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https://www.scribd.com/document/310361571/UJIAN-PENGESANAN-AHKIR-TAHUN-PAPER-2-1-pdf | [
"You are on page 1of 18\n\nhttp://sahatmozac.blogspot.\n\ncom\nSULIT\n\n1449/2\n\n1449/2\nMathematics\nPaper 2\nNOV\n2008\n\nName :.\nForm :..\n\nUJIAN PENGESANAN PERTENGAHAN TAHUN\n\nSIJIL PELAJARAN MALAYSIA 2008\n\nFull\nMarks\n3\n\n4. Working step must be written clearly.\n\n10\n\n11\n\n12\n\n12\n\n13\n\n12\n\n14\n\n12\n\n15\n\n12\n\nMATHEMATICS\nPaper 2\nTwo hour and thirty minutes\n\nSection\n\nQuestion\n\nDO NOT OPEN THIS QUESTION PAPER UNTIL YOU\n\nARE TOLD TO DO SO\n1. 1. This question paper consist of two section: Section A and Section\nB. Answer all question in Section A and Section B.\n\n5. Diagrams given are not according to scale unless stated.\n\n6. Marks for each question are given in bracket.\n7. A list of formulae is given in pages 2 and 3.\n\n8. Non programmable scientific calculator are allowed.\n\n9 This question paper must be hand up at the end of the exam.\n\nMark\nObtain\n\nTotal\n\n1449/22008\n\nHak Cipta JPM\n\nhttp://mathsmozac.blogspot.com\n\n[Lihat sebelah\nSULIT\n\nhttp://sahatmozac.blogspot.com\nSULIT\n\n1449/2\n\nMATHEMATICAL FORMULAE\nThe following formulae may be helpful in answering the questions. The symbols given are the\nones commonly used..\nRELATIONS\n1\n\na a a\n\na m a n a m n\n\n( a m ) n a mn\n\n1 d b\n\nA 1\n\nc\na\n\nn( A)\nP( A)\nn\nS\n\nP( A' ) 1 P ( A)\n\nDistance ( x 2 x1 ) 2 ( y 2 y1 ) 2\n\n8\n9\n\nx x 2 y1 y 2\nMidpoint , ( x, y ) 1\n,\n\n2\n2\ndistance travelled\nAverage speed =\ntime taken\nsum of data\nnumber of data\n\n10\n\nMean =\n\n11\n\nMean\n\n12\n\nm n\n\nsum of(class mark frequency)\n\nsum of frequencie s\n\nPythagoras Theorem\nc 2 a 2 b 2\n\n13\n\ny y 1\nm 2\nx 2 x1\n\n14\n\ny int ercept\nm\nx int ercept\n\n1449/2@2008\n\nHak Cipta JPM\n\nhttp://mathsmozac.blogspot.com\nSULIT\n\nhttp://sahatmozac.blogspot.com\nSULIT\n\n1449/2\n\nSHAPES AND SPACE\n\n1\nsum of parallel sides height\n2\nCircumference of circle = d 2r\n\nArea of circle = r 2\n\nVolume of right prism = cross sectional area length\n\nVolume of cylinder = j 2h\n\nVolume of cone =\n\nVolume of sphere =\n\n10\n\n11\n\nSum of interior angles of a polygon = (n 2) 180\n\n12\n13\n\nArea of trapezium =\n\n1 2\nr h\n3\n4 3\nr\n3\n1\nbase area height\n3\n\narc length\nangle subtended at center\n\no\ncircumfere nce of circle\n360\narea of sector angle subtended at centre\n\no\narea of circle\n360\nPA'\nPA\n\n14\n\nScale factor, k =\n\n15\n\n1449/22008 Hak Cipta JPM\n\nhttp://mathsmozac.blogspot.com\n\n[Lihat sebelah\nSULIT\n\nhttp://sahatmozac.blogspot.com\nSULIT\n\n1449/2\n\nBahagian A\n[52 markah]\nAnswer all questions in this section.\n1\n\nThe Venn diagram in the answer space shows sets P, Q and R such that the\nuniversal set, = P Q R. On the diagrams in the answer space, shade\n(a) the set P' R' ,\n(b) the set (P Q ) R.\n[ 3 marks ]\n\n(a)\nQ\nR\nP\n\n(b)\n\nQ\nR\nP\n\n1449/22008 Hak Cipta JPM\n\nhttp://mathsmozac.blogspot.com\n\n[Lihat sebelah\nSULIT\n\nhttp://sahatmozac.blogspot.com\nSULIT\n\n2.\n\n1449/2\n\nDiagram 1 shows a cuboid with a horizontal base ABCD. E, F, G, H are the\n\nmidpoints of AD , JM , KL and BC respectively .\n.\n\nD\nF\n\nH\nK\n2 cm\nB\n\n10 cm\n\nJ\nDIAGRAM 1\n5 cm\n\nIdentify and calculate the angle between the plane KLE and the plane ABCD.\n[ 4 marks ]\n\n1449/22008 Hak Cipta JPM\n\nhttp://mathsmozac.blogspot.com\n\n[Lihat sebelah\nSULIT\n\nhttp://sahatmozac.blogspot.com\nSULIT\n3\n\nSolve the quadratic equation 9x (x 3) = 21x + 15\n\n1449/2\n\n[4 marks].\n\nCalculate the values of x and y that satisfy the following simultaneous linear\nequations:\nx\n\n3y = 9\n\n4\nx y = 6\n3\n\n[ 4 marks]\n\n1449/22008 Hak Cipta JPM\n\nhttp://mathsmozac.blogspot.com\n\n[Lihat sebelah\nSULIT\n\nhttp://sahatmozac.blogspot.com\nSULIT\n5\n\n1449/2\n\nDiagram I shows a cylindrical solid of height h cm with a base radius of 7 cm. A\n\nconical cavity PQR of the same height is hollowed out. The volume of the remaining\nsolid is 2772 cm3.\nDiagram I\nP\n\n22\n\nUse\n7\n\n[ 5 marks ]\n\n1449/22008 Hak Cipta JPM\n\nhttp://mathsmozac.blogspot.com\n\n[Lihat sebelah\nSULIT\n\nhttp://sahatmozac.blogspot.com\nSULIT\n6\n\n1449/2\n\n(a)\n\nState whether the following statement is true or false ?\n\n2\nk 3k = 2k or 8 64\n\n(b)\n\nWrite down two implications based on the following sentence :.\n\nm + 3 = 10 if and only if m = 7\n\n(c)\n\n1 = 2(1)2 1\n7 = 2(2)2 1\n17 = 2(3)2 1\n.\n.\nBased on the information above, make a general conclusion by induction\n[5 marks]\n\n(a)\n\n..\n\n(b)\n\nImplication 1 :\n\nImplication 2 :\n.\n\n(c)\n\n..\n\nA box contains 4 red balls,6 black balls and 2 yellow balls.\n\n(a)\n\nIf a ball is picked at random from the box, state the probability that it is red in\ncolour.\n\n(b)\n\nThe ball that was picked is returned into the box and 4 black balls are added into\nthe box. If a ball is picked at random from the box, state the probability that it is\nblack in colour\n[ 4 marks]\n\n(a)\n\n1449/22008 Hak Cipta JPM\n\nhttp://mathsmozac.blogspot.com\n\n[Lihat sebelah\nSULIT\n\nhttp://sahatmozac.blogspot.com\nSULIT\n\n1449/2\n\n(b)\n\n8. Diagram 3 shows two sectors, PQR and TUV, with the same centre O. The angle of\neach sector is 270 . OSR is a semicircle with centre V. PTO is a straight line and\nOP = 14 cm.\n\nDiagram 3\nUsing =\n\n22\n, calculate\n7\n\n(a) the perimeter, in cm, of the whole diagram,\n\n(b) the area, in cm 2 , of the shaded region.\n[ 6 marks ]\n(a)\n\n(b)\n\n1449/22008 Hak Cipta JPM\n\nhttp://mathsmozac.blogspot.com\n\n[Lihat sebelah\nSULIT\n\nhttp://sahatmozac.blogspot.com\nSULIT\n9\n\n10\n\n1449/2\n\nTable 1 shows the ages 200 of participants in the charity run.\n\nAge (year)\n15 19\n20 24\n25 - 29\n30 34\n35 39\n40 44\n45 - 50\n\nFrequency\n25\n37\n58\n32\n20\n17\n11\n\nMidpoint\n\nTable 1\n(a)\n\nState the class size and modal class based on the grouped data in table1.\n[2 marks]\n\n(b)\n\nComplete the table 1 and find the mean age of the participants.\n\n[4 marks]\n(a)\n\nClass size\n\n= .\n\nModal class = .\n\n(b)\nAge (year)\n15 19\n20 24\n25 - 29\n30 34\n35 39\n40 44\n45 - 50\n\nFrequency\n25\n37\n58\n32\n20\n17\n11\n\nMidpoint\n\nMean age = ..\n\n1449/22008 Hak Cipta JPM\n\nhttp://mathsmozac.blogspot.com\n\n[Lihat sebelah\nSULIT\n\nhttp://sahatmozac.blogspot.com\nSULIT\n\n11\n\n1449/2\n\ny\n10\nT\n\nQ ( 6, c)\n\nR ( 2, -3 )\nDIAGRAM 4\n\nIn Diagram 4, straight line QP is parallel to RT. QR is parallel to x-axis . P and S is\n\nlocated on the x-axis. Gradient of QP is -3.\nCalculate\n(a)\n\nthe value of c\n\n(b)\n\nthe RT equations\n\n(c)\n\nx - intercept of RT\n[ 5 mark ]\n\n(a)\n\n(b)\n\n(c)\n\n1449/22008 Hak Cipta JPM\n\nhttp://mathsmozac.blogspot.com\n\n[Lihat sebelah\nSULIT\n\nhttp://sahatmozac.blogspot.com\nSULIT\n11.\n\n12\n(a)\n\n1449/2\n\nH\n\nThe angle of depression of peak D from peak H is 550. The length of\n\nEF is 13 m and the height of DE is 7 m. Calculate the height in m, of\nthe pole of FSH.\n[ 3 marks ]\n(b)\nE\nF\nD\n\nThe diagram shows a prism with horizontal base ABCD. Given that\nAB = 6 ,BC = 8 and AF = 5. Calculate the angle between line FC and\nthe plane ABCD.\n[ 3 marks ]\n\nAnswer: ( a )\n\n(b)\n\n1449/22008 Hak Cipta JPM\n\nhttp://mathsmozac.blogspot.com\n\n[Lihat sebelah\nSULIT\n\nhttp://sahatmozac.blogspot.com\nSULIT\n\n13\n\n1449/2\n\nSection B\n[48 marks]\nAnswer all questions from this section.\n12\n\n(a)\n\nGiven that sin = sin 59 and 90\n\n360\n, find the value of .\n\n(b)\n\nGiven that x is an angle in quadrant III and sin x = -0.2924, find\n\nthe value of tan x correct to two decimal places.\n\nIn the diagram above, ABC is a straight line. Calculate the\n\n(i) length of AD,\n(ii) value of cos y.\n(d)\n\nGiven that x = sin 270, y = cos 240 and z = tan 360, find the\nvalue of\n(i)\n(ii)\n(iii)\n(iv)\n\ny x z,\nz y x,\nx (y + z),\nz\nx + y z.\n\n( 12 marks)\n\n(a)\n\n(b)\n\n1449/22008 Hak Cipta JPM\n\nhttp://mathsmozac.blogspot.com\n\n[Lihat sebelah\nSULIT\n\nhttp://sahatmozac.blogspot.com\nSULIT\n\n14\n\n1449/2\n\n(c) ( i )\n\n( ii )\n\n(d) ( i )\n\n( ii )\n\n( iii )\n\n( iv )\n\n1449/22008 Hak Cipta JPM\n\nhttp://mathsmozac.blogspot.com\n\n[Lihat sebelah\nSULIT\n\nhttp://sahatmozac.blogspot.com\nSULIT\n\n15\n\n1449/2\n\n13. (a) Given the universal set = W X Y. Where\n\nSet W = {a, b, c }\nSet Y = {c, d, e, f, g}\nSet X = {g}\nI. In the Venn diagram below, fill in set W, set Y and Set X\nII.\nList all the elements of (W X)\nIII.\nFind n( X Y)\n(b) Given that the universal set = {x: 20x 50}\nSet K = { x: x is multiple of 5}\nSet L = { x: x is prime number}\nSet H = { x: x is greater that 30}\nI. Write down all the elements of set K and set L\nII.\nWrite down the elements of (K L) H\nIII. Find n(L H)\n(a) I\nW\n\nY\nX\n\nII .\n\nIII .\n\n(b)\n\nI .\n\nII .\n\nIII. .\n\n1449/22008 Hak Cipta JPM\n\nhttp://mathsmozac.blogspot.com\n\n[Lihat sebelah\nSULIT\n\nhttp://sahatmozac.blogspot.com\nSULIT\n\n14\n\n16\n\n1449/2\n\nTable 2 shows the distribution of the weight of 40 students in a class.\n\nWeight ( kg ) Frequency\n35 39\n2\n40 44\n5\n45 49\n8\n50 54\n10\n55 59\n7\n60 64\n4\n65 69\n3\n70 74\n1\n(a)\n\nTABLE 2\nState the modal class for the data in the Table 2\n\n(b)\n\n(i)\n\nConstruct a cumulative frequency table for the data in Table 2\n\n[3 marks]\n\n(ii)\n\nUsing the scale of 2cm to 5cm on the x-axis and 2cm to 5 students to\ny-axis, draw an ogive for the data.\n[5 marks]\n\n(c)\n\n[1 mark]\n\nFrom the graph, find\n\n(i)\n(ii)\n\nmedian\ninterquartile range\n[3 marks]\n\n(a)\n(b)\nWeight ( kg ) Frequency\n35\n40\n45\n50\n55\n60\n65\n70\n(c)\n\n39\n44\n49\n54\n59\n64\n69\n74\n\nUpper Boundary\n\nCumulative\nFrequency\n\n2\n5\n8\n10\n7\n4\n3\n1\n\n(i)\n(ii)\n\n1449/22008 Hak Cipta JPM\n\nhttp://mathsmozac.blogspot.com\n\n[Lihat sebelah\nSULIT\n\nhttp://sahatmozac.blogspot.com\nSULIT\n15\n\n17\n\n1449/2\n\n( a ) Find the equation of the straight line that passes through the point ( 2, -5 ) and\nis parallel to y = 4 3x.\n[2 marks]\n\nB ( 1, h )\n\nA( 0 , 2 )\nC\nx\n6\n\n( b ) In the diagram above, the gradient of the straight line AB is 2. Find\n\n( i ) the value of h,\n\n[1 marks]\n\n[2 marks]\n\n[1 marks]\n\n( iv ) the y-intercept of the straight line of BC.\n\n[2 marks]\n\n(c)\n\nIf M is the midpoint of AC, find the equation of the straight line BM. Hence,\nstate its y-intercept.\n[4 marks]\n\n(a)\n\n(b)\n\n(i)\n\n(ii)\n\n1449/22008 Hak Cipta JPM\n\nhttp://mathsmozac.blogspot.com\n\n[Lihat sebelah\nSULIT\n\nhttp://sahatmozac.blogspot.com\nSULIT\n\n18\n\n1449/2\n\n(iii)\n\n(iv)\n\n(c)\n\n1449/22008 Hak Cipta JPM\n\nhttp://mathsmozac.blogspot.com\n\n[Lihat sebelah\nSULIT"
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http://www.foo.be/docs-free/applied-cryptography/ch14/14-06.html | [
"Applied Cryptography, Second Edition: Protocols, Algorthms, and Source Code in C (cloth)\n(Publisher: John Wiley & Sons, Inc.)\nAuthor(s): Bruce Schneier\nISBN: 0471128457\nPublication Date: 01/01/96\n\nThese subkeys must all be calculated before encryption or decryption.\n\nTo encrypt a 1024-byte block X:\n\n(1) Divide X into 256 32-bit sub-blocks: X0, X1, X2,..., X255.\n(2) Permute the sub-blocks of X according to P.\n(3) For r = 0 to 3\nFor g = 0 to 63\nA = X(4g)<<<2r\nB = X(4g+1)<<<2r\nC = X(4g+2)<<<2r\nD = X(4g+3)<<<2r\nFor step s = 0 to 7\nA = A ⊕ (B + fr(B,C,D) + S512r+8 g+s)\nTEMP = D\nD = C\nC = B\nB = A <<< 5\nA = TEMP\nX(4g)<<<2r = A\nX(4g+1)<<<2r = B\nX(4g+2)<<<2r = C\nX(4g+3)<<<2r = D\n(4) Recombine X0, X1, X2,..., X255 to form the ciphertext.\n\nThe functions fr(B,C,D) are similar to those used in MD5:\n\nf0(B,C,D) = (B Λ C) ν ((¬ B) Λ D)\nf1(B,C,D) = (B Λ D) ν (C Λ (¬ D))\nf2(B,C,D) = BCD\nf3(B,C,D) = C ⊕ (B ν (¬ D))\n\nDecryption is the reverse process.\n\nGenerating the subkeys is a large task. Here is how the permutation array, P, could be generated from an 80-bit key, K.\n\n(1) Initialize K0, K1, K2,..., K9 with the 10 bytes of K.\n(2) For i = 10 to 255\nKi = Ki - 2Ki - 6Ki - 7Ki - 10\n(3) For i = 0 to 255, Pi = i\n(4) m = 0\n(5) For j = 0 to 1\nFor i = 256 to 1 step -1\nm = (K256 - i + K257 - i) mod i\nK257 - i = K257 - i <<< 3\nSwap Pi and Pi - 1\n\nThe S-array of 2048 32-bit words could be generated in a similar manner, either from the same 80-bit key or from another key. The authors caution that these details should “be viewed as motivational; there may very well be alternative schemes which are both more efficient and offer improved security” .\n\nCrab was proposed as a testbed of new ideas and not as a working algorithm. It uses many of the same techniques as MD5. Biham has argued that a very large block size makes an algorithm easier to cryptanalyze . On the other hand, Crab may make efficient use of a very large key. In such a case, “easier to cryptanalyze” might not mean much.\n\n### 14.7 SXAL8/MBAL\n\nThis is a 64-bit block algorithm from Japan . SXAL8 is the basic algorithm; MBAL is an expanded version with a variable block length. Since MBAL does some clever things internally, the authors claim that they can get adequate security with only a few rounds. With a block length of 1024 bytes, MBAL is about 70 times faster than DES. Unfortunately, shows that MBAL is susceptible to differential cryptanalysis, and shows that it is susceptible to linear cryptanalysis.\n\n### 14.8 RC5\n\nRC5 is a block cipher with a variety of parameters: block size, key size, and number of rounds. It was invented by Ron Rivest and analyzed by RSA Laboratories [1324,1325].\n\nThere are three operations: XOR, addition, and rotations. Rotations are constant-time operations on most processors and variable rotations are a nonlinear function. These rotations, which depend on both the key and the data, are the interesting operation.\n\nRC5 has a variable-length block, but this example will focus on a 64-bit data block. Encryption uses 2r + 2 key-dependent 32-bit words—S0, S1, S2,..., S2r + 1—where r is the number of rounds. We’ll generate those words later. To encrypt, first divide the plaintext block into two 32-bit words: A and B. (RC5 assumes a little-endian convention for packing bytes into words: The first byte goes into the low-order bit positions of register A, etc.) Then:\n\nA = A + S0\nB = B + S1\nFor i = 1 to r:\nA = ((AB) <<< B) + S2i\nB = ((BA) <<< A) + S2i + 1\n\nThe output is in the registers A and B.\n\nDecryption is just as easy. Divide the plaintext block into two words, A and B, and then:\n\nFor i = r down to 1:\nB = ((BS2i + 1) >>> A) ⊕ A\nA = ((AS2i) >>> B) ⊕ B\nB = BS1\nA = AS0\n\nThe symbol “>>>” is a right circular shift. Of course, all addition and subtraction are mod 232.\n\nCreating the array of keys is more complicated, but also straightforward. First, copy the bytes of the key into an array, L, of c 32-bit words, padding the final word with zeros if necessary. Then, initialize an array, S, using a linear congruential generator mod 232:\n\nS0 = P\nfor i = 1 to 2(r + 1) – 1:\nSi = (Si - 1 + Q) mod 232\n\nP = 0xb7e15163 and Q = 0x9e3779b9; these constants are based on the binary representation of e and phi.\n\nFinally, mix L into S:\n\ni = j = 0\nA = B = 0\ndo 3n times (where n is the maximum of 2(r + 1) and c):\nA = Si = (Si + A + B) <<< 3\nB = Lj = (Lj + A + B) <<< (A + B)\ni = (i + 1) mod 2(r + 1)\nj = (j + 1) mod c\n\n[an error occurred while processing this directive]"
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http://vshibo.cn/Static/docs.html | [
"1. 体验版\n2. 0元 10小时直播券\n3. 同时在线人数300以下\n• 直播综合管理\n• 开设子代理账户\n• 界面管理\n• 直播回看\n• 页面装修自定义\n• 广告位设置\n• 活动LOGO自主添加\n• 宣传片预热\n• 播放状态设置\n• APP同步\n• 官网同步\n• 授权观看设置\n• 弹幕设置\n• 分享送红包设置\n• 评论送红包设置\n• 评论管理\n• 观看用户列表\n• 黑名单\n• 红包管理\n• 添加微信红包\n• 修改微信红包\n• 互动数据\n• 授权观看\n• 打赏红包\n• 打赏榜\n• 邀约榜\n• 分享红包\n• 群发红包\n• 评论中奖\n• 摇一摇中奖\n• 刮刮卡中奖\n• 大转盘中奖\n• 虚拟礼物\n• 数据导出\n• 专属运维服务\n• 专属客户经理\n• CNN智能调度,2000+分发节点\n• 微信公众号绑定\n• 独立域名、logo\n• 独立服务器\n• 子账户管理\n1. 企业版\n2. 18000元/年 无时长限制\n3. 同时在线人数5000人以下\n• 直播综合管理\n• 开设子代理账户\n• 界面管理\n• 直播回看\n• 页面装修自定义\n• 广告位设置\n• 活动LOGO自主添加\n• 宣传片预热\n• 播放状态设置\n• APP同步\n• 官网同步\n• 授权观看设置\n• 弹幕设置\n• 分享送红包设置\n• 评论送红包设置\n• 评论管理\n• 观看用户列表\n• 黑名单\n• 红包管理\n• 添加微信红包\n• 修改微信红包\n• 互动数据\n• 授权观看\n• 打赏红包\n• 打赏榜\n• 邀约榜\n• 分享红包\n• 群发红包\n• 评论中奖\n• 摇一摇中奖\n• 刮刮卡中奖\n• 大转盘中奖\n• 虚拟礼物\n• 数据导出\n• 专属运维服务\n• 专属客户经理\n• CNN智能调度,2000+分发节点\n• 微信公众号绑定\n• 独立域名、logo\n• 独立服务器\n• 子账户管理\n1. 定制版\n2. 30000元/年 无时长限制\n3. 同时在线人数10000人以下\n• 直播综合管理\n• 开设子代理账户\n• 界面管理\n• 直播回看\n• 页面装修自定义\n• 广告位设置\n• 活动LOGO自主添加\n• 宣传片预热\n• 播放状态设置\n• APP同步\n• 官网同步\n• 授权观看设置\n• 弹幕设置\n• 分享送红包设置\n• 评论送红包设置\n• 评论管理\n• 观看用户列表\n• 黑名单\n• 红包管理\n• 添加微信红包\n• 修改微信红包\n• 互动数据\n• 授权观看\n• 打赏红包\n• 打赏榜\n• 邀约榜\n• 分享红包\n• 群发红包\n• 评论中奖\n• 摇一摇中奖\n• 刮刮卡中奖\n• 大转盘中奖\n• 虚拟礼物\n• 数据导出\n• 专属运维服务\n• 专属客户经理\n• CNN智能调度,2000+分发节点\n• 微信公众号绑定\n• 独立域名、logo\n• 独立服务器\n• 子账户管理\n1. 市级独家代理\n2. 首年5万元—150万元\n\n代理费依城市而定，可发展下级代理\n\n次年起，仅需5万/年系统维护费\n\n3. 流量不限,并发量不限,时长不限\n• 直播综合管理\n• 开设子代理账户\n• 界面管理\n• 直播回看\n• 页面装修自定义\n• 广告位设置\n• 活动LOGO自主添加\n• 宣传片预热\n• 播放状态设置\n• APP同步\n• 官网同步\n• 授权观看设置\n• 弹幕设置\n• 分享送红包设置\n• 评论送红包设置\n• 评论管理\n• 观看用户列表\n• 黑名单\n• 红包管理\n• 添加微信红包\n• 修改微信红包\n• 互动数据\n• 授权观看\n• 打赏红包\n• 打赏榜\n• 邀约榜\n• 分享红包\n• 群发红包\n• 评论中奖\n• 摇一摇中奖\n• 刮刮卡中奖\n• 大转盘中奖\n• 虚拟礼物\n• 数据导出\n• 专属运维服务\n• 专属客户经理\n• CNN智能调度,2000+分发节点\n• 微信公众号绑定\n• 独立域名、logo\n• 独立服务器\n• 子账户管理",
null,
"QQ"
]
| [
null,
"http://vshibo.cn/Public/Home/assets/img/index/footer.jpg",
null
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https://www.colorhexa.com/0e1a11 | [
"# #0e1a11 Color Information\n\nIn a RGB color space, hex #0e1a11 is composed of 5.5% red, 10.2% green and 6.7% blue. Whereas in a CMYK color space, it is composed of 46.2% cyan, 0% magenta, 34.6% yellow and 89.8% black. It has a hue angle of 135 degrees, a saturation of 30% and a lightness of 7.8%. #0e1a11 color hex could be obtained by blending #1c3422 with #000000. Closest websafe color is: #003300.\n\n• R 5\n• G 10\n• B 7\nRGB color chart\n• C 46\n• M 0\n• Y 35\n• K 90\nCMYK color chart\n\n#0e1a11 color description : Very dark (mostly black) cyan - lime green.\n\n# #0e1a11 Color Conversion\n\nThe hexadecimal color #0e1a11 has RGB values of R:14, G:26, B:17 and CMYK values of C:0.46, M:0, Y:0.35, K:0.9. Its decimal value is 924177.\n\nHex triplet RGB Decimal 0e1a11 `#0e1a11` 14, 26, 17 `rgb(14,26,17)` 5.5, 10.2, 6.7 `rgb(5.5%,10.2%,6.7%)` 46, 0, 35, 90 135°, 30, 7.8 `hsl(135,30%,7.8%)` 135°, 46.2, 10.2 003300 `#003300`\nCIE-LAB 7.882, -7.281, 4.087 0.652, 0.873, 0.664 0.298, 0.399, 0.873 7.882, 8.349, 150.692 7.882, -3.297, 3.157 9.341, -3.895, 2.322 00001110, 00011010, 00010001\n\n# Color Schemes with #0e1a11\n\n• #0e1a11\n``#0e1a11` `rgb(14,26,17)``\n• #1a0e17\n``#1a0e17` `rgb(26,14,23)``\nComplementary Color\n• #111a0e\n``#111a0e` `rgb(17,26,14)``\n• #0e1a11\n``#0e1a11` `rgb(14,26,17)``\n• #0e1a17\n``#0e1a17` `rgb(14,26,23)``\nAnalogous Color\n• #1a0e11\n``#1a0e11` `rgb(26,14,17)``\n• #0e1a11\n``#0e1a11` `rgb(14,26,17)``\n• #170e1a\n``#170e1a` `rgb(23,14,26)``\nSplit Complementary Color\n• #1a110e\n``#1a110e` `rgb(26,17,14)``\n• #0e1a11\n``#0e1a11` `rgb(14,26,17)``\n• #110e1a\n``#110e1a` `rgb(17,14,26)``\n• #171a0e\n``#171a0e` `rgb(23,26,14)``\n• #0e1a11\n``#0e1a11` `rgb(14,26,17)``\n• #110e1a\n``#110e1a` `rgb(17,14,26)``\n• #1a0e17\n``#1a0e17` `rgb(26,14,23)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #050906\n``#050906` `rgb(5,9,6)``\n• #0e1a11\n``#0e1a11` `rgb(14,26,17)``\n• #172b1c\n``#172b1c` `rgb(23,43,28)``\n• #203b27\n``#203b27` `rgb(32,59,39)``\n• #294c32\n``#294c32` `rgb(41,76,50)``\nMonochromatic Color\n\n# Alternatives to #0e1a11\n\nBelow, you can see some colors close to #0e1a11. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #0e1a0e\n``#0e1a0e` `rgb(14,26,14)``\n• #0e1a0f\n``#0e1a0f` `rgb(14,26,15)``\n• #0e1a10\n``#0e1a10` `rgb(14,26,16)``\n• #0e1a11\n``#0e1a11` `rgb(14,26,17)``\n• #0e1a12\n``#0e1a12` `rgb(14,26,18)``\n• #0e1a13\n``#0e1a13` `rgb(14,26,19)``\n• #0e1a14\n``#0e1a14` `rgb(14,26,20)``\nSimilar Colors\n\n# #0e1a11 Preview\n\nThis text has a font color of #0e1a11.\n\n``<span style=\"color:#0e1a11;\">Text here</span>``\n#0e1a11 background color\n\nThis paragraph has a background color of #0e1a11.\n\n``<p style=\"background-color:#0e1a11;\">Content here</p>``\n#0e1a11 border color\n\nThis element has a border color of #0e1a11.\n\n``<div style=\"border:1px solid #0e1a11;\">Content here</div>``\nCSS codes\n``.text {color:#0e1a11;}``\n``.background {background-color:#0e1a11;}``\n``.border {border:1px solid #0e1a11;}``\n\n# Shades and Tints of #0e1a11\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, #000000 is the darkest color, while #f3f8f4 is the lightest one.\n\n• #000000\n``#000000` `rgb(0,0,0)``\n• #070d09\n``#070d09` `rgb(7,13,9)``\n• #0e1a11\n``#0e1a11` `rgb(14,26,17)``\n• #152719\n``#152719` `rgb(21,39,25)``\n• #1c3422\n``#1c3422` `rgb(28,52,34)``\n• #23402a\n``#23402a` `rgb(35,64,42)``\n• #294d32\n``#294d32` `rgb(41,77,50)``\n• #305a3b\n``#305a3b` `rgb(48,90,59)``\n• #376743\n``#376743` `rgb(55,103,67)``\n• #3e734b\n``#3e734b` `rgb(62,115,75)``\n• #458054\n``#458054` `rgb(69,128,84)``\n• #4c8d5c\n``#4c8d5c` `rgb(76,141,92)``\n• #539a64\n``#539a64` `rgb(83,154,100)``\n• #5aa66d\n``#5aa66d` `rgb(90,166,109)``\n``#67ad78` `rgb(103,173,120)``\n• #73b483\n``#73b483` `rgb(115,180,131)``\n• #80bb8f\n``#80bb8f` `rgb(128,187,143)``\n• #8dc19a\n``#8dc19a` `rgb(141,193,154)``\n• #9ac8a5\n``#9ac8a5` `rgb(154,200,165)``\n• #a6cfb0\n``#a6cfb0` `rgb(166,207,176)``\n• #b3d6bc\n``#b3d6bc` `rgb(179,214,188)``\n• #c0ddc7\n``#c0ddc7` `rgb(192,221,199)``\n• #cde4d2\n``#cde4d2` `rgb(205,228,210)``\n• #d9ebde\n``#d9ebde` `rgb(217,235,222)``\n• #e6f2e9\n``#e6f2e9` `rgb(230,242,233)``\n• #f3f8f4\n``#f3f8f4` `rgb(243,248,244)``\nTint Color Variation\n\n# Tones of #0e1a11\n\nA tone is produced by adding gray to any pure hue. In this case, #131513 is the less saturated color, while #00280a is the most saturated one.\n\n• #131513\n``#131513` `rgb(19,21,19)``\n• #111713\n``#111713` `rgb(17,23,19)``\n• #101812\n``#101812` `rgb(16,24,18)``\n• #0e1a11\n``#0e1a11` `rgb(14,26,17)``\n• #0c1c10\n``#0c1c10` `rgb(12,28,16)``\n• #0b1d0f\n``#0b1d0f` `rgb(11,29,15)``\n• #091f0f\n``#091f0f` `rgb(9,31,15)``\n• #08200e\n``#08200e` `rgb(8,32,14)``\n• #06220d\n``#06220d` `rgb(6,34,13)``\n• #05230c\n``#05230c` `rgb(5,35,12)``\n• #03250c\n``#03250c` `rgb(3,37,12)``\n• #02260b\n``#02260b` `rgb(2,38,11)``\n• #00280a\n``#00280a` `rgb(0,40,10)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #0e1a11 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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http://swmath.org/software/4634 | [
"# azove\n\nOn threshold BDDs and the optimal variable ordering problem Many combinatorial optimization problems can be formulated as $0/1$ integer programs ($0/1$ IPs). The investigation of the structure of these problems raises the following tasks: count or enumerate the feasible solutions and find an optimal solution according to a given linear objective function. All these tasks can be accomplished using binary decision diagrams (BDDs), a very popular and effective datastructure in computational logics and hardware verification. par We present a novel approach for these tasks which consists of an $output-sensitive$ algorithm for building a BDD for a linear constraint (a so-called threshold BDD) and a parallel AND operation on threshold BDDs. In particular our algorithm is capable of solving knapsack problems, subset sum problems and multidimensional knapsack problems. par BDDs are represented as a directed acyclic graph. The size of a BDD is the number of nodes of its graph. It heavily depends on the chosen variable ordering. Finding the optimal variable ordering is an NP-hard problem. We derive a $0/1$ IP for finding an optimal variable ordering of a threshold BDD. This $0/1$ IP formulation provides the basis for the computation of the variable ordering spectrum of a threshold function. par We introduce our new tool azove 2.0 as an enhancement to azove 1.1 which is a tool for counting and enumerating $0/1$ points. Computational results on benchmarks from the literature show the strength of our new method.\n\n##",
null,
"Keywords for this software\n\nAnything in here will be replaced on browsers that support the canvas element\n\n## References in zbMATH (referenced in 11 articles , 2 standard articles )\n\nShowing results 1 to 11 of 11.\nSorted by year (citations)"
]
| [
null,
"http://swmath.org/media/img/minus.png",
null
]
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https://www.ams.org/journals/tran/1972-166-00/S0002-9947-1972-0294557-9/ | [
"",
null,
"",
null,
"",
null,
"ISSN 1088-6850(online) ISSN 0002-9947(print)\n\nExtending congruences on semigroups\n\nAuthor: A. R. Stralka\nJournal: Trans. Amer. Math. Soc. 166 (1972), 147-161\nMSC: Primary 22A15\nDOI: https://doi.org/10.1090/S0002-9947-1972-0294557-9\nMathSciNet review: 0294557\nFull-text PDF Free Access\n\nAbstract: The two main results are: (1) Let S be a semigroup which satisfies the relation $abcd = acbd$, let A be a subsemigroup of Reg S which is a band of groups and let $[\\varphi ]$ be a congruence on A. Then $[\\varphi ]$ can be extended to a congruence on S. (2) Let S be a compact topological semigroup which satisfies the relation $abcd = acbd$, let A be a closed subsemigroup of Reg S and let $[\\varphi ]$ be a closed congruence on A such that $\\dim \\varphi (A)|\\mathcal {H} = 0$. Then $[\\varphi ]$ can be extended to a closed congruence on S.\n\n[Enhancements On Off] (What's this?)\n\nRetrieve articles in Transactions of the American Mathematical Society with MSC: 22A15\n\nRetrieve articles in all journals with MSC: 22A15"
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https://themagedclo.tk/ | [
"# e-book Ordinary Differential Equations and Operators\n\nLecture 1 Play Video. Lecture 2 Play Video. Solving Differential Equations: Two Distinct Real Roots This is a video in the series for solving systems of ordinary differential equations.\n\n### mathematics and statistics online\n\nLecture 3 Play Video. Lecture 4 Play Video. Lecture 5 Play Video. Lecture 6 Play Video. Test For Stability of the Origin This video addresses the stability of the origin on a phase portrait of a 2X2 system of ordinary differential equations. Lecture 7 Play Video. Lecture 8 Play Video. Lecture 9 Play Video.",
null,
"Lecture 10 Play Video. Introduction to Differential Equations This is an introduction to differential equations. Lecture 11 Play Video. Lecture 12 Play Video. Lecture 13 Play Video. Lecture 14 Play Video. Lecture 15 Play Video. Derivation a General Solution and Integrating Factor for a Linear Differential Equation This video defines total differential, exact equations and uses clairiots theorem to derive the form of the integrating factor for a First order linear ODE.\n\nLecture 16 Play Video.\n\n1. Select a Web Site.\n2. A Tribute to F.V. Atkinson. Proceedings of a Symposium held at Dundee, Scotland, March - July 1982.\n3. Select a Web Site;\n4. Differential equations.\n5. An Introduction to Family Therapy.\n6. Domain of linear ordinary differential operators - MuPAD?\n7. Environmental UV Photobiology.\n\nLinear Differental Equations: Example I This the first example of two on how to solve a linear differential equation using and integrating factor. Lecture 17 Play Video. Linear Differential Equations: Example II This the second example of two on how to solve a linear differential equation using and integrating factor.\n\nLecture 18 Play Video. Using Differential Operators A differential operator acts on a function.\n\n## Differential Equations\n\nLecture 19 Play Video. Solving Linear DE's with Two Distinct Real Roots One way to solve homogeneous linear differential equations is by using differential operators and characteristic equations. Lecture 20 Play Video. Solving Linear DE's with Repeated Real Roots One way to solve homogeneous linear differential equations is by using differential operators and characteristic equations.\n\nLecture 21 Play Video. Solving Linear DE's with Complex Roots One way to solve homogeneous linear differential equations is by using differential operators and characteristic equations. Lecture 22 Play Video.\n\n### Recommended\n\nFundamental Solution Set for Linear DE's Three criteria for a fundamental set of solutions to a differential equation must be satisfied. Lecture 23 Play Video. Factoring Operators When dealing with differential operators with constant coefficients then the operators are factorable and do factor like polynomials.\n\nLecture 24 Play Video. Annihilator Method I This is the first example of using the annihilator method for solving non-homogeneous linear differential equations. Lecture 25 Play Video. Writing a Differential Equation as a System This video is about writing a differential equation as a system of differential equations.\n\nLecture 26 Play Video. Existence and Uniqueness Linear D. Lecture 27 Play Video. Lecture 28 Play Video. Runge-Kutta Method The video is about Runge-Kutta method for approximating solutions of a differential equation using a slope field. Lecture 29 Play Video. The Wronskian and a Test for Independence The Wronskian is a fun name to say and it is not hard to calculate.\n\nLecture 30 Play Video. Euler Method The video is about Euler method for approximating solutions of a differential equation using a slope field. A definiteness result for determinantal operators Pages Binding, Paul et al. On second-order left-definite boundary value problems Pages Bennewitz, C. A von Neumann factorization of some selfadjoint extensions of positive symmetric differential operators and its application to inequalities Pages Brown, Richard C. Inclusion theorems for solutions of differential equations with aid of pointwise or vector monotonicity Pages Collatz, Prof.\n\nThe liouville-green asymptotic theory for second-order differential equations: A new approach and some extensions Pages Eastham, M. Non-self-adjoint operators and their essential spectra Pages Evans, W. A concept of adjointness and symmetry of differential expressions based on the generalised Lagrange identity and Green's formula Pages Everitt, W.\n\nOn quadratic integral inequalities associated with second-order symmetric differential expressions Pages Everitt, W. Nonoscillation theorems for differential equations with general deviating arguments Pages Fink, A. Self-adjoint 4-th order boundary value problem in the limit - 4 case Pages Fulton, Charles T.\n\nDifferential equation introduction - First order differential equations - Khan Academy\n\nFactorization and the Friedrichs extension for ordinary differential operators Pages Kauffman, Robert M. Boundedness criteria for hyperbolic characteristic initial value problems Pages Kreith, K. Matrix riccati inequality and oscillation of second order differential systems Pages Kwong, Man Kam. A new approach to second order linear oscillation theory Pages Kwong, Man Kam et al. Small amplitude limit cycles of polynomial differential equations Pages Lloyd, N.\n\nRegular growth and zero-tending solutions Pages Macki, Jack W.\n\n• Course summary?\n• Ordinary differential equation - Wikipedia!\n• From Gene to Protein: Translation Into Biotechnology;\n• X-Men - The Brotherhood of Monsters."
]
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https://math.libretexts.org/Courses/Misericordia_University/MTH_226%3A_Calculus_III/Chapter_12%3A_Vectors_and_the_Geometry_of_Space/12.6E%3A_Exercises_for_Quadric_Surfaces | [
"Skip to main content\n$$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$\n\n# 12.6E: Exercises for Quadric Surfaces\n\n$$\\newcommand{\\vecs}{\\overset { \\rightharpoonup} {\\mathbf{#1}} }$$ $$\\newcommand{\\vecd}{\\overset{-\\!-\\!\\rightharpoonup}{\\vphantom{a}\\smash {#1}}}$$$$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$\n\nFor exercises 1 - 6, sketch and describe the cylindrical surface of the given equation.\n\n1) [T] $$x^2+z^2=1$$\n\nAnswer:\n\nThe surface is a cylinder with the rulings parallel to the y-axis.",
null,
"2) [T] $$x^2+y^2=9$$\n\n3) [T] $$z=cos(\\frac{π}{2}+x)$$\n\nAnswer:\n\nThe surface is a cylinder with rulings parallel to the y-axis.",
null,
"4) [T] $$z=e^x$$\n\n5) [T] $$z=9−y^2$$\n\nAnswer:\n\nThe surface is a cylinder with rulings parallel to the x-axis.",
null,
"6) [T] $$z=ln(x)$$\n\nFor exercises 7 - 10, the graph of a quadric surface is given.\n\na. Specify the name of the quadric surface.\n\nb. Determine the axis of symmetry of the quadric surface.\n\n7)",
null,
"Answer:\na. Cylinder; b. The $$x$$-axis\n\n8)",
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"9)",
null,
"Answer:\na. Hyperboloid of two sheets; b. The $$x$$-axis\n\n10)",
null,
"For exercises 11 - 16, match the given quadric surface with its corresponding equation in standard form.\n\na. $$\\frac{x^2}{4}+\\frac{y^2}{9}−\\frac{z^2}{12}=1$$\n\nb. $$\\frac{x^2}{4}−\\frac{y^2}{9}−\\frac{z^2}{12}=1$$\n\nc. $$\\frac{x^2}{4}+\\frac{y^2}{9}+\\frac{z^2}{12}=1$$\n\nd. $$z^2=4x^2+3y^2$$\n\ne. $$z=4x^2−y^2$$\n\nf. $$4x^2+y^2−z^2=0$$\n\n11) Hyperboloid of two sheets\n\nAnswer:\nb.\n\n12) Ellipsoid\n\n13) Elliptic paraboloid\n\nAnswer:\nd.\n\n14) Hyperbolic paraboloid\n\n15) Hyperboloid of one sheet\n\nAnswer:\na.\n\n16) Elliptic cone\n\nFor exercises 17 - 28, rewrite the given equation of the quadric surface in standard form. Identify the surface.\n\n17) $$−x^2+36y^2+36z^2=9$$\n\nAnswer:\n$$−\\frac{x^2}{9}+\\frac{y^2}{\\frac{1}{4}}+\\frac{z^2}{\\frac{1}{4}}=1,$$ hyperboloid of one sheet with the $$x$$-axis as its axis of symmetry\n\n18) $$−4x^2+25y^2+z^2=100$$\n\n19) $$−3x^2+5y^2−z^2=10$$\n\nAnswer:\n$$−\\frac{x^2}{\\frac{10}{3}}+\\frac{y^2}{2}−\\frac{z^2}{10}=1,$$ hyperboloid of two sheets with the $$y$$-axis as its axis of symmetry\n\n20) $$3x^2−y^2−6z^2=18$$\n\n21) $$5y=x^2−z^2$$\n\nAnswer:\n$$y=−\\frac{z^2}{5}+\\frac{x^2}{5},$$ hyperbolic paraboloid with the $$y$$-axis as its axis of symmetry\n\n22) $$8x^2−5y^2−10z=0$$\n\n23) $$x^2+5y^2+3z^2−15=0$$\n\nAnswer:\n$$\\frac{x^2}{15}+\\frac{y^2}{3}+\\frac{z^2}{5}=1,$$ ellipsoid\n\n24) $$63x^2+7y^2+9z^2−63=0$$\n\n25) $$x^2+5y^2−8z^2=0$$\n\nAnswer:\n$$\\frac{x^2}{40}+\\frac{y^2}{8}−\\frac{z^2}{5}=0,$$ elliptic cone with the $$z$$-axis as its axis of symmetry\n\n26) $$5x^2−4y^2+20z^2=0$$\n\n27) $$6x=3y^2+2z^2$$\n\nAnswer:\n$$x=\\frac{y^2}{2}+\\frac{z^2}{3},$$ elliptic paraboloid with the $$x$$-axis as its axis of symmetry\n\n28) $$49y=x^2+7z^2$$\n\nFor exercises 29 - 34, find the trace of the given quadric surface in the specified plane of coordinates and sketch it.\n\n29) [T] $$x^2+z^2+4y=0,z=0$$\n\nAnswer:\n\nParabola $$y=−\\frac{x^2}{4},$$",
null,
"30) [T] $$x^2+z^2+4y=0,\\quad x=0$$\n\n31) [T] $$−4x^2+25y^2+z^2=100,\\quad x=0$$\n\nAnswer:\n\nEllipse $$\\frac{y^2}{4}+\\frac{z^2}{100}=1,$$",
null,
"32) [T] $$−4x^2+25y^2+z^2=100,\\quad y=0$$\n\n33) [T] $$x^2+\\frac{y^2}{4}+\\frac{z^2}{100}=1,\\quad x=0$$\n\nAnswer:\n\nEllipse $$\\frac{y^2}{4}+\\frac{z^2}{100}=1,$$",
null,
"34) [T] $$x^2−y−z^2=1,\\quad y=0$$\n\n35) Use the graph of the given quadric surface to answer the questions.",
null,
"a. Specify the name of the quadric surface.\n\nb. Which of the equations—$$16x^2+9y^2+36z^2=3600,9x^2+36y^2+16z^2=3600,$$ or $$36x^2+9y^2+16z^2=3600$$ —corresponds to the graph?\n\nc. Use b. to write the equation of the quadric surface in standard form.\n\nAnswer:\na. Ellipsoid\nb. The third equation\nc. $$\\frac{x^2}{100}+\\frac{y^2}{400}+\\frac{z^2}{225}=1$$\n\n36) Use the graph of the given quadric surface to answer the questions.",
null,
"a. Specify the name of the quadric surface.\n\nb. Which of the equations—$$36z=9x^2+y^2,9x^2+4y^2=36z$$, or $$−36z=−81x^2+4y^2$$ —corresponds to the graph above?\n\nc. Use b. to write the equation of the quadric surface in standard form.\n\nFor exercises 37 - 42, the equation of a quadric surface is given.\n\na. Use the method of completing the square to write the equation in standard form.\n\nb. Identify the surface.\n\n37) $$x^2+2z^2+6x−8z+1=0$$\n\nAnswer:\na. $$\\frac{(x+3)^2}{16}+\\frac{(z−2)^2}{8}=1$$\nb. Cylinder centered at $$(−3,2)$$ with rulings parallel to the $$y$$-axis\n\n38) $$4x^2−y^2+z^2−8x+2y+2z+3=0$$\n\n39) $$x^2+4y^2−4z^2−6x−16y−16z+5=0$$\n\nAnswer:\na. $$\\frac{(x−3)^2}{4}+(y−2)^2−(z+2)^2=1$$\nb. Hyperboloid of one sheet centered at $$(3,2,−2),$$ with the $$z$$-axis as its axis of symmetry\n\n40) $$x^2+z^2−4y+4=0$$\n\n41) $$x^2+\\frac{y^2}{4}−\\frac{z^2}{3}+6x+9=0$$\n\nAnswer:\na. $$(x+3)^2+\\frac{y^2}{4}−\\frac{z^2}{3}=0$$\nb. Elliptic cone centered at $$(−3,0,0),$$ with the $$z$$-axis as its axis of symmetry\n\n42) $$x^2−y^2+z^2−12z+2x+37=0$$\n\n43) Write the standard form of the equation of the ellipsoid centered at the origin that passes through points $$A(2,0,0),B(0,0,1),$$ and $$C(12,\\sqrt{11},\\frac{1}{2}).$$\n\nAnswer:\n$$\\frac{x^2}{4}+\\frac{y^2}{16}+z^2=1$$\n\n44) Write the standard form of the equation of the ellipsoid centered at point $$P(1,1,0)$$ that passes through points $$A(6,1,0),B(4,2,0)$$ and $$C(1,2,1)$$.\n\n45) Determine the intersection points of elliptic cone $$x^2−y^2−z^2=0$$ with the line of symmetric equations $$\\frac{x−1}{2}=\\frac{y+1}{3}=z.$$\n\nAnswer:\n$$(1,−1,0)$$ and $$(\\frac{13}{3},4,\\frac{5}{3})$$\n\n46) Determine the intersection points of parabolic hyperboloid $$z=3x^2−2y^2$$ with the line of parametric equations $$x=3t,y=2t,z=19t$$, where $$t∈R.$$\n\n47) Find the equation of the quadric surface with points $$P(x,y,z)$$ that are equidistant from point $$Q(0,−1,0)$$ and plane of equation $$y=1.$$ Identify the surface.\n\nAnswer:\n$$x^2+z^2+4y=0,$$ elliptic paraboloid\n\n48) Find the equation of the quadric surface with points $$P(x,y,z)$$ that are equidistant from point $$Q(0,2,0)$$ and plane of equation $$y=−2.$$ Identify the surface.\n\n49) If the surface of a parabolic reflector is described by equation $$400z=x^2+y^2,$$ find the focal point of the reflector.\n\nAnswer:\n$$(0,0,100)$$\n\n50) Consider the parabolic reflector described by equation $$z=20x^2+20y^2.$$ Find its focal point.\n\n51) Show that quadric surface $$x^2+y^2+z^2+2xy+2xz+2yz+x+y+z=0$$ reduces to two parallel planes.\n\n52) Show that quadric surface $$x^2+y^2+z^2−2xy−2xz+2yz−1=0$$ reduces to two parallel planes passing.\n\n53) [T] The intersection between cylinder $$(x−1)^2+y^2=1$$ and sphere $$x^2+y^2+z^2=4$$ is called a Viviani curve.",
null,
"a. Solve the system consisting of the equations of the surfaces to find the equation of the intersection curve. (Hint: Find $$x$$ and $$y$$ in terms of $$z$$.)\n\nb. Use a computer algebra system (CAS) or CalcPlot3D to visualize the intersection curve on sphere $$x^2+y^2+z^2=4$$.\n\nAnswer:\n\na. $$x=2−\\frac{z^2}{2},y=±\\frac{z}{2}\\sqrt{4−z^2},$$ where $$z∈[−2,2];$$\n\nb.",
null,
"54) Hyperboloid of one sheet $$25x^2+25y^2−z^2=25$$ and elliptic cone $$−25x^2+75y^2+z^2=0$$ are represented in the following figure along with their intersection curves. Identify the intersection curves and find their equations (Hint: Find y from the system consisting of the equations of the surfaces.)",
null,
"55) [T] Use a CAS or CalcPlot3D to create the intersection between cylinder $$9x^2+4y^2=18$$ and ellipsoid $$36x^2+16y^2+9z^2=144$$, and find the equations of the intersection curves.\n\nAnswer:\n\ntwo ellipses of equations $$\\frac{x^2}{2}+\\frac{y^2}{\\frac{9}{2}}=1$$ in planes $$z=±2\\sqrt{2}$$",
null,
"56) [T] A spheroid is an ellipsoid with two equal semiaxes. For instance, the equation of a spheroid with the z-axis as its axis of symmetry is given by $$\\frac{x^2}{a^2}+\\frac{y^2}{a^2}+\\frac{z^2}{c^2}=1$$, where $$a$$ and $$c$$ are positive real numbers. The spheroid is called oblate if $$c<a$$, and prolate for $$c>a$$.\n\na. The eye cornea is approximated as a prolate spheroid with an axis that is the eye, where $$a=8.7mm$$ and $$c=9.6mm$$.Write the equation of the spheroid that models the cornea and sketch the surface.\n\nb. Give two examples of objects with prolate spheroid shapes.\n\n57) [T] In cartography, Earth is approximated by an oblate spheroid rather than a sphere. The radii at the equator and poles are approximately $$3963$$mi and $$3950$$mi, respectively.\n\na. Write the equation in standard form of the ellipsoid that represents the shape of Earth. Assume the center of Earth is at the origin and that the trace formed by plane $$z=0$$ corresponds to the equator.\n\nb. Sketch the graph.\n\nc. Find the equation of the intersection curve of the surface with plane $$z=1000$$ that is parallel to the xy-plane. The intersection curve is called a parallel.\n\nd. Find the equation of the intersection curve of the surface with plane $$x+y=0$$ that passes through the z-axis. The intersection curve is called a meridian.\n\nAnswer:\n\na. $$\\frac{x^2}{3963^2}+\\frac{y^2}{3963^2}+\\frac{z^2}{3950^2}=1$$\n\nb.",
null,
"c. The intersection curve is the ellipse of equation $$\\frac{x^2}{3963^2}+\\frac{y^2}{3963^2}=\\frac{(2950)(4950)}{3950^2}$$, and the intersection is an ellipse.\nd. The intersection curve is the ellipse of equation $$\\frac{2y^2}{3963^2}+\\frac{z^2}{3950^2}=1.$$\n\n58) [T] A set of buzzing stunt magnets (or “rattlesnake eggs”) includes two sparkling, polished, superstrong spheroid-shaped magnets well-known for children’s entertainment. Each magnet is $$1.625$$ in. long and $$0.5$$ in. wide at the middle. While tossing them into the air, they create a buzzing sound as they attract each other.\n\na. Write the equation of the prolate spheroid centered at the origin that describes the shape of one of the magnets.\n\nb. Write the equations of the prolate spheroids that model the shape of the buzzing stunt magnets. Use a CAS or CalcPlot3D to create the graphs.\n\n59) [T] A heart-shaped surface is given by equation $$(x^2+\\frac{9}{4}y^2+z^2−1)^3−x^2z^3−\\frac{9}{80}y^2z^3=0.$$\n\na. Use a CAS or CalcPlot3D to graph the surface that models this shape.\n\nb. Determine and sketch the trace of the heart-shaped surface on the xz-plane.\n\nAnswer:\n\na.",
null,
"b. The intersection curve is $$(x^2+z^2−1)^3−x^2z^3=0.$$",
null,
"60) [T] The ring torus symmetric about the z-axis is a special type of surface in topology and its equation is given by $$(x^2+y^2+z^2+R^2−r^2)^2=4R^2(x^2+y^2)$$, where $$R>r>0$$. The numbers $$R$$ and $$r$$ are called are the major and minor radii, respectively, of the surface. The following figure shows a ring torus for which $$R=2$$ and $$r=1$$.",
null,
"a. Write the equation of the ring torus with $$R=2$$ and $$r=1$$, and use a CAS or CalcPlot3D to graph the surface. Compare the graph with the figure given.\n\nb. Determine the equation and sketch the trace of the ring torus from a. on the xy-plane.\n\nc. Give two examples of objects with ring torus shapes.\n\n## Contributors\n\nGilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.\n\nExercises and LaTeX edited by Paul Seeburger"
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"https://math.libretexts.org/@api/deki/files/3523/CNX_Calc_Figure_12_06_223.jpg",
null,
"https://math.libretexts.org/@api/deki/files/3524/CNX_Calc_Figure_12_06_206.jpg",
null,
"https://math.libretexts.org/@api/deki/files/3525/CNX_Calc_Figure_12_06_209.jpg",
null,
"https://math.libretexts.org/@api/deki/files/3526/CNX_Calc_Figure_12_06_226.jpg",
null,
"https://math.libretexts.org/@api/deki/files/3527/CNX_Calc_Figure_12_06_227.jpg",
null,
"https://math.libretexts.org/@api/deki/files/3528/CNX_Calc_Figure_12_06_228.jpg",
null,
"https://math.libretexts.org/@api/deki/files/3529/CNX_Calc_Figure_12_06_229.jpg",
null,
"https://math.libretexts.org/@api/deki/files/3530/CNX_Calc_Figure_12_06_230.jpg",
null,
"https://math.libretexts.org/@api/deki/files/3531/CNX_Calc_Figure_12_06_231.jpg",
null,
"https://math.libretexts.org/@api/deki/files/3532/CNX_Calc_Figure_12_06_233.jpg",
null,
"https://math.libretexts.org/@api/deki/files/3533/CNX_Calc_Figure_12_06_237.jpg",
null,
"https://math.libretexts.org/@api/deki/files/3534/CNX_Calc_Figure_12_06_238.jpg",
null,
"https://math.libretexts.org/@api/deki/files/3535/CNX_Calc_Figure_12_06_239.jpg",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.76908016,"math_prob":1.0000074,"size":10774,"snap":"2021-04-2021-17","text_gpt3_token_len":4005,"char_repetition_ratio":0.17520891,"word_repetition_ratio":0.16572858,"special_character_ratio":0.4069055,"punctuation_ratio":0.11001642,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.000009,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-19T02:55:21Z\",\"WARC-Record-ID\":\"<urn:uuid:2458cbb3-f6c3-4db4-994a-17ed45528701>\",\"Content-Length\":\"112204\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e47f2b87-b297-4b93-bb52-0638454322f1>\",\"WARC-Concurrent-To\":\"<urn:uuid:6584c3cb-a26d-47e6-b054-b7f9dee0c80e>\",\"WARC-IP-Address\":\"99.86.230.16\",\"WARC-Target-URI\":\"https://math.libretexts.org/Courses/Misericordia_University/MTH_226%3A_Calculus_III/Chapter_12%3A_Vectors_and_the_Geometry_of_Space/12.6E%3A_Exercises_for_Quadric_Surfaces\",\"WARC-Payload-Digest\":\"sha1:B4PVFKZ3XWNDFYVKSYDXDRGSGSEH6WC3\",\"WARC-Block-Digest\":\"sha1:2ZTIHIZFR5ADSLYDOCTDR5XKSBPEOPSL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038863420.65_warc_CC-MAIN-20210419015157-20210419045157-00602.warc.gz\"}"} |
https://byjus.com/question-answer/which-of-the-following-has-sum-of-its-zeroes-as-zero-18/ | [
"",
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"Question\n\n# Which of the following graph represents a quadratic polynomial which has sum of its zeroes is zero?\n\nA\n\nA)",
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"No worries! We‘ve got your back. Try BYJU‘S free classes today!\nB\n\nB)",
null,
"No worries! We‘ve got your back. Try BYJU‘S free classes today!\nC\n\nC)",
null,
"No worries! We‘ve got your back. Try BYJU‘S free classes today!\nD\n\nD)",
null,
"Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses\nOpen in App\nSolution\n\n## The correct option is D D)",
null,
"We know that, the values at which a quadratic polynomial cuts the x-axis are its zeroes. Option A: Polynomial cuts the x-axis at : -3, -2 ; sum of zeroes: -5 Option B: Polynomial cuts the x-axis at : 2, 3; sum of zeroes: 5 Option C: Polynomial cuts the x-axis at : 2, 3; sum of zeroes: 5 Option D: Polynomial cuts the x-axis at : -2, 2; sum of zeros: 0 Hence, the answer is D.",
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"Similar questions",
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"Zeroes of a Polynomial\nMATHEMATICS\nWatch in App",
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| [
null,
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https://research.wur.nl/en/publications/significant-non-linearity-in-nitrous-oxide-chamber-data-and-its-e | [
"# Significant non-linearity in nitrous oxide chamber data and its effect on calculated annual emissions\n\nP.C. Stolk, C.M.J. Jacobs, E.J. Moors, A. Hensen, G.L. Velthof, P. Kabat\n\n### Abstract\n\nChambers are widely used to measure surface fluxes of nitrous oxide (N2O). Usually linear regression is used to calculate the fluxes from the chamber data. Non-linearity in the chamber data can result in an underestimation of the flux. Non-linear regression models are available for these data, but are not commonly used. In this study we compared the fit of linear and non-linear regression models to determine significant non-linearity in the chamber data. We assessed the influence of this significant non-linearity on the annual fluxes. For a two year dataset from an automatic chamber we calculated the fluxes with linear and non-linear regression methods. Based on the fit of the methods 32% of the data was defined significant non-linear. Significant non-linearity was not recognized by the goodness of fit of the linear regression alone. Using non-linear regression for these data and linear regression for the rest, increases the annual flux with 21% to 53% compared to the flux determined from linear regression alone. We suggest that differences this large are due to leakage through the soil. Macropores or a coarse textured soil can add to fast leakage from the chamber. Yet, also for chambers without leakage non-linearity in the chamber data is unavoidable, due to feedback from the increasing concentration in the chamber. To prevent a possibly small, but systematic underestimation of the flux, we recommend comparing the fit of a linear regression model with a non-linear regression model. The non-linear regression model should be used if the fit is significantly better. Open questions are how macropores affect chamber measurements and how optimization of chamber design can prevent this.\nOriginal language English 115-141 Biogeosciences 6 https://doi.org/10.5194/bgd-6-115-2009 Published - 2009\n\n### Keywords\n\n• nitrous oxide\n• emission\n• data analysis\n• regression analysis"
]
| [
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https://lionessays.com/i-need-some-math-assistance/ | [
"# I need some Math assistance\n\nI need some Math assistance\n\nPlease complete the follow problems neatly. Clearly label each problem, show all work, and use Microsoft Word’s equation editor to properly format all mathematics.\n\n1.\n\n1. Y= 3x^3\n\n3.\n\n4.\n\na.\n\n1. Find the slope (-1,6) and (3,-2)\n2. Find the equation of the line that is parallel 2x+3y=5 and passes through (-8,3)\n3. A manuscript translator charges a standing fee of \\$50 plus \\$2.50 per page translated. Write a linear equation for the amount earned for translating pages.\n4. In section 2.1 you read about how we can model certain situations using linear equations. In particular, example 8 shows how you can create an equation to predict sales of a company\n\nPick a company whose sales you want to model and predict. Look up their sales information and pick 2 years of data. Using that information, create a linear equation that can be used to predict the sales of the company. Once you have your equation, predict the sales 2 years from now.\n\nMake sure you include in your post the links where you found the relevant information. Clearly show your work. This includes naming the company, sharing the ordered pairs, defining variables, finding the slope, determining the equation, and then using that equation to predict.\n\n9.\n\nThe post I need some Math assistance appeared first on Lion Essays.\n\n## “Looking for a Similar Assignment? Get Expert Help at an Amazing Discount!”",
null,
"I need some Math assistance was first posted on April 20, 2019 at 8:17 pm."
]
| [
null,
"https://i2.wp.com/Lion Essays.com/wp-content/uploads/2018/08/.",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.90461767,"math_prob":0.96371806,"size":1716,"snap":"2022-05-2022-21","text_gpt3_token_len":387,"char_repetition_ratio":0.11974299,"word_repetition_ratio":0.013745705,"special_character_ratio":0.23484848,"punctuation_ratio":0.12427746,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9874248,"pos_list":[0,1,2],"im_url_duplicate_count":[null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-27T00:23:43Z\",\"WARC-Record-ID\":\"<urn:uuid:8991f277-7f1e-4cb9-ab69-0dfcaa1cf124>\",\"Content-Length\":\"57596\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2fecffd3-e019-49f8-85d4-7ec4828ac37f>\",\"WARC-Concurrent-To\":\"<urn:uuid:49cada88-e488-4a0c-be26-c33e3b23df4e>\",\"WARC-IP-Address\":\"162.0.227.207\",\"WARC-Target-URI\":\"https://lionessays.com/i-need-some-math-assistance/\",\"WARC-Payload-Digest\":\"sha1:B3BCNIFDTOHPEUWGOBI5XAQRFI5U6KF6\",\"WARC-Block-Digest\":\"sha1:HPXHMP3IC5KTTLQY6QCZRDKIUOWKBKPD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662627464.60_warc_CC-MAIN-20220526224902-20220527014902-00082.warc.gz\"}"} |
https://en.m.wikiversity.org/wiki/QB/AstroWikipSolSys2 | [
"# QB/AstroWikipSolSys2\n\n< QB\n\nThe enrollment key for each course is 123. They are all is set to practice mode, giving students unlimited attempts at each question. Instructors can also print out copies of the quiz for classroom use. If you have any problems leave a message at user talk:Guy vandegrift.\n\n• Quizbank now resides on MyOpenMath at https://www.myopenmath.com (although I hope Wikiversity can play an important role in helping students and teachers use these questions!)\n• At the moment, most of the physics questions have already been transferred. To see them, join myopenmath.com as a student, and \"enroll\" in one or both of the following courses:\n• Quizbank physics 1 (id 60675)\n• Quizbank physics 2 (id 61712)\n• Quizbank astronomy (id 63705)\n\nCurrentID: -\n\nSee special:permalink/1863372 for a wikitext version of this quiz.\n\n### LaTexMarkup begin\n\nRequired images: [[file:Wikiversity-logo-en.svg|45px]]\n%This code creates both the question and answer key using \\newcommand\\mytest\n%%% EDIT QUIZ INFO HERE %%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\newcommand{\\quizname}{QB/AstroWikipSolSys2}\n\n\\newcommand{\\quiztype}{conceptual}%[[Category:QB/conceptual]]\n%%%%% PREAMBLE%%%%%%%%%%%%\n\\newif\\ifkey %estabkishes Boolean ifkey to turn on and off endnotes\n\n\\documentclass[11pt]{exam}\n\\RequirePackage{amssymb, amsfonts, amsmath, latexsym, verbatim,\nxspace, setspace,datetime}\n\\RequirePackage{tikz, pgflibraryplotmarks, hyperref}\n\\usepackage[left=.5in, right=.5in, bottom=.5in, top=.75in]{geometry}\n\\usepackage{endnotes, multicol,textgreek} %\n\\usepackage{graphicx} %\n\\singlespacing %OR \\onehalfspacing OR \\doublespacing\n\\parindent 0ex % Turns off paragraph indentation\n% BEGIN DOCUMENT\n\\begin{document}\n\\title{AstroWikipSolSys2}\n\\author{The LaTex code that creates this quiz is released to the Public Domain\\\\\nAttribution for each question is documented in the Appendix}\n\\maketitle\n\\begin{center}\n\\includegraphics[width=0.15\\textwidth]{666px-Wikiversity-logo-en.png}\n\\\\Latex markup at\\\\\n\\end{center}\n\\begin{frame}{}\n\\begin{multicols}{3}\n\\tableofcontents\n\\end{multicols}\n\\end{frame}\n\\pagebreak\\section{Quiz}\n\\keytrue\n\\begin{questions}\\keytrue\n\n\\question In astrophysics, what is accretion? \\ifkey\\endnote{ placed in Public Domain by Guy Vandegrift: {\\url{https://en.wikiversity.org/wiki/special:permalink/1863372}}}\\fi\n\\begin{choices}\n\\CorrectChoice the growth of a massive object by gravitationally attracting more matter\n\\choice the growth in size of a massive star as its outer atmosphere expands\n\\choice the growth of a comet's tail as it comes close to the Sun\n\\choice the increase in temperature and pressure of a star as it collapses from its own gravity\n\\choice the condensation of volatiles as a gas cools\n\\end{choices}\n\n\\question Dwarf planets are defined as objects orbiting the Sun and smaller than planets, that? \\ifkey\\endnote{ placed in Public Domain by Guy Vandegrift: {\\url{https://en.wikiversity.org/wiki/special:permalink/1863372}}}\\fi\n\\begin{choices}\n\\CorrectChoice have been rounded by their own gravity\n\\choice possess an atmosphere\n\\choice lack an atmosphere\n\\choice are too far from the Sun to be planets\n\\choice lie in the asteroid belt\n\\end{choices}\n\n\\question Dwarf planets have no natural satellites, \\ifkey\\endnote{ placed in Public Domain by Guy Vandegrift: {\\url{https://en.wikiversity.org/wiki/special:permalink/1863372}}}\\fi\n\\begin{choices}\n\\choice true\n\\CorrectChoice false\n\\end{choices}\n\n\\question Pluto is classified as \\ifkey\\endnote{ placed in Public Domain by Guy Vandegrift: {\\url{https://en.wikiversity.org/wiki/special:permalink/1863372}}}\\fi\n\\begin{choices}\n\\CorrectChoice a dwarf planet and a trans-Neptunian object.\n\\choice an asteroid belt object\n\\choice a dwarf planet with no natural satellites\n\\choice a natural satellite of Neptune\n\\choice a natural satellite of Uranus\n\\end{choices}\n\n\\question How many of the outer planets have rings? \\ifkey\\endnote{ placed in Public Domain by Guy Vandegrift: {\\url{https://en.wikiversity.org/wiki/special:permalink/1863372}}}\\fi\n\\begin{choices}\n\\CorrectChoice 4\n\\choice 3\n\\choice 2\n\\choice 1\n\\end{choices}\n\n\\question Currently there are approximately 8 billion people on Earth. For every person on Earth there will are approximately \\_\\_\\_ stars in the Milky Way galaxy. \\ifkey\\endnote{ placed in Public Domain by Guy Vandegrift: {\\url{https://en.wikiversity.org/wiki/special:permalink/1863372}}}\\fi\n\\begin{choices}\n\\CorrectChoice 20\n\\choice 2\n\\choice 200\n\\choice 2000\n\\end{choices}\n\n\\question The revolution of Haley's comet around the Sun is nearly circular. \\ifkey\\endnote{ placed in Public Domain by Guy Vandegrift: {\\url{https://en.wikiversity.org/wiki/special:permalink/1863372}}}\\fi\n\\begin{choices}\n\\choice true\n\\CorrectChoice false\n\\end{choices}\n\n\\question The revolution of Haley's comet around the Sun is opposite that of the 8 planets.\\ifkey\\endnote{ placed in Public Domain by Guy Vandegrift: {\\url{https://en.wikiversity.org/wiki/special:permalink/1863372}}}\\fi\n\\begin{choices}\n\\CorrectChoice true\n\\choice false\n\\end{choices}\n\n\\question The frost line is situated approximately \\ifkey\\endnote{ placed in Public Domain by Guy Vandegrift: {\\url{https://en.wikiversity.org/wiki/special:permalink/1863372}}}\\fi\n\\begin{choices}\n\\CorrectChoice 5 times as far from the Sun as the Earth is from the Sun\n\\choice 10 times as far from the Sun as the Earth is from the Sun\n\\choice 5 times as far from the Earth as the Earth's surface is from its center\n\\choice 10 times as far from the Earth as the Earth's surface is from its center\n\\end{choices}\n\n\\end{questions}\n\\newpage"
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https://jwcn-eurasipjournals.springeropen.com/articles/10.1186/1687-1499-2013-283 | [
"# Joint source-channel coding and optimization for mobile video streaming in heterogeneous wireless networks\n\n## Abstract\n\nThis paper investigates mobile video delivery in a heterogeneous wireless network from a video server to a multi-homed client. Joint source-channel coding (JSCC) has proven to be an effective solution for video transmission over bandwidth-limited, error-prone wireless networks. However, one major problem with the existing JSCC approaches is that they consider the network between the server and the client as a single transport link. The situation becomes more complicated in the context of multiple available links because involving a low-bandwidth, highly lossy, or long-delay wireless network in the transmission will only degrade the video quality. To address the critical problem, we propose a novel flow rate allocation-based JSCC (FRA-JSCC) approach that includes three key phases: (1) forward error correction redundancy estimation under loss requirement, (2) source rate adaption under delay constraint, and (3) dynamic rate allocation to minimize end-to-end video distortion. We present a mathematical formulation of JSCC to optimize video quality over multiple wireless channels and provide comprehensive analysis for channel distortion. We evaluate the performance of FRA-JSCC through emulations in Exata and compare it with the existing schemes. Experimental results show that FRA-JSCC outperforms the competing models in improving the video peak signal-to-noise ratio as well as in reducing the end-to-end delay.\n\n## 1 Introduction\n\nIn the past few years, mobile video streaming (e.g., Youtube and Hulu ) has become one of the most popular applications, and video traffic headed for handheld devices (e.g., smart cell phones and iPad) has experienced explosive growth. According to the Cisco Visual Index report, video streaming accounts for 57% of mobile data usage in 2012 and will reach 69% by the year 2017. Global mobile data is expected to increase 13-fold between 2012 and 2017. Furthermore, high-definition video has surpassed the standard definition video by the end of 2012 and will comprise 79% of video traffic by 2016.\n\nAlthough the proliferation of wireless infrastructures has offered the users with many access options (e.g., cellular networks, wireless local area network (WLAN), and Worldwide Interoperability for Microwave Access (WiMAX)), it is still a challenging problem to efficiently provide mobile video streaming services due to performance limitations of single wireless networks. Current WLAN systems cannot provide satisfactory quality of video streaming services due to the small coverage and relatively limited bandwidth as the number of mobile users increases [4, 5]. Even worse, WLAN systems are not robust enough to sustain user mobility [6, 7]. On the other hand, cellular networks, e.g., Universal Mobile Telecommunications System (UMTS) and HSDPA, can provide more robust wireless connections to mobile users. However, their bandwidth is not adequate to support high-quality video streaming with stringent bandwidth requirements . Although 4G LTE and WiMAX can provide a much higher peak data rate and extended coverage, they are not widely deployed yet and the bandwidth limitation will still become a problem as the wireless spectrum is shared by many users . The performance limitations of single wireless networks naturally turn research attentions to aggregate the bandwidth of heterogeneous wireless networks, and it has already attracted considerable research attentions . Conventionally, these bandwidth aggregation algorithms are designed for dynamically allocating video flows with seldom considerations in inherent channel errors and fluctuations, which can significantly impact on the streaming video quality [6, 11].\n\nTo address the challenging problems, joint source-channel coding (JSCC) has proven to be an effective solution in designing error-resilient wireless video transmission systems [12, 13]. However, one major problem with the existing JSCC approaches (e.g., [14, 15]) is that they consider the network between the server and the client as a single transport link . The problem becomes more complicated in the context of integrated heterogeneous wireless networks, in which multiple access networks may be simultaneously available. In , Jurca et al. studied on the physical path selection and source rate allocation for video streaming over multi-path networks, and experimental results show that video streaming through only certain reliable wireless networks gives better video quality than that through all possible wireless networks. The problem statement is presented in Figure 1, and it can be illustrated that involving an unreliable wireless access network in the transmission during the client mobility will only degrade the user-perceived video quality.\n\nMotivated by optimizing the JSCC for mobile video delivery in heterogeneous wireless networks, we propose a flow rate allocation-based JSCC (FRA-JSCC) approach in this work. By the term ‘flow rate allocation’, we mean dynamically picking the appropriate wireless access networks and assigning the transmission rates to each of them. First, the video source rate adaption scheme is designed to satisfy the delay requirements of real-time video applications. Second, forward error correction (FEC) redundancy estimation is performed to meet the tolerable loss rate. Third, a simple but effective search algorithm for flow rate allocation is proposed to minimize end-to-end video distortion. Specifically, the contributions of this paper can be summarized in the following:\n\n• An efficient end-to-end video delivery scheme in integrated heterogeneous wireless networks that uses JSCC in conjunction with flow rate allocation in order to improve the perceived video quality.\n\n• A mathematical model of JSCC to minimize the end-to-end video distortion over multiple wireless channels. The channel distortion is comprehensively analyzed with both transmission and overdue loss.\n\n• Extensive semi-physical emulations in Exata with the real-time H.264 video streaming. Experimental results show that (1) FRA-JSCC improves the average video peak signal-to-noise ratio (PSNR) by up to 3.5, 8.45, and 11 dB compared to the fountain code-based virtual path (FCVP) , joint multimedia-FEC rate (JMFR) , and dynamic multi-path (DMP) ; (2) FRA-JSCC reduces the average end-to-end delay by up to 20.8, 11.5, and 40.3 ms compared to the FCVP, JMFR, and DMP; (3) FRA-JSCC mitigates the effective loss rate by up to 6.05%, 10.5%, and 15.5% compared to the FCVP, JMFR, and DMP.\n\nThe remainder of this paper is organized as follows: in Section 2, we briefly discuss the related work. Section 3 presents the system model and problem formulation. In Section 4, we describe the design of the proposed FRA-JSCC in detail. The performance evaluation is provided in Section 5. Conclusion remarks are given out in Section 6. The basic notations used throughout this paper are listed in Table 1.\n\n## 2 Related work\n\nThe related work to this paper can be generally categorized into two branches: joint source-channel coding and video delivery in heterogeneous wireless networks. We will discuss on each topic respectively in this section.\n\n### 2.1 Joint source-channel coding\n\nIn summary, the JSCC problem includes joint coding and optimal rate calculation for video coding and channel coding, which provides various protection level to the video data according to its level of importance and channel conditions. Most of the related work in video transmission focus on (1) finding an optimal bit rate for video coding and channel coding, e.g., [19, 20]; (2) designing the video coding mechanism to achieve the target source rate under given channel conditions, e.g., ; (3) designing the channel coding to achieve the required reliability, e.g., low-density parity check , turbo , Reed-Solomon (RS) , and fountain codes; (4) designing joint optimization framework, including all available error control components together with error concealment and transmission control, to improve global system performance, e.g., . The authors of deal with the optimal allocation of MPEG-2 encoding and media-independent forward error correction rates under the total given bandwidth. They define optimality in terms of minimum perceptual distortion given a set of video and network parameters. They compute the network error parameters after FEC decoding and derive the global set of equations that lead to optimal dynamic rate allocation. In a more recent work , Ji et al. studied on the optimization approach of JSCC for layered video broadcasting to heterogeneous devices. The objective is to achieve maximum overall receiving quality of the heterogeneous quality of service (QoS) receivers.\n\nAll these works consider the network as a single transport link between the server and the client. They do not address multi-path streaming scenarios, where more than one network path is allocated to the application. Different from previous JSCC approaches, Jurca et al. researched on the optimal FEC scheme and layer selection in multi-path scenario. This approach uses a multi-layer coded video stream, and the base-layer stream is protected by duplicated transmission using multiple physical paths during the handoff. However, the major flaw is that it is generally under the assumption that all the wireless networks are reliable for improving the overall video quality and thus lacks effective network selection algorithm.\n\n### 2.2 Video delivery in heterogeneous wireless networks\n\nVideo delivery in heterogeneous wireless networks has recently attracted much attention, and the general review can be referred to [27, 28]. In the Earliest Delivery Path First algorithm, it takes into account the available bandwidth, propagation delay, and video frame size to estimate the arrival time and aims to find an earliest path to deliver the video packet. The load balancing algorithm (LBA) performs stream adaption in response to varying network status by only transmitting those packets which are estimated to arrive at the client within the decoding deadline and conserves bandwidth by dropping packets that cannot be decoded because they rely on previous packets that have been dropped. A packet prioritization scheme in LBA gives a higher weight to I frames over B and P frames and also to base layer packets over enhancement layer packets. The LBA scheduler sorts packets according to priority weighting and sacrifices lower priority packets to ensure the delivery of those with a higher priority. Song et al. propose a probabilistic multi-path transmission (PMT) scheme, which sends video traffic bursts over multiple available channels based on a probability generation function of packet delay. PMT is not robust to client mobility as it does not dynamically adjust the flow splitting probability according to time-varying channel status. Han et al. proposed an end-to-end virtual path construction system over heterogeneous wireless networks based on fountain code. The goal of this system is to maximize the encoding bit rate on the basis of aggregate bandwidth as well as overcoming the channel loss. However, the big block size of fountain code will lead to a long delay, which is not appropriate for real-time video streaming over the bandwidth-limited and time-varying wireless networks.\n\nBesides, encoded multi-path streaming (EMS) and multi-path loss tolerant (MPLOT) are typical protocols exploiting path diversity in wired/wireless multi-path networks based on erasure code. EMS scheme splits traffic loads over multiple paths according to the path loss rate and dynamically adjusts FEC redundancy. However, EMS was generally under the assumption that all the available paths could be beneficial for the transmission as in . MPLOT is a transport protocol that aims at maximizing the throughput of the upper layer application. However, MPLOT cannot guarantee real-time video delivery as it does not address tight delay constraints.\n\n## 3 System model and problem formulation\n\nThe system model for the proposed FRA-JSCC is depicted in Figure 2. We consider a scenario of a heterogeneous wireless network integrating",
null,
"wireless connections from a single video server to a single destination node. This system involves the models for network path, end-to-end video distortion, and forward error correction. Parts of this section describe each of them.\n\n### 3.1 Network model\n\nThe end-to-end connection from the video server to the wireless interface of the mobile client is considered as an independent physical path which includes the wired and wireless domains. It is well known that the wireless access is most likely to be the bottleneck link for the end-to-end transmission due to the limited bandwidth and time-varying channel status. The transmission data packets may encounter loss due to buffer overflow in immediate routers or erasures caused by channel fading in the error-prone wireless channels. In order to simplify the discussion, we generally consider a packet to be lost due to the link fault either in the wired/wireless packet switching networks. Each physical path P r is associated with the following metrics:\n\n• Available bandwidth μ r (expressed in Kbps). μ r captures the variation of background traffic and bandwidth fluctuation.\n\n• Propagation delay t r which includes the link delays of the wired and wireless networks.\n\n• Average loss probability ${\\pi }_{B}^{r}\\in \\left[0,1\\right]$, assumed to be an i.i.d process and independent of the video streaming rate.\n\nWe model the burst loss behavior on each physical path by the continuous-time Gilbert model. It is a two-state stationary continuous time Markov chain. The state ${\\mathcal{X}}_{r}\\left(t\\right)$ assumes one of two values: G (good) or B (bad). If a packet is sent at time t with ${\\mathcal{X}}_{r}\\left(t\\right)=G$, then the packet can be successfully delivered. Otherwise, when ${\\mathcal{X}}_{r}\\left(t\\right)=B$, then the packet is lost.\n\nWe denote by ${\\pi }_{G}^{r}$ and ${\\pi }_{B}^{r}$ the stationary probabilities that P i is good or bad. Let ${\\xi }_{B}^{r}$ and ${\\xi }_{G}^{r}$ represent the transition probability from G to B and B to G, respectively. In this work, we adopt two system-dependent parameters to specify the continuous-time Markov chain packet loss model: (1) the average loss rate ${\\pi }_{B}^{r}$ and (2) the average loss burst length $1/{\\xi }_{B}^{r}$. Then, we can have\n\n${\\pi }_{G}^{r}=\\frac{{\\xi }_{B}^{r}}{{\\xi }_{B}^{r}+{\\xi }_{G}^{r}}\\text{and}{\\pi }_{B}^{r}=\\frac{{\\xi }_{G}^{r}}{{\\xi }_{B}^{r}+{\\xi }_{G}^{r}}.$\n(1)\n\nThe available bandwidth and propagation delay of each wireless network can be estimated by packet probing mechanisms (e.g., the pathChirp algorithm employed in this work) over each interface of the mobile client. The loss parameters ${\\pi }_{B}^{r}$ and ${\\xi }_{B}^{r}$ can be sensed through control protocols or delay measurements .\n\n### 3.2 Video distortion model\n\nIn this subsection, we introduce a generic video distortion model . The end-to-end distortion (Dtotal), perceived by the end user, can generally be computed as the sum of the source distortion (Dsrc) and the channel distortion (Dchl). Overall, the end-to-end distortion can thus be written as\n\n${D}_{\\text{total}}={D}_{\\text{src}}+{D}_{\\text{chl}}.$\n(2)\n\nThe video quality depends on both the distortion due to a lossy encoding of the media information and the distortion due to losses experienced in the network. Dsrc is mostly determined by the video source rate and the video sequence parameters (e.g., for the same encoding bit rate, the more complex the sequence, the higher the source distortion). The source distortion decays with increasing encoding rate; the decay is quite steep for low bit rate values, but it becomes very slow at high bit rate. The channel distortion is dependent on the effective loss rate ${\\pi }_{B}^{\\ast }$, which is caused by the transmission loss and expired arrivals of video packets. It is roughly proportional to the number of video frames that cannot be decoded correctly. Hence, we can explicitly formulate Dtotal (in units of mean square error) as\n\n${D}_{\\text{total}}=\\underset{{D}_{\\text{src}}}{\\underset{⏟}{{D}_{0}+\\frac{\\alpha }{V-{V}_{0}}}}+\\underset{{D}_{\\text{chl}}}{\\underset{⏟}{\\beta ×{\\pi }_{B}^{\\ast }}}\\phantom{\\rule{0.3em}{0ex}},$\n(3)\n\nin which α, V0, D0, and β are constants for a specific video codec and video sequence. These parameters can be estimated from three or more trial encodings using nonlinear regression techniques. To allow fast adaptation of the flow rate allocation to abrupt changes in the video content, these parameters can be updated for each group of pictures (GOP) in the encoded video sequence . Since this model takes into account the effects of intra-coding and spatial loop filtering, it provides accurate estimates for end-to-end distortion .\n\n### 3.3 Forward error correction\n\nIn this work, we use the systematic RS block erasure code for video data protection against channel losses. Generically, a FEC block of n data packets contains k source packets and n-k redundant packets. Usually, the receiver can fully reconstruct the original k data packets if at least k packets of the FEC block are correctly received. In FEC (n,k) code, for every k source packets, (n-k) redundant data packets are introduced to make up a codeword of packets. As long as a client receives at least k out of the n data packets, it can recover all the source packets. If the number of received packet is less than k, the arrival source packets can still be used to contribute to the video decoding process because they have been kept intact by the RS encoding process. In general, for the same code rate k/n, increasing the value of n would enhance the performance of RS code. The FEC code rate n/k needs to be dynamically chosen based on the loss requirement and channel status.\n\nPractically, the frame-level , GOP-level , or sub-GOP-level FEC coding is often applied for video data protection. In this work, we implement the GoP-level FEC coding (see Figure 3) in order to seamlessly integrate with the source rate adaption mechanism.\n\n### 3.4 Effective loss rate\n\nThe effective loss rate ${\\pi }_{B}^{\\ast }$ represents the combined rates of the lost packets due to channel losses and expired arrivals, i.e.,\n\n${\\pi }_{B}^{\\ast }={\\pi }_{\\text{tran}}^{\\ast }+\\left(1-{\\pi }_{\\text{tran}}^{\\ast }\\right)×{\\pi }_{\\text{over}}^{\\ast }.$\n(4)\n\nFor real-time video applications, each video frame is associated with a decoding deadline. This deadline sets a maximum delay bound for a frame to be successfully delivered to the client in order to contribute to the decoding process. Next, we will provide a comprehensive analysis for the transmission and overdue loss, respectively.\n\n#### 3.4.1 Transmission loss rate\n\nLet c denote a n-tuple which represents a particular failure configuration. If the i th FEC data packet is lost during the transmission, then c i = B. By taking into account all the possible configurations, we can compute the transmission loss rate as\n\n${\\pi }_{\\text{tran}}^{\\ast }=\\frac{1}{k}\\sum _{\\text{all}c}\\mathcal{ℒ}\\left(c\\right)×\\mathbb{P}\\left(c\\right),$\n(5)\n\nin which $0<\\mathcal{ℒ}\\left(c\\right) is the number of lost source packets for a given c. For the systematic FEC(n,k) we can have\n\n$\\mathcal{ℒ}\\left(c\\right)=\\left\\{\\begin{array}{ll}0& \\text{if}{\\sum }_{i=1}^{k}{1}_{{c}^{i}=B}\\le n-k,\\\\ {\\sum }_{i=1}^{k}{1}_{{c}^{i}=B}& \\text{otherwise}.\\end{array}\\right\\$\n(6)\n\nAs the",
null,
"physical paths to the multi-homed client are independent of each other, we can compute $\\mathbb{P}\\left(c\\right)$ as follows:\n\n$\\mathbb{P}\\left(c\\right)=\\prod _{r=1}^{\\mathcal{R}}\\left({\\varphi }_{r}×\\mathbb{P}\\left({c}^{r}\\right)\\right),$\n(7)\n\nwhere $\\mathbb{P}\\left({c}^{r}\\right)$ is the probability of a failure configuration cron P r and ϕ r is an element of the selection vector ($\\Phi ={\\varphi }_{1},\\dots ,{\\varphi }_{\\mathcal{R}}$) for wireless access networks. ϕ r is defined by\n\n${\\varphi }_{r}=\\left\\{\\begin{array}{ll}1& \\text{if the}r\\text{th wireless access network is picked},\\\\ 0& \\text{otherwise}.\\end{array}\\right\\$\n\nLet ${p}_{i,j}^{r}\\left(\\theta \\right)$ denote the probability of the transition from state i to j on P r in time θ, then we can have\n\n${p}_{i,j}^{r}\\left(\\theta \\right)=\\mathbb{P}\\left[{X}_{r}\\left(\\theta \\right)=j|{X}_{r}\\left(0\\right)=i\\right].$\n(8)\n\nFor the classic Markov chain analysis, we can have\n\n$\\begin{array}{l}{p}_{G,G}^{r}\\left(\\theta \\right)={\\pi }_{G}^{r}+{\\pi }_{B}^{r}×\\kappa ,\\phantom{\\rule{1em}{0ex}}{p}_{G,B}^{r}\\left(\\theta \\right)={\\pi }_{B}^{r}-{\\pi }_{B}^{r}×\\kappa ,\\\\ {p}_{B,G}^{r}\\left(\\theta \\right)={\\pi }_{G}^{r}-{\\pi }_{G}^{r}×\\kappa ,\\phantom{\\rule{1em}{0ex}}{p}_{B,B}^{r}\\left(\\theta \\right)={\\pi }_{B}^{r}+{\\pi }_{G}^{r}×\\kappa ,\\end{array}$\n(9)\n\nin which $\\kappa =exp\\left[-\\left(\\underset{B}{\\overset{r}{\\xi }}+{\\xi }_{G}^{r}\\right)×\\theta \\right]$. We assume that each element in the vector $N=\\left\\{{n}_{1},{n}_{2},\\dots ,{n}_{\\mathcal{R}}\\right\\}$, $\\sum _{r}{n}_{r}=n$ represents the number of packets dispatched onto each physical path. Now, the value of $\\mathbb{P}\\left({c}^{r}\\right)$ can be computed as follows\n\n$\\mathbb{P}\\left({c}^{r}\\right)={\\varphi }_{r}×{\\pi }_{{c}_{1}^{r}}^{r}\\prod _{i=1}^{{n}_{r}-1}{p}_{{c}_{i}^{r},{c}_{i+1}^{r}}^{r}\\left({\\theta }^{r}\\right).$\n(10)\n\nAfter a sequence of algebraic computations, we can obtain\n\n${\\pi }_{\\text{tran}}^{\\ast }=\\frac{1}{k}\\sum _{\\text{all}c}\\mathcal{ℒ}\\left(c\\right)\\prod _{r=1}^{\\mathcal{R}}{\\pi }_{{c}_{1}^{r}}^{r}\\prod _{i=1}^{{n}_{r}-1}{p}_{{c}_{i}^{r},{c}_{i+1}^{r}}^{r}\\left({\\theta }^{r}\\right).$\n(11)\n\nThe above equation allows us to compute the transmission loss rate of a specific scheduling approach. Based on the Equation 11 in , we can obtain the expected value of ${\\pi }_{\\text{tran}}^{\\ast }$ as in Equation 12,\n\n$\\begin{array}{l}\\begin{array}{l}\\phantom{\\rule{1em}{0ex}}\\phantom{\\rule{0.3em}{0ex}}{\\pi }_{\\text{tran}}^{\\ast }=\\frac{1}{k}\\phantom{\\rule{0.3em}{0ex}}\\sum _{j=n-k+1}^{n}\\phantom{\\rule{0.3em}{0ex}}\\sum _{\\begin{array}{l}0\\le {j}_{1},\\phantom{\\rule{0.6em}{0ex}}.\\phantom{\\rule{0.3em}{0ex}}.\\phantom{\\rule{0.3em}{0ex}},{j}_{\\mathcal{N}}\\le j\\\\ {j}_{1}+.\\phantom{\\rule{0.3em}{0ex}}.+{j}_{\\mathcal{N}}=j\\end{array}}\\left(\\prod _{r=1}^{\\mathcal{R}}{\\varphi }_{r}\\left({\\pi }_{G}^{r}×\\mathbb{P}\\left(\\left[\\begin{array}{l}{n}_{r}-1\\\\ {j}_{r}\\end{array}\\right]|G\\right)+{\\pi }_{B}^{r}×\\mathbb{P}\\left(\\left[\\begin{array}{l}{n}_{r}-1\\\\ {j}_{r}-1\\end{array}\\right]|B\\right)\\right)\\right)×\\\\ \\left(\\sum _{r=1}^{\\mathcal{R}}{\\varphi }_{r}×\\sum _{i=0}^{{k}_{r}}i×\\frac{{\\pi }_{G}^{r}×\\mathbb{P}\\left(\\left[\\begin{array}{l}{k}_{r}-1\\\\ i\\end{array}\\right]|G\\right)×\\mathbb{P}\\left(\\left[\\begin{array}{l}{n}_{r}-{k}_{r}\\\\ {j}_{r}-i\\end{array}\\right]|G\\right)+{\\pi }_{B}^{r}×\\mathbb{P}\\left(\\left[\\begin{array}{l}i-1\\\\ {k}_{r}-1\\end{array}\\right]|B\\right)×\\mathbb{P}\\left(\\left[\\begin{array}{l}{n}_{r}-{k}_{r}\\\\ {j}_{r}-i\\end{array}\\right]|B\\right)}{{\\pi }_{G}^{r}×\\mathbb{P}\\left(\\left[\\begin{array}{l}{n}_{r}-1\\\\ {j}_{r}\\end{array}\\right]|G\\right)+{\\pi }_{B}^{r}×\\mathbb{P}\\left(\\left[\\begin{array}{l}{n}_{r}-1\\\\ {j}_{r}-1\\end{array}\\right]|B\\right)}\\right),\\end{array}\\end{array}$\n(12)\n\nin which $\\mathbb{P}\\left(\\left[\\begin{array}{l}{n}_{r}-1\\\\ {j}_{r}-1\\end{array}\\right]|q\\right),\\phantom{\\rule{0.3em}{0ex}}q\\in \\left\\{G,B\\right\\}$ denotes that any b out of a consecutive packets are lost given that this block is preceded by a packet which is state q. The detailed computations of $\\mathbb{P}\\left(\\left[\\begin{array}{l}{n}_{r}-1\\\\ {j}_{r}-1\\end{array}\\right]|q\\right),\\phantom{\\rule{0.3em}{0ex}}q\\in \\left\\{G,B\\right\\}$ can be referred to .\n\n#### 3.4.2 Overdue loss rate\n\nThe end-to-end packet delay over a single wireless network (d r ) is dominated by the queueing delay at the bottleneck link, and it can be approximated by an exponential distribution , i.e.,\n\n$\\mathbb{P}\\left\\{{d}_{r}>\\mathcal{T}\\right\\}\\approx \\frac{1}{\\sqrt[2\\pi ]{}}exp\\left(-\\frac{\\mathcal{T}}{{d}_{r}}\\right),$\n(13)\n\nin which",
null,
"denotes the maximum delay constraint that prevents the playback buffer starvation. Now, we calculate the value of d r . It can be obtained with the following equation:\n\n${d}_{r}={t}_{r}+\\frac{{\\mu }_{r}\\left(1-{\\pi }_{r}\\right)}{{n}_{r}×S},$\n(14)\n\nwhere μ r (1 - π r ) represents the ‘loss-free’ bandwidth of P r and S represents the packet payload size. Then, the probability for expired arrival of packets can be obtained with\n\n$\\mathbb{P}\\left\\{{d}_{r}>\\mathcal{T}\\right\\}\\approx \\frac{1}{\\sqrt[2\\pi ]{}}exp\\left\\{\\frac{\\mathcal{T}×{n}_{r}×S}{{t}_{r}×{n}_{r}×S+{\\mu }_{r}\\left(1-{\\pi }_{r}\\right)}\\right\\}.$\n(15)\n\nThe overdue loss rate can be obtained with the equation of\n\n$\\begin{array}{ll}{\\pi }_{\\text{over}}^{\\ast }=\\phantom{\\rule{0.3em}{0ex}}& \\frac{\\sum _{r=1}^{\\mathcal{R}}{n}_{r}×{\\varphi }_{r}×\\mathbb{P}\\left\\{{d}_{r}>\\mathcal{T}\\right\\}}{n},\\\\ =\\phantom{\\rule{0.3em}{0ex}}\\frac{1}{\\sqrt[2\\pi ]{}×n}\\sum _{r=1}^{\\mathcal{R}}{n}_{r}×{\\varphi }_{r}\\\\ \\phantom{\\rule{1em}{0ex}}×exp\\left\\{\\frac{\\mathcal{T}×{n}_{r}×S}{{t}_{r}×{n}_{r}×S+{\\mu }_{r}\\left(1-{\\pi }_{r}\\right)}\\right\\}.\\end{array}$\n(16)\n\n### 3.5 Problem formulation\n\nWe are now ready to formulate the problem of flow rate allocation combining the JSCC for video delivery in heterogeneous wireless networks. Note that it is not practical for the video encoder to trace the frequent variation in source rate. Therefore, we adapt the source rate in units of GOP, based on the channel status, FEC code rate, and delay requirements. To allow fast adaptation of the source rate to abrupt changes in the video content, this parameter is updated for each GOP in the encoded video sequence, typically once every 0.25 s (with J = 8 frames, F = 30 fps). The objective is to minimize the summation of the total distortion Dtotal subject to loss, delay and bandwidth constraints:\n\n$\\begin{array}{l}\\text{For each GOP, determine the value of}\\Phi ,\\phantom{\\rule{0.3em}{0ex}}\\Omega ,\\phantom{\\rule{0.3em}{0ex}}V,\\phantom{\\rule{0.3em}{0ex}}n\\\\ \\text{tominimize}\\\\ {D}_{\\text{total}}=\\underset{{D}_{\\text{src}}}{\\underset{⏟}{{D}_{0}+\\frac{\\alpha }{V-{V}_{0}}}}+\\underset{{D}_{\\text{chl}}}{\\underset{⏟}{\\underset{{D}_{\\text{tran}}}{\\underset{⏟}{\\beta ×{\\pi }_{\\text{tran}}^{\\ast }}}+\\underset{{D}_{\\text{over}}}{\\underset{⏟}{\\beta ×\\left(1-{\\pi }_{\\text{tran}}^{\\ast }\\right)×{\\pi }_{\\text{over}}^{\\ast }}}}},\\end{array}$\n$\\text{subject to:}\\left\\{\\begin{array}{l}V×n/k×\\frac{{\\omega }_{r}}{{\\sum }_{i=1}^{\\mathcal{R}}{\\omega }_{i}}<{\\mu }_{r},\\phantom{\\rule{0.3em}{0ex}}\\text{for}1\\le r<\\mathcal{R},\\\\ V×n/k\\le {\\sum }_{r=1}^{\\mathcal{R}}{\\mu }_{r},\\\\ {\\pi }_{\\text{tran}}^{\\ast }+\\left(1-{\\pi }_{\\text{tran}}^{\\ast }\\right)×{\\pi }_{\\text{over}}^{\\ast }<\\Delta ,\\\\ {\\pi }_{\\text{tran}}^{\\ast }=\\text{Equation 12},\\\\ {\\pi }_{\\text{over}}^{\\ast }=\\text{Equation 16}.\\end{array}\\right\\$\n(17)\n\nThis is a nonlinear optimization problem with linear constraints. With regard to the computational cost and convergence, it is impractical to derive the exact solution for the minimal video distortion. In the next section, we will show how to resolve this optimization problem in the design of the proposed FRA-JSCC.\n\n## 4 Design of flow rate allocation-based joint source-channel coding\n\nIn this section, we describe the overall design of the proposed FRA-JSCC and outline the functionality of its major components. The system design is presented in Figure 4, and it includes components implemented in both the server and client side, respectively. In order to solve the optimization problem (17), the proposed FRA-JSCC performs the following working steps at the server side: (1) FEC redundancy estimation, (2) source rate adaption, and (3) flow rate allocation. Specifically, the value of FEC redundancy ((n - k)/k) and video source rate (V) is based on the rate allocation vector (Φ). The input and feedback information (e.g., the loss, delay constraints, and channel status) is necessary for the computation steps. The loss and delay requirement is imposed by the video application in order to achieve the required QoS. The encoded video streaming is split among multiple available wireless networks at the weighted round robin distributor, and the packet transmitter is responsible for dispatching the FEC data packets onto different channels.\n\nAt the client side, the video frames will be stored in the playback buffer after the FEC decoding process. The inter-frame resequence step aims at reordering the video frames in case they arrive at the client out-of-order. As each video frame is associated with a decoding deadline, the overdue frames will be discarded and concealed by copying from the last received ones. Next, we will describe the key components in the system design and their working steps.\n\n### 4.1 FEC redundancy estimation\n\nFor estimating the FEC redundancy, we model the multiple wireless networks as a single virtual link with effective loss rate ${\\pi }_{B}^{\\ast }$. Consider the transmission of k FEC packets (each of size S) over the virtual link from the source to the destination. Let (n - k)/k denote the redundancy (i.e., the fraction of redundant FEC packets in the FEC block). There is an inherent tradeoff between FEC redundancy and its error correction power . With more redundant packets, the receiver can recover from more severe losses, at the cost of larger end-to-end delays and higher loads imposed on networks. Therefore, in the design of FRA-JSCC, the goal is to use ‘just enough’ FEC redundancy to meet the video application’s loss requirement (Δ). With this objective, the FEC adaption policy can be derived under fairly general assumptions by simply bounding the loss tail probability.\n\nTherefore, the FEC redundancy estimation problem can be stated as\n\n$n=\\text{arg min}\\left\\{\\text{diff}\\left(\\Delta -\\underset{B}{\\overset{\\ast }{\\pi }}\\right)\\right\\},$\n(18)\n\nin which\n\n$\\text{diff}\\left(\\Delta -{\\pi }_{B}^{\\ast }\\right)=\\left\\{\\begin{array}{ll}\\Delta -{\\pi }_{B}^{\\ast }& \\text{if}{\\pi }_{B}^{\\ast }<\\Delta ,\\\\ \\infty & \\text{otherwise},\\end{array}\\right\\$\n(19)\n\nand ${\\pi }_{B}^{\\ast }$ can be estimated using Equation 12. Therefore, the FEC redundancy can be obtained i.f.f Φ is determined.\n\nAccording to the information theory , video source distortion can be minimized by increasing the effective encoding rate. On the other hand, the increasing encoding rate will lead to higher transmission rate which imposes heavier load on channels. If the imposed load exceeds the network capacity, it will in turn result in longer delay and packet loss due to network congestion. There is an inherent conflict between the source and channel distortion. Therefore, the critical point in source rate adaption is to find the upper bound under application and channel constraints. The constraint imposed by video applications is the delay requirements. In real-time video applications, delay plays a vital role in enhancing streaming video quality. If a video frame arrives at a destination past the decoding deadline, it is considered lost. In this paper, we propose a source rate adaption algorithm under delay requirements, taking into account FEC redundancy carried out in the last subsection.\n\nThe maximum number of packets that can be transmitted through the r th wireless network in the tolerable maximum delay",
null,
"is calculated by\n\n${\\omega }_{r}=⌊\\frac{{\\mu }_{r}×\\left(1-{\\pi }_{r}\\right)×\\left(\\mathcal{T}-{t}_{r}\\right)}{S}⌋,\\phantom{\\rule{0.3em}{0ex}}\\text{for}1\\le r\\le \\mathcal{R.}$\n(20)\n\nwhere x denotes the largest integer less than x. Now, we set the weighting factor for the r th wireless channel of the weighted round robin distributor to ω r × ϕ r for $1\\le r\\le \\mathcal{R}$. The proposed joint source and FEC control scheme calculates the FEC decoding failure rate based on the effective loss rate to determine the code rate. First, we define the maximum number of packets which can be transmitted using the constructed virtual link by\n\n$\\Theta =\\sum _{r=1}^{\\mathcal{R}}{\\varphi }_{r}×{\\omega }_{r}.$\n(21)\n\nThen, the duration of a GOP to be displayed at the client side can be obtained with J/F, in which J is the number of frames in a GOP and F is the video frame rate (in terms of frames per second). The number of bytes within a GOP after being encoded within the duration is Θ × S × k/n, where k/n denotes the FEC code rate. Consequently, the resulting maximum video source rate V for a FEC block is determined by the equation\n\n$V=\\frac{\\Theta ×S×k/n}{J/F}.$\n(22)\n\n### 4.3 Flow rate allocation\n\nThe source packets together with the redundancy packets consist of the ‘flow’ mentioned throughout this paper. The goal of the flow rate allocation is to select appropriate wireless access networks out of all the candidates so as to minimize end-to-end video distortion.\n\n$\\phantom{\\rule{-12.0pt}{0ex}}\\text{Minimize:}{D}_{\\text{total}}\\left(\\Phi \\right)={D}_{0}+\\frac{\\alpha }{V\\left(\\Phi \\right)-{V}_{0}}+\\beta ×{\\pi }_{B}^{\\ast }\\left(\\Phi \\right).$\n(23)\n\nUntil now, we have obtained the expressions of n and V. According to the Theorem 1 in , the optimal flow rate allocation solution takes the form of a consecutive series of 1’s, followed by a consecutive series of 0’s, i.e., Φ = [1,1,…,1,0,0,…,0]. Indeed, the inclusion of a wireless access network with high loss rate, long propagation delay, or low bandwidth can theoretically increase Dtotal because more FEC redundancy may be required to compensate for the increased uncertainty. In order to find the optimal solution, we first rank all the available wireless networks according to their ‘loss-free’ bandwidth (μ r (1 - π r )), which has proven to be a good indicator of the network path quality . Then, the optimal flow rate allocation vector can be obtained with a simple but effective search algorithm, i.e.,",
null,
"Practically, a mobile device has a small number of network interfaces due to the limited battery life, mobility, cost, etc. Thus, the computational complexity required for the proposed flow rate allocation algorithm is negligible although the full search method is used.\n\n### 4.4 Channel status monitoring\n\nEstimating channel status information based on end-to-end monitoring has been attracting research attention for years. Over heterogeneous wireless networks, it is very important to identify the physical characteristics of each wireless channel in order to utilize channel resources efficiently. The available bandwidth, propagation delay, and channel loss rate are especially important properties for a high-quality video streaming service. So far, numerous algorithms have been proposed to estimate the available bandwidth over wired/wireless networks in the literature [31, 41, 42]. In this paper, the pathChirp algorithm is employed to estimate the available bandwidth through each wireless network with high accuracy and efficiency. During the transient state, a server sends some probe packets with exponentially distributed intervals through each wireless network interface when a video request arrives from a client. Based on the probe packet arrival intervals, the client estimates the available bandwidth using the pathChirp algorithm (for detailed descriptions, please refer to ). In the steady state, video data packets are transmitted at a fixed interval, and the client continuously monitors the packet arrival intervals in a sliding window and estimates the available bandwidth based on these intervals. We can easily calculate the propagation delay by the time stamp in each packet header. Now, we can obtain the following information\n\n$\\begin{array}{l}\\mathbit{\\mu }=\\left\\{{\\mu }_{1},{\\mu }_{2},\\dots ,{\\mu }_{\\mathcal{R}}\\right\\},\\phantom{\\rule{0.3em}{0ex}}\\mathbit{\\pi }=\\left\\{{\\pi }_{B}^{1},{\\pi }_{B}^{2},\\dots ,{\\pi }_{B}^{\\mathcal{R}}\\right\\},\\\\ \\text{and}\\phantom{\\rule{0.3em}{0ex}}\\mathbit{t}=\\left\\{{t}_{1},{t}_{2},\\dots ,{t}_{\\mathcal{R}}\\right\\}.\\end{array}$\n\nA client periodically reports information on each physical path to a parameter control unit of a server through the most reliable uplink channel. This information is used to determine the results of FEC parameter tuning, source rate adaption, and source rate allocation in the system design. The procedures of the proposed FRA-JSCC are presented in Algorithm 1.\n\n#### Algorithm 1 Flow Rate Allocation based Joint Source Channel Coding.",
null,
"## 5 Performance evaluation\n\nIn this section, we evaluate the efficacy of the proposed FRA-JSCC by comparing it with the existing schemes for video delivery over heterogeneous wireless networks. We first describe the emulation methodology that includes the emulation setup, reference schemes, performance metrics, and emulation scenario.\n\n### 5.1 Emulation methodology\n\n#### 5.1.1 Emulation setup\n\nWe adopt the Exata and Joint Scalable Video Model (JSVM) as the network emulator and video codec, respectively. The architecture of evaluation system is presented in Figure 5, and the main configurations are set as follows:\n\n• Exata 2.1 is used as the network emulator. Exata is an advanced edition of QualNet in which we can perform semi-physical emulations. In order to implement the real video streaming-based emulations, we integrate the source code of JSVMa (as Objective File Library (.LIB)) with Exata and develop an application layer protocol of ‘Video Transmission’. The detailed descriptions of the development steps could be referred to Exata Programmer’s Guide . In the emulation topology, the video server has one wired network interface and the mobile client has three wireless network interfaces, i.e., cellular, WLAN, and WiMAX. We can construct an end-to-end connection to a specific wireless network interface by binding a pair of IP addresses from the server and the client. The configurations of the emulated background traffic in the wired networks are listed in Table 2. The server and client are mapped to real computers, which are connected to the emulation server through the Exata Connection Manager. The IEEE 802.11b is adopted as the WLAN protocol. The configurations of heterogeneous wireless networks are summarized in Table 3[4, 5, 45].\n\n• H.264/SVC reference software JSVM 9.18 is adopted as the video encoder. The generated video streaming is encoded at 30 frames per second and a GOP consists of 8 frames. The test video sequences are Foreman, Mother & Daughter, Hall, and Container in QCIF (quarter common interchange format) with 300 frames. Each of the sequences features a different pattern of temporal motion and spatial characteristics which is reflected in their corresponding video quality versus encoding rate dependencies. We concatenate them 10 times to be 3,000 frames long in order to obtain statistically meaningful results. The loss requirement (Δ) and delay constraint (",
null,
") are set to 1% and 250 ms, respectively.\n\n#### 5.1.2 Reference schemes\n\nWe compare the performance of FRA-JSCC with the following schemes for video delivery in heterogeneous wireless networks:\n\n• FCVP []. As the system proposed in aims at exploiting the path diversity in heterogeneous wireless networks based on fountain code, we name it fountain code-based virtual path construction system. In the implementation of FCVP, the control parameters were updated for every 0.5 s. The symbol and packet size is set to be 8 and 512 bytes, respectively.\n\n• JMFR []. The joint multimedia-FEC rate allocation scheme computes the optimal source and FEC rate for scalable video over multi-path networks based on the utility algorithm. The number of video layers is set to be 1 in all the emulations.\n\n• DMP []. The dynamic multi-path streaming utilizes multiple paths by maintaining a transmission control protocol (TCP) connection on each path. The sender puts the data packets in a single sender queue. At any time, only one TCP connection can gain the access to the sender queue. The winning TCP connection will keep sending data until the connection is blocked. Another available TCP connection will then gain the access to the sender queue and continue sending data. In order to fairly compare the performance with other competing models, we dynamically adjust the video encoding rate based on the aggregate bandwidth of the available links.\n\n#### 5.1.3 Performance metrics\n\nWe adopt the following performance metrics to evaluate the proposed approach against the above competing approaches:\n\n• PSNR. Peak signal-to-noise ratio is a standard metric of video quality and is a function of the mean square error between the original and the received video frames. If a video frame is lost or past the deadline, it is considered lost but may be concealed by copying from the last received frame before it.\n\n• Average end-to-end delay. The end-to-end delay of a video frame consists of delay in the network and the resequencing time at the client. It is counted from the generation time of a video frame to the time when it can be decoded.\n\n• Effective loss rate. As introduced in Section 3.4, the effective loss rate ${\\pi }_{B}^{\\ast }$ includes the transmission and overdue loss. PSNR measures video quality after error concealment for the lost video frames. We measure the effective loss rate to testify the competing models in mitigating the packet loss.\n\n#### 5.1.4 Emulation scenario\n\nWe conduct all the emulations in the mobile scenario with trajectories indexed from 1 to 4 as shown in Figure 5. The four mobile trajectories represent the different access options for the mobile user in the integrated heterogeneous wireless networks, e.g., the user could simultaneously access the UMTS and WiMAX while moving along the first trajectory. The mobile client requests to the server through a wireless interface and constructs the connection whenever it moves in the coverage. The moving speed of the client is set to be 2 m/s in all the emulations. In all the emulations, the components of FRA-JSCC are working at the GOP level, i.e., every 0.25 s. It is necessary to update the JSCC parameters for each GOP due to the time-varying wireless channel status. However, with regard to the coding efficiency, it is impractical to trace the rate variation at the video frame level.\n\nFor the confidence results, we repeat each set of emulations with different video sequences more than five times and obtained the average results with a 95% confidence interval. The microscopic and mobility results were presented with the measurements of finer granularity.\n\n### 5.2 Evaluation results\n\nBefore showing the experimental results of the performance metrics in detail, we first present the channel status information, which is the feedback with a 0.25-s period from the client. Figure 6 plots the available bandwidth of different wireless access networks during the client mobility along mobile trajectory 3. It can be observed that the available bandwidth of both WLAN and WiMAX experiences fluctuations due to the injected background traffic and client mobility. The instantaneous loss rates are shown in Figure 7. Due to the lack of space, we do not present all the channel status information during the experimentations in this section.\n\n#### 5.2.1 PSNR\n\nAs shown in Figure 8, FRA-JSCC achieves higher PSNR values and lower variations than the other competing models. The average video PSNR in trajectory 2 is lower than that in trajectory 1 as the WLAN is less stable than the WiMAX. The results verify the instance in Figure 1 and the conclusions in related work [6, 7]. Besides, the superiority of FRA-JSCC and FCVP over the other two schemes is larger in trajectories 3 and 4 as more wireless access networks are available. The substantial improvements in video quality confirm the importance of JSCC in conjunction with flow rate allocation in heterogeneous wireless networks. FRA-JSCC outperforms the FCVP as the Reed-Solomon code is more appropriate than the fountain code for the real-time video and thus reduce the erasure-coding-induced delays. In order to have a microscopic view of the results, we also depict the mean values and standard deviations (Stddev) of mobile trajectory 4 in Table 4. The per frame video PSNR during the interval of [ 0, 20] s is presented in Figure 9. It can be observed that FRA-JSCC maintains the PSNR values at a relatively higher range. In the mobile trajectory 1, the superiority of FRA-JSCC over the JMFR becomes more obvious and is due to the increase number of access options.\n\n#### 5.2.2 Average end-to-end delay\n\nFigure 10 plots the average end-to-end delays as well as the confidence intervals. FRA-JSCC achieves the lowest delay of all the competing models. The delay performance of FCVP is inferior to that of FRA-JSCC and JMFR due to the large block size of fountain and the coding inefficiency. The results indicate the Reed-Solomon code is more suitable for real-time video applications than the fountain code. Figure 11a depicts the cumulative distribution function of the end-to-end video frame delay from a single experiment. We can see that the per-frame delay is significantly lower here than that of the other three reference schemes. Although the FEC encoding is not employed in the DMP, the lost video frames need to be retransmitted, and thus, the end-to-end delay will be increased. As each video frame is associated with a decoding deadline in real-time applications, we plot the ratio of video frames past the decoding deadline of 200 ms in Figure 11b.\n\n#### 5.2.3 Effective loss rate\n\nFigure 12 depicts the effective loss rates of all the competing schemes under different mobile trajectories. The pattern is very similar to the results presented in Figure 8 as the PSNR is generally proportional to the ratio of lost video frames. FRA-JSCC significantly outperforms the reference schemes as it takes into account both the loss and delay requirements. However, different from the results in end-to-end delay, FCVP outperforms JMFR and DMP in minimizing the effective loss rate as it includes a physical path selection algorithm in the system design. Thus, the transmission loss is substantially decreased.\n\n## 6 Conclusions\n\nIn this paper, we have presented a flow rate allocation-based JSCC approach for mobile video delivery in heterogeneous wireless networks. Through modeling and analysis, we have developed solutions for FEC redundancy adaption, video source rate adaption, and flow rate allocation. Experimental results show that the proposed FRA-JSCC is able to dynamically select the appropriate wireless access networks out of all candidates and significantly improve the video PSNR. As future work, we will consider (1) designing a seamless vertical handoff algorithm for optimal-quality video in the integrated WLAN, WiMAX, and cellular networks. The work in formulates the heterogeneous wireless networks as restless bandit systems. However, it does not provide in-depth analysis on the physical characteristics (e.g., the coverage and received signal strength) of each wireless network. We would also consider (2) including an optimal path interleaving mechanism with the FRA-JSCC to overcome the burst loss.\n\n## Endnotes\n\na We choose the JSVM in convenience for the source code integration as both Exata and JSVM are developed using the C++ code, while the H.264/AVC JM (http://iphome.hhi.de/suehring/tml/) software is developed using C language.\n\n## References\n\n1. Hurley C, Chen S, Karim J, YouTube 2005.http://www.youtube.com/ . Accessed 1 Sept 2013\n\n2. Hulu: NBC Universal & New Corp. Los Angeles, CA; 2007. . Accessed 1 Sept 2013 http://www.hulu.com/\n\n3. Cisco: Cisco visual networking index: Global mobile data traffic forecast update, 2012–2017, May 2013 (online). Available: http://www.cisco.com/en/US/solutions/collateral/ns341/ns525/402ns537/ns705/ns827/white_paper_c11-520862.html. Accessed 29 Nov 2013\n\n4. Oliveira T, Mahadevan S, Agrawal DP: Handling network uncertainty in heterogeneous wireless networks. In INFOCOM, 2011 Proceedings IEEE. IEEE Piscataway; 2011:2390-2398.\n\n5. Si P, Ji H, Yu FR: Optimal network selection in heterogeneous wireless multimedia networks. Wireless Netw 2009, 16(5):1277-1288.\n\n6. Han S, Joo H, Lee D, Song H: An end-to-end virtual path construction system for stable live video streaming over heterogeneous wireless networks. IEEE J. Select. Areas Commun 2011, 29(5):1032-1041.\n\n7. Yooon J, Zhang H, Banerjee S, Rangarajan S: MuVi: a multicast video delivery scheme for 4G cellular networks. In Proceedings of the 18th Annual International Conference on Mobile Computing and Networking (Mobicom ’12). ACM, New York; 2012:209-220.\n\n8. Chebrolu K, Rao R: Bandwidth aggregation for real-time applications in heterogeneous wireless networks. IEEE Trans. Mobile Comput 2006, 5(4):388-403.\n\n9. Song W, Zhuang W: Performance analysis of probabilistic multipath transmission of video streaming traffic over multi-radio wireless devices. IEEE Trans. Wireless Commun 2012, 11(4):1554-1564.\n\n10. Jurca D, Frossard P: Video packet selection and scheduling for multipath streaming. IEEE Trans. Multimedia 2007, 9(3):629-641.\n\n11. Khalek AA, Heath RW, Caramanis C: A cross-layer design for perceptual optimization Of H.264/SVC with unequal error protection. IEEE J. Select. Areas Commun 2012, 30(7):1157-1171.\n\n12. Zhang Y, Gao W, Lu Y, Huang Q, Zhao D: Joint source-channel rate-distortion optimization for H.264 video coding over error-prone networks. IEEE Trans. Multimedia 2007, 9(3):445-454.\n\n13. Ji W, Li Z, Chen Y: Joint source-channel coding and optimization for layered video broadcasting to heterogeneous devices. IEEE Trans. Multimedia 2012, 14(2):443-455.\n\n14. Frossard P, Verscheure O: Joint source/FEC rate selection for quality-optimal MPEG-2 video delivery. IEEE Trans. Image Process 2001, 10(2):1815-1825.\n\n15. Ahmad S, Hamzaoui R, Akaidi MA: Adaptive unicast video streaming with rateless codes and feedback. IEEE Trans. Circuits Syst. Video, Technol 2010, 20(2):275-285.\n\n16. Jurca D, Frossard P, Jovanovic A: Forward error correction for multipath media streaming. IEEE Trans. Circuits Syst. Video, Technol 2009, 19(9):1315-1326.\n\n17. Jurca D, Frossard P: Media flow rate allocation in multipath networks. IEEE Trans. Multimedia 2007, 9(6):1227-1240.\n\n18. Wang B, Wei W, Guo Z, Towsley D: Multipath live streaming via TCP: scheme, performance and benefits. TOMCCAP 2009., 5(3): doi:10.1145/1556134.1556142\n\n19. Bystrom M, Modestino JW: Combined source-channel coding schemes for video transmission over an additive white gaussian noise channel. IEEE J. Select. Areas Commun 2000, 18(6):880-890.\n\n20. Cernea DC, Munteanu A, Alecu A, Cornelis J, Schelkens P: Scalable joint source and channel coding of meshes. IEEE Trans. Multimedia 2008, 10(3):503-513.\n\n21. He Z, Cai J, Chen CW: Joint source channel rate-distortion analysis for adaptive mode selection and rate control in wireless video coding. IEEE Trans. Circuits Syst. Video, Technol 2002, 12(6):511-523. 10.1109/TCSVT.2002.800313\n\n22. Duyck D, Capirone D, Boutros J, Moeneclaey M: Analysis and construction of full-diversity joint network-LDPC codes for cooperative communications. EURASIP J. Wireless Commun. Netw 2010, 2010: 805216. 10.1155/2010/805216\n\n23. Jaspar X, Guillemot C, Vandendorpe L: Joint source-channel turbo techniques for discrete-valued sources: from theory to practice. Proc. IEEE 2007, 95(6):1345-1361.\n\n24. Qian L, Jones DL, Ramchandran K, Appadwedula S: A general joint source-channel matching method for wireless video transmission. In Data Compression Conference (DCC ’99), Snowbird, 29–31 Mar 1999. IEEE Piscataway; 1999:414-423.\n\n25. Xu Q, Stankovic V, Xiong Z: Distributed joint source-channel coding of video using raptor codes. IEEE Trans. Circuits Syst. Video, Technol 2007, 25(4):851-861.\n\n26. Zhai F, Eisenberg Y, Pappas TN, Berry R, Katsaggelos AK: Rate-distortion optimized hybrid error control for real-time packetized video transmission. IEEE Trans. Image Process 2006, 15(1):40-53.\n\n27. Apostolopoulos J, Trott M: Path diversity for enhanced media streaming. IEEE Commun. Mag 2004, 42: 80-87.\n\n28. Ramaboli A, Falowo O, Chan A: Bandwidth aggregation in heterogeneous wireless networks: a survey of current approaches and issues. J. Netw. Comput. Appl 2012, 35(6):1674-1690. 10.1016/j.jnca.2012.05.015\n\n29. Chow ALH, Yang H, Xia CH, Kim M, Liu Z, Lei H: EMS: Encoded multipath streaming for real-time live streaming applications. In 17th IEEE International Conference on Network Protocols (ICNP 2009), Princeton 13–16 Oct 2009. IEEE, Piscataway; 2010:233-243.\n\n30. Sharma V, Kar K, Ramakrishnan KK, Kalyanaraman S: A transport protocol to exploit multipath diversity in wireless networks. IEEE/ACM Trans. Netw 2012, 20(4):1024-1039.\n\n31. Ribeiro V, Riedi R, Baraniuk R, Navratil J, Cottrell L: pathChirp: efficient available bandwidth estimation for network paths. In Proceedings of Passive and Active Measurement Workshop. La Jol; 6–8 April 2003.\n\n32. Paxson V, Almes G, Mahdavi J, Mathis M: Framework for IP performance metrics. IETF Technical Report, RFC 2330 1998.\n\n33. Stuhlmüller K, Färber N, Link M, Girod B: Analysis of video transmission over lossy channels. IEEE J. Select. Areas Commun 2000, 18(6):1012-1032.\n\n34. Zhu X, Agrawal P, Singh JP, Alpcan T, Girod B: Distributed rate allocation policies for multihomed video streaming over heterogeneous access networks. IEEE Trans. Multimedia 2009, 11(4):752-764.\n\n35. Thomos N, Argyropoulos S, Boulgouris N, Strintzis M: Robust transmission of h.264/avc video using adaptive slice grouping and unequal error protection. In IEEE International Conference on Multimedia and Expo, Toronto, 9–12 July 2006. IEEE, Piscataway; 2006:593-596.\n\n36. Baccaglini E, Tillo T, Olmo G: Slice sorting for unequal loss protection of video streams. IEEE Signal Process. Lett 2008, 15: 581-584.\n\n37. Xiao J, Tillo T, Lin C, Zhao Y: Dynamic sub-GOP forward error correction code for real-time video applications. IEEE Trans. Multimedia 2012, 14(4):1298-1308.\n\n38. Kurant M: Exploiting the path propagation time differences in multipath transmission with FEC. IEEE J. Select. Areas Commun 2011, 29(5):1021-1031.\n\n39. Kompella S, Mao S, Hou YT, Sherali HD: On path selection and rate allocation for video in wireless mesh networks. IEEE/ACM Trans. Netw 2009, 17(1):212-224.\n\n40. Sun M, Reibman A: Compressed Video Over Networks. Marcel Dekker Inc., New York; 2000.\n\n41. Jain M, Dovrolis C: Pathload: a measurement tool for end-to-end available bandwidth. In Proceedings of Passive and Active Measurement Workshop. Fort Collins; 25–27 Mar 2002.\n\n42. Zhou A, Liu M, Song Y, et al.: A new method for end-to-end available bandwidth estimation. In Proceedings of IEEE GLOBECOM, New Orleans, 30 Nov–4 Dec 2008. IEEE, Piscataway; 2008:1-5.\n\n43. Exata: (SCALABLE Network Technologies, Inc., Culver City, 2008). . Accessed 29 Nov 2013 http://www.scalable-networks.com/exata\n\n44. QualNet: (SCALABLE Network Technologies, Inc., Culver City, 2008). . Accessed 29 Nov 2013 http://www.scalable-networks.com/qualnet\n\n45. Song W, Cheng Y, Zhuang W: Improving voice and data services in cellular/WLAN integrated networks by admission control. IEEE Trans. Wireless Commun 2007, 6(11):4025-4037.\n\n## Acknowledgements\n\nThis research is supported by the National Grand Fundamental Research 973 Program of China under grant nos. 2011CB302506, 2012CB315802, and 2013CB329102; Research Program of Chongqing Municipal Education Commission (grant no. KJ130523); CQUPT Research Fund for Young Scholars (grant no. A2012-79); National Key Technology Research and Development Program of China ‘Research on the mobile community cultural service aggregation supporting technology’ (grant no. 2012BAH94F02); Novel Mobile Service Control Network Architecture and Key Technologies (2010ZX03004-001-01); National High-tech R &D Program of China (863 Program) under grant no. 2013AA102301; National Natural Science Foundation of China under grant nos. 61003067, 61171102, 61001118, and 61132001; Program for New Century Excellent Talents in University (grant no. NCET-11-0592); Project of New Generation Broadband Wireless Network under grant no. 2011ZX03002-002-01; and Beijing Nova Program under grant no. 2008B50. The authors would like to express their gratitude to the anonymous reviewers who provided comments to improve the paper quality.\n\n## Author information\n\nAuthors\n\n### Corresponding author\n\nCorrespondence to Jiyan Wu.\n\n### Competing interests\n\nThe authors declare that they have no competing interests.\n\n## Authors’ original submitted files for images\n\nBelow are the links to the authors’ original submitted files for images.\n\n## Rights and permissions\n\nReprints and Permissions\n\nWu, J., Shang, Y., Huang, J. et al. Joint source-channel coding and optimization for mobile video streaming in heterogeneous wireless networks. J Wireless Com Network 2013, 283 (2013). https://doi.org/10.1186/1687-1499-2013-283\n\n• Accepted:\n\n• Published:\n\n• DOI: https://doi.org/10.1186/1687-1499-2013-283\n\n### Keywords\n\n• Mobile video streaming\n• Heterogeneous wireless networks\n• Multi-homing\n• Joint source-channel coding\n• Flow rate allocation",
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https://marketplace.corporatefinanceinstitute.com/vendor/corporate-finance-institute-cfi/page/5/?add-to-cart=496 | [
"# Templates by Corporate Finance Institute®\n\n65 to 80 of 256 Results\nCost of Equity Excel Calculator\n(0)\n7,885\n1093\nThis Cost of Equity Excel Calculator can be a useful tool for prospective investors. The cost of equity is the required…\nQuick Ratio Excel Calculator\n(0)\n8,244\n1106\nThis is a ready-to-use Quick Ratio Excel template, which helps to indicate how well a company can pay off its…\nCost-Volume-Profit (CVP) Analysis Excel Template\n(0)\n10,030\n1107\nCost-volume-profit analysis (CVP analysis), also commonly referred to as breakeven analysis, is a method to evaluate how profitability is impacted…\nDiscount Factor Excel Template\n(0)\n8,625\n1059\nThis discount factor excel template can be a useful tool for analysts when creating financial models. The discount factor is…\nAccumulated Depreciation Excel Template\n(1)\n10,328\n1031\nThis Accumulated Depreciation Excel Template is a useful tool to calculate the total depreciation of an asset when given the…\nAccounts Payable Excel Template\n(0)\n10,288\n1046\nThis Accounts Payable Excel Template, is an educational resource that highlights where you would find the A/P account on a…\nDebt/Equity Ratio Excel Template\n(0)\n8,086\n1107\nThis Debt to Equity Excel Template Ratio is an educational resource that shows you an example on how to calculate…\nEquity Beta and Asset Beta Conversion Excel Model Template\n(1)\n10,001\n952\nUnlevered beta (also referred to as asset beta) is the company’s beta score without taking into account the debt that…\nProperty Plant and Equipment Schedule Template\n(0)\n9,050\n994\nThis Property Plant and Equipment Schedule Template is a simple and easy to use, one-sheet worksheet that will allow you…\nWorking Capital Cycle Template\n(0)\n7,639\n1001\nThe working capital cycle is the amount of time it takes a company to convert its working capital into cash.…\nYear over Year Analysis (YoY) Excel Template\n(0)\n8,529\n1035\nThis Year over Year (YOY) excel template is a great tool to analyze time-series data. This type of analysis is…\nCapital Expenditure (CapEx) Excel Calculator\n(0)\n8,034\n981\nCapital expenditure (CapEx) is the amount of money invested by a business in the acquisition, maintenance, and improvement of fixed…"
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http://www.lmfdb.org/EllipticCurve/Q/158d3/ | [
"Show commands for: Magma / SageMath / Pari/GP\n\nMinimal Weierstrass equation\n\nmagma: E := EllipticCurve([1, 0, 1, -47, 118]); // or\nmagma: E := EllipticCurve(\"158d3\");\nsage: E = EllipticCurve([1, 0, 1, -47, 118]) # or\nsage: E = EllipticCurve(\"158d3\")\ngp: E = ellinit([1, 0, 1, -47, 118]) \\\\ or\ngp: E = ellinit(\"158d3\")\n\n$$y^2 + x y + y = x^{3} - 47 x + 118$$\n\nMordell-Weil group structure\n\n$$\\Z/{3}\\Z$$\n\nTorsion generators\n\nmagma: TorsionSubgroup(E);\nsage: E.torsion_subgroup().gens()\ngp: elltors(E)\n\n$$\\left(4, -2\\right)$$\n\nIntegral points\n\nmagma: IntegralPoints(E);\nsage: E.integral_points()\n\n$$\\left(4, -2\\right)$$\n\nNote: only one of each pair $\\pm P$ is listed.\n\nInvariants\n\n magma: Conductor(E); sage: E.conductor().factor() gp: ellglobalred(E) Conductor: $$158$$ = $$2 \\cdot 79$$ magma: Discriminant(E); sage: E.discriminant().factor() gp: E.disc Discriminant: $$316$$ = $$2^{2} \\cdot 79$$ magma: jInvariant(E); sage: E.j_invariant().factor() gp: E.j j-invariant: $$\\frac{11134383337}{316}$$ = $$2^{-2} \\cdot 7^{3} \\cdot 11^{3} \\cdot 29^{3} \\cdot 79^{-1}$$ Endomorphism ring: $$\\Z$$ (no Complex Multiplication) Sato-Tate Group: $\\mathrm{SU}(2)$\n\nBSD invariants\n\n magma: Rank(E); sage: E.rank() Rank: $$0$$ magma: Regulator(E); sage: E.regulator() Regulator: $$1$$ magma: RealPeriod(E); sage: E.period_lattice().omega() gp: E.omega Real period: $$5.0558812156$$ magma: TamagawaNumbers(E); sage: E.tamagawa_numbers() gp: gr=ellglobalred(E); [[gr[i,1],gr[i]] | i<-[1..#gr[,1]]] Tamagawa product: $$2$$ = $$2\\cdot1$$ magma: Order(TorsionSubgroup(E)); sage: E.torsion_order() gp: elltors(E) Torsion order: $$3$$ magma: MordellWeilShaInformation(E); sage: E.sha().an_numerical() Analytic order of Ш: $$1$$ (exact)\n\nModular invariants\n\nModular form158.2.a.b\n\nmagma: ModularForm(E);\nsage: E.q_eigenform(20)\ngp: xy = elltaniyama(E);\ngp: x*deriv(xy)/(2*xy+E.a1*xy+E.a3)\n\n$$q - q^{2} + q^{3} + q^{4} + 3q^{5} - q^{6} - q^{7} - q^{8} - 2q^{9} - 3q^{10} + q^{12} + 5q^{13} + q^{14} + 3q^{15} + q^{16} + 2q^{18} + 2q^{19} + O(q^{20})$$\n\nFor more coefficients, see the Downloads section to the right.\n\n magma: ModularDegree(E); sage: E.modular_degree() Modular degree: 120 $$\\Gamma_0(N)$$-optimal: no Manin constant: 3\n\nSpecial L-value\n\nmagma: Lr1 where r,Lr1 := AnalyticRank(E: Precision:=12);\nsage: r = E.rank();\nsage: E.lseries().dokchitser().derivative(1,r)/r.factorial()\ngp: ar = ellanalyticrank(E);\ngp: ar/factorial(ar)\n\n$$L(E,1)$$ ≈ $$1.12352915902$$\n\nLocal data\n\nmagma: [LocalInformation(E,p) : p in BadPrimes(E)];\nsage: E.local_data()\ngp: ellglobalred(E)\nprime Tamagawa number Kodaira symbol Reduction type Root number ord($$N$$) ord($$\\Delta$$) ord$$(j)_{-}$$\n$$2$$ $$2$$ $$I_{2}$$ Non-split multiplicative 1 1 2 2\n$$79$$ $$1$$ $$I_{1}$$ Split multiplicative -1 1 1 1\n\nGalois representations\n\nThe 2-adic representation attached to this elliptic curve is surjective.\n\nmagma: [GaloisRepresentation(E,p): p in PrimesUpTo(20)];\nsage: rho = E.galois_representation();\nsage: [rho.image_type(p) for p in rho.non_surjective()]\n\nThe mod $$p$$ Galois representation has maximal image $$\\GL(2,\\F_p)$$ for all primes $$p$$ except those listed.\n\nprime Image of Galois representation\n$$3$$ B.1.1\n\n$p$-adic data\n\n$p$-adic regulators\n\nsage: [E.padic_regulator(p) for p in primes(3,20) if E.conductor().valuation(p)<2]\n\nAll $$p$$-adic regulators are identically $$1$$ since the rank is $$0$$.\n\nIwasawa invariants\n\n$p$ Reduction type $\\lambda$-invariant(s) 2 3 79 nonsplit ordinary split 2 0 1 0 0 0\n\nAll Iwasawa $\\lambda$ and $\\mu$-invariants for primes $p\\ge 5$ of good reduction are zero.\n\nIsogenies\n\nThis curve has non-trivial cyclic isogenies of degree $$d$$ for $$d=$$ 3 and 9.\nIts isogeny class 158d consists of 3 curves linked by isogenies of degrees dividing 9.\n\nGrowth of torsion in number fields\n\nThe number fields $K$ of degree up to 7 such that $E(K)_{\\rm tors}$ is strictly larger than $E(\\Q)_{\\rm tors}$ $\\cong \\Z/{3}\\Z$ are as follows:\n\n$[K:\\Q]$ $K$ $E(K)_{\\rm tors}$ Base-change curve\n3 3.3.6241.1 $$\\Z/9\\Z$$ Not in database\n3.3.316.1 $$\\Z/6\\Z$$ Not in database\n6 6.0.16826434992.1 $$\\Z/3\\Z \\times \\Z/3\\Z$$ Not in database\n6.0.2696112.3 $$\\Z/9\\Z$$ Not in database\n6.6.31554496.1 $$\\Z/2\\Z \\times \\Z/6\\Z$$ Not in database\n\nWe only show fields where the torsion growth is primitive. For each field $K$ we either show its label, or a defining polynomial when $K$ is not in the database."
