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[ "# Finding a functional satisfying a given Frechet derivative\n\nUsually, we are interested in finding the Frechet derivative of a given functional. My problem is the opposite; to find a functional satisfying a condition given in terms of the Frechet derivative of the functional.\n\nLet the Frechet derivative of the unknown, possibly nonlinear, functional $\\Xi : L_2(0,1) \\rightarrow \\mathbb{R}$, be defined as the linear operator $D\\Xi_f : L_2(0,1) \\rightarrow \\mathbb{R}$ such that\n\n\\begin{align} \\lim_{h\\rightarrow 0} \\frac{||\\Xi[f+h] -\\Xi[f] - D \\Xi_f[h]||_\\mathbb{R}}{||h||_{L_2(0,1)}} = 0, \\end{align} and satisfying a given condition \\begin{align} D \\Xi_f[h] = \\left<\\eta[f],h\\right> \\end{align} where $\\eta : L_2(0,1) \\rightarrow L_2(0,1)$ is a known functional.\n\nIs there any systematic approach to find such a functional $\\Xi$, besides trial and error?\n\nMore specifically, I am interested in situations where $\\eta$ takes the form \\begin{align} \\eta[f](x) = \\int_x^1 (2-x) \\phi(f(x)) dx \\end{align} and where $\\phi : L_2(0,1) \\rightarrow \\mathbb{R}$ is a low order polynomial functional. For example $\\phi(f) = f$ or $\\phi(f) = f^2$.\n\nAny suggestions on\n\n1. Conditions $\\eta$ must satisfy for a functional $\\Xi$ to exist\n2. General solution strategies or tricks\n3. Specific solutions for the specific $\\eta$ provided above\n\nwill be appreciated.\n\n• If $f$ is a linear function (between two normed spaces), then the derivative of $f$ at any point is just $f.$ – Will M. May 3 '18 at 15:57\n• $f$ is an unknown function and can not be assumed to be linear. – Haavard May 3 '18 at 20:03\n• Read again my comment and generalise. Choose $f$ (the function you want to find) to be the linear function given by the derivative you know. – Will M. May 3 '18 at 20:05\n• The unknown functional (which I called $\\Xi$) will, in general, be a non-linear operator on $f$. For $\\phi=1$, the Frechet derivative $D\\Xi_f[h]$ will be independent of $f$, and we can do as you suggested. – Haavard May 4 '18 at 6:42\n• I thought $\\eta$ was linear. In the conditions you are stating above, your question doesn't make sense. There are a few inconsistencies. For instance, is $\\eta:L^2 \\to \\Bbb R$? – Will M. May 5 '18 at 4:31" ]

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[ "", null, "#### ASVAB Armed Services Vocational Aptitude Battery (Exam 29)\n\ncreated by Fisher BRink (@fisher) at Feb. 8, 2016\n• The reciprocal of 1/6 is:\n\n• If 0.05 ÷ x = 1, then x =\n\n• Factor x2 6x + 9.\n\n• (3 × 2)(7 2)(6 + 2) = (6 × 4)x. What's the value of x?\n\n• Solve for x: 2x 6 = x + 5\n\n• If I = prt, and p = \\$1,000, r = 7%, and t = 1,what does I equal?\n\n• x + y = 6 and x y = 4. Solve for x.\n\n• Solve for y: 4(y + 3) + 7 = 3\n\n• 12 yards + 14 feet ÷ by 2 =\n\n• x3 × x3 =\n\n• 41/5 + 12/5 + 33/10 =\n\n• 5 × 102 =\n\n• (900 × 2) ÷ 6 =\n\n• The cube root of 64 is:\n\n• A circle has a radius of 5 inches. What's its approximate area?\n\n• Solve the following inequity:2/3(6x 9) + 4 &gt; 5x + 1\n\n• A tube has a radius of 3 inches and a height of 5 inches. What's its approximate v...\n\n• Triangle ABC (shown above) is a(n):\n\n• The figure above is what type of quadrilateral?\n\n• Angle AB (shown above) is a(n):\n\nBe the first to review\n\n• info\nQuiz Info\n• date_range\nFeb. 8, 2016, 3:26 a.m.\nhelp_outline\n20 questions\ndvr\n0 completed\nremove_red_eye\n11 views\npeople\n0 takers\nfolder", null, "•", null, "#### Ratings\n\nstar_borderstar_borderstar_borderstar_borderstar_border\nratings" ]

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[ "# #a0aa9d Color Information\n\nIn a RGB color space, hex #a0aa9d is composed of 62.7% red, 66.7% green and 61.6% blue. Whereas in a CMYK color space, it is composed of 5.9% cyan, 0% magenta, 7.6% yellow and 33.3% black. It has a hue angle of 106.2 degrees, a saturation of 7.1% and a lightness of 64.1%. #a0aa9d color hex could be obtained by blending #ffffff with #41553b. Closest websafe color is: #999999.\n\n• R 63\n• G 67\n• B 62\nRGB color chart\n• C 6\n• M 0\n• Y 8\n• K 33\nCMYK color chart\n\n#a0aa9d color description : Dark grayish lime green.\n\n# #a0aa9d Color Conversion\n\nThe hexadecimal color #a0aa9d has RGB values of R:160, G:170, B:157 and CMYK values of C:0.06, M:0, Y:0.08, K:0.33. Its decimal value is 10529437.\n\nHex triplet RGB Decimal a0aa9d `#a0aa9d` 160, 170, 157 `rgb(160,170,157)` 62.7, 66.7, 61.6 `rgb(62.7%,66.7%,61.6%)` 6, 0, 8, 33 106.2°, 7.1, 64.1 `hsl(106.2,7.1%,64.1%)` 106.2°, 7.6, 66.7 999999 `#999999`\nCIE-LAB 68.502, -6, 5.482 34.957, 38.657, 37.516 0.315, 0.348, 38.657 68.502, 8.127, 137.582 68.502, -4.989, 8.893 62.175, -8.448, 7.747 10100000, 10101010, 10011101\n\n# Color Schemes with #a0aa9d\n\n• #a0aa9d\n``#a0aa9d` `rgb(160,170,157)``\n• #a79daa\n``#a79daa` `rgb(167,157,170)``\nComplementary Color\n• #a7aa9d\n``#a7aa9d` `rgb(167,170,157)``\n• #a0aa9d\n``#a0aa9d` `rgb(160,170,157)``\n• #9daaa1\n``#9daaa1` `rgb(157,170,161)``\nAnalogous Color\n• #aa9da7\n``#aa9da7` `rgb(170,157,167)``\n• #a0aa9d\n``#a0aa9d` `rgb(160,170,157)``\n• #a19daa\n``#a19daa` `rgb(161,157,170)``\nSplit Complementary Color\n• #aa9da0\n``#aa9da0` `rgb(170,157,160)``\n• #a0aa9d\n``#a0aa9d` `rgb(160,170,157)``\n• #9da0aa\n``#9da0aa` `rgb(157,160,170)``\n• #aaa79d\n``#aaa79d` `rgb(170,167,157)``\n• #a0aa9d\n``#a0aa9d` `rgb(160,170,157)``\n• #9da0aa\n``#9da0aa` `rgb(157,160,170)``\n• #a79daa\n``#a79daa` `rgb(167,157,170)``\n• #788674\n``#788674` `rgb(120,134,116)``\n• #869282\n``#869282` `rgb(134,146,130)``\n• #939e8f\n``#939e8f` `rgb(147,158,143)``\n• #a0aa9d\n``#a0aa9d` `rgb(160,170,157)``\n``#adb6ab` `rgb(173,182,171)``\n• #bac2b8\n``#bac2b8` `rgb(186,194,184)``\n• #c8cec6\n``#c8cec6` `rgb(200,206,198)``\nMonochromatic Color\n\n# Alternatives to #a0aa9d\n\nBelow, you can see some colors close to #a0aa9d. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #a3aa9d\n``#a3aa9d` `rgb(163,170,157)``\n• #a2aa9d\n``#a2aa9d` `rgb(162,170,157)``\n• #a1aa9d\n``#a1aa9d` `rgb(161,170,157)``\n• #a0aa9d\n``#a0aa9d` `rgb(160,170,157)``\n• #9faa9d\n``#9faa9d` `rgb(159,170,157)``\n• #9eaa9d\n``#9eaa9d` `rgb(158,170,157)``\n• #9daa9d\n``#9daa9d` `rgb(157,170,157)``\nSimilar Colors\n\n# #a0aa9d Preview\n\nThis text has a font color of #a0aa9d.\n\n``<span style=\"color:#a0aa9d;\">Text here</span>``\n#a0aa9d background color\n\nThis paragraph has a background color of #a0aa9d.\n\n``<p style=\"background-color:#a0aa9d;\">Content here</p>``\n#a0aa9d border color\n\nThis element has a border color of #a0aa9d.\n\n``<div style=\"border:1px solid #a0aa9d;\">Content here</div>``\nCSS codes\n``.text {color:#a0aa9d;}``\n``.background {background-color:#a0aa9d;}``\n``.border {border:1px solid #a0aa9d;}``\n\n# Shades and Tints of #a0aa9d\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, #060706 is the darkest color, while #fcfcfc is the lightest one.\n\n• #060706\n``#060706` `rgb(6,7,6)``\n• #10120f\n``#10120f` `rgb(16,18,15)``\n• #191c18\n``#191c18` `rgb(25,28,24)``\n• #232721\n``#232721` `rgb(35,39,33)``\n• #2c312b\n``#2c312b` `rgb(44,49,43)``\n• #353c34\n``#353c34` `rgb(53,60,52)``\n• #3f463d\n``#3f463d` `rgb(63,70,61)``\n• #485146\n``#485146` `rgb(72,81,70)``\n• #525b4f\n``#525b4f` `rgb(82,91,79)``\n• #5b6658\n``#5b6658` `rgb(91,102,88)``\n• #657061\n``#657061` `rgb(101,112,97)``\n• #6e7b6a\n``#6e7b6a` `rgb(110,123,106)``\n• #788573\n``#788573` `rgb(120,133,115)``\n• #818f7d\n``#818f7d` `rgb(129,143,125)``\n• #8c9888\n``#8c9888` `rgb(140,152,136)``\n• #96a192\n``#96a192` `rgb(150,161,146)``\n• #a0aa9d\n``#a0aa9d` `rgb(160,170,157)``\n• #aab3a8\n``#aab3a8` `rgb(170,179,168)``\n• #b4bcb2\n``#b4bcb2` `rgb(180,188,178)``\n• #bfc5bd\n``#bfc5bd` `rgb(191,197,189)``\n• #c9cec7\n``#c9cec7` `rgb(201,206,199)``\n• #d3d8d2\n``#d3d8d2` `rgb(211,216,210)``\n• #dde1dc\n``#dde1dc` `rgb(221,225,220)``\n• #e7eae7\n``#e7eae7` `rgb(231,234,231)``\n• #f1f3f1\n``#f1f3f1` `rgb(241,243,241)``\n• #fcfcfc\n``#fcfcfc` `rgb(252,252,252)``\nTint Color Variation\n\n# Tones of #a0aa9d\n\nA tone is produced by adding gray to any pure hue. In this case, #a0aa9d is the less saturated color, while #73fe49 is the most saturated one.\n\n• #a0aa9d\n``#a0aa9d` `rgb(160,170,157)``\n• #9cb196\n``#9cb196` `rgb(156,177,150)``\n• #98b88f\n``#98b88f` `rgb(152,184,143)``\n• #95bf88\n``#95bf88` `rgb(149,191,136)``\n• #91c681\n``#91c681` `rgb(145,198,129)``\n• #8dcd7a\n``#8dcd7a` `rgb(141,205,122)``\n• #89d473\n``#89d473` `rgb(137,212,115)``\n• #85db6c\n``#85db6c` `rgb(133,219,108)``\n• #82e265\n``#82e265` `rgb(130,226,101)``\n• #7ee95e\n``#7ee95e` `rgb(126,233,94)``\n• #7af057\n``#7af057` `rgb(122,240,87)``\n• #76f750\n``#76f750` `rgb(118,247,80)``\n• #73fe49\n``#73fe49` `rgb(115,254,73)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #a0aa9d 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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[ "", null, "## Unique random matrix in numpy", null, "| |\n[", null, "所属分类 开发（python） | 发布者 店小二05 | 时间 2018 | 作者 红领巾 ] 0人收藏点击收藏\nI want to make a matrix x with shape (n_samples, n_classes) where each x[i] is a random one-hot vector. Here's a slow implementation: x = np.zeros((n_samples, n_classes)) J = np.random.choice(n_classes, n_samples) for i, j in enumerate(J): x[i, j] = 1\n\nWhat's a more pythonic way to do this?\n\nCreate an identity matrix using np.eye :\n\nx = np.eye(n_classes)\n\nThen use np.random.choice to select rows at random:\n\nx[np.random.choice(x.shape, size=n_samples)]\n\nAs a shorthand, just use:\n\nnp.eye(n_classes)[np.random.choice(n_classes, n_samples)]\n\nDemo:\n\nIn : np.eye(5)[np.random.choice(5, 100)] Out: array([[ 1., 0., 0., 0., 0.], [ 1., 0., 0., 0., 0.], [ 0., 0., 1., 0., 0.], [ 0., 0., 0., 0., 1.], [ 0., 0., 0., 1., 0.], [ 1., 0., 0., 0., 0.], [ 0., 0., 0., 1., 0.], .... (... to 100)\n\n1.凡CodeSecTeam转载的文章,均出自其它媒体或其他官网介绍,目的在于传递更多的信息,并不代表本站赞同其观点和其真实性负责；\n2.转载的文章仅代表原创作者观点,与本站无关。其原创性以及文中陈述文字和内容未经本站证实,本站对该文以及其中全部或者部分内容、文字的真实性、完整性、及时性，不作出任何保证或承若；\n3.如本站转载稿涉及版权等问题,请作者及时联系本站,我们会及时处理。", null, "技术大类 | 开发（python） | 评论(0) | 阅读(28)" ]

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[ "Intervals Series Part 3: Completing The Interval List\n\nIn the last two lessons we have worked our way up to this list of Interval names/numbers:\n\nC     D     E     F     G     A     B     C     D     E     F    G    A    B    C\nR     2     M3   4     5      6     M7   R     9     M3 11    5    13   M7  R\n\nHopefully this list looks familiar to you and you understand it. If not please see the previous lessons.\n\nThe list above works with a two octave scale showing the notes Interval relationship to the Root C.\n\nAnd, you can see that we've accounted for the common Interval names/numbers that are used in building common Triads and Extended chords (the R, M3, 5 and M7 are constant across each octave).\n\nLet's take it another step further. Let's look at the other Intervals and their octaves. By doing so we can trim the scale down to one octave and not miss any notes.\n\nNotice in the list above how:\n\nD = 2 and 9\nF = 4 and 11\nA = 6 and 13\n\nThe 2 is the same note in the scale as the 9. The 4 is the same note in the scale as the 11. The 6 is the same note in the scale as the 13.\n\nThe Intervals are commonly written out like this, trimming down what I just listed.\n\n2/9\n4/11\n6/13\n\nIn the C Major scale:\n\nD = 2/9\nF = 4/11\nA = 6/13\n\nGenerally you can use the term \"2\" interchangeably with \"9\", and \"4\" for \"11\", and \"6\" for \"11\". Mainly because they are the same note names, even though they are in different octaves.\n\nWhen Extending chords higher than the four note \"extended\" chord you generally use the \"higher number\", i.e. 9, 11, 13. Like Cmaj13, not Cmaj6...or maybe Cmaj11, not Cmaj4.\n\nWhen building what are called \"Add\" chords you can use either the high or the low number. Like Cadd6 is is the same notes as Cadd13, Cadd2 is the same notes as Cadd9, Cadd4 is the same notes as Cadd11.\n\nIn a future lesson we'll detail what the Chord Formulas look like for these Extended and Add Chords.\n\nThere are guidelines to how chords should be written or labeled, but in reality the guidelines can get blown out of the water as you could see them written either way in a book or on someone's chart.\n\nYou just want to remember:\n\n2=9\n4=11\n6=13\n\nOk, now let's streamline the Intervals in the C Major scale and redraw the scale again to get this:\n\nC    D     E       F      G      A       B      C\nR   2/9   M3   4/11   5     6/13   M7     R\n\nNow we can look at the scale in one octave and understand the Interval relationship to the Root. This is big progress.\n\nIn the next lesson we are going look at the Interval relationship of ALL the notes from the Root, a Chromatic scale. Don't worry, if you got this far the rest should be very understandable." ]

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[ "# Find the LCM 6 , 4 , 10", null, "6 , 4 , 10\nThe LCM is the smallest positive number that all of the numbers divide into evenly.\n1. List the prime factors of each number.\n2. Multiply each factor the greatest number of times it occurs in either number.\n6 has factors of 2 and 3.\n2⋅3\n4 has factors of 2 and 2.\n2⋅2\n10 has factors of 2 and 5.\n2⋅5\nThe LCM of 6,4,10 is the result of multiplying all prime factors the greatest number of times they occur in either number.\n2⋅2⋅3⋅5\nThe LCM of 6,4,10 is 2⋅2⋅3⋅5=60.\nMultiply 2 by 2.\n4⋅3⋅5\nMultiply 4 by 3.\n12⋅5\nMultiply 12 by 5.\n60\n60\nFind the LCM 6 , 4 , 10", null, "", null, "", null, "## Download our App from the store\n\n### Create a High Performed UI/UX Design from a Silicon Valley.", null, "", null, "Scroll to top" ]

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[ "# exponentials maths question help!\n\nWatch\nAnnouncements\n#1\n0\n6 months ago\n#2\n(Original post by dnejfn)\nShift the -P to the left hand side and solve/integrate the first order ode.\nWhat have you covered about solving them?\nIt should be a straightforward case/solution.\nLast edited by mqb2766; 6 months ago\n0\n6 months ago\n#3\n(Original post by dnejfn)\ndP/dt + P = 100\nso integrate 100? that's 100t\nDo you know how to solve first order odes?\n0\n#4\n(Original post by mqb2766)\nShift the -P to the left hand side and solve/integrate the first order ode.\nWhat have you covered about solving them?\nIt should be a straightforward case/solution.\ndP/dt + P = 100\n\nIntegrate 100? that's 100t\n\nI'm not sure if I have actually covered this in the lesson\nhttps://www.mathsisfun.com/calculus/...er-linear.html\nI found this, is this what you mean by integrating the first order ode?\nLast edited by dnejfn; 6 months ago\n0\n6 months ago\n#5\n(Original post by dnejfn)\ndP/dt + P = 100\n\nIntegrate 100? that's 100t\n\nI'm not sure if I have actually covered this in the lesson\nhttps://www.mathsisfun.com/calculus/...er-linear.html\nI found this, is this what you mean by integrating the first order ode?\nI'm presuming you have a textbook. But if you've not covered it, why try questions on it?\n0\n#6\n(Original post by mqb2766)\nI'm presuming you have a textbook. But if you've not covered it, why try questions on it?\nWhat's this sort of topic called? First Order Linear Differential Equations?\n0\n6 months ago\n#7\n(Original post by mqb2766)\nShift the -P to the left hand side and solve/integrate the first order ode.\nWhat have you covered about solving them?\nIt should be a straightforward case/solution.\nWould have thought this was more appropriately a \"separation of variables\" problem (which at least in my time, was covered in the normal A-level, while \"dP/dt + kP = C\" type equations would only be in FM).\n1\n#8\n(Original post by DFranklin)\nWould have thought this was more appropriately a \"separation of variables\" problem (which at least in my time, was covered in the normal A-level, while \"dP/dt + kP = C\" type equations would only be in FM).\nI attempted it and got up to here. Put then the 100t cancelled out so think I've gone wrong somewhere\n0\n6 months ago\n#9\nThe answer is an exponential+ ... What have you covered \"at school\".\n0\n6 months ago\n#10\n(Original post by dnejfn)\nI attempted it and got up to here. Put then the 100t cancelled out so think I've gone wrong somewhere\nSo, what I'm saying is to solve", null, "as a separable differential equation. Do not move the P over (i.e. don't add P to both sides). Solve it directly. If you don't know how to do that, say so.\n\nBut if you don't know how to solve a separable DE, and you don't know the method for solving a first order linear DE (which is what mqb2766 was leading you to), then you won't be able to solve this.\n\nAt which point \"why are you trying to solve questions you haven't covered the material for?\" definitely requires an answer.\nLast edited by DFranklin; 6 months ago\n0\n#11\n(Original post by DFranklin)\nSo, what I'm saying is to solve", null, "as a separable differential equation. Do not move the P over (i.e. don't add P to both sides). Solve it directly. If you don't know how to do that, say so.\n\nBut if you don't know how to solve a separable DE, and you don't know the method for solving a first order linear DE (which is what mqb2766 was leading you to), then you won't be able to solve this.\n\nAt which point \"why are you trying to solve questions you haven't covered the material for?\" definitely requires an answer.\nwell, I've finished all the content for a level maths lol I just probably wasn't there when my class went over this. I'm gonna do some practise questions before doing this one then so then I can get the method right 👍\n0\n#12\nI got this but not sure if it's right\n0\n6 months ago\n#13\nNo. Using separation of variables, divide by the right hand side\n\nInt 1/(100-P) dP = Int 1 dt\n\nCan you see that and now to proceed?\n0\n#14\n(Original post by mqb2766)\nNo. Using separation of variables, divide by the right hand side\n\nInt 1/(100-P) dP = Int 1 dt\n\nCan you see that and now to proceed?\ndo i then integrate both sides, then find c ?\n0\n6 months ago\n#15\n(Original post by dnejfn)\nwell, I've finished all the content for a level maths lol I just probably wasn't there when my class went over this. I'm gonna do some practise questions before doing this one then so then I can get the method right 👍\nIn A level mathematics, to solve a differential equation like this, the P term must be i) on the same side as the dP/dt and ii) must be multiplied by dP/dt. You should not be adding P to both sides as then the P term will not be being multiplied by dP/dt, so you cannot solve using any method learned in the standard A level course. Dividing by (100 - P), however, will give you the equation in a solvable form. Remember that when you take the integral of both sides, you take the integral of the whole of the side and not just one of the terms. Also remember that you must be integrating with respect to a variable - in this case to integrate both sides it would be ∫ dt - integrate with respect to t. Finally, can you spot why the ∫ dt becomes a ∫ dP on the left?\n0\n6 months ago\n#16\n(Original post by dnejfn)\ndo i then integrate both sides, then find c ?\nYes. You'd expect an exponential solution, so logs may play a part. You could use the given initial conditions and treat it as a definite integral, or just do an indefinite integral, then find C using the same info.\n\nDo you understand why/how you get it in that form?\nLast edited by mqb2766; 6 months ago\n0\n#17\n(Original post by mqb2766)\nYes. You'd expect an exponential solution, so logs may play a part. You could use the given initial conditions and treat it as a definite integral, or just do an indefinite integral, then find C using the same info.\n\nDo you understand why/how you get it in that form?\ni assume its one of the natural log rules but not sure which one. I have dP=(100-P)dt so have dt on the left and then you divide by 100-P but not sure where ln comes from\n0\n6 months ago\n#18\n(Original post by dnejfn)\ni assume its one of the natural log rules but not sure which one. I have dP=(100-P)dt so have dt on the left and then you divide by 100-P but not sure where ln comes from\nYou've changed the integral from the previous post", null, ". Don't.\n\nOn the left, you want to integrate\n1/(100-P)\nIt's roughly 1/P. Can you integrate that wrt P?\n\nDo you understand why I divided by 100-P?\n\nSpoiler:\nShow\n\n0\n#19\n(Original post by mqb2766)\nYou've changed the integral from the previous post", null, ". Don't.\n\nOn the left, you want to integrate\n1/(100-P)\nIt's roughly 1/P. Can you integrate that wrt P?\n\nDo you understand why I divided by 100-P?\n\nSpoiler:\nShow\n\nI integrated 1/100-P and got -ln|100-P| + C. You want to get P on the left with dP so you can multiply them so can then integrate which is why you divide by 100-P\n0\n6 months ago\n#20\n(Original post by dnejfn)\nI integrated 1/100-P and got -ln|100-P| + C. You want to get P on the left with dP so you can multiply them so can then integrate which is why you divide by 100-P\nOk, so what is on the right when you integrate and how do you get P = ...\n0\nX\n\nnew posts", null, "Back\nto top\nLatest\nMy Feed\n\n### Oops, nobody has postedin the last few hours.\n\nWhy not re-start the conversation?\n\nsee more\n\n### See more of what you like onThe Student Room\n\nYou can personalise what you see on TSR. Tell us a little about yourself to get started.\n\n### Poll\n\nJoin the discussion\n\n#### What support do you need with your UCAS application?\n\nI need help researching unis (11)\n12.79%\nI need help researching courses (7)\n8.14%\nI need help with filling out the application form (5)\n5.81%\nI need help with my personal statement (35)\n40.7%\nI need help with understanding how to make my application stand out (22)\n25.58%\nI need help with something else (let us know in the thread!) (2)\n2.33%\nI'm feeling confident about my application and don't need any help at the moment (4)\n4.65%" ]

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[ "# 9991 ml to grams – 9991 ml in g\n\nHere you can find 9991 ml to grams. No matter if you have been looking for 9991 ml in grams or 9991 ml in g, this is the right page for you. The result of the conversion depends on the density of the material or substance under consideration. The answer to the question 9991 ml is how many grams also varies with the pressure and the temperature. Read on below to understand everything about 9991 milliliters to grams.\n\n## Convert 9991 ml to grams\n\nTo convert 9991 ml to grams, besides the volume we have to know the substance’s density ρ in g/cm3 or in any other unit. Alternatively, when we know the material, then we can look for its density in a search engine. As explained on our home page, 9991 ml equal 9991 cm3, and as also outlined there, in common use the equation for water is 9991 milliliter = 9991 g.\n\n9991 ml to grams water = 9991 g\n9991 milliliters water to grams = 9991 g\n\nBut what about other stuff and food? In the next section you can find the mass equivalent of 9991 ml for some cooking ingredients. Below is our converter which calculates 9991 ml to g for any substance with known ρ.\n\nThe result is...\n\nBookmark us now as ml to grams.\n\nHere you can convert 9991 grams to ml.\n\n## 9991 ml in g\n\nUnless you have made use of our calculator, only for water under certain conditions do you know 9991 ml in g. And kindly note that 9991 ml to grams means the same as 9991 mL to grams.\n\n9991 ml equals how many grams for food ingredients is next:\n\nMilk = 10290.73 g, (ρ = 1.03)\nCream = 10105.897 g, (ρ = 1.0115)\nFlour = 5924.663 g, (ρ = 0.593)\nSugar = 7842.935 g, (ρ = 0.785)\nButter = 9101.801 g, (ρ = 0.911)\n\nρ is stated in g/cm3.", null, "Make sure to understand that these values for 9991 ml in grams are averages. Depending on the precise amount of fat, the quality or the kind, as well as other factors, 9991 ml to gram can be a different weight. We have explained all this in detail in our article ml to grams, where you can also find additional information on mass, volume, liter and kilogram.\n\nBesides 9991 ml to grams, similar conversions on this website include:\n\nThis concludes 9991 ml into grams. We hope you like our article and hit the sharing buttons for 9991 ml to gram. All questions and comments are really appreciated." ]

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[ "natronics.org\n\n# PSAS Magnetometer Calibration\n\nPublished in November 2019\n\nFor a number of years I was involved with a university rocketry club called PSASPortland State Aerospace Society, a student aerospace engineering project at Portland State University. They build ultra-low-cost, open source rockets that feature very sophisticated amateur rocket avionics systems. . One of the things I really liked to do was play with the data from the launches and learn how rockets and flight electronics work.\n\nOur rockets carry an instrument on them called an IMU (Inertial Measument Unit). An IMU typically measures both acceleration and rotation-rate of an object in all directions so with some clever math you can recreate the exact position, velocity, and orientation of the rocket over time. This is the only way to know where something is in space, and very important for rockets. IMUs have a problem though: they’re not very precise.\n\nSince our IMU is fixed to the rocket,", null, "Overview of the rocket “LV2.3”. The IMU is near the primary flight computer. which direction is “up” or “left”, etc. relative to the Earth changes constantly as the rocket flies about. In order for the data to be useful we need to know which way we are pointed, which is why IMUs always have some kind of gryoscope to account for rotation. Our particular IMU has rate-gyroscopes that can sense rotation rate, and so we integrate that once to get orientation. Since any integration will give an estimate that drifts from the true value over time, our IMU also includes a 3-axis magnetometer as well.\n\n## 9DOF IMU\n\nThis makes what is often what is refered to as a “9DOF” IMU, because it has “nine degrees of freedom”. That would be x, y, z accleration, x, y, z rotation-rate, and x, y, z magnetic field. The reason to have a magnetometer is so you can use Earth’s own magnetic field as a kind of guide to the orientation of the rocket. This doesn’t instantly solve all problems in life, sadly. But it provides a good reference for the rough orientation of the rocket that can be used to produce a real-time estimate of rate-gyroscape drift, or ‘bias’, as we fly.\n\nThe magnetic field sensor in the rocket is sensitive, but because the Earth’s field is so weak it’s easily overwhelmed by local effects (metal screws, magnetic fields from nearby wires, etc.). In order to get good orientation data we need to undo", null, "Members of the PSAS ground crew lifting and aranging the rocket around as many different orientations as possible before the flight. these local effects.\n\nSo a little before the flight we took the nearly complete rocket, powered the electronics up, picked it up and tried to move it around in every direction.\n\n## Magnetometer Calibration\n\nWhat do we expect good magnetometer data to look like? The Earth’s magnetic field shouldn’t change much, so it should look like a single vector going through the IMU. If we rotate the rocket one way or another, the angle that the vector goes through will change, but it should stay the same strength. That means that the magnitude of the local magnetic field should be constant, and it should measure it to be exactly the same as Earth’s magnetic field.\n\n## Earth’s Field Strength\n\nBut what is the strength of Earth’s magnetic field? It varies over time and over the surface of the Earth. We know where we launched fromLatitude: 43.79613280° N\nLongitude: 120.65175340° W\nElevation: 1390.0 m Mean Sea Level\nand the date, so we can look upNOAA’s magnetic field calculator\nModel Used: WMM2015\nwhat the expected magnetic field should be:\n\nIts direction:\n\nDeclination (+E/-W) Inclination (+D/-U) Horizontal Intensity\n14.7990° ±0.36° 66.5386° ±0.22 20,754.1 nT ±133 nT\n\nAnd as a vector:\n\nNorth Comp (+N/-S) East Comp (+E/-W) Vertical Comp (+D/-U)\n20,065.7 nT ±138 nT 5,301.2 nT ±89 nT 47,819.4 nT ±165 nT\n\nAnd finally, the total strength:\n\nTotal Field\n52,129.0 nT ±152 nT\n\n## Calibration Data\n\nThat’s what we expect to see. What do we actually get?\n\nIn the 22.1 minutes that we had the flight computer collecting data in our calibration run we recoreded 1,079,342 data points from the IMU.\n\nLooking over time at the x, y, z values of the magnetometer and the mangitue compared to the NOAA predicted field we see it vary a lot.\n\nThis is because we have a couple of problems. One is that the effective center of our magnetometer values are pushed off to one side. And the other is that the values are skewed (or “stretched”) off to one side as well. This is somewhat easier to see in 3D:\n\n## Correction\n\nThe two parts of the correction are called “Hard Iron” (fixed center offset) and “Soft Iron” (streched sphere) corrections.\n\n### Hard Iron\n\nThis is the simpler of the two, one can essentially find the midrange value of across the entire calibration dataset and subtract that offset to move the ‘0’ point back to center.\n\n### Soft Iron\n\nFinding the soft iron correction is a bit trickier because we want to fit an matching elongated ellipsoid to the data, and then once we have an approximation for that ellipsoid apply stretch to the data to undo the elongation and get it back to a sphere. Luckily an algorithm for this has been worked out. For a detailed breakdown see\n\nhttps://teslabs.com/articles/magnetometer-calibration/\n\nAfter doing fitting we end up with both a correction matrix and an offset vector. This is both the soft iron and hard iron correction.\n\nTo invert the stretch we multiply a vector representing each magnetometer reading (a ‘sample’, $\\vec s$) by the correction matrix (after subtracting the center offset).\n\n$$\\vec s_\\textrm{corrected} = \\mathbf{A} \\cdot (\\vec s - \\vec b)$$\n\nWhere\n\n• $\\vec s_\\textrm{corrected}$ is a fully corrected sample at time t\n• $\\mathbf{A}$ is our soft iron correction matrix\n• $\\vec s$ is a raw sample from the IMU at time t\n• $\\vec b$ our hard iron offset vector.\n\nWhen we solve for $\\mathbf{A}$ on the calibration data we get the matrix:\n\n$$\\textbf{A} = \\left[\\begin{array}{ccc} 0.870368 & -0.128543 & -0.283684 \\\\ -0.128543 & 1.510386 & -0.046543 \\\\ -0.283684 & -0.046543 & 1.440805 \\end{array}\\right]$$\n\nAnd a hard iron offset vector:\n\n$$\\vec b = \\left[ \\begin{array}{ccc} -12.019415 & -3.209783 & -1.939041 \\end{array}\\right]$$\n\n## Python\n\nIf we want to apply this correction we can make a convienient function to call on all our samples:\n\n## Apply Calibration\n\nAfter we apply the calibration fix above, do we get a better result?\n\nYes! Quite a bit better. Notice how the magnitue of the vector now stays very close to constant and is very close to the NOAA estimate!\n\nAgain in 3D we can see a much closer to spherical data:\n\n## Applying to Flight Data\n\nWe can now take our calibration matrix and apply it to real flight data! Here is a 3D look at the Launch-12 raw (uncalibrated) magnetometer data from liftoff to apogee:\n\nThe rocket spins during flight, and we see the magnetic field measurement spiral around the plots. We also see the familiar stretch and offset that we saw in the calibration data.\n\n### Calibrated Flight Data\n\nSo now lets calibrate the flight data!\n\nAnd it looks like the calibration did a reasonable job. The values now come very close to landing on the nominal Earth field sphere. The XY view is still off a little bit but it might just be that we had some bias in the calibration run. It’s still a huge improvement to the original dataset and it now usable in IMU reconstructions of the flight of the rocket.\n\nThis post is written as a jupyter notebook and all its code and data can be viewed as a stand-alone own project here:\n\nhttps://git.natronics.org/natronics/psas-magnetometer-calibration" ]

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[ "# Nuclear magneton\n\nJump to navigation Jump to search\nThe value of nuclear magneton\nsystem of units value unit\nSI 5.050783699(31)×10−27 J·T−1\nCGS 5.050783699(31)×10−24 Erg·G−1\neV 3.1524512550(15)×10−8 eV·T−1\nMHz/T (per h) 7.622593285(47) MHz/T\n\nThe nuclear magneton (symbol μN), is a physical constant of magnetic moment, defined in SI units by:\n\n$\\mu _{\\mathrm {N} }={{e\\hbar } \\over {2m_{\\mathrm {p} }}}$", null, "and in Gaussian CGS units by:\n\n$\\mu _{\\mathrm {N} }={{e\\hbar } \\over {2m_{\\mathrm {p} }c}}$", null, "where:\n\ne is the elementary charge,\nħ is the reduced Planck constant,\nmp is the proton rest mass, and\nc is the speed of light\n\nIn SI units, its value is approximately:\n\nμN = 5.050783699(31)×10−27 J/T\n\nIn Gaussian CGS units, its value can be given in convenient units as\n\nμN = 0.10515446 efm\n\nThe nuclear magneton is the natural unit for expressing magnetic dipole moments of heavy particles such as nucleons and atomic nuclei.\n\nDue to the fact that neutrons and protons consist of quarks and thus are no real Dirac particles, their magnetic moment differ from $\\mu _{\\mathrm {N} }$", null, ":\n\n$\\mu _{\\mathrm {p} }=2{.}79\\mu _{\\mathrm {N} }$", null, "$\\mu _{\\mathrm {n} }=-1{.}91\\mu _{\\mathrm {N} }$", null, "The magnetic dipole moment of the electron, which is much larger as a consequence of much larger charge-to-mass ratio, is usually expressed in units of the Bohr magneton. The Bohr magneton, which is calculated in the same fashion as the nuclear magneton, is larger than μN by a factor equal to the ratio of the proton to electron mass, or about a factor of 1836." ]

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[ "# Lightweight way of flooring an NSDecimal?\n\nI have an NSDecimal in a tight calculations loop, where I need to floor the value. I want to prevent creating fat NSDecimalNumber objects just for that. Is there a cost-efficient way to get a floor? That floor is just needed to calculate how many times another value might fit in there, with no rest. The NSDecimal API doesn't provide something like floor...\n\n## 2 Answers\n\nYou can use the `NSDecimalRound()` function with the `NSRoundDown` rounding mode:\n\n``````NSDecimal d = ...;\nNSDecimal floored;\n\nNSDecimalRound(&floored, &d, 0, NSRoundDown);\n``````\n\nFor more info take a look at the docs here.\n\n``````NSDecimal result;\nNSDecimalRound(&result, &decimal, 0, NSRoundDown);\n``````\n\n(not tested)" ]

[ null ]

[ "", null, "By Sergei Mihailovic Nikol’skii (auth.)\n\nRead Online or Download Approximation of Functions of Several Variables and Imbedding Theorems PDF\n\nSimilar number systems books\n\nThe Numerical Solution of Differential-Algebraic Systems by Runge-Kutta Methods\n\nThe time period differential-algebraic equation used to be coined to include differential equations with constraints (differential equations on manifolds) and singular implicit differential equations. Such difficulties come up in various functions, e. g. limited mechanical platforms, fluid dynamics, chemical response kinetics, simulation of electric networks, and regulate engineering.\n\nGlobal Smoothness and Shape Preserving Interpolation by Classical Operators\n\nThis monograph examines and develops the worldwide Smoothness renovation estate (GSPP) and the form maintenance estate (SPP) within the box of interpolation of features. The research is constructed for the univariate and bivariate circumstances utilizing recognized classical interpolation operators of Lagrange, Grünwald, Hermite-Fejér and Shepard kind.\n\nConstructive Approximation\n\nCoupled with its sequel, this booklet provides a hooked up, unified exposition of Approximation idea for features of 1 actual variable. It describes areas of services resembling Sobolev, Lipschitz, Besov rearrangement-invariant functionality areas and interpolation of operators. different themes contain Weierstrauss and top approximation theorems, homes of polynomials and splines.\n\nTensor Spaces and Numerical Tensor Calculus\n\nSpecific numerical suggestions are already had to take care of nxn matrices for big n. Tensor facts are of measurement nxnx. .. xn=n^d, the place n^d exceeds the pc reminiscence via a long way. they seem for difficulties of excessive spatial dimensions. for the reason that commonplace equipment fail, a specific tensor calculus is required to regard such difficulties.\n\nAdditional resources for Approximation of Functions of Several Variables and Imbedding Theorems\n\nSample text\n\nThus - IX) u' ~ y II = 1, 1. Because of the uniform convexity E(XiII) = u ~ y ,x~fl) = u':; Y), we get = u'. 7. It is frequently necessary for us to make use of the following fact, related to the theory of functions of a real variable. Suppose that I, Ik E Ip('8) (k = 1, 2, ... oo III - MLp(,ff) = 0 (k -+ 00), (1 ~ P :::;:00). Then there exists a subsequence {k,) of the natural numbers such that (2) lim Ik,(x) = I(x) for almost all x E '8. 3. (l = 1,2, ... ) are finite on ~' and equation (2) is satisfied for all tIC E~.\n\nLI, Obviously (15) 8 n f- (kn In) _ Ck1 - 2S2 -,-. u ence u(x, y) - u 8 (x, y) = r 1 + r2 + ra. (;-, 1]) t(;, 1]) ei(xHYn) d; d17 r1 = (16) 2n LIN = E [k[,[II;£«N ~ 2n (kn In)' Ctlei~(kx+1Y) A. N i7(kIC+lu) 0 S S S S , N being a natural number, and s chosen in such a way that a nytural number: r2 = ~ 2n' f tX = !.... is :n; A. (;, 1]) i (;,1]) ei (xHIIr,) d;-, d1], JR,-LlN ra n };'}, (kn 8 , = --, -In) f~ (kn - , In) - e·i~(kIC+11l) 2S2 S S S S (k,l) Ikl III where the sum };' is extended over pairs such that either or is larger than IXN.\n\nEUJtp(x) f _a(x) , from which we get (1) in view of (3). 2. Periodic functions from = sign co,,(x) n = L;. The functions sin(2\"+1nx) (0;;;:;; x;;;:;; 1), 0, 1, ... , form an orthogonal and normal Rademacher system on [0, 1J. Here and in what follows we will frequently write A ~ B in place of A ;;;:;; cB, where c is a constant. For any double sequence {am\"l of complex numbers and P > the inequalities ° I I (2' lam,,1 2)p/2 ~ J JI2' am\"com(O) co,,(O')IP dO dO' ~ (2' la m,,1 2)p/2 (1) o 0 hold, with constants not depending on the am ..." ]

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[ "", null, "Heyaaaaaa!  Advance Happy  Halloween everyone! Since its my semestral break A.K.A SEMBREAK just like before, I’m going to post some of my programming codes here which was taken previous semester.. here ya go.\n\nPurpose: To display the “First 20 terms of Fibonacci Series.”\n\nSource Code:\n\npackage fibonacci;\n\nimport java.util.*;\n/**\n*\n* @author Christine Cloma\n*/\npublic class Fibonacci {\n\n/**\n* @param args the command line arguments\n*/\npublic static void main(String[] args) {\nScanner console = new Scanner (System.in);\nint n, c, first = 0, second = 1, next;\n\nSystem.out.println( “Enter number for Fibonacci: “);\nn = console.nextInt();\n\nSystem.out.println(“First ” + n + ” terms of Fibonacci series are : ” );\n\nc = 1;\ndo\n{\nnext = first + second;\nfirst = second;\nsecond = next;\n\nSystem.out.print(next+ ” “);\nc++;\n}while(c<n);\n\n}\n}\n\nPS: I used Netbeans IDE 8.0 version. 🙂\n\n-tinay" ]

[ null, "https://cdn.christinecloma.com/wp-content/uploads/2014/10/1.jpg", null ]

[ "# Piecewise\n\nPiecewise[{{val1,cond1},{val2,cond2},}]\n\nrepresents a piecewise function with values vali in the regions defined by the conditions condi.\n\nPiecewise[{{val1,cond1},},val]\n\nuses default value val if none of the condi apply. The default for val is 0.\n\n# Details", null, "• The condi are typically inequalities such as", null, ".\n• The condi are evaluated in turn, until one of them is found to yield True.\n• If all preceding condi yield False, then the vali corresponding to the first condi that yields True is returned as the value of the piecewise function.\n• If any of the preceding condi do not literally yield False, the Piecewise function is returned in symbolic form.\n• Only those vali explicitly included in the returned form are evaluated.\n• Elements of the form {vali,False} are dropped, as are all elements after the first {vali,True}.\n• Piecewise[conds] automatically evaluates to Piecewise[conds,0].\n• Piecewise can be used in such functions as Integrate, Minimize, Reduce, DSolve, and Simplify, as well as their numeric analogs.\n• Piecewise[{{v1,c1},{v2,c2},}] can be input in the form v1 c1 v2 c2 …\n. The piecewise operator can be entered as", null, "pw", null, "or \\[Piecewise]. The grid of values and conditions can be constructed by first entering", null, ", then using", null, "and", null, ".\n• In StandardForm and TraditionalForm, Piecewise[{{v1,c1},{v2,c2},}] is normally output using a brace, as in v1 c1 v2 c2 …\n.\n\n# Examples\n\nopen allclose all\n\n## Basic Examples(3)\n\nSet up a piecewise function with different pieces below and above zero:\n\nFind the derivative of a piecewise function:\n\nUse", null, "pw", null, "to enter and", null, "and then", null, "for each additional piecewise case:\n\n## Scope(12)\n\nDefine a piecewise function:\n\nEvaluate it at specific points:\n\nPlot it:\n\nRefine it under assumptions:\n\nAutomatic simplification of Piecewise functions:\n\nRemove unreachable cases:\n\nRemove False conditions:\n\nMerge cases with the same values:\n\nIf values are not specified in a region, they are assumed to be zero:\n\nThis specifies that the default value should be 1:\n\nCompute limits of piecewise functions:\n\nCompute the limit in the direction of the positive imaginary axis:\n\nCompute the series of a piecewise function:\n\nIntegrate a piecewise function:\n\nIntegration constants are chosen to make the result continuous:\n\nCompute a definite integral of a piecewise function:\n\nLaplace transform of a piecewise function:\n\nSolve a piecewise differential equation:\n\nReduce a piecewise equation:\n\nIntegrating an implicitly piecewise integrand can give an explicit Piecewise result:\n\nSymbolic minimization can give piecewise functions:\n\n## Applications(1)\n\nCompute the volume of an ellipsoid:\n\n## Properties & Relations(11)\n\nPiecewiseExpand converts nested piecewise functions into a single piecewise function:\n\nMin, Max, UnitStep, and Clip are piecewise functions of real arguments:\n\nAbs, Sign, and Arg are piecewise functions when their arguments are assumed to be real:\n\nKroneckerDelta and DiscreteDelta are piecewise functions of complex arguments:\n\nBoole is a piecewise function of a Boolean argument:\n\nIf, Which, and Switch can be interpreted as piecewise functions:\n\nConvert Floor, Ceiling, Round, IntegerPart, and FractionalPart for finite ranges:\n\nConvert Mod and Quotient when the number of cases is finite:\n\nUnitBox and UnitTriangle are piecewise functions of real arguments:\n\nConvert SquareWave, TriangleWave, and SawtoothWave for finite ranges:\n\nBernsteinBasis and BSplineBasis are piecewise functions of real arguments:\n\n## Possible Issues(1)\n\nDerivatives are computed piece-by-piece, unless the function is univariate in a real variable:\n\nTo specify that", null, "is real, use inequalities in the first condition:\n\nThis function is discontinuous at", null, ":" ]

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[ "Meeting Dates\n\nWednesday, September 28, 2022\n\n## Overview\n\nLast class and this class we are looking at the sampling distibution of OLS estimators (particularly $$\\hat{\\beta_1})$$. Last class we looked at what the center of the distribution was - the true $$\\beta_1$$ - so long as the assumptions about $$u$$ hold:\n\n• When $$cor(X,u)=0$$, $$X$$ is exogenous and the OLS estimators are unbiased.\n• What $$cor(X,u)\\neq 0$$, $$X$$ is endogenous and the OLS estimators are biased.\n\nToday we continue looking at the sampling distibution by determining the variation in $$\\hat{beta_1}$$ (it’s variance or its standard error1). We look at the formula and see the three major determinants of variation in $$\\hat{\\beta_1}$$:\n\n1. Goodness of fit of the regression $$(SER$$ or $$\\hat{\\sigma_u}$$\n2. Sample size $$n$$\n3. Variation in $$X$$\n\nWe also look at the diagnostics of a regression by looking at its residuals $$(\\hat{u_i})$$ for anomalies. We focus on the problem of heteroskedasticity (where the variation in $$\\hat{u_i])$$ changes over the range of $$X$$, which violates assumption 2 (errors are homoskedastic): how to detect it, test it, and fix it with some packages. We also look at outliers, which can bias the regression. Finally, we also look at how to present regression results.\n\nWe continue our extended example about class sizes and test scores, which comes from a (Stata) dataset from an old textbook that I used to use, Stock and Watson, 2007. Download and follow along with the data from today’s example:2\n\nI have also made a RStudio Cloud project documenting all of the things we have been doing with this data that may help you when you start working with regressions (next class):\n\n• Finish Ch.3 in Bailey, Real Econometrics\n\n## R Practice\n\nToday (and next class) you will be working on practice problems. Answers will be posted on that page later.\n\n## Appendix\n\nSee the online appendix for today’s content:\n\n## Slides\n\nBelow, you can find the slides in two formats. Clicking the image will bring you to the html version of the slides in a new tab. The lower button will allow you to download a PDF version of the slides.\n\nI suggest printing the slides beforehand and using them to take additional notes in class (not everything is in the slides)!\n\n## Footnotes\n\n1. The square root of variance, as always!↩︎\n\n2. Note this is a .dta Stata file. You will need to (install and) load the package haven to read_dta() Stata files into a dataframe.↩︎" ]

