourgues 1998, Lesgourgues 2000]. Spectral features can also be generated by resonant production of particles during inflation [Chung et al. 2000].\n\nSilk & Gawiser (2000) have noted independently that it is difficult for a CDM model to fit LSS and CMB observations simultaneously without considering departures from scale-invariance; they advocate adding a ‘bump’ at to a primordial H-Z spectrum.888Griffiths, Silk & Zaroubi (2000) have very recently shown that this can also fit the recent CMB and LSS data; however they choose to discount the APM data in favour of the (less precise) power spectrum inferred from the PSCz survey [Hamilton & Tegmark 2000] which does not show a ‘shoulder’ at . We would point out that our models in Table 4 also fit the PSCz power spectrum. Einasto et al. (1999) have also concluded from an exhaustive analysis of LSS data that a scale-free primordial spectrum is excluded. In particular it has been noted that the power spectrum of clusters is better fitted if the primordial spectrum has a step-like feature at [Gramann & Suhhonenko 1999]. The implications of broken scale-invariance for accurate determination of cosmological parameters using forthcoming LSS and CMB data have been investigated [Wang, Spergel & Strauss 1999, Hannestad 2000].\n\nIt is clear that the notion of a scale imprinted on the primordial density perturbation is favoured by a number of observational considerations as well as having a natural physical interpretation in the context of realistic inflationary models. In this paper we have demonstrated that the recent small angular-scale CMB anisotropy data favour a deviation from scale-invariance in the range independently of the similar previous indication from studies of LSS, in particular the APM survey. This is reassuring since the latter rests on the assumption (for which arguments are given in Appendix A) that there is no scale-dependent bias between APM galaxies and dark matter.\n\nAllowing the primordial spectrum to depart from scale-invariance has important consequences for the deductions that can be made about cosmological parameters from LSS and CMB data. The standard interpretation of LSS data (e.g. APM and from cluster abundances) favours a low universe assuming a COBE-normalised H-Z spectrum. However as we have shown these observations can be satisfactorily accounted for with much larger values of if we allow for a break in the primordial spectrum. Our best fit models in Table 4, which obey the BBN constraint (5) on the baryon density, prefer higher values of (and lower values of ). The required amplitude of the break decreases with increasing but one cannot do without such a break altogether. In particular the value favoured by the SN Ia data [Perlmutter et al. 1999, Riess et al. 1998] is not permitted, while it is quite possible to have a universe with zero cosmological constant and . Of course the high observed fraction of baryons in clusters, combined with the BBN value of the baryon density (5), implies an independent limit on the matter density [White et al. 1993]; a recent analysis quotes at 95% c.l. [Mohr, Haiman & Holder 2000]. There are many assumptions made in such analyses, viz. that clusters are spherical and in hydrostatic equilibrium with an isothermal temperature profile, that the measured baryon fraction reflects the universal value, that there is no preheating before or energy injection after cluster formation, etc. It is important to reassess these issues in the light of our increasing understanding of galaxy and cluster formation to quantify the systematic uncertainties better.