]
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https://math.stackexchange.com/questions/3272449/how-do-we-approach-with-these-kind-of-probability-problems | [
"# How do we approach with these kind of probability problems?\n\nA step by step thinking approach to these types of problem.\n\nQ: Given Player A,B,C, and D randomly lined up. first two players in the line play games, winner of the game plays with the person third in line, further the winner plays with the fourth in the line. Finding probability that A wins the tournament, given A wins each game it plays with probability p.\n\nIf $$A$$ is number $$1$$ on the line (probability on that is $$\\frac14$$) then he must win $$3$$ consecutive games (probability on that $$p^3$$).\n\nIf $$A$$ is number $$2$$ on the line (probability on that is $$\\frac14$$) then he must win $$3$$ consecutive games (probability on that $$p^3$$).\n\nIf $$A$$ is number $$3$$ on the line (probability on that is $$\\frac14$$) then he must win $$2$$ consecutive games (probability on that $$p^2$$).\n\nIf $$A$$ is number $$4$$ on the line (probability on that is $$\\frac14$$) then he must win $$1$$ game (probability on that $$p$$).\n\nCan you take it from here?\n\n• Yes. You solved the problem. I went blank when I saw problem, my intention is to know how should I approach whenever I encounter such problems. Your explanation made this problem so simple, and I wonder why I haven't thought it this way. Wanting to develop this thinking process. – Roopesh Singh Jun 24 at 8:43\n• Just keep going in your tries. Then mathematical maturity will surely grow. It needs time though ;). – drhab Jun 24 at 8:47\n\nFirst calculate the probability that $$A$$ is in each position, i.e. first second third and last. Then, for each position, calculate the probability that $$A$$ wins the tournament. Then, multiply and sum these probabilities, i.e.\n\n$$P(A\\mbox{ wins}) = P(A\\mbox{ wins |} A\\mbox{ is first})P(A\\mbox{ is first}) + \\ldots + P(A\\mbox{ wins |} A\\mbox{ is last})P(A\\mbox{ is last})$$"
]
| [
null
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https://octave.org/doc/v4.0.1/Example-Code.html | [
"Next: , Previous: , Up: Diagonal and Permutation Matrices [Contents][Index]\n\n### 21.4 Examples of Usage\n\nThe following can be used to solve a linear system `A*x = b` using the pivoted LU factorization:\n\n``` [L, U, P] = lu (A); ## now L*U = P*A\nx = U \\ L \\ P*b;\n```\n\nThis is one way to normalize columns of a matrix X to unit norm:\n\n``` s = norm (X, \"columns\");\nX /= diag (s);\n```\n\n``` s = norm (X, \"columns\");\nX ./= s;\n```\n\nThe following expression is a way to efficiently calculate the sign of a permutation, given by a permutation vector p. It will also work in earlier versions of Octave, but slowly.\n\n``` det (eye (length (p))(p, :))\n```\n\nFinally, here’s how to solve a linear system `A*x = b` with Tikhonov regularization (ridge regression) using SVD (a skeleton only):\n\n``` m = rows (A); n = columns (A);\n[U, S, V] = svd (A);\n## determine the regularization factor alpha\n## alpha = …\n## transform to orthogonal basis\nb = U'*b;\n## Use the standard formula, replacing A with S.\n## S is diagonal, so the following will be very fast and accurate.\nx = (S'*S + alpha^2 * eye (n)) \\ (S' * b);\n## transform to solution basis\nx = V*x;\n```"
]
| [
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https://demonstrations.wolfram.com/KarushKuhnTuckerKKTConditionsForNonlinearProgrammingWithIneq/ | [
"",
null,
"# Karush-Kuhn-Tucker (KKT) Conditions for Nonlinear Programming with Inequality Constraints\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\nThis Demonstration explores a constrained nonlinear program in which the objective is to minimize a function",
null,
"subject to a single inequality constraint",
null,
". The top-left box shows the level sets of",
null,
"as gray contours, the level sets of",
null,
"as blue contours and the feasible region as a shaded blue area. The optimal feasible solution is shown as a red dot. Clicking anywhere inside the top-left box selects a point as a candidate solution; the radio buttons adjust the location of the feasible region. All three boxes display visual representations of the Karush–Kuhn–Tucker conditions, along with a green checkmark or a red Χ to indicate that the candidate does or does not satisfy the condition, respectively. In particular, the top two boxes display",
null,
"as a black arrow and",
null,
"as a blue arrow, while the bottom left box displays a point corresponding to the KKT multiplier",
null,
"(provided it exists, which requires that either",
null,
"or that",
null,
"and",
null,
"are collinear) and the value of",
null,
". In order for a solution to be the gobal optimum, it is necessary to satisfy all of the conditions simultaneously.\n\nContributed by: Adam Rumpf (July 2018)\nOpen content licensed under CC BY-NC-SA\n\n## Snapshots",
null,
"",
null,
"",
null,
"## Details\n\nThe Karush–Kuhn–Tucker conditions (a.k.a. KKT conditions or Kuhn–Tucker conditions) are a set of necessary conditions for a solution of a constrained nonlinear program to be optimal . The KKT conditions generalize the method of Lagrange multipliers for nonlinear programs with equality constraints, allowing for both equalities and inequalities. In this Demonstration, we consider only inequality constraints. Specifically, the problem used for this Demonstration has the form",
null,
"such that",
null,
"where",
null,
"and",
null,
"are both real-valued, differentiable functions on",
null,
"that satisfy the regularity conditions needed for the KKT conditions to apply . For this simple problem, the KKT conditions state that a solution",
null,
"is a local optimum if and only if there exists a constant",
null,
"(called a KKT multiplier) such that the following four conditions hold:\n\n1. Stationarity:",
null,
"2. Primal feasibility:",
null,
"3. Dual feasibility:",
null,
"4. Complementary slackness:",
null,
"There are two possibilities for the optimal solution: it can occur either on the boundary of the feasible set (where",
null,
") or on the interior (where",
null,
"). If it occurs on the boundary, then we are left with the equivalent of an equality constraint, in which case the simple method of Lagrange multipliers applies. The preceding stationarity condition is identical to the one for Lagrange multipliers, and it captures instances for which either",
null,
"(meaning that",
null,
"becomes flat on the boundary) or",
null,
"(meaning that the level sets of",
null,
"and",
null,
"lie tangent to each other). The first case forces",
null,
", while the second forces",
null,
".\n\nIf, on the other hand, the optimum occurs on the interior, then the constraint has no effect on the calculation of the optimum. This can only occur where",
null,
", but the stationarity condition guarantees simply that at least one out of",
null,
"and",
null,
"is true, with no guarantees about which one holds. Additional conditions are needed to ensure that",
null,
"for any candidate optimum on the interior. The snapshots include various examples of solutions that are not locally optimal and satisfy some (but not all) of the KKT conditions.\n\nDual feasibility ensures that the optimum occurs on the correct side of the feasible boundary by ensuring that",
null,
"and",
null,
"point in opposite directions. Complementary slackness ensures that the correct restriction is enforced. The condition itself forces at least one of",
null,
"and",
null,
"to vanish. On the interior of the feasible set, we have",
null,
", in which case complementary slackness enforces",
null,
"and thus",
null,
". On the boundary, we have",
null,
", in which case",
null,
"is unrestricted.\n\nThe three boxes contained in this Demonstration graphically illustrate each of these conditions. Primal feasibility is satisfied exactly when the chosen point lies inside the feasible region. The top two boxes of the Demonstration display",
null,
"as a black arrow and",
null,
"as a blue arrow. The bottom-left box shows a point on the",
null,
"versus",
null,
"axis corresponding to the current values of",
null,
"and",
null,
". The positive",
null,
"and negative",
null,
"axes are highlighted to indicate the coordinates that simultaneously satisfy dual feasibility and complementary slackness. Note that",
null,
"is only defined when",
null,
"or when",
null,
"and",
null,
"are collinear, and that no point is plotted when",
null,
"does not exist.\n\nSnapshot 1: an optimal solution on the boundary of the feasible region. Here we do not have",
null,
", so the optimal solution is a stationary point on the boundary. Note that",
null,
"(the black arrow) and",
null,
"(the blue arrow) point in opposite directions. In this case, we have",
null,
"and",
null,
"; therefore, dual feasibility and complementary slackness are also satisfied.\n\nSnapshot 2: a clearly nonoptimal solution on the interior of the feasible region. Stationarity is not satisfied, therefore there exists no value",
null,
"such that",
null,
". With no KKT multiplier, dual feasibility and complementary slackness are meaningless, and so there is nothing to report in the bottom-left box.\n\nSnapshot 3: a nonoptimal solution that does not satisfy complementary slackness. This means that",
null,
"(so",
null,
") and",
null,
"(so we are not on the boundary), which contradicts the only possible interior solutions coinciding with",
null,
"being flat.\n\nSnapshot 4: a nonoptimal solution that satisfies only stationarity. This indicates that if the level sets of",
null,
"were extended outside of the feasible region, then one would lie tangent to a level set of",
null,
"at this point, but the point is obviously not feasible.\n\nSnapshot 5: an optimal solution inside the feasible region. As is necessary for such a solution, we have",
null,
".\n\nSnapshot 6: a nonoptimal solution that does not satisfy dual feasibility. This is identical to the example from Snapshot 5, but the point has been shifted to the right and away from where",
null,
"is flat. Here",
null,
"and",
null,
"point in the same direction and",
null,
", which indicates that moving further toward the interior would decrease",
null,
"; therefore, the current solution cannot be optimal.\n\nReferences\n\n D. P. Bertsekas, Nonlinear Programming (2nd ed.), Belmont, MA: Athena Scientific, 1999.\n\n R. G. Eustáquio, E. W. Karas and A. A. Ribeiro, Constraint Qualifications for Nonlinear Programming, Working paper, Department of Mathematics, Federal University of Paraná, Brazil, 2007. www.researchgate.net/publication/230663810_Constraint _Qualifications _for _Nonlinear _Programming."
]
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https://dsp.stackexchange.com/questions/56510/convolution-of-two-impulse-signals | [
"# Convolution of Two Impulse Signals\n\nI have encountered convolution of two different impulse signals.\n\nx[n] = (1/2)^n . u[n-2] * u[n]\nx[n] = u[n] * [n]\n\nu[n] = discrete impulse signal\n. = product operation\n* = convolution operation\n\n\nFor the first one, I found this solution:\n\nx[n] = 1/4 if n = 2\nx[n] = 0 if n != 2\n\n\nFor the second one, I found impulse signal itself\n\nEdit: Are my answers are true ? My professor told me that the answer for the first one is wrong, but he did not say the correct answer.\n\n• what is the question ? – Ahmad Bazzi Apr 7 '19 at 15:49\n• I cannot make sure that my answers are true or not – Goktug Apr 7 '19 at 16:00\n• My answer is specified above. – Goktug Apr 7 '19 at 16:01\n• $u[n]$ is generally used to denote the unit step function, not the unit impulse function which is usually denoted $\\delta[n]$. Please don't introduce new notation unnecessarily. – Dilip Sarwate Apr 7 '19 at 22:38\n\nBut you better use the standard notation as Dilip Sarwate already indicated; $$u[n]$$ is the unit-step and $$\\delta[n]$$ is the unit impulse. Then\n$$0.5^n \\delta[n-2] \\star \\delta[n] = 0.5^2 \\delta[n-2] = \\begin{cases} { 0.25 ~~~, ~~~n= 2 \\\\ 0.00 ~~~,~~~n \\neq 2 } \\end{cases}$$\nThis is basically an exercise to test the student's understanding of the concept that the unit impulse is effectively the unit in convolutions, that is: $$\\delta \\star x = x$$ for all signals $$x$$. Perhaps a systems explaination might help. If an LTI system has impulse response $$h$$, then we know that the output of the system when $$x$$ is the input is $$y = h \\star x$$. So, $$\\delta \\star x$$ can be thought of as the output of an LTI system with impulse response $$\\delta$$ when the input to the LTI system is $$x$$. What LTI system has output $$\\delta$$ when its input is the unit impulse $$\\delta$$?? It is just the canonical straight wire with (no) gain that audio enthusiasts dream about! And so, $$\\delta \\star x = x$$ for all $$x$$.\nWith this, it it is easy to verify that the OP's answer to the first question is correct (but maybe his professor wanted to see the answer as $$\\left(\\frac 12\\right)^n \\delta[n-2]$$ or $$\\left(\\frac 12\\right)^2 \\delta[n-2]$$ instead of what the OP wrote) while the second answer $$\\delta[n]\\star [n] = \\delta[n]$$ is incorrect, it should be $$[n]$$."
]
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https://de.scribd.com/document/148858357/Mod-ALL | [
"Sie sind auf Seite 1von 51\n\n# Ph. D.\n\nFeb. 2013\n\nDo\n\nout of\n\nproblems\n\n## Problem 1: (40 points) The specific Gibbs function of a gas is given by\n\ng P RT ln \u0010 AP P 0\n\nwhere is a function of . Find expressions for: (a) the equation of state [15 points] (b) the specific entropy [10 points] (c) the specific Helmhotz function [15 points]\n\n## Problem 2: (40 points)\n\nThe angle of deviation, , is defined as the angle between the direction of the incident beam and the direction of the beam transmitted by the prism. (a) When the incident beam strikes the prism at an angle such that the angle of deviation is minimized, this minimum angle of deviation, , gives a simple relation for the index of refraction of the prism. This occurs when the angle of incidence for the 1st interface ( )is equal to the angle of refraction for the 2nd interface () . Find the refractive index of refraction, , of the prism in terms of the angles and for this situation. [25 points] (b) Find the angle of deviation, , for the general case (shown above) when it is not minimized. Find this angle in terms of the index of refraction, , of the prism, the incident angle of the 1st interface (), and . [15 points]\n\nProblem 3: (40 points) (a) Derive an approximate formula for the strength of the magnetic field at the hydrogen nucleus produced by an electron in the first Bohr orbit (using the Bohr model of the atom). [20 points] (b) Give a rough order of magnitude estimate of the strength of the magnetic field (in Teslas). Below are useful physical constants that you can use: [10 points] permeability of free space: N/A2 permittivity of free space: C2/Nm2 charge of the electron: C mass of the electron: kg 3ODQFNV\u0003FRQVWDQW\u001d\u0003\u0003\u0003\u0003\u0003\u0003\u0003\u0003\u0003\u0003\u0003\u0003 Jsec (c) If the nuclear Bohr mageton is J/Tesla give an order of magnitude estimate of the hyperfine splitting (in eV) of spectral lines. [10 points]\n\nProblem 4: (40 points) Solve for the threshold energy for the following examples of the production of an electron-positron pair. (a) Show the Feynman diagram for the process [5 points] (b) What is the threshold energy for the photon (expressed in terms of electron masses) for the process in Part (a)? [5 points] (c) What is the threshold energy for the photon (expressed in terms of electron masses) for the process where is a free electron? [15 points] (d) Why are the threshold energies in Parts (b) and (c) different? [5 points] (e) In intergalactic space, protons can interact with the photons in the cosmic microwave background via Outline the steps (but don't calculate) needed to obtain the proton threshold energy for this process. Assume the CMB photons all have the same energy ( degrees Kelvin). [10 points]\n\nProblem 5: (40 points) Measurements of the entropy of a certain paramagnetic salt, as a function of the temperature (in degrees Kelvin) and of the magnetic field ( in gauss), have led to the following table (in units of per mole):\n\n\u0014\u0011 \u0014\u0011\u0017\u0018 \u0014\u0011 \u0013 \u0012\u0011 \u0012\u0011 \u0012\u0011 \u0018 \u0011 \u0011\u0015\u0018 \u0011\u0015 \u0011\u0015\n\n\u0014\u0011 \u0014\u0011 \u0013 \u0014\u0011\u0017\u0016 \u0014\u0011 \u0016\u0012 \u0012\u0011 \u0019\u0012 \u0012\u0011 \u0016 \u0012\u0011 \u0016\u0013 \u0012\u0011 \u0016 \u0012\u0011 \u0013 \u0012\u0011 \u0018\n\n\u0014\u0011 \u0019 \u0014\u0011 \u0014\u0011\u0017 \u0014\u0011\u0017\u0018 \u0014\u0011 \u0012 \u0014\u0011 \u0018\u0012 \u0012\u0011 \u0019\u0012 \u0012\u0011 \u0013 \u0012\u0011 \u0016\n\n\u0010 \u0012\u0011 \u0018 \u0012\u0011 \u0012\u0011 \u0018\u0018 \u0012\u0011 \u0019\u0018 \u0012\u0011 \u0016\u0013 \u0012\u0011\u0015 \u0012\u0011\u0015 \u0012\u0011 \u0013!\n\n\u0012\u0011 \u0013 \u0012\u0011\u0015\u0016 \u0012\u0011 \u0019 \u0012\u0011 \u0018 \u0014\u0011 \u0019 \u0012\u0011 \u0019\u0013 \u0012\u0011 \u0016 \u0012\u0011 \u0016\u0019 \u0012\u0011 \u0012\u0011 \u0013\n\n\u0012\u0011 \u0013 \u0012\u0011 \u0013 \u0012\u0011 \u0011\u0015\u0018 \u0011 \u0018 \u0011 \u0018 \u0011 \u0018\u0019 \u0011 \u0018 \u0011 \u0018\u0013 \u0011 \u0018\n\nIt is proposed to use this salt to produce very low temperatures by adiabatic demagnetization. The sample can be pre-cooled to 0.8 Kelvin by pumping on liquid He. The biggest available field is 10,000 gauss. What is the lowest temperature which can be reached? Each step in your reasoning should be explained very carefully (this is an essay question: numerical computations are not necessary)\n\n## Ph. D. Qualifying Exam Modern and Statistical Physics\n\nDo\n\nSept. 2012\n\nout of\n\nproblems\n\nProblem 1: (40 points) 1. Classical thermodynamics, Gay-Lussac-Joule experiment. (a) For any reversible process, calculate the dependence of the specific !\" internal energy on volume, , from the First law of Thermodynamics. Here, the internal energy = , . Here u and v are defined as u = U/n and v = V/n, where U, V, and n are the internal energy, volume, and a number of kilomoles. (hint: utilize the condition of an exact differential.) [25 points] (b) For an ideal gas, find the dependence of on by calculating [15 points] Problem 2: (40 points) In a two-slit Young interference, the aperture-to-screen distance is and the wavelength is !\"# . (a) What slit separation, , is required to produce a fringe spacing of at the screen? [15 points] (b) Assume a thin glass plate of thickness and index of refraction is placed over one of the slits (see figure below). The glass plate causes the entire fringe pattern to laterally shift up or down on the screen. Calculate the lateral fringe displacement. [25 points]\n!\" !\" ! !\" !\n\n## Screen Glass plate\n\nProblem 3: (40 points) A meson of mass ! decays at rest into a muon (mass ! ) and a neutrino of negligible mass. Express your answers below in terms of the masses ! and ! . (a) What is the momentum of the muon? [15 points] (b) What is the kinetic energy ! of the muon? [15 points] (c) The muon decays, too. Let ! be the mean lifetime of the muon in its rest frame. What is the mean distance traveled by the muon in the rest frame of the meson? [10 points]\n\nProblem 4: (40 points) A beam of either positively or negatively charged muons is stopped in a piece of Pb. Answer the following questions separately for both a positive muon and a negative muon. Hint: consider atom formation and the energy levels of a hydrogen-like atom. (a) Where is the most likely place for the muon to be after it stops in the Pb for ! and for ! ? [10 points] (b) What type of particles are emitted after the muon stops for ! and for ! ? [10 points] (c) What are the energy ranges of the particles emitted in item (b)? Give the answer in terms of eV or MeV for ! and for ! . [10 points] (d) What is the lifetime of the muon in its own frame compared to it decaying in vacuum? Do not calculate a number but state if it is the same, a shorter, or a longer time and why for ! and for ! . [10 points] masses: muon 105 MeV proton 938 MeV electron 0.5 MeV neutron 940 MeV pion 135 and 140 MeV photon and neutrino 0\n\nProblem 5: (40 points) Classical Thermodynamics of a Two-state System: A (very small) discrete system has only two states 1 and 2 with energies ! = ! and ! = ! , respectively. This could, for instance, be a spin 1 2 particle in an external magnetic field. Since this system contains only one particle, the different thermodynamic ensembles (of the systems described below) do not provide equivalent descriptions of the physics. We want to explore this difference for this simplest possible system. (a) If the system is isolated from the environment which are the possible values for the internal energy of the system? [5 points] (b) In the following we assume that the system is not isolated any more but instead interacting with a heat bath of temperature . Using the canonical distribution of classical thermodynamics, what are the probabilities ! to find the system in each of the two states in this case? [10 points] (c) Find the internal energy as a function of the temperature of the heat bath (express it in terms of a familiar hyperbolic function). [15 points] (d) Using the limiting values of the expression in (c), what are the possible values for the internal energy of the system when coupled to the heat bath? [10 points]\n\n## Ph. D. Qualifying Exam M odern and Statistical Physics\n\nFeb. 2012\n\nDo\n\nout of\n\nproblems\n\nProblem 1: (40 points) 1. Phase diagram, the critical point: The equation of state of van der\na Waals gas is P \u0010 2 v \u0010 b RT , where a and b are characteristic v\n\nconstants for a given gas, and P, v, T and R are pressure, specific volume, temperature and gas constant, respectively. When we regard the above van der Waals equation as P = P(v, T) (a) what are the three conditions at the critical point in the critical isotherm (T = TC) on a P-v diagram? [15 points] (b) Express the critical specific volume, vC, the critical temperature, TC, and the critical pressure, PC, at the critical point, in terms of the constants a and b. [25 points]\n\nProblem 2: (40 points) A metal ring is dipped into a soapy solution (index of refraction ) and held in a vertical plane so that a wedge-shaped film formed under the influence of gravity. At near-normal illumination with blue-green light (wavelength ) from an argon laser, one can see fringes per cm. Determine the wedge angle of the soap film. (Note: assume that the wedge angle is very small).\n\nProblem 3: (40 points) 2. A pi-mu atom consists of a pion and a muon bound in a Hydrogen-like atom. (a) What are the energy levels for such an atom compared to those for Hydrogen expressed in terms of the electron, pion, and muon masses? [20 points] (b) The pi-mu atoms are produced in decays pi-mu atom neutrino. If the has , what are the minimum and maximum energies of the pi-mu atom in the moving frame of the ? (express these energies in terms of the particle masses) [20 points]\n\nProblem 4: (40 points) Briefly explain or describe 4 of the following 6 phenomena (in no more than 200 words for each phenomena): [10 points each] (a) Electromagnetic structures of the neutron and proton (b) The laser (c) C, P, T, and CP symmetry (and any possible well known violations) (d) The transistor (e) 7KH\u0003-\u0012\u0003SDUWLFOH (f) Superconductivity\n\nProblem 5: (40 points) A system consisting of N (a very large number) identical weakly interacting particles is in equilibrium with a heat bath. The total number of individual states available to each particle is 2N. Of these, N states are degenerate with energy 0, and N states are degenerate with energy .\n\nN states\n\nN states\n\nIt is found by observation that the total energy of the system is . Find the temperature of the heat bath under three different assumptions: (a) That the particles are bosons. [10 points] (b) That the particles are fermions. [10 points] (c) That the particles obey the (unphysical) Boltzmann distribution. [10 points] (d) You should find from Parts (a), (b), and (c) that Explain why this is so. [10 points]\n\n## Modern and Statistical Physics Fall 2011\n\n9/24/2011 Do 3 out of 5 problems\n1. Classical Thermodynamics Ideal Gas [40 points] Assume that the earths atmosphere is an isothermal ideal gas in a gravitational field. Consider a thin layer of the ideal gas at height z and of thickness dz. There is a difference of pressure across this thin layer, dP, which must just balance the gravitational force on the mass. If the pressure at z = 0 is P0 , determine the pressure as a function of height z.\n\nFig.1 2. The Kinetic theory of Gas [40 points] The ditribution of particle speeds of a certain hypothetical gas is given by\nN ( v ) dv = Ave v v0 dv ,\n\nwherer A and v0 are constants. a. Determine A so that f ( v ) N ( v ) N is a true probability density function; i.e.,\n\nf ( v ) dv = 1 .\n0\n\n[10 points]\n\nb. Find the mean speed, v , and the root mean square speed, vrms , in terms of v0. [10 points] c. Find the most probable speed vm. [10 points] d. What is the standard deviation of the speeds from the mean, , in this case? [10 points] 3. Solid state physics, thermodynamics, statistical mechanics [40 points] There is a semiconductor with two bands: The lower band is described by E k , the upper band by E0 E k . The two bands are separated by an energy gap of 2. (Hint: Use electron-hole\n\n( )\n\n( )\n\nsymmetry. This is also evident from the way the upper band is defined.) a. Calculate the chemical potential as a function of the temperature. [20 points]\n\nb. Examine whether the system can be treated using classical statics at low temperatures. [20 points] 4. High energy physics [40 points] a. A particle of mass m is produced with energy E and decays after traveling a distance l. How long did the particle live in its own rest frame? [8 points] b. Use the result of part (a) to determine the spatial separation between the production and decay vertices (a.k.a. decay length) of a B0 meson carrying an energy of 65 GeV, if it lived 1.5 ps in its own rest frame. The mass of the B0 meson is 5.3 GeV. [8 points] c. Determine the maximum energy that can be carried off by any one of the decay particles, when a particle of mass m0 at rest decays into three particles with masses m1, m2, and m3. [24 points] 5. Diffraction grating (Multiple slit diffraction) [40 points] Consider diffraction of an array of multiple slits (N equal slits of width d and common spacing of a, numbered from 0 to N - 1) , or diffraction grating, illuminated by monochromatic plane wave normal to the array (Fig 2). We assume that the length of the slit is long enough, and the obervation point P is far enough from the slit. a. Obtain the expression for the intensity of the multiple slit diffraction pattern at a point P. [20 points] b. Now using the above diffraction grating, we perform an experiment to determine the wavelength of light. When the wavelength of the light changes to + d, the diffraction pattern shifts. If this shifts is smaller than the width of a bright band of the diffraction pattern, the wavelength and + d cannot be distinguished one another. Determine the accuracy of the measurement of the wavelength by this experimental method. [20 points]\n\nFig.2\n\n## Modern and Statistical Physics Spring 2011\n\n2/26/2011 Do 3 out of 5 problems\n1. High energy physics [40 points] Consider the reaction p + gamma --> neutron + pi+. a. Assume that the gammas are the cosmic blackbody radiation with average kT = 3 10 4 eV. What is the threshold energy that the proton must have for the reaction to occur? [20 points] mp ~ mn ~ 1000 MeV, m = 140 MeV, m = 105 MeV, m = 0 eV. b. The pion then decays to + . In the pion rest frame, what is the neutrino energy? What is the neutrino energy in the \"lab\" frame? This is the source of the highest energy neutrinos in the universe. Give answers in eV units. [20 points] 2. Relativity C-O white dwarf [40 points] a. A C-O white dwarf has radius R and mass M. Assuming the electrons are degenerate, what is the average energy of the electrons in terms of M, R and fundamental constants assuming the electrons are non-relativistic? [15 points] b. Repeat assuming the electrons are relativistic. [15 points] c. Now assume that the radius shrinks by a factor of 2 so R' = R/2 while the mass remains the same. What is the relative change in the gravitational and electron energies (assuming both relativistic and non-relativistic) between R and R'? [5 points] d. Is the star more stable (less likely to collapse) if the electrons are relativistic or nonrelativistic? Why? [5 points] 3. 2D harmonic oscillator [40 points] Give a 2D harmonic oscillator with Hamiltonian, H =\n\n2 p x 2m\n\n2 p y 2m\n\n1 m 2 x 2 + y 2 + kmxy . 2\n\na. For k=0, what are the energies of the ground state and first and second excited states? What are the degeneracies of each state? [20 points] b. For k>0, using first order perturbation theory, what are the energy shifts of the ground state and the first excited states? [20 points]\n\n## Classical mechanics Van der Waals Gas [40 points]\n\nThe Van der Waals equation of state for one mole of a non-ideal gas reads\n\na p + 2 (V b ) = RT . V\n[Note: part (d) of this problem can be done independently of part (a) to (c).] a. Sketch four isotherms of the Van derWaals gas in the p-V plane (V along the horizontal axis, p along the vertical axis). Identify the critical point. [10 points] b. Evaluate the dimensionless ratio pV/RT at the critical point. [10 points] c. In a portion of the p-V plane below the critical point, the liquid and gas phases can coexist. In this region the isotherms given by the Van der Waals equation are unphysical and must be modified. The physically correct isotherms in this region are lines of constant pressure, p0(T). Maxwell proposed that p0(T) should be chosen so that the area under the modified isotherm should equal the area under the original Van der Waals isotherm. Draw a modified isotherm and explain the idea behind Maxwell's construction. [10 points] d. Show that the heat capacity at constant volume of a Van der Waals gas is a function of temperature alone (i.e., independent of V). [10 points] 5. Statistical mechanics [40 points] There is a system of two identical particles, which may occupy any of the three energy levels 0 = 0, 1 = , 2 = 3 . The system is in thermal equilibrium at Temperature T, which means that the system is a canonical ensemble. For each of the following cases, determine the partition function and the energy and carefully enumerate the configurations. a. The particles are Fermions. [10 points] b. The particles are Bosons. [10 points] c. The particles obey Boltzmann statistics and now they are distinguishable. [10 points] d. Discuss the conditions under which Fermions or Bosons may be treated as Boltzmann particles. [10 points]\n\nModernandStatisticalPhysics.September/October2010 Do3of5problems\n1.AtopquarkdecaystoabottomquarkandaWboson.TheWbosonthendecaystoapositronanda neutrino.Giveyouranswersbelowintermsofthemassesmt,mb,mWofthetopquark,thebottomquark, andtheWbosonrespectively.Youmaytreatthepositronandtheneutrinoasmassless.Youare encouragedtosetc=1. (a)(14points)Whatisthemomentumofthebottomquark,intherestframeofthetopquark? (b)(14points)Whatisthemaximumpossiblemomentumofthepositron,intherestframeofthetop quark? (c)(12points)Supposethetopquarkhasspeed0.5cinthelabframe,beforeitdecays.Whatisthe maximumpossiblemomentumofthepositroninthelabframe?\n\n## (a)(10points)Ifthereisnoenergydifferencebetweenthetwostates,whatistheaveragelengthofthe chain? (b)(25points)Fixthechainatoneendandhangweightfromtheotherend,supplyingaforceFasshown. DeterminetheaveragelengthofthechainatanytemperatureTandshowthatitisequivalentto\n\nL T =\n.\n\naNe Fa / 2 kT e Fa / 2 kT e Fa / 2 kT\n\n(c)(5points)InwhichtemperaturelimitistheextensionproportionaltoF(Hooke'slaw)?Calculatethe constantofproportionality(thatisthekfromHookeslaw).\n\n3.Artificialmaterials(ormetamaterials)canbeengineeredtoprovideanegativeindexofrefractionn<0. Inthisproblemweexploretheamazingpropertiesoftwoarrangementsofmetamaterialsdescribedinthe Figurebelow. 1. (6points)Consideranincomingopticalrayattheinterfacefromvacuumtoametamaterial.Write downtheexpressionfortherefractedangleanddrawtheincomingandrefractedrayforn=1. Indicatethedifference(s)withtheresultobtainedforconventionalmaterials. 2. (17points)WenowconsideraslabofmetamaterialwiththicknesstasdrawnintheFigure(a) withn=1. a. (6points)Letapointlikesourceobjectbeatadistanced=tupstreamoftheslab,show thatthereisanimageandgivethelocationofthisimage. b. (6points)Samequestioniftheobjectisatadistanced=1/2t. c. (5points)Commentonthepropertiesofasimpleslabwithnegativeindexofrefraction comparedtoastandard(n>0)slab. 3. (17points)ConsidertheconfigurationshowninFigure(b). a. (6points)Viageometricalconstruction,showthatthesourceisimagedonthelower rightquadrant. b. (6points)Showthatraysemittedbythesourcearealsoimagedonthesourceitself. c. (5points)Discusspossibleapplicationsofsuchimaging/recirculatingconfigurations.\n\n4.\n\na.\n\nu = U V = aT 4 .[20points]\nb. FindthetotalentropyofaphonongasasafunctionofitstemperatureT,volumeVand theconstants k , h, c .[20points]\n\n5.Aflashlightiscollimatedsothatitsbeamgoesoffina90degreeconeinitsownrestframe.\n\n## Modem and Statistical Physics. February 2010 Do 3 of 5 problems\n\n1. Consider the reaction nubar\n\n*p)\n\n## inMeV. b. If the reaction occurs off Carbon\n\nc. Assuming a Fermi gas model, what is the Fermi energy for protons nC 12? What is the average proton kinetic energy? Give in MeV. d. Up to what antineutrino energy will suppression in C 12 still be a factor?\n\n12 (with proton density of 0.1/F^3), the reaction will be suppressed (hint, note the two lowest spectroscopic states in nuclei.) why. Explain energies. low antineutrino at\n\ng3g.6MeYlc2 hbar*c mass neutron mass proton: 938.3 MeV/c2 mass electron: 0.5 MeV/c2\n\n197\n\nMeV*F\n\nhc\n\n1240 MeV*F\n\n2.\nT:\n\nForthis problem, do not calculate any exponentials or roots but put in all the other numbers for the final\na.\n\nFor a H atom, what is the ratio of the probability to be in the n:2Ievel compared to the n:l level at 3000 degrees K? E(lS) -- -13.6 eV and kT: .025 eV at T : 300 degrees K b. For a metal with a Fermi energy equal to 4 eV and at T : 300 degrees K determine the following assuming a Fermi gas model: i. Wfrat is the density for conduction electrons? hc: 1240 eV*nm mass electron : 0.5 MeV/c2 ii. What is the ratio of the density of states between 4.4 eY and 4.0 eV? iii. What is the ratio of the probability to have the energy be 4.4 eY compared to 4.0 eV? c. For a photon in a Bosonic gas at T : 300 degrees K: i. What is the ratio of the density of states between 4.4 eY and 4.0 eV? ii. What is the ratio of the probability to have the energy be 4.4 eV compared to 4.0 eV?\nis traveling towards the Earth at 0.6c. It hres a projectile of mass M with a velocity of 0.8c frame. What is the energy of the projectile in the Earth's frame? If the rocket ship is rocketship's in the traveling perpendicular to a line from the ship to the Earth and again fires a projectile at 0.8c in the ship's frame and uf 90 d.gr\"\"s from the direction of the ship's motion (in the rocketship's frame), what is the energy of the projectile in the Earth's frame? What is the tangent of the angle of the projectile relative to the ship's motion (don't calculate but give as the ratio of two components)?\n\n3. A rocketship\n\n4. Optical mirage: We explore the propagation of an optical ray in the (Oxz) plane; see Figure. We assume the refractive index n obeys\nthe relation\n\na.\n\n## z = an2 * 6 where a andb are constants. where\n\nn(z)cosd:floaosd,o=A b.\nconstant.\n\nA is\n\nray trajectory\n\n(#)' =(1)'-\n\nl and show\n\nthat\n\nc.\n\n## the resulting ray trajectory is a pardbola.\n\n(20 points) We now wish that the model discussed in (f and (2) canbe used to explain the formation of optical mirages. We consider the (Oxz) plane to be filled with an ideal gas. We assume the gas to be in mechanical equilibrium (so that the change of pressure dP depends on the height variation dz as dP=-pgdz where g is the\n\ngravitational constant and pthe gas density). We consider a temperature gradient along z of the form T(z) : T0(l - kz)where k is a constant and firther assume the refractive index and the gas\ndensity satisff the relation trn'z\n\n-tIt p = const\n\n## (15 points) Show that the refractive index is related to\n\nz via z : an2 -l b\n\nand explicitly\n\ndetermine the constants a and b. (5 points) Qualitatively describe the formation of a \"warm\" mirage due to a hot floor.\n\n5. Partition function of\n\nan Einsteinsolid is\n\n7_ e \"' t - t-'-en'r\nwhere\n\n'\n\na. b. c. d.\n\nof 3N distinguishable oscillators.\nCalculate the Helmholtz function F. [10 points] Calculate the Enhopy,S. [10 points] Show that the entropy approaches zero as the temperature goes to absolute zero. [10\n\nft\n\n## Modern and Statistical Physics. September 2009 Do 3 of 5 problems\n\ncubic (fcc) is the most dense and the simple cubic (sc) is the least dense of the three cubic Bravais lattices. Suppose identical solid sphere are distributed through space in such a way that their centers lie on the points ofeach ofthese three structures, and spheres on neighboring pointsjust touch, without overlapping. (Such an arrangement of sphere is called a close-packing arrangement) a). Assuming that the spheres have unit density, show that the density ofa set ofclose-packed spheres on\nI . The face-centered\n\n## eachofthetbreestructures(the\"packingfraction\") ,\" centered cubic (bcc), and n l6 p 0.52 for sc.\n\ntlinl6x0.l4\n\nforfcc,\n\n\",linl8r0.68\n\nforbody-\n\nb). Show that for a fcc Bravais lattice the free electron Fermi sphere for valence 3 extends beyond point W of the first Brillouin zone (see figure), so that the first Brillouin zone is completely frlled fHint: prove u kF W = (1296 ll25 o')t' = 1.0081\n\nlf\n\n2. Considerthe reactiony+ e ) r| +n- +ewiththe electroninitiallyfree andatrest (withthis frame designated as the lab frame). a) What is the minimum photon energy for this reaction to proceed? b) Assuming that the photon energy is the minimum determined in a), what is the velocity of either the pions after the reaction?\n\nof\n\nc)Assumethepionthendecaystoamuonandneutrinon)V+v.Inthelabframe,whatisthe\nmaximum energy of the muon produced in the pion decay assuming the photon energy determined in a). Give your answers in terms of the pion, muon, and electron masses while setting the photon and neutrino masses to 0.\n3. A 55 year old man can focus objects clearly from 100 cm to 300 cm. Representing the eye as a simple lens 2 cm from the retina, a) what is the focal length of the lens at the far point (focused at 300 cm)? b) what is the focal length of the lens at the ner point (focused at 100 cm)? c) what strength lens (focal length) must he wear in the lower part of his bifocal eyeglasses to focus at 25 cm?\n\n4. A smooth vertical tube having two different sections is open from both ends and equipped with two pistons of different areas. Each piston slides within its respective tube section. One mole of ideal gas is inclosed between the pistons. The pistons are connected by a non-stretchable rod. The outside air pressure is 1 atm. The total mass of the pistons is M. The cross sectional area of the larger upper piston A1, and the lower piston A2 are related by 4 : A\" + AA. Find the rise in the temperature of the gas between the\npistons required to lift the piston assembly by a distance L?\n\n5. Assume that the neutron density in a neutron star is 0. l/frn3 lthat is 0. I neutron per cubic Fermi). Assuming T:0, ignoringany gravitational forces, and using a Fermi gas model with uniform density determine\n\na) b) c) d)\n\nthe average energy ofthe neutrons the average energy ofelectrons ifthe electron density is 1% ofthe neutron density show two reactions, one which can convert a neutron to a proton and one which can convert a proton\n\nto a neutron\n\ndetermine the neutron to proton ratio. Hint, consider the Fermi energies of the neutrons, protons and electrons at equilibrium Give answers in i) and b) in MeV using hc : 1240 MeVfrn, the mass of the neutron: 1000 MeV/c2, and the mass of the electron : 0.5 MeV/c2. You will need to decide if the particles are relativistic or nonrelativistic. The answer for d) can be given in terms of the particle masses.\n\nFebruary 2009. Stat/Modern/Thermo PhD. Qualifier. Pick 3 questions out of 5. 1. a. (20) Show using conservation of energy and momentum that it is not possible for a free electron moving through vacuum to emit a photon. b. (20) Now consider the related problem of an electron moving through superfluid helium. Show that in this case the electron can emit a phonon as long as it moves with a velocity exceeding a critical velocity vc and find vc . The excitation spectrum of the phonons in superfluid helium is given by E = up where u is the velocity of sound and p is the momentum of the phonon. You may do this part in the nonrelativistic limit. 2. a. (10) A CO white dwarf has a radius = R, mass = M and is assumed to have constant density. Assuming the electrons are degenerate and nonrelativistic find the average energy of the electrons in terms of M, R and fundamental constants. b. (10) Repeat part a) assuming the electrons are relativistic. c. (10) Assume the radius shrinks by a factor of 2 so R = R / 2 while the mass remains the same. Find the relative change in the gravitational and electron energies (assuming both relativistic and nonrelativistic) between R and R. d. (10) Discuss whether the star is more stable (less likely to collapse) if the electrons are relativistic or if they are nonrelativistic. 3. When a large number of atoms come together to produce a solid each atomic level broadens into a band. Draw a picture of simplified band structure, define the valence and conduction bands, and describe the temperaturedependent behavior of conductivity for: a. (10) Metals (define the term Fermi energy) b. (10) Insulators (define the term band gap) c. (10) Intrinsic semiconductors d. (10) The ntype and ptype semiconductors 4.\n\nOne mole of a monatomic ideal gas initially at temperature T0 expands from volume V0 to 2V0 . Calculate the work of expansion and the heat absorbed by the gas for the case of expansion at: a. (20) Constant temperature b. (20) Constant pressure\n\n## 5. a. (20) Consider a large number of N localized particles in an external magnetic\n\nfield H (directed along the z direction). Each particle has a spin s = 1/2. Find the number of states accessible to the system as a function of M s , the z component of the total spin of the system. Determine the value of M s for which the number of states is maximum.\n\nb. (20) Define the absolute zero of the thermodynamic temperature. Explain the\n\nmeaning of negative absolute temperature, and give a concrete example to show how the negative absolute temperature can be reached.\n\n## September 2008. Stat/Modern/Thermo PhD. Qualifier. Pick 3 questions out of 5.\n\n1. The electronic structure of the atom is described by four quantum numbers. a. Define these quantum numbers and describe the values they can acquire. b. Write down the electronic configurations for elements with atomic numbers: 8 (Oxygen), 19 (Potassium), 29 (Copper). c. Define the term atomic valence and find the valence for the elements O, K, and Cu. d. List four main types of bonding found in materials and give examples of the elements and/or compounds for each type of bonding. 2. Consider the reaction + p e + + n . a. Find the threshold energy for the if the reaction occurs off Hydrogen. b. Explain why the reaction will be suppressed due to Pauli Exclusion if it occurs off 12 C (with proton density of 0.1/F3). c. Assuming a Fermi gas model, find the approximate suppression as a function of neutrino energy. ( mn = 940MeV/c 2 , =c = 197 MEV F, hc = 1240 MeV F)\n\n3. A parallel beam of electrons is directed through a narrow slit of width a and a screen is placed at a distance d from the slit. The first diffraction minimum is observed at a distance y from the central maximum. a. Find the velocity of the electrons in terms of a , d , y , and me assuming the electrons are non-relativistic. b. Find the velocity assuming the electrons are relativistic. c. Estimate the spread of the electron beam through the slit using the Heisenberg uncertainty principle and show that this is consistent with the spread estimated based on the width of the central maximum.\n\n4. We have an ensemble of N spins that have a Zeeman splitting described by the Hamiltonian\nH = B B .\ni =1 N\n\n(1)\n\nHere B is the magnetic field, B is the Bohr magneton, and = 1 is the direction of the spin. a. Show that the partition function is given by B Z = (2 cosh B ) N . (2) k BT b. Show that the energy in terms of the partition function is given by ln Z (3) E = k BT 2 T and the magnetization by ln Z M = k BT (4) B c. Determine E and M for the partition function in Eqn. (2) and sketch them as a function of temperature.\n\n5. Chemical potential of an ideal gas. a. Starting with TdS = dU + PdV show that the chemical potential of an ideal gas can be written in terms of the temperature T and the volume V as :\n\n## = cPT cV T ln(T ) RT ln(V ) S0T + constant .\n\nHere, cP is the specific heat constant pressure, cv is the specific heat at constant volume and S0 is a constant. b. Starting from TdS = dH VdP find a similar expression for the chemical potential but now as a function of T and P. c. Show that the chemical potential at the fixed temperature T varies with pressure: P = 0 + RT ln . P0 Here, 0 is the value of at the reference point (P0, T)\n\n## Stat/Modern/Thermo PhD. candidacy examination. March 2008 Pick 3 questions out of 5.\n\n1) Consider a string of length L , mass density and string tension . The string is\n\n## maintained at finite temperature T .\n\na. Find the average amplitude of a mode of the string of wavelength = 2 L n in the\n\n## limit of kB T >> hf , with f the frequency of the mode.\n\nb. Explain why it is necessary to state that kB T >> hf . 2) Assume that the crystal lattice structure of solid comprising N atoms can be treated as an\n\nof 3N distinguishable one-dimensional oscillators (Einstein solid). assembly a. What is the partition function Z? Use the Einstein temperature E ( h/k, where is natural frequency). b. Calculate the Helmholtz function F. c. Calculate the entropy S. d. Show that the entropy approaches zero as the temperature goes to absolute zero. The variation of the internal energy U as a function of entropy S is predicted by classical equilibrium thermodynamics to have a functional form of U U 0 = ( S S0 ) where is a constant for a fixed value of volume.\na. Show that the temperature as a function of entropy in the case of constant volume\nn\n\n3)\n\nis given by T = n ( S S0 )\n\nn 1\n\n1 n 1 S S0 1 T = b. Show that Cv = . n 1 n 1 n\n\nT 2U At some point it may be useful to derive the relationship = 2 . S V S V 4) Silicon has Z=14. a. Relative to a hydrogen atom, what are the energy and radius of an electron in the 1s shell? b. In the 3p shell? c. Assume that there are 2 electrons in the 3p shell. What are the allowed spectroscopic states for this 2-electron system? (Give S, L, J). d. What is the energy ordering and why?\n5) A charged kaon (at rest) decays by K + + + 0 and then 0 + . a. In terms of the masses of the particles what is the 0 s energy? b. What is the maximum photon energy?\n\nModern/Optics/Thermodynamics September 2007 Pick 4 out of 6 problems. 1 In inertial frame O a rod of length l is oriented along the x-axis and moving with velocity u in the positive y direction. This rod is then viewed from an inertial reference frame O moving with velocity v in the positive x direction. a) What is the length of the rod in O? [10 points] b) What angle does the rod make with respect to the x axis? [30 points] 2 A cubic box (with sides of length L) holds diatomic H2 gas at temperature T. Each H2 molecule consists of two hydrogen atoms with mass of m each separated by distance d. Assume that the gas behaves like an ideal gas. Ignore the vibrational degree of freedom. a. What is the average velocity of the molecules? [10 points] b. What is the average velocity of rotation of the molecules around an axis which is the perpendicular bisector of the line joining the two atoms (assuming each atom as a point mass)? [10 points] c. Derive the expressions expected for the molar heat capacities Cp and Cv for such a gas. [10 x 2 points] 3. a. What is the threshold kinetic energy for the proton for p+ n p+ p+ assuming the neutron is at rest? [20 points] b. If now the neutron is in a Carbon nuclei of size 60 F3 (with 6 protons and 6 neutrons), what is the Fermi energy of the neutron, and what is the threshold kinetic energy in this case? [20 points] mn = mp = 1000 MeV, m = 140 MeV, hc = 200 MeVF 4. A smooth vertical tube having two different sections is open from both ends and equipped with two pistons of different areas. Each piston slides within its respective tube section. One mole of ideal gas is enclosed between the pistons. The pistons are connected by a non-stretchable rod. The outside air pressure is 1 atm. The total mass of the pistons is M. The cross sectional area of the larger upper piston A1, and the lower piston A2 are related by A1= A2+A. How much (in Kelvin) must the inner gas (between the pistons) be heated in order to lift the piston assembly by L=5cm?\n\n## M=5kg p0=1atm A1-A2=A=10cm2 CONSTANTS: g=9.8 m/s2 R=8.3J/(oKmole) kboltz=1.38x1023J/oK\n\np0\n\n5. Eight non-interacting neutrons are confined to a 3D square well of size D= 5 F (10-15 m) such that V = -50 MeV for 0 < x < D, 0 < y < D, 0<z<D and V = 0 everywhere else. a. How many energy levels are there in this well? [10 points] b. What is the degeneracy of each energy level? [10 points] c. What is the approximate Fermi energy for this system? [10 points] d. What is the relative probability to be in the lowest energy state to the fourth lowest energy state at kT= 10 MeV? Just write down the ratio (don't calculate the value). [10 points] mn = 1000 MeV, hc = 200 MeVF 6. Interference by a biprism. A plane wave (wavelength ) enters perpendicular to the biprism (the prism angle , refractive index n) as shown in the figure. The wave transmitted through both sides of the biprism is bent (refracted) and overlap at the viewing screen S (parallel to the biprism) where an interference pattern can be observed. a. Find the refracted angle, . [20 points] b. Find the interval of adjacent interference lines. [20 points]\n\nbiprism\n\nStatistical and Modern. Spring 2007. Pick 4 out of 6. 1) a) Starting with the first law of thermodynamics and the definition of c p and cv , show that\nU V c p cv = p + . V T T P Here c p and cv are the specific heat capacities per mole at constant pressure and volume\n\nrespectively, and U and V are the energy and volume of one mole. b) Use the above result plus the expression\nU p p + =T V T T V to find c p cv for a van der Waals gas with equation of state\n\n## a p + 2 (V b ) = RT . V Here a and b are constants.\n\nc) Use this result to show that as V at constant p , you obtain the ideal gas result for c p cv . 2) The rotational motion of a diatomic molecule is specified by two angular variables and and the corresponding canonical conjugate momenta, p , p . Assuming the form of the kinetic energy of the rotational motion to be\nrot = 1 2 1 p + p 2 2 2I 2 I sin ( ) r (T ) = 2 IkT h2\n\na) Derive the classical formula for the rotational partition function , r(T ),\n\nb) Calculate the Helmholtz free energy Frot . c) Calculate the corresponding entropy and specific heat. The following may be helpful\n\nax\n\ndx =\n\nsin\n\ndx = cot( ax ) / a 2 (ax )\n\n3) Assume that the neutron density in a neutron star is 0.1/fm 3 (that is 0.1 neutron per cubic Fermi). Assuming T=0 and ignoring any gravitational forces calculate the ratio of neutrons to protons to electrons.\n\nHint: determine their Fermi energy. The electron, neutron and protons masses are .511 MeV/c 2 , 939.6 MeV/c 2 and 938.3 MeV/c 2 . The constant hc = 1240MeV fm . You should be able to work out \"by hand\" an approximate value. 4) A atom consists of a pion and a muon bound in a Hydrogen-like atom. a) What are the energy levels for such an atom compared to those for Hydrogen? b) atoms are produced in KL decays ( K L + ). If the KL has = 0.8 what are the minimum and maximum energies of the atom expressed in terms of the K , and masses with m = 0 ? c) Approximately what fraction of KL decays will produce a atom (hint: use the Heisenberg uncertainty principle)?\n5) a) You are familiar with the quarter-wave thin film coating that acts as a reflectionreducer. For the moment, let us look at a simpler thin filmthe air gap between two pieces of glass such as you would find in a Newtons rings experiment. Why do we get constructive interference in the reflected when the thickness is one-fourth of the wavelength of light or some odd multiple of a quarter wavelength? Why isnt it constructive at one-half wavelength of the light? For assistance, I present two of the Fresnel equations (in two forms) for reflected light.\n\nrP = r =\n\nnt cos i ni cos t tan(i t ) = nt cos i + ni cos t tan( i + t ) ni cos i nt cos t sin( i t ) = ni cos i + nt cos t sin( i + t )\n\nWhere the parallel and perpendicular symbols refer to the plane of incidence, and i,t refer to incident and transmitted media, s are angles of incidence and transmission, and n s are indices of refraction. b) In light of the previous, to get destructive reflection in a thin film-i.e.-a quarter-wave film, such as the one illustrated below, what condition must prevail among the indices of refraction for the three media (n0 may be taken as = 1.0 for air.) c) The destructive interference described in part b) will generally not be complete. Find the value n0 of n1 as a function of n2 which gives completely destructive interference at normal incidence.\n\nn1\n\nn2\n\n6) In a big-bang theory of the universe, the radiation energy initially confined in a small region adiabatically expands in a spherically symmetric manner. Here the radiation (photon) pressure is expressed as p = U 3V , and the black body radiation energy density is u = U V = aT 4 The radiation cools down as it expands. a) Derive a relation between the temperature T and the radius R of the spherical volume of radiation, based purely on thermodynamic considerations. b) For the above problem, show the total entropy of a photon gas is expressed as\nS= 4 3 aT V . 3\n\nModern Physics, Optics, Statistical Mechanics and Thermodynamics September, 2006 Pick 4 out of 6 problems 1) A flashlight, in its own rest frame, is directed at an angle of 45 degrees to the x-axis in the xy plane. What angle will the light beam appear to make to an observer moving towards the flashlight along the x-axis at velocity ? a. Now consider a relativistic electron traveling in a circular orbit. Explain why the radiation from the electron will be confined to a narrow cone, and calculate the opening angle of the cone. b. 2) A simple model of a rubber band is a one-dimensional (horizontal) chain consisting of N ( N \u0015 1 ) linked segments, as shown schematically in the diagram.\n\nEach segment has two possible states: horizontal with length a, or vertical, contributing nothing to the length. The segments are linked such that they cannot come apart. The chain is in thermal contact with a reservoir at temperature T. a. If there is no energy difference between the two states, what is the average length of the chain? b. Fix chain at one end and hang weight from the other end, supplying a force F as shown. Determine the average length of the chain at any temperature T . Find the length in the limits T 0 and T .\n\nc. In which temperature limit is the extension proportional to F (Hookelaw)? Calculate the constant of proportionality.\n\n3) The complete formula for multislit diffraction pattern in Fraunhofer diffraction is given by the expression\n\nwhere\n\n## I 0 is the flux density in the = 0 direction for one slit,\n\nkb sin , and b is the slit width, 2 ka = sin , and a is the slit separation, and 2 2 is the propagation number. k=\n\na. Show that I (0) = N 2 I 0 , where N is the number of slits. b. For N 3 there are secondary maxima between the principal maxima. Deduce the rule for the number of minima between principal maxima, and, then the number of secondary maxima. Actually deduce this, dont just recite it if you happen to remember it! c. The result in a indicates that the flux density at = 0 varies as N 2 . Explain how that can be true. After all, e.g., if you have three slits, then there is three times a much light. How can the flux density be nine times as great? 4) Consider a system of two types of charge carriers in the Drude model. The two carriers have different densities (n1 and n2) and opposite charge (e and e), and their masses and relaxation times m1, m2 and 1, 2, respectively. Determine the Hall coefficient RH of this system.\n\n5) Assume that the cross section ratio for the two body reactions given below depends only on phase space of the final state particlesLarine o . Calculate the ratio.\n\n+ e + e\n\n+ p + p\n6). Consider an infinitely long cylinder that has been cut in half, with the two halves thermally insulated from each other (see figure). The top half is held at temperature T=30C, and the bottom half at T=10C. Find the temperature T(x,y) inside the cylinder. Hint: Use the conformal mapping z +1 w = ln z 1 z = x + iy\n\nModern Physics, Optics, Statistical Mechanics and Thermodynamics February, 2006 Pick 4 out of 6 problems 1. (Bohn) Consider an electron beam of circular cross section moving down the z-axis and passing through a system of focusing magnets. Suppose the beam density remains uniform, and its initial radius is R0 . Suppose further that the transverse components of the electrons momenta stay uniformly distributed over a circle in the momentum space , and the initial radius of this circle is P0 . a. If the focusing system reduces the beam radius from R0 to R1 , how does the distribution of transverse momenta change? b. Part (a) is highly idealized. Suppose, instead, that the electron beam is bunched, and each electron bunch moves non- relativistically through the focusing system. Suppose further that Coulomb self-forces affect the beam, and that the focusing system imparts an external magnetic field Bext. Write down the Vlasov-Poisson kinetic equation for the distribution function of electrons in the six-dimensional phase space of a single electron. c. The Vlasov-Poisson equations are, in general, notoriously difficult to solve. Why? d. Suppose, now, that the Hamiltonian for the beam bunch is independent of time and that the distribution function is a function only of the Hamiltonian (again, with regards to the six-dimensional phase space of a single electron). Is the beam in equilibrium? Explain your answer. T 1 T 2. (Ito) The Joule coefficient may be written = P , and the Joule v u c v T v Thomson coefficient may be written = (T 1). Here u is the internal P h c P energy, the expansivity, and the isothermal compressibility. Using these two coefficients, a. Find and for a van der Waals gas and b. Show that both are zero for an ideal gas. 3. (Ito) Consider a two-level system with an energy 2 separating upper and lower states. Assume that the energy splitting is the result of an external magnetic field B. Given that the total magnetic energy is U B = N tanh , show that the associated kT 2 2 kT 2 e heat capacity is CB = Nk 2. 2 kT ( e kT + 1) 4. (Benbow) Suppose you have large, long-focal length achromatic lens to use as the objective of an astronomical telescope. We are told in geometrical optics that if parallel rays are incident on a lens, the rays will converge to a point at the focal length\n\nof the lens.\n\nIf that is so, how can we expect to obtain an image of the moon (certainly not a point source) if we place a screen at the focal point of the lens? Explain, using words, not equations. 5. (Brown) Although a photon has no rest mass, it nevertheless interacts with electrons as though it has the inertial mass p p m= = c where the velocity of the photon is = c . a. Compute the photon mass (in units of eV) for photons of wavelengths 5000 (x-ray region). Compare to the mass of an electron (visible region) and 1.0 (which is 0.511 MeV). (Note: h = 4136 . 1015 eV sec ). b. When we drop a stone of mass m from a height H near the earths surface, the gravitational pull of the earth accelerates it as it falls and the stone gains the energy mgH on the way to the ground. A photon of frequency that falls in a gravitational field gains energy, just as a stone does. Using the photon mass relation, determine the new frequency, , of the fallen photon, and thus the frequency shift, , suffered by the photon. c. A passenger in an airplane flying at 20,000 feet aims a laser beam of red light ) towards the ground. What is the shift, , in the laser beam ( = 7000 wavelength an observer on the ground would measure? 6. (Hedin) Consider the two reactions of an elastic scatter of a neutrino off an electron at rest for muon-type and electron-type neutrinos: a) + e + e b) e + e e + e a. Which will have the larger cross section for E = 1 GeV? b. Explain (best if you show the first order Feynman diagrams). c. For E =1 GeV, what is the maximum angle a scattered electron will have with respect to the incoming neutrino?\n\nModern Physics: Pick 4 of 6 1) The potential energy of gas molecules in a certain central field depends on the distance r from the fields center as U ( r ) = r 2 where is a positive constant. The gas temperature is T, the concentration of molecules at the center of the filed is n0. a) Find the number of molecules dN, located at the distances between r and r+dr from the center of the field. b) Find the most probable distance separating the molecules from the center of the field. c) How many times will the concentration of molecules in the center of the field will change if the temperature decreases by (Tnew = T). (Hint, how does dN/N, that is, the fraction of molecules located in spherical layer between r and r+dr, behave at the center) d) Find the number of molecules dN, whose potential energy lies within the interval from U to U+dU. FORMULA SHEET for PROBLEM 1 ,n=0 1, n = 0 2 1 ,n= 1 2, n =1 n x2 n x 2 2 x e dx = x e dx = 0 0 1, n = 1 ,n=2 4 2, n = 2 1 ,n=3 2\n\n2) Consider a particle of mass m undergoing Brownian motion in one dimension. The particle is under the influence of a viscous friction force mv, an oscillatory driving force masin(t) and a random force mA(t). Its equations of motion are \u0005 = v, v \u0005 + v = a sin (t ) + A ( t ) . x a) Suppose A(t ) = 0 and A(t ) A( ) = ( t ) , where is a constant and (t) is the Dirac delta function. Suppose the initial conditions are q(0) =q0, v(0) = /. Find the mean speed v(t ) . Note:\nd e sin( ) =\n0 t\n\ne t ( sin(t ) cos(t ) ) +\n\n2 + 2\n\nb) Evaluate (by whatever means you chose) v(t ) in the limit 0, and provide a physical interpretation of your result. 3) The density of states for a free particle of momentum p confined within a box of volume is given by dn p 2 = dp 2 =3 a) Calculate the density of states per unit energy for a non-relativistic electron and for a massless neutrino. Assume each is confined within a box of volume but otherwise free of interactions. b) Assume that the transition probability for beta decay is dominated by the density of states term. Take the electron to be non-relativistic and the neutrino massless. In terms of the total decay energy, Etot, calculate the most likely energy for the emitted beta particle? 4) An energetic proton strikes a proton at rest. A K+ is produced. Write down a reaction for producing a K+ showing all the final state particles. What is the minimum kinetic energy for the incoming proton to produce this final state (express in terms of the masses of the particles)? 5) An interstellar proton interacts with the 3 degree cosmic microwave background (CMB) and produces an electron-positron pair via p+photon -> p+electron+positron. Assume the CMB is monoenergtic with E=kT= .002 eV. What is the minimum proton energy to produce this reaction? What is the electron energy at this threshold proton energy? The proton mass is 938 MeV/c2 and the electron mass is 0.5 MeV/c2.\n\n6)\n\n## Half Silvered Mirror\n\nDetector\n\nA Michaelson interferometer has two arms, the first of length L and the second of length L+d. The interferometer is illuminated by a polychromatic source of light with a frequency distribution given by 2 I ( ) = I 0 / 2 2 exp ( 0 ) / 2 2 a) Describe qualitatively how you expect the interference pattern for the case with polydisperse light to differ from the case where the interferometer is illuminated with a single frequency of light.\n\nb) Calculate the intensity of light observed as a function of the arm offset d. Does this result confirm your prediction from part a? You may want to use the following integral.\n\n( x y )2\n2 a2\n\ncos 2 ( kx ) dx =\n\n2 1 + cos ( 2ky ) e2 k\n2 2\n\n## Statistical Mechanics: Pick 2 of\n\n3.\n\nFebruary,2005\nsystem of identical stars for which the\n\n## distribution function in the phase space of a single star is\n\n:c.lv?)-I,, ro.vr!,, 2 L \\/ 2 |\nI\n=o\nfor Y\n\n## f (r,v) = CIH o - H (', O)]*''' an-3t2 t-\n\n<!r'\n2\n\nHere Y = V(t) = Ho-Q(r) denotes a relative potential, with O(r)being the actual potential, and ydenotes speed- Using Poisson's equation V'?@(r) = 4trGp(r) , find:\na) The density distribution function\n\nfr)\n\n## conesponding to the index n = 5.\n\nb) Are the stellar orbits in this system chaotic? Why or why not? Clearly explain your reasoning.\n\n2. T\\e equation for the distribution of free electrons in a metal in the vicinity of absolute\nzero is\n\n. ,l2m/2 d\"=:FJEat .\nMaking use of this equation, find at T=0K a) The velocity distribution of free electrons.\n\n3/\n\nb) The ratio of the mean velocity of free electrons to their maximum velocityc) What is the functional dependence of the Ferrni Energy on the density of electrons? 3. The distribution functions for identical particles (indistinguishable or distinguishable) can be represented by the equation:\n\nNr81\n\n*o Where a = 1 for Fermi-Dirac statistics, a= 'Gt-r)'tr -1 for Bose-Einstein statistics and a=0 for Maxwell-B oltzmann statistics.\ng, us. (e, - a) t nf for all three distriburions. b) For the Fermi-Dirac distribution, sketch N , I g, vs. e, for ?\"= 0 and I slightly greater\na) Sketch and compare N , I than zero.\n\nc) Show that for a system of a large number, N, of bosons at very low temperature (such that they are all in nondegenerate lowest energy state t= 0), the chemical potential varies with temperature according to:\n\np-+-kTlNasI-+0.\nModern Physics: Pick 2 out of 4\n\n1. Consider an assembly of identical two state atoms in equilibrium with a radiation field. The two states of the atom are labeled 1 and2 and their energy difference is hvstate 2 has the higher enersy. Lnt\n\np(v)\n\n=,yj\\)denote\n\n## the energv densitv\n\nof the radiation field per unit frequency interval. The probability for stimulated emission from state 2 to I and from state 1 to 2 are given by\n\nW|r= Br,rP(v)\nWrt.r-- Br,rP(v)\n\n## The probability of spontaneous emission from state 2 is given by, W),t-- A.\n\na) What is the state 2.\n\nratio\n\nN|/ Nz\n\n## of the number of atoms in state 1, to the number of atoms in\n\nb) Show thatB2l = Br,z and use this to find the rutio A/B' c) Explain briefly why it is necessary to have a non-thermal population distribution between two states in order to achieve laser action.\n\ndecay via\n\nKL\n\nin KL\n\n## pion-muon atom + neutrino.\n\na) What are the energy levels of the pion-muon atom relative to the hydrogen atom?\n\nb) What is the lifetime of the pion-muon atom in terms of the pion and muon lifetimes? c) If the KL is at rest when it decays, what are the momentum and kinetic energy of the pi-mu atom?\nPlease state all answers in terms of the masses and lifetimes of the particles involved (pion, muon, kaon, proton, or electron). 3. Find the energy Q, expressed in terms of the masses of the atoms Mp, M.' and the electron m\", liberated in B- and B+ decays, and in K-electron-capture if the masses of the parent atom is Mn, the daughter atom M6, and an electron me are known. (Assume electron binding energies are negligible)-\n\n4.\n\nwaves are to interfere, stipulations are that they must be traveling in the same direction in the same region of space, and that their oscillatoryplanes be parallel. ff this last condition is not met, something else will occur.\n\nIf two\n\na.\n\nSuppose\n\nEx = Eoicos(kz- rn)\n\nE, = Eoicos(kz-m)\nWhat happens here? What do you see if the wave is traveling toward you? b. Suppose\n\nEt\n\nEolcos(kz-\n\na)\n\nEv =\n\nEolsrntkz- ax1\n\n'\n\nWhathappens now? What do you see if the wave is traveling toward you?\n\nPh.D. Qualifying Exam. statistical Mechanics, Thermodynamics and Mo dem Physics. September 2004\nPart\n\n## Statistical Mechanics Thermodynamics. @ick 2 out of 3) If you answer all thrree\n\nwill\n\n1. A Carnot engine\n\nis operated between two heat reseryoirs at temperatures of 400 K to 300 K. (a) If the engine receives 1200 kilocalories from the reservoir at400K in each cycle, how much heat does it reject to the reservoir at 300 K? (b) If the engine is operated as a refrigerator (i.e., in reverse) and receives 1200 kilocalories from the reservoir at 300 K, how much heat does it deliver to the reservoir at 400 K? (c) How much work is done by the engine in each case? (d) What is the efiiciency of the engine in (a) and the coefficient of perfonnance in (bX\n\n## 2. An ideal monatomic gas consists\n\nisothermally to\n\nfill\n\na volume\n\nofN atoms in a volume V. The gas is allowed to expend 2v. showthatthe enhopy change is-Nkln2.\n\nJ. A mercury atom moves in a cubical box whose edge is I m\\ong. Its kinetic energJ is equal to the average kinetic energy of an atom of an ideal gas at 1000K. If the quantum numbers n\", r7y, and n\" are all equal to z, calculafe z. Note that the atomic mass of mercury is 200.6 g/mol (hint calculate the mass of an individual atom).\n\nJK:8.617 x 10-5 eV/K Stefan-Boltzmann constan! o =A[o=Ta :5.670x l0{ JK4m-2s-r ' l1h3ctPlanck's constant, h:6.62 x 10-3a Js Speed of lighg c:2.99792458 x 108 ms-r R: 8.34 x 103 J kmole-r K-r\n10\n\nk:\n\n-23\n\n0-23\n\nJ/K ,\n\nPart\n1..\n\nII.\n\n## Modern Physics and Optics (Pick 2 out of 4)\n\nAssume that the mu-neutrino vrandthe tau neutrino y.are composed of a mixture of two mass\n\neigenstates\n\n## -sin(e)[v,\\ f\",) _fcos(o) \\\",/ \\sin(o) cos(o) /\\v,/\n\nIn free space, the states v1 andv2 ovolve according to\n\n## \" fl\",(\",t) ,)(lv r(x, t)\n\nr\\\n\n_ _*.,,\n\n^(\"-'\"\"'o\n\n[v,\n\n(o) t\n\n\\\"-*\"\"p,(o)\n\n'/\n\na) Show tlrat the transition probability for a mu-neufino to turn into a tau neutrino is given by:\n\nPAt*\n\n\")=\n\n## b) Show that in the extreme relativistic limit this result becomes:\n\nPQt-- t) = sin'(2o1\"rr'l(*l\n\n\"\n\nHerq\n\n## I is the distance taveled by the neutrino, and E is the total energy.\n\n't)st1 \\,, L o* l\n\n## F 5 F (10^-15 m) suchthat for0<xcD V:-50MeV\n\nand\na-\n\nV:0\n\no<y<D\n\neverywhere else.\n\nHowmany energy levels are there inthis well? b. What is the degeneracy of each energy level? c. What is the approximate energy of the lowest energJ state? (and at least outline how one can improve estimating the energy).\nmass\n\nproton:\n\n938\n\nMeV/c^2\n\n## Hbar*c :197 MeV*F\n\n3. An electron with p 1 GeV/c strikes a proton at rest. Their masses are 0.51 Meylc^2 and 93g MeV/c\"2\nthe exiting proton can have? b. what is the maximum total energy that the exiting proton can have? c. What energy would the electron need to have to mate a rho meson, with mass of 770 MeV/C2? That is e+F>sFprrho.\na. What is the ma:rimtrm momenfum transverse to the incoming elecfion's direction that\n\n4.\n\n## For the waveform faveling in a dense glass\n\nE(r,t) =Xt(+i.3il(u/*)\"os[o.oa+rr\n\nx t06e\n\n+t.rtlnx ldsr]\n\nb. Detennine the wavelength ofthe light in air. c. Resolve the\" Eo\" into an amplifude times a unitvector. d. Determine which directionthe wave is propagating.\n\n## a^ Determine the ildex ofrefraction of the glass.\n\nPh.D. Qualifying Exam. Mechanics, Thermodynamics and Modern Physics. Statistical January 2044\nPart\n\n## note that only the first two\n\n1)\n\nCalculate the total electomagnetic energy inside an oven of volume 1 m3 heated to a temperature of 600 degrees Fahrenheit.\n\n2)\n\n## Anideal monatomic gas undergoes specific volume vz.\n\na reversible expansion\n\n## (a) (b) (c)\n\n3)\n\nCalculate the change in specific entropy As if the expansion is isobaric. Calculate As if the process is isothermal. Whioh is larger? By how much?\n\n## For N distinguishable coins the thermodynamic probability\n\ni, ,\n\n=ffi,\n\nwhere\n\nNr is the number of heads andN-Nr ttre number of tails. (a) Assume thatN is large enough that Stirling's approximation (lnn!-nlnn valid. Show that lno is ma:rimum forNr : N/2.\n\n-n) is\n\n(b)\n\n## Show that @^o-eNro2.\n\nPossible useful information -23 1.38 x 10 Boltzmann constant, J/K: 8.617 x 10-5 eV/K Stefan-Boltzmann constan! o:5.670 x 104 JK4m-2s-1 Planck's constanL h:6.62x 10-34 Js Speed of l^ight, c:2.99792458 x 108 ms-l R: 8-.34 x 103J knole-r K-l dU: TdS - PdV\n\nk:\n\nH:U+PV F : U.TS\nG:\nF+PV\n\nPart\n\n## II. Modern Physics and Optics\n\n1) A proton is confined\n\n(Pick 2 out of 4)\n\n## to a 2D square well of size D: 5 F (10\"t m) such that\n\nV:\nand\n\n-50 MeV\n\nfor0<x(D,0<y<D,\neverywhere else.\n\nv:0\na-\n\nHow many enerry levels are there in this well? b. What is the degeneracy of each energy level? c. What is the approximate energy of the lowest energy state? Give both the zeroth and first order approximation.\nMass proton:938\n\n## MeV/c2 h*c:197 MeV*F\n\n2)\n\nConsider a neufon star of mass M and radius R. Assume T: 0 K. Ignoring gravitational energy and assuming a simple Fermi gas model, what is the ma:<imum energy a neutron can have? What is the average energy of the neutrons? Do not calculate actual values but leave in the form using the mass of neufion, mo, and other constants.\n\n3)\n\nA photon from the cosmic microwave backgfound with energy 1.2x10-3 eV (corresponding to a wavelength of I mm) collides head-on with a relativistic electron having kinetiCenerry 1 Ge\\l The photon is backscattered i.e., scattered by 180o, after which the electron continues to move in its initial direction. Find the energy of\nthe photon after tlre collision. What is the wavelength of the backscattered photon? Consider a plane electromagnetic wave incident in the normal direction on a system of nnarrowslits each separated by a distance d. Derive a formula for the intensity of the soattering as a function of angle and wavelength in the Fraunhoffer approximation.\n\n4)\n\nO-*J?r\"\n\nStat-Mech Pick2 of 3.\n\nProblem 1:\nConsider the Fokker-Planck equation for the distribution function f:\n\nurhere 0 denotes a gradient with respect to velocity dynamical friction and dimlsion:\nanA\n\n## #*y, +q.ff = Q.(40*O(D.A),\n\nv.\nTake a simple model\n\nof\n\nof y.\n\n1is\n\n## the unit vector in the direction\n\nab. c.\n\nDescribe the overall effect of the right-hand side on the evolution of the systern How is the coefFrcient Brelated to the force fluctuations a constituent particle Deduce from the Fokker-Planck equation\n\ne4periences?\n\n## bw B afr D are related for a system in\n\nthermal equilibriurn\n\nProblem2.\nConsider the simplest model for desorption of atoms of mass m from a surface: one that ignored the tanslational degrees of freedonl and takes the possible energy states of the atom as -e (when adsorbed on the surfrce) and 0 when in the 'vacuum' at terperature\n\n*, b.\n\nT.\n\nUse the canonical ensemble for Natoms to obtain an eqrression for the number\n\nof\n\n## atdms inthe vacuum. Wbat is the total energy\n\nofthe system\n\nas a function oftemperature?\n\nProblem 3.\nConduction electrons are confined in a metal at T= 300 K. The Fermi energy is 4 eV. What is the relative density of states N(E) for energy of 4.4 eV compared to 3.6 eV? What is the relative probability for anelectronto have enErgy of 4.4 eV compared to 3.6 eV?\n\n## F'- il 't c'6i\n\nModem/Optics Pick 2 of 4\nProblem 4.\nDerive Malus's law: Malus's law relates the intensityof light passing tbrotr\\$ttwo ideal polarizers with their polarizing directions at an angle A with each other. Polarizers act on the electric field vector (amplitude) of the light wave traversing thern An ideal plafiznr passes 50% ofpurely unpolarized light incident on its surface, the remainder being now polarized parallel to the plarizng direction Now, armed with Malus's law, consider two ideal polarizers arranged so that a beam of light passing through both finds the second one at an angle of 45'to the fust. Then a 3'd ideal polarizer is inserted between them and rotated at angular frequency ar, Derive an expression for the intensity ofthe light emitted from the system as a function oftime.\n\nProblem 5.\nTwo neutrinos leave a supenrcva explosion at the same time. They arrive at earth a distance L away separated in time by At. The first neutrino to arrive is found to bave energy Et andthe second energy Ez. Findthe mass ofthe neutrino.\n\nProblem 6.\nThe disintegration constant )\" of aradioactive nucleus is defined as the fraction of nuclei that decay per unit time. Let N(t) bethe number of nuclei present at time /. Derive the decay law, N(t): N(0)exp(-tr t) What is the relation between half-life Tn afr ,L2 In an excitation experiment, a parent nucleusp is produced at arate.R, and decays with a disintegration constant 2 p,to a daughter nucleus 4 whose disintegration constants is La. At H, the number ofparent nuclei isifi0):N6oand the number of daughter nuclei is ,va(0):0. Find nrlO and ila(Q for t>0, and show that after a long time, an equilibrium is attained.\n\na. b. c.\n\nProblem 7.\nAn electron is in the spin state\n\nb.\n\na.\n\nDetermine the normal Find the expectation values ^rr\"^'\"1\"!r!^i)o of S*, Sr, and Sr.\n\nNIU Ph.D./Master qualifier examination 2003 Spring Statistical Physics and Modern Physics\nSolve 4 problems. Choose 2 problems from I, II and from V and VI, choose 1 problem from IV and VII.\n\nL\ne,'\n\n## Consider a nonrelativistic free particle in a cubical container of edge length\n\n-L\n\nand volume\n\nr3\n\na b\nc\n\nEach quantum state r of this particle has a corresponding kinetic energy on Z What is t,(V)?\n\nwhich depends\n\n## Find the contribution to the gas pressure p,\n\n^c crdll(J ^ ^-) vL f/ v\n.\n\np of any ideal\nE\ngas\n\n## Use this result to show that the mean pressure\n\nof weakly interacting\n\nAV V\n\n=1rlr,\n\nirrespective\n\n## of whether the gas obeys classical, Fermi-Dirac, or Bose-trinstein statistics.\n\nII\n\nConsider a system of two atoms, each having only 3 quantum states of energies 0, e and 2e . The system is n contact with aheatreservoir at temperature T. Write down the partition function Z far the system if the particles obey\n\nA. B.\nC. D.\n\nClassical statistics and are distinguishable. Classical statistics and are indistinguishable. Fermi-Dirac statistics. Bose-Einstein statistics.\ngas obeys the equation\n\nlll\n\nA certain\n\noistate P(v - b) = RT\" where r? is the universal gas constant, andb is a constant suchthat 0 <b <Vfot allV.\n\na b . V\n\nDerive an expression for the work obtained from a reversible isothermal expression one mole of this gas from an initial volume Vito aftnalvohsne V7.\n\nof\n\n## same process produce more or less work?\n\nIVFindthethresholdenergyfortheprocess,y+p-uo*p,inwhichasingle'zrmesonis\nproduced when a photon strikes a proton at rest. The rest energy of the pion is 135 MeV and the nroton 938 MeV. Take an ideal monatomic gas (y: 5/3) around the Carnot cycle, wherc point I atthe beginning of the adiabatic compression has pressure Pr : Pa (atmospheric presswe), volume Vt:13liters, andtemperatureTt:300 K. Point 3 has pressureP::2Po andvolume V3:26 liters. Calculate the values of volume and pressure at all four points of the Carnot cycle.\n\nW An ideal\n\ninitially at temperature, Ti, pressure, Pi and volume, V has its pressure reduced to a final value Pi< Pi via one of the following processes: (1) isochoric; (2) isothermal; (3) adiabatic.\ngas\n\na b c d\n\n## Sketch each process schematically on a P - V diagram.\n\nIn which process is he work done on the gas zero? In which process is the work done on the gas greatest?\nShow that the ratio the absolute magnitudes of the heat transfers, Qi*t1,.r\".-u1 to Qiro\"1,o6\" is given by\n\nleu,,*^^l=\n\nlril=4Eilt'tp lYisochnncl\n'ul*\n\n*e l,,lg)l\nll \\'/ ll\n\ntt\n\ne VII\n\nIn which process is the change in internal energy of the gas the greatest? Explain your\nreasoning.\n\nThe k-alpha radiation from copper tZ19) occurs via the following process- An incident high energy electron removes an electron from the n:1 orbital and subsequently an electron from the n:2 orbital falls down to the n:l orbital releasing an x-ray.\n\na b c d\n\nIgnoring electron-electron interactions calculate, in units of the Rydberg constant R, the energy of the Cu k-alpha x-ray. More realistically, the second 1S electron will shield the nuclear charge. Assuming perfect shielding, calculate a better estimate of the Cu k-alpha energy.\nDiscuss, in qualitative terms, how inter-electron interactions might modifu the Cu k-alpha energy, beyond just the perfect shielding approximation. The energy of the Cu k-alpha x-ray is slightly modified by the electron's spin-orbit interaction. This leads to a splitting of the k-alpha line into a k-alpa-l and k-alpha-2.\n\ni ii iii\n\nList all possible values of total (orbital + spin) angular momentum for the n:1 and\n\nn:2\n\nstates.\n\nList all transitions from states with n:2 to states with selection rules for atomic transitions.\n\n## n:l that are permitted\n\nby the\n\nExplain the observ'ed ratio of 2:1 for the Cu k-alpha-l to k-alpha-2 fluorescence yields."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8525164,"math_prob":0.9857346,"size":59144,"snap":"2019-35-2019-39","text_gpt3_token_len":15504,"char_repetition_ratio":0.16428813,"word_repetition_ratio":0.06094289,"special_character_ratio":0.24673678,"punctuation_ratio":0.11789702,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.996171,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-23T16:47:26Z\",\"WARC-Record-ID\":\"<urn:uuid:ca0ef94a-0d49-44e7-a580-6ecbff2bdf99>\",\"Content-Length\":\"420948\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d036ac1f-e6e2-4288-bd24-9521fc3f5a57>\",\"WARC-Concurrent-To\":\"<urn:uuid:66f04862-1cb4-4b39-b89f-402ed87ae6de>\",\"WARC-IP-Address\":\"151.101.250.152\",\"WARC-Target-URI\":\"https://de.scribd.com/document/148858357/Mod-ALL\",\"WARC-Payload-Digest\":\"sha1:IBKBZTJRL22THJTKFU6W2UAKMIIHGSQT\",\"WARC-Block-Digest\":\"sha1:7AXTTC7TSBDMV5ZFNYAVWQV7Y4OGTI52\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514577363.98_warc_CC-MAIN-20190923150847-20190923172847-00520.warc.gz\"}"} |
https://statisticsglobe.com/convert-vector-from-rcpparmadillo-to-rcpp-r | [
"# Convert Vector from RcppArmadillo to Rcpp & Vice Versa in R (2 Examples)\n\nAs R users, we use the package Rcpp to make our R code much more efficient by integrating C++ code. We can define objects in the C++ code as standard Rcpp or RcppArmadillo objects.\n\nIn this post we will show you how to convert a vector from RcppArmadillo to standard Rcpp. And we also show you the opposite way: How to convert a standard Rcpp vector to an Armadillo vector.\n\nTake a look at our blog posts about Rcpp and RcppArmadillo first if both terms are new to you.\n\nThis post has the following structure:\n\nLet’s do some vector conversion.\n\n## Example 1: Convert Vector from RcppArmadillo to Rcpp\n\n```if (!require('Rcpp', quietly = TRUE)) { install.packages('Rcpp') } library('Rcpp') # Load package 'Rcpp'```\n\nAs an example, we create a function “f_example” which interprets its input vector as an Armadillo vector and transforms it into an Rcpp vector.\n\n```cppFunction(depends = \"RcppArmadillo\", ' Rcpp::NumericVector f_example( arma::vec vec_arma ) { // Transform Armadillo vector \"vec_arma\" into a standard Rcpp vector // Here are 2 possibilities to do that Rcpp::NumericVector vec_Rcpp = as<NumericVector>(wrap(vec_arma)); Rcpp::NumericVector vec_Rcpp_2 = NumericVector(vec_arma.begin(), vec_arma.end()); // Take a look at the the different vectors Rcout << \"vec_arma:\\\\n\" << vec_arma << std::endl; Rcout << \"vec_Rcpp:\\\\n\" << vec_Rcpp << \"\\\\n\" << std::endl; Rcout << \"vec_Rcpp_2:\\\\n\" << vec_Rcpp_2 << \"\\\\n\" << std::endl; return vec_Rcpp_2; } ')```\n\nWe use Rcout to print certain objects in the console. It is pretty useful for writing C++ code and taking a look at some intermediate results. Both lines of code, as(wrap(vec_arma)) and NumericVector(vec_arma.begin(), vec_arma.end()) transform the Armadillo vector into an Rcpp vector. We can run an example to see the different versions of the vector.\n\n```f_example(1:5) # vec_arma: # 1.0000 # 2.0000 # 3.0000 # 4.0000 # 5.0000 # # vec_Rcpp: # 1 2 3 4 5 # # vec_Rcpp_2: # 1 2 3 4 5 # # 1 2 3 4 5```\n\n## Example 2: Convert Vector from Rcpp to RcppArmadillo\n\nNow, we do it the other way around. We create a function “f_example_2” which interprets its input vector as an Rcpp vector and transforms it into an Armadillo vector.\n\n```cppFunction(depends = \"RcppArmadillo\", ' arma::vec f_example_2( Rcpp::NumericVector vec_rcpp ) { // Transform Rcpp vector \"vec_rcpp\" into an Armadillo vector arma::vec vec_arma = as<arma::vec>(wrap(vec_rcpp)); // Take a look at the the different vectors Rcout << \"vec_rcpp:\\\\n\" << vec_rcpp << \"\\\\n\" << std::endl; Rcout << \"vec_arma:\\\\n\" << vec_arma << \"\\\\n\" << std::endl; return vec_arma; } ')```\n\nTake a look at the function for example values 1 to 5.\n\n```f_example_2(1:5) # vec_rcpp: # 1 2 3 4 5 # # vec_arma: # 1.0000 # 2.0000 # 3.0000 # 4.0000 # 5.0000 # # # [,1] # [1,] 1 # [2,] 2 # [3,] 3 # [4,] 4 # [5,] 5```\n\nYou see that in RcppArmadillo, vectors are interpreted as column vectors.\n\n## Video & Further Resources\n\nFor more information, we recommend you to take a look at the CRAN information on the Rcpp and RcppArmadillo package. Furthermore, the Rcpp homepage and the Armadillo homepage are very useful.\n\nIf you want to learn more, you might want to take a look at the following video about templates from CppCon 2021, uploaded by the CppCon channel.\n\nPlease accept YouTube cookies to play this video. By accepting you will be accessing content from YouTube, a service provided by an external third party.",
null,
"If you accept this notice, your choice will be saved and the page will refresh.\n\nWe have more videos on https://statisticsglobe.com/ which you might want to take a look at:\n\nWe showed you how to convert Rcpp vectors into RcppArmadillo vectors and vice versa. For questions or comments, please use the section below.\n\nThis page was created in collaboration with Anna-Lena Wölwer. Have a look at Anna-Lena’s author page to get more information about her academic background and the other articles she has written for Statistics Globe.\n\nSubscribe to the Statistics Globe Newsletter"
]
| [
null,
"https://statisticsglobe.com/wp-content/uploads/2020/09/YouTube-Tutorial-Preload-Thumbnail-R-Programming-Video.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.65405035,"math_prob":0.55929625,"size":5392,"snap":"2022-27-2022-33","text_gpt3_token_len":1692,"char_repetition_ratio":0.1722346,"word_repetition_ratio":0.49046016,"special_character_ratio":0.33308604,"punctuation_ratio":0.18208092,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9895086,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-29T22:21:08Z\",\"WARC-Record-ID\":\"<urn:uuid:c96e2b21-4502-4ba2-b57a-5b37443cdb80>\",\"Content-Length\":\"136501\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:32ba86e8-12dc-48f6-a197-3021d9254491>\",\"WARC-Concurrent-To\":\"<urn:uuid:93e8bedf-888d-48bd-ada8-20867563894a>\",\"WARC-IP-Address\":\"217.160.0.159\",\"WARC-Target-URI\":\"https://statisticsglobe.com/convert-vector-from-rcpparmadillo-to-rcpp-r\",\"WARC-Payload-Digest\":\"sha1:M4BALGRVHFJPVPG76TUPUFKQUKVVF6DE\",\"WARC-Block-Digest\":\"sha1:KJ2QJAGFDKKR5TPPVYAUCI6ZHBIACZMI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103645173.39_warc_CC-MAIN-20220629211420-20220630001420-00634.warc.gz\"}"} |
https://qwng.github.io/teaching/fall2018/stat225/slides/lecture1.html | [
"## Introduction to Probability Models\n\nLecture 1\n\nQi Wang, Department of Statistics\n\nAug 20, 2018\n\n### About the Instrcutor\n\n• Course Instructor: Qi Wang(Pronounced as Chee Waung)\n• Email: [email protected]\n• Homepage: http://www.stat.purdue.edu/~wang2047/\n• Office: MATH G143\n\n## Syllabus\n\n### Course Info\n\n• Textbook: Introduction to Probability by Mark Daniel Ward and Ellen Gundlach, $1_{st}$ edition, W.H. Freeman\n• Course website:\n• http://www.stat.purdue.edu/~cfurtner/stat225\n• user name: stat225\n• password: fall2018\n• Session website: http://www.stat.purdue.edu/~wang2047/teaching/fall2018/stat225\n\n### Grading\n\n Homework 22.0% 125pts Quizzes 18.4% 105pts Exam 1 17.5% 100pts Exam 2 17.5% 100pts Final Exam 22.0% 125pts Class Participation 2.6% 15pts Total 100% 570pts\n\n### Grades\n\n• There will be NO curving of individual exam grades\n• A student must earn a minimum of 60% on AT LEAST ONE of the 3 exams in order to pass this class.\n\n### Quiz\n\n• 8 are scheduled\n• Close book and close notes\n• The lowest quiz will be dropped\n• Make-up quiz\n• Official documented University business or a documented illness\n• Contact your instructor at least TWO DAYS in advance\n\n### Homework\n\n• 5 assignments\n• Due at the begining of class\n• Late homework will NOT be accepted\n• Must be handwritten or typed using mathematical notation.\n• Each homework is worth 25 points, NO homeworks are dropped.\n\n### Exams\n\n• Two evening exams from 8:00 -- 9:30 pm\n• Exam 1: Tuesday, 9/25/2018\n• Exam 2: Tuesday, 10/30/2018\n• A final exam, during the day during final exam week\n• Close book and close notes\n• Items allowed\n• pencils\n• erasers\n• a scientific calculator (must not have capability to do integration)\n• one-page cheat sheet for mid-terms and two-page for the final\n• Show a photo ID to your instructor\n\n### Cheat sheet\n\n• $8 \\frac{1}{2} \\times 11$\n• Handwritten in your own writing\n• Both sides\n• Handing in your cheat sheet at the end of the exam is required\n• Use of printed or photocopied material on a cheat sheet is prohibited and considered cheating in this course\n\n### Emergency\n\n• If you hear a fire alarm inside, proceed outside\n• If you hear a siren outside, proceed inside\n• Fire emergency:\n• immediately suspend class, evacuate the building, and proceed outdoors\n• do not use the elevator\n• meet outside by fountain near John Purdue’s grave\n• Tornado warning/servere weather event\n• suspend class and shelter in interior hallway on $1_{st}$ floor\n• Shelter in place\n• suspend class and shelter in the classroom\n• shutting the door and turning off the lights\n\n• A gambler's dispute in 1654 led to the creation of a mathematical theory of probability by two famous French mathematicians, Blaise Pascal and Pierre de Fermat\n•",
null,
"",
null,
"• The game consisted in throwing a pair of dice 24 times, the problem was to decide whether or not to bet even money on the occurrence of at least one \"double six\" during the 24 throws\n\n## Basic Terminology\n\n### Basic Terminology\n\nThis course is a course on Probability. The following terminology will assist us in the study of Probability.\n\n• Element: a single item(outcome), typically denoted by $\\omega$\n• Set: a collection of elements\n• For example, $A = \\{x, y, z\\}$\n• $x \\in A$, x is a memeber of set A\n• $j \\notin B$, j is not a smember of set B\n• Population: the collection of all individuals or items under consideration\n• Random Experiment: an action whose outcome cannot be predicted with certianty beforehand\n• Sample Space: the set of all possible outcomes for a random experiment, typically denoted by $\\Omega$, the textbook uses $S$\n• Event a result that may or may not occur, a subset of $\\Omega$\n• Subset: a set in which every element is contained in another set\n• Notation: $A \\subset B$, A is a subset of B\n• Complement: a set that contains all of the elements in $\\Omega$ that aren't in the original set\n• Notation: $A^c$ is the complement of $A$\n• Empty Set: the set with no element in it, denoted by $\\emptyset$ or $\\{\\}$\n\n### Example 1\n\nWe check whether the Standard and Poor’s 500 Index at the end of the day shows an increase, a decrease or remains the same as the previous days ending index.\n\n1. For one day, what is the sample space for this scenario?\n2. What is the sample space for two days?\n3. Define event A: the S & P decreases at least one day. List the outcomes in A.\n4. What are the outcomes in the complement of A?\n5. What do you notice if we combine A and its complement?"