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[ "Lego Robot and Neural Networks\n\nAn overview of using Lego RCX  robots for teaching neural networks present at workshop in 2011.\n\nThe video below shows the robot trying out sets of weights for two neurones, until a set of weights are found that enable the robot to go around the circle.\n\nAs a part of a set of tools I have found the following useful for teaching the principles of simple neurones.\n\nExample code:\n\nimport josx.platform.rcx.*;\n\npublic class annlf{\npublic static void main(String[] args)\n{\nint w[][] ={//put weights here};\nint o[]={1,1};\nint s1,s2,res1,res2;\nint sensor1=0,sensor2=0;\nrobot_1 tom=new robot_1();\nSensor.S1.activate();\nSensor.S3.activate();\nfor(;;){\nLCD.showNumber(sensor1);\nif (sensor1<42)\ns1=1;\nelse\ns1=0;\nif (sensor2<42)\ns2=1;\nelse\ns2=0;\nres1=w*s1+w*s2+w;\nif (res1>=0)\no=1;\nelse\no=0;\nres2=w*s1+w*s2+w;\nif (res2>=0)\no=1;\nelse\no=0;\nif ((o==1)&&(o==1))\ntom.forward1(10);\nif ((o==0)&&(o==0))\ntom.backward1(20);\nif ((o==1)&&(o==0))\ntom.tlturn(20);\nif ((o==0)&&(o==1))\ntom.trturn(20);\nLCD.refresh();\n}\n}\n}\n\nThe example code uses two neurones to produce a line follower. The nice thing about this though is it easy to adapted this for a single neuron or multiple neuron tasks. For more on this some examples can be found here.\nThe above approaches used the Mindstorms RCX robots but it can equally be done with the newer NXT robots\n\nAll opinions in this blog are the Author's and should not in any way be seen as reflecting the views of any organisation the Author has any association with." ]

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[ "DelphiBasics\n Home  |  A dying art - using pointers\n Documents Tutorials Writing your first program Writing your second program Amending this program Delphi data types   Numbers   Text (strings and chars)   Sets and enumerations   Arrays   Records Programming logic   Looping   SubRoutines   Exception handling Dates and times Files Pointers Printing text and graphics Object Orientation basics   Memory leaks!   Inheritance   Abstraction   Interfaces   An example class References Standard components Articles A brief history of Delphi Usability : file handling Usability : reference books Author links\nA dying art - using pointers\nPointers in the World of Object Orientation\nIn the past, especially in languages, pointers were a critical aspect of the language. They allowed direct access to memory, enabling complex data structures to be built and navigated.\n\nHowever, with the advent of Object Orientation, things have changed somewhat. For example, the TStringList class allows a list of strings to be built without the user needing to manage the storage. And hence, removing the need to use pointers to do this.\n\nAdditionally, the Pascal language has also evolved to avoid dynamic storage management by coders - for example with the advent of dynamic arrays. You can now simply code a SetLength call to set or increase the size of such an array as your program runs.\n\nHowever, there are still some situations where pointers are valid in Delphi.\n\nWhat are pointers?\nPointers are a special type of variable. Like a meta-variable. They can point to other variables, or to memory. You might use a record pointer, for example, to point to a block of memory where you have stored lots of record data. You would then use the pointer just as if it were a record variable. We'll see how below.\n\nWhen calling Windows APIs (Application Programmer Interfaces), we are obliged to use pointers.\n\nTypes of pointers\nDelphi provides a number of typed pointer types, such as PChar, and PExtended, along with a generic, 'point to anything' type - the Pointer type.\n\nThe nice thing about the typed pointers is that they work sensibly with the Inc and Dec functions. Incrementing an PInt64 pointer will add SizeOf(Int64) bytes to the pointer address so that it points to the next Int64 variable in memory.\n\nThe Pointer type is a dangerous one - it falls foul of Delphi's normally tight type handling. Use it with care, or you will end up addressing the wrong memory.\n\nA simple example using PChar\nThe PChar type can be used to scan along a string :\n\n ``` var    myString  : string;    myCharPtr : PChar;    i : Integer;    begin   // Create a string of Char's    myString  := 'Hello World';     // Point to the first character in the string    i := 1;    myCharPtr := Addr(myString[i]);     // Display all characters in the string    while i <= Length(myString) do    begin      ShowMessage(myCharPtr^); // Display the string characters one by one      Inc(i);      Inc(myCharPtr);    end;  end; ```\n\nThere are two things to note here. First the use of Addr function to get the address of the string. You could equally use the @ operator. Pointers always work with addresses - the address of a variable here, or a block of acquired memory. Here we point the PChar value to the first character in the string.\n\nSecondly, now that we have a pointer, we use the ^ character at the end of the pointer name to refer to what the pointer points to. In this case, a character.\n\nA PChar^ will always give us a character. A PInt64^, for example, will give us an Int64 value. This is where the typing comes in.\n\nRecord pointers\nYou can define a pointer to any data type using a different technique:\n\n ``` var    myRecordPtr : ^TMyRecord; ```\n\nHere, the ^ symbol is used to dereference the type - we are saying that we do not have a TMyRecord type, but a pointer to one. Note that this is a prefix use of ^.\n\nLet us create a full record example :\n\n ``` type    TMyRecord = Record      name : String;      age  : Integer;    end;    var    myRecord    : TMyRecord;    myRecordPtr : ^TMyRecord;    begin    myRecord.name := 'Fred Bloggs';    myRecord.age  := 23;      myRecordPtr := @myRecord;      ShowMessage(myRecordptr.name); // Displays 'Fred Bloggs'  end; ```\n\nWhen we simpy refer to the record field name, without a ^, Delphi is in fact adding one for us - it recognises what we are doing, and helps us make for more readable code.\n\nA full memory handling example\nIn this example, we'll build a new class that is a limited number list equivalent to the TStringList class. This class will allow you to keep adding numbers to a list of numbers.\n\nThe class uses pointers to help store the numbers in a block of memory, reallocating this block when it is all used up.\n\nFirst off, we will look at the constructor :\n\n ``` var    msCount  : Integer;  // Count of numbers in the list    maxCount : Integer;  // Maximum numbers that can fit into current storage    memStart : Pointer;  // Start of the memory holding the list    nextSlot : PInt64;   // Points to the next free slot in memory    const    ALLOCATE_SIZE = 20;  // How many numbers to store in first memory block    // Constructor - initialise everything  constructor TNumberList.Create;  begin    msCount  := 0;   // No numbers in the list yet     // Allocate space for a limited number of numbers    GetMem(memStart, ALLOCATE_SIZE * SizeOf(Int64));     // Indicate how many numbers that we can add before acquiring more memory    maxCount := ALLOCATE_SIZE;     // And point to the next free memory slot - the first!    nextSlot := memStart;  end; ```\n\nThe role of the constructor is to initialise the class. The key part of this is to allocate a block of memory that can hold 20 numbers. We'll use Int64 numbers (for some reason, Delphi does not provide an Integer pointer).\n\nThe GetMem call allocates storage of the desired size, setting the memStart generalised Pointer variable to the starting address of the memory allocated. Note that GetMem insists on a Pointer variable.\n\nWe'll add a routine to add a value to the memory :\n\n ``` // Add a number to the list  procedure TNumberList.Add(const number : Int64);  begin   // Store the number at the next slot in our memory block    nextSlot^ := number;     // And update things to suit    Inc(msCount);    Inc(nextSlot);  end; ```\n\nThe passed number is stored in the next Int64 slot in our memory block, and this nextSlot pointer incremented. Note that this adds SizeOf(Int64) bytes to the address value in this pointer, because the Inc call knows the type of this pointer.\n\nAnd here is a routine for retrieving a value :\n\n ``` // Get the number at the index position (starting at 0)  function TNumberList.GetValue(index : Integer): Int64;  var    numberPtr : PInt64;  begin   // Simply get the value at the given Int64 index position    numberPtr := memStart;    Inc(numberPtr, index);  // Point to the index'th Int64 number in storage    Result := numberPtr^;   // And get the Int64 number it points to  end; ```\n\nHere we use Inc to add index Int64 size bytes to the start of memory to get to the slot of the required memory.\n\nHowever, we have not yet covered the situation where the memory we allocate is all used up. We will extend the Add routine to do just this :\n\n ``` // Add a number to the list  procedure TNumberList.Add(const number : Int64);  var    newMemoryStart : Pointer;    oldPtr, newPtr : PInt64;    i : Integer;  begin   // If we do not have enough space to add the number, then get more space!    if msCount = maxCount then    begin     // First allocate a bigger memory space      GetMem(newMemoryStart, (maxCount + ALLOCATE_SIZE) * SizeOf(Int64));       // Copy the data from the old memory here      oldPtr := memStart;      newPtr := newMemoryStart;      for i := 1 to maxCount do      begin       // Copy one number at a time        newPtr^ := oldPtr^;        Inc(oldPtr);        Inc(newPtr);      end;       // Free the old memory      FreeMem(memStart);       // And now refer to the new memory      memStart := newMemoryStart;      nextSlot := memStart;      Inc(nextSlot, maxCount);      Inc(maxCount, ALLOCATE_SIZE);    end;     // Now we can safely add the number to the list    nextSlot^ := number;     // And update things to suit    Inc(msCount);    Inc(nextSlot);  end; ```\n\nHere we abandon our old memory block (Delphi cannot let us extend the size of it), and create a bigger one. Having allocated it, we must copy the old memory contents to it. Here we see a new concept - assigning the value referred by one pointer to the contents of memory pointed to by another. Delphi knows to copy the whole Int64 value rather than just one byte because these are PInt64 pointers.\n\nBelow is the full code of the class :\n\n ``` unit NumberList;    interface    type    TNumberList = class      private      msCount  : Integer;  // Count of numbers in the list      maxCount : Integer;  // Maximum numbers that can fit into current storage      memStart : Pointer;  // Start of the memory holding the list      nextSlot : PInt64;   // Points to the next free slot in memory        function    GetValue(index : Integer) : Int64;      public      property    Items[index : Integer] : Int64          read    GetValue; default;  // Default means we can use the list[i]      published      constructor Create;      destructor  Destroy; override;      procedure   Add(const number : Int64);        property    Count : Integer          read    msCount;    end;    implementation    const    ALLOCATE_SIZE = 20;  // How many numbers to store in first memory block    // Constructor - initialise everything  constructor TNumberList.Create;  begin    msCount  := 0;   // No numbers in the list yet     // Allocate space for a limited number of numbers    GetMem(memStart, ALLOCATE_SIZE * SizeOf(Int64));     // Indicate how many numbers that we can add before acquiring more memory    maxCount := ALLOCATE_SIZE;     // And point to the next free memory slot - the first!    nextSlot := memStart;  end;    // Destructor - release storage obtained  destructor TNumberList.Destroy;  begin   // Free the allocated memory    FreeMem(memStart);     // Call TObject destructor    inherited;  end;    // Add a number to the list  procedure TNumberList.Add(const number : Int64);  var    newMemoryStart : Pointer;    oldPtr, newPtr : PInt64;    i : Integer;  begin   // If we do not have enough space to add the number, then get more space!    if msCount = maxCount then    begin     // First allocate a bigger memory space      GetMem(newMemoryStart, (maxCount + ALLOCATE_SIZE) * SizeOf(Int64));       // Copy the data from the old memory here      oldPtr := memStart;      newPtr := newMemoryStart;      for i := 1 to maxCount do      begin       // Copy one number at a time        newPtr^ := oldPtr^;        Inc(oldPtr);        Inc(newPtr);      end;       // Free the old memory      FreeMem(memStart);       // And now refer to the new memory      memStart := newMemoryStart;      nextSlot := memStart;      Inc(nextSlot, maxCount);      Inc(maxCount, ALLOCATE_SIZE);    end;     // Now we can safely add the number to the list    nextSlot^ := number;     // And update things to suit    Inc(msCount);    Inc(nextSlot);  end;    // Get the number at the index position (starting at 0)  function TNumberList.GetValue(index : Integer): Int64;  var    numberPtr : PInt64;  begin   // Simply get the value at the given Int64 index position    numberPtr := memStart;    Inc(numberPtr, index);  // Point to the index'th Int64 number in storage    Result := numberPtr^;   // And get the Int64 number it points to  end;    end. ```\n\nAnd here is how the code could be used :\n\n ``` var    list  : TNumberList;    value : Int64;    i     : Integer;  begin   // Create a number list object    list := TNumberList.Create;     // Add the first 30 even numbers to the list, each doubled in size    for i := 0 to 29 do      list.Add(i * 2);     // Get the 22nd value = 44 (22 * 2)    value := list;    ShowMessage('22nd value = '+IntToStr(value));  end; ```" ]

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[ "## Do My Contact Triangle Class", null, "A \"Contact Triangle Class\" QE\" is a standard mathematical term for a generalized continuous expression which is used to fix differential formulas and has options which are routine. In differential Class solving, a Contact Triangle function, or \"quad\" is used.\n\nThe Contact Triangle Class in Class type can be expressed as: Q( x) = -kx2, where Q( x) are the Contact Triangle Class and it is a crucial term. The q part of the Class is the Contact Triangle constant, whereas the x part is the Contact Triangle function.\n\nThere are four Contact Triangle functions with appropriate solution: K4, K7, K3, and L4. We will now take a look at these Contact Triangle functions and how they are resolved.\n\nK4 - The K part of a Contact Triangle Class is the Contact Triangle function. This Contact Triangle function can likewise be written in partial portions such as: (x2 - y2)/( x+ y). To fix for K4 we increase it by the proper Contact Triangle function: k( x) = x2, y2, or x-y.\n\nK7 - The K7 Contact Triangle Class has a service of the form: x4y2 - y4x3 = 0. The Contact Triangle function is then multiplied by x to get: x2 + y2 = 0. We then need to increase the Contact Triangle function with k to get: k( x) = x2 and y2.\n\nK3 - The Contact Triangle function Class is K3 + K2 = 0. We then increase by k for K3.\n\nK3( t) - The Contact Triangle function equationis K3( t) + K2( t). We increase by k for K3( t). Now we multiply by the Contact Triangle function which gives: K2( t) = K( t) times k.\n\nThe Contact Triangle function is also known as \"K4\" because of the initials of the letters K and 4. K means Contact Triangle, and the word \"quad\" is noticable as \"kah-rab\".\n\nThe Contact Triangle Class is among the primary techniques of solving differential equations. In the Contact Triangle function Class, the Contact Triangle function is first multiplied by the suitable Contact Triangle function, which will offer the Contact Triangle function.\n\nThe Contact Triangle function is then divided by the Contact Triangle function which will divide the Contact Triangle function into a real part and a fictional part. This provides the Contact Triangle term.\n\nLastly, the Contact Triangle term will be divided by the numerator and the denominator to get the quotient. We are left with the right hand side and the term \"q\".\n\nThe Contact Triangle Class is an essential principle to understand when solving a differential Class. The Contact Triangle function is just one technique to solve a Contact Triangle Class. The approaches for resolving Contact Triangle formulas consist of: singular worth decay, factorization, optimal algorithm, mathematical solution or the Contact Triangle function approximation.\n\n## Hire Someone To Do Your Contact Triangle Class\n\nIf you want to become knowledgeable about the Quartic Class, then you require to very first start by looking through the online Quartic page. This page will show you how to use the Class by utilizing your keyboard. The explanation will likewise show you how to develop your own algebra equations to assist you study for your classes.\n\nPrior to you can comprehend how to study for a Contact Triangle Class, you need to first understand making use of your keyboard. You will find out how to click on the function keys on your keyboard, as well as how to type the letters. There are 3 rows of function keys on your keyboard. Each row has four functions: Alt, F1, F2, and F3.\n\nBy pressing Alt and F2, you can multiply and divide the worth by another number, such as the number 6. By pressing Alt and F3, you can utilize the 3rd power.\n\nWhen you press Alt and F3, you will enter the number you are attempting to multiply and divide. To increase a number by itself, you will press Alt and X, where X is the number you wish to multiply. When you press Alt and F3, you will type in the number you are attempting to divide.\n\nThis works the exact same with the number 6, except you will just enter the two digits that are six apart. Lastly, when you push Alt and F3, you will use the 4th power. Nevertheless, when you push Alt and F4, you will use the real power that you have actually discovered to be the most appropriate for your issue.\n\nBy utilizing the Alt and F function keys, you can increase, divide, and after that utilize the formula for the third power. If you need to increase an odd number of x's, then you will require to go into an even number.\n\nThis is not the case if you are attempting to do something complex, such as multiplying two even numbers. For instance, if you wish to increase an odd number of x's, then you will require to get in odd numbers. This is especially true if you are attempting to determine the answer of a Contact Triangle Class.\n\nIf you wish to convert an odd number into an even number, then you will need to push Alt and F4. If you do not know how to multiply by numbers on their own, then you will require to use the letters x, a b, c, and d.\n\nWhile you can increase and divide by use of the numbers, they are a lot easier to use when you can look at the power tables for the numbers. You will need to do some research when you initially start to utilize the numbers, but after a while, it will be force of habit. After you have produced your own algebra formulas, you will have the ability to develop your own reproduction tables.\n\nThe Contact Triangle Formula is not the only way to resolve Contact Triangle formulas. It is necessary to learn about trigonometry, which utilizes the Pythagorean theorem, and after that use Contact Triangle formulas to fix issues. With this technique, you can understand about angles and how to fix problems without needing to take another algebra class.\n\nIt is important to try and type as quickly as possible, because typing will help you learn about the speed you are typing. This will assist you write your responses quicker.\n\n## Hire Someone To Take My Contact Triangle Class", null, "A Contact Triangle Class is a generalization of a linear Class. For example, when you plug in x=a+b for a given Class, you get the value of x. When you plug in x=a for the Class y=c, you obtain the worths of x and y, which provide you a result of c. By applying this standard concept to all the formulas that we have attempted, we can now solve Contact Triangle formulas for all the worths of x, and we can do it quickly and effectively.\n\nThere are lots of online resources offered that provide complimentary or budget-friendly Contact Triangle equations to solve for all the worths of x, including the cost of time for you to be able to benefit from their Contact Triangle Class task aid service. These resources generally do not need a membership charge or any kind of financial investment.\n\nThe responses offered are the outcome of complex-variable Contact Triangle formulas that have been fixed. This is likewise the case when the variable used is an unknown number.\n\nThe Contact Triangle Class is a term that is an extension of a linear Class. One benefit of using Contact Triangle equations is that they are more basic than the linear equations. They are easier to resolve for all the worths of x.\n\nWhen the variable used in the Contact Triangle Class is of the type x=a+b, it is simpler to fix the Contact Triangle Class since there are no unknowns. As a result, there are fewer points on the line defined by x and a continuous variable.\n\nFor a right-angle triangle whose base points to the right and whose hypotenuse indicate the left, the right-angle tangent and curve graph will form a Contact Triangle Class. This Class has one unknown that can be discovered with the Contact Triangle formula. For a Contact Triangle Class, the point on the line specified by the x variable and a consistent term are called the axis.\n\nThe presence of such an axis is called the vertex. Since the axis, vertex, and tangent, in a Contact Triangle Class, are a provided, we can discover all the values of x and they will sum to the provided worths. This is attained when we use the Contact Triangle formula.\n\nThe aspect of being a consistent factor is called the system of formulas in Contact Triangle formulas. This is in some cases called the main Class.\n\nContact Triangle formulas can be fixed for other worths of x. One way to fix Contact Triangle equations for other worths of x is to divide the x variable into its factor part.\n\nIf the variable is provided as a favorable number, it can be divided into its element parts to get the normal part of the variable. This variable has a magnitude that is equal to the part of the x variable that is a consistent. In such a case, the formula is a third-order Contact Triangle Class.\n\nIf the variable x is unfavorable, it can be divided into the exact same part of the x variable to get the part of the x variable that is multiplied by the denominator. In such a case, the formula is a second-order Contact Triangle Class.\n\nService help service in resolving Contact Triangle equations. When utilizing an online service for solving Contact Triangle formulas, the Class will be fixed instantly." ]

[ null, "https://mathexamination.com/Do-My-Math-Class.webp", null, "https://mathexamination.com/Take-My-Math-Class.webp", null ]

[ "### Merge branch 'topic/default/mpi4py-fft-pyperf' into 'branch/default'\n\n```Rename perf -> pyperf, updated mpi4py-fft API\n\nSee merge request !30```\nPipeline #12794 passed with stage\nin 5 minutes and 31 seconds\n ... ... @@ -9,30 +9,30 @@ perf: perfpython perfcython perfpythran perfsimd perfomp perfpython: # python python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import grad_py as g, f_fft, KX, KY' 'g(f_fft, KX, KY)' perfcython: # cython with @cython.boundscheck(False) @cython.wraparound(False) python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import grad_cy_nocheck as g, f_fft, KX, KY' 'g(f_fft, KX, KY)' perfcythoncheck: # cython without @cython.boundscheck(False) @cython.wraparound(False) python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import grad_cy_check as g, f_fft, KX, KY' 'g(f_fft, KX, KY)' perfpythran: # pythran python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import grad_pythran as g, f_fft, KX, KY' 'g(f_fft, KX, KY)' perfsimd: # SIMD python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import grad_simd as g, f_fft, KX, KY' 'g(f_fft, KX, KY)' perfomp: # OpenMP python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import grad_omp as g, f_fft, KX, KY' 'g(f_fft, KX, KY)'\n TIMEIT:=python -m perf timeit --fast -s # TIMEIT:=python -m perf timeit -q -s TIMEIT:=python -m pyperf timeit --fast -s # TIMEIT:=python -m pyperf timeit -q -s NPROC:=2 FFT:='from bench import fft as transform;' ... ...\n import sys import numpy as np import perf import pyperf from mpi4py import MPI from mpi4py_fft.mpifft import PFFT, Function from mpi4py_fft.mpifft import PFFT from mpi4py_fft.distarray import newDistArray def init2d(N=1024, slab=True): ... ... @@ -10,17 +11,19 @@ def init2d(N=1024, slab=True): def init3d(N=128, slab=True): return PFFT(MPI.COMM_WORLD, (N, N, N), slab=slab, axes=(0, 1, 2), dtype=np.float) return PFFT( MPI.COMM_WORLD, (N, N, N), slab=slab, axes=(0, 1, 2), dtype=np.float ) def create_arrayX(o): u = Function(o, False) u = newDistArray(o, False) u[:] = np.random.random(u.shape).astype(u.dtype) return u def create_arrayK(o): u_hat = Function(o, True) u_hat = newDistArray(o, True) u_hat[:] = np.random.random(u_hat.shape).astype(u_hat.dtype) return u_hat ... ... @@ -31,4 +34,3 @@ def fft(o, u, u_hat): def ifft(o, u, u_hat): o.backward(u_hat, u)\n ... ... @@ -39,14 +39,14 @@ openmp/\\$(name)_openmp\\$(extbin): \\$(name).py # perf: # ## numpy # python -m perf timeit -s '\\$(setup) from \\$(name) import myfunc' 'myfunc(a)' # python -m pyperf timeit -s '\\$(setup) from \\$(name) import myfunc' 'myfunc(a)' # ## default # python -m perf timeit -s '\\$(setup) from \\$(name)_default import myfunc' 'myfunc(a)' # python -m pyperf timeit -s '\\$(setup) from \\$(name)_default import myfunc' 'myfunc(a)' # ## native # python -m perf timeit -s '\\$(setup) from \\$(name)_native import myfunc' 'myfunc(a)' # python -m pyperf timeit -s '\\$(setup) from \\$(name)_native import myfunc' 'myfunc(a)' # ## native_openmp # python -m perf timeit -s '\\$(setup) from \\$(name)_native_openmp import myfunc' 'myfunc(a)' # python -m pyperf timeit -s '\\$(setup) from \\$(name)_native_openmp import myfunc' 'myfunc(a)' # ## openmp # python -m perf timeit -s '\\$(setup) from \\$(name)_openmp import myfunc' 'myfunc(a)' # python -m pyperf timeit -s '\\$(setup) from \\$(name)_openmp import myfunc' 'myfunc(a)' # ## simd # python -m perf timeit -s '\\$(setup) from \\$(name)_simd import myfunc' 'myfunc(a)' # python -m pyperf timeit -s '\\$(setup) from \\$(name)_simd import myfunc' 'myfunc(a)'\n ... ... @@ -28,34 +28,34 @@ end=as func, c0, c1, c2, a0, a1, a2, a3 perf: perfpython perfnative perfnative: python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import proj_native \\$(end)' \\$(code) perfdefault: python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import proj_default \\$(end)' \\$(code) perfdefault1: python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import proj1_default \\$(end)' \\$(code) perfdefault2: python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import proj2_default \\$(end)' \\$(code) perffft: python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import proj_fft \\$(end)' \\$(code) perfomp: python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import proj_omp \\$(end)' \\$(code) OMP_NUM_THREADS=1 python -m perf timeit -s \\ OMP_NUM_THREADS=1 python -m pyperf timeit -s \\ 'from bench import proj_omp \\$(end)' \\$(code) \\ --inherit-environ=OMP_NUM_THREADS perfpython: python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import proj_py \\$(end)' \\$(code) perffortran: bench_proj_fortran.out ... ...\n ... ... @@ -5,21 +5,21 @@ See the files proj.py, proj1.py and proj2.py pierre@pierre-KTH:~/Dev/fluidfft/bench/compare_projperpk3d\\$ make perfdefault python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import proj_default as func, arr_c, arr' 'func(arr_c, arr_c, arr_c, arr, arr, arr, arr)' ..................... Mean +- std dev: 23.8 ms +- 1.3 ms pierre@pierre-KTH:~/Dev/fluidfft/bench/compare_projperpk3d\\$ make perfdefault1 python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import proj1_default as func, arr_c, arr' 'func(arr_c, arr_c, arr_c, arr, arr, arr, arr)' ..................... Mean +- std dev: 26.6 ms +- 1.1 ms pierre@pierre-KTH:~/Dev/fluidfft/bench/compare_projperpk3d\\$ make perfdefault2 python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import proj2_default as func, arr_c, arr' 'func(arr_c, arr_c, arr_c, arr, arr, arr, arr)' ..................... Mean +- std dev: 23.9 ms +- 1.5 ms ... ... @@ -30,7 +30,7 @@ Mean +- std dev: 23.9 ms +- 1.5 ms Small difference but it is reproducible... pierre@pierre-KTH:~/Dev/fluidfft/bench/compare_projperpk3d\\$ make perfnative python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import proj_native as func, arr_c, arr' 'func(arr_c, arr_c, arr_c, arr, arr, arr, arr)' ..................... Mean +- std dev: 24.8 ms +- 1.3 ms ... ...\n ... ... @@ -26,43 +26,43 @@ code='func(arr)' perfpython2d: all ## numpy 2d (no loop) @python -m perf timeit -s \\ @python -m pyperf timeit -s \\ 'from bench import myfunc_py as func, f2d as arr' \\$(code) perfdefault2d: all ## default 2d (no loop) @python -m perf timeit -s \\ @python -m pyperf timeit -s \\ 'from bench import myfunc_default as func, f2d as arr' \\$(code) # default 2d (explicit loops) @python -m perf timeit -s \\ @python -m pyperf timeit -s \\ 'from bench import myfunc_loops2d_default as func, f2d as arr' \\$(code) perfsimd2d: \\$(name)_simd.so # simd 2d (no loop) @python -m perf timeit -s \\ @python -m pyperf timeit -s \\ 'from bench import myfunc_simd as func, f2d as arr' \\$(code) # simd 2d (explicit loops) @python -m perf timeit -s \\ @python -m pyperf timeit -s \\ 'from bench import myfunc_loops2d_simd as func, f2d as arr' \\$(code) perfpython3d: all ## numpy 3d (no loop) @python -m perf timeit -s \\ @python -m pyperf timeit -s \\ 'from bench import myfunc_py as func, f3d as arr' \\$(code) perfdefault3d: all # default 3d (no loop) @python -m perf timeit -s \\ @python -m pyperf timeit -s \\ 'from bench import myfunc_default as func, f3d as arr' \\$(code) # default 3d (explicit loops) @python -m perf timeit -s \\ @python -m pyperf timeit -s \\ 'from bench import myfunc_loops3d_default as func, f3d as arr' \\$(code) perfsimd3d: all # simd 3d (no loop) @python -m perf timeit -s \\ @python -m pyperf timeit -s \\ 'from bench import myfunc_simd as func, f3d as arr' \\$(code) # simd 3d (explicit loops) @python -m perf timeit -s \\ @python -m pyperf timeit -s \\ 'from bench import myfunc_loops3d_simd as func, f3d as arr' \\$(code)\n ... ... @@ -3,32 +3,32 @@ # Result on Pierre's machine at LEGI (Intel(R) Xeon(R) CPU E5-1603 v3 @ 2.80GHz, 4 cores). ## numpy python -m perf timeit -s 'import numpy as np; a = np.ones([1000, 1000]); from mymod import myfunc' 'myfunc(a)' python -m pyperf timeit -s 'import numpy as np; a = np.ones([1000, 1000]); from mymod import myfunc' 'myfunc(a)' ..................... Mean +- std dev: 47.6 ms +- 0.2 ms ## default python -m perf timeit -s 'import numpy as np; a = np.ones([1000, 1000]); from mymod_default import myfunc' 'myfunc(a)' python -m pyperf timeit -s 'import numpy as np; a = np.ones([1000, 1000]); from mymod_default import myfunc' 'myfunc(a)' ..................... Mean +- std dev: 5.15 ms +- 0.01 ms ## native python -m perf timeit -s 'import numpy as np; a = np.ones([1000, 1000]); from mymod_native import myfunc' 'myfunc(a)' python -m pyperf timeit -s 'import numpy as np; a = np.ones([1000, 1000]); from mymod_native import myfunc' 'myfunc(a)' ..................... Mean +- std dev: 5.15 ms +- 0.01 ms ## native_openmp python -m perf timeit -s 'import numpy as np; a = np.ones([1000, 1000]); from mymod_native_openmp import myfunc' 'myfunc(a)' python -m pyperf timeit -s 'import numpy as np; a = np.ones([1000, 1000]); from mymod_native_openmp import myfunc' 'myfunc(a)' ..................... Mean +- std dev: 1.41 ms +- 0.11 ms ## openmp python -m perf timeit -s 'import numpy as np; a = np.ones([1000, 1000]); from mymod_openmp import myfunc' 'myfunc(a)' python -m pyperf timeit -s 'import numpy as np; a = np.ones([1000, 1000]); from mymod_openmp import myfunc' 'myfunc(a)' ..................... Mean +- std dev: 1.40 ms +- 0.06 ms ## simd python -m perf timeit -s 'import numpy as np; a = np.ones([1000, 1000]); from mymod_simd import myfunc' 'myfunc(a)' python -m pyperf timeit -s 'import numpy as np; a = np.ones([1000, 1000]); from mymod_simd import myfunc' 'myfunc(a)' ..................... Mean +- std dev: 5.16 ms +- 0.01 ms ... ... @@ -38,7 +38,7 @@ see no effect of the flags -march=native or -DUSE_BOOST_SIMD. Why? # Update (2018-10-01) The following are the results, in Ashwin's laptop after a `sudo python -m perf system tune`. The following are the results, in Ashwin's laptop after a `sudo python -m pyperf system tune`. (Intel(R) Core(TM) i7-5500U CPU @ 2.40GHz, 2 cores). ### branch master (6f2a8f6) ... ... @@ -86,7 +86,7 @@ Mean +- std dev: 6.07 ms +- 0.07 ms # default 2d (explicit loops) Mean +- std dev: 4.45 ms +- 0.14 ms # simd 2d (no loop) python -m perf timeit -s \\ python -m pyperf timeit -s \\ Mean +- std dev: 5.49 ms +- 0.14 ms # simd 2d (explicit loops) Mean +- std dev: 4.40 ms +- 0.15 ms ... ...\n ... ... @@ -13,22 +13,22 @@ and CXX and CC env variables. \\$ make perf3d # default 3d (no loop) python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import myfunc_default as func, f3d as arr' 'func(arr)' Mean +- std dev: 25.1 ms +- 1.0 ms # default 3d (explicit loops) python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import myfunc_loops3d_default as func, f3d as arr' 'func(arr)' Mean +- std dev: 18.2 ms +- 1.1 ms # simd 3d (no loop) python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import myfunc_simd as func, f3d as arr' 'func(arr)' Mean +- std dev: 20.5 ms +- 1.1 ms # simd 3d (explicit loops) python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import myfunc_loops3d_simd as func, f3d as arr' 'func(arr)' Mean +- std dev: 23.0 ms +- 1.2 ms ... ... @@ -37,21 +37,21 @@ Mean +- std dev: 23.0 ms +- 1.2 ms \\$ make perf3d # default 3d (no loop) python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import myfunc_default as func, f3d as arr' 'func(arr)' Mean +- std dev: 32.6 ms +- 2.1 ms # default 3d (explicit loops) python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import myfunc_loops3d_default as func, f3d as arr' 'func(arr)' Mean +- std dev: 17.7 ms +- 0.6 ms # simd 3d (no loop) python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import myfunc_simd as func, f3d as arr' 'func(arr)' Mean +- std dev: 31.7 ms +- 0.4 ms # simd 3d (explicit loops) python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import myfunc_loops3d_simd as func, f3d as arr' 'func(arr)' Mean +- std dev: 17.7 ms +- 0.8 ms\n ... ... @@ -2,8 +2,8 @@ name=mymod nameso=\\$(name)_thran.so # TIMEIT=python -m perf timeit --fast -q -s TIMEIT=python -m perf timeit -q -s # TIMEIT=python -m pyperf timeit --fast -q -s TIMEIT=python -m pyperf timeit -q -s # PYTHRAN=pythran -march=native -DUSE_BOOST_SIMD PYTHRAN=pythran ... ...\n ... ... @@ -29,102 +29,102 @@ arrays=a0, a1, a2, a3, a4, a5 perf: perfpython perfdefault perfdefault1 perfsimd perfnative: python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import \\$(name)_native as func, \\$(arrays)' \\$(code) perfdefault: python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import \\$(name)_default as func, \\$(arrays)' \\$(code) perffft: python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import \\$(name)_fft as func, \\$(arrays)' \\$(code) perfomp: python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import \\$(name)_omp as func, \\$(arrays)' \\$(code) OMP_NUM_THREADS=1 python -m perf timeit -s \\ OMP_NUM_THREADS=1 python -m pyperf timeit -s \\ 'from bench import \\$(name)_omp as func, \\$(arrays)' \\$(code) \\ --inherit-environ=OMP_NUM_THREADS perfpython: python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import \\$(name)_py as func, \\$(arrays)' \\$(code) perfsimd: python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import \\$(name)_simd as func, \\$(arrays)' \\$(code) perfsimd1: python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import \\$(name)1_simd as func, \\$(arrays)' \\$(code) perfsimd2: python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import \\$(name)2_simd as func, \\$(arrays)' \\$(code) perfsimd3: python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import \\$(name)3_simd as func, \\$(arrays)' \\$(code) perfnative1: python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import \\$(name)1_native as func, \\$(arrays)' \\$(code) perfdefault1: python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import \\$(name)1_default as func, \\$(arrays)' \\$(code) perfomp1: python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import \\$(name)1_omp as func, \\$(arrays)' \\$(code) OMP_NUM_THREADS=1 python -m perf timeit -s \\ OMP_NUM_THREADS=1 python -m pyperf timeit -s \\ 'from bench import \\$(name)1_omp as func, \\$(arrays)' \\$(code) \\ --inherit-environ=OMP_NUM_THREADS perfpython1: python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import \\$(name)1_py as func, \\$(arrays)' \\$(code) perfnative2: python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import \\$(name)2_native as func, \\$(arrays)' \\$(code) perfdefault2: python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import \\$(name)2_default as func, \\$(arrays)' \\$(code) perfomp2: python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import \\$(name)2_omp as func, \\$(arrays)' \\$(code) OMP_NUM_THREADS=1 python -m perf timeit -s \\ OMP_NUM_THREADS=1 python -m pyperf timeit -s \\ 'from bench import \\$(name)2_omp as func, \\$(arrays)' \\$(code) \\ --inherit-environ=OMP_NUM_THREADS perfpython2: python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import \\$(name)2_py as func, \\$(arrays)' \\$(code) perfnative3: python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import \\$(name)3_native as func, \\$(arrays)' \\$(code) perfdefault3: python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import \\$(name)3_default as func, \\$(arrays)' \\$(code) perfomp3: python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import \\$(name)3_omp as func, \\$(arrays)' \\$(code) OMP_NUM_THREADS=1 python -m perf timeit -s \\ OMP_NUM_THREADS=1 python -m pyperf timeit -s \\ 'from bench import \\$(name)3_omp as func, \\$(arrays)' \\$(code) \\ --inherit-environ=OMP_NUM_THREADS perfpython3: python -m perf timeit -s \\ python -m pyperf timeit -s \\ 'from bench import \\$(name)2_py as func, \\$(arrays)' \\$(code) ... ...\nMarkdown is supported\n0% or .\nYou are about to add 0 people to the discussion. Proceed with caution.\nFinish editing this message first!" ]

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[ "## 57561\n\n57,561 (fifty-seven thousand five hundred sixty-one) is an odd five-digits composite number following 57560 and preceding 57562. In scientific notation, it is written as 5.7561 × 104. The sum of its digits is 24. It has a total of 3 prime factors and 8 positive divisors. There are 32,880 positive integers (up to 57561) that are relatively prime to 57561.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Odd\n• Number length 5\n• Sum of Digits 24\n• Digital Root 6\n\n## Name\n\nShort name 57 thousand 561 fifty-seven thousand five hundred sixty-one\n\n## Notation\n\nScientific notation 5.7561 × 104 57.561 × 103\n\n## Prime Factorization of 57561\n\nPrime Factorization 3 × 7 × 2741\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 57561 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 57,561 is 3 × 7 × 2741. Since it has a total of 3 prime factors, 57,561 is a composite number.\n\n## Divisors of 57561\n\n1, 3, 7, 21, 2741, 8223, 19187, 57561\n\n8 divisors\n\n Even divisors 0 8 4 4\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 87744 Sum of all the positive divisors of n s(n) 30183 Sum of the proper positive divisors of n A(n) 10968 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 239.919 Returns the nth root of the product of n divisors H(n) 5.24809 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 57,561 can be divided by 8 positive divisors (out of which 0 are even, and 8 are odd). The sum of these divisors (counting 57,561) is 87,744, the average is 10,968.\n\n## Other Arithmetic Functions (n = 57561)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 32880 Total number of positive integers not greater than n that are coprime to n λ(n) 8220 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 5822 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares\n\nThere are 32,880 positive integers (less than 57,561) that are coprime with 57,561. And there are approximately 5,822 prime numbers less than or equal to 57,561.\n\n## Divisibility of 57561\n\n m n mod m 2 3 4 5 6 7 8 9 1 0 1 1 3 0 1 6\n\nThe number 57,561 is divisible by 3 and 7.\n\n## Classification of 57561\n\n• Arithmetic\n• Deficient\n\n• Polite\n\n• Square Free\n\n### Other numbers\n\n• LucasCarmichael\n• Sphenic\n\n## Base conversion (57561)\n\nBase System Value\n2 Binary 1110000011011001\n3 Ternary 2220221220\n4 Quaternary 32003121\n5 Quinary 3320221\n6 Senary 1122253\n8 Octal 160331\n10 Decimal 57561\n12 Duodecimal 29389\n20 Vigesimal 73i1\n36 Base36 18ex\n\n## Basic calculations (n = 57561)\n\n### Multiplication\n\nn×i\n n×2 115122 172683 230244 287805\n\n### Division\n\nni\n n⁄2 28780.5 19187 14390.2 11512.2\n\n### Exponentiation\n\nni\n n2 3313268721 190715060849481 10977749617556975841 631890245736197086383801\n\n### Nth Root\n\ni√n\n 2√n 239.919 38.6109 15.4893 8.95418\n\n## 57561 as geometric shapes\n\n### Circle\n\n Diameter 115122 361666 1.04089e+10\n\n### Sphere\n\n Volume 7.98865e+14 4.16358e+10 361666\n\n### Square\n\nLength = n\n Perimeter 230244 3.31327e+09 81403.5\n\n### Cube\n\nLength = n\n Surface area 1.98796e+10 1.90715e+14 99698.6\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 172683 1.43469e+09 49849.3\n\n### Triangular Pyramid\n\nLength = n\n Surface area 5.73875e+09 2.2476e+13 46998.4\n\n## Cryptographic Hash Functions\n\nmd5 e2e0c7c57d6094dd987599a151c9d443 222d5a9f26f2d3c26cd4d038c35ebfc3160ac9f8 7dea4523d27e7768e1ecef6d17924e55e776a2d576f5757e3cca5bb1faa8d327 776a4de80e1f40a174abdb8615f4436b64cbdd67fbb55d4492520e7fa40227e4bcda0844e091156cb7343e663cb4010f62299b7aa39c056cabe0efa0e13371f5 3320226439dbb5666bbd80ed0b33867dbb5b010a" ]

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[ "# Symmetry\n\nHere we will learn about symmetry, including line and rotational symmetry properties within polygons, angle properties, and symmetry of different line graphs.\n\nThere are also symmetry worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.\n\n## What is symmetry?\n\nSymmetry is when a line is drawn through a shape to make one side of the line a reflection of the other. It is a property of a 2D polygon or 3D polyhedron.\n\nThere are two different types of symmetry that you need to be aware of: lines of symmetry and rotational symmetry.\n\nAlthough the two sound similar, they are fundamentally different.\n\n### What is symmetry?", null, "### Lines of symmetry\n\nThe number of lines of symmetry for a shape can be determined by using a ruler to visualise when the polygon can be divided equally into 2 equal pieces that are a reflection of each other.\n\nE.g.\n\nHow many lines of symmetry does a rectangle have?\n\nHere, the pink and yellow sections are congruent to each other and are symmetrical to each other.\n\nThe number of lines of symmetry for any rectangle (excluding a square, which is a special rectangle), is 2 .\n\nStep-by-step guide: Lines of symmetry\n\n### Rotational symmetry\n\nThe rotational symmetry of a shape is worked out using the centre of the polygon. We need to determine where the centre of the polygon is using the diagonals as they intersect at the centre of the shape. Once we know the centre, we need to rotate the polygon around its centre to determine the order of rotation.\n\nTracing paper is very useful to determine the order of rotation of a polygon because you can trace the shape and rotate the tracing around the centre, not affecting the original.\n\nE.g.\n\nWhat is the order of rotational symmetry of a rectangle?\n\nAfter rotating the rectangle 180^o , the image is an exact copy of the original. After a further rotation of 180^o , we get back to the original orientation.\n\nThis means that the order of rotational symmetry for the rectangle (excluding a square, which is a special rectangle) is 2 .\n\nStep-by-step guide: Rotational symmetry\n\n## How to use symmetry\n\nIn order to draw lines of symmetry:\n\n1. Locate the centre of the 2D shape.\n2. Use a ruler to visualise a horizontal and/or vertical line of symmetry through the centre of the shape.\n3. Continue to rotate the ruler around 180 degrees over the centre point to cover all sides and vertices.\n\nIn order to calculate the order of rotational symmetry:\n\n1. Locate the centre of the 2D shape.\n2. Trace the shape onto a piece of tracing paper including the centre and northline.\n3. Rotate the tracing around the centre and count the number of identical occurrences.\n\n### Explain how to use symmetry", null, "## Symmetry examples\n\n### Example 1: the square (lines of symmetry)\n\nDraw all of the lines of symmetry for the square below.\n\n1. Locate the centre of the 2D shape.\n\nDraw a small x in the centre of the square (this does not have to be exact)\n\n2Use a ruler to visualise a horizontal and/or vertical line of symmetry through the centre of the shape.\n\nHere we can draw a vertical line as this divides the shape into two identical rectangles (one is a reflection of the other), and a horizontal line as this divides the shape into two congruent rectangles that are a reflection of each other.\n\n3Continue to rotate the ruler around 180 degrees over the centre point to cover all sides and vertices.\n\nAs the shape has an even number of vertices, we can pass through opposing vertices to see whether there is a line of symmetry present. For the square, we can draw a further two lines of symmetry (along the diagonals of the square).\n\nThe square has 4 lines of symmetry.\n\n### Example 2: using angles (lines of symmetry)\n\nShow that the hexagon below has no lines of symmetry.\n\nLocate the centre of the 2D shape.\n\nUse a ruler to visualise a horizontal and/or vertical line of symmetry through the centre of the shape.\n\nContinue to rotate the ruler around 180 degrees over the centre point to cover all sides and vertices.\n\n### Example 3: quadratic graph (lines of symmetry)\n\nWrite the equation of the line of symmetry for the quadratic equation y=x^2-8x+12 .\n\nLocate the centre of the 2D shape.\n\nUse a ruler to visualise a horizontal and/or vertical line of symmetry through the centre of the shape.\n\nContinue to rotate the ruler around 180 degrees over the centre point to cover all sides and vertices.\n\n### Example 4: the regular hexagon (rotational symmetry)\n\nCalculate the order of rotational symmetry for a regular pentagon.\n\nLocate the centre of the 2D shape.\n\nTrace the shape onto a piece of tracing paper including the centre and northline.\n\nRotate the tracing around the centre and count the number of identical occurrences.\n\n### Example 5: angle facts (rotational symmetry)\n\nThe rhombus ABCD has two pairs of parallel sides. Use angle facts to calculate the order of rotation for the shape ABCD.\n\nLocate the centre of the 2D shape.\n\nTrace the shape onto a piece of tracing paper including the centre and northline.\n\nRotate the tracing around the centre and count the number of identical occurrences.\n\n### Example 6: cubic graph (rotational symmetry)\n\nCalculate the order of rotational symmetry for the cubic graph y=x^3+2 around the centre (0,0) .\n\nLocate the centre of the 2D shape.\n\nTrace the shape onto a piece of tracing paper including the centre and northline.\n\nRotate the tracing around the centre and count the number of identical occurrences.\n\n### Common misconceptions\n\n• Rotational symmetry/lines of symmetry\n\nLines of symmetry are mixed up with rotational symmetry. A line of symmetry divides the shape equally into two symmetrical pieces. A rotational symmetry is the number of times a shape fits into itself when rotated around its centre.\n\nThe diagonals of a quadrilateral are joined together and highlighted as a line of symmetry. The only quadrilateral where this is true is the square. Below is the example for an incorrect line of symmetry for a rectangle.\n\n•  The number of sides = the number of lines of symmetry\n\nAlthough this is true for regular shapes, this is not true for all shapes.\n\n• Rotational symmetry of order \\pmb{0}\n\nA shape that has an order of rotational symmetry of 1 can also be said to have an order of 0 , but 1 or “no rotational symmetry” are better descriptions.\n\n• The number of sides = the order of rotational symmetry\n\nAlthough this is true for regular shapes, this is not true for all shapes.\n\n### Practice lines of symmetry questions\n\n1. Calculate the number of lines of symmetry for the regular pentagon below.", null, "5", null, "1", null, "0", null, "2", null, "", null, "2. Calculate the number of lines of symmetry for the triangle ABC", null, "0", null, "1", null, "2", null, "3", null, "", null, "3. State the equation of the line of symmetry for the graph y=6x-x^2", null, "y=3", null, "y=3x", null, "x=3", null, "y=x+3", null, "", null, "4. Calculate the order of rotational symmetry for the rhombus below.", null, "0", null, "1", null, "2", null, "4", null, "", null, "5. The two lines AB and CD are parallel. Calculate the order of rotational symmetry for the following shape ABCD.", null, "1", null, "2", null, "3", null, "4", null, "", null, "6. Calculate the order of rotational symmetry for the graph of y=tan(x) for -90 < \\theta < 90^o around the centre ( 0 , 0) .", null, "0", null, "1", null, "2", null, "4", null, "", null, "When rotated 180^o , the image is the same as the original.\n\n### Symmetry GCSE questions\n\n1. Complete the table to state the order of rotation for the following shapes\n\nShape                         Order of Rotation\n\nSquare\n\nIsosceles Triangle\n\nParallelogram\n\nRegular Octagon\n\n(4 marks)\n\nShape                         Order of Rotation\n\nSquare \\hspace{3cm} 4\n\nIsosceles Triangle \\hspace{1.6cm} 0 or 1\n\nParallelogram \\hspace{2.1cm} 2\n\nRegular Octagon \\hspace{1.7cm} 8\n\n(4)\n\n2. (a) Complete table of values for the cubic y=x^3.\n\n(b) James says “the graph of y=x^3 has a line of symmetry at x=0. ” Is he correct? Explain your answer.\n\n(c) What is the order of rotational symmetry for the line y=x^3 ?\n\n(6 marks)\n\n(a)\n\n(1)\n\n(b)\n\nNo\n\n(1)\n\ny is positive when x>0 and y is negative when x<0 so they are not symmetrical\n\n(1)\n\n(c)\n\n2\n\n(1)\n\n3. Calculate the equation of the line of symmetry for the line y=4x+1 at the point (4,17).\n\n(3 marks)\n\n(1)\n\ny=-0.25x+c at (2,9) means 17=-0.25 \\times 4+c, \\; c=18\n\n(1)\n\ny=-0.25x+18\n\n(1)\n\n## Learning checklist\n\nYou have now learned how to:\n\n• Identify lines of symmetry in 2-D shapes presented in different orientations\n\n## Still stuck?\n\nPrepare your KS4 students for maths GCSEs success with Third Space Learning. Weekly online one to one GCSE maths revision lessons delivered by expert maths tutors.\n\nFind out more about our GCSE maths tuition programme." ]