\n\nThe hypothesis of a primordial density perturbation with broken scale-invariance, although radical, is eminently falsifiable. The ongoing 2DF and SDSS redshift surveys can confirm or rule out such a feature in the power spectrum of galaxy clustering, while the forthcoming MAP and Planck missions will determine whether all the secondary acoustic peaks in the CMB angular spectrum are indeed suppressed as expected. By contrast the alternative hypothesis of a baryon density higher than the BBN value predicts a boosted third peak (e.g. Tegmark & Zaldariagga 2000), so this would be a definitive test distinguishing between the two possibilities. We note that in either case there are important implications for the physics of the early universe. To increase the baryon-to-photon ratio after BBN by requires a decrease in the comoving entropy; the only mechanism suggested which can achieve this proposes that the photons in our universe become cooled through their coupling to a ‘shadow’ universe [Bartlett & Hall 1991]. However this possibility, recently invoked by Kaplinghat and Turner (2000), was shown to be severely constrained, if not ruled out altogether, from considerations of the concommitant spectral distortion in the CMB together with the bound from Supernova 1987a on new weakly interacting particles [Birkel & Sarkar 1997]. By contrast broken scale-invariance has a natural explanation in a phase transition occuring during inflation as expected in supersymmetric theories. If established this would provide the first direct connection between astronomical data and physics at very high energies.\n\n## Acknowledgments\n\nWe thank Pedro Ferreira, Joe Silk, Graham Ross and Román Scoccimarro for stimulating discussions. J.B. and E.G. acknowledge grants from IEEC/CSIC and DGES(MEC) (Spain) — project PB96-0925 and Acción Especial ESP1998-1803-E; J.B. would also like to thank INAOE for their warm hospitality. The work of M.G.S. was supported by the Fundação para a Ciencia e a Tecnologia (Portugal) under program PRAXIS XXI/BD/18305/98.\n\n## Appendix A The effects of biasing\n\nIndependently of how or where galaxies form they must eventually fall into the dominant gravitational wells and thus trace out the underlying mass distribution (see Peebles 1980, Fry 1996). However, in principle, the galaxy distribution may be ‘biased’ with respect to the underlying mass fluctuations (e.g. Bardeen et al. 1986). There are several observations indicating that this effect is in fact small. The theoretical predictions for the first few connected moments, based on this hypothesis [Juszkiewicz, Bouchet & Colombi 1993, Bernardeau 1994a, Bernardeau 19994b] are in good agreement with the APM data [Gaztañaga 1994, Gaztañaga & Frieman 1994, Frieman & Gaztañaga 1999]. The current precision of this higher order correlation test is 20% and expected to improve with further data. The APM angular correlation function shows an inflection point near very similar to the long awaited ‘shoulder’ in , generated by gravitational dynamics [Gott & Rees 1975]. The agreement between the two characteristic scales can be used to constrain the linear biasing factor for the APM catalogue to be within 20% of unity [Gaztañaga & Juszkiewicz 2000]. We summarise these arguments [Gaztañaga & Baugh 1998] below for completeness since the assumption of small or no bias is crucial for the present study.\n\nLet us assume that the (smoothed) galaxy fluctuations are related to those in the mass by a local transformation, , which can be expanded as a Taylor series: . Then the 2-point function on the scale will be:\n\n ξg2(r) = b21ξm2(r)+b1b2⟨δm(x)δm(x+r)2⟩ (14) +b1b2⟨δ2m(x)δm(x+r)⟩+…,\n\nwhere all further terms are of order 4 or higher in and correspond therefore to either higher order correlations, with , or higher powers in . If is Gaussian or hierarchical (as is the case for evolution through gravity), the higher order correlations are at most of order . This means that at large scales where , the first term is dominant so that only the amplitude of the 2-point statistics, but not its shape, may be altered by biasing. This has been confirmed in N-body simulations and galaxy biasing models [Mo, Jing & White 1997, Scoccimarro et al 2000].", null, "Figure 9: The ratio of the non-linear to linear power spectrum illustrating non-linear effects on P(k). Long and short-dashed lines show non-linear effects in the mass while the other lines show non-linear effects in the final galaxy distribution for various halo models (from Scoccimarro et al 2000).