]
| [
null,
"https://qwng.github.io/teaching/fall2018/stat225/image/blaise-pascal.jpg",
null,
"https://qwng.github.io/teaching/fall2018/stat225/image/pierre_de_fermat.jpg",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.85241795,"math_prob":0.65436137,"size":4340,"snap":"2019-35-2019-39","text_gpt3_token_len":1149,"char_repetition_ratio":0.07818266,"word_repetition_ratio":0.015831135,"special_character_ratio":0.27096775,"punctuation_ratio":0.103658535,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9548988,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-19T01:23:07Z\",\"WARC-Record-ID\":\"<urn:uuid:f233418f-6487-4343-8a47-bda6ef126ea5>\",\"Content-Length\":\"11162\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0dc87e6e-8265-4988-8bc1-0ce902627e70>\",\"WARC-Concurrent-To\":\"<urn:uuid:7ac13ee5-db87-45dd-a960-b3105df5a22a>\",\"WARC-IP-Address\":\"185.199.111.153\",\"WARC-Target-URI\":\"https://qwng.github.io/teaching/fall2018/stat225/slides/lecture1.html\",\"WARC-Payload-Digest\":\"sha1:LSPUISJ55DNOQDNJOE5IGVZWLHDHT66C\",\"WARC-Block-Digest\":\"sha1:56657QSXUA64IP52752BY7WVATIFRXEF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027314638.49_warc_CC-MAIN-20190819011034-20190819033034-00017.warc.gz\"}"} |
https://journals.plos.org/ploscompbiol/article?id=10.1371/journal.pcbi.1002536 | [
"# Fast Coding of Orientation in Primary Visual Cortex\n\n• Oren Shriki ,\n\[email protected]\n\nAffiliations Department of Physiology and Neurobiology, Ben-Gurion University of the Negev, Be'er-Sheva, Israel, Laboratory of Systems Neuroscience, National Institute of Mental Health, Bethesda, Maryland, United States of America\n\nAffiliations Dom Purpura Department of Neuroscience, Albert Einstein College of Medicine, New York, New York, United States of America, Department of Ophthalmology and Visual Sciences, Albert Einstein College of Medicine, New York, New York, United States of America\n\n• Maoz Shamir\n\nAffiliations Department of Physiology and Neurobiology, Ben-Gurion University of the Negev, Be'er-Sheva, Israel, Department of Physics, Ben-Gurion University of the Negev, Be'er-Sheva, Israel\n\n# Fast Coding of Orientation in Primary Visual Cortex\n\n• Oren Shriki,\n• Maoz Shamir",
null,
"x\n\n## Abstract\n\nUnderstanding how populations of neurons encode sensory information is a major goal of systems neuroscience. Attempts to answer this question have focused on responses measured over several hundred milliseconds, a duration much longer than that frequently used by animals to make decisions about the environment. How reliably sensory information is encoded on briefer time scales, and how best to extract this information, is unknown. Although it has been proposed that neuronal response latency provides a major cue for fast decisions in the visual system, this hypothesis has not been tested systematically and in a quantitative manner. Here we use a simple ‘race to threshold’ readout mechanism to quantify the information content of spike time latency of primary visual (V1) cortical cells to stimulus orientation. We find that many V1 cells show pronounced tuning of their spike latency to stimulus orientation and that almost as much information can be extracted from spike latencies as from firing rates measured over much longer durations. To extract this information, stimulus onset must be estimated accurately. We show that the responses of cells with weak tuning of spike latency can provide a reliable onset detector. We find that spike latency information can be pooled from a large neuronal population, provided that the decision threshold is scaled linearly with the population size, yielding a processing time of the order of a few tens of milliseconds. Our results provide a novel mechanism for extracting information from neuronal populations over the very brief time scales in which behavioral judgments must sometimes be made.\n\n## Author Summary\n\nHow can humans and animals make complex decisions on time scales as short as 100 ms? The information required for such decisions is coded in neural activity and should be read out on a very brief time scale. Traditional approaches to coding of neural information rely on the number of electrical pulses, or spikes, that neurons fire in a certain time window. Although this type of code is likely to be used by the brain for higher cognitive tasks, it may be too slow for fast decisions. Here, we explore an alternative code which is based on the latency of spikes with respect to a reference signal. By analyzing the simultaneous responses of many cells in monkey visual cortex, we show that information about the orientation of visual stimuli can be extracted reliably from spike latencies on very short time scales.\n\n## Introduction\n\nFiring rates of many primary visual cortical cells are tuned to the orientation of visual stimuli . This dependence of neuronal firing rates on the stimulus implies that information about the stimulus can be decoded from the spike count. The trial to trial variability of firing limits the accuracy with which a stimulus can be estimated from the neuronal spike count . To decrease this variability and increase the accuracy of the rate code, studies have typically used responses measured over several hundred milliseconds , , . However, increasing evidence indicates that the central nervous system can process complex information on very short time scales.\n\nVisual psychophysical and evoked potential studies have shown that human subjects can classify natural scenes or emotional facial expressions on the basis of 100–150 ms of processing . Evidence for fast processing of visual stimuli also exists from behavioral and electrophysiological experiments in monkeys . A recent study by Stanford et al. shows that monkeys can make perceptual decisions regarding the color of stimuli after about 30 ms of processing time. Evidence for fast coding also exists for the auditory system , and the somatosensory system , . The overall theme deriving from these studies is that sensory systems are able to process the gist of a scene rapidly .\n\nIt has been suggested that the temporal structure of the neuronal response and in particular, response latency, is the source of fast decisions in the brain , , . However, the accuracy of codes based on these responses has not been studied in the visual system systematically.\n\nA common approach to measuring response latency is to define it as the transition from spontaneous firing to stimulus-dependent firing, e.g., by detecting the time at the which the PSTH (Post Stimulus Time Histogram) reaches half of its maximal firing rate . This attempts to estimate the ‘pure’ latency component of the response, but it involves defining that quantity by a different number of spikes for each condition. For instance, latency might be defined by the time to the first ten spikes at the preferred stimulus and to the first spike at a non-preferred stimulus. Thus, in this approach the criterion for neural response time depends on the stimulus, making it impractical for decoding: the readout parameters cannot scale in a stimulus dependent manner, as that requires the readout to know the stimulus in order to estimate it.\n\nRecently, we proposed a simple spike latency code readout , the temporal Winner-take-all (tWTA). The tWTA determines the external stimulus by the label, e.g. preferred orientation, of the cell that fired the first spike in the population. It avoids attempting to estimate ‘pure’ onset latency and instead takes a pragmatic approach in which each cell's time to first spike will depend both on its latency and the strength of its response.\n\nFormally, consider a population of N neurons coding for the orientation of a visual stimulus,",
null,
". Let us denote by",
null,
"the time of the first spike of neuron i, with preferred orientation",
null,
", following some reference signal tref. The tWTA algorithm estimates the orientation of the external stimulus as the preferred orientation of the neuron which fired first with respect to tref:",
null,
"This definition can be generalized to estimate the stimulus by the preferred orientation of the cell that fired the nth spike first, or to incorporate a competition between groups, ‘columns’ of cells (see below). Here we investigate neural coding on brief time scales by applying the tWTA to simultaneously recorded populations of neurons in the primary visual cortex of macaque monkeys responding to the orientation of visual stimuli.\n\n## Results\n\nResponses of multiple neurons were measured in primary visual cortex (V1) of anesthetized monkeys using electrode arrays. The stimuli were drifting sinusoidal gratings of varying orientations. The duration of each stimulus was 300–400 ms and each stimulus was repeated 200–400 times. Details about stimulus parameters and numbers of recorded units in each dataset appear in Table 1 (see Materials and Methods). The recorded cells consisted of well-isolated single units and small multiunit clusters.\n\n### Tuning of spike latencies\n\nWe first investigated the tuning of first spike times to stimulus orientation. Figure 1A presents eight raster plots showing the response of the same V1 neuron to eight different orientations of the visual stimulus. Qualitatively, both response strength and response latency seem tuned to the stimulus. Measuring latency by simply calculating the mean time to the first spike is problematic because stimuli that evoke weak responses may result in no spikes on some trials. A more principled approach is to incorporate both response time and probability of firing by computing the probability density function and the corresponding cumulative distribution function of the first spike latency.\n\nFigure 1. Orientation tuning of spike latencies.\n\n(A) Raster plot for of a sample cell in the data (taken from dataset 1 in Table 1). For each orientation, 100 randomly chosen trials (out of 400) are shown. For clarity, only the first 120 ms after stimulus onset are shown. Stimulus duration was 400 ms. (B) Cumulative distribution functions of first, second and third spike latencies (n denotes the spike number) for the same neuron. Each row corresponds to a different stimulus orientation and the gray levels represent the probability of the spike occurring before the time indicated on the abscissa. (C) Tuning curves of first, second and third spike latencies, computed as level curves of the corresponding cumulative distributions at 0.5. Cosine fits are shown as solid lines and are also shown as dashed lines in (A). (Error bars were calculated according to the method described in Materials and Methods, but are often smaller than the marker size). (D) Rate tuning curve for the same cell over the entire stimulus duration (black circles) and a fitted von-Mises function (solid line). (Error bars were calculated using the standard error of the mean, but are smaller than the marker size).\n\nhttps://doi.org/10.1371/journal.pcbi.1002536.g001\n\nFigure 1B (upper panel) shows the cumulative distribution function,",
null,
"; i.e., the probability of firing the first spike before time",
null,
"for a given orientation",
null,
"(",
null,
"is measured with respect to the onset of the external stimulus). It is convenient to think of the level curves of this function,",
null,
", as tuning curves of the neuron. For instance, Figure 1C shows the",
null,
"level curve (red circles, fits shown by the solid red line and the dashed line in Figure 1B), which indicates the time at which there was a 50% chance that the neuron had fired its first spike, for each orientation. Typically, the level curves have unimodal orientation tuning, with a single minimum which we define as the latency-based preferred orientation of the cell. Note that although the choice of the 0.5 level curve is arbitrary, similar results were obtained for other criteria. For comparison, the conventional rate-tuning curve of the same neuron is shown in Figure 1D (black circles represent mean firing rates over the entire response, solid curve represents fitted von-Mises function, stimulus duration was 400 ms; see Materials and Methods). The rate tuning is also characterized by a unimodal curve that peaks at the rate-based preferred orientation.\n\nFigure 2 shows three additional examples of V1 responses in each column. Eight raster plots for eight orientations are depicted at the top row for each cell. The stimulus dependence of the temporal structure of neural response can be seen from the PSTHs at the second row. The latency tuning curve, in terms of 0.5 level curve of first spike time cumulative distribution, is shown on the third row, and the conventional rate tuning curve appears on the fourth row for comparison. Examining the PSTHs of each cell, one can see that response strength has a considerable contribution to first spike latency, in our definition. For example, in cell B it is mainly the firing rate that is tuned to stimulus orientation. Nevertheless, due to the high firing rate near the preferred orientation, the first spike times tend to be shorter near that orientation. It is also evident that the temporal structure of the response is tuned to the stimulus as well. The modulation of the entire temporal structure (and not a simple temporal shift) limits the ability to extract the ‘pure’ latency tuning. However, as mentioned above, it is the distribution of the nth spike time that governs the tWTA readout accuracy; hence, the definition of spike latency used here.\n\nFigure 2. Additional examples of spike latency tuning.\n\nEach column, (A)–(C), corresponds to data from a different unit. First row: Raster plot for each of the 8 orientations. For each orientation, 100 randomly chosen trials are shown for 120 ms after stimulus onset. Second row: PSTH (Post Stimulus Time Histogram) for the same time window. Third row: Tuning curve of first spike latency. Cosine fit is shown as a solid line. Fourth row: Rate tuning curve (black circles) and a fitted von-Mises function (solid line). The cell in (A) is taken from dataset 1 in Table 1, in which stimulus duration was 400 ms and the number of trials was 400. The cells in (B) and (C) are taken from dataset 5 in Table 1, in which stimulus duration was 300 ms and the number of trials was 300. These are the same 3 cells as in Figure 5.\n\nhttps://doi.org/10.1371/journal.pcbi.1002536.g002\n\nThe middle and bottom panels of Figure 1B depict the cumulative distribution function for the second and third spike times, respectively; the green and blue traces in Figure 1C show the corresponding latency tuning curves (level curves at 0.5). The level curve for the cumulative distribution of the nth spike time indicates a tradeoff: the curves are delayed in time as n increases, but tuning becomes more pronounced. To quantify this behavior we characterized each tuning curve by a ‘DC’ component, denoted by A, which represents the mean latency across all orientations, and by the ‘modulation amplitude’, denoted by B (see Materials and Methods). Figures 3A and B show the dependence of the mean (A) and the modulation amplitude (B) of the spike-time tuning curve as a function of n, averaged across the population (dataset 3 in Table 1). The delay is evident from the linear increase of A with the spike number, while the increase of tuning amplitude, B, indicates that the tuning becomes more pronounced as n increases. A scatter plot showing the mean latency of the first spike against the tuning modulation of the first spike indicates that they are correlated (Figure 3C; correlation coefficient 0.85). This is a manifestation of an empirical result that the first spike latency at the preferred orientation (AB) is approximately constant, and thus neurons with larger modulation amplitudes also have larger mean latencies. Note, that because (A–B) is the fitted latency at the preferred orientation and is expected to be positive, we would expect that in general A will be larger than B. We find that, typically, the rate-based preferred orientation is very close to the latency-based preferred orientation. Figure 3D shows the distribution of the difference (in absolute value) between the rate and the latency preferred orientations of cells with a tuned first spike latency. In about 90% of the cells this difference is less than 20°.\n\nFigure 3. Population statistics of latency tuning.\n\nThe tuning curves were fitted using a cosine function,",
null,
", where θ is the stimulus orientation and φ is the latency preferred orientation. (A) Dependence of the mean DC component, A, on spike number (averaged over the population). (B) Dependence of the modulation amplitude, B, on spike number (averaged over the population). Error bars in (A) and (B) represent ±onestandard error of the mean. (C) A scatter plot of A vs. B for first spike latency (each point represents one unit; correlation coefficient 0.85). The cells that are marked in red are onset detectors (see text and Figure 4). The statistical analyses in panels (A)–(C) were performed using dataset 3 in Table 1 (159 cells). Similar results were obtained for the other 4 datasets. The correlation coefficients between A and B were, in decreasing order: 0.88, 0.84, 0.76 and 0.66 (D) Histogram of the difference between the first spike latency-based preferred orientation and the conventional rate-based preferred orientation. In order to avoid artifacts from poorly tuned cells, the histogram shows only cells for which the modulation, B, of the first spike latency tuning curve was larger than 15 ms (∼50% of the cells from datasets 1 to 5 in Table 1).\n\nhttps://doi.org/10.1371/journal.pcbi.1002536.g003\n\nIn summary, the latency to the first spike is stimulus dependent: it is shortest for the same orientation that evokes the highest firing rate in the cell. Defining response latency by the first two or three spikes, rather than the first single spike, results in tuning with the same preference but with deeper modulation. Thus, spike latency appears to contain useful information about stimulus orientation.\n\n### Generating a reference signal to measure spike latencies\n\nBecause the brain does not have direct access to information about when a stimulus was presented, a reference signal is required to extract information about stimulus orientation from the first spike latency. Such a reference signal can be reported by neurons which are sensitive to the mere onset of the stimulus. An ideal onset neuron is expected not only to have a uniform spike time latency for all orientations, but also a low spontaneous firing rate, to prevent false alarms. In fact, several neurons in the data showed weak orientation tuning of their first spike latency as well as a low spontaneous firing rate. Figure 4A shows a scatter plot of the spontaneous firing rate against the modulation amplitude, B, of the latency tuning curve for a single dataset (dataset 3 in Table 1). We categorized neurons as onset detectors if their modulation amplitude was less than 15 ms and their spontaneous firing rate was less than 5 spks/sec (gray shading in Figure 4A). Typically, we had 10–25 onset detectors in a dataset (10–25% of the population ; see Table 1). Because the parameters A and B are correlated, these neurons also tend to have an earlier latency (Figure 3C, red dots).\n\nFigure 4. Onset detection.\n\n(A) Mean spontaneous firing rate vs. the modulation amplitude, B, of the first spike latency tuning curves. Each point corresponds to a single neuron. All neurons in this figure were taken from the same dataset (dataset 3 in Table 1). Neurons were categorized as onset detectors if the modulation was smaller than 15 ms and the spontaneous rate was below 5 spks/sec (shaded box in the lower left corner, 25 cells). (B) ‘ROC’ curve of the onset detection mechanism for a time window of 20 ms. The inset shows the false alarm rate as a function of detection threshold in standard deviations. (C) Mean onset time as a function of detection threshold. The gray band represents ±1 standard deviation. The black circles in (B) and (C) mark the detection threshold of 4 standard deviations above baseline, which we use throughout the paper. (D) Distribution of onset times. The number of spikes was required to be 4 standard deviations above the mean number of spikes in a 20 ms window during spontaneous firing.\n\nhttps://doi.org/10.1371/journal.pcbi.1002536.g004\n\nIn a given trial, onset time was determined using a simple coincidence detection mechanism. Stimulus presence was detected if the group of onset cells fired at least m spikes during a time interval of T ms, and stimulus onset was estimated by the first crossing time of this threshold. A high value of the threshold m results in a very low false-alarm rate but compromises the probability of hit, whereas a low value of m increases the hit probability but also the false-alarm rate. By varying the m criterion we can quantify the Receiver-Operating Characteristic (ROC) curve of this onset detection mechanism; i.e., the dependence of the hit probability on the false alarm rate (Figure 4B). Note that, in contrast to standard two alternative forced choice tasks, in a detection task there are no well-defined trials of ‘no stimulus’, and the stimulus may be absent over a wide range of time intervals. The mean number of false alarms will scale linearly with the duration in which they are counted. Hence, in a detection task, false alarm is measured in rate of occurrence and not in probability. Unless otherwise stated, throughout this paper we use the following parameters for onset detection: a time window of T = 20 ms, with a criterion of 4 standard deviations above the mean number of spikes in this time interval during spontaneous firing. This choice takes into account the need for a fast detection of the onset (Figure 4C) while maintaining a high hit probability and a low false-alarm rate. The distribution of estimated onset times (relative to stimulus onset) with this criterion is depicted in Figure 4D. Because the detection of stimulus onset involves a simple integration of spikes emitted by onset detectors, it can be realized in a straightforward way in an integrate-and-fire neuron, producing a similar distribution of onset times (Figure S1).\n\nWe have shown that first spike latency contains information about stimulus orientation and that there is a distinct subset of neurons whose responses can be used as a timing reference signal. To read out the information embedded in the neural response latencies, we used a temporal Winner-Take-All (tWTA) mechanism, with respect to the above onset mechanism . The complete definition of the method used to compute tWTA performance is provided in Materials and Methods.\n\nThe performance of the tWTA is affected by the spontaneous firing rates of the neurons, since the mechanism can erroneously identify a spontaneous spike as an informative one. This effect is reduced by taking a more general readout, the n-tWTA, in which the identity of the stimulus is determined by the cell or group of cells that fired the first n spikes with respect to the reference signal. This may come at the expense of the time it takes to make a decision. However, if the number of spikes, n, is less than or equal to the group size, N, then the mean decision time of the n-tWTA will be less than the mean first spike time of a single cell, keeping the mechanism fast.\n\n### Discrimination accuracy based on single cell responses\n\nAs a first test of the tWTA accuracy we quantified how well it can discriminate between two orientations based on single cell responses. We consider the case where one of the orientations is the cell's preferred orientation θ0 (as defined by its latency tuning curve) and the other orientation is θ0+Δθ. The tWTA decision rule is to associate the shorter latency with the stimulus at the preferred orientation of the cell and the longer latency with the other stimulus. The probability of correct discrimination, PC, using the n-tWTA was calculated from the probability density function, fn(θ,t), of the n'th spike latency, as estimated from the data with time relative to the external stimulus onset (see Materials and Methods). Similar to psychometric curves in psychophysical experiments, the curve that describes the probability of a correct response as a function of the orientation difference Δθ is termed the neurometric curve of the cell.\n\nFigures 5A, C and E show the neurometric curves of 3 single cells. The red, green and blue curves correspond to the n-tWTA readout for n = 1, 2, and 3, respectively. For comparison, we show the neurometric curve of the conventional rate code readout in black (the firing rate was estimated from the total number of spikes fired by the cell during the entire response). Typically, as n increases, the performance improves and approaches that of the rate code. Figures 6A and B compare the accuracy of the first spike latency code, in terms of probability of correct discrimination, and the rate code, for a relatively fine discrimination task (Figure 6A; 22.5 deg) and for a coarse one (Figure 6B; 90 deg). Latency and rate code accuracy are correlated and, for the coarse discrimination task, the latency code performance is often comparable to that of the rate code. The cumulative distributions of the accuracy of the different codes in these two tasks are shown in Figure 6C.\n\nFigure 5. Orientation discrimination using single cell spike latencies and firing rates.\n\n(A) Neurometric curves for a single cell using the first spike latency (red), second spike latency (green), third spike latency (blue) and the firing rate (black). These curves represent the probability of correct discrimination in a 2AFC paradigm where one stimulus is at the cell's preferred orientation, PO, and the other at PO±Δθ. (Error bars represent the standard error of the mean, but are often smaller than the marker size). (B) Neurometric curves for 90° discrimination as a function of decision time. The black curve represents probability of correct discrimination based on firing rate for different time windows starting at stimulus onset (the curve starts at 60 ms because deviation from spontaneous activity starts at about this time). (The gray band represents ± standard error of the mean). The horizontal line represents the asymptotic performance using firing rate from the full response (black circle at 90° in A). The filled circles represent decisions using first, second and third spike latencies with the same color code as in (A) (error bars are smaller than the marker size). Each circle is plotted at the corresponding mean decision time. This cell was taken from dataset 1 in Table 1. (C)–(F) The same as (A) and (B) for two other cells from dataset 5 in Table 1. (The 3 cells in this figure are the same cells as in Figure 2).\n\nhttps://doi.org/10.1371/journal.pcbi.1002536.g005\n\nFigure 6. Statistics of orientation discrimination using single cell spike latencies and firing rates.\n\n(A)–(B) Pc (probability of correct discrimination) using first spike latency vs. Pc using the spike count of the entire response. Each point corresponds to a single cell. The identity line is shown for comparison (solid black line). (A) Comparison of performance at a fine resolution discrimination task, Δθ = 22.5°. (B) Comparison of performance at a coarse resolution discrimination task, Δθ = 90°. (C) Proportion of cells above a given performance level. The dashed curves correspond to a 22.5° discrimination task and the solid curves to a 90° discrimination task. Different curves correspond to first spike latency (red), second spike latency (green), third spike latency (blue) and firing rate from the entire response (black). (D) Comparison of latency and rate performance at a given decision time. The abscissa is the difference between Pc using the n'th spike latency and Pc of the conventional rate code readout, where the rate is estimated from the spike count in the time window from stimulus onset to the mean decision time using the n'th spike latency. These differences correspond to the vertical distances between the circles and the solid black curve in the right panels of Figure 5. The curves show the proportion of cells above a given difference. The color code is the same as in (C). The data for all panels are from the tuned cells (B>15 ms) in datasets 1, 2, 4 and 5 in Table 1 (244 cells).\n\nhttps://doi.org/10.1371/journal.pcbi.1002536.g006\n\nFigures 5B, D and F show the accuracy of the rate code as a function of the time used for the discrimination for three example cells (same cells as in Figure 5A, C and E). For comparison we plot the accuracy of the n'th spike latency code readout at its mean decision time (see Materials and Methods). On brief timescales, the latency code readout is superior to that of the conventional rate code. To quantify this effect, we show in Figure 6D the cumulative distribution of the difference between the accuracy of the n-tWTA and the accuracy of the rate code, as computed at the mean decision time using the n'th spike latency. As is clear from the figure, this difference is always positive, emphasizing the superiority of the latency code on brief timescales.\n\nThe responses we measured were evoked by drifting gratings. We also recorded and analyzed additional data using flashed static gratings of brief (50 ms) and long (300 ms) durations. These data provided qualitatively similar results (see Figure S2).\n\n### Discrimination accuracy based on population responses\n\nDecisions in the central nervous system are expected to involve large numbers of cells. In large populations, the n-tWTA discrimination in a two alternative forced choice paradigm can be thought of as a competition between two ‘columns’ towards a threshold of firing n spikes. To study the dependence of n-tWTA accuracy on the population size we divided the tuned neurons (B>15 ms) into artificial columns of equal orientation width according to the latency-based preferred orientation of the cells (see Materials and Methods). For each pair of columns, we measured the probability of correct discrimination as a function of the number of cells in the population (see Materials and Methods). Importantly, unless stated otherwise, the spike latencies in each trial were measured with respect to the onset detection mechanism described above. Thus, the analysis uses only information that is present in the brain, and, in principle, can be performed by an appropriate neuronal mechanism (see Discussion).\n\nFigures 7A, B and C show the n-tWTA probability of correct discrimination for three representative pairs of columns as a function of the number of cells in each column, N. The pairs of columns differ in terms of the difference between the preferred orientations,",
null,
". The blue curve depicts the performance of the naïve tWTA (n = 1) readout. Initially, for small N, tWTA performance increases with N. However, beyond a critical size of",
null,
", tWTA performance saturates. Theory has shown that two factors may limit tWTA performance. The first is correlations in the first spike latencies of different cells and the second is the spontaneous firing of the cells . We find that although first spike latency is correlated (Figure S3), its effect on tWTA accuracy is negligible (Figure S4; Text S1). The dominant factor that limits accumulation of information from large populations is the spontaneous firing. Clearly, adding more cells also results in adding more spontaneous spikes which interfere with informative spikes (see for a detailed analysis). This effect can be reduced by increasing the decision threshold criterion; i.e., by increasing n.\n\nFigure 7. Orientation discrimination using the n-tWTA readout in populations of neurons.\n\n(A–C) Probability of correct discrimination (Pc) as a function of population size (N) for two populations that differ in preferred orientation by 45° (A), 67.5° (B) and 80° (C). Different curves correspond to different values of n (see legend). (D–F) Probability of correct discrimination using the optimal value of n for each N (for the above pairs of populations). The inset shows the optimal n for each N. (G–I) Mean decision times relative to the onset signal for the neurometric curves in the top panels. (Decision times larger than 200 ms are not shown. Error bars represent ± standard error of the mean). The black circles mark the decision times when n = N; i.e., when the number of spikes used for the decision is equal to the group size. Note that the data for the left two columns are from dataset 5 in Table 1 whereas the data for the right column are from dataset 3. These datasets had different levels of spontaneous and evoked firing, which are responsible for the differences in the optimal n and in the decision times.\n\nhttps://doi.org/10.1371/journal.pcbi.1002536.g007\n\nWe next analyzed the performance of the n-tWTA readout, which takes the winning group to be the first to fire n spikes. Different curves in Figure 7A, B and C correspond to different values of the decision threshold, n, in the n-tWTA readout. In this regime, as n is increased the maximal performance is also increased. Figures 7D, E, and F show the performance of the best n-tWTA for each N (that is, the value of the uppermost curve in a vertical cross-section above this N). The inset shows the corresponding value of n,",
null,
", as a function of the population size N. As the population size, N, grows, it pays to consider more spikes in the readout. Moreover, for these values of population size we obtain that",
null,
"is approximately linear in N.\n\nFigures 7G, H, and I show the mean decision time of the n-tWTA readout, relative to the onset signal (decision times higher than 200 ms are truncated). As expected, for a given decision threshold, n, increasing the number of neurons reduces the decision time significantly. The important point is that the average waiting time for the nth spike in a population of N∼n cells is around the average waiting time for the first spike of a single cell (black filled circles), which is typically in the range of 40–80 ms. Thus, considering both more spikes and more neurons (N∼n) can substantially improve reliability without compromising the decision time.\n\nIn the preceding analysis we measured response timing relative to an internal stimulus onset detection mechanism. We wondered whether performance could be improved by making use of the absolute timing of stimulus onset. In principle, this could decrease the detrimental effect of spontaneous firing . To evaluate this we used an artificial reference signal (i.e. not based on neural responses) which varied from 0 to 120 ms relative to the external stimulus onset. Spike times were then measured relative to this reference signal (spikes before the signal were ignored). Figure 8 shows the accuracy of the tWTA readout (n = 1) as a function of the onset time. Estimating the onset too early causes the readout mechanism to consider more spontaneous spikes which only contribute noise. Overestimating the onset time results in a loss of informative spikes. The performance is thus non-monotonic. Since most cells start responding about 60 to 90 ms following stimulus onset, tWTA accuracy peaks at about this time, at a performance level comparable to that achieved using the internal onset detection signal. For comparison, Figure 8 also shows the mean time (±1 standard deviation) of our onset detection mechanism for the same dataset. As can be seen, the onset detection mechanism matches the range of times that produce optimal performance. We conclude that the speed and accuracy of our decoding is similar to that which would be achieved by making use of absolute information as to when the stimulus was presented.\n\nFigure 8. Effect of onset time on orientation discrimination using the first spike latency.\n\nTo investigate the effect of onset time, we measured the spike times relative to an artificial reference signal. The curves show the probability of correct discrimination (Pc) as a function of onset time for two populations that differ in preferred orientation by 90°. Each curve corresponds to a different population size, N (see legend). The black vertical line and the gray band represent the mean onset time ±1 standard deviation using the onset detection mechanism.\n\nhttps://doi.org/10.1371/journal.pcbi.1002536.g008\n\n### Discriminating multiple alternatives\n\nWe next studied the issue of tWTA accuracy in a multiple (M)-alternative-forced-choice task using the following setting. All the tuned neurons (B>15 ms) in each dataset were divided into M ‘columns’ according to their preferred orientation, as depicted in Figure 9A (see Materials and Methods). Note that the number of cells in different groups is not identical and that dividing them into many groups may result in some that contain no cells. The orientation label of each column was defined as the center of that column. The n-tWTA decision in a competition among M columns was defined as the orientation label of the first column to reach a threshold of n spikes. The resolution of this decision is inversely related to the number of alternatives,",
null,
". Figure 9b shows the probability of correct discrimination of the n-tWTA as a function of",
null,
"in one of the datasets. Different curves correspond to different values of n. The dashed line represents chance value, which is inversely proportional to the number of alternatives. As the decision threshold, n, is increased, n-tWTA performance improves. This improvement is more significant for coarse discrimination tasks; i.e., for large",
null,
".\n\nFigure 9. Discrimination among multiple alternatives using the n-tWTA in populations of neurons.\n\n(A) The tuned neurons in one of the datasets (dataset 3 in Table 1) were divided according to their preferred orientations into M groups of equal orientation width, Δθ = 180°/M. To illustrate this division, each point on the circle represents a neuron (the angle is twice the preferred orientation). The left plot illustrates division into M = 4 groups of width Δθ = 45° and the right plot illustrates division into M = 9 groups of width Δθ = 20°. Each group is labeled by the orientation of its center. The lengths of the blue bars are proportional to the number of neurons in each group. (B) Probability of correct discrimination of the n-tWTA as a function of group width. The different curves correspond to n = 1,2,3,4,5 and 20. (C–D) Distribution of errors for group width of Δθ = 1° for n = 1 (C) and n = 2 (D).\n\nhttps://doi.org/10.1371/journal.pcbi.1002536.g009\n\nTo gain more insight, Figure 9C depicts the distribution of errors in a fine discrimination task (",
null,
") using the tWTA (n = 1). The error distribution is very broad and there are relatively many large errors. These large errors are related to spontaneous firing and reflect the fact that discrimination at fine resolutions involves a competition among many groups (180 in this case). In a substantial fraction of the trials the winning group is the first to fire a spontaneous spike, which carries no information about the stimulus; hence errors in these cases are distributed uniformly. Using the n-tWTA readout with n = 2 decreases this effect and makes the distribution narrower, as depicted in Figure 9d. Nevertheless, the decision is still based on a competition between one “correct” group and many (M−1 = 179) “incorrect” groups. The chances that one of the “incorrect” groups will fire its first two spikes before the “correct” group are still high and the distribution of errors is still relatively wide. With larger groups of neurons in each bin (i.e. with more samples than that provided by our microelectrode arrays), the decision threshold, n, could be increased so as to improve performance for these more difficult discriminations. Nevertheless, for our dataset, we can conclude that n-tWTA can perform coarse discriminations remarkably quickly and with high accuracy.\n\n## Discussion\n\nWe performed a quantitative analysis of spike latency coding of orientation in primary visual cortex. We found that spike time latency is tuned to the orientation of visual stimuli. Surprisingly, for many neurons, the performance of a WTA decoder based on spike latency was comparable to the performance based on the total spike count during the entire response. This decoding could be performed by measuring latency relative to a reference signal in cortex, namely the pooled responses of a subset of neurons with low spontaneous rates and poor latency-based selectivity for orientation. Performance of the decoder could in principle be improved by using larger populations of neurons. We found that spontaneous firing limits the ability to accumulate information from the spike time latencies of large cell populations, but this can be overcome by scaling the decision threshold linearly with the population size.\n\n### Tuning of spike latencies\n\nCoding of visual attributes by spike latencies was studied previously in the context of contrast processing , , where it was demonstrated that higher stimulus contrast results in shorter response latency. However, some confusion exists in the literature as to the tuning of first spike latency to the orientation of visual stimuli. Whereas Celebrini et al reported tuning of spike latency of V1 neurons to orientation, Gawne et al. claimed that stimulus orientation mainly modulates response strength and only weakly affects response latency .\n\nWe have shown that first spike latencies of V1 neurons are tuned to the orientation of external stimuli. This tuning is typically unimodal and the minimal latency is close to the orientation that evokes the maximal firing in the cell. The apparent discrepancy with Gawne et al. is due to different definitions of response latency. In their study, Gawne et al. defined response latency to be the time at which the PSTH reaches half of its peak. The utility of this measure is that it attempts to estimate changes in the ‘pure latency’ in a manner that is unaffected by the changes in the firing rate of the cell. However, since firing rate is modulated by orientation, this definition may measure the latency to a single spike at the null orientation and the latency to ten spikes at the preferred orientation. Hence, using this definition should result in flatter latency tuning curves. Indeed, when applying this definition to our data, we found little modulation of latency with orientation (Figure S5). Moreover, since response strength, the temporal structure of the PSTH, and response latency itself may all be modulated by the stimulus, it is very difficult to obtain a reliable estimate of ‘pure latency’ tuning based on finite amounts of data.\n\nHere we took a more pragmatic approach. Since we are interested in the issue of decoding neural responses on brief time scales, we studied latency tuning using the probability density function of first spike time, which is the quantity that governs tWTA accuracy. Our results thus hold regardless of whether differences in first spike latency arise entirely from differences in response strength, or whether there is some tendency for neurons' absolute latency to vary with stimulus conditions.\n\n### Onset estimation\n\nTo extract the information embedded in spike latencies, a reference signal is required. Note that a reference signal is also required for decisions based on spike count in order to determine the start of the counting window. In the general case of latency coding, the onset signal gives a natural reference for measuring latency. However, in our case we do not use the absolute response time, but instead only use relative timing, i.e., who fired first. In this case, an important feature of the onset signal is to filter out spontaneous spikes that are not stimulus dependent and hence carry no information (see Figure 8).\n\nIn the case of ‘active sensing’, the intrinsic signal of the motor command can, in principle, serve as the onset signal. However, in the case of ‘passive sensing’; e.g., when a child suddenly jumps in front of your car, the onset signal must be estimated from the responses of sensory neurons. Here we suggested a principle by which stimulus onset is estimated by the group of cells that are not tuned to the information that must be processed rapidly. We showed that a simple summation of the responses of ‘onset’ neurons during short time intervals can provide a reliable reference signal, with sufficient accuracy to allow for accurate identification of stimulus orientation. The onset cells were characterized by weak first spike latency tuning, to limit stimulus dependent bias of the estimated onset time, and low spontaneous firing rates to reduce the false alarm rate. Because the tuning modulation and the mean latency are correlated (Figure 3C), these cells also tend to have an early response. However, even if the onset signal arrives slightly after the tuned neurons started to fire, the performance is only mildly decreased (Figure 8). In terms of the identity of the onset cells, one possibility is that these are inhibitory interneurons, which are known to be responsive but poorly tuned , . Since these neurons do not project downstream, this would imply that onset detection is performed locally. A similar approach has been applied in the past for the estimation of the onset of auditory stimuli by Chase & Young . The main differences are twofold. One, Chase & Young used a ‘pseudo population’ signal whereas we use simultaneous recordings of real neural populations. Two, we used the responses of a distinguished subclass of cells with weakly tuned first spike latency for our onset signal, whereas Chase & Young pooled the responses of all the cells.\n\n### A fast and simple readout mechanism in the brain\n\nIt remains an open question whether the brain employs a latency-based readout like the tWTA. Nevertheless, the utility of the tWTA in our study has been to enable us to investigate and quantify the information embedded in spike time latency. Let us consider, for example, the case of a two alternative forced choice discrimination task, based on a competition between two neurons. At the time of the first spike the tWTA decision is identical to that of the conventional rate-based readout. The advantage of a latency-based readout is clear when both neurons fired one spike in the counting window. In those cases the latency based readout can extract information from the temporal structure of the response, whereas there is no information in the total spike count. A rate code readout will perform better when more spikes were fired, but this results in a slower readout. A recent study reported that the minimal processing time required for visual perceptual decisions in the monkey is about 30 ms . This brief time scale is on par with the processing time of the latency readout, i.e the mean decision time following the internal onset signal (see e.g. Figure 7I).\n\nTo test more directly if a candidate readout mechanism is used by the brain one would need to correlate the behavior of animals with the relevant aspects of neural activity. In a recent study , activity of single neurons in monkey V1 was measured together with reaction times for visually guided saccades. It was shown that first spike latency was correlated with behavior whereas firing rate was not, suggesting that spike latency may indeed serve as a source of information for fast decisions in the brain.\n\n### Implementation of the tWTA readout in the brain\n\nAs noted above, the implementation of the n-tWTA readout requires an integration process and a threshold decision mechanism. In this sense, n-tWTA competition is very similar to the ‘race to threshold’ mechanism suggested by Mazurek et al , in which the decision in a two alternative task is determined by integrating ‘evidence’ (spikes) for the two competing alternatives to reach a decision threshold (n spikes). The decision mechanism involves a winner-take-all type competition, which is an algorithm that others have also used to decode neural response . Winner-take-all competition can be implemented using reciprocal inhibition between the integrators that represent the different alternatives , (Figure 10). Each inhibitory neuron accumulates evidence for the corresponding alternative and fires when it crosses a threshold. Higher threshold values reflect a stricter decision criterion and correspond to higher values of n in the n-tWTA readout. The integration time constant of the neurons should be on the order of the relevant time scale for decisions (∼10–30 ms).\n\nFigure 10. Neuronal architectures for implementing the tWTA readout for a two-alternative task.\n\nThe figure describes in a qualitative manner neuronal architectures that can implement the tWTA readout in the context of a two-alternative task. The inputs from population A and population B represent the two alternatives. Both architectures rely on reciprocal inhibition for implementing a ‘race to threshold’ competition but they differ in the implementation of the gating mechanism (see Discussion for details). (A) Implementation of the gating mechanism using NMDA synapses. (B) Implementation of the gating mechanism using disinhibition.\n\nhttps://doi.org/10.1371/journal.pcbi.1002536.g010\n\nThe circuit also requires a gating mechanism that triggers the integration process based on the reference signal. One qualitative way to implement such a gating mechanism is using NMDA synapses for the tuned inputs (Figure 10A). The inputs from the onset cells are first integrated by a coincidence detector, which in turn excites the inhibitory cells through AMPA synapses (as shown in Figure S1, such a coincidence detector can be implemented using a simple integrate-and-fire neuron). Only when this detector is active, the inhibitory cells become depolarized and the magnesium block of the NMDA synapses is removed, allowing for integration of the tuned inputs. When the onset cells are silent, the NMDA synapses do not allow inputs from the tuned populations to be integrated. The gating mechanism can also be implemented using a disinhibition pathway (Figure 10B). In this case the onset cells are assumed to be inhibitory. Their inputs are integrated by a neuron which inhibits the competing neurons. Thus, the competing neurons are released from inhibition only when the onset cells are active, allowing the ‘race to threshold’ to begin.\n\nPrevious studies have proposed more sophisticated mechanisms to combine information from the first spikes of different neurons in a large population. These methods include rank order , and synfire chains . The utility of tWTA is that its simplicity enables statistical analysis of its accuracy, whereas sophisticated readout mechanisms that rely on specific combinations of firing orders cannot be tested with finite data on the order of a few hundred repetitions per stimulus condition. Furthermore, these readouts may be more difficult to implement in biological circuits.\n\nRecently first spike latency code has been analyzed in the framework of fast discrimination of sound source location in the auditory system . There are several interesting similarities and differences worth noting. In both systems, many cells exhibit tuning of their first spike latency to the stimulus. Tuned cells are typically characterized by a unimodal latency tuning curve that peaks close to the preferred stimulus of the cell, as defined by the rate tuning curve. In addition, the accuracy of first spike latency readout is typically comparable though somewhat inferior to the accuracy of the conventional rate code in single tuned cells in both systems. The main differences between the systems are the higher spontaneous firing rates in visual cortex and the poorer performance of V1 neurons for orientation discrimination. To overcome the detrimental effect of spontaneous spikes, we developed here a novel onset detection mechanism, based on pooling the responses from a set of simultaneously recorded neurons. The use of simultaneous data from array recordings rather than single units also enabled us to investigate the accuracy of latency coding at the population level without the use of artificial pseudo populations of neurons.\n\nIn summary, our study demonstrates that the orientation tuning of first spike latencies enables accurate discrimination of orientations on brief time scales. Spontaneous firing limits the resolution of the decision. However, larger populations can afford better resolution. Furthermore, in many cases when fast decisions are essential, it is important that the probability of correct response will be high but coarse resolution may suffice. This may be a general principle used by the nervous system when fast decisions are essential. For example, when an object suddenly appears on the road while we are driving, all we need to know is its rough location. In most cases we react before we realize whether this object is a child, a dog or just a plastic bag. These finer details can be sorted out later as more spikes are accumulated using readout mechanisms that take into account the entire neural response.\n\n## Materials and Methods\n\n### Ethics statement\n\nAll procedures were approved by the Institutional Animal Care and Use Committee at the Albert Einstein College of Medicine of Yeshiva University, and were in compliance with the guideline set forth in the United States Public Health Service Guide for the Care and Use of Laboratory Animals.\n\n### Experimental procedures\n\nThe methods we use to record from neural populations have been described in detail . In short, we recorded from anesthetized (sufentanil citrate, typically 6–18 microg/kg/hr, adjusted as needed for each animal), paralyzed (vecuronium bromide, 0.1 mg/kg/h) macaque monkeys (macaca fascicularis). Vital signs were monitored continuously to assure adequate anesthesia and the well-being of the animal. The pupils were dilated with topical atropine and the corneas protected with gas-permeable hard contact lenses. Supplementary lenses were used to bring the retinal image into focus.\n\nNeural activity was recorded using the Cyberkinetics “Utah” Array (Cyberkinetics Neurotechnology Systems), using methods reported previously ,. The array consists of a 10×10 grid of silicon microelectrodes (1 mm in length) spaced 400 µm apart, thus covering 12.96 mm2. The array was inserted roughly 0.6 mm into cortex using a pneumatic insertion device , resulting in recordings confined mostly to layers 2–3. Signals from each microelectrode were amplified and bandpass filtered (250 Hz to 7.5 kHz). Waveform segments that exceeded a threshold (periodically adjusted using a multiple of the rms noise on each channel) were digitized (30 kHz) and sorted off-line. Sorted units included both well-isolated single units and small multiunit clusters. Neuronal receptive fields were roughly 2–5° from the fovea.\n\nVisual stimuli were displayed at a resolution of 1024×768 pixels and a video frame rate of 100 Hz on a calibrated CRT monitor. Stimuli were oriented drifting gratings presented in a circular aperture surrounded by a gray field of average luminance (8 orientations in 4 datasets and 36 orientations in one dataset). Stimuli were presented binocularly, for 300–400 ms, and separated by 500–800 ms intervals during which we presented an isoluminant gray screen. Stimulus orientation was block randomized, and each stimulus was presented 200–400 times (see Table 1 for details). In 4 datasets the initial phase of the drifting grating was identical across trials. To test whether our results were skewed by this, we collected and analyzed additional data using initial phases that were randomized across trials. We obtained similar results from this dataset (see Figure S6). To verify that our results also generalize to static images, we collected and analyzed responses to static gratings presented for 50 or 300 ms (dataset 6 in Table 1). We obtained similar results from this dataset (see Figure S2).\n\n### Rate tuning\n\nThe rate tuning curves represent the mean firing rate across all trials at each orientation. The firing rate in a trial was calculated using a time window from stimulus onset to 300 ms after stimulus offset. The tuning curves are well fitted using the Von-Mises function:",
null,
"where θ is the stimulus orientation and φ is the rate-preferred orientation of the cell.\n\n### Latency tuning\n\nTo generate latency tuning curves for a neuron we first estimate the probability density function of the first spike latency of this neuron, f1(θ,t). This is done by computing the histogram of the first spike times over trials and then normalizing it. Note that because in some trials there may be no spikes, the integral of the probability density function may not be 1 but slightly below. The spike times are measured with respect to the external stimulus onset and the histogram is generated using bins of 10 ms from time 0 to 300 ms after stimulus termination. The corresponding cumulative distribution, F1(θ,t), is generated by direct numerical integration of the density function. A similar procedure is applied to obtain the nth spike time probability density, fn(θ,t), and cumulative distribution, Fn(θ,t), for general n.\n\nThe latency tuning curve of the n'th spike is defined as the level curve at 0.5 of the corresponding cumulative distribution function. These level curves are fitted using a cosine function of the form:",
null,
"where θ is the stimulus orientation and φ is termed the latency preferred orientation of the cell. Parameter A represents the mean latency and B represents the modulation of the tuning.\n\nHere, the reference time is chosen to be the onset of the external stimulus, but in principle other external reference times can be used, e.g. 20 ms after stimulus onset. We note that in the cosine fit, changing the reference time will change the value of A but not B. The arbitrary choice of the reference is also why a simple cosine function is more appropriate here than the von-Mises function. Choosing the reference such that at some orientations the latency is zero requires parameter k at the von-Mises function to diverge to infinity. In addition, if the latency is negative with respect to the reference at some orientations, the von-Mises function will not fit at all, as it is purely positive.\n\nBecause in some trials there may be no spikes, error bars for the latency tuning curves cannot be simply calculated from the standard error of the mean associated with the spike times. In order to generate error bars, we first calculated the standard errors of the mean for the cumulative distribution, F. This can be done by noting that F is the mean of a Bernoulli variable and thus its variance is",
null,
". The standard error of the mean is therefore:",
null,
", where K is the number of trials. We then calculated the level curves at 0.5 for F+SEM(F) and for F-SEM(F), and used them to generate lower and upper error bars, respectively. These error bars are depicted in Figure 1C and in subsequent plots of spike latency tuning.\n\n### Onset detection\n\nIn each dataset we identify a group of cells that can serve for the detection of stimulus onset. These cells are characterized by poor tuning and low spontaneous firing rates. The spontaneous firing rates are estimated from the recordings during the inter-stimulus interval (ISI) after each stimulus. From each ISI we remove the first 300 ms, assuming that after this period the cell returned to its spontaneous rate (i.e. any post-response adaptation of spontaneous rate would have dissipated). The tuning is characterized by the modulation amplitude, B, of the cosine fit to the first spike latency tuning curve. In each dataset, the cells with a spontaneous rate lower than 5 spks/sec and with a modulation lower than 15 ms, were labeled as onset detectors. Using this definition, the number of onset detectors in a dataset is roughly 10–25% of the population (see Table 1).\n\nThe onset signal in each trial is generated using coincidence detection. We used a running time window of T ms and looked for the first time in which there were at least m spikes in this window (but see also Figure S1). The onset time is then defined as the end of this window. To set m, we first estimated the mean and standard deviation of the number of spikes that these cells fire in a time window T during spontaneous firing. We then set the threshold m to be Nv standard deviations above this baseline value. By varying Nv for a given T we generated ROC curves for the onset detection process. In subsequent analyses we used T = 20 ms and Nv = 4 standard deviations. This onset signal was used as the reference time tref for measuring spike latencies in the tWTA.\n\n### Discrimination accuracy based on single cell responses\n\nThe discrimination accuracy of single cells is computed in the context of a Two-Interval 2-Alternative-Forced-Choice paradigm. We assume that the cell is presented with two stimuli, one at orientation θ1 and the other at orientation θ2, where θ1 is the preferred orientation of the cell. The probability that the tWTA will yield the correct response is the probability that the latency of the response to θ1 will be shorter than the latency of the response to θ2. To find this probability, we multiply the probability that the neuron first fired at time t in response to θ1 by the probability that it did not fire before t in response to θ2, and then we sum over all possible times, t (the time is measured with respect to the onset of the external stimulus). Formally, this is given by the following integral:",
null,
"However, recording time is finite. Our data contains only 300–400 ms of stimulus presence and the following 700–800 ms of inter-stimulus time; hence, in some cases the decision threshold is not reached during our recording time. In practice we assume that after time T0, that contains the stimulus presence time and the initial 300 ms of the following inter stimulus period, the neuron returns to its spontaneous firing rate. Assuming Poisson firing with mean rate λ after time T0, we obtain:",
null,
"It is also important to note that f and F are estimated from the data using time bins of Δt. The spikes from the responses to θ1 and θ2 may fall within the same time bin, leading to correct discrimination at chance level. Correcting for this effect we obtain:",
null,
"Finally, for general n, the correction that stems from the spontaneous firing after response termination is more complicated due to all the combinations of spike trains that have to be taken into account. The general expression is then:",
null,
"where the coefficients",
null,
"are given by:",
null,
"and",
null,
"is the probability that neuron i fired m spikes up to time T in response to stimulus θi.\n\nThe probability of correct response Pc is the mean of a Bernoulli variable and the corresponding standard error of the mean can be calculated as",
null,
", where K is the number of trials.\n\nTo prevent possible interaction between the discrimination accuracy analysis and the latency tuning analysis, we separated each dataset into a training and test set, each consisting of half of the trials (randomly chosen). The training set was used for estimating the latency preferred orientation of the cell. The test set was then used for constructing the neurometric curve, based on the preferred orientation from the training set.\n\nTo calculate the mean decision time we first compute the probability that decision will be made between t and t+ Δt,",
null,
"and then compute its mean.\n\n### Discrimination accuracy based on population responses\n\nTo study the dependence of n-tWTA accuracy on the population size we divided the neurons into several artificial columns of equal orientation width (for datasets with 8 orientations we divided into 8 columns of 22.5° width and for the dataset with 36 orientations (dataset 3 in Table 1) we divided into 9 columns of 20° width). Each neuron was assigned to the column with the closest orientation to its own preferred orientation (the number of neurons in such a column ranged from 1 to 14). For each pair of columns, we then constructed a neurometric curve, which measures the probability of correct response as a function of the number of neurons, N.\n\nGiven two subsets of N cells from each column, we simply went over all trials with the orientation of the first column and then over all trials with the orientation of the second. In each trial, the subset that first fired the n'th spike after the onset signal from the onset neurons was the n-tWTA. If the time of the n'th spike was the same for both subsets we tested whether one of the subsets fired additional spikes in the same bin and took the winner as the subset that had more spikes. The average number of correct responses using the n-tWTA gave an estimate of the probability of correct response for these two subsets of cells. For a given N we averaged this value over 1000 realizations of the subsets of neurons. The decision time in a given trial was the time relative to the onset signal and we calculated its mean and standard error of the mean across all trials.\n\n### Discrimination among multiple alternatives\n\nTo investigate discrimination among multiple alternatives, the neurons were divided according to their preferred orientation into M groups of equal orientation width, Δθ. For convenience, we set one group to be centered at the stimulus orientation (e.g., if M = 18 and the stimulus orientation is 45°, the centers will be at 5°, 15°, 25°,…, 175°). On a given trial, the group that was first to fire n spikes was the n-tWTA. If several groups fired the n'th spike at the same time we chose among them in a random manner. The error in the trial was the (signed) difference between the orientation of the winning group and the stimulus orientation. The probability of correct response was calculated as the average number of times in which the correct group was the winner.\n\n## Supporting Information\n\n### Figure S1.\n\nOnset detection using an integrate-and-fire neuron. In addition to the onset detection mechanism described in the main text, we investigated an implementation of onset detection in a more biologically plausible architecture, namely by performing the coincidence detection using a leaky integrate-and-fire neuron. The neuron integrates the spikes from the onset neurons with equal synaptic weights until a threshold is reached. For simplicity, the integration process was set such that each spike increases the membrane potential by one unit. The voltage then decays exponentially with a time constant of 20 ms. (A) Illustration of the integration process by a leaky integrate-and-fire neuron. The top trace shows a series of spikes from all onset neurons relative to stimulus onset and the bottom trace show the membrane potential of the integrate-fire-neuron (in arbitrary units) as it integrates these spike. The accumulation of spikes around 45–65 ms after stimulus onset causes the neuron to first cross the specified threshold and an onset is detected. (B) Mean onset time as a function of threshold. The mean is calculated over all trials in one dataset. The gray stripe represents ±one standard deviation. The horizontal line represents the mean onset time using the running window approach which was used in the main text (the time window was 20 ms and the threshold was set to 4 standard deviations above spontaneous firing, which corresponds here to m = 6 spikes). The vertical line represents the required threshold for the integrate-and-fire neuron (3.3) to achieve the same mean onset time. (C) Distribution of the difference in onset times between the two mechanisms across all trials. The threshold for the integrate-and-fire neuron is the one which achieves the same mean onset time as the running window method (3.3).\n\nhttps://doi.org/10.1371/journal.pcbi.1002536.s001\n\n(TIF)\n\n### Figure S2.\n\nLatency tuning and orientation discrimination for static gratings. (A) First spike latency tuning curves for one cell. Different colors denote different stimulus durations (50 ms and 300 ms; see legend). (B) The corresponding neurometric curves for the same cell. For each stimulus duration, the solid curve corresponds to the first spike latency neurometric curve and the dashed curve to the conventional rate neurometric curve. (C)–(D) Same as (A)–(B) for a different cell. (E) Proportion of cells above a given level of the mean latency, A, for the two stimulus durations. (F) Proportion of cells above a given level of the modulation tuning, B, for the two stimulus durations. (G)–(H) Statistics of orientation discrimination. (G) Proportion of cells above a given performance level for the 300 ms stimulus. The dashed curves correspond to a 22.5° discrimination task and the solid curves to a 90° discrimination task. Different curves correspond to first spike latency (red), second spike latency (green), third spike latency (blue) and firing rate from the entire response (black). (H) Same as (G) for the 50 ms stimulus. The analyses were performed using dataset 6 in Table 1 (98 tuned cells for the 300 ms stimulus and 89 tuned cells for the 50 ms stimulus).\n\nhttps://doi.org/10.1371/journal.pcbi.1002536.s002\n\n(TIF)\n\n### Figure S3.\n\nPairwise correlations of response latencies among neurons. (A) Trial to trial fluctuations of first spike latencies for a pair of cells. For each neuron we calculated a normalized measure of its first spike latency by subtracting the mean latency and dividing by the standard deviation. These cells had similar latency preferred orientations, 139° and 136°, and their normalized first spike latencies are shown for the first 25 trials in which the stimulus orientation was 135°. The first spike latency of the two cells fluctuates from trial to trial around its mean. However, typically, when one cell fires sooner than its mean latency the other does as well, and similarly when the response is delayed, resulting in a positive correlation coefficient of 0.45. (B) Distribution of first spike latency correlation coefficients. For each pair of cells with latency tuning (B>15 ms) a correlation coefficient is calculated separately for each stimulus orientation. The mean of this distribution is 0.07 and its standard deviation is 0.04. (C) Dependence of correlations on the difference in preferred orientations (POs). The cell pairs were divided into six groups according to the difference between their preferred orientations, ΔPO. Each point represents the mean correlation coefficient for all pairs of neurons at a given range of ΔPO and the error bars represent the corresponding standard errors. The solid line represents the linear regression and its slope is −10−4±2•10−5 (deg-1). These results indicate that there is a very weak dependence of the first spike latency correlation on the difference between the latency preferred orientations of the cells.\n\nhttps://doi.org/10.1371/journal.pcbi.1002536.s003\n\n(TIF)\n\n### Figure S4.\n\nEffect of pairwise correlations on tWTA accuracy. Correlations in the trial to trial fluctuations of spike latencies can affect the utility of pooling information from groups of cells. To quantify the effect of correlations we compared the discrimination performance using the original simultaneous data to the performance using shuffled data with no correlations. In the shuffled data, the responses of different cells on a given trial were taken from different trials in the original data. The onset signal from the onset neurons was also shuffled among trials. For each shuffled version of the data we found the corresponding neurometric curve, and then averaged the result from 50 shuffles. (A–C) Probability of correct discrimination (Pc) as a function of population size (N) for two populations that differ in preferred orientation by 45° (A), 67.5° (B) and 80° (C) (same pairs of populations as in Figure 7). (D) Probability of correct discrimination in a model of two columns (see Text S1). (E–H) The effect of shuffling the responses of neurons in different trials. The curves show the difference in Pc between the original and the shuffled data (Pc(original)-Pc(shuffled)) for each pair of columns on the left. Although the performance is typically better in the original correlated data, the overall size of the effect is relatively small, on the order of 1%. For small populations, the correlations increase tWTA accuracy. However, as the population size increases, this difference decays to zero. The simplified model (H) captures the behavior of the data (see Text S1).\n\nhttps://doi.org/10.1371/journal.pcbi.1002536.s004\n\n(TIF)\n\n### Figure S5.\n\nEffect of different definitions of response latency. Gawne et al. recorded responses to oriented bars in V1 of behaving monkeys and reported that the effect of stimulus orientation on response latency is relatively weak. The difference with our results can be attributed to different definitions of response latency. Gawne et al. defined response latency to be the time at which the PSTH reaches its half peak (in cases where the peak was less than twice the spontaneous activity, latency was left undefined). We used the cumulative distribution function of first spike times and defined the first spike latency tuning curve as a level curve of this distribution. (A–C) First-spike latency tuning curves as computed by first spike (cumulative) distribution level curves (red) and the corresponding latency tuning curves for the same cells using the time at which the PSTH reaches half of its peak (black). (Lower and upper error bars for the halfmax definition were calculated using the times at which the PSTH reaches half the peak of the PSTH minus and plus its standard error of the mean). B denotes the modulation amplitude using our latency definition and B′ using the half max definition. For the cells in (A) and (B), the latency tuning curve using the half max definition (black) is relatively flat, whereas the first spike latency tuning curve using our definition (red) shows a strong modulation. For the cell in (C) both definitions show pronounced tuning. (D) Scatter plot of the modulation amplitudes, B and B′, for each cell. Notably, the tuning amplitudes using the halfmax definition are relatively small, whereas our definition results in many cells with significant modulation.\n\nhttps://doi.org/10.1371/journal.pcbi.1002536.s005\n\n(TIF)\n\n### Figure S6.\n\nEffect of stimulus phase on spike latency tuning. The panels on the left side (A,C,E,G) depict results from a dataset in which all stimuli of the same orientation had identical initial phases; the panels on the right side (B,D,F,H) depict results from a dataset in which the initial phases were random. The two datasets were recorded in the same animal using the same electrode array. (A–B) PSTHs of two single cells. The stimuli were near the preferred orientation of the cells and the PSTH was constructed from 300 repetitions. For fixed initial phase, the PSTH is characterized by a periodic modulation whereas for random phases there is no such modulation. (C–D) First spike latency tuning curves of the neurons in (A) and (B). (E–F) Distribution of latency tuning modulation amplitude, B, in the two datasets. The substantial tuning of the response latency in the random phase dataset cannot be attributed to the stimulus initial phase. The inset in (F) shows the two cumulative distribution functions (black for the fixed phase dataset and blue for the random phase dataset). The similarity of the two distributions indicates that there is no significant difference between the orientation tuning levels of first spike latencies in the two datasets. (G–H) Mean neurometric curves using the first spike latency (red), second spike latency (green), third spike latency (blue) and the firing rate (black). Error bars represent the standard deviation and are shown only for the rate and the first spike latency neurometric curves. The fact that performance is similar indicates that our results reflect the tuning of the first spike latency to stimulus orientation and not the initial phase of the stimulus.\n\nhttps://doi.org/10.1371/journal.pcbi.1002536.s006\n\n(TIF)\n\n### Text S1.\n\nA simple model for studying the effect of pairwise correlations on tWTA accuracy.\n\nhttps://doi.org/10.1371/journal.pcbi.1002536.s007\n\n(DOC)\n\n## Author Contributions\n\nConceived and designed the experiments: OS AK MS. Performed the experiments: AK. Analyzed the data: OS. Contributed reagents/materials/analysis tools: MS. Wrote the paper: OS AK MS.\n\n## References\n\n1. 1. Hubel DH, Wiesel TN (1968) Receptive fields and functional architecture of monkey striate cortex. J Physiol 195: 215–243.\n2. 2. Zohary E, Shadlen MN, Newsome WT (1994) Correlated Neuronal Discharge Rate and its Implications for Psychophysical Performance. Nature 370: 140–143.\n3. 3. 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"https://journals.plos.org/ploscompbiol/article/file",
null,
"https://journals.plos.org/ploscompbiol/article/file",
null,
"https://journals.plos.org/ploscompbiol/article/file",
null,
"https://journals.plos.org/ploscompbiol/article/file",
null,
"https://journals.plos.org/ploscompbiol/article/file",
null,
"https://journals.plos.org/ploscompbiol/article/file",
null,
"https://journals.plos.org/ploscompbiol/article/file",
null,
"https://journals.plos.org/ploscompbiol/article/file",
null,
"https://journals.plos.org/ploscompbiol/article/file",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.9150885,"math_prob":0.9232267,"size":81881,"snap":"2019-43-2019-47","text_gpt3_token_len":17585,"char_repetition_ratio":0.18703665,"word_repetition_ratio":0.058300022,"special_character_ratio":0.2155323,"punctuation_ratio":0.10320403,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.95436895,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-12T09:27:10Z\",\"WARC-Record-ID\":\"<urn:uuid:eaa38a00-6460-49d7-94d5-cb7313c87f5b>\",\"Content-Length\":\"245614\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ea2f0631-f4a3-4762-9006-12624b4feeca>\",\"WARC-Concurrent-To\":\"<urn:uuid:7a43d9bd-0bb4-48a1-b99e-ca503d473204>\",\"WARC-IP-Address\":\"216.74.38.76\",\"WARC-Target-URI\":\"https://journals.plos.org/ploscompbiol/article?id=10.1371/journal.pcbi.1002536\",\"WARC-Payload-Digest\":\"sha1:ZWAOHY232LHB5OTBVZSK32F7XAKWHYQ6\",\"WARC-Block-Digest\":\"sha1:PIZ5P6Q55AHXWVSVGKAMFRTYO7WOWU22\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496664808.68_warc_CC-MAIN-20191112074214-20191112102214-00178.warc.gz\"}"} |
https://crypto.stackexchange.com/questions/50955/why-is-eax-not-a-generic-composition | [
"# Why is EAX not a generic composition?\n\nIn the document http://web.cs.ucdavis.edu/~rogaway/papers/eax.pdf (p. 7) Rogaway et. al. state that EAX is not a generic composition of an encryption and an authentication method.\n\nThe EAX algorithm is given in the following Image:",
null,
"(the image is taken from p. 7 of the referenced paper and simplified by omitting the associated data)\n\nSo, the plaintext M is encrypted by a CTR block cipher mode and the resulting ciphertext is authenticated by an OMAC tag. So far, this looks pretty much like an encrypt-then-authenticate based generic composition. The only thing particular here is, that the tag $\\mathcal{C}$ isn't the final tag but is XORed by the tagged nonce $\\mathcal{N}$.\n\nMy questions are:\n\n1. Why isn't EAX considered to be a generic composition ?\n2. Would EAX be a generic composition of the block cipher mode encrytion and the OMAC authentication, if $\\mathcal{C}$ would be used as authentication tag T ?\n\nAdded: Actually, I think the answer depends on the point of view.\n\n(1) Proof of security: Security proofs for generic compositions assume that two different, independet keys are used for encryption and authentication. As EAX uses the same key for both, security results for generic compositions don't apply to EAX. So in this point of view EAX is not a generic composition.\n\n(2) Design principle: When one tries to classify AE algorithms, one can look how they are designed. By design, a generic composition is an algorithm that runs an encryption and an authentication algorithm in a (generic) order that is independent from the primitives: E.g. Encrypt-then-Authenticate, Authenticate-then-Encrypt, etc. In this point of view I would consider EAX (as well as CCM) as a generic composition.\n\nWhen reading the paper, it seems that there are three or so properties that make a generic authenticated encryption method:\n\n1. a generic encryption method and MAC are used;\n2. two different keys are used for both;\n3. two pass encryption is used (generally encrypt-than-MAC).\n\nEAX uses two well defined algorithms: AES and CMAC. So it seems the definition fails on (1). As EAX uses only 1 key it will always fail on (2) as well.\n\nBut yes, if you do not consider the special way $\\mathcal{N}$ is constructed then EAX outputting just $\\mathcal{C}$ could be considered a relatively generic construction when it comes to (3).\n\nAdding it up, I personally think this should be a big NO, EAX is not a generic construction as defined in the paper; and just outputting $\\mathcal{C}$ only makes a tiny difference.\n\n• Thanks for your answer. I agree with (2): The authors certainly require a generic composition to use two different, independet keys for encryption and authentication, what EAX does not do. However, I disagree on (1). The algorithms for CTR and OMAC in Fig. 1 (p. 6) of the paper are specified for any block cipher E, not just for a particular block cipher like AES. – user120513 Aug 22 '17 at 1:40\n• Ah, sorry, that could be a difference between the NIST version and the org. paper. Or it is that a confuse it with a different scheme. I'll try and re-read that section and adjust accordingly. – Maarten Bodewes Aug 22 '17 at 8:20"
]
| [
null,
"https://i.stack.imgur.com/EM2fT.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.91641515,"math_prob":0.8959294,"size":1699,"snap":"2019-51-2020-05","text_gpt3_token_len":384,"char_repetition_ratio":0.17168142,"word_repetition_ratio":0.014814815,"special_character_ratio":0.2130665,"punctuation_ratio":0.12462006,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96409225,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-27T03:48:39Z\",\"WARC-Record-ID\":\"<urn:uuid:2b69056f-37dc-4260-a5d2-d6e0cd87904c>\",\"Content-Length\":\"134272\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9955b072-a6e9-44ae-bfd6-44ae10c9f4a3>\",\"WARC-Concurrent-To\":\"<urn:uuid:e7bc7517-bcaa-4083-a8e2-23515ce8dd1e>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://crypto.stackexchange.com/questions/50955/why-is-eax-not-a-generic-composition\",\"WARC-Payload-Digest\":\"sha1:UFIGXUHN7AWW727VQYTSY4QLCVHSB55X\",\"WARC-Block-Digest\":\"sha1:DVVOD7YDWEHITSU3PRKWIGPEJHWPV7DX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251694176.67_warc_CC-MAIN-20200127020458-20200127050458-00400.warc.gz\"}"} |
https://www.quantumstudy.com/category/physics-solutions/kinetic-theory-of-gases/ | [
"## As shown schematically in the figure , two vessels contain water solution (temperature T) of potassium permanganate ….\n\nQ: As shown schematically in the figure , two vessels contain water solution (temperature T) of potassium Permanganate (KMnO4) of different concentration n1 and n2 (n1 >n2) molecules per unit volume with Δn =(n1-n2) << n1 . When they are connected by a tube of small length l and cross-sectional area S , KMnO4 starts to diffuse from left to right vessel through the tube . Consider the collection of molecules to behave as dilute ideal gases and the difference in their partial pressure in the two vessels causing the diffusion . The speed v of molecules is limited by the viscous force -β v on each molecule , where β is constant . Neglecting all terms of the order (Δn)^2 , Which of the following is/are correct ? (kB is the Boltzmann constant)",
null,
"(a) the force causing the molecules to move across the tube is Δn kB T S\n\n(b) force balance implies n1βvl = Δn kB T\n\n(c) total number of molecules going across the tube per sec is $(\\frac{\\Delta n}{l}) (\\frac{k_B T}{\\beta})S$\n\n(d) rate of molecules getting transferred through the tube does not change with time\n\nAns: (a,b,c)\n\nSolution: n1 >> (n1-n2) = Δn\n\n$p_1 = \\frac{n_1 R T}{N_A}$ and $p_2 = \\frac{n_2 R T}{N_A}$\n\n$F = (n_1 -n_2)k_B T S = \\Delta n k_B T S$\n\n$V = \\frac{\\Delta n k_B T S}{\\beta}$\n\n$\\Delta n k_B T S = l n_1 S \\beta v$\n\n$n_1 \\beta v l = \\Delta n k_B T$\n\nTotal number of molecules/second $= \\frac{(n_1 v dt)S}{dt}$\n\n$= n_1 v S = \\frac{\\Delta n K_B T v S}{\\beta v l}$\n\n$= (\\frac{\\Delta n}{l}) (\\frac{k_B T}{\\beta})S$\n\nAs Δn will decrease with time , therefore rate of molecules getting transfer decreases with time\n\n## Consider a collection of a large number of particles each with speed v. The direction of velocity is randomly distributed in the collection….\n\nQ: Consider a collection of a large number of particles each with speed v. The direction of velocity is randomly distributed in the collection. What is the magnitude of the relative velocity between a pairs in the collection\n\n(a) 2V/π\n\n(b) V/ π\n\n(c) 8V/ π\n\n(d) 4V/ π\n\nAns: (d)\n\n## The absolute zero is the temperature at which\n\nQ: The absolute zero is the temperature at which\n\n(a) Water freezes\n\n(b) All substance exist in solid state\n\n(c) Molecular motion ceases\n\n(d) None of the above\n\nAns: (c)\n\n## The temperature at which the r.m.s. speed of hydrogen molecules is equal to escape velocity on earth surface, will be\n\nQ: The temperature at which the r.m.s. speed of hydrogen molecules is equal to escape velocity on earth surface, will be\n\n(a) 1060 K\n\n(b) 5030 K\n\n(c) 8270 K\n\n(d) 10063 K\n\nAns: (d)\n\n## The root mean square speed of the molecules of a diatomic gas is v. When the temperature is doubled ….\n\nQ: The root mean square speed of the molecules of a diatomic gas is v. When the temperature is doubled, the molecules dissociate into two atoms. The new root mean square speed of the atom is\n\n(a) √2 v\n\n(b) v\n\n(c) 2 v\n\n(d) 4 v\n\nAns: (c)\n\n## N molecules each of mass m of gas A and 2 N molecules each of mass 2 m of gas B are contained in the same vessel at temperature T….\n\nQ: N molecules each of mass m of gas A and 2 N molecules each of mass 2 m of gas B are contained in the same vessel at temperature T. The mean square of the velocity of molecules of gas B is v2 and the mean square of x component of the velocity of molecules of gas A is w2 . The ratio w2/v2 is\n\n(a) 1\n\n(b) 2\n\n(c) 1/3\n\n(d) 2/3\n\nAns: (d)\n\n## The mean free path of molecules of a gas, (radius r) is inversely proportional to\n\nQ: The mean free path of molecules of a gas, (radius r) is inversely proportional to\n\n(a) r\n\n(b) √r\n\n(c) r3\n\n(d) r2\n\nAns: (d)\n\n## Which one of the following is not an assumption of kinetic theory of gases\n\nQ: Which one of the following is not an assumption of kinetic theory of gases\n\n(a) The volume occupied by the molecules of the gas is negligible\n\n(b) The force of attraction between the molecules is negligible\n\n(c) The collision between the molecules are elastic\n\n(d) All molecules have same speed\n\nAns: (d)\n\n## At room temperature, the r.m.s. speed of the molecule certain diatomic gas is found to be 1930 m/s. The gas is\n\nQ: At room temperature, the r.m.s. speed of the molecule certain diatomic gas is found to be 1930 m/s. The gas is\n\n(a) H2\n\n(b) F2\n\n(c) O2\n\n(d) Cl2\n\nAns: (a)\n\n## A cubical box with porous walls containing an equal number of O2 and H2 molecules is placed in a large evacuated chamber….\n\nQ: A cubical box with porous walls containing an equal number of O2 and H2 molecules is placed in a large evacuated chamber. The entire system is maintained at constant temperature T. The ratio of vrms of O2 molecules to that of the vrms, of H2 molecules, found in the chamber outside the box after a short interval is\n\n(a) 1/2√2\n\n(b) 1/4\n\n(c)1/√2\n\n(d) √2"
]
| [
null,
"https://www.quantumstudy.com/wp-content/uploads/qaphy1/fm47.png",
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]
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http://seis.geus.net/software/seisan/node187.html | [
"## 27.1.2 Program operation\n\nThe program first reads the parameter file, default codaq.par which must be in your current directory. It then reads the codaq.inp file with the events to analyze (also in current directory). The S-file names given here can, as shown in the examples above, be in the database or elsewhere, e.g. in your local directory. In the S-file, the name of the waveform file is given. If more than one waveform file is given, all files will be searched for the specified station and component. The program will first look in the current directory, and then in WAV and thereafter in the WAV database and other directories as given in the SEISAN.DEF file in DAT. The program can therefore work without moving the data from the database, however you can also move both the S- files and waveform files to your local directory. Remember that the S-files must be updated in order to have origin time, since the program uses the origin time and P arrival times from the S-files.\n\nRunning the program:\n\n```Type codaq, the program asks about output:\n\n0: Only q is calculated, ente ris 0 for default\n1: Q is calculated and a plot on the tek screen is shown\n2: Q ,\nand at the same time hard copy plots are made.\n3: Q is calculated and hard copy plots are made, but\nno screen plot.\n\nParameter file, name codaq.par is default (return)\nJust hit return if default file, otherwise give name.\n\nFile with event stations, codaq.inp is default (return)\nJust hit return if default file, otherwise give name.\n\n```\n\nThe program will now start to run. Alterantively, the progran can be started with arguments on the promt line:\n\ncodaq n parameter-file data-file\n\nor alternatively when doing channel averages\n\ncodaq -c n parameter-file data-file\n\nand no questions are asked. n is the choice 0 to 3 above.\n\nIf no plot is chosen, one line will appear on the screen for each station used and one for each frequency. The program will start a new page for each new event. If you are plotting on the screen, you will therefore have to hit return to get the next plot. The screen might not have been filled out if there are few data.\n\nAll questions will appear in the text window. At the end, a summary is given, which is the same as logged in the output file codaq.out.\n\nThe abbreviations are:\n\n H: Focal depth M: Magnitude TP: P travel time TC: Start time of coda window relative to origin time F: Frequency Q: Corresponding coda q, if 0 value is",
null,
"10000 or negative S/N: Signal to noise ratio AV Q: Average q SD: Standard deviation for average NT: Total number of q values at all frequencies N: Number of q values at given frequency q: Average of q values 1/q: q is calculated as 1/q averages, probably the best to use f:1/q: Q values calculated using the relation derived from the 1/q averages q = q0*f**v obtained with the average 1/q-values cq0: Constant q0 obtained using the fixed user selected v v: Constant v determined cor: Correlation coefficient on determining q vs f corr: Average correlation coefficients of individual codaq calculations when fitting the envelope, both average and standard deviation is given\n\nIf a station is not present or no P is read, a message will be given. The program will search for the first P arrival time in the S-file. If several are present for the same station, it will use the first.\n\nPeter Voss : Tue Jun 8 13:38:42 UTC 2021"
]
| [
null,
"http://seis.geus.net/software/seisan/img33.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.83353496,"math_prob":0.8457521,"size":2379,"snap":"2021-21-2021-25","text_gpt3_token_len":601,"char_repetition_ratio":0.112842105,"word_repetition_ratio":0.037914693,"special_character_ratio":0.24422026,"punctuation_ratio":0.13163064,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9761336,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-17T08:48:39Z\",\"WARC-Record-ID\":\"<urn:uuid:6ff6c56e-217f-405c-a41e-5f9f228b44f6>\",\"Content-Length\":\"9676\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:269974ba-243b-43ad-b71e-7e80019c2fc1>\",\"WARC-Concurrent-To\":\"<urn:uuid:f555a960-7e17-450e-9164-8498b7a9e3bd>\",\"WARC-IP-Address\":\"152.115.61.167\",\"WARC-Target-URI\":\"http://seis.geus.net/software/seisan/node187.html\",\"WARC-Payload-Digest\":\"sha1:5LNLE46WPLZAJNCLTNVFV2LWS22VNGQE\",\"WARC-Block-Digest\":\"sha1:HRCMTI4YFPEMENOWKB5U3O3X2OUDA5YX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487629632.54_warc_CC-MAIN-20210617072023-20210617102023-00204.warc.gz\"}"} |
https://prepinsta.com/mettl-coding-questions-2/ | [
"# Mettl Coding Questions – 2\n\n## Longest Prefix Suffix\n\nGiven a string of character, find the length of longest proper prefix which is also a proper suffix.\nExample:\nS = abab\nlps is 2 because, ab.. is prefix and ..ab is also a suffix.\n\nInput:\nFirst line is T number of test cases. 1<=T<=100.\nEach test case has one line denoting the string of length less than 100000.\n\nExpected time compexity is O(N).\n\nOutput:\nPrint length of longest proper prefix which is also a proper suffix.\n\nExample:\nInput:\n2\nabab\naaaa\n\nOutput:\n2\n3\n\n### C++\n\n#include < bits / stdc++.h >\nusing namespace std;\nint lps(string);\nint main() {\n//code\nint T;\ncin >> T;\ngetchar();\nwhile (T–) {\nstring s;\ncin >> s;\nprintf(“%d\\n”, lps(s));\n}\nreturn 0;\n}\nint lps(string s) {\nint n = s.size();\nint lps[n];\nint i = 1, j = 0;\nlps = 0;\nwhile (i < n) {\nif (s[i] == s[j]) {\nj++;\nlps[i] = j;\ni++;\n} else {\nif (j != 0)\nj = lps[j – 1];\nelse {\nlps[i] = 0;\ni++;\n}\n}\n}\nreturn lps[n – 1];\n}\n\nJAVA\nimport java.util.*;\nimport java.lang.*;\nimport java.io.*;\nclass PreSuf {\npublic static void main(String[] args) {\n//code\nScanner s = new Scanner(System.in);\nint t = s.nextInt();\nfor (int i = 0; i < t; i++) {\nString s1 = s.next();\nint j = 1, k = 0, l = s1.length(), max = 0, len = 0;\nint lps[] = new int[l];\nwhile (j < l) {\nif (s1.charAt(j) == s1.charAt(len)) {\nlen++;\nlps[j] = len;\nj++;\n} else {\nif (len != 0) {\nlen = lps[len – 1];\n} else {\nlps[j] = 0;\nj++;\n}\n}\n}\n\nSystem.out.println(lps[l – 1]);\n}\n}\n}\n\n### One comment on “Mettl Coding Questions – 2”\n\n•",
null,
"PrepInsta\n\ncomment",
null,
"0"
]
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null,
"https://lh3.googleusercontent.com/a-/AOh14GgK2AnsnrsEqGgioK8Pi9PhjL8a3LUBdxLpIWJhxvU=s96-c",
null,
"data:image/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==",
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https://answers.everydaycalculation.com/simplify-fraction/1979-2520 | [
"Solutions by everydaycalculation.com\n\n## Reduce 1979/2520 to lowest terms\n\n1979/2520 is already in the simplest form. It can be written as 0.785317 in decimal form (rounded to 6 decimal places).\n\n#### Steps to simplifying fractions\n\n1. Find the GCD (or HCF) of numerator and denominator\nGCD of 1979 and 2520 is 1\n2. Divide both the numerator and denominator by the GCD\n1979 ÷ 1/2520 ÷ 1\n3. Reduced fraction: 1979/2520\nTherefore, 1979/2520 simplified is 1979/2520\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",
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https://unix.stackexchange.com/questions/537164/tcsh-search-entire-array-for-a-variable | [
"# tcsh: search entire array for a variable\n\nI am modifying a suite of scripts written in tcsh. Unfortunately, these scripts can't be rewritten because the program they interface with is also written in tcsh. I just want to get out of the way the idea that if I could rewrite these in BASH or literally any other language, I would.\n\nAs part of these scripts many of our variables are saved in arrays. My question is: Is there a way in tcsh where you can query if the content of a variable is present in an array (regardless of its position in the array) without using loops?\n\nFor example, I have the following array of valid ROI names:\n\n``````set shortROI = {\"lAmy_\",\"mAmy_\",\"A32p_\",\"A32sg_\"}\n``````\n\nWhen the user executes the script they have an option to use a flag to say they want to redo processing for one or more specific ROIs from the options above. So like, if the flag is `-remake_ROI`, as part of calling the script they could add `-remake_ROI lAmy` or `-remake_ROI lAmy A32p_` to remake those ROIs only. I have a very large `while` loop currently processing all possible flags but the section relevant here looks like this:\n\n``````set ac = 1\nset redo_rois = 0\nset roi_proc_list =\n\nwhile ( \\$ac <= \\$#argv )\nforeach roi_opt ( \\$shortROI )\nif ( \"\\$argv[\\$ac]\" == \\$roi_opt )\n@ redo_rois ++\nset roi_proc_list = ( \\$roi_proc_list \\$roi_opts )\nendif\nend\n@ ac ++\nend\n``````\n\n`ac` is the command line argument count, `redo_rois` is the number of ROIs to re-process and `roi_proc_list` is the actual list to redo. I was wondering instead of the many loops, if there was something like `if ( \"\\$argv[\\$ac]\" == \"\\$shortROI[*]\" )` that would get the job done.\n\nThank anyone for their help.\n\n``````if ( \" \\$shortROI \" =~ *\" \\$argv[\\$ac] \"* ) then\n...\nendif\n``````\n\nExample:\n\n``````set list = (foo bar baz)\n\nforeach a (\\$*)\nif(\" \\$list \" =~ *\" \\$a \"*) then\necho \"\\$a in list\"\nelse\necho \"\\$a NOT in list\"\nendif\nend\n``````\n\nThis may incorrectly return true if either the variable on the right side or any of the words from the list contain spaces. If you can decide on a character which cannot appear in either side (eg. `@`), you can replace the spaces with it in each word:\n\n``````foreach a (\\$*:q)\nif(\" \\$list:q:gas/ /@/ \" =~ *\" \\$a:q:as/ /@/ \"*) then\n...\n``````\n• Thank you for this - it's just what I was looking for. I appreciate the caveats about the spaces as well. Aug 27, 2019 at 21:22\n• Great idea for how to handle spaces! Using octal, you can substitute with a non-printable character instead of '@'. I successfully used \" \\$list:q:gas/ /\\0177/ \", for example. Mar 24, 2021 at 2:56"
]
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https://www.colorhexa.com/e5ebec | [
"# #e5ebec Color Information\n\nIn a RGB color space, hex #e5ebec is composed of 89.8% red, 92.2% green and 92.5% blue. Whereas in a CMYK color space, it is composed of 3% cyan, 0.4% magenta, 0% yellow and 7.5% black. It has a hue angle of 188.6 degrees, a saturation of 15.6% and a lightness of 91.2%. #e5ebec color hex could be obtained by blending #ffffff with #cbd7d9. Closest websafe color is: #ccffff.\n\n• R 90\n• G 92\n• B 93\nRGB color chart\n• C 3\n• M 0\n• Y 0\n• K 7\nCMYK color chart\n\n#e5ebec color description : Light grayish cyan.\n\n# #e5ebec Color Conversion\n\nThe hexadecimal color #e5ebec has RGB values of R:229, G:235, B:236 and CMYK values of C:0.03, M:0, Y:0, K:0.07. Its decimal value is 15068140.\n\nHex triplet RGB Decimal e5ebec `#e5ebec` 229, 235, 236 `rgb(229,235,236)` 89.8, 92.2, 92.5 `rgb(89.8%,92.2%,92.5%)` 3, 0, 0, 7 188.6°, 15.6, 91.2 `hsl(188.6,15.6%,91.2%)` 188.6°, 3, 92.5 ccffff `#ccffff`\nCIE-LAB 92.632, -1.814, -1.189 77.159, 82.13, 91.14 0.308, 0.328, 82.13 92.632, 2.169, 213.245 92.632, -3.389, -1.51 90.626, -6.62, 3.812 11100101, 11101011, 11101100\n\n# Color Schemes with #e5ebec\n\n• #e5ebec\n``#e5ebec` `rgb(229,235,236)``\n• #ece6e5\n``#ece6e5` `rgb(236,230,229)``\nComplementary Color\n• #e5ecea\n``#e5ecea` `rgb(229,236,234)``\n• #e5ebec\n``#e5ebec` `rgb(229,235,236)``\n• #e5e8ec\n``#e5e8ec` `rgb(229,232,236)``\nAnalogous Color\n• #eceae5\n``#eceae5` `rgb(236,234,229)``\n• #e5ebec\n``#e5ebec` `rgb(229,235,236)``\n• #ece5e8\n``#ece5e8` `rgb(236,229,232)``\nSplit Complementary Color\n• #ebece5\n``#ebece5` `rgb(235,236,229)``\n• #e5ebec\n``#e5ebec` `rgb(229,235,236)``\n• #ece5eb\n``#ece5eb` `rgb(236,229,235)``\n• #e5ece6\n``#e5ece6` `rgb(229,236,230)``\n• #e5ebec\n``#e5ebec` `rgb(229,235,236)``\n• #ece5eb\n``#ece5eb` `rgb(236,229,235)``\n• #ece6e5\n``#ece6e5` `rgb(236,230,229)``\n• #b9c9cc\n``#b9c9cc` `rgb(185,201,204)``\n• #c8d4d6\n``#c8d4d6` `rgb(200,212,214)``\n• #d6e0e1\n``#d6e0e1` `rgb(214,224,225)``\n• #e5ebec\n``#e5ebec` `rgb(229,235,236)``\n• #f4f6f7\n``#f4f6f7` `rgb(244,246,247)``\n• #ffffff\n``#ffffff` `rgb(255,255,255)``\n• #ffffff\n``#ffffff` `rgb(255,255,255)``\nMonochromatic Color\n\n# Alternatives to #e5ebec\n\nBelow, you can see some colors close to #e5ebec. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #e5eceb\n``#e5eceb` `rgb(229,236,235)``\n• #e5ecec\n``#e5ecec` `rgb(229,236,236)``\n• #e5ecec\n``#e5ecec` `rgb(229,236,236)``\n• #e5ebec\n``#e5ebec` `rgb(229,235,236)``\n• #e5eaec\n``#e5eaec` `rgb(229,234,236)``\n• #e5eaec\n``#e5eaec` `rgb(229,234,236)``\n• #e5e9ec\n``#e5e9ec` `rgb(229,233,236)``\nSimilar Colors\n\n# #e5ebec Preview\n\nThis text has a font color of #e5ebec.\n\n``<span style=\"color:#e5ebec;\">Text here</span>``\n#e5ebec background color\n\nThis paragraph has a background color of #e5ebec.\n\n``<p style=\"background-color:#e5ebec;\">Content here</p>``\n#e5ebec border color\n\nThis element has a border color of #e5ebec.\n\n``<div style=\"border:1px solid #e5ebec;\">Content here</div>``\nCSS codes\n``.text {color:#e5ebec;}``\n``.background {background-color:#e5ebec;}``\n``.border {border:1px solid #e5ebec;}``\n\n# Shades and Tints of #e5ebec\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, #060808 is the darkest color, while #fcfcfd is the lightest one.\n\n• #060808\n``#060808` `rgb(6,8,8)``\n• #0e1313\n``#0e1313` `rgb(14,19,19)``\n• #161d1f\n``#161d1f` `rgb(22,29,31)``\n• #1f282a\n``#1f282a` `rgb(31,40,42)``\n• #273335\n``#273335` `rgb(39,51,53)``\n• #2f3e41\n``#2f3e41` `rgb(47,62,65)``\n• #38494c\n``#38494c` `rgb(56,73,76)``\n• #405457\n``#405457` `rgb(64,84,87)``\n• #485f63\n``#485f63` `rgb(72,95,99)``\n• #506a6e\n``#506a6e` `rgb(80,106,110)``\n• #597579\n``#597579` `rgb(89,117,121)``\n• #618085\n``#618085` `rgb(97,128,133)``\n• #698a90\n``#698a90` `rgb(105,138,144)``\n• #749499\n``#749499` `rgb(116,148,153)``\n• #7f9da1\n``#7f9da1` `rgb(127,157,161)``\n• #8aa5aa\n``#8aa5aa` `rgb(138,165,170)``\n• #96aeb2\n``#96aeb2` `rgb(150,174,178)``\n• #a1b7ba\n``#a1b7ba` `rgb(161,183,186)``\n• #acbfc3\n``#acbfc3` `rgb(172,191,195)``\n• #b8c8cb\n``#b8c8cb` `rgb(184,200,203)``\n• #c3d1d3\n``#c3d1d3` `rgb(195,209,211)``\n``#cedadb` `rgb(206,218,219)``\n• #dae2e4\n``#dae2e4` `rgb(218,226,228)``\n• #e5ebec\n``#e5ebec` `rgb(229,235,236)``\n• #f0f4f4\n``#f0f4f4` `rgb(240,244,244)``\n• #fcfcfd\n``#fcfcfd` `rgb(252,252,253)``\nTint Color Variation\n\n# Tones of #e5ebec\n\nA tone is produced by adding gray to any pure hue. In this case, #e8e9e9 is the less saturated color, while #d4f7fd is the most saturated one.\n\n• #e8e9e9\n``#e8e9e9` `rgb(232,233,233)``\n• #e7eaea\n``#e7eaea` `rgb(231,234,234)``\n• #e5ebec\n``#e5ebec` `rgb(229,235,236)``\n• #e3ecee\n``#e3ecee` `rgb(227,236,238)``\n• #e2edef\n``#e2edef` `rgb(226,237,239)``\n• #e0eff1\n``#e0eff1` `rgb(224,239,241)``\n• #def0f3\n``#def0f3` `rgb(222,240,243)``\n• #dcf1f5\n``#dcf1f5` `rgb(220,241,245)``\n• #dbf2f6\n``#dbf2f6` `rgb(219,242,246)``\n• #d9f4f8\n``#d9f4f8` `rgb(217,244,248)``\n• #d7f5fa\n``#d7f5fa` `rgb(215,245,250)``\n• #d5f6fc\n``#d5f6fc` `rgb(213,246,252)``\n• #d4f7fd\n``#d4f7fd` `rgb(212,247,253)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #e5ebec is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population"
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https://nbviewer.jupyter.org/github/jump-dev/Convex.jl/blob/v0.9.0/examples/binary_knapsack.ipynb | [
"# Binary (or 0-1) knapsack problem¶\n\nGiven a knapsack of some capacity $C$ and $n$ objects with object $i$ having weight $w_i$ and profit $p_i$, the goal is to choose some subset of the objects that can fit in the knapsack (i.e. the sum of their weights is no more than $C$) while maximizing profit.\n\nThis can be formulated as a mixed-integer program as: $$\\begin{array}{ll} \\mbox{maximize} & x' p \\\\ \\mbox{subject to} & x \\in \\{0, 1\\} \\\\ & w' x <= C \\\\ \\end{array}$$\n\n$x$ is a vector is size $n$ where $x_i$ is one if we chose to keep the object in the knapsack, 0 otherwise.\n\nIn :\n# Data taken from http://people.sc.fsu.edu/~jburkardt/datasets/knapsack_01/knapsack_01.html\nw = [23; 31; 29; 44; 53; 38; 63; 85; 89; 82]\nC = 165\np = [92; 57; 49; 68; 60; 43; 67; 84; 87; 72];\nn = length(w)\n\nOut:\n10\nIn :\nusing Convex, GLPKMathProgInterface\nx = Variable(n, :Bin)\nproblem = maximize(dot(p, x), dot(w, x) <= C)\nsolve!(problem, GLPKSolverMIP())"
]
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http://dougspeed.com/single-predictor-analysis/ | [
"# Single-Predictor Analysis\n\nHere we explain how to perform one-predictor-at-a-time analysis, using linear regression (either classical or mixed-model) or logistic regression (only classical). When analysing a binary phenotype, it is theoretically better to use logistic regression, but in practice, linear regression usually suffices (and is substantially faster).\n\nAlways read the screen output, which suggests arguments and estimates memory usage.\n_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _\n\nLinear regression:\n\nThe main argument is --linear <outfile>.\n\nThis requires the options\n\n--bfile/--gen/--sp/--speed <datastem> - to specify the genetic data files (see File Formats).\n\n--pheno <phenofile> - to specify phenotypes (in PLINK format). Samples without a phenotype will be excluded. If <phenofile> contains more than one phenotype, specify which should be used with --mpheno <integer>.\n\nBy default, LDAK will perform classical linear regression. To instead perform mixed-model linear regression, you should first Calculate Kinships, then provide the kinship matrix using --grm <kinfile>. Typically, this kinship matrix is computed assuming a thinned version of the GCTA Model, that restricts to SNPs in approximate linkage equilibrium. Note that it is common to use leave-one-chromosome-out (LOCO) analysis to avoid proximal contamination (explained in these papers by Lippert et al. and Yang et al.). To perform LOCV, you should use --chr <integer> to analyse one chromosome at a time, then provide a kinship matrix calculated across all other chromosomes. For more details, see the example below.\n\nYou can use --covar <covarfile> to provide covariates (in PLINK format) as fixed effect in the regression. If the data contains predictors that are definitely associated with the phenotype, you can include these as fixed effects using --top-preds <toppredslist>. This is useful if wishing to perform a conditional analysis (i.e., to find secondary associations).\n\nTo perform weighted linear regression, use --sample-weights <sampleweightfile> (this is not possible if providing a kinship matrix). The file <sampleweightfile> should have three columns, where each row provides two sample IDs followed by a positive float. Note that by default, LDAK will use the sandwich estimator of the effect size variance (see this page for an explanation); to instead revert to the standard estimator of variance, add --sandwich NO.\n\nIf the phenotype is binary, you can use --prevalence <float> to specify the population prevalence. LDAK will then additionally report estimates on the liability scale.\n\nIf you add --permute YES - the phenotypic values will be shuffled. This is useful if wishing to perform permutation analysis to see the distribution of p-values or test statistics when there is no true signal.\n\nThe main output files is <outfile>.assoc, which provides full results for each predictor. <outfile>.summaries provides the results necessary for performing analyses using SumHer, while <outfile>.pvalues contains just the p-values. <outfile>.coeff provides estimates of the fixed effects, while <outfile>.score contains prediction models corresponding to six different p-value thresholds.\n_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _\n\nLogistic regression:\n\nThe main argument is --logistic <outfile>.\n\nThis requires the options\n\n--bfile/--gen/--sp/--speed <datastem> - to specify the genetic data files (see File Formats).\n\n--pheno <phenofile> - to specify phenotypes (in PLINK format). The phenotype must be binary (either cases 1 and controls 0, or cases 2, controls 1). Samples without a phenotype will be excluded. If <phenofile> contains more than one phenotype, specify which should be used with --mpheno <integer>.\n\nYou can use --covar <covarfile> to provide covariates (in PLINK format) as fixed effect in the regression. If the data contains predictors that are definitely associated with the phenotype, you can include these as fixed effects using --top-preds <toppredslist>. This is useful if wishing to perform a conditional analysis (i.e., to find secondary associations).\n\nIf you add --permute YES - the phenotypic values will be shuffled. This is useful if wishing to perform permutation analysis to see the distribution of p-values or test statistics when there is no true signal.\n\nThe main output files is <outfile>.assoc, which provides full results for each predictor. <outfile>.summaries provides the results necessary for performing analyses using SumHer, while <outfile>.pvalues contains just the p-values. <outfile>.coeff provides estimates of the fixed effects, while <outfile>.score contains prediction models corresponding to six different p-value thresholds.\n_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _\n\nExample:\n\nHere we use the binary PLINK files human.bed, human.bim and human.fam, the phenotypes quant.pheno and binary.pheno, and the covariates covar.covar from the Test Datasets.\n_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _\n\n1 - Classical linear regression.\n\nWe regress quant.pheno on each SNP in the genetic data by running\n\n./ldak.out --linear single --bfile human --pheno quant.pheno\n\nThe main results are saved in single.assoc. To repeat this including the covariates, we run\n\n./ldak.out --linear single2 --bfile human --pheno quant.pheno --covar covar.covar\n\nThe main results are saved in single2.assoc. The most significant SNP is 21:15603999 (P=5e-17 without covariates, 2e-14 with covariates). We can perform a conditional analysis including this SNP as a covariate by running\n\necho 21:15603999 > top.txt\n./ldak.out --linear single3 --bfile human --pheno quant.pheno --top-preds top.txt\n\nThe main results are saved in single3.assoc.\n_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _\n\n2 - Mixed-model linear regression.\n\nWe need to create a kinship matrix. We recommend doing this assuming a thinned version of the GCTA Model, for which we must obtain a list of predictors in approximate linkage equilibrium\n\n./ldak.out --thin le --bfile human --window-prune .05 --window-cm 1\n./ldak.out --calc-kins-direct le --bfile human --ignore-weights YES --power -1 --extract le.in\n\nThe list of thinned predictors is saved in le.in. We perform mixed-model linear regression by running\n\n./ldak.out --linear single4 --bfile human --pheno quant.pheno --grm le\n\nThe main results are saved in single4.assoc.\n\nTo perform LOCO analysis, we must first create a kinship matrix for each chromosome which is constructed using only predictors on other chromosomes (a complementary kinship matrix). We can do this using the following script\n\n#First we create per-chromosome kinship matrices\nfor j in {21..22}; do\n./ldak.out --calc-kins-direct le\\$j --bfile human --ignore-weights YES --power -1 --extract le.in --chr \\$j\ndone\n\n#Then we join these to obtain a genome-wide kinship matrix\nrm list.All\n\nfor j in {21..22}; do echo \"le\\$j\" >> list.All; done\n\n#Finally, we subtract the per-chromosome kinship matrices from the genome-wide matrix\nfor j in {21..22}\n\ndo echo \"leAll\nle\\$j\" > list.\\$j\n./ldak.out --sub-grm leN\\$j --mgrm list.\\$j\ndone\n\nThe complementary kinship matrices are saved with stems leN21 and leN22. Note that in this script, we loop from 21 to 22, because our example dataset contains only these two chromosomes; usually you would loop from 1 to 22. Now we can perform the mixed-model linear regression for each chromosome in turn by running\n\nfor j in {21..22}\ndo ./ldak.out --linear loco\\$j --bfile human --pheno quant.pheno --grm leN\\$j --chr \\$j\ndone\n\nThe main results are saved in locov21.assoc and locov22.assoc.\n_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _\n\n3 - Classical logistic regression.\n\nWe regress binary.pheno on each SNP in the genetic data by running\n\n./ldak.out --logistic single5 --bfile human --pheno binary.pheno\n\nThe main results are saved in single5.assoc."