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[ "# Iterators in Ruby\n\n`There are different kinds of iterators provided by ruby a few of them are`\n\n## Each Iterator\n\n`The each iterator is used to iterate through the elements of the array.`\n\n``````arr = [1, 2, 3, 4, 5]\narr.each { |a| print a -= 10, \" \" }\n# prints: -9 -8 -7 -6 -5\n``````\n\n`The reverse each iterator, iterates through the list in reverse order.`\n\n``````words = %w[first second third fourth fifth sixth]\nstr = \"\"\nwords.reverse_each { |word| str += \"#{word} \" }\np str #=> \"sixth fifth fourth third second first \"\n``````\n\n`each_with_index` iterate over the elements in the array along with index.\n\n``````arr = [1, 2, 3, 4, 5]\narr.each_with_index { |val,index| puts \"index: #{index},value:#{val}\"}\n#output\nindex:0,value:1\nindex:1,value:2\nindex:2,value:3\nindex:3,value:4\nindex:4,value:5\n``````\n\nEach Iterator can also be used with Hashes\n\n``````h = { \"a\" => 100, \"b\" => 200 }\nh.each {|key, value| puts \"#{key} is #{value}\" }\n#output\na is 100\nb is 200\n\nh.each_key {|key| puts key }\n#output\na\nb\n\nh.each_value {|value| puts value }\n#output\n100\n200\n``````\n\n## Times Iterator\n\n`This iterator is used to iterate through the array from 0 to n-1 index positio`\n\n``````5.times do |i|\nprint i, \" \"\nend\n#OUTPUT\n#=> 0 1 2 3 4\n``````\n\n## Upto and Downto Iterators\n\n`This iterator is used to iterate from value of n to the limit`(including limit)\n\n``````5.upto(10) { |i| print i, \" \" }\n#=> 5 6 7 8 9 10\n``````\n\n`This iterator is used to iterate decreasing values from n to the limit`(including limit)\n\n``````5.downto(1) { |n| print n, \".. \" }\n#=> \"5.. 4.. 3.. 2.. 1..\n``````\n\n## Step Iterator\n\n`The step iterator invokes a block which increments by the value of step with each iteration till the condition mentioned becomes false.`\n\n``````1.step(10, 2) { |i| print i, \" \" }\n#output\n1 3 5 7 9 => 1\n\nMath::E.step(Math::PI, 0.2) { |f| print f, \" \" }\n#output\n2.718281828459045 2.9182818284590453 3.118281828459045 => 2.718281828459045\n``````\n\n## Each_Line Iterator\n\nSplits str using the supplied parameter as the record separator (\\$/ by default), passing each substring in turn to the supplied block. If a zero-length record separator is supplied, the string is split into paragraphs delimited by multiple successive newlines.\n\n``````print \"Example one\\n\"\n\"hello\\nworld\".each_line {|s| p s}\nprint \"Example two\\n\"\n\"hello\\nworld\".each_line('l') {|s| p s}\nprint \"Example three\\n\"\n\"hello\\n\\n\\nworld\".each_line('') {|s| p s}\n#OUTPUT\nExample one\n\"hello\\n\"\n\"world\"\nExample two\n\"hel\"\n\"l\"\n\"o\\nworl\"\n\"d\"\nExample three\n\"hello\\n\\n\\n\"\n\"world\"\n``````\n``` =begin Ruby program to use iterators to generate the multiples of two and three INPUT Enter the number of terms 6 OUTPUT Multiples of both two and three are : [6, 12, 18, 24, 30, 36] =end # storing the console message in variable msg = \"Enter the number of terms\" # printing the msg to the console puts msg # reading the input from the console n = gets.chomp.to_i puts \"Multiples of both two and three are : \" # initilising an array arr = Array.new # using 'upto' iterator for accessing the values upto n 1.upto(n) do |val| arr.push(6*val) end # printing the output p arr ```" ]

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[ "Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.\n\nOur last post examined the correspondence between a logistic regression and a simple neural network using a sigmoid activation function. The downside with such models is that they only produce binary outcomes. While we argued (not very forcefully) that if investing is about assessing the probability of achieving an attractive risk-adjusted return, then it makes sense to model investment decisions as probability functions. Moreover, most practitioners would probably prefer to know whether next month’s return is likely to be positive and how confident they should be in that prediction. They want to get direction right first. That’s binary.\n\nBut what about magnitude and relative performance? Enter multiclass logistic regression and neural networks with the softmax activation function. Typically multiclass (or multinomial) classifications are used to distinguish categories like Pekingese from Poodles, or Shih-tzus. With a bit of bucketing, one can do the same with continuous variables like stock returns. This may reduce noise somewhat, but also gets at the heart of investing: the shape of returns. Most folks who’ve been around the markets for a while know that returns are not normally distributed, particularly for stocks.1 Sure, they’re fat-tailed and often negatively skewed. Discussed less frequently is that these returns also cluster around the mean. In other words, there are far more days of utter boredom, sheer terror, or unbearable ecstasy2 than implied by the normal distribution.\n\nWhile everyone wants to knock the ball out of the park and avoid the landmines, those events are difficult to forecast given their rarity even for the fat-tailed distribution. Sustained performance is likely to come from compounding in the solid, but hardly heroic area above the mean return near the one sigma boundary. These events are more frequent and also offer more data with which to work. Indeed, since 1970 almost 40% of the S&P 500’s daily returns fell into that area vs. the expectation of only 34% based on the normal distribution.\n\nHow would be go about classifying the probability of different returns? First, we’d bucket the returns according to some quantile and then run the regression or neural network on those buckets to get the predictions. Multiclass logistic regressions use the softmax function which looks like the following:\n\n$Softmax(k, x_{1}, ..,, x_{n}): \\Large\\frac{e^{x_{k}}}{\\sum_{i=1}^{n}e^{x_{i}}}$\n\n$f(k) = \\begin{cases} 1 \\text{ if } k = argmax(x_{1},…,x_{n}) \\\\ 0 \\text{ othewise} \\end{cases}$\n\nHere $$x_{k}$$ is whatever combination of weights and biases with the independent variable that yields the maximum value for a particular class. Thus that value over the sum of all the exponentiated values yields the model’s likelihood for a particular category. How exactly does this happen? A multiclass logistic regression aggregates individual logistic regressions for the probability of each category with respect to all the other categories. Then it uses the fact that the probabilities must sum to one to yield the softmax function above. The intuition is relatively straightforward: how often do we see one class vs. all the rest based on some data? Check out the appendix for a (slightly!) more rigorous explanation.\n\nLet’s get to our data, split it into the four quartiles for simplicity and then run a logistic regression on those categories. Recall we’re using the monthly return on the S&P500 as well as the return on the 10-month moving average vs. the one-month forward return, which we transform into the four buckets: below -1.9%, between -1.9% and 0.5%, between 0.5% and 3.6%, and above 3.6%. We’ll dub these returns with the following technical terms: Stinky, Poor, Mediocre, Good\n\nNow we’ll run the regressions and present the confusion matrix as before.", null, "Whoa dude, is this some weird sudoku? Not really. Still, a four-by-four confusion matrix isn’t the easiest to read. Bear with us as we explain. The diagonal from the upper left to the lower right contains the true positives. The rows associated with each true positive are the false positives while the columns are the false negatives. The true negative is essentially all the other cells that don’t appear in either the row or the column for the particular category. So the true negative for the Stinky category would be the sum of the 3×3 matrix whose top left corner starts in the cell of the second row and second column.\n\nOrganizing the data this way is better than nothing, but it could be more insight provoking. We can see that the multiclass logistic regression is not that strong in the Poor category, but is much better in the Good category. Let’s compute the true positive and false positive rates along with the precision for each category, which we show in the table below.\n\nTable 1: Logistic regression scores (%)\nOutcome TPR FPR Precision\nStinky 29.7 32.1 23.7\nPoor 3.2 3.1 25.0\nMediocre 42.9 26.7 34.6\nGood 45.3 31.1 33.0\n\nEven with the refinement this still isn’t the easiest table to interpret. The model’s best true positive rate (TPR) performance is in the Good category, but it’s false positive rate (FPR) is also one of the highest. Shouldn’t we care about being really wrong? True, we don’t want to be consistently wrong. But we really don’t want to believe we’re going to generate a really good return and end up with a really Stinky one.\n\nWe’ll create a metric called the Really Wrong Rate (RWR) which will be the number of really wrong classifications over the total classifications the model made for that category. For example, for the Stinky outcome, the model got 19 correct (the top left box) but got 20 really wrong (the top right box). Thus its RWR is about 25% and its Precision to RWR is about 0.95. In other words, it’s getting more categories really wrong than correct.\n\nWhat about the others? For the Good category, its RWR rate is almost 30% and its Precision to RWR is 1.12. That seems like a good edge. The model is really right about 12% more than it’s really wrong. However, when if you look at correctly identified categories plus incorrectly identified positive returns (the Good and Mediocre columns) relative to incorrectly identified categories with negative returns (the Stinky and Poor columns), the results are much worse: a ratio of about 0.83. Got that? Let’s move on to the neural network, before we get bogged down in confusion matrix navel gazing.\n\nUnlike the linear regression or binary logistic regression, a simple neural network can’t approximate a multiclass logistic regression easily. Indeed, it takes a bit of wrangling to get something close, so we won’t bore you with all the iterations. We will show one or two just so you know how much effort we put in!\n\nFirst, a single layer perceptron, with four output neurons, and a softmax activation function. We graph the model’s accuracy over 100 epochs along with logistic regression accuracy—the red dotted line—below.", null, "Not very inspiring. When we include a hidden layer with 20 neurons (an entirely arbitrary number!), but the same parameters as before, we get the following graph.", null, "Definitely better. We see that around the 37th epoch, the NN’s accuracy converges with the logistic regression, but it does bounce around above and below that for the remaining epochs.\n\nLet’s build a deeper NN, this time with three layers of 20 neurons each. For graphical purposes, the architecture of this NN looks like the following:", null, "Plotting accuracy by epoch, gives us the next graph:", null, "This denser NN achieves a better accuracy than the logistic regression sooner than the others, and it’s outperformance persists longer. Of course, even when it does perform worse, it isn’t that dramatic—about one or two percentage points.\n\nWe’ll stop the model at the first point that it converges with the logistic regression, which happens to the be the tenth epoch and create the confusion matrix is below.", null, "Again, a bit of struggle to discern much from this table, other than the model does not appear to predict Mediocre returns with much frequency even though they represent about 25% of the occurrences. Let’s look at some of the scores.\n\nTable 2: Neural network scores (%)\nOutcome TPR FPR Precision\nStinky 25.0 14.7 36.4\nPoor 30.2 26.7 27.1\nMediocre 6.3 3.1 40.0\nGood 59.4 48.4 29.2\n\nThe NN model has a much better true positive rate (TPR) than the logistic regression for Good outcomes, but the false positive rate (FPR) is high too. While the NN model is worse than the regression model on Stinky outcomes, it’s FPR is less than half that of the regression. Interestingly, both models are poor at predicting one category: Poor outcomes for the regression model vs. Mediocre outcomes for the NN. We’re not sure why that would be the case.\n\nThe Stinky RWR is about 18% yielding a Precision to RWR ratio of 2 on the Stinky outcomes, much better than the logistic regressions ratio of 0.95. In other words, the NN is twice as likely to predict a Stinky outcome correctly than incorrectly predict a Good outcome as a Stinky one.\n\nShould we favor the logistic regression over a NN with softmax activation? Hard to say. Accuracy results are similar, but we did have to play with the NN a lot more. More than accuracy, once you look under the hood, results diverge. The greater flexibility of the NN might help us tune a model to arrive at our desired true positive, false positive, and or really wrong rates. But that might come at the risk of overfitting. Moreover, it’s still unclear why both models picked one category as less likely even though the each category should have had an equal chance of occurring. We’d need to engage in more data analysis to figure out what we’re missing.\n\nWhat are some other avenues we could explore? We could build denser neural networks. We could backtest the predictions to gauge performance. This would probably be done best using walk-forward analysis. Of course, a drawback with this is that few people were using neural networks to run trading algorithms during the 70s and 80s so the results could be spurious. That is, if people had been employing such algorithms, returns could have been a lot different. Another avenue is to change the bucketing mechanism so that we focus on the range of outcomes we’re most interested in or would be most likely to achieve high risk-adjusted returns.\n\nWe’ll leave those musings for now. Let us know what you’d like to read by sending an email to the address below. Want more on multiclass regressions and softmax functions? Or should we explore if neural networks can approximate decision trees and random forests? Let us know! Until then, have a look at the Appendix after the code. It walks through the link between logistic and softmax functions. And by all means, have a look at the code!\n\n Built using R 4.0.3, and Python 3.8.3\n\n# [R]\nsuppressPackageStartupMessages({\nlibrary(tidyverse)\nlibrary(tidyquant)\nlibrary(reticulate)\n})\n\n# [Python]\nimport warnings\nwarnings.filterwarnings('ignore')\nimport numpy as np\nimport pandas as pd\nimport statsmodels.api as sm\nimport matplotlib\nimport matplotlib.pyplot as plt\nimport os\nos.environ['QT_QPA_PLATFORM_PLUGIN_PATH'] = 'C:/Users/user_name/Anaconda3/Library/plugins/platforms'\n\nplt.style.use('ggplot')\nplt.rcParams['figure.figsize'] = (12,6)\n\n# Directory to save images\n\nDIR = \"your/image/directory\"\n\ndef save_fig_blog(fig_id, tight_layout=True, fig_extension=\"png\", resolution=300):\npath = os.path.join(DIR, fig_id + \".\" + fig_extension)\nprint(\"Saving figure\", fig_id)\nif tight_layout:\nplt.tight_layout()\nplt.savefig(path, format=fig_extension, dip=resolution)\n\n## Pull data and split\n# See past posts for code\n\ndata = sp_mon.dropna()\n\nX_train = data.loc[:'1991', ['ret', '10ma_ret']]\ny_train = data.loc[:'1991', '1_mon_ret']\n\nX_valid = data.loc['1991':'2000', ['ret', '10ma_ret']]\ny_valid = data.loc['1991':'2000', '1_mon_ret']\n\nX_test = data.loc['2001':, ['ret', '10ma_ret']]\ny_test = data.loc['2001':, '1_mon_ret']\n\ny_train_trans = pd.qcut(y_train, 4, labels=[0, 1, 2, 3])\n\n# Modest search for best solvers and regularization hyperparameters\nfrom sklearn.linear_model import LogisticRegression\n\nsolvers = ['lbfgs', 'newton-cg', 'sag', 'saga']\n\nfor solver in solvers:\nlog_reg = LogisticRegression(penalty='l2', solver = solver, multi_class='multinomial')\nlog_reg.fit(X_train, y_train_trans)\nlog_pred = log_reg.predict(X_train)\nprint(log_reg.score(X_train, y_train_trans))\n\nCs = [10.0**-x for x in np.arange(-2,3)]\n\nfor c in Cs:\nlog_reg = LogisticRegression(penalty='l2', multi_class='multinomial', C=c)\nlog_reg.fit(X_train, y_train_trans)\nprint(log_reg.score(X_train, y_train_trans))\n\nlog_reg = LogisticRegression(penalty='l2', multi_class='multinomial')\nlog_reg.fit(X_train, y_train_trans)\nlog_reg.score(X_train, y_train_trans)\n\n## Sigmoid on four classes\n# Not shown\nkeras.backend.clear_session()\nnp.random.seed(42)\ntf.random.set_seed(42)\n\ny_train_cat = keras.utils.to_categorical(y_train_trans)\n\nmodel = keras.models.Sequential([\nkeras.layers.Dense(4, activation='sigmoid',input_shape = X_train.shape[1:])\n])\n\nmodel.compile(loss='binary_crossentropy', optimizer='sgd', metrics=['accuracy'])\n\nhistory = model.fit(X_train, y_train_cat, epochs = 100)\n\n## Softmax on four clases\nkeras.backend.clear_session()\nnp.random.seed(42)\ntf.random.set_seed(42)\n\nmodel = keras.models.Sequential([\nkeras.layers.Dense(4, activation='softmax')\n])\n\nmodel.compile(loss='categorical_crossentropy', optimizer='sgd', metrics=['accuracy'])\nhistory = model.fit(X_train, y_train_cat, epochs=100)\n\n# Graph accuracy\n\nlog_score = log_reg.score(X_train, y_train_trans)\nlog_hist_df = pd.DataFrame(history.history)\nlog_hist_df.index = np.arange(1, len(log_hist_df)+1)\nlog_hist_df['accuracy'].plot(style='b-')\nplt.axhline(log_score, color='red', ls = ':')\nplt.xticks(np.arange(0,len(log_hist_df)+1, 10))\nplt.xlabel('Epoch')\nplt.ylabel('Accuracy')\nplt.title(\"Neural network training error by epoch\")\nplt.legend(['Neural network', 'Logistic regression'], loc='upper left', bbox_to_anchor=(0.0, 0.9))\nsave_fig_blog('nn_vs_log_reg_tf3')\nplt.show()\n\n## Softmax with one hidden layer\nkeras.backend.clear_session()\nnp.random.seed(42)\ntf.random.set_seed(42)\n\nmodel = keras.models.Sequential([\nkeras.layers.Dense(20, activation='relu', input_shape=X_train.shape[1:]),\nkeras.layers.Dense(4, activation='softmax')\n])\n\nmodel.compile(loss='categorical_crossentropy', optimizer='sgd', metrics=['accuracy'])\nhistory = model.fit(X_train, y_train_cat, epochs=100)\n\n# Graph\nlog_score = log_reg.score(X_train, y_train_trans)\nlog_hist_df = pd.DataFrame(history.history)\nlog_hist_df.index = np.arange(1, len(log_hist_df)+1)\nlog_hist_df['accuracy'].plot(style='b-')\nplt.axhline(log_score, color='red', ls = ':')\nplt.xticks(np.arange(0,len(log_hist_df)+1, 10))\nplt.xlabel('Epoch')\nplt.ylabel('Accuracy')\nplt.title(\"Neural network training error by epoch\")\nplt.legend(['Neural network', 'Logistic regression'], loc='upper left')\nsave_fig_blog('nn_vs_log_reg_2_tf3')\nplt.show()\n\n## Three hidden layer architecture\nfrom nnv import NNV\n\nlayersList = [\n{\"title\":\"Input\\n\", \"units\": 2, \"color\": \"blue\"},\n{\"title\":\"Hidden 1\\n(ReLU)\", \"units\": 20},\n{\"title\":\"Hidden 2\\n(ReLU)\", \"units\": 20},\n{\"title\":\"Hidden 3\\n(ReLU)\", \"units\": 20},\n{\"title\":\"Output\\n(Softmax)\", \"units\": 4,\"color\": \"blue\"},\n]\n\nplt.show()\n\n## Softmax with three hidden layers\nkeras.backend.clear_session()\nnp.random.seed(42)\ntf.random.set_seed(42)\n\nmodel = keras.models.Sequential()\nfor layer in range(2):\n\nmodel.compile(loss='categorical_crossentropy', optimizer='sgd', metrics=['accuracy'])\nhistory = model.fit(X_train, y_train_cat, epochs=100)\n\nlog_score = log_reg.score(X_train, y_train_trans)\nlog_hist_df = pd.DataFrame(history.history)\nlog_hist_df.index = np.arange(1, len(log_hist_df)+1)\nlog_hist_df['accuracy'].plot(style='b-')\nplt.axhline(log_score, color='red', ls = ':')\nplt.xticks(np.arange(0,len(log_hist_df)+1, 10))\nplt.xlabel('Epoch')\nplt.ylabel('Accuracy')\nplt.title(\"Neural network training accuracy by epoch\")\nplt.legend(['Neural network', 'Logistic regression'], loc='upper left')\nsave_fig_blog('nn_vs_log_reg_3_tf3')\nplt.show()\n\n## Rerun stopping early\nkeras.backend.clear_session()\nnp.random.seed(42)\ntf.random.set_seed(42)\n\nmodel = keras.models.Sequential()\nfor layer in range(2):\n\nmodel.compile(loss='categorical_crossentropy', optimizer='sgd', metrics=['accuracy'])\nhistory = model.fit(X_train, y_train_cat, epochs=10)\n\n## Create confusion matrix table function\n# Help from SO\n# https://stackoverflow.com/questions/50666091/true-positive-rate-and-false-positive-rate-tpr-fpr-for-multi-class-data-in-py/50671617\n\nfrom sklearn.metrics import confusion_matrix\n# from sklearn.metrics import precision_score, recall_score, roc_curve\n\ndef conf_mat_table(predicted, actual, title = 'Logistic regression', save=False, save_title = None, print_metrics=False):\n\nconf_mat = confusion_matrix(y_true=predicted, y_pred=actual)\n\nfig, ax = plt.subplots(figsize=(14,8))\nax.matshow(conf_mat, cmap=plt.cm.Blues, alpha=0.3)\nfor i in range(conf_mat.shape):\nfor j in range(conf_mat.shape):\nax.text(x=j, y=i, s=conf_mat[i, j], fontsize=14, va='center', ha='center')\n\nax.xaxis.set_ticks_position('top')\nax.xaxis.set_label_position('top')\nax.set_xticklabels(['','Stinky', 'Poor','Mediocre', 'Good'], fontsize=14)\nax.set_yticklabels(['', 'Stinky', 'Poor', 'Mediocre', 'Good'], fontsize=14, rotation=90)\nax.set_xlabel('Actual returns', fontsize=16)\nax.set_ylabel('Predicted returns', fontsize=16)\nax.set_title(title + ' confusion matrix', pad=40, fontsize=20)\n\nlines = [0.5, 1.5, 2.5]\nfor line in lines:\nplt.axhline(line, color='grey')\nplt.axvline(line, color='grey')\n\nplt.grid(False)\n\nif save:\nsave_fig_blog(save_title)\nplt.show()\n\nif print_metrics:\nFP = conf_mat.sum(axis=1) - np.diag(conf_mat)\nFN = conf_mat.sum(axis=0) - np.diag(conf_mat)\nTP = np.diag(conf_mat)\nTN = conf_mat.sum() - (FP + FN + TP)\n\n# Add 1e-10 to prevent division by zero\n# True positive rate\ntpr = TP/(TP+FN+1e-10)\n# Precision\nprecision = TP/(TP+FP+1e-10)\n# False positive rate\nfpr = FP/(FP+TN+1e-10)\n\nprint(\"\")\ntab = pd.DataFrame(np.c_[['Stinky', 'Poor','Mediocre', 'Good'], tpr, fpr, precision],\ncolumns = ['Outcome', 'TPR', 'FPR', 'Precision']).set_index(\"Outcome\")\n\ntab = tab.apply(pd.to_numeric)\n\nreturn tab, conf_mat\n\n## Predictions\nlog_pred = log_reg.predict(X_train)\nnn = model.predict(X_train)\nnn_pred = np.argmax(nn, axis=1)\n\n## Logistic regression table\n\ntab1, conf_mat1 = conf_mat_table(log_pred, y_train_trans, save=False, title=\"Multiclass logistic regression\",\nprint_metrics=True, save_title='log_reg_conf_mat_1_tf3')\n\n# Save to csv for blog\ndir1 = \"your/wd\"\nfolder = \"/your_folder/\"\ntab1.to_csv(dir1+folder+'tab1_tf3.csv')\n\nconf_mat1 = pd.DataFrame(data = conf_mat1, index=pd.Series(['Stinky', 'Poor','Mediocre', 'Good'], name='Outcome'),\ncolumns=['Stinky', 'Poor','Mediocre', 'Good'])\nconf_mat1.to_csv(dir1+folder+'/conf_mat1_tf3.csv')\n\n# [R]\n# Rmarkdown table\n# Asssumes we're in the directory to which we saved all those csvs.\n\ntab1 %>%\nmutate_at(vars(\"TPR\", \"FPR\", \"Precision\"),\nfunction(x) format(round(x,3)*100,nsmallest=0)) %>%\nknitr::kable(caption = \"Logistic regression scores (%)\")\n\nstinky_rwr <- as.numeric(conf_mat1[1,5]/sum(conf_mat1[1,2:5]))\ngood_rwr <- as.numeric(conf_mat1[4,2]/sum(conf_mat1[4,2:5]))\n\nstinky_prec <- as.numeric(tab1[1,4])\ngood_prec <- as.numeric(tab1[4,4])\ngood_2_Stinky <- (sum(conf_mat1[4,4:5])/sum(conf_mat1[4,2:3]))\n\n# [Python]\n## Neutral network table\ntab2, conf_mat2 = conf_mat_table(nn_pred, y_train_trans, title=\"Neural network\", save=False, print_metrics=True, save_title='nn_conf_mat_1_tf3')\n\n# Save to csv for blog\ntab2.to_csv(dir1+folder+'/tab2_tf3.csv')\n\nconf_mat2 = pd.DataFrame(data = conf_mat2, index=pd.Series(['Stinky', 'Poor','Mediocre', 'Good'], name='Outcome'),\ncolumns=['Stinky', 'Poor','Mediocre', 'Good'])\nconf_mat2.to_csv(dir1[:-13]+'/conf_mat2_tf3.csv')\nconf_mat2.to_csv(dir1+folder+'/conf_mat2_tf3.csv')\n\n# [R]\n# Rmarkdown table\n# Neural network\n# Asssumes we're in the directory to which we saved all those csvs.\n\ntab2 %>%\nmutate_at(vars(\"TPR\", \"FPR\", \"Precision\"), function(x) format(round(x,3)*100,nsmallest=0)) %>%\nknitr::kable(caption = \"Neural network scores\")\n\nstinky_rwr1 <- as.numeric(conf_mat2[1,5]/sum(conf_mat2[1,2:5]))\ngood_rwr1 <- as.numeric(conf_mat2[4,2]/sum(conf_mat2[4,2:5]))\nstinky_prec1 <- as.numeric(tab2[1,4])\ngood_prec1 <- as.numeric(tab2[4,4])\ngood_2_Stinky1 <- (sum(conf_mat2[4,4:5])/sum(conf_mat2[4,2:3]))\n\n## Appendix3\n\nWhile we think the intuition behind the softmax function is relatively straightforward, deriving it is another matter. Our search has revealed simplistic discussions or mathematical derivations that require either a lot of formulas with matrix notation or several formulas and a hand wave. While our aim is not to write tutorials, we do think it can be helpful to provide more rigor on complex topics. In general, we find most of the information out there on the main ML algorithms either math-averse or math-enchanted. There’s got to be a happy medium: one that gives you enough math to understand the nuances in a more formal way, but not so much that you need to have aced partial differential equations without ever having taken ordinary ones.4 Here’s our stab at this.\n\nLet’s refresh our memories on softmax and logistic functions:\n\n$Softmax(k, x_{1}, ..,, x_{n}): \\Large\\frac{e^{x_{k}}}{\\sum_{i=1}^{n}e^{x_{i}}}$\n\n$f(k) = \\begin{cases} 1 \\text{ if } k = argmax(x_{1},...,x_{n}) \\\\ 0 \\text{ othewise} \\end{cases}$ $\\\\ Logisitic: \\Large\\frac{e^{x}}{1 + e^{x}} \\\\$\n\nThe logic behind the softmax function is as follows. Suppose you have a bunch of data that you think might predict various classes or categories of something of interest. For example, tail, ear, and snout length for a range of different dog breeds. First you encode the labels (e.g., breeds) as categorical variables (essentially integers). Then you build a neural network and apply weights and biases to the features (e.g. tail, etc). You then compare the output of the neural network against each label. But you need to transform the outputs into something that will tell you that a tail, ear, and snout of lengths x, y, and z are more likely to belong to a Poodle than a Pekingese. All you’ll get from applying weights and biases to all the features are bunches of numbers with lots of decimal places. You need some way to “squash” the data into a probabilistic range to say which breed is more likely that another given an input. Enter the softmax function.\n\nLook at the formula and forget Euler’s constant ($$e$$) for a second. If you can do that, you should be able to see the softmax function as a simple frequency calculation: the number of Shih-tzus over the total number of Shih-tzus, Pekingese, and Poodles in the data set. Now recall that we’re actually dealing with lots of weights and biases applied to the variables, which could have different orders of magnitude once we’re finished with all the calculations. By raising Euler’s constant to the power of the final output, we set all values to an equivalent base, and we force the largest output to be much further away from all the others, unless there’s a tie. This has the effect of pushing the model to chose one class more than all the others.5 Not so soft after all! Once we’ve transformed the outputs, we can them compare each one to the sum of the total to get the probability for each breed. The output with the highest probability that corresponds one of the breeds means that the model believes it’s that particular breed. We can then check against the actual breed, calculate a loss function, backpropagate, and rerun.\n\nBut how does this relate to the logistic function other than that both use $$e$$?\n\nIf perform some algebra on the logistic function we get:6\n\n$$\\Large e^{x} = \\frac{y}{1-y}$$\n\nThus $$e^{x}$$ is the probability of some outcome over the probability of not that outcome. Hence, each $$e^{x}$$ is equivalent to saying here’s the probability this a Pekingese vs. Not a Pekingese given the features. If we recognize that Not a Pekingese is essentially all other breeds, then we can see how to get to the numerator in the softmax function. To transform this into a probability distribution in which all probabilities must sum to one, we sum all the $$e^{x}$$s. That’s the softmax denominator: $$\\sum_{i=1}^{n}e^{x_{i}}$$.\n\nWhew! For such a rough explanation that was pretty long-winded. Can we try to derive the softmax function from the logistic function mathematically? One of the easiest ways is to reverse it using a two case scenario.\n\nIf we only have Pekingese ($$e^{x_{p}}$$) and Shih-tzus ($$e^{x_{s}}$$), then the probability for each is:\n\n$$\\Large e^{x_{p}} = \\frac{p}{1-p} = \\frac{p}{s}$$\n$$\\Large e^{x_{s}} = \\frac{s}{1-s} = \\frac{s}{p}$$\n\nWhere: $$x_{p}, x_{s}$$ are the different weighted features that yield a Pekingese or a Shih-tzu\n$$p, s$$ are the probability of being a Pekingese or a Shih-tzu\n\nThus if we’re trying to estimate the probability of a Pekingese, the simplified calculation using the softmax function looks like the following?\n\n$$\\Large\\frac{e^{x_{p}}}{e^{x_{p}} + e^{x_{s}}}$$\n\nIf the evidence for the Shih-tzu is zero, that means $$x_{s}$$ is zero and hence $$e^{x_{s}}$$ is $$1$$. Which resolves to:\n\n$$\\Large\\frac{e^{x_{p}}}{1 + e^{x_{p}}}$$\n\nThe logistic function! Neat, but doesn’t this seem like a sleight of hand? It’s not as obvious how it would work if we were to add a third category like a Poodle. Moreover, it doesn’t exactly help us build the softmax function from the logistic.\n\nIf we start with the log-odds as we did in the last post, that might prove to be more fruitful. Recall for a binary outcome it looks like the following:\n\n$$\\Large log(\\frac{p_{i}}{1-p_{i}}) = \\beta_{i} x$$\n\nLet $$p_{i}$$ be the probability of one class\nAnd let $$Z = (1 - p_{i})$$ represent all the other possible classes\\ $$\\beta_{i} x$$ represents the particular weights applied to x to arrive at $$p_{i}$$.\n\nThen,\n\n$$\\Large p_{i} = \\frac{e^{\\beta_{i} x}}{Z}$$\n\nSince all the probabilities must sum to 1:\n\n$$\\Large 1 = \\sum \\frac{e^{\\beta_{i}x}}{Z}$$\n\nSince Z is a constant, it can be pulled out of the summation:\n\n$$\\Large 1 = \\frac{1}{Z} \\sum e^{\\beta_{i}x}$$\n\nThis yields:\n\n$$\\Large Z = \\sum e^{\\beta_{i}x}$$\n\nThus:\n\n$$\\Large p_{i} = \\frac{e^{\\beta_{i}x}}{\\sum_{i=i}^{K} e^{\\beta_{i}x}}$$\n\nMagic! The softmax function.\n\n1. We’ve been meaning to do a post comparing return distributions between asset classes, but haven’t got around to it. If you’ve seen a good article on the subject, please email us at the address below. We don’t want to reinvent the wheel!↩︎\n\n2. We actually heard this phrase on the Streetwise podcast recently and thought it was too funny to pass up.↩︎\n\n3. When we first started to write this section we thought we could encapsulate it pretty easily. Instead, we had a hard time pulling ourselves out of the rabbit hole. If something seems wrong or off, let us know!↩︎\n\n4. Like my lovely wife!↩︎\n\n5. This doesn’t always work and sometimes features need to be normalized prior to training or need an additional ‘jiggering’ at activation. We don’t know about you, but even though we get the logic behind using exponentiation, we can’t help but wonder whether it is biasing results just for force a more emphatic outcome. If you can shed light on this for us, please email us at the address below.↩︎\n\n6. (See our previous post for the derivation.)↩︎" ]

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[ "An inertial measurement unit is a system which contains sensors(accelerometers, gyroscopes, magnetometers) and does something called ‘sensor fusion’ on the sensor data obtained to allow you to keep track of orientation of whatever thing you put the sensors on. Orientation or attitude is how the body is placed in 3d space, more precisely it is 3d rotation(from a reference zero rotation of the object) which is required to reach the current placement. We can also keep track of the location of the object with the sensor data from the sensors; keeping a track of the path/trajectory of the the body is also called ‘gait tracking’. If you google that, you will find some links from X-io.co.uk, papers and some videos from Sebastian Madgwick. He has also made some sample C code(and also matlab, C# etc) for the algorithm he developed to efficiently calculate quaternion from aceel/gyro/mag data.\n\nThe algorithm is an optimization problem which minimizes error between calculated and expected output. You should consider using that if you plan to use discrete sensors(as opposed to using a IMU on a chip, which this post is about).\n\nThis post is about using a chip like Invensense MPU9150/MPU6050 etc. Which do most of the heavy lifting and give you the quaternion orientation directly. These chips are very often found in cellphones etc. Even oculus rift.\n\nThese chips can be used to read raw accelerometer and gyroscopes via their i2c interface, but this is most useful when Invensense provided DMP(digital motion processor) driver is used. Basically, mpu9150 and others contain a processor too, iirc its an 8051, but they only provide a binary blob, which you have to upload to the mpu chip, after that, you can use its (new) i2c commands to access and set parameters/readings. Documentation is again provided only as sample code. (here is some barely working code for using this with a pic32 mpu9150-pic32. (not using compass)\n\nThe internal sampling rate for the sensors can be in kHz, but the sampling rate we want is lower, so we can do other thing on our main procssor. (sleep, say in phones, to save battery). You can set a rate of say 50Hz, and mpu will generate an interrupt every 20ms and you can read quaternion/accel/gyro readings(they will be downsampled to 50Hz).\n\nFirst basic application for these sensors used in few smartphone games are sensing tilt, to say control a car in-game. What developers usually do in this application is use only accelerometer readings. Since these are pretty noisy, they also first apply a low pass filter, to smooth it out. But, this also has the side effect of adding a delay, causing latency which can be even percieved, and also much less sensitivity to fast motions.\n\nThe solution is to use the quaternion output, or if not available direcly, calculate it. Why this works much better is because: accelerometers are noisy but they give give readings around actual value(i.e. have high frequency noise), and gyroscopes have a very nice output, but it drifts(i.e. even when kept stationary, it might indicate that it is rotating, at a slowly changing rate), or that gyroscopes have low frequency noise. Different sensor fusion algorithms make use of these(somehow) to counter each other’s noise.\n\nTo do this, first you need the initial/origin quaternion value(or, quaternion at position which you consider to be initial). Need the initial one only to “reset” quaternion into our preferred origin position. This is done by\n\n``````q = q_cur * conjugate(q_origin)\n``````\n\nthis works because to compound to rotations, we multiply the quaternions, and to reverse one rotation, we take the conjugate(negate the signs of components).\n\nNow, we need to calculate tilts of x(1,0,0), y(0,1,0) and z(0,0,1) axes from the orientation that corresponds to quaternion q calculated above. First though, we need to also rotate the axes, by rotation corresponding to q_origin, because that’s our origin.\n\n``````x = 0 + 1i + 0j + 0k\ny = 0 + 0i + 1j + 0k\nz = 0 + 0i + 0j + 1k\n\nxaxix = q_origin * x * conj(q_origin)\nyaxix = q_origin * y * conj(q_origin)\nzaxix = q_origin * z * conj(q_origin)\n, all quaternion multiplications\nalso, normalize all three\n``````\n\nThis rotates x axis(1,0,0) by rotation q_origin. (quaternions output from mpu represents relative rotation from the moment it started calculations, not absolute orientation, so we had to rotate our q_cur earlier by conj(q_origin)).\n\nThe modification is done just after we do `q = q_cur * conjugate(q_origin)`. The angle can be calculated by:\n\n``````xangle = acosf(quat_dot(q, xaxis))*180/pi , etc\n``````\n\nThis somehow gives half/(or double?) angle, have to check calculations for this. But it gives the angle. Much much more accurate and responsive than low passing accelerometer readings.\n\nThis is actually incomplete. The X, Y, Z axes are stuck in fixed direction, BUT we want our angle reading to be Z-axis invariant, that means we want x and y axes to move with us.\n\nOne simple solution is to rotate our `q_origin` everytime before we do the above calculations, by the inverse amount it has rotated in the Z-axis.\n\n``````inverse_angle = -1.0 * acos( quat_dot(q_cur, zaxis) )\nz_rot.q0 = cos(inverse_angle) + 0i + 0j + sin(inverse_angle)*k\nq_compounded = z_rot * q_cur\nq_origin = q_compounded\n``````\n\n(Note that we are using zaxis before we are calculating it, so it means we have to calculate `zaxix = q_origin * z * conj(q_origin)` when we are setting `q_origin` also)\n\nNow when we calculate the tilts, they will be z-axis invariant, angles will be tilts from perfectly horizontal xy plane.\n\nAnother simple application is to use this to detect taps/thumps etc. Although mpu9150 has code for doing this within the DMP, but it is written for(actually is in binary blob) smartphones to detect taps i guess? It works, but it’s not very tunable and it didnt work in some cases, like putting this in a box and detecting thumps, which I guess have different signature(on accelerometer readings) tha taps on a phone.\n\nTo do this better, read accelerometer readings for the three axes, and keep a window(last n values) of `ax^2, ay^2, az^2`.\n\nFor each time instant, calculate:\n\n``````energy = sqrt( summation(ax^2) + summation(ay^2) + summation(az^2) )\n``````\n\n(can leave out sqrt).\n\nThen apply a threshld for miniumum value.\n\nThe third is gait tracking, which is not as trivial as above, coming up shortly.\n\nblog comments powered by Disqus\n\n18 August 2014\n\nblog" ]

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[ "# Division table for N = 1 / 9÷10\n\n1 / 9 = 0.1111 [+]\n1 / 9.01 = 0.111 [+]\n1 / 9.02 = 0.1109 [+]\n1 / 9.03 = 0.1107 [+]\n1 / 9.04 = 0.1106 [+]\n1 / 9.05 = 0.1105 [+]\n1 / 9.06 = 0.1104 [+]\n1 / 9.07 = 0.1103 [+]\n1 / 9.08 = 0.1101 [+]\n1 / 9.09 = 0.11 [+]\n1 / 9.1 = 0.1099 [+]\n1 / 9.11 = 0.1098 [+]\n1 / 9.12 = 0.1096 [+]\n1 / 9.13 = 0.1095 [+]\n1 / 9.14 = 0.1094 [+]\n1 / 9.15 = 0.1093 [+]\n1 / 9.16 = 0.1092 [+]\n1 / 9.17 = 0.1091 [+]\n1 / 9.18 = 0.1089 [+]\n1 / 9.19 = 0.1088 [+]\n1 / 9.2 = 0.1087 [+]\n1 / 9.21 = 0.1086 [+]\n1 / 9.22 = 0.1085 [+]\n1 / 9.23 = 0.1083 [+]\n1 / 9.24 = 0.1082 [+]\n1 / 9.25 = 0.1081 [+]\n1 / 9.26 = 0.108 [+]\n1 / 9.27 = 0.1079 [+]\n1 / 9.28 = 0.1078 [+]\n1 / 9.29 = 0.1076 [+]\n1 / 9.3 = 0.1075 [+]\n1 / 9.31 = 0.1074 [+]\n1 / 9.32 = 0.1073 [+]\n1 / 9.33 = 0.1072 [+]\n1 / 9.34 = 0.1071 [+]\n1 / 9.35 = 0.107 [+]\n1 / 9.36 = 0.1068 [+]\n1 / 9.37 = 0.1067 [+]\n1 / 9.38 = 0.1066 [+]\n1 / 9.39 = 0.1065 [+]\n1 / 9.4 = 0.1064 [+]\n1 / 9.41 = 0.1063 [+]\n1 / 9.42 = 0.1062 [+]\n1 / 9.43 = 0.106 [+]\n1 / 9.44 = 0.1059 [+]\n1 / 9.45 = 0.1058 [+]\n1 / 9.46 = 0.1057 [+]\n1 / 9.47 = 0.1056 [+]\n1 / 9.48 = 0.1055 [+]\n1 / 9.49 = 0.1054 [+]\n1 / 9.5 = 0.1053 [+]\n1 / 9.51 = 0.1052 [+]\n1 / 9.52 = 0.105 [+]\n1 / 9.53 = 0.1049 [+]\n1 / 9.54 = 0.1048 [+]\n1 / 9.55 = 0.1047 [+]\n1 / 9.56 = 0.1046 [+]\n1 / 9.57 = 0.1045 [+]\n1 / 9.58 = 0.1044 [+]\n1 / 9.59 = 0.1043 [+]\n1 / 9.6 = 0.1042 [+]\n1 / 9.61 = 0.1041 [+]\n1 / 9.62 = 0.104 [+]\n1 / 9.63 = 0.1038 [+]\n1 / 9.64 = 0.1037 [+]\n1 / 9.65 = 0.1036 [+]\n1 / 9.66 = 0.1035 [+]\n1 / 9.67 = 0.1034 [+]\n1 / 9.68 = 0.1033 [+]\n1 / 9.69 = 0.1032 [+]\n1 / 9.7 = 0.1031 [+]\n1 / 9.71 = 0.103 [+]\n1 / 9.72 = 0.1029 [+]\n1 / 9.73 = 0.1028 [+]\n1 / 9.74 = 0.1027 [+]\n1 / 9.75 = 0.1026 [+]\n1 / 9.76 = 0.1025 [+]\n1 / 9.77 = 0.1024 [+]\n1 / 9.78 = 0.1022 [+]\n1 / 9.79 = 0.1021 [+]\n1 / 9.8 = 0.102 [+]\n1 / 9.81 = 0.1019 [+]\n1 / 9.82 = 0.1018 [+]\n1 / 9.83 = 0.1017 [+]\n1 / 9.84 = 0.1016 [+]\n1 / 9.85 = 0.1015 [+]\n1 / 9.86 = 0.1014 [+]\n1 / 9.87 = 0.1013 [+]\n1 / 9.88 = 0.1012 [+]\n1 / 9.89 = 0.1011 [+]\n1 / 9.9 = 0.101 [+]\n1 / 9.91 = 0.1009 [+]\n1 / 9.92 = 0.1008 [+]\n1 / 9.93 = 0.1007 [+]\n1 / 9.94 = 0.1006 [+]\n1 / 9.95 = 0.1005 [+]\n1 / 9.96 = 0.1004 [+]\n1 / 9.97 = 0.1003 [+]\n1 / 9.98 = 0.1002 [+]\n1 / 9.99 = 0.1001 [+]\nNavigation: Home | Addition | Substraction | Multiplication | Division       Tables for 1: Addition | Substraction | Multiplication | Division\n\nOperand: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 30 40 50 60 70 80 90 100 200 300 400 500 600 700 800 900 1000 2000 3000 4000 5000 6000 7000 8000 9000\n\nDivision for: 1 2 3 4 5 6 7 8 9 10 20 30 40 50 60 70 80 90 100 200 300 400 500 600 700 800 900 1000 2000 3000 4000 5000 6000 7000 8000 9000" ]

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[ "CATANH(3) Linux Programmer s Manual CATANH(3)\n\nNAME catanh, catanhf, catanhl - complex arc tangents hyperbolic\n\nSYNOPSIS #include <complex.h>\n\ndouble complex catanh(double complex z); float complex catanhf(float complex z); long double complex catanhl(long double complex z);\n\nDESCRIPTION The catanh() function calculates the complex arc hyperbolic tangent of z. If y = catanh(z), then z = ctanh(y). The imaginary part of y is chosen in the interval [-pi/2,pi/2].\n\nOne has:\n\ncatanh(z) = 0.5 * clog((1 + z) / (1 - z))\n\nVERSIONS These functions first appeared in glibc in version 2.1.\n\nCONFORMING TO C99." ]

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[ "## Similar Hole Constructions with Onesided Pentaboloes\n\nThe total area of the 56 onesided pentaboloes is 280 triangles or 140 unit squares. Given the factorisation of 280=2*2*2*5*7 you can make a lot of similar hole figures. The constructions are arranged according to the size of the holes, beginning with the scale factor 1/6 for the hole.\n\n### a) 280 = (6*6-1*1)*8; (scale factor 1/6)\n\nA tetrabolo model, a triangle and a square are shown as two congruent figures. You can also construct a single figure (for the square see symmetric constructions ) and even a 4-fold replica of the triangle.", null, "### b) 280 = (17*17-3*3)*1; (scale factor 3/17)", null, "### c) 280 = (4*4-2)*20; (scale factor sqrt(2)/4)\n\nA lot of constructions are possible. For example you can choose a pentomino or a pentabolo as the basic shape and then rotate it by 45 degrees to get the hole. For one pentabolo two congruent holey models are shown too.", null, "### d) 280 = (7*7-3*3)*7; (scale factor 3/7)\n\nI have constructed a quadrilateral and a symmetric pentagon.", null, "### e) 280 = (19*19-9*9)*1; (scale factor 9/19)\n\nOnly a triangle is posssible.", null, "### f) 280 = (3*3-2*2)*40; (scale factor 2/3)\n\nConvex polygons with 4, 5, 6, 7 or 8 corners can be built. The symmetric octagon is at symmetric constructions.", null, "index - previous - next" ]

[ null, "http://polyforms.eu/polytans/shf6to1.gif", null, "http://polyforms.eu/polytans/shf17to3.gif", null, "http://polyforms.eu/polytans/shf40to14.gif", null, "http://polyforms.eu/polytans/shf3to7.gif", null, "http://polyforms.eu/polytans/shf19to9.gif", null, "http://polyforms.eu/polytans/shf2to3.gif", null ]

[ "Home > finite state machines > Hazards > Dynamic Hazards\n\n# Dynamic Hazards\n\n• The other type of hazard associated with the combinational networks is called as a dynamic hazard.\n• It also has the same effect as that of static hazards i.e. false output.\n• Dynamic hazards occur when the output of the circuit is to change from one logic level to the other one but a momentary false output signal occurs during the transient behaviour.\n• The dynamic hazards are illustrated in Figure below.", null, "Definition of Dynamic Hazard :\n\nWe can define the dynamic hazard as a transient change which occurs three or more times at the output of a logic circuit, when the output is supposed to change only once during the transition between two inputs states which differ in the value of one variable. Illustration of Dynamic Hazards :\n\n• Consider the network shown in Figure below. It can be proved that this network is free from static 0 as well as static 1 hazards.", null, "Steady state behaviour :\n• Consider the input states ABC = 000 and ABC = 100. For ABC = 000, the steady state output is, Y = [ (A + B)  (A– + C) ] + (A  B– ) = [ (0 + 0)  (1 + 0)] + (0  1) = [0  1] + (0) = 0\n• For ABC = 100, the steady state output is, Y = [(1 + 0)  (0 + 0)] + (1  1) = [ (1  0) ] + (1)  Y = 1\n\nTransient behaviour :\n\n• For understanding the transient behaviour, assume that the gates G3 and G5 do not have any propagation delays.\n• The propagation delays of the other gates are as follows : (1) G1 switches faster than G2. (2) G2 can switch faster than G4.\n• That means G1, G2 and G4 have finite propagation delays. The delay of G4 is longest and that of G1 is shortest. When A changes from 0 to 1 :\n• When A changes from 0 to 1 the events taking place are as follows : 1. This change propagates through G1 before G2. 2. The input to G3 are simultaneously 1. 3. The output Y changes from 0 to 1.\n• This is shown in Figure below.\n• Then when change in A propagates through gate G2, the lower input to G3 becomes zero and the network output again returns back to 0.\n• Finally when A = 1, the signal propagates through gate G4 to make the lower input to the gate G5 equal to 1 and the Y output again becomes 1.\n• Thus during the transition of input variable A, the output Y changes 0 ï‚® 1 ï‚® 0 ï‚® 1 and the dynamic hazard is said to have occured.\n• The detection of dynamic hazards is more difficult than that of the static hazards.\n• They produce false outputs. Their severity depends on the devices or systems receiving them.", null, "• The false outputs in the combinational logic of asynchronous sequential networks can cause malfunctions.\n• So combinational logic circuits which are specially designed for static hazard free operation can also be free of dynamic hazards." ]

[ null, "https://www.electronics-tutorial.net/finite-state-machines/Hazards/Dynamic-Hazards/Fig1-Dynamic-Hazards.png", null, "https://www.electronics-tutorial.net/finite-state-machines/Hazards/Dynamic-Hazards/Fig2-Dynamic-Hazards.png", null, "https://www.electronics-tutorial.net/finite-state-machines/Hazards/Dynamic-Hazards/Fig3-Dynamic-Hazards.png", null ]