\n\nThus from the above arguments, the small variance on large scales () means that it is reasonable to assume that the shape of for galaxies at scales coincides with the shape of the underlying linear matter power spectrum. This argument, which is based simply on the smallness of the variance and the assumption of hierarchical clustering, can also be applied to gravity, since the leading contribution to the correlation functions in perturbation theory is indeed given by a local transformation (see Fosalba & Gaztañaga 1997). This is clearly illustrated in Figure 11 of Gaztañaga and Baugh (1998). By comparing the linear and non-linear shape of , one can see that it has not been changed significantly through gravitational evolution on scales where the rms fluctuations are small, i.e. for .\n\nHow important are non-linearities for , i.e. for ? Figure 9 shows the ratio of the non-linear to linear power spectrum , normalised to be unity at small in order to emphasize the non-linear effects. The long-dashed line corresponds to the APM-like for dynamics (from N-body simulations in Baugh & Gaztañaga 1996). The short-dashed line corresponds to a , CDM model (using the non-linear mass fitting formulae). The dotted, continuous and dot-dashed lines correspond to different non-linear models for galaxy formation based on halo models (with linear bias for ) from Figure 6 in Scoccimarro et al (2000). In all cases non-linear effects become significant () only at and are of order unity by . Non-linear effects seem smaller for galaxies than for the dark matter (in agreement with the observed in Baugh & Gaztañaga 1996). One could imagine constructing galaxy formation models with stronger non-linear effects, but it will then be difficult to account for the observed galaxy higher-order correlations. As pointed out in Scoccimarro et al (2000), galaxy clustering in the framework of the halo models is strongly constrained by the measured galaxy skewness. Thus, given current observational constraints and present understanding of galaxy formation one can conclude that non-linear effects cannot have altered the linear by a factor exceeding , up to ." ]
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https://lkml.org/lkml/2001/8/23/213
[ "", null, "", null, "`I tested the following under linux-2.4.8-ac8, linux-2.4.8pre4 and2.4.5pre4, all had similar behaviour.I have written a webserver that serves many large files, and thus, thedisks are the bottleneck. To get around the problem of blocking reads(this killed thttpd's performance totally, for example) I can start one ormore reader threads. And strace of them under load looks like this: 0.000400 read(6, \"\\300g#\\v\", 4) = 4 <0.000091> 0.000315 lseek(1129, 70376488, SEEK_SET) = 70376488 <0.000088> 0.000287 read(1129, \"\\212\\266\\233^\\250\\23\\256D\\21E\\'#c\\242\\351pp`:Q[\\22/:\\27\"..., 65536) = 65536 <0.047408> 0.051059 write(9, \"\\300g#\\v\", 4) = 4 <0.000052> 0.000222 read(6, \"@\\3755\\r\", 4) = 4 <0.000036> 0.000187 lseek(946, 26180056, SEEK_SET) = 26180056 <0.000035> 0.000162 read(946, \"\\0\\20\\24\\0\\330\\6\\30\\264\\345\\263\\247\\213\\264\\0\\274\\4\\340\"..., 65536) = 65536 <0.029500> 0.029976 write(9, \"@\\3755\\r\", 4) = 4 <0.000098> 0.000331 read(6, \"\\300\\373j\\r\", 4) = 4 <0.000088> 0.000309 lseek(944, 33816188, SEEK_SET) = 33816188 <0.000090> 0.000287 read(944, \"\\210]\\360C\\340\\200\\315\\363@\\205\\203\\250\\316\\256\\\"\\34,E\"..., 65536) = 65536 <0.043455> 0.043885 write(9, \"\\300\\373j\\r\", 4) = 4 <0.000042> 0.000200 read(6, \"\\0\\227M\\r\", 4) = 4 <0.000035> 0.000191 lseek(315, 1310720, SEEK_SET) = 1310720 <0.000034> 0.000162 read(315, \"\\7\\17\\376\\250\\37\\312m\\210\\24\\215s\\257v\\246\\354\\272\\253\"..., 65536) = 