]
| [
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.72471076,"math_prob":0.68888295,"size":7917,"snap":"2023-14-2023-23","text_gpt3_token_len":1957,"char_repetition_ratio":0.15000632,"word_repetition_ratio":0.44512662,"special_character_ratio":0.26588354,"punctuation_ratio":0.11786834,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9806744,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-02T01:53:45Z\",\"WARC-Record-ID\":\"<urn:uuid:e33c4596-6f90-4de0-b173-1a8b15cb35a2>\",\"Content-Length\":\"57400\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e900fca9-f63a-4b81-a263-063fedd9174b>\",\"WARC-Concurrent-To\":\"<urn:uuid:3534e59c-0e01-490a-83e6-9d42624a207f>\",\"WARC-IP-Address\":\"35.214.77.229\",\"WARC-Target-URI\":\"http://dougspeed.com/single-predictor-analysis/\",\"WARC-Payload-Digest\":\"sha1:SP5NXU4KOIPZDF5VIIQSPHBGDEYZI6MF\",\"WARC-Block-Digest\":\"sha1:5HBXJ7EVYGJ4O5MN2Y64NWJFLLQNYCXV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224648245.63_warc_CC-MAIN-20230602003804-20230602033804-00160.warc.gz\"}"} |
https://www.learnbymarketing.com/tutorials/rpart-decision-trees-in-r/ | [
"# Decision Trees in R\n\nThis tutorial covers the basics of working with the rpart library and some of the advanced parameters to help with pre-pruning a decision tree.\n\nIf you’re not already familiar with the concepts of a decision tree, please check out this explanation of decision tree concepts to get yourself up to speed. Fortunately, R’s rpart library is a clear interpretation of the classic CART book.\n\nLibraries Needed: rpart, rpart.plot\n\nData Needed: http://archive.ics.uci.edu/ml/datasets/Bank+Marketing (bank.Zip)\n\n```rpart(y~., data, parms=list(split=c(\"information\",\"gini\")),\ncp = 0.01, minsplit=20, minbucket=7, maxdepth=30)\n```\n\n## Data Understanding",
null,
"When dealing with a classification problem, you’re looking for opportunities to exploit a pattern in the data. It’s also very important to note that, for decision trees, you’re looking for linear patterns.\n\nThe bank balance is also something to look into.",
null,
"## Modeling\n\nWe’ll explore a few different ways of using rpart and we’ll explore the different parameters you can apply.\n\n### Basic Tree With Default Parameters",
null,
"```default.model <- rpart(y~., data = train)\ninfo.model <- rpart(y~., data = train, parms=list(split=\"information\"))\n```\n• The default splitting method for classification is “gini”.\n• To define the split criteria, you use `parms=list(split=\"...\")`\n\nYou can see that just changing the split criteria has already created a very different looking tree. Explaining the differences between gini index and information gain is beyond this short tutorial. But I have written a quick intro to the differences between gini index and information gain elsewhere.\n\nChoosing between the gini index and information gain is an analysis all in itself and will take some experimentation.\n\n### Minsplit, Minbucket, Maxdepth\n\n```overfit.model <- rpart(y~., data = train,\nmaxdepth= 5, minsplit=2,\nminbucket = 1)\n```\n\nOne of the benefits of decision tree training is that you can stop training based on several thresholds. For example, a hypothetical decision tree splits the data into two nodes of 45 and 5. Probably, 5 is too small of a number (most likely overfitting the data) to have as a terminal node. Wouldn’t it be nice to have a way to stop the algorithm when it encounters this situation?\n\nThe option minbucket provides the smallest number of observations that are allowed in a terminal node. If a split decision breaks up the data into a node with less than the minbucket, it won’t accept it.\n\nThe minsplit parameter is the smallest number of observations in the parent node that could be split further. The default is 20. If you have less than 20 records in a parent node, it is labeled as a terminal node.\n\nFinally, the maxdepth parameter prevents the tree from growing past a certain depth / height. In the example code, I arbitrarily set it to 5. The default is 30 (and anything beyond that, per the help docs, may cause bad results on 32 bit machines).\n\nYou can use the maxdepth option to create single-rule trees. These are examples of the one rule method for classification (which often has very good performance).",
null,
"```one.rule.model <- rpart(y~., data=train, maxdepth = 1)\nrpart.plot(one.rule.model, main=\"Single Rule Model\")\n```\n\nAll of these options are ways preventing a model from overfitting via pre-pruning. However, there’s one more parameter you may need to adjust…\n\n### cp: Complexity Parameter\n\nThe complexity parameter (cp) in rpart is the minimum improvement in the model needed at each node. It’s based on the cost complexity of the model defined as…",
null,
"• For the given tree, add up the misclassification at every terminal node.\n• Then multiply the number of splits time a penalty term (lambda) and add it to the total misclassification.\n• The lambda is determined through cross-validation and not reported in R.\n• The cp we see using `printcp()` is the scaled version of lambda over the misclassifcation rate of the overall data.\n\nThe cp value is a stopping parameter. It helps speed up the search for splits because it can identify splits that don’t meet this criteria and prune them before going too far.\n\nIf you take the approach of building really deep trees, the default value of 0.01 might be too restrictive.",
null,
"```super.overfit.model <- rpart(y~., data = train, minsplit=2,\nminbucket = 1, cp = 0.0001)\nrpart.plot(super.overfit.model, main = \"Really Overfit\")\n```\n\n### Using a Loss Matrix\n\nThe final option we’ll discuss is using the loss matrix from the parms list. Imagine you’re running a marketing campaign that offers some incredible discount for high-value customers you expect to leave.\n\nGetting this prediction is very important. Not only for the obvious reason of wanting to keep these high-value customers but also due to the fact that your prediction has a real cost to it.\n\nIf you mistake a normal customer for one that is about to leave, your discount is throwing away money. So rpart offers a loss matrix.\n\n```cost.driven.model <- rpart(y~.,data=train,\nparms=list(\nloss=matrix(c(0,1,5,0),\nbyrow=TRUE,\nnrow=2))\n)\n#The Matrix looks like...\n# [,1] [,2]\n#[1,] 0 1\n#[2,] 5 0\n```\n\nThe loss matrix is structured with actuals on the rows and predictions on the columns. When dealing with multiple classes, your matrix will look slightly different. Here’s what the confusion matrix looks like for my example.\n\n``` Predicted\nActual Class1 Class2\nClass1 TP FN\nClass2 FP TN\n**Assuming Class1 is the positive class.\n```\n\nSo for this given situation, it’s 5 times worse to generate a false positive than a false negative. This will have the effect of making Class1 predicted less frequently. This makes sense since your loss matrix says “Watch out for being wrong on Class1!! I’d rather you said Class2 if you’re not really certain it’s Class1…”\n\nA few notes on working with the Loss Matrix in R:\n\n• The order of the loss matrix depends on the order of your factor variable in R.\n• If you have multiple classes, think of it in terms of rows. “How much does it cost us if this real class is marked as something else”.\n• The diagonal should always be zero and the non-diagonal values should be greater than zero.\n• For example, for a four class problem, if you want to make Class2 the “positive” variable, you should mark all non-zero entries for that row with some cost.\n``` [,1] [,2] [,3] [,4]\n[1,] 0 1 1 1\n[2,] 5 0 5 5\n[3,] 1 1 0 1\n[4,] 1 1 1 0\n```\n\n## Interpreting RPart Output",
null,
"So you’ve built a few model by now. Let’s get analyzing with a few key functions.\n\n• `rpart.plot` from the rpart.plot package prints very nice decision trees.\n• `print(rpart_model)` Produces a simple summary of your model at each split. Shows…\n• Split criteria\n• # rows in this node\n• # Misclassified\n• Predicted Class\n• % of rows in predicted class for this node.\n• `printcp(rpart_model)` the cp table showing the improvement in cost complexity at each node\n• `summary(rpart_model)` the most descriptive output, providing…\n• CP Table\n• Variable Importance\n• Description of the Node and Split (including # going left or right and even surrogate splits.\n• Can be very verbose, so print with caution\n• `predict(rpart_model, newdata, method=\"class\")` lets you apply the model to new data. Using the “class” method, it will return the most likely class label.\n\nThere’s still more to the rpart function! There are prior probabilities and weighting observations. There are variable costs (and not just classification costs). There are poisson and anova splitting methods. However, this tutorial has covered the basics and some of the advanced parameters.\n\nKeep experimenting and pushing your decision tree knowledge!"
]
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"https://www.learnbymarketing.com/wp-content/uploads/2016/05/dtree-cost-complexity.png",
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"https://www.learnbymarketing.com/wp-content/uploads/2016/05/dtree-r-vis04-tut-1-300x181.png",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8636745,"math_prob":0.9014737,"size":7413,"snap":"2023-40-2023-50","text_gpt3_token_len":1712,"char_repetition_ratio":0.104332566,"word_repetition_ratio":0.014913008,"special_character_ratio":0.23027115,"punctuation_ratio":0.1266256,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97329444,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,3,null,3,null,3,null,3,null,3,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-28T17:50:04Z\",\"WARC-Record-ID\":\"<urn:uuid:f0f98bb7-2ee2-447d-a07a-62f0ab105488>\",\"Content-Length\":\"44829\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e846a55b-c46a-4464-b24b-a220b00266ea>\",\"WARC-Concurrent-To\":\"<urn:uuid:90bb56fe-2da8-467b-a59b-b78584f82763>\",\"WARC-IP-Address\":\"205.196.221.110\",\"WARC-Target-URI\":\"https://www.learnbymarketing.com/tutorials/rpart-decision-trees-in-r/\",\"WARC-Payload-Digest\":\"sha1:LWI7YGJJITXDKRQFUEMTBEUDIQTVAYHY\",\"WARC-Block-Digest\":\"sha1:EDWLE4SDLTDSDY75OTEHLGIVKMWJAZKR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510427.16_warc_CC-MAIN-20230928162907-20230928192907-00840.warc.gz\"}"} |
https://www.beilstein-journals.org/bjnano/articles/10/127 | [
"On the relaxation time of interacting superparamagnetic nanoparticles and implications for magnetic fluid hyperthermia\n\nNational Institute of Materials Physics, P.O. Box MG 7, 077125, Magurele, Romania\n\n1. Corresponding author email\n\nAssociate Editor: P. Leiderer\nBeilstein J. Nanotechnol. 2019, 10, 1280–1289. doi:10.3762/bjnano.10.127\nReceived 15 Oct 2018, Accepted 24 May 2019, Published 24 Jun 2019\n\n• Full Research Paper\n\nAbstract\n\nA critical discussion on the presently available models for the relaxation time of magnetic nanoparticles approaching the superparamagnetic regime in the presence of interparticle dipolar interactions is considered. The direct implications of such interactions for magnetic fluid hyperthermia therapy through susceptibility loss mechanisms give rise to an indirect method for their study via specific absorption rate measurements performed on ferrofluids of different volume fractions. The theoretical support for the specific evolution of the relaxation time constant and the anisotropy energy barrier versus the interparticle interactions in a perturbation approach of the simple Néel expression for the relaxation time is provided via static and time-dependent micromagnetic simulations.\n\nKeywords: magnetic hyperthermia; magnetic nanoparticles; magnetic relaxation time; micromagnetic simulation\n\nIntroduction\n\nMagnetic relaxation phenomena in nanoparticulate systems are under intensive investigation today, especially due to their implications in various fields of nanotechnology such as biomedicine, magnetic data storage and sensors [1-6]. Concerning the biomedical applications, the magnetic relaxation of nanoparticles is of key interest in magnetic resonance imaging (through the influence of the relaxation time of the nanoparticulate contrasting agents on proton relaxivity [7-9]) and cancer therapy (through magnetic fluid hyperthermia therapy [10,11]). The efficiency of the magnetic nanoparticles (MNPs) in a colloidal system to convert the energy of AC magnetic fields into temperature increments is of high importance for magnetic hyperthermia therapy. Usually the system consists of MNPs dispersed in an aqueous medium, known also as a ferrofluid. The heat transfer from the specifically configured AC field (with biologically compatible amplitude and frequency) to the tissue loaded with suitably functionalized MNPs can be performed by different mechanisms, depending on type and size of the MNPs. In the case of magnetic oxide nanoparticles (usually ferrites) with an average size of less than 30 nm (considered at this moment as the most suitable for such purposes), two heat transfer mechanisms must be considered [12,13]: (i) hysteretic losses and (ii) magnetic relaxation processes. The effective mechanism depends on the relationship between the magnetic relaxation time, τ, and the inverse of the AC field frequency, 1/f, defining the time window, τM, of the magnetic excitation (τM may also represent a measuring time window). In this respect, two regimes have to be mentioned here : (i) a static regime, corresponding to τ >> τM, where the main heat transfer mechanism is through hysteretic losses and (ii) the dynamic regime, corresponding to τ << τM, where the main heat transfer mechanism is through magnetic relaxation phenomena. The relaxation time depends on the volume of the MNPs, and in the case of nanoparticulate systems of finite size, also on the polydispersity. Both mechanisms can contribute, with the first, for nanoparticles larger than a critical size, and the second, for finer nanoparticles. It is thus very important that both the size distribution of the MNPs and the relaxation time corresponding to each volume are well characterized.\n\nThe heat transfer is reflected in a power deposition term, qp. Among other parameters, this term depends on the volume fraction of the nanoparticles in the ferrofluid, φ, which is defined as φ = VMNPs/(VMNPs + VFF) where VMNPs is the total volume of the MNPs and VFF is the total volume of the ferrofluid. In turn, qp, which quantifies the energy transfer from the magnetic AC field to the ferrofluid (or specific tissue loaded by MNPs), is related to the specific absorption rate (SAR). Actually the SAR represents the power absorbed per mass unit of ferrofluid (W/kg). Accordingly, SAR = P/m = c × ΔTt = qp/ρ, where P is the power dissipated in the mass m of ferrofluid. Terms c and ρ are the specific heat and the density of the ferrofluid respectively whereas ΔTt is the rate of the temperature increment. The above relation is valid only for adiabatic systems where the absorbed energy is completely transformed into heat. The SAR can be simply evaluated if the rate of the temperature increment is experimentally estimated in adiabatic-like systems. In order to reach such conditions, the experiments have to be done on low volume ferrofluid systems (simulating as much as possible the loaded tissue) with suitable isolation aimed to diminish the heat losses. However, even extremely careful experimental caution cannot completely prevent the heat transfer from the actuated ferrofluid. Therefore, new methodologies to properly compensate for the heat losses have been proposed over time. The simplest procedure of taking the initial heating slope [11,15] has been extended to more sophisticated cases of correcting the heating curves through suitably exploited cooling curves [16,17] or alternative calorimetric calibrations . Once an experimental methodology for a suitable estimation of the SAR is reasonably achieved, two opportunities appear: (i) the possibility of building a SAR database for nanoparticulate systems obtained by different processing routes as a function of particle size, volume fraction and dispersing liquid parameters and (ii) the possibility to investigate the mechanisms involved in the heat transfer under the AC field actuation as thoroughly as possible. The first option is directly connected to realistic numerical solutions of the bio-heat-transfer equation which is of key interest in the optimization of cancer therapy by magnetic hyperthermia therapy . The second option may be effective in exploiting the theoretical models proposed for linking the experimental SAR dependencies to the physical mechanisms. From the theoretical point-of-view, the heat transfer is only due to the actuated nanoparticles, and therefore, the power, P, absorbed by the ferrofluid is in fact the power absorbed by the nanoparticles. Accordingly, an additional coefficient SARNP counting for the power dissipated by the mass of the nanoparticles can be defined, the link between the two coefficients being expressed as:\n\nIn the above relation, P/VNP = P* represents the absorbed power per unit volume of MNPs which can be directly related to the evolution of the magnetization (or susceptibility) of the system (magnetic moment per unit volume).\n\nFor nanoparticles in the static regime, P* can be simply expressed by multiplying the area of the hysteresis loop developed under the amplitude of the AC field with its frequency. However, according to Equation 1, the experimentally obtained SAR values should increase linearly with φ under the condition that P* does not depend explicitly on φ. However, both experimental evidence and Monte Carlo modeling have been provided in to understand the influence of the dipolar interparticle interactions on P*, leading to a reduction of the heating power and consequently on the hyperthermia effects of ferromagnetic nanoparticles.\n\nFor noninteracting MNPs in the dynamic magnetic regime (superparamagnetic regime), it has been shown by Rosensweig in 2002 that only a susceptibility loss mechanism has to be considered (P* ≈ χ’’ where χ’’ is the out of phase component of the magnetic susceptibility). Furthermore, earlier more sophisticated theories have been considered in order to investigate the role of the interparticle interactions on the magnetic relaxation phenomena or on the behavior of the magnetic susceptibility. The most successful of these approaches is related to the first order and second order modified mean field theory [22,23]. It should be noted that the mean field theory (providing means to include the effective field experienced by a particle due to the action of the neighboring nanoparticles) at various level of approximation has been tested with respect to experimental dependencies. This led to available approaches for describing the more complex cases related to particle size distribution and different concentrations. The influence of the dipolar interactions on the frequency-dependent magnetic susceptibility has been more recently extensively studied by Ivanov et al. [24,25]. For example, in the Debye theory of polar relaxation was applied in the case of MNPs undergoing only Brownian motion. This was further extended to a first order modified mean field theory leading to a relatively simple expression of the out-of-phase component of the susceptibility in ferrofluids with dipolar interparticle interactions. Accordingly, the out-of-plane component of the susceptibility for interacting MNPs is proportional to the out-of-phase component of susceptibility of similar noninteracting MNPs modulated by a factor which depends on the in phase component of susceptibility of noninteracting MNPs. In spite of its simplicity, this approach cannot be easily applied to hyperthermia applications, where the nanoparticles are usually vectorially bound to the tissue and Brownian relaxation can be neglected. However, even in the case of usual ferrofluids consisting of dispersed and motion-free nanoparticles, the Brownian relaxation mechanism becomes dominant over the Néel mechanism only for nanoparticles larger than a critical radius (e.g., 8 nm in the case of magnetite nanoparticles) . In such conditions, the heat transfer mechanism might also be completed by a hysteretic loss (the nanoparticle size can be large enough to open a hysteresis loop), so it becomes really difficult to compare experimental results on SAR with the above-mentioned approach. On the other hand, according to the above-mentioned theory, the out-of-phase component of the susceptibility of interacting nanoparticles can be expressed only as a function of components of noninteracting nanoparticles, which gives support for the validity of a perturbational approach towards an upper limit of volume fraction, with the possibility to extend such reasoning to the expression of the relaxation time. In fact, the first experimental support for the specific influence of the magnetic dipolar interactions on different parameters in the relaxation law of nanoparticles in the dynamic magnetic regime has been provided in . Accordingly, it has been experimentally proven that the effect of the nanoparticle interaction is not only to increase the anisotropy energy barrier per nanoparticle but also to decrease the specific time constant in the simplest Néel–Brown expression of the relaxation time, which is usually considered only as a material dependent parameter. In this context, the threefold aim of this paper is: (i) to critically discuss the presently available models for the relaxation time of nanoparticles in the magnetic dynamic regime with focus on the effect of the interparticle interactions, (ii) to provide a specific approach for the experimental investigation of the interparticle interactions through SAR measurements and (iii) to provide theoretical support for the specific evolution of the relaxation time constant and the anisotropy energy barrier in the Néel expression as function of interparticle interaction through a suitable exploitation of static and time-dependent micromagnetic simulations.\n\nResults and Discussion\n\nRelaxation time and interparticle interactions: a critical survey\n\nThe first expression of the relaxation time of magnetic monodomain particles (of Stoner–Wohlfarth-type) with the thermal excitation energy (kT) approaching the particle anisotropy barrier energy (e.g., ΔE = KV in the case of uniaxial anisotropy) has been provided by Néel under the assumption that the particle macrospin behaves as a gyroscopic system :\n\nIn Equation 2, τ0N is a time characteristic depending slightly on temperature and other material parameters such as magnetization, gyromagnetic ratio, Young’s modulus, etc. An improved model for the magnetic relaxation was performed by Brown who supposed that the orientation of the particle macrospin may be described by the Gilbert equation. Accordingly, the relaxation time was introduced as a specific parameter in a Fokker–Planck-type equation devoted to the probability density of orientation . Brown derived different formula for the relaxation time (in different approximations for the ΔE/kT ratio) which were further refined by Aharoni . However, their form did not significantly differ from Equation 2, the only difference being in the expression of the time factor τ0 (hence the general name of Néel–Brown for Equation 2). Other expressions of the relaxation time have been further obtained by solving the Fokker–Planck equations in different approximations, as excellently summarized in . As a general behavior, the exponential dependence of the relaxation time versus the ratio ΔE/kT remains valid as soon as this ratio is higher than unity.\n\nDifferent attempts to consider the influence of the interparticle dipolar interactions on the relaxation mechanisms have been provided by Mørup et al. [31,32] and Dormann et al. . The first idea in a perturbation theory due to a mean field effect of the dipolar interactions is to express the magnetic energy of the system by a superposition of uniparticle energies, with the anisotropy energy modified by an additional Zeeman term due to the presence of the particle magnetic moment in the effective field created by the neighboring particles. Such an approach has been used in , resulting in the relaxation time (Equation 2) in the new corrected form:\n\nwith ΔE* > ΔE (e.g., ΔE = KV for uniaxial symmetry and lack of interparticle interactions) and τ0* < τ0 (e.g., τ0 is the relaxation time constant for noninteracting MNPs), where both starred (*) parameters depend on the number and arrangement of the neighboring MNPs. According to Dormann’s model, the interparticle dipolar interaction will always increase the anisotropy energy barrier and will decrease the time constant with respect to the case of noninteracting particles, but in a such a way that the relaxation time will remain larger in the first case. On the contrary, Mørup et al. have reported, starting from temperature-dependent Mössbauer spectroscopy data obtained on maghemite based ferrofluids, a decrease of the relaxation time with increasing volume fraction. An intriguing explanation based on the decrease of the anisotropy energy per nanoparticle due to the dipolar interactions for the specific kT/KV ratio has been provided in a subsequent paper . As a consequence, the effect of the interparticle interactions on the magnetic relaxation time, which is of key importance in magnetic hyperthermia applications, has remained an open issue to be investigated both experimentally and theoretically.\n\nExperimental investigation of the relaxation time through SAR measurements\n\nAccording to the above-mentioned issues, the proposed approach for the investigation of the relaxation time through SAR measurements is to express the power absorbed per unit volume of nanoparticles, P*, within Rosensweig’s model specific to monodisperse noninteracting nanoparticles of a given volume, V. Nanoparticles that are smaller than a critical size that lead to the predominance of the Néel relaxation mechanism over the Brownian one are considered. Taking into account only a perturbative effect of the interparticle interactions on the magnetic relaxation mechanism, it will be implicitly assumed that the above model may be used up for an upper limit of volume fraction, φ, with the only effect of the perturbation reflected in the expression of the relaxation time. Accordingly, the dissipated power is\n\nwhere τ is the effective relaxation time (only the Néel component), f and H are the frequency and the amplitude of the applied AC magnetic field, µ0 is the permittivity of air and χ0 is the equilibrium susceptibility (χ0",
null,
"µ0ηMs2VM/3kBT where Ms and VM are the spontaneous magnetization and the magnetic volume of the nanoparticle, respectively). The simple Néel–Brown expression,\n\nwas considered, where τ0 is a time constant (supposed to depend only on the material, because the temperature variation in hyperthermia is of only a few degrees) and ΔE* is an anisotropy energy barrier (supposed to be modified by interparticle interactions).\n\nAccording to the Equation 1 and the definition of SAR(T), we find\n\nThe time increment of the ferrofluid temperature, T(t), can be evaluated versus parameters specific to P* by the numerical integration of the last part of Equation 6. On the other hand, T(t) can be also experimentally obtained by heating the ferrofluid under a completely characterized AC magnetic field and using specific methodologies to minimize the heat losses [16,17]. The investigation of the relaxation mechanism can be therefore performed by the simulation of the experimentally obtained T(t) curves (adiabatic-like) through the theoretical T(t) dependence obtained by integration of Equation 6, under the condition that the material-dependent parameters for the magnetite nanoparticle based ferrofluids of different volume fractions were independently derived by alternative techniques [26,33]. Detailed experimental and methodological aspects involving a very diluted ferrofluid as reference are described in . The adiabatic-like curve obtained in the case of a concentrated ferrofluid (φ = 0.094) consisting of quasi-ellipsoidal magnetite nanoparticles of average magnetic volume of 4.3 × 10−25 m3 dispersed in transformer oil, with a spontaneous magnetization Ms = 4.5 × 105 A m−2, as determined by DC low temperature magnetometry and an effective anisotropy energy barrier ΔE* = 7.2 × 10−21 J, as estimated by Mössbauer spectroscopy, is presented in Figure 1 (red open circles). The amplitude of the AC field was 21 kA/m, as numerically estimated from the current amplitude and geometrical characteristics of the RF induction coil, by a finite element method.",
null,
"Figure 1: (a) Red open circles – adiabatic curve obtained by a proper experimental methodology in the case of a ferrofluid with φ = 0.094, black open stars – theoretical curve obtained from the Rosensweig model with input parameters specific to the ferrofluid with φ = 0.094 (e.g., KV = 7.2 × 10−20 and H0 = 21 kA/m), but considering for the relaxation time constant the value obtained from a more diluted sample (τ0 = 3.4 × 10−9 s, φ = 0.005), filled blue stars – the same theoretical curve adjusted to τ0 at 1.2 × 10−9 s for the best fit of the adiabatic curve. (b) Volume fraction dependence of τ0* and anisotropy energy barrier (ΔE*) as experimentally obtained. Figure 1: (a) Red open circles – adiabatic curve obtained by a proper experimental methodology in the case of... Jump to Figure 1\n\nFigure 1 also shows two theoretical simulations according to the numerical solution of Equation 6 involving the Rosensweig model for power P* (Equation 4) with the above-mentioned input parameters and for two different time constants: τ0 = 3.4 × 10−9 s and 1.2 × 10−9 s. The first time constant corresponds to the best fit of the experimental curve of a similar system but for a very low volume fraction (φ = 0.005). It can be observed that this value (typical for noninteracting nanoparticles) cannot conveniently fit the experimental curve for the concentrated system (with φ = 0.094), for which the best fit of the initial slope is obtained for a lower value of τ0 (1.2 × 10−9 s in this case). A direct proof for the dependence of τ0 on the strength of the dipolar interactions is provided by the above example.\n\nA more extended experimental study on the influence of the volume fraction and external AC field amplitude on the SAR values has been previously performed and partially reported in . The work was performed on similar ferrofluids with four different volume fractions, where the ferrofluid of the lowest volume fraction (φ = 0.005) is considered as the reference for noninteracting nanoparticles. All samples were magnetically excited by a radiofrequency magnetic field with a constant frequency of 235 kHz (single radiofrequency inductor) and at four amplitude values: 14 kA m−1, 21 kA m−1, 28 kA m−1 and 35 kA m−1. As expected, the experimental SAR values increase with the volume fraction (in a more saturated manner as compared to the predicted linear dependence), the increment being larger for higher amplitudes of the AC field (e.g., for 14 kA m−1 the SAR increases from 0.3 W/g in the ferrofluid with φ = 0.005 to 2.8 W/g in the ferrofluid with φ = 0.16 whereas for 28 kA m−1 the SAR increases from 1.1 W/g in the ferrofluid with φ = 0.005 to 4.0 W/g in the ferrofluid with φ = 0.16). According to the above reported values (expressed in W/g of ferrofluid), the SAR increases with the square of the field amplitude (as theoretically expected) in the case of very diluted ferrofluids, and much more slowly in the case of concentrated ferrofluids (e.g., for φ = 0.16, SAR increases less than two times when the field amplitude doubles). Note that any comparison between the SAR values obtained under different experimental conditions and on different ferrofluids (even formed by a same type of nanoparticles) is not conclusive. In order to avoid the effect of the dilution, the SAR values expressed in W/g of magnetic material can be used. Based on the definition of the volume fraction and taking a density of roughly 5.6 g/cm3 for magnetite nanoparticles and 0.9 g/cm3 for the transformer oil, one can compute a quantity of 0.03 g of magnetic material in 1 g of ferrofluid in the case of the sample with φ = 0.005 and of 0.54 g of magnetic material in 1 g of ferrofluid in the case of the sample with φ = 0.16. Furthermore, the SAR values of about 37 W/g of magnetic material and 7.5 W/g of magnetic material are deduced for the sample with φ = 0.005 and for the sample with φ = 0.16, respectively (under the same experimental excitations with the field frequency of 235 kHz and the field amplitude of 28 kA m−1). Nevertheless, such a strong reduction of the heating power due to only interparticle interactions should be taken into account in hyperthermia biomedical applications, which has also been reported in other previous studies [34,35].\n\nTheoretical investigation of the relaxation time by micromagnetic simulations\n\nThe dynamic magnetic behavior of parallelepiped nanoparticles with specific configurations (reflecting different volume fractions) has been simulated using the object oriented micromagnetic framework , an open source software developed by the National Institute of Standards and Technology. The core software uses numerical methods (Euler or Runge–Kutta) in order to solve a system of Landau–Lifshitz–Gilbert (LLG) equations describing the time evolution of each magnetic moment (only one magnetic moment is assigned to each monodomain-like nanoparticle). For the specific purpose of this work, i.e., the observation of the magnetic behavior under thermal excitations, an extension of the program, as developed by Oliver Lemke, was used . The influence of temperature on the particulate magnetic moments, m, has been modeled by altering the original LLG equation with a strongly fluctuating term hfluct. Thus, the LLG equation becomes a Langevin-type stochastic differential equation [28,37]:\n\nThe values of hfluct at each magnetic moment are provided according to a Gaussian distribution which is characterized by two parameters, the mean and the variance, taking values of 0 and",
null,
"respectively, where γ is the gyro-magnetic ratio and α the damping constant (the usual value of 0.5 was considered). Ms and V are the spontaneous magnetization and the volume of the magnetic material/MNP, respectively, and Heffect is an effective field originating from the total magnetic energy of the system (with anisotropy energy, demagnetization energy, exchange energy and Zeeman energy, as components).\n\nA group of thirteen magnetic nanoparticles (each of size 4.4 × 4 × 4 nm in order to assure the uniaxial magnetic anisotropy) with no magneto-crystalline anisotropy, a stiffness constant (A) of 1.3 × 10−11 J/m3 and a spontaneous magnetization (Ms) of 8.5 × 105 A/m have been arranged in a bidimensional geometry as in Figure 2.",
null,
"Figure 2: The specific geometry of magnetic nanoparticles in the micromagnetic simulations. Different configurations may be obtained for different couples of d1 and d2 distances. Figure 2: The specific geometry of magnetic nanoparticles in the micromagnetic simulations. Different configu... Jump to Figure 2\n\nNote that the magnetic parameters of realistic order of magnitude have been chosen in order to reflect real experiments on magnetite and also to assure the magnetic monodomain-like behavior for the nanoparticles. However, the considered nanoparticles differ in size/shape and bidimensional arrangement from those leading to the results reported in Figure 1, and therefore, rather a qualitative agreement should be expected between simulation and experiment. In this particular case, the volume fraction (as a measure of the particle density) will be computed as the volume occupied by touching nanoparticles relative to the total volume occupied by the 13 nanoparticles distributed in only one layer, being noted accordingly by φ*. The particles are considered to be magnetic monodomain and consequently defined by their associated magnetic moment (or cell magnetization), considering a fixed nanoparticle volume in each occupied cell (thus the terms MNPs and magnetic moments will be further used interchangeably). Time-dependent simulations have been performed for three different densities of nanoparticles, reflected by the following interparticle distances d1 and d2: high density (volume fraction) with d1 = 8 nm and d2 = 8.8 nm (i.e., 1 free cell between neighboring MNPs), medium density with d1 = 12 nm and d2 = 13.2 nm (i.e., 2 free cells between neighboring MNPs) and low density with d1 = 16 nm and d2 = 17.6 nm (i.e., 3 free cells between neighboring MNPs). For static simulations, a special case was considered, namely a purely noninteracting case of a very low density of MNPs (7 free cells spacing between neighboring MNPs, with d1 = 32 nm and d2 = 35.2 nm).\n\nThe equivalent volume fraction, φ*, corresponding to the interparticle spacing of ncells = 1, 2 and 3 cells, respectively, can be straightforwardly estimated to be 1/(1 + ncells)2, resulting in approximate values of 0.25 for high density packing, 0.11 for medium density packing and 0.06 for low density packing.\n\nBoth time-dependent and time-independent simulations have been performed, each of them being briefly discussed with respect to the provided results and limitations.\n\nTime-independent approach: simulation and data analysis\n\nRegarding the time-independent analysis, a monotonously increasing magnetic field ranging from −100 mT up to 100 mT has been applied along the Oy axis (perpendicular to the MNP length). Considering the uniaxial anisotropy (as due to only the shape anisotropy), defined along Ox axis (along the MNP length), it is possible to calculate both the magnitude of the anisotropy energy barrier of each particle (e.g., as in Figure 3a,b) or the anisotropy energy of the whole system (e.g., as in Figure 3c) for all the assumed three configurations by simply scanning the magnetic field orthogonal to the anisotropy axis and computing the demagnetization energy.",
null,
"Figure 3: Demagnetization energy profiles and anisotropy energy barriers for MNP P1, for low and high volume fractions, respectively (a and b) and for the whole considered nanoparticulate system for the same configurations (c). Figure 3: Demagnetization energy profiles and anisotropy energy barriers for MNP P1, for low and high volume ... Jump to Figure 3\n\nFigure 3a shows the demagnetization energy profiles of particle P1 as function of the applied magnetic field in the case of the low volume fraction configuration. The fact that the energy profiles are independent of the particle surroundings (i.e., number of first order neighbors) gives support for noninteracting particle behavior for the chosen low volume fraction. Another observation that shows the independent nature of the NPs for the low volume fraction case is the relation between the overall potential barrier and the potential barrier of individual particles, i.e., the 1.3 × 10−20 J energy barrier (Figure 3c, top) is the sum of 13 one particle energy barriers of 1 × 10−21 J (Figure 3a).\n\nOn the other hand, Figure 3b and Figure 3c (bottom), corresponding to the case of a high volume fraction, shows a strong influence of the particle surroundings on the particle energy barrier. For example, according to Figure 3c, the energy barrier of MNP P1 is much higher in the case of a high volume fraction configuration as compared to the case of a low volume fraction configuration. Thus, a strong interaction between the nanoparticles of the system can be concluded for this case, leading to an overall energy barrier significantly different from the sum of the energy barriers belonging to each particle in the configuration. Note also that the outer particles, with a less symmetrical surrounding, present a demagnetization energy profile (not shown here) which differs from a Stoner–Wohlfarth-like behavior, in contrast to the case of the inner nanoparticle P1 for which this behavior is roughly respected (see Figure 3b). However, in the real experimental cases dealing with a very high number of MNPs, the inner ones (with a complete number of neighbors) are by far dominant over the outer ones. Therefore, the time-dependent analysis will be performed only for such inner nanoparticles with Stoner–Wohlfarth-like behavior, sensing the interparticle dipolar interactions only as a perturbation of their own anisotropy energy. It is worth mentioning that the demagnetization energy profile of the whole system of 13 particles is also influenced by the volume fraction (and hence by interparticle interactions). So, at a first glance, time-independent simulations already suggest that anisotropy energy barriers and hence the switching time provided by Equation 5 may be altered by the volume fraction of the system of monodisperse (identical) nanoparticles. However, any information on the characteristic time constant τ0 is out of reach with this approach and this aspect will be investigated further by time-dependent simulations.\n\nTime-dependent approach: simulation and data analysis\n\nThe magnetic switching of each nanoparticle in all three geometric configurations described above has been investigated via time-dependent (dynamic) micromagnetic simulations (numerical solution of Equation 7) performed at different temperatures (e.g., from 60 K to 140 K). The time dependence of the magnetic moment of one particle (e.g., P1 at 60 K) along the Ox axis (easy magnetization axis) is presented in Figure 4a. Large files consisting of such time dependences of the normalized magnetic moment (magnetization) have been obtained for each MNPs P1 at the mentioned temperatures and volume fractions. The following methodology was applied for a reliable evaluation of both the anisotropy energy barrier and relaxation time constant for any MNP of type P1, for a specific volume fraction.",
null,
"Figure 4: Temporal evolution of normalized magnetization along the Ox axis for particle P1 within a high volumetric fraction configuration, at 60 K (a). Counting method for the random switching of normalized magnetization (b). Arrhenius plot for the same particle, in the same configuration for temperatures ranging from 60 K to 140 K (c). Figure 4: Temporal evolution of normalized magnetization along the Ox axis for particle P1 within a high volu... Jump to Figure 4\n\nThe relaxation time, τ, was properly estimated (see Figure 4b) for each temperature from the time dependencies of the magnetic moment of P1 (similar to the one presented in Figure 4a). Then a plot of ln τ vs (1/T) according to Equation 5 was considered. Such an Arrhenius plot (logarithmic representation of Equation 5) is shown in Figure 4c for MNP P1, in the case of a high volume fraction configuration. The reasonable linear dependence suggests on one hand the validity of the perturbation-like treatment of the relaxation time even for the high concentration case, and on the other hand, it provides both the anisotropy energy barrier ΔE* and the relaxation time constant τ0* assumed to be affected by the interparticle interactions.\n\nIt is worth noting that the above-mentioned methodology depends critically on the reliability in the determination of the relaxation time (or switching frequency) from Figure 4a like dependencies. In this respect, a C++ program has been developed for automatically computing the number of magnetic switching events over a certain period (e.g., during 100 ns) from such time-dependent magnetization curves. Given a noise of 0.2 in Figure 4a like curves, any jump of the normalized magnetization between a maximum of 0.8 and a minimum of −0.8 was regarded as a switching event. Each intersection of the plots with the 0.8 and −0.8 lines triggers an H and L flag, respectively. A switching event is counted when H and L are both triggered. The schematic representation of the counting method is shown in Figure 4b. The relaxation time is provided by dividing the total simulation time by the total number of switching events.\n\nThe values obtained for the anisotropy energy EB = ΔE* and the time constant τ0* for the inner MNP P1 (considered as representative in assemblies of a large number of MNPs) for different volume fractions are shown in Figure 5.",
null,
"Figure 5: Relaxation time constant (a) and anisotropy energy barrier (b) of an inner MNP versus the volume fraction in %, as estimated from the theoretical simulations. Figure 5: Relaxation time constant (a) and anisotropy energy barrier (b) of an inner MNP versus the volume fr... Jump to Figure 5\n\nIt is worth noting that the equality between the anisotropy energy barrier per nanoparticle obtained in the case of static and dynamic (temperature dependent) simulations is not straightforward due to the fact that in the first case the external excitation (i.e., the magnetic field) has been applied strictly along the Ox axis whereas in the second case, the excitations, including the fluctuating field due to temperature activation of the surrounding magnetic moments (described by hfluct) act on all three axes. It should also be mentioned that both the anisotropy energy and relaxation time constant depend not only on the volume fraction (average distances between MNPs) but also on the number of magnetic neighbors, being different for outer or inner MNPs, but also for a bidimensional or tridimensional configurations of neighbors. Given the extension of a real system of MNPs relative to the present simulations, the most representative values describing the magnetic relaxation regime of ferrofluids are those corresponding to inner MNPs, as above mentioned, whereas the tridimensional arrangement (equivalent to an increased number of neighbors at a similar volume fraction and hence, to an increased interaction) would be reflected by an amplification of the specific behaviors observed from the bidimensional configuration of the MNPs. According to Figure 5, the relaxation time constant decreases almost exponentially and the anisotropy energy barrier of the nanoparticle increases almost linearly with the volume fraction. This result is in reasonable qualitative agreement with the experimental results presented in Figure 1.\n\nConclusion\n\nThe direct implications of the magnetic relaxation phenomena of interacting superparamagnetic nanoparticles on the specific absorption rate of ferrofluids, through susceptibility loss mechanisms, are discussed. Various models supporting the influence of interparticle interactions on the relaxation time are reviewed, leading to the conclusion that they are controversial, not versatile enough, and are limited to particular cases. A more general perturbation approach of the simple Néel expression is introduced with both the relaxation time constant and the anisotropy energy barrier depending on interparticle interactions, as suggested by specific absorption rate measurements performed on ferrofluids of different volume fractions. A theoretical investigation of such parameters versus interparticle interactions was performed via static and time dependent micromagnetic simulations on different configurations of nanoparticles. It has been proven that a monotonous increase of the anisotropy energy barrier and a quasi-exponential decrease of the relaxation time constant versus interparticle interactions, even up to the highest experimentally reachable volume fractions of ferrofluids is present. The mentioned evolutions of the two parameters (anisotropy energy barrier and relaxation time constant) in relation to interparticle interactions lead to a significant decrease in the SAR values in ferrofluids of high volume fraction, which should be taken into account in hyperthermia therapy applications.\n\nAcknowledgements\n\nThis work was supported by the Core Program PN18-110101/2018 and the grant of the Romanian Ministry of Research and Innovation, CCCDI-UEFISCDI, project number PN-III-P1.2-PCCDI-2017-0871/2018 and POC P_37_697 REBMAT.\n\nInteresting articles\n\nAn adapted Coffey model for studying susceptibility losses in interacting magnetic nanoparticles\n\nMihaela Osaci and Matteo Cacciola\n\nAntitumor magnetic hyperthermia induced by RGD-functionalized Fe3O4 nanoparticles, in an experimental model of colorectal liver metastases\n\nOihane K. Arriortua, Eneko Garaio, Borja Herrero de la Parte, Maite Insausti, Luis Lezama, Fernando Plazaola, Jose Angel García, Jesús M. Aizpurua, Maialen Sagartzazu, Mireia Irazola, Nestor Etxebarria, Ignacio García-Alonso, Alberto Saiz-López and José Javier Echevarria-Uraga\n\nOptimizing qPlus sensor assemblies for simultaneous scanning tunneling and noncontact atomic force microscopy operation based on finite element method analysis\n\nOmur E. Dagdeviren and Udo D. Schwarz\n\nNews\n\nBeilstein Archives – the preprint server for the Beilstein Journals is online!",
null,
"Register for this Beilstein Nanotechnology Symposium!",
null,
"The most widely used journal bibliometrics are presented on this page.",
null,
"© 2019 Kuncser et al.; licensee Beilstein-Institut.\nThis is an Open Access article under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0). Please note that the reuse, redistribution and reproduction in particular requires that the authors and source are credited.\nThe license is subject to the Beilstein Journal of Nanotechnology terms and conditions: (https://www.beilstein-journals.org/bjnano)"
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http://nd.ics.org.ru/nd190411/ | [
"0\n2013\nImpact Factor\n\n# Semi-Invariant Form of Equilibrium Stability Criteria for Systems with One Cosymmetry\n\n2019, Vol. 15, no. 4, pp. 525-531\n\nAuthor(s): Kurakin L. G., Kurdoglyan A. V.\n\nThe systems of differential equations with one cosymmetry are considered . The ordinary object for such systems is a one-dimensional continuous family of equilibria. The stability spectrum changes along this family, but it necessarily contains zero. We consider the nondegeneracy condition, thus the boundary equilibria separate the family on linearly stable and instable areas. The stability of the boundary equilibria depends on nonlinear terms of the system.\nThe stability problem for the systems with one cosymmetry is studied in . The general problem is to apply the stability criteria one needs to compute coefficients of the model system. It is especially difficult if the system has a large dimension, while a number of critical variables may be small. A method for calculating coefficients is proposed in .\nIn this work the expressions for the known stability criteria are proposed in a form convenient for calculation. The explicit formulas of the coefficients of the model system are given in semi-invariant form. They are expressed using the generalized eigenvectors of the linear matrix and its conjugate matrix.\nKeywords: stability, critical case, neutral manifold, cosymmetry, semi-invariant form\nCitation: Kurakin L. G., Kurdoglyan A. V., Semi-Invariant Form of Equilibrium Stability Criteria for Systems with One Cosymmetry, Rus. J. Nonlin. Dyn., 2019, Vol. 15, no. 4, pp. 525-531\nDOI:10.20537/nd190411",
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https://programmer.help/blogs/617505da09a97.html | [
"# Computer base and base conversion\n\n### Multiple bases of the computer:\n\nToday, I'd like to share the conversion between binary, decimal and hexadecimal. It's full of dry goods.",
null,
"First of all, I'll give you a brief introduction to the common hexadecimal systems in the computer field: binary, octal, decimal and hexadecimal.\n\nBinary:\n\nEvery two into one, there are only 0 and 1 in the number.",
null,
"Where S represents one digit, k is the position of the digit, and the base is 2.\n\noctal number system:\n\nEvery eight into one, the number contains 0, 1, 2, 3, 4, 5, 6 and 7.",
null,
"Where S represents a number, k is the position of the number, and the base number is 8.\n\ndecimal system:\n\nEvery decimal one, the number contains 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9, where S represents 1 number, k is the position of the number, and the base is 10.",
null,
"Where S represents a number, k is the position of the number, and the base number is 10.\n\nEvery 16 into one, because 10-15 with 16 as the base cannot be represented by A single number, it is replaced by English letters. 10 is represented by A, 11 by B, 12 by C, 13 by D and 14 by F. So hex contains: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A,B,C,D,F.",
null,
"After we have a simple understanding of various hexadecimals, the problem of how to convert them to each other has emerged. Next, we use c language code to realize the conversion between them one by one. There are 12 mutual transformations between them. (four common ones are divided here)\n\n### Code implementation (body):\n\n##### Binary to decimal:\n\nThe calculation method is to multiply the N-power of 2 from right to left, n starts from zero, and the ^ symbol represents the power.\n\nFor example: 111 (omitting the previous 0), its decimal expression is 22 + 21 + 2 ^ 0 = 7\n\nWe only need to find the symbol '1' in the array arr and calculate the power of 2 according to its position. The symbol '0' does not need to be found and does not participate in the calculation.\n\nCode implementation:\n\n```#include<stdio.h>\n#include<stdlib.h>\n#include<string.h>\nint main()\n{\nchar arr;//Create an array char arr to receive the number of hexadecimals to be converted. (note that it is an array of char type)\nwhile (gets(arr) != NULL)\n{\nint len, i, sum = 0, num, j;//Initialize variables. num is the value of each bit, and sum is the last sum of each bit.\n\nlen = strlen(arr);//len is the length of the input character array.\nfor (i = 0; i < len; i++)\n{\nnum = 1;\nif (arr[i] == '1')//If the bit is 1, then the power operation is performed. If it is 0, then it doesn't matter (0 doesn't participate in the calculation)\n{\nfor (j = 1; j <= len - i - 1; j++)//j is the number of powers of 2 on each bit.\n{\nnum = num * 2;\n}\nsum = sum + num;//sum is the last decimal value.\n}\n}\nprintf(\"%d\\n\", sum);\n}\nreturn 0;\n}\n\n```\n##### Decimal to binary:\n\nDecimal to binary is the inverse process of binary to decimal.\n\nTake 10 as an example.\n\n10 / 2 = 5 (remainder is 0)\n\n5 / 2 = 2 (the remainder is 1)\n\n2 / 2 = 1 (remainder is 0)\n\n1 / 2 = 0 (the remainder is 1) ends.\n\nSo the final 1010 is the binary expression of 10.\n\nCode implementation:\n\n```#include<stdio.h>\n#include<stdlib.h>\n#include<string.h>\nint main()\n{\nint n = 0;\nscanf(\"%d\", &n);//Get a decimal number\nint i = 0;\nint arr;//Binary is represented by an integer array\nwhile (n)//The calculation can continue as long as n is not 0\n{\ni++;\narr[i] = n % 2;//Assign a value to each bit of the array\nn = n / 2;//One bit is automatically eliminated after assignment\n}\nfor (int j = i; j > 0; j--)\n{\nprintf(\"%d\", arr[j]);\n}\nreturn 0;\n}\n\n```\n##### Hex to decimal\n\nMultiply by the n-th power of 16 from right to left, and N starts from zero.\n\nExample: 32\n\n3x161+2x160=50\n\nCode implementation:\n\n```#include<stdio.h>\n#include<stdlib.h>\n#include<string.h>\n#Include < math. H > / / don't forget to reference this library\nint main()\n{\nint b = { 0 };//Convert hexadecimal number to int type\nint i, j, sum=0;//Don't forget to initialize the sum here.\nint c = 0;//Final decimal number\ngets(a);\n//Convert it from char type to int type and store it in array b [].\nwhile (a[sum] != '\\0')\n{\nif ((a[sum] >= 'a') && (a[sum] <= 'f'))\n{\nb[sum] = a[sum] - 'a' + 10;\nsum++;\ncontinue;\n}\nif ((a[sum] >= 'A') && (a[sum] <= 'F'))\n{\nb[sum] = a[sum] - 'A' + 10;\nsum++;\ncontinue;\n}\nb[sum] = a[sum] - '0';\nsum++;\n}\nfor(i = 0; i < sum; i++)Decimal to hexadecimal and hexadecimal to decimal are reciprocal\n{\nb[sum - 1 - i] = b[sum - 1 - i] * pow(16, i);\n}\n//Direct accumulation\nfor (j = 0;j<sum;j++)\n{\nc = c + b[j];\n}\nprintf(\"%d\", c);\nreturn 0;\n}\n\n```\n\nFor example: 50\n\n50 / 16 = 3 (the remainder is 2)\n\n3 / 16 = 0 (the remainder is 3)\n\nSo its decimal system is 32\n\nCode implementation:\n\n```#include<stdio.h>\nint main()\n{\nint a = 0;\nint arr = { 0 };//Put the converted hexadecimal number into the array arr.\nint y = 0;\nscanf(\"%d\", &a);//Gets a decimal number\nwhile (a != 0)\n{\ny++;\narr[y] = a % 16;\na = a / 16;\nif (arr[y] > 9)\n{\narr[y] = 'A' + (arr[y] - 10);\n}\nelse\n{\narr[y] = '0' + arr[y];\n}\n}\nfor (int i = y; i > 0; i--)\n{\nprintf(\"%c\", arr[i]);\n}\nreturn 0;\n}\n//The hexadecimal number stored in the array is reversed. When printing, you can directly print it backwards. What you print out is the hexadecimal number.\n//It is the same as the decimal to binary conversion above.\n\n```\n\nIf you think it's helpful, you can praise it and collect it.",
null,
"Tags: C\n\nPosted on Sun, 24 Oct 2021 03:06:39 -0400 by suresh1"
]
| [
null,
"https://programmer.help/images/blog/43e74cd2537925a01c09ecd1ff807c2f.jpg",
null,
"https://programmer.help/images/blog/6f8df168643d8271ac6a4661750495c3.jpg",
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"https://programmer.help/images/blog/16f25798df0cbcd62e2dcb090535401c.jpg",
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"https://programmer.help/images/blog/dd31f38bb0058b29da020a4d0f8adcc2.jpg",
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"https://programmer.help/images/blog/86aa9b76685884626caf260988073f57.jpg",
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"https://programmer.help/images/blog/1fb96d1a0e21ae142cb62538bff0e7b8.jpg",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8122842,"math_prob":0.9982109,"size":5398,"snap":"2023-40-2023-50","text_gpt3_token_len":1621,"char_repetition_ratio":0.14386356,"word_repetition_ratio":0.07692308,"special_character_ratio":0.34864765,"punctuation_ratio":0.16570008,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998722,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-29T06:51:39Z\",\"WARC-Record-ID\":\"<urn:uuid:4ce85410-1441-45f7-a855-ce8786bbd6f1>\",\"Content-Length\":\"12839\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bdfa1405-e50b-4c76-a05f-c7798f22b6a4>\",\"WARC-Concurrent-To\":\"<urn:uuid:97fc5dbc-3bdf-4cfb-ade8-6cbabcad0c5c>\",\"WARC-IP-Address\":\"178.238.237.47\",\"WARC-Target-URI\":\"https://programmer.help/blogs/617505da09a97.html\",\"WARC-Payload-Digest\":\"sha1:VAUNTBZPQS326A6FNFZGZBVGAOLTCD46\",\"WARC-Block-Digest\":\"sha1:YD6VP34SKTK6C4QIJ25ZJ24F6IUFBEJF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510498.88_warc_CC-MAIN-20230929054611-20230929084611-00854.warc.gz\"}"} |
https://byjus.com/jee-questions/if-a-b-c-are-in-gp-and-x-y-are-the-arithmetic-means-between-a-b-and-b-c-respectively-then-a-by-x-plus-c-y-is-equal-to/ | [
"",
null,
"Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :\n\n# If a, b, c are in GP and x,y are the arithmetic means between a,b and b,c respectively, then a/x + c/y is equal to\n\n(1) 0\n\n(2) 1\n\n(3) 2\n\n(4) ½\n\nSolution:\n\nSince a, b, c are in GP\n\nb2 = ac\n\nSince x is the arithmetic mean between a and b\n\nx = (a+b)/2\n\nAlso y = (b+c)/2\n\na/x + c/y = 2a/(a+b) + 2c/(b+c)\n\n= [2a(b+c) + 2c(a+b)]/(a+b)(b+c)\n\n= (2ab + 2ac+ 2ac + 2bc)/(ab + b2 + ac + bc)\n\n= 2(ab + 2ac + bc)/(ab+ 2ac+ bc) ( since b2 = ac)\n\n= 2\n\nHence option (3) is the answer.",
null,
"(11)",
null,
"(3)"
]
| [
null,
"https://www.facebook.com/tr",
null,
"https://cdn1.byjus.com/wp-content/uploads/2021/08/upvote-lineart.svg",
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"https://cdn1.byjus.com/wp-content/uploads/2021/08/downvote-lineart.svg",
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https://math.stackexchange.com/questions/tagged/decision-theory | [
"# Questions tagged [decision-theory]\n\nFor questions regarding formal decision problems. In contrast, questions involving strategic aspects (where the solution depends on the behavior of others) are discussed in game theory.\n\n216 questions\nFilter by\nSorted by\nTagged with\n35 views\n\n### How to find a utility function\n\nThe choices are of the form $(x; y)$ where $x$ represents the amount of time you have left to live, say anywhere from $0$ to $50$ years, and $y$ represents the amount of time you have left to work, ...\n23 views\n\n### Creating a Majority Graph from multiple preference orders\n\nI can't find much on voting theory on this exceptional site. I am trying to find a way of constructing a majority graph based on a few preference. When I try to construct one, I end up breaking the ...\n35 views\n\n56 views\n\n32 views\n\n### Compute conditional probability\n\nI have to solve this problem for a decision analysis network: \"There is a genetic trait present in 20% of the population that makes that, for men over 65 years, the probability of suffering the ...\n136 views\n\n### Deriving of optimal decision boundary of two Gaussians\n\nGiven two Gaussians with the same variance $\\sigma$ and means $\\mu_1$ and $\\mu_2$, where each Gaussian represents a class $C_1$ and $C_2$ with the same prior probabilities, i.e. $p(C_1) = p_(C_2)$, we ...\n16 views\n\n### Finding optimum time for change of equipment based on shift pattern available resources\n\nI am trying to sort out a relatively simple problem, where I believe an algorithm or solving technique may already exist (Hungarian Algorithm?). I'd like to solve it using an methodology rather than ...\n49 views\n\n36 views\n\n### What approaches can be used to calculating a fair weighted score?\n\nI want to calculate a weighted score in a fair way, for a soccer substitution algorithm. There might not be a clear answer to this question, but I am looking for guidance on how to approach computing ...\n56 views\n\n### Likelihood ratio test between two hypothesis\n\nI have two hypothesis as below: Under $H_1$, $f_x(x) = 3/2 * x^2$ where $x \\epsilon (-1,1)$ Under $H_0$, $f_x(x) =$ Uniformly distributed between $x \\epsilon (-1,1)$ What is the maximum likelihood ...\n327 views\n\n### Value of the game from payoff matrix\n\nI am absolutely new to decision theory . I came across this following payoff matrix in the book.(Math. Stats : John E Freund). ..."
]
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http://www.lutsko.com/publication/00030 | [
"# On the multifractal properties of the energy dissipation derived from turbulence data\n\nE. Aurell, U. Frisch, J. Lutsko, and M. Vergassola, \"On the multifractal properties of the energy dissipation derived from turbulence data\", J. Fluid Mech, 238, 467 (1992)\n\n## Abstract\n\nVarious difficulties can be expected in trying to extract from experimental data the distribution of singularities, the f(α) function, of the energy dissipation. One reason is that the multifractal model of turbulence implies a dependence of the viscous cutoff on the singularity exponent. It is an open question if current hot-wire probes can resolve the scales implied by exponents a significantly less than 1. Two exactly soluble models are used to show how spurious scaling can occur, due to finite Reynolds number effects. In the Gaussian model the true velocity signal is replaced by independent Gaussian random variables. The dissipation, defined as the square of the difference of successive variables, has trivial scaling in so far as all the moments of spatial averages of the dissipation behave asymptotically as a uniform dissipation. Still, contamination by subdominant terms requires that scaling exponents for high-order moments be identified over an increasingly large range of scales. If the available range is limited by the Reynolds number, scaling exponents for high orders will be systematically underestimated and spurious intermittency will be inferred. Burgers’ model is used to highlight further problems. At finite Reynolds numbers, regions with no small-scale activity (away from shocks) have a residual dissipation which contributes a spurious point (α = 1,f(α) = 1). In addition, when the f(α) function is obtained by Legendre transform techniques, convex hull effects generate an entire spurious segment. Finally, Burgers’ model also indicates that the relation between exponents of structure functions and exponents of local dissipation moments, which goes back to Kolmogorov's (1962) work, leads to an inconsistency for structure functions of low positive order."
]
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null
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https://enjoyphysics.cn/Article694 | [
"",
null,
"## 景深（Depth of Field）",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"// 模糊程度( 即提取当前像素的周围像素的范围)\nfloat BlurDistance = 0.003f;\n// This will use the texture bound to the object( like from the sprite batch ).\nsampler ColorMapSampler : register(s0);\n\nfloat4 PixelShader(float2 Tex: TEXCOORD0) : COLOR\n{\nfloat4 Color;\n\n// 使用修改过的纹理坐标从ColorMapSampler采样颜色，并将颜色叠加。\nColor = tex2D( ColorMapSampler, float2(Tex.x+BlurDistance, Tex.y+BlurDistance));\nColor += tex2D( ColorMapSampler, float2(Tex.x-BlurDistance, Tex.y-BlurDistance));\nColor += tex2D( ColorMapSampler, float2(Tex.x+BlurDistance, Tex.y-BlurDistance));\nColor += tex2D( ColorMapSampler, float2(Tex.x-BlurDistance, Tex.y+BlurDistance));\n// 我们需要将颜色除以叠加次数（本例中为4）以获得平均颜色\nColor = Color / 4;\n\n// 返回经过模糊的颜色\nreturn Color;\n}\n\nfloat Distance;\nfloat Range;\nfloat Near;\nfloat Far;\n\nsampler SceneSampler : register(s0);\ntexture D1M;\nsampler D1MSampler = sampler_state\n{\nTexture = ;\nMinFilter = Linear;\nMagFilter = Linear;\nMipFilter = Linear;\n};\n\ntexture BlurScene;\nsampler BlurSceneSampler = sampler_state\n{\nTexture = ;\nMinFilter = Linear;\nMagFilter = Linear;\nMipFilter = Linear;\n};\n\nfloat4 PixelShader(float2 Tex: TEXCOORD0) : COLOR\n{\n// 获取正常场景的颜色\nfloat4 NormalScene = tex2D(SceneSampler, Tex);\n\n// 获取模糊过的场景的颜色\nfloat4 BlurScene = tex2D(BlurSceneSampler, Tex);\n\n// 获取深度纹理\nfloat fDepth = tex2D(D1MSampler, Tex).r;\n\n// 颠倒深度，这样背景为白色，近处为黑色\nfDepth = 1 - fDepth;\n\n// Calculate the distance from the selected distance and range on our DoF effect, set from the application\nfloat fSceneZ = ( -Near * Far ) / ( fDepth - Far);\nfloat blurFactor = saturate(abs(fSceneZ-Distance)/Range);\n\n// Based on how far the texel is from \"distance\" in Distance, stored in blurFactor, mix the scene\nreturn lerp(NormalScene,BlurScene,blurFactor);\n}\n\nfloat fSceneZ = ( -Near * Far ) / ( fDepth - Far);\nfloat blurFactor = saturate(abs(fSceneZ-Distance)/Range);\n\nreturn lerp(NormalScene,BlurScene,blurFactor);\n\nvoid SetShaderParameters( float fD, float fR, float nC, float fC )\n{\nfocusDistance = fD;\nfocusRange = fR;\nnearClip = nC;\nfarClip = fC;\nfarClip = farClip / ( farClip - nearClip );\neffectPostDoF.Parameters[\"Distance\"].SetValue(focusDistance);\neffectPostDoF.Parameters[\"Range\"].SetValue(focusRange);\neffectPostDoF.Parameters[\"Near\"].SetValue(nearClip);\neffectPostDoF.Parameters[\"Far\"].SetValue(farClip);\n}\n\n2006 - 2023，推荐分辨率1024*768以上，推荐浏览器Chrome、Edge等现代浏览器，截止2021年12月5日的访问次数：1872万9823 站长邮箱"
]
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null,
"https://enjoyphysics.cn/images/soft/Hlsl/Tutorial20_pic1.jpg",
null,
"https://enjoyphysics.cn/images/soft/Hlsl/Tutorial20_pic2.jpg",
null,
"https://enjoyphysics.cn/images/soft/Hlsl/Tutorial20_pic3.jpg",
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"https://enjoyphysics.cn/images/soft/Hlsl/Tutorial20_pic4.jpg",
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"https://enjoyphysics.cn/images/soft/Hlsl/Tutorial20_pic6.jpg",
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"https://enjoyphysics.cn/images/soft/Hlsl/Tutorial20_pic5.jpg",
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https://dotnetbasic.com/2019/11/python-pandas-interview-questions.html | [
"# python pandas interview questions",
null,
"We will Know “python pandas interview questions”.Pandas is a Python library that is used for data manipulation and data analysis. It provides highly optimized performance. It is written in Python. Also, it is free software released under the three-clause BSD license.\n\n1). Define the Pandas/Python pandas?\nAns:- Pandas is a Python library package that has robust features. It is very fast, flexible, and expressive data structures designed to make working with “relational” or “labeled” data both easy.\n\n2). Define Series in Pandas?\nAns:- It is a feature that is a one-dimensional labeled array capable of holding data of any type (integer, string, float, python objects, etc.).It is nothing but a column in an excel sheet.\n\n3). Define DataFrame in Pandas?\nAns: It is a pandas feature which is a 2-dimensional labeled data structure with columns of potentially different types. As an example, it is like a spreadsheet or SQL table, or a dict of Series objects.\n\n4). What are the major features of the pandas Library?\nAns: Follow is the key feature of the panda’s library.\n\n• Data Alignment\n• Memory Efficient\n• Reshaping\n• Merge and join\n• Time Series\n\n5). What is the name of pandas library tools used to create a scatter plot matrix?\nAns: Scatter_matrix\n\n6). What is pylab?\nAns: PyLab is a package that contains NumPy, SciPy, and Matplotlib into a single namespace.\n\n7). range () vs and xrange () functions in Python?\nAns: The range() function returns a list but the xrange () function returns an object that works as an iterator for generating numbers on-demand in python.\n\n8). what is the concept of monkey patching?\nAns: Monkey patching is the programming technique used to modify or extend other codes during runtime. Best practices to use in testing purposes, but not use in a production environment as debugging the code could become difficult.\n\n9). What is the map function in Python?\nAns: Map function executes the function given as the first argument on all the elements of the iterable given as the second argument. If the function given takes in more than 1 argument, then many iterables are given.\n\n10). Which library is used for Machine Learning in Python?\nAns: SciKit-Learn\n\nConclusion\n\n1.",
null,
"bun hairstyle says:"
]
| [
null,
"https://dotnetbasic.com/wp-content/uploads/2019/11/python.png",
null,
"https://secure.gravatar.com/avatar/da136170f0aaf0c52498b8a09cceb054",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.8636245,"math_prob":0.72407854,"size":2561,"snap":"2020-45-2020-50","text_gpt3_token_len":550,"char_repetition_ratio":0.11967149,"word_repetition_ratio":0.0,"special_character_ratio":0.21788365,"punctuation_ratio":0.13772455,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95717025,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,2,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-24T18:39:16Z\",\"WARC-Record-ID\":\"<urn:uuid:08bd9d95-5685-47ab-9c74-aaddbd1faa60>\",\"Content-Length\":\"66359\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:98b4c990-ac17-4c2e-8fb6-7e91468786c7>\",\"WARC-Concurrent-To\":\"<urn:uuid:54673b32-9f51-4fde-80de-fd8c53500c92>\",\"WARC-IP-Address\":\"160.153.61.33\",\"WARC-Target-URI\":\"https://dotnetbasic.com/2019/11/python-pandas-interview-questions.html\",\"WARC-Payload-Digest\":\"sha1:YYXSNPJXLTNZ4KDH43JA4L75C6LLX33Q\",\"WARC-Block-Digest\":\"sha1:S2RQRSG6E46COM4MTSUFCVRJZQOZPJ6O\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141176922.14_warc_CC-MAIN-20201124170142-20201124200142-00627.warc.gz\"}"} |
https://stackoverflow.com/questions/14732277/quadratic-algorithm-for-4-sum/14737071 | [
"# Quadratic algorithm for 4-SUM\n\nThe 4-SUM is as follows: Given an array of N distinct integers find 4 integers a,,b,c,d such that a+b+c+d = 0. I could come up with a cubic algorithm using quadratic algorithm for 3-SUM problem. Can we do better than cubic for 4-SUM?\n\n## 3 Answers\n\nYes you can. Go over all pairs of numbers and store their sum(and also store which numbers give that sum). After that for each sum check if its negation is found among the sums you have. Using a hash you can reach quadratic complexity, using std::map, you will reach `O(n^2*log(n))`.\n\nEDIT: to make sure no number is used more than once it will better to store indices instead of the actual numbers for each sum. Also as a given sum may be formed by more than one pair, you will have to use a hash multimap. Having in mind the numbers are different for a sum `X = a1 + a2` the sum `-X` may be formed at most once using `a1` and once using `a2` so for a given sum `X` you will have to iterate over at most 3 pairs giving `-X` as sum. This is still constant.\n\n• How do you make sure you don't choose the same element twice? – amit Feb 6 '13 at 15:16\n• @amit `\"also store which numbers give that sum\"`. If you found a number chosen twice, just ignore that sum. I imagine there's a simple improvement, but not in the big-O complexity. – Dukeling Feb 6 '13 at 15:17\n• But this sum might be generated in a different form as well, You should store a list or something per sum, and look for the first element in that list that does not contain one of the two elements. Can you prove it is O(1)? (looking in this list) – amit Feb 6 '13 at 15:19\n• @amit in fact I was considering using a hash multimap for that very reason. – Ivaylo Strandjev Feb 6 '13 at 15:22\n• Please extend your answer itself to contain how exactly is it making sure you do not use the same element twice. – amit Feb 6 '13 at 15:23\n\nThere is also a O(N^2) algorithm for this problem using O(N^2) extra memory.\n\n1- Generate all the pairwise sums in O(N^2) and store the pair (a_i, a_j) in a hash table and use the absolute value of their sum as the key of the hash table (a_i and a_j are two distinct number of the input array)\n\n2- iterate over the table and find a key which has both negative and positive sum with four distinctive elements and return is as the answer\n\nThere is an alternative if you prefer not using hash table. Since your numbers are integers, you can sort the list of all the sum in a linear time of the elements in the sum list using something like a Radix sort (there are O(N^2) elements in the sum list).\n\n• Codes speak loader. – Han XIAO Aug 18 '18 at 13:46\n• For step 2 you are iterating over a table of size O(N^2) (all possible pairs). You want to find a key with 2 elements and another with 2 elements. i.e. you need to check all possible key-pairs in the table of size O(N^2). This will require looping over all O(N^2) keys twice to find the pair. This makes the algorithm O(N^2 * N^2) = O(N^4) which is no better than brute force (except you're also using O(N^2) space) or am I missing something? – Frikster Feb 10 at 19:26\n• I am an idiot who had a momentary lapse in judgement. You obviously just check your first key and then check for the existence of the negation of that key in the hash in O(1). This means you iterate once over the dictionary with O(N^2) keys meaning the final algorithm is still O(N^2) just like you say. – Frikster Feb 10 at 19:30\n\nfor each index i and j, adding (S[i]+S[j]) in multimap(multimap bcz sum can be duplicate from different integers) and checking whether the multimap contains the sum -(S[i]+S[j]). If it does then add to the resultant set.\n\n``````void fourSUM(vector<int> array)\n{\nunordered_set<vector<int>> result; // set to avoid duplicate results\nmultimap<int,pair<int,int>> twosum_map;\n\nfor(unsigned int i=0; i< array.size();i++)\n{\nfor(unsigned int j=i+1; j< array.size();j++)\n{\n// insert sum in multimap\ntwosum_map.insert(make_pair(array[i]+array[j],make_pair(i,j)));\n\n// look for -ve sum in multimap\nint lookfor = -(array[i]+array[j]);\nstd::pair <std::multimap<int,pair<int,int>>::iterator, std::multimap<int,pair<int,int>>::iterator> ret;\nret = twosum_map.equal_range(lookfor);\n\nfor(std::multimap<int,pair<int,int>>::iterator it = ret.first; it != ret.second; it++)\n{\nvector<int> oneresult;\noneresult.push_back(array[i]);\noneresult.push_back(array[j]);\noneresult.push_back(array[it->second.first]);\noneresult.push_back(array[it->second.second]);\nresult.insert(oneresult); // this avoids duplicate results\n}\n}\n}\n\n// Display Result\nfor(set<vector<int>>::iterator it=result.begin();it != result.end();it++)\n{\nfor( vector<int>::const_iterator vit = it->begin(); vit != it->end(); vit++)\n{\ncout << *vit << \" \";\n}\ncout << endl;\n}\n}\n``````\n\ninsertion/find in multimap is log(n). insertion in set is log(n).\n\ncomplexity O(n^2 * log(n)).\n\n• HOW CAN `unordered_set<vector<int>>` avoid duplicates? – Han XIAO Aug 18 '18 at 13:19\n• You code won't even compile. `unordered_set<vector<int>>` is wrong. – Han XIAO Aug 18 '18 at 13:31\n• @HanXIAO who cares? beneficial as pseudo code nonetheless – naor.z Dec 15 '18 at 18:43"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.810071,"math_prob":0.9712351,"size":4897,"snap":"2019-35-2019-39","text_gpt3_token_len":1322,"char_repetition_ratio":0.10831801,"word_repetition_ratio":0.012004802,"special_character_ratio":0.2852767,"punctuation_ratio":0.13031423,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99768525,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-18T05:53:41Z\",\"WARC-Record-ID\":\"<urn:uuid:a2ec5677-6996-490c-ab43-ae6c2e112a40>\",\"Content-Length\":\"157520\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e2ef5cfb-d388-49f0-a3d2-efdbd93f5ef1>\",\"WARC-Concurrent-To\":\"<urn:uuid:586e943e-bbfe-4a00-93b7-6bf0b7f44123>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://stackoverflow.com/questions/14732277/quadratic-algorithm-for-4-sum/14737071\",\"WARC-Payload-Digest\":\"sha1:KSQBOGC5NLXDH4X6VX3QICV4IYU5R4MG\",\"WARC-Block-Digest\":\"sha1:IGXYRXEWYG2ULRB64AWCPKBBSWACXU7E\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027313617.6_warc_CC-MAIN-20190818042813-20190818064813-00034.warc.gz\"}"} |
https://hub.jmonkeyengine.org/t/get-the-height-of-a-boundingcapsule/11689 | [
"# Get the height of a BoundingCapsule\n\nI wondered if anyone can tell me how to compute the height of a BoundingCapsule.\n\nI tried like this, but it is not getting the correct results (its not a cylinder after all:)):\n\n`capsule.getVolume()/FastMath.pow(FastMath.PI*capsule.getRadius(),2);`\n\nSorry, found it out myself. Did it like this (maybe someone has an idea involving less computing?\n\n``` float radius=capsule.getRadius(); float volume=capsule.getVolume(); //subtract the volume of the round edges (a sphere with radius radius) volume-= ( ((4.0f/3.0f) * FastMath.PI ) * FastMath.pow(radius,3) ); //compute the height of the remaining cylinder float height=(volume/(FastMath.PI*FastMath.pow(radius,2))); //add the diameter of the sphere we too off height+=(radius*2); ```"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.6464828,"math_prob":0.8852383,"size":965,"snap":"2023-14-2023-23","text_gpt3_token_len":244,"char_repetition_ratio":0.15608741,"word_repetition_ratio":0.48,"special_character_ratio":0.2476684,"punctuation_ratio":0.1919192,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99749804,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-04-01T23:51:36Z\",\"WARC-Record-ID\":\"<urn:uuid:b00aa875-3319-4b34-ab29-47f8046067bb>\",\"Content-Length\":\"18465\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:da831994-35b9-4d23-8f67-3379c2071816>\",\"WARC-Concurrent-To\":\"<urn:uuid:265dc53f-22ec-4508-8813-f3b5764c6a7d>\",\"WARC-IP-Address\":\"172.67.140.141\",\"WARC-Target-URI\":\"https://hub.jmonkeyengine.org/t/get-the-height-of-a-boundingcapsule/11689\",\"WARC-Payload-Digest\":\"sha1:VYPY6WJEC34EAFNG4LRNVKRQITC3VE5O\",\"WARC-Block-Digest\":\"sha1:TPIZBXW6V7ZNUCVYRK7BLWN7EDIWA2HE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296950363.89_warc_CC-MAIN-20230401221921-20230402011921-00516.warc.gz\"}"} |
https://markov-bases.de/search.php?perpage=20&sortby=state%20space%20card.&revsort=revsort | [
"",
null,
"# The Markov Bases Database\n\n## 298 entries found\n\nDidn't find what you were looking for? Write us an email.\n\n1. N is the number of random variables for those models where the state space is a product space.\n2. dim is the dimension of the model. dim+1 is the rank of the sufficient statistics matrix.\n3. d is the cardinality of the state space.\n4. The properties in the table are:\n• binary: binary variables\n• hierarchical: is it hierarchical\n• graph: is it a graph-model\n• graphical: is it a graphical (clique) model\n• normal: is the semigroup normal\n• unique: is the Markov Basis unique\n• reducible: is the model reducible; that is, does the simplicial complex have a complete separator?\n5. deg is the Markov degree, the highest degree of an element in any minimial Markov basis.\n6. The size of a minimal Markov basis.\n7. The number of symmetry classes (usually the maximal symmetry)."
]
| [
null,
"https://markov-bases.de/img/MBDB5.png",
null
]
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https://mathematica.stackexchange.com/questions/152874/regarding-input-to-nettrain-function/152881 | [
"# Regarding input to NetTrain function\n\nEssentially I have used the SemanticImport[] function on a csv file with headers for each column. Now I understand that each of these headers is referred to 'key' in mathematica. *\n\nI have no problem doing Predict[]. But I understand that NetTrain requires a different input format I was looking under Association, but have not found clarity in doing this.\n\nI ran the following lines:\n\nnet = NetChain[{5, 1}, \"Input\" -> 8, \"Output\" -> \"Scalar\"];\ntrained = NetTrain[net, inputdata]\n\n\nBut an error is shown:\n\nNetTrain::invdataset: Datasets provided to NetTrain must consist of a list of associations with fixed keys.\n\n\nUPDATE: solved by xslittlegrass. Thank you\n\nThis seems to work. Set up some data:\n\nassocdata =\nTable[Association[\"Latitude\" -> RandomInteger[{-90, 90}],\n\"Longitude\" -> RandomInteger[{-180, 180}],\n\"Temperature\" -> RandomReal[{-40, 49}]], 5]\n\n\n{<|\"Latitude\" -> 64, \"Longitude\" -> 25, \"Temperature\" -> -24.8889|>, <|\"Latitude\" -> -49, \"Longitude\" -> -101, \"Temperature\" -> -28.0145|>, <|\"Latitude\" -> 9, \"Longitude\" -> -112, \"Temperature\" -> 22.6383|>, <|\"Latitude\" -> -65, \"Longitude\" -> 150,\n\"Temperature\" -> 13.6052|>, <|\"Latitude\" -> 25, \"Longitude\" -> 110,\n\"Temperature\" -> 29.0704|>}\n\nThen\n\nlistdata = {#[], #[]} -> #[] & /@ assocdata\n\n\ngives\n\n{{64, 25} -> -24.8889, {-49, -101} -> -28.0145, {9, -112} -> 22.6383, {-65, 150} -> 13.6052, {25, 110} -> 29.0704}\n\nIs that what you meant?\n\n• yes, your answer gives me the correct format I am looking for, but somehow NetTrain doesn't like it. Could you have a look at my updated question and would appreciate your advice. Thank you – Corse Aug 3 '17 at 3:26\n• @Corse Unfortunately I can't play around with it since I'm on 10.4 and NetTrain is 11+. The only guess I can make is that it wants an input of the form Association[\"Lattiude\" -> assocdata[[;; , 1]], \"Longitude\" -> assocdata[[;; , 2]], \"Temperature\" -> assocdata[[;; , 3]]]' (where assocdata has the same form as in my answer). Not convinced it'll work, though, since it's not a \"list of associations with fixed keys\". Not even sure what a fixed key is. Is it perhaps interpreting your latitude, etc as variable names? Sorry, but I'm just guessing now. – aardvark2012 Aug 3 '17 at 4:52\n• I would trawl through the documentation on all the different *Layers (Linear, Elementwise, etc), play around with them and see if you can figure out what they're doing. – aardvark2012 Aug 3 '17 at 4:53\n• @Corse No problem. Thanks for the accept. – aardvark2012 Aug 3 '17 at 7:04\n\nI think aardvark2012's answer is correct, this is just a comment on your follow-up questions. Your follow-up questions can be addressed by providing NetTrain with the compatible training data formats.\n\nYou can either use a Dataset or a list as training data. If a Dataset is used, then the format should be like Dataset[{<|\"Input\" -> {64, 25}, \"Output\" -> -24.8889|>, <|\"Input\" -> {-49, -101}, \"Output\" -> -28.0145|>}]. For example\n\ninputdata =\nDataset@Table[\nAssociation[\"Input\" -> RandomReal[{0, 1}, {2}],\n\"Output\" -> RandomReal[{0, 1}]], {10}]",
null,
"then you can train the network like\n\nnet = NetChain[{5, 1}, \"Input\" -> 2, \"Output\" -> \"Scalar\"];\ntrained = NetTrain[net, inputdata]\n\n\nIf using a list, then the training data should be like {{64, 25} -> -24.8889, {-49, -101} -> -28.0145, {9, -112} -> 22.6383, {-65, 150} -> 13.6052, {25, 110} -> 29.0704}. For example\n\ninputdata =\nTable[RandomReal[{0, 1}, {2}] -> RandomReal[{0, 1}], {10}];\n\nnet = NetChain[{5, 1}, \"Input\" -> 2, \"Output\" -> \"Scalar\"];\ntrained = NetTrain[net, inputdata]\n\n\nYou can also specify the input and output ports with NetGraph, which then allow you to train on a Dataset with named keys directly. For example\n\ninputdata = Dataset[Association[Thread[header -> #]] & /@ data]",
null,
"net = NetGraph[{ReshapeLayer[{1}], ReshapeLayer[{1}], CatenateLayer[],\nLinearLayer, LinearLayer}, {NetPort[\"lat\"] -> 1 -> 3,\nNetPort[\"lon\"] -> 2 -> 3 -> 4 -> 5 -> NetPort[\"temperature\"]}]",
null,
"NetTrain[net, inputdata, Method -> {\"ADAM\", \"LearningRate\" -> 0.01}]\n\n\nHope it helps.\n\nHere is an example of how to read in and train on a csv dataset\n\npath =\nTable[{RandomReal[{-90, 90}], RandomReal[{-180, 180}],\nRandomReal[{-50, 50}]}, {100}]]\n\ninputdata = #[[1 ;; 2]] -> #[] & /@ Import[path, \"Data\"];\n\nnet = NetChain[{5, 1}, \"Input\" -> 2, \"Output\" -> \"Scalar\"];\ntrained =\nNetTrain[net, inputdata, Method -> {\"ADAM\", \"LearningRate\" -> 0.01}]\n\n\nanother example using dataset\n\npath =\nStringRiffle[\nPrepend[Table[\nStringRiffle[\nToString[NumberForm[#, {3, 4}]] & /@ {RandomReal[{-90, 90}],\nRandomReal[{-180, 180}], RandomReal[{-50, 50}]},\n\"\\t\"], {100}],\nStringRiffle[{\"lat\", \"lon\", \"temperature\"}, \"\\t\"]], \"\\n\"]]\n\ntmp = Import[path, \"Table\"];\n\ninputdata =\nAssociation[\"Input\" -> #[[1 ;; 2]], \"Output\" -> #[]] & /@\nRest@tmp;\n\nnet = NetChain[{5, 1}, \"Input\" -> 2, \"Output\" -> \"Scalar\"];\ntrained =\nNetTrain[net, inputdata, Method -> {\"ADAM\", \"LearningRate\" -> 0.01}]\n`\n• thank you for the clarification. I refer to this: Association[\"Input\" -> RandomReal[{0, 1}, {2}], \"Output\" -> RandomReal[{0, 1}]], {10}]. How could I replace the RandomReal part using the Keys in my dataset? say I have header as \"lat\", \"lon\" and \"temperature\" from the CSV file. – Corse Aug 3 '17 at 14:55\n• @Corse See the updates. – xslittlegrass Aug 3 '17 at 16:16\n• this works well and that was very enlightening and helpful for my understanding on the input formats of NetTrain/NetGraph. Thank you very much! – Corse Aug 4 '17 at 2:10\n• a minor question to seek your advice: for the purposes of regression/prediction, is it normally recommended to put a BatchNormalizationLayer[] as the first hidden layer? Does it have the effect of scaling the values of each variable and is it at all necessary in Mathematica? – Corse Aug 4 '17 at 2:33\n• one more query if you don't mind, if i have an additional column (input variable) that is of a class vector, how could I modify the part: inputdata = #[[1 ;; 2]] -> #[] & /@ Import[path, \"Data\"]; to utilize the NetEncoder function for 'UnitVector'? – Corse Aug 4 '17 at 3:46"
]
| [
null,
"https://i.stack.imgur.com/s0eX6.png",
null,
"https://i.stack.imgur.com/l6dGM.png",
null,
"https://i.stack.imgur.com/l1ewN.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.617474,"math_prob":0.8503692,"size":2506,"snap":"2020-24-2020-29","text_gpt3_token_len":825,"char_repetition_ratio":0.13629097,"word_repetition_ratio":0.17045455,"special_character_ratio":0.42657623,"punctuation_ratio":0.2462203,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.983394,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-05-31T20:25:53Z\",\"WARC-Record-ID\":\"<urn:uuid:ce6eb126-8e3b-4bdf-bf88-c857dda16997>\",\"Content-Length\":\"161283\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ac6cfcdd-01f6-4ff0-8be1-43a89e78f0a3>\",\"WARC-Concurrent-To\":\"<urn:uuid:a316bd3f-9567-4496-9c13-145fa0f0b70a>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://mathematica.stackexchange.com/questions/152874/regarding-input-to-nettrain-function/152881\",\"WARC-Payload-Digest\":\"sha1:MXQSVL675KWXZT566HEYRAZEEIZU7NVI\",\"WARC-Block-Digest\":\"sha1:LEUEHSNOCLKNNRJUGGRMWMHETPCCAC5O\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347413624.48_warc_CC-MAIN-20200531182830-20200531212830-00274.warc.gz\"}"} |
https://www.embedded.com/tracking-demodulation/ | [
"",
null,
"Tracking Demodulation - Embedded.com\n\n# Tracking Demodulation\n\nHigh resolution encoders are expensive, but alternatives exist. You can develop a high-resolution system using a sinusoidal encoder or resolver.\n\nLast month, we covered some of the basics of using cosinusoids to determine position in a motion control application. We translated the magnitudes of two voltages-one representing the sine and the other representing the cosine of position-into an angle indicating position. We obtained these voltages using a tracking device called a resolver. As the motion control industry (and industry in general) reaches for tighter specifications, higher performance, and lower cost, this procedure, called interpolation, is an increasingly popular alternative.\n\nThis month, I'll show you how to develop high-resolution position feedback with two cosinusoids that are in quadrature, meaning 90 degrees out of phase. We'll examine two applications in particular (there are more): sinusoidal encoders and resolvers. We will concentrate on the algorithmic techniques common to both, and save the differences for next month. First, I would like to address the interpolation scheme we will use, as well as some possible enhancements.\n\nVoltages to angles",
null,
"Figure 1 shows two cosinusoids in quadrature. They represent the feedback from a sinusoidal encoder or resolver. We label each of the two cosinusoids according to their phase:",
null,
"is zero voltage at zero degrees.",
null,
"is equal to 1V at zero degrees, which makes it 90 degrees out of phase with the first cosinusoid. Recall that sine and cosine represent ratios based on the length of sides of a triangle as captured within the unit circle. The ratio of these two ratios is equal to another function, called the tangent:",
null,
"By taking the arctangent of the two inputs, we can find the angular position of the shaft of the encoder:",
null,
"This information will help us track our position and, thereby, velocities in the servo system.\n\nA high resolution arctangent evaluation is a long and difficult calculation for an integer-based processor. Sine and cosine, however, are relatively easy. This, combined with the fact that most of these applications will require a filter of some sort to remove noise (as much as possible) anyway, makes a tracking demodulator an excellent alternative. It's fast, it's lean-it's even possible to put it in an FPGA and offload the processing altogether. Resolution\n\nThe resolution of our interpolator depends on a number of factors. Most significantly, it depends on the error in the analog-to-digital converter (ADC).\n\nIf we choose a reasonably good 12-bit ADC, we can consider the error equal to its least significant bit for quantization purposes, which is 2-12. This neglects other errors in the system, which I cannot estimate here. However, assuming good layout and adequate electronics, we should be able to account for the absent error by doubling that of the ADC.If we spread the error evenly throughout a single rotation, we have:",
null,
"or about 0.014 degrees. Because this is the error, the resolution in a revolution must be approximately 213.65, or 12,855.\n\nA basic system\n\nIn its simplest and most direct form, an interpolating system accepts the cosinusoidal inputs at the ADC through suitable buffering and filtering, performs an arctangent, and determines the angle within the necessary resolution.\n\nIn many applications, the noise from the integrated gate bipolar transistor's (IGBT) switching, which is transmitted either through proximity or by direct connection, can cause problems in the conversions. To counter this, we could high-pass filter the input. The filter's corner frequency (-3dB point) has to be far enough from our operating frequency to keep the phase flat, but low enough to remove most of the unwanted noise. If your pulse width modulation (PWM) frequency is, say, 16kHz, you could have a second-order filter with corner frequency at 1kHz, or even 2kHz, that would put the stop band between 48dB and 36dB, without sacrificing performance.The choice of the corner frequency-and the form of the filter-is application dependent. For our needs, we'll use a simple Butterworth filter, which combines a reasonably flat pass band and minimal phase distortion. We'll also pick a Q of 0.707 for minimal error at the turnover frequency (In this case, Q is the ratio of the turnover frequency to the 3dB point. The higher the Q, the more distortion at the turnover frequency).\n\nSometimes, however, this is not the optimal implementation. If board space is critical and the DSP is already working hard, or if memory is at a premium, you might want to go another way.\n\nTracking devices\n\nDepending on the application, tracking devices can make up for deficiencies in other areas by providing a means to perform complex functions in only a few steps, or with slower elements. A good example of such a device is the tracking ADC. Built with a comparator and a counter, it produces analog-to-digital conversions at a higher speed with slower, less expensive parts.\n\nThe heart of this converter is a counter driven by a constant high-speed clock; the output of the counter is connected to a digital-to-analog converter (DAC). When the counter counts up, the magnitude of the voltage at the output of the DAC rises. When the counter goes down, so does the magnitude. To make a tracking ADC, a comparator is introduced ahead of the counter. The comparator has two inputs. If the input from the external circuitry is greater than the output of the DAC, the counter is told to count up; when it is less, the counter is instructed to count down.\n\nImagine performing analog-to-digital conversion on a sine wave. Depending on the sample rate, the difference between each sample can be very small. Here, it provides a good result quickly, but you need to use it carefully. If you were trying to convert a square wave with a tracking ADC, each edge would require the counter to count all the way up or all the way down, which could result in a very slow system.\n\nA tracking demodulator\n\nThe tracking demodulator is similar to the tracking ADC because it operates off the differences between a predicted and an actual value to compute the next value. This algorithm also incorporates a second-order filter in the processing. We will specify the sample rate for this algorithm at 8kHz. Another controller might use a different rate, depending on the update period. The tracking demodulator, like the tracking ADC, compares the input from an analog cosinusoidal device-either an encoder or resolver-with an estimated value derived from previous values. In the block diagram in Figure 2, we subtract our predicted position (the one we generate) from the actual position (the voltages in quadrature returned from the encoder or resolver) using the formula for the difference of two angles:",
null,
"",
null,
"After scaling the differences by a value representing the corner frequency of the filter, we put them through an accumulator. (An accumulator sums: the predicted value is equal to the sum of all past samples plus the latest input.) When accelerating from a stop, the differences between predicted and actual values add up quickly, but as soon as velocity stabilizes, the differences fall to zero. At this point, the output of our first accumulator will be a constant that represents the predicted velocity, in terms of a position displacement per sample cycle.The output of this accumulator is then summed in the second accumulator, which produces position. We also include a scaled addition of velocity (output of the first accumulator) to this final position sum as a correction factor-this is the damping in the second-order transfer function.\n\nAn example\n\nIn the following example, I disregard the second-order filter function because it isn't necessary for an understanding of the algorithm.Suppose that we step onto a seat on a Ferris wheel at the point closest to the ground. Our speed and position are both zero. Both accumulators are at zero. We will be checking our position every second.\n\nThe Ferris wheel operator moves our chair up one increment to seat the next paying guest. This takes him one second. At this point, we compare our actual position with our predicted position. Our predicted position is zero; we had been sitting motionless near the ground and all our accumulators had decayed to zero. So on our first comparison, we use the previous formula to calculate a difference. That first difference is placed in the first accumulator and subsequently added to the second accumulator. The output of the second accumulator is our new position.\n\nOne second later, the Ferris wheel moves again, and a new passenger gets on. We take the sine and cosine of our estimated position and compare it to our sine and cosine inputs. The difference is added to the first accumulator, which is then added to the second and stored as our new position.\n\nAnother passenger gets on. Again, we difference the expected position with the actual, but because in each increment we have moved the same distance in the same amount of time, we find that the expected position is greater. A negative difference is added to the first accumulator to reduce the integral. This smaller value is added to the second accumulator and a new position is produced.\n\nWe continue this process with the Ferris wheel moving the same amount each second. Soon the expected position is the same as the actual. Now, we add the first accumulator to the second to estimate position. When we compare, we see no difference. At a constant velocity, the value in the first accumulator stabilizes and does not change. (This is just as it is in a car moving at 60mph. To know how far you have gone you add 88 feet for every second of travel.) The first accumulator integrates acceleration and now represents incremental velocity. Integrating these increments in the second accumulator gives us a reliable position. Because no differences show up between expected and actual position, we are adding zero to the first accumulator and subsequently adding a constant to the second for incremental position change.\n\nNow, the ride is over and the Ferris wheel is slowing. Once again a difference appears; our predicted position is greater than the real position. We are decelerating. This is reflected in the first accumulator, which is reduced, and then in the second accumulator, which shows smaller and smaller position changes.\n\nWhen we again come to a stop, the differences will eventually clear the first accumulator and, depending on actual position, may clear the second.\n\nThis is how you track position with differences between predicted and actual position.\n\nThe filter\n\nThe filter/demodulator consists of a second-order filter constructed with accumulators. The input samples are scaled to control where the accumulator overflows-if it ever does-to fit the function to applications requiring integration.\n\nWe start by determining what value can be added to the accumulator during each sample period to result in an overflow in one second, the amount of time in which the rotor moves. In other words, we are setting up the accumulator for a time constant of one second.\n\nTo do this, we must set some constraints on the system. We choose a sample rate for the accumulator of 8kHz, a common value for many applications. We also choose a 16-bit word size, another common value.If 2p = 216 we can know instantly that the value that will overflow our accumulator is:",
null,
"This means that 8.192 = 1Hz. This relationship is also important when using this technique to determine velocity.\n\nThe scaling factors\n\nThe real magic of this technique is that it can be combined with a second-order filter to help remove some of the transient noise from the switching of the IGBTs. What I have been calling scaling factors are really the coefficients of this filter, and they are very easy to figure out. Because you may opt to implement this on any processor, I'm omitting processor-specific details.\n\nTwo scaling functions are involved in this process, both derived directly from the transfer function for a second-order low pass filter. One depends on the corner frequency. The other depends on the Q, or damping, factor. The transfer function is:",
null,
"Both will be developed as fractions so that we can multiply rather than divide them, which would be cumbersome and noisy. The first scaling factor is:",
null,
"This will be the inverse of the square of frequency in Hertz. We'll call it fr. The next is the damping factor:",
null,
"We will call this damp. You see them noted in the block diagram. They occur as simple multiplications.Next month, we'll take the tracking demodulator concept and see how it applies to systems using analog cosinusoids in quadrature to indicate position.\n\nDon Morgan is senior engineer at Ultra Stereo Labs and a consultant with 25 years of experience in signal processing, embedded systems, hardware, and software. Morgan recently completed a book about numerical methods, featuring multi-rate signal processing and wavelets, called Numerical Methods for DSP Systems in C. He is also the author of Practical DSP Modeling, Techniques, and Programming in C, published by John Wiley & Sons, and Numerical Methods for Embedded Systems from M&T. Don's e-mail address is ."
]
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https://federalprism.com/what-is-the-multiple-game/ | [
"# What is the multiple game?\n\n## What is the multiple game?\n\nThe game prepares a student for multiplication by practicing how to recognize which numbers are multiples of five. The numbers 5, 10, 15, and so on generate scores when they come up in the course of the game.\n\n## How do you play the multiple game?\n\nThe number must be a factor or multiple of the first number. Players continue to take it in turns to cross out numbers, at each stage choosing a number that is a factor or multiple of the number just crossed out by the other player. The first person who is unable to cross out a number loses.\n\nWhat are multiples math is fun?\n\nA multiple is the result of multiplying a number by an integer (not a fraction).\n\n### What Is factors and multiples in maths?\n\nA factor is one of two or more numbers that divides a given number without a remainder. A multiple of a number is a number that can be divided evenly by another number without a remainder. Factors and multiples are inverse concepts. A number sentence can help us to understand factors.\n\n### What are factors maths?\n\nfactor, in mathematics, a number or algebraic expression that divides another number or expression evenly—i.e., with no remainder. For example, 3 and 6 are factors of 12 because 12 ÷ 3 = 4 exactly and 12 ÷ 6 = 2 exactly. The other factors of 12 are 1, 2, 4, and 12.\n\nAre factors and multiples the same thing?\n\n## Do all multiples of 9 add up to 9?\n\nThe first 10 multiples of 9 have digits that always sum to make 9, which can make them easier to memorise. For example, when looking at the numbers of 9, 18, 27 and 36, we have 9 = 9, 1 + 8 = 9, 2 + 7 = 9 and 3 + 6 = 9. This pattern continues all the way up to 90, which has 9 + 0 = 9.\n\n## What are multiples 7?\n\nMultiples of 7 are: 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, …\n\nWhat is the multiples game?\n\nThe game challenges students to solve a set of problems on finding multiples. These engaging problems encourage them to apply their prior knowledge of the topic and find the answer. Students will select the correct option to mark their responses.\n\n### How to learn multiplication tables?\n\nHere you can learn the multiplication tables in an interactive way. Adding a game element with the free multiplication games makes it more fun to practice. This is a good variation on the speed test, the tables diploma, and exercises at school with a reinforcing learning effect.\n\n### How many math games are there in 2029?\n\nSign up and get access to our entire library of 2029 Math games & 51 Math courses The game challenges students to solve a set of problems on finding multiples. These engaging problems encourage them to apply their prior knowledge of the topic and find the answer.\n\nHow do you find all the multiples in space?\n\nRace down the intergalactic highway and collect all the space multiples. The more your collect, the faster you will go! Click on a lane to steer your spaceship. Click again to jump. Apply properties of operations as strategies to multiply. Find all factor pairs for a whole number in the range 1-100."