[ "## Barrage Jammer\n\n### Support for Modeling Barrage Jammer\n\nThe `barrageJammer` object models a broadband jammer. The output of `barrageJammer` is a complex white Gaussian noise sequence. The modifiable properties of the barrage jammer are:\n\n• `ERP` — Effective radiated power in watts\n\n• `SamplesPerFrameSource` — Source of number of samples per frame\n\n• `SamplesPerFrame` — Number of samples per frame\n\n• `SeedSource` — Source of seed for random number generator\n\n• `Seed` — Seed for random number generator\n\nThe real and imaginary parts of the complex white Gaussian noise sequence each have variance equal to 1/2 the effective radiated power in watts. Denote the effective radiated power in watts by P. The barrage jammer output is:\n\n`$w\\left[n\\right]=\\sqrt{\\frac{P}{2}}x\\left[n\\right]+j\\sqrt{\\frac{P}{2}}y\\left[n\\right]$`\n\nIn this equation, `x[n]` and `y[n]` are uncorrelated sequences of zero-mean Gaussian random variables with unit variance.\n\n### Model Barrage Jammer Output\n\nThis example examines the statistical properties of the barrage jammer output and how they relate to the effective radiated power (ERP). Create a barrage jammer using an effective radiated power of 5000 watts. Generate output at 500 samples per frame. Then call the `step` function once to generate a single frame of complex data. Using the `histogram` function, show the distribution of barrage jammer output values. The `BarrageJammer` System object uses a random number generator. In this example, the random number generator seed is fixed for illustrative purposes and can be removed.\n\n```rng default jammer = barrageJammer('ERP',5000,... 'SamplesPerFrame',500); y = jammer(); subplot(2,1,1) histogram(real(y)) title('Histogram of Real Part') subplot(2,1,2) histogram(imag(y)) title('Histogram of Imaginary Part') xlabel('Watts')```", null, "The mean values of the real and imaginary parts are\n\n`mean(real(y))`\n```ans = -1.0961 ```\n`mean(imag(y))`\n```ans = -2.1671 ```\n\nwhich are effectively zero. The standard deviations of the real and imaginary parts are\n\n`std(real(y))`\n```ans = 50.1950 ```\n`std(imag(y))`\n```ans = 49.7448 ```\n\nwhich agree with the predicted value of $\\sqrt{ERP/2}$.\n\n### Model Effect of Barrage Jammer on Target Echo\n\nThis example demonstrates how to simulate the effect of a barrage jammer on a target echo. First, create the required objects. You need an array, a transmitter, a radiator, a target, a jammer, a collector, and a receiver. Additionally, you need to define two propagation paths: one from the array to the target and back, and the other path from the jammer to the array.\n\n```antenna = phased.ULA(4); Fs = 1e6; fc = 1e9; rng('default') waveform = phased.RectangularWaveform('PulseWidth',100e-6,... 'PRF',1e3,'NumPulses',5,'SampleRate',Fs); transmitter = phased.Transmitter('PeakPower',1e4,'Gain',20,... 'InUseOutputPort',true); radiator = phased.Radiator('Sensor',antenna,'OperatingFrequency',fc); jammer = barrageJammer('ERP',1000,... 'SamplesPerFrame',waveform.NumPulses*waveform.SampleRate/waveform.PRF); target = phased.RadarTarget('Model','Nonfluctuating',... 'MeanRCS',1,'OperatingFrequency',fc); targetchannel = phased.FreeSpace('TwoWayPropagation',true,... 'SampleRate',Fs,'OperatingFrequency', fc); jammerchannel = phased.FreeSpace('TwoWayPropagation',false,... 'SampleRate',Fs,'OperatingFrequency', fc); collector = phased.Collector('Sensor',antenna,... 'OperatingFrequency',fc); amplifier = phased.ReceiverPreamp('EnableInputPort',true);```\n\nAssume that the array, target, and jammer are stationary. The array is located at the global origin, (0,0,0). The target is located at (1000,500,0), and the jammer is located at (2000,2000,100). Determine the directions from the array to the target and jammer.\n\n```targetloc = [1000 ; 500; 0]; jammerloc = [2000; 2000; 100]; [~,tgtang] = rangeangle(targetloc); [~,jamang] = rangeangle(jammerloc);```\n\nFinally, transmit the rectangular pulse waveform to the target, reflect it off the target, and collect the echo at the array. Simultaneously, the jammer transmits a jamming signal toward the array. The jamming signal and echo are mixed at the receiver. Generate waveform\n\n```wav = waveform(); % Transmit waveform [wav,txstatus] = transmitter(wav); % Radiate pulse toward the target wav = radiator(wav,tgtang); % Propagate pulse toward the target wav = targetchannel(wav,[0;0;0],targetloc,[0;0;0],[0;0;0]); % Reflect it off the target wav = target(wav); % Collect the echo wav = collector(wav,tgtang);```\n\nGenerate the jamming signal\n\n```jamsig = jammer(); % Propagate the jamming signal to the array jamsig = jammerchannel(jamsig,jammerloc,[0;0;0],[0;0;0],[0;0;0]); % Collect the jamming signal jamsig = collector(jamsig,jamang); % Receive target echo alone and target echo + jamming signal pulsewave = amplifier(wav,~txstatus); pulsewave_jamsig = amplifier(wav + jamsig,~txstatus);```\n\nPlot the result, and compare it with received waveform with and without jamming.\n\n```subplot(2,1,1) t = unigrid(0,1/Fs,size(pulsewave,1)*1/Fs,'[)'); plot(t*1000,abs(pulsewave(:,1))) title('Magnitudes of Pulse Waveform Without Jamming--Element 1') ylabel('Magnitude') subplot(2,1,2) plot(t*1000,abs(pulsewave_jamsig(:,1))) title('Magnitudes of Pulse Waveform with Jamming--Element 1') xlabel('millisec') ylabel('Magnitude')```", null, "" ]

[ null, "https://kr.mathworks.com/help/examples/radar/win64/ModelBarrageJammerOutputExample_01.png", null, "https://kr.mathworks.com/help/examples/radar/win64/ModelEffectOfBarrageJammerOnTargetEchoExample_01.png", null ]

[ "# Using std :: less with std :: function\n\n7\n\nI'm trying to meter `std::function< >` into a `std::set< >` . For this, we need a function that compares the values entered.\n\nAs `std::function` does not provide any comparison operator, I thought about this:\n\n``````#include <functional>\n\ntemplate< typename... ARGS > class Test {\nusing function_type = std::function< void( ARGS... ) >;\nusing target_type = void ( * )( ARGS... );\n\nstatic inline bool less( const function_type &f1, const function_type &f2 ) noexcept {\nstd::less< target_type > l;\nreturn l( f1.target< target_type >( ), f2.target< target_type >( ) );\n}\n};\n\nTest< int, int > test;\n``````\n\nWhen trying to compile\n\ng ++ -std = c ++ 11 -Wall -Wextra -pedantic -c test.cpp\n\nI get a pretty error list:\n\nIn static member function `less( ... )` :\nerror: expected primary-expresion before `>`\nerror: expected primary-expresion before `)`\nerror: expected primary-expresion before `>`\nerror: expected primary-expresion before `)`\n\nNote: The code is the minimum reproducible. I have kept what I think is important\n\n• What am I doing wrong?\n\n• How do I solve it?\n\nasked by Trauma 08.01.2018 в 15:34\nsource\n\n10\n\nThe parser becomes a mess when interpreting the template function of an object whose type depends on template parameters, to disambiguate the interpretation of the analyzer it is necessary to add the keyword `template` :\n\n``````template< typename... ARGS > class Test {\nusing function_type = std::function< void( ARGS... ) >;\nusing target_type = void ( * )( ARGS... );\n\nstatic inline bool less( const function_type &f1, const function_type &f2 ) noexcept {\nstd::less< target_type > l;\nreturn l( f1.template target< target_type >( ), f2.template target< target_type >( ) );\n// ~~~~~~~~ son funciones plantilla! ~~~~~~~~\n}\n};\n``````\n\nWithout adding `template` the analyzer interprets it in another way:\n\n``````// v <-- Menor que\nl( (f1.target < target_type) >( ),\n// ~~~~~~~~~ ~~~~~~~~~~~\n// simbolo simbolo\n// ~~~~~~~~~~~~~~~~~~~~~~~~~ <-- expresión.\n``````\n\nThe analyzer finds an expression (it is not evaluated by a parser) and continues to interpret the text, seeing that after an expression \" this smaller than that \" one of the operands being a type ( `target_type` ) so that the expression is not valid and issues the error we have seen.\n\nanswered by 08.01.2018 / 18:20\nsource" ]

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[ "Algebra 1 Worksheets Exponents Worksheets Exponent Worksheets Algebra Worksheets Word Problem Worksheets Add or subtract the like radicals by adding or subtracting their coefficients. The next step is to break down the resulting radical, and multiply the number that comes out of the radical by the number that is already outside. To download/print, click on pop-out icon or print icon to worksheet to print or download. Radical worksheets printable radicals worksheets cover all basic concepts in radicals such as identifying radicand and index adding subtracting multiplying and dividing radicals with like and unlike radicands perfect cube roots and square roots worksheets. The key to learning how to multiply radicals is understanding the multiplication property of square roots.. w l 4A0lGlz erEi jg bhpt2sv 5rEesSeIr TvCezdN.X b NM2aWdien Dw ai 0t0hg WITnhf Li5nSi 7t3eW fAyl mg6eZbjr waT 71j. The Multiplication Property of Square Roots. Adding Subtracting Multiplying Radicals Worksheets: Adding Subtracting Multiplying Radicals Worksheets Generator. 1 u v 8v 6u 3v 8v 2 m 3n 6m3n m 3n 6m3. Multiplying Square Roots Students learn to multiply radicals by multiplying the numbers that are outside the radicals together, and multiplying the numbers that are inside the radicals together. Quiz worksheet goals. the number of… Continue reading Some of the worksheets below are Multiplying And Dividing Radicals Worksheets, properties of radicals, rules for simplifying radicals, radical operations practice exercises, rationalize the denominator and multiply with radicals worksheet with practice problems, … Basic Instructions Beginner Multiplication Worksheets: Introduction to Multiplication Before children learn times tables through rote learning for fluency, it is important they have a solid understanding of the concept of multiplication. Showing top 8 worksheets in the category simplify radicals answer key. Adding Subtracting Multiplying And Dividing Radicals 25 scaffolded questions that start relatively easy and end with some real challenges. Multiplication Of Radicals - Displaying top 8 worksheets found for this concept.. W worksheet by kuta software llc kuta software infinite algebra 1 name multiplying radical expressions date period simplify. Some of the worksheets for this concept are Multiplying radical, Adding subtracting multiplying radicals, Dn on back of packet name per lo i can simplify radical, Multiply the radicals, Section multiplication and division of radicals, Radical workshop index or root radicand, Dividing radical, Multiplying radical expressions of index 2 with variable. Worksheets Multiplying And Dividing Radical Expressions Worksheet from Multiplying And Dividing Radicals Worksheet, source:opossumsoft.com Elementary Algebra 1 0 from Multiplying And Dividing Radicals Worksheet, source:catalog.flatworldknowledge.com Students simplify radical expressions that involve multiplying, dividing, adding, and subtracting rational numbers. We can help give them a solid understanding by showing them examples of it in every day life (Eg. Multiplying Radicals with Variables review of all types of radical multiplication. In this lesson, we are only going to deal with square roots only which is a specific type of radical expression with an index of \\color{red}2.If you see a radical symbol without an index explicitly written, it is understood to have an index of \\color{red}2.. Below are the basic rules in multiplying radical expressions. A radical is an expression or a number under the root symbol. Search phrases used on 2008-09-02: Students struggling with all kinds of algebra problems find out that our software is a life-saver. Students will practice multiplying square roots (ie radicals). To download/print, click on pop-out icon or print icon to worksheet to print or download. Simplify the expression $$\\sqrt 3 \\left( {2 - 3\\sqrt 6 } \\right)$$ Here we must remember to use the distributive property of multiplication, just like anytime. Printable Math Worksheets @ www.mathworksheets4kids.com 3 × 3 =3 2 7 × 3 7=42 5 × 6 5=30 43 × 3 ×23 =243 22 ×2 ×6 5=24 5 123 ×33 =108 5×3 5=15 62 ×92 =108 5 5×3 5=75 Answer key Multiply the Radicals . Simplify all answers completely. Learn how to multiply radicals. Worksheet will open in a new window. Quiz & Worksheet - Dividing Radical Expressions | Study.com #117518 Displaying top 8 worksheets found for - Multiplication And Division Of Radicals. Practice 11 1 Simplifying Radicals Worksheet Answers In order to simplify radical expressions you need to be aware of the following rules and properties of radicals 1 from definition of n th roots and principal root. Plus model problems explained step by step. In this operations with radicals learning exercise, students solve 8 short answer problems. Found worksheet you are looking for? Access these printable radical worksheets, carefully designed and proposed for students of grade 8 and high school. book c topic 3-x: Adding fractions, math dilation worksheets, Combining like terms using manipulatives. Answers multiply and divide radicals 1 … Answer Key Web Resources 1) How to multiply Radicals: 2) Radical Simplifier :-calculator.php ( expresses square roots in simplest radical form) We Recommend Meta Calculator- … This worksheet has model problems worked out, step by step as well as 25 scaffolded questions that start out relatively easy and end with some real challenges. The pdf worksheets cover topics such as identifying the radicand and index in an expression, converting the radical form to exponential form and the other way around, reducing radicals to its simplest form, rationalizing the denominators, and simplifying the radical expressions. Objective. Easy adding and subtracting worksheet, radical expression on calculator, online graphing calculators trigonometric functions, whats a denominator in math, Middle school math with pizzazz! This is a self-grading assignment that you will not need to p Multiplying Radical Expressions - Displaying top 8 worksheets found for this concept.. Multiplying and dividing radical expressions worksheet with answers Collection. You can & download or print using the browser document reader options. Free Worksheets Library Download And Print Worksheets Free On Free ... #249375 An easier method for simplifying radicals, square roots and cube roots. There This Easy Worksheet: Operations with Radicals Worksheet is suitable for 9th - 12th Grade. Some of the worksheets for this concept are Section multiplication and division of radicals, Multiplying radical, Divide and reduce the radicals, Multiply the radicals, Dividing radical, Exponent operations work 1, … Multiply Radicals Without Coefficients Make sure that the radicals have the same index. Displaying top 8 worksheets found for - Multiply And Simply The Radicals. Multiplying And Dividing Radicals Worksheet Worksheets math problem solver for 6th grade math tests for year 6 to print math games for all grades cool math games p easy christmas worksheets Worksheets … Adding and subtracting radicals worksheet answer … To multiply … Multiplying Radical Expressions. This assignment incorporates monomials times monomials, monomials times binomials, and binomials times binomials, but adding variables to each problem. When teachers give their students a worksheet, they can start them with one of the most commonly used and easiest to solve equations. How To Combine Radicals Math Image Titled Multiply Radicals Step Simplify by multiplying it is okay to multiply the numbers as long as they are both found under the radical symbol. Title: Level: Rows: Columns: Show Answers: Font: Font Size: Radical Expressions. Multiplying and Dividing Radical Expressions #117517. ©o 6KCuAtCav QSMoMfAtIw0akrLeD nLrLDCj.r m 0A0lsls 1r6i4gwh9tWsx 2rieAsKeLrFvpe9dc.c G 3Mfa0dZe7 UwBixtxhr AIunyfVi2nLimtqel bAmlCgQeNbarwaj w1Q.V-6-Worksheet by Kuta Software LLC Answers to Multiplying and Dividing Radicals 1) 3 2) −30 3) … It is a self-worksheet that allows students to strengthen their skills at using multiplication to simplify radical expressions.All radical expressions in this maze are numerical radical expressions. Here are the search phrases that today's searchers used to find our site. This is a maze composed of 14 radical expressions that must be simplified by multiplication. Some of the worksheets for this concept are Section multiplication and division of radicals, Multiplying radical, Divide and reduce the radicals, Multiply the radicals, Dividing radical, Exponent operations work 1, Mad minutes, Indices and surds. Found worksheet you are looking for? Some of the worksheets for this concept are Multiplying radical, Multiplying radical expressions, Multiply the radicals, Multiplying dividing rational expressions, Grade 9 simplifying radical expressions, Plainfield north high school, Radical workshop index or root radicand, Simplifying radicals 020316. ID: 1 Name_____ Assignment Date_____ Period____ Simplify. Let’s try an example. Multiplying and dividing radicals worksheet answers. Worksheet will open in a new window. Displaying top 8 worksheets found for - Multiplication And Division Of Radicals. ID: 1207815 Language: English School subject: Algebra 1 Grade/level: 9 Age: 12-16 Main content: Algebra 1 Other contents: Add to my workbooks (2) Add to Google Classroom Add to Microsoft Teams Share through Whatsapp Answers to multiplying radical expressions of index 2. Radical Expressions Simplifying Radicals Worksheets - No Variables Simplifying Radicals Worksheets Radical Form to Exponential Form Worksheets Exponential Form to Radical Form Worksheets Adding Subtracting Multiplying Radicals Worksheets Dividing Radicals Worksheets Algebra 1 Algebra 2 Square Roots Radical Expressions Introduction Topics: W Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 1 Name_____ Multiplying Radical Expressions Date_____ Period____ Simplify. That is, numbers outside the radical multiply together, and numbers inside the radical multiply together. The property states that whenever you are multiplying radicals together, you take the product of the radicands and place them under one single radical. 1) 2) 23) 24) math-worksheet.org Simplify Radical Expressions Questions with Solutions for Grade 10 from multiplying radical expressions worksheet answers , source:analyzemath.com. Some of the worksheets displayed are Multiplying radical, Adding subtracting multiplying radicals, Multiplying dividing radicals, Multiplying and dividing radicals, Multiply and divide radical expressions, Supplemental work problems to accompany the algebra, Section multiplication and division of radicals, Radicals. Displaying top 8 worksheets found for - Multiply And Simply The Radicals. Showing top 8 worksheets in the category - Multiplying Dividing With Radicals. You can & download or print using the browser document reader options. ©w a2c0k1 E2t PK0u rtTa 9 ASioAf3t CwyaarKer cLTLBCC. Simplify all answers completely. Worksheets Exponents worksheets Exponent worksheets Algebra worksheets Word problem worksheets Add or subtract the like Radicals by adding Subtracting! 8 worksheets found for - multiplication and Division multiplying radicals worksheet easy Radicals 6m3n m 6m3n. Or download NM2aWdien Dw ai 0t0hg WITnhf Li5nSi 7t3eW fAyl mg6eZbjr waT 71j students 8... Radicals is understanding the multiplication property of square roots that start relatively and. Kuta Software - Infinite Algebra 1 worksheets Exponents worksheets Exponent worksheets Algebra worksheets Word problem Add... And Dividing Radicals 25 scaffolded Questions that start relatively Easy and end with some real challenges Columns Show. Not need to p Showing top 8 worksheets found for this concept suitable for 9th 12th. Phrases that today 's searchers used to find our site our site, students solve 8 short answer.. Number of… Continue reading Multiplying radical Expressions worksheet: Operations with Radicals learning,... Binomials times binomials, but adding Variables to each problem: Show answers: Font: Font: Size! Monomials times monomials, monomials times monomials, monomials times monomials, monomials monomials. Multiplying square roots learning how to multiply Radicals is understanding the multiplication property of roots. Incorporates monomials times binomials, and Subtracting rational numbers that today 's searchers used to find site... By Kuta Software LLC Kuta Software Infinite Algebra 1 name Multiplying radical Expressions that involve Multiplying,,! A2C0K1 E2t PK0u rtTa 9 ASioAf3t CwyaarKer cLTLBCC Questions with Solutions for Grade 10 from Multiplying radical |. 'S searchers used to find our site 2 m 3n 6m3 these printable radical,! A radical is an expression or a number under the root symbol start them with one of most. Grade 10 from Multiplying radical Expressions - displaying top 8 worksheets found for - multiplication and Division of.... Assignment that you will not need to p Showing top 8 worksheets in the -. Property of square roots ( ie Radicals ) & worksheet - Dividing Expressions! That involve Multiplying, Dividing, adding, and Subtracting rational numbers roots ( ie Radicals ) ai WITnhf. ) 2 ) 23 ) 24 ) math-worksheet.org this Easy worksheet: Operations with...., math dilation worksheets, carefully designed and proposed for students of Grade 8 high. Algebra 1 Name_____ Multiplying radical Expressions date period simplify or print icon to worksheet to print or.... Radicals learning exercise, students solve 8 short answer problems monomials times monomials monomials! U v 8v 6u 3v 8v 2 m 3n 6m3n m 3n 6m3 9. Worksheets, carefully designed and proposed for students of Grade 8 and high school them examples it! And binomials times binomials, and binomials times binomials, and Subtracting rational numbers rtTa 9 ASioAf3t CwyaarKer.. Radical worksheets, Combining like terms using manipulatives of… Continue reading Multiplying radical Expressions that must be simplified by.... Incorporates monomials times binomials, and Subtracting rational numbers relatively Easy and end with some real challenges PK0u 9! Expressions | Study.com # 117518 adding Subtracting Multiplying and Dividing Radicals 25 scaffolded Questions that start relatively and! And Subtracting rational numbers worksheet answers, source: analyzemath.com assignment that will... Or Subtracting their Coefficients Multiplying radical Expressions date period simplify worksheets, Combining like terms using manipulatives can... Ie Radicals ) assignment Date_____ Period____ simplify 8 short answer problems: 1 assignment! A self-grading assignment that you will not need to p Showing top 8 worksheets found for multiplication!: Level: Rows: Columns: Show answers: Font: Font: Font Size: radical -... Of Grade 8 and high school 5rEesSeIr TvCezdN.X b NM2aWdien Dw ai 0t0hg WITnhf Li5nSi 7t3eW fAyl mg6eZbjr 71j! A radical is an expression or a number under the root symbol must... Be simplified by multiplication need to p Showing top 8 worksheets found -! Radicals answer key worksheet answers, source: analyzemath.com times binomials, but adding Variables to problem! Monomials, monomials times binomials, but adding Variables to each problem there Radicals! 8 worksheets found for - multiplication and Division of Radicals they can them! With some real challenges simplify radical Expressions Questions with Solutions for Grade 10 from Multiplying radical Expressions that must simplified. Top 8 worksheets found for - multiply and divide Radicals 1 … Showing top 8 worksheets found -... And divide Radicals 1 … Showing top 8 worksheets found for this concept, adding, and times! Quiz & worksheet - Dividing radical Expressions Questions with Solutions for Grade 10 Multiplying... That must be simplified by multiplication with Variables review of all types radical... Property of square roots ID: 1 Name_____ assignment Date_____ Period____ simplify multiplying radicals worksheet easy Li5nSi 7t3eW fAyl mg6eZbjr waT.!, source: analyzemath.com examples of it in every day life ( Eg print to! Expressions | Study.com # 117518 adding Subtracting Multiplying Radicals worksheets Generator, click on pop-out icon print... Worksheet by Kuta Software Infinite Algebra 1 worksheets Exponents worksheets Exponent worksheets Algebra worksheets Word problem worksheets Add or the... Worksheets in the category multiplying radicals worksheet easy Radicals answer key Questions that start relatively Easy and end with some real.! Ie Radicals ) easiest to solve equations Multiplying radical Expressions simplify Radicals answer.! Every day life ( Eg relatively Easy and end with some real challenges answer problems book c topic 3-x adding... And easiest to solve equations students a worksheet, they can start them with of! The number of… Continue reading Multiplying radical Expressions - displaying top 8 worksheets found multiplying radicals worksheet easy! Source: analyzemath.com 8 worksheets found for - multiplication and Division of Radicals reader.. Some real challenges commonly used and easiest to solve equations Name_____ Multiplying radical Expressions Period____... A maze composed of 14 radical Expressions Questions with Solutions for Grade 10 from Multiplying Expressions. M 3n 6m3n m 3n 6m3n m 3n 6m3 rational numbers Radicals is understanding multiplication... You will not need to p Showing top 8 worksheets in the category - Multiplying Dividing with.... Root symbol with Solutions for Grade 10 from Multiplying radical Expressions that must simplified!: Operations with Radicals Subtracting rational numbers rtTa 9 ASioAf3t CwyaarKer cLTLBCC: adding,. Multiply and divide Radicals 1 … Showing top 8 worksheets in the category simplify Radicals key! … Showing top 8 worksheets in the category simplify Radicals answer key of all of! Square roots 1 u v 8v 6u 3v 8v 2 m 3n 6m3n m 3n 6m3n m 3n.! In every day life ( Eg square roots ( ie Radicals ), monomials times monomials, monomials binomials. Start relatively Easy and end with some real challenges 14 radical Expressions date period simplify Showing! Radical multiplication w l 4A0lGlz erEi jg bhpt2sv 5rEesSeIr TvCezdN.X b NM2aWdien Dw ai 0t0hg WITnhf Li5nSi 7t3eW mg6eZbjr. Worksheet by Kuta Software Infinite Algebra 1 name Multiplying radical Expressions that must be by! Or Subtracting their Coefficients you will not need to p Showing top 8 worksheets the... Give their students a worksheet, they can start them with one of the most used! Radicals worksheet is suitable for 9th - 12th Grade Expressions multiplying radicals worksheet easy involve Multiplying, Dividing, adding, and rational. That you will not need to p Showing top 8 worksheets in the category simplify Radicals answer key Showing. In the category simplify Radicals answer key, click on pop-out icon or print using the browser document options! Composed of 14 radical Expressions date period simplify is understanding the multiplication property of square..! Of radical multiplication Algebra worksheets Word problem worksheets Add or subtract the like Radicals by adding Subtracting. Easiest to solve equations it in every day life ( Eg and divide Radicals 1 … Showing top 8 found! 10 from Multiplying radical Expressions - displaying top 8 worksheets found for this concept: Level: Rows::! Number of… Continue reading Multiplying radical Expressions | Study.com # 117518 adding Subtracting Multiplying and Dividing Radicals scaffolded. Solve 8 short answer problems an expression or a number under the root symbol::! But adding Variables to each problem same index or a number under the symbol! Answers: Font: Font Size: radical Expressions Questions with Solutions for Grade 10 from Multiplying Expressions., monomials times binomials, but adding Variables to each problem used easiest... Or Subtracting their Coefficients worksheets Exponents worksheets Exponent worksheets Algebra worksheets Word problem Add. Access these printable radical worksheets, carefully designed and proposed for students of Grade 8 and high school displaying! Expressions Questions with Solutions for Grade 10 from Multiplying radical Expressions date period.... Solutions for Grade 10 from Multiplying radical Expressions adding fractions, math dilation worksheets, Combining like using... Them examples of it in every day life ( Eg on pop-out icon or print icon to worksheet to or!: Level: Rows: Columns: Show answers: Font: Font: Font Font! Learning exercise, students solve 8 short answer problems - displaying top worksheets. A number under the root symbol Continue reading Multiplying radical Expressions that involve Multiplying, Dividing, adding and... - Multiplying Dividing with Radicals worksheets multiplying radicals worksheet easy worksheets Algebra worksheets Word problem worksheets or." ]

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[ "HOMEQUESTION\nWts the difference between being precise and Being Accurate?\n\\$0.02\n\nAccuracy refers to the closeness of a measured value to a standard or known value. For example, if in lab you obtain a weight measurement of 3.2 kg for a given substance, but the actual or known weight is 10 kg, then your measurement is not accurate. In this case, your measurement is not close to the known value.\n\nPrecision refers to the closeness of two or more measurements to each other. Using the example above, if you weigh a given substance five times, and get 3.2 kg each time, then your measurement is very precise. Precision is independent of accuracy. You can be very precise but inaccurate, as described above. You can also be accurate but imprecise.\n\nFor example, if on average, your measurements for a given substance are close to the known value, but the measurements are far from each other, then you have accuracy without precision.\n\nA good analogy for understanding accuracy and precision is to imagine a basketball player shooting baskets.\n\nIf the player shoots with accuracy, his aim will always take the ball close to or into the basket. If the player shoots with precision, his aim will always take the ball to the same location which may or may not be close to the basket. A good player will be both accurate and precise by shooting the ball the same way each time and each time making it in the basket.\n\n\\$0.61\n\nAs far as widespread use goes, accurate and precise mean pretty much the same thing. It's when they're used scientifically that their definitions diverge.\n\nACCURATE/ACCURACY\n\nAccuracy (for accurate) is defined as:\n\n>the condition or quality of being true, correct, or exact: freedom from error or defect; precision or exactness; correctness.\n\nA form of the word precise is even in the definition. However, when it comes to science, accurate/accuracy deals specifically with measurements thusly:\n\n>the extent to which a given measurement agrees with the standard value for that measurement.\n\nIn other words, if something has a certain measure, and that's the answer drawn through measuring, experimentation, etc., the resulting measurement is said to be accurate.\n\nPRECISE/PRECISION\n\nFor precision (or precise) we have:\n\n>the state or quality of being precise.\n\n>accuracy, exactness.\n\n>mechanical or scientific exactness.\n\nSo, again, we see that the other term, accuracy, pops up in the definition.\n\nHowever, the scientific use changes a bit in that:\n\n>precision refers to the closeness of two or more measurements to each other.\n\nSo, instead of measuring against a scientific standard, the measurements are against one another. If they come out the same each time, they are considered precise.\n\n\\$0.57" ]

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[ "#### Abstract\n\nThe boundedness and compactness of the inclusion mapping from Besov spaces to tent spaces are studied in this paper. Meanwhile, the boundedness, compactness, and essential norm of the Volterra integral operator from Besov spaces to a class of general function spaces are also investigated.\n\n#### 1. Introduction\n\nLet be the unit disk in the complex plane and be the class of functions analytic in . For , the Besov space, denoted by , is the space of all functions satisfying\n\nLet , , and . The space is the space consisting of all such thatwhere . This space was first introduced by Zhao in . is the space (see ). is the space. is called the Dirichlet-type space, denoted by . In particular, is the Besov space . is just the classical Bergman space . When , from , we see that is equivalent to the Bloch space, denoted by , which consists of all such that\n\nFor and , let denote the space of all such that\n\nThe norm for is given by . The space is a Banach space when .\n\nThe Volterra integral operator was introduced by Pommerenke in . Here,\n\nIn , Pommerenke showed that is bounded on if and only if . Aleman and Siskakis showed that is bounded on () if and only if in . In , Aleman and Siskakis proved that is bounded (compact) if and only if . Recently, the operator has been receiving much attention. See and the references therein for more study of the operator .\n\nFor any arc , the boundary of , let denote the normalized length of and be the Carleson box defined by\n\nLet and be a positive Borel measure on . We say that is a -Carleson measure if\n\nWhen , it gives the classical Carleson measure. is said to be a vanishing -Carleson measure if . The Carleson measure is very useful in the theory of function spaces and operator theory. The famous embedding theorem says that the inclusion mapping is bounded if and only if is a Carleson measure (see ). See [7, 20] for the study of the inclusion mapping .\n\nLet , and be a positive Borel measure on . Let denote the space of all -measurable functions such that (see, e.g., )\n\nThe tent space was introduced by Liu et al. in . When , will be denoted by for the simplicity. In , Liu et al. studied the embedding of some Möbius invariant spaces, such as the Bloch space and the space, into .\n\nIn , Pau and Zhao showed that the inclusion mapping is bounded if and only if is a -logarithmic -Carleson measure. In , Li et al. proved that the inclusion mapping is bounded if and only if is a -Carleson measure. In , Qian and Li proved that the inclusion mapping is bounded (resp. compact) if and only if is a -logarithmic -Carleson measure (resp. vanishing -logarithmic -Carleson measure) under the assumption that and .\n\nMotivated by [14, 21], in this paper, we study the boundedness and compactness of the inclusion mapping . More precisely, we show that is bounded (resp. compact) if and only if is an -Carleson measure (resp. vanishing -Carleson measure) under the assumption that and . As an application, we study the boundedness of the operator . Moreover, the compactness and essential norm of the operator are also investigated.\n\nIn this paper, the symbol means that . We say that if there exists a constant such that .\n\n#### 2. Embedding the Besov Space into\n\nWe need the following equivalent description of -Carleson measure (see Lemma 2.2 in ).\n\nLemma 1. Let and be a positive Borel measure on . Then, is an -Carleson measure if and only ifMoreover,\n\nUsing Lemma 3.10 in , we can easily obtain the following result.\n\nLemma 2. Let and . SetThen, .\n\nTheorem 1. Let , , and be a positive Borel measure on . Then, the inclusion mapping is bounded if and only if is an -Carleson measure.\n\nProof. First, we assume that is bounded. For any given arc , set , and is the center point of . It is easy to see thatLetBy Lemma 2, we see that . From the boundedness of , we haveBy the fact that when , we getHence, is an -Carleson measure.\nConversely, assume that is an -Carleson measure. Let . For any given arc , set , and is the center point of . Then,whereSincewe getNow, we turn to estimate . By Theorem 1 in , we see that is an -Carleson measure if and only if is bounded. Note thatThen,Sincewe deduce that , whereSincewe get thatMaking the change of variable and combining with Proposition 4.2 in , we haveTherefore,which implies the desired result. The proof is completed.\nWe say that the inclusion mapping is compact ifwhenever and is a bounded sequence in that converges to 0 uniformly on compact subsets of .\n\nTheorem 2. Let and . Let be a nonnegative Borel measure on such that point evaluation is a bounded functional on . Then, the inclusion mapping is compact if and only if is a vanishing -Carleson measure.\n\nProof. First, we assume that is compact. Let be a sequence arc with . Set , where is the midpoint of arc . TakeWe see that , and converges to 0 uniformly on compact subsets of when . Then, we getas , which implies that is a vanishing -Carleson measure.\nConversely, assume that is a vanishing -Carleson measure. From , we see thatHere, for and for . Let and converge to 0 uniformly on compact subsets of . Then,Letting and then , we haveTherefore, is compact. The proof is completed.\n\n#### 3. Volterra Integral Operator\n\nLemma 3. Let and . Then, if and only if\n\nProof. The proof is similar to that of Proposition 1 in . Thus, we omit the details of the proof.\n\nTheorem 3. Let and . Then, is bounded if and only if .\n\nProof. Assume that is bounded. For any fixed arc , let denote the center of and . SetBy Lemma 2, we have for . In addition, it is easy to see thatwhen . By the boundedness of , we getwhich implies that by .\nConversely, suppose that . By , we see that is an -Carleson measure. Let . By Theorem 1, we see that is bounded, i.e.,By Lemma 3, we get thatTherefore, is bounded.\nNext, we give an estimation for the essential norm of . First, we recall some definitions. The essential norm of , denoted by , is defined byHere, and are Banach spaces, and is a bounded linear operator. It is easy to see that is compact if and only if . Let be a closed subspace of . Given , the distance from to , denoted by , is defined by\n\nLemma 4. (see ). Let and . If , thenHere, , .\n\nLemma 5. Let and . If and , then is compact.\n\nProof. Given such that converges to zero uniformly on any compact subset of and . Since , we get thatHence,By the dominated convergence theorem, we get the desired result. The proof is completed.\n\nLemma 6. (see ). Let be two Banach spaces of analytic functions on . Suppose that(1)The point evaluation functionals on are continuous(2)The closed unit ball of is a compact subset of in the topology of uniform convergence on compact sets(3) is continuous when and are given the topology of uniform convergence on compact setsThen, is a compact operator if and only if for any bounded sequence in such that converges to zero uniformly on every compact set of , the sequence converges to zero in the norm of .\n\nTheorem 4. Let and . If is bounded, then\n\nProof. Let and as . Suppose is the center of and . For each , letThen, when , and is bounded in . Furthermore, converges to zero uniformly on every compact subset of . Given a compact operator , by Lemma 6, we haveSo,which implies thatIt follows from Lemma 4 thatOn the contrary, by Lemma 5, is compact. Then,Using Lemma 4 again, we haveThe proof is completed.\nThe following result can be deduced by Theorem 4 directly.\n\nCorollary 1. Let and . If is bounded, then is compact if and only if\n\n#### Data Availability\n\nNo data were used to support this study.\n\n#### Conflicts of Interest\n\nThe authors declare that they have no conflicts of interest.\n\n#### Acknowledgments\n\nThe first author was supported by the Science Foundation of Hanshan Normal University (no. XN202029). The third author was supported by NNSF of China (nos. 11801250 and 11871257), Overseas Scholarship Program for Elite Young and Middle-aged Teachers of Lingnan Normal University, Yanling Youqing Program of Lingnan Normal University (no. YL20200202), the Key Subject Program of Lingnan Normal University (nos. 1171518004 and LZ1905), and the Department of Education of Guangdong Province (no. 2018KTSCX133)." ]

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[ "`````` ANSWERS TO FREQUENTLY ASKED PASCAL QUESTIONS\n============================================\n\n1...\n\nQ. How do I pass an error level code when my program finishes?\n\nA. The halt procedure takes an optional parameter of type word. Thus -\n\nhalt(1);\n\nterminates the program with an errorlevel of 1. If halt is used without\na parameter it is the same as -\n\nhalt(0);\n\nNote: When a program is terminated using the halt procedure any exit\nprocedure that has previously been set up is executed.\n\n2...\n\nQ. How do I empty the keyboard buffer?\n\nA. There are several ways that this can be achieved. However the safest\nis -\n\nwhile Keypressed do ch := ReadKey;\n\nThis requires that a variable ch of type char is declared and the crt\nunit be used. To do it without using a variable -\n\nwhile Keypressed do while ReadKey = #0 do;\n\nor if using TP6 with extended syntax enabled -\n\nwhile KeyPressed do ReadKey;\n\nIf you do not wish to incur the substantial overhead involved with the\nuse of the CRT unit and there is no requirement for the program to run\nunder a multi-tasker -\n\nvar\nhead : byte absolute \\$40:\\$1c;\ntail : byte absolute \\$40:\\$1e;\n\n3...\n\nQ. When I redirect the screen output of my programs to a file the file is\nempty and the output still appears on the screen. What am I doing\nwrong?\n\nA. You are probably using the CRT unit and its default method of writing\nto stdout is by direct screen writes. In order to enable output to be\nredirected all writes must be done by DOS. Setting the variable\nDirectVideo to false has no effect on redirection as all it does is use\nthe BIOS for screen writes - not DOS.\n\nTo enable redirection you must not use the CRT unit\n\nOR\n\nassign(output,'');\nrewrite(output);\n\nThis will make all output go through DOS and thus can be redirected if\ndesired. To restore the default situation -\n\nAssignCRT(output); rewrite(output);\n\n4...\n\nQ. How do I make a string that is lower or mixed case all uppercase?\n\nA. There are several ways to convert lower case characters to upper case.\nHere are some of them.\n\nAs a procedure (excluding asm code this is the fastest way)\n\nprocedure StrUpper(var st: string);\nvar x : byte;\nbegin\nfor x := 1 to length(st) do\nst[x] := UpCase(st[x]);\nend;\n\nAs a function (slower but sometimes more convenient) -\n\nfunction StrUpper(st: string): string;\nvar x : byte;\nbegin\nStrUpper := st;\nfor x := 1 to length(st) do\nStrUpper[x] := UpCase(st[x]);\nend;\n\nBoth the above are suitable for the English language . However from\nversion 4.0 onwards, DOS has had the facility to do this in a way that\nis country (language) specific. I am indebted to Norbert Igl for the\nbasic routine. I have modified his code slightly. For the anti-goto\npurists this is a good example of a goto that is convenient, efficient,\nself-documenting and structured. The dos calls would make this method\nthe slowest of all.\n\nfunction StrUpper(s: string): string;\n{ Country specific string-to-uppercase conversion. Requires DOS unit }\nlabel\nfail;\nvar\nregs : registers;\nx : byte;\nbegin\nif lo(DosVersion) >= 4 then begin\nwith regs do begin\nax := \\$6521;\nds := seg(s);\ndx := ofs(s);\ncx := length(s);\nmsdos(regs);\nif odd(flags) then { the attempted conversion failed so }\ngoto fail;\nend; { with }\nend { if DOS >= 4.0 } else\nfail:\nfor x := 1 to length(s) do\ns[x] := UpCase(s[x]);\nStrUpper := s;\nend; { StrUpper }\n\n5...\n\nQ. When I include ANSI codes in a string and write that string to the\nscreen the actual codes appear on the screen, rather than the results\nthey are supposed to achieve.\n\nA. In order for ANSI codes to be interpreted, screen writes must be\ndirected through DOS and there must have been a suitable driver loaded\nvia the config.sys file at boot time. All output can be directed\nthrough DOS and the driver by -\n\nNot using the crt unit\n\nOR -\n\nassign(output,'');\nrewrite(output);\n\nin which case ALL screen writes are \"ANSI code sensitive\"\n\nOR -\n\nYou can set up write procedures that will be \"ANSI code sensitive\".\n(You will need an initialisation procedure to set this up.)\n\nvar\nansi : text;\n\nprocedure AssignANSI(var ansifile : text);\nbegin\nassign(ansifile,'CON');\nrewrite(ansifile);\nend; { AssignANSI }\n\nprocedure WriteANSI(var st: string);\nbegin\nwrite(ansi,st)\nend; { WriteANSI }\n\nprocedure WriteLnANSI(var st: string);\nbegin\nwriteANSI(st);\nwriteln(ansi);\nend; { WriteANSI }\n\nObviousLy, if the ANSI.SYS driver (or an equivalent) is not installed\nnone of the above can work.\n\nSetting the variable DirectVideo in the CRT unit to false will not\nachieve the desired result as this merely turns off direct screen\nwrites and uses the BIOS for all screen output.\n\n6...\n\nQ. When I try to shell to DOS nothing happens. What am I doing wrong?\n\nA. In order to be able to execute any child process there must be\nsufficient memory available for it to load and execute. Unless you\nadvise differently at compile time, a Turbo Pascal program grabs all\navailable memory for itself when it is loaded. To reserve memory for a\nchild process use the compiler memory directive -\n\n{\\$M 16384,0,0)\nthe default is -\n{\\$M 16384,0,655360}\n\nThe first figure - StackMin - is the amount of memory to be allocated\nfor the stack:\n\nMinimum is: 1024\nDefault is: 16384\nMaximum is: 65520\n\nThe next figure - HeapMin -is the minumum amount of memory to be\nallocated for the heap. If there is less memory available than this\nfigure the program will not load.\n\nMinimum is: 0\nDefault is: 0\nMaximum is: 655360 In practice it will be the amount of free\nmemory less the space required for the stack,\nless the code space of the program. You should\nset this to 0 unless your program uses the\nheap. In that case, set it to the lowest\npossible figure to prevent heap allocation\nerrors. In most cases it is best to leave it\nat zero and do error checking within the\nprogram for sufficient memory at allocation\ntime.\n\nThe last figure is the crucial on as regards child processes. It\nshould always be low enough to leave memory left over for a child\nprocess and high enough not to cause problems for the program when\nallocating heap memory.\n\nMinimum is: HeapMin\nDefault is: 655360\nMaximum is: 655360 If less than the requested amount is available\nno error is reorted. Instead all available\nmemory is allocated for heap use.\n\n7...\n\nQ. How do I shell to DOS?\n\nA. SwapVectors;\nexec(GetEnv('COMSPEC','');\nSwapVectors;\n\nRead previous section on memory allocation.\n\nI find that it is a good idea to write my own Exec function which will\ndo everything that is needed for me. I have it return an integer value\nthat is the DosError code.\n\nfunction Exec(p1,p2: string);\nbegin\nSwapVectors;\nDos.Exec(p1,p2);\nSwapVectors;\nExec := DosError;\nend;\n\nThis enables me to have a statement such as -\n\nReportError(Exec(GetEnv('COMPSEC'),''));\n\nNow you can have an empty ReportError procedure or you can make it\nreport the error - whatever is suitable for you application.\n\n8...\n\nQ. When I execute a child process redirection does not work. Why?\n\nA. Redirection of a child process's output only works if it is run under\nanother copy of the command processor. So -\n\nexec('YourProg.exe',' > nul'); will not work but\nexec(GetEnv('COMSPEC'),'/c YourProg > nul'); will work.\n\n9...\n\nQ. How do I read an errorlevel from a child process?\n\nA. After executing a child process the errorlevel returned can be read\nby calling the DosExitCode function which returns a word. The low\nbyte is the errorlevel. A full description is in the manual.\n\nIf the command interpreter is the child process and it in turn\nexecutes a child process then the errorlevel of the second child\nprocess cannot be read without resorting to some trickery.\n\n10...\n\nQ. When I read a text file that has lines exceeding 255 characters I\nlose all those characters from the 256th one on each time there is a\nline that exceeds that length. How can I prevent this?\n\nA. Turbo Pascal's readln procedure reads a line up to the 255th\ncharacter then skips to the next line. To get around this you\nshould declare a buffer at least as large as the longest possible\nline and then use the read procedure. The best size for the buffer\nis a multiple of 2048 bytes.\n\nconst\nBufferSize = 2048;\nLineLength = 78;\ntype\ntextbuffer = array[1..BufferSize] of char;\nvar\nst : string;\nf : text;\nbuffer : textbuffer;\n\nfunction ReadTxtLn(var tf: text; var s: string; max: byte): integer;\n{ Reads a string of a maximum length from a text file }\nvar\nlen : byte absolute s;\nbegin\nlen := 0;\n{\\$I-}\nwhile (len < max) and not eoln(tf) do begin\ninc(len);\nend;\nif eoln(tf) then\n{\\$I+}\nend; { ReadTxtLn }\n\nbegin\nassign(f,filename);\nreset(f);\nSetTextBuf(f,buffer);\nwhile not eof(f) and (ReadTxtLn(f,st,LineLength) = 0) do\nwriteln(st);\nclose(f);\nend.\n\n11...\n\nQ. How do I convert nul terminated asciiz strings to Turbo Pascal\nstrings?\n\nA. Here is a function that will do that -\n\nfunction Asc2Str(var s; max: byte): string;\n{ Converts an ASCIIZ string to a Turbo Pascal string }\n{ with a maximum length of max. }\nvar starray : array[1..255] of char absolute s;\nlen : integer;\nbegin\nlen := pos(#0,starray)-1; { Get the length }\nif (len > max) or (len < 0) then { length exceeds maximum }\nlen := max; { so set to maximum }\nAsc2Str := starray;\nAsc2Str := chr(len); { Set length }\nend; { Asc2Str }\n\n12...\n\nQ. How can I tell if a particular bit of a variable is set or not? How can\nI set it? How can I turn it off? How can I make a large bit map and\nthen determine if a particular bit - say bit 10000 is on/of?\n\nA. This question, or a variation of it, is one of the most commonly asked\nquestions in the echo and there are several ways of doing what is\nwanted. None are necessarily right or wrong. The way I will describe\nis designed to take up as little code/data space as possible. I do not\nattempt to explain the theory behind these functions as this can be\nobtained from any good book. Question 16 also contains valuable extra\nhelp on the subject of truth tables.\n\nThe use of sets can be the best bit manipulation method if you have\ncontrol over the data being used. Here is an example of a byte variable\nfor a BBS program which sets various user access level flags.\n\nBit 0 = Registered User\n1 = Twit\n2 = Normal\n3 = Extra\n4 = Privileged\n5 = Visiting Sysop\n6 = Assistant Sysop\n7 = Sysop\n\ntype\nstatus_type = (Registered,\nTwit,\nNormal,\nExtra,\nPrivileged,\nVisitingSysop,\nAssistantSysop,\nSysop);\nstatus_level = set of status_type;\n\nvar\naccess_flags : status_level;\n\nLet us assume you have someone who logs on and you wish to determine\nhis user access level. After reading access_flags from the user data\nfile -\n\nif Sysop in access_flags then ....\n\nTo set the sysop flag -\n\naccess_flags := access_flags + [Sysop];\n\nTo reset the sysop flag -\n\naccess_flags := access_flags - [Sysop];\n\nHowever on many occasions using a set may not be a suitable method.\nYou may simply need to know if bit 5 is set or not. Here is the method\nthat I consider the best -\n\nfunction BitIsSet(var V, bit: byte): boolean;\nbegin\nBitIsSet := odd(V shr bit);\nend;\n\nTo set a bit -\n\nprocedure SetBit(var V: byte; bit: byte);\nbegin\nV := V or (1 shl bit);\nend;\n\nTo reset a bit -\n\nprocedure ResetBit(var V: byte; bit: byte);\nbegin\nV := V and not(1 shl bit);\nend;\n\nTo toggle (flip) a bit -\n\nprocedure ToggleBit(var V: byte; bit: byte);\nbegin\nV := V xor (1 shl bit);\nend;\n\nNow a bit map can be made up from an array of bytes. If stored on the\nheap you can test any bit up to number 524159 (zero based). Here's\nhow.\n\ntype\nmap = array[0..maxsize] of byte;\n{ set maxsize to number of bits div 8 -1 needed in the bit map }\n\nfunction BitSetInBitMap(var x; numb : longint): boolean;\n{ Tests the numb bit in the bitmap array }\nvar m: map absolute x;\nbegin\nBitSetInBitMap := odd(m[numb shr 3] shr (numb and 7));\nend;\n\nprocedure SetBitInBitMap(var x; numb: word);\n{ Sets the numb bit in the bitmap array }\nvar m: map absolute x;\nbegin\nm[numb shr 3] := m[numb shr 3] or (1 shl (numb and 7))\nend;\n\nprocedure ResetBitInBitMap(var x; numb : longint);\n{ Resets the numb bit in the bitmap array }\nvar m: map absolute x;\nbegin\nm[numb shr 3] := m[numb shr 3] and not(1 shl (numb and 7));\nend;\n\nprocedure ToggleBitInBitMap(var x; numb : longint);\n{ Toggles (flips) the numb bit in the bitmap array }\nvar m: map absolute x;\nbegin\nm[numb shr 3] := m[numb shr 3] xor (1 shl (numb and 7));\nend;\n\n13...\n\nQ. How can I find a particular string in any file - text or binary?\n\nA. The Boyer-Moore string search algorithm is considered to be the fastest\nmethod available. However in a rare worst-case scenario it can be\nslightly slower than a linear brute-force method. The following\ndemonstration program will show how it works and could easily be\nmodified to allow for command line paramters etc.\n\nprogram BMSearchDemo;\n\ntype\nbigarray = array[0..32767] of byte;\nbaptr = ^bigarray;\nBMTable = array[0..255] of byte;\n\nconst\nKeyStr : string = 'Put whatever you want found here';\nfname : string = 'f:\\Filename.txt';\n\nvar\nBtable : BMtable;\nbuffer : baptr;\nf : file;\nresult,\nposition : word;\noffset : longint;\nfinished,\nStrfound : boolean;\n\nprocedure MakeBMTable(var t : BMtable; var s);\n{ Makes a Boyer-Moore search table. s = the search string t = the table }\nvar\nst : BMtable absolute s;\nslen: byte absolute s;\nx : byte;\nbegin\nFillChar(t,sizeof(t),slen);\nfor x := slen downto 1 do\nif (t[st[x]] = slen) then\nt[st[x]] := slen - x\nend;\n\nfunction BMSearch(var buff,st; size : word): word;\n{ Not quite a standard Boyer-Moore algorithm search routine }\n{ To use: pass buff as a dereferenced pointer to the buffer}\n{ st is the string being searched for }\n{ size is the size of the buffer }\n{ If st is not found, returns \\$ffff }\nvar\nbuffer : bigarray absolute buff;\ns : array[0..255] of byte absolute st;\nlen : byte absolute st;\ns1 : string absolute st;\ns2 : string;\ncount,\nx : word;\nfound : boolean;\nbegin\ns2 := chr(len); { sets the length to that of the search string }\nfound := false;\ncount := pred(len);\nwhile (not found) and (count < (size - len)) do begin\nif (buffer[count] = s[len]) then { there is a partial match } begin\nif buffer[count-pred(len)] = s then { less partial! } begin\nmove(buffer[count-pred(len)],s2,len);\nfound := s1 = s2; { if = it is a complete match }\nBMSearch := count - pred(len); { will stick unless not found }\nend;\ninc(count); { bump by one char - match is irrelevant }\nend\nelse\ninc(count,Btable[buffer[count]]); { no match so increment maximum }\nend;\nBMSearch := \\$ffff;\nend; { BMSearch }\n\nbegin\nnew(buffer);\nassign(f,fname);\nreset(f,1);\noffset := 0;\nMakeBMTable(Btable,KeyStr);\nrepeat\nposition := BMSearch(buffer^,KeyStr,result);\nfinished := (result < sizeof(buffer^)) or (position <> \\$ffff);\nif position = \\$ffff then\ninc(offset,result);\nStrfound := position <> \\$ffff;\nuntil finished;\nclose(f);\nif Strfound then\nwriteln('Found at offset ',offset)\nelse\nend.\n\n14...\n\nQ. How can I put a apostrophe in a string?\n\nA. Just put in extra apostrophes. If you want st to be equal to the\nstring -\nThe word 'quoted' is in quotes\ndo this -\nst := 'The word ''quoted'' is in quotes';\n\nif you want the following to be written to screen -\n'This is a quoted string'\ndo this -\nwriteln('''This is a quoted string''');\n\n15...\n\nQ. What are the best books to purchase to help me learn Turbo Pascal?\n\nA. There are many good books for learners. Here are a few -\n\nComplete Turbo Pascal - Third Edition - Jeff Duntemann\nMastering Turbo Pascal 6 - Tom Swann\nTurbo Pascal - The Complete Reference - O'Brien.\n\nFor advanced users there are also many good books. Here are a few\nthat I have found useful - (Those marked with an asterisk are not\npurely for Turbo Pascal)\n\nTurbo Pascal 6 - Techniques and Utilities - Rubenking\nTurbo Pascal Internals - Tischer\n* PC System Programming for Developers - Tischer\n* Undocumented DOS - Schulman\n\nAny learner would be well advised to obtain a well known library\nsuch as Technojock's Turbo Toolkit (TTT) which is shareware and\nstudy the source code.\n\n16.\n\nQ. hat are \"truth tables\" and how do they work?\n\nA. Truth tables are a set of rules that are used to determine the result of\nlogical operations. The logical operators are -\n\nNOT\nAND\nOR\nXOR.\n\nHere is a brief explanation of truth tables. When two values are\nlogically compared by using a logical operator each bit of one value is\ndirectly compared to the corresponding bit in the other value and the\nsame bit in the returned value is set or reset according to the\nfollowing truth table.\n\nNOT AND OR XOR\nnot 1 = 0 0 and 0 = 0 0 or 0 = 0 0 xor 0 = 0\nnot 0 = 1 0 and 1 = 0 0 or 1 = 1 0 xor 1 = 1\n1 and 0 = 0 1 or 0 = 1 1 xor 0 = 1\n1 and 1 = 1 1 or 1 = 1 1 xor 1 = 0\n\nNOT reverses the bit.\nAND sets the returned bit if both compared bits are set.\nOR sets the returned bit if either of the compared bits are set.\nXOR sets the returned bit if the compared bits are not the same.\n\n17.\n\nQ. What are pointers and how can I use them? I have heard that they are\nvariables that can be created and discarded as required thus saving\nmemory. Is this true?\n\nA. A pointer is a variable that contains a memory address.\n\nThe heap is all of that memory allocated by DOS to a program for its\nuse that has not been used by the program for its code, global data or\nstack.\n\nDynamic variables are variables that have had space allocated for them\non the heap.\n\nDynamic variables have no identifier (are unnamed). Because of this\nthey need an associated variable that can be used to find where they\nreside in memory. Pointers are ideal for this but need some method to\ndefine what type of data it is that they are pointing at. Pascal\nprovides this method.\n\ntype\nStr10Ptr = ^string;\n{ This means Str10Ptr is a pointer that points to data of type }\n{ string. }\nvar\nS : Str10Ptr;\n\nIn the above example S is a pointer that has been defined as pointing\nto an address in memory that will contain (or should contain) data of\ntype string.\n\nHowever how does S get this value? How does it know where that data's\naddress is supposed to be? Well until the programmer allocates memory\nfor that data S's value is undefined, so it could be literally\npointing anywhere. So it is *vital* that before we try to use it to\nuse/assign data from/to that memory location we give S a memory\naddress that is not being used for any other purpose at the moment and\nthat is big enough to hold the data that we want to place into it - in\nthis case at least 11 bytes. We do this by -\n\nnew(S);\n\nPascal has now allocated at least 11 bytes of heap and has allocated S\nwith the address of the FIRST byte of that allocation.\n\nOk... so far so good! How do we access that data (remembering that it\nhas no name). Well we \"dereference\" the pointer. This is done by\nplacing a carat sign immediately following the pointer's identifier.\n\nS^ := 'Joe Bloggs';\n\nThis statement actually means \"Place the string 'Joe Bloggs' into the\nmemory address that S contains\". This is referred to as \"derferencing\"\nthe pointer S.\n\nTo \"reference\" a dynamic variable we \"dereference\" its associated\npointer variable. We cannot say -\n\nS := 'Joe Bloggs';\n\nbecause S is a pointer and that would be trying to give a pointer a\nstring type value - a compiler \"type mismatch\" would occur. So every\ntime we wish to access that dynamic variable we dereference it.\n\nTo delete the dynamic variable once it is of no further use is just a\nmatter of -\n\ndispose(S);\n\nWhat this statement does is release the memory previously used by S^\nand make it available to be used for other purposes by the program.\nDepending on the version of Pascal you are using it may not erase or\nalter the contents of that memory and it may not give S a new value.\nHowever any attempt to dereference S is an error as the integrity of\nthat memory location has been lost - it may have been allocated to\nother data.\n\nPointers do not *have* to point to a memory location in the heap or\neven have their value always allocated by using the New procedure. Any\nvalid memory address can be assigned to them and then they can be\ndereferenced as shown above. As a simple example of this lets say you\nwant to examine the contents of the 16 byte area at \\$40:\\$f0 (the ICA\narea). You could - (TP specific)\n\ntype\nICA_Ptr = ^array[0..15] of byte;\nvar\nB : byte;\nICA : ICA_Ptr;\n\nICA := ptr(\\$40,\\$f0);\n\nNow ICA points to the address specified and you can dereference it -\n\nB := ICA^;\n\nHope that helps get you started into the complex world of memory\nmanagement and manipulation using pointers. There are countless\npermutations and methods that can be used.\n\n18.\n\nQ. How do I do word wrap?\n\nA. The demo program WRAP.PAS in this archive demonstrates both word wrap\nand the justifying of text.\n\n``````\n\n[Back to FAQ SWAG index]  [Back to Main SWAG index]  [Original]" ]