65536 <0.025821> 0.026236 write(9, \"\\0\\227M\\r\", 4) = 4 <0.000040>filehandles 6 and 9 are request / result filehandles, i.e. the thread(s)read the request on filehandle 6, execute it (e.g. lseek/read) and thenwrite the result back to fh 9.as you can see, the read-request/lseek/write-result syscalls are veryfast, while the read-from-file-call totally dominates the runtime, whichis sensible, since the disks can only seek so-and-so-many times persecond. under this config the server steadily delivers about 23mbits/s.Now I thought I'd start more threads to give the kernel elevator morechances to optimize reads, however, it gets much slower. when I start 128threads, the strace of one thread now looks like this: 0.002181 read(6, \"\\300\\311\\205\\f\", 4) = 4 <0.000091> 0.000351 lseek(857, 12798664, SEEK_SET) = 12798664 <0.000087> 0.000292 read(857, \"^0\\260\\274\\363\\3078\\314\\373\\343\\254h\\360\\276\\347\\332\\305\"..., 65536) = 65536 <28.176472> 28.177062 write(9, \"\\300\\311\\205\\f\", 4) = 4 <0.000039> 0.000200 read(6, \"@\\2\\265\\10\", 4) = 4 <0.000033> 0.000480 lseek(471, 6553600, SEEK_SET) = 6553600 <0.000033> 0.000182 read(471, \"\\244\\t*\\\\\\225`+\\270@\\210\\206\\367\\10\\261\\4m\\32\\206\\377x\"..., 65536) = 65536 <36.023682> 36.025902 write(9, \"@\\2\\265\\10\", 4) = 4 <0.000107> 0.000422 read(6, \"\\0A\\235\\r\", 4) = 4 <0.000107> 0.000335 lseek(1239, 7143424, SEEK_SET) = 7143424 <0.000088> 0.000309 read(1239, \"\\'\\213\\315\\372\\331+x\\271\\234Bx\\255\\274\\\"G\\202\\264L+>\\266\"..., 65536) = 65536 <0.154575> 0.155760 write(9, \"\\0A\\235\\r\", 4) = 4 <0.000093> 0.000337 read(6, \"@G\\235\\r\", 4) = 4 <0.000387> 0.000676 lseek(998, 4989944, SEEK_SET) = 4989944 <0.000093> 0.000357 read(998, \"\\214P\\325|k\\226\\260\\31\\351\\260\\10\\10:\\23d`\\271Tu\\32\\252\"..., 65536) = 65536 <36.512863> 36.513243 write(9, \"@G\\235\\r\", 4) = 4 <0.002407>as you can see, read request now can take about ten times longer than theyshould (30 / 128 = 0.2). as a result, webserver thruput decreases to amere 4mbits/s.I would have expected a slight slowdown at most (more processing, maybecache effects), but not that much. any ideas why this is so and what couldbe done against it?-- -----==- | ----==-- _ | ---==---(_)__ __ ____ __ Marc Lehmann +-- --==---/ / _ \\/ // /\\ \\/ / [email protected] |e| -=====/_/_//_/\\_,_/ /_/\\_\\ XX11-RIPE --+ The choice of a GNU generation | |-To unsubscribe from this list: send the line \"unsubscribe linux-kernel\" inthe body of a message to [email protected] majordomo info at http://vger.kernel.org/majordomo-info.htmlPlease read the FAQ at http://www.tux.org/lkml/`", null, "", null, "", null, "" ]
[ null, "https://lkml.org/images/toprowlk.gif", null, "https://lkml.org/images/icornerl.gif", null, "https://lkml.org/images/icornerr.gif", null, "https://lkml.org/images/bcornerl.gif", null, "https://lkml.org/images/bcornerr.gif", null ]
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https://mathoverflow.net/questions/252302/extending-sacks-forcing
[ "# Extending Sacks forcing\n\nSacks forcing allows us to build a model $V[G]$, such that there is no \"intermediate model\" between $V$ and $V[G]$, meaning if $V \\subseteq W \\subseteq V[G]$ is a model of ZFC then either $W = V$ or $W = V[G]$.\n\nMy question is:\n\n1. Whether we know how to force a model to have exactly $1$ intermediate model?\n2. Assuming the answer is Yes, do we know how to extend this result arbitrarily long? Meaning, do we know how to force a model where all the inner models (i.e standard, transitive classes that contain all the ordinals) are well ordered and have a $1-1$ correspondence with the ordinals?\n\nThis would necessate to start with $V_0 = L$, and then possibly take $V_1 = V_0[G]$ (where this would be a regular Sacks forcing) and somehow continue to create $V_\\alpha$ for all $\\alpha \\in ORD$, without 'accidentally' creating any more inner models." ]
[ null ]