]
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http://marcspages.co.uk/rtc/0221.htm | [
"",
null,
"",
null,
"Wiring Analogues as Digitals Analogue Input as Digital InputWith MODBUS now being a well established standard, most systems follow suit and relay most, if not all internal information using a 16-bit format. With MODBUS the ON state is when the value is equal to or above 32768 and OFF when 0 to 32767. If the analogue input on the device in question uses the 16-bit word to convey the analogue level such that the mid voltage/current range (e.g. 12 mA on a 4-20mA signal) equals 32768 or 8000H, then using the analogue as a digital input merely becomes a matter of ensuring there is sufficient current through the analogue input when ON is desired, and conversely, significantly low current to equal the OFF state. In the following examples it is assumed that the analogue input reads 4-20mA with 12mA being represented by a value of 32768 (8000H). To test if the input will satisfy this requirement inject 4mA into the input and read the output value (in our example it was very near 16384 or 4000H) and then inject 20mA and read the value (in our example this was very near 49152 or C000H). This gives the span as used by the input, in our case 4-20mA was spanned over 8000H counts, centred around 8000H - in other words, perfect.",
null,
"In order to ensure the input exceeds 32768 (8000H) it is advised to use 16mA rather than rely on 12mA as any small errors will make the input unreliable. Using Ohm's Law (V ÷ I = R) we can determine the value of the resistor. In our example we used the supply voltage of 24 volts therefore 24 ÷ 0.016 = 1500 (or otherwise known as 1.5k). It must be borne in mind that the maximum of 20mA must also not be exceeded and this minimum resistance is again calculated with Ohm's Law and in our example 24 ÷ 0.02 = 1200, (or 1.2k).",
null,
"The same method can be adopted when using open collector outputs as indicated in the accompanying diagram. The only criteria is to ensure the output is able to sink the whole 20mA with some capacity spare so as to ensure long working life of the output. It will be noticed in both the above examples that the current limiting resistor is placed between the +24V supply and the high (+) input to the analogue channel, with the low (-) input being used as the equivalent digital input. This is done for safety in that should the line from the input be brought to earth no damage is done to the analogue channel. If the resistor were in between the low (-) input and the contact and a short to ground occurred ahead of the resistor the full potential of the supply would be found across the analogue channel which would damage it.",
null,
"Certain outputs, especially older solid state types, may not be able to sink this level of current, however, all is not lost. Reducing the current the output has to carry is done by using a \"bleed\" resistor which then carries some of the current. We will assume, for the sake of an example, our output is not allowed to sink more than 10mA. We want to draw 16mA to safely assume the value transferred is high enough so we need the bleed resistor to draw the remaining 6mA the output is not allowed to carry, therefore, 24 ÷ 0.006 = 4000 (or 4k). Such a value is not available but 3.9k is and this is then used. To calculate the sink resistor to draw 10mA, 24 ÷ 0.01 = 2400 (or 2.4k). In the above example the 6mA drawn through the analogue input by the bleed resistor when the output is off is low enough to ensure the value is well below the transition value of 32768 (8000H). Analogue Output as Digital OutputAnalogue outputs are not that easy to interface to digital inputs but with a little trickery they can, in most cases, be successfully interconnected. Some outputs work from 0-20mA making interfacing extremely easy while others only operate from 4-20mA, the lowest current being more than most digital inputs require. A further problem is some analogue outputs are 'alarmed'. If the full current is not being drawn it could force them to either shut down or bombard the controlling system with a continuous warning. As most digital inputs are optically isolated interfacing is simply achieved using a resistor across the input on a 0-20mA circuit, the value being: (VAo - 5) ÷ (.02 - IDi) where:VAo = Voltage feeding the analogue loop andIDi the current taken by the digital input. The reason the voltage is reduced by 5 volts is the current regulating circuit within the analogue output requires a little voltage for everything to work. The full 20mA is reduced by the portion of current fed through the digital input. In our example we will assume an analogue loop feed voltage of 24 volts and a digital input current of 3mA and the result would therefore be 19 ÷ 0.17 = 1118 ohms. Always use the next lowest common value resistor, in this case 1000 ohms (this ensures the analogue output will not alarm). If the analogue output is a 4-20mA type, or the digital inputs are fed from separate voltage rails, an additional diode and zener diode are required. We will require the voltages of both the digital input rail and the analogue loop feed in order to complete the calculations. In our next example our digital input system uses 12 volts on the opto-isolator, it draws the same 3mA, and the analogue is fed from 24 volts. Calculating the resistor is done using using the same formula as above. Where things change is we now need to calculate the voltage swing across the resistor relative to the digital input. This is achieved by first calculating the 'off' and 'on' voltage levels: Voff = 24 - (.004 X R) and in our case = 20VVon = 24 - ((.02 - IDi) X R) and is = 7V The swing is VDi - Voff to VDi - Von and. in our example, is -8v to +5. The first value also determines the zener diode value, but in this example the value being negative indicates we don't require a zener diode and may be removed from the circuit. As they say, \"simple when you know how\". | | Ask a Question | 03.11.00"
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http://allreactions.com/index.php/group-4a/silicon | [
"### SILICON - Si - [silicium]\n\nProperties of Silicon:\n\nNonmetal. Large crystals - dark gray, with metal gloss, very firm, very fragile, opaque, the semiconductor at the room temperature. Amorphous in the form of very small crystals - white (without impurity) or brown (with impurity). Melts with reduction of volume. It is steady on air (formation of a protective oxidic film). In a crystal form - small reactionary ability. Doesn't react with water, acids (including also fluoric acid), hydrogen.\n\nIn an amorphous form - more active. Reacts with the concentrated fluoric acid, alkalis (it is partially transferred to solution even in the alkalescent environment), absorbs significant amounts of various gases (including hydrogen). Is oxidized by oxygen and halogens. Reacts with hydrogen halides, ammonia, hydrogen sulfide, metals sulfides by heating. It is extremely active in the melted state, reacts with alkalis, alkaline earth metals and other metals. Alloyed (but doesn't react) with beryllium, aluminum, gallium, indium, tin, antimony, zinc, silver, gold. An alloy with iron - ferrosilicium (12-90% of Si) is industrially important. The second for prevalence (after oxygen) an element in Earth lithosphere.\n\n Molar mass g/mol 28.086 Density g/cm3 2.33 Melting point °C 1415 Boiling point °C 3250\n\nMethods for the preparation of Silicon:\n\nSiH4 = Si + 2H2 (400-1000°C).\n\nSiO2 + 2Mg = 2MgO + Si (800-900°C, in atm. of argon).\n\nSiO2 →(air, Mg, -MgO, -Mg3N2) → Si, Mg2Si (700-900°C).\n\nSiO2 + 5C(coke) + CaO = Si + CaC2 + 3CO (800-1000°C).\n\nSiCl4 + 2H2 = Si + 4HCl (800°C).\n\nSiCl4 + Li[AlH4] = Si + LiCl + AlCl3 + 2H2 (over 450°C).\n\nSiCl4 + 4M = Si(amorphous) + 4MCl (M = Na, K; 600-700°C).\n\n3Na2[SiF6] + 4Al = 3Si + 2Na3[AlF6] + 2AlF3 (700°C).\n\nNa2[SiF6] → Electrolysis → Si↓(on cathode) + 2F2↑(on anode) + 2NaF (in the liquid NaF).\n\nСhemical reactions with Silicon:\n\nSi(amorphous) + 2H2O(steam) = SiO2 + 2H2 (400-500°C).\n\nSi(amorphous) + 4NaOH(conc.) = Na4SiO4 + 2H2↑.\n\nSi(amorphous) + 6HF(conc.) = H2[SiF6] + 2H2↑.\n\nSi + 4HF(gas) = SiF4 + 2H2 (40-100°C).\n\n3Si + 18HF(conc.) + 4HNO3(conc.) = 3H2[SiF6] + 4NO↑ + 8H2O.\n\n3Si + 18HF(conc.) + 2KClO3 = 3H2[SiF6] + 2KCl + 6H2O.\n\nSi + 6HF(conc.) + KNO3 = H2[SiF6] + 2KNO2 + 2H2O.\n\nSi + O2 = SiO2 (1200-1300°C).\n\nSi + 2F2 = SiF4 (normal temp., burning in the fluorine).\n\nSi + 2Cl2 = SiCl4 (340-420°C, under argon).\n\nSi + 2Br2 = SiBr4 (620-700°C, under argon).\n\nSi + 2I2 = SiI4 (750-810°C, under argon).\n\nSi + 4HI = SiI4 + 2H2 (400-500°C).\n\nSi + S = SiS (650-700°C, pressure).\n\nSi + 2S = SiS2 (250-600°C).\n\nSi + 2E = SiE2(800°C, E = Se, Te; under argon).\n\n3Si + 2N2 = Si3N4 (1200-1500°C).\n\nSi + C(graphite) = SiC (1200-1300°C).\n\nSi + M = MSi (by alloying, M = Na, K, Rb, Cs).\n\nSi + 2M = M2Si (by alloying, M = Mg, Ca).\n\nSi + M = MSi, MSi2 (by alloying, M = Ca, Sr, Ba).\n\n2Si + M = MSi2 (by alloying, M = La, Th, Ti, Cr, Mo, Mn, Fe).\n\n3Si + 4NH3 = Si3N4 + 6H2 (1300-1500°C).\n\nSi + 2H2S = SiS2 + 2H2 (1200-1300°C).\n\nDid you know?\n\nGroup 1 and 2 metals are so reactive that they are found in nature as compounds."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.7388805,"math_prob":0.99732465,"size":2940,"snap":"2020-45-2020-50","text_gpt3_token_len":1155,"char_repetition_ratio":0.1328338,"word_repetition_ratio":0.0038095238,"special_character_ratio":0.4085034,"punctuation_ratio":0.1910299,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9686269,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-29T22:27:39Z\",\"WARC-Record-ID\":\"<urn:uuid:e1f766f2-133a-4051-af04-f602bcf8bfdf>\",\"Content-Length\":\"17412\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0d8812bc-3610-41f8-9727-a963df554dd0>\",\"WARC-Concurrent-To\":\"<urn:uuid:3cf8ba05-36af-4f1c-ace9-b34ee3cd77de>\",\"WARC-IP-Address\":\"185.15.211.107\",\"WARC-Target-URI\":\"http://allreactions.com/index.php/group-4a/silicon\",\"WARC-Payload-Digest\":\"sha1:B665LO5POYSCDXESWXDWYJ4X6DI5XPII\",\"WARC-Block-Digest\":\"sha1:AIKP5SZL5T7PBRGV53E4Q2E7EFOPHJ5O\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141203418.47_warc_CC-MAIN-20201129214615-20201130004615-00496.warc.gz\"}"} |
https://answers.everydaycalculation.com/multiply-fractions/60-28-times-56-9 | [
"Solutions by everydaycalculation.com\n\n## Multiply 60/28 with 56/9\n\n1st number: 2 4/28, 2nd number: 6 2/9\n\nThis multiplication involving fractions can also be rephrased as \"What is 60/28 of 6 2/9?\"\n\n60/28 × 56/9 is 40/3.\n\n#### Steps for multiplying fractions\n\n1. Simply multiply the numerators and denominators separately:\n2. 60/28 × 56/9 = 60 × 56/28 × 9 = 3360/252\n3. After reducing the fraction, the answer is 40/3\n4. In mixed form: 131/3\n\nMathStep (Works offline)",
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"Download our mobile app and learn to work with fractions in your own time:"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.87175107,"math_prob":0.98511475,"size":414,"snap":"2019-43-2019-47","text_gpt3_token_len":150,"char_repetition_ratio":0.12926829,"word_repetition_ratio":0.0,"special_character_ratio":0.41062802,"punctuation_ratio":0.097826086,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9763339,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-14T14:40:43Z\",\"WARC-Record-ID\":\"<urn:uuid:f586fdcc-8ce3-4418-bf2b-7aeaca610bed>\",\"Content-Length\":\"7059\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e32a7a2c-52d7-4e28-a05f-a033e01adf4c>\",\"WARC-Concurrent-To\":\"<urn:uuid:44ba43e5-7f13-4ff8-8fe0-468dc5106b57>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/multiply-fractions/60-28-times-56-9\",\"WARC-Payload-Digest\":\"sha1:KR4QMXBRMI7BE324JV5SWF6FOJWTRBJB\",\"WARC-Block-Digest\":\"sha1:AQQBTVKW5BAJDUHR2VSCKJBBOSLLDVG6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496668525.62_warc_CC-MAIN-20191114131434-20191114155434-00258.warc.gz\"}"} |
https://aboutdev.wordpress.com/2009/12/19/excelcification-brain-teaser-code/ | [
"## Excelcification: Brain Teaser Code\n\nProblem: Someone at work recently asked me how you would go about converting excel header rows into integers. In the image below, you see each column and its corresponding integer value.\n\nThus, A = 0, B = 1, C = 2 and so on. Since I have no better word to describe this process, I’m going to call it Excelcification.\n\nDef: Excelcification. The act of converting a alpha column into its numeric representation.\n\nHowever, Excel cells go from A to Z and then become AA, AB, AC etc. So if Z = 25, AA = 26, AB = 27 and so on.\n\nYour mission is to create a method that takes a string that corresponds to the excel column (don’t have to worry about spaces etc for now) and change that to its integer value.\n\nIn case you didn’t know, the largest column in Excel 2010 that I could see was XFD.\n\nSo what is the Excelcification of “XFD”?\n\n[Jeopardy theme music plays…]\n\n[Some time later…]\n\n[Some more time later…]\n\nAre you done? How did you do it?\n\nMy process was to recognize that this is going to be Base 26 arithmetic, also know as Hexavigesimal. But you didn’t need to know that. If you know how base 2 (binary) works, you can extrapolate it to work for base 26. So how does base 2 work. If you remember your truth tables from high school/college, you would know that in binary:\n\n01 = 2^0\n\n10 = 2^1\n\n11 = 2^1 + 2^0\n\nNow extrapolate this for base 26 and realize that in binary, your digits are only 0 and 1 whereas in hexavigesimal, your digits are 0 through 25. (See the link there to A through Z).\n\nThus,\n\nA = (26^0) * Numeric_Value(A)\n\nB = (26^0) * Numeric_Value(B)\n\netc.\n\nSo if you create a method for this your code should be:\n\n``` public static class ExtensionMethods\n{\n/// <summary>\n/// Converts a string into its hexavigesical (base 26) representation.\n/// </summary>\n/// <param name=\"sxCol\">Input string of letters.</param>\n/// <returns>-1 if input is null or empty, base 26 integer representation of input otherwise.</returns>\npublic static int Excelcify(this string sxCol)\n{\nint result = -1;\nsxCol = sxCol.ToUpper();\n\nif (string.IsNullOrEmpty(sxCol))\nreturn result;\n\nfor (int i = sxCol.Length; i > 0; i--)\n{\nchar _c = sxCol[i-1];\n//\n// Function => (26 ^ reversed_char_index) * char_value\n// A = 1 ------ Z = 26 ------ AA = 27 ------ AZ = 54\n// 64 because there 'A' starts at index 65 and we want to give 'A' the value 1.\nresult += Math.Pow(26, sxCol.Length - i).ToSafeInt() * (_c.ToSafeInt() - 64);\n}\n\nreturn result;\n}\n} ```\n\nAnd you would call your code as such:\n\n``` int iResult = “A”.Excelcify();\nDebug.Assert(res == 0);\niResult = \"Z\".Excelcify();\nDebug.Assert(res == 25);\niResult = \"AA\".Excelcify();\nDebug.Assert(res == 26);\n```\n\n#### Thus “AAA” = 702, and “XFD” = 16383.\n\nThat is exactly 2^14 cells for those binary folk!\n\n## 6 thoughts on “Excelcification: Brain Teaser Code”\n\n1.",
null,
"Aaron Wood says:\n\nI assume that the ToSafeInt() method does the conversion of letters to an int? I.e. if you put ‘C’ into the ToSafeInt method, you get back 3?\n\n•",
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"aboutdev says:\n\nYep ToSafeInt is an extension method to convert the base class types to integer values.\n\nASCII tables show the values of the letters (http://en.wikipedia.org/wiki/ASCII). I’m glad you took a stab at it Aaron. I like little teasers like this to keep my brain moving.\n\n•",
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"Aaron Wood says:\n\nThat was definitely a good teaser. Thanks for posting it.\n\n2.",
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"Aaron Wood says:\n\nNevermind, I see that the math function handles characters apparently.\n\n3.",
null,
"Aaron Wood says:\n\nOh wait. Now I see. Char has an index value, but it starts at 65 (for ‘A’). Wheee, I finally get it now. It’s like I’m back at my interview, lol.\n\nNice, elegant solution. Mine was not looking quite so elegant, lol.\n\n4. […] Excelcification: Brain Teaser Code « Aboutdev's Weblog […]"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8476043,"math_prob":0.87391865,"size":3492,"snap":"2021-04-2021-17","text_gpt3_token_len":956,"char_repetition_ratio":0.09202982,"word_repetition_ratio":0.0,"special_character_ratio":0.3024055,"punctuation_ratio":0.14779006,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98782,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-16T08:37:28Z\",\"WARC-Record-ID\":\"<urn:uuid:bd79d9f6-d927-4dba-802a-87a218e352a1>\",\"Content-Length\":\"94695\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bb20ebf5-73b9-4df4-8a40-f8e29fcec25a>\",\"WARC-Concurrent-To\":\"<urn:uuid:e9a30627-00c1-41f7-b9ae-b1d674d26494>\",\"WARC-IP-Address\":\"192.0.78.12\",\"WARC-Target-URI\":\"https://aboutdev.wordpress.com/2009/12/19/excelcification-brain-teaser-code/\",\"WARC-Payload-Digest\":\"sha1:WSQHME55CUFGPN4PNO5Z7BDBPNZCUPB3\",\"WARC-Block-Digest\":\"sha1:DY7ZIH7GK2BBLTJUSMATLE7L6OBLXEZ7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703505861.1_warc_CC-MAIN-20210116074510-20210116104510-00703.warc.gz\"}"} |
http://answers.dataiku.com/961/perform-analysis-notebook-dataset-without-loading-dataset | [
"Hi,\n\nI have a dataset representing transactions for all the products.\n\nI would like to perform a loop for on product in a python notebook for loading the transaction for this product, then perform analysis and write the results in a dataset.\n\nHow can I load only a partition of my dataset from a python Notebook ?\n\nretagged\n\nHi,\n\nI assume you use the get_dataframe() method and then work with a pandas dataframe. (Let me know if you do something different).\n\nHere is what you can do:\n\n1) Get only a sample of a dataset with my_dataset.get_dataframe(sampling='head', limit=10000)\n\n2) Load the dataset by chunks with my_dataser.iter_dataframes(chunksize=10000)\n\n``````my_dataset = dataiku.Dataset(\"name_dataset\")\nfor partial_dataframe in my_dataset.iter_dataframes(chunksize=10000):\n# Insert here applicative logic on each partial dataframe.\npass``````\n\nYou can read more in the documentation."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.91178286,"math_prob":0.6896509,"size":328,"snap":"2019-35-2019-39","text_gpt3_token_len":64,"char_repetition_ratio":0.13580246,"word_repetition_ratio":0.0,"special_character_ratio":0.19512194,"punctuation_ratio":0.078125,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9956826,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-18T16:21:55Z\",\"WARC-Record-ID\":\"<urn:uuid:c49736ef-6358-400f-86ab-10b467dc0290>\",\"Content-Length\":\"32541\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d54d1faa-16aa-49a8-82d2-09990cbae2c0>\",\"WARC-Concurrent-To\":\"<urn:uuid:88891d7c-43cc-4b2d-a880-f98c998cb4b3>\",\"WARC-IP-Address\":\"104.25.142.109\",\"WARC-Target-URI\":\"http://answers.dataiku.com/961/perform-analysis-notebook-dataset-without-loading-dataset\",\"WARC-Payload-Digest\":\"sha1:ONBYJ6KDOZENLQOYU5XI6SX2F257IWPT\",\"WARC-Block-Digest\":\"sha1:OEHOWYTMR24J2H2COWUGJJ4UXGS4SPD3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514573309.22_warc_CC-MAIN-20190918151927-20190918173927-00552.warc.gz\"}"} |
https://www.proprofs.com/quiz-school/story.php?title=meter-usage-and-circuit-diagnosis | [
"# Quiz On Meter Usage And Circuit Diagnosis\n\n25 Questions | Attempts: 8886",
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"",
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"Settings",
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"How good are you at understanding electrical engineering and electrical measuring tools? Check out this quiz based on meter usage and circuit diagnosis to test your memory for the same. An electrical or energy meter is a measuring tool that measures the total power consumed over time. Identifying the components or connections responsible for malfunctioning a defective circuit board is called the circuit diagnosis. Here, we will ask questions about insulators, electrical circuits, and Ohm's law. Can you pass this quiz? Play it out and test yourself.\n\n• 1.\nWhat materials make a good insulator?\n• A.\n\nCopper and aluminum\n\n• B.\n\nCeramic and plastic\n\n• C.\n\nBoth A and B\n\n• D.\n\nNone of the above\n\n• 2.\nTech A says AC stands for alternating voltage. Tech B says DC is used in electric motors on most hybrid vehicles. Who is correct?\n• A.\n\nTech A\n\n• B.\n\nTech B\n\n• C.\n\nBoth Techs A and B\n\n• D.\n\nNeither Tech A nor B\n\n• 3.\nTech A says high resistance causes an increase in current flow. Tech B says a voltage drop is another name for high resistance. Who is correct?\n• A.\n\nTech A\n\n• B.\n\nTech B\n\n• C.\n\nBoth Techs A and B\n\n• D.\n\nNeither Tech A nor B\n\n• 4.\nTech A says parallel circuits are like links in a chain. Tech B says total current in a parallel circuit equals the sum of the current flowing in each branch of the circuit. Who is correct?\n• A.\n\nTech A\n\n• B.\n\nTech B\n\n• C.\n\nBoth Techs A and B\n\n• D.\n\nNeither Tech A nor B\n\n• 5.\nTech A says relays are turned on and off by a small amount of current. Tech B says there are two types of relays: normally closed (NC) and normally open (NO) types. Who is correct?\n• A.\n\nTech A\n\n• B.\n\nTech B\n\n• C.\n\nBoth Techs A and B\n\n• D.\n\nNeither Tech A nor B\n\n• 6.\nWhat does DVOM stand for?\n• A.\n\nDigital volt–hour meter\n\n• B.\n\nDigital volt-ohmmeter\n\n• C.\n\nDigital valve ohmmeter\n\n• D.\n\nDigital variable oval meter\n\n• 7.\nWhen reading a digital volt-ohmmeter (DVOM), you have a reading of 2168 mV, which is the same as:\n• A.\n\n2168 millivolts.\n\n• B.\n\n2.168 volts.\n\n• C.\n\n1000 mV.\n\n• D.\n\nBoth A and B\n\n• 8.\nWhen probing wires, you should do all these EXCEPT:\n• A.\n\nBack probe when possible.\n\n• B.\n\nUse caution when piercing wiring not to damage the wire internally.\n\n• C.\n\nReinsulate the hole with room temperature vulcanizing silicone.\n\n• D.\n\nNever use excessive force.\n\n• 9.\nTech A says you can measure up to 100 amps directly through the meter. Tech B says when checking high volts, you need to use a volt clamp so you don't damage the digital volt-ohmmeter (DVOM). Who is correct?\n• A.\n\nTech A\n\n• B.\n\nTech B\n\n• C.\n\nBoth Techs A and B\n\n• D.\n\nNeither Tech A nor B\n\n• 10.\nTech A says most digital volt-ohmmeters (DVOMs) have both auto range and a manual range setting. Tech B says OL means overload and indicates the voltage being read is higher than the maximum. Who is correct?\n• A.\n\nTech A\n\n• B.\n\nTech B\n\n• C.\n\nBoth Techs A and B\n\n• D.\n\nNeither Tech A nor B\n\n• 11.\nWhen doing a voltage drop test, the reading is 2.7 volts across the circuit. How many volts are going to the load?\n• A.\n\n12 volts\n\n• B.\n\n9.3 volts\n\n• C.\n\n8.3 volts\n\n• D.\n\n3.7 volts\n\n• 12.\nTech A says testing across the battery will show you better voltage. Tech B says when checking across a switch, it has a 2-volt drop, which means only about 10 volts are going to the light. Who is correct?\n• A.\n\nTech A\n\n• B.\n\nTech B\n\n• C.\n\nBoth Techs A and B\n\n• D.\n\nNeither Tech A nor B\n\n• 13.\nTech A says when doing a voltage drop test, the voltmeter needs to be set to ohm. Tech B says current has to be flowing to do a voltage drop test. Who is correct?\n• A.\n\nTech A\n\n• B.\n\nTech B\n\n• C.\n\nBoth Techs A and B\n\n• D.\n\nNeither Tech A nor B\n\n• 14.\nTech A says that in a relay, current is sent through a resistor and a magnetic field is produced. Tech B says to check the amps in a circuit, you need to connect the leads parallel to the circuit. Who is correct?\n• A.\n\nTech A\n\n• B.\n\nTech B\n\n• C.\n\nBoth Techs A and B\n\n• D.\n\nNeither Tech A nor B\n\n• 15.\nTech A says when checking resistance in a circuit, you need to have power connected to the component. Tech B says ideally, you should have a component disconnected. Who is correct?\n• A.\n\nTech A\n\n• B.\n\nTech B\n\n• C.\n\nBoth Techs A and B\n\n• D.\n\nNeither Tech A nor B\n\n• 16.\nTech A says Ohm's law will show how many amps are needed in a circuit with 110 ohm resistor and a 12-volt power supply. The amps needed are:\n• A.\n\n12 amps.\n\n• B.\n\n1.2 amps.\n\n• C.\n\n0.12 amps.\n\n• D.\n\n0.012 amps.\n\n• 17.\nTech A says there are three ways to check a circuit with Ohm's law. Tech B says if you know two of the measurements, Ohm's law will show you the third. Who is correct?\n• A.\n\nTech A\n\n• B.\n\nTech B\n\n• C.\n\nBoth Techs A and B\n\n• D.\n\nNeither Tech A nor B\n\n• 18.\nWhen discussing a circuit, you know the voltage is 0.202 volts and the current draw is 0.00202 amp. What is the resistance in the circuit?\n• A.\n\n1\n\n• B.\n\n10\n\n• C.\n\n100\n\n• D.\n\n10000\n\n• 19.\nWhen doing a voltage drop test, you should do all these EXCEPT:\n• A.\n\nSelect auto range volts DC.\n\n• B.\n\nConnect the red lead to volt/ohm.\n\n• C.\n\nRemove the positive battery cable.\n\n• D.\n\nEnsure the current is flowing.\n\n• 20.\nTech A says you can calculate total circuit current with this formula I = V ÷ R. Tech B says you can determine the total voltage drop using Ohm's law. Who is correct?\n• A.\n\nTech A\n\n• B.\n\nTech B\n\n• C.\n\nBoth Techs A and B\n\n• D.\n\nNeither Tech A nor B\n\n• 21.\nTech A says voltage drops can be measured as long as current is flowing. Tech B says voltage drops can be measured across components, connectors, or cables. Who is correct?\n• A.\n\nTech A\n\n• B.\n\nTech B\n\n• C.\n\nBoth Techs A and B\n\n• D.\n\nNeither Tech A nor B\n\n• 22.\nWhen calculating current flow in a parallel circuit that has 4 resistors, you would:\n• A.\n\nDivide the total current flow by the volts.\n\n• B.\n\nMultiply the current flow by 4.\n\n• C.\n\nAdd the current flow for each resistor together.\n\n• D.\n\nSubtract the volts from amps and add 2 volts.\n\n• 23.\nA variable resistor:\n• A.\n\nHas a fixed resistance.\n\n• B.\n\nChanges with the amount of voltage applied.\n\n• C.\n\nHas a movable arm that swipes across a coil that has less resistance at one end and more at the other.\n\n• D.\n\nAll of the above\n\n• 24.\nTech A says checking a variable resistor is almost the same as a regular resistor. Tech B says you have to have a different type of digital volt-ohmmeter (DVOM) to read a variable resistor. Who is correct?\n• A.\n\nTech A\n\n• B.\n\nTech B\n\n• C.\n\nBoth Techs A and B\n\n• D.\n\nNeither Tech A nor B\n\n• 25.\nTech A says an oscilloscope is commonly referred to as a lab scope or just scope. Tech B says lab scopes are not very important test instruments in a modern automobile. Who is correct?\n• A.\n\nTech A\n\n• B.\n\nTech B\n\n• C.\n\nBoth Techs A and B\n\n• D.\n\nNeither Tech A nor B\n\n## Related Topics",
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https://au.mathworks.com/matlabcentral/answers/546825-cross-product-of-by-an-array?s_tid=prof_contriblnk | [
"# cross product of by an array\n\n11 views (last 30 days)\nOday Shahadh on 11 Jun 2020\nCommented: Oday Shahadh on 11 Jun 2020\nI have (a )which is an array as below:\na =\na{1} = [185x3 double]\na{2} = [185x3 double]\na{3} = [185x3 double]\na{4} = [185x3 double]\na{5} = [185x3 double]\nAlso I have (L)which is (5,3) matrix\nI need to make cross product as below\n(0,0,-2) x a{1} = [185x3 double]\n0,0,-1 x a{2} = [185x3 double]\n0,0,0) x a{3} = [185x3 double]\n0,0,1) x a{4} = [185x3 double]\n0,0,2) x a{5} = [185x3 double]\nthis should result another vectors which I need to quiver3 them\n\nTodd on 11 Jun 2020\nYou can represent a cross product as a matrix product\na X b = tilde(a)*b\nwhere tilde(a) is a skew-symmetric matrix defined by\ntilde = @(v)[\n0 -v(3) v(2)\nv(3) 0 -v(1)\n-v(2) v(1) 0\n];\nFor example\na = [0;0;-1];\nb = [1;2;3];\nc = tilde(a)*b\nb could be a 3xm matrix of column vectors, in which case tilde(a)*b is a crossed with each column of b.\na = [0;0;-1];\nb = rand(3,185);\nc = tilde(a)*b\nAlso note that tilde(a)*b = -tilde(b)*a.\nMy example assumes column vectors, so you'll have to transpose things to apply it to your data.\nOday Shahadh on 11 Jun 2020\nthanks Todd, sorry I made a mistake in typing my question\nthe range is from(0,0,-2) to (0,0,2) as seen above"
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https://www.colorhexa.com/42c26b | [
"# #42c26b Color Information\n\nIn a RGB color space, hex #42c26b is composed of 25.9% red, 76.1% green and 42% blue. Whereas in a CMYK color space, it is composed of 66% cyan, 0% magenta, 44.8% yellow and 23.9% black. It has a hue angle of 139.2 degrees, a saturation of 51.2% and a lightness of 51%. #42c26b color hex could be obtained by blending #84ffd6 with #008500. Closest websafe color is: #33cc66.\n\n• R 26\n• G 76\n• B 42\nRGB color chart\n• C 66\n• M 0\n• Y 45\n• K 24\nCMYK color chart\n\n#42c26b color description : Moderate cyan - lime green.\n\n# #42c26b Color Conversion\n\nThe hexadecimal color #42c26b has RGB values of R:66, G:194, B:107 and CMYK values of C:0.66, M:0, Y:0.45, K:0.24. Its decimal value is 4375147.\n\nHex triplet RGB Decimal 42c26b `#42c26b` 66, 194, 107 `rgb(66,194,107)` 25.9, 76.1, 42 `rgb(25.9%,76.1%,42%)` 66, 0, 45, 24 139.2°, 51.2, 51 `hsl(139.2,51.2%,51%)` 139.2°, 66, 76.1 33cc66 `#33cc66`\nCIE-LAB 70.036, -53.982, 33.692 24.191, 40.801, 20.51 0.283, 0.477, 40.801 70.036, 63.633, 148.03 70.036, -53.862, 52.763 63.876, -44.182, 25.676 01000010, 11000010, 01101011\n\n# Color Schemes with #42c26b\n\n• #42c26b\n``#42c26b` `rgb(66,194,107)``\n• #c24299\n``#c24299` `rgb(194,66,153)``\nComplementary Color\n• #59c242\n``#59c242` `rgb(89,194,66)``\n• #42c26b\n``#42c26b` `rgb(66,194,107)``\n• #42c2ab\n``#42c2ab` `rgb(66,194,171)``\nAnalogous Color\n• #c24259\n``#c24259` `rgb(194,66,89)``\n• #42c26b\n``#42c26b` `rgb(66,194,107)``\n• #ab42c2\n``#ab42c2` `rgb(171,66,194)``\nSplit Complementary Color\n• #c26b42\n``#c26b42` `rgb(194,107,66)``\n• #42c26b\n``#42c26b` `rgb(66,194,107)``\n• #6b42c2\n``#6b42c2` `rgb(107,66,194)``\n• #99c242\n``#99c242` `rgb(153,194,66)``\n• #42c26b\n``#42c26b` `rgb(66,194,107)``\n• #6b42c2\n``#6b42c2` `rgb(107,66,194)``\n• #c24299\n``#c24299` `rgb(194,66,153)``\n• #2d8b4b\n``#2d8b4b` `rgb(45,139,75)``\n• #339e55\n``#339e55` `rgb(51,158,85)``\n• #39b160\n``#39b160` `rgb(57,177,96)``\n• #42c26b\n``#42c26b` `rgb(66,194,107)``\n• #55c87a\n``#55c87a` `rgb(85,200,122)``\n• #69ce89\n``#69ce89` `rgb(105,206,137)``\n• #7cd598\n``#7cd598` `rgb(124,213,152)``\nMonochromatic Color\n\n# Alternatives to #42c26b\n\nBelow, you can see some colors close to #42c26b. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #42c24b\n``#42c24b` `rgb(66,194,75)``\n• #42c256\n``#42c256` `rgb(66,194,86)``\n• #42c260\n``#42c260` `rgb(66,194,96)``\n• #42c26b\n``#42c26b` `rgb(66,194,107)``\n• #42c276\n``#42c276` `rgb(66,194,118)``\n• #42c280\n``#42c280` `rgb(66,194,128)``\n• #42c28b\n``#42c28b` `rgb(66,194,139)``\nSimilar Colors\n\n# #42c26b Preview\n\nThis text has a font color of #42c26b.\n\n``<span style=\"color:#42c26b;\">Text here</span>``\n#42c26b background color\n\nThis paragraph has a background color of #42c26b.\n\n``<p style=\"background-color:#42c26b;\">Content here</p>``\n#42c26b border color\n\nThis element has a border color of #42c26b.\n\n``<div style=\"border:1px solid #42c26b;\">Content here</div>``\nCSS codes\n``.text {color:#42c26b;}``\n``.background {background-color:#42c26b;}``\n``.border {border:1px solid #42c26b;}``\n\n# Shades and Tints of #42c26b\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, #010402 is the darkest color, while #f4fbf6 is the lightest one.\n\n• #010402\n``#010402` `rgb(1,4,2)``\n• #06130a\n``#06130a` `rgb(6,19,10)``\n• #0b2112\n``#0b2112` `rgb(11,33,18)``\n• #10301a\n``#10301a` `rgb(16,48,26)``\n• #143f22\n``#143f22` `rgb(20,63,34)``\n• #194e2a\n``#194e2a` `rgb(25,78,42)``\n• #1e5d32\n``#1e5d32` `rgb(30,93,50)``\n• #236c3a\n``#236c3a` `rgb(35,108,58)``\n• #287a42\n``#287a42` `rgb(40,122,66)``\n• #2c894a\n``#2c894a` `rgb(44,137,74)``\n• #319852\n``#319852` `rgb(49,152,82)``\n• #36a75a\n``#36a75a` `rgb(54,167,90)``\n• #3bb662\n``#3bb662` `rgb(59,182,98)``\n• #42c26b\n``#42c26b` `rgb(66,194,107)``\n• #51c777\n``#51c777` `rgb(81,199,119)``\n• #60cc82\n``#60cc82` `rgb(96,204,130)``\n• #6ed08e\n``#6ed08e` `rgb(110,208,142)``\n• #7dd599\n``#7dd599` `rgb(125,213,153)``\n• #8cdaa5\n``#8cdaa5` `rgb(140,218,165)``\n• #9bdfb1\n``#9bdfb1` `rgb(155,223,177)``\n• #aae4bc\n``#aae4bc` `rgb(170,228,188)``\n• #b9e8c8\n``#b9e8c8` `rgb(185,232,200)``\n• #c7edd4\n``#c7edd4` `rgb(199,237,212)``\n• #d6f2df\n``#d6f2df` `rgb(214,242,223)``\n• #e5f7eb\n``#e5f7eb` `rgb(229,247,235)``\n• #f4fbf6\n``#f4fbf6` `rgb(244,251,246)``\nTint Color Variation\n\n# Tones of #42c26b\n\nA tone is produced by adding gray to any pure hue. In this case, #7c8880 is the less saturated color, while #08fc56 is the most saturated one.\n\n• #7c8880\n``#7c8880` `rgb(124,136,128)``\n• #72927c\n``#72927c` `rgb(114,146,124)``\n• #689c79\n``#689c79` `rgb(104,156,121)``\n• #5fa575\n``#5fa575` `rgb(95,165,117)``\n• #55af72\n``#55af72` `rgb(85,175,114)``\n• #4cb86e\n``#4cb86e` `rgb(76,184,110)``\n• #42c26b\n``#42c26b` `rgb(66,194,107)``\n• #38cc68\n``#38cc68` `rgb(56,204,104)``\n• #2fd564\n``#2fd564` `rgb(47,213,100)``\n• #25df61\n``#25df61` `rgb(37,223,97)``\n• #1ce85d\n``#1ce85d` `rgb(28,232,93)``\n• #12f25a\n``#12f25a` `rgb(18,242,90)``\n• #08fc56\n``#08fc56` `rgb(8,252,86)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #42c26b is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population"
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https://www.intechopen.com/books/rational-fitting-techniques-for-the-modeling-of-electric-power-components-and-systems-using-matlab-environment/rational-fitting-techniques-for-the-modeling-of-electric-power-components-and-systems-using-matlab-e | [
"Open access peer-reviewed compact\n\n# Rational Fitting Techniques for the Modeling of Electric Power Components and Systems Using MATLAB Environment\n\nBy Eduardo Salvador Bañuelos-Cabral, José Alberto Gutiérrez-Robles and Bjørn Gustavsen\n\nSubmitted: May 26th 2017Reviewed: September 29th 2017Published: December 6th 2017\n\nDOI: 10.5772/intechopen.71358\n\n## Abstract\n\nThis book provides a detailed description of some of the most widely used rational fitting techniques for approximation of frequency domain responses. The techniques are: Bode’s asymptotic approximation, the Levy method, iteratively reweighted least squares, the Sanathanan-Koerner method, the Noda method, Vector Fitting, the Levenberg-Marquardt method, and the Damped Gauss-Newton method. A MATLAB routine for each technique is presented. These techniques are tested by approximating synthetic frequency domain responses. Then, they are applied to the rational approximation of the frequency-dependent parameters corresponding to a single-phase transmission line. The effect of the rational function-based models is evaluated, considering transients in three cases: Open-ended, short-circuited, and perfectly matched lines. The error levels obtained in time domain simulations are consistent with the fitting deviations of the frequency-dependent parameters. The book concludes by showing main advantages and disadvantages for each technique.\n\n### Keywords\n\n• rational approximation\n• least squares\n• weighted least squares\n• Bode’s asymptotic approximation\n• vector fitting\n\n## 1. Introduction\n\nOne of the main problems encountered in the modeling of power system components is the inclusion of frequency-dependent effects in a time domain simulation . In practice, these effects are described by discrete frequency domain responses that are obtained via calculations or measurements . For the time domain simulation, implementation of numerical convolutions is possible but is computationally inefficient. The frequency-dependent effects are usually performed in the frequency domain via complex-curve fitting processes, leading to rational function-based models which can be expressed in pole-zero form, pole-residue form, and polynomial form . Once obtained, these models must be converted into a form suitable for representation in an electromagnetic Transient Program (EMTP-type) circuit simulator . Commonly, such representations can be in the form of network synthesis or in ordinary differential equations (ODEs). Some methods have been adapted to particular problems due to the difficulty present in the development of a general methodology. These rational fitting techniques have been developed since the 1950s for model synthesis based on frequency response data. Some antecedents are presented below.\n\nIn 1959, Levy presented a mathematical procedure of linearized Least Squares (LS) for model synthesis based on a polynomial form. Some years later, in 1963, Sanathanan and Koerner improved the Levy method by introducing an iterative weighted method to reduce the biasing in the approximation caused by the linearization. Also, based on the Levy method, Lawrence and Rogers presented in 1979 a sequential algorithm that allows the fitting progress of a transfer function point-to-point without matrix inversion.\n\nIn addition, a technique based on pole-zero form that quickly became popular for the modeling of overhead lines was presented by Marti in 1982. This method is a numerical implementation of the well-known Bode diagrams technique. Recently, in 2017, Marti and Tavighi presented an investigation based in the same rational approximation technique for the modeling of transmission lines. In this work, this fitting technique will be referred to as Asymptotic Approximation or Bode.\n\nIn 1993, Soysal and Semlyen used the Gauss–Seidel optimization method to improve the results given by Levy and, 6 years later Gustavsen and Semlyen developed the Vector Fitting method which has become one of the most widely used techniques.\n\nIn 2006, Gustavsen proposed a modification of his algorithm in order to improve the ability of VF to relocate poles to better positions. This is achieved by replacing the high-frequency asymptotic requirement of the VF scaling function with a more relaxed condition. He called this method Relaxed VF (RVF).\n\nThe VF method is closely related to the Universal Line Model (ULM) . The ULM is formulated in terms of rational approximation of the line parameters through VF. A recent publication, in 2016, Bañuelos et al. propose the use of only real poles and zeroes in the ULM to improve its numerical efficiency.\n\nIn 2005, Noda presented an iterative algorithm that partitions the entire frequency range in order to avoid ill-conditioning of the system when Levy method is used. Recently, in 2015, Bañuelos-Cabral et al. propose the implementation of Damped Gauss-Newton (DGN) to increase the accuracy of this technique.\n\nThe aim of this book is to provide a MATLAB algorithm and a detailed description of the rational approximation techniques that are most commonly used to approximate functions in frequency domain. These techniques are: Bode’s Asymptotic Approximation (Bode), the Levy (Levy or LS) method, Iteratively Reweighted Least Squares (IRLS), the Sanathanan-Koerner (SK) method, the Noda (Noda) method, Vector Fitting (VF), the Levenberg–Marquardt (LM) method and the Damped Gauss-Newton (DGN) method.\n\nIn this chapter, the methodology for determining the state-space (SS) representation in concordance with the model used in the approximation is presented first. Next, the abovementioned fitting techniques are described in detail. Then, the techniques are tested by approximating a synthetic frequency domain response. The techniques are implemented in MATLAB environment. Finally, advantages of the methods are demonstrated in the rational approximation of the frequency-dependent parameters corresponding to a single-phase transmission line.\n\n## 2. State-space model from a transfer function\n\nThe SS representation of a given system can be determined from its frequency response on rational form. Basically, there are three different rational forms (models) that can represent a measured or calculated frequency response: pole-zero form, polynomial form, and pole-residue form. In this section, a methodology is presented for determining the SS representation according to the given model type.\n\n### 2.1. Pole-zero form\n\nUsage of pole-zero form or series realization leads to a rational function-based model of nth-order given by (1), which is a ratio between products of first-order transfer functions,\n\nFsksz1sz2sznsp1sp2spm.E1\n\nThe generalized graphic representation of pole-zero form with n = m is shown in Figure 1 .\n\nFrom Figure 1, it is possible to obtain the state equations and the output equation as\n\nẋ1=p1x1+uẋ2=p2x2+z1x1+ẋ1ẋ3=p3x3+z2x2+ẋ2ẋn1=pn1xn1+zn2xn2+ẋn2ẋn=pnxn+zn1xn1+ẋn1E2\ny=kznxn+ẋn.E3\n\nUsing algebraic manipulation in (2) and (3) to remove ẋnfrom the right side, the state equations and output equation in matrix form are\n\nẋ1ẋ2ẋn1ẋn=p1000ĉ1p200ĉ1ĉ2pn10ĉ1ĉ2ĉn1pnx1x2xn1xn+1111uE4\ny=ĉ1ĉ2ĉn1ĉnx1x2xn1xn+kuE5\n\nwhere ĉn=kznpn. There is a direct relation between the SS coefficients in (4) and (5) with the transfer function coefficients in (1) for pole-zero form realization. Thus, the SS description can be obtained directly from the transfer function by inspection.\n\n### 2.2. Polynomial form\n\nIn this case, polynomial form or direct form realization leads to a rational function-based model of nth-order given by (6), which is a ratio of two polynomials,\n\nFsa0+a1s+a2s2++ansnb0+b1s+b2s2++bmsm.E6\n\nA graphical representation of the polynomial form with bm = 1 and n = m is shown in Figure 2 .\n\nFrom Figure 2, it is possible to obtain the state equations and output equation as\n\nẋ1=x2ẋ2=x3ẋn1=xnẋn=b0x1b1x2bn1xn+uE7\ny=a0x1+a1x2++an1xn+anẋn.E8\n\nTo remove ẋnfrom (8) the last equation of (7) can be used. These equations can be written more conveniently as\n\nẋ1ẋ2ẋn1ẋn=010000010000001b0b1b2bn2bn1x1x2xn1xn+0001uE9\ny=â0â1ân1x1x2xn+anu,E10\n\nwhere âi=aianbi. Also, there is a direct relation between the SS coefficients in (9) and (10) with the transfer function coefficients in (6). The matrices of the SS description can be obtained directly from the transfer function through inspection.\n\n### 2.3. Pole-residue form\n\nFinally, pole-residue form or parallel realization leads to rational function-based model of nth-order given by (11), which is a sum of partial fractions\n\nFsc1sp1+c2sp2++cnspn+d.E11\n\nThe generalized graphic representation of the pole-residue form is shown in Figure 3 .\n\nFrom this Figure 3, it is possible to obtain the state equations and output equation as\n\nẋ1=p1x1+uẋ2=p2x2+uẋn=pnxn+uE12\ny=c1x1+c2x2++cnxn+du.E13\n\nIn matrix form it yields\n\nẋ1ẋ2ẋn=p1000p2000pnx1x2xn+111uE14\ny=c1c2cnx1x2xn+du.E15\n\nSimilarly to the pole-zero form and the polynomial form, the SS representation can be obtained through inspection. It should be mentioned that the pole-residue form produces an uncoupled SS system.\n\n## 3. Fitting methods\n\nThis section provides a description of the most widely used techniques for rational approximation of frequency domain responses within the power systems area.\n\n### 3.1. Asymptotic approximation (Bode) technique\n\nThe Asymptotic Approximation technique consists of a numerical implementation of the graphical technique known as Bode diagrams . The Asymptotic Approximation was first implemented by Marti to include the frequency dependence in transmission line modeling. This technique considers the pole-zero form with real poles and zeros only,\n\nFs=ksz1sz2sznsp1sp2spm.E16\n\nConsidering s = , we get\n\nFs=kz1z2znp1p2pm.E17\n\nEq. (17) can be expressed in a standard form as\n\nF=kz11z21zn1z1z2znp11p21pm1p1p2pmE18\n\nand in polar form\n\nF=K0z11α1z21α2zn1αnp11β1p21β2pm1βmE19\n\nwhere K0 = kz1z2zn/p1p2pm. The Asymptotic Approximation technique approximates only the magnitude of the frequency response, which can be expressed as\n\nF=K0z11z21zn1p11p21pm1.E20\n\nFinally, by using logarithmic properties in Eq. (20), it is obtained:\n\nlog10F=log10K0+log10/z11++log10/zn1log10/p11log10/pm1.E21\n\nIn Bode diagrams, each term of (21) is plotted individually in accordance with its already known asymptotic behavior and combined in order to obtain the desired diagram.\n\nIn the numerical implementation, the function to be fitted is compared with the sum of the line segments given by the addition of a pole or a zero in the model (21); the precision of the fitting depends on the sensitivity to locate these poles and zeros into the model.\n\n### 3.2. Levy (Levy) method\n\nThis method was introduced by Levy for complex-curve fitting. The Levy method makes the identification of the polynomial coefficients (22) in a LS sense. It is also known simply as Lest Squares (LS).\n\nFsNsDs=a0+a1s+a2s2++ansn1+b1s+b2s2++bmsm.E22\n\nThe numerical difference between the frequency response to be fitted and the model represents the error in the approximation, that is,\n\nεs=FsNsDs.E23\n\nMultiplying both sides of Eq. (23) by D(s), we obtain\n\nεs=εsDs=FsDsNs.E24\n\nConsidering that ε′ tends to zero, (24) can be expressed as\n\nFs1+b1s+b2s2++bmsma0+a1s+a2s2++ansn=0.E25\n\nThe unknown coefficients in (22) can now be solved by formulating (25) as a LS problem (Ax = b). Frequently, the sample size of the frequency response is greater than the number of polynomial coefficients to be calculated, so an overdetermined system is obtained, as follows:\n\nAk=1sksknskFskskmFskE26\nx=a0a1anb1b2bmTE27\nb=Fs1FskT,E28\n\nwhere k denotes the k-th data sample. The objective function to be minimized is\n\nminxAxb2.E29\n\nThis is called a linear least squares problem. The solution x satisfies the normal equations :\n\nATAx=ATb.E30\n\nFinally, the LS solution is obtained by\n\nx=ATA1ATb.E31\n\n### 3.3. Weighted least squares (WLS)\n\nThe LS solution of a given system equation assumes that each equation (row in A) is equally important. However, the system equation itself may be biased, e. g. in the Levy method which is biased due to the multiplication of F(s) with D(s) in (24). In the rational approximation of frequency domain responses, Weighted Least Squares (WLS) is used to mitigate the biasing by giving more weight to specific equations in order to overcome the own deficiency of the used technique. Iteratively Reweighted Least Squares (IRLS), the Sanathanan-Koerner (SK) method and the Noda (Noda) method are techniques that implement the concept of WLS. These techniques are described below.\n\n#### 3.3.1. Iteratively reweighted least squares (IRLS) iteration\n\nA robust regression procedure is an alternative to LS solution when data are contaminated with outliers or influential observations (measurement and/or computation errors) [16, 17]. The idea of robust regression is to weight (less) these observations differently based on the proposal of weighting functions. IRLS can be considered as a robust regression procedure.\n\nIt is proposed to use IRLS for the rational approximation of frequency domain responses by using the weighting functions in inverse form, namely, data are not considered to be contaminated with outliers or influential observations; the error in the fitting in each iteration is used to weight (more) the frequency response data that have not been approximated correctly.\n\nIn WLS, the objective function to be minimized is\n\nminxWbAx2E32\n\nwhere W is a diagonal weighting matrix. Eq. (32) is called a linear weighted least squares problem and x the linear weighted least squares solution of the system. This solution satisfies the normal equations :\n\nATWAx=ATWb.E33\n\nThen, the WLS solution is given by\n\nx=ATWA1ATWb.E34\n\nIt is possible to solve (34) through an iterative process (35), where i is the iteration number.\n\nxi+1=ATWiA1ATWib.E35\n\nEq. (35) is the IRLS method. According to (35), W must be updated iteratively; the error in the approximation is used for this purpose, which can be expressed as\n\nεsx=FsF̂sx,E36\n\nwhere F̂sxis the model to be fitted. Each element of W is updated according to\n\nwki=wki1ψεki1,E37\n\nwith ψ(ε) being the weighting function. In the beginning of the process, W is an identity matrix.\n\nTable 1 shows a list of weighting functions proposed to use in the implementation of IRLS. Weighting functions 1, 2, and 3 are the inverse of the functions used in robust regression [16, 17], and functions 4, 5, and 6 are proposed in this work.\n\nNumberψ(ε)\n1|ε|\n21+ε21\n31εp2,p=1.2\n4ε21+ε2\n5ε21+ε2\n6εpp,p=1.2\n\n### Table 1.\n\nWeighting functions.\n\nUltimately, in Figure 4 the behavior of the weighting functions with ε = [−2, 2] is shown.\n\n#### 3.3.2. Sanathanan-Koerner (SK) iteration\n\nThis method is based on the polynomial form:\n\nFsNsDs=a0+a1s+a2s2++ansn1+b1s+b2s2++bmsm.E38\n\nThe objective is to identify the coefficients for the polynomials N(s) and D(s) in (38) that minimize the error function\n\nεs=FsDsNs.E39\n\nSanathanan and Koerner proposed an iterative procedure where (39) is divided by the denominator from the previous iteration; thus, (39) can be expressed as\n\nεs=1Dsl1FsDslNsl.E40\n\nThe subscript l denotes the iteration number, and D(s) is considered 1 in the first iteration.\n\n#### 3.3.3. Noda (Noda) iteration\n\nThis algorithm proposed by Noda for the rational fitting identification is based on the function:\n\nFsNsDs=a0+a1s+a2s2++ansn1+b1s+b2s2++bmsm.E41\n\nThis technique corresponds to IRLS using the weighting function number 1 in Table 1. Additionally, Noda also proposed a procedure to prevent the ill-conditioning of the system by partitioning the given frequency response data into sections along the frequency axis. The technique is then applied to each section of the frequency response in order to identify the poles, and finally, the corresponding residues are obtained by means of a standard LS procedure, using the entire frequency response.\n\n### 3.4. Vector Fitting (VF) method\n\nVector Fitting performs the fitting by replacing a set of heuristically calculated initial poles with a set of relocated ones through an iterative procedure . VF works in two stages: First, it improves the position of the initial poles iteratively. Second, it calculates the residues in one step. This method is based on the pole-zero form\n\nFsn=1Ncnspn+d+sh,E42\n\nwhere cn are the residues, pn are the poles, d is the constant term and h the proportional part.\n\n#### 3.4.1. Pole identification\n\nConsidering the initial poles as an, and multiplying F(s) by an auxiliary function σ(s) gives\n\nσsFsσFfits=n=1Ncnsan+d+shE43\n\nwhere σ(s) is defined as\n\nσsσfits=n=1Nc˜nsan+1,E44\n\nwith σFfit(s) being the fitting of σ(s)F(s) and σfit(s) being the fitting of σ(s). From (43) one can obtain,\n\nFsσFfitsσs.E45\n\nSubstituting (43) and (44) into (45) we obtain,\n\nFshn=1N+1szn/n=1Nsann=1Nsz˜n/n=1Nsan=hn=1N+1sznn=1Nsz˜n.E46\n\nEq. (46) indicates that the zeros of σ(s) are an approximation of the poles of F(s). To obtain the zeros one multiplies (44) by F(s) which results in\n\nσsFsn=1Nc˜nsan+1Fs.E47\n\nEquating Eqs. (43) and (47) yields\n\nn=1Ncnsan+d+sh=n=1Nc˜nsan+1Fs.E48\n\nAlgebraic manipulation in (48) gives\n\nn=1Ncnsan+d+shn=1Nc˜nFssan=Fs.E49\n\nEq. (49) can now be formulated as a LS problem (Ax = b). Usually, the sample size of the frequency response is greater than the number of coefficients to be calculated, so an overdetermined system is obtained:\n\nAk=1ska11skaN1skFskska1FskskaNE50\nx=c1cNdhc˜1c˜NTE51\nb=Fs1FskTE52\n\nThe LS solution of the system (Ax = b) delivers the residues of σfit(s); the zeros of this function are the poles an for the next iteration, which correspond to the eigenvalues of ,\n\nH=Gkc˜T,E53\n\nwhere G is a diagonal matrix containing the poles and k is a unit column vector.\n\n#### 3.4.2. Residue identification\n\nOnce the poles pn for the rational approximation in (42) have been obtained, the residues can be found by solving (42) as a LS problem. The new overdetermined system is solved similarly as (50), (51) and (52).\n\n#### 3.4.3. Relaxation (RVF)\n\nThe scaling function σ(s) (44) is observed to approach unity at high frequencies. This asymmetry with respect to the frequency gives a tendency to relocate poles in the direction of lower frequencies. In addition to this biasing, it also reduces the convergence speed and often leads to a reduced accuracy for the final model. This problem is effectively alleviated by the introduction of Relaxed Vector Fitting (RVF) , where the scaling function (44) is replaced by\n\nσsσfits=n=1Nc˜nsan+d˜E54\n\nwith d˜as a real variable. To obtain a non-trivial solution, a single line is added to the LS problem, where the sum of the real part of σ(s) over the frequency samples is fixed. This relaxed criterion allows σ(s) to freely vary in shape.\n\n### 3.5. Levenberg-Marquardt (LM) method\n\nThe Levenberg–Marquardt (LM) method is a technique used to improve iteratively parameter values in order to reduce the sum of the squares of the errors between the model and the calculated or measured data [18, 19]. This method is actually a combination of two minimization methods: The gradient descent (GD) method and the Gauss-Newton (GN) method.\n\nConsider the model F̂sx, which is a function of s and n unknown parameters contained in vector x. The model is to be fitted to a set of frequency response data F(s) as follows:\n\nFsF̂sx,E55\n\nwhere x = [x1, x2, ⋯xn] is a vector that contains the coefficients to be fitted. The LS error between the fitting of the model and the frequency response data is\n\nγx=12k=1NsF̂skxFsk2E56\nγx=12F̂sxFsTF̂sxFsE57\nγx=12F̂sxTF̂sxF̂sxTFs+12FsTFs.E58\n\nThe aim of LM method is to find iteratively a perturbation h to modify the parameters x that reduces the objective function γ(x) .\n\n#### 3.5.1. Gradient descent (GD) method\n\nThe gradient of a function indicates the direction of the maximum rate of change in a specific point of the function. GD updates the parameter values in the opposite direction to the gradient of the objective function. Thus, the gradient of γ(x) is:\n\nxγx=F̂sxFsTxF̂sxFsE59\nxγx=F̂sxFsTJ,E60\n\nwhere J is the Jacobian matrix which represents the local sensitivity of the model F̂sxto variation in the parameters x. Finally, the perturbation h that moves the parameters in the direction of steepest descent is given by\n\nhGD=JTF̂sxFsE61\n\n#### 3.5.2. Gauss-Newton (GN) method\n\nThe Gauss-Newton (GN) method supposes that the objective function γ(x) is approximately quadratic in the parameters near the optimal solution. Approximating the model with a first-order Taylor series,\n\nF̂sx+hF̂sx+F̂sxxh=F̂sx+Jh.E62\n\nSubstituting (62) into (57) gives\n\nγx+h=12F̂TF̂F̂TF+12FTF+F̂FTJh+12hTJTJh.E63\n\nThe perturbation h that minimizes γ(x) is reached when\n\nγx+hh=0.E64\n\nTaking the derivative for (63) gives\n\nγx+hh=F̂sxFsTJ+hTJTJE65\n\nand applying condition (64), we obtain the perturbation h calculated by the GN method\n\nhGN=JTJ1JTF̂sxFs.E66\n\n#### 3.5.3. Levenberg-Marquardt method (LM)\n\nLM works more like the GD method when the parameters are far from their optimal value, and more like the GN method when the parameters are close to their optimal value. Levenberg and Marquardt proposed this method:\n\nJTJ+λIhLM=JTF̂sxFs,E67\n\nwhere λ is known as the Levenberg–Marquardt parameter. Clearly, for large values of λ, the LM method results in a GD update; large distance from the function. The GD method is utilized to provide steady and convergent progress toward the solution. As the solution approaches the minimum, λ is adaptively decreased; then, the LM method approaches the GN method, and the solution typically converges rapidly to the local minimum.\n\nThe following modification has also been suggested:\n\nJTJ+λdiagJTJhLM=JTF̂sxFsE68\n\nThere are different methods for update the Levenberg-Marquardt parameter λ. A complete review is presented in Ref. . Nevertheless, a simple method consists of reducing this parameter as the solution approaches the optimal value. The numerical implementation consists iteratively of:\n\n1. Calculate the objective function γ(x), (57).\n\n2. Calculate the Jacobian J, taking into account the implemented model (1), (6) or (11).\n\n3. Calculate the perturbation h, (67).\n\n4. If γ(x + h) ≤ γ(x), then x is replaced by x + h and λ is reduced.\n\n5. If γ(x + h) > γ(x), then x is not replaced and λ is increased.\n\n6. Convergence is achieved when the parameters reach a tolerance value or when the iteration count exceeds a pre-specified limit.\n\n#### 3.5.4. Damped gauss-Newton (DGN)\n\nThe DGN method can be applied to any of the alternative function-based models: pole-zero form (1), polynomial form (6), and pole-residue form (11). In the DGN method, a damping factor α is introduced\n\nhDGN=αJTJ1JTF̂sxFs.E69\n\nThe DGN method (69) always takes the descendent direction that satisfies a linear search method. It has slow convergence when the parameters to be fitted are far from the solution, but the method maximizes its advantages when it is close to the solution . For that reason, DGN should be used when the parameters are close to their optimal values.\n\nClearly, α defines a fraction of the step taken from GN to update the x parameters. In this work, we select α by a backtracking strategy, whereby α is reduced from unity until an acceptable value for γ(x) is found.\n\n## 4. MATLAB algorithms and applications\n\nIn this section, algorithms for the rational approximation of frequency domain responses in MATLAB environment are provided. Rational fitting of synthetics functions through Bode, DGN, LS or Levy, IRLS, LM and VF are presented. Finally, the techniques are applied to the rational approximation of the frequency-dependent parameters corresponding to a single-phase transmission line.\n\n### 4.1. Bode and DGN algorithms\n\nBode and DGN routines are presented in this section. The concept is simple, first Bode is implemented for the curve fitting process of a synthetic function, then DGN is used to improve the accuracy of the rational approximation given by Bode. A synthetic function is selected in the main program “Fitting_Bode_DGN.m” presented in Table 2. After application of the Bode routine “Bode_process.m,” shown in Table 3, a set of poles, residues and a constant term are obtained. These results are the initial values for the DGN routine “Damped_Gauss_Newton.m” presented in Table 4.\n\n %================================================% Main program of BODE-DGN% Authors: Eduardo Salvador Bañuelos Cabral% José Alberto Gutiérrez Robles% Bjørn Gustavsen%===============================================clcclear allclose all%% Initial SettingsNs = 400; % Number of samplesf = logspace(-2,10,Ns); % Frequency (Hz)w = 2.*pi.*f; % Frequency (rad/seg)s = 1j*w; % Complex Frequency%% Synthetic functionschoice = menu('CHOOSE A FUNCTION','Function F1(s)',… 'Function F2(s)','Function F3(s)','Function F4(s)',… 'Function F5(s)','Function F6(s)');if choice == 1 Fs = (110.*s./((s+10).*(s+100))); tol = 2.2; % Tolerance in decibels to set a new pole and/or new zero ite = 40; % Number of iteration in DGN methodendif choice == 2 Fs = ((s+10).*(s+100))./((s+14580).*(s+10355550)); tol = 2.2; % Tolerance in decibels to set a new pole and/or new zero ite = 40; % Number of iteration in DGN methodendif choice == 3 Fs = ((s+10655).*(s+10))./((s+148450).*(s+198545852).*(s+155222220)); tol = 2.2; % Tolerance in decibels to set a new pole and/or new zero ite = 40; % Number of iteration in DGN methodendif choice == 4 Fs = (10124).*s.*(s+35.7).*(s+88.9)./((s+100.5).*(s+220.7).*(s+5900).*(s+1370).*(s+21000)); tol = 2; % Tolerance in decibels to set a new pole and/or new zero ite = 50; % Number of iteration in DGN methodendif choice == 5 Fs = ((s+79).*(s+1045))./((s+1458).*(s+103555).*(s+127710355).*(s+1244103555)); tol = 1.8; % Tolerance in decibels to set a new pole and/or new zero ite = 40; % Number of iteration in DGN methodendif choice == 6 Fs = ((s+64518).*(s+8451629).*(s+312).*(s+54841216192))./((s+456).*(s+7852).* (s+982365).*…(s+93256888).*(s+79325684588536)); tol = 2; % Tolerance in decibels to set a new pole and/or new zero ite = 40; % Number of iteration in DGN methodend%% Bode method[P,Z,k] = Bode_process(Fs,f,Ns,tol); % Bode subroutineas = k.*poly(Z); bs = poly(P); % Polynomials[r,p,ks] = residue(as,bs); % Poles, residues and constant termTF=isempty(ks);if(TF==1);ks=0;end%% Damped Gauss-NewtonXmcp = [r; p; ks]; % Poles, residues and constant term from BodeNp = length(p); % Number of poles[KGN,RGN,PGN,Ff] = Damped_Gauss_Newton(Np,Ns,Fs,s,Xmcp,ite);Fs_fitDGN = zeros(1,Ns);for k = 1:length(PGN) Fs_fitDGN = Fs_fitDGN + (RGN(k)./(s.' - PGN(k))).';endFs_fitDGN = Fs_fitDGN + KGN;error = abs(Fs - Fs_fitDGN);%% Plotsfigure(3)loglog(f,abs(Fs),'k',f,abs(Fs_fitDGN),'r‐‐',f,error,'‐‐b','LineWidth',2);xlabel('Frequency [Hz]'); ylabel('Magnitude [p.u.]');legend('Data','DGN','Deviation',2)figure (4)semilogy(Ff,'‐‐b','LineWidth',2)xlabel('Iteration count'); ylabel('RMS-error');\n\n### Table 2.\n\n“Fitting_Bode_DGN.m”.\n\n %================================================% BODE PROCESS PROGRAM% Authors: Eduardo Salvador Bañuelos Cabral% José Alberto Gutiérrez Robles% Bjørn Gustavsen%================================================% Inputs % ‐‐‐ Fs, function to be fitted % ‐‐‐ f, frequency (Hz) % ‐‐‐ Ns, number of samples % ‐‐‐ tol, tolerance% Outputs % ‐‐‐ P, Poles % ‐‐‐ Z, Zeros % ‐‐‐ k, Constant termfunction [P,Z,k] = Bode_process(Fs,f,Ns,tol)P = []; Z = []; % Initialize P and Z like empty matricesFsdb = 20*log10(abs(Fs)); % Function in decibelsk0db = Fsdb(1); % k0 in decibelsk0 = 10.^(k0db./20); % k0 in magnitudeFsdb1 = ones(1,Ns).*k0db; % Constant term% Plot of the function in decibelsfigure(1)semilogx(f,Fsdb,'k',f,Fsdb1,'r‐‐','LineWidth',2)ylabel('Decibels'), xlabel('Frequency [Hz]'), hold onlegend('Data','Bode',2)c = 1; h = 1; r = 0; % CountersFsdb2 = zeros(1,Ns); % Initialize Fsdb2while (r < Ns) % Bode process algorithm Fsfitdb = Fsdb1 + Fsdb2; semilogx(f,Fsfitdb,'r‐‐','LineWidth',2) pause(0.5) for r = 1:1:Ns error = abs((Fsdb(r))-(Fsfitdb(r))); % Deviation if error >= (tol) % Tolerance % New zero if (Fsfitdb(r)< Fsdb(r)); Z(h)=f(r); h=h+1; break; end % New pole if (Fsfitdb(r)> Fsdb(r)); P(c)=f(r); c=c+1; break; end end end % Contribution of each zero in decibels Arg1 = 0; for kp = 1:length(Z) Arg1 = Arg1 + 20.*log10(abs(1+1i*f/Z(kp))); end % Contribution of each pole in decibels Arg2 = 0; for kp = 1:length(P) Arg2 = Arg2 - 20.*log10(abs(1+1i*f/P(kp))); end Fsdb2 = Arg1 + Arg2;end% Poles and zerosP = -P*2*pi;Z = -Z*2*pi;% Construct the constant termNum=1;for k=1:length(P);Num=Num*P(k);endDen=1;for k=1:length(Z);Den=Den*Z(k);endk = abs(Num)*k0/abs(Den);% Plotsfigure(2)semilogx(f,Fsdb,'k',f,Fsfitdb,'‐‐r','LineWidth',2);xlabel('Frequency [Hz]'); ylabel('Decibels')legend('Data','Bode',2)\n\n### Table 3.\n\n“Bode_process.m”.\n\n %================================================% DAMPED GAUSS NEWTON METHOD% Authors: Eduardo Salvador Bañuelos Cabral% José Alberto Gutiérrez Robles% Bjørn Gustavsen%===============================================% Inputs % ‐‐‐ Np, number of poles % ‐‐‐ Ns, Number of samples % ‐‐‐ Fs, function to be fitted % ‐‐‐ s, complex frequency (rad/s) % ‐‐‐ X, initial poles, zeros and constant term % ‐‐‐ ite,iterations% Outputs % ‐‐‐ Ks, constant term % ‐‐‐ Rs, residues % ‐‐‐ Ps, poles % ‐‐‐ Ff, objective function (deviation)function [Ks,Rs,Ps,Ff] = Damped_Gauss_Newton(Np,Ns,Fs,s,X,ite)a = ones(Ns,1); % Vector columnJ1 = zeros(Ns,Np); % Matrix size (J1)J2 = zeros(Ns,Np); % Matrix size (J2)Jn = zeros(2*Ns,2*Np+1); % Matrix size (Jn)At = zeros(2*Np+1,2*Np+1); % Matrix size (At)en = zeros(2*Ns,1); % Vector size (en)epn = zeros(2*Ns,1); % Vector size (epn)Ff = zeros(ite,1); % Vector size (Ff)Euclidian = zeros(1,1+2*Np); % Vector sizefor ki = 1:ite % Damped Gauss Newton methodology R = X(1:Np); % Residues P = X(Np+1:2*Np); % Poles K = X(2*Np+1); % Constant term Fa = zeros(1,Ns); % Set the approximation to the function for k = 1:length(P) Fa = Fa + (R(k)./(s.' - P(k))).'; end Fa = Fa + K; % Construct the approximation to the function error = (Fa.' - Fs.'); % Deviation for n = 1:1:Np % Loop to construct the Jacobian J1(1:Ns,n) = 1./(s-P(n)); J2(1:Ns,n) = R(n)./((s-P(n)).^2); end J = [J1 J2 a]; % Jacobian [Xmax Ymax] = size(J); % Matrix size (J) Jr = real(J); % Real part of vector J Ji = imag(J); % Imaginary part of vector J er = real(error); % Real part of vector error ei = imag(error); % Imaginary part of vector error km = 1; % Counter for k = 2:2:2*Xmax % Interleaved Jn(k-1,:) = Jr(km,:); Jn(k,:) = Ji(km,:); en(k-1,1) = er(km); en(k,1) = ei(km); km = km+1; end F = ((norm(en,2)^2)); % Objective function Ff(ki,1) = F; % Storage the objective function must tend to zero [Q,R] = qr(Jn); % Matrix Q and R of Jn Jn = R; % It updates matrix Jn Gf = (Jn.'*Q.')*en; % Gradient (QR decomposition) Hess = Jn.'*Jn; % Hessian (QR decomposition) for col = 1:Ymax % Euclidian norm Euclidian(col) = norm(Hess(:,col),2); At(:,col) = Hess(:,col)./Euclidian(col); end h = (At)\\-Gf; % Solution for the system (Ax = b) h = h./Euclidian.'; % Real solution stop = 1; % Variable to stop the next loop al = 1; % Variable to weigh the approximation while (stop == 1) Xp = X + al*h; % New coefficients (without updating x) Rp = Xp(1:Np); % Residues Pp = Xp(Np+1:2*Np); % Poles Kp = Xp(2*Np+1); % Constant term Fap = zeros(1,Ns); % Set the new approximation of the function for k = 1:length(P) Fap = Fap + (Rp(k)./(s.' - Pp(k))).'; end Fap = Fap + Kp; %New approximation of the function ep = (Fap.' - Fs.'); % Deviation epr = real(ep); % Real part of vector ep epi = imag(ep); % Imaginary part of vector ep km = 1; % Counter for k = 2:2:2*Xmax % Interleaved epn(k-1,1) = epr(km); epn(k,1) = epi(km); km = km+1; end Fp = ((norm(epn,2)^2)); % Objective function % Updating process Erel = 1e-4*al*h.'*Gf; % Relative error to stop the while process if (Fp < F + Erel) X = X + al*h; % Final approximation stop = 0; % Go out the while else al = al*0.9; % Weigh to update X if (al < 1e-30) stop = 0; else stop = 1; end end endendRs = X(1:Np); % ResiduesPs = X(Np+1:2*Np); % PolesKs = X(2*Np+1); % Constant term\n\n### Table 4.\n\n“Damped_Gauss_Newton.m”.\n\nIn Figure 5 the asymptotic behavior of the Bode routine is shown, and in Figure 6 the final rational approximation given by Bode is presented. Following, DGN routine improves the curve fitting process, this is presented in Figure 7. Moreover, the RMS-error on each iteration is shown in Figure 8.\n\n### 4.2. LS or Levy algorithm\n\nIn this section the LS method or Levy method routine is introduced. In the main program “Fitting_OLS.m,” Table 5, a synthetic function can be selected by the user. Then the LS routine “Least_Squares_Method.m,” Table 6, is implemented for the rational approximation of this function.\n\n %====================================================% Main program of ORDINARY LEAST SQUARES% Authors: Eduardo Salvador Bañuelos Cabral% José Alberto Gutiérrez Robles% Bjørn Gustavsen%====================================================clcclear allclose all%% Initial SettingsNs = 400; % Number of samplesf = logspace(-2,6,Ns); % Frequency \"Hertz\"w = 2.*pi.*f; % Frequency \"rad/seg\"s = 1i*w; % Complex Frequency%% Synthetic functionschoice = menu('CHOOSE A FUNCTION','Function F1(s)','Function F2(s)','Function F3(s)');if choice == 1Np = 2;Fs = 0.01+(210*(s)./((s+10).*(s+100)));endif choice == 2 Np = 5; Fs = (-81+18e6*s+27e6*s.^2+62e3*s.^3+42*s.^4+5e-7*s.^5)./… (1+54e5*s+36e7*s.^2+83e4*s.^3+90*s.^4+36e-3*s.^5);endif choice == 3 Np = 6; Fs =(100./(s+200))+(250./(s+2000))+(1500./(s+(10-1j*100)))+(1500./(s+(10+1j*100)))+… (800./(s+(1000+1j*30000)))+(800./(s+(1000-1j*30000)));end%% Ordinary Least Squares[Ks,Rs,Ps,x,A1,an,bn] = Least_Squares_Method(Np,Ns,Fs,s);% Fitting using partial fractionsFs_fit = zeros(1,Ns);for i = 1:Np Fs_fit = Fs_fit + (Rs(i)./(s.' - Ps(i))).';endFs_fit = Fs_fit + Ks;error = abs((Fs.' - Fs_fit.' ));%% Plotsfigure(1)loglog(f,abs(Fs),'-k',f,abs(Fs_fit),'‐‐r',f,error,'-.b','LineWidth',2);legend('Data','Ordinary Least Squares','Deviation',3);xlabel('Frequency [Hz]'); ylabel('Magnitude [p.u.]')\n\n### Table 5.\n\n“Fitting_OLS.m”.\n\n %==========================================================% Function to implement ORDINARY LEAST SQUARES% Authors: Eduardo Salvador Bañuelos Cabral% José Alberto Gutiérrez Robles% Bjørn Gustavsen%==========================================================% Inputs % ‐‐‐ Np, number of poles % ‐‐‐ Ns, number of samples % ‐‐‐ Fs, function to be fitted % ‐‐‐ s, complex frequency (rad/s)% Outputs % ‐‐‐ Ks, constant term % ‐‐‐ Rs, residues % ‐‐‐ Ps, poles % ‐‐‐ x, vector of coefficients (polynomials) % ‐‐‐ A1, matrix to evaluate the fitting % ‐‐‐ an, coefficients of the numerator % ‐‐‐ bn, coefficients of the denominatorfunction [Ks,Rs,Ps,x,A1,an,bn] = Least_Squares_Method(Np,Ns,Fs,s)a = ones(Ns,1); % Column vectora1 = zeros(Ns,Np); % Matrix size (s^n)a2 = zeros(Ns,Np); % Matrix size (Fs*s^n)An = zeros(2*Ns,1+2*Np); % Matrix size (real and imaginary part)Bn = zeros(2*Ns,1); % Vector size (real and imaginary part)Euclidian = zeros(1,1+2*Np); % Vector size% Construction of vector b.b = Fs.';% Construction of matrix A.for n = 1:1:Np a1(1:Ns,n) = s.^((Np+1)-n); a2(1:Ns,n) = -b.*a1(1:Ns,n);endA1 = [a1 a]; % Matrix to evaluate the fittingA = [a1 a a2]; % Matriz of the system[Xmax Ymax] = size(A); % Size for matrix AAr = real(A); % Real part of matrix AAi = imag(A); % Imaginary part of matrix Abr = real(b); % Real part of vector Bbi = imag(b); % Imaginary part of vector Bkm = 1; % Construction of matrix An and vector Bn (interleaved)for k = 2:2:2*Xmax An(k-1,:) = Ar(km,:); An(k,:) = Ai(km,:); Bn(k-1,1) = br(km); Bn(k,1) = bi(km); km = km+1;end[Q,R] = qr(An,0); % QR decomposition - Matrix Q and R of AnAt = R; % It updates the matrix AnB = Q.'*Bn; % It updates the vector Bnfor col = 1:Ymax Euclidian(col) = norm(At(:,col),2); % Euclidian norm At(:,col) = At(:,col)./Euclidian(col);endx = At\\B; % Solution for the system (Ax = b)x = x./Euclidian.'; % Real solutionan = x(1:Np+1); % Numerator coefficientsbn = [x(Np+2:2*Np+1);1]; % Denominator coefficients% Poles, residues and constant term[Rs,Ps,Ks] = residue(an,bn);TF = isempty(Ks);if (TF==1) Ks = 0;end\n\n### Table 6.\n\n“Least_Squares_Method.m”.\n\nFigure 9 shows the synthetic frequency response data and the approximation given by the LS routine, additionally, the deviation in terms of absolute error is presented.",
null,
"Figure 9.Synthetic frequency response data and the approximation given by the LS routine.\n\n### 4.3. IRLS algorithm\n\nBecause in practice the Noda method and the SK method correspond to a specific weighting function in the IRLS technique, only the latter is presented. In the beginning of the implementation a synthetic function is selected in the main program “Fitting_IRLS.m,” presented in Table 7. Afterwards, the algorithm of the method “IRLS.m,” Table 8, is implemented for the rational approximation of the selected function.\n\n %=======================================================% Main program of ITERATIVELY REWEIGHTED LEAST SQUARES% Authors: Eduardo Salvador Bañuelos Cabral% José Alberto Gutiérrez Robles% Bjørn Gustavsen%=======================================================clcclear allclose all%% Initial SettingsNs = 500; % Number of samplesf = logspace(-2,6,Ns); % Frequency (Hz)w = 2.*pi.*f; % Frequency (rad/seg)s = 1i*w; % Complex Frequency%% Synthetic functionschoice = menu('CHOOSE A FUNCTION','Function F1(s)',… 'Function F2(s)','Function F3(s)','Function F4(s)',… 'Function F5(s)','Function F6(s)');if choice == 1 Np = 12; Fs =(0.83*s.^12-0.35*s.^11+0.32*s.^10+60000*s.^9+600.36*s.^8+650.23*s.^7+… 54.5*s.^6+45.6*s.^5 +0.1*s.^4-1240*s.^3+80025*s.^2+5547.*s+5008)./… (s.^12-0.89*s.^11-0.23*s.^10-4565.*s.^9+34569.02*s.^8-55450*s.^7+… 60*s.^6+1005*s.^5+8050*s.^4+7502*s.^3+10*s.^2+1014.*s-50);endif choice == 2 Np = 9; Fs = (6000*s.^9+600*s.^8+650*s.^7+54*s.^6+457*s.^5+100*s.^4+1240*s.^3+8000*s.^2+… 55400.*s+5000)./(450*s.^9+3500*s.^8+500*s.^7-600*s.^6+1000*s.^5+700*s.^4-… 7500*s.^3+1000*s.^2+500.*s);endif choice == 3 Np = 7;Fs =(514*s.^7-364.25*s.^6-0.35*s.^5+635*s.^4-20802*s.^3+5304*s.^2+4520.*s-… 18020.025 )./(241*s.^7-32.02*s.^6+538.23*s.^5-2588*s.^4-22560*s.^3-… 604*s.^2+4150.*(s)-21052.31 );endif choice == 4 Np = 3; Ps = [-0.52; -0.12; -0.04]; Rs = [-2.18; -7192.25; 20.58]; Ks = 5; Fs = zeros(1,Ns);for k = 1:length(Ps) Fs = Fs + (Rs(k)./(s.' - Ps(k))).';endFs = Fs + Ks;endif choice == 5 Np = 11; Ps = 1.0e+004.*[-.88; -5.42; - 27.28; 0.88; -9665; 235.7; … -0.26; -3.87; 46.32; -0.13; -3756 ]; Rs = 1.0e+005.*[834; 22593; 893; 2653; 654; 32; … 44.14; 6405; 136.79; 0.12; 125]; Ks = 0; Fs = zeros(1,Ns); for k = 1:length(Ps) Fs = Fs + (Rs(k)./(s.' - Ps(k))).'; end Fs = Fs + Ks;endif choice == 6 Np = 18; Ps = [-4500; -41000; -100+1i*5000; -100-1i*5000; -120+1i*15000; -120-1i*15000; … -3000+1i*35000; -3000-1i*35000; -200+1i*45000; -200-1i*45000; -1500+1i*45000;… -1500-1i*45000; -500+1i*70000; -500-1i*70000; -1000+1i*73000; -1000-1i*73000;… -2000+1i*90000; -2000-1i*90000]; Rs = [-3000; -83000; -5+1i*7000; -5-1i*7000; -20+1i*18000; -20-1i*18000;… 6000 + 1i*45000; 6000-1i*45000; 40+1i*60000; 40-1i*60000; 90+1i*10000;… 90-1i*10000; 50000+1i*80000; 50000-1i*80000; 1000+1i*45000; 1000-1i*45000;… -5000+1i*92000; -5000-1i*92000]; Ks = 0.2; Fs = zeros(1,Ns); for k = 1:length(Ps) Fs = Fs + (Rs(k)./(s.' - Ps(k))).';endFs = Fs + Ks;end%% Iteratively Reweighted Least Squaresfw = 1; % weighting function (1-6)[Ks,Rs,Ps,x,A1,an,bn,k1,Err] = IRLS(Np,Ns,Fs,s,fw);% Fitting using partial fractionsFs_fit = zeros(1,Ns);for i = 1:Np Fs_fit = Fs_fit + (Rs(i)./(s.' - Ps(i))).';endFs_fit = Fs_fit + Ks ;e = abs((Fs.' - Fs_fit.')); % Deviation of the fitting process%% Plotsfigure(1)semilogy(Err(1:length(Err)),'-b','LineWidth',2)xlabel('Itetation count');ylabel('RMS-error')figure(2)loglog(f,abs(Fs),'-k',f,abs(Fs_fit),'‐‐r',f,e,'‐‐b','LineWidth',2),legend('Data','IRLS','Deviation')xlabel('Frequency [Hz]'); ylabel('Magnitude [p.u.]');\n\n### Table 7.\n\n“Fitting_IRLS.m”.\n\n %=====================================================% ITERATIVELY REWEIGHTED LEAST SQUARES METHOD% Authors: Eduardo Salvador Bañuelos Cabral% José Alberto Gutiérrez Robles% Bjørn Gustavsen%=====================================================% Inputs % ‐‐‐ Np, number of poles % ‐‐‐ Ns, number of samples % ‐‐‐ Fs, function to be fitted % ‐‐‐ s, complex frequency (rad/s) % ‐‐‐ fw, weighting function% Outputs % ‐‐‐ Ks, constant term % ‐‐‐ Rs, residues % ‐‐‐ Ps, poles % ‐‐‐ x, coefficients vector (polynomials) % ‐‐‐ A1, matrix to evaluate the fitting % ‐‐‐ an, numerator coefficients % ‐‐‐ bn, denominator coefficients % ‐‐‐ k1, minimum error % ‐‐‐ Error, RMS-errorfunction [Ks,Rs,Ps,x,A1,an,bn,k1,Error] = IRLS(Np,Ns,Fs,s,fw)a = ones(Ns,1); % Vector columna1 = zeros(Ns,Np); % Matrix size (s^n)a2 = zeros(Ns,Np); % Matrix size (Fs*s^n)An = zeros(2*Ns,1+2*Np); % Matrix size considering real and imaginary partBn = zeros(2*Ns,1); % Vector size considering real and imaginary partEuclidian = zeros(1,1+2*Np); % Vector sizeW = eye(length(Ns*2)); % Weighting matrixb = Fs.'; % Construction of vector b.for n = 1:1:Np % Construction of matrix A. a1(1:Ns,n) = s.^((Np+1)-n); a2(1:Ns,n) = -b.*a1(1:Ns,n);endA1 = [a1 a]; % Matrix to evaluate the fittingA = [a1 a a2]; % Matriz of the system[Xmax Ymax] = size(A); % Size for matrix AAr = real(A); % Real part of matrix AAi = imag(A); % Imaginary part of matrix Abr = real(b); % Real part of vector Bbi = imag(b); % Imaginary part of vector B% Construction of matrix An and vector Bn (interleaved)km = 1;for k = 2:2:2*Xmax An(k-1,:) = Ar(km,:); An(k,:) = Ai(km,:); Bn(k-1,1) = br(km); Bn(k,1) = bi(km); km = km+1;end[Q,R] = qr(An,0);% Matrix Q and R of AnAt = R; % It updates matrix AnB = Q.'*Bn; % It updates vector Bnfor col = 1:Ymax % Applying the Euclidian norm to At Euclidian(col) = norm(At(:,col),2); At(:,col) = At(:,col)./Euclidian(col);endxn(:,1) = At\\B; % Solution for the system (Ax = B)xn(:,1) = xn(:,1)./Euclidian.'; % Real solution% weighting iterativeError(1) = 1e2; % Initial errorDisc = 1e2; % Discriminatingn=1; % Counter to storage de errork0=1; % Counter of iterationswhile Disc < 0 || Error(n) > 1e-3 || k0 < 13 an = xn(1:Np+1,k0); % Numerator coefficients bn = [xn(Np+2:2*Np+1,k0);1]; % Denominator coefficients Fsfit = (A1*an)./(A1*bn); % Fitting evaluation vres = (Fsfit - b); % Direct error n=n+1; % Counter to storage de error Error(n)=sum(abs(vres)); % Error Disc = Error(n)-Error(n-1); % Discriminating vres_r = real(vres); % Real part of the vector vres vres_i = imag(vres); % Imaginary part of the vector vres km = 1; %(interleaved for k = 2:2:2*Xmax res(k-1,1) = vres_r(km); res(k,1) = vres_i(km); km = km+1; end Ds = std(res); % Standard deviation res = res./Ds; % Vector \"res\" with standard deviation % WEIGHING FUNCTIONS if fw ==1, w = diag(W).*abs(res) + 1e-130; end if fw ==2, w = diag(W).*(sqrt(1+((res.^2)))-1) + 1e-130; end if fw ==3, w = diag(W).*1./(abs(res).^(1.2-2)) + 1e-130; end if fw ==4, w = diag(W).*((res).^2)./(1+(res).^2) + 1e-130; end if fw ==5, w = diag(W).*((res).^2)./sqrt(1+(res).^2)+ 1e-130; end if fw ==6, w = diag(W).*(abs(res).^(1.2))./1.2 + 1e-130; end W = diag(w); % It updates the matrix W whit weighting functions Bn_p = W*Bn; % Weighting An An_p = W*An; % Weighting Bn [Q,R] = qr(An_p,0); % Matrix Q and R of An_p At = R; % It updates the matrix An_p B = Q.'*Bn_p; % It updates the matrix Bn_p for col = 1:Ymax % Euclidian norm Euclidian(col) = norm(At(:,col),2); At(:,col) = At(:,col)./Euclidian(col); end k0=k0+1; % Number of iterations xn(:,k0) = At\\B; % Solution for the system (Ax = b) xn(:,k0) = xn(:,k0)./Euclidian.'; % Real solution if k0 > 20, break, end % End the while, if k0 > 20end[k1,k2]=min(Error); % Position of the minimum errorx=xn(:,k2); % Coefficients with minimum error% Poles, residues and constant terman = x(1:Np+1); % Numerator coefficientsbn = [x(Np+2:2*Np+1);1]; % Denominator coefficients[Rs,Ps,Ks] = residue(an,bn);TF = isempty(Ks);if (TF==1) Ks = 0;end\n\n### Table 8.\n\n“IRLS.m”.\n\nIn Figure 10 the synthetic frequency response behavior and the rational approximation given by IRLS is presented, together with its deviation in terms of absolute error. Moreover, the RMS-error on each iteration is shown in Figure 11.",
null,
"Figure 10.Rational approximation given by the IRLS method for the synthetic function.",
null,
"Figure 11.RMS-error on each iteration given by the IRLS method.\n\n### 4.4. LM algorithm\n\nIn this section the LM routine is introduced. First, a synthetic function can be selected in the main program “Fitting_LM_Polynomials.m,” presented in Table 9. Afterwards, the algorithm of the method “LM_method.m,” Table 10, is implemented for the rational approximation of the selected function.\n\n %=======================================================% Main program of LEVENBERG-MARQUARDT (Polynomials)% Authors: Eduardo Salvador Bañuelos Cabral% José Alberto Gutiérrez Robles% Bjørn Gustavsen%=======================================================clcclear allclose all%% Initial SettingsNs = 500; % Number of samplesf = logspace(-2,8,Ns); % Frequency \"Hertz\"w = 2.*pi.*f; % Frequency \"rad/seg\"s = 1i*w; % Complex Frequency%% Synthetic functionschoice = menu('CHOOSE A FUNCTION','Function F1(s)','Function F2(s)',…'Function F3(s)','Function F4(s)','Function F5(s)');if choice == 1 ite = 150; % Number of iterations Np = 7; % Order of the proposed function Fs = (-364*s.^6-s.^5+635*s.^4-20802*s.^3+5304*s.^2+4520.*s-18020)./… (241*s.^7-32*s.^6+538*s.^5-2588*s.^4-22560*s.^3-604*s.^2+4150.*s-21052);endif choice == 2 ite = 50; % Number of iterations Np = 12; % Order of the proposed function Fs = (s.^12-0.4*s.^11+0.3*s.^10+60003*s.^9+600*s.^8+650*s.^7+54*s.^6+45*s.^5+… 0.1*s.^4-1240*s.^3+80025*s.^2+5547*s+5008)./(s.^12-0.9*s.^11-0.2*s.^10-… 4565*s.^9+34569*s.^8-55450*s.^7+60*s.^6+1005*s.^5+8050*s.^4+7502*s.^3+… 10*s.^2+1014*s-50);endif choice == 3 ite = 50; % Number of iterations Np = 14; % Order of the proposed function Fs = 2.*pi*(652*s.^14-0.8*s.^13-0.8*s.^12-0.4*s.^11+25487*s.^10+600003*s.^9+… 6040*s.^8-60050*s.^7+54*s.^6+45*s.^5+0.1*s.^4-1240*s.^3+8025*s.^2+5547*s+508)./… (2365*s.^14-8362*s.^13-8*s.^12-0.9*s.^11-65547*s.^10-4565*s.^9+34569*s.^8-… 55450*s.^7+60*s.^6-1005*s.^5+850*s.^4-7502*s.^3-10*s.^2-1014*s+75230);endif choice == 4 ite = 50; % Number of iterations Np = 3; % Order of the proposed function Ps = [-0.5; -0.1; -0.05]; Rs = [-2.2; -7192; 20]; % Process to obtain Fs from poles and residues Ks = 5; Fs = zeros(1,Ns); for k = 1:length(Ps) Fs = Fs + (Rs(k)./(s.' - Ps(k))).'; end Fs = Fs + Ks;endif choice == 5 ite = 100; % Number of iterations Np = 11; % Order of the proposed function Ps = 1.0e4*[-0.9;-5.4;-27.3;0.9;-9665;235;-0.3;-3.9;46.3;-0.1;-3756]; Rs = 1.0e5*[834;22593;893;2653;654;32;44;6405;136;0.1;125]; % Process to obtain Fs from poles and residues Ks = 0; Fs = zeros(1,Ns); for k = 1:length(Ps) Fs = Fs + (Rs(k)./(s.' - Ps(k))).'; end Fs = Fs + Ks;end%% Levenberg Marquardt[Ks,Rs,Ps,x,A1,an,bn,Ff] = LM_method(Np,Ns,Fs,s,ite);% Fitting using partial fractionsFs_fit = zeros(1,Ns);for i = 1:Np Fs_fit = Fs_fit + (Rs(i)./(s.' - Ps(i))).';endFs_fit = Fs_fit + Ks ;% Deviation in the fitted processerror = abs((Fs.' - Fs_fit.' ));%% Plotsfigure(1)loglog(f,abs(Fs),'-k',f,abs(Fs_fit),'-.r',f,error,'-.b','LineWidth',2);legend('Data','LM (Polynomials)','Deviation');xlabel('Frequency [Hz]'); ylabel('Magnitude [p.u.]');figure(2)semilogy(Ff,'‐‐b','LineWidth',2);xlabel('Iteration count');ylabel('RMS-error');\n\n### Table 9.\n\n“Fitting_LM_Polynomials.m”.\n\n %==========================================================% Function to implement LEVENBERG-MARQUARDT (Polynomials)% Authors: Eduardo Salvador Bañuelos Cabral% José Alberto Gutiérrez Robles% Bjørn Gustavsen%==========================================================% Inputs % ‐‐‐ Np, number of poles % ‐‐‐ Ns, number of samples % ‐‐‐ Fs, function to be fitted % ‐‐‐ s, complex frequency (rad/s) % ‐‐‐ ite, number of iterations% Outputs % ‐‐‐ Ks, constant term % ‐‐‐ Rs, residues % ‐‐‐ Ps, poles % ‐‐‐ x, vector of coefficients (polynomials) % ‐‐‐ A1, matrix to evaluate the fitting % ‐‐‐ an, coefficients of the numerator % ‐‐‐ bn, coefficients of the denominator % ‐‐‐ Ff, objective function (deviation)function [Ks,Rs,Ps,x,A1,an,bn,Ff] = LM_method(Np,Ns,Fs,s,ite)x = ones((2*Np)+1,1); % Initial valuesa = ones(Ns,1); % Vector columna1 = zeros(Ns,Np); % Matrix size (s^n)J1 = zeros(Ns,Np+1); % Matrix size (J1)J2 = zeros(Ns,Np); % Matrix size (J1)At = zeros(2*Np+1,2*Np+1); % Matrix size (At)en = zeros(2*Ns,1); % Vector size (en)epn = zeros(2*Ns,1); % Vector size (enp)Ff = zeros(ite,1); % Vector size (Ff)Euclidian = zeros(1,1+2*Np); % Euclidian norm% Construction of the matrix A1for n = 1:1:Np a1(1:Ns,n) = s.^((Np+1)-n);endA1 = [a1 a];v = 2; % LM coefficient (m = m*v)u = 5; % LM coefficient (m = m/u)% Levenberg-Marquardt methodfor ki = 1:ite % Main loop x1 = x(1:Np+1); % Coefficients of the numerator x2 = [x(Np+2:2*Np+1);1]; % Coefficients of the denominator Yg = (A1*x1)./(A1*x2); % Fitting evaluation error = (Yg - Fs.'); % Deviation % Process to obtain the Jacobian (J) for k = 1:1:Np J1(:,k) = (s.^((Np+1)-k))./((A1*x2).'); J2(:,k) = (-(A1*x1).'.*s.^((Np+1)-k))./((A1*x2).^2).'; end J1(:,Np+1) = 1./((A1*x2).'); J = [J1 J2]; [Xmax Ymax] = size(J); % Matrix size (J) Jr = real(J); % Real part of vector J Ji = imag(J); % Imaginary part of vector J er = real(error); % Real part of vector e ei = imag(error); % Imaginary part of vector e km = 1; % Interleaved for k = 2:2:2*Xmax Jn(k-1,:) = Jr(km,:); Jn(k,:) = Ji(km,:); en(k-1,1) = er(km); en(k,1) = ei(km); km = km+1; end F = ((norm(en,2)^2)); % Objective function tends to zero to converge Ff(ki,1) = F; % Storage the RMS-error (objective function) [Q,R] = qr(Jn); % Matrix Q and R of Jn Jn = R; % It updates matrix Jn Gf = (Jn.'*Q.')*en; % Gradient (QR decomposition) Hess = Jn.'*Jn; % Hessian (QR decomposition) if ki == 1 % Inicial value for m m = max(diag(Hess)); end paro = 0; % Variable to enter into the while while (paro==0) I = diag(diag(Hess)); % Construction of matrix I, diagonal of Hessian Hess_mod = (Hess + m*I); % Modified Hessian for col = 1:Ymax % Loop to normalize At Euclidian(col) = norm(Hess_mod(:,col),2); % Euclidian norm At(:,col) = Hess_mod(:,col)./Euclidian(col); end h = (At)\\-Gf; % Solution for the system (Ax = b) h = h./Euclidian.'; % Real solution xnew = x + h; % New coefficients (without updating x) % == Updated coefficients are evaluated, if it meets the assessment then x is updated == % x1new = xnew(1:Np+1); % Coefficients of the numerator x2new = [xnew(Np+2:2*Np+1);1]; % Coefficients of the denominator Ygnew = (A1*x1new)./(A1*x2new); % Fitting evaluation ep = (Ygnew - Fs.'); % Deviation epr = real(ep); % Real part of vector ep epi = imag(ep); % Imaginary part of vector ep km = 1; % Interleaved for k = 2:2:2*Xmax epn(k-1,1) = epr(km); epn(k,1) = epi(km); km = km+1; end Fp = ((norm(epn,2)^2)); % Objective function % Updating of m if (F < Fp) m = m*v; end if (F >= Fp) % Condition to go out the while x = x + h; m = m/u; break; end endendan = x(1:Np+1); % Coefficients of the numeratorbn = [x(Np+2:2*Np+1);1]; % Coefficients of the denominator% Poles, residues and constant term[Rs,Ps,Ks] = residue(an,bn);TF = isempty(Ks);if (TF==1) Ks = 0;end\n\n### Table 10.\n\n“LM_method.m”.\n\nIn Figure 12 the synthetic frequency response behavior and the rational approximation given by IRLS is presented, together with its deviation in terms of absolute error. Moreover, the RMS-error on each iteration is shown in Figure 13.\n\n### 4.5. VF algorithm\n\nVF has become one of the main methodologies for the rational approximation of frequency domain responses, the MATLAB routine is available online , “vectfit3.m.” Here, we present an example of its application to the rational approximation of a synthetic function.\n\nIn the beginning of the implementation a synthetic function is selected in the main program “Fitting_VF.m,” presented in Table 11. Afterwards, initial poles for the fitting must be established “InitialPoles.m,” Table 12, for the VF routine. Finally in “VF.m,” VF is used (via vectfit3.m) to perform the rational approximation of the selected function. In VF.m routine, Table 13, the parameter opts.relax is set to 1. This implies that “relaxation” is being used.\n\n %=====================================================================% Main program of VECTOR FITTING% (vectfit3 available in:https://www.sintef.no/projectweb/vectfit/% Author: Bjørn Gustavsen)% Authors: Eduardo Salvador Bañuelos Cabral% José Alberto Gutiérrez Robles% Bjørn Gustavsen%====================================================================clcclear allclose all%% Initial SettingsNs = 500; % Number of samplesf = logspace(-2,6,Ns); % Frequency (Hz)w = 2.*pi.*f; % Frequency (rad/seg)s = 1i*w; % Complex Frequency%% Synthetic functionschoice = menu('CHOOSE A FUNCTION','Function F1(s)',… 'Function F2(s)','Function F3(s)','Function F4(s)');if choice == 1 Np = 12; Fs =(0.83*s.^12-0.35*s.^11+0.32*s.^10+60000*s.^9+600.36*s.^8+650.23*s.^7+… 54.5*s.^6+45.6*s.^5 +0.1*s.^4-1240*s.^3+80025*s.^2+5547.*s+5008)./… (s.^12-0.89*s.^11-0.23*s.^10-4565.*s.^9+34569.02*s.^8-55450*s.^7+60*s.^6+… 1005*s.^5+8050*s.^4+7502*s.^3+10*s.^2+1014.*s-50);endif choice == 2 Np = 9; Fs = (6000*s.^9+600*s.^8+650*s.^7+54*s.^6+457*s.^5+100*s.^4+1240*s.^3+8000*s.^2+… 55400.*s+5000 )./(450*s.^9+3500*s.^8+500*s.^7-600*s.^6+1000*s.^5+… 700*s.^4-7500*s.^3+1000*s.^2+500.*s);endif choice == 3 Np = 7; Fs =(514*s.^7-364.25*s.^6-0.35*s.^5+635*s.^4-20802*s.^3+5304*s.^2+4520.*s-… 18020.025 )./(241*s.^7-32.02*s.^6+538.23*s.^5-2588*s.^4-22560*s.^3-… 604*s.^2+4150.*(s)-21052.31 );endif choice == 4 Np = 18; Ps = 2*pi*[-4500; -41000; -100+1i*5000; -100-1i*5000; -120+1i*15000; … -120-1i*15000; -3000+1i*35000; -3000-1i*35000; -200+1i*45000; -200-1i*45000;… -1500+1i*45000; -1500-1i*45000;-500+1i*70000; -500-1i*70000; -1000+1i*73000;… -1000-1i*73000; -2000+1i*90000; -2000-1i*90000]; Rs = [-3000; -83000; -5+1i*7000; -5-1i*7000; -20+1i*18000; -20-1i*18000;… 6000+1i*45000; 6000-1i*45000; 40+1i*60000; 40-1i*60000; 90+1i*10000; … 90-1i*10000; 50000+1i*80000; 50000-1i*80000; 1000+1i*45000; 1000-1i*45000;… -5000+1i*92000; -5000-1i*92000]; Ks = 0.2; Fs = zeros(1,Ns); for k = 1:length(Ps) Fs = Fs + (Rs(k)./(s.' - Ps(k))).'; end Fs = Fs + Ks ;end%% Vector Fitting Method[Ps]=InitialPoles(f,Np); %Initial poles subroutine[P,C,D,E,fVF] = VF(Ps,Ns,Fs,s,20,3); % Vector Fitting MethodFsfit = zeros(1,Ns);for k = 1:length(P) Fsfit = Fsfit + (C(k)./(s.' - P(k))).';endFsfit = Fsfit + D+ E.*s; % Fitting resulteVF = abs(Fs - Fsfit); % Deviationfigure(1)loglog(f,abs(Fs),'-k',f,abs(Fsfit),'‐‐r',f,abs(eVF),'b','LineWidth',2);xlabel('Frequency [Hz]'); ylabel('Magnitude [p.u.]')legend('Data','VF','Deviation')\n\n### Table 11.\n\n“Fitting_VF.m”.\n\n %==========================================================% Function to implement INITIAL POLES for Vector Fitting% Authors: Eduardo Salvador Bañuelos Cabral% José Alberto Gutiérrez Robles% Bjørn Gustavsen%==========================================================% Inputs % ‐‐‐ f, frequency % ‐‐‐ Npol, number of poles% Outputs % ‐‐‐ Ps, polesfunction [Ps] = InitialPoles(f,Npol)% Set the initial poles (even, odd)even = fix(Npol/2);p_odd = Npol/2 - even;disc = p_odd ~= 0;% Set a real pole in case of disc == 1if disc == 0 % Even initial poles pols = [];else % Odd initial poles pols = [(max(f)-min(f))/2];end% Initial complex polesbet = linspace(min(f),max(f),even);for n=1:length(bet) alf=-bet(n)*1e-2; pols=[pols (alf-1j*bet(n)) (alf+1j*bet(n)) ];endPs = pols.'; % Column vector of initial poles\n\n### Table 12.\n\n“InitialPoles.m”.\n\n %=====================================================================% Function to use VECTOR FITTING% (vectfit3.m available in:https://www.sintef.no/projectweb/vectfit/% Author: Bjørn Gustavsen)% Authors: Eduardo Salvador Bañuelos Cabral% José Alberto Gutiérrez Robles% Bjørn Gustavsen%=====================================================================% Inputs % ‐‐‐ Ps, initial poles % ‐‐‐ Ns, number of samples % ‐‐‐ Fs, function to be fitted % ‐‐‐ s, complex frequency (rad/s) % ‐‐‐ ite, iterations % ‐‐‐ Ka, kind of fitting% Outputs % ‐‐‐ P, poles % ‐‐‐ C, residues % ‐‐‐ D, constant term % ‐‐‐ E. % ‐‐‐ RMS.function [P,C,D,E,RMS] = VF(Ps,Ns,Fs,s,Ni,Ka)weight=ones(1,Ns); % Vector of weightsopts.relax=1; % Use vector fitting with relaxed non-triviality constraintopts.stable=0; % Enforce stable polesopts.asymp=Ka; % Include both D, E in fittingopts.skip_pole=0; % Do NOT skip pole identificationopts.skip_res=0; % DO skip identification of residues (C,D,E)opts.cmplx_ss=1; % Create real-only state space modelopts.spy1=0; % No plotting for first stage of vector fittingopts.spy2=0; % Create magnitude plot for fitting of f(s)opts.logx=1; % Use linear abscissa axisopts.logy=1; % Use logarithmic ordinate axisopts.errplot=1; % Include deviation in magnitude plotopts.phaseplot=0; % Do NOT produce plot of phase angleopts.legend=1; % Include legends in plotsfor k = 1:Ni % Loop to make N-iterations [SER,Ps,rmserr,fit] = vectfit3(Fs,s,Ps,weight,opts); RMS(k) = rmserr;endP = Ps; % PolesC = SER.C; % ResiduesD = SER.D; % Constant termE = SER.E; % Proportional term\n\n### Table 13.\n\n“VF.m”.\n\nFigure 14 shows the synthetic frequency response behavior and the rational approximation given by VF, together with its deviation in terms of absolute error.",
null,
"Figure 14.Synthetic frequency response data and the rational approximation given by VF.\n\n### 4.6. Single-phase transmission line modeling\n\nBode, Levy or LS, IRLS, VF, and LM are applied to the rational approximation of the frequency-dependent parameters corresponding to a single-phase transmission line. In Figure 15a, the diagram of the equivalent circuit for the test with L = 70 mH and Ri = 1.058 Ω is shown. The current source is considered as an ideal 60 Hz. The line length (l) is 100 km, with a diameter of 2.7 cm. The conductor is placed horizontally at a height of 20 m, with a 100 Ωm resistivity. The termination impedance at the end of line (Rl) is placed as 1e6 Ω, 1e-6 Ω, and 462 Ω for evaluation of the different line end cases.",
null,
"Figure 15.(a) Diagram of the equivalent circuit for the test, (b) behavior of Z(ω) and Y(ω).\n\nThe modeling consists of calculating rational approximations for the line series impedance Z and line shunt admittance Y by applying the aforementioned rational fitting techniques. Figure 15b shows the behavior of Z and Y as a function of frequency, calculated with 16,384 linearly spaced samples and Figure 16 shows a flowchart for the implementation of this test case.\n\nFigure 17a shows the results for the fitting of Y with Np = 3, in terms of model deviations (relative error) by Bode, Levy, IRLS, VF, and LM. It can be observed that the best approximations are obtained with Levy, IRLS, LM and VF which converge to practically the same deviation.",
null,
"Figure 17.(a) Relative errors for the rational approximations of Y(ω) and (b) relative errors for the rational approximations of Z(ω).\n\nFigure 17b shows the results for the fitting of Z with Np = 14, in terms of the model deviations (relative error) by Bode, Levy, IRLS, VF, and LM. It can be seen that the best approximation is obtained with VF, IRLS, and LM.\n\nThen, the effect of the modeling errors in Y and Z on a time domain response is evaluated. The line terminal nodal admittance matrix Yn is established with respect to the two line ends 2 and 3 of Figure 15a as\n\nI2I3=YnV2V3E70\n\nwhere Yn is given by\n\nYn=YcAYcBYcBYcAE71\n\nand Yc, A and B by\n\nYcω=Zω1ZωYωE72\nAω=cothZωYωlandBω=cschZωYωlE73\n\nBy using the Numerical inversion of Laplace Transform (NLT) [2, 12], the voltage responses on time domain (v1, v2 and v3) are calculated for the cases of open-ended, short-circuited, and perfectly matched lines end. These results are considered as a reference solution.\n\nThe rational function-based models obtained with Bode, Levy, IRLS, VF, and LM are used to calculate Z and Y. Using a similar procedure as mentioned for each technique, the voltage responses (v1, v2 and v3) are calculated through the NLT.\n\nThe objective is to calculate the error in the time domain introduced by the rational approximations of Z and Y in frequency domain, taking into account the reference solution.\n\nThe comparative results are shown in Figures 1820 for the cases of short-circuited, perfectly matched, and open-ended lines end, respectively. The absolute errors are consistent with the deviations for the rational approximation of Z.",
null,
"Figure 18.(a) Time domain simulation of the voltage response at the short-circuited line end, (b) absolute error with respect to the reference solution (v3).",
null,
"Figure 19.(a) Time domain simulation of the voltage response at the matched line end, (b) absolute error with respect to the reference solution (v3).",
null,
"Figure 20.(a) Time domain simulation of the voltage response at the open line end, (b) absolute error with respect to the reference solution (v3).\n\n## 5. Conclusions\n\nIn this book, Bode, Levy, Noda, SK, IRLS, VF, LM, and DGN methods have been described in detail for complex-curve fitting, and a routine implemented in MATLAB environment presented for each technique. Rational approximation of synthetic frequency responses have used to show the operation of the programs. Moreover, the techniques are used to approximate the series line impedance (Z) and the shunt line admittance (Y) corresponding to a single-phase transmission line. The main conclusions are:\n\n1. Asymptotic Approximation or Bode has proved to be a reliable technique for the modeling of overhead transmission lines. Also, this method can realize the fitting on the magnitude of the function and uses only real poles and zeros. However, the error level in the approximation can be substantial and it depends on the sensitivity criterion used when inserting new poles and zeros in the fitting.\n\n2. The Levy method is pioneered for complex-curve fitting. Many techniques have been developed based on this methodology. Nevertheless, the method is biased, because it weights the fitting on high frequencies too much; this fact is demonstrated in the test cases.\n\n3. The IRLS method is an accurate technique for the rational fitting of frequency responses. This method permits the implementation of different weighting functions in order to improve the level error in the approximation. Its disadvantage lies in the numerical ill-conditioning encountered in approximations with a wide frequency range.\n\n4. The VF method is a robust and accurate technique. This fact has positioned this methodology as one of the most important in this field.\n\n5. The LM method is a technique that shows good results in terms of level error. An advantage is that it can be implemented in pole-zero form, pole-residue form, and polynomial form.\n\n6. The main disadvantage for the optimization techniques like LM and DGN is that they can get stuck in a local minimum.\n\n7. The rational approximation for Z and Y in the single-phase transmission line modeling shows that the same technique does not always deliver the same result. Levy, LM and IRLS delivers more accurate result for the fitting of Y while VF reach the best result for the fitting of Z. The error levels obtained in time domain simulations are consistent with the fitting of Z, because in the modeling of overhead transmission lines, these parameters are more relevant, therefore VF reach the best level error.\n\ncompact PDF\nCitations in RIS format\nCitations in bibtex format\n\n## More\n\n© 2017 The Author(s). Licensee IntechOpen. This compact is distributed under the terms of the Creative Commons Attribution-NonCommercial 4.0 License, which permits use, distribution and reproduction for non-commercial purposes, provided the original is properly cited.\n\n## How to cite and reference\n\n### Cite this compact Copy to clipboard\n\nEduardo Salvador Bañuelos-Cabral, José Alberto Gutiérrez-Robles and Bjørn Gustavsen (December 6th 2017). Rational Fitting Techniques for the Modeling of Electric Power Components and Systems Using MATLAB Environment, Rational Fitting Techniques for the Modeling of Electric Power Components and Systems Using MATLAB Environment, Eduardo Salvador Banuelos- Cabral, Jose Alberto Gutierrez-Robles and Bjorn Gustavsen, IntechOpen, DOI: 10.5772/intechopen.71358. Available from:\n\n### Related Content\n\nNext compact\n\n#### Rational Fitting Techniques for the Modeling of Electric Power Components and Systems Using MATLAB Environment\n\nBy Eduardo Salvador Bañuelos-Cabral, José Alberto Gutiérrez-Robles and Bjørn Gustavsen\n\nFirst chapter\n\n#### Synthesis Processes for Li-Ion Battery Electrodes – From Solid State Reaction to Solvothermal Self-Assembly Methods\n\nBy Verónica Palomares and Teófilo Rojo\n\nWe are IntechOpen, the world's leading publisher of Open Access books. Built by scientists, for scientists. Our readership spans scientists, professors, researchers, librarians, and students, as well as business professionals. We share our knowledge and peer-reveiwed research papers with libraries, scientific and engineering societies, and also work with corporate R&D departments and government entities."