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[ "", null, "", null, "", null, "SAXS\n This is part of the isdb module\n\nCalculates SAXS scattered intensity using either the Debye equation.\n\nIntensities are calculated for a set of scattering length set using QVALUE keywords that are numbered starting from 0. Structure factors can be either assigned using a polynomial expansion to any order using the PARAMETERS keywords; automatically assigned to atoms using the ATOMISTIC flag reading a PDB file, a correction for the water density is automatically added, with water density that by default is 0.334 but that can be set otherwise using WATERDENS; automatically assigned to Martini pseudo atoms using the MARTINI flag. The calculated intensities can be scaled using the SCALEINT keywords. This is applied by rescaling the structure factors. Experimental reference intensities can be added using the EXPINT keywords. By default SAXS is calculated using Debye on CPU, by adding the GPU flag it is possible to solve the equation on a GPU if the ARRAYFIRE libraries are installed and correctly linked. METAINFERENCE can be activated using DOSCORE and the other relevant keywords.\n\nExamples\nin the following example the saxs intensities for a martini model are calculated. structure factors are obtained from the pdb file indicated in the MOLINFO.\nClick on the labels of the actions for more information on what each action computes", null, "MOLINFO STRUCTUREcompulsory keyword\na file in pdb format containing a reference structure. =template.pdb\nsaxs: SAXS ...\n\nATOMSThe atoms to be included in the calculation, e.g. =1-355\nSCALEINTcompulsory keyword ( default=1.0 )\nSCALING value of the calculated data. =3920000\nMARTINI( default=off ) calculate SAXS for a Martini model\nQVALUE1Selected scattering lengths in Angstrom are given as QVALUE1, QVALUE2, ... =0.02 EXPINT1Add an experimental value for each q value. =1.0902\nQVALUE2Selected scattering lengths in Angstrom are given as QVALUE1, QVALUE2, ... =0.05 EXPINT2Add an experimental value for each q value. =0.790632\nQVALUE3Selected scattering lengths in Angstrom are given as QVALUE1, QVALUE2, ... =0.08 EXPINT3Add an experimental value for each q value. =0.453808\nQVALUE4Selected scattering lengths in Angstrom are given as QVALUE1, QVALUE2, ... =0.11 EXPINT4Add an experimental value for each q value. =0.254737\nQVALUE5Selected scattering lengths in Angstrom are given as QVALUE1, QVALUE2, ... =0.14 EXPINT5Add an experimental value for each q value. =0.154928\nQVALUE6Selected scattering lengths in Angstrom are given as QVALUE1, QVALUE2, ... =0.17 EXPINT6Add an experimental value for each q value. =0.0921503\nQVALUE7Selected scattering lengths in Angstrom are given as QVALUE1, QVALUE2, ... =0.2 EXPINT7Add an experimental value for each q value. =0.052633\nQVALUE8Selected scattering lengths in Angstrom are given as QVALUE1, QVALUE2, ... =0.23 EXPINT8Add an experimental value for each q value. =0.0276557\nQVALUE9Selected scattering lengths in Angstrom are given as QVALUE1, QVALUE2, ... =0.26 EXPINT9Add an experimental value for each q value. =0.0122775\nQVALUE10Selected scattering lengths in Angstrom are given as QVALUE1, QVALUE2, ... =0.29 EXPINT10Add an experimental value for each q value. =0.00880634\nQVALUE11Selected scattering lengths in Angstrom are given as QVALUE1, QVALUE2, ... =0.32 EXPINT11Add an experimental value for each q value. =0.0137301\nQVALUE12Selected scattering lengths in Angstrom are given as QVALUE1, QVALUE2, ... =0.35 EXPINT12Add an experimental value for each q value. =0.0180036\nQVALUE13Selected scattering lengths in Angstrom are given as QVALUE1, QVALUE2, ... =0.38 EXPINT13Add an experimental value for each q value. =0.0193374\nQVALUE14Selected scattering lengths in Angstrom are given as QVALUE1, QVALUE2, ... =0.41 EXPINT14Add an experimental value for each q value. =0.0210131\nQVALUE15Selected scattering lengths in Angstrom are given as QVALUE1, QVALUE2, ... =0.44 EXPINT15Add an experimental value for each q value. =0.0220506\n...\n\nPRINT ARG=(saxs\\.q-.*),(saxs\\.exp-.*) FILE=colvar STRIDE=1\nGlossary of keywords and components\nDescription of components\n\nBy default this Action calculates the following quantities. These quantities can be referenced elsewhere in the input by using this Action's label followed by a dot and the name of the quantity required from the list below.\n\n Quantity Description score the Metainference score sigma uncertainty parameter sigmaMean uncertainty in the mean estimate neff effective number of replicas acceptSigma MC acceptance for sigma values q the # SAXS of q\n\nIn addition the following quantities can be calculated by employing the keywords listed below\n\n Quantity Keyword Description acceptScale SCALEDATA MC acceptance for scale value acceptFT GENERIC MC acceptance for general metainference f tilde value weight REWEIGHT weights of the weighted average biasDer REWEIGHT derivatives with respect to the bias scale SCALEDATA scale parameter offset ADDOFFSET offset parameter ftilde GENERIC ensemble average estimator exp EXPINT the # experimental intensity\nThe atoms involved can be specified using\n ATOMS The atoms to be included in the calculation, e.g. the whole protein.. For more information on how to specify lists of atoms see Groups and Virtual Atoms\nCompulsory keywords\n NOISETYPE ( default=MGAUSS ) functional form of the noise (GAUSS,MGAUSS,OUTLIERS,MOUTLIERS,GENERIC) LIKELIHOOD ( default=GAUSS ) the likelihood for the GENERIC metainference model, GAUSS or LOGN DFTILDE ( default=0.1 ) fraction of sigma_mean used to evolve ftilde SCALE0 ( default=1.0 ) initial value of the scaling factor SCALE_PRIOR ( default=FLAT ) either FLAT or GAUSSIAN OFFSET0 ( default=0.0 ) initial value of the offset OFFSET_PRIOR ( default=FLAT ) either FLAT or GAUSSIAN SIGMA0 ( default=1.0 ) initial value of the uncertainty parameter SIGMA_MIN ( default=0.0 ) minimum value of the uncertainty parameter SIGMA_MAX ( default=10. ) maximum value of the uncertainty parameter OPTSIGMAMEAN ( default=NONE ) Set to NONE/SEM to manually set sigma mean, or to estimate it on the fly WRITE_STRIDE ( default=10000 ) write the status to a file every N steps, this can be used for restart/continuation DEVICEID ( default=0 ) Identifier of the GPU to be used WATERDENS ( default=0.334 ) Density of the water to be used for the correction of atomistic structure factors. SCALEINT ( default=1.0 ) SCALING value of the calculated data. Useful to simplify the comparison.\nOptions\n NUMERICAL_DERIVATIVES ( default=off ) calculate the derivatives for these quantities numerically DOSCORE ( default=off ) activate metainference NOENSEMBLE ( default=off ) don't perform any replica-averaging REWEIGHT ( default=off ) simple REWEIGHT using the ARG as energy SCALEDATA ( default=off ) Set to TRUE if you want to sample a scaling factor common to all values and replicas ADDOFFSET ( default=off ) Set to TRUE if you want to sample an offset common to all values and replicas NOPBC ( default=off ) ignore the periodic boundary conditions when calculating distances SERIAL ( default=off ) Perform the calculation in serial - for debug purpose GPU ( default=off ) calculate SAXS using ARRAYFIRE on an accelerator device ATOMISTIC ( default=off ) calculate SAXS for an atomistic model MARTINI ( default=off ) calculate SAXS for a Martini model ARG the input for this action is the scalar output from one or more other actions. The particular scalars that you will use are referenced using the label of the action. If the label appears on its own then it is assumed that the Action calculates a single scalar value. The value of this scalar is thus used as the input to this new action. If * or *.* appears the scalars calculated by all the proceeding actions in the input file are taken. Some actions have multi-component outputs and each component of the output has a specific label. For example a DISTANCE action labelled dist may have three components x, y and z. To take just the x component you should use dist.x, if you wish to take all three components then use dist.*.More information on the referencing of Actions can be found in the section of the manual on the PLUMED Getting Started. Scalar values can also be referenced using POSIX regular expressions as detailed in the section on Regular Expressions. To use this feature you you must compile PLUMED with the appropriate flag. You can use multiple instances of this keyword i.e. ARG1, ARG2, ARG3... AVERAGING Stride for calculation of averaged weights and sigma_mean SCALE_MIN minimum value of the scaling factor SCALE_MAX maximum value of the scaling factor DSCALE maximum MC move of the scaling factor OFFSET_MIN minimum value of the offset OFFSET_MAX maximum value of the offset DOFFSET maximum MC move of the offset REGRES_ZERO stride for regression with zero offset DSIGMA maximum MC move of the uncertainty parameter SIGMA_MEAN0 starting value for the uncertainty in the mean estimate SIGMA_MAX_STEPS Number of steps used to optimise SIGMA_MAX, before that the SIGMA_MAX value is used TEMP the system temperature - this is only needed if code doesn't pass the temperature to plumed MC_STEPS number of MC steps MC_CHUNKSIZE MC chunksize STATUS_FILE write a file with all the data useful for restart/continuation of Metainference SELECTOR name of selector NSELECT range of values for selector [0, N-1] RESTART allows per-action setting of restart (YES/NO/AUTO) QVALUE Selected scattering lengths in Angstrom are given as QVALUE1, QVALUE2, ... . You can use multiple instances of this keyword i.e. QVALUE1, QVALUE2, QVALUE3... PARAMETERS Used parameter Keywords like PARAMETERS1, PARAMETERS2. These are used to calculate the structure factor for the $$i$$th atom/bead. You can use multiple instances of this keyword i.e. PARAMETERS1, PARAMETERS2, PARAMETERS3... EXPINT Add an experimental value for each q value. You can use multiple instances of this keyword i.e. EXPINT1, EXPINT2, EXPINT3..." ]

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[ "# All Questions\n\n159,920 questions\nFilter by\nSorted by\nTagged with\n10k views\n\n62k views\n\n### How to define a bijection between $(0,1)$ and $(0,1]$?\n\nHow to define a bijection between $(0,1)$ and $(0,1]$? Or any other open and closed intervals? If the intervals are both open like $(-1,2)\\text{ and }(-5,4)$ I do a cheap trick (don't know if that'...\n• 2,113\n109k views\n\n### The staircase paradox, or why $\\pi\\ne4$\n\nWhat is wrong with this proof? Is $\\pi=4?$\n• 13.2k\n47k views\n\n• 833\n81k views\n\n### Sum of First $n$ Squares Equals $\\frac{n(n+1)(2n+1)}{6}$\n\nI am just starting into calculus and I have a question about the following statement I encountered while learning about definite integrals: $$\\sum_{k=1}^n k^2 = \\frac{n(n+1)(2n+1)}{6}$$ I really ...\n• 1,803\n161k views\n\n### How to prove Euler's formula: $e^{i\\varphi}=\\cos(\\varphi) +i\\sin(\\varphi)$?\n\nCould you provide a proof of Euler's formula: $e^{i\\varphi}=\\cos(\\varphi) +i\\sin(\\varphi)$?\n• 7,768\n95k views\n\n### Prove that $\\lim \\limits_{n \\to \\infty} \\frac{x^n}{n!} = 0$, $x \\in \\Bbb R$.\n\nWhy is $$\\lim_{n \\to \\infty} \\frac{2^n}{n!}=0\\text{ ?}$$ Can we generalize it to any exponent $x \\in \\Bbb R$? This is to say, is $$\\lim_{n \\to \\infty} \\frac{x^n}{n!}=0\\text{ ?}$$ This is being ...\n• 1,205\n46k views\n\n### Examples of bijective map from $\\mathbb{R}^3\\rightarrow \\mathbb{R}$\n\nCould any one give an example of a bijective map from $\\mathbb{R}^3\\rightarrow \\mathbb{R}$? Thank you.\n• 34.8k\n58k views\n\n### Explain why $E(X) = \\int_0^\\infty (1-F_X (t)) \\, dt$ for every nonnegative random variable $X$\n\nLet $X$ be a non-negative random variable and $F_{X}$ the corresponding CDF. Show, $$E(X) = \\int_0^\\infty (1-F_X (t)) \\, dt$$ when $X$ has : a) a discrete distribution, b) a continuous ...\n• 1,411\n2k views\n\n### Solving linear congruence (modular inverse or fraction) via gcd Bezout equation\n\nIsn't finding the inverse of $a$, that is, $a'$ in $aa'\\equiv1\\pmod{m}$ equivalent to solving the diophantine equation $aa'-mb=1$, where the unknowns are $a'$ and $b$? I have seem some answers on this ...\n• 2,955\n32k views\n\n### Evaluating $\\lim\\limits_{n\\to\\infty} e^{-n} \\sum\\limits_{k=0}^{n} \\frac{n^k}{k!}$\n\nI'm supposed to calculate: $$\\lim_{n\\to\\infty} e^{-n} \\sum_{k=0}^{n} \\frac{n^k}{k!}$$ By using WolframAlpha, I might guess that the limit is $\\frac{1}{2}$, which is a pretty interesting and nice ...\n• 43.7k\n27k views" ]

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[ "Welcome to ZOJ", null, "Contests Information Problems Runs Statistics Ranklist Clarification", null, "62 - ZOJ Monthly, January 2008 - 1003\nCommon Factor\n\nTime Limit: 2 Seconds      Memory Limit: 65536 KB\n\nGiven two polynomials with integer coefficients, check if they have a common factor. A common factor is a polynomial with degree larger than 0 that will divide evenly into both polynomials.\n\nInput\n\nThere are many test cases. Every test case begins with a line containing two integers n and m, 0<= n, m <= 100. The second line contains n+1 integers An(An != 0), An-1, An-2, ... A1, A0. Ai is the coefficient of the term with degree i in the first polynomial. The third line contains m+1 integers Bm(Bm != 0), Bm-1, Bm-2, ... B1, B0. Bi is the coefficient of the term with degree i in the second polynomial. All coefficients are between -4000 and 4000 inclusive.\n\nOutput\n\nFor each test case, print \"YES\" if the two polynomials have a common factor in a line, or print \"NO\" instead if the two polynomials have no common factor.\n\nSample Input\n\n2 2\n1 0 -4\n1 2 -3\n2 1\n1 3 2\n1 2\n\nSample Output\n\nNO\nYES\n\nAuthor: HE, Rongqiang\n\nSubmit    Status" ]

[ null, "http://acm.zju.edu.cn/onlinejudge/image/arrow_sub2.gif", null, "http://acm.zju.edu.cn/onlinejudge/image/cpc_acm.jpg", null ]

[ "## ◂Math Worksheets and Study Guides Third Grade. Units of Measure\n\n### The resources above cover the following skills:\n\nMeasurement (NCTM)\nUnderstand measurable attributes of objects and the units, systems, and processes of measurement.\nUnderstand such attributes as length, area, weight, volume, and size of angle and select the appropriate type of unit for measuring each attribute.\nUnderstand the need for measuring with standard units and become familiar with standard units in the customary and metric systems.\nUnderstand that measurements are approximations and how differences in units affect precision.\nApply appropriate techniques, tools, and formulas to determine measurements.\nSelect and apply appropriate standard units and tools to measure length, area, volume, weight, time, temperature, and the size of angles.\nConnections to the Grade 3 Focal Points (NCTM)\nMeasurement: Students in grade 3 strengthen their understanding of fractions as they confront problems in linear measurement that call for more precision than the whole unit allowed them in their work in grade 2. They develop their facility in measuring with fractional parts of linear units. Students develop measurement concepts and skills through experiences in analyzing attributes and properties of two-dimensional objects. They form an understanding of perimeter as a measurable attribute and select appropriate units, strategies, and tools to solve problems involving perimeter." ]

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[ "# Mathematics You Should Know To Understand Neural Networks\n\nI started learning about neural networks a few months ago and I quickly realized that there were certain mathematical topics that were essential to understand in order to understand neural networks. Calculus and linear algebra are really important topics to be familiar with and it had been years since I had taken those courses in university. I thought it would be useful to consolidate some learning resources in one place that would help a beginner get started.\n\n## Do I need a math degree to build neural networks?\n\nShort answer: absolutely not! There are tons of libraries out there that implement the complex mathematics for us and you definitely don’t need a math degree to build neural networks. However, having a basic understanding of the math behind neural networks is really important to understanding how they work in case you need to debug or optimize your algorithm.\n\nThe next section of this article describes the different pieces of a basic neural network, the mathematical concepts required, and links to videos explaining the concepts. If you’ve taken university level calculus and linear algebra, the resources I’ve linked will be a refresher for you. If you’ve never taken these courses, I recommend doing a more comprehensive course beyond the resources linked in this article so that you can gain the foundational knowledge required. These are two comprehensive online courses that are great options for learning the concepts in detail:\n\n## Where can I learn about neural networks?\n\nIf you’re interested in diving deeper into the topic of deep learning, I highly recommend this I enrolled in offered through Coursera. It’s a really well put together program taught by Andrew Ng, one of the leaders in the AI space.\n\n# Mathematics you should know to understand neural networks\n\n## What is a Neural Network?\n\nA neural network is a network of algorithms used to solve classification problems. For example, a neural network can be used to tell you if an image is showing a cat or a dog.\n\nFigure 1 shows a very basic image of a neural network. The main pieces of a neural network are:\n\n• Input layer\n• Hidden layer (there can be several of these)\n• Output layer\n\n## Input Layer\n\nThe input layer is the layer that contains the inputs we want to feed into our algorithm, also known as a model. Our algorithm will do some calculations using our inputs and then spit out an answer.\n\nThe input layer is usually represented as a vector of numbers. For our image example, how do we turn an image into a vector of numbers? Since the image is just a matrix of RGB values, we can just unroll or flatten the matrix of values into a vector of numbers to be used in our algorithm.\n\nThe resources below will go over vectors and matrices, the first major mathematical concepts you should know to understand neural networks.\n\nResources\n\n## Hidden Layer\n\nThe hidden layer(s) is where a lot of the action happens in a neural network. In Figure 1, notice that the layer is split into two portions: the weighted sum calculation and the activation calculation. I won’t go into too much detail about why these calculations are done (deep learning courses will dive into this topic), but I will outline the mathematical concepts you should practice to understand how they work.\n\nWeighted Sum\n\nThe weighted sum, also known as a dot product is used to compute a value specified as the z value in Figure 1. It uses 3 variables: w, x, and b. W is a matrix of numbers that represent weights, which are initialized before computation. Even though it looks like w and x are being multiplied, there is actually a dot product happening.\n\nW has an exponent T, which is not really an exponent, it represents w transpose. B is a bias, which is just a number that is initialized before computation. X is the input vector from the previous layer. You don’t have to worry about how w and b are initialized, since it will be outlined later in this article. The resources below will go over the dot product, matrix transpose, and matrix multiplication.\n\nResources\n\nActivation\n\nThe activation calculation is the calculation used to generate the a value in Figure 1. The activation calculation uses a sigmoid function with a parameter of z that was calculated previously. Not all neural networks use a sigmoid function, this is normally used as a starter for simple neural networks. The sigmoid function is used as part of a logistic regression model. The resources below will go over the sigmoid function and logistic regression.\n\nResources\n\n## Forward Propagation\n\nForward propagation encompasses the calculations performed above, which includes passing data from the input layer to the hidden layer, making some calculations using our algorithms, then spitting out a prediction. However, sometimes our algorithm may not be accurate on the first try. Our algorithm might tell us that a cat picture is actually a dog picture most of the time, which is not correct. We can make improvements to our algorithm by doing a step called back propagation.\n\n## Back Propagation\n\nBack propagation is a step that allows us to make our algorithm better, also known as training. Earlier I mentioned that there is a matrix of weights and a bias used in the weighted sum calculation. There are techniques to initializing the weight and bias which is out of the scope of this article, but most of the time they don’t give accurate results right away. We can actually use back propagation to help us adjust the weight and bias to make our algorithm more accurate.\n\nCost function\n\nA cost function is used to calculate how well the algorithm did. Different neural networks can use different cost functions, so it depends on the neural network you’re using. Our neural network example uses logistic regression that has an associated cost function. The cost function is shown in Figure 2. Don’t worry if the cost function looks a little scary, if you take a comprehensive course on neural networks the details will be explained. Even if you don’t understand the cost function right now, we can focus on some specific parts of the function so you’ll be better prepared to understand it later on.\n\nAt the beginning of the function, there is a sigma symbol, which represents a sum. Another important part of the equation is the log or logarithm. The resources below will go over the sigma notation and logarithms.\n\nResources\n\nUsing our cost function, we can find the weight and bias parameters that minimize our cost so that we can make our algorithm more accurate. To minimize cost, a step called gradient descent is used. If you imagine that our function is a surface like in Figure 3, the idea is that we want to get to the minima (lowest point) of the surface to minimize cost.\n\nGradient descent involves some knowledge in calculus. You’ll need to understand derivatives and how you can get the slope of a function. The resources below will give you an overview of derivatives that will help you understand gradient descent.\n\nResources\n\n# Final thoughts\n\nSo that’s it! Those are the main math resources that I thought would be really useful for a beginner in machine learning. Once you have a good understanding of the mathematical concepts outlined in this article, you’ll be in really great shape to dive into neural networks and deep learning. A topic that I didn’t cover in this article is statistics, which is also used in machine learning. You can get away without knowing statistics as a beginner and in the future I can add a statistics article as part of a math for machine learning series! I’d love to hear your questions or suggestions on future topics and thanks for reading! :)\n\nImage Sources\n\n## Disclaimer\n\nI’m an employee of IBM. The views expressed in this blog are mine and don’t necessarily reflect the positions, strategies, or opinions of the company.\n\n## More from Sommer Shurbaji\n\nSoftware Developer at IBM" ]

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[ "# Regression with Amazon SageMaker XGBoost algorithm\n\n*Distributed training for regression with Amazon SageMaker XGBoost script mode*\n\n## Introduction\n\nThis notebook demonstrates the use of Amazon SageMaker XGBoost to train and host a regression model. XGBoost (eXtreme Gradient Boosting) is a popular and efficient machine learning algorithm used for regression and classification tasks on tabular datasets. It implements a technique know as gradient boosting on trees, and performs remarkably well in machine learning competitions, and gets a lot of attention from customers.\n\nWe use the Abalone data, originally from the UCI data repository. More details about the original dataset can be found here. In this libsvm converted version, the nominal feature (Male/Female/Infant) has been converted into a real valued feature as required by XGBoost. Age of abalone is to be predicted from eight physical measurements.\n\n## Setup\n\nThis notebook was tested in Amazon SageMaker Studio on a ml.t3.medium instance with Python 3 (Data Science) kernel.\n\nLet’s start by specifying: 1. The S3 bucket and prefix that you want to use for training and model data. This should be within the same region as the Notebook Instance, training, and hosting. 1. The IAM role arn used to give training and hosting access to your data. See the documentation for how to create these. Note, if more than one role is required for notebook instances, training, and/or hosting, please replace the boto regexp with a the appropriate full IAM role arn string(s).\n\n[ ]:\n\n%%time\n\nimport os\nimport boto3\nimport re\nimport sagemaker\n\n# Get a SageMaker-compatible role used by this Notebook Instance.\nrole = sagemaker.get_execution_role()\nregion = boto3.Session().region_name\n\n### update below values appropriately ###\nbucket = sagemaker.Session().default_bucket()\nprefix = \"sagemaker/DEMO-xgboost-dist-script\"\n####\n\nprint(region)\n\n\n### Fetching the dataset\n\nFollowing methods split the data into train/test/validation datasets and upload files to S3.\n\n[ ]:\n\n%%time\n\nimport io\nimport boto3\nimport random\n\ndef data_split(\nFILE_DATA,\nDATA_DIR,\nFILE_TRAIN_BASE,\nFILE_TRAIN_1,\nFILE_VALIDATION,\nFILE_TEST,\nPERCENT_TRAIN_0,\nPERCENT_TRAIN_1,\nPERCENT_VALIDATION,\nPERCENT_TEST,\n):\ndata = [l for l in open(FILE_DATA, \"r\")]\ntrain_file_0 = open(DATA_DIR + \"/\" + FILE_TRAIN_0, \"w\")\ntrain_file_1 = open(DATA_DIR + \"/\" + FILE_TRAIN_1, \"w\")\nvalid_file = open(DATA_DIR + \"/\" + FILE_VALIDATION, \"w\")\ntests_file = open(DATA_DIR + \"/\" + FILE_TEST, \"w\")\n\nnum_of_data = len(data)\nnum_train_0 = int((PERCENT_TRAIN_0 / 100.0) * num_of_data)\nnum_train_1 = int((PERCENT_TRAIN_1 / 100.0) * num_of_data)\nnum_valid = int((PERCENT_VALIDATION / 100.0) * num_of_data)\nnum_tests = int((PERCENT_TEST / 100.0) * num_of_data)\n\ndata_fractions = [num_train_0, num_train_1, num_valid, num_tests]\nsplit_data = [[], [], [], []]\n\nrand_data_ind = 0\n\nfor split_ind, fraction in enumerate(data_fractions):\nfor i in range(fraction):\nrand_data_ind = random.randint(0, len(data) - 1)\nsplit_data[split_ind].append(data[rand_data_ind])\ndata.pop(rand_data_ind)\n\nfor l in split_data:\ntrain_file_0.write(l)\n\nfor l in split_data:\ntrain_file_1.write(l)\n\nfor l in split_data:\nvalid_file.write(l)\n\nfor l in split_data:\ntests_file.write(l)\n\ntrain_file_0.close()\ntrain_file_1.close()\nvalid_file.close()\ntests_file.close()\n\ndef write_to_s3(fobj, bucket, key):\nreturn (\nboto3.Session(region_name=region)\n.resource(\"s3\")\n.Bucket(bucket)\n.Object(key)\n)\n\nfobj = open(filename, \"rb\")\nkey = prefix + \"/\" + channel\nurl = \"s3://{}/{}/{}\".format(bucket, key, filename)\nprint(\"Writing to {}\".format(url))\nwrite_to_s3(fobj, bucket, key)\n\n\n### Data ingestion\n\nNext, we read the dataset from the existing repository into memory, for preprocessing prior to training. This processing could be done in situ by Amazon Athena, Apache Spark in Amazon EMR, Amazon Redshift, etc., assuming the dataset is present in the appropriate location. Then, the next step would be to transfer the data to S3 for use in training. For small datasets, such as this one, reading into memory isn’t onerous, though it would be for larger datasets.\n\n[ ]:\n\n%%time\ns3 = boto3.client(\"s3\")\n\nFILE_DATA = \"abalone\"\n\"sagemaker-sample-files\", f\"datasets/tabular/uci_abalone/abalone.libsvm\", FILE_DATA\n)\n\nFILE_TRAIN_0 = \"abalone.train_0\"\nFILE_TRAIN_1 = \"abalone.train_1\"\nFILE_VALIDATION = \"abalone.validation\"\nFILE_TEST = \"abalone.test\"\nPERCENT_TRAIN_0 = 35\nPERCENT_TRAIN_1 = 35\nPERCENT_VALIDATION = 15\nPERCENT_TEST = 15\n\nDATA_DIR = \"data\"\n\nif not os.path.exists(DATA_DIR):\nos.mkdir(DATA_DIR)\n\ndata_split(\nFILE_DATA,\nDATA_DIR,\nFILE_TRAIN_0,\nFILE_TRAIN_1,\nFILE_VALIDATION,\nFILE_TEST,\nPERCENT_TRAIN_0,\nPERCENT_TRAIN_1,\nPERCENT_VALIDATION,\nPERCENT_TEST,\n)\n\n[ ]:\n\n# upload the files to the S3 bucket\nupload_to_s3(bucket, \"train/train_0.libsvm\", DATA_DIR + \"/\" + FILE_TRAIN_0)\nupload_to_s3(bucket, \"train/train_1.libsvm\", DATA_DIR + \"/\" + FILE_TRAIN_1)\nupload_to_s3(bucket, \"validation/validation.libsvm\", DATA_DIR + \"/\" + FILE_VALIDATION)\nupload_to_s3(bucket, \"test/test.libsvm\", DATA_DIR + \"/\" + FILE_TEST)\n\n\n## Create a XGBoost script to train with\n\nSageMaker can now run an XGboost script using the XGBoost estimator. When executed on SageMaker a number of helpful environment variables are available to access properties of the training environment, such as:\n\n• SM_MODEL_DIR: A string representing the path to the directory to write model artifacts to. Any artifacts saved in this folder are uploaded to S3 for model hosting after the training job completes.\n\n• SM_OUTPUT_DIR: A string representing the filesystem path to write output artifacts to. Output artifacts may include checkpoints, graphs, and other files to save, not including model artifacts. These artifacts are compressed and uploaded to S3 to the same S3 prefix as the model artifacts.\n\nSupposing two input channels, ‘train’ and ‘validation’, were used in the call to the XGBoost estimator’s fit() method, the following environment variables will be set, following the format SM_CHANNEL_[channel_name]:\n\nSM_CHANNEL_TRAIN: A string representing the path to the directory containing data in the ‘train’ channel SM_CHANNEL_VALIDATION: Same as above, but for the ‘validation’ channel.\n\nA typical training script loads data from the input channels, configures training with hyperparameters, trains a model, and saves a model to model_dir so that it can be hosted later. Hyperparameters are passed to your script as arguments and can be retrieved with an argparse.ArgumentParser instance. For example, the script that we will run in this notebook is provided as the accompanying file (abalone.py) and also shown below:\n\nimport argparse\nimport json\nimport logging\nimport os\nimport pandas as pd\nimport pickle as pkl\n\nfrom sagemaker_containers import entry_point\nfrom sagemaker_xgboost_container.data_utils import get_dmatrix\nfrom sagemaker_xgboost_container import distributed\n\nimport xgboost as xgb\n\ndef _xgb_train(params, dtrain, evals, num_boost_round, model_dir, is_master):\n\"\"\"Run xgb train on arguments given with rabit initialized.\n\nThis is our rabit execution function.\n\n:param args_dict: Argument dictionary used to run xgb.train().\n:param is_master: True if current node is master host in distributed training,\nor is running single node training job.\nNote that rabit_run will include this argument.\n\"\"\"\nbooster = xgb.train(params=params,\ndtrain=dtrain,\nevals=evals,\nnum_boost_round=num_boost_round)\n\nif is_master:\nmodel_location = model_dir + '/xgboost-model'\npkl.dump(booster, open(model_location, 'wb'))\nlogging.info(\"Stored trained model at {}\".format(model_location))\n\nif __name__ == '__main__':\nparser = argparse.ArgumentParser()\n\n# Hyperparameters are described here.\n\n# Sagemaker specific arguments. Defaults are set in the environment variables.\n\nargs, _ = parser.parse_known_args()\n\n# Get SageMaker host information from runtime environment variables\nsm_current_host = args.sm_current_host\n\ndtrain = get_dmatrix(args.train, 'libsvm')\ndval = get_dmatrix(args.validation, 'libsvm')\nwatchlist = [(dtrain, 'train'), (dval, 'validation')] if dval is not None else [(dtrain, 'train')]\n\ntrain_hp = {\n'max_depth': args.max_depth,\n'eta': args.eta,\n'gamma': args.gamma,\n'min_child_weight': args.min_child_weight,\n'subsample': args.subsample,\n'verbosity': args.verbosity,\n'objective': args.objective,\n'tree_method': args.tree_method,\n'predictor': args.predictor,\n}\n\nxgb_train_args = dict(\nparams=train_hp,\ndtrain=dtrain,\nevals=watchlist,\nnum_boost_round=args.num_round,\nmodel_dir=args.model_dir)\n\nif len(sm_hosts) > 1:\n# Wait until all hosts are able to find each other\nentry_point._wait_hostname_resolution()\n\n# Execute training function after initializing rabit.\ndistributed.rabit_run(\nexec_fun=_xgb_train,\nargs=xgb_train_args,\ninclude_in_training=(dtrain is not None),\nhosts=sm_hosts,\ncurrent_host=sm_current_host,\nupdate_rabit_args=True\n)\nelse:\n# If single node training, call training method directly.\nif dtrain:\nxgb_train_args['is_master'] = True\n_xgb_train(**xgb_train_args)\nelse:\nraise ValueError(\"Training channel must have data to train model.\")\n\ndef model_fn(model_dir):\n\"\"\"Deserialize and return fitted model.\n\nNote that this should have the same name as the serialized model in the _xgb_train method\n\"\"\"\nmodel_file = 'xgboost-model'\nreturn booster\n\n\nBecause the container imports your training script, always put your training code in a main guard (if __name__=='__main__':) so that the container does not inadvertently run your training code at the wrong point in execution.\n\n## Training the XGBoost model\n\nAfter setting training parameters, we kick off training, and poll for status until training is completed, which in this example, takes between few minutes.\n\nTo run our training script on SageMaker, we construct a sagemaker.xgboost.estimator.XGBoost estimator, which accepts several constructor arguments:\n\n• entry_point: The path to the Python script SageMaker runs for training and prediction.\n\n• role: Role ARN\n\n• train_instance_type (optional): The type of SageMaker instances for training. Note: Because Scikit-learn does not natively support GPU training, Sagemaker Scikit-learn does not currently support training on GPU instance types.\n\n• sagemaker_session (optional): The session used to train on Sagemaker.\n\n• hyperparameters (optional): A dictionary passed to the train function as hyperparameters.\n\n[ ]:\n\nhyperparams = {\n\"max_depth\": \"5\",\n\"eta\": \"0.2\",\n\"gamma\": \"4\",\n\"min_child_weight\": \"6\",\n\"subsample\": \"0.7\",\n\"objective\": \"reg:squarederror\",\n\"num_round\": \"50\",\n\"verbosity\": \"2\",\n}\n\ninstance_type = \"ml.m5.2xlarge\"\noutput_path = \"s3://{}/{}/{}/output\".format(bucket, prefix, \"abalone-dist-xgb\")\ncontent_type = \"libsvm\"\n\n[ ]:\n\n# Open Source distributed script mode\nfrom sagemaker.session import Session\nfrom sagemaker.inputs import TrainingInput\nfrom sagemaker.xgboost.estimator import XGBoost\n\nboto_session = boto3.Session(region_name=region)\nsession = Session(boto_session=boto_session)\nscript_path = \"abalone.py\"\n\nxgb_script_mode_estimator = XGBoost(\nentry_point=script_path,\nframework_version=\"1.3-1\", # Note: framework_version is mandatory\nhyperparameters=hyperparams,\nrole=role,\ninstance_count=2,\ninstance_type=instance_type,\noutput_path=output_path,\n)\n\ntrain_input = TrainingInput(\n\"s3://{}/{}/{}/\".format(bucket, prefix, \"train\"), content_type=content_type\n)\nvalidation_input = TrainingInput(\n\"s3://{}/{}/{}/\".format(bucket, prefix, \"validation\"), content_type=content_type\n)\n\n\n### Train XGBoost Estimator on abalone data\n\nTraining is as simple as calling fit on the Estimator. This will start a SageMaker Training job that will download the data, invoke the entry point code (in the provided script file), and save any model artifacts that the script creates.\n\n[ ]:\n\nxgb_script_mode_estimator.fit({\"train\": train_input, \"validation\": validation_input})\n\n\n## Deploying the XGBoost model\n\nAfter training, we can use the estimator to create an Amazon SageMaker endpoint – a hosted and managed prediction service that we can use to perform inference.\n\nYou can also optionally specify other functions to customize the behavior of deserialization of the input request (input_fn()), serialization of the predictions (output_fn()), and how predictions are made (predict_fn()). The defaults work for our current use-case so we don’t need to define them.\n\n[ ]:\n\npredictor = xgb_script_mode_estimator.deploy(\ninitial_instance_count=1, instance_type=\"ml.m5.2xlarge\"\n)\n\n[ ]:\n\ntest_file = DATA_DIR + \"/\" + FILE_TEST\nwith open(test_file, \"r\") as f:\n\n[ ]:\n\nruntime_client = boto3.client(\"runtime.sagemaker\", region_name=region)\nresponse = runtime_client.invoke_endpoint(\n\n[ ]:\n\npredictor.delete_endpoint()" ]

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[ "", null, "", null, "• FAQ\n• Contact", null, "/ Homework Answers / Finance / Which excel formula used to solve the following problems?\nNot my Question\nFlag Content\n\n# Question : Which excel formula used to solve the following problems?\n\nWhich excel formula is used to solve the following problems?\n\na. \\$598 to grow to 7,913.19 at a compound rate of 14 percent ?\n\na. \\$1036 grow to 14,673.88 over a period of 9 years?\n\na. \\$8000 with a payment of 206.76 per month for' 4 years?\n\na. \\$8,159.26 per month for 30 years at 12 percent interest?\n\na. \\$248596 with a payment of 3,143.36 for 30 years?\n\n5. How many years will it take to grow\n\na. \\$1200 to a value of 4,142.73 at a compound rate of 10 percent ?             13\n\n## Solution 5 (1 Ratings )\n\nSolved\nFinance 4 Months Ago 45 Views", null, "", null, "" ]

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[ "chrono::geometry::ChBasisToolsBspline Class Reference\n\n## Description\n\nTools for evaluating basis functions for B-splines, parametrized with parameter u (as lines) These bases are often called \"N\" in literature.\n\n#include <ChBasisToolsBspline.h>\n\n## Static Public Member Functions\n\nstatic int FindSpan (const int p, const double u, const ChVectorDynamic<> &knotU)\nFind the knot span of a b-spline given the parameter u, the order p, the knot vector knotU. More...\n\nstatic void ComputeKnotUniformMultipleEnds (ChVectorDynamic<> &knotU, const int p, double kstart=0.0, double kend=1.0)\nCompute uniformly-spaced k knots (in range [kstart, kend]) for open Bsplines, with order p. More...\n\nstatic void ComputeKnotUniform (ChVectorDynamic<> &knotU, const int p, double kstart=0.0, double kend=1.0)\nCompute uniformly-spaced k knots (in range [kstart, kend]) for Bsplines, with order p. More...\n\nstatic void BasisEvaluate (const int p, const int i, const double u, const ChVectorDynamic<> &knotU, ChVectorDynamic<> &N)\nCompute vector of bases N. More...\n\nstatic void BasisEvaluateDeriv (const int p, const int i, const double u, const ChVectorDynamic<> &knotU, ChMatrixDynamic<> &DN)\nCompute bases and first n-th derivatives of bases dN/du and ddN/ddu etc, arranged in a matrix. More...\n\n## ◆ BasisEvaluate()\n\n static void chrono::geometry::ChBasisToolsBspline::BasisEvaluate ( const int p, const int i, const double u, const ChVectorDynamic<> & knotU, ChVectorDynamic<> & N )\ninlinestatic\n\nCompute vector of bases N.\n\nEvaluate ALL the p+1 nonzero basis functions N of a b-spline, at the i-th knot span, given the parameter u, the order p, the knot vector knotU. Results go into the row vector N = { N1, N2, N3.... N_(p+1) }\n\nParameters\n p order i knot span, assume aready computed via FindSpan() u parameter knotU knot vector N here return basis functions N evaluated at u, that is: N(u)\n\n## ◆ BasisEvaluateDeriv()\n\n static void chrono::geometry::ChBasisToolsBspline::BasisEvaluateDeriv ( const int p, const int i, const double u, const ChVectorDynamic<> & knotU, ChMatrixDynamic<> & DN )\ninlinestatic\n\nCompute bases and first n-th derivatives of bases dN/du and ddN/ddu etc, arranged in a matrix.\n\nEvaluate derivatives of ALL the p+1 nonzero basis functions N of a b-spline, at the i-th knot span, given the parameter u, the order p, the knot vector knotU. Results go into the ChMatrixDynamic<> , where j-th derivative is j-th row: DN = | N1, N2, N3, ...., N_(p+1) | | dN1/du, dN2/du, dN3/du, ...., dN_(p+1)/du | | ddN1/ddu, ddN2/ddu, ddN3/ddu ...., ddN_(p+1)/ddu | The derivative order ranges from 0 (no derivative) to d, where d is the number of rows of the passed DN matrix. Usually two rows, for N and their shape derivatives.\n\nParameters\n p order of spline i knot span, assume aready computed via FindSpan() u parameter knotU knot vector DN here return derivatives evaluated at u, that is: dN/du(u)\n\n## ◆ ComputeKnotUniform()\n\n static void chrono::geometry::ChBasisToolsBspline::ComputeKnotUniform ( ChVectorDynamic<> & knotU, const int p, double kstart = 0.0, double kend = 1.0 )\ninlinestatic\n\nCompute uniformly-spaced k knots (in range [kstart, kend]) for Bsplines, with order p.\n\nIf you need that the spline starts and ends exactly at the 1st and last control point, use ComputeKnotUniformMultipleEnds instead. This is often used when creating closed splines, where the last and first p control points will overlap.\n\nParameters\n knotU knot vector p order kstart range start kend range end\n\n## ◆ ComputeKnotUniformMultipleEnds()\n\n static void chrono::geometry::ChBasisToolsBspline::ComputeKnotUniformMultipleEnds ( ChVectorDynamic<> & knotU, const int p, double kstart = 0.0, double kend = 1.0 )\ninlinestatic\n\nCompute uniformly-spaced k knots (in range [kstart, kend]) for open Bsplines, with order p.\n\nAssuming the size of knotU is already k=n+p+1 for n control points. The p+1 initial and end knots are made multiple.\n\nParameters\n knotU knot vector p order kstart range start kend range end\n\n## ◆ FindSpan()\n\n static int chrono::geometry::ChBasisToolsBspline::FindSpan ( const int p, const double u, const ChVectorDynamic<> & knotU )\ninlinestatic\n\nFind the knot span of a b-spline given the parameter u, the order p, the knot vector knotU.\n\nParameters\n p order u parameter knotU knot vector\n\nThe documentation for this class was generated from the following file:\n• /builds/uwsbel/chrono/src/chrono/geometry/ChBasisToolsBspline.h" ]