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https://docs.djangoproject.com/en/1.8/_modules/django/contrib/gis/gdal/envelope/
[ "• fr\n• Language: en\n• 1.10\n• dev\n• Documentation version: 1.8\n\n# Source code for django.contrib.gis.gdal.envelope\n\n```\"\"\"\nThe GDAL/OGR library uses an Envelope structure to hold the bounding\nbox information for a geometry. The envelope (bounding box) contains\ntwo pairs of coordinates, one for the lower left coordinate and one\nfor the upper right coordinate:\n\n+----------o Upper right; (max_x, max_y)\n| |\n| |\n| |\nLower left (min_x, min_y) o----------+\n\"\"\"\nfrom ctypes import Structure, c_double\n\nfrom django.contrib.gis.gdal.error import GDALException\n\n# The OGR definition of an Envelope is a C structure containing four doubles.\n# http://www.gdal.org/ogr/ogr__core_8h-source.html\nclass OGREnvelope(Structure):\n\"Represents the OGREnvelope C Structure.\"\n_fields_ = [(\"MinX\", c_double),\n(\"MaxX\", c_double),\n(\"MinY\", c_double),\n(\"MaxY\", c_double),\n]\n\n[docs]class Envelope(object):\n\"\"\"\nThe Envelope object is a C structure that contains the minimum and\nmaximum X, Y coordinates for a rectangle bounding box. The naming\nof the variables is compatible with the OGR Envelope structure.\n\"\"\"\n\ndef __init__(self, *args):\n\"\"\"\nThe initialization function may take an OGREnvelope structure, 4-element\ntuple or list, or 4 individual arguments.\n\"\"\"\n\nif len(args) == 1:\nif isinstance(args, OGREnvelope):\n# OGREnvelope (a ctypes Structure) was passed in.\nself._envelope = args\nelif isinstance(args, (tuple, list)):\n# A tuple was passed in.\nif len(args) != 4:\nraise GDALException('Incorrect number of tuple elements (%d).' % len(args))\nelse:\nself._from_sequence(args)\nelse:\nraise TypeError('Incorrect type of argument: %s' % str(type(args)))\nelif len(args) == 4:\n# Individual parameters passed in.\n# Thanks to ww for the help\nself._from_sequence([float(a) for a in args])\nelse:\nraise GDALException('Incorrect number (%d) of arguments.' % len(args))\n\n# Checking the x,y coordinates\nif self.min_x > self.max_x:\nraise GDALException('Envelope minimum X > maximum X.')\nif self.min_y > self.max_y:\nraise GDALException('Envelope minimum Y > maximum Y.')\n\ndef __eq__(self, other):\n\"\"\"\nReturns True if the envelopes are equivalent; can compare against\nother Envelopes and 4-tuples.\n\"\"\"\nif isinstance(other, Envelope):\nreturn (self.min_x == other.min_x) and (self.min_y == other.min_y) and \\\n(self.max_x == other.max_x) and (self.max_y == other.max_y)\nelif isinstance(other, tuple) and len(other) == 4:\nreturn (self.min_x == other) and (self.min_y == other) and \\\n(self.max_x == other) and (self.max_y == other)\nelse:\nraise GDALException('Equivalence testing only works with other Envelopes.')\n\ndef __str__(self):\n\"Returns a string representation of the tuple.\"\nreturn str(self.tuple)\n\ndef _from_sequence(self, seq):\n\"Initializes the C OGR Envelope structure from the given sequence.\"\nself._envelope = OGREnvelope()\nself._envelope.MinX = seq\nself._envelope.MinY = seq\nself._envelope.MaxX = seq\nself._envelope.MaxY = seq\n\n[docs] def expand_to_include(self, *args):\n\"\"\"\nModifies the envelope to expand to include the boundaries of\nthe passed-in 2-tuple (a point), 4-tuple (an extent) or\nenvelope.\n\"\"\"\n# We provide a number of different signatures for this method,\n# and the logic here is all about converting them into a\n# 4-tuple single parameter which does the actual work of\n# expanding the envelope.\nif len(args) == 1:\nif isinstance(args, Envelope):\nreturn self.expand_to_include(args.tuple)\nelif hasattr(args, 'x') and hasattr(args, 'y'):\nreturn self.expand_to_include(args.x, args.y, args.x, args.y)\nelif isinstance(args, (tuple, list)):\n# A tuple was passed in.