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http://mymathforum.com/algebra/47553-equation-has-me-stumped.html | [
"User Name Remember Me? Password\n\n Algebra Pre-Algebra and Basic Algebra Math Forum\n\n November 2nd, 2014, 09:53 PM #1 Newbie Joined: Jun 2014 From: U.S.A Posts: 13 Thanks: 0 This equation has me stumped",
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"This is what I basically did: 25-x^2=(x+5)(x-5) which means multiply 6 by x-5 = 6x - 30. Multiply -9 by X+5 = -9x - 45. Combine them gets me -75-3x=8x. Add 3 gets 11x/-75. giving me a final answer of -75/11 What did i do wrong?!?! Last edited by HelpMeNow; November 2nd, 2014 at 10:14 PM.",
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"November 2nd, 2014, 10:17 PM #2\nSenior Member\n\nJoined: Jan 2012\nFrom: Erewhon\n\nPosts: 245\nThanks: 112\n\nQuote:\n Originally Posted by HelpMeNow",
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"",
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"This is what I basically did: 25-x^2=(x+5)(x-5) which means multiply 6 by x-5 = 6x - 30. Multiply -9 by X+5 = -9x - 45. Combine them gets me -75-3x=8x. Add 3 gets 11x/-75. What did i do wrong?!?!\n$$\\frac{6}{x+5}-\\frac{9}{x-5}=\\frac{6(x-5)-9(x+5)}{x^2-25}=\\frac{-3x-75}{x^2-25}$$\n\nSo:\n$$\\frac{-3x-75}{x^2-25}=\\frac{8x}{25-x^2}=\\frac{-8x}{x^2-25}$$\n\nSo if $x^2-25 \\ne 0$, you have $-3x-75=-8x$\n\nCB",
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"November 2nd, 2014, 10:45 PM #3\nNewbie\n\nJoined: Jun 2014\nFrom: U.S.A\n\nPosts: 13\nThanks: 0\n\nQuote:\n Originally Posted by CaptainBlack",
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"$$\\frac{6}{x+5}-\\frac{9}{x-5}=\\frac{6(x-5)-9(x+5)}{x^2-25}=\\frac{-3x-75}{x^2-25}$$ So: $$\\frac{-3x-75}{x^2-25}=\\frac{8x}{25-x^2}=\\frac{-8x}{x^2-25}$$ So if $x^2-25 \\ne 0$, you have $-3x-75=-8x$ CB\nWhat?",
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"November 2nd, 2014, 11:13 PM #4\nSenior Member\n\nJoined: Apr 2013\n\nPosts: 425\nThanks: 24\n\nQuote:\n Originally Posted by HelpMeNow",
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"What?\nWhat \"What\"????",
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"",
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"November 3rd, 2014, 01:38 AM #5\nSenior Member\n\nJoined: Jan 2012\nFrom: Erewhon\n\nPosts: 245\nThanks: 112\n\nQuote:\n Originally Posted by HelpMeNow",
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"What?\nYour explanation of what you did is incomprehensible, as far as I can tell you lost a minus sign in the wash and you did not observe that you had to assume that $x^2\\ne 5$ as otherwise both sides of the equation you are trying to solve are undefined.\n\nCB",
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"November 3rd, 2014, 02:35 AM #6\nMath Team\n\nJoined: Apr 2010\n\nPosts: 2,780\nThanks: 361\n\nQuote:\n Originally Posted by CaptainBlack you had to assume that $\\displaystyle x^2\\ne 5$\n(Presumably, you meant $\\displaystyle x^2\\ne 25$)",
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"November 3rd, 2014, 04:56 AM #7\nSenior Member\n\nJoined: Jan 2012\nFrom: Erewhon\n\nPosts: 245\nThanks: 112\n\nQuote:\n Originally Posted by Hoempa",
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"(Presumably, you meant $\\displaystyle x^2\\ne 25$)\nYes, it was/is correct in the original response.\n\nCB",
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"November 3rd, 2014, 05:00 AM #8\nSenior Member\n\nJoined: Apr 2014\nFrom: Glasgow\n\nPosts: 2,157\nThanks: 732\n\nMath Focus: Physics, mathematical modelling, numerical and computational solutions\nQuote:\n Originally Posted by HelpMeNow",
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"What?\n$\\displaystyle \\frac{8x}{25 - x^2} = \\frac{8x}{-(x^2 - 25)} = \\frac{-8x}{x^2 - 25}$\n\nThis is so that the denominators on the LHS and RHS are exactly equivalent, so you can equate the numerators. Consequently, you have $\\displaystyle -3x - 75 = -8x$ instead of $\\displaystyle -3x-75 = 8x$.\n\nLast edited by Benit13; November 3rd, 2014 at 05:02 AM.",
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https://ask.sagemath.org/question/9172/plot-series-of-3d-direction-vectors-not-all-from-origen/?answer=13840 | [
"# Plot Series of 3D Direction Vectors (Not All from Origen)\n\nHi,\n\nI'm trying to plot a series of vectors. I've pasted what I have below. The problem is that I want to plot the second vector, starting from the end of the first vector. If I run the commands below, I get two vectors plotted as I want (with arrows) but they are both from the origin.\n\nI tried looking at MATPLOTLIB to see what to do but I got stuck.\n\nIs there an easy way to plot a series of vectors in the way that I want?\n\na=vector([1,1,1])\nb=vector([2,2,3])\naPlot=plot(a, legend_label='Vector a')\nbPlot=plot(b, legend_label='Vector b')\nAllPlot=aPlot+bPlot\nAllPlot.show()\n\nedit retag close merge delete\n\nSort by » oldest newest most voted\nsage: a1=arrow((0,0,0), (1,1,1))\nsage: a2=arrow((1,1,1), (2,2,3))\nsage: a3=arrow((2,2,3), (3,3,1))\nsage: a1+a2+a3\n\nmore\n\nThe point being of course that a vector \"always\" starts at the origin, though some books would disagree.\n\nYou can use the start keyword to give a starting point or vector:\n\na=vector([1,1,1])\nb=vector([-1,2,3])\nplot(a)+plot(b, start=a)\n\nmore"
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https://casaguides.nrao.edu/index.php?title=Self-Calibration_Template&oldid=21932 | [
"# Self-Calibration Template\n\nThis guide continues with products created in Image the Continuum Template. All imaging parameters are set in the previous guide. You will need calibrated_final_cont.ms to proceed. Commands for this guide can be found in scriptForImaging_template.py available on github.\n\n## Self-Calibration on the Continuum (optional)\n\nIf you have a high signal-to-noise detection of your source, you can use the source itself to calibrate the phases and potentially the amplitudes of the visibilities as a function of time. This technique is called self-calibration and takes advantage of the fact that interferometer data is over-constrained (we have more equations than we have solutions). The recommended signal to noise depends on the number of antennas. For an array of 40 antennas, you should have a S/N of at least 45 in your continuum image to attempt self-calibration.\n\nSelf-calibration is an iterative process. You start by generating a model of your source using clean. Then you use this model to determine the gains as a function of time using the gaincal task. Finally you apply these solutions to the data and re-image. These steps are repeated until you are happy with your model or the solution interval is too short to reach the necessary signal-to-noise. In general, you start with phase-only self-calibration and only do amplitude calibration at the end of the self-calibration process if there are amplitude-based gain artifacts in the data (see the Imaging and Deconvolution talk from the Synthesis Imaging Summer School for more detail. Amplitude calibration should be used with caution because it has the potential to change the fluxes of sources in your data.\n\nSelf-calibration can either be performed with line or continuum data. Here we demonstrate how to do it with continuum data. The principle is the same for line data, except you form your image using the brightest line emission. For an example of self-calibration with line data, see the VLA high frequency spectral line tutorial for IRC+10216.\n\nFor a more detailed explanation of self-calibration, please see the NRAO Live! presentation When, why, and how to do self-calibration.\n\nThe first thing you should do when beginning self-calibration is to save the original flags in your data set. This step is necessary because when you apply the calibration to a data set it flags data that is not associated with a solution. Applying a bad calibration table to your data can result in the excessive flagging of your data. Saving your original flags allows you to restore the original state of your data easily. We also recommend you save your flags after each applycal command. This will allow you to go back to a given point in the self-cal process without starting from the beginning. See Self_Calibration_Template#Restart_Self-Calibration if you need to start over or go back a step in self-cal.\n\n```# in CASA\nflagmanager(vis=contvis,mode='save',versionname='before_selfcal',merge='replace')\n```\n\nNow we can set up the parameters for the self calibration. Begin by setting the parameters for your ms and continuum image name. The ms calibrated_final_cont.ms was created in the previous part of this guide: Image the Continuum Template.\n\n```# in CASA\ncontvis = 'calibrated_final_cont.ms'\ncontimagename = 'calibrated_final_cont'\n```\n\nIf you run the self-calibration process more than once, you will need to remove any models that could be left in the image header; which can be accomplished by using the delmod task as well as the clearcal task (which will need to be uncommented in the code below).\n\n```# in CASA\n# delmod(vis=contvis,field=field,otf=True)\n# clearcal(vis=contvis)\n```\n\nYou should use the same reference antenna that was used during calibration. For pipeline reductions, this can be found in the stage hif_refant. For manual reductions, the reference antenna will be stated near the top of *.ms.scriptForCalibration.py. If there are multiple executions, make sure to use an antenna that has good solutions and is present in all executions. The Analysis Utilities task au.commonAntennas can help you find antennas that meet this criteria. If you haven't installed Analysis Utilities, see Obtaining Analysis Utilities for instructions. Tasks such as plotants or listobs/vishead can also be used to find antennas in all executions.\n\n```# in CASA\nrefant = 'DV09' # pick a reference antenna.\n```\n\nIn the example below, we combine the signal from all spectral windows to improve the signal-to-noise for our gain solution. When combining the spectral windows, you need to map the solution from the combined spectral window (0) to the individual spectral windows using the spwmap parameter. This parameter is a list where the index of the element in the list indicates the spectral window and the value for that index the window that it is mapped to. For example, if we have three spectral windows in the original data set and use combine='spw' for our gain solution, we set spwmap=[0,0,0] to map spw=0 to spw=0, spw=1 to spw=0, and spw=2 to spw=0.\n\n```spwmap = [0,0,0] # mapping self-calibration solutions to individual spectral windows. Generally an array of n zeroes, where n is the number of spectral windows in the data sets.\n```\n\nYou then begin the process of shallowly cleaning your continuum data to create an initial model for your data in the model column of your data set. Usually you only should do at most a few hundred iterations on the brightest source(s) in the field.\n\n```# in CASA\n# shallow clean on the continuum\n\nrmtables(contimagename + '_p0'+ ext)\n\nclean(vis=contvis,\nimagename=contimagename + '_p0',\nfield=field,\n# phasecenter=phasecenter, # uncomment if mosaic.\nmode='mfs',\npsfmode='clark',\nimsize = imsize,\ncell= cell,\nweighting = weighting,\nrobust=robust,\nniter=niter,\nthreshold=threshold,\ninteractive=True,\nusescratch=True, # needed for 4.3 and 4.4 (and maybe 4.5)\nimagermode=imagermode)\n\n# Note number of iterations performed.\n```\n\nNext you take that model and use it to determine the per-antenna phase solutions as a function of time using gaincal. We use the rmtables command here to completely eradicate any previous solution table from CASA's memory. Note that we start here with a fairly long solution interval that is the length of a scan (solint='inf').\n\n```# per scan solution\nrmtables('pcal1')\ngaincal(vis=contvis,\ncaltable='pcal1',\nfield=field,\ngaintype='T',\nrefant=refant,\ncalmode='p',\ncombine='spw',\nsolint='inf',\nminsnr=3.0,\nminblperant=6)\n```\n\nInspect the logging messages output by gaincal to see how many solutions were expected/attempted/succeeded. If you have a large number of failed solutions, do not proceed further! This usually means that you do not have enough signal-to-noise on your source to proceed with self-calibration. Note that if you apply a calibration table with many failed solutions to the data, it will flag the data associated with these solutions. You may try changing the minsnr and minblperant parameters in the gaincal call above to find more solutions, but this option should be used with caution because these parameters are already low. This is especially relevant for 7-meter datasets, where achieving 6 baselines per antenna may be difficult. If you choose to modify these parameters, make sure to examine the results in detail.\n\n```Calibration solve statistics per spw: (expected/attempted/succeeded):\nSpw 0: 235/235/235\nSpw 1: 235/235/235\nSpw 2: 235/235/235\n```\n\n<figure id=\"Plotcal_image.png\">\n\n</figure>\n\nYou should also check the solutions (above) that were obtained using plotcal. The solutions should vary smoothly with time and there should not be any large outliers.\n\n```In CASA\n# Check the solution\nplotcal(caltable='pcal1',\nxaxis='time',\nyaxis='phase',\ntimerange='',\niteration='antenna',\nsubplot=421,\nplotrange=[0,0,-180,180])\n```\n\nIf you are satisfied with the solutions, apply them to the ms. Note: this step usually takes a while, so go ahead and get a cup of coffee.\n\n<figure id=\"Pcal1.png\">\n\n</figure>\n\n```# in CASA\n# apply the calibration to the data for next round of imaging\napplycal(vis=contvis,\nfield=field,\nspwmap=spwmap,\ngaintable=['pcal1'],\ngainfield='',\ncalwt=False,\nflagbackup=False,\ninterp='linearperobs')\n```\n\nSave the flags in case you need to go back to this step.\n\n```# in CASA\nflagmanager(vis=contvis,mode='save',versionname='after_pcal1')\n```\n\nUsing our new gain solutions, we can generate an improved model for our source using clean. During this second clean iteration, you should clean a bit deeper than previously, but not all the way down to the noise. Remember the goal here is to build up a good model for your source, so you don't want to include things that are not real emission.\n\n```# clean deeper\nrmtables(contimagename + '_p1'+ ext)\n\nclean(vis=contvis,\nfield=field,\n# phasecenter=phasecenter, # uncomment if mosaic.\nimagename=contimagename + '_p1',\nmode='mfs',\npsfmode='clark',\nimsize = imsize,\ncell= cell,\nweighting = weighting,\nrobust=robust,\nniter=niter,\nthreshold=threshold,\ninteractive=True,\nusescratch=True, # needed for 4.3 and 4.4 (and maybe 4.5)\nimagermode=imagermode)\n\n# Note number of iterations performed.\n```\n\nYou should inspect the new image in the viewer to assess how well self-calibration worked. In general, the noise should be lower and the signal-to-noise ratio higher than in the original image and there should be no major new artifacts in the image. If the signal-to-noise ratio does not increase, you may not have enough signal-to-noise on your target to run self-calibration. If you don't see an improvement in the image, clear the calibration with clearcal (see commands above) and proceed to continuum subtraction (if you are working on a line project).\n\nIf the image has improved you can proceed with the self-cal and generate a new gain solution table. The next gaincal call uses a shorter solution interval (solint) to generate solutions on a shorter time interval. Note that we always want to generate our model based on the latest gaincal solutions. The calibration tables, however, should be generated by comparing the original visibilities to the model. This prevents bad solutions from propagating through different iterations of self-calibration.\n\n```# in CASA\n# shorter solution\nrmtables('pcal2')\ngaincal(vis=contvis,\nfield=field,\ncaltable='pcal2',\ngaintype='T',\nrefant=refant,\ncalmode='p',\ncombine='spw',\nsolint='30.25s', # solint=30.25s gets you five 12m integrations, while solint=50.5s gets you five 7m integration\nminsnr=3.0,\nminblperant=6)\n\n# Check the solution\nplotcal(caltable='pcal2',\nxaxis='time',\nyaxis='phase',\ntimerange='',\niteration='antenna',\nsubplot=421,\nplotrange=[0,0,-180,180])\n\n# apply the calibration to the data for next round of imaging\napplycal(vis=contvis,\nspwmap=spwmap,\nfield=field,\ngaintable=['pcal2'],\ngainfield='',\ncalwt=False,\nflagbackup=False,\ninterp='linearperobs')\nflagmanager(vis=contvis,mode='save',versionname='after_pcal2')\n```\n\nNow we can repeat the clean/gaincal/applycal process until we start seeing many failed solutions and only small improvements to the data. Remember to clean a bit deeper each time to improve your model and be cautious about what emission you are including in your clean mask. You should also decrease the solution interval each time to better model the gain variations with time. Note that it generally only take 3-4 rounds of phase self-calibration to produce a good solution.\n\n```# clean deeper\nrmtables(contimagename + '_p2'+ ext)\n\nclean(vis=contvis,\nimagename=contimagename + '_p2',\nfield=field,\n# phasecenter=phasecenter, # uncomment if mosaic.\nmode='mfs',\npsfmode='clark',\nimsize = imsize,\ncell= cell,\nweighting = weighting,\nrobust=robust,\nniter=niter,\nthreshold=threshold,\ninteractive=True,\nusescratch=True, # needed for 4.3 and 4.4 (and maybe 4.5)\nimagermode=imagermode)\n\n# Note number of iterations performed.\n\n# shorter solution\nrmtables('pcal3')\ngaincal(vis=contvis,\nfield=field,\ncaltable='pcal3',\ngaintype='T',\nrefant=refant,\ncalmode='p',\ncombine='spw',\nsolint='int',\nminsnr=3.0,\nminblperant=6)\n\n# Check the solution\nplotcal(caltable='pcal3',\nxaxis='time',\nyaxis='phase',\ntimerange='',\niteration='antenna',\nsubplot=421,\nplotrange=[0,0,-180,180])\n\n# apply the calibration to the data for next round of imaging\napplycal(vis=contvis,\nspwmap=spwmap,\nfield=field,\ngaintable=['pcal3'],\ngainfield='',\ncalwt=False,\nflagbackup=False,\ninterp='linearperobs')\n\nflagmanager(vis=contvis,mode='save',versionname='after_pcal3')\n\nrmtables(contimagename + '_p3'+ ext)\n\nclean(vis=contvis,\nimagename=contimagename + '_p3',\nfield=field,\n# phasecenter=phasecenter, # uncomment if mosaic.\nmode='mfs',\npsfmode='clark',\nimsize = imsize,\ncell= cell,\nweighting = weighting,\nrobust=robust,\nniter=niter,\nthreshold=threshold,\ninteractive=True,\nusescratch=True, # needed for 4.3 and 4.4 (and maybe 4.5)\nimagermode=imagermode)\n\n# Note number of iterations performed.\n```\n\n<figure id=\"Apcal.png\">\n\n</figure>\n\nThe next section performs an amplitude calibration. This is an optional part of self-calibration. If you are happy with the results of your phase calibration, you can stop here. However, if you see amplitude-based artifacts, you can attempt to improve the situation using amplitude self-calibration. Since amplitude solutions are inherently less constrained than phase solutions, we use a longer solution interval only here. If you have a complex source with lots of extended emission, you may set a uvrange limit on the data to avoid downweighting the large scale emission\n\nWhile the phase calibration involved iterating with different solution intervals, the amplitude self-calibration will only use an infinite solution interval.\n\n```# in CASA\nrmtables('apcal')\ngaincal(vis=contvis,\nfield=field,\ncaltable='apcal',\ngaintype='T',\nrefant=refant,\ncalmode='ap',\ncombine='spw',\nsolint='inf',\nminsnr=3.0,\nminblperant=6,\n# uvrange='>50m', # may need to use to exclude extended emission\ngaintable='pcal3',\nspwmap=spwmap,\nsolnorm=True)\n\nplotcal(caltable='apcal',\nxaxis='time',\nyaxis='amp',\ntimerange='',\niteration='antenna',\nsubplot=421,\nplotrange=[0,0,0.2,1.8])\n\napplycal(vis=contvis,\nspwmap=[spwmap,spwmap], # select which spws to apply the solutions for each table\nfield=field,\ngaintable=['pcal3','apcal'],\ngainfield='',\ncalwt=False,\nflagbackup=False,\ninterp='linearperobs')\n\nflagmanager(vis=contvis,mode='save',versionname='after_apcal')\n\n# Make amplitude and phase self-calibrated image.\nrmtables(contimagename + '_ap'+ ext)\n\nclean(vis=contvis,\nimagename=contimagename + '_ap',\nfield=field,\n# phasecenter=phasecenter, # uncomment if mosaic.\nmode='mfs',\npsfmode='clark',\nimsize = imsize,\ncell= cell,\nweighting = weighting,\nrobust=robust,\nniter=niter,\nthreshold=threshold,\ninteractive=True,\nusescratch=True, # needed for 4.3 and 4.4 (and maybe 4.5)\nimagermode=imagermode)\n```\n\nCompare the final image to the initial image and see if the image dynamic range (ratio betwen peak flux and rms noise) has improved and phase- and amplitude-based errors have improved.\n\nFinally, you should split out the results of your self-calibration for safe-keeping.\n\n```# in CASA\nsplit(vis=contvis,\noutputvis=contvis+'.selfcal', datacolumn='corrected')\n```\n\n## Restart Self-Calibration\n\nIf you would like to revert to a certain point in the self-cal process, use the following commands to do so.\n\n```# uncomment the following to revert to a given point in the iterative process (here after pcal2 has been applied)\n# flagmanager(vis=contvis, mode='restore',versionname='after_pcal2')\n# clearcal(contvis)\n# delmod(contvis,field=field,otf=True)\n```\n\nThen return to the applycal step that applied the desired calibration table. In the example code above, you would return to where the table pcal2 was applied.\n\nIf you are unhappy with the self-calibration, use the clearcal and delmod tasks and restore the original flags to return your ms to it’s original pre-self-cal state.\n\n```# in CASA\n# uncomment the following to revert to pre self-cal ms\n# flagmanager(vis=contvis, mode='restore',versionname='before_selfcal')\n# clearcal(contvis)\n# delmod(contvis,field=field,otf=True)\n```\n\nIf you have line data, you will subtract the continuum and apply the self-calibration results to the line data on the next page, Spectral Line Imaging Template."
]
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https://math.stackexchange.com/questions/tagged/exponential-function?sort=votes&pageSize=50 | [
"# Questions tagged [exponential-function]\n\nFor question involving exponential functions and questions on exponential growth or decay.\n\n5,192 questions\n15answers\n19k views\n\n### The math behind Warren Buffett's famous rule – never lose money\n\nThis is a question about a mathematical concept, but I think I will be able to ask the question better with a little bit of background first. Warren Buffett famously provided 2 rules to investing: ...\n9answers\n15k views\n\n### Proof that $C\\exp(x)$ is the only set of functions for which $f(x) = f'(x)$\n\nI was wondering the following. And I probably know the answer already: NO. Is there another number with similar properties as $e$? So that the derivative of $\\exp(x)$ is the same as the function ...\n13answers\n91k views\n\n### Do factorials really grow faster than exponential functions? [closed]\n\nHaving trouble understanding this. Is there anyway to prove it?\n25answers\n14k views\n\n### Simplest or nicest proof that $1+x \\le e^x$\n\nThe elementary but very useful inequality that $1+x \\le e^x$ for all real $x$ has a number of different proofs, some of which can be found online. But is there a particularly slick, intuitive or ...\n11answers\n10k views\n\n### Comparing $\\pi^e$ and $e^\\pi$ without calculating them\n\nHow can I calculate, without calculator or similar device, the values of $\\pi^e$ and $e^\\pi$ in order to compare them?\n2answers\n24k views\n\n### Why is it hard to prove whether $\\pi+e$ is an irrational number?\n\nFrom this list I came to know that it is hard to conclude $\\pi+e$ is an irrational? Can somebody discuss with reference \"Why this is hard ?\" Is it still an open problem ? If yes it will be helpful ...\n6answers\n6k views\n\n### What's so “natural” about the base of natural logarithms?\n\nThere are so many available bases. Why is the strange number $e$ preferred over all else? Of course one could integrate $\\frac{1}x$ and see this. But is there more to the story?\n5answers\n8k views\n\n### If $a+b=1$ so $a^{4b^2}+b^{4a^2}\\leq1$\n\nLet $a$ and $b$ be positive numbers such that $a+b=1$. Prove that: $$a^{4b^2}+b^{4a^2}\\leq1$$ I think this inequality is very interesting because the equality \"occurs\" for $a=b=\\frac{1}{2}$ and ...\n13answers\n52k views\n\n### How to prove that exponential grows faster than polynomial?\n\nIn other words, how to prove: For all real constants $a$ and $b$ such that $a > 1$, $$\\lim_{n\\rightarrow\\infty}\\frac{n^b}{a^n} = 0$$ I know the definition of limit but I feel that it's not ...\n3answers\n955 views\n\n### Is $\\lfloor n!/e\\rfloor$ always even for $n\\in\\mathbb N$?\n\nI checked several thousand natural numbers and observed that $\\lfloor n!/e\\rfloor$ seems to always be an even number. Is it indeed true for all $n\\in\\mathbb N$? How can we prove it? Are there any ...\n2answers\n1k views\n\n2answers\n1k views\n\n### From $e^n$ to $e^x$\n\nSolve for $f: \\mathbb{R}\\to\\mathbb{R}\\ \\ \\$ s.t. $$f(n)=e^n \\ \\ \\forall n\\in\\mathbb{N}$$ $$f^{(y)}(x)>0 \\ \\forall y\\in\\mathbb{N^*} \\ \\forall x\\in\\mathbb R$$ Could you please prove that ...\n8answers\n3k views\n\n### What's the value of $\\sum\\limits_{k=1}^{\\infty}\\frac{k^2}{k!}$?\n\nFor some series, it is easy to say whether it is convergent or not by the \"convergence test\", e.g., ratio test. However, it is nontrivial to calculate the value of the sum when the series converges. ...\n17answers\n4k views\n\n14answers\n2k views\n\n### Intuitive proofs that $\\lim\\limits_{n\\to\\infty}\\left(1+\\frac xn\\right)^n=e^x$\n\nAt this link someone asked how to prove rigorously that $$\\lim_{n\\to\\infty}\\left(1+\\frac xn\\right)^n = e^x.$$ What good intuitive arguments exist for this statement? Later edit: . .&...\n3answers\n1k views\n\n3answers\n3k views\n\n### How to verify if a curve is exponential by eyeballing?\n\nA plane curve is printed on a piece of paper with the directions of both axes specified. How can I (roughly) verify if the curve is of the form $y=a e^{bx}+c$ without fitting or doing any quantitative ...\n7answers\n4k views\n\n1answer\n721 views\n\n### Peculiar locations of the root and the maximum of $(x+1)^{x+1}-x^{x+2}$\n\nRelated to some other problems, I got interested in this function: $$(x+1)^{x+1}-x^{x+2}$$ Its root is very close to $\\pi$. This is Mathematica code that finds the root: ...\n3answers\n680 views"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.84397453,"math_prob":0.9965657,"size":13699,"snap":"2019-13-2019-22","text_gpt3_token_len":4445,"char_repetition_ratio":0.13428259,"word_repetition_ratio":0.0085227275,"special_character_ratio":0.3260822,"punctuation_ratio":0.11103448,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9997842,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-22T21:25:51Z\",\"WARC-Record-ID\":\"<urn:uuid:071150c8-dc05-48be-be7e-3f066ce1f888>\",\"Content-Length\":\"225156\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:26598beb-974f-428d-ad08-f271afe212e4>\",\"WARC-Concurrent-To\":\"<urn:uuid:6c3baac7-fb58-4add-acd2-5948e41ce81a>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/tagged/exponential-function?sort=votes&pageSize=50\",\"WARC-Payload-Digest\":\"sha1:4V2FF5H2HAVVB3MHW2ADWGHNLODRHABT\",\"WARC-Block-Digest\":\"sha1:U4IO3DOWBZ7RHJ7ILKJA2ZA27RCR2YZK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232256958.53_warc_CC-MAIN-20190522203319-20190522225319-00458.warc.gz\"}"} |
https://www.esaral.com/q/mark-against-the-correct-answer-84521 | [
"",
null,
"# Mark (✓) against the correct answer:\n\nQuestion:\n\nMark (✓) against the correct answer:\n\nThe parallel sides of a trapezium are 54 cm and 26 cm and the distance between them is 15 cm. The area of the trapezium is\n\n(a) 702 cm2\n\n(b) 810 cm2\n\n(c) 405 cm2\n\n(d) 600 cm2\n\nSolution:\n\n(d) $600 \\mathrm{~cm}^{2}$\n\nArea of the trapezium $=\\left\\{\\frac{1}{2} \\times(54+26) \\times 15\\right\\} \\mathrm{cm}^{2}$\n\n$=\\left(\\frac{1}{2} \\times 80 \\times 15\\right) \\mathrm{cm}^{2}$\n\n$=600 \\mathrm{~cm}^{2}$"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.5558929,"math_prob":0.99990356,"size":434,"snap":"2023-14-2023-23","text_gpt3_token_len":177,"char_repetition_ratio":0.15116279,"word_repetition_ratio":0.0,"special_character_ratio":0.4423963,"punctuation_ratio":0.04597701,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9941737,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-03T20:38:07Z\",\"WARC-Record-ID\":\"<urn:uuid:6edd70e1-edff-4116-83fa-30c5e5f3e1e5>\",\"Content-Length\":\"24720\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a367ed22-9462-4bd8-9700-e3bfbfb15f51>\",\"WARC-Concurrent-To\":\"<urn:uuid:0c74d5d4-4587-4a00-8e5c-c28e81ec5a2d>\",\"WARC-IP-Address\":\"172.67.213.11\",\"WARC-Target-URI\":\"https://www.esaral.com/q/mark-against-the-correct-answer-84521\",\"WARC-Payload-Digest\":\"sha1:L6JQKICWVARUMYH57WMAXETNZF7YUL52\",\"WARC-Block-Digest\":\"sha1:ZCB6PR5OQB2D62F4GSHBMRDB6GG63V6N\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224649343.34_warc_CC-MAIN-20230603201228-20230603231228-00139.warc.gz\"}"} |
https://www.kopykitab.com/blog/rd-sharma-solutions-class-12-maths-chapter-19-exercise-19-1/ | [
"RD Sharma Solutions Class 12 Maths Chapter 19 Exercise 19.1 (2021)",
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"RD Sharma Solutions Class 12 Maths Chapter 19 Exercise 19.1: Solutions PDF is an important reference guide to help students score well in the Class 12 exam. By solving exercise problems from RD Sharma Solutions on a daily basis, you can improve problem solving and logical thinking skills.\n\nDownload RD Sharma Solutions Class 12 Maths Chapter 19 Exercise 19.1 PDF:\n\nRD Sharma Solutions For Class 12th Maths All Chapters\n\nAccess to the Solutions for RD Sharma Solutions Class 12 Maths Chapter 19 Exercise 19.1\n\nIndefinite Integrals Ex 19.1 Q1",
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"",
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"Indefinite Integrals Ex 19.1 Q2",
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"Indefinite Integrals Ex 19.1 Q4",
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"Indefinite Integrals Ex 19.1 Q5",
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"Indefinite Integrals Ex 19.1 Q6",
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"RD Sharma Solutions Class 12 Maths Chapter 19 Exercise 19.1: Important Topics From The Chapter\n\nSome of the essential topics covered in chapter are listed below.\n\n• Definition of primitive or antiderivative\n• Definition and meaning of indefinite integral\n• Fundamental integration formulae\n• Some standard results on integration along with the corollary\n• Integration of trigonometric functions\n• Integration of exponential functions\n• Miscellaneous problems\n• Geometrical interpretation of indefinite integral\n• Comparison between differentiation and integration\n• Methods of integration\n• Integration by substitution\n• Some standard results\n• Evaluation of integrals by using trigonometric substitutions\n• Some special integrals\n• Integration by parts\n• Some important integrals along with theorems\n• Integration of rational algebraic functions by using partial fractions\n• When the denominator is expressible as a product of distinct linear factors\n• When the denominator contains some repeating linear factors\n• The denominator contains irreducible quadratic factors\n• Integration of some special irrational algebraic functions\n\nWe have included all the information regarding CBSE RD Sharma Solutions Class 12 Maths Chapter 19 Exercise 19.1. If youa hve any query feel free to ask in the comments ection.\n\nFAQ: RD Sharma Solutions Class 12 Maths Chapter 19 Exercise 19.1\n\nYes, you can download RD Sharma Class 12 Maths Solutions Chapter 19 Exercise 19.1 PDF free.\n\nAre these solutions a reliable source of studying for board exams?\n\nYes, RD Sharma Class 12 Maths Solutions Chapter 19 Exercise 19.1 is a reliable source of studying for board exams.\n\nWhat are the essential topics included in RD Sharma Solutions Class 12 Maths Chapter 19 Exercise 19.1?\n\nYou can refer to the above article to know more about the important topics included in RD Sharma Solutions Class 12 Maths Chapter 19 Exercise 19.1.\n\nIs PDF for Class 12 RD Sharma Solution open in smartphones?\n\nYes, you can open PDFs on any device.\n\nWhat will be taught in RD Sharma Solutions Class 12 Maths Chapter 19 Exercise 19.1?\n\nIt states some standard results on integration with fundamental integration formulas."
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http://bablefishfx.comwww.tug.org/pipermail/metapost/2005-February/000234.html | [
"# [metapost] Re: d\\'ej\\`a vu or turningnumber, area et caetera\n\nDaniel H. Luecking luecking at uark.edu\nFri Feb 4 20:28:08 CET 2005\n\n```On Fri, 4 Feb 2005, Boguslaw Jackowski wrote:\n\n> The problem of such paths can be easily solved. Let a, b, c, d denote the\n> nodes of the B\\'ezier segment a .. controls b and c .. d.\n>\n> A more general definition of `postdir' (and, for completeness, `predir'),\n> given below, is based on the following observation, being the consequence of\n> the de l'H\\^ospital's rule: normally the vector a-->b determines the ``post''\n> direction at node a; if b coincides with a, then the vector a-->c determines\n> the direction; if c also coincides coincides with a, then the last resort is\n> vector c-->d; if d also coincides with a, the B\\'ezier segment is\n> degenerated, and can be removed.\n\nMfpic's grafbase.mf (version 0.7) has postdirection and predirection.\nIt is used to determine the orientation of arrowheads at the end of\npaths. Version 0.8 (coming Real Soon Now) uses a different approach, but\nnow uses these to create parallel paths. The code follows.\n\n% Note: \"post[n](f)\" = \"postcontrol n of f\", etc. Also, the time is\n% not expected to be integral, and if a trivial segment (all\n% four controls coinciding) occurs, the next segment is examined.\n\nvardef predirection@# (expr p) =\n- postdirection[length p - @#] (reverse p)\nenddef;\n\nvardef postdirection@# (expr p) =\nsave _n; _n := length (p);\nsave v; pair v;\nv:= __dir (subpath (@#, @# + _n) of p);\nif v = origin : % use predirection as a fallback\nv := - __dir (subpath (@#, @# - _n) of p);\nfi\nv\nenddef;\n\nvardef __dir (expr p) =\nsave v, w; pair v, w; w := pnt0 (p);\nv := origin;\nfor n = 1 upto length (p) :\nv := post[n-1] (p) - w; exitif v <> origin;\nv := pre [ n ] (p) - w; exitif v <> origin;\nv := pnt [ n ] (p) - w; exitif v <> origin;\nendfor\nv if v<>(0,0): /abs(v) fi\nenddef;\n\nvardef trivial primary p = __dir (p) = origin enddef;\n\n> Now, `turn_ang' works quite efficiently (speed-up is some 30% in\n> comparison with Werner's `is_clockwise') and yields ``reasonable''\n> results for ``reasonable'' paths. It fails for paths with cusps,\n> loops, and with segments turning by the angle >=180. Note that\n> cusps and large turning angles are indifferent for the `Area' function.\n>\n> Ps. Note that the `direction' operator, defined in MP/MF as follows:\n>\n> vardef direction expr t of p =\n> postcontrol t of p - precontrol t of p enddef;\n>\n> can be improved using the functions `predir' and `postdir'\n> defined above:\n>\n> vardef gendir expr t of p =\n> if round(t)=t:\n> predir t of p + postdir t of p\n> else:\n> gendir 1 of\n> ((subpath (floor t, t) of p) & (subpath (t, ceiling t) of p))\n> fi\n> enddef;\n\nI personally think that at an integer t, a user should specify either\npostdir or predir. There is little use in obtaining some intermediate\nangle when they differ.\n\nDan\n\n--\nDan Luecking\nDept. of Mathematical Sciences\nUniversity of Arkansas\nFayetteville, AR 72101\n\n```"
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https://cses.fi/alon/task/1750 | [
"CSES - Planets Queries I\n• Time limit: 1.00 s\n• Memory limit: 512 MB\nYou are playing a game consisting of $n$ planets. Each planet has a teleporter to another planet (or the planet itself).\n\nYour task is to process $q$ queries of the form: when you begin on planet $x$ and travel through $k$ teleporters, which planet will you reach?\n\nInput\n\nThe first input line has two integers $n$ and $q$: the number of planets and queries. The planets are numbered $1,2,\\dots,n$.\n\nThe second line has $n$ integers $t_1,t_2,\\dots,t_n$: for each planet, the destination of the teleporter. It is possible that $t_i=i$.\n\nFinally, there are $q$ lines describing the queries. Each line has two integers $x$ and $k$: you start on planet $x$ and travel through $k$ teleporters.\n\nOutput\n\nPrint the answer to each query.\n\nConstraints\n• $1 \\le n, q \\le 2 \\cdot 10^5$\n• $1 \\le t_i \\le n$\n• $1 \\le x \\le n$\n• $0 \\le k \\le 10^9$\nExample\n\nInput:\n4 3\n2 1 1 4\n1 2\n3 4\n4 1\n\nOutput:\n1\n2\n4"
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https://onlineclasshandlers.com/page/294/ | [
"Home » Page 294\n\n## Score Better",
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"### PHY 1004 MDC Physics Mass Weight and Reading on The Scale Lab Report\n\nPHY 1004 MDC Physics Mass Weight and Reading on The Scale Lab Report custom answer.\n\n### PHY 222 Kalamazoo Valley Community College Magnetic Field Analysis Lab Report\n\nPHY 222 Kalamazoo Valley Community College Magnetic Field Analysis Lab Report custom answer.\n\n### PHYS 211 KFUPM Physics Force Table & Simple Pendulum Lab Experiments\n\nPHYS 211 KFUPM Physics Force Table & Simple Pendulum Lab Experiments custom answer.\n\n### PHY 130 Cuyamaca College Energy Conservation Questions\n\nPHY 130 Cuyamaca College Energy Conservation Questions custom answer.\n\n### Saudi Electronic University Electrodynamic Application Report\n\nSaudi Electronic University Electrodynamic Application Report custom answer.\n\n### Physics Worksheet\n\nPhysics Worksheet custom answer.\n\n### PHYSICS 2049 Palm Beach State College Physics Lab Report\n\nPHYSICS 2049 Palm Beach State College Physics Lab Report custom answer.\n\n### Grossmont College Inductor Lab Worksheet\n\nGrossmont College Inductor Lab Worksheet custom answer.\n\n### Centenary College of Louisiana Physics Worksheet\n\nCentenary College of Louisiana Physics Worksheet custom answer.\n\n### University of Toronto Energy and Momentum Physics Worksheet\n\nUniversity of Toronto Energy and Momentum Physics Worksheet custom answer.\n\n#### Solve problem 4 using MATLAB. And paste code and output screenshot here: Solution of problem 3 is also attached at end to solve problem 4 problem3 Solution\n\nSolve problem 4 using MATLAB. And paste code and output\nscreenshot here:\nSolution of problem 3 is also attached at end to solve\nproblem 4\n\nproblem3\nSolution\n\ncustom answer show all workings plsexcel sheet attached 3. A university wants to know what proportion of students are regular bike riders so that they can install an appropriate number of bike racks. The spreadsheet bike. csv contains the result of survey conducted in 20 classes. Here N is a class size and n is\n\ncustom answer write a brief overview of a proposed theory. The proposal is on how management affects students performance in UGBS Online Class Handlers: Advanced Math, Advanced Math questions and answers, homework help, Math\n\ncustom answer please Help me Instructions: Below are some typical answers students have submitted in the past. Describe the mistake(s) in your own words. What grade (percentage) would you give a student making this mistake? In deciding the grade, consider the following: how ‘big’ is the mistake, how does the mistake compare to the others\n\n#### An economist wants to determine whether average price/earnings (P/E) ratios differ for firms in three industries. Independent samples of five firms in each industry produced the following results after conducting a one-way ANOVA. SSB (sum of squares between groups) = 258.82 and SST (sum of squares total) = 424.04. Based on this information, what would be the\n\nAn economist wants to determine whether average price/earnings\n(P/E) ratios differ for firms in three industries. Independent\nsamples of five firms in each industry produced the following\nresults after conducting a one-way ANOVA. SSB (sum of squares\nbetween groups) = 258.82 and SST (sum of squares total) = 424.04.\nBased on this information, what would be the\n\n#### Why would a t-test be used for statistical analysis of surveys?\n\nWhy would a t-test be used for statistical analysis of surveys?\n\n#### A financial security generates a cash flow of \\$25,000 every five years forever with the first cash flow occurring in 3 years’ time. The appropriate opportunity cost is 12% p.a. compounded annually. What should be the security’s price today?\n\nA financial security generates a cash flow of \\$25,000 every five\nyears forever with the first cash flow occurring in 3 years’ time.\nThe appropriate opportunity cost is 12% p.a. compounded annually.\nWhat should be the security’s price today?\n\n#### 6. Find vectors ( mathbf{v}, mathbf{w} in mathbb{C}^{1} ) so that ( mathbf{w} ) is a complex scalar multiple of ( mathbf{v} ) but not a real scalar multiple.\n\n6. Find vectors ( mathbf{v}, mathbf{w} in mathbb{C}^{1} ) so that ( mathbf{w} ) is a complex scalar multiple of ( mathbf{v} ) but not a real scalar multiple.\n\n#### find the root mean square as shown in the formula of the given data\n\nfind the root mean square as shown in the formula of the given data\n\n#### Was wondering how to do a where statement for months between 4 and 11 and also help with the question after\n\nWas wondering how to do a where statement\nfor months between 4 and 11\nand also help with the question after"
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https://www.bankersadda.com/quadratic-equations-for-bank-exam/ | [
"",
null,
"Latest Banking jobs » Quadratic Equations For Bank Exam\n\n# Quadratic Equations For Bank Exam\n\nQuadratic equation is one such topic which is asked in almost all the banking exams. Approximately five questions are asked from this topic which carries 5 marks and you can easily fetch those five marks if you solve this topic with right approach. In this article, we will be telling you the correct way of solving the questions from this topic and if you want to make sure that you master this topic then readout this article.\n\nPractice with,\n\nIn quadratic equation, students are expected to find the relation between two variable given in the two equation. The related can be of Smaller to, Greater to, Greater than, Smaller than and Equal to.\n\nLinear Equation– In this type, when a student solve the equation then he/she will only get one value for X and for Y.\n\nfor ex- 2X+3Y=5, X+2Y=6\n\n2X+3Y=5—–(eq.1), (X+2Y=6)X2——(eq.2)\n\nSolving these two equations, we will get\n\nWe get, X=-8 & Y= 7\n\nHence Y>X, this will be our final answer\n\nSquares Equation– In this type of question, we have to find the sqaure root of the given below and we end up getting two values each for X and Y but one is negative while other is positive. The best thing about such type of questions is that the answer to such questions are always Cannot be Determined.\n\nFor e.q., X= 1600 & Y=2500\n\nOn finding their square root we will get X=+40, -40 & Y= +50, -50\n\nHence the answer will be Cannot be determined.\n\nSqaures and Sqaure root equation– In this type of question, one is sqaure while the other one is square root and we know that sqaure root always gives positive value when we solve it.\n\nFor e.g. X=1600 & Y= 2500\n\nOn solving them we will get, X= +40, -40 & Y=+50\n\nHence, our answer will be Y>X\n\nCube Cases– In this type of questions, the cube will be given and you have to answer the relation between those cubes.\n\nFor e.g. X= 1331 & Y= 729\n\nWhen we solve these two equation, we will get X= 11 & Y= 9\n\nHence, X>Y. The trick to solve these questions is that whichsoever cube will be greater when you will solve it that cube will remain greater.\n\n### Table Method to Solve Quadratic Equation\n\n Sign of coefficient ‘x’ Sign of coefficient ‘y’ Sign of roots + + – – + – – + – + + + – – + –\n\nThis is one of the best methods to solve Quadratic Equation questions. You can answer the question without actually solving the equations.\n\nFor e.g. x2 – 7x + 10 = 0, y2 + 8y + 15 = 0\n\nWe can solve this equation without actually solving the equation. Let’s see\n\n• First look at the sign in equation.1 which are -, + means its roots will be +,+ as per our table.\n• Secondly, look at the other equation, we will see the signs are +, + which means that our roots will be -, -.\n• Hence X>Y as it has both the roots positive.\n\n### Different cases in Quadratic Equations\n\n CASES Roots of Eq.1 Roots of Eq. 2 Conclusion CASE-2 +, + +, + Easy CASE-3 +, + -, – Eq 1>Eq 2 CASE-4 +, – +, – Cannot be defined CASE-5 -, – -, – Easy CASE-6 -, – +,+ Eq1< Eq2 CASE-7 +, – +,+ Solve the Equations\n\nQ. How many questions from this topic are asked in Banking exams?\n\nAns. Approximately five questions are asked in Banking exams.\n\nQ. How many marks does this topic carry?\n\nAns. Usually five marks are allocated for this topic.\n\nQ. Which is the easiest method to solve Quadratic Equation?\n\nAns. Table Method is the easiest method to solve Quadratic Equation.\n\nPractice with,\n\n×\n\nThank You, Your details have been submitted we will get back to you.",
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" Monotonic Vector Space Model(II): Application in System Capability Indicator Analysis and Design\n\nAmerican Journal of Operations Research\nVol.04 No.01(2014), Article ID:42349,9 pages\n10.4236/ajor.2014.41005\n\nMonotonic Vector Space Model(II): Application in System Capability Indicator Analysis and Design\n\nJianwen Hu\n\nNational University of Defense Technology, Changsha, China\n\nEmail: [email protected]\n\nABSTRACT\n\nReceived December 7, 2013; revised January 7, 2014; accepted January 14, 2014\n\nThe application of some MVS operations (each dimension of which corresponds to a capability indicator) in sys- tem capability analysis and design, including capability indicator requirement analysis, effectiveness analysis, sensibility analysis, fuzzy analysis, stability analysis, capability optimization design, etc., is discussed in the second paper of this series of papers. And some MVS-based models and algorithms for capability analysis and design are put forward. Finally, an example of capability analysis and optimization design is given for explaining the usages of related models and algorithms.\n\nKeywords:\n\nMonotonic Vector Space; System Capability Indicators; Analysis; Optimization Design\n\n1. Introduction\n\nSystem capability indicator analysis and design are very important and also difficult for system development and system engineering. Many practical methods have been proposed and will be introduced in the following.\n\n1.1. The Representative System Capability Indicator Analysis Methods\n\n・ Index method. It is a traditional method for weapon system capability indicator analysis. This method is suit- able for synthesis and easy to understood. However, the index formula and weight influenced by analyzer are difficult to be determined. And the analysis result of index method is to some extent ambiguous.\n\n・ ADC model . Based on the principle of division and transformation of system state, this method proposed by WEISIC is able to give clear quantitative analysis result. However, the calculation overload will exponen- tially grow with the increasing of system dimension number. Furthermore, other models and algorithms are required to get the capability vector C of ADC model.\n\n・ Weight-based multi-attributes analysis method [2,3]. This type of method such as AHP , ELECTRE, LINAMP, etc., determines weight of each indicator firstly. Then, all indicators are synthesized according to each of their weights in a linear way. Obviously, they are regardless of the mutual influence among indicators and so fail to reveal the nonlinearity of complex system. And furthermore their analysis results are also unclear.\n\n・ SEA method . This method compares system capabilities with mission requirements in a common attribute space and emphasizes the whole characteristic of complex system. This comparison leads to the evaluation of partial measures of effectiveness that are then combined to yield a global measure. But system mission requirement loci used by this method are difficult to acquire.\n\n1.2. System Capability Indicator Design Method\n\n・ Linera programming methods. This kind of method is traditional and neat. Many complex problems (espe- cialy with nolineartity and uncertainty) have to be over trivialized so as to be solved by these neat methods. Therefore, these methods are not directly effective for many complex problems.\n\n・ Capability-based Planning (CPB) . CPB method emphasizes the flexibility, adaptiveness and robustness of capability and usually adopts exploratory analysis strategy. Compared with programming method, CPB, as a flexible and non-neat method, is more suitable for complex problems.\n\nMonotonic vector space (MVS) model is used for system capability indicator analysis and design in this paper. Since each dimension of MVS corresponds to one capability indicator, the MVS, therefore, is also called Mo- notonic Indicator Space (MIS) here. And likewise there exist some monotonic mappings between capability in- dicators and requirement measures in MIS.\n\n1.3. Assumptions Adopted in This Paper\n\nA monotonic relationship between the system capability indicators and the system requirement measures is assumed in this paper. It must be noted that system capability indicators are different from system parameters. For example, the velocity of a certain aircraft is not a capability indicator, but the maximal (or minimal) velocity of the aircraft is a system capability indicator. This assumption is usually in accordance with the practical situation in many cases. For example, in reference , there exists a monotonic relationship between the capability indi- cators (including reliability indicator",
null,
", survivability indicator",
null,
", delay time indicator",
null,
", and link capacity indicator",
null,
") and the requirement measure, i.e. the ratio of kill to loss. In reference , there also exists a mo- notonic relationship between the capability indicators (including link reliability indicator and delay time indicator) and the requirement measure (the kill probability). Actually, similar monotonic relationship between the capability indicators and the requirement measure can be found in many practical cases introduced by references [8-12].\n\nIn addition, the weak capability is assumed to be a subset of the stronger one. Thus, if a system with weak capability meets some certain mission requirements, the one with stronger capability will be certain to meet these mission requirements. For example, if an aircraft with the maximal cruise velocity of",
null,
"can ar- rive at a certain location in one hour (a requirement measure), the one with maximal cruise velocity of 3000 km/h will definitely be able to arrive at the same location in one hour as long as the other capability indicators of these two aircrafts are equal.\n\nThe contents of system capability indicator analysis and design based on MIS include indicator requirement analysis, effectiveness analysis, sensitivity analysis, fuzzy analysis, relativity analysis, finance-based optimiza- tion design and time-based optimization design. The mutual relationships among these contents are shown in Figure 1. Obviously, system indicator analysis is the start point of system capability indicator analysis and de- sign. And in the following sections, the related models and algorithms for system capability indicator analysis and design will be dealt with in detail.\n\n2. MIS-Based System Capability Indicator Analysis\n\n2.1. System Capability Indicator Requirement Analysis\n\nThe partitioning operation in MVS introduced by the first part of this series paper is used to produce the system capability requirement locus. For example, the dimensions of a monotonic indicator space",
null,
"for a ground-to-air missile system may include the searching radar detecting capability indicator dimension",
null,
", delay time indicator dimension",
null,
", and tracking radar detecting capability indicator dimension",
null,
". So, the equation\n\nFigure 1.MVS-based system capability indicators analysis and design.",
null,
"describes the relationship between these three system capability indicators (i.e.,",
null,
",",
null,
", and",
null,
") and one requirement measure",
null,
"(the kill probability). The function",
null,
"may adopt several different forms, each of which corresponds to a different context. For example,",
null,
"and",
null,
"correspond to the different contexts in which the enemy stealth aircraft is respectively type A and type B. Function",
null,
"is obviously monotonic, that is to say, the value of u (the kill probability) will monotonically increase or decrease with one of three system capability indicators changing monotonically while the others maintaining to be constants. Because the mono- tonic relationship between capability indicators (",
null,
"and) and the requirement measure (the kill probability) is different from that between capability indicator and the kill probability, so, and are called monotonic increasing indicators and called monotonic decreasing indicator.\n\nEvery dimension in MIS can be continuous or discrete type. System capability indicators are usually of con- tinuous type as all the above mentioned ones. For discrete-type indicators, if its range is large enough, it still can be considered as continuous type just like many integer programming problems can be solved by general linear programming model. In this paper, all the dimensions of MIS are assumed to be continuous. The purpose of system capability indicator requirement analysis is to find out the MVRL or MIRL (monotonic indicator re- quirement locus) which can then be acquired with the algorithm proposed by the first part of this series papers. Figure 2 shows the MIRL for the above-mentioned ground-to-air missile system. There are 3 MIS dimensions, each of which corresponds to a capability indicator. The requirement measure (the kill probability for enemy invading stealth aircraft of type A and type B) is assumed as. The simulation calculation realizes the mapping from capability indicators to requirement measure. The MIRLs are produced by the partitioning algo- rithm at first and then synthesizing operation is used to produce the intersection of these multiple MIRLs which correspond to multiple requirements. According to the algorithm for producing MIRL, the algorithm output is a heap tree consisting of MIHs and MEHs. The MIHs compose the MIRL.\n\n2.2. Effectiveness Analysis for the Indicators of Stochastic Type\n\nDue to the uncertainty of the system capability indicators, it is usually difficult to represent them precisely. So the probability distribution function is applied to represent the system capability indicators. We represent a sys- tem capability indicator with such equation as, where and respectively denote the ex- pected value and a stochastic variable. Furthermore, a union probability density function where is the vector in the MIS is used to describe the system capability indicators. The system effectiveness is denoted as\n\nand its expression is, where is the MIRL. It can be easily known that the range of\n\nis. As the system effectiveness, reflects the extent to which system meets the requirements. If the\n\nMIRL is approximated by the hyperboxes, i.e., where is the jth hyperbox and is the number of hyperboxes, the expression of will change into.\n\n2.3. Sensitivity and Relativity Analysis for the Indicators of Stochastic Type\n\nThe purpose of getting the indicator’s sensitivity includes: 1) acquire the weightiness of indicator and determine\n\n(a) (b) (c)\n\nFigure 2.(a) MIRL1 for the stealth aircraft of type A; (b) MIRL2 for the type B; (c) the intersection of MIRL1 and MIRL2 acquired by the approaching method.\n\nwhich indicator is critical; 2) more carefully deal with the uncertain and sensitive context or system capability indicators; and 3) optimize system.\n\nThe sensitivity analysis is realized by the perturbation analysis. This paper uses the analytic method to acquire the indicator’s sensitivity. In addition, every indicator is normalized within a common range for simplicity.\n\nDefinition 1. Denote the sensitivity of indicator as and its definition is\n\n,\n\nwhere and is the expected value of the system capability indicator (stochastic variable)\n\nand also the perturbation variable.\n\nDefinition 2. Denote the relativity between indicator and as.\n\nIf;\n\nIf;\n\nDefinition 3. Possibility measure . is assumed as a non-empty set and as the power set of.\n\nIf meets three axioms, i.e., , , and, it is\n\ncalled possibility measure.\n\nDefinition 4. System effectiveness based on the fuzzy indicators and possibility measure. The ex- pression of is, where is multi-dimensional fuzzy indicator and is the re- quirement locus. The value of can be acquired by fuzzy simulation method which is detailed as follows.\n\nStep 1 Set and is a very small estimate of.\n\nStep 2 According to subordinating fuction of, uniformly produce from set and let.\n\nStep 3 Set.\n\nStep 4 If and, set\n\nStep 5 Repeat step 2 to 4 times.\n\nStep 6 If, return.\n\nCredibility, which reveals the possibility and impossibility simultaneously, is a novel mode describing fuzzy relation. So, as a measure, credibility is more reasonable than possibility. The definition of credibility measure is stated as follows.\n\nDefinition 5. Credibility measure . With the assumption that is a possibility space and is the element of power set, credibility measure is defined as\n\n.\n\nDefinition 6. System effectiveness based on the fuzzy indicators and credibility measure. The ex-\n\npression of can be stated as, where is multi-\n\ndimensional fuzzy indicator and is the requirement locus. Similarly, it can be acquired by fuzzy simulation method.\n\nStep 1 Uniformly produce from and let if, where is a very small number.\n\nStep 2 Let\n\nand then return.\n\nFor the above two algorithms, it is important to judge whether the is within or not. And based on the tree structure output of partitioning algorithms introduced in the first one of this series papers, it can be quickly judged whether the is within or not. The judging procedure is detailed as follows. Firstly, judge whether the is within the top MEH (root of the tree) or not. If it is not, it is not within either and the judging procedure ends. If is within the MEH, judge whether is within the next-level MIH or not. If is within the MIH, it is also within and the judging procedure ends. If it is not, judge whether is within one certain MEH of this level or not. If is contained by one MEH, continue the recursive searching proce- dure from this MEH. If searching reaches the bottom of the tree and no one MIH containing is found, it means that this is not within.\n\n2.4. Effectiveness Analysis Based on Fuzzy MIRL\n\nDefinition 7. Fuzzy monotonic indicator requirement locus (FMIRL). is a fuzzy set on a cer- tain domain; is the l-cut set of; and is the effective range in P (MIS) which meets the re- quirement. These concepts are now illustrated by an example. For an air defense system, the mission re- quirement measure (fuzzy set) is killing the enemy aircraft. The domain is the kill probability. Set a l to de- termine the. is a requirement measure of kill probability that must be larger than a certain value.\n\nThe fuzzy system capability indicator (assumed to be stochastic) effectiveness analysis model is introduced as follows. can be determined according to l. If, we can get the corresponding conclusion, i.e., and according to the monotonic assumption.\n\nand;\n\nwhen. So,. And can be solved rapidly by\n\nthe binary method. The fuzzy effectiveness analysis model is not only suitable for the fuzzy requirement but also able to decrease misclassification rate of the boundary points especially when the system capability indicator values are distributed near the boundary of MIRL.\n\n3. MIS-Based System Capability Indicator Optimization Design\n\n3.1. Budget-Oriented Optimization Design\n\nBudget-oriented capability indicator optimization design will be discussed in this section. In order to make a system meet a certain requirement, there are usually many development “paths” with different cost for system capability indicators. Figure 3 gives three linking paths (i.e., OA, OB and OC) between point O representing the initial system capability and the requirement locus. And point O is assumed to be out of the requirement locus. Certainly, besides these three paths, there exist countless other linking paths between point O and requirement locus. The purpose of capability indicator optimization design is to find out the path with the minimal budget or least time cost. In this paper, the optimization model is used to get the optimized expectation values of the statis- tical capability indicators according to practical problems. For simplicity, all the capability indicators are as- sumed to be mutually independent and have fixed variances (i.e., every capability indicator can be developed independently). And the optimization design model is:\n\nIn this model: is union probability density function and, as the optimization variable, repre-\n\nFigure 3. The paths to monotonic indicators requirement locus.\n\nsents the expectation value of;, , and respectively represent the MIRL (monotonic indicators requirement locus), the effectiveness value, and the capability indicator effectiveness requirement threshold; represents the budget-cost function of capability indicator varying from to. The which meets the effectiveness requirement and has the least budget can be found through optimization. The corresponding duality programming is\n\n.\n\nThis model will be used to get the maximized capability indicator effectiveness value with the budget being less than.\n\n3.2. Time-Oriented Optimization Design\n\nCapability development is usually a time-consuming work. The following optimization model is able to find the path from the original capability point to MIRL with the least time.\n\n.\n\nIn this model: represents time-cost function of the capability indicator varying from to and represents the total number of capability indicators. This model can also be transformed into the following programming model,\n\n.\n\nSimilarly, the dual programming can be changed into\n\n.\n\n4. A Notional Example: Number Optimization of Missiles Available\n\nIn this section, some models and algorithms mentioned above are used to program missile available numbers of six types so as to meet the mission requirement with the least budget cost.\n\n4.1. Problem Context\n\nTen types of opponent targets (i.e., T1, T2, T3, T4, T5, T6, T7, T8, T9, and T10) are assumed to be destroyed to win the air superiority. Target-attacking missile types are assumed to be type A, type B, type C, type D, type E, and type F.\n\nBecause missile production cost is usually very large, so the missile number available of each type is an im- portant capability indicator. The capability indicator optimization design will be carried out in a MIS with six dimensions (i.e., six missile available number indicators). The required number of one certain type missile, when only this type of missile is used to destroy one target of a certain type, is presented in the following table. “No” in the table means that type of missile can not be assigned to attack the corresponding type of target.\n\nIf there exists solution meeting the following equality and inequalities group, it means that the missile number available of each type can satisfy the mission requirement.\n\nIn the above equation and inequalities group, , , and respectively represent the number of the type of missile assigned to the type of target, the required number of the type missile to destroy type target, and the number available of type missile.\n\nIf the number available of each type of missiles all meet the mission, the requirement measure is set to 1, otherwise, set to 0. And the function is obviously monotonic.\n\n4.2. Capability Indicator Analysis and Optimization Design\n\nThe MIS-based capability indicator analysis and optimization design is performed according to the following steps.\n\n1) Construct the MIS, i.e., select all the MIS dimensions and standardize every dimension value range into the range [0,100].\n\n2) Create the monotonic function representing the relation between the capability indicators and requirement measure. Actually, the function is already put forward in section 4.1.\n\n3) Apply the partitioning algorithm to acquire the MIRL, i.e., accomplish capability indicators requirement analysis.\n\n4) Use the capability indicators effectiveness model (introduced in section 2.2) to judge whether the present capability meet the requirement or not. If not, go to step5 to perform optimization design, else stop capability indicator analysis and optimization design.\n\n5) Trade off missile number for each type with the optimization model introduced in section 3.1.\n\nBased on the data of Table 1, the MIRL shown in Figure 4 is acquired by the partitioning algorithm. The numbers of six types of missiles (i.e., type A, B, C, D, E, and F) available is respectively assumed to be 14 - 26 number unit, 14 - 26 number unit, 12 - 29 number unit, 14 - 26 number unit, 14 - 26 number unit, and 12 - 29 number unit.\n\nAll the numbers of six types of missiles available are assumed to be subject to normal distribution and according to principle, the corresponding distribution functions can be approximately stated as follows:\n\nFurthermore, the budget costs of six types of incremental missiles are assumed to be evaluated with the fol- lowing functions:\n\n;;;\n\n;;.\n\nIn these functions: respectively represent the incremental missile numbers of the type A, B, C, D, E, and F; respectively represent the budget costs of the incremental missiles of type A, B, C, D, E, and F.\n\nIf the effectiveness value (introduced in section 2.2) is required to be larger than 0.9, the optimized in- cremental numbers of six types of missiles can be acquired through conducting the following programming model.\n\nTable 1. The required number of a certain type missile to destroy a certain type target.\n\n(a) (b)\n\nFigure 4. (a) The MIRL of missile type A, B, and C; (b) The MIRL of missile type D, E, and F.\n\nIn this model:\n\nrespectively represent the probability density functions of missile numbers of type A, B, C, D, E, and F; respectively represent the expectation values of six distribution functions; respectively represent the incremental missile numbers of type A, B, C, D, E, and F. The initial numbers of six types of missiles are all and variances of six distribu- tion functions are 2 or 3.\n\nThrough solving the programming model mentioned above, we get such conclusions that the optimized in- cremental missile numbers of type A, B, C, D, E, and F should be 28 number unit, 28.2 number unit, 27.9 num- ber unit, 25.7 number unit, 36.5 number unit, and 35.9 number unit and the optimized least budget cost will be 3275 (cost unit).\n\n5. Conclusion and Perspective\n\nThis series of papers propose and introduce a novel model, i.e., MVS (monotonic vector space) model, in which there exist some monotonic mappings. Many MVS operations (such as partitioning operation, synthesizing op- eration, sampling operation, etc.) can be directly used to solve related practical problems, which have been ela- borated in the first one of this series papers. Based on MVS, some system capability indicator analysis and op- timization models (e.g., requirement analysis model and effectiveness analysis model) are put forward and used to perform capability indicator analysis and optimization design in the second one of this series of papers. The future work for MVS model includes: 1) improving the present operation algorithms and developing other types of operations (such as partitioning operation in non-deterministic discrete or non-deterministic continuous MVS), which is of great significance to practical problems; 2) extending the application of the MVS model in other fields such as imaging processing and multi-object programming, etc.\n\nFunding\n\nSupported by NSFC, China, PRC, Grant No 70871120.\n\nReferences\n\n1. WSEIAC Report, “AFSC TR-65-2, Vol. 1-3, Final Report of Task Group 2, Preciction Measurement,” 1965.\n2. C. L. Hwang, et al., “Multiple Attribute Decision Making,” Springger-Verlag, Berlin, 1981. http://dx.doi.org/10.1007/978-3-642-48318-9\n3. W. D. Cook and M. A. Kress, “A Multiple-Criteria Composite Index Model for Quantitative and Quantitative Data,” European Journal of Operation Research, Vol. 78, No. 3, 1994, pp. 367-379. http://dx.doi.org/10.1016/0377-2217(94)90046-9\n4. T. L. Saaty, “The Analytic Hierarchy Process,” McGraw-Hill, 1980.\n5. V. Bouthonnier and A. H. Levis, “System Effectiveness Analysis of C3 Systems,” IEEE Transaction on Systems, Man, and Cybernetics, Vol. 14, No. 1, 1984, pp. 48-54. http://dx.doi.org/10.1109/TSMC.1984.6313268\n6. P. K. Davis, “Analytic Architecture for Capabilities-Based Planning, Mission-System Analysis, and Transformation,” RAND MR-1513-OSD, 2002.\n7. M. C. Bohner, “Computer Graphics for System Effectiveness Analysis,” M.S. Thesis, Massachusetts Institute of Technology, Dept. of Electrical Engineering and Computer Science, 1986.\n8. X. F. Wu, “SEA Methods and Its Application in C3I System(4)-System Effectiveness Analysis,” System Engineering Theory and Practice, Vol. 2, No. 2, 1999, pp. 44-49.\n9. B. A. Nixon, “Management of Performance Requirements for Information Systems,” IEEE Trans on Software Engineering, Vol. 26, No. 12, 2000..\n10. P. H. Cothier and A. H. Levis, “Timeliness and Measures of Effectiveness in Command and Control,” In: S. J. Andriole and S. M. Halpin, Eds., Information Technology for Command and Control, IEEE Press, New York, 1991.\n11. N. M. Paul, “Evolution Equations of C3I: Toward a New Understanding of Military of Combat,” In: Application of Artificial Intelligence to Command & Control Systems IEE Computing Series 13, Peter Peregrinus Ltd..\n12. L.Washington, “Effectiveness Analysis of Flexible Manufacturing Systems LIDS-TH-1430,” 1985.\n13. B. Liu, “Uncertain Programming with Applications,” Tsinghua Springer Press, Beijing, 2003."
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https://alexanderpruss.blogspot.com/2013/04/grounding-and-category-theory.html?showComment=1365539672482 | [
"## Tuesday, April 9, 2013\n\n### Grounding and category theory\n\nI've been searching for the right kind of mathematical structure to think about the phenomenon of grounding or partial grounding with. The orthodoxy is that the right structure is a partial ordering. That the axioms of partial ordering are satisfied by partial grounding has been challenged and defended. But even the critics have tended to take partial grounding to be something like[note 1] a single relation, or perhaps a small collection of related relations, between pairs of propositions or between a proposition and a set of propositions. I've offered two suggestions (first and second) on how to model grounding using graphs. But I now think all of these approaches abstract away too much of the structure of grounding and/or are unable to capture all the prima facie possibilities that a theory of grounding should recognize.\n\nFor instance, the relational approach loses sight of the structural fact that one can sometimes have two different grounding relations between a pair of propositions. Let W be the proposition that Smith is drinking water and let H be the proposition that Smith is drinking H2O. Let D be the disjunction of W and H: the proposition that Smith is drinking water or drinking H2O. If we think of grounding as a relation, we can certainly say that H grounds D. But we want to be able to say that there are two groundings between H and D: H grounds D directly by being one of its disjuncts and indirectly by grounding W which is another disjunct. And this structure is not captured by the orthodoxy. The graph approach nicely captures this sort of thing, but it does not adequately capture the compositional structure which is that the indirect grounding that H provides for D is a composition between a grounding by H of W and a grounding by W of D. There are ways to make it capture this, say by identifying composition of grounding with sequences of arrows in the directed graph, but this won't work for infinite sequences of arrows, something that we should not rule out in the formalism. I realized this when trying to finish a paper on grounding and fundamentality.\n\nI am now wondering if the right structure isn't that of a category. Maybe the objects are true propositions. The arrows are groundings, i.e., token grounding-like relations. Every arrow is at least a partial weak grounding (weak, because there are identity arrows). Some arrows may be full groundings. There is a nice associative compositional structure.\n\nThere will be further structure in the category. For instance, perhaps, every family of true propositions will have a coproduct, which is the conjunction of the propositions in that family. The canonical injections are the partial groundings that conjuncts give to the conjunction, and the universal property of the coproduct basically says that when a proposition is weakly partially grounded by each member of a family of propositions, then there is a coproduct weak partial grounding arrow from the conjunction of that family to the proposition. This is very nice.\n\nWe might also consider a move to a category where the objects are all propositions, but that creates the challenge that we need to keep track of which propositions are true and which groundings are actual. For that p partially grounds q is, in general, a contingent matter.[note 2] Truth and actuality of grounding respects the category structure. Actual groundings only obtain between truths, and compositions of actual groundings are actual.\n\nThe category structure is going to yield transitivity of weak partial grounding. There are apparent counterexamples to transitivity. But it is my hope that when we keep track of the additional structure, and think of the token groundings as central rather than the relation of there being a token grounding, the result is not going to be problematic.\n\nIt is now interesting to investigate what different category theoretic phenomena occur in the grounding category, and how they connect up with metaphysical phenomena. One thing I'd like to see is if there is a neat category theoretic characterization of full, as opposed to partial, grounding.\n\nI have an intuitive worry about the above approaches. Intuition would suggest that if conjunctions are coproducts, then disjunctions would be products. But they're not. For in general there is no grounding from a disjunction to disjuncts. This could be related to another worry, that because categories include identity arrows, I had to take the arrows to be weak groundings—i.e., I had to allow each proposition to count as having a grounding between itself and itself.\n\nI do not know if category theory will in the end provide a good mathematical home to grounding structures, but I am hopeful.\n\nBrian Cutter said...\n\nVery cool! I'd also be interested to see whether full grounding, as opposed to partial grounding, could be given a neat category-theoretic treatment. (I don't know enough about category theory to even speculate about this.) If not, then that may be a problem for the whole category-theoretic approach to grounding. As others have pointed out, it does not seem possible to take partial grounding as primitive and define full grounding in terms of it. The worry stems from the fact that two truths might have the same collection of partial grounds while differing in their full grounds. (So, e.g., if A and B are each fundamental truths, then A&B and AvB are both partially grounded in A and partially grounded in B, and not partially grounded in anything else. But AvB, unlike A&B, is fully grounded in A, as well as in B.) On the other hand, it is straightforward to define partial grounding in terms of full grounding, at least on the standard conception of full grounding as a relation between a set/plurality of facts/propositions and a fact/proposition. (q is a partial ground of p iff q is a member of some set/plurality which is a full ground of p.)\n\nAlexander R Pruss said...\n\nI was hoping for this to be an alternative to my relativization of grounding to graphs. But if it's not an alternative to it but an extension, then I will say that grounding is always relative to a grounding-category.\n\nI can then define full grounding more or less as I do on graphs. Here's how I do on graphs. These are directed graphs, and each arrow means \"directly partially grounds\". (I am worried about the \"directly\", which was why I switched to categories.) A set S of nodes fully grounds a node x relative to a graph G provided that every maximal ancestral lineage of x meets S. S is an ancestral lineage of x iff S is a set of ancestors of x and S is totally ordered under the transitive closure (in the full graph) of the arrow relation.\n\nHere's how this takes care of the disjunction problem. There is no grounding overdetermination within a single grounding graph. Thus, A or B, in the case where both A and B are true, has two grounding graphs, one of them ending with A→(A or B) and the other ending with B→(A or B). In the first graph every maximal ancestral lineage of (A or B) intersects {A}. In the second graph every maximal ancestral lineage of (A or B) intersects {B}.\n\nOn the other hand, any grounding graph for (A and B) ends with two arrows going to (A and B), one from A and the other from B. Thus, {A,B} will be a full grounding for (A and B) (as will {A,B1,B2,B3} if B1, B2 and B3 are the only three nodes that have arrows pointing to B, etc.)\n\nNow, back to the Category Theory version. (Which I am very rusty on. About two decades ago, I passed a comp on Category Theory, but have not used it almost at all in between.) Define a partial ordering on the category by y<=x iff there is an arrow from y to x. An ancestral lineage of x is a set of objects less than x that are totally ordered by <=. A set S of objects is then a full grounding of x provided that S intersects every maximal ancestral lineage of x.\n\nThe primitive concept on this view is: C is a grounding category. Intuitively, a grounding category represents a complete, consistent, correct and non-redundant grounding story for every object (i.e., proposition) in the category. The arrows are weak partial groundings relative to those stories.\n\nThere is a lot to work out. I've only been thinking about categories in this context since this morning, in light of technical troubles with the graph-theoretic approach.\n\nI may want to weaken the category axioms to remove the identity arrows in some cases.\n\nAlexander R Pruss said...\nThis comment has been removed by the author.\nAlexander R Pruss said...\n\nI've also been playing with the idea of using a similar framework for causation and for explanation.\n\nAlexander R Pruss said...\n\n1. In my long comment, <= will only be a partial ordering if there are no loops.\n\n2. Another option for a category-based approach is to make the objects in the category be labeled by propositions, without the objects actually being propositions. This would mean there is a map from objects to propositions. We could then suppose the categories to have no loops, which would give us something like irreflexivity. The objects would then be \"grounding nodes\" rather than propositions. The same proposition can, perhaps, be found at more than one grounding node. For instance, consider the grounding loop: <I should respect you> → <I should keep promises to you> → <I should respect you>, when I promised to respect you. We might want to have two separate nodes for <I should respect you>--one, a node that grounds the duty to keep promises to you, and the other the node that is grounded by the duty to keep promises to you.\n\nI don't know what the nodes would correspond to in the metaphysics of the world. Maybe states of affairs? For maybe there is more than one state of affairs of my being obligated to respect you. One such state of affairs grounds my duty to keep my promises and the other is grounded by my duty to keep my promises.\n\nThis would get rid of the duplication between propositions and states of affairs on some ontologies, like Plantinga's.\n\nPropositions could even be taken to be equivalence classes of states of affairs.\n\nAnd we could then make sense of the odd locution \"It is doubly true that p.\" It is doubly true that p iff there are two states of affairs, each labeled with p.\n\nThis may be nuts.\n\nBrian Cutter said...\n\nVery interesting. A lot here to think about. I'm pretty satisfied with the graph-relative definition of full grounding in terms of partial grounding. (Though I'm also worried about defining partial grounding in terms of direct partial grounding, perhaps for the same reasons you are. A lot of folks in the grounding literature think that where F is a determinate of G, and a is F, then [Fa] grounds [Ga]. Where we have a determinable associated with an infinite quality space, e.g. height, we might get collections of propositions, e.g. a collection of increasingly determinable true propositions about my height, which can be densely ordered by the grounding relation, where none of the propositions in the collection can be said to \"directly\" ground any of the others.)\nI also like the second point in the last comment as a way of dealing with grounding loops. Another case which comes to mind in which we might have a similar grounding loop, which might be handled similarly: Assume two (apparently widely accepted) principles: (i) If a is F, then [a is F] grounds [something is F]. (ii) Given any true proposition p, p grounds [p is true]. Now, let Q be the proposition that something is true. Q grounds [Q is true] by the second principle. And [Q is true] grounds Q by the first principle.\n\nAlexander R Pruss said...\n\nThe height case doesn't bother me, because I don't think the less determinate height fact is grounded in a more determinate height fact, unless the latter is maximally determinate.\nBut we can come up with a case where we have an infinite conjunction p1 and p2 and ...., where pk is grounded in qk. Then the whole conjunction is grounded in q1 and p2 and ..., which is grounded in q1 and q2 and p3 and .... But then we want all of this sequence to be grounded in q1 and q2 and q3 and ....\nYes, the truth case is another case in the vicinity, but it's also next door to Patrick Grim style paradoxes about \"all propositions\" so I am cautious. But then again my own motivating cases are related to truth glut paradoxes like the truthteller, so I am not in much better shape.\nIf you think these approaches are worthwhile, maybe we can write something together. You know the grounding literature better than I.\n\nAlexander R Pruss said...\n\nBy the way, I think these infinite cases might rather neatly correspond to the concept of a limit in a category.\n\nAlexander R Pruss said...\n\nHow should we interpret the identity arrows that go from p to p? One way is that a special case of the concept of weak partial grounding is the concept of grounding equivalence. For instance (p and q) is grounding-equivalent to (q and p). There will be a grounding arrow from (p and q) to (q and p) and one going back. Moreover, these two arrows will be inverses of each other.\n\nWe can now say that f:p→q is a grounding-equivalence iff f is an isomorphism, i.e., there is a g:p→q such that fg and gf are both identity arrows.\n\nGiving grounding-equivalences, we will be able to say that when f is the obvious grounding by p of (p and q) and g is the obvious grounding by p of (q and p), these aren't completely separate groundings. Rather, they are intimately related: f = hg, where h is the grounding-equivalence by (q and p) of (p and q).\n\nAlexander R Pruss said...\n\nMy account of full grounding in terms of partial grounding has this counterexample. Let RLJ be the proposition that Romeo loves Juliet. Let R be the proposition that Romeo exists. Let J be the proposition that Juliet exists. Suppose love is a fundamental relation. Then RLJ is partially grounded in R as well as in J. But it is not fully grounded in {R,J}, of course.",
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"Anonymous said...\n\nLogics via category theory - Toposes"
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https://or.meta.stackexchange.com/questions/110/what-makes-a-good-help-me-with-my-model-question?noredirect=1 | [
"# What makes a good “help me with my model” question?\n\nI suspect that OR.SE will see a good number of questions that ask for help with an optimization model that's not working as desired. Here are a few examples of such questions, from other sites:\n\nStack Overflow:\n\nmath.SE:\n\nWe've also had one or two on OR.SE, if I'm not mistaken. Many of these questions are well crafted and insightful and make it easy for readers to understand the question and develop answers. Others are sloppy and non-intuitive and make me not want to invest time to answer them. (I'm not calling these specific questions out as good or bad examples, just as examples of the kinds of questions I'm referring to in this post.)\n\nI'm sure many on this site don't really want to answer \"help me with my model\" questions, no matter how well written they are, and that's fine. Others (myself included) enjoy answering questions like that from time to time.\n\nOne way or another, I am sure that questions like this will drive a lot of traffic our way, and we should welcome them.\n\nI also think that one of the selling points of OR.SE is that we can provide guidance on all aspects of the modeling process, from algebraic formulation through coding and analysis. Right now, theory questions tend to be asked on math.SE, coding/implementation questions on SO, modeling questions tend to get split between them, and there are many other exceptions of course. On OR.SE, one can ask questions both that include math models (MathJax!) and programming (code blocks!) in the same post.\n\nI think we should provide some guidance to askers about how to ask a \"help me with my model\" question. This will make it easier on both the askers and the answerers, and will (I hope) help jump-start the quality of these questions. My idea is that at some point this guidance would be part of our help page, and we can also link to it when responding to questions that need some work before they are answerable.\n\nI'll post my thoughts in an answer below. I hope others will comment and/or post their own answers.\n\n# How to ask for help with your model\n\nWe love mathematical models and are happy to help you with yours. Before you ask your question, please read these suggestions about how to ask a good question. Doing so will make it easier for us to provide good answers.\n\n1. Use a descriptive title. Include (some of) the relevant details and be specific. Avoid vague, general titles like \"Why isn't my model working?\" Good titles will catch the reader's eye and make us want to read the question.\n\n2. Include your algebraic model. Include your entire algebraic formulation, or at least the relevant portion of it. Write your formulation using MathJax. Explain all of the notation and what the objective function and constraints are doing, if it's not obvious.\n\n3. Include minimal, reproducible code. If you are including code (AMPL, GAMS, Python, etc.), use a minimal, complete, verifiable example. That is, give us enough code so that we can run it, reproduce your problem/error, and try to debug it. Don't post very long code listings that contain a lot of stuff that's irrelevant to the problem.\n\n4. Be specific about what's not working. Tell us exactly what's not working the way you want it to. Are you getting error messages? Include them. Are you getting results that don't seem logical or numerically correct? Include them, and tell us what you think is wrong with them.\n\n5. Keep it self-contained. Your question should be as self-contained as possible. Avoid posts that require us to click on external images just to see your formulation, or to read a journal article to understand what you want your model to do. External links are good if they provide additional context, but the question should be able to stand on its own even if we don't click on the links.\n\n6. Use the tag. This tag will let users know that your question is one that needs help with a modeling problem.\n\n## A good example\n\nTitle: Why does AMPL/CPLEX give me \"nonlinear\" error for a knapsack-type problem?\n\nQuestion: I am trying to modify a 0–1 knapsack problem to require that at least half of the items chosen are \"priority\" items. But I'm getting an error message in AMPL that I don't understand.\n\nHere is my model. We have $$n$$ items, each with a weight $$w_i$$ and a value $$v_i$$. For item $$i$$, $$p_i$$ = 1 if the item is a \"priority\" item and 0 otherwise. The knapsack has a capacity of $$W$$. (These are all parameters.) The decision variables are $$x_i$$, which equals 1 if we choose the item, and 0 otherwise. The integer programming formulation is:\n\n\\begin{alignat}{2} \\text{maximize} \\quad & \\sum_{i=1}^n v_ix_i && \\\\ \\text{subject to} \\quad & \\sum_{i=1}^n w_ix_i \\le W &\\quad& \\forall i=1,\\ldots,n \\\\ & \\frac{\\sum_{i=1}^n p_ix_i}{\\sum_{i=1}^n x_i} \\ge 0.5 \\\\ & x_i \\in \\{0,1\\} && \\forall i=1,\\ldots,n \\end{alignat}\n\nThe first and third constraints are standard knapsack constraints. The second constraint says at least half of the items chosen have to be priority items.\n\nI implemented my model in AMPL. Here is my *.mod file.\n\nparam n; # number of items\nset ITEMS = 1..n; # set of items\n\nparam weight{ITEMS}; # weight of each item\nparam value{ITEMS}; # value of each item\nparam is_priority{ITEMS} binary; # 1/0 if item is priority/not\n\nparam capacity; # knapsack capacity\n\nvar x{ITEMS} binary; # do we choose the item?\n\nmaximize TotalValue:\nsum {i in ITEMS} value[i] * x[i];\n\nsubj to Capacity:\nsum {i in ITEMS} weight[i] * x[i] <= capacity;\n\nsubj to HalfPriority:\n(sum {i in ITEMS} is_priority[i] * x[i]) / (sum {i in ITEMS} x[i]) >= 0.5;\n\n\nHere is a simplified version of my *.dat file. (My real *.dat file has n > 500.)\n\nparam n := 4;\n\nparam: weight value is_priority :=\n1 20 50 1\n2 15 40 0\n3 15 55 0\n4 20 30 1 ;\n\nparam capacity := 40;\n\n\nI am using CPLEX as the solver. When I solve the model, I get the following error message:\n\nCPLEX 12.8.0.0: Constraint _scon is a nonquadratic nonlinear constraint.\n\n\nWhat am I doing wrong?\n\n(Spoiler: The \"priority\" constraint is nonlinear. Just multiply both sides by the denominator.)\n\n## A bad example\n\nTitle: Question about solver\n\nQuestion: I want the knapsack problem to include a constraint that half the items are priority items.\n\nHere is the constraint in AMPL:\n\nsubj to HalfPriority:\n(sum {i in ITEMS} is_priority[i] * x[i]) / (sum {i in ITEMS} x[i]) >= 0.5;\n\n\nBut AMPL gives me an error. What am I doing wrong?"
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https://www.colorhexa.com/012b0f | [
"# #012b0f Color Information\n\nIn a RGB color space, hex #012b0f is composed of 0.4% red, 16.9% green and 5.9% blue. Whereas in a CMYK color space, it is composed of 97.7% cyan, 0% magenta, 65.1% yellow and 83.1% black. It has a hue angle of 140 degrees, a saturation of 95.5% and a lightness of 8.6%. #012b0f color hex could be obtained by blending #02561e with #000000. Closest websafe color is: #003300.\n\n• R 0\n• G 17\n• B 6\nRGB color chart\n• C 98\n• M 0\n• Y 65\n• K 83\nCMYK color chart\n\n#012b0f color description : Very dark (mostly black) cyan - lime green.\n\n# #012b0f Color Conversion\n\nThe hexadecimal color #012b0f has RGB values of R:1, G:43, B:15 and CMYK values of C:0.98, M:0, Y:0.65, K:0.83. Its decimal value is 76559.\n\nHex triplet RGB Decimal 012b0f `#012b0f` 1, 43, 15 `rgb(1,43,15)` 0.4, 16.9, 5.9 `rgb(0.4%,16.9%,5.9%)` 98, 0, 65, 83 140°, 95.5, 8.6 `hsl(140,95.5%,8.6%)` 140°, 97.7, 16.9 003300 `#003300`\nCIE-LAB 14.223, -22.094, 13.901 0.963, 1.769, 0.743 0.277, 0.509, 1.769 14.223, 26.103, 147.823 14.223, -12.626, 12.435 13.299, -10.353, 5.999 00000001, 00101011, 00001111\n\n# Color Schemes with #012b0f\n\n• #012b0f\n``#012b0f` `rgb(1,43,15)``\n• #2b011d\n``#2b011d` `rgb(43,1,29)``\nComplementary Color\n• #082b01\n``#082b01` `rgb(8,43,1)``\n• #012b0f\n``#012b0f` `rgb(1,43,15)``\n• #012b24\n``#012b24` `rgb(1,43,36)``\nAnalogous Color\n• #2b0108\n``#2b0108` `rgb(43,1,8)``\n• #012b0f\n``#012b0f` `rgb(1,43,15)``\n• #24012b\n``#24012b` `rgb(36,1,43)``\nSplit Complementary Color\n• #2b0f01\n``#2b0f01` `rgb(43,15,1)``\n• #012b0f\n``#012b0f` `rgb(1,43,15)``\n• #0f012b\n``#0f012b` `rgb(15,1,43)``\n• #1d2b01\n``#1d2b01` `rgb(29,43,1)``\n• #012b0f\n``#012b0f` `rgb(1,43,15)``\n• #0f012b\n``#0f012b` `rgb(15,1,43)``\n• #2b011d\n``#2b011d` `rgb(43,1,29)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #001206\n``#001206` `rgb(0,18,6)``\n• #012b0f\n``#012b0f` `rgb(1,43,15)``\n• #024418\n``#024418` `rgb(2,68,24)``\n• #025d20\n``#025d20` `rgb(2,93,32)``\n• #037629\n``#037629` `rgb(3,118,41)``\nMonochromatic Color\n\n# Alternatives to #012b0f\n\nBelow, you can see some colors close to #012b0f. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #012b04\n``#012b04` `rgb(1,43,4)``\n• #012b08\n``#012b08` `rgb(1,43,8)``\n• #012b0c\n``#012b0c` `rgb(1,43,12)``\n• #012b0f\n``#012b0f` `rgb(1,43,15)``\n• #012b13\n``#012b13` `rgb(1,43,19)``\n• #012b16\n``#012b16` `rgb(1,43,22)``\n• #012b1a\n``#012b1a` `rgb(1,43,26)``\nSimilar Colors\n\n# #012b0f Preview\n\nThis text has a font color of #012b0f.\n\n``<span style=\"color:#012b0f;\">Text here</span>``\n#012b0f background color\n\nThis paragraph has a background color of #012b0f.\n\n``<p style=\"background-color:#012b0f;\">Content here</p>``\n#012b0f border color\n\nThis element has a border color of #012b0f.\n\n``<div style=\"border:1px solid #012b0f;\">Content here</div>``\nCSS codes\n``.text {color:#012b0f;}``\n``.background {background-color:#012b0f;}``\n``.border {border:1px solid #012b0f;}``\n\n# Shades and Tints of #012b0f\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, #000502 is the darkest color, while #f0fff5 is the lightest one.\n\n• #000502\n``#000502` `rgb(0,5,2)``\n• #011808\n``#011808` `rgb(1,24,8)``\n• #012b0f\n``#012b0f` `rgb(1,43,15)``\n• #013e16\n``#013e16` `rgb(1,62,22)``\n• #02511c\n``#02511c` `rgb(2,81,28)``\n• #026523\n``#026523` `rgb(2,101,35)``\n• #03782a\n``#03782a` `rgb(3,120,42)``\n• #038b30\n``#038b30` `rgb(3,139,48)``\n• #049e37\n``#049e37` `rgb(4,158,55)``\n• #04b13e\n``#04b13e` `rgb(4,177,62)``\n• #05c444\n``#05c444` `rgb(5,196,68)``\n• #05d84b\n``#05d84b` `rgb(5,216,75)``\n• #05eb52\n``#05eb52` `rgb(5,235,82)``\n• #0af95a\n``#0af95a` `rgb(10,249,90)``\n• #1efa67\n``#1efa67` `rgb(30,250,103)``\n• #31fa74\n``#31fa74` `rgb(49,250,116)``\n• #44fb81\n``#44fb81` `rgb(68,251,129)``\n• #57fb8e\n``#57fb8e` `rgb(87,251,142)``\n• #6afc9b\n``#6afc9b` `rgb(106,252,155)``\n• #7dfca8\n``#7dfca8` `rgb(125,252,168)``\n• #91fcb5\n``#91fcb5` `rgb(145,252,181)``\n• #a4fdc2\n``#a4fdc2` `rgb(164,253,194)``\n• #b7fdce\n``#b7fdce` `rgb(183,253,206)``\n• #cafedb\n``#cafedb` `rgb(202,254,219)``\n• #ddfee8\n``#ddfee8` `rgb(221,254,232)``\n• #f0fff5\n``#f0fff5` `rgb(240,255,245)``\nTint Color Variation\n\n# Tones of #012b0f\n\nA tone is produced by adding gray to any pure hue. In this case, #151716 is the less saturated color, while #012b0f is the most saturated one.\n\n• #151716\n``#151716` `rgb(21,23,22)``\n• #141815\n``#141815` `rgb(20,24,21)``\n• #121a15\n``#121a15` `rgb(18,26,21)``\n• #101c14\n``#101c14` `rgb(16,28,20)``\n• #0f1d14\n``#0f1d14` `rgb(15,29,20)``\n• #0d1f13\n``#0d1f13` `rgb(13,31,19)``\n• #0b2112\n``#0b2112` `rgb(11,33,18)``\n• #092312\n``#092312` `rgb(9,35,18)``\n• #082411\n``#082411` `rgb(8,36,17)``\n• #062611\n``#062611` `rgb(6,38,17)``\n• #042810\n``#042810` `rgb(4,40,16)``\n• #032910\n``#032910` `rgb(3,41,16)``\n• #012b0f\n``#012b0f` `rgb(1,43,15)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #012b0f is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population"
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https://cs.stackexchange.com/questions/115079/algorithm-for-solving-system-of-congruences | [
"# Algorithm for solving system of congruences\n\nI'm trying to understand an algorithm shown in my textbook for solving a system of congruences. The problem states:\n\nLet $$r$$ and $$m$$ be arrays of size $$n$$, such that every two numbers (modulos) from $$m$$ are coprime. Find $$x$$ such that\n\n$$x \\bmod m_0=r_0\\\\x \\bmod m_1=r_1\\\\...\\\\x \\bmod m_{n-1}=r_{n-1}$$\n\nSo yeah, the classic congruence system problem. The naiive approach is shown first and then comes this:\n\nLet $$M = m_0 \\cdot m_1 \\cdot ... \\cdot m_{n-1}$$. Let's assume we can find the numbers $$w_0, w_1, ...w_{n-1}$$ such that $$w_i \\bmod m_i = 1$$ and $$w_i \\bmod m_j = 0$$ for $$i \\neq j$$. So we're finding numbers $$w_j$$ such that it is divisible by every number except for $$m_j$$ where the remainder is $$1$$. Then the solution x can be constructed as: $$x = r_0\\cdot w_0 + r_1\\cdot w_1 + ...+r_{n-1}\\cdot w_{n-1} (\\bmod M)$$\n\nWell I guess my question is simply - why? I can't seem to reason this out. It seems to me that $$x$$ is always going to be zero. If we take the sum given above and break it up into sums $$(r_0\\cdot w_0)\\bmod M + (r_1\\cdot w_1)\\bmod M + ...$$\n\nit seems like everything should be zero? Let's take the first expression $$(r_0 \\cdot w_0) \\bmod M$$. We've got $$r_0$$ times something that's a multiple of every $$m$$ except for $$m_0$$ (because of the way the numbers $$w$$ were defined), meaning that whole thing will be a multiple of every $$m$$ except for $$m_0$$. And then we're taking a modulo $$M$$ which is also a multiple of every $$m$$.\n\nHow does this give us a solution? The textbook is pretty vague about this and doesn't explain why it works this way. So, any ideas?\n\nThanks.\n\nIt is often less confusing to refrain from using congruences and simply write what it means in extension. I.e. when the textbook says to take an $$x$$ which is equal modulo $$M$$ to $$r_0\\cdot w_0 + \\dots + r_{n-1}\\cdot w_{n-1}$$, it means that any $$x$$ of the form $$x=r_0\\cdot w_0 + \\dots + r_{n-1}\\cdot w_{n-1} + k\\dot M$$, with $$k\\in \\mathbb{Z}$$, is a solution. Under this form it should be clearer that $$x$$ is indeed a solution since $$\\forall i\\in \\mathbb{N}_n,x\\equiv_{(m_i)}r_i\\cdot w_i + \\sum_{j\\neq i}r_j\\cdot w_j+ k\\cdot M\\equiv_{(m_i)}r_i$$ (as $$m_i|M$$, $$w_i\\equiv_{(m_i)}1$$, and for $$j\\neq i$$ $$m_i|w_j$$)."