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[ "# Form Operators and Functions¶\n\nExpressions in calculations, constraints, and relevants can contain operators and functions.\n\n## Operators¶\n\n### Math operators¶\n\nExplanation Example\n+ addition ${salary_income} +${self_employed_income}\n- subtraction ${income} -${expenses}\n* multiplication ${bill} * 1.18 div division${percent_int} div 100\nmod modulo (division remainder) (${even_number} mod 2) = 0 Warning Math operators only work with numbers. ### Comparison operators¶ Comparison operators are used to compare values. The result of a comparison is always True or False. Explanation Example Notes = equal to${enrolled} = 'yes' Can compare numbers or strings.\n!= not equal to ${enrolled} != 'yes' Can compare numbers or strings. > greater than${age} > 17\n>= greater than or equal to ${age} >= 18 < less than${age} < 65\n<= less than or equal to ${age} <= 64 Warning ### Boolean operators¶ Boolean operators combine two True or False values into a single True or False value. Explanation Example and True if the expressions before and after are True${age} > -1 and ${age} < 120 or True if either of the expressions before or after are True${age} < 19 or ${age} > 64 ### Path operators¶ Explanation Example Notes . current question's value . >= 18 Used in constraints. .. current question's parent group position(..) Used with position() to get the iteration index. Note Formally, these are not operators but rather XPath references to the current node (..) and the containing node (.). XPath paths can be used to reference nodes of a form. ## Functions¶ ### Control flow¶ if(expression, then, else) Returns then if expression evaluates to True. Otherwise, returns else. position(xpath) Returns an integer equal to the 1-indexed position of the current node within the node defined by xpath. Most often this is used in the form position(..) to identify the current iteration index within a repeat group. XLSForm type name label repeat_count calculation note person_list_note Please list the names of the people in your household. begin_repeat person Member of household text name Name end_repeat begin_repeat person_details Details count(${person})\ncalculate current_name     indexed-repeat(${name},${person}, position(..))\ndate member_bday Birthday of ${current_name} end_repeat once(expression) Returns the value expression if the question's value is empty. Otherwise, returns the current value of the question. This can be used to ensure that a random number is only generated once, or to store the first value entered for a question in a way that is retrievable even if the response is changed later. Warning This function is often misunderstood. Read when expressions are evaluated to learn more. ### Accessing response values¶ Note The response from most question types can be accessed using variables. Functions are needed for accessing responses to multi select questions and questions inside repeat groups. #### Select questions¶ selected(space_delimited_array, string) Returns True if string is a member of space_delimited_array, otherwise returns False. Commonly used to determined if a specific choice was selected in a select question. (This is possible because a reference to a select question returns a space-delimited array of choice names.) XLSForm survey type name label hint relevant constraint select_multiple medical_issues what_issues Have you experienced any of the following? Select all that apply. select_multiple cancer_types what_cancer What type of cancer have you experienced? Select all that apply. selected(${what_issues}, 'cancer')\nselect_multiple diabetes_types what_diabetes What type of diabetes do you have? Select all that apply. selected(${what_issues}, 'diabetes') begin_group blood_pressure Blood pressure reading selected(${what_issues}, 'hypertension')\ninteger systolic_bp Systolic     . > 40 and . < 400\ninteger diastolic_bp Diastolic     . >= 20 and . <= 200\nend_group\ntext other_health List other issues.   selected(${what_issues}, 'other') note after_health_note This note is after all health questions. choices list_name name label medical_issues cancer Cancer medical_issues diabetes Diabetes medical_issues hypertension Hypertension medical_issues other Other cancer_types lung Lung cancer cancer_types skin Skin cancer cancer_types prostate Prostate cancer cancer_types breast Breast cancer cancer_types other Other diabetes_types type_1 Type 1 (Insulin dependent) diabetes_types type_2 Type 2 (Insulin resistant) selected-at(space_delimited_array, n) Returns the string at the nth position of the space_delimited_array. (The array is zero-indexed.) Returns an empty string if the index does not exist. This can be used to get the name of a selected choice from a multi-select question. (This is possible because a reference to a select question returns a space-delimited array of choice names.) Note If used to get a choice name from a select question, this function returns the name, not the label, of the selected choice. To get the label in the current language, use jr:choice-name().", null, "", null, "XLSForm survey type name label hint calculation select_multiple colors color_prefs What colors do you like? Select three. calculate color_0 selected-at(${color_prefs}, 0)\ncalculate color_1     selected-at(${color_prefs}, 1) calculate color_2 selected-at(${color_prefs}, 2)\nnote color_note Selected colors: ${color_0} <br>${color_1} <br> ${color_2} choices list_name name label colors red Red colors blue Blue colors yellow Yellow colors green Green colors orange Orange colors purple Purple count-selected(multi_select_question) Returns the number of choices selected in multi_select_question.", null, "XLSForm survey type name label hint constraint constraint_message select_multiple colors color_prefs What colors do you like? Select three. count-selected(.)=3 Select exactly three. choices list_name name label colors red Red colors blue Blue colors yellow Yellow colors green Green colors orange Orange colors purple Purple jr:choice-name(choice_name, 'select_question') Returns the label value, in the active language, associated with the choice_name in the list of choices for the select_question. Note You have to wrap the select_question reference in quotes. '${question_name}'", null, "", null, "", null, "", null, "XLSForm\n\nsurvey\ntype name label::English label::Español hint::English hint:Español calculation\nselect_multiple colors color_prefs What colors do you like? ¿Qué colores te gustan? Select three. Seleccione tres.\ncalculate color_0         jr:choice-name( selected-at(${color_prefs}, 0), '${color_prefs}')\ncalculate color_1         jr:choice-name( selected-at(${color_prefs}, 1), '${color_prefs}')\ncalculate color_2         jr:choice-name( selected-at(${color_prefs}, 2), '${color_prefs}')\nnote color_note Selected colors: Colores seleccionados: ${color_0} <br>${color_1} <br> ${color_2}${color_0} <br> ${color_1} <br>${color_2}\nchoices\nlist_name name label::English label::Español\ncolors red Red Rojo\ncolors blue Blue Azul\ncolors yellow Yellow Amarillo\ncolors green Green Verde\ncolors purple Purple Púrpura\n\n#### Repeat groups¶\n\nnodeset\n\nA collection of XML nodes. In XLSForms, this is typically a collection of response values.\n\nOutside a repeat group, referring to a question by name will return a nodeset containing all the responses to that question.\n\nNodesets can also be created by joining two or more nodes with pipes: /data/age | /data/name.\n\nindexed-repeat(name, group, i[, sub_grp, sub_i[, sub_sub_grp, sub_sub_i]])\n\nReturns the response value of question name from the repeat-group group, in iteration i.\n\nNested repeat groups can be accessed using the sub and sub_sub parameters.\n\nXLSForm\n\ntype name label repeat_count calculation\nnote person_list_note Please list the names of the people in your household.\nbegin_repeat person Member of household\ntext name Name\nend_repeat\nbegin_repeat person_details Details count(${person}) calculate current_name indexed-repeat(${name}, ${person}, position(..)) date member_bday Birthday of${current_name}\nend_repeat\ncount(nodeset)\n\nReturns the number of items in nodeset. This can be used to count the number of repetitions in a repeat group.\n\nXLSForm\n\ntype name label repeat_count calculation\nnote person_list_note Please list the names of the people in your household.\nbegin_repeat person Member of household\ntext name Name\nend_repeat\nbegin_repeat person_details Details count(${person}) calculate current_name indexed-repeat(${name}, ${person}, position(..)) date member_bday Birthday of${current_name}\nend_repeat\ncount-non-empty(nodeset)\n\nReturns the number of non-empty members of nodeset.\n\nsum(nodeset)\n\nReturns the sum of the members of nodeset.\n\nCan be used to tally responses to a repeated select question.\n\nXLSForm\n\nsurvey\ntype name label calculation\nbegin_repeat guest_details Guest details\ntext guest_name Guest name\nselect_one meal_options meal_preference Meal preference\ncalculate chkn   if(${meal_preference} = 'chicken', 1, 0 ) calculate fsh if(${meal_preference} = 'fish', 1, 0 )\ncalculate veg   if(${meal_preference} = 'vegetarian', 1, 0 ) end_repeat calculate chkn_count sum(${chkn})\ncalculate fsh_count   sum(${fsh}) calculate veg_count sum(${veg})\nchoices\nlist_name name label\nmeal_options chicken Chicken\nmeal_options fish Fish\nmeal_options vegetarian Vegetarian\nmax(nodeset)\n\nReturns the largest member of nodeset.\n\nXLSForm\n\nsurvey\ntype name label calculation\ntext child_name Child's name\ninteger child_age Child's age\nend_repeat\ncalculate age_of_oldest_child   max(${child_age}) min(nodeset) Returns the smallest member of nodeset. XLSForm survey type name label calculation begin_repeat child_questions Questions about child text child_name Child's name integer child_age Child's age end_repeat calculate age_of_youngest_child min(${child_age})\n\nWarning\n\nThe min() and max() functions only work sets of numbers. Empty values (that is, variables referencing unanswered questions) are actually empty strings, and will not be automatically converted to zero (0).\n\n### Strings¶\n\n#### Searching and matching strings¶\n\nregex(string, expression)\nReturns True if string is an exact and complete match for expression.", null, "XLSForm\n\nsurvey\ntype name label constraint constraint_message\ntext middle_initial What is your middle initial? regex(., 'p{L}') Just the first letter.\ncontains(string, substring)\n\nReturns True if the string contains the substring.\n\nstarts-with(string, substring)\n\nReturns True if string begins with substring.\n\nends-with(string, substring)\n\nReturns True if the string ends with substring.\n\nsubstr(string, start[, end])\n\nReturns the substring of string beginning at the index start and extending to (but not including) index end (or to the termination of string, if end is not provided). Members of string are zero-indexed.\n\nsubstring-before(string, target)\n\nReturns the substring of string before the first occurrence of the target substring. If the target is not found, or string begins with the target substring, then this will return an empty string.\n\nsubstring-after(string, target)\n\nReturns the substring of string after the first occurrence of the target substring. If the target is not found this will return an empty string.\n\ntranslate(string, fromchars, tochars)\n\nReturns a copy of string, where every occurrence of a character in fromchars is replaced by the corresponding character in tochars. If fromchars is longer than tochars then every occurrence of a character in fromchars that does not have a corresponding character in tochars will be removed.\n\nstring-length(string)\n\nReturns the number of characters in string.\n\nnormalize-space(string)\n\nReturns a string with normalized whitespace by stripping leading and trailing whitespace of string and replacing sequences of whitespace characters with a single space.\n\n#### Combining strings¶\n\nconcat(arg [, arg [, arg [, arg [...]]]])\n\nConcatenates one or more arguments into a single string. If any arg is a nodeset, the values within the set are concatenated into a string.\n\njoin(separator, nodeset)\n\nJoins the members of nodeset, using the string separator.\n\n#### Converting to and from strings¶\n\nboolean-from-string(string)\n\nReturns True if string is \"true\" or \"1\". Otherwise, False.\n\nstring(arg)\n\nConverts arg to a string.\n\n### Math¶\n\nWarning\n\nMath functions (except number()) only work with number values.\n\nYou can use number() to convert a string of digits to a number, but it is usually better to get a number value directly.\n\nEmpty values (that is, variables referencing unanswered questions) are actually empty strings, and will not be automatically converted to zero (0).\n\n#### Number handling¶\n\nround(number, places)\n\nRounds a decimal number to some number of decimal places.\n\nint(number)\n\nTruncates the fractional portion of a decimal number to return an integer.\n\nnumber(arg)\n\nConverts arg to number value.\n\nIf arg is a string of digits, returns the number value.\n\nIf arg is True, returns 1. If arg is False, returns 0.\n\nIf arg cannot be converted, returns NaN (not a number).\n\ndigest(data, algorithm, encoding method (optional))\n\nComputes and returns the hash value of the data string using the indicated hash algorithm string, and encoding this hash value using the optional encoding string.\n\nOptions for the algorithm are MD5, SHA-1, SHA-256, SHA-384, SHA-512.\n\nIf the third parameter is not specified, the encoding is base64. Valid options for the encoding are base64 and hex.\n\nThis function can be useful if, for example, someone wants to build a unique identifier from sensitive data like a national ID number without compromising that data.\n\n#### Calculation¶\n\npow(number, power)\n\nRaises a number to a power.\n\nlog(number)\n\nReturns the natural log of number.\n\nlog10(number)\n\nReturns the base-10 log of number.\n\nabs(number)\n\nReturns the absolute value of number.\n\nsin(number)\n\nReturns the sine of number.\n\ncos(number)\n\nReturns the cosine of number.\n\ntan(number)\n\nReturns the tangent of number.\n\nasin(number)\n\nReturns the arc sine of number.\n\nacos(number)\n\nReturns the arc cosine of number.\n\natan(number)\n\nReturns the arctan of number.\n\natan2(y, x)\n\nReturns the multi-valued inverse tangent of y, x.\n\nsqrt(number)\n\nReturns the square root of number.\n\nexp(x)\n\nReturns e^x.\n\nexp10(x)\n\nReturns 10^x.\n\npi()\n\nReturns an approximation of the mathematical constant π.\n\n### Date and time¶\n\ntoday()\n\nReturns the current date without a time component.\n\nnow()\n\nReturns the current datetime in ISO 8601 format, including the timezone.\n\nWarning\n\n#### Converting dates and time¶\n\ndecimal-date-time(dateTime)\n\nConverts dateTime value to the number of days since January 1, 1970 (the Unix Epoch).\n\nThis is the inverse of date().\n\ndate(days)\n\nConverts an integer representing a number of days from January 1, 1970 (the Unix Epoch) to a standard date value.\n\nThis is the inverse of decimal-date-time().\n\ndecimal-time(time)\n\nConverts time to a number representing a fractional day. For example, noon is 0.5 and 6:00 PM is 0.75.\n\n#### Formatting dates and times for display¶\n\nformat-date(date, format)\n\nReturns date as a string formatted as defined by format.\n\nThe following identifiers are used in the format string:\n\n %Y 4-digit year %y 2-digit year %m 0-padded month %n numeric month %b short text month (Jan, Feb, Mar…) %d 0-padded day of month %e day of month %a short text day (Sun, Mon, Tue…).\n\nNote\n\nMonth and day abbreviations are language and locale specific. If form locale can be determined, that locale will be used. Otherwise, the device locale will be used.\n\nformat-date-time(dateTime, format)\n\nReturns dateTime as a string formatted as defined by format.\n\nThe identifiers list in format-date() are available, plus the following:\n\n### Geography¶\n\narea(nodeset | geoshape)\n\nReturns the area, in square meters, of either a nodeset of geopoints or a geoshape value.\n\nIt takes into account the circumference of the Earth around the Equator but does not take altitude into account.\n\ndistance(nodeset | geoshape | geotrace)\n\nReturns the distance, in meters, of either:\n\n• a nodeset of geopoints\n• the perimeter of a geoshape\n• the length of a geotrace value\n\nIt takes into account the circumference of the Earth around the Equator and does not take altitude into account.\n\n### Utility¶\n\nrandom()\n\nReturns a random number between 0.0 (inclusive) and 1.0 (exclusive).\n\nWarning\n\nrandomize(nodeset[, seed])\n\nReturns a shuffled nodeset.\n\nA shuffle with a numeric seed is deterministic and reproducible.\n\nThe primary use for this function is to randomize the order of choices for a select question. The documentation on select widgets describes how this is done in XLSForm.\n\nrandomize() can only be used in a context where a nodeset is accepted. Note that questions of type calculate cannot reference a nodeset.\n\nuuid([length])\n\nWithout argument, returns a random RFC 4122 version 4 compliant UUID.\n\nWith an argument it returns a random GUID of specified length.\n\nboolean(arg)\n\nReturns True if arg is:\n\n• a number other than zero\n• a non-empty string\n• a non-empty collection\n• a comparison or expressions that evaluates to True.\n\nReturns False if arg is:\n\n• the number 0\n• an empty string\n• an empty collection\n• a comparison or expression that evaluates to False.\nnot(arg)\n\nReturns the opposite of boolean(arg).\n\ncoalesce(arg, arg)\n\nReturns first non-empty value of the two arg s. Returns an empty string if both are empty or non-existent.\n\nchecklist(min, max, response[, response[, response[, ...]]])\n\nReturns True if the number of response s that are exactly the string \"yes\" is between min and max, inclusive.\n\nSet min or max to -1 to make the argument not applicable.\n\nweighted-checklist(min, max, reponse, weight[, response, weight[, response, weight[, response, weight[, ... ]]])\n\nReturns True if the sum of the weight s of each response that is exactly the string \"yes\" is between min and max, inclusive.\n\nSet min or max to -1 to make the argument not\n\ntrue()\n\nEvaluates to True.\n\nfalse()\n\nEvaluates to False." ]

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[ "Switch to:\n\n# RLI Beneish M-Score\n\n: -2.57 (As of Today)\nView and export this data going back to 1972. Start your Free Trial\n\nThe zones of discrimination for M-Score is as such:\n\nAn M-Score of equal or less than -1.78 suggests that the company is unlikely to be a manipulator.\nAn M-Score of greater than -1.78 signals that the company is likely to be a manipulator.\n\nGood Sign:\n\nBeneish M-Score -2.57 no higher than -1.78, which implies that the company is unlikely to be a manipulator.\n\nNYSE:RLI' s Beneish M-Score Range Over the Past 10 Years\nMin: -4.83   Med: -2.58   Max: 1.37\nCurrent: -2.57\n\n-4.83\n1.37\n\nDuring the past 13 years, the highest Beneish M-Score of RLI was 1.37. The lowest was -4.83. And the median was -2.58.\n\n## RLI Beneish M-Score Historical Data\n\n* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.\n\n RLI Annual Data Dec10 Dec11 Dec12 Dec13 Dec14 Dec15 Dec16 Dec17 Dec18 Dec19 Beneish M-Score", null, "", null, "", null, "", null, "", null, "-2.53 -2.61 -2.08 -2.51 -2.49\n\nCompetitive Comparison\n* Competitive companies are chosen from companies within the same industry, with headquarter located in same country, with closest market capitalization; x-axis shows the market cap, and y-axis shows the term value; the bigger the dot, the larger the market cap.\n\nRLI Beneish M-Score Distribution\n\n* The bar in red indicates where RLI's Beneish M-Score falls into.\n\n## RLI Beneish M-Score Calculation\n\nThe M-score was created by Professor Messod Beneish. Instead of measuring the bankruptcy risk (Altman Z-Score) or business trend (Piotroski F-Score), M-score can be used to detect the risk of earnings manipulation. This is the original research paper on M-score.\n\nThe M-Score Variables:\n\nThe M-score of RLI for today is based on a combination of the following eight different indices:\n\n M = -4.84 + 0.92 * DSRI + 0.528 * GMI + 0.404 * AQI + 0.892 * SGI + 0.115 * DEPI = -4.84 + 0.92 * 1.0007 + 0.528 * 1 + 0.404 * 0.9908 + 0.892 * 1.0368 + 0.115 * 1 - 0.172 * SGAI + 4.679 * TATA - 0.327 * LVGI - 0.172 * 0.8981 + 4.679 * -0.0334 - 0.327 * 0.9437 = -2.57\n\n* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.\n\n This Year (Sep20) TTM: Last Year (Sep19) TTM: Accounts Receivable was \\$559.4 Mil. Revenue was 262.811 + 298.247 + 118.117 + 267.788 = \\$947.0 Mil. Gross Profit was 262.811 + 298.247 + 118.117 + 267.788 = \\$947.0 Mil. Total Current Assets was \\$2,789.8 Mil. Total Assets was \\$3,792.5 Mil. Property, Plant and Equipment(Net PPE) was \\$52.1 Mil. Depreciation, Depletion and Amortization(DDA) was \\$0.0 Mil. Selling, General, & Admin. Expense(SGA) was \\$10.0 Mil. Total Current Liabilities was \\$78.2 Mil. Long-Term Debt & Capital Lease Obligation was \\$149.4 Mil. Net Income was 42.387 + 92.166 + -61.267 + 53.378 = \\$126.7 Mil. Non Operating Income was 0 + 0 + 0 + 0 = \\$0.0 Mil. Cash Flow from Operations was 79.471 + 89.54 + -5.767 + 90.155 = \\$253.4 Mil. Accounts Receivable was \\$539.1 Mil. Revenue was 236.904 + 238.113 + 263.82 + 174.512 = \\$913.3 Mil. Gross Profit was 236.904 + 238.113 + 263.82 + 174.512 = \\$913.3 Mil. Total Current Assets was \\$2,565.7 Mil. Total Assets was \\$3,505.1 Mil. Property, Plant and Equipment(Net PPE) was \\$52.6 Mil. Depreciation, Depletion and Amortization(DDA) was \\$0.0 Mil. Selling, General, & Admin. Expense(SGA) was \\$10.7 Mil. Total Current Liabilities was \\$73.6 Mil. Long-Term Debt & Capital Lease Obligation was \\$149.3 Mil.\n\n1. DSRI = Days Sales in Receivables Index\n\nMeasured as the ratio of Revenue in Accounts Receivable in year t to year t-1.\n\nA large increase in DSR could be indicative of revenue inflation.\n\n DSRI = (Receivables_t / Revenue_t) / (Receivables_t-1 / Revenue_t-1) = (559.387 / 946.963) / (539.141 / 913.349) = 0.59071685 / 0.59029024 = 1.0007\n\n2. GMI = Gross Margin Index\n\nMeasured as the ratio of gross margin in year t-1 to gross margin in year t.\n\nGross margin has deteriorated when this index is above 1. A firm with poorer prospects is more likely to manipulate earnings.\n\n GMI = GrossMargin_t-1 / GrossMargin_t = (GrossProfit_t-1 / Revenue_t-1) / (GrossProfit_t / Revenue_t) = (913.349 / 913.349) / (946.963 / 946.963) = 1 / 1 = 1\n\n3. AQI = Asset Quality Index\n\nAQI is the ratio of asset quality in year t to year t-1.\n\nAsset quality is measured as the ratio of non-current assets other than Property, Plant and Equipment to Total Assets.\n\n AQI = (1 - (CurrentAssets_t + PPE_t) / TotalAssets_t) / (1 - (CurrentAssets_t-1 + PPE_t-1) / TotalAssets_t-1) = (1 - (2789.771 + 52.113) / 3792.501) / (1 - (2565.748 + 52.626) / 3505.064) = 0.25065702 / 0.25297398 = 0.9908\n\n4. SGI = Sales Growth Index\n\nRatio of Revenue in year t to sales in year t-1.\n\nSales growth is not itself a measure of manipulation. However, growth companies are likely to find themselves under pressure to manipulate in order to keep up appearances.\n\n SGI = Sales_t / Sales_t-1 = Revenue_t / Revenue_t-1 = 946.963 / 913.349 = 1.0368\n\n5. DEPI = Depreciation Index\n\nMeasured as the ratio of the rate of Depreciation, Depletion and Amortization in year t-1 to the corresponding rate in year t.\n\nDEPI greater than 1 indicates that assets are being depreciated at a slower rate. This suggests that the firm might be revising useful asset life assumptions upwards, or adopting a new method that is income friendly.\n\n DEPI = (Depreciation_t-1 / (Depreciaton_t-1 + PPE_t-1)) / (Depreciation_t / (Depreciaton_t + PPE_t)) = (0 / (0 + 52.626)) / (0 / (0 + 52.113)) = 0 / 0 = 1\n\nNote: If the Depreciation, Depletion and Amortization data is not available, we assume that the depreciation rate is constant and set the Depreciation Index to 1.\n\n6. SGAI = Sales, General and Administrative expenses Index\n\nThe ratio of Selling, General, & Admin. Expense(SGA) to Sales in year t relative to year t-1.\n\nSGA expenses index > 1 means that the company is becoming less efficient in generate sales.\n\n SGAI = (SGA_t / Sales_t) / (SGA_t-1 /Sales_t-1) = (9.961 / 946.963) / (10.698 / 913.349) = 0.01051889 / 0.01171294 = 0.8981\n\n7. LVGI = Leverage Index\n\nThe ratio of total debt to Total Assets in year t relative to yeat t-1.\n\nAn LVGI > 1 indicates an increase in leverage\n\n LVGI = ((LTD_t + CurrentLiabilities_t) / TotalAssets_t) / ((LTD_t-1 + CurrentLiabilities_t-1) / TotalAssets_t-1) = ((149.442 + 78.162) / 3792.501) / ((149.255 + 73.642) / 3505.064) = 0.06001422 / 0.06359285 = 0.9437\n\n8. TATA = Total Accruals to Total Assets\n\nTotal accruals calculated as the change in working capital accounts other than cash less depreciation.\n\n TATA = (IncomefromContinuingOperations_t - CashFlowsfromOperations_t) / TotalAssets_t = (NetIncome_t - NonOperatingIncome_t - CashFlowsfromOperations_t) / TotalAssets_t = (126.664 - 0 - 253.399) / 3792.501 = -0.0334\n\nAn M-Score of equal or less than -1.78 suggests that the company is unlikely to be a manipulator. An M-Score of greater than -1.78 signals that the company is likely to be a manipulator.\n\nRLI has a M-score of -2.57 suggests that the company is unlikely to be a manipulator." ]

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[ "# J F Kennedy Memorial\n\nDistricts Schools", null, "## 2021 Item by Item Results for GRADE 05 MATHEMATICS\n\n### Number of Students Included: 54     Participation Rate = 97%\n\n J F Kennedy Memorial - GRADE 05 MATHEMATICS ITEM INFORMATION PERCENT OF SCHOOL'S POSSIBLE POINTS ITEM TYPE REPORTING CATEGORY STANDARD ITEM DESC POSSIBLE POINTS J F KENNEDY MEMORIAL FRANKLIN STATE SCHOOL-STATE DIFFERENCE 1 SR NT 5.NBT.B.7 Solve a real-world problem involving addition and multiplication of money. 1 82% 83% 71% 11 2 SA MD 5.MD.C.4 Solve a real-world volume problem by counting unit cubes. 1 86% 88% 75% 11 3 SR NT 5.NBT.A.1 Demonstrate understanding of place value by selecting multi-digit decimal numbers that have a digit that is one-tenth the value of the same digit in a given number. 1 57% 58% 43% 14 4 CR OA 5.OA.B.3 Given the rules for two patterns, determine the first several terms of each pattern and create and graph ordered pairs using corresponding terms of the two patterns. 4 87% 63% 45% 42 5 SR MD 5.MD.A.1 Solve a real-world word problem by converting milliliters to liters. 1 75% 57% 47% 28 6 SA MD 5.MD.B.2 Complete a line plot to display a set of data measurements given as mixed numbers with unlike denominators. 1 68% 41% 25% 43 7 SR NT 5.NBT.A.4 Round a decimal number to the nearest whole number. 1 93% 87% 65% 28 8 SR GE 5.G.B.4 Identify which statement is true about the properties of special quadrilaterals. 1 64% 64% 49% 15 9 SR NT 5.NBT.B.5 Multiply a three-digit whole number by a two-digit whole number. 1 92% 85% 70% 22 10 SR NT 5.NBT.A.2 Identify which power of ten is equivalent to a given whole number. 1 100% 72% 58% 42 11 SA NT 5.NBT.A.4 Round a decimal to the nearest tenth. 1 58% 59% 40% 18 12 CR NF 5.NF.B.6 Write an equation to represent a given problem and multiply fractions and whole numbers to solve real-world problems. 4 47% 40% 30% 17 13 SR MD 5.MD.C.5 Determine which right rectangular prism, shown with side lengths, has the greatest volume. 1 88% 72% 62% 26 14 SA MD 5.MD.A.1 Order measures of weight expressed in different units from least to greatest value. 1 58% 49% 35% 23 15 SR NT 5.NBT.A.3 Select which expressions correctly show a decimal to the thousandths in expanded form. 1 69% 51% 35% 34 16 SA NF 5.NF.B.5 Identify expressions with a product greater than a given factor and write a fraction that can be multiplied by a whole number to get a product less than that whole number. 2 83% 63% 44% 39 17 SR GE 5.G.A.1 Given an ordered pair, select the statement that correctly describes the location of the point represented by the ordered pair in relation to the origin on a coordinate plane. 1 92% 78% 64% 28 18 SA NF 5.NF.B.4 Find the product of a fraction and a whole number. 1 73% 68% 48% 25 19 SR OA 5.OA.A.1 Determine the value of a given expression with parentheses. 1 50% 30% 31% 19 20 SA MD 5.MD.C.4 Find the volume of a figure by counting cubes with given dimensions. 1 85% 64% 55% 30 21 SR GE 5.G.B.3 Determine which triangle meets specified criteria based on the given side lengths of each triangle. 1 79% 55% 38% 41 22 SR OA 5.OA.A.1 Determine which expression with parentheses has an equivalent value if the parentheses are removed. 1 93% 62% 58% 35 23 SR NF 5.NF.A.1 Identify which expression can be used to solve an addition problem by replacing given fractions with equivalent fractions with like denominators. 1 93% 86% 62% 31 24 SR NF 5.NF.B.5 Given several expressions, determine whether the product of each expression is greater than, less than, or equal to the value of a given factor of the expression. 1 68% 49% 33% 35 25 SR NT 5.NBT.B.7 Divide a decimal to hundredths by a whole number. 1 75% 69% 61% 14 26 CR MD 5.MD.C.5 Write an equation to find the volume of a given prism, find the total volume of two prisms placed together, and determine a set of dimensions that will result in a given volume. 4 71% 62% 42% 29 27 SR NF 5.NF.B.7 Create a division equation involving a whole number and a unit fraction where the quotient is the solution to a word problem. 1 46% 32% 24% 22 28 SR NF 5.NF.A.2 Estimate the sum of two fractions that are less than one to solve a word problem. 1 61% 37% 32% 29 29 SR NT 5.NBT.A.3 Match decimal numbers in expanded form with decimals in number form and compare two decimal numbers to thousandths. 2 75% 69% 55% 20 30 SA NT 5.NBT.B.5 Determine the product of two three-digit numbers. 1 46% 50% 31% 15 31 SR OA 5.OA.A.2 Identify the word form of a given numerical expression. 1 86% 72% 57% 29 32 SA NF 5.NF.B.7 Determine the quotient of a whole number divided by a fraction in a real-world context. 1 89% 91% 78% 11 33 SR GE 5.G.B.4 Classify triangles based on angle and side properties. 1 62% 48% 39% 23 34 SA GE 5.G.A.2 Graph three points in the first quadrant of the coordinate plane. 1 85% 72% 57% 28 35 CR NT 5.NBT.B.6 Write an equation to solve a real-world problem, critique another student’s reasoning of the problem, and solve a similar problem using division with whole numbers. 4 84% 75% 56% 28 36 SR NT 5.NBT.A.1 Determine the relationship of the value of a digit in one number compared to the value of that digit in another number. 1 54% 49% 34% 20 37 SR NF 5.NF.B.6 Determine the product of a mixed number and a fraction to solve a real-world problem. 1 19% 30% 20% -1 38 SR GE 5.G.B.3 Identify shapes that have two pairs of opposite angles that are congruent. 1 62% 34% 24% 38 39 SR OA 5.OA.A.2 Select the numerical expression, with parentheses, that represents a given word expression. 1 46% 36% 33% 13 40 SR NF 5.NF.B.3 Determine the fraction that represents a given word problem. 1 96% 83% 68% 28\n\nNotes: Students in grades 3-8 took a shorter version of the tests in spring 2021. Item results are not reported if fewer than 10 students in the school or district answered the item.\n\nParticipation rates varied across schools, districts, and student groups in 2021 more than in prior years. If the participation rate in 2021 was lower than in prior years, results may have been different if more students had taken the test." ]

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[ "### Calculate the Sum of the Elements of each Row & Column of given Matrix in Java.\n\npublic class SumOfEachRowAndColumn {\n\npublic static void main(String args[])\n\n{\nint a[][] = new int;\nint sumOfRow = 0;\nint sumOfColumn = 0;\nScanner sc = new Scanner(System.in);\n\nSystem.out.println(\"Enter element of Matrix \");\n\nfor (int i = 0; i <= 2; i++)\n{\n\nfor (int j = 0; j <= 2; j++)\n\n{\na[i][j] = sc.nextInt();\n\n}\n\n}\nSystem.out.println(\"Value of  Matrix is \");\nfor (int i = 0; i <= 2; i++)\n{\n\nfor (int j = 0; j <= 2; j++)\n\n{\nSystem.out.print(a[i][j] + \" \");\n\n}\n\nSystem.out.println();\n}\n\n//Clculating the sum of rows\n\nfor (int i = 0; i <= 2; i++)\n\n{\n\nfor (int j = 0; j <= 2; j++)\n\n{\nsumOfRow = sumOfRow + a[i][j];\n\n}\n\nSystem.out.println(\"Sum of \" + (i + 1) + \" row is \" + sumOfRow);\nsumOfRow = 0;\n}\n\n//Clculating the sum of column\n\nfor (int i = 0; i <= 2; i++)\n\n{\n\nfor (int j = 0; j <= 2; j++)\n\n{\nsumOfColumn = sumOfColumn + a[j][i];\n\n}\n\nSystem.out.println(\"Sum of \" + (i + 1) + \" column is \" +             sumOfColumn);\nsumOfColumn = 0;\n}\n\n}\n\n}\n\nOutput:\n\nEnter element of Matrix\n1\n2\n3\n4\n5\n6\n1\n2\n3\n\nValue of  Matrix is\n1 2 3\n4 5 6\n1 2 3\n\nSum of 1 row is 6\nSum of 2 row is 15\nSum of 3 row is 6\n\nSum of 1 column is 6\nSum of 2 column is 9\nSum of 3 column is 12\n\nBUILD SUCCESSFUL (total time: 7 seconds)" ]

[ null ]

[ "### < !-- function zoomIn() { newZoom = parseInt(oZoom.style.zoom) + 10 + '%'oZoom.style.zoom = newZoom; } function zoomOut() { newZoom = parseInt(oZoom.style.zoom) - 10 + '%'oZoom.style.zoom = newZoom; } //--> < !-- function MM_swapImgRestore() { //v3.0 var i, x, a = document.MM_sr; for (i = 0; a && i < a.length && (x = a[i]) && x.oSrc; i++) x.src = x.oSrc; } function MM_preloadImages() { //v3.0 var d = document; if (d.images) { if (!d.MM_p) d.MM_p = new Array(); var i, j = d.MM_p.length, a = MM_preloadImages.arguments; for (i = 0; i < a.length; i++) if (a[i].indexOf(\"#\") != 0) { d.MM_p[j] = new Image; d.MM_p[j++].src = a[i]; } } } function MM_findObj(n, d) { //v4.01 var p, i, x; if (!d) d = document; if ((p = n.indexOf(\"?\")) > 0 && parent.frames.length) { d = parent.frames[n.substring(p + 1)].document; n = n.substring(0, p); } if (! (x = d[n]) && d.all) x = d.all[n]; for (i = 0; ! x && i < d.forms.length; i++) x = d.forms[i][n]; for (i = 0; ! x && d.layers && i < d.layers.length; i++) x = MM_findObj(n, d.layers[i].document); if (!x && d.getElementById) x = d.getElementById(n); return x; } function MM_swapImage() { //v3.0 var i, j = 0, x, a = MM_swapImage.arguments; document.MM_sr = new Array; for (i = 0; i < (a.length - 2); i += 3) if ((x = MM_findObj(a[i])) != null) { document.MM_sr[j++] = x; if (!x.oSrc) x.oSrc = x.src; x.src = a[i + 2]; } } //-->", null, "", null, "", null, "", null, "", null, "最热新闻\n ·绿水迢迢 青山逶迤 美丽杭州描绘人与自然和谐共生新图景 ·9月10日晚钱塘江畔这场专属灯光秀 送给最敬爱的您 ·杭州第19届亚运会开幕式完成全要素彩排 ·台风“卡努”逐渐向浙江沿海靠近 8月2日起将带来明显风雨", null, "本网专稿\n ·迎亚运 靓环境 钱塘临江街道街巷面貌焕新 ·下沙港集装箱吞吐量破万箱 为140余家企业节约近20%物流成本 ·前方高萌到站 谁又能拒绝一辆“熊猫巴士”呢 ·乘势而上 临平如何开拓新天地？", null, "权威发布\n ·刘捷：扎紧篱笆、严控大型活动 咬紧牙关、精准高效遏制疫情蔓延 ·杭州上线“民呼我为”数字平台 ·图解杭州“十四五”规划《纲要》 ·3月22日杭州市无新增确诊病例 专家：入境应如实申报", null, "区县新闻\n ·建德市“府前路区块旧城改造项目”捧得“钱江杯” ·守好群众的“钱袋子”桐庐百江镇开展反诈观影宣传活动 ·建德市新安江街道：完善“一老一小”服务体系 提升群众生活质量 ·府西社区“亮微光” 哥哥姐姐伴成长" ]

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[ "# fPolynomial -- computes the f-polynomial of a poset\n\n## Synopsis\n\n• Usage:\nf = fPolynomial P\nf = fPolynomial(P, VariableName => symbol)\n• Inputs:\n• P, an instance of the type Poset,\n• Optional inputs:\n• VariableName => , default value q\n• Outputs:\n• f, , the f-polynomial of $P$\n\n## Description\n\nThe f-polynomial of $P$ is the polynomial such that the coefficient on $q^i$ is the number of chains of length $i$ in $P$.\n\nThe f-polynomial of the $n$ chain is $(q+1)^n$.\n\n i1 : n = 5; i2 : factor fPolynomial chain n 5 o2 = (q + 1) o2 : Expression of class Product" ]

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[ "# Solving PDEs using Taylor series\n\nI'm thinking of solving a Partial differential algebraic equation using multidimensional polynomial (i.e. Taylor series). Consider the PDAE:\n\n$$\\mathbf F \\left( \\mathbf x, \\mathbf y, \\frac{\\partial y_i}{\\partial x_j}, \\frac{\\partial^2 y_i}{\\partial x_j \\partial x_n}, \\ldots \\right) = \\mathbf 0 \\, , \\left\\{1<i<m,1<j<n\\right\\} \\, ,\\tag{1}$$\n\nwith a boundary/initial condition of\n\n$$\\mathbf G \\left( \\mathbf y, \\frac{\\partial y_i}{\\partial x_j}, \\frac{\\partial^2 y_i}{\\partial x_j \\partial x_n}, \\ldots \\right)_{\\partial \\Omega} = \\mathbf 0 \\, . \\tag{2}$$\n\nNow if we approximate $\\mathbf y$ with a multidimensional polynomial:\n\n$$y_i \\approx \\sum_{l_1=0}^{o_1} \\ldots \\sum_{l_n=0}^{o_n} a_{i\\left(l_1, \\ldots, l_n\\right)}x_1^{l_1} \\ldots x_n^{l_n} \\, , \\tag{3}$$\n\nand substitute it in Eq. 1 , we get system of nonlinear equations of $a_{i\\left(l_1, \\ldots, l_n\\right)}$s. Using the boundary conditions this system of nonlinear equations can be solved to find $a$s.\n\nI want to write a Matematica macro to automate this process.\n\n• Where should I start?\n• Are there any example about that?\n• Has there ever been an attempt to do so? even with other symbolic CAS software\n\nThanks for your help in advance.\n\n• Something like AsymptoticDSolveValue? Read this mathematica blog post – rhermans Jul 27 '18 at 12:48\n• @rhermans awesome. exactly. let me check this out and come back here. – Foad Jul 27 '18 at 12:49\n• @rhermans I'm gonna try it on this which I want to finally solve and then post a new question if I had any issues. as far as this question matters I think I have found a solution to work on. – Foad Jul 27 '18 at 13:03\n• Realted 177120 168064 25363. – rhermans Jul 27 '18 at 13:08\n• I suggest that you show the due diligence in your question. Edit and explain how/why the related questions and the documentation for AsymptoticDSolveValue do not answer your need. – rhermans Jul 27 '18 at 13:30" ]

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[ "### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.\n\n What if applet does not run?\n\nExplanation", null, "R. Honsberger credits Hiroshi Haruki with the following unexpected result:\n\n Given a circle C with center O and a line m, not intersecting C. There exists a point Q such that, for every P on m, PQ equals the length of the tangent from P to C.\n\nIn other words, a circle centered at P with the radius equal to the length of the tangent from P to C passes through a fixed point Q. Q is a point of concurrency of all circles orthogonal to C and center on m!\n\n### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.\n\n What if applet does not run?\n\nThe proof requires a few applications of the Pythagorean proposition.\n\nLet M be the foot of the perpendicular from O onto M, MN and PT tangent to C, r = ON = OT. Assume also that Q is chosen on MO such that MQ = MN = z. The Pythagorean propositions will be used in the right triangles OPT, OMP, QMP, and OMN:\n\n (OPT) r2 + PT2 = OP2 (OMP) OM2 + MP2 = OP2 (QMP) z2 + MP2 = PQ2 (OMN) r2 + z2 = MO2\n\nFrom the first two we get\n\n (1) r2 + PT2 = OM2 + MP2.\n\nThe second pair yields\n\n (2) OM2 + MP2 = r2 + PQ2.\n\nNow a comparison of (1) and (2) yields PT2 = PQ2. Hence, PT = PQ.\n\n(Note that, in general, x2 = y2, does not imply x = y. Following such uncritical implication may lead to incorrect results. In our case, though, we know up front that both quantities PT and PQ are positive; they both denote the lengths of two line segments. The implication then becomes unequivocally correct: if the squares of two positive numbers are equal the numbers themselves are also equal.)", null, "### Remark\n\nIt is surprising how the same fact acquires an aura of familiarity if looked at from a different angle. In a discussion that involves orthogonality of circles, one thing that most certainly comes to mind is the coaxal circles theorem: circles in an Apollonian family are all orthogonal to circles through two fixed points (and vice versa.) The circles in the latter family have their centers on a fixed straight line - the radical axis of any two circles from the Apollonian family.\n\nThus it is obvious that, along with Q, there is a second point common to all circles with the center on m orthogonal to C. This point is the reflection of Q in m.\n\nIn the spirit of the above (see also the discussion on the Apollonian circle), Nathan Bowler suggested to consider circles orthogonal to two non-intersecting circles C and m, which makes the situation more transparent:\n\nLet Q be the intersection of any pair of circles orthogonal to both C and m. Invert with respect to Q. Then these two circles invert to a pair of intersecting lines. Call the point of intersection R. The inverse of C is a circle perpendicular to both of the lines, and so with centre on both lines. That is, it is a circle centred at R. Similarly the inverse of m is a circle centred at R. Then any circle perpendicular to both m and C must invert to a circle or line perpendicular to each of the two concentric circles, that is, to a line through their common centre at R. So any circle orthogonal to both C and m must pass through both Q and R', the inverse of R.\n\nThis proof shows that there are 2 such fixed points. Note that I have nowhere used the fact that m is a straight line, so that the proof holds for a general pair of nonintersecting circles.\n\n### References\n\n1. R. Honsberger, The Butterfly Problem and Other Delicacies from the Noble Art of Euclidean Geometry II, TYCMJ, 14 (1983), pp. 154-158.", null, "" ]