\nif len(args) == 2:\nreturn self.expand_to_include((args, args, args, args))\nelif len(args) == 4:\n(minx, miny, maxx, maxy) = args\nif minx < self._envelope.MinX:\nself._envelope.MinX = minx\nif miny < self._envelope.MinY:\nself._envelope.MinY = miny\nif maxx > self._envelope.MaxX:\nself._envelope.MaxX = maxx\nif maxy > self._envelope.MaxY:\nself._envelope.MaxY = maxy\nelse:\nraise GDALException('Incorrect number of tuple elements (%d).' % len(args))\nelse:\nraise TypeError('Incorrect type of argument: %s' % str(type(args)))\nelif len(args) == 2:\n# An x and an y parameter were passed in\nreturn self.expand_to_include((args, args, args, args))\nelif len(args) == 4:\n# Individual parameters passed in.\nreturn self.expand_to_include(args)\nelse:\nraise GDALException('Incorrect number (%d) of arguments.' % len(args))\n\n@property\ndef min_x(self):\n\"Returns the value of the minimum X coordinate.\"\nreturn self._envelope.MinX\n\n@property\ndef min_y(self):\n\"Returns the value of the minimum Y coordinate.\"\nreturn self._envelope.MinY\n\n@property\ndef max_x(self):\n\"Returns the value of the maximum X coordinate.\"\nreturn self._envelope.MaxX\n\n@property\ndef max_y(self):\n\"Returns the value of the maximum Y coordinate.\"\nreturn self._envelope.MaxY\n\n@property\ndef ur(self):\n\"Returns the upper-right coordinate.\"\nreturn (self.max_x, self.max_y)\n\n@property\ndef ll(self):\n\"Returns the lower-left coordinate.\"\nreturn (self.min_x, self.min_y)\n\n@property\ndef tuple(self):\n\"Returns a tuple representing the envelope.\"\nreturn (self.min_x, self.min_y, self.max_x, self.max_y)\n\n@property\ndef wkt(self):\n\"Returns WKT representing a Polygon for this envelope.\"\n# TODO: Fix significant figures.\nreturn 'POLYGON((%s %s,%s %s,%s %s,%s %s,%s %s))' % \\\n(self.min_x, self.min_y, self.min_x, self.max_y,\nself.max_x, self.max_y, self.max_x, self.min_y,\nself.min_x, self.min_y)\n```" ]
[ null ]
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https://gadgetspidy.com/insertion-sort-2/
[ "# Insertion sort\n\nQUESTION\n\nSorting \\nOne common task for computers is to sort data. For example, people might want to see all their files on a computer sorted by size. Since sorting is a simple problem with many different possible solutions, it is often used to introduce the study of algorithms.\\n\\nInsertion Sort \\nThese challenges will cover Insertion Sort, a simple and intuitive sorting algorithm. We will first start with an already sorted list.\\n\\nInsert element into sorted list \\n\\nGiven a sorted list with an unsorted number in the rightmost cell, can you write some simple code to insert into the array so that it remains sorted?\\n\\nPrint the array every time a value is shifted in the array until the array is fully sorted. The goal of this challenge is to follow the correct order of insertion sort.\\n\\nGuideline: You can copy the value of e to a variable and consider its cell \\”empty\\”. Since this leaves an extra cell empty on the right, you can shift everything over until v can be inserted. This will create a duplicate of each value, but when you reach the right spot, you can replace it with e .\\n\\nInput Format \\n\\nThere will be two lines of input:\\nsize – the size of the array\\nArr – the unsorted array of integers\\n\\nOutput Format \\nOn each line, output the entire array every time an item is shifted in it.\\n\\nConstraints \\n\\n1<=size<=1000 \\n-10000<=e<=10000, e belongs to arr.\n\n``````#include <stdio.h>\n#include <string.h>\n#include <math.h>\n#include <stdlib.h>\n#include <assert.h>\n\nvoid insertionSort(int ar_size, int * ar)\n{\nint i,j,k,m;\nfor(i=1;i<ar_size;i++)\n{\nfor(j=0;j<i;j++)\n{\nif(ar[i]<ar[j])\n{\nar[i] = ar[i] + ar[j];\nar[j] = ar[i] - ar[j];\nar[i] = ar[i] - ar[j];\n}\n}\nfor(j=0;j<ar_size;j++)\n{\nprintf(\"%d \",ar[j]);\n}\nprintf(\"\\n\");\n}\n}\nint main(void)\n{\nint _ar_size;\nscanf(\"%d\", &_ar_size);\nint _ar[_ar_size], _ar_i;\nfor(_ar_i = 0; _ar_i < _ar_size; _ar_i++)\n{\nscanf(\"%d\", &_ar[_ar_i]);\n}\ninsertionSort(_ar_size, _ar);\nreturn 0;\n}", null, "", null, "" ]
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