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https://wiki.fis-ski.com/index.php/Contour_line | [
"# Contour line",
null,
"Contour line (also isoline or isarithm) of a function of two variables is a curve along which the function has a constant value. In cartography, a contour line (often just called a \"contour\") joins points of equal elevation (height) above a given level, such as mean sea level.\n\nContour line (also isoline or isarithm) of a function of two variables is a curve along which the function has a constant value. In cartography, a contour line (often just called a \"contour\") joins points of equal elevation (height) above a given level, such as mean sea level.\n\nContour line for a function of two variables is a curve connecting points where the function has the same particular value. The gradient of the function is always perpendicular to the contour lines.\n\nWhen the lines are close together the magnitude of the gradient is large: the variation is steep. A level set is a generalization of a contour line for functions of any number of variables.\n\n## Reference;\n\n1. See Wikipedia Contour lines"
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"https://wiki.fis-ski.com/images/thumb/350px-Countour_lines_topmap.JPG",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.797072,"math_prob":0.9868735,"size":995,"snap":"2020-45-2020-50","text_gpt3_token_len":221,"char_repetition_ratio":0.1493441,"word_repetition_ratio":0.049689442,"special_character_ratio":0.20100503,"punctuation_ratio":0.065476194,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9928891,"pos_list":[0,1,2],"im_url_duplicate_count":[null,9,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-01T14:35:46Z\",\"WARC-Record-ID\":\"<urn:uuid:6c4bb967-1957-460b-b9c4-80152a906a61>\",\"Content-Length\":\"14785\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3b8bd469-8caf-45e6-9f21-0b5969b1e00a>\",\"WARC-Concurrent-To\":\"<urn:uuid:95734732-88be-4540-b842-c07707a4b1a2>\",\"WARC-IP-Address\":\"94.103.99.161\",\"WARC-Target-URI\":\"https://wiki.fis-ski.com/index.php/Contour_line\",\"WARC-Payload-Digest\":\"sha1:HB6BXO7VJEVCUXEY2LV2XEV3LKH2JIRY\",\"WARC-Block-Digest\":\"sha1:5WDFEC7NRQLZ6I33F5KIWKDUOPH47BL3\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141674594.59_warc_CC-MAIN-20201201135627-20201201165627-00210.warc.gz\"}"} |
https://studyres.com/doc/24206836/q---indiastudychannel | [
"Survey\n\n* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project\n\nDocument related concepts\n\nOptical aberration wikipedia, lookup\n\nBirefringence wikipedia, lookup\n\nMagnetic circular dichroism wikipedia, lookup\n\nChristiaan Huygens wikipedia, lookup\n\nAiry disk wikipedia, lookup\n\nRetroreflector wikipedia, lookup\n\nDiffraction grating wikipedia, lookup\n\nThomas Young (scientist) wikipedia, lookup\n\nFourier optics wikipedia, lookup\n\nNonlinear optics wikipedia, lookup\n\nDiffraction wikipedia, lookup\n\nTranscript\n```DIFFRACTION\nDiffraction is the penetration of light wave towards the geometrical shadow region around a sharp obstacle\nsuch as the edge of a slit, sharp aperture etc.\n(a) In Fresnel class of diffraction, the source and/or screen are at a finite distance from the aperture. The\nwave front is either spherical or cylindrical in shape\n(b) In Fraunhoffer class of diffraction, the source and screen are at infinite distance from the diffracting\naperture. The wave front is plane.\nSingle slit Fraunhoffer diffraction\nP\nA slit of width ‘a’ is divided into N parallel strips of width x. Each strip acts as a\nradiator of Huygens’s wavelets and produces a characteristic wave disturbance at\nP, whose position on the screen for a particular arrangement of apparatus can be\ndescribed by the angle .\n\na\nIf the strips are narrow enough – which we assume – all points on a given strip\nhave essentially the same optical path length to P, and therefore all the light from the strip will have the same\nphase when it arrives at P. The amplitudes E0 of the wave disturbances at P from the various strips may be\ntaken as equal if is not too large.\nThe wave disturbances from adjacent strips have a constant phase difference between them at P, given by\nphase difference path difference\n\n2\n\n2\n \n\n x sin \n\nThus at P, N vectors with the same amplitude E0, the same\n\nfrequency and the same phase difference DF between adjacent\n\nmembers combine to produce a result disturbance. In the\nR\nlimiting case (N → ∞), → 0\nR\nFrom geometry\nE\n \nE 2R sin , m\nR\n2\nE\n \nE m .sin \n\n /2\n2\nsin \n\n, where \nOr E E m .\n\n2\nEn\n\nEm\nAs\n\n2\n\na sin \n sin \nAs I E I I m \n\n \nIn brief\n sin \nE E m \n;\n \n a sin \n\n2\n2\n sin \nI I m \n ;\n \n a sin \n\n2\n\n2\nSingle slit diffraction formula\nMinima occur when, a = np\nn – 1, 2, 3…….\n a sin \n x sin x\n\n\nn\n.\nIf q is small, min \nCondition for minima\na\nFor n 3, it can be assumed that the nth secondary maximum is midway between the adjacent minima.\nIf we know the shape of front at t = 0 then from Huygens principle allows us to determine the shape of wave\nfront at any time t. Let us consider a diverging wave originating from point 0. f1 f2 represent a portion of the\nspherical wave front at t = 0 Now according to Huygens principle, each point of the wave front is the source\nof a secondary disturbance and the wavelets emanating from these points spread out in all dandies with the\nspeed of the wave. These wavelets emanating from the wave front are usually referred to as secondary\nwavelets and if we draw a common tangent to all these spheres we obtain the new position of the wave front\nat dater time. Thus if we wish to determine the shape of the wave front at t = we draw the sphere of radius\nv from each point on the spherical wave front. The common tangent draw to all these spheres gives the new\nposition of the wave front C1 C2 with center at O. we also have a back wave which is shown as 1 2.\nAccording to Huygens the amplitude of the secondary wavelength in forward direction and zero in the\nbackward direction\nSandeep tiwari\n```\nRelated documents"
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https://practicaldev-herokuapp-com.global.ssl.fastly.net/mathewchan/python-exercise-21-list-comprehensions-1a09 | [
"## DEV Community\n\nHHMathewChan\n\nPosted on • Originally published at rebirthwithcode.tech\n\n# Python Exercise 21: list comprehensions\n\n## Question\n\nYou are given three integers (x,y,z) representing the dimensions of a cuboid along with an integer n .\n\n• Print a list of all possible coordinates\n• given by (i, j , k) in a 3D grid\n• where the sum of i+j+k is not equal to n.\n• i.e. can be greater than or smaller than n\n• Here 0<=i<=x; 0<=j<=y; 0<=k<=z\n• Print the list in lexicographic increasing order.\n\n## Example\n\nx = 1\ny = 1\nz = 2\nn = 3\n\n• All permutations of (i, j, k) are:\n``````[[0, 0, 0], [0, 0, 1], [0, 0, 2], [0, 1, 0], [0, 1, 1], [0, 1, 2], [1, 0, 0], [1, 0, 1], [1, 0, 2], [1, 1, 0], [1, 1, 1], [1, 1, 2]]\n``````\n• Print an array of the elements that sum of i, j, k does not equal to n (The required output)\n``````[[0, 0, 0], [0, 0, 1], [0, 0, 2], [0, 1, 0], [0, 1, 1], [1, 0, 0], [1, 0, 1], [1, 1, 0], [1, 1, 2]]\n``````\n\n## My solution\n\n• code\n`````` # make list of all x,y and z possible number\nx = int(input())\ny = int(input())\nz = int(input())\nn = int(input())\nx = [i for i in range(x+1)]\ny = [i for i in range(y+1)]\nz = [i for i in range(z+1)]\n# for each combination of i,j,k will form a sublist in the final_list, and sum of i,j,k will not equal to n\nfinal_list = [[i,j,k]for i in x for j in y for k in z if (i+j+k)!=n]\nprint(final_list)\n``````\n\n## Other solution\n\n• a shorter version\n``````x = int(input())\ny = int(input())\nz = int(input())\nn = int(input())\n# for each combination of i,j,k will form a sublist in the final_list, and sum of i,j,k will not equal to n\nfinal_list = [[i,j,k]for i in range(x+1) for j in range(y+1) for k in range(z+1) if (i+j+k)!=n]\nprint(final_list)\n``````\n\n## My reflection\n\n• The first time to try problem need to obtain variable form standard input\n• good exercise.\n• The first time to use multiple for loop in list comprehension.\n\n## Credit\n\nchallenge on hackerrank"
]
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https://fbfmnigeria.org/and-pdf/999-kiran-gupta-classical-mechanics-of-particles-and-rigid-bodies-pdf-243-797.php | [
"and pdfWednesday, June 2, 2021 6:38:41 AM5\n\n# Kiran Gupta Classical Mechanics Of Particles And Rigid Bodies Pdf\n\nFile Name: kiran gupta classical mechanics of particles and rigid bodies .zip\nSize: 2860Kb\nPublished: 02.06.2021",
null,
"",
null,
"## Classical mechanics of particles and rigid bodies\n\nVariational Principle and Lagrange's Equations: Some techniques of the calculus of variations- Derivations of Lagrange's equations-from Hamilton's principle-extension of Hamilton's principle to nonholonomic systems-advantages of variational principle formulation- conservation theorems and symmetry properties.\n\nTwo body Central force Problems ; Reduction to the equivalent one-body problem-the equations of motion and first integrals-the equivalent one-dimensional problems and classification of orbits-the virial theorem-the differential equation of orbit and integrable power-law potentials-conditions for closed orbits bertrand's theorem -the Kepler Problem Inverse square law of force-the motion in time in the Kepler problem-the Laplace-Runge-Lenz vector-scattering in a central force field, Transformation of the scattering problem to the laboratory co-ordinates.\n\nI The Kinematics of Rigid Body Motion: The independent co-ordinates of a rigid body-orthogonal transformation-formal properties of the transformation matrix, The Euler Angles, Euler's theorem on the motion of a rigid body-finite rotations-infinitesimal rotations-rate of change of vector-the Coriolis force. The Rigid Body Equations of Motion: Angular momentum and kinetic energy of motion about a point- Tensor and dyadics-the inertia tensor and the momentum of inertia-the eigen values of the inertia tensor and the principal axis transformation-methods of solving rigid body problems and the Euler equations of motion Torque- Free motion of a rigid body-the heavy symmetrical top with one point fixed-precession of the equinoxes and of satellite orbits-precession of system of changes in a magnetic field.\n\nElasticity: Introduction, Displacement vector and the strain tensor, Stress tensor, Strain energy, Possible forms of free energy and stress tensor for isotropic solids, Elastic moduli for Isotropic solids, Elastic properties of general solids: Hooke's law and stiffness constants, Elastic properties of isotropic solids, propagation of elastic waves in isotropic elastic media.\n\nClassical Mechanics of particles and Rigid body-kiran C. Gupta, New age Publishers 2. Classical Mechanics-J. Uppadaya 3. Classical mechanics S. Gupta, Meenakshi prakashan, , New Delhi. Introduction to classical mechanics R. Takwall and P.\n\nAn Introduction to Continuum Mechanics-M. Gurtin, Academic Press. Linear Algebra: Various types of matrices, rank of matrix, Types of linear equation, Linear dependence and independence of vectors, eigen values and eigen vectors, Cayley Hamilton Theorem, Digonalisation of matrices, Elementary ideas about Tensors, Types of tensors, Transformation properties, Introductory group theory, Generators of continuous groups.\n\nFourier Series, Fourier and Laplace Transforms. I Ordinary Differential Equation: Differential equation of the First order and First Degree, variable separation, Homogeneous differential equation, Linear Differential Equation, Exact differential equation, Equation of first order and Higher degree, Method of finding the complementary function and particular integrals, Series solution- Frobenous method.\n\nBessel's differential equation and its solution, Bessel's functions, Recurrence formula, Generating function, Legendre equation and its solution, Legendre's Polynomial, Rodrigue's formula, laguerre's differential equation, Laguerre's functions, Hermite polynomials. Wave equation, Heat equation, Possion equation. Mathematical Methods for Physicist: G. Arfken, Hans. Weber,-Academic Press 2.\n\nMathematical Physics: H. Dass, Rama Verma-S. Chand and Company Ltd. Matrices and tensors: A. Joshi 2. Xavier-New Age International Publishers 3. Mathematical Physics: B. Rajput 4. Mathematical Physics: Satyaprakash 5.\n\nIntroduction Mathematical Physics: Charlie Harper. Common elementary computer science:programming instructions, simple algorithms and computational methods.\n\nOverview of C: Introduction, Sample C programs,basic structure of C program,executing a \"C\" Program,Constants, Variable and Data types, Operators and Expressions, Control statements while, do-while, for statements, nested loops, if-else, switch, break, continue statements.\n\nC Functions: Defining and accessing a function passing arguments to a function, function prototypes. Physics of Semiconductor Devices- S. SZE 2. Semiconductor Optoelectronic devices:- P. Bhattacharya PHI 3. Digital Electronics and Computer Design: M.\n\nMano PHI 4. Electronics Fundamentals and Applications: J. Ryder 5. Computer Oriented Numerical Methods V. Numerical Methods in Science and Engineering M. Xavier-New Age International Publishers. Physics of Semiconductor Devices- D. Gun Barrel and Tube Launcher: Theory of thin cylinders, use of plastic region of the material and its application to the pre-stressing tube, theory of failure, approximate Mises Hencky criteria, principle of monoblock non auto-fr ettaged and auto-frettaged barrels.\n\nGun Design: Gun design rules, Design of combustion chambers, Rifling profile and stress due to rifling, Gun tube acoustics, Gun erosion I Ordnance: Obturation, Muzzle brake, fume extractor, Firing mechanism, functions and characteristics of saddle, cradle, traversing and elevating gears, balancing gears. Recoil System: Elements of recoil mechanism, small arms, method of obtaining automatic fire, factors affecting recoil gas system, feed mechanism, trigger mechanism, sights.\n\nElements of Ordnance, , T. Hayes, John Wiley, New York 1. Measurement of velocity of liquid by Ultrasonic Interferometer. Measurement of band gap energy of different semiconductors including LED. Implementation of Adder using Universal Logic gates. To study Multivibrator circuits. Study of different types of Flip Flops. Realization of Boolean expression by using K- Maps. Study of BCD to 7 segment Display. Study of Left register, Right register and Ring counter Study of R-C coupled Amplifier Study of Solar cell KIT.\n\nCanonical ensemble and energy fluctuation, Grand canonical ensemble and density fluctuation, Equivalence of canonical and grand canonical ensemble. I Quantum Statistical Mechanics: The density matrix, Ensembles in quantum statistical mechanics, Ideal gas in micro canonical and grand-canonical ensemble, Equation of state for Ideal Fermi gas, Theory of white dwarf stars.\n\nPhase Transitions: Ideal Bose gas, Photons and Planck s Law, Phonons, Bose-Einstein condensation, Thermodynamic description of phase transition, Phase transitions of second kind, Discontinuity of specific heat, Change in symmetry in a phase transition of second kind 1. Statistical Mechanics-K: Huang 1. Elementary Statistical Physics- C Kittel 2. Statistical Mechanics-F: Mohling 3. Statistical Mechanics-Landau and Lifshitz.\n\nPhysics Transitions and Critical Phenomena-H. Stanley 5. Thermal Physics-C. Kittel 6. Fundamentals of Statistical and Thermal Physics-F. Mathematical Basics: Expansion Theorem, Completeness and Closure property of the basis set, Coordinate and Momentum representation, Compatible and incompatible observables, Commutator algebra, Uncertainty relation as a consequence of noncommutability, Minimum uncertainty wave packet Quantum Dynamics: Time evolution of quantum states, time evolution operator and its properties, Schroedinger picture, Heisenberg picture, Interaction picture, Equation of motion, Operator method of solution of Harmonic Oscillator, Matrix representation and time evolution of creation and annihilation operator.\n\nMotion in Spherical Symmetric Field: Hydrogen atom, Reduction of two body problem to equivalent one body problem, Radial equation, Energy eigenvalues and eigenfunctions, Degeneracy, radial probability distribution. Free particle problem incoming and outgoing spherical waves, Expansion of plane waves in terms of spherical waves, bound states of a 3-D square well, particle in a sphere.\n\nApproximation Methods: Time independent perturbation theory and application, variational method, WKB approximation, Time dependent perturbation theory and Fermi s Golden rule, selection rules. Scatterings: Elementary theory of scattering, Phase shifts, Partial waves, Born approximations.\n\nQuantum Mechanics-Joichan 1. Quantum Mechanics- Gasorowicz. Bernoulli s equation along a stream line and in rotational flow, Bernoulli s equation from thermodynamics,static and dynamics pressure, Losses due to geometric changes:-sudden expansion and contraction Venturimetre.\n\nI Viscous Effect: Normal stress shear stress, Navier-Stokes theorem, Flow through a parallel channel, Flow past a sphere, Terminal velocity order of magnitude analysis, Approximation of the Navier-Stokes equations. Boundary layer concepts:- Momentum integral equation, velocity profile, Boundary layer thickness, Skin Friction effecter, Transverse component of velocity, Displacement thickness, momentum thickness.\n\nDrag:- Bluff badles, Aerofoil, Boundary layer control, entrance region. Compressible flow: Perfection gas Relations:- Speed of propagation in gas, in isothermal and adiabatic condition, Mach number, Limits of incompressibility. Isentropic flow:- Laws of conservation, Static and stagnation values, flow through a duct of varying crosssection, mass flow rate, choking a converging passage, constant area adiabatic flow and Fanno like, constant area frictionless flow and Raleigh line.\n\nFluid Mechanics, A. Mohanty, PHI 2. Fluid Dynamics, R. Mises, Springer 1. Foundation of Fluid Mechanics, S. Yuan, PHI 2. Text Book of Fluid Mechanics, R. Khurmi, S. Chand 3. Perspective in Fluid Dynamics, Batchelor, Cambridge. Chemical Thermodynamics of Gun Propellant: Introduction to gun, ammunition, projectiles and missiles- Energetic of gun propellants-composition of gaseous products- Corrections to the basic calculationsprediction of propellant performance-the ratio of heat capacities.",
null,
"## Classical Mechanics By Gupta Kumar Sharma.pdf\n\nUpdating results WorldCat is the world's largest library catalog, helping you find library materials online. Don't have an account? Your Web browser is not enabled for JavaScript. Some features of WorldCat will not be available.\n\nClassical mechanics of particles and rigid bodies by Kiran C. Gupta, , New Age International edition, in English - 2nd ed.\n\n## Elementary Analytical Dynamics Of A Particle\n\n#### ISBN 10: 0470209585\n\nClassical mechanics book by gupta kumar sharma pdf download Download link of classical mechanics by dr jc upadhyaya pdf for msc physics msc be net gate and. Gupta kumar sharma classical mechanics are almost similar but special relativity is given in upadhyay.\n\nVariational Principle and Lagrange's Equations: Some techniques of the calculus of variations- Derivations of Lagrange's equations-from Hamilton's principle-extension of Hamilton's principle to nonholonomic systems-advantages of variational principle formulation- conservation theorems and symmetry properties. Two body Central force Problems ; Reduction to the equivalent one-body problem-the equations of motion and first integrals-the equivalent one-dimensional problems and classification of orbits-the virial theorem-the differential equation of orbit and integrable power-law potentials-conditions for closed orbits bertrand's theorem -the Kepler Problem Inverse square law of force-the motion in time in the Kepler problem-the Laplace-Runge-Lenz vector-scattering in a central force field, Transformation of the scattering problem to the laboratory co-ordinates. I The Kinematics of Rigid Body Motion: The independent co-ordinates of a rigid body-orthogonal transformation-formal properties of the transformation matrix, The Euler Angles, Euler's theorem on the motion of a rigid body-finite rotations-infinitesimal rotations-rate of change of vector-the Coriolis force. The Rigid Body Equations of Motion: Angular momentum and kinetic energy of motion about a point- Tensor and dyadics-the inertia tensor and the momentum of inertia-the eigen values of the inertia tensor and the principal axis transformation-methods of solving rigid body problems and the Euler equations of motion Torque- Free motion of a rigid body-the heavy symmetrical top with one point fixed-precession of the equinoxes and of satellite orbits-precession of system of changes in a magnetic field.\n\nОн снова ответил Да. Мгновение спустя компьютер подал звуковой сигнал. СЛЕДОПЫТ ОТОЗВАН Хейл улыбнулся.\n\nНе может быть! - сказала она по-испански. У Беккера застрял комок в горле. Росио была куда смелее своего клиента. - Не может быть? - повторил он, сохраняя ледяной тон. - Может, пройдем, чтобы я смог вам это доказать.\n\nТрюк, старый как мир. Никуда я не звонил. ГЛАВА 83 Беккеровская веспа, без сомнения, была самым миниатюрным транспортным средством, когда-либо передвигавшимся по шоссе, ведущему в севильский аэропорт. Наибольшая скорость, которую она развивала, достигала 50 миль в час, причем делала это со страшным воем, напоминая скорее циркулярную пилу, а не мотоцикл, и, увы, ей не хватало слишком много лошадиных сил, чтобы взмыть в воздух. В боковое зеркало заднего вида он увидел, как такси выехало на темное шоссе в сотне метров позади него и сразу же стало сокращать дистанцию.\n\nБеккер двинулся по едва освещенному коридору. Все здесь напоминало зловещую декорацию к голливудскому фильму ужасов. В воздухе стоял тяжелый запах мочи. Лампочки в конце коридора не горели, и на протяжении последних двадцати метров можно было различать только смутные силуэты. Женщина с кровотечением… плачущая молодая пара… молящаяся маленькая девочка.\n\nАлгоритм создает шифр, который кажется абсолютно стойким, а ТРАНСТЕКСТ перебирает все варианты, пока не находит ключ. Стратмор ответил ей тоном учителя, терпеливого и умеющего держать себя в руках: - Да, Сьюзан, ТРАНСТЕКСТ всегда найдет шифр, каким бы длинным он ни. - Он выдержал длинную паузу. - Если только… Сьюзан хотела что-то сказать, но поняла, что сейчас-то Стратмор и взорвет бомбу. Если только - .",
null,
"1. ## Celnihovsli\n\n04.06.2021 at 15:46\n\n3. Classical mechanics of particles and rigid bodies. by Kiran C Gupta · Classical mechanics of particles and rigid bodies. by Kiran C Gupta. Print book: Juvenile.\n\n2. ## Withdbirdflatfac\n\n07.06.2021 at 00:48\n\nOnline shopping from a great selection at Books Store.\n\n3. ## Pompeo H.\n\n07.06.2021 at 20:10\n\nWritten in English.\n\n4. ## Mothcontscanib\n\n08.06.2021 at 13:30"
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https://collegephysicsanswers.com/openstax-solutions/find-frequency-range-visible-light-given-it-encompasses-wavelengths-380-760-nm | [
"Question\nFind the frequency range of visible light, given that it encompasses wavelengths from 380 to 760 nm.\n\n$3.95 \\times 10^{14} \\textrm{ Hz} \\leq f_{visible} \\leq 7.89 \\times 10^{14} \\textrm{ Hz}$\n\nSolution Video\n\n# OpenStax College Physics Solution, Chapter 24, Problem 9 (Problems & Exercises) (0:54)",
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https://cmrcet.ml/railway-stations/ | [
"## Question\n\nGiven schedule of n trains with respective arrival and stoppage time at a Railway Station, find minimum number of platforms needed.\n\n# Solution by Nikhil Mohan.\n\n## Algorithm\n\nSort the arrival and departure time of trains.\nCreate two pointers i=0, and j=0 and a variable to store ans and current count plat\nRun a loop while i<n and j<n and compare the ith element of arrival array and jth element of departure array.\nif the arrival time is less than or equal to departure then one more platform is needed so increase the count, i.e. plat++ and increment i\nElse if the arrival time greater than departure then one less platform is needed so decrease the count, i.e. plat++ and increment j\nUpdate the ans, i.e ans = max(ans, plat).\n\nImplementation: This doesn’t create a single sorted list of all events, rather it individually sorts arr[] and dep[] arrays, and then uses merge process of merge sort to process them together as a single sorted array.\n\n## Solution\n\n``````#include<bits/stdc++.h>\nusing namespace std;\ntypedef long long ll;\n\n// Returns minimum number of platforms reqquired\nint solve(ll a[],ll b[],int n)\n{ // Sort arrival and departure arrays\nsort(a,a+n);\nsort(b,b+n);\n\n// p indicates number of platforms needed at a time\nint p=1,result=1;\n\n// Similar to merge in merge sort to process\n// all events in sorted order\nfor(int i=1,j=0; i<n && j<n;)\n{\n// If next event in sorted order is arrival,\n//increment count of platforms needed\nif (a[i]<=b[j])\n{p++;\ni++;}\n// Else decrement count of platforms needed\nelse if (a[i]>b[j])\n{ p--;\nj++;}\n\n// Update result\nresult=max(result,p);\n}\n//returning the result to main function\nreturn result;\n}\n\nvoid test()\n{\nll n;\nscanf(\"%lld\",&n);\n\nll a[n]; // arival array\nll b[n]; // departure array\nll s,t;\n\n//scanning the input given\nfor(int i=0;i<n;i++)\n{scanf(\"%lld%lld\",&s,&t);\na[i]=s;\nb[i]=s+t; // calculated departure time\n}\n\n//printing output,solve return the minimum stations required\ncout<<solve(a,b,n);\n\n}\n\n//driver code\nint main()\n{\ntest();\n}``````"
]
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https://forum.arduino.cc/index.php?amp;topic=598486.0 | [
"Go Down\n\nTopic: Serial communication Matlab-Arduino (Read 1 time)previous topic - next topic\n\nm_dal",
null,
"Feb 18, 2019, 12:23 pm\nHi everyone, I am having the following problem: I calculate a matrix on Matlab and I need to send this matrix to Arduino in order to move my stepper motors. The flow I implemented is the following:\nI transform the matrix into a string in Matlab;\nI saved the string as a .txt file;\nI use Hyperterminal to send the file on the Serial port;\nI read the string and I newly transform it into a matrix on the Arduino Sketch.\n\nEverything is working well, but I need some code improvements to reduce its own complexity. Have you got any suggestions? I know, it's possible to write on the serial port using Matlab but the problem is this one: when I try to print every single value of the Matrix just simply using a for cycle and Serial.print() command I cannot open the serial monitor to verify if the values have been correctly received, because the Serial port has just been opened to begin the communication between Matlab and Arduino. Feel free to ask me any clarifications if you don't understand my problem. Thank you all.\n\nm_dal",
null,
"#1\nFeb 18, 2019, 12:41 pm\nHi everyone, I am having the following problem: I calculate a matrix on Matlab and I need to send this matrix to Arduino in order to move my stepper motors. The flow I implemented is the following:\nI transform the matrix into a string in Matlab;\nI saved the string as a .txt file;\nI use Hyperterminal to send the file on the Serial port;\nI read the string and I newly transform it into a matrix on the Arduino Sketch.\n\nEverything is working well, but I need some code improvements to reduce its own complexity. Have you got any suggestions? I know, it's possible to write on the serial port using Matlab but the problem is this one: when I try to print every single value of the Matrix just simply using a for cycle and Serial.print() command I cannot open the serial monitor to verify if the values have been correctly received, because the Serial port has just been opened to begin the communication between Matlab and Arduino. Feel free to ask me any clarifications if you don't understand my problem. Thank you all.\n\nMetallor",
null,
"#2\nFeb 18, 2019, 12:57 pm\nOne suggestion for you, post your code. We have no idea how complex your code is if we can't see it. And remember to use code tags.\n\ngroundFungus",
null,
"#3\nFeb 18, 2019, 01:32 pm\nAs you have seen, only one device at a time can be connected to a serial port. A solution is to set up a software serial port and use a USB to TTL converter to connect the soft serial port to serial monitor for debug while using the hardware serial to talk to Matlab.\n\nUKHeliBob",
null,
"#4\nFeb 18, 2019, 01:41 pm\nWhat pins on the Arduino are you using to receive the serial data ? Not 0 and 1 perhaps ?\nPlease do not send me PMs asking for help. Post in the forum then everyone will benefit from seeing the questions and answers.\n\nm_dal",
null,
"#5\nFeb 18, 2019, 02:47 pm\nOk I understood; there can be a more efficient way to send data to Arduino? I want to to avoid the transformation of the matrix into a string and then re-transforming the string into a matrix. Thank you\n\nPaulS",
null,
"#6\nFeb 18, 2019, 03:04 pm\nQuote\nI want to to avoid the transformation of the matrix into a string and then re-transforming the string into a matrix.\nI can't see how to apply a matrix to a stepper motor. So, I have to wonder why you want to send a matrix to the Arduino. The PC can do a better job of determining how many steps a motor should take, and just that/those value(s) to the Arduino.\nThe art of getting good answers lies in asking good questions.\n\nm_dal",
null,
"#7\nFeb 18, 2019, 03:16 pmLast Edit: Feb 18, 2019, 03:20 pm by m_dal\nI use AccelStepper and MultiStepper libraries because i have six different stepper motors to be controlled and to be moved simultaneously: the motors move a robotic arm. The matrix has a number of rows equal to the number of position to be assumed by the arm and each column represent a single motor (six motors so six columns). I use a for cycle to send each row of the matrix to the function moveTo() provided by the library to move each motor: the function takes an array as an input which is represented by a single row of the matrix and when all the motors have reached their target, the for loop allow to go to the following line of the matrix and move the motors to the new target. here is a part of the code:\n\nif (conta5==1788){ //once B has been completed store each row in a 1-D array and move the robot (1788 represents the numbers of the matrix 298 rows 6 columns)\nlong positions ;\nfor (int i=0;i<298;i++){\nfor (int j=0;j<6;j++){\npositions [j]=B [j];\n}\nsteppers.moveTo(positions);\nsteppers.runSpeedToPosition();\n}\n\nm_dal",
null,
"#8\nFeb 18, 2019, 03:25 pm\nhere is the entire code: note that in Matlab I transformed the matrix into a string and saved it into a .txt file: the string is like this one:\n2 3 4;5 6 8; 12923 3475 -2] for example\n(I used ; char to understand when to fill the following row of the matrix and ] as a string terminator)\n\nI have six motors representing the joints of my robotic arm and I use accelstepper and multi stepper libraries to have a simultaneous movement of all the motors.\n\nString stringa= \"\"; //create a string to store what comes from serial\nchar c; //create a variable char to store each character from serial\nint B ; //matrix that must be filled with data read from serial\nint count=0; //counter that increased if a ';' is found\nint conta4=0; //counter to store the length of the string\nint conta5=0; //counter that increases if a value is added to B\nint conta2=0; //counter that increases if a numeric value is found on the string\nint conta3=0; //counter that increases if a value is added to B\n//(it is set to 0 if a row of B has been entirely filled)\nint num; //variable that stores the value of x(char array) into integer\nbool s=false; //to repeat the loop only once\nchar q=';';\nchar w=' ';\nchar y=']';\nchar x;//variable char to store values until ';' ' ' or ']' will be found\n#include <AccelStepper.h>\n#include <MultiStepper.h>\n\n// Define some steppers and the pins the will use (first pin for the number of wires,\n// second one for the number of steps,\n// and third one for direction)\nAccelStepper stepper1(1,9,8);\nAccelStepper stepper2(1,2,3);\nAccelStepper stepper3(1,10,11);\nAccelStepper stepper4(1,22,23);\nAccelStepper stepper5(1,24,25);\nAccelStepper stepper6(1,26,27);\n\nMultiStepper steppers;\n\nvoid setup() {\nSerial.begin(4800);\n\nstepper1.setMaxSpeed(500.0);\nSerial.print(\"stepper1 in posizione:\");\nSerial.println(stepper1.currentPosition());\n\nstepper2.setMaxSpeed(500.0);\nSerial.print(\"stepper2 in posizione:\");\nSerial.println(stepper2.currentPosition());\n\nstepper3.setMaxSpeed(500.0);\nSerial.print(\"stepper3 in posizione:\");\nSerial.println(stepper3.currentPosition());\n\nstepper4.setMaxSpeed(500.0);\nSerial.print(\"stepper4 in posizione:\");\nSerial.println(stepper3.currentPosition());\n\nstepper5.setMaxSpeed(500.0);\nSerial.print(\"stepper5 in posizione:\");\nSerial.println(stepper3.currentPosition());\n\nstepper6.setMaxSpeed(500.0);\nSerial.print(\"stepper6 in posizione:\");\nSerial.println(stepper3.currentPosition());\n\ndelay(3000);\n\n}\nvoid loop() {\nif (Serial.available() > 0) { //read from serial\n\nstringa = stringa + c; // add Char to string\nconta4=conta4+1; //increase counter to store the length of the string\n}\nif(s==false){\nif(stringa.charAt(conta4-1)==y){//enter in if only is the string has been acquired\nfor ( int i=0;i<conta4-1;i++)//go through the i-char of the string\n{\nif(stringa.charAt(i)==q) //if ';' is found\n{\ncount=count+1; //increase count (go to the following row of the matrix)\n\n}\n\nif (stringa.charAt(i)!= q && stringa.charAt(i)!= w) //(a number has been found)\n{\nx[conta2]=stringa.charAt(i); //add number to x\nconta2=conta2+1; //increase the counter if a number or a sign has been found\n}\nif(stringa.charAt(i)==w || stringa.charAt(i+1)==q|| stringa.charAt(i+1)==y){ //the number\n//to be stored in the matix\nnum= atoi(x); //convert the char array x into a number\nif (conta3==6){ //the row of B has been completed\nconta3=0;\n}\nfor (int i=0; i<conta2;i++){\nx=' ';//set all the values of char array x to ' ' once x has been converted to integer\n}\nB[count][conta3 % 6]=num; //add num to B\ns=true; //exit from loop once the for cycle is over\nconta2=0; //set to 0 the index of x\nconta3=conta3+1; //increase counter because a number has been added to the matrix B\nconta5=conta5+1;//increase counter because a number has been added to the matrix B\n\n}\n}\nif (conta5==1788){ //once B has been completed store each row in a 1-D array and move the robot\nlong positions ;\nfor (int i=0;i<298;i++){\nfor (int j=0;j<6;j++){\npositions [j]=B[j];\n}\nsteppers.moveTo(positions);\nsteppers.runSpeedToPosition();\n}\n}\n}\n\n}\n}\n\nUKHeliBob",
null,
"#9\nFeb 18, 2019, 04:09 pm\nDo you see those italic characters in your code ?\n\nYou did not use code tags when you posted it so some of your code has been interpreted as HTML commands. Please read read this before posting a programming question and follow the instructions about posting code\nPlease do not send me PMs asking for help. Post in the forum then everyone will benefit from seeing the questions and answers.\n\nPaulS",
null,
"#10\nFeb 18, 2019, 04:23 pm\nQuote\nThe matrix has a number of rows equal to the number of position to be assumed by the arm\nDoesn't matter. You should be sending data for ONE position at a time, and only when the arm is at the last position sent.\n\nQuote\neach column represent a single motor\nSo, you need to send 6 values, not a matrix. Much simpler.\n\nWhen WILL you learn to use code tags?\nThe art of getting good answers lies in asking good questions.\n\nm_dal",
null,
"#11\nFeb 18, 2019, 04:41 pm\nSorry I am new, later I will post my code in the correct way. I have a trajectory made of several points (around 300) to be reached by the arm; so each row of the matrix represents a single point of the trajectory. I just tried to send via serial one row of the matrix at a time (so an array of six values) but the problem is to syncronize the movement of all the motors with the time of sending values to serial: for example while the motors are completing their movements it's possible that different rows are sent via serial and I miss one because motors are still moving: that's why I prefer to send the entire matrix and then using a for loop to go through each row of the matrix.\n\nPaulS",
null,
"#12\nFeb 18, 2019, 05:27 pm\nQuote\nfor example while the motors are completing their movements it's possible that different rows are sent via serial and I miss one because motors are still moving:\nYou need to STOP matlab from spamming the Arduino. Only send a new row when the Arduino is ready for you to send a new row. That doesn't necessarily mean when it has completed a move. It could be working on a move while receiving data for the next move.\nThe art of getting good answers lies in asking good questions.\n\nRobin2",
null,
"#13\nFeb 18, 2019, 07:00 pm\nIt is not a good idea to use the String (capital S) class on an Arduino as it can cause memory corruption in the small memory on an Arduino. This can happen after the program has been running perfectly for some time. Just use cstrings - char arrays terminated with '\\0' (NULL).\n\nHave a look at the examples in Serial Input Basics - simple reliable ways to receive data. There is also a parse example to illustrate how to extract numbers from the received text.\n\nThe technique in the 3rd example will be the most reliable. It is what I use for Arduino to Arduino communication.\n\nYou can send data in a compatible format with code like this (or the equivalent in any other programming language)\nCode: [Select]\nSerial.print('<'); // start marker\nSerial.print(value1);\nSerial.print(','); // comma separator\nSerial.print(value2);\nSerial.println('>'); // end marker\n\nAnd to make it easy for people to help you please modify your post and use the code button </>",
null,
"Code: [Select]\nso your code looks like this and is easy to copy to a text editor. See How to use the Forum\n\nYour code is too long for me to study quickly without copying to my text editor. The text editor shows line numbers, identifies matching brackets and allows me to search for things like all instances of a particular variable or function.\n\nAlso please use the AutoFormat tool to indent your code for easier reading.\n\n...R\nTwo or three hours spent thinking and reading documentation solves most programming problems.\n\nm_dal",
null,
"#14\nFeb 20, 2019, 10:44 am\nYou need to STOP matlab from spamming the Arduino. Only send a new row when the Arduino is ready for you to send a new row. That doesn't necessarily mean when it has completed a move. It could be working on a move while receiving data for the next move.\nYes it's clear but the risk is the one I explained: a new row is sent on the serial and the robot is still completing the previous movement, there could be the possibility that another row is sent on the serial so I lose one position. The problem is to synchronize the time in which a row is sent on serial and the movement of the robot itself: there would be perfect that a new row arrive immediately after a movement is completed but I don't know how to make this synchronization. So it could be easier to receive the entire matrix and then move row by row once the matrix has been entirely received.\n\nGo Up"
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http://cloud.originlab.com/doc/en/Origin-Help/Graph-Type-Data-Req | [
"# 31.2 Data Selection Requirements for Origin Graph Types\n\nThe data requirements for each of the built-in graph types are listed below. The data selection can be either columns or selected ranges from columns as shown below:",
null,
"Additionally rows can be selected from different parts of the sheet by using CTRL + select, releasing and repeating as shown below:",
null,
"When this option is available, the Description field refers to the \"CTRL key being depressed.\"\n\nGraph Type Selected Data Requirements Description\n\n2 Point segment\n\n2D Waterfall\n\n3D Ribbons\n\n3D Walls\n\n3D Waterfall\n\n3 Point segment\n\n4 Panel\n\n9 Panel\n\nArea\n\nBar\n\nColumn\n\nHorizontal 2 panel\n\nHorizontal step\n\nLine\n\nLine+symbol\n\nPolar\n\nScatter\n\nSpline connected\n\nStack\n\nStacked bar\n\nStacked column\n\nVertical 2 panel\n\nVertical drop line\n\nVertical step\n\nXYY 3D Bars\n\nZoom\n\nOne or more columns.\n\nIf one column is selected: The column supplies the Y values. The data is plotted versus row number.\n\nIf more than one column is selected: The leftmost column supplies the X values. All other columns supply the Y values. The data is plotted versus the X values.\n\nIf more than one column is selected and the CTRL key is depressed: All the columns supply the Y values. The data is plotted versus row number.\n\nBox chart\n\nOne or more columns.\n\nEach selected column supplies Y values for a separate box. Column names supply the associated X values.\n\nDouble Y Axis\n\nTwo or more columns.\n\nIf two columns are selected: Both columns supply Y values. The data is plotted versus row number.\n\nIf three columns are selected: The leftmost column supplies X values. Other columns supply Y values. The data is plotted versus X values.\n\nFill area\n\nTwo or three columns.\n\nIf two columns are selected: Both columns supply Y values. The data is plotted versus row number.\n\nIf three columns are selected: The leftmost column supplies X values. The other two columns supply Y values. The data is plotted versus X values.\n\nFloating bar\n\nFloating column\n\nTwo or more columns.\n\nIf two columns are selected and the CTRL key is depressed: The leftmost column supplies the starting Y values. The second column supplies the ending Y values. The data is plotted versus row number.\n\nIf three or more columns are selected: The leftmost column supplies the X values. The second column supplies the starting Y values. The next column supplies the intermediate Y values (etc.). The rightmost column supplies the ending Y values. The data is plotted versus X values.\n\nIf three or more columns are selected and the CTRL key is depressed: The leftmost column supplies the starting Y values. The next column supplies the intermediate Y values (etc.). The rightmost column supplies the ending Y values. The data is plotted versus row number.\n\nHigh-Low-Close\n\nThree columns or four columns.\n\nIf three columns are selected: The leftmost column supplies the high values, the next column supplies the low values, and the last column supplies the closing values. The data is plotted versus row number.\n\nIf four columns are selected: The leftmost column supplies the X values. The second column supplies the high values. The third column supplies the low values. The rightmost column supplies the closing values. The data is plotted versus the X values.\n\nHistogram\n\nOne or more columns.\n\nA histogram graph is created containing an interlaced histogram for each of the selected columns.\n\nHistogram + Probabilities\n\nOne column.\n\nA histogram is created for the selected column. The cumulative counts are also displayed.\n\nIndexed size (Bubble) and color map\n\nThree columns or four columns.\n\nIf three columns are selected: The leftmost column supplies the Y values. The second column supplies the size values. The rightmost column supplies the color values. The data is plotted versus row number.\n\nIf four columns are selected: The leftmost column supplies the X values. The second column supplies the Y values. The third column supplies the size values. The rightmost column supplies the color values. The data is plotted versus the X values.\n\nColor map, Indexed size (Bubble)\n\nTwo columns or three columns.\n\nIf two columns are selected: The leftmost column supplies the Y values. The second column supplies the size or color values. The data is plotted versus row number.\n\nIf three columns are selected: The leftmost column supplies the X values. The next column supplies the Y values. The rightmost column supplies the size or color values. The data is plotted versus the X values.\n\nLine series\n\nTwo or three columns.\n\nIf two or three columns are selected: The columns supply the Y values. Each series of row values comprises a line + symbol data plot. The data plot's X values are determined by the selected Y column number (1, 2, or 3). The data plot's Y values are determined by the actual cell values in the selected column.\n\nPie\n\nOne column.\n\nThe selected column values are summed, and the percentage of the total is determined for each selected value. The pie chart displays the percentage of the total for each selected value as a pie section.\n\nQC (X bar R)\n\nOne or more columns.\n\nIf one column is selected: The column supplies the Y values. The X values are determined by the subgroup number (defined by the subgroup size and the number of column values). The data is plotted versus X values.\n\nIf more than one column is selected: The selected columns supply the Y values. The X values are determined by the row number. The data is plotted versus X values.\n\nTernary\n\nThree columns.\n\nThe leftmost column supplies the X values. The second column supplies the Y values. The rightmost column supplies the Z values. The X, Y, Z data is plotted.\n\nXYAM Vector\n\nThree columns or four columns.\n\nIf three columns are selected: The leftmost column supplies the Y values. The second column supplies the angle values. The rightmost column supplies the magnitude values. The data is plotted versus row number.\n\nIf four columns are selected: The leftmost column supplies the X values. The second column supplies the Y values. The third column supplies the angle values. The rightmost column supplies the magnitude values. The data is plotted versus the X values.\n\nXYXY Vector\n\nFour columns.\n\nThe leftmost column supplies the X start values. The second column supplies the Y start values. The third column supplies the X end values. The rightmost column supplies the Y end values."
]
| [
null,
"http://d2mvzyuse3lwjc.cloudfront.net/doc/en/UserGuide/images/Graph_Type_Data_Requirements/Column_Selection.png",
null,
"http://d2mvzyuse3lwjc.cloudfront.net/doc/en/UserGuide/images/Graph_Type_Data_Requirements/CTRL_Selection.png",
null
]
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https://docs.wso2.com/display/CEP400/Inbuilt+Aggregate+Functions | [
"This documentation is for WSO2 Complex Event Processor 4.0.0. View documentation for the latest release.\nWSO2 Complex Event Processor is succeeded by WSO2 Stream Processor. To view the latest documentation for WSO2 SP, see WSO2 Stream Processor Documentation.\n||\nGo to start of banner\n\n# Inbuilt Aggregate Functions\n\nFollowing are the supported inbuilt aggregate functions of Siddhi\n\n#### sum\n\n`<long|double> sum(<int|long|double|float>`` arg)`\n\n• Extension TypeAggregate Function\n• Description: Sums all the events.\n• Parameterarg: The value that need to be summed.\n• Return Type: Returns long if the input parameter type is int or long and Returns double if the input parameter type is float or double.\n• Examples`sum(20)` returns sum of 20s as a long value for each event arrival and expiry. `sum(temp)` returns the sum of all temp attributes based on each event arrival and expiry.\n\n#### avg\n\n`<double> avg(<int|long|double|float> arg)`\n\n• Extension Type: Aggregate Function\n• Description: Calculates the average for all the events.\n• Parameter: arg: The value that need to be averaged.\n• Return Type: Returns calculated average value as a double.\n• Examples: avg(temp) returns the average temp value for all the events based on their arrival and expiry.\n\n#### max\n\n`<int|long|double|float> max(<int|long|double|float> arg)`\n\n• Extension Type: Aggregate Function\n• Description: Returns the max value for all the events.\n• Parameter: arg: The value that need to be compared to find the max value.\n• Return Type: Returns max value in same type as the input.\n• Examples: max(temp) returns the maximum temp value recorded for all the events based on their arrival and expiry.\n\n#### min\n\n`<int|long|double|float> min(<int|long|double|float> arg)`\n\n• Extension Type: Aggregate Function\n• Description: Returns the min value for all the events.\n• Parameter: arg: The value that need to be compared to find the min value.\n• Return Type: Returns min value in same type as the input.\n• Examples: min(temp) returns the minimum temp value recorded for all the events based on their arrival and expiry.\n\n#### count\n\n`<long> count()`\n\n• Extension Type: Aggregate Function\n• Description: Returns the count of all the events.\n• Return Type: Returns event count as a long.\n• Examples: count() will return the count of all the events.\n\n#### stddev\n\n`<double> stddev(<int|long|double|float> arg)`\n\n• Extension Type: Aggregate Function\n• Description: Returns the calculated standard deviation for all the events.\n• Parameter: arg: The value that should be used in calculating the standard deviation.\n• Return Type: Returns calculated standard deviation value as a double.\n• Examples: stddev(temp) returns the calculated standard deviation of temp, for all the events based on their arrival and expiry.\n\n• No labels"
]
| [
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https://www.directhardwaresupply.com/store/page/16/?filter_product-group=hard-disk | [
"Showing 241–256 of 3173 results\n\n';if(form!=''){form.closest('form').append(alert_box);}else{\\$('#form-5d835a8175a9e').append(alert_box);}} function addConditionClass(field_id,cond_class,form_fields_wrapper){\\$(field_id).each(function(){if(\\$(this).is(':input')||\\$(this).is('select')) \\$(this).addClass('cond_filler_'+cond_class);\\$(this).children().each(function(){addConditionClass(\\$(this),cond_class,form_fields_wrapper);})});return false;} function compareRule(objs,cmp_operator,cmp_value,cmp_id,\\$form_part_0){var comp_res=false;var areOperandsCb=false;switch(cmp_operator){case'is':if(cmp_value.startsWith('Checkbox_')){test=objs.closest('#form_part_0').find('#'+cmp_value+' :input:checked');areOperandsCb=cmp_id.startsWith('Checkbox_')?true:false;if(areOperandsCb&&objs.length!=test.length){break;}}else{test=objs.closest('#form_part_0').find('#'+cmp_value+' :input');} \\$(objs).each(function(){if(areOperandsCb){comp_res=false;} \\$cmp1=\\$(this).val();\\$(test).each(function(){\\$cmp2=\\$(this).val();if(\\$cmp1==\\$cmp2){comp_res=true;if(!areOperandsCb){return;}}});if(areOperandsCb&&false==comp_res){return;}});break;case'is-not':if(cmp_value.startsWith('Checkbox_')){test=\\$form_part_0.find('#'+cmp_value+' :input:checked');areOperandsCb=cmp_id.startsWith('Checkbox_')?true:false;if(areOperandsCb&&objs.length!=test.length){return true;}}else{test=objs.closest('#form_part_0').find('#'+cmp_value+' :input');} for(let objsElement of objs){comp_res=false;\\$cmp1=\\$(objsElement).val();for(let testElement of test){\\$cmp2=\\$(testElement).val();if(\\$cmp1!=\\$cmp2){comp_res=true;}else if(areOperandsCb){comp_res=false;break;}} if(areOperandsCb&&true==comp_res){break;}} break;case'less-than':\\$(objs).each(function(){if(!isNaN(cmp_value)){cmp_value=Number(cmp_value);} if(\\$(this).val()cmp_value){comp_res=true;return;}});break;case'starts-with':\\$(objs).each(function(){if(\\$(this).val().indexOf(cmp_value)==0){comp_res=true;return;}});break;case'contains':\\$(objs).each(function(){if(\\$(this).val().indexOf(cmp_value)!=-1){comp_res=true;return;}});break;case'ends-with':\\$(objs).each(function(){indexPoint=(\\$(this).val().length-cmp_value.length);if(indexPoint>=0&&\\$(this).val().indexOf(cmp_value,indexPoint)==indexPoint){comp_res=true;return;}});break;default:comp_res=false;break;} return comp_res;} function applyRule(field_id){\\$('.cond_filler_'+field_id).each(function(){var this_conditions=\\$('#'+field_id).attr('data-cond-fields').split('|');var this_action=\\$('#'+field_id).attr('data-cond-action').split(':');var cmp_res=this_action=='all'?true:false;for(i=0;i\n';if(form!=''){form.closest('form').append(alert_box);}else{\\$('#form-5d835a8177dda').append(alert_box);}} function addConditionClass(field_id,cond_class,form_fields_wrapper){\\$(field_id).each(function(){if(\\$(this).is(':input')||\\$(this).is('select')) \\$(this).addClass('cond_filler_'+cond_class);\\$(this).children().each(function(){addConditionClass(\\$(this),cond_class,form_fields_wrapper);})});return false;} function compareRule(objs,cmp_operator,cmp_value,cmp_id,\\$form_part_0){var comp_res=false;var areOperandsCb=false;switch(cmp_operator){case'is':if(cmp_value.startsWith('Checkbox_')){test=objs.closest('#form_part_0').find('#'+cmp_value+' :input:checked');areOperandsCb=cmp_id.startsWith('Checkbox_')?true:false;if(areOperandsCb&&objs.length!=test.length){break;}}else{test=objs.closest('#form_part_0').find('#'+cmp_value+' :input');} \\$(objs).each(function(){if(areOperandsCb){comp_res=false;} \\$cmp1=\\$(this).val();\\$(test).each(function(){\\$cmp2=\\$(this).val();if(\\$cmp1==\\$cmp2){comp_res=true;if(!areOperandsCb){return;}}});if(areOperandsCb&&false==comp_res){return;}});break;case'is-not':if(cmp_value.startsWith('Checkbox_')){test=\\$form_part_0.find('#'+cmp_value+' :input:checked');areOperandsCb=cmp_id.startsWith('Checkbox_')?true:false;if(areOperandsCb&&objs.length!=test.length){return true;}}else{test=objs.closest('#form_part_0').find('#'+cmp_value+' :input');} for(let objsElement of objs){comp_res=false;\\$cmp1=\\$(objsElement).val();for(let testElement of test){\\$cmp2=\\$(testElement).val();if(\\$cmp1!=\\$cmp2){comp_res=true;}else if(areOperandsCb){comp_res=false;break;}} if(areOperandsCb&&true==comp_res){break;}} break;case'less-than':\\$(objs).each(function(){if(!isNaN(cmp_value)){cmp_value=Number(cmp_value);} if(\\$(this).val()cmp_value){comp_res=true;return;}});break;case'starts-with':\\$(objs).each(function(){if(\\$(this).val().indexOf(cmp_value)==0){comp_res=true;return;}});break;case'contains':\\$(objs).each(function(){if(\\$(this).val().indexOf(cmp_value)!=-1){comp_res=true;return;}});break;case'ends-with':\\$(objs).each(function(){indexPoint=(\\$(this).val().length-cmp_value.length);if(indexPoint>=0&&\\$(this).val().indexOf(cmp_value,indexPoint)==indexPoint){comp_res=true;return;}});break;default:comp_res=false;break;} return comp_res;} function applyRule(field_id){\\$('.cond_filler_'+field_id).each(function(){var this_conditions=\\$('#'+field_id).attr('data-cond-fields').split('|');var this_action=\\$('#'+field_id).attr('data-cond-action').split(':');var cmp_res=this_action=='all'?true:false;for(i=0;i\n';if(form!=''){form.closest('form').append(alert_box);}else{\\$('#form-5d835a81797bd').append(alert_box);}} function addConditionClass(field_id,cond_class,form_fields_wrapper){\\$(field_id).each(function(){if(\\$(this).is(':input')||\\$(this).is('select')) \\$(this).addClass('cond_filler_'+cond_class);\\$(this).children().each(function(){addConditionClass(\\$(this),cond_class,form_fields_wrapper);})});return false;} function compareRule(objs,cmp_operator,cmp_value,cmp_id,\\$form_part_0){var comp_res=false;var areOperandsCb=false;switch(cmp_operator){case'is':if(cmp_value.startsWith('Checkbox_')){test=objs.closest('#form_part_0').find('#'+cmp_value+' :input:checked');areOperandsCb=cmp_id.startsWith('Checkbox_')?true:false;if(areOperandsCb&&objs.length!=test.length){break;}}else{test=objs.closest('#form_part_0').find('#'+cmp_value+' :input');} \\$(objs).each(function(){if(areOperandsCb){comp_res=false;} \\$cmp1=\\$(this).val();\\$(test).each(function(){\\$cmp2=\\$(this).val();if(\\$cmp1==\\$cmp2){comp_res=true;if(!areOperandsCb){return;}}});if(areOperandsCb&&false==comp_res){return;}});break;case'is-not':if(cmp_value.startsWith('Checkbox_')){test=\\$form_part_0.find('#'+cmp_value+' :input:checked');areOperandsCb=cmp_id.startsWith('Checkbox_')?true:false;if(areOperandsCb&&objs.length!=test.length){return true;}}else{test=objs.closest('#form_part_0').find('#'+cmp_value+' :input');} for(let objsElement of objs){comp_res=false;\\$cmp1=\\$(objsElement).val();for(let testElement of test){\\$cmp2=\\$(testElement).val();if(\\$cmp1!=\\$cmp2){comp_res=true;}else if(areOperandsCb){comp_res=false;break;}} if(areOperandsCb&&true==comp_res){break;}} break;case'less-than':\\$(objs).each(function(){if(!isNaN(cmp_value)){cmp_value=Number(cmp_value);} if(\\$(this).val()cmp_value){comp_res=true;return;}});break;case'starts-with':\\$(objs).each(function(){if(\\$(this).val().indexOf(cmp_value)==0){comp_res=true;return;}});break;case'contains':\\$(objs).each(function(){if(\\$(this).val().indexOf(cmp_value)!=-1){comp_res=true;return;}});break;case'ends-with':\\$(objs).each(function(){indexPoint=(\\$(this).val().length-cmp_value.length);if(indexPoint>=0&&\\$(this).val().indexOf(cmp_value,indexPoint)==indexPoint){comp_res=true;return;}});break;default:comp_res=false;break;} return comp_res;} function applyRule(field_id){\\$('.cond_filler_'+field_id).each(function(){var this_conditions=\\$('#'+field_id).attr('data-cond-fields').split('|');var this_action=\\$('#'+field_id).attr('data-cond-action').split(':');var cmp_res=this_action=='all'?true:false;for(i=0;i\n';if(form!=''){form.closest('form').append(alert_box);}else{\\$('#form-5d835a817ae80').append(alert_box);}} function addConditionClass(field_id,cond_class,form_fields_wrapper){\\$(field_id).each(function(){if(\\$(this).is(':input')||\\$(this).is('select')) \\$(this).addClass('cond_filler_'+cond_class);\\$(this).children().each(function(){addConditionClass(\\$(this),cond_class,form_fields_wrapper);})});return false;} function compareRule(objs,cmp_operator,cmp_value,cmp_id,\\$form_part_0){var comp_res=false;var areOperandsCb=false;switch(cmp_operator){case'is':if(cmp_value.startsWith('Checkbox_')){test=objs.closest('#form_part_0').find('#'+cmp_value+' :input:checked');areOperandsCb=cmp_id.startsWith('Checkbox_')?true:false;if(areOperandsCb&&objs.length!=test.length){break;}}else{test=objs.closest('#form_part_0').find('#'+cmp_value+' :input');} \\$(objs).each(function(){if(areOperandsCb){comp_res=false;} \\$cmp1=\\$(this).val();\\$(test).each(function(){\\$cmp2=\\$(this).val();if(\\$cmp1==\\$cmp2){comp_res=true;if(!areOperandsCb){return;}}});if(areOperandsCb&&false==comp_res){return;}});break;case'is-not':if(cmp_value.startsWith('Checkbox_')){test=\\$form_part_0.find('#'+cmp_value+' :input:checked');areOperandsCb=cmp_id.startsWith('Checkbox_')?true:false;if(areOperandsCb&&objs.length!=test.length){return true;}}else{test=objs.closest('#form_part_0').find('#'+cmp_value+' :input');} for(let objsElement of objs){comp_res=false;\\$cmp1=\\$(objsElement).val();for(let testElement of test){\\$cmp2=\\$(testElement).val();if(\\$cmp1!=\\$cmp2){comp_res=true;}else if(areOperandsCb){comp_res=false;break;}} if(areOperandsCb&&true==comp_res){break;}} break;case'less-than':\\$(objs).each(function(){if(!isNaN(cmp_value)){cmp_value=Number(cmp_value);} if(\\$(this).val()cmp_value){comp_res=true;return;}});break;case'starts-with':\\$(objs).each(function(){if(\\$(this).val().indexOf(cmp_value)==0){comp_res=true;return;}});break;case'contains':\\$(objs).each(function(){if(\\$(this).val().indexOf(cmp_value)!=-1){comp_res=true;return;}});break;case'ends-with':\\$(objs).each(function(){indexPoint=(\\$(this).val().length-cmp_value.length);if(indexPoint>=0&&\\$(this).val().indexOf(cmp_value,indexPoint)==indexPoint){comp_res=true;return;}});break;default:comp_res=false;break;} return comp_res;} function applyRule(field_id){\\$('.cond_filler_'+field_id).each(function(){var this_conditions=\\$('#'+field_id).attr('data-cond-fields').split('|');var this_action=\\$('#'+field_id).attr('data-cond-action').split(':');var cmp_res=this_action=='all'?true:false;for(i=0;i\n';if(form!=''){form.closest('form').append(alert_box);}else{\\$('#form-5d835a817c8e3').append(alert_box);}} function addConditionClass(field_id,cond_class,form_fields_wrapper){\\$(field_id).each(function(){if(\\$(this).is(':input')||\\$(this).is('select')) \\$(this).addClass('cond_filler_'+cond_class);\\$(this).children().each(function(){addConditionClass(\\$(this),cond_class,form_fields_wrapper);})});return false;} function compareRule(objs,cmp_operator,cmp_value,cmp_id,\\$form_part_0){var comp_res=false;var areOperandsCb=false;switch(cmp_operator){case'is':if(cmp_value.startsWith('Checkbox_')){test=objs.closest('#form_part_0').find('#'+cmp_value+' :input:checked');areOperandsCb=cmp_id.startsWith('Checkbox_')?true:false;if(areOperandsCb&&objs.length!=test.length){break;}}else{test=objs.closest('#form_part_0').find('#'+cmp_value+' :input');} \\$(objs).each(function(){if(areOperandsCb){comp_res=false;} \\$cmp1=\\$(this).val();\\$(test).each(function(){\\$cmp2=\\$(this).val();if(\\$cmp1==\\$cmp2){comp_res=true;if(!areOperandsCb){return;}}});if(areOperandsCb&&false==comp_res){return;}});break;case'is-not':if(cmp_value.startsWith('Checkbox_')){test=\\$form_part_0.find('#'+cmp_value+' :input:checked');areOperandsCb=cmp_id.startsWith('Checkbox_')?true:false;if(areOperandsCb&&objs.length!=test.length){return true;}}else{test=objs.closest('#form_part_0').find('#'+cmp_value+' :input');} for(let objsElement of objs){comp_res=false;\\$cmp1=\\$(objsElement).val();for(let testElement of test){\\$cmp2=\\$(testElement).val();if(\\$cmp1!=\\$cmp2){comp_res=true;}else if(areOperandsCb){comp_res=false;break;}} if(areOperandsCb&&true==comp_res){break;}} break;case'less-than':\\$(objs).each(function(){if(!isNaN(cmp_value)){cmp_value=Number(cmp_value);} if(\\$(this).val()cmp_value){comp_res=true;return;}});break;case'starts-with':\\$(objs).each(function(){if(\\$(this).val().indexOf(cmp_value)==0){comp_res=true;return;}});break;case'contains':\\$(objs).each(function(){if(\\$(this).val().indexOf(cmp_value)!=-1){comp_res=true;return;}});break;case'ends-with':\\$(objs).each(function(){indexPoint=(\\$(this).val().length-cmp_value.length);if(indexPoint>=0&&\\$(this).val().indexOf(cmp_value,indexPoint)==indexPoint){comp_res=true;return;}});break;default:comp_res=false;break;} return comp_res;} function applyRule(field_id){\\$('.cond_filler_'+field_id).each(function(){var this_conditions=\\$('#'+field_id).attr('data-cond-fields').split('|');var this_action=\\$('#'+field_id).attr('data-cond-action').split(':');var cmp_res=this_action=='all'?true:false;for(i=0;i\n';if(form!=''){form.closest('form').append(alert_box);}else{\\$('#form-5d835a817e3a0').append(alert_box);}} function addConditionClass(field_id,cond_class,form_fields_wrapper){\\$(field_id).each(function(){if(\\$(this).is(':input')||\\$(this).is('select')) \\$(this).addClass('cond_filler_'+cond_class);\\$(this).children().each(function(){addConditionClass(\\$(this),cond_class,form_fields_wrapper);})});return false;} function compareRule(objs,cmp_operator,cmp_value,cmp_id,\\$form_part_0){var comp_res=false;var areOperandsCb=false;switch(cmp_operator){case'is':if(cmp_value.startsWith('Checkbox_')){test=objs.closest('#form_part_0').find('#'+cmp_value+' :input:checked');areOperandsCb=cmp_id.startsWith('Checkbox_')?true:false;if(areOperandsCb&&objs.length!=test.length){break;}}else{test=objs.closest('#form_part_0').find('#'+cmp_value+' :input');} \\$(objs).each(function(){if(areOperandsCb){comp_res=false;} \\$cmp1=\\$(this).val();\\$(test).each(function(){\\$cmp2=\\$(this).val();if(\\$cmp1==\\$cmp2){comp_res=true;if(!areOperandsCb){return;}}});if(areOperandsCb&&false==comp_res){return;}});break;case'is-not':if(cmp_value.startsWith('Checkbox_')){test=\\$form_part_0.find('#'+cmp_value+' :input:checked');areOperandsCb=cmp_id.startsWith('Checkbox_')?true:false;if(areOperandsCb&&objs.length!=test.length){return true;}}else{test=objs.closest('#form_part_0').find('#'+cmp_value+' :input');} for(let objsElement of objs){comp_res=false;\\$cmp1=\\$(objsElement).val();for(let testElement of test){\\$cmp2=\\$(testElement).val();if(\\$cmp1!=\\$cmp2){comp_res=true;}else if(areOperandsCb){comp_res=false;break;}} if(areOperandsCb&&true==comp_res){break;}} break;case'less-than':\\$(objs).each(function(){if(!isNaN(cmp_value)){cmp_value=Number(cmp_value);} if(\\$(this).val()cmp_value){comp_res=true;return;}});break;case'starts-with':\\$(objs).each(function(){if(\\$(this).val().indexOf(cmp_value)==0){comp_res=true;return;}});break;case'contains':\\$(objs).each(function(){if(\\$(this).val().indexOf(cmp_value)!=-1){comp_res=true;return;}});break;case'ends-with':\\$(objs).each(function(){indexPoint=(\\$(this).val().length-cmp_value.length);if(indexPoint>=0&&\\$(this).val().indexOf(cmp_value,indexPoint)==indexPoint){comp_res=true;return;}});break;default:comp_res=false;break;} return comp_res;} function applyRule(field_id){\\$('.cond_filler_'+field_id).each(function(){var this_conditions=\\$('#'+field_id).attr('data-cond-fields').split('|');var this_action=\\$('#'+field_id).attr('data-cond-action').split(':');var cmp_res=this_action=='all'?true:false;for(i=0;i\n';if(form!=''){form.closest('form').append(alert_box);}else{\\$('#form-5d835a817fd26').append(alert_box);}} function addConditionClass(field_id,cond_class,form_fields_wrapper){\\$(field_id).each(function(){if(\\$(this).is(':input')||\\$(this).is('select')) \\$(this).addClass('cond_filler_'+cond_class);\\$(this).children().each(function(){addConditionClass(\\$(this),cond_class,form_fields_wrapper);})});return false;} function compareRule(objs,cmp_operator,cmp_value,cmp_id,\\$form_part_0){var comp_res=false;var areOperandsCb=false;switch(cmp_operator){case'is':if(cmp_value.startsWith('Checkbox_')){test=objs.closest('#form_part_0').find('#'+cmp_value+' :input:checked');areOperandsCb=cmp_id.startsWith('Checkbox_')?true:false;if(areOperandsCb&&objs.length!=test.length){break;}}else{test=objs.closest('#form_part_0').find('#'+cmp_value+' :input');} \\$(objs).each(function(){if(areOperandsCb){comp_res=false;} \\$cmp1=\\$(this).val();\\$(test).each(function(){\\$cmp2=\\$(this).val();if(\\$cmp1==\\$cmp2){comp_res=true;if(!areOperandsCb){return;}}});if(areOperandsCb&&false==comp_res){return;}});break;case'is-not':if(cmp_value.startsWith('Checkbox_')){test=\\$form_part_0.find('#'+cmp_value+' :input:checked');areOperandsCb=cmp_id.startsWith('Checkbox_')?true:false;if(areOperandsCb&&objs.length!=test.length){return true;}}else{test=objs.closest('#form_part_0').find('#'+cmp_value+' :input');} for(let objsElement of objs){comp_res=false;\\$cmp1=\\$(objsElement).val();for(let testElement of test){\\$cmp2=\\$(testElement).val();if(\\$cmp1!=\\$cmp2){comp_res=true;}else if(areOperandsCb){comp_res=false;break;}} if(areOperandsCb&&true==comp_res){break;}} break;case'less-than':\\$(objs).each(function(){if(!isNaN(cmp_value)){cmp_value=Number(cmp_value);} if(\\$(this).val()cmp_value){comp_res=true;return;}});break;case'starts-with':\\$(objs).each(function(){if(\\$(this).val().indexOf(cmp_value)==0){comp_res=true;return;}});break;case'contains':\\$(objs).each(function(){if(\\$(this).val().indexOf(cmp_value)!=-1){comp_res=true;return;}});break;case'ends-with':\\$(objs).each(function(){indexPoint=(\\$(this).val().length-cmp_value.length);if(indexPoint>=0&&\\$(this).val().indexOf(cmp_value,indexPoint)==indexPoint){comp_res=true;return;}});break;default:comp_res=false;break;} return comp_res;} function applyRule(field_id){\\$('.cond_filler_'+field_id).each(function(){var this_conditions=\\$('#'+field_id).attr('data-cond-fields').split('|');var this_action=\\$('#'+field_id).attr('data-cond-action').split(':');var cmp_res=this_action=='all'?true:false;for(i=0;i\n';if(form!=''){form.closest('form').append(alert_box);}else{\\$('#form-5d835a818155f').append(alert_box);}} function addConditionClass(field_id,cond_class,form_fields_wrapper){\\$(field_id).each(function(){if(\\$(this).is(':input')||\\$(this).is('select')) \\$(this).addClass('cond_filler_'+cond_class);\\$(this).children().each(function(){addConditionClass(\\$(this),cond_class,form_fields_wrapper);})});return false;} function compareRule(objs,cmp_operator,cmp_value,cmp_id,\\$form_part_0){var comp_res=false;var areOperandsCb=false;switch(cmp_operator){case'is':if(cmp_value.startsWith('Checkbox_')){test=objs.closest('#form_part_0').find('#'+cmp_value+' :input:checked');areOperandsCb=cmp_id.startsWith('Checkbox_')?true:false;if(areOperandsCb&&objs.length!=test.length){break;}}else{test=objs.closest('#form_part_0').find('#'+cmp_value+' :input');} \\$(objs).each(function(){if(areOperandsCb){comp_res=false;} \\$cmp1=\\$(this).val();\\$(test).each(function(){\\$cmp2=\\$(this).val();if(\\$cmp1==\\$cmp2){comp_res=true;if(!areOperandsCb){return;}}});if(areOperandsCb&&false==comp_res){return;}});break;case'is-not':if(cmp_value.startsWith('Checkbox_')){test=\\$form_part_0.find('#'+cmp_value+' :input:checked');areOperandsCb=cmp_id.startsWith('Checkbox_')?true:false;if(areOperandsCb&&objs.length!=test.length){return true;}}else{test=objs.closest('#form_part_0').find('#'+cmp_value+' :input');} for(let objsElement of objs){comp_res=false;\\$cmp1=\\$(objsElement).val();for(let testElement of test){\\$cmp2=\\$(testElement).val();if(\\$cmp1!=\\$cmp2){comp_res=true;}else if(areOperandsCb){comp_res=false;break;}} if(areOperandsCb&&true==comp_res){break;}} break;case'less-than':\\$(objs).each(function(){if(!isNaN(cmp_value)){cmp_value=Number(cmp_value);} if(\\$(this).val()cmp_value){comp_res=true;return;}});break;case'starts-with':\\$(objs).each(function(){if(\\$(this).val().indexOf(cmp_value)==0){comp_res=true;return;}});break;case'contains':\\$(objs).each(function(){if(\\$(this).val().indexOf(cmp_value)!=-1){comp_res=true;return;}});break;case'ends-with':\\$(objs).each(function(){indexPoint=(\\$(this).val().length-cmp_value.length);if(indexPoint>=0&&\\$(this).val().indexOf(cmp_value,indexPoint)==indexPoint){comp_res=true;return;}});break;default:comp_res=false;break;} return comp_res;} function applyRule(field_id){\\$('.cond_filler_'+field_id).each(function(){var this_conditions=\\$('#'+field_id).attr('data-cond-fields').split('|');var this_action=\\$('#'+field_id).attr('data-cond-action').split(':');var cmp_res=this_action=='all'?true:false;for(i=0;i\n';if(form!=''){form.closest('form').append(alert_box);}else{\\$('#form-5d835a8182c77').append(alert_box);}} function addConditionClass(field_id,cond_class,form_fields_wrapper){\\$(field_id).each(function(){if(\\$(this).is(':input')||\\$(this).is('select')) \\$(this).addClass('cond_filler_'+cond_class);\\$(this).children().each(function(){addConditionClass(\\$(this),cond_class,form_fields_wrapper);})});return false;} function compareRule(objs,cmp_operator,cmp_value,cmp_id,\\$form_part_0){var comp_res=false;var areOperandsCb=false;switch(cmp_operator){case'is':if(cmp_value.startsWith('Checkbox_')){test=objs.closest('#form_part_0').find('#'+cmp_value+' :input:checked');areOperandsCb=cmp_id.startsWith('Checkbox_')?true:false;if(areOperandsCb&&objs.length!=test.length){break;}}else{test=objs.closest('#form_part_0').find('#'+cmp_value+' :input');} \\$(objs).each(function(){if(areOperandsCb){comp_res=false;} \\$cmp1=\\$(this).val();\\$(test).each(function(){\\$cmp2=\\$(this).val();if(\\$cmp1==\\$cmp2){comp_res=true;if(!areOperandsCb){return;}}});if(areOperandsCb&&false==comp_res){return;}});break;case'is-not':if(cmp_value.startsWith('Checkbox_')){test=\\$form_part_0.find('#'+cmp_value+' :input:checked');areOperandsCb=cmp_id.startsWith('Checkbox_')?true:false;if(areOperandsCb&&objs.length!=test.length){return true;}}else{test=objs.closest('#form_part_0').find('#'+cmp_value+' :input');} for(let objsElement of objs){comp_res=false;\\$cmp1=\\$(objsElement).val();for(let testElement of test){\\$cmp2=\\$(testElement).val();if(\\$cmp1!=\\$cmp2){comp_res=true;}else if(areOperandsCb){comp_res=false;break;}} if(areOperandsCb&&true==comp_res){break;}} break;case'less-than':\\$(objs).each(function(){if(!isNaN(cmp_value)){cmp_value=Number(cmp_value);} if(\\$(this).val()cmp_value){comp_res=true;return;}});break;case'starts-with':\\$(objs).each(function(){if(\\$(this).val().indexOf(cmp_value)==0){comp_res=true;return;}});break;case'contains':\\$(objs).each(function(){if(\\$(this).val().indexOf(cmp_value)!=-1){comp_res=true;return;}});break;case'ends-with':\\$(objs).each(function(){indexPoint=(\\$(this).val().length-cmp_value.length);if(indexPoint>=0&&\\$(this).val().indexOf(cmp_value,indexPoint)==indexPoint){comp_res=true;return;}});break;default:comp_res=false;break;} return comp_res;} function applyRule(field_id){\\$('.cond_filler_'+field_id).each(function(){var this_conditions=\\$('#'+field_id).attr('data-cond-fields').split('|');var this_action=\\$('#'+field_id).attr('data-cond-action').split(':');var cmp_res=this_action=='all'?true:false;for(i=0;i\n';if(form!=''){form.closest('form').append(alert_box);}else{\\$('#form-5d835a8184333').append(alert_box);}} function addConditionClass(field_id,cond_class,form_fields_wrapper){\\$(field_id).each(function(){if(\\$(this).is(':input')||\\$(this).is('select')) \\$(this).addClass('cond_filler_'+cond_class);\\$(this).children().each(function(){addConditionClass(\\$(this),cond_class,form_fields_wrapper);})});return false;} function compareRule(objs,cmp_operator,cmp_value,cmp_id,\\$form_part_0){var comp_res=false;var areOperandsCb=false;switch(cmp_operator){case'is':if(cmp_value.startsWith('Checkbox_')){test=objs.closest('#form_part_0').find('#'+cmp_value+' :input:checked');areOperandsCb=cmp_id.startsWith('Checkbox_')?true:false;if(areOperandsCb&&objs.length!=test.length){break;}}else{test=objs.closest('#form_part_0').find('#'+cmp_value+' :input');} \\$(objs).each(function(){if(areOperandsCb){comp_res=false;} \\$cmp1=\\$(this).val();\\$(test).each(function(){\\$cmp2=\\$(this).val();if(\\$cmp1==\\$cmp2){comp_res=true;if(!areOperandsCb){return;}}});if(areOperandsCb&&false==comp_res){return;}});break;case'is-not':if(cmp_value.startsWith('Checkbox_')){test=\\$form_part_0.find('#'+cmp_value+' :input:checked');areOperandsCb=cmp_id.startsWith('Checkbox_')?true:false;if(areOperandsCb&&objs.length!=test.length){return true;}}else{test=objs.closest('#form_part_0').find('#'+cmp_value+' :input');} for(let objsElement of objs){comp_res=false;\\$cmp1=\\$(objsElement).val();for(let testElement of test){\\$cmp2=\\$(testElement).val();if(\\$cmp1!=\\$cmp2){comp_res=true;}else if(areOperandsCb){comp_res=false;break;}} if(areOperandsCb&&true==comp_res){break;}} break;case'less-than':\\$(objs).each(function(){if(!isNaN(cmp_value)){cmp_value=Number(cmp_value);} if(\\$(this).val()cmp_value){comp_res=true;return;}});break;case'starts-with':\\$(objs).each(function(){if(\\$(this).val().indexOf(cmp_value)==0){comp_res=true;return;}});break;case'contains':\\$(objs).each(function(){if(\\$(this).val().indexOf(cmp_value)!=-1){comp_res=true;return;}});break;case'ends-with':\\$(objs).each(function(){indexPoint=(\\$(this).val().length-cmp_value.length);if(indexPoint>=0&&\\$(this).val().indexOf(cmp_value,indexPoint)==indexPoint){comp_res=true;return;}});break;default:comp_res=false;break;} return comp_res;} function applyRule(field_id){\\$('.cond_filler_'+field_id).each(function(){var this_conditions=\\$('#'+field_id).attr('data-cond-fields').split('|');var this_action=\\$('#'+field_id).attr('data-cond-action').split(':');var cmp_res=this_action=='all'?true:false;for(i=0;i\n';if(form!=''){form.closest('form').append(alert_box);}else{\\$('#form-5d835a8185960').append(alert_box);}} function addConditionClass(field_id,cond_class,form_fields_wrapper){\\$(field_id).each(function(){if(\\$(this).is(':input')||\\$(this).is('select')) \\$(this).addClass('cond_filler_'+cond_class);\\$(this).children().each(function(){addConditionClass(\\$(this),cond_class,form_fields_wrapper);})});return false;} function compareRule(objs,cmp_operator,cmp_value,cmp_id,\\$form_part_0){var comp_res=false;var areOperandsCb=false;switch(cmp_operator){case'is':if(cmp_value.startsWith('Checkbox_')){test=objs.closest('#form_part_0').find('#'+cmp_value+' :input:checked');areOperandsCb=cmp_id.startsWith('Checkbox_')?true:false;if(areOperandsCb&&objs.length!=test.length){break;}}else{test=objs.closest('#form_part_0').find('#'+cmp_value+' :input');} \\$(objs).each(function(){if(areOperandsCb){comp_res=false;} \\$cmp1=\\$(this).val();\\$(test).each(function(){\\$cmp2=\\$(this).val();if(\\$cmp1==\\$cmp2){comp_res=true;if(!areOperandsCb){return;}}});if(areOperandsCb&&false==comp_res){return;}});break;case'is-not':if(cmp_value.startsWith('Checkbox_')){test=\\$form_part_0.find('#'+cmp_value+' :input:checked');areOperandsCb=cmp_id.startsWith('Checkbox_')?true:false;if(areOperandsCb&&objs.length!=test.length){return true;}}else{test=objs.closest('#form_part_0').find('#'+cmp_value+' :input');} for(let objsElement of objs){comp_res=false;\\$cmp1=\\$(objsElement).val();for(let testElement of test){\\$cmp2=\\$(testElement).val();if(\\$cmp1!=\\$cmp2){comp_res=true;}else if(areOperandsCb){comp_res=false;break;}} if(areOperandsCb&&true==comp_res){break;}} break;case'less-than':\\$(objs).each(function(){if(!isNaN(cmp_value)){cmp_value=Number(cmp_value);} if(\\$(this).val()cmp_value){comp_res=true;return;}});break;case'starts-with':\\$(objs).each(function(){if(\\$(this).val().indexOf(cmp_value)==0){comp_res=true;return;}});break;case'contains':\\$(objs).each(function(){if(\\$(this).val().indexOf(cmp_value)!=-1){comp_res=true;return;}});break;case'ends-with':\\$(objs).each(function(){indexPoint=(\\$(this).val().length-cmp_value.length);if(indexPoint>=0&&\\$(this).val().indexOf(cmp_value,indexPoint)==indexPoint){comp_res=true;return;}});break;default:comp_res=false;break;} return comp_res;} function applyRule(field_id){\\$('.cond_filler_'+field_id).each(function(){var this_conditions=\\$('#'+field_id).attr('data-cond-fields').split('|');var this_action=\\$('#'+field_id).attr('data-cond-action').split(':');var cmp_res=this_action=='all'?true:false;for(i=0;i"
]
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https://www.litscape.com/word_analysis/edgewise | [
"# Definition of edgewise\n\n## \"edgewise\" in the adverb sense\n\n### 1. edgeways, edgewise\n\nwith the edge forward or on, by, or toward the edge\n\n\"he sawed the board edgeways\"\n\n\"held it edgewise\"\n\n### 2. edgewise, edgeways\n\nas if by an edge barely\n\n\"I could not get a word in edgewise\"\n\nSource: WordNet® (An amazing lexical database of English)\n\nWordNet®. Princeton University. 2010.\n\n# edgewise in Scrabble®\n\nThe word edgewise is playable in Scrabble®, no blanks required. Because it is longer than 7 letters, you would have to play off an existing word or do it in several moves.\n\nEDGEWISE\n(153)\n\nwedgies\n\nEDGEWISE\n(153)\nEDGEWISE\n(126)\nEDGEWISE\n(102)\nEDGEWISE\n(84)\nEDGEWISE\n(52)\nEDGEWISE\n(52)\nEDGEWISE\n(45)\nEDGEWISE\n(45)\nEDGEWISE\n(42)\nEDGEWISE\n(42)\nEDGEWISE\n(42)\nEDGEWISE\n(42)\nEDGEWISE\n(42)\nEDGEWISE\n(36)\nEDGEWISE\n(34)\nEDGEWISE\n(34)\nEDGEWISE\n(32)\nEDGEWISE\n(32)\nEDGEWISE\n(30)\nEDGEWISE\n(30)\nEDGEWISE\n(30)\nEDGEWISE\n(30)\nEDGEWISE\n(30)\nEDGEWISE\n(30)\nEDGEWISE\n(30)\nEDGEWISE\n(28)\nEDGEWISE\n(28)\nEDGEWISE\n(28)\nEDGEWISE\n(28)\nEDGEWISE\n(28)\nEDGEWISE\n(26)\nEDGEWISE\n(26)\nEDGEWISE\n(26)\nEDGEWISE\n(26)\nEDGEWISE\n(23)\nEDGEWISE\n(19)\nEDGEWISE\n(19)\nEDGEWISE\n(19)\nEDGEWISE\n(19)\nEDGEWISE\n(17)\nEDGEWISE\n(17)\nEDGEWISE\n(17)\nEDGEWISE\n(17)\nEDGEWISE\n(16)\nEDGEWISE\n(16)\nEDGEWISE\n(15)\n\nEDGEWISE\n(153)\nEDGEWISE\n(126)\nEDGEWISE\n(102)\nWEDGIES\n(98 = 48 + 50)\nWEDGIES\n(98 = 48 + 50)\nWEDGIES\n(92 = 42 + 50)\nWEDGIES\n(92 = 42 + 50)\nWEDGIES\n(92 = 42 + 50)\nWEDGIES\n(89 = 39 + 50)\nWEDGIES\n(89 = 39 + 50)\nWEDGIES\n(89 = 39 + 50)\nWEDGIES\n(89 = 39 + 50)\nWEDGIES\n(86 = 36 + 50)\nWEDGIES\n(86 = 36 + 50)\nEDGEWISE\n(84)\nWEDGIES\n(82 = 32 + 50)\nWEDGIES\n(82 = 32 + 50)\nWEDGIES\n(82 = 32 + 50)\nWEDGIES\n(78 = 28 + 50)\nWEDGIES\n(78 = 28 + 50)\nWEDGIES\n(78 = 28 + 50)\nWEDGIES\n(78 = 28 + 50)\nWEDGIES\n(78 = 28 + 50)\nWEDGIES\n(76 = 26 + 50)\nWEDGIES\n(76 = 26 + 50)\nWEDGIES\n(76 = 26 + 50)\nWEDGIES\n(76 = 26 + 50)\nWEDGIES\n(76 = 26 + 50)\nWEDGIES\n(74 = 24 + 50)\nWEDGIES\n(74 = 24 + 50)\nWEDGIES\n(74 = 24 + 50)\nWEDGIES\n(74 = 24 + 50)\nWEDGIES\n(74 = 24 + 50)\nWEDGIES\n(72 = 22 + 50)\nWEDGIES\n(69 = 19 + 50)\nWEDGIES\n(68 = 18 + 50)\nWEDGIES\n(68 = 18 + 50)\nWEDGIES\n(67 = 17 + 50)\nWEDGIES\n(66 = 16 + 50)\nWEDGIES\n(66 = 16 + 50)\nWEDGIES\n(65 = 15 + 50)\nWEDGIES\n(65 = 15 + 50)\nWEDGIES\n(65 = 15 + 50)\nWEDGIES\n(65 = 15 + 50)\nWEDGIES\n(64 = 14 + 50)\nWEDGIES\n(64 = 14 + 50)\nEDGEWISE\n(52)\nEDGEWISE\n(52)\nEDGEWISE\n(45)\nEDGEWISE\n(45)\nWEDGES\n(45)\nEDGEWISE\n(42)\nEDGEWISE\n(42)\nEDGEWISE\n(42)\nWEDGE\n(42)\nEDGEWISE\n(42)\nEDGEWISE\n(42)\nWEDGES\n(39)\nWIDES\n(39)\nWISED\n(39)\nWEDGES\n(39)\nWEEDS\n(39)\nWEDGES\n(38)\nWEDGE\n(36)\nWEDS\n(36)\nWEED\n(36)\nWEDGE\n(36)\nWEDGES\n(36)\nWIGS\n(36)\nWEDGES\n(36)\nWEDGES\n(36)\nEDGEWISE\n(36)\nWIDE\n(36)\nEDGEWISE\n(34)\nWEEDS\n(34)\nWISED\n(34)\nWIDES\n(34)\nEDGEWISE\n(34)\nWISE\n(33)\nWEDGE\n(33)\nWEDGES\n(33)\nSEWED\n(33)\nWEEDS\n(33)\nWISED\n(33)\nWEDGES\n(33)\nWEDGE\n(33)\nEDGEWISE\n(32)\nEDGEWISE\n(32)\nSEWED\n(30)\nSEWED\n(30)\nSEWED\n(30)\nSWIG\n(30)\nSIEGED\n(30)\nSIEGED\n(30)\nWEDGES\n(30)\nWEDGE\n(30)\nWEDGE\n(30)\nWEDGE\n(30)\nDEWS\n(30)\nWEDGES\n(30)\nWISED\n(30)\nEDGEWISE\n(30)\nWIDES\n(30)\nEDGEWISE\n(30)\nEDGEWISE\n(30)\nWISED\n(30)\nEDGEWISE\n(30)\nWIDES\n(30)\nWIDES\n(30)\nEDGEWISE\n(30)\nEDGEWISE\n(30)\nWEEDS\n(30)\nWEEDS\n(30)\nWEED\n(30)\nEDGEWISE\n(30)\nEDGEWISE\n(28)\nEDGEWISE\n(28)\nWEDGE\n(28)\nEDGEWISE\n(28)\nEDGEWISE\n(28)\nEDGEWISE\n(28)\nWEDGE\n(28)\nSIEGED\n(27)\nSEWED\n(27)\nWISED\n(27)\nSEWED\n(27)\nSEWED\n(27)\nSWIG\n(27)\nWISED\n(27)\nSIEGED\n(27)\nSIEGED\n(27)\nSIEGED\n(27)\nEDGES\n(27)\nWISED\n(27)\nWIDES\n(27)\nWEEDS\n(27)\nWEEDS\n(27)\nWEEDS\n(27)\nWIDE\n(27)\nWIDES\n(27)\nWEDS\n(27)\nWIDES\n(27)\nWIGS\n(27)\nDEWS\n(27)\nSEDGE\n(27)\nWEDGES\n(26)\nWIDES\n(26)\nWISED\n(26)\nWIDES\n(26)\nWISED\n(26)\nWEEDS\n(26)\nSEWED\n(26)\nWISED\n(26)\nWEEDS\n(26)\nWEDGES\n(26)\nWEDGES\n(26)\nEDGEWISE\n(26)\nEDGEWISE\n(26)\nEDGEWISE\n(26)\nEDGEWISE\n(26)\nSWIG\n(24)\nSIEGED\n(24)\nWEED\n(24)\nSWIG\n(24)\nWEED\n(24)\nDEWS\n(24)\nWEED\n(24)\nWEED\n(24)\nWEED\n(24)\nWEDGES\n(24)\nSIEGE\n(24)\nWIGS\n(24)\nWEDGES\n(24)\nGEESE\n(24)\nWIDE\n(24)\nSIEGED\n(24)\nWEDGES\n(24)\nWIDE\n(24)\nWIDE\n(24)\nWIDE\n(24)\nWIDE\n(24)\nWISE\n(24)\nSIEGED\n(24)\nSWIG\n(24)\nSEDGE\n(24)\nDIGS\n(24)\nWIGS\n(24)\nSWIG\n(24)\nEDGES\n(24)\nEDGES\n(24)\nWIGS\n(24)\nWIGS\n(24)\nEWES\n(24)\nEDGES\n(24)\nDEWS\n(24)\nDEWS\n(24)\nWEDGE\n(24)\nEWES\n(24)\nWEDS\n(24)\nDEWS\n(24)\n\n# edgewise in Words With Friends™\n\nThe word edgewise is playable in Words With Friends™, no blanks required. Because it is longer than 7 letters, you would have to play off an existing word or do it in several moves.\n\nEDGEWISE\n(132)\n\nwedgies\n\nEDGEWISE\n(132)\nEDGEWISE\n(96)\nEDGEWISE\n(84)\nEDGEWISE\n(72)\nEDGEWISE\n(66)\nEDGEWISE\n(60)\nEDGEWISE\n(60)\nEDGEWISE\n(60)\nEDGEWISE\n(56)\nEDGEWISE\n(56)\nEDGEWISE\n(54)\nEDGEWISE\n(54)\nEDGEWISE\n(54)\nEDGEWISE\n(48)\nEDGEWISE\n(44)\nEDGEWISE\n(40)\nEDGEWISE\n(36)\nEDGEWISE\n(34)\nEDGEWISE\n(32)\nEDGEWISE\n(32)\nEDGEWISE\n(32)\nEDGEWISE\n(32)\nEDGEWISE\n(32)\nEDGEWISE\n(32)\nEDGEWISE\n(30)\nEDGEWISE\n(30)\nEDGEWISE\n(28)\nEDGEWISE\n(28)\nEDGEWISE\n(28)\nEDGEWISE\n(28)\nEDGEWISE\n(28)\nEDGEWISE\n(28)\nEDGEWISE\n(28)\nEDGEWISE\n(28)\nEDGEWISE\n(24)\nEDGEWISE\n(22)\nEDGEWISE\n(21)\nEDGEWISE\n(20)\nEDGEWISE\n(20)\nEDGEWISE\n(20)\nEDGEWISE\n(20)\nEDGEWISE\n(19)\nEDGEWISE\n(19)\nEDGEWISE\n(18)\nEDGEWISE\n(18)\nEDGEWISE\n(18)\nEDGEWISE\n(18)\nEDGEWISE\n(18)\nEDGEWISE\n(17)\nEDGEWISE\n(17)\nEDGEWISE\n(17)\nEDGEWISE\n(17)\nEDGEWISE\n(16)\nEDGEWISE\n(16)\nEDGEWISE\n(16)\nEDGEWISE\n(16)\nEDGEWISE\n(15)\nEDGEWISE\n(15)\n\nEDGEWISE\n(132)\nWEDGIES\n(110 = 75 + 35)\nWEDGIES\n(98 = 63 + 35)\nWEDGIES\n(98 = 63 + 35)\nWEDGIES\n(98 = 63 + 35)\nEDGEWISE\n(96)\nWEDGIES\n(92 = 57 + 35)\nWEDGIES\n(92 = 57 + 35)\nWEDGIES\n(87 = 52 + 35)\nWEDGIES\n(87 = 52 + 35)\nWEDGIES\n(87 = 52 + 35)\nWEDGIES\n(86 = 51 + 35)\nEDGEWISE\n(84)\nWEDGIES\n(80 = 45 + 35)\nWEDGIES\n(80 = 45 + 35)\nWEDGIES\n(80 = 45 + 35)\nWEDGIES\n(77 = 42 + 35)\nEDGEWISE\n(72)\nWEDGES\n(72)\nWEDGIES\n(69 = 34 + 35)\nWEDGIES\n(69 = 34 + 35)\nEDGEWISE\n(66)\nWEDGIES\n(65 = 30 + 35)\nWEDGIES\n(65 = 30 + 35)\nWEDGIES\n(65 = 30 + 35)\nWEDGIES\n(65 = 30 + 35)\nWEDGIES\n(65 = 30 + 35)\nWEDGIES\n(63 = 28 + 35)\nWEDGIES\n(63 = 28 + 35)\nWEDGIES\n(63 = 28 + 35)\nWEDGIES\n(63 = 28 + 35)\nWEDGIES\n(61 = 26 + 35)\nWEDGIES\n(61 = 26 + 35)\nWEDGIES\n(61 = 26 + 35)\nWEDGIES\n(61 = 26 + 35)\nWEDGIES\n(61 = 26 + 35)\nWEDGIES\n(61 = 26 + 35)\nWEDGIES\n(61 = 26 + 35)\nWEDGES\n(60)\nEDGEWISE\n(60)\nWEDGES\n(60)\nEDGEWISE\n(60)\nEDGEWISE\n(60)\nWEDGIES\n(58 = 23 + 35)\nWEDGIES\n(57 = 22 + 35)\nWEDGE\n(57)\nSIEGED\n(57)\nEDGEWISE\n(56)\nEDGEWISE\n(56)\nWEDGIES\n(55 = 20 + 35)\nWEDGIES\n(55 = 20 + 35)\nWEDGIES\n(54 = 19 + 35)\nEDGEWISE\n(54)\nEDGEWISE\n(54)\nEDGEWISE\n(54)\nWEDGIES\n(54 = 19 + 35)\nWEDGIES\n(54 = 19 + 35)\nWEDGES\n(54)\nWEDGIES\n(54 = 19 + 35)\nWEDGIES\n(53 = 18 + 35)\nWEDGIES\n(53 = 18 + 35)\nWEDGIES\n(52 = 17 + 35)\nWEDGIES\n(52 = 17 + 35)\nWEDGIES\n(52 = 17 + 35)\nWEDGIES\n(52 = 17 + 35)\nWEDGIES\n(52 = 17 + 35)\nWEDGIES\n(51 = 16 + 35)\nWEDGIES\n(51 = 16 + 35)\nWEDGIES\n(51 = 16 + 35)\nWEEDS\n(51)\nWIGS\n(51)\nWEDGE\n(51)\nWISED\n(51)\nWIDES\n(51)\nWEDGIES\n(51 = 16 + 35)\nWEDGIES\n(50 = 15 + 35)\nWEDGIES\n(50 = 15 + 35)\nWEDGIES\n(50 = 15 + 35)\nWEDGIES\n(49 = 14 + 35)\nWEDGIES\n(49 = 14 + 35)\nWEDGIES\n(49 = 14 + 35)\nWIDE\n(48)\nEDGEWISE\n(48)\nWEDGES\n(48)\nWEDGES\n(48)\nWEDGES\n(48)\nWEED\n(48)\nWEDS\n(48)\nWEDGIES\n(48 = 13 + 35)\nWISE\n(45)\nSWIG\n(45)\nSIEGED\n(45)\nWEDGE\n(44)\nEDGEWISE\n(44)\nWEDGES\n(42)\nSEDGE\n(42)\nWEDGES\n(42)\nWEDGES\n(42)\nWEDGES\n(40)\nEDGEWISE\n(40)\nWEDGE\n(39)\nGEESE\n(39)\nSEWED\n(39)\nSIEGE\n(39)\nSIEGED\n(39)\nWISED\n(39)\nSIEGED\n(39)\nWEDGE\n(39)\nWEEDS\n(39)\nWEDGE\n(38)\nSIEGED\n(36)\nDEWS\n(36)\nWEED\n(36)\nEDGEWISE\n(36)\nWEDGES\n(36)\nWEDGES\n(36)\nSIEGED\n(36)\nSEWED\n(36)\nGEES\n(36)\nWISED\n(36)\nEDGES\n(36)\nWEEDS\n(36)\nWIDES\n(36)\nWISED\n(34)\nWIDES\n(34)\nEDGEWISE\n(34)\nWEEDS\n(34)\nWIDES\n(33)\nSIEGED\n(33)\nWIDES\n(33)\nSIEGED\n(33)\nDIGS\n(33)\nSIEGED\n(33)\nWEDGE\n(33)\nSEWED\n(33)\nSEWED\n(33)\nSEWED\n(33)\nWIDES\n(33)\nWISED\n(33)\nSWIG\n(33)\nSIEGED\n(33)\nWEDGE\n(33)\nWEEDS\n(33)\nWEEDS\n(33)\nWIGS\n(33)\nWEDGE\n(33)\nWISED\n(33)\nEDGES\n(32)\nWEDGES\n(32)\nSEDGE\n(32)\nEDGEWISE\n(32)\nEDGEWISE\n(32)\nEDGEWISE\n(32)\nEDGEWISE\n(32)\nEDGEWISE\n(32)\nEDGEWISE\n(32)\nEDGEWISE\n(30)\nEDGES\n(30)\nEDGES\n(30)\nEDGES\n(30)\nDEWS\n(30)\nSEDGE\n(30)\nSEDGE\n(30)\nSEDGE\n(30)\nWIDE\n(30)\nWEDS\n(30)\nEDGEWISE\n(30)\nWEDGE\n(30)\nWEDGES\n(30)\nWEDGE\n(28)\nGEESE\n(28)\nWEDGES\n(28)\nWEDGES\n(28)\nEDGEWISE\n(28)\nWEDGES\n(28)\nSIEGE\n(28)\nWEDGES\n(28)\nEDGEWISE\n(28)\nEDGEWISE\n(28)\nEDGEWISE\n(28)\nEDGEWISE\n(28)\nEDGEWISE\n(28)\nEDGEWISE\n(28)\nEDGEWISE\n(28)\nDIES\n(27)\nSIEGE\n(27)\nWISE\n(27)\nSIEGE\n(27)\nSWIG\n(27)\nDEES\n(27)\nSIEGED\n(27)\nEWES\n(27)\nSIEGED\n(27)\nDIGS\n(27)\nWIGS\n(27)\nSWIG\n(27)\nWIDES\n(27)\nWIDES\n(27)\nEWES\n(27)\nEDGE\n(27)\n\n# Word Growth involving edgewise\n\nedge\n\nis wise\n\n## Longer words containing edgewise\n\n(No longer words found)"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.7458847,"math_prob":0.9995937,"size":393,"snap":"2022-27-2022-33","text_gpt3_token_len":118,"char_repetition_ratio":0.18508998,"word_repetition_ratio":0.0,"special_character_ratio":0.24173027,"punctuation_ratio":0.14864865,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9977547,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-06T16:30:28Z\",\"WARC-Record-ID\":\"<urn:uuid:3b586dc4-62cf-4f31-bb9a-b0a672374b8b>\",\"Content-Length\":\"142183\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1e86ba01-1e11-4b0c-a151-02a744c9ebf1>\",\"WARC-Concurrent-To\":\"<urn:uuid:fee2fae0-0195-446f-ab7d-8095f32f8b9c>\",\"WARC-IP-Address\":\"104.21.48.183\",\"WARC-Target-URI\":\"https://www.litscape.com/word_analysis/edgewise\",\"WARC-Payload-Digest\":\"sha1:BHBSROGJDPI7ADK7TN3XRNCQ26MGPF2M\",\"WARC-Block-Digest\":\"sha1:VHJVF2INDW534WQCBM2IOOW34XNZ6MU5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104675818.94_warc_CC-MAIN-20220706151618-20220706181618-00069.warc.gz\"}"} |
https://ssconlineexam.com/onlinetest/ssc-cgl-tier-1/general-intelligence-and-reasoning/gir-test-63 | [
"# ssc cgl tier 1 :: general intelligence and reasoning :: gir test 63\n\n## Home ssc cgl tier 1 / general intelligence and reasoning Questions and Answers\n\n1 . In question, choose the correct alternative from the given ones that will complete the series.\nGHIK, HIJL, IJKM, JKLN, ?\nBMJH\nJHLM\nKJIM\nKLMO",
null,
"View Answer",
null,
"Discuss in Forum\n\n2 . In question, choose the correct alternative from the given ones that will complete the series.\nHJK, MOP, RTU, ?\nWXZ\nWXY\nWYZ\nWYX",
null,
"View Answer",
null,
"Discuss in Forum\n\n3 . In question, choose the correct alternative from the given ones that will complete the series.\n6, 13, 28, ?, 122\n29\n59\n34\n93",
null,
"View Answer",
null,
"Discuss in Forum\n\n4 .\nFind the wrong number in the given number series\n\n2, 17, 47, 77, 122, 177, 242\n177\n122\n2\n47",
null,
"View Answer",
null,
"Discuss in Forum\n\n5 .\nFrom the following alternatives select the word which cannot be formed by using the letters of the given word\n\n$RESIDENTIAL$\nALTERS\nDENTAL\nARDENT\nDIGITAL",
null,
"View Answer",
null,
"Discuss in Forum\n\n6 .\nIf A = 6, B = 7, C = 8, D = 9, then the numbers 7, 23, 20, 25, 13, 10, 23 will give the word\nNEITHER\nBROTHER\nBOUSTER\nBRAWLER",
null,
"View Answer",
null,
"Discuss in Forum\n\n7 .\nIn a certain code INTRODUCE is written as RTNIOECUD. In the same code which alternative will be written for KNOWLEDGE?\nGDELOWNKE\nWONKLEGDE\nWKONLEGDE\nWOKNLEGDE",
null,
"View Answer",
null,
"Discuss in Forum\n\n8 .\nIf a means plus, b means minus, c means multiplication, and d means division what will come at question mark (?) in the given equation?\n\n12 c 8 a 9 b 48 d 6 = ?\n96\n97\n105\n87",
null,
"View Answer",
null,
"Discuss in Forum\n\n9 .\nSelect the correct combination of mathematical signs to replace * and balance the given expression :\n\n40 * 20 * 10 * 2 * 4\n- $\\times$ $\\div$ =\n- = $\\div$ $\\times$\n= $\\times$ - $\\div$\n$\\div$ $\\times$ - =",
null,
"View Answer",
null,
"Discuss in Forum\n\n10 .In question select the missing number from the given responses",
null,
"3\n5\n93\n89",
null,
"View Answer",
null,
"Discuss in Forum"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.7152942,"math_prob":0.9619065,"size":1738,"snap":"2020-10-2020-16","text_gpt3_token_len":480,"char_repetition_ratio":0.24855825,"word_repetition_ratio":0.31360948,"special_character_ratio":0.30897585,"punctuation_ratio":0.21802326,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9790044,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,3,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-18T13:57:56Z\",\"WARC-Record-ID\":\"<urn:uuid:5a207910-7572-4b52-8cf1-6ab08b55a226>\",\"Content-Length\":\"50286\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1b995422-db76-4d82-ac43-a0447c8f55f1>\",\"WARC-Concurrent-To\":\"<urn:uuid:3f270633-e005-4c24-a5f9-36059e9b7664>\",\"WARC-IP-Address\":\"35.200.86.105\",\"WARC-Target-URI\":\"https://ssconlineexam.com/onlinetest/ssc-cgl-tier-1/general-intelligence-and-reasoning/gir-test-63\",\"WARC-Payload-Digest\":\"sha1:Q25DOLOKXSOT5XJSKBRYQWYUUX4OY3UH\",\"WARC-Block-Digest\":\"sha1:334MQWMF65D5JIK3RJJPNHUAUOR65HOL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875143695.67_warc_CC-MAIN-20200218120100-20200218150100-00370.warc.gz\"}"} |
https://mathematica.stackexchange.com/questions/219816/multiply-an-array-by-an-especific-number-at-an-specific-position-in-the-array | [
"# Multiply an array by an especific number at an specific position in the array\n\nIf I have a table of let's say 2000 values in column 1 and 2, how can I multiply only an specific part of that table by a certain number?. In particular, I want to multiply the second column only from row 3 to row 2000 (not including row 1 and 2) by a number (let's say 5) and leaving column 1 intact. I would very much appreciate your help.\n\nI am aware of Elegant operations on matrix rows and columns but I could not find a solution for this in particular there.\n\n• table[[3;;2000,2]]=5 table[[3;;2000,2]]?\n– kglr\nApr 18, 2020 at 0:57\n• try also MapAt[5 #&, table, {3;;2000,2}]\n– kglr\nApr 18, 2020 at 0:59\n• Thank you kglr. The first method works but I cannot access column 1. I would like to have column 2 multiplied by the number while at the same time column 1 does not dissapear. The second method does not work: If I have table = Table[{1, 2}, 10]; and then I use MapAt[5 # &, table, {3 ;; 10, 2}], then I get {{1, 2}, {1, 2}, {78125, 3906250}, {78125, 3906250}, {78125, 3906250}, {78125, 3906250}, {78125, 3906250}, {78125, 3906250}, {78125, 3906250}, {78125, 3906250}}\n– John\nApr 18, 2020 at 1:16\n\ntable = Table[{1, 2}, 10];\n\n\nTo create new tables without changing table:\n\ntable2 = Module[{t = #}, t[[2 ;; 10, 2]] = 5 t[[2 ;; 10, 2]]; t] & @ table;\ntable3 = Module[{t = #}, t = MapAt[5 # &, t, {2 ;; 10, 2}]] & @ table;\ntable4 = Module[{t = #}, t[[2 ;; 10, 2]] *= 5; t] & @ table;\n\n{MatrixForm /@ {table, table2, table3, table4},\n{\"table\", \"table2\", \"table3\", \"table4\"}}], Spacer]",
null,
"• Thank you very much kglr! This works perfect!\n– John\nApr 18, 2020 at 1:56\n\nMultiplying the row range $$a_1$$ to $$a_2$$ and column range $$b_1$$ to $$b_2$$ by the scalar $$c$$ in the matrix $$A$$:\n\nA[[a1;;a2, b1;;b2]] *= c\n\n• Unfortunately, this method does not seem to work. It gives a multiplication by 5, every single time you hit shift + Enter. In other words, at first it multiplies by 5 and then you do it again and now is the last by 5 and then the last by 5 again and so on, saving the last calculation.\n– John\nApr 18, 2020 at 1:18\n• That's what any line of code that depends on the previous state of $A$ will do. If you want a line of code that fixes $A$ no matter what, you will have to hardcode the value into the line A[[a1;;a2, b1;;b2]] = c * hard_coded_A[[a1;;a2, b1;;b2]]. This is bad practice. If you write your notebook sequentially assuming that line will be hit only once, it will work. It will be more readable that way too. Apr 18, 2020 at 1:22\n• I understand but the code does not seem to work very well either in any case. For instance, if I have table = Table[{1, 2}, 10]; and then I use your code as table[[3 ;; 10]] *= 5, in the very first iteration it will give {{5, 50}, {5, 50}, {5, 50}, {5, 50}, {5, 50}, {5, 50}, {5, 50}, {5, 50}}... which is obviously wrong, as it seems to multiply by 5 the first row and by 25 the second.\n– John\nApr 18, 2020 at 1:30\n• I don't follow. The following code m = Table[{1,2},10]; m[[3;;10]]*=5; outputs {{5, 10}, {5, 10}, {5, 10}, {5, 10}, {5, 10}, {5, 10}, {5, 10}, {5, 10}}. Apr 18, 2020 at 1:35\n• Correct for the first iteration. But remember that I would like to have the first row unchanged and in that code both rows are multiplied. Is there any way to have only the second row multiplied by 5 and not both rows?\n– John\nApr 18, 2020 at 1:39"
]
| [
null,
"https://i.stack.imgur.com/0MEeW.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.87334424,"math_prob":0.973295,"size":559,"snap":"2022-40-2023-06","text_gpt3_token_len":160,"char_repetition_ratio":0.113513514,"word_repetition_ratio":0.0,"special_character_ratio":0.32200357,"punctuation_ratio":0.14728682,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99631643,"pos_list":[0,1,2],"im_url_duplicate_count":[null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-05T08:16:43Z\",\"WARC-Record-ID\":\"<urn:uuid:c0a1f2bc-7ccc-4b55-a71b-6b2950e06c99>\",\"Content-Length\":\"246923\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:235ac7c9-8432-4b65-9cce-9ddf5b7106ec>\",\"WARC-Concurrent-To\":\"<urn:uuid:bcd0f764-c793-4cfe-a3e1-3bb2a0d9a169>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://mathematica.stackexchange.com/questions/219816/multiply-an-array-by-an-especific-number-at-an-specific-position-in-the-array\",\"WARC-Payload-Digest\":\"sha1:AI6Y62CJJA6BW6L362IAFCOBJW6XAJMQ\",\"WARC-Block-Digest\":\"sha1:27RCHWHPXP77Q2J4RSKZNTUWBEPSEQB3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337595.1_warc_CC-MAIN-20221005073953-20221005103953-00414.warc.gz\"}"} |
https://math.hecker.org/2011/05/15/linear-algebra-and-its-applications-exercise-1-6-14/ | [
"## Linear Algebra and Its Applications, Exercise 1.6.14\n\nExercise 1.6.14. For any m x n matrix A, prove that",
null,
"$AA^T$ and",
null,
"$A^TA$ are symmetric matrices. Provide an example where these matrices are not equal.\n\nAnswer: Per Equation 1M(i) on page 47 we have",
null,
"$(AB)^T = B^TA^T$\n\nSubstituting",
null,
"$A^T$ for B we have",
null,
"$(AA^T)^T = (A^T)^TA^T = AA^T$\n\nSince",
null,
"$AA^T$ is equal to its own transpose it is a symmetric matrix.\n\nSimilarly we have",
null,
"$(A^TA)^T = A^T(A^T)^T = A^TA$\n\nSince",
null,
"$A^TA$ is equal to its own transpose it too is a symmetric matrix.\n\nNext, we pick an example 2 x 2 matrix",
null,
"$A = \\begin{bmatrix} 1&2 \\\\ 1&3 \\end{bmatrix}$\n\nWe then have",
null,
"$AA^T = \\begin{bmatrix} 1&2 \\\\ 1&3 \\end{bmatrix} \\begin{bmatrix} 1&1 \\\\ 2&3 \\end{bmatrix} = \\begin{bmatrix} 5&7 \\\\ 7&10 \\end{bmatrix}$\n\nand",
null,
"$A^TA = \\begin{bmatrix} 1&1 \\\\ 2&3 \\end{bmatrix} \\begin{bmatrix} 1&2 \\\\ 1&3 \\end{bmatrix} = \\begin{bmatrix} 2&5 \\\\ 5&13 \\end{bmatrix}$\n\nSo in general",
null,
"$AA^T$ does not equal",
null,
"$A^TA$.\n\nNOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition",
null,
"by Gilbert Strang.\n\nIf you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition",
null,
", Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition",
null,
"and the accompanying free online course, and Dr Strang’s other books",
null,
".\n\nThis entry was posted in Uncategorized. Bookmark the permalink."