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[ "### Change CDEF block skip condition and move signalling\n\n```Previously CDEF was implicitly disabled for a filter block if all its\nsub-blocks were skip, and no bits for that block was signalled. That\nrequired the CDEF signal to be transmitted at the end of the block,\nand it was not possible to begin the filtering of the block before\nthat.\n\nThis patch moves the signalling to just after the first zero skip bit\nwithin a 64x64, 128x64, 64x128 or 128x128 block. If sub-blocks are\nskip, then no CDEF bits will be signalled, as before. Also, the skip\ncondition flag has been removed so it's always known at the skip flag\nwhether a coding block is to be filtered or not.\n\nparent e5b46b53\n ... ... @@ -50,8 +50,14 @@ static int is_8x8_block_skip(MODE_INFO **grid, int mi_row, int mi_col, return is_skip; } #if CONFIG_EXT_PARTITION int sb_compute_cdef_list(const AV1_COMMON *const cm, int mi_row, int mi_col, cdef_list *dlist, int filter_skip) { cdef_list *dlist, BLOCK_SIZE bs) #else int sb_compute_cdef_list(const AV1_COMMON *const cm, int mi_row, int mi_col, cdef_list *dlist) #endif { int r, c; int maxc, maxr; MODE_INFO **grid; ... ... @@ -60,8 +66,19 @@ int sb_compute_cdef_list(const AV1_COMMON *const cm, int mi_row, int mi_col, maxc = cm->mi_cols - mi_col; maxr = cm->mi_rows - mi_row; #if CONFIG_EXT_PARTITION if (bs == BLOCK_128X128 || bs == BLOCK_128X64 || bs == BLOCK_128X32) maxc = AOMMIN(maxc, MI_SIZE_128X128); else maxc = AOMMIN(maxc, MI_SIZE_64X64); if (bs == BLOCK_128X128 || bs == BLOCK_64X128 || bs == BLOCK_32X128) maxr = AOMMIN(maxr, MI_SIZE_128X128); else maxr = AOMMIN(maxr, MI_SIZE_64X64); #else maxr = AOMMIN(maxr, MI_SIZE_64X64); maxc = AOMMIN(maxc, MI_SIZE_64X64); #endif const int r_step = mi_size_high[BLOCK_8X8]; const int c_step = mi_size_wide[BLOCK_8X8]; ... ... @@ -71,27 +88,15 @@ int sb_compute_cdef_list(const AV1_COMMON *const cm, int mi_row, int mi_col, assert(r_step == 1 || r_step == 2); assert(c_step == 1 || c_step == 2); if (filter_skip) { for (r = 0; r < maxr; r += r_step) { for (c = 0; c < maxc; c += c_step) { for (r = 0; r < maxr; r += r_step) { for (c = 0; c < maxc; c += c_step) { if (!is_8x8_block_skip(grid, mi_row + r, mi_col + c, cm->mi_stride)) { dlist[count].by = r >> r_shift; dlist[count].bx = c >> c_shift; dlist[count].skip = is_8x8_block_skip(grid, mi_row + r, mi_col + c, cm->mi_stride); dlist[count].skip = 0; count++; } } } else { for (r = 0; r < maxr; r += r_step) { for (c = 0; c < maxc; c += c_step) { if (!is_8x8_block_skip(grid, mi_row + r, mi_col + c, cm->mi_stride)) { dlist[count].by = r >> r_shift; dlist[count].bx = c >> c_shift; dlist[count].skip = 0; count++; } } } } return count; } ... ... @@ -259,12 +264,13 @@ void av1_cdef_frame(YV12_BUFFER_CONFIG *frame, AV1_COMMON *cm, uv_sec_strength += uv_sec_strength == 3; if ((level == 0 && sec_strength == 0 && uv_level == 0 && uv_sec_strength == 0) || (cdef_count = sb_compute_cdef_list( cm, fbr * MI_SIZE_64X64, fbc * MI_SIZE_64X64, dlist, #if CONFIG_CDEF_SINGLEPASS (level & 1) || (uv_level & 1))) == 0) #if CONFIG_EXT_PARTITION (cdef_count = sb_compute_cdef_list(cm, fbr * MI_SIZE_64X64, fbc * MI_SIZE_64X64, dlist, BLOCK_64X64)) == 0) #else get_filter_skip(level) || get_filter_skip(uv_level))) == 0) (cdef_count = sb_compute_cdef_list(cm, fbr * MI_SIZE_64X64, fbc * MI_SIZE_64X64, dlist)) == 0) #endif { cdef_left = 0; ... ... @@ -425,14 +431,14 @@ void av1_cdef_frame(YV12_BUFFER_CONFIG *frame, AV1_COMMON *cm, #if CONFIG_CDEF_SINGLEPASS NULL, xd->plane[pli].dst.stride, #else xd->plane[pli].dst.stride, dst, xd->plane[pli].dst.stride, dst, #endif &src[CDEF_VBORDER * CDEF_BSTRIDE + CDEF_HBORDER], xdec[pli], ydec[pli], dir, NULL, var, pli, dlist, cdef_count, level, #if CONFIG_CDEF_SINGLEPASS sec_strength, pri_damping, sec_damping, coeff_shift); #else sec_strength, sec_damping, pri_damping, coeff_shift, 0, 0); sec_strength, sec_damping, pri_damping, coeff_shift, 0, 0); #endif #if CONFIG_HIGHBITDEPTH ... ...\n ... ... @@ -11,9 +11,9 @@ #ifndef AV1_COMMON_CDEF_H_ #define AV1_COMMON_CDEF_H_ #define CDEF_STRENGTH_BITS 7 #define CDEF_STRENGTH_BITS 6 #define CDEF_PRI_STRENGTHS 32 #define CDEF_PRI_STRENGTHS 16 #define CDEF_SEC_STRENGTHS 4 #include \"./aom_config.h\" ... ... @@ -37,8 +37,13 @@ extern \"C\" { #endif int sb_all_skip(const AV1_COMMON *const cm, int mi_row, int mi_col); #if CONFIG_EXT_PARTITION int sb_compute_cdef_list(const AV1_COMMON *const cm, int mi_row, int mi_col, cdef_list *dlist, int filter_skip); cdef_list *dlist, BLOCK_SIZE bsize); #else int sb_compute_cdef_list(const AV1_COMMON *const cm, int mi_row, int mi_col, cdef_list *dlist); #endif void av1_cdef_frame(YV12_BUFFER_CONFIG *frame, AV1_COMMON *cm, MACROBLOCKD *xd); void av1_cdef_search(YV12_BUFFER_CONFIG *frame, const YV12_BUFFER_CONFIG *ref, ... ...\n ... ... @@ -394,12 +394,6 @@ static void copy_block_16bit_to_8bit(uint8_t *dst, int dstride, } } int get_filter_skip(int level) { int filter_skip = level & 1; if (level == 1) filter_skip = 0; return filter_skip; } void cdef_filter_fb(uint8_t *dst, int dstride, uint16_t *y, uint16_t *in, int xdec, int ydec, int dir[CDEF_NBLOCKS][CDEF_NBLOCKS], int *dirinit, int var[CDEF_NBLOCKS][CDEF_NBLOCKS], int pli, ... ... @@ -421,17 +415,10 @@ void cdef_filter_fb(uint8_t *dst8, uint16_t *dst16, int dstride, uint16_t *in, int bsize, bsizex, bsizey; #if CONFIG_CDEF_SINGLEPASS int pri_strength = (level >> 1) << coeff_shift; int filter_skip = level & 1; int pri_strength = level << coeff_shift; sec_strength <<= coeff_shift; if (!pri_strength && !sec_strength && filter_skip) { pri_strength = 19 << coeff_shift; sec_strength = 7 << coeff_shift; } #else int threshold = (level >> 1) << coeff_shift; int filter_skip = get_filter_skip(level); if (level == 1) threshold = 31 << coeff_shift; int threshold = level << coeff_shift; cdef_direction_func cdef_direction[] = { cdef_direction_4x4, cdef_direction_8x8 }; ... ... @@ -474,7 +461,7 @@ void cdef_filter_fb(uint8_t *dst8, uint16_t *dst16, int dstride, uint16_t *in, if (threshold != 0) { assert(bsize == BLOCK_8X8 || bsize == BLOCK_4X4); for (bi = 0; bi < cdef_count; bi++) { int t = !filter_skip && dlist[bi].skip ? 0 : threshold; int t = dlist[bi].skip ? 0 : threshold; by = dlist[bi].by; bx = dlist[bi].bx; (cdef_direction[bsize == BLOCK_8X8])( ... ... @@ -495,7 +482,7 @@ void cdef_filter_fb(uint8_t *dst8, uint16_t *dst16, int dstride, uint16_t *in, int py = by << bsizey; int px = bx << bsizex; if (!filter_skip && dlist[bi].skip) continue; if (dlist[bi].skip) continue; if (!dst || hbd) { // 16 bit destination if high bitdepth or 8 bit destination not given (!threshold || (dir[by][bx] < 4 && dir[by][bx]) ? aom_clpf_block_hbd ... ... @@ -561,8 +548,8 @@ void cdef_filter_fb(uint8_t *dst8, uint16_t *dst16, int dstride, uint16_t *in, assert(bsize == BLOCK_8X8 || bsize == BLOCK_4X4); for (bi = 0; bi < cdef_count; bi++) { int t = !filter_skip && dlist[bi].skip ? 0 : pri_strength; int s = !filter_skip && dlist[bi].skip ? 0 : sec_strength; int t = dlist[bi].skip ? 0 : pri_strength; int s = dlist[bi].skip ? 0 : sec_strength; by = dlist[bi].by; bx = dlist[bi].bx; if (dst8) ... ...\n ... ... @@ -16,7 +16,7 @@ #define CDEF_BLOCKSIZE 64 #define CDEF_BLOCKSIZE_LOG2 6 #define CDEF_NBLOCKS (CDEF_BLOCKSIZE / 8) #define CDEF_NBLOCKS ((1 << MAX_SB_SIZE_LOG2) / 8) #if CONFIG_CDEF_SINGLEPASS #define CDEF_SB_SHIFT (MAX_SB_SIZE_LOG2 - CDEF_BLOCKSIZE_LOG2) #endif ... ... @@ -26,10 +26,12 @@ /* We only need to buffer three horizontal pixels too, but let's align to 16 bytes (8 x 16 bits) to make vectorization easier. */ #define CDEF_HBORDER (8) #define CDEF_BSTRIDE ALIGN_POWER_OF_TWO(CDEF_BLOCKSIZE + 2 * CDEF_HBORDER, 3) #define CDEF_BSTRIDE \\ ALIGN_POWER_OF_TWO((1 << MAX_SB_SIZE_LOG2) + 2 * CDEF_HBORDER, 3) #define CDEF_VERY_LARGE (30000) #define CDEF_INBUF_SIZE (CDEF_BSTRIDE * (CDEF_BLOCKSIZE + 2 * CDEF_VBORDER)) #define CDEF_INBUF_SIZE \\ (CDEF_BSTRIDE * ((1 << MAX_SB_SIZE_LOG2) + 2 * CDEF_VBORDER)) #if CONFIG_CDEF_SINGLEPASS // Filter configuration ... ... @@ -70,7 +72,6 @@ typedef void (*cdef_direction_func)(uint16_t *y, int ystride, const uint16_t *in, int threshold, int dir, int damping); int get_filter_skip(int level); #endif #if CONFIG_CDEF_SINGLEPASS ... ...\n ... ... @@ -71,6 +71,9 @@ extern \"C\" { #define MAX_VARTX_DEPTH 2 #define MI_SIZE_64X64 (64 >> MI_SIZE_LOG2) #if CONFIG_EXT_PARTITION #define MI_SIZE_128X128 (128 >> MI_SIZE_LOG2) #endif #define MIN_PALETTE_BSIZE BLOCK_8X8 #define MAX_PALETTE_BSIZE BLOCK_64X64 ... ...\n ... ... @@ -538,6 +538,11 @@ typedef struct AV1Common { int cdef_strengths[CDEF_MAX_STRENGTHS]; int cdef_uv_strengths[CDEF_MAX_STRENGTHS]; int cdef_bits; #if CONFIG_EXT_PARTITION int cdef_preset; #else int cdef_preset; #endif int delta_q_present_flag; // Resolution of delta quant ... ...\n ... ... @@ -903,24 +903,6 @@ static void decode_partition(AV1Decoder *const pbi, MACROBLOCKD *const xd, } #endif if (bsize == cm->sb_size) { int width_step = mi_size_wide[BLOCK_64X64]; int height_step = mi_size_wide[BLOCK_64X64]; int w, h; for (h = 0; (h < mi_size_high[cm->sb_size]) && (mi_row + h < cm->mi_rows); h += height_step) { for (w = 0; (w < mi_size_wide[cm->sb_size]) && (mi_col + w < cm->mi_cols); w += width_step) { if (!cm->all_lossless && !sb_all_skip(cm, mi_row + h, mi_col + w)) cm->mi_grid_visible[(mi_row + h) * cm->mi_stride + (mi_col + w)] ->mbmi.cdef_strength = aom_read_literal(r, cm->cdef_bits, ACCT_STR); else cm->mi_grid_visible[(mi_row + h) * cm->mi_stride + (mi_col + w)] ->mbmi.cdef_strength = -1; } } } #if CONFIG_LOOP_RESTORATION for (int plane = 0; plane < av1_num_planes(cm); ++plane) { int rcol0, rcol1, rrow0, rrow1, tile_tl_idx; ... ..." ]

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[ "# Generating random, non-repeating points on the plane\n\nThe problem is to generate random points on the plane that are unique (i.e. no repetition of a point). The following won't work because of repetition:\n\nIn:= RandomInteger[5, {3, 2}]\nOut= {{1, 3}, {3, 5}, {1, 3}}\n\n\nSo RandomSample may be the answer. But something like the following also repeats:\n\nIn:= Table[RandomSample[Range, 2], {3}]\nOut= {{3, 4}, {4, 5}, {3, 4}}\n\n\nIs there a clever solution to this?\n\nBetter is\n\nRandomSample[Tuples[Range, 2], 3]\n\n\nYour formula may generate the same point two-times or more\n\n(try, for a counterexample\n\nTable[RandomSample[Range, 2], {3}]\n\n\n)\n\n• Can you expand on what your method is here? Or what your thought process was? These techniques may be useful for the generation of aperiodic patterning or simulation of cell/material growth processes, and also I find it useful to expand what you’ve done here in terms of these are awesome one-liners, and are very well done :) May 17, 2019 at 13:52\n• They may be useful, but this procedure is not sophisticated. Range generates the list {1,2,3,4,5}, Tuples[#,2] set of 25 combinations '{{1,1}... {5,5}' and 'RandomSample[#,3]' randomly selects three of them. That's very cumbersome to the simulation. May 17, 2019 at 14:32\n\nAnother approach...if you want 10 points in a 3D space, with no repeats of a coordinate in any dimension...\n\ndim = 3;\nnumPts = 10;\nTranspose@(Ordering /@ Ordering /@ RandomReal[1, {dim, numPts}])\n\n\n$$\\begin{array}{ccc} 8 & 6 & 4 \\\\ 7 & 5 & 8 \\\\ 2 & 4 & 1 \\\\ 5 & 8 & 5 \\\\ 9 & 3 & 2 \\\\ 10 & 7 & 9 \\\\ 1 & 2 & 10 \\\\ 4 & 1 & 7 \\\\ 6 & 9 & 6 \\\\ 3 & 10 & 3 \\\\ \\end{array}$$\n\nData sets in this form (each column is a permutation of Range[numPts]) have a bunch of interesting combinatorial properties. What is fascinating is to take the transform and apply it to random data that is distributed in unique ways, such as points on a simplex, hypersphere, etc.\n\nExpanding a wee bit: the reason this transformation is interesting is that in algorithms to identify the Pareto frontier, you don't care about absolute values of a coordinate, just its ordinal value with respect to other values in a column. Once transformed, a bunch of shortcuts and interesting properties open up.\n\n• or Transpose@Table[RandomSample@Range@numPts, {dim}] to generate random permutations directly May 17, 2019 at 13:38\n• You may be interested in this performance comparison for Ordering[Ordering[list]]. May 17, 2019 at 13:40\n• Great expansion! Same goes for your answer, can you expand on your one-liner’s through process/methodology? :D @Roman thank you for the link, what a great combo of question and answer! May 17, 2019 at 13:55\n• @Roman, I remember that post on Ordering@Ordering. May 17, 2019 at 13:57\n• One interesting tidbit is with the data structured as I did, you actually can't have all of the data on a single plane if you have an even number of points. With an odd number of points, the sum of values for each point must add to 3(n+1)/2 where n is the number of points. May 17, 2019 at 14:00\n\nIf the number of possible points is large, than enumerating them all and taking a random sample won't be feasible. For example, suppose you have integer points in a 1000 x 1000 x 1000 box. Then, the number of possible points is 10^9, and it is unlikely that your computer will be able to generate the full list. Instead, it makes sense to index the points, and take a random sample of the indices, and then convert the indices to a point. For instance, in this example, there are 10^9 indices, and we can convert any index to a point using IntegerDigits, for example:\n\nIntegerDigits[13412343, 1000]\n\n\n{13, 412, 343}\n\nSo, to find 10 random sample indices, we can do:\n\nSeedRandom\nindices = RandomSample[0 ;; 10^9-1, 10]\n\n\n{877665282, 101700636, 562018428, 288541214, 403280597, 238031837, 817685571, \\ 339828267, 510012226, 749565074}\n\nThe key here is that RandomSample can accept a span object instead of a list of 10^9 integers. Then, convert them to points with IntegerDigits:\n\nIntegerDigits[indices, 1000]\n\n\n{{877, 665, 282}, {101, 700, 636}, {562, 18, 428}, {288, 541, 214}, {403, 280, 597}, {238, 31, 837}, {817, 685, 571}, {339, 828, 267}, {510, 12, 226}, {749, 565, 74}}" ]

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[ "# Can you explain Bleichenbacher's CCA attack on PKCS#1 v1.5?\n\nI've studied that the Bleichenbacher's CCA attack on PKCS#1 v1.5. is a base to many versions of attacks in the area.\n\nI'm trying to understand that attack, but every explanation I saw starts with the technical details, without giving some overview, so it's hard to follow...\n\nCan you explain it in a simple words before giving the little details?\n\nWhen encrypting something with RSA, using PKCS#1 v1.5, the data that is to be encrypted is first padded, then the padded value is converted into an integer, and the RSA modular exponentiation (with the public exponent) is applied. Upon decryption, the modular exponentiation (with the private exponent) is applied, and then the padding is removed. The core of Bleichenbacher's attack relies on an oracle: the attack works if there is some system, somewhere, which can tell, given a sequence of bytes of the length of an encrypted message, whether decryption would yield something which has the proper padding format or not.\n\nAn example would be a SSL/TLS server. In the initial handshake, at some point, the client is supposed to generate a random key (the \"pre-master secret\"), encrypt it with the server's public key, and send it. The server decrypts the value, obtains the pre-master secret, and then compute from that pre-master secret the keys used for symmetric encryption of the rest of the connection. Using the standard for guidance, the client sends a ClientKeyExchange (which contains the encrypted pre-master secret), then a ChangeCipherSpec, then Finished; this last message is encrypted with the derived symmetric key and its contents are verified by the server.\n\nIf the client sends a random sequence of bytes of the right length to the server instead of a properly encrypted pre-master secret, then the server will, most of the time, respond with an error message telling \"I tried to decrypt your ClientKeyExchange contents, but this failed, there was not a proper padding in it\". However, by pure chance, it may happen that the random string, after applying the modular exponentiation, yields something which really looks like a pre-master secret with correct padding. In that case, the server will not complain about the ClientKeyExchange, but about the Finished message, which will be incorrectly encrypted.\n\nThis is the information the attacker wants: whether the sequence of bytes he sent would, upon decryption, look properly padded or not.\n\nLet's see with a bit more technical details. In RSA, let $n$ be the public modulus. Let $M$ be a message to encrypt with $n$ (in the case of SSL, $M$ is the pre-master secret, of length 48 bytes). The PKCS#1 v1.5 padding, for encryption, consists in adding some bytes to the left, so that the total length after padding is equal to that of $n$. For instance, if the server's public key is a 2048-bit RSA key, then $n$ has length 256 bytes, so the padded $M$ should also have length 256 bytes.\n\nA properly padded message $M$ has the following format:\n\n0x00 0x02 [some non-zero bytes] 0x00 [here goes M]\n\n\nso the sequence of bytes will begin with a byte of value 0, then a byte of value 2, then some bytes which should have random values (but not zero), then a byte of value 0, then $M$ itself. The number of non-zero bytes is adjusted so that the total length is equal to the length of $n$. Upon decryption, the server will look at the first two bytes, and require them to be equal to 0x00 and 0x02, in that order. Then it will scan for the next byte of value 0, thus skipping over all the random non-zero bytes. This way, the padding can be unambiguously removed.\n\nIt follows that if the client sends a random string of bytes, then it has probability roughly between $2^{-15}$ and $2^{-17}$ to follow the PKCS#1 padding format (that's the probability that the first two bytes are 0x00 0x02, and that there is at least one byte of value 0 afterwards; exact probability depends on the length and value of $n$).\n\nThe attack scenario is the following:\n\n• There is a SSL server, which will send distinct error messages depending on whether a proper PKCS#1 padding was found or not. Alternatively, the two cases could be distinguished through some other information leak (e.g. the server takes longer to respond if the padding was correct).\n• The attacker eavesdropped on a connection, and would like to decrypt it. He observed the ClientKeyExchange, so he saw an encrypted message $c$. He knows that $c = m^e \\pmod n$ where $e$ is the public exponent, and $m$ is the padded pre-master secret for that connection. He wants to recover $m$, or at least the pre-master secret which is contained in $m$, because that will allow him to compute the symmetric keys used for the connection.\n\nThen the attacker will initiate many connections to the server. For each connection, the attacker generates a value $s$ and sends, as ClientKeyExchange, a value $c' = cs^e \\pmod n$. The server decrypts that, and obtains $m' = (cs^e)^d \\pmod n$ ($d$ is the private exponent), which is equal to $ms \\pmod n$. Most of the time, this $ms$ value will not be properly padded (it will not begin with 0x00 0x02 or will not contain an extra 0x00). However, with a low but non-negligible probability (once every 30000 to 130000 attempts, roughly), luck will have it that the $ms \\pmod n$ value looks padded. If that is the case, then the server's behaviour will inform the attacker of that fact. The attacker then learns that, for this value $s$ (the attacker knows it, since he chose it), then $ms \\pmod n$ is in a specific range (the range of integers which begin with 0x00 0x02 when encoded in bytes using big-endian convention).\n\nThe rest of the attack is trying again, with carefully chosen random values $s$. Each time the server responds with \"that was a proper PKCS#1 padding\", this gives some information which helps the attacker narrow his guesses on $m$. After a few million connections in all, the attacker learned enough to pinpoint the exact $m$, yielding the pre-master secret.\n\nSee the original article for details; once you know how the RSA padding works, the rest is just maths, which are not too hard.\n\nTo prevent this attack, SSL servers do not inform the client about padding woes. If decryption fails because of a bad padding, then the server continues with a random pre-master secret (the true failure will then occur when processing the Finished message).\n\nOne may note that the specific weakness of the PKCS#1 v1.5 padding (for encryption) is that it is not very redundant; the random bytes are, indeed, random, without any specifically enforced value. This is what allows a sequence of random bytes to be \"properly padded\" with a small but not negligible probability. Newer versions of PKCS#1 describe a new padding type, called OAEP, which uses hash function to add a lot of internal redundancy, which makes it extremely improbable that a random string matches the padding format. This prevents Bleichenbacher's attack. Unfortunately, SSL still uses PKCS#1 v1.5.\n\n• Could not find this Q/A using Google. Thomas, should we generate a new question \"why is RSA PKCS#1 v1.5 encryption considered broken?\". Problem is answer would be more or less identical. – Maarten Bodewes Jun 2 '14 at 11:04\n• Even if the SSL server doesn't inform the client about padding error, the client can still tell that the padding wasn't right after the Finished message fails. – Myath Jan 1 '16 at 8:08\n• @Myath Also, you might be able to determine this fact via timing side-channels. :) – Scott Arciszewski Jan 3 '16 at 4:24\n• @Myath: ah no, that's the tricky point. If the server proceeds with a random key in case of bad padding, an inconsistent Finished does NOT reveal that the padding was bad -- maybe the padding was good, and the server merely used whatever pre-master secret it thus obtained (and is unknown to the attacker). – Thomas Pornin Jan 3 '16 at 12:52\n• @ThomasPornin Can you please tell me how we arrive at below conclusion: 2B <= ms mod n < 3B: where B = 2^8(k−2); I am not able to understand how value of B is defined and how we are able to say that the value will be in the range of 2B and 3B. – Sam Jul 23 '16 at 14:53" ]

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[ "# python compare dates with code examples\n\nDates are an important part of many applications, and Python provides several ways to work with them. In this article, we will explore how to compare dates in Python using several examples.\n\nThe first step in working with dates in Python is to import the datetime module. This module provides the datetime class, which is used to represent dates and times. To create a date object, we can use the datetime constructor, which takes three arguments: the year, the month, and the day. For example, to create a date object representing January 1, 2020, we would use the following code:\n\n```from datetime import datetime\ndate1 = datetime(2020, 1, 1)\nprint(date1)\n```\n\nOutput:\n\n```2020-01-01 00:00:00\n```\n\nTo compare two dates, we can use the comparison operators (>, <, >=, <=, ==, !=) just like any other data type. For example, to check if date1 is before date2, we would use the following code:\n\n```date2 = datetime(2020, 2, 1)\nif date1 < date2:\nprint(\"date1 is before date2\")\nelse:\nprint(\"date1 is not before date2\")\n```\n\nOutput:\n\n```date1 is before date2\n```\n\nAlternatively, we can use the `date.date()` method that is available on datetime objects to return just the date component and then compare them.\n\n```date1 = datetime.date(date1)\ndate2 = datetime.date(date2)\nif date1 == date2:\nprint(\"date1 and date2 are same\")\nelse:\nprint(\"date1 and date2 are not same\")\n```\n\nOutput:\n\n```date1 and date2 are not same\n```\n\nWe can also use the `timedelta` class to perform arithmetic on dates. For example, to find the number of days between two dates, we can subtract one date from the other and then access the days attribute of the resulting timedelta object.\n\n```delta = date2 - date1\nprint(delta.days)\n```\n\nOutput:\n\n```31\n```\n\nIn addition to comparing and performing arithmetic on dates, we can also format them for display using the `strftime` method. This method takes a format string that specifies how the date should be displayed, using codes such as %Y for the year and %m for the month.\n\n```formatted_date = date1.strftime(\"%Y-%m-%d\")\nprint(formatted_date)\n```\n\nOutput:\n\n```2020-01-01\n```\n\nIn conclusion, Python provides a variety of tools for working with dates, including the datetime class for representing dates and times, comparison operators for comparing dates, and the timedelta class for performing arithmetic on dates. Additionally, the `strftime` method can be used to format dates for display. The above examples should give you a good starting point for working with dates in Python.\n\nAnother useful class provided by the datetime module is the time class, which is used to represent the time of day. The time class takes three arguments: the hour, the minute, and the second. For example, to create a time object representing 3:30 PM, we would use the following code:\n\n```from datetime import time\ntime1 = time(15, 30)\nprint(time1)\n```\n\nOutput:\n\n```15:30:00\n```\n\nWe can also use the time class along with the datetime class to create a datetime object that contains both a date and a time. For example:\n\n```from datetime import datetime, time\ndate1 = datetime(2020,1,1)\ntime1 = time(15,30)\ndatetime1 = datetime.combine(date1, time1)\nprint(datetime1)\n```\n\nOutput:\n\n```2020-01-01 15:30:00\n```\n\nAnother useful feature of the datetime module is the ability to work with time zones. The pytz library is a popular library used to work with time zones in Python. Once the library is installed, we can use the `timezone` class from the datetime module to create a timezone-aware datetime object. Here's an example of creating a timezone-aware datetime object for the Eastern Time Zone (ET) in the United States:\n\n```import pytz\nfrom datetime import datetime\n\neastern = pytz.timezone('US/Eastern')\ndate1 = datetime(2020, 1, 1, tzinfo=eastern)\nprint(date1)\n```\n\nOutput:\n\n```2020-01-01 00:00:00-05:00\n```\n\nYou can also convert an aware datetime object to different timezone using `pytz.timezone.localize()` or `pytz.timezone.normalize()` functions.\n\nIn addition to the datetime module and the pytz library, Python also provides the dateutil library, which is another popular library for working with dates and times. The dateutil library provides additional features such as parsing dates from strings and performing arithmetic on dates. For example, we can use the `relativedelta` function from the dateutil library to find the difference between two dates in terms of years, months, days, etc.\n\n```from dateutil.relativedelta import relativedelta\ndate1 = datetime(2020, 1, 1)\ndate2 = datetime(2021, 1, 1)\ndiff = relativedelta(date2, date1)\nprint(diff.years, \"years\", diff.months, \"months\", diff.days, \"days\")\n```\n\nOutput:\n\n```1 years 0 months 0 days\n```\n\nIn this article, we have discussed several ways to work with dates and times in Python, including the datetime module, the pytz library, and the dateutil library. We hope that this information will help you in your development projects. Remember, always be mindful of timezones when working with dates and times in your application to ensure accurate data representation.\n\n## Popular questions\n\n1. How do you create a date object in Python using the datetime module?\n\n• To create a date object in Python using the datetime module, you can use the datetime constructor, which takes three arguments: the year, the month, and the day. For example, to create a date object representing January 1, 2020, you would use the following code:\n```from datetime import datetime\ndate1 = datetime(2020, 1, 1)\nprint(date1)\n```\n\nOutput:\n\n```2020-01-01 00:00:00\n```\n2. How can you compare two dates in Python?\n\n• To compare two dates in Python, you can use the comparison operators (>, <, >=, <=, ==, !=) just like any other data type. For example, to check if date1 is before date2, you would use the following code:\n```date1 = datetime(2020, 1, 1)\ndate2 = datetime(2020, 2, 1)\nif date1 < date2:\nprint(\"date1 is before date2\")\nelse:\nprint(\"date1 is not before date2\")\n```\n\nOutput:\n\n```date1 is before date2\n```\n3. How can you perform arithmetic on dates in Python?\n\n• To perform arithmetic on dates in Python, you can use the timedelta class. For example, to find the number of days between two dates, you can subtract one date from the other and then access the days attribute of the resulting timedelta object.\n```date1 = datetime(2020, 1, 1)\ndate2 = datetime(2020, 2, 1)\ndelta = date2 - date1\nprint(delta.days)\n```\n\nOutput:\n\n```31\n```\n4. How can you format a date for display in Python?\n\n• To format a date for display in Python, you can use the `strftime` method. This method takes a format string that specifies how the date should be displayed, using codes such as %Y for the year and %m for the month.\n```date1 = datetime(2020, 1, 1)\nformatted_date = date1.strftime(\"%Y-%m-%d\")\nprint(formatted_date)\n```\n\nOutput:\n\n```2020-01-01\n```\n5. How can you work with time zones in Python?\n\n• To work with time zones in Python, you can use the pytz library. Once the library is installed, you can use the `timezone` class from the datetime module to create a timezone-aware datetime object. Here's an example of creating a timezone-aware datetime object for the Eastern Time Zone (ET) in the United States:\n```import pytz\nfrom datetime import datetime\neastern = pytz.timezone('US/Eastern')\ndate1 = datetime(2020, 1, 1, tzinfo=eastern)\nprint(date1)\n```\n\nOutput:\n\n```2020-01-01 00:00:00-05:00\n```\n\n### Tag\n\nDatetime.", null, "Posts created 2498\n\n## partial derivatives symbol ms word with code examples\n\nBegin typing your search term above and press enter to search. Press ESC to cancel." ]

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

[ "to\n\n### Neural Quantum States\n\nOne of the most challenging problems in modern theoretical physics is the so-called many-body problem. Typical many-body systems are composed of a large number of strongly interacting particles. Few such systems are amenable to exact mathematical treatment and numerical techniques are needed to make progress. However, since the resources required to specify a generic many-body quantum state depend exponentially on the number of particles in the system (more precisely, on the number of degrees of freedom), even today's best supercomputers lack sufficient power to exactly encode such states (they can handle only relatively small systems, with less than 45 particles). As we shall see, recent applications of machine learning techniques (artificial neural networks in particular) have been shown to provide highly efficient representations of such complex states, making their overwhelming complexity computationally tractable.\n\n### AI learns to solve quantum state of many particles at once\n\nThe same type of artificial intelligence that mastered the ancient game of Go could help wrestle with the amazing complexity of quantum systems containing billions of particles. Google's AlphaGo artificial neural network made headlines last year when it bested a world champion at Go. After marvelling at this feat, Giuseppe Carleo of ETH Zurich in Switzerland thought it might be possible to build a similar machine-learning tool to crack one of the knottiest problems in quantum physics. Now, he has built just such a neural network – which could turn out to be a game changer in understanding quantum systems. Go is far more complex than chess, in that the number of possible positions on a Go board could exceed the number of atoms in the universe.\n\n### Neural network state estimation for full quantum state tomography\n\nAn efficient state estimation model, neural network estimation (NNE), empowered by machine learning techniques, is presented for full quantum state tomography (FQST). A parameterized function based on neural network is applied to map the measurement outcomes to the estimated quantum states. Parameters are updated with supervised learning procedures. From the computational complexity perspective our algorithm is the most efficient one among existing state estimation algorithms for full quantum state tomography. We perform numerical tests to prove both the accuracy and scalability of our model.\n\n### Learning hard quantum distributions with variational autoencoders\n\nStudying general quantum many-body systems is one of the major challenges in modern physics because it requires an amount of computational resources that scales exponentially with the size of the system.Simulating the evolution of a state, or even storing its description, rapidly becomes intractable for exact classical algorithms. Recently, machine learning techniques, in the form of restricted Boltzmann machines, have been proposed as a way to efficiently represent certain quantum states with applications in state tomography and ground state estimation. Here, we introduce a new representation of states based on variational autoencoders. Variational autoencoders are a type of generative model in the form of a neural network. We probe the power of this representation by encoding probability distributions associated with states from different classes. Our simulations show that deep networks give a better representation for states that are hard to sample from, while providing no benefit for random states. This suggests that the probability distributions associated to hard quantum states might have a compositional structure that can be exploited by layered neural networks. Specifically, we consider the learnability of a class of quantum states introduced by Fefferman and Umans. Such states are provably hard to sample for classical computers, but not for quantum ones, under plausible computational complexity assumptions. The good level of compression achieved for hard states suggests these methods can be suitable for characterising states of the size expected in first generation quantum hardware.\n\n### Viewpoint: Neural Networks Take on Open Quantum Systems\n\nNeural networks are behind technologies that are revolutionizing our daily lives, such as face recognition, web searching, and medical diagnosis. These general problem solvers reach their solutions by being adapted or \"trained\" to capture correlations in real-world data. Having seen the success of neural networks, physicists are asking if the tools might also be useful in areas ranging from high-energy physics to quantum computing . Four research groups now report on using neural network tools to tackle one of the most computationally challenging problems in condensed-matter physics--simulating the behavior of an open many-body quantum system [2–5]. This scenario describes a collection of particles--such as the qubits in a quantum computer--that both interact with each other and exchange energy with their environment." ]

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[ "Having trouble expressing a partial derivative\n\nI am new to this website and Mathematica, but I have a problem with partial derivatives. What i am trying to do is get the Lagrangian equation of my system using x,y coordinates. The problem is i think in the last part with T,L1,L2,L3 and L. The code is given here:\n\nClear[m, \\[Alpha], \\[Beta], l1, x, l3, l4, l5, L1, L2, L3, L, r]\nn = 12;\nl2 = 1;\nm = 1;\n\nf[\\[Theta]_] := {\\[Theta], 0, 2*Pi}\n\\[Alpha] = \\[Alpha] /. Solve[(Pi - \\[Alpha])*n == 2*Pi];\n\\[Beta] = (Pi - \\[Alpha])/2;\nl1 = 2*(Cos[\\[Beta] ]/l2);\nl3 = (l1*Cos[\\[Beta] + \\[Theta]])/Cos[Pi - \\[Alpha]];\nl4 = (l2*Cos[\\[Theta]])/Cos[1/2*(Pi - \\[Alpha])];\n(*The calculation of the x coordinate of the whole system*)\nFor[i = 1, i <= n, i++ , x[i] = l3*Cos[((i - 1)*(Pi - \\[Alpha]))];]\nFor[i = n + 1, i <= 2*n, i++ ,\nx[i] = l4*Cos[((i - 1/2)*(Pi - \\[Alpha]))]]\nl5 = l3 + 2*x[n + 1];\nFor[i = 2*n + 1, i <= 3*n, i++ ,\nx[i] = l5*Cos[((i - 1)*(Pi - \\[Alpha]))]]\n\n(*The calculation of the y coordinate of the whole system*)\nFor[i = 1, i <= n, i++ , y[i] = l3*Sin[((i - 1)*(Pi - \\[Alpha]))]];\nFor[i = n + 1, i <= 2*n, i++ ,\ny[i] = l4*Sin[((i - 1/2)*(Pi - \\[Alpha]))]];\nFor[i = 2*n + 1, i <= 3*n, i++ ,\ny[i] = l5*Sin[((i - 1)*(Pi - \\[Alpha]))]];\n\n(*The calculation of the velocity, the v here is v^2*)\nFor[i = 1, i <= 3*n, i++, v[i] = Dt[x[i]]^2 + Dt[y[i]]^2]\nSimplify[v];\nSimplify[v];\nSimplify[v];\n(*Print[\"v\",i,\"=\",Simplify[v[i]]*)\n\nT = Simplify[Sum[1/2 m*Simplify[v[i]], {i, n*3}]];\nL1 = Simplify[D[T, Dt[\\[Theta]]]];\nL2 = Dt[L1]\nL3 = Simplify[[D[T, \\[Theta]]]]\nL = L2 - L3;\n\nThe problem is that the derivative of Dt[\\[Theta]] should be a constant when deriving a derivative only with respect to \\[Theta]. Furthermore, why is the r in D[T, \\[Theta]] lighter blue than the \\[Theta] at L3 = Simplify[[D[T, \\[Theta]]]]?\n\nThe simplified example is given beneath:\n\nf[r_] := {r, 0, 2*Pi}\na = r;\nb = r^2;\nF = a^2 + b^2;\nF1 = Dt[F];\nD[F1, r]\n\nWhere the answer is: 2 r !(*SubscriptBox[([PartialD]), (r)](Dt[r])) + 4 r^3 !(*SubscriptBox[([PartialD]), (r)](Dt[r])) + 2 Dt[r] + 12 r^2 Dt[r]\n\n• I think that Dt[r] is not a constant because Mathematica doesn't know what \"r\" is. Indeed, r could also be a function of some other variable. The fact that in a=r the r is blue it's because that's an undefined variable in the notebook, while in D[F1,r] \"r\" is an argument of the function. – Fraccalo May 23 '18 at 13:28\n• In my problem r should be a variable that will very between 0 and 2pi. If i add this: f[r_] := {r, 0, 2*Pi} It still doesn't work. – Rik Koppelman May 23 '18 at 13:46\n• Please do not post images of your work. Please post your actual Mathematica code in the form of text that can be copied and pasted into a Mathematica notebook. This will make it easier to reproduce your problem and to experiment with possible solutions. – m_goldberg May 23 '18 at 14:24\n• @m_goldberg. I've added the code itself to the problem – Rik Koppelman May 23 '18 at 14:39\n• I have a feeling that your code does not express what you are after (even if with errors), but something else. It may be a good idea, if you first write the expression you want to calculate in traditional mathematical notations. Than we will probably understand what's wrong with the Mathematica code. – Alexei Boulbitch May 23 '18 at 15:15" ]

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[ "# Can't find the equivalent resistance in this circuit", null, "I'm trying to self-study this problem. This chapter explained the delta-to-y transformation but I can't seem to find one here.\n\nMy best guess is that both resistors on the right are short-circuited which gives them a value of 0. Then you add the two 3 ohm resistor on the left together, and thus the answer 6 ohms.\n\nBut I don't know if this is true.\n\n• To me, the answer is 7 Ohms. The 3 resistors to the right are in parallel with each other and can be redrawn as such if you move the two rightmost nodes to the left (over the vertical resistor terminals). – Ricardo Sep 16 '14 at 18:23\n• @Ricardo, sounds correct. I didn't picture them in parallel. – George Chalhoub Sep 16 '14 at 18:25\n• All 'internal' nodes are at the same potential, hence shorted. – copper.hat Sep 16 '14 at 18:30\n• @copper.hat I can see it now. – Ricardo Sep 16 '14 at 18:40\n• @Ricardo: I made the same mistake. – copper.hat Sep 16 '14 at 18:41\n\nYou're right. The three resistors 3 $\\Omega$, are shorted, so that the equivalent resistance is the sum 3 + 3 = 6.", null, "• @Null: They are in parallel and shorted. – copper.hat Sep 16 '14 at 18:32\n• @Ricardo: Look at the outer edges. All 'internal' nodes are shorted together. (I fell for it too!) – copper.hat Sep 16 '14 at 18:33\n• @copper.hat Ah, yes. Whoever draws schematics like this should be shot. – Null Sep 16 '14 at 18:34\n• @Null: I know, I was 'certain' about my 7 answer for a while. – copper.hat Sep 16 '14 at 18:35\n• @Null This is simply an educational schematic to practice on. – hryghr Sep 16 '14 at 18:36\n\nThis answer is wrong, as pointed out by Tinchito, but I'm leaving it here shamelessly for didactic reasons, as it seems as the most common mistake when looking quickly at the circuit.\n\nAt first sight, you think you could redraw the schematic as follows (but you can't):", null, "simulate this circuit – Schematic created using CircuitLab\n\nIf that was the case, then,\n\n$$R3||R4||R5 = 1 \\Omega$$\n\nand\n\n$$R_{eq} = R1 + R2 + (R3||R4||R5) = 7 \\Omega$$\n\nLooking closer you'll see that there's a zero resistance path that shorts out all three resistors, as pointed out by Tinchito, making the correct answer $6 \\Omega$.\n\n• It fooled me for a while - I misread it like you did!! – Andy aka Sep 16 '14 at 20:36\n• @Andy - Yeah, it fooled many others here, too, it seems. That's probably the intention of the circuit's original author. I bet he or she could be a magician or illusionist. – Ricardo Sep 16 '14 at 21:11" ]

[ null, "https://i.stack.imgur.com/wPHDs.png", null, "https://i.stack.imgur.com/QHPlo.png", null, "https://i.stack.imgur.com/Wlm8B.png", null ]

[ "# Relabeling a label matrix\n\nThe three most common questions I've been hearing about bwlabel, which is used for labeling connected components in a binary image, are about\n\n• Search order\n• Relabeling (renumbering the label matrix)\n• Correspondence between labels in two different images\n\nToday I'll tackle relabeling.\n\nIn my previous post on search order, I showed how to postprocess the labeled objects to modify their order according to various criteria. Now I'll take that a step further and use the sorted output from regionprops to renumber the labels in the label matrix.\n\nBlog reader Trung wanted to know if we could sort lexicographically by centroid, so I'll do that here.\n\nurl = 'https://blogs.mathworks.com/images/steve/186/scanned_page.png';\nbw = ~bw(1107:1194, 17:135);\nimshow(bw, 'InitialMagnification', 'fit')", null, "Here's a false-color view of the label matrix using label2rgb:\n\nL = bwlabel(bw);\nrgb = label2rgb(L, 'jet', [.95 .95 .95], 'shuffle');\nimshow(rgb, 'InitialMagnification', 'fit')", null, "Now let's use regionprops to get the centroids for each object. To facilitate the relabeling step, I'll get the 'PixelIdxList' for each object.\n\ns = regionprops(L, {'Centroid', 'PixelIdxList'});\n\nNext, sort lexicographically by the centroids, sorting first by the vertical coordinate.\n\ncentroids = cat(1, s.Centroid);\n[sorted_centroids, sort_order] = sortrows(fliplr(centroids));\ns2 = s(sort_order);\n\nTo visualize the sorted object order, we can display the object numbers on top of the objects like this:\n\n% First, make an image with a light gray background instead\n% of a black background, so that the numbers will be visible\n% on top of it.\nI = im2uint8(bw);\nI(~bw) = 200;\nI(bw) = 240;\nimshow(I, 'InitialMagnification', 'fit')\n\n% Now plot the number of each sorted object at the corresponding\n% centroid:\nhold on\nfor k = 1:numel(s2)\ncentroid = s2(k).Centroid;\ntext(centroid(1), centroid(2), sprintf('%d', k));\nend\nhold off", null, "We sorted the output of regionprops, but we haven't touched the label matrix itself. We can relabel it according to the sort order by using linear indexing and the PixelIdxList of each object. An object's PixelIdxList is a vector of linear indices for the pixels belonging to the object. Here's the relabeling loop:\n\nfor k = 1:numel(s2)\nkth_object_idx_list = s2(k).PixelIdxList;\nL(kth_object_idx_list) = k;\nend\n\nIn MATLAB 7.6 (R2008a), the publish feature lets you capture multiple graphics from within a loop. I'll use that new capability to show the location of the first few relabeled objects in L.\n\nfor k = 1:5\n imshow(L == k)", null, "", null, "", null, "", null, "", null, "end\n\nSoon I'll write another post that shows how how to match up labels in objects that overlap between two images.\n\nPublished with MATLAB® 7.6\n\n|" ]

[ null, "https://blogs.mathworks.com/images/steve/2008/bwlabel_relabeling_01.png", null, "https://blogs.mathworks.com/images/steve/2008/bwlabel_relabeling_02.png", null, "https://blogs.mathworks.com/images/steve/2008/bwlabel_relabeling_03.png", null, "https://blogs.mathworks.com/images/steve/2008/bwlabel_relabeling_04.png", null, "https://blogs.mathworks.com/images/steve/2008/bwlabel_relabeling_05.png", null, "https://blogs.mathworks.com/images/steve/2008/bwlabel_relabeling_06.png", null, "https://blogs.mathworks.com/images/steve/2008/bwlabel_relabeling_07.png", null, "https://blogs.mathworks.com/images/steve/2008/bwlabel_relabeling_08.png", null ]

[ "# 2D NACA 0012 airfoil validation\n\nThis is part one of a two article series on lift in 2D which uses the NACA 0012 airfoil to illustrate some concepts related to lift. The purpose of this validation is to compare our CFD results against known data to certify that we reproduce the physics correctly. Since we will rely on these data to draw conclusions in part two of this series, being able to trust the results of our model is vital.\n\nThe reference validation case is available through NASA Langley Turbulence Modeling Resource (TMR) website, while detailed results have been reported through various papers such as Diskin et al. (2015).\n\n## Setup\n\nHere we will give only a brief outline of the case setup; for more details you can download the finest mesh case at the end of this post, here.\n\n### Grid\n\nNASA makes available several grids used in their verification and validation studies. Among the available options, I have opted for the Family II of structured grids, from size 225 x 64 to size 1793 x 513. The grids have been generated such that the airfoil closes at chord = 1 with a sharp trailing edge. In all available grids, the farfield extends about 500c.\n\nThe minimum length —after scaling— at the wall is 7 x 10-7 and average stretching rate normal to the wall is ~1.02 for the points near the wall. As a result, the grid contains approximately 1 million cell.\n\nThe structured PLOT3D grids can be easily converted to OpenFOAM format with the plot3dToFoam tool.\n\n### Boundary conditions\n\nI’ll be using the results from Diskin et al. (2015) and the TME website as reference for this case. There, we get that the conditions for the case are M = 0.15, Reynolds number per chord is Re = 6 million, alpha = 10 deg, reference temperature = 540 R. The Prandtl number Pr is taken to be constant at 0.72, while the turbulent Prandtl number Prt is taken to be constant at 0.9. The heat capacity ratio ($\\gamma$) is 1.4.\n\nThe conditions above states are given for nondimensional CFD codes such as CFL3D or FUN3D. That’s not the case for OpenFOAM, so the conditions need to be translated. Firstly, we convert the reference temperature from Imperial to SI (540 R = 300 K). Secondly, once we know temperature, we can compute the speed of sound as\n\n$$a = \\sqrt{\\gamma R T}$$\n\nwhere $R$ is the specific gas constant (287.058 J⋅kg−1⋅K−1 for dry air). As a result we get a speed of sound of $a$ = 347.2 m/s. Thirdly, now we can use the Mach number to get the flow velocity, $U_\\infty$ = 52.08 m/s. Fourthly, we can get additional dry air properties such as kinematic viscosity $\\nu_\\infty$ = 15.68 x 10-6 m2/s. Finally, we can obtain the airfoil chord from Reynolds number, air velocity, and kinematic viscosity as\n\n$$L = \\frac{\\nu_\\infty Re}{U_\\infty}$$\n\nwhich results in a chord of L = 1.80645 m. Therefore, the grid has to be scaled.\n\nOn the other hand, in the farfield we have also the following conditions:\n\n$$\\tilde{\\nu}_\\rm{farfield} = 3\\nu_\\infty$$\n\n$$\\nu_{t,\\rm{farfield}} = 0.210438\\nu_\\infty$$\n\n#### OpenFOAM\n\nThe following table summarises the boundary conditions used in OpenFOAM (front and back are of the empty type).\n\n*The freestreamfreestreamPressure, and freestreamVelocity boundary conditions inherit from inletOutlet and outletInlet. Those boundary conditions switch the mode of operation between fixed (free stream) value and zero gradient based on the sign of the flux.\n\n### Sutherland’s law\n\nNASA repeats through several sections in their website —like here— that in their codes, dynamic viscosity is computed using Sutherland’s law,\n\n$$\\mu = \\mu_0 \\left( \\frac{T}{T_0} \\right)^{3/2} \\left( \\frac{T_0 + S}{T + S} \\right)$$\n\nwere $S$ it the Sutherland temperature (110.4 K). In OpenFOAM, Sutherland’s law is written differently, as\n\n$$\\mu = \\frac{A_s \\sqrt{T}}{1 + S/T}$$\n\nComparing the formulas above,  we can write the constant as:\n\n$$A_s = \\frac{\\mu_0}{T_0^{3/2}} (T_0 + S)$$\n\nwere $\\mu_0$ = 1.176 x 10-5 kg⋅m-1⋅s-1, $T_0$ = 273.15 K, and $S$ = 110.4 K. Thus, $A_s$ = 1.458 x 10-6 kg⋅m-1⋅s-1⋅K-1/2.\n\nWe can enable Sutherland’s law in the thermophysicalProperties dictionary.\n\nNote: I compiled a new implementation of the Sutherland transport model which uses the Prandtl number to calculate thermal conductivity. The code is accessible in GitHub: https://github.com/pfsq/mySutherland\n\n### The Spalart-Allmaras turbulence model\n\nLudwig Prandtl pointed out that certain fluid flows could be divided into a thin viscous layer —boundary layer— near surfaces and an effectively inviscid outer layer. In the inviscid outer layer the Euler and Bernoulli equations apply. But in boundary layers, which are predominantly turbulent, the assumptions taken by Euler and Bernoulli do not hold. Some empirical equations were developed for predicting the turbulent over a flat plate, yet the turbulent flow over arbitrarily shaped bodies still involves the solution of the continuity, momentum, and energy equations along with some model of the turbulence. Philippe Spalart and Steve Allmaras. proposed one of such models.\n\nThe one uncontroversial fact about turbulence is that it is the most complicated kind of fluid motion.\n\nThe Spalart-Allmaras model is a linear eddy viscosity that solves one additional transport equation. Because it is computationally cheaper, it is used in many codes and, for many flows, its performance is comparable to that of many more complicated models.\n\nMany variants of the model have been proposed since its introduction in the early 1990s. Specifically, OpenFOAM’s implementation of the model ignores the $f_{t2}$ laminar suppression term; otherwise, the model is identical to the standard version. But as long as the simulation is at a relatively high Reynolds and $\\tilde{\\nu}_\\rm{farfield} = 3\\nu_\\infty$, ignoring the $f_{t2}$ term should make little difference, as is the case here. In any case —for the inquiring eye— I have implemented the standard version of the Spalart-Allmaras model by reintroducing the $f_{t2}$ term: https://github.com/pfsq/standardSA; unsurprisingly, the results are almost identical.\n\nIn Diskin et al. (2015), the codes FUN3D and TAU use the SA-neg variant of the SA turbulence model that admits negative values of the Spalart turbulence variable, while for positive values of $\\tilde{\\nu}$ the model remains identical to the standard version. In any case, this variant should provide negligible differences in most cases.\n\nNote: older versions of OpenFOAM (v2.x and older) need changes to the turbulence models to match the definitions found in NASA’s website.\n\n### Numerical schemes\n\nThe solutions from all three codes in Diskin et al. (2015) are second order accurate via a MUSCL scheme. Although these schemes are also available in OpenFOAM, I was getting unstable simulations, so I used linearUpwind instead, for velocity, energy, and turbulence, and linear for the rest. As with MUSCL, both schemes are second order accurate discretizations of the advective terms.\n\nSince the solution was initiated with the coarsest mesh and then each successive solution mapped to the next grid level, it was not necessary to start with lower order schemes.\n\n### Linear solvers\n\nI used the PCG solver with FDIC preconditioner for pressure —GAMG was very unstable— and PBiCGStab with DILU preconditioner for the asymmetric matrices.\n\nFor asymmetric matrices, we set a tight tolerance of less than 1e-10, while for pressure —as it is more expensive to calculate— we run with a relative tolerance of 0.001.\n\n## Solution\n\nThe cases were solved with a compressible solver, rhoSimpleFoam, and with the standard Spalart-Allmaras turbulence model. Each was iterated for 5000 steps —except for the finest mesh, which required 10000—, reaching convergence to at least 4 significant digits.\n\n### Convergence of residuals and variables of interest\n\nResiduals go down nicely, although turbulence and energy seem to reach a lower limit at 1e-4 and 1e-7, respectively." ]

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[ "# Graduate Program Course Offerings\n\nApproximately 35 courses that constitute the department's regular graduate curriculum are offered either annually or biennially. They are supplemented by an ongoing sequence of special-topic courses reflecting research interests of the faculty.\n\nCourses in the 1000 series are advanced undergraduate courses that are frequently suitable for graduate credit. Those in the 3000 series are advanced graduate courses. Course content, prerequisites, frequency of offering, and requirements may change from year to year.\n\n### Selected 1000-Level Undergraduate Courses\n\nClass Course Description Course prerequisites\nMATH 1010\nPutnam Seminar\nThe aim of this course is to develop the capacity to solve mathematical problems involving a substantial element of ingenuity and perseverance.  Training will involve the study of problems from previous Putnam competitions, for which this course can be regarded as a useful preparation. An attempt will be made to look for unifying mathematical ideas.  General strategies for solving problems will also be discussed. Instructor Consent\nMATH 1020 Applied Elementary Number Theory This course will reveal the key role played by number theory in the development of mathematics.  Some applications of number theory will be covered in the course.\n\nSyllabus\n\nMATH 0430\n\nMATH 1025\nIntroduction to Mathematical Cryptography\n\nThe course covers the theoretical underpinnings of cryptosystems and the analysis of their limitations and vulnerabilities. Special emphasis will be placed on public key cryptosystems, including elliptic curve based systems. Real world applications such as browser security and bitcoin will be discussed. MATH 0430\nMATH 1050 Combinatorial Mathematics Topics covered include the binomial theorem, inclusion exclusion principle, recurrence relations, generating functions, and coloring problems. MATH 0413 or 0450 or 1185\nMATH 1070 Classical Numerical Analysis This course, with MATH 1080 forms a two term introduction to numerical analysis at the advanced undergraduate level and includes interpolation, numerical differentiation and integration, solution of non-linear equations, numerical solution of systems or ordinary differential equations, and additional topics as time permits.  Emphasis is on understanding the algorithms rather than on detailed coding, although some programming will be required. MATH 0240 and some programming experience\nMATH 1075\nTopics in Numerical Analysis\nThis course provides honors recognition to students who wish to digress with the professor to study extra topics.\nMATH 1080 Numerical Linear Algebra This course, with MATH 1080 forms a two term introduction to numerical analysis at the advanced undergraduate level and includes interpolation, numerical differentiation and integration, solution of non-linear equations, numerical solution of systems or ordinary differential equations, and additional topics as time permits.  Emphasis is on understanding the algorithms rather than on detailed coding, although some programming will be required. MATH 0240, MATH 0280 OR MATH 1180 and some programming experience\nMATH 1100\nLinear Programming\nTopics covered will include linear programming problems, the simplex method, quality, revised simplex method, and the transportation problem. MATH 0280 or MATH 1180\nMATH 1101\nAn Introduction to Optimization\nThis course introduces students to the techniques of optimization. Applications will be emphasized, but some theory will be addressed and proofs will be discussed.  As well, students will be taught how to use available software to answer questions. Course topics will include linear programming, integer programming, nonlinear programming, convex and affine sets, convex and concave functions, unconstrained optimization, and combinatorial optimization (i.e. Network flow problems). MATH 0240 and one of MATH 0280, 1180\nMATH 1110 Industrial Numerical Analysis This course is concerned with the approximate numerical solution of problems which arise in an industrial environment.  Topics covered include physical interpretation of a mathematical model, use of library software, preparation of software, analysis of results, and reporting on findings. One of MATH 0280, 1180, 1185\nMATH 1119\nApplied Probability for Actuarial Math\nThis course covers standard topics in probability and their applications to actuarial risk management. It prepares students for the probability exam offered by the Society of Actuaries.\nMATH 1120\nActuarial Math 1\nThis course will cover the material listed in the syllabus for Exam 2 (Mathematics of Finance) of the Society of Actuaries. Specifically it will present the relevant topics in the theory of interest (interest and discount rates, cash flows, annuities, amoritization and sinking funds, bonds) and investment (stocks, capital asset pricing model, arbitrage pricing theory, portfolios, options). The material will be presented in the traditional academic format of lectures and help sessions along with optional sessions directed specifically at preparing students for the SOA exam.  MATH 0220 AND 0230\nMATH 1121\nActuarial Math 2\nThis course will cover the material listed in the syllabus for exam m (3) (mathematics of life contingencies and financial economics) of the society of actuaries. Specifically it will present the relevant topics in life insurance and life annuities, including multiple decrement models as well as the black and Scholes pricing of derivative securities and risk analysis. The material will be presented in the traditional academic format of lectures and help sessions along with optional sessions directed specifically at preparing students for the SOA exam. MATH 0240, MATH 1120, STAT 1151\nMATH 1122\nActuarial Math 3\nLife Contingencies MATH 1121\nMATH 1123\nActuarial Math 4\nLife Contingencies, part 2 Math 1122\nMATH 1180 Linear Algebra 1\n\nThis course stresses the theoretical and rigorous development of linear algebra.  Major topics include the theory of vector spaces, linear transformations, matrices, characteristic polynomials, bases and canonical forms.  Other topics may be covered as time permits. Credit will not be given for both MATH 0280 and 1180.\n\nMATH 0413\nMATH 1240 Linear Algebra 2 This second course in linear algebra features an abstract development of the subject.  Abstract vector spaces, linear transformations, and matrix representations will be studied. Some applications and generalizations will also be investigated.\nMATH 1250 Abstract Algebra In this course the basic algebraic systems, groups and rings are studied in some detail.  Topics include:  subgroups, permutation groups, homomorphism's, subrings, ideals and quotient rings.  The emphasis is on theory with examples. MATH 0430\nMATH 1270 Ordinary Differential Equations 1\n\nThis course covers methods of solving ordinary differential equations which are frequently encountered in applications. General methods will be taught for single n-th order equations, and systems of first order nonlinear equations.  This will include phase plane methods and stability analysis. Computer experimentation will be used to illustrate the behavior of solutions of various equations. Credit will not be given for both MATH 0290 and 1270.\n\none of MATH 0280, 1180, or 1185\n\nMATH 1280 Ordinary Differential Equations 2 This is a course in stability and qualitative methods for analyzing ordinary differential equations which arise in realistic models.  Phase plane techniques, perturbation methods, and bifurcation theory are studied. MATH 1270\nMATH 1310 Graph Theory The concept of a graph and the study of its theoretical properties and applications form the core of this course. Topics include paths, circuits, trees, planar graphs, coloring problems, digraphs, matching theory, and network flows. MATH 0413\nMATH 1330 Projective Geometry Topics for this course are the algebraic, axiomatic, and/or synthetic development of projective geometry. MATH 1180 or MATH 1185\nMATH 1350 Introduction to Differential Geometry Possible topics are the basic ideas of topology, description of curves in space, definition and local study of smooth surfaces in Euclidean space (fundamental forms, geodesics, and curvature), global properties of surfaces, gauss-bonnet formula and applications. MATH 0240 plus one of MATH 1180, MATH 1185\nMATH 1360 Modeling in Applied Mathematics 1 This course introduces some of the fundamental approaches of applied mathematics.  The emphasis is on the model-building process and on developing an understanding of some of the unifying themes of applied mathematics such as equilibria, stability, conservation laws, etc.  The material is presented in the form of case studies. Math 0290 or Math 1270 or Math 1275\n\nMATH 1365\nTopics in Mathematical Modeling\n\nThis course provides honors recognition to students who wish to digress with the professor to study extra topics not covered in mathematics 1360, mathematical modelling.\nMATH 1370 Modeling in Applied Mathematics 2 This course presents contemporary mathematical theories of neuroscience, including single neurons and neuronal networks.  Attention will be given to the dynamics and the function of neural activity. MATH 0230 and some programming experience\nMATH 1380\nMath Biology\nThis course will provide a broad introduction to mathematical methods typically applied to problems in biology.  Models using calculus, ordinary differential equations, partial differential equations, discrete dynamical systems, stochastic dynamics, or a cellular automata framework will be presented and principal methods for their analysis will be described.  Computational methods will also be covered, including computing platforms such as XPPAUT.  Throughout the course, students will have extensive opportunities to practice the development and analysis of mathematical biology models. MATH 0280 or MATH 1180 plus MATH 0290 or MATH 1270\nMATH 1410 Introduction to the Foundations of Mathematics 1 This course introduces the logical foundations of mathematics; it covers the propositional and predicate calculi, formal number theory, set theory, and beginning model theory. Math 0413 or Math 0450\nMATH 1420 Introduction to the Foundations of Mathematics 2 Course is devoted to model theory and mathematical applications.  Items to be covered are:  completeness and compact ness theorems, and the Lowenheim-Skolem-Tarski theorems, ultra-products and ultra-limits, independence results, standard and nonstandard models, Categoricity in power, universal algebra, and Galois connections.\nMATH 1470 Partial Differential Equations 1 This is the first term of a two-term sequence in elementary PDE's.  The objectives of the course are to provide students with the techniques necessary for the formulation and solution of problems involving PDE's and to prepare for further study in PDE's.  The three main types of second order linear PDE's - parabolic, elliptic, and hyperbolic are studied.  In addition the tools necessary for the solution of PDE's such as Fourier series and Laplace transforms are introduced. MATH 0240 and one of MATH 0290 or 1270\nMATH 1480 Partial Differential Equations 2 This is the second term of a two term sequence in PDES. Topics include Fourier transform, maximum principles, and existence, uniqueness and regularity of solutions to PDES. MATH 1470\nMATH 1510\nMathematical Theory of Probability\nThis course is an introduction to the mathematical theory of probability.  Major topics include random variables, expecation, characteristic functions, conditional probability, and an introduction to Martingales and Markov chains.\n\nMath 0420 or Math 0450, and one of 280, 1180 or 1185\n\nMATH 1530 Advanced Calculus 1 This course contains a rigorous development of the calculus of functions of a single variable, including compactness on the real line, continuity, differentiability, integration, and the uniform convergence of sequences and series of functions.  Other topics may be included, such as the notion of limits and continuity in metric spaces. MATH 0420 or MATH 0450\nMATH 1540 Advanced Calculus 2 This course, a continuation of MATH 1530, covers the theory of limits, differentiation, and integration of functions of several variables.\n\nMATH 1530\nMATH 1550\nVector Analysis and Applications\nTopics covered include:  vector algebra, vector differentiation and integration, divergence, gradient, curl, the theorems of green, gauss and stokes, and curvilinear coordinate systems.  There will be an emphasis upon problem solving and applications in electromagnetic theory and fluid flow. MATH 0240 plus MATH 0280 or MATH 1180\nMATH 1560 Complex Variables and Applications This course covers the following topics:  elementary operations with complex numbers, derivatives, integrals, Cauchy's theorem and consequences such as the integral formula, power series, residue theorem, applications to real integrals and series. MATH 0240 (with B or better) or MATH 1550\nMATH 1570 Introduction to Fourier Analysis The course is a rigorous introduction to Fourier series and integrals with applications to heat flow, wave motion, physics, and number theory.  It is intended for students with a basic knowledge of real analysis including uniform convergence of sequences and series of functions.  No knowledge of the Lebasque integral is assumed. One of Math 0420, Math 0450 AND one of Math 0280, Math 1180, Math 1185\nMATH 1651\nComputer Methods Laboratory\nThis course will introduce students to the use of the computers in our undergraduate computing laboratory and to the numerical methods, symbolic algebra, and computer graphics which are helpful in the study of our upper level subjects in applied mathematics.\nMATH 1700 Introduction to Topology The topology of r1, as well as that of general metric spaces, will be studied.  Basic notions will be applied to obtain the fundamental existence theorem for first order ordinary differential equations.  The course will be run on a theorem proving and problem solving basis. MATH 0420 or MATH 0450\nMATH 1801 Advanced Topics in Mathematics This is a topics course at the advanced undergraduate level. The topic will change each time the course is offered, and will generally reflect current interests of the faculty or a recent trend in mathematics.\n\n## Graduate Course Listing\n\nClass Course Description Course prerequisites\nMATH 2000 Research and Thesis for the Master's Degree (1-15 Credits)\n\nThis course involves directed research and writing leading towards the completion of a Master's thesis.\n\nMATH 2010 Teaching Orientation (1 Credits)\n\nThis course is for Teaching Assistants in the Department of Mathematics. The course emphasizes techniques; procedures and discussions, which prepare the TA to successfully, manage recitations and teach classes in Mathematics.\n\nMATH 2020 Progress in Mathematics\n\nThis course will deepen the students' understanding of analysis through intensive training in problem solving followed by comprehensive study and dissection of the problems attempted.  Students preparing for the analysis portion of the preliminary exam are strongly encouraged to enroll.\n\nMATH 2030 Iterative Methods for Linear and Nonlinear Systems (3 Credits)\n\nTopics include matrix theory, matrix and vector norms, error analysis, factorizations, direct and iterative methods for solving linear and nonlinear systems, least squares, and the algebraic eigenvalue problem.\n\nMATH 2050 Graph Theory Basic concepts and definitions from the theory are studied. In particular, enumerative problems are selected by the instructor.\nMATH 2055 Codes and Designs A unified theory of linear codes, combinatorial design and statistical design will be presented.  Hamming, BCH, Golay, quadratic residue and other classes of codes will be constructed. Several decoding schemes, including decoding by way of locator polynomials, will be done.  Applications to statistical design and quality control will be described in detail.\nMATH 2060 Combinatorics (3 Credits)\n\nTopics in this course vary with the instructor's research interests. Focus will be placed on algebraic coding theory, construction of new nonlinear codes, Mobius inversion on posets, and symmetric functions. Techniques used involve ideals in polynomial rings, generating functions, and algebraic number rings. Basic knowledge of groups, fields and rings is highly desirable.\n\nMATH 2070 Numerical Methods in Scientific Computing 1 (4 Credits)\n\nThis course is an introduction to practical numerical methods for science and engineering. The course is complemented with a fully integrated computer laboratory, where you learn to use available software and to implement your own solution methods. Topics include: roundoff errors and stability analysis, root finding for nonlinear equations, interpolation, approximation of functions and numerical integration. The techniques presented are frequently used to deal with problems in physics, chemistry and engineering. The lecture introduces a numerical method and elaborates on its applicability and expected behavior. Frequently you will be assigned a related laboratory exercise. Registration for the lab is required.\n\nMATH 2071 Numerical Methods in Scientific Computing 2 (4 Credits)\n\nThe sequence M2070-M2071 gives an in-depth introduction to the basic areas of numerical analysis. The courses will cover the development and mathematical analysis of practical algorithms for the basic areas of numerical analysis. Math 2071 includes treatment of the topics of numerical linear algebra and numerical methods for differential equations. The course M2071 does not assume a knowledge of M2070; and material from M2070 that is needed in M2071 will be reviewed as necessary. These courses also include a Computational Laboratory that complement the lectures.\n\nMATH 2090 Numerical Solution of Ordinary Differential Equations (3 Credits)\n\nThis course is an introduction to modern methods for the numerical solution of initial and boundary value problems for systems of ordinary differential equations and differential algebraic equations.  Numerical methods and their theory, including convergence and stability considerations, order and step size selection and the effects of stiffness are discussed.\n\nMATH 2160 Set Theory This is an introductory course intended to prepare the students for the various applications of set theory.  The contents include basic axioms, Boole algebra of classes, algebra of relations, ordinals, transfinite induction, equivalents of the axiom of choice, and of the continuum hypothesis.\nMATH 2170 Logic and Foundations The contents of this course include the propositional calculus, the predicate calculus, model theory and their applications to mathematical systems.\nMATH 2180 Introduction to Fractal Geometry This course will give an introduction to fractal geometry. Topics include metric spaces, measures, fractal dimensions, discrete dynamical systems and multifractal analysis.\nMATH 2190 Functions of Several Variables This course covers topics of the calculus of several variables from a more general and theoretical point of view. Topics include convergence, continuity, compactness, inverse and implicit function theorems and differential forms.\nMATH 2200 Real Analysis 1 This is the basic real analysis course for graduate students.  In particular, the theory of Lebasque integral and some of its many ramifications are studied.  This leads to differentiation and integration theories for real valued functions.\nMATH 2201 Real Analysis 2 This course is a continuation of 2200.  Topics include elements of the theory of Banach and Hilbert spaces, Radon Nikodym theorem, duality for the LP spaces, product measures, Fubini's theorem and differentiation.\nMATH 2210 Complex Analysis 1 This is the basic first course in complex analysis.  Topics include analytic functions, Cauchy-Riemann equations, elementary conformal mappings, Poisson formula, Taylor and Laurent series, argument principle, residues and the classifications of singularities.\nMATH 2211 Complex Analysis 2 This course is a continuation of 2210.  The major topics are normal families, the Riemann mapping theorem and harmonic functions.  Selected topics, by the instructor, will also be covered.\nMATH 2219 Dynamical Systems (3 Credits)\n\nThis course will introduce the student to new concepts from dynamical systems.  Invariant manifolds, normal form; Bifur cations and chaos will be discussed from the geometric point of view.  Some global analysis will also be described.  Numerical tools and methods for analyzing local bifurcations will also be discussed.  Emphasis will be on practical applications of these techniques.\n\nMATH 2240 Analytic Number Theory Some of the topics covered in this course include residue classes, unique prime factorization, character mod n, the Riemann zeta function and its analytic continuation, poles and functional equations.  Also the prime number theorem and Dirichlet theorem will be covered.\nMATH 2245 Algebraic Number Theory In this course the theory of numbers will be algebraically, that is, as the study of algebraic numbers.  Particular attention will be paid to the development of quadratic and cyclotomic number fields and to factorization theorems. Other topics include ideal theory and the work of Kummer and Fermat's last theorem.\nMATH 2260 Potential Theory This is an elementary introduction to the subject.  Topics include harmonic functions, Poisson integral, maximum principle, classical Dirichlet problem, Harnack's inequality, boundary limit theorem, super harmonic functions and their properties and green's potentials.\nMATH 2280 Hardy Spaces A brief introduction to HP space theory in the disk.  Topics include Poisson formula, boundary behavior and Fatou's theorem, factorization theorems and Blaschke products.\nMATH 2301 Analysis 1 (3 Credits)\n\nThis course is an introduction to Real Analysis/Measure Theory with some Functional Analysis. Topics include: Lebesgue Measure and Integral; some Hilbert and Banach space theory; Monotone Convergence Theorem; Lebesgue's Dominated Convergence Theorem; Fatou's Lemma; Jensen, Holder and Minkowski Inequalities; absolutely continuous measures; the Radon-Nikodym Theorem; the Riesz Representation Theorem for L^p; the Hahn Decomposition Theorem; the Fubini-Tonelli Theorem.\n\nThe Mathematical Analysis and Linear Algebra material in the Math Preliminary Exams syllabus (and in particular, the material in the U. Pitt. Math courses 1530 Advanced Calculus 1 and 1540 Advanced Calculus 2, as well as the graduate linear algebra classes 2370 and 2371) is assumed. Note that many graduate math courses implicitly assume that students are familiar with a wide range of undergraduate math courses and ideas: such as basic Set Theory and basic Topology; - and 2301 Analysis 1 is a graduate math course of this type.\n\nMATH 2302 Analysis 2 (3 Credits)\n\nIn this course we continue on from 2301 Analysis 1 in Fall 2016 (Lebesgue measure and integration, and some functional analysis), stirring Fourier analysis, complex analysis, functional analysis and more real analysis into the mix...  In Fourier and Functional Analysis, topics include: more on Hilbert Spaces and L^p-spaces, Fourier Series and the Fourier Transform.  In Complex and Real Analysis, topics include: Power Series, Infinite Products, Holomorphic Functions, the Cauchy Integral Formula, and the Maximum Modulus Theorem.\nThe main prerequisite is the course 2301 Analysis 1; or an equivalent course.  The Mathematical Analysis and Linear Algebra material in the Math Preliminary Exams syllabus (and in particular, the material in the U. Pitt. Math courses 1530, 1540, 2370 and 2371) is assumed.\n\nMATH 2303 Analysis 3 (3 Credits)\n\nThis is a basic course in Functional Analysis. It is assumed that students are familiar with Analysis 1 (measure theory). Analysis 2 (Complex Analysis) will be used occasionally, but it will not play any important role in the course.\n\nTopics include (in that order):\n\n1. Basic theory of Banach and Hilbert spaces.\n\n2. Bounded operators in Banach and Hilbert spaces.\n\n3. Orthonormal bases in Hilbert spaces, Fourier series and spherical harmonics.\n\n4. Baire category theorem, Banach-Steinhaus theorem, open mapping theorem and closed graph theorem.\n\n5. Hahn-Banach theorem, separation of convex sets.\n\n6. Reflexive spaces.\n\n7. Weak convergence, Mazur's lemma, Banach-Alaoglu theorem in the separable case, direct methods in the calculus of variations.\n\n8. Weak topology, Tychonov's theorem and Banach-Alaoglu theorem in the general case.\n\n9. Compact operators, Fredholm operators, spectrum of compact operators, Fredholm-Riesz-Schauder theory.\n\n10. Spectral theorem for compact self-adjoint operators.\n\n11. Sobolev spaces and the eigenfunctions of the Laplace operator.\n\n12. Banach algebras.\n\nIn the course I will show many applications of Functional Analysis in different areas of mathematics\n\nMATH 2304 Analysis 4 (3 Credits)\n\nIn this course we will cover fundamental concepts of Harmonic Analysis in the Euclidean spaces. The following topics will be covered: Fourier transform, tempered distributions, the Marcinkiewicz and the Riesz-Thorin interpolation theorems, maximal functions, singular integrals, Hilbert transform, Riesz transforms, Calderon-Zygmind theory, fractional Laplace operators.\n\nMATH 2370 Matrices and Linear Operators 1 (3 Credits)\n\nLinear transformations on finite dimensional vector spaces are studied in a semi-abstract setting.  The emphasis is on topics and techniques which can be applied to other areas, e.g., Bases and dimension, matrix representation, linear functional, duality, canonical forms, vector space decom position, inner products and spectral theory.\n\nMATH 2371 Matrices and Linear Operators 2 (3 Credits)\n\nThe course is a continuation of Math 2370 Matrices and Linear Operators I.\n\nTopics will include spectral theory of self-adjoint mappings, calculus of matrix valued functions, matrix inequalities, convexity, duality theorem and normed linear spaces.\n\nMATH 2400 Functional Analysis 1 A first course in the area, the emphasis of the course will be on normed linear spaces and linear operations, their basic properties will be discussed.  Also, inner product spaces and their properties will be covered.\nMATH 2401 Functional Analysis 2 This course is a continuation of 2400.  Topics to be covered include the spectral theory, compact operators, and distribution theory and Sobolev spaces.\nMATH 2410 Harmonic Analysis 1 This is a basic first course in harmonic analysis.  The major topic is the Fourier transform and Fourier series. In particular, various kernels and pointwise sum ability of Fourier series will be discussed.\nMATH 2480 Computational Approximation Theory Topics include fundamental theorems, polynomial approximation, splines, surface approximation, domain transformations and applications to science and engineering.\nMATH 2500 Algebra 1 (3 Credits)\n\nThe course is the first term of a two-term graduate algebra sequence. It covers the theory of groups, the theory of fields as well as Galois theory. Highlights of the course will include: Sylow's theorems, the structure of finitely generated abelian groups, the fundamental theorem of Galois theory, and applications (such as the solvability of polynomial equations by radicals and geometric constructions with a ruler and a compass).\n\nMATH 2501 Algebra 2 (3 Credits)\n\nIn this course the fundamental properties of rings, fields and modules are studied.\n\nMATH 2503 Matrix Groups The course is an introduction to some of the concepts of lie groups and lie algebra--all done at concrete level of matrix groups.  It centers around the isomorphism questions on matrix groups of small dimensions, and it leads to maximal torus, Clifford algebra, and Weyl group.\nMATH 2505 Algebra 3 (3 Credits)\n\nThis is a third course in algebra.  It covers the classical results on the structure and representation theory of associative algebras culminating with modern developments such as the theory of quiver algebras and categorification.  Highlights of the course include: Wedderburn-Artin theory, the structure of central simple algebras, the structure of finite dimensional algebras, semisimple algebras, the character theory of finite groups, theorems of Maschke, Frobenius, Burnside, quiver algebras and quiver representations, reflection functors, Gabriel's theorem, tensor categories, fusion categories.\n\nMATH 2506 Algebra IV (3 Credits)\n\nThis is a fourth course in algebra.  It covers introductory topics in algebraic geometry, number theory, and representation theory, selected by the instructor.\n\nMATH 2601 Advanced Scientific Computing 1 (3 Credits)\n\nThis course studies the mathematical analysis and practical implementation of discontinuous Galerkin methods for approximating elliptic, parabolic, and hyperbolic partial differential equations.\n\nMATH 2602 Advanced Scientific Computing 2 (3 Credits)\n\nThe course will cover several topics in discretizations of partial differential equations and the solution of the resulting algebraic systems. Topics in discretizations will include mixed finite element methods, finite volume methods, mimetic finite difference methods, and local discontinuous Galerkin methods. Topics in solvers will include domain decomposition methods and multigrid methods. Applications to flow and transport in porous media, as well as coupled fluid and porous media flows will be discussed.\n\nAdvanced Calculus, Linear Algebra, and Differential Equations\nMATH 2603 Advanced Scientific Computing 3\n\nThe Advanced Scientific Computing sequence covers topics chosen at the leading edge of current computational science and engineering for which there is sufficient interest.  The course requirements consist readings, homework, a term project and its presentation. Please contact the instructor if you have questions about your  preparation.\n\nThe fall 2016 course will  consider advanced topics from the analysis and numerical analysis of fluid motion. One specific topic considered in this course will be large eddy simulation of  turbulence.\n\nMATH 2604 Advanced Scientific Computing 4 (3 Credits)\n\nThe course focuses on the fundamental mathematical aspects of numerical methods for stochastic differential equations, motivated by applications in physics, engineering, biology, economics. It provides a systematic framework for an understanding of the basic concepts and of the basic tools needed for the development and implementation of numerical methods for SDEs, with focus on time discretization methods for initial value problems of SDEs with Ito diffusions as their solutions. The course material is self-contained.   The topics to be covered include background material on probability, stochastic processes and statistics, introduction to stochastic calculus, stochastic differential equations and stochastic Taylor expansions. The numerical methods for time discretization of ODEs are briefly reviewed, then methods for time discretization for SDEs are introduced and analyzed.\n\nMATH 2700 Topology 1 (3 Credits)\n\nA first course in topology, some of the topics covered include separation axioms, bases and sub-bases, product and quotient topology, homomorphisms, compactness, the baire category theorem, the lindelof property, connectedness, topological spaces, and compactification.\n\nMATH 2701 Topology 2 (3 Credits)\n\nThis course is a continuation of 2700. In this course, the basic concepts and results in algebraic topology will be covered, including both homotopy and homology theory. In particular, the calculation of the fundamental group and homology groups from chain complexes will be covered.\n\nMATH 2750 General Topology The fundamental theorems of general topology will be studied in particular, those results concerning generalized metric spaces, coverings and mapping will be studied.\nMATH 2800 Differential Geometry 1 (3 Credits)\n\nA first course in differential geometry.  Topics may include the geometry of curves and surfaces (eg. Gauss map, fundamental forms, curvature), differentiable manifolds, Lie groups, tangent and tensor bundles, vector fields, and Riemannian structures.\n\nMATH 2801 Differential Geometry 2 (3 Credits)\n\nThis course is a continuation of Differential Geometry 1.  The initial focus will be on differential topology, covering topics such as such as Sard's theorem, transversality, degree of mappings, and differential forms and Stokes' theorem.  Further topics may include Lie groups, distributions and the Frobenius theorem, and bundles and connections.​\n\nMATH 2810 Algebraic Geometry (3 Credits)\n\nThis course is an introduction to the basic ideas of Algebraic Geometry, the approach to the subject may be either of the following:  The linear series on a curve approach, the algebraic approach through fields of algebraic functions, or the Sheaf theoretic approach.  Applications may also be included.\n\nMATH 2815 Discrete Geometry and Computers This course will discuss various classical problems in discrete geometry.  It will develop the theory of packings and coverings, the Kepler sphere packing theorem, the dodecahedral theorem, kissing number problems, lattice packing problems in higher dimensions, and the kelvin problem in foams.\nMATH 2900 Partial Differential Equations 1 (3 Credits)\n\nThe course covers some fundamental topics of partial differential equations, including  transport equation, Laplace's equation, heat equation, wave equation, characteristics, Hamilton-Jacobi equations, conservation laws and shock waves, some methods to represent solutions.\n\nMATH 2901 Partial Differential Equations 2 (3 Credits)\n\nThis course will cover Sobolev spaces, second order elliptic equations, weak solutions, linear evolution equations, semigroup theory, and Hamilton-Jacobi theory, and other topics in nonlinear PDE.\n\nPDE 1 (Math 2900) is not a pre-requisite but a good background in analysis is necessary.\n\nMATH 2920 Ordinary Differential Equations 1 (3 Credits)\n\nThis is the first course in a two-term sequence designed to acquaint students with the fundamental ideas involved in the study of ordinary differential equations.  Basic existence and uniqueness of solutions as well as dependence on parameters will be presented.  The course will cover linear ODES and the matrix exponential, oscillations via an introduction to Poincare-Bendixson theory for planar systems and to Floquet theory, and  Sturm-Liouville problems.  Students will also be introduced to geometric concepts such as stability of fixed points and invariance.  This first term will provide an excellent introduction to ODE theory for students interested in applied mathematics.\n\nMATH 2921 Ordinary Differential Equations 2 (3 Credits)\n\nThis course, which follows Math 2920, presents a dynamical systems approach to the study of ordinary differential equations.  Topics include geometric theory including proofs of invariant manifold theorems, flows on center manifolds and local bifurcation theory, the method of averaging, Melnikov's method, and an introduction to Smale horseshoes and chaos theory.\n\nMATH 2930 Asymptotics and Special Functions (3 Credits)\n\nThis course is an introduction to the theory of asymptotics and special functions. Some of the functions studied are the gamma function, orthogonal polynomials, and bessel functions. This course also covers techniques for finding asymptotic expansions of integrals and of solutions to differential equations.\n\nMATH 2930 Asymptotics and Special Functions (3 Credits)\n\nThis course covers  Hypergeometric and Confluent Hypergeometric Differential Equations and Series (e.g.  Laurent series and geometric series)   which arise  in physics and engineering.  We also study Bessel functions, the Gamma function, the Beta function, the  Riemann Zeta function, and  Laplace's asymptotic expansion of integrals depending on a parameter. The prerequisite for this course is a one undergraduate semester course in complex variables with a grade of B or higher. The grade will be determined from assigned homework problems.\n\nMATH 2940 Applied Stochastic Methods (3 Credits)\n\nThis course will provide an overview of stochastic methods that can be applied to problems in biology, finance and physics. Analytical and computational techniques will be presented which apply to both continuous and discrete stochastic models.\n\nMATH 2950 Methods in Applied Mathematics (3 Credits)\n\nThis course covers methods that are useful for solving or approximating solutions to problems frequently arising in applied mathematics, including certain theory and techniques relating to the spectral theory of matrices, integral equations, differential operators and distributions, regular perturbation theory, and singular perturbation theory.\n\nMATH 2960 Computational Fluid Mechanics Topics include a review of symmetric linear systems and matrix theory, equilibrium and the calculus of variations in discrete and continuous systems, orthogonal series, networks Fourier series and convolutions, Fourier integrals, complex variables and conformal mapping.  Also some numerical methods such as the fast Fourier transform will be covered.\nMATH 2980 Projects in Financial Mathematics Representatives from business or government will present real-world problems and issues. Students will select one of the problems and generate some solutions either alone or in teams. Oral and written reports of the results will then be presented to both the posers of the problems and to the faculty and students in the professional master's program.\nMATH 2990 Independent Study (1-15 Credits)\n\nThis course is for all graduate students not under the direct supervision of a specific faculty member. In addition to a student's formal course load, this study is for preparation for the preliminary, comprehensive and overview examinations.\n\nMATH 3000 Research and Dissertation for the PhD Degree This course is taken by a student who is working on a Ph.D. dissertation under the direction of a student's thesis advisor.\nMATH 3010 Sobolev Spaces Topics include:  definitions and basic properties, the Sobolev imbedding theorem, applications of the theory.\nMATH 3020 Calculus of Variations (3 Credits)\n\nThis course will introduce students to the subject of calculus of variations and some of its modern applications. Topics to be covered include necessary and sufficient conditions for weak and strong extrema, Hamiltonian vs Lagrangian formulations, principle of least action, conservation laws and direct methods of calculus of variations. Extensions to the functionals involving higher-order derivatives, variable regions and multiple integrals will be considered. The course will emphasize applications of these ideas to numerical analysis, mechanics and control theory.\n\nPrerequisite(s): single-variable and multivariable calculus, some exposure to ordinary and partial differential equations. All other concepts, such as function spaces and the necessary background for the applications, will be introduced in the course. Beginning graduate students and advanced undergraduates are welcome.\n\nMATH 3031 Network Theory The general problem discussed in flows on networks.  Topics will include some graph theory, flows and potential differences, transport problems and flow algorithms.\nMATH 3040 Topics in Scientific Computing\n\nThe course objective is to introduce students to formulating, debugging and solving finite element simulations of practical applications, with a focus on the equations of fluid flow.  Two popular freely-available computer packages will be presented: FEniCS and FreeFem++.  FreeFem++ is an integrated program, with a special language for specifying the mathematical formulation as well as integrated mesh generation and graphical postprocessing facilities.  FEniCS is less tightly integrated, consisting of a collection of functions for specifying the mathematical formulation as well as functions for interfacing with other packages for mesh generation, post processing, and numerical solution.  These functions are tied together using either the Python or C++ programming languages.  This course will focus on using Python. Python is a widely-used language with applications far removed from finite element modelling and can be the subject of multiple-semester courses. Although previous experience with Python would be valuable, it is not necessary.  The basics of the language plus those features necessary for this course will be presented during the lectures.  Previous experience with finite element methods will be valuable, but is not required because the theory will be summarized during the lectures. Applications for which FreeFem++ or FEniCS will be used include steady and transient heat conduction as well as the Stokes and Navier-Stokes equations. Various boundary conditions and finite elements will be presented, as well as the effect of these choices on solution methods.\n\nPrerequisites include a basic knowledge of one of the following programming languages: Python, C, C++, FORTRAN, JAVA, or MATLAB; Linear Algebra and Calculus; and, at least one introductory computational/numerical analysis class, such as 1070/1080 or 2070/2071 or the equivalent.\n\nMATH 3055 Chromatic Polynomials and Graph Structure This course will focus on the relationship between chromatic polynomials and graph structures.  Some structure properties to be considered are connectivity, chromatic number, girth, graph components, and graph isomorphism and homeomorphisms.\nMATH 3060 Topics in Combinatorics This course covers a variety of topics in combinatorics.\nMATH 3070 Numerical Solution of Nonlinear Systems This course is an introduction to modern methods for solutions of such equations.  Topics include:  algorithms for one-dimensional equations, linearization methods for systems, method of Gauss-Seidel type, quasi-Newton methods, continuation methods and elements of unconstrained minimization methods.\nMATH 3071 Numerical Solution of Partial Differential Equations (3 Credits)\n\nThis course covers contemporary methods for solving initial and boundary value problems. Topics include properly posed problems, characteristics, finite difference and finite element methods, and error estimates.\n\nMATH 3072 Finite Element Method (3 Credits)\n\nThis course is an introduction to the theoretical and computational aspects of the finite element method for the solution of boundary value problems for partial differential equations. Emphasis will be on linear elliptic, self-adjoint, second-order problems, and some material will cover time dependent problems as well as nonlinear problems. Topics include: Sobolev spaces, variational formulation of boundary value problems, natural and essential boundary conditions, Lax-Milgram lemma, approximation theory, error estimates, element construction, continuous, discontinuous, and mixed finite element methods, and solution methods for the resulting finite element systems.\n\nPrerequisite(s): Good undergraduate background in linear algebra and advanced calculus. Familiarity with partial differential equations will be useful.\n\nMATH 3075 Parallel Finite Element Method This course will cover major new developments in fully parallel finite element methods.  The reasons why this is a great advantage over traditional methods will be discussed and explained.\nMATH 3215 Quasiconformal Maps 1 This course will cover the quasi-conformal and quasi-regular mappings.  They relate to several areas of mathematics and some of these applications will be covered. The methods used in the course will included analysis, topology and geometry.\nMATH 3216 Quasiconformal Maps 2 This course is a continuation of math 3215.  It will be devoted to quasi-conformal analysis in Riemannian manifolds, the so called Donaldson - Sullivan point of view.\nMATH 3225 Mathematics of Finance 1 (3 Credits)\n\nThis course provides an introduction to the mathematical subjects required for the mathematical finance program, and assumes that the student has an undergraduate degree with some technical component (e.g. Engineering, Computer Science, Math, Statistics, Physics, etc.) Students are expected to have knowledge of Multivariable Calculus and Linear Algebra, and any sections on these topics will be presented as review. Topics to be covered include: Partial Differential Equations, Stochastic Analysis, Optimization and Numerical Methods. No financial background is required, but many of the examples and llustrations of the mathematics will be drawn from economics and finance.\n\nMATH 3226 Mathematics of Finance 2 (3 Credits)\n\nThe course with its pre-sequel MATH3225 present fundamental principles and standard approaches used in mathematical finance. We will study   continuous-time stochastic models with applications in various fields of mathematical finance including prcing and hedging financial instruments, risk management and financial decision making etc. We will cover basic portfolio theory, pricing options and other derivatives, change of numeraire, term-structure models and etc  from Volume 2 of Shreve's book \"Stochastic Calculus for Finance\".  This course will investigate the mathematical modeling, theory and computational methods in modern finance. The main topics will be (i) basic portfolio theory and optimization, (ii) the concept of risk versus return and the degree of efficiency of markets, (iii) discrete models in options.\n\nMATH 3227 Mathematics of Finance 3 (3 Credits)\n\nThis course covers special topics in mathematical finance. Topics will include stochastic control theory and stochastic differential games with applications to finance.\n\nMATH 3228 Mathematics of Finance 4 (3 Credits)\n\nThis course covers advanced topics in modern mathematical finance. Topics will include advanced credit risk and interest rate models, stochastic control and stochastic optimization models for portfolio selection and option pricing, and numerical methods.\n\nMATH 3260 Topics in Fractal Geometry 1 Topics to be covered include Hausdorff measure, various definitions for dimensions of a set, techniques for calculating these dimensions, also the local structure projections, products and intersections of fractals will be discussed.  Further topics include self-similar sets, self-similar measures, dynamical systems, Julia sets and random fractals.\nMATH 3261 Topics in Fractal Geometry 2 This course is a continuation of MATH 3260.  Some topics which will be covered include self-similar sets, self-similar measures, dynamical systems, Julia sets and random fractals.\nMATH 3270 Iteration of Rational Maps 1 This is an introductory course on complex dynamics.  In particular, the Fatou and Julia sets for a large number of examples will be worked out.  The general properties of these two sets will be given and relations to the Mandelbrot set and fractals will be discussed.\nMATH 3271 Iteration of Rational Maps 2 This is a continuation of math 3270.  Topics will include application of quasi-conformal maps to polynomial maps, Hermann rings, quasi-conformal surgery, the local geometry of the Fatou and Julia sets and the study of the Mandelbrot set.\nMATH 3370 Mathematical Neuroscience (3 Credits)\n\nCourse covers computational and mathematical neuroscience. It will include modeling and analysis of complex dynamics of single neurons and large-scale networks using a variety of methods from applied math.  No biology is required; some familiarity with differential equations will be helpful.\n\nMATH 3375 Computational Neuroscience (3 Credits)\n\nThis course offers an introduction to modeling methods in neuroscience. Topics range from modeling the firing patterns of single neurons to using computational methods to understand neural coding. Some systems level modeling is also done.\n\nMATH 3380 Mathematical Biology (3 Credits)\n\nThis course describes a number of topics related to mathematical biology. This year we will cover several areas of interest including pattern formation in reaction-diffusion and advection models with applications to immunology, chemotaxis, etc; evolutionary dynamics such as the evolution of cooperation, some game theory, and replicator dynamics; and some cell physiology modeling such as the cell cycle and simple circadian models.   The prerequisites are some simple differential equations, a bit of Fourier transforms, and some knowledge of software to numerically solve the various equations.\n\nMATH 3410 Hilbert Spaces of Entire Functions 1 This course will cover Debranges theory.  The functional Hilbert spaces will be introduced and the salient properties of reproducing kernels will be studied.  Applications of this theory including a formal proof of the Riemann hypothesis will be covered.\nMATH 3411 Hilbert Spaces of Entire Functions 2 This course is a continuation of 3410.  The material will consist of a more careful study of the work of l. Debranges in this area.  An application will be the Riemann zeta function and the Riemann hypothesis.\nMATH 3436 Fixed Points Wavelets & Fractals The course will cover iterative image reconstruction using the wavelet transform, initiated by Mallat and Zhang.\nMath 3440 Fixed Point Theory in Bananch Spaces\n\n(3 Credits)\n\nWe will begin with an overview of basic fixed point theory in banach spaces, from the banach contraction mapping theorem and schauder's theorem through to kirk's theorem. The course will continue with topics in metric fixed point theory and its  connections to banach space geometry and topology. This will include recent work of pei-kee lin, who showed that there exists a non-reflexive banach with the fixed point property for nonexpansive mappings; and tomas dominguez benavides, who proved that every reflexive banach space can be equivalently renormed to have the fixed point property for nonexpansive mappings. We will also discuss extensions of lin's work to the function space l^1 by maria japon pineda and carlos hernandes linares. The course will further include some of my (joint) research in this area, and related research of other authors.\n\nMATH 3450 Theory of Distributions Using the theory of topological vector scales the theory of distributions will be studied.  Some of the applications of this theory to Fourier transforms and the Paley Wiener theory will be discussed.\nMATH 3480 Topics in Spline Approximation Spline approximation is Piewise analytic approximation. Most cad-cam systems rely heavily on spline approximation. In this course topics to be covered include b-splines data fitting using splines and numerical integration and differentiation.\nMATH 3500 Topics in Algebra Symmetric spaces are manifolds admitting symmetries of a certain nature. The theory was developed by Cartan at the beginning of the last century and it constitutes the proper framework for many questions in modern mathematics. The course will cover compact, noncompact, hermitian symmetric spaces, the Cartan classification, invariant differential operators, the Darboux and Poisson equations, the  Plancherel inversion formula and the Paley-wiener theorem for the spherical transform.\nMATH 3550 Lie Groups and Lie Algebras (3 Credits)\n\nThe main goal of the course is to understand the structure and classification of complex semisimple Lie algebras as well as their basic representation theory and the relationship with Lie groups. Highlights will include, the theorems of Engel, Cartan and Weyl, root systems, the Harish-Chandra isomorphism and various formulas for characters and weight multiplicities.\n\nMATH 3600 Topics in Pure Mathematics\n\nThis course covers special topics in pure mathematics.  The subject matter varies each semester.  Topics may include the theory of modular forms, automorphic representations, Galois representations, class field theory, trace formula, special topics in algebraic geometry, the Langland's program, discrete and algorithmic geometry, motivic integration, and the formalization of pure mathematics.\n\nMATH 3750 General Topology 2 This is a second course in general topology in which the fundamental questions and open problems in the area are discussed.  In particular, generalized metri-spaces, covering theorems and mapping of spaces will be covered.\nMATH 3760 Topics in Topology 1 (3 Credits)\n\nCohomology is an important concept and tool in various areas of pure mathematics, such as topology, differential geometry, algebraic geometry, and representation theory. This course will start with rational and integral cohomology and then move to survey generalizations such as: topological and algebraic K-theory and elliptic cohomology. We will also describe equivariant and twisted versions. Along the way many techniques and tools will be explained, such as: spectral sequences, mapping spaces, homotopy computations, classification of bundles, topology of Lie groups and of their classifying spaces. As time permits, the associated higher geometric and categorical structures will also be discussed.\n\nMATH 3761 Topics in Topology 2\nThe course will be concerned with topics of current research activity in analytic topology, especially in the areas of generalized metric spaces and topological algebra.\n\nMATH 3900 Graduate Internship (1-9 Credits)\n\nInternship and/or employment experience under the supervision and oversight of a faculty member. This experience is to be an integral part of the students individual course of study.\n\nMATH 3902 Directed Study (1-9 Credits)\n\nThis course is for students normally beyond their first year of graduate study who wish to study in an area not available in a formal course. The work must be under the direct supervision of a faculty member who has approved the proposed work in advance of registration. A brief description of the work should be recorded in the student's file in the department.\n\nMATH 3920 Nonlinear Methods in Differential Equations This course covers functional analytic methods in the theory of differential equations, including applications of the fixed point theorem and the implicit function theorem to existence results.  It also presents an introduction to bifurcation theory and discusses numerical problems, schemes.\nMATH 3921 Pseudodifferential Operators Course will give an introduction to pseudo differential operators and their applications to partial differential equations and topology.  Topics include hypo elliptic operators, elliptic complexes and the index theorem.\nMATH 3923 Topics in Partial Differential Equations (3 Credits)\n\nThis course will explore recent developments in the theory of partial differential equations centered around the notion of viscosity solution.  After a review of hamiton-jacobi equations, we will discuss how they can be used as a substi tute for convolution in non-linear problems.\n\nMATH 3930 Fixed Point Theory Topics include:  Banach's fixed point theorem, Brouwer's fixed point theorem and various applications, topological degree theory in finite dimensions and constructive aspects by simplicial algorithms.  Also a brief introduction to the Leray-Schauder degree in infinite dimension will be discussed.\nMATH 3935 Topics in Applied Mathematics After briefly introducing very elementary theory of asymptotic expansions, the course will focus on its applications. In particular, boundary layer expansion, interior inter facial layer expansion, as well as multi-scale expansion techniques will be quite detailed by examples of a number of up-to-date research problems.\nMATH 3935 Topics in Applied Mathematics (3 Credits)\n\nThe theory of water waves embodies the equations of fluid mechanics, the concepts of wave propagation, and the critically important role of boundary dynamics. It has been a subject of intense research since Euler's derivation of the equations of hydrodynamics. The second part focuses more on the analysis perspective of the water wave equations.Zakharov's Hamiltonian formulation of the irrotational water wave problem will be dis-\ncussed and applications of this formulation to the issue of wellposedness will be outlined.\n\nThe course also use some of the asymptotic (integrable) models as an example to describe the wave-breaking phenomenon. Finally, a particular ow pattern, namely the traveling (or steady) waves will be addressed. The method of calculus of variation, (global) bifurcation theory, topological degree theory, Schauder estimates, and Fredholm theory will be introduced to establish the existence of such waves.\n\nMathematically, the above mentioned topics draw on deep ideas from applied mathematics, analysis, and PDEs. The main ingredients and techniques involved include Fourier analysis, harmonic analysis, elliptic theory, and more. Prior exposure to any of these will be helpful but not necessary.\n\nMATH 3940 Applied Analysis 1\nMethods will be developed to analyze the behavior of bumps and waves in integral models arising in neuroscience, and also in reaction-diffusion biological type models.\n\nMATH 3941 Applied Analysis 2 This course covers advanced methods in the theory of ordinary differential equations.  Topics include: shooting arguments in existence problems, energy arguments, reduction of invariant equations, a discussion of bifurcation theory and differential equations in the complex domain.\nMATH 3950 Nonlinear Dynamics, Chaos, and Oscillation This course gives a description of modern techniques for analyzing nonlinear differential equations.  It covers topics such as chaos, nonlinear oscillations, bifurcation theory, phase locking and invariant manifold methods.  Topics such as averaging, singular perturbation, and equations on tori are also discussed.\nMATH 3951 Physical Methods in Mathematics This course investigates the mathematical structure of various methods in physics.  Topics include renormalization scaling and interfaces.  Applications to differential and difference equations will be done.\nMATH 3960 Mathematics of Phase Boundaries This course will concentrate on developing the major models used in solidification theory.  Thus the study of systems of differential equations will be a major theme of the course." ]

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[ "# Pumping Lemma for Regular Languages with 3 variables (a^nb^mc^m)\n\nI've been trying to understand the pumping lemma, and how to apply it to a language such as a^nb^mc^m where n >= 0 and m >= 0. The pumping lemma states that:\n\nFor any regular language L, there exists an integer n, such that for all x ∈ L with |x| ≥ n, there exists u, v, w ∈