]
| [
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"http://www.assoc-amazon.com/e/ir",
null,
"http://www.assoc-amazon.com/e/ir",
null,
"http://www.assoc-amazon.com/e/ir",
null,
"https://www.assoc-amazon.com/e/ir",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.8797338,"math_prob":0.99973804,"size":895,"snap":"2022-40-2023-06","text_gpt3_token_len":200,"char_repetition_ratio":0.099887766,"word_repetition_ratio":0.09150327,"special_character_ratio":0.2122905,"punctuation_ratio":0.10555556,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998393,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-31T21:06:15Z\",\"WARC-Record-ID\":\"<urn:uuid:2515d846-9c08-435c-a80e-c4687d521b13>\",\"Content-Length\":\"83018\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fdc52704-591f-4e73-b72d-32c8d52ae977>\",\"WARC-Concurrent-To\":\"<urn:uuid:7df1c86c-76d3-44dd-b37a-4d92d14689a6>\",\"WARC-IP-Address\":\"192.0.78.12\",\"WARC-Target-URI\":\"https://math.hecker.org/2011/05/15/linear-algebra-and-its-applications-exercise-1-6-14/\",\"WARC-Payload-Digest\":\"sha1:NNGVWWWJGLSPMTGEWEXJ3NHTYAFZJUW2\",\"WARC-Block-Digest\":\"sha1:W7PVKOYFK7KSUA427EIQZBDU7AJKCCHU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499890.39_warc_CC-MAIN-20230131190543-20230131220543-00215.warc.gz\"}"} |
https://practicaldev-herokuapp-com.global.ssl.fastly.net/tomasforsman/why-length-is-9-2i4l | [
"## DEV Community",
null,
"# Why (! + [] + [] + ![]).length is 9\n\nEdit: At the bottom I've added what I've figured out since I wrote this.\n\nToday I sat down and figured out why you get this result in JavaScript\n\n``````(! + [] + [] + ![]).length\n> 9\n``````\n\nSo I sat down with the js console and tried to figure things out. There were a lot of this going on:\n\n``````(!\"\")\n>true\n\n(![])\n>false\n\n(!+[])\n>true\n\n(!+[]+[])\n>\"true\"\n\n(+[]+\"2\")\n>\"02\"\n\n([]+\"2\")\n>\"2\"\n``````\n\nWhat I have come up with, and I might be totally off on some of these so please correct anything that doesn't look right, is the following.\n\n• JavaScript does operations from right to left.\n• The + operand first tries to add the left to the right, if that isn't possible it tries to convert the operand on the right to a number and if it can't do that it converts it to a string. If there is something to the left of the + it adds them together, and if it can't it sticks around. If there is nothing to the left it goes away.\n• The ! operand converts the operand to the right to a boolean and then reverts it. Only the following values are the same as false as booleans:\n``````false\n0\n\"\"\nnull\nNaN\nundefined\n``````\n\nAll other values is not false and thus true.\n\nSo now it goes something like this:\n\n``````// We start with\n(! + [] + [] + ![]).length // Remember that we are going from right to left starting\n// with the operation before doing -length\n![] === false // ! converts the content of [] to true, since it's not part of\n// the false group above, and then reverts it.\n+ === + // the + is lonely so far with no operand to the right but sticks\n// around since it has something on the left.\n[]+[] === \"\" // The contents of [] can't be added or converted to a number so\n// the right operand becomes a string. There's still something to\n// the left so the + stays.\n!+[] === true // This the same as !+0 since +[] is 0\n\n// Now we got:\n(\"false\" + \"\" + false).length\n\"\"+false === \"false\" // + adds \"\" to false. It then sticks around.\n\"true\" + \"\" === \"true\" // The + stays after adding \"true\" to \"\"\n\n// ---\n(\"true\"++\"false\")\n+\"false\" ==== \"false\" // + has nothing but an operand to the left so it goes away.\ntrue + \"\" === \"true\" // + adds them together and stays\n\n// ---\n(\"true\"+\"false\")\n\"true\" + \"false\" === \"truefalse\" // + still stays\n\n// ---\n(+\"truefalse\")\n+\"truefalse\" === \"truefalse\" // + has nothing to do after this so it goes away\n\n// ---\n(\"truefalse\") // The operation is done so now we are left with what's outside.\n\n\"truefalse\".length === 9\n\n``````\n\nYes, I did go through every step, even those that seems pointless. I'm not at all sure this is how it works but it is what seems to happen to me.\n\nThoughts?\n\nEdit:\nAfter comments and looking at the documentation this is now how I figure things going.\n((!(+[]))+[]+(![]))\nUnary operators are going right to left thus !+[] becomes +[] -> !0 === true.",
null,
"JavaScript does operations from right to left.\n\nIn some cases it does, but not in this case. The reason you see behaviour that may appear to be right-to-left evaluation is because of operator precedence, where the unary `!` has a higher precedence than the binary `+`.\n\n`+` actually evaluates from left to right, so more verbosely parenthesized, it looks like this:\n\n``````(((!(+[])) + []) + (![])).length\n``````\n\nEvaluating this parenthesized form step-by-step, we get this:\n\n``````/* 1. */ (((!(+[])) + []) + (![])).length // parethesized form.\n/* 2. */ (((!0) + []) + (![])).length // `+[]` is evaluated to `0`.\n/* 3. */ (((!false) + []) + (![])).length // `0` is coerced to the boolean `false`.\n/* 4. */ ((true + []) + (![])).length // `!false` is evaluated to `true`.\n/* 5. */ ((\"true\" + \"\") + (![])).length // `[]` is coerced as a primitive to the string `\"\"`, which means `true` is also coerced to a string.\n/* 6. */ (\"true\" + (![])).length // `\"true\" + \"\"` is string concatenation, which evaluates to `\"true\"`.\n/* 7. */ (\"true\" + (!true)).length // `[]` is coerced to the boolean `true`.\n/* 8. */ (\"true\" + false).length // `!true` evaluates to `false`.\n/* 9. */ (\"true\" + \"false\").length // `false` is coerced to a string because `\"true\"` is a string.\n/* 10. */ \"truefalse\".length // `\"true\" + \"false\"` is evaluated as string concatenation to `\"truefalse\".\n/* 11. */ 9 // the length of `\"truefalse\"` is `9`.\n``````",
null,
"Eugene Karataev\n\nIf I'm not sure what's going on with the JS code, I use ASTexplorer to look at the code with the compiler's eyes. It converts a code string into the tree of instructions to be executed by the compiler step by step.\nBut sometimes text representation of a tree is not expressive enough, so I built a little tool which visualizes an AST tree.\nFor `(! + [] + [] + ![]).length` AST in graph form looks like this:",
null,
"In runtime calculations starts from the bottom left and flow to the top.",
null,
"Jacob Paris • Edited\n``````(({})[[]]+[]).length\n``````\n\nThis is also equal to 9 but for a very different reason",
null,
"Ben Sinclair\n``````(({})[[]]+[])+(({})[[]]+[])+(({})[[]]+[])+(({})[[]]+[])\n``````\n\nStanding on the shoulders of giants.",
null,
"Jacob Paris\n\nYou can stand on your own shoulders this was perfect",
null,
"Ain't that the truef.",
null,
"Sagar Bhattacharya • Edited\n\nIt is a great observation , typecasting in JS is weird. This is the reason I love this platform.",
null,
"nabbisen\n\nTotally greatly interesting.\nThank you for your great post.\nThe behaviors seem a bit strange and are beautiful 😆\n\nDEV Community"
]
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null,
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null
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https://convertlive.com/u/convert/milliwatts/to/megawatts | [
"# Milliwatts to Megawatts\n\nMilliwatts = Megawatts\n\nPrecision: decimal digits\n\nConvert from Milliwatts to Megawatts. Type in the amount you want to convert and press the Convert button.\n\nBelongs in category\nPower\n\n 1 Milliwatts = 1.0×10-9 Megawatts 10 Milliwatts = 1.0×10-8 Megawatts 2500 Milliwatts = 2.5×10-6 Megawatts 2 Milliwatts = 2.0×10-9 Megawatts 20 Milliwatts = 2.0×10-8 Megawatts 5000 Milliwatts = 5.0×10-6 Megawatts 3 Milliwatts = 3.0×10-9 Megawatts 30 Milliwatts = 3.0×10-8 Megawatts 10000 Milliwatts = 1.0×10-5 Megawatts 4 Milliwatts = 4.0×10-9 Megawatts 40 Milliwatts = 4.0×10-8 Megawatts 25000 Milliwatts = 2.5×10-5 Megawatts 5 Milliwatts = 5.0×10-9 Megawatts 50 Milliwatts = 5.0×10-8 Megawatts 50000 Milliwatts = 5.0×10-5 Megawatts 6 Milliwatts = 6.0×10-9 Megawatts 100 Milliwatts = 1.0×10-7 Megawatts 100000 Milliwatts = 0.0001 Megawatts 7 Milliwatts = 7.0×10-9 Megawatts 250 Milliwatts = 2.5×10-7 Megawatts 250000 Milliwatts = 0.00025 Megawatts 8 Milliwatts = 8.0×10-9 Megawatts 500 Milliwatts = 5.0×10-7 Megawatts 500000 Milliwatts = 0.0005 Megawatts 9 Milliwatts = 9.0×10-9 Megawatts 1000 Milliwatts = 1.0×10-6 Megawatts 1000000 Milliwatts = 0.001 Megawatts\n\nEmbed this unit converter in your page or blog, by copying the following HTML code:"
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https://patents.justia.com/patent/20110037878 | [
"# ZOOM LENS SYSTEM AND IMAGE PICKUP APPARATUS INCLUDING THE SAME\n\n- Canon\n\nA zoom lens including in order from an object side to an image side: a positive first lens unit which does not move for zooming; a negative second lens unit for magnification, and a rear lens group including two or more lens units, in which: the first lens unit includes a first subunit not moving for focusing and a second subunit moving for focusing, and the first subunit includes two or more negative lenses and one or more positive lenses; and an average Abbe number and an average partial dispersion ratio of materials of the negative lenses included in the first subunit, average Abbe number and average partial dispersion ratio of materials of the one or more positive lenses included in the first lens unit, combined focal length of the negative lenses included in the first subunit, and focal length of the zoom lens at telephoto end are appropriately set.\n\n## Latest Canon Patents:\n\nDescription\nBACKGROUND OF THE INVENTION\n\n1. Field of the Invention\n\nThe present invention relates to a zoom lens system which is suitable for use in a broadcasting television (TV) camera, a video camera, a digital still camera, and a silver-halide camera, and also to an image pickup apparatus including the zoom lens.\n\n2. Description of the Related Art\n\nIn recent years, there have been demanded a zoom lens system having a high zoom ratio and high optical performance for image pickup apparatus such as a television (TV) camera, a silver-halide camera, a digital camera, and a video camera. A positive lead and telephoto type four-unit zoom lens system in which four lens units are provided in total and one of the lens units located closest to an object side has a positive refractive power has been known as the zoom lens system having a high zoom ratio. For instance, there is known a four-unit zoom lens system which includes a first lens unit having a positive refractive power for focusing, a second lens unit having a negative refractive power for magnification, a third lens unit having a negative refractive power for correcting image plane variation, and a fourth lens unit having a positive refractive power for image formation. As to this four-unit zoom lens system, there is known a four-unit zoom lens system in which an optical material having an extraordinary dispersion characteristic is used, so that chromatic aberration is corrected appropriately and a high optical performance is provided (see U.S. Pat. Nos. 6,825,990 and 6,141,157).\n\nU.S. Pat. No. 6,825,990 discloses a structure in which a first sub lens unit which is fixed during focusing in a first lens unit which is fixed during zooming includes two negative lenses. As to the two negative lenses, a low dispersion material is used as a material of the negative lens disposed on the side closest to an object, so that lateral chromatic aberration is corrected appropriately mainly at the wide angle end. U.S. Pat. No. 6,141,157 discloses a structure in which a thin resin layer having a meniscus shape and a negative refractive power is formed on a lens surface of a second or third positive lens counted from the object side in a first lens unit which is fixed during zooming. Utilizing this resin layer, higher order lateral chromatic aberration is corrected appropriately mainly on the wide angle side without an increase in size or weight.\n\nA positive lead type four-unit zoom lens system having a structure described above may support a high zoom ratio relatively easily. In order to obtain high optical performance in this four-unit zoom lens system, it is important to correct lateral chromatic aberration at the wide angle end and longitudinal chromatic aberration at the telephoto end appropriately. It is easy to correct the chromatic aberration appropriately including the lateral chromatic aberration and the longitudinal chromatic aberration if an optical material having an extraordinary dispersion characteristic is used. However, it is difficult to correct the chromatic aberration appropriately by simply using a lens made of the optical material having an extraordinary dispersion characteristic. In particular, in order to obtain high optical performance over the entire zoom range in the four-unit zoom lens system described above, it is an important factor to set appropriately a material of each lens included in the first lens unit which does not move for zooming. For instance, in order to correct a secondary spectrum of the longitudinal chromatic aberration appropriately on the telephoto side, it is important to set an appropriate difference between dispersions of materials of the positive lens and the negative lens in the first lens unit. If this difference is set inappropriately, it is difficult to suppress higher order aberration mainly.\n\nSUMMARY OF THE INVENTION\n\nA zoom lens system according to the present invention includes in order of from an object side to an image side: a first lens unit (F) having a positive refractive power, which does not move for zooming; a second lens unit (V, V1) having a negative refractive power for a magnification action; and a rear lens group including two or more lens units, characterized in that: the first lens unit (F) includes a first sub lens unit (1a) which does not move for focusing and a second sub lens unit (1b) which moves for focusing, and the first sub lens unit includes two or more negative lenses and one or more positive lenses; and the following conditional expressions are relative,\n\n−1.193×10−3<(θpa−θna)/(νpa−νna)<−0.904×10−3;\n\n40<νpa−νna<55; and\n\n−0.465<f1ns/ft,\n\nwhere νna and θna represent an average Abbe number and an average partial dispersion ratio of materials of the two or more negative lenses included in the first sub lens unit (1a), respectively, νpa and θpa represent an average Abbe number and an average partial dispersion ratio of materials of the one or more positive lenses included in the first lens unit (F), respectively, fins represents a combined focal length of the two or more negative lenses included in the first sub lens unit (1a), and ft represents a focal length of the zoom lens system at a telephoto end.\n\nFurther features of the present invention will become apparent from the following description of exemplary embodiments with reference to the attached drawings.\n\nBRIEF DESCRIPTION OF THE DRAWINGS\n\nFIG. 1 is a cross sectional view of a lens system of Numerical Embodiment 1 at the wide angle end in infinity focus.\n\nFIG. 2A is an aberration diagram of Numerical Embodiment 1 at the wide angle end in focus at an objective distance of 2.5 m.\n\nFIG. 2B is an aberration diagram of Numerical Embodiment 1 at the intermediate zoom position in focus at an objective distance of 2.5 m.\n\nFIG. 2C is an aberration diagram of Numerical Embodiment 1 at the telephoto end in focus at an objective distance of 2.5 m.\n\nFIG. 3 is a cross sectional view of a lens system of Numerical Embodiment 2 at the wide angle end in infinity focus.\n\nFIG. 4A is an aberration diagram of Numerical Embodiment 2 at the wide angle end in focus at an objective distance of 2.5 m.\n\nFIG. 4B is an aberration diagram of Numerical Embodiment 2 at the intermediate zoom position in focus at an objective distance of 2.5 m.\n\nFIG. 4C is an aberration diagram of Numerical Embodiment 2 at the telephoto end in focus at an objective distance of 2.5 m.\n\nFIG. 5 is a cross sectional view of a lens system of Numerical Embodiment 3 at the wide angle end in infinity focus.\n\nFIG. 6A is an aberration diagram of Numerical Embodiment 3 at the wide angle end in focus at an objective distance of 2.5 m.\n\nFIG. 6B is an aberration diagram of Numerical Embodiment 3 at the intermediate zoom position in focus at an objective distance of 2.5 m.\n\nFIG. 6C is an aberration diagram of Numerical Embodiment 3 at the telephoto end in focus at an objective distance of 2.5 m.\n\nFIG. 7 is a cross sectional view of a lens system of Numerical Embodiment 4 at the wide angle end in infinity focus.\n\nFIG. 8A is an aberration diagram of Numerical Embodiment 4 at the wide angle end in focus at an objective distance of 2.8 m.\n\nFIG. 8B is an aberration diagram of Numerical Embodiment 4 at the intermediate zoom position in focus at an objective distance of 2.8 m.\n\nFIG. 8C is an aberration diagram of Numerical Embodiment 4 at the telephoto end in focus at an objective distance of 2.8 m.\n\nFIG. 9 is a schematic diagram illustrating an achromatic positive lens unit and remaining secondary spectrum.\n\nFIG. 10 is a schematic graph of a distribution of an Abbe number ν and a partial dispersion ratio θ of optical materials.\n\nFIG. 11 is a schematic diagram of a main part of an image pickup apparatus according to the present invention.\n\nDESCRIPTION OF THE EMBODIMENTS\n\nAn object of the present invention is to provide a zoom lens system that may correct the secondary spectrum of the longitudinal chromatic aberration appropriately on the telephoto side, and may easily realize high zoom ratio, a small size and light weight, and an image pickup apparatus including the zoom lens system.\n\nHereinafter, an embodiment of the present invention is described in detail with reference to the attached drawings. A zoom lens system according to the present invention includes a first lens unit having a positive refractive power which does not move for zooming, a second lens unit having a negative refractive power for magnification action, and a rear lens group including two or more lens units, arranged in this order from an object side to an image side. Here, “the lens unit does not move for zooming” means that the lens unit is not driven for a purpose of zooming, but the lens unit may move for focusing if zooming and focusing are performed simultaneously.\n\nFIG. 1 is a cross sectional view of a zoom lens system of Embodiment 1 (Numerical Embodiment 1) of the present invention at the wide angle end (short focal length end) in the focus state at the infinite objective distance. FIGS. 2A, 2B and 2C are aberration diagrams of Numerical Embodiment 1 in focus at an objective distance of 2.5 m at the wide angle end, at the intermediate zoom position (focal length f=33.3 mm), and at the telephoto end (long focal length end), respectively. Note that the focal length and the objective distance are values when the values of Numerical Embodiments are expressed in the unit of mm. The same is true for each of the following embodiments. FIG. 3 is a cross sectional view of a zoom lens system of Embodiment 2 (Numerical Embodiment 2) of the present invention at the wide angle end in the focus state at the infinite objective distance. FIGS. 4A, 4B and 4C are aberration diagrams of Numerical Embodiment 2 in focus at an objective distance of 2.5 m at the wide angle end, at the intermediate zoom position (focal length f=33.3 mm), and at the telephoto end, respectively.\n\nFIG. 5 is a cross sectional view of a zoom lens system of Embodiment 3 (Numerical Embodiment 3) of the present invention at the wide angle end in the focus state at the infinite objective distance. FIGS. 6A, 6B and 6C are aberration diagrams of Numerical Embodiment 3 in focus at an objective distance of 2.5 m at the wide angle end, at the intermediate zoom position (focal length f=35 mm), and at the telephoto end, respectively. FIG. 7 is a cross sectional view of a zoom lens system of Embodiment 4 (Numerical Embodiment 4) of the present invention at the wide angle end in the focus state at the infinite objective distance. FIGS. 8A, 8B and 8C are aberration diagrams of Numerical Embodiment 4 in focus at an objective distance of 2.8 m at the wide angle end, at the intermediate zoom position (focal length f=29.5 mm), and at the telephoto end, respectively.\n\nIn the cross sectional views in FIGS. 1, 3 and 5, a front lens unit (first lens unit) F having a positive refractive power does not move for zooming. A sub-unit (first sub lens unit) 1a is fixed during focusing (does not move for focusing) and is disposed closest to the object side in the first lens unit F. A sub-unit (second sub lens unit) 1b having a positive refractive power moves during focusing and is disposed on the image side in the first lens unit F. A variator (second lens unit) V having a negative refractive power for magnification is moved monotonously on the optical axis toward the image plane side, so that magnification (zooming) is performed from the wide angle end to the telephoto end. A compensator (third lens unit) C having a negative refractive power moves on the optical axis non-linearly in association with a movement of the second lens unit so as to correct an image plane variation accompanying the magnification. The variator V and the compensator C constitute a magnification-varying system. A stop (aperture stop) SP is disposed on the image side of the third lens unit C. A relay lens unit (fourth lens unit) R having a positive refractive power is fixed so as to have an image formation action. A glass block P in the diagrams includes a color separation prism, an optical filter, and the like. An image pickup surface I corresponds to an image pickup surface of a solid state image pickup element (photoelectric transducer). Note that the compensator C and the relay lens unit R constitute a rear lens group.\n\nIn FIG. 7, a front lens unit (first lens unit) F having a positive refractive power does not move for zooming. A sub-unit (first sub lens unit) 1a does not move for focusing and is disposed closest to the object side in the first lens unit F. A sub-unit (second sub lens unit) 1b having a positive refractive power moves during focusing and is disposed on the image side in the first lens unit F. A first variator (second lens unit) V1 having a negative refractive power for magnification is moved on the optical axis toward the image plane side, so that magnification (zooming) is performed from the wide angle end to the telephoto end. A second variator (third lens unit) V2 having a negative refractive power works for magnification in the same manner as the first variator V1. A fixed lens unit (fourth lens unit) R having a positive refractive power is fixed. A compensator (fifth lens unit) C having a positive refractive power moves on the optical axis non-linearly so as to correct an image plane variation accompanying the magnification. The first variator V1, the second variator V2, and the compensator C constitute the magnification system. A stop (aperture stop) SP is disposed on the image side of the second variator V2. A glass block P in the diagram includes a color separation prism, an optical filter, and the like. An image pickup surface I corresponds to an image pickup surface of a solid state image pickup element. Note that the second variator V2, the fixed lens unit R, and the compensator C constitute a rear lens group.\n\nIn each embodiment, the first lens unit F includes the first sub lens unit 1a which does not move for focusing and the second sub lens unit 1b which moves for focusing. The first sub lens unit 1a includes two or more negative lenses and one or more positive lenses. An average Abbe number of materials of the negative lenses included in the first sub lens unit 1a and an average partial dispersion ratio are denoted by νna and θna, respectively. An average Abbe number of materials of the positive lenses included in the first lens unit F and an average partial dispersion ratio are denoted by νpa and θpa, respectively. A combined focal length of the negative lenses included in the first sub lens unit 1a is denoted by fins, and a focal length of the zoom lens system at the telephoto end is denoted by ft. Then, the following conditions are satisfied.\n\n−1.193×10−3<(θpa−θna)/(νpa−νna)<−0.904×10−3 (1)\n\n40<νpa−νna<55 (2)\n\n−0.465<f1ns/ft (3)\n\nHere, the combined focal length fins is expressed by the following equation:\n\n$1 f 1 ns = 1 f 1 + 1 f 2 + ... 1 fn$\n\nwhere n represents a number of negative lenses included and fi represents a focal length of the i-th negative lens.\n\nIn each embodiment, the conditions including the lens configuration of the first lens unit F and the dispersion characteristic of the lens material are defined by the conditional expressions (1) to (3), so that the secondary spectrum of the longitudinal chromatic aberration at the telephoto end is corrected appropriately. The conditional expression (1) shows the condition for reducing a remaining amount of the secondary spectrum of the longitudinal chromatic aberration in the first lens unit F so that the secondary spectrum of the longitudinal chromatic aberration at the telephoto end is appropriately corrected. The general outline of this condition is described with reference to FIGS. 9 and 10.\n\nFIG. 9 is a schematic diagram illustrating an achromatic lens unit LP having a positive refractive power and remaining secondary spectrum. FIG. 10 is a schematic diagram of a distribution of an Abbe number ν and a partial dispersion ratio θ of existing optical materials. Here, the Abbe number ν and the partial dispersion ratio θ are respectively expressed by the following equations:\n\nν=(Nd−1)/(NF−NC) (a)\n\nθ=(Ng−NF)/(NF−NC) (b)\n\nwhere Ng represents a refractive index with respect to the g-line, NF represents a refractive index with respect to the F-line, Nd represents a refractive index with respect to the d-line, and NC represents a refractive index with respect to the C-line.\n\nAs illustrated in FIG. 10, the existing optical material is distributed in the region having narrow partial dispersion ratio θ with respect to the Abbe number ν, and there is a tendency that the partial dispersion ratio θ increases as the Abbe number ν decreases. A correction condition of chromatic aberration of a thin lens system constituted of two lenses G1 and G2 having refracting powers φ1 and φ2 and Abbe numbers of materials ν1 and ν2, respectively is expressed by the following equation.\n\nφ1/ν1+φ2/ν2=0 (c)\n\nHere, combined refracting power φ is expressed as follows.\n\nφ=φ1+φ2 (d)\n\nIf the equation (c) is satisfied, an imaging position is identical between the C-line and the F-line as illustrated in FIG. 9.\n\nIn this case, the refracting powers φ1 and φ2 are expressed by the following equations.\n\nφ1=φ×ν1/(ν1−ν2) (e)\n\nφ2=−φ×ν2/(ν1−ν2) (f)\n\nIn FIG. 9, as to the achromatic lens unit LP having a positive refractive power, a material having a large Abbe number ν1 is used for the positive lens G1, and a material having a small Abbe number ν2 is used for the negative lens G2. Therefore, the positive lens G1 has a small partial dispersion ratio θ1, and the negative lens G2 has a large partial dispersion ratio θ2 as illustrated in FIG. 10. As illustrated in FIG. 9, if the chromatic aberration is corrected at the F-line and the C-line, the imaging point at the g-line is shifted to the image side. This deviation amount defined as a secondary spectrum amount Δ is expressed by the following equation.\n\nΔ=−(1/φ)×(θ1−θ2)/(ν1−ν2) (g)\n\nHere, the secondary spectrum amounts of the first sub lens unit 1a, the second sub lens unit 1b, and the lens unit in the image side of the magnification systems V and C are denoted by Δ1a, Δ1b, and ΔZ, respectively. Imaging magnification factors of the second sub lens unit 1b and the lens unit in the image side of the magnification systems V and C are denoted by β1b and βZ, respectively. Then, the secondary spectrum amount Δ of the entire lens system is expressed by the following equation.\n\nΔ=Δ1ax×α1b2×βZ2+Δ1b×(1−β1b)×βZ2+ΔZ×(1−βZ) (h)\n\nThe secondary spectrum amount Δ becomes significant in the first sub lens unit 1a and the second sub lens unit 1b in which an axial marginal light beam passes through at a high position on the telephoto side. Therefore, the longitudinal chromatic aberration secondary spectrum amount Δ may be reduced on the telephoto side by suppressing the sum of the secondary spectrum amounts Δ1a and Δ1b of the longitudinal chromatic aberration generated in the first sub lens unit 1a and the second sub lens unit 1b.\n\nWhen the upper limit of the conditional expression (1) is exceeded, dispersions of the positive lens and the negative lens naturally become close to each other in existing general optical materials. Then, each lens power in the first lens unit F increases so that a curvature of the lens surface increases. As a result, it becomes difficult to correct various aberrations, in particular, a higher order aberration. In addition, because the lens becomes thick, it becomes difficult to reduce size and weight of the first lens unit F. If the lower limit of the conditional expression (1) is exceeded, the sum of the secondary spectrum amounts (Δ1a1b) of the first sub lens unit 1a and the second sub lens unit 1b increases. Therefore, it becomes difficult to correct the longitudinal chromatic aberration appropriately on the telephoto end.\n\nThe conditional expression (2) defines a relative relationship between material dispersions of the positive lens and the negative lens for the first lens unit F to have an appropriate achromatic effect. If the upper limit of the conditional expression (2) is exceeded, it becomes difficult for the first lens unit F including the positive lens and the negative lens for the achromatic effect to have an appropriate refractive power. Then, it becomes difficult to push the rear principal point position of the first lens unit F toward the image side. As a result, it becomes difficult to realize wide angle and to reduce size and weight of the first lens unit F. If the lower limit of the conditional expression (2) is exceeded, each lens power in the first lens unit F increases so that a curvature of the lens surface increases. As a result, it becomes difficult to correct, in particular, higher order aberration. In addition, the lens becomes thick, so that it becomes difficult to reduce size and weight of the first lens unit F.\n\nThe conditional expression (3) defines the condition to satisfy both appropriate correction of the chromatic aberration and reduction of size and weight of the first lens unit F when a zoom lens system having a large magnification ratio (zoom ratio) is obtained. If the lower limit of the conditional expression (3) is exceeded, the negative refractive power included in the first sub lens unit 1a is decreased, so that it becomes difficult to obtain the appropriate achromatic effect at the telephoto end of the zoom lens. In addition, it becomes difficult to push the rear principal point position of the first lens unit F toward the image side, and it becomes difficult to realize a wide angle and to reduce size and weight of the first lens unit F. Further, it is preferred that numerical value ranges of the conditional expressions (1) to (3) be set as follows.\n\n−1.191×10−3<(θpa−θna)/(νpa−νna)<−0.910×10−3 (1a)\n\n41.0<νpa−νna<52.0 (2a)\n\n−0.465<f1ns/ft<−0.400 (3a)\n\nThe technical meanings of the conditional expressions (1a) and (2a) are the same as the technical meanings of the conditional expressions (1) and (2) described above. If the upper limit of the conditional expression (3a) is exceeded, the negative refractive power included in the first sub lens unit 1a increases, so that it becomes difficult to correct various aberrations, in particular, the higher order aberration. In addition, a curvature of the lens surface of the negative lens increases, so that it becomes difficult to reduce size and weight of the first lens unit F.\n\nWith the structure described above, it is possible to provide a zoom lens system having a high zoom ratio in which the secondary spectrum of the longitudinal chromatic aberration is appropriately corrected at the telephoto end, and in which reduction of size and weight is achieved. In each embodiment, it is more preferred that one or more of the following conditions be satisfied,\n\n15.0<νna<30.3 (4)\n\n0.350<f1/ft<0.425 (5)\n\n15<ft/fw (6)\n\nwhere f1 represents a focal length of the first lens unit F, fw represents a focal length of the zoom lens system at the wide angle end.\n\nThe conditional expression (4) defines the average Abbe number νna of materials of negative lenses included in the first sub lens unit 1a that are appropriate for correcting the secondary spectrum of the longitudinal chromatic aberration of the zoom lens system at the telephoto side more appropriately.\n\nIf the upper limit of the conditional expression (4) is exceeded, the refracting powers of the two negative lenses in the first sub lens unit 1a increase, so that it becomes difficult to correct various aberrations, in particular, higher order aberration. In addition, a curvature of the lens surface of the lens increases, so that the lens becomes thick. As a result, it becomes difficult to reduce size and weight of the first sub lens unit 1a. If the lower limit of the conditional expression (4) is exceeded, the two negative lenses in the first lens unit F may not have appropriate refracting powers, so that it becomes difficult to push a principal point interval toward the image side. As a result, it becomes difficult to realize a wide angle and to reduce size and weight of the first lens unit F.\n\nThe conditional expression (5) defines an optimal range of the focal length of the first lens unit F with respect to the focal length at the telephoto end for realizing high magnification-varying factor (high zoom ratio) while the longitudinal chromatic aberration is corrected appropriately. If the lower limit of the conditional expression (5) is exceeded, the refractive power of the first lens unit F increases, so that it becomes difficult to correct mainly spherical aberration at the telephoto end and higher order components of the off-axial aberration appropriately. If the upper limit of the conditional expression (5) is exceeded, the refracting power of the first lens unit F is decreased, so that it becomes difficult to realize both the high zooming factor (high zoom ratio) and downsizing of the first lens unit F.\n\nThe conditional expression (6) is for correcting the chromatic aberration appropriately at the telephoto end while obtaining an optimal zoom ratio. If the zoom ratio exceeds the lower limit of the conditional expression (6), it is easy to realize reduction of size and weight without deteriorating the longitudinal chromatic aberration at the telephoto end and various aberrations even by the conventional lens structure. It is more preferred that the numerical value ranges in the conditional expressions (4) to (6) be set as follows.\n\n20.0<νna<30.0 (4a)\n\n0.360<f1/ft<0.423 (5a)\n\n16<ft/fw<25 (6a)\n\nWhen the zoom lens system of the present invention is used for an image pickup apparatus including a photoelectric transducer such as a CCD, it is preferred that the condition expressed by the following expression be satisfied,\n\n0.6<fw/IS (7)\n\nwhere IS represents a diagonal length of an effective surface of the photoelectric transducer.\n\nThe conditional expression (7) defines an optimal relative relationship between the focal length at the wide angle end of a zoom lens system and the diagonal length of the image pickup element when the zoom lens system of the present invention is used for the image pickup apparatus. If the lower limit of the conditional expression (7) is exceeded, the zoom lens system becomes to have an excessively wide field angle, so that it becomes difficult to correct a distortion or off-axial aberration such as lateral chromatic aberration. It is more preferred that the numerical value of the conditional expression (7) be set as follows.\n\n0.62<fw/IS<0.80 (7a)\n\nNext, the configuration of the first lens unit F in each embodiment is described. First, the configuration of the first lens unit F in Embodiment 1 illustrated in FIG. 1 is described. The first lens unit F corresponds to the first to the twelfth lens surfaces, and is constituted by the first sub lens unit 1a including the first to the sixth lens surfaces and the second sub lens unit 1b including the seventh to the twelfth surfaces. The first sub lens unit 1a is constituted by a negative lens, a negative lens, and a positive lens arranged in this order from the object side. The second sub lens unit 1b is constituted by three positive lenses. Table 1 shows values corresponding to the individual conditional expressions in Embodiment 1. Numerical Embodiment 1 satisfies every conditional expression, so that the longitudinal chromatic aberration is appropriately corrected at the telephoto end while the high zoom ratio (high magnification factor) and reduction in size and weight of the entire system are achieved.\n\nNext, the configuration of the first lens unit F in Embodiment 2 illustrated in FIG. 3 is described. The first lens unit F corresponds to the first to the fourteenth lens surfaces, and is constituted by the first sub lens unit 1a including the first to the eighth lens surfaces and the second sub lens unit 1b including the ninth to the fourteenth surfaces. The first sub lens unit 1a is constituted by a negative lens, a positive lens, a negative lens, and a positive lens arranged in this order from the object side. The second sub lens unit 1b is constituted by three positive lenses. Table 1 shows values corresponding to the individual conditional expressions in Embodiment 2. Numerical Embodiment 2 satisfies every conditional expression, so that the longitudinal chromatic aberration is appropriately corrected at the telephoto end while the high zoom ratio and reduction in size and weight of the entire system are achieved.\n\nNext, the configuration of the first lens unit F in Embodiment 3 illustrated in FIG. 5 is described. The first lens unit F corresponds to the first to the fourteenth lens surfaces, and is constituted by the first sub lens unit 1a including the first to the eighth lens surfaces and the second sub lens unit 1b including the ninth to the fourteenth surfaces. The first sub lens unit 1a is constituted by a negative lens, a positive lens, a negative lens, and a positive lens arranged in this order from the object side. The second sub lens unit 1b is constituted by three positive lenses. Table 1 shows values corresponding to the individual conditional expressions in Embodiment 3. Numerical Embodiment 3 satisfies every conditional expression, so that the longitudinal chromatic aberration is appropriately corrected at the telephoto end while the high zoom ratio and reduction in size and weight of the entire system are achieved.\n\nNext, the first lens unit F in Embodiment 4 illustrated in FIG. 7 is described. The first lens unit F corresponding to the first to the eleventh lens surfaces and is constituted by the first sub lens unit 1a including the first to the seventh lens surfaces and the second sub lens unit 1b including the eighth to the eleventh surfaces. The first sub lens unit 1a is constituted by a negative lens, a cemented lens constituted by a negative lens and a positive lens, and a positive lens arranged in this order from the object side. The second sub lens unit 1b is constituted by two positive lenses. Table 1 shows values corresponding to the individual conditional expressions in Embodiment 4. Numerical Embodiment 4 satisfies every conditional expression, so that the longitudinal chromatic aberration is appropriately corrected at the telephoto end while the high zoom ratio and the reduction in size and weight are achieved. As described above, the preferred embodiments of the present invention are described. However, it goes without saying that the present invention is not limited to these embodiments, which may be modified and changed variously within the spirit thereof.\n\nFIG. 11 is a principal schematic diagram illustrating an image pickup apparatus (television (TV) camera system) using the zoom lens system according to each of the Embodiments 1 to 3 as an image pickup optical system. In FIG. 11, a zoom lens system 101 according to any one of Embodiments 1 to 3 and a camera 124 are provided. The zoom lens system 101 is detachably attached to the camera 124. An image pickup apparatus 125 has a structure in which the zoom lens system 101 is attached to the camera 124. The zoom lens system 101 includes a first lens unit F, a magnification-varying section LZ, and a fourth lens unit R for imaging. The first lens unit F includes a focusing lens unit. The magnification section LZ includes a second lens unit which moves on the optical axis so as to vary the magnification and a third lens unit which moves on the optical axis so as to correct an image plane variation accompanying the magnification.\n\nThe zoom lens system 101 includes the aperture stop SP. The fourth lens unit R includes a lens unit (magnification optical system) IE which may be inserted onto or removed from the optical path. The lens unit IE is provided to shift the focal length range of the entire system of the zoom lens system 101. Drive mechanisms 114 and 115 such as helicoids or cams drive the first lens unit F and the magnification section LZ, respectively, in the optical axis direction. Motors (drive units) 116 to 118 are provided to electrically drive the drive mechanisms 114 and 115 and the aperture stop SP. Detectors 119 to 121 such as encoders, potentiometers, or photosensors detect positions of the first lens unit F and the magnification section LZ on the optical axis and a stop diameter of the aperture stop SP.\n\nThe camera 124 includes a glass block 109 corresponding to an optical filter or a color separation prism, and a solid-state image pickup element (photoelectric transducer) 110 such as a CCD sensor or a CMOS sensor, for receiving a subject image formed by the zoom lens system 101. CPUs 111 and 122 perform various drive controls of the camera 124 and the main body of the zoom lens system 101, respectively. When the zoom lens system according to the present invention is applied to the TV camera system as described above, the image pickup apparatus having high optical performance is realized. Note that the zoom lens system according to the Embodiment 4 of the present invention may also be applied to a TV camera as in the case of Embodiments 1 to 3.\n\nHereinafter, Numerical Embodiments 1 to 4 corresponding to Embodiments 1 to 4 of the present invention are described. In the respective numerical embodiments, a surface number “i” is counted from the object side. In addition, ri indicates a curvature radius of an i-th surface counted from the object side, and di indicates an interval between the i-th surface and an (i+1)-th surface which are counted from the image side. Further, ndi and νdi indicate a refractive power and an Abbe number of an i-th optical material, respectively. Last three surfaces correspond to a glass block such as a filter. Assume that the optical axis direction is an X-axis, a direction perpendicular to the optical axis is an H axis, and a light traveling direction is positive. In this case, when R denotes a paraxial curvature radius, k denotes a conic constant, and A3, A4, A5, A6, A7, A8, A9, A10, A11, and A12 denote aspherical coefficients, an aspherical surface shape is expressed by the following expression. For example, “e−Z” indicates “×10−Z”. The mark “*” indicates the aspherical surface. Table 1 shows a correspondence relationship between the respective embodiments and the conditional expressions described above.\n\n$X = H 2 / R 1 + 1 - ( 1 + k ) ( H / R ) 2 + A 4 H 4 + A 6 H 6 + A 8 H 8 + A 10 H 10 + A 12 H 12 + A 3 H 3 + A 5 H 5 + A 7 H 7 + A 9 H 9 + A 11 H 11 Equation 2$\n\nNumerical Embodiment 1 Surface data Surface Effective number r d nd νd diameter 1 −332.526 1.80 1.90200 25.1 76.53 2 524.636 5.84 75.14 3 −272.645 1.80 1.72047 34.7 74.42 4 119.340 0.12 73.71 5 119.387 13.93 1.43875 95.0 73.78 6 −118.750 8.04 74.03 7 1839.222 5.81 1.43875 95.0 73.20 8 −156.962 0.15 73.09 9 129.687 8.48 1.59240 68.3 71.91 10 −369.251 0.15 71.70 11 62.991 8.08 1.77250 49.6 67.86 12 163.572 (Variable) 66.72 13* 228.519 0.70 1.88300 40.8 26.83 14 16.139 5.93 22.17 15 −162.360 6.59 1.80518 25.4 21.70 16 −15.005 0.70 1.75500 52.3 21.31 17 27.315 0.68 19.68 18 22.172 5.61 1.60342 38.0 19.88 19 −39.636 0.88 19.33 20 −24.853 0.70 1.83481 42.7 19.21 21 −134.693 (Variable) 19.68 22 −28.312 0.70 1.74320 49.3 22.52 23 46.740 2.80 1.84666 23.8 24.99 24 −2634.956 (Variable) 25.48 25 (Stop) 1.30 29.38 26 363.155 4.88 1.65844 50.9 30.81 27 −34.292 0.15 31.18 28 78.344 3.20 1.51823 58.9 31.81 29 −1293.355 0.15 31.71 30 79.771 7.00 1.51633 64.1 31.49 31 −31.873 1.80 1.83400 37.2 31.21 32 −345.747 35.20 31.35 33 52.089 5.88 1.48749 70.2 30.29 34 −49.909 1.67 30.00 35 −88.953 1.80 1.83481 42.7 28.00 36 28.616 6.25 1.51742 52.4 27.30 37 −108.754 3.17 27.53 38 89.432 6.93 1.48749 70.2 27.77 39 −29.788 1.80 1.83400 37.2 27.59 40 −123.485 0.18 28.20 41 53.971 4.90 1.51633 64.1 28.39 42 −82.923 4.50 28.13 43 30.00 1.60342 38.0 40.00 44 16.20 1.51633 64.2 40.00 45 (Variable) 40.00 Image plane Aspherical Surface data Thirteenth surface K = 8.58860e+000 A4 = 7.05382e−006 A6 = −1.80303e−008 A8 = 7.49637e−011 A10 = −8.01854e−013 A12 = 5.80206e−015 A3 = −4.50041e−007 A5 = 1.66019e−008 A7 = −8.87373e−010 A9 = 1.99340e−011 A11 = −1.17115e−013 Various data Zoom ratio 20.00 Focal length 8.20 16.40 33.29 109.33 164.00 F number 1.80 1.80 1.80 1.87 2.73 Field angle 33.85 18.54 9.38 2.88 1.92 Image height 5.50 5.50 5.50 5.50 5.50 Total lens length 267.97 267.97 267.97 267.97 267.97 BF 41.33 41.33 41.33 41.33 41.33 d12 1.17 22.84 37.62 51.83 54.42 d21 55.34 30.57 13.32 2.39 4.78 d24 4.40 7.50 9.97 6.69 1.71 d45 7.50 7.50 7.50 7.50 7.50 Entrance pupil 48.84 93.36 167.10 410.71 541.88 position Exit pupil 348.87 348.87 348.87 348.87 348.87 position Front principal 57.24 110.54 203.64 555.06 784.67 point position Rear principal −0.70 −8.90 −25.80 −101.84 −156.50 point position Zoom lens unit data Front Rear Lens principal principal First Focal structure point point Unit surface length length position position 1 1 67.24 54.20 37.71 5.78 2 13 −13.70 21.79 2.66 −11.33 3 22 −42.20 3.50 −0.07 −1.98 4 25 64.97 136.94 74.63 −153.22\n\nNumerical Embodiment 2 Surface data Surface Effective number r d nd νd diameter 1 −321.282 1.80 1.80000 29.8 77.09 2 155.978 0.27 74.59 3 148.778 5.00 1.43387 95.1 74.52 4 939.166 1.64 74.28 5 7215.908 1.80 1.80000 29.8 73.96 6 142.263 1.56 72.80 7 181.560 11.04 1.43387 95.1 72.86 8 −127.642 10.88 72.96 9 −2147.703 4.17 1.43387 95.1 71.20 10 −253.410 0.15 71.03 11 146.659 8.29 1.59240 68.3 71.37 12 −222.853 0.15 71.31 13 59.292 7.70 1.77250 49.6 67.36 14 152.159 (Variable) 66.69 15* 228.519 0.70 1.88300 40.8 27.77 16 16.263 5.43 22.82 17 −10264.248 6.59 1.80518 25.4 22.54 18 −15.549 0.70 1.75500 52.3 22.13 19 25.590 0.68 20.01 20 20.925 5.61 1.60342 38.0 20.17 21 −53.448 1.38 19.48 22 −24.853 0.70 1.83481 42.7 19.20 23 −134.693 (Variable) 19.56 24 −28.312 0.70 1.74320 49.3 22.53 25 46.740 2.80 1.84666 23.8 25.00 26 −2634.956 (Variable) 25.48 27 (Stop) 1.30 29.39 28 394.928 5.18 1.65844 50.9 30.80 29 −33.348 0.15 31.26 30 74.155 3.20 1.51823 58.9 31.80 31 3654.252 0.15 31.67 32 83.980 7.00 1.51633 64.1 31.46 33 −31.528 1.50 1.83400 37.2 31.16 34 −336.280 35.20 31.30 35 38.818 6.68 1.48749 70.2 30.21 36 −49.746 1.67 29.82 37 −75.132 1.80 1.83481 42.7 27.57 38 27.229 6.15 1.51742 52.4 26.68 39 −399.526 2.67 26.88 40 180.001 6.83 1.48749 70.2 27.19 41 −24.173 1.80 1.83400 37.2 27.22 42 −51.301 0.18 28.26 43 58.679 4.80 1.51633 64.1 28.27 44 −104.601 4.50 27.91 45 30.00 1.60342 38.0 40.00 46 16.20 1.51633 64.2 40.00 47 (Variable) 40.00 Image plane Aspherical Surface data Fifteenth surface K = 8.58860e+000 A4 = 7.05382e−006 A6 = −1.80303e−008 A8 = 7.49637e−011 A10 = −8.01854e−013 A12 = 5.80206e−015 A3 = −4.50041e−007 A5 = 1.66019e−008 A7 = −8.87373e−010 A9 = 1.99340e−011 A11 = −1.17115e−013 Various data Zoom ratio 20.00 Focal length 8.20 16.40 33.29 109.33 164.00 F number 1.80 1.80 1.80 1.87 2.73 Field angle 33.85 18.54 9.38 2.88 1.92 Image height 5.50 5.50 5.50 5.50 5.50 Total lens length 268.50 268.50 268.50 268.50 268.50 BF 41.33 41.33 41.33 41.33 41.33 d14 0.53 22.20 36.98 51.19 53.78 d23 56.24 31.47 14.23 3.29 5.68 d26 4.40 7.50 9.97 6.69 1.71 d47 7.50 7.50 7.50 7.50 7.50 Entrance pupil 49.27 93.79 167.53 411.14 542.31 position Exit pupil 332.44 332.44 332.44 332.44 332.44 position Front principal 57.68 111.02 204.24 557.26 789.08 point position Rear principal −0.70 −8.90 −25.80 −101.84 −156.50 point position Zoom lens unit data Front Rear Lens principal principal First Focal structure point point Unit surface length length position position 1 1 67.24 54.47 38.14 5.59 2 15 −13.70 21.79 3.10 −10.42 3 24 −42.20 3.50 −0.07 −1.98 4 27 65.91 136.94 76.20 −155.53\n\nNumerical Embodiment 3 Surface data Surface Effective number r d nd νd diameter 1 −156.710 2.00 1.72047 34.7 88.52 2 126.311 3.19 82.79 3 148.734 5.82 1.43875 95.0 82.48 4 567.137 0.40 82.08 5 276.593 2.00 2.10446 17.1 81.43 6 116.332 0.00 79.53 7 116.332 16.37 1.49700 81.5 79.53 8 −110.799 8.99 79.71 9 155.999 5.15 1.59240 68.3 78.64 10 626.005 0.15 78.33 11 121.931 10.16 1.70000 48.1 76.82 12 −263.541 0.15 76.32 13 49.218 6.20 1.83481 42.7 63.39 14 85.248 (Variable) 62.65 15* 143.379 1.00 1.88300 40.8 28.33 16 14.990 5.56 21.61 17 −67.291 7.22 1.80809 22.8 21.49 18 −12.199 0.75 1.88300 40.8 20.38 19 28.461 0.60 18.47 20 23.739 5.13 1.63980 34.5 18.61 21 −36.481 2.69 18.09 22 −16.019 0.75 1.81600 46.6 16.93 23 −24.425 (Variable) 17.23 24 −20.647 0.75 1.75500 52.3 16.28 25 154.228 2.93 1.84649 23.9 20.90 26 −78.885 (Variable) 21.60 27 (Stop) 1.34 25.28 28 538.417 4.44 1.67003 47.2 25.74 29 −31.174 0.20 27.10 30 62.452 3.88 1.50127 56.5 27.90 31 −226.896 0.15 27.80 32 125.628 5.71 1.50127 56.5 27.50 33 −29.646 1.20 1.88300 40.8 27.10 34 −1686.338 46.64 27.10 35 57.525 7.01 1.51633 64.1 26.90 36 −38.749 0.47 23.54 37 −69.664 1.40 1.83400 37.2 22.69 38 20.495 7.23 1.48749 70.2 22.13 39 227.801 0.61 23.38 40 37.446 5.48 1.50127 56.5 24.56 41 −38.416 1.40 1.83481 42.7 24.66 42 −103.245 0.15 25.15 43 40.048 4.54 1.51633 64.1 25.55 44 −97.950 4.00 25.31 45 33.00 1.60859 46.4 23.57 46 13.20 1.51633 64.2 40.00 47 (Variable) 40.00 Image plane Aspherical Surface data Fifteenth surface K = −4.98064e+002 A4 = 3.14859e−005 A6 = −3.27331e−007 A8 = −1.45370e−009 A10 = −1.93067e−011 A12 = 4.84910e−014 A3 = 8.91291e−006 A5 = 1.87330e−006 A7 = 9.07873e−010 A9 = 4.49930e−010 A11 = −8.01597e−013 Various data Zoom ratio 20.00 Focal length 7.00 17.50 35.00 63.00 140.00 F number 1.90 1.90 1.90 1.90 2.43 Field angle 38.16 17.45 8.93 4.99 2.25 Image height 5.50 5.50 5.50 5.50 5.50 Total lens length 268.86 268.86 268.86 268.86 268.86 BF 38.96 38.96 38.96 38.96 38.96 d14 0.96 21.96 31.72 37.19 41.38 d23 40.45 15.87 5.20 2.00 7.23 d26 8.65 12.24 13.15 10.88 1.46 d47 5.79 5.79 5.79 5.79 5.79 Entrance pupil 48.53 104.16 180.24 287.22 525.01 position Exit pupil 84.00 84.00 84.00 84.00 84.00 position Front principal 56.16 125.57 230.90 400.96 915.61 point position Rear principal −1.21 −11.71 −29.21 −57.21 −134.21 point position Zoom lens unit data Front Rear Lens principal principal First Focal structure point point Unit surface length length position position 1 1 53.00 60.59 36.96 4.17 2 15 −12.16 23.72 2.53 −13.62 3 24 −40.86 3.68 −0.90 −2.97 4 27 149.70 142.05 236.08 −325.18\n\nNumerical Embodiment 4 Surface data Surface Effective number r d nd νd diameter 1 −155.833 2.20 1.80100 35.0 78.36 2 245.932 6.67 75.10 3 13317.833 2.20 1.84666 23.8 73.82 4 112.751 12.47 1.43875 95.0 72.35 5 −172.062 0.15 72.60 6 396.107 9.96 1.43387 95.1 73.22 7 −105.572 5.39 73.26 8 114.110 9.04 1.72916 54.7 69.30 9 −256.562 0.15 68.71 10 54.825 6.05 1.78800 47.4 62.45 11 103.504 (Variable) 61.61 12 51.789 1.00 1.83481 42.7 31.55 13 15.910 6.61 24.93 14 −400.917 6.84 1.80809 22.8 24.62 15 −17.072 0.75 1.83481 42.7 23.76 16 31.672 0.30 21.17 17 19.845 6.28 1.60342 38.0 21.08 18 −32.634 1.00 20.28 19 −25.408 0.75 1.83489 42.6 18.99 20 102.231 (Variable) 18.28 21 −22.336 0.75 1.75500 52.3 15.30 22 31.032 2.52 1.84649 23.9 16.72 23 470.386 (Variable) 17.28 24 (Stop) 1.34 24.01 25 −513.754 4.44 1.67003 47.2 24.91 26 −29.491 0.15 25.66 27 55.051 3.88 1.51633 64.1 26.31 28 186.478 0.15 26.12 29 61.856 5.71 1.50127 56.5 26.09 30 −32.834 1.20 1.88300 40.8 25.82 31 −336.685 (Variable) 25.98 32 55.061 6.20 1.48749 70.2 26.60 33 −42.524 0.15 26.23 34 −106.356 1.40 1.83400 37.2 25.37 35 23.054 4.74 1.51633 64.2 24.28 36 −199.026 0.86 24.30 37 44.732 6.26 1.51742 52.4 24.92 38 −22.898 1.40 1.88300 40.8 24.88 39 −243.333 0.15 25.85 40 66.393 4.71 1.51742 52.4 26.38 41 −36.700 (Variable) 26.42 42 33.00 1.60859 46.4 40.00 43 13.20 1.51633 64.2 40.00 44 (Variable) 40.00 Image plane Various data Zoom ratio 16.85 Focal length 8.00 13.61 29.55 59.74 134.79 F number 1.90 1.90 1.90 1.90 2.50 Field angle 34.51 22.00 10.54 5.26 2.34 Image height 5.50 5.50 5.50 5.50 5.50 Total lens length 250.44 250.44 250.44 250.44 250.44 BF 42.92 41.73 41.02 41.29 40.39 d11 0.70 14.94 29.31 37.91 43.74 d20 41.44 25.03 8.90 1.98 3.57 d23 8.55 10.72 12.48 10.79 3.38 d31 33.00 34.19 34.90 34.63 35.54 d41 5.26 4.07 3.35 3.63 2.72 d44 8.50 8.50 8.50 8.50 8.50 Entrance pupil 48.74 76.75 139.36 232.64 409.07 position Exit pupil 259.33 231.24 217.46 222.51 206.70 position Front principal 57.00 91.19 173.08 309.06 635.53 point position Rear principal 0.50 −5.11 −21.05 −51.25 −126.29 point position Zoom lens unit data Front Rear Lens principal principal First Focal structure point point Unit surface length length position position 1 1 56.83 54.28 36.40 7.74 2 12 −14.39 23.53 5.67 −8.29 3 21 −30.56 3.27 0.06 −1.72 4 24 34.74 16.87 2.69 −8.22 5 32 47.76 25.87 12.68 −6.01 6 42 46.20 14.58 −14.58\n\nTABLE 1 Values corresponding to individual conditional expressions in Numerical Embodiments 1 to 4 Conditional expression Conditional Numerical Numerical Numerical Numerical number expression Embodiment 1 Embodiment 2 Embodiment 3 Embodiment 4 (1) (θpa − θna)/ −1.190 × 10−3 −1.17 × 10−3 −1.160 × 10−3 −0.924 × 10−3 (vpa − vna) (2) vpa − vna 47.07 50.80 41.23 43.66 (3) f1ns/ft −0.461 −0.460 −0.448 −0.464 (4) vna 29.90 29.84 25.90 29.38 (5) f1/ft 0.410 0.410 0.379 0.422 (6) Z 20 20 20 17 (7) fw/IS 0.745 0.745 0.636 0.727\n\nAccording to this embodiment described above, it is possible to provide a zoom lens system in which high zoom ratio may easily be realized, a secondary spectrum of the longitudinal chromatic aberration may be appropriately corrected at the telephoto side, and reduction of size and weight may be realized easily.\n\nWhile the present invention has been described with reference to exemplary embodiments, it is to be understood that the invention is not limited to the disclosed exemplary embodiments. The scope of the following claims is to be accorded the broadest interpretation so as to encompass all such modifications and equivalent structures and functions.\n\nThis application claims the benefit of Japanese Patent Application No. 2009-188463, filed on Aug. 17, 2009, which is hereby incorporated by reference herein in its entirety.\n\n## Claims\n\n1. A zoom lens system, comprising in order of from an object side to an image side: where νna and θna represent an average Abbe number and an average partial dispersion ratio of materials of the two or more negative lenses included in the first sub lens unit, respectively, νpa and θpa represent an average Abbe number and an average partial dispersion ratio of materials of the one or more positive lenses included in the first lens unit, respectively, fins represents a combined focal length of the two or more negative lenses included in the first sub lens unit, and ft represents a focal length of the zoom lens system at a telephoto end.\n\na first lens unit having a positive refractive power, which does not move for zooming;\na second lens unit having a negative refractive power for a magnification action; and\na rear lens group including two or more lens units,\nwherein the first lens unit includes a first sub lens unit which does not move for focusing and a second sub lens unit which moves for focusing, and the first sub lens unit includes two or more negative lenses and one or more positive lenses; and\nwherein the following conditional expressions are relative, −1.193×10−3<(θpa−θna)/(νpa−νna)<−0.904×10−3; 40<νpa−νna<55; and −0.465<f1ns/ft,\n\n2. A zoom lens system according to claim 1, wherein the average Abbe number (νna) of the material of the negative lens included in the first sub lens unit satisfies the following condition:\n\n15.0<νna<30.3.\n\n3. A zoom lens system according to claim 1, wherein the following expression is satisfied, when f1 represents a focal length of the first lens unit.\n\n0.350<f1/ft<0.425,\n\n4. A zoom lens system according to claim 1, wherein the following expression is satisfied, where fw represents a focal length of the zoom lens system at a wide angle end.\n\n15<ft/fw,\n\n5. A zoom lens system according to claim 1, wherein the rear lens group comprises in order of from the object side to the image side:\n\na third lens unit having a negative refractive power, which moves on an optical axis in association with a movement of the second lens unit; and\na fourth lens unit having a positive refractive power, which does not move for zooming.\n\n6. A zoom lens system according to claim 1, wherein the rear lens group comprises in order of from the object side to the image side:\n\na third lens unit having a negative refractive power, which moves for zooming;\na fourth lens unit having a positive refractive power, which does not move for zooming; and\na fifth lens unit having a positive refractive power, which moves for zooming.\n\n7. An image pickup apparatus, comprising:\n\nthe zoom lens system according to any one of claims 1 to 6; and\na photoelectric transducer for receiving an optical image formed by the zoom lens system so as to perform photoelectric conversion of the optical image.\n\n8. An image pickup apparatus according to claim 7, wherein the following expression is satisfied, wherein fw represents a focal length of the zoom lens system at a wide angle end and IS represents a diagonal length of an effective surface of the photoelectric transducer.\n\n0.6<fw/IS,\nPatent History\nPublication number: 20110037878\nType: Application\nFiled: Aug 2, 2010\nPublication Date: Feb 17, 2011\nPatent Grant number: 8223440\nApplicant: CANON KABUSHIKI KAISHA (Tokyo)\nInventors: Tsuyoshi Wakazono (Utsunomiya-shi), Masao Hori (Utsunomiya-shi)\nApplication Number: 12/848,329\nClassifications\nCurrent U.S. Class: Optical Zoom (348/240.3); With Mechanical Compensation (359/683); + - - + Arrangement (359/688); 348/E05.055\nInternational Classification: G02B 15/14 (20060101); H04N 5/262 (20060101);"
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https://byjus.com/question-answer/all-cuboids-are-cube-false-true/ | [
"",
null,
"",
null,
"Question\n\n# All cuboids are cube.False True\n\nSolution\n\n## The correct option is A False The given statement is false. A cuboid becomes cube only when all sides of a cube are equal. However the converse is true, i.e. all cubes are cuboid. Cube can be considered as a special cuboid having equal length, breadth, and height.",
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"",
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"Suggest corrections",
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"data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7",
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"data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7",
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"data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7",
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https://chemistry.stackexchange.com/feeds/question/55253 | [
"Why does dissolving a salt in water get a high pH? - Chemistry Stack Exchange most recent 30 from chemistry.stackexchange.com 2019-10-17T20:44:14Z https://chemistry.stackexchange.com/feeds/question/55253 https://creativecommons.org/licenses/by-sa/4.0/rdf https://chemistry.stackexchange.com/q/55253 3 Why does dissolving a salt in water get a high pH? Hamze https://chemistry.stackexchange.com/users/32067 2016-07-19T23:04:49Z 2017-09-21T06:28:31Z <p>I dissolved some sodium carbonate in water and I measured the $\\mathrm{pH}$. It turned out to be 11. I don't really understand why dissolving it in water increased the $\\mathrm{pH}$. I mean, I know that $\\mathrm{pH}$ of water was 7, so it must have something to do with the $\\ce{H+}$ ions. We just covered acid and bases today so please bear with me.</p> <p>Here are my thoughts. My salt had some $\\ce{OH-}$ salt and this combined with the $\\ce{H+}$ to produce $\\ce{H2O}$ thereby decreasing the amount of $\\ce{H+}$ ion. Or maybe one of my salt ions combine with the dissolved $\\ce{H+}$, again thereby decreasing the $\\ce{H+}$ concentration and increasing the $\\mathrm{pH}$, so I'd have something like this could have something like:</p> <p>$$\\ce{Na+ + H+ -> NaH2+}$$</p> <p>or maybe</p> <p>$$\\ce{CO3^2- + H+ -> HCO3-}$$</p> <p>Which one do you think happened?</p> https://chemistry.stackexchange.com/questions/55253/-/55254#55254 0 Answer by sixtytrees for Why does dissolving a salt in water get a high pH? sixtytrees https://chemistry.stackexchange.com/users/31921 2016-07-19T23:22:50Z 2017-09-21T06:28:31Z <p><strong>TL;DR</strong>: Because there is an equilibrium between anion of $\\ce{CO3^2-}$ and $\\ce{HCO3^-}$ that depletes $\\ce{H+}$ and that causes more water to dissociate and form free $\\ce{OH-}$.</p> <hr> <p>You should start by reading about hydrolysis and buffers (chemistry). Briefly, you have these equilibria:</p> <p>\\begin{align} \\ce{M+ + OH- &<=> MOH}& &\\text{(limiting cases: $\\ce{M+}$ = $\\ce{Na+}$ or $\\ce{NH4+}$)} \\\\ \\ce{H+ + A- &<=> HA}& &\\text{(limiting cases: $\\ce{A-}$ = $\\ce{I-}$ or $\\ce{CH3COO-}$)} \\\\ \\ce{H+ + OH- &<=> H2O} & \\end{align}</p> <p>Here $[\\ce{Y}]$ means concentration of $\\ce{Y}$. Note that $\\frac{[\\ce{MOH}]}{[\\ce{M+}][\\ce{OH-}]} = K_\\mathrm{b}$, $\\frac{[\\ce{HA}]}{[\\ce{H+}][\\ce{A-}]} = K_\\mathrm{a}$ and $[\\ce{H+}][\\ce{OH-}] = 10^{-14}$.</p> <p>Depending on the $K_\\mathrm{a}$ and $K_\\mathrm{b}$ you have several situations. Acids and bases are called strong if they favor dissociation on ions and weak if they favor association of ions.</p> <p>If both your acid and base is strong, then they both dissociate and the released $\\ce{H+}$ and $\\ce{OH-}$ are recombining to form water. If acid is strong and a base is weak then acid dissociates fully and yield a lot of $\\ce{H+}$, but base doesn't dissociate fully and yield less $\\ce{OH-}$. As a result of recombination of $\\ce{H+}$ and $\\ce{OH-}$, you get excess of $\\ce{H+}$ and acidic condition. If the acid is weak and the base is strong you have more free $\\ce{OH-}$ and basic reaction much by the same logic. If both acid and base are weak $\\mathrm{pH}$ depends on relative strengths on acid and base. </p> <p>In your case acid is weak ($\\ce{CO2}$ forms slightly acidic solution in water) and base is strong ($\\ce{NaOH}$ is a strong base). Thus, you have a basic solution.</p> <p>This story is overly simplified because you have an acid with two stages of dissociation:</p> <p>\\begin{align} \\ce{H2CO3 &<=> H+ + HCO3-} \\\\ \\ce{HCO3- &<=> H+ + CO3^2-} \\end{align}</p> <p>This adds one equation to the system but similar logic applies.</p> https://chemistry.stackexchange.com/questions/55253/-/55255#55255 9 Answer by IanC for Why does dissolving a salt in water get a high pH? IanC https://chemistry.stackexchange.com/users/31978 2016-07-19T23:25:03Z 2016-07-20T20:22:55Z <p>It's a little more complicated than that.. When you dissolved $\\ce{Na2CO3}$ in water you had a solution of ions: $\\ce{Na+}$, $\\ce{CO3^2-}$, $\\ce{H3O+}$ and $\\ce{OH-}$ (the later are present in water even when there is no salt, since it dissociates too, however they are in equal quantity in pure water so the pH is 7).</p> <p>Now when you mix all those ions, the equilibrium isn't only the equilibrium of water dissociation ($\\ce{H2O + H2O <=> H3O+ + OH-}$). Now you have a whole series of equilibria occurring in the solution simultaneously:</p> <p>1) Water dissociation: $\\ce{H2O + H2O <=> H3O+ + OH-}$</p> <p>2) Salt dissociation: $\\ce{Na2CO3 <=> 2Na+ + CO3^2-}$</p> <p>3) Base equilibrium*: $\\ce{NaOH <=> Na+ + OH-}$</p> <p>4) Acid equilibrium*: $\\ce{CO3^2- + 2 H3O+ <=> H2CO3 + 2 H2O}$</p> <p>The salt dissociation doesn't affect the pH directly. We have two equilibria though that do unbalance the proportion of $\\ce{H3O+}$ and $\\ce{OH-}$ ions, which are the \"base equilibrium\" and \"acid equilibrium\". We don't think about them at first sight because we didn't add any $\\ce{NaOH}$ or $\\ce{H2CO3}$ to the solution, but everything needed to form them is right there.</p> <p>$\\ce{NaOH}$ is a strong base, however $\\ce{H2CO3}$ is a weak acid. So it's likely (and you witnessed it in experience) that more $\\ce{H3O+}$ will be consumed in comparison with the $\\ce{OH-}$, making the solution more alkaline.</p> https://chemistry.stackexchange.com/questions/55253/-/55306#55306 2 Answer by porphyrin for Why does dissolving a salt in water get a high pH? porphyrin https://chemistry.stackexchange.com/users/30424 2016-07-20T17:21:13Z 2017-09-21T05:53:36Z <p>You have some detailed answers, but the principles are quite simple to grasp. </p> <p>A 'strong' acid or base is one in which there is complete, i.e. 100% dissociation of the molecule when dissolved in water. Examples are $\\ce{HCl}$ which dissociates into $\\ce{H+}$ and $\\ce{Cl-}$, and NaOH into $\\ce{Na+}$ and $\\ce{OH-}$. If 1 mole of each is added to pure water the solution remains neutral ($\\mathrm{pH = 7}$) because there are equal numbers of $\\ce{H+}$ and $\\ce{OH-}$.</p> <p>A 'weak' acid or 'weak' base is one which does not dissociate completely, so, for example, some fraction of $\\ce{H2CO3}$ molecules remain in solution and the rest dissociate producing $\\ce{H+}$ and $\\ce{CO3^2-}$ ions. Because some protons remain bound in the $\\ce{H2CO3}$ when 1 mole of strong base is added to 1 mole of the acid, there is an excess of $\\ce{OH-}$ and the solution is alkaline. A similar argument applies to a weak base and strong acid.</p>"
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https://www.easycalculation.com/formulas/true-porosity.html | [
"# True Porosity Formula\n\nThe page shows the True porosity formula to calculate the bulk density and porosity relationship of any material. It can be computed based on the apparent solid specific gravity and true specific gravity. According to the given Porosity equation bulk density, just divide the true specific gravity from apparent solid specific gravity. Subtract the obtained value from 1. Now multiply the resultant value with 100 to calculate the value in percentage.\n\n# Porosity Equation Bulk Density\n\n#### Formula:\n\nTrue porosity = ( 1 - ( Ga / Gt ) ) x 100 )\n\nWhere,\n\nGa - Apparent Solid Specific Gravity\nGt - True Specific Gravity\n\n#### Related Calculator:\n\nYou can also use our online calculator developed as per the true porosity formula mentioned above to cross-check your manually computed true porosity percentage in few fraction of seconds."
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8984908,"math_prob":0.9687125,"size":640,"snap":"2020-10-2020-16","text_gpt3_token_len":115,"char_repetition_ratio":0.14779875,"word_repetition_ratio":0.0,"special_character_ratio":0.1734375,"punctuation_ratio":0.06481481,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.998052,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-06T21:17:34Z\",\"WARC-Record-ID\":\"<urn:uuid:c0bdd546-d58e-4007-a516-7ed8b9c30bf4>\",\"Content-Length\":\"21311\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b963cacd-d252-46e9-8e48-d21178aa5198>\",\"WARC-Concurrent-To\":\"<urn:uuid:7a27612c-03d3-49c2-8762-7f5e41595ddf>\",\"WARC-IP-Address\":\"50.116.14.108\",\"WARC-Target-URI\":\"https://www.easycalculation.com/formulas/true-porosity.html\",\"WARC-Payload-Digest\":\"sha1:BWCACQAWAGE27AQXCYQB3DHSTEG53QUF\",\"WARC-Block-Digest\":\"sha1:OIHORDPIWGV75C5GKDBE7QYZTD2M3AIV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585371660550.75_warc_CC-MAIN-20200406200320-20200406230820-00329.warc.gz\"}"} |
https://harmonicsofnature.com/tag/music-based-on-the-rotation-of-earth/ | [
"## Music and the fabric of time\n\nWell, I just came across this article where the author goes back to the ancient Hebrew divisions of time and overlays musical frequencies in keeping with these time divisions and – LP and behold – comes up with the exact same frequencies for the complete musical scale that I did, based on the phenomena of the “resonant still points” which I demonstrate on my homepage.\n\nhttps://ethnographicsblog.wordpress.com/2019/04/19/a-horological-and-mathematical-defense-of-philosophical-pitch/\n\nThe Hebrew measure of time was the “helek” which equates to 3.333333 (recurring) seconds. Gives you more time to think, I suppose.\n\nThis measure of time was devised by dividing each of the 360 degrees by which the Earth turns every day into 72 parts to give a total of 25,920 helakim (plural of helek) per day.\n\nFirst off, my frequency (and his) for C is 259.2. (By the way, each time you multiply a frequency by 10 you’re getting the Major Third of the original – so 25,920 also suggests a frequency for G# (C-D-E, E-F#-G#).\n\nAs the author points out, 25,920 also equates to the number of years in the Great Year – the time it takes for the world’s axial “wobble” to precess through 360 degrees, going through the twelve Ages – one for each of the Astrological signs. And it takes 72 years for the equinox to precess by one degree.\n\nTwo hours is 25,920/12 = 2160 helakim, and 2,160 years is the length of an Age. And this suggests a microcosm/macrocosm thing where we enjoy a tiny Age, or change of astrological sign, every two hours of the day.\n\nAnd every day is, in effect, a mini Great Year as our position on the planet passes through all 12 astrological signs.\n\nThe Earth rotates 1 degree on its axis every 4 minutes (72 helakim x 3.333 seconds = 240 seconds = 4 minutes.) (3.333 recurring is a favourite number of the Free Masons but perhaps their big secret is simply the Helek. By the way, this video beautifully presents much of this.)\n\nEvery day, we turn 360 degrees, 4 minutes per degree = 1,440 minutes which is a number the author equates to F# as 1,440 Hz, which is an octave of 360 Hz – which is also the frequency I’ve found for F#.\n\nDividing the hour into seconds suggests B at 60 Hz.\n\nHe makes the root frequency for his scale 108 Hz because 360 degrees divided by 3.3333 seconds per helek = 108. 360 represents the full daily and annual rotation of the planet on its axis and around the zodiac. Now, 108 Hz is a sub-octave of 216 and 432 Hz – so that’s the frequency for A. Same as mine. So he sort of encompasses a whole year into A as the root note.\n\nHe then derives the harmonic series from this A using the “5-Limit” harmonic approach (which is simply deriving 5ths (multiply the frequency by 3) and the major third (multiply by 5).\n\nThe author also considers beats per minute as a starting point – so that the rhythm of the music and the music itself are aligned with the fabric of time.\n\nHe works into this a division of time (in seconds or helakim) by whole numbers: 2, 3, 5, 7, 9.\n\nSo he divides the day into seconds (60 seconds times 60 minutes times 12 hours) = 86,400 seconds, (a number which relates to A=432 Hz). He also starts with a notion of C at one cycle per second where its octaves would be 2 Hz, 4, 8, 16, 32, 64, 128, 256 Hz, etc.\n\nAnd with one helek = 3.333-seconds, a minute is 18 helakim – which relates to D at 9, 18, 36, 72, 144, 288 Hz\n\nAs he says, “Using 5 Limit Tuning with the root set to A (at 216) rather than C, the frequencies of notes C4 (256), G4 (384), E4 (320), D4 (288), and B4 (240) are reducible to, respectively: 1, 3, 5, 9, and 15“. Meaning that 1, 3, 5, 9 are sub-octaves of the given frequencies, e.g. 9 x 2 x 2 x 2 x 2 x 2 = 288 Hz = D.\n\nIf, as I believe, the phenomenon I can demonstrate on my tone generator is indeed the phenomenon of sound interacting with the fabric of the universe, then what more powerful evidence than to find that these frequencies all tie back to a natural way of measuring time – at least on our Earth. Do check it out: https://ethnographicsblog.wordpress.com/2019/04/19/a-horological-and-mathematical-defense-of-philosophical-pitch/"
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https://converter.ninja/length/yards-to-meters/355-yd-to-m/ | [
"# 355 yards in meters\n\n## Conversion\n\n355 yards is equivalent to 324.612 meters.\n\n## Conversion formula How to convert 355 yards to meters?\n\nWe know (by definition) that: $1\\mathrm{yd}=0.9144\\mathrm{m}$\n\nWe can set up a proportion to solve for the number of meters.\n\n$1 yd 355 yd = 0.9144 m x m$\n\nNow, we cross multiply to solve for our unknown $x$:\n\n$x\\mathrm{m}=\\frac{355\\mathrm{yd}}{1\\mathrm{yd}}*0.9144\\mathrm{m}\\to x\\mathrm{m}=324.612\\mathrm{m}$\n\nConclusion: $355 yd = 324.612 m$",
null,
"## Conversion in the opposite direction\n\nThe inverse of the conversion factor is that 1 meter is equal to 0.00308060084038791 times 355 yards.\n\nIt can also be expressed as: 355 yards is equal to $\\frac{1}{\\mathrm{0.00308060084038791}}$ meters.\n\n## Approximation\n\nAn approximate numerical result would be: three hundred and fifty-five yards is about three hundred and twenty-four point six one meters, or alternatively, a meter is about zero times three hundred and fifty-five yards.\n\n## Footnotes\n\n The precision is 15 significant digits (fourteen digits to the right of the decimal point).\n\nResults may contain small errors due to the use of floating point arithmetic."
]
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"https://converter.ninja/images/355_yd_in_m.jpg",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8988686,"math_prob":0.9984768,"size":843,"snap":"2022-27-2022-33","text_gpt3_token_len":184,"char_repetition_ratio":0.12038141,"word_repetition_ratio":0.0,"special_character_ratio":0.23487544,"punctuation_ratio":0.1,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99854803,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-30T04:28:21Z\",\"WARC-Record-ID\":\"<urn:uuid:aca9e916-2a2c-44e6-8038-3e257014edfb>\",\"Content-Length\":\"17818\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:09243c5e-f14a-4142-9a51-41071f2c6a7c>\",\"WARC-Concurrent-To\":\"<urn:uuid:67fe468e-d495-4eb6-a3f4-0f46348a62f0>\",\"WARC-IP-Address\":\"172.67.167.118\",\"WARC-Target-URI\":\"https://converter.ninja/length/yards-to-meters/355-yd-to-m/\",\"WARC-Payload-Digest\":\"sha1:3KLEQ4WWZ2DVVECRZJQBGNH6DSJHYQBW\",\"WARC-Block-Digest\":\"sha1:PZYAFIXAWNF24BXZB5BQTIKSHMQJUOPU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103661137.41_warc_CC-MAIN-20220630031950-20220630061950-00313.warc.gz\"}"} |
https://biggboss3.net/qa/how-many-hours-minutes-and-seconds-are-in-3-days.html | [
"",
null,
"# How Many Hours Minutes And Seconds Are In 3 Days?\n\n## How many minutes are in 3 hour?\n\n180 minutesIn 3 h there are 180 min .\n\nWhich is the same to say that 3 hours is 180 minutes..\n\n## What is 6.75 hours in hours and minutes?\n\n6.75 hours with the decimal point is 6.75 hours in terms of hours. 6:75 with the colon is 6 hours and 75 minutes. . 75 = fractional hours.\n\n## How many minutes are in hour?\n\n60 minutesThere are 60 minutes in 1 hour. To convert from minutes to hours, divide the number of minutes by 60. For example, 120 minutes equals 2 hours because 120/60=2.\n\n## How many minutes are in a day?\n\n1440 minutesThere is total 1440 minutes in a day. You can determine it in easy calculation. One hour is consist of 60 minutes. One day is consist of 24 hours.\n\n## What is 2 hours and 45 minutes expressed in just minutes?\n\nWhat is 2 hours and 45 minutes expressed in just minutes? 2 × 60 = 120. 120 + 45 = 165. The answer is 165 minutes.\n\n## How many seconds is 3 minutes?\n\n180 secondsIn 3 min there are 180 s . Which is the same to say that 3 minutes is 180 seconds.\n\n## How many days hours minutes and seconds are in 3 years?\n\n3 Years – 36 Months – 1,095 Days, 26,280 Hours – 1,576,800 Minutes – 94,608,000 Seconds…\n\n## How many seconds is 2 hours?\n\nPlease share if you found this tool useful:Conversions Table1 Hours to Seconds = 360070 Hours to Seconds = 2520002 Hours to Seconds = 720080 Hours to Seconds = 2880003 Hours to Seconds = 1080090 Hours to Seconds = 3240004 Hours to Seconds = 14400100 Hours to Seconds = 36000011 more rows\n\n## Is 1.25 an hour and 15 minutes?\n\n25 x 60 = 15 minutes. Here you can convert another time in terms of hours to hours and minutes. What is 1.26 decimal hours in hours and minutes?\n\n## How many seconds are there in a day of 24 hours?\n\nThere are 60 seconds in a minute, 60 seconds in an hour, 3600 seconds in a hour, 24 hours in a day, 7 days in a week, 52 weeks in a year, 365 days in a year, 10 years in a decade, 100 years in a century, 10 decades in a century, 100000 years in an eon.\n\n## How much is 100 years in days?\n\nThis conversion of 100 years to days has been calculated by multiplying 100 years by 365 and the result is 36,500 days.\n\n## How many minutes is 2 by 3 an hour?\n\nTo find 2/3 of an hour first we will divide 1 hour that is 60 minutes into 3 parts. So we will get 20 minutes. So we 2parts out of 3 parts of an hour so we get 20X2=40minutes.\n\n## How many days is 27 years?\n\n9,855 daysThis conversion of 27 years to days has been calculated by multiplying 27 years by 365 and the result is 9,855 days.\n\n## How many hours makes 3 days?\n\n72 HoursDays to Hours Conversion Examples 3 Days = 72 Hours. 4 Days = 96 Hours. 5 Days = 120 Hours. 6 Days = 144 Hours.\n\n## What is .25 of an hour?\n\nConversion Chart – Minutes to Hundredths of an Hour Enter time in Oracle Self Service as hundredths of an hour. For example 15 minutes (¼ hour) equals . 25, 30 minutes (½ hour) equals . 5, etc.\n\n## How many seconds are in 2 days?\n\n›› More information from the unit converter You can view more details on each measurement unit: days or seconds The SI base unit for time is the second. 1 days is equal to 86400 second.\n\n## What is 3.9 hours in hours and minutes?\n\nThis conversion of 3.9 hours to minutes has been calculated by multiplying 3.9 hours by 60 and the result is 234 minutes.\n\n## What is 4.8 hours in hours and minutes?\n\nThis conversion of 4.8 hours to minutes has been calculated by multiplying 4.8 hours by 60 and the result is 288 minutes.\n\n## How many hours is a full day?\n\nThis is called a sidereal day. On Earth, a sidereal day is almost exactly 23 hours and 56 minutes.\n\n## How do you convert days to hours minutes and seconds?\n\nEasy d to hr conversion. A day is the approximate time it takes for the Earth to complete one rotation. It is defined as exactly 86,400 seconds. An hour is a unit of time equal to 60 minutes, or 3,600 seconds….Convert Days to Hours.dhr0.00010.00240.00020.00480.00030.00720.00040.0096238 more rows\n\n## What is .8 of an hour?\n\nBilling Increment Chart—Minutes to Tenths of an HourMinutesTime25-30.531-36.637-42.743-48.86 more rows\n\n## How do you convert 3 hours to 20 minutes in hours?\n\nHere we have to convert 3 hours 20 minutes into hours. Hence, 3 hours 20 minutes is equal to 3.333 hours.\n\n## How many days are in 100 years in leap year?\n\nSo every 4th year we add an extra day (the 29th of February), which makes 365.25 days a year. This is fairly close, but is wrong by about 1 day every 100 years. So every 100 years we don’t have a leap year, and that gets us 365.24 days per year (1 day less in 100 year = -0.01 days per year).\n\n## How do I calculate seconds?\n\nDivide the number of minutes by 60 again to convert seconds to hours.Example: How many hours is 7200 seconds?7200 / 60 = 120 minutes.120 / 60 = 2 hours.Answer: 7200 seconds equals 2 hours.\n\n## How many minutes is 5 hour?\n\nHours to Minutes Conversion TableHoursMinutes3 Hours180 Minutes4 Hours240 Minutes5 Hours300 Minutes6 Hours360 Minutes20 more rows\n\n## What is 4.5 hours in hours and minutes?\n\nThis conversion of 4.5 hours to minutes has been calculated by multiplying 4.5 hours by 60 and the result is 270 minutes.\n\n## How much is in a century?\n\nA century is a period of 100 years. Centuries are numbered ordinally in English and many other languages. The word century comes from the Latin centum, meaning one hundred."
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"https://mc.yandex.ru/watch/68550892",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.92353296,"math_prob":0.92940617,"size":6223,"snap":"2021-21-2021-25","text_gpt3_token_len":1748,"char_repetition_ratio":0.24087474,"word_repetition_ratio":0.23289903,"special_character_ratio":0.33070865,"punctuation_ratio":0.13515407,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99093616,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-13T12:16:13Z\",\"WARC-Record-ID\":\"<urn:uuid:9ac194c2-110b-4427-9d75-e07c9c2514c0>\",\"Content-Length\":\"41139\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7e8b264d-1356-4706-80d1-e1b84faac372>\",\"WARC-Concurrent-To\":\"<urn:uuid:2ce0a60e-7ae0-40ed-a050-9b5c58f84ed8>\",\"WARC-IP-Address\":\"87.236.16.33\",\"WARC-Target-URI\":\"https://biggboss3.net/qa/how-many-hours-minutes-and-seconds-are-in-3-days.html\",\"WARC-Payload-Digest\":\"sha1:NBLEMIPUFKJSFBDG2FPHMDTUNKMQIADA\",\"WARC-Block-Digest\":\"sha1:74MKY4RG4CGSMZA2SOP3CLQLTYQNAOVY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487608702.10_warc_CC-MAIN-20210613100830-20210613130830-00524.warc.gz\"}"} |
https://tailieuthamkhao.com/linking-grammar-model-21-155 | [
"# Vietnamese linking grammar model - 21\n\n3.3.1.2. Training Algorithm\n\nAs discussed in Chapter 1 with a context-free grammar, probabilities acting as parameters can initially be generated randomly, and then updated every time a new sentence is analyzed and added to the set. corpus. The training algorithm proposed by aims to recalculate the parameter value after processing the input sentence. Like the context-free grammar, this algorithm relies on two parameters, the inner probability and the outer probability.\n\nThe probability in PrI ( L, R, l, r ) is the probability that words from L to R can be linked together such that the connections l and r are satisfied.\n\nThe outer probability Pro ( L, R, l, r ) is the probability that words outside the range L to R can be associated with each other such that the outer join requirements l and r are satisfied.\n\nThe inner probability is calculated recursively according to the relations:\n\nAccording to the parsing algorithm in Figure 3.4, it is clear that PI ( wi , wi+1, NIL, NIL ) = 1 with 0 ≤ i ≤ n-1.\n\nFor example, With the linking grammar and the sentence “I bought a flower” mentioned above,\n\nMaybe you are interested!\n\nPrI ( 1, 4, NIL, NcNt3 ) = Pr (3, (McN)(NcNt3),→ | 1, 4, NIL, NcNt3 ) ×\n\nPrI ( 1, 3, NIL, McN ) × PrI ( 3, 4, NIL, NIL )\n\nwith the values of the probabilities given in (3.1) :\n\nPrI( 1, 3, NIL, McN) = Pr(2, ( )(McN), → | 1, 3, NIL, McN) ×\n\nPrI (1, 2, NIL,NIL) × PrI (2, 3, NIL, NIL)\n\n= 0.06 × 1 × 1 = 0.06\n\nPr ( 3, (McN)(NcNt3),→ | 1, 4, NIL, NcNt3 ) = 0.05\n\nso, PrI (buy, flower, NIL, NcNt3) = 0.05 × 0.06 = 0.003 (3.5)\n\nThe probabilities outside PrO are calculated recursively: initially, for each d ∈ D(W0) there is left[d] = NIL, set\n\nThe probability is added to the 4 possible cases in the previous step (then R and L also play the role of W):\n\nFigure 3.22. Describe how to calculate probability Pr0 ⊲left(L, W, l ⊳, ⊲ left[D])\n\nAccording to , Counts are calculated in the formulas from (3.6) to (3.9) below:\n\nThe count(L, R, l, r) value is calculated in the analysis algorithm:\n\nwhere δ is a function that takes 1 if l = NIL, 0 otherwise, match takes 1 if the two matches match, 0 otherwise. Notice match(c,NIL) = match(NIL,c) = 0.\n\nThe Pr(S) value stated in the above formulas is calculated according to the following formula:\n\nThe values Count(L, R, l, r), Count(W, l, r) and Count(d, l, r) are calculated directly according to the connections and selections that appear in the corpus.\n\nDate published: 01/11/2021\nRating:\n4.0/5 (1 ratings)"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.86039543,"math_prob":0.9966155,"size":2694,"snap":"2023-14-2023-23","text_gpt3_token_len":782,"char_repetition_ratio":0.13197026,"word_repetition_ratio":0.07070707,"special_character_ratio":0.30178174,"punctuation_ratio":0.18644068,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99925566,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-05-29T22:18:59Z\",\"WARC-Record-ID\":\"<urn:uuid:29550d4e-5beb-4d84-8281-c6f787b1a440>\",\"Content-Length\":\"55188\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:213f4cfa-2145-44f8-bc4b-6e329d7a9590>\",\"WARC-Concurrent-To\":\"<urn:uuid:3aa3d916-aa7d-41bc-8db9-919f020020d7>\",\"WARC-IP-Address\":\"209.95.52.84\",\"WARC-Target-URI\":\"https://tailieuthamkhao.com/linking-grammar-model-21-155\",\"WARC-Payload-Digest\":\"sha1:E3ZDCP3VJPC5Z3Q4VXWLJJJKWXOCRAV7\",\"WARC-Block-Digest\":\"sha1:YZVP43IWJOCYXK5SMHS2VK5XSOJWDNSO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224644913.39_warc_CC-MAIN-20230529205037-20230529235037-00362.warc.gz\"}"} |
http://tasks.illustrativemathematics.org/content-standards/4/NF/A | [
"# 4.NF.A\n\nExtend understanding of fraction equivalence and ordering.\n\n## Standards\n\n4.NF.A.1 Explain why a fraction $a/b$ is equivalent to a fraction $(n \\times a)/(n \\times b)$ by using...\n4.NF.A.2 Compare two fractions with different numerators and different denominators, e.g., by creating...\n\n## Tasks aligned to this cluster\n\nRunning Laps\nMoney in the piggy bank"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.76582336,"math_prob":0.981474,"size":354,"snap":"2022-40-2023-06","text_gpt3_token_len":89,"char_repetition_ratio":0.13428572,"word_repetition_ratio":0.0,"special_character_ratio":0.2457627,"punctuation_ratio":0.23170732,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9906069,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-28T13:20:52Z\",\"WARC-Record-ID\":\"<urn:uuid:4a58dbda-c11d-4b96-b5e0-6d35c047dc81>\",\"Content-Length\":\"20160\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:20ba1791-76b9-4f02-acf4-4034df8a63ee>\",\"WARC-Concurrent-To\":\"<urn:uuid:63d6b38b-97ef-43c9-940f-767845dca2d9>\",\"WARC-IP-Address\":\"54.204.143.11\",\"WARC-Target-URI\":\"http://tasks.illustrativemathematics.org/content-standards/4/NF/A\",\"WARC-Payload-Digest\":\"sha1:UMXRBFWQGDFLQS3VLXAI7DEDMRKUHCRU\",\"WARC-Block-Digest\":\"sha1:KRTK2AGB3A3DZM3IXGGCXL64IVRZP3DI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030335254.72_warc_CC-MAIN-20220928113848-20220928143848-00582.warc.gz\"}"} |
https://www.math.upenn.edu/events/spanning-clusters-and-subcritical-connectivity-high-dimensional-percolation | [
"### Probability and Combinatorics\n\nTuesday, September 14, 2021 - 3:30pm\n\n#### Jack Hanson\n\nCity College of NY\n\nLocation\n\nTemple University\n\nWachman Hall 617\n\nIn their study of percolation, physicists have proposed scaling hypotheses'' relating the behavior of the model in the critical ($p = p_c$) and subcritical ($p < p_c$) regimes. We show a version of such a scaling hypothesis for the one-arm probability $\\pi(n;p)$ -- the probability that the open cluster of the origin has Euclidean diameter at least $n$.\n\nAs a consequence of our analysis, we obtain the correct scaling of the lower tail of cluster volumes and the chemical (intrinsic) distances within clusters. We also show that the number of spanning clusters of a side-length $n$ box is tight on scale $n^{d-6}$. A new tool of our analysis is a sharp asymptotic for connectivity probabilities when paths are restricted to lie in half-spaces."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.91309315,"math_prob":0.9731024,"size":746,"snap":"2022-05-2022-21","text_gpt3_token_len":166,"char_repetition_ratio":0.11725067,"word_repetition_ratio":0.0,"special_character_ratio":0.22654155,"punctuation_ratio":0.05925926,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98296034,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-29T05:29:26Z\",\"WARC-Record-ID\":\"<urn:uuid:8417cd9e-ac16-458a-b376-608e2aaf9ac5>\",\"Content-Length\":\"42547\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:05b804ef-1042-4681-ac08-c91b10ace7d7>\",\"WARC-Concurrent-To\":\"<urn:uuid:a8ea43a4-684c-435f-89e5-59f4c880ff38>\",\"WARC-IP-Address\":\"23.185.0.2\",\"WARC-Target-URI\":\"https://www.math.upenn.edu/events/spanning-clusters-and-subcritical-connectivity-high-dimensional-percolation\",\"WARC-Payload-Digest\":\"sha1:3PPJYFMBBDMKBGFXGZV2RTM2RVLWYMTJ\",\"WARC-Block-Digest\":\"sha1:PQAK7TFZUQRP735K3AERX7KCRC6F7BC4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652663039492.94_warc_CC-MAIN-20220529041832-20220529071832-00318.warc.gz\"}"} |
https://research-explorer.app.ist.ac.at/record/2731 | [
"# Linear Boltzmann equation as the weak coupling limit of a random Schrödinger equation\n\nL. Erdös, H. Yau, Communications on Pure and Applied Mathematics 53 (2000) 667–735.\n\nNo fulltext has been uploaded. References only!\n\nJournal Article | Published\nAuthor\nAbstract\nWe study the time evolution of a quantum particle in a Gaussian random environment. We show that in the weak coupling limit the Wigner distribution of the wave function converges to a solution of a linear Boltzmann equation globally in time. The Boltzmann collision kernel is given by the Born approximation of the quantum differential scattering cross section.\nPublishing Year\nDate Published\n2000-06-01\nJournal Title\nCommunications on Pure and Applied Mathematics\nVolume\n53\nIssue\n6\nPage\n667 - 735\nIST-REx-ID\n\n### Cite this\n\nErdös L, Yau H. Linear Boltzmann equation as the weak coupling limit of a random Schrödinger equation. Communications on Pure and Applied Mathematics. 2000;53(6):667-735. doi:10.1002/(SICI)1097-0312(200006)53:6<667::AID-CPA1>3.0.CO;2-5\nErdös, L., & Yau, H. (2000). Linear Boltzmann equation as the weak coupling limit of a random Schrödinger equation. Communications on Pure and Applied Mathematics, 53(6), 667–735. https://doi.org/10.1002/(SICI)1097-0312(200006)53:6<667::AID-CPA1>3.0.CO;2-5\nErdös, László, and Horng Yau. “Linear Boltzmann Equation as the Weak Coupling Limit of a Random Schrödinger Equation.” Communications on Pure and Applied Mathematics 53, no. 6 (2000): 667–735. https://doi.org/10.1002/(SICI)1097-0312(200006)53:6<667::AID-CPA1>3.0.CO;2-5.\nL. Erdös and H. Yau, “Linear Boltzmann equation as the weak coupling limit of a random Schrödinger equation,” Communications on Pure and Applied Mathematics, vol. 53, no. 6, pp. 667–735, 2000.\nErdös L, Yau H. 2000. Linear Boltzmann equation as the weak coupling limit of a random Schrödinger equation. Communications on Pure and Applied Mathematics. 53(6), 667–735.\nErdös, László, and Horng Yau. “Linear Boltzmann Equation as the Weak Coupling Limit of a Random Schrödinger Equation.” Communications on Pure and Applied Mathematics, vol. 53, no. 6, Wiley-Blackwell, 2000, pp. 667–735, doi:10.1002/(SICI)1097-0312(200006)53:6<667::AID-CPA1>3.0.CO;2-5.\n\n### Export\n\nMarked Publications\n\nOpen Data IST Research Explorer"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.7191694,"math_prob":0.861641,"size":2305,"snap":"2020-45-2020-50","text_gpt3_token_len":724,"char_repetition_ratio":0.13472404,"word_repetition_ratio":0.375,"special_character_ratio":0.3140998,"punctuation_ratio":0.21370968,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9596148,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-29T14:46:25Z\",\"WARC-Record-ID\":\"<urn:uuid:06572f35-ccb5-4824-b76e-8ffc95b7ffd7>\",\"Content-Length\":\"27572\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:96924975-c2ce-4fd2-95ff-93afbe10fe94>\",\"WARC-Concurrent-To\":\"<urn:uuid:1c707ac1-6723-4454-b59f-ba30b5ff829a>\",\"WARC-IP-Address\":\"81.223.84.196\",\"WARC-Target-URI\":\"https://research-explorer.app.ist.ac.at/record/2731\",\"WARC-Payload-Digest\":\"sha1:XOODMPN24B2OPRPDL2LOZT3OZL7QMUBT\",\"WARC-Block-Digest\":\"sha1:ZIW6MPZYAPD4DCDU7ZUGKZLSIZ5EKPHG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107904287.88_warc_CC-MAIN-20201029124628-20201029154628-00067.warc.gz\"}"} |
https://www.examrace.com/CSIR/CSIR-Free-Study-Material/CSIR-Physical-Sciences/Physics-Questions/Physics-Objective-Questions-Part-20.html | [
"# CSIR NET JRF Physics Objective questions Part 20\n\nQ-1. Magnetic field is measured in\n\n(a) Weber\n\n(b) Hennery\n\n(c)\n\n(d)\n\nQ-2. The dimensions of Plank’s constant are\n\n(a)\n\n(b)\n\n(c)\n\n(d)\n\nQ-3. The speed of boat is 5 km/hr in still water. It crosses a river of width 1 km along the shortest possible path in 15 minutes. Then velocity of river is\n\n(a) 4 km/hr\n\n(b) 3 km/hr\n\n(c) 2 km/hr\n\n(d) 1 km/hr\n\nQ-4. A bullet is dropped from the same height when another bullet is fired horizontally. They will hit the ground\n\n(a) Simultaneously\n\n(b) One after the other\n\n(c) Depends on the observer\n\n(d) Depends up on mass of bullet\n\nQ-5.The displacement of a particle moving in a straight line depends on time(t) as: x=\n\nThe ratio of its initial acceleration to its initial velocity depends\n\n(a) Only on\n\n(b) Only on\n\n(c) Only on\n\n(d) Only on\n\nQ-6. If a cyclist moving with a speed of 4.9 m/s on a level road can take a sharp circular turn of radius 4 m, then coefficient of friction between the cycle tires and road is\n\n(a) 0.41\n\n(b) 0.51\n\n(c) 0.61\n\n(d) 0.71\n\nQ-7. A body of mass 5 kg is moving in a circle of radius 1 m with an angular velocity of 2 radian/sec. The centripetal force acting on the body is\n\n(a) 10 N\n\n(b) 20 N\n\n(c) 30 N\n\n(d) 40 N\n\nQ-8. A bullet of mass 25 g moving with a velocity of 200 m/s is stopped within 5 cm of the target. The average resistance offered by the target is\n\n(a) 10 KN\n\n(b) 20 KN\n\n(c) 30 KN\n\n(d) 40 KN\n\nQ-9. A machine delivering power movers a body along a straight line. The distance moved by the body in time is proportional to\n\n(a) t\n\n(b)\n\n(c)\n\n(d)\n\nQ-10. If the radius of earth is reduced by 1% without changing the mass, then change in the acceleration due to gravity will be\n\n(a) 1% decrease\n\n(b) 1% increase\n\n(c) 2% increase\n\n(d) 2% decrease\n\nQ-11. If the spinning speed of earth is increased, then weight of the body at the equator\n\n(a) Increases\n\n(b) Decreases\n\n(c) doubles\n\n(d) Does not change\n\nQ-12. The ratio of energy required to raise a satellite to a height ‘h’ above the earth’s surface to that required to put it into the orbit is\n\n(a)\n\n(b)\n\n(c)\n\n(d)\n\nQ-13. A circular disc is rotating with angular velocity If a man standing at the edge of the disc walks towards its centre, then angular velocity of the disc will\n\n(a) Decrease\n\n(b) Increase\n\n(c) Be halved\n\n(d) Not change\n\nQ-14. For a gas, if the ratio of specific heats at constant pressure and volume is then the value of degree of freedom is\n\n(a)\n\n(b)\n\n(c)\n\n(d)\n\nQ-15. A life is ascending with acceleration equal to g/3. What will be the time-period of a simple pendulum suspended from its time-period in stationary life is T?\n\n(a) T/2\n\n(b)\n\n(c)\n\n(d)\n\nQ-16. If the equation of a sound wave is given as: y=0.0015 then wavelength of this wave is\n\n(a) 0.4 unit\n\n(b) 0.3 unit\n\n(c) 0.2 unit\n\n(d) 0.1 unit\n\nQ-17. A simple pendulum of length ‘I’ has a maximum angular displacement The maximum kinetic energy of the bob of mass m is.\n\n(a) Mgl\n\n(b) 0.5 mgl\n\n(c) Mgl sin\n\n(d) Mgl\n\nQ-18. A standing wave is represented by: y=a sin(100 t). cos(0.01 x); where t in seconds and x in meters. The velocity of wave is\n\n(a)\n\n(b)\n\n(c)\n\n(d)\n\nQ-19. The amplitude of the vibrating particle due to superposition of two simple harmonic motion of is\n\n(a) 1\n\n(b)\n\n(c)\n\n(d) 2\n\nQ-20. In a sinusoidal wave, the time required for a particular point to move from maximum displacement is 0.17 sec. The frequency of the wave is\n\n(a) 0.36 Hz\n\n(b) 0.73 Hz\n\n(c) 1.47 Hz\n\n(d) 2.94 Hz\n\nQ-21. When a current flows in a wire. There exists an electric field in the direction of\n\n(a) Flow of current\n\n(b) Opposite to the flow of current\n\n(c) Perpendicular to the flow of current\n\n(d) At an angle of to the flow of current\n\nQ-22. Two identical mercury drops, each of radius r are charged to the same potential V. if the mercury drop coalesce to form a big drop of radius R, then potential of the combined drop will be\n\n(a)\n\n(b)\n\n(c)\n\n(d)\n\nQ-23. The energy stored in a capacitor is actually stored\n\n(a) Between the plates\n\n(b) On the positive plate\n\n(c) On the negative plate\n\n(d) On the outer surface of the plates\n\nQ-24. In the given figure, the capacitors have a capacitance of 4 each If the capacitor has a capacitance between A and B is\n\n(a)\n\n(b)\n\n(c)\n\n(d)\n\nQ-25. A 100 W, 200 V bulb is connected to a 160 volts supply. The actual power consumption would be\n\n(a)\n\n(b)\n\n(c)\n\n(d)\n\nQ-26. To convert a galvanometer in a voltmeter. We must connect a\n\n(a) Low resistance in series\n\n(b) High resistance in series\n\n(c) Low resistance in parallel\n\n(d) High resistance in Parallel\n\nQ-27. A galvanometer of 100 resistance gives full scale deflection with 0.01 A current. How much resistance should be connect to convert into an ammeter of range 10 A?\n\n(a) 0.2 in series\n\n(b) 0.2 in Parallel\n\n(c) 0.1 in series\n\n(d) 0.1 in Parallel\n\nQ-28 The potential difference between two electrodes of a galvanic cell, in an open circuit, is known as\n\n(a) Current\n\n(b) Impedance\n\n(c) Potential difference\n\n(d) Electromotive force\n\nQ-29. The magnetic field due to a current carrying circular loop of radius 12 cm at its centre is The magnetic field due to this loop at a point on the ax is at a distance of 5 cm from the centre is\n\n(a)\n\n(b)\n\n(c)\n\n(d)\n\nQ-30. An e.m.f. of 15 volt is applied in a circuit containing 5H inductance and 10 resistance. The ration of currents at t = and at t=1 sec is\n\n(a)\n\n(b)\n\n(c)\n\n(d)\n\nQ-31. Two magnets of magnetic moment M and 2M are placed in a vibration magnetometer, with identical poles in the same direction. The time-period of vibration of the combination is if the same magnets are placed with opposite poles together and vibrate with time period\n\n.\n\n(a)\n\n(b)\n\n(c)\n\n(d)\n\nQ-32. Which of the following waves have the maximum wavelength?\n\n(a) X-rays\n\n(c) UV rays\n\n(d) IR rays\n\nQ-33. At what angle, a ray of light will be incident on face of an equilateral prism, so that the emergent ray may graze the second surface of the\n\n(a)\n\n(b)\n\n(c)\n\n(d)\n\nQ-34. A paper, with two marks having separation d, is held normal to the line of right of an observer at distance of 50 cm. The diameter of the, eyes-lens of the observer is 2 mm. Which of the following is the least value of d, so that the marks can be seen as separate? (mean wavelength of visible light may be taken 5000 A)\n\n(a)\n\n(b)\n\n(c)\n\n(d)\n\nQ-35. How many images will be formed if two mirrors are fitted on adjacent wall and one mirror on roof?\n\n(a) 2\n\n(b) 5\n\n(c) 7\n\n(d) 10\n\nQ-36. An optician prescribes spectacles to a patient with a combination of a convex lens of focal length 40 cm and concave lens 25 cm. The power of spectacles is\n\n(a)\n\n(b)\n\n(c)\n\n(d)\n\nQ-37. The velocity of an electron in the inner-most orbit of an atom is\n\n(a) Zero\n\n(b) Mean\n\n(c) Lowest\n\n(d) Highest\n\nQ-38. The hydrogen atom can give spectral lines in the Lyman, Ballmer and Paschal series. Which of the following statement is correct?\n\n(a) Paschal series is in visible region\n\n(b) Blamer series is in visible region\n\n(c) Lyman series is in infra-red region\n\n(d) Blamer series is in ultra violet-region\n\nQ-39. A sample of a radioactive substance contains 2828 atoms. If its half-life is 2 days, how many atoms will be left intact in the sample after one day?\n\n(a) 1414\n\n(b) 707\n\n(c) 2000\n\n(d) 1000\n\nQ-40. In a nuclear reactor, the fast moving neutrons are showered down by passing them through\n\n(a) oil\n\n(b) Vacuum\n\n(c) Heavy water\n\n(d) Kerosene\n\nDeveloped by:"
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https://www.asknumbers.com/lbs-to-oz/16-lbs-to-oz.aspx | [
"# How Many Ounces in 16 Pounds?\n\n16 Pounds to ounces (lbs to oz) converter. How many ounces in 16 pounds?\n\n16 Lbs equal to 256 oz or there are 256 ounces in 16 pounds.\n\n←→\nstep\nRound:\nEnter Pound\nEnter Ounce\n\n## How to convert 16 pounds to ounces?\n\nThe conversion factor from pound to ounce is 16. To convert any value of pounds to ounces, multiply the pound value by the conversion factor.\n\nTo convert 16 lbs to oz, multiply 16 by 16, that makes 16 lbs equal to 256 oz.\n\n16 lbs to oz formula\n\noz = lbs value * 16\n\noz = 16 * 16\n\noz = 256\n\nCommon conversions from 16.x lbs to oz:\n(rounded to 3 decimals)\n\n• 16 lbs = 256 oz\n• 16.1 lbs = 257.6 oz\n• 16.2 lbs = 259.2 oz\n• 16.3 lbs = 260.8 oz\n• 16.4 lbs = 262.4 oz\n• 16.5 lbs = 264.0 oz\n• 16.6 lbs = 265.6 oz\n• 16.7 lbs = 267.2 oz\n• 16.8 lbs = 268.8 oz\n• 16.9 lbs = 270.4 oz\n\nWhat is a Pound?\n\nPound (lb) is an Imperial system mass unit. 1 Pound = 16 Ounces. 1 Troy pound = 12 Troy ounces. The symbol is \"lb\".\n\nWhat is a Ounce?\n\nOunce is an Imperial system mass unit. 1 oz = 0.0625 lb. 1 troy oz = 0.08334 troy lb. The symbol is \"oz\".\n\nCreate Conversion Table\nClick \"Create Table\". Enter a \"Start\" value (5, 100 etc). Select an \"Increment\" value (0.01, 5 etc) and select \"Accuracy\" to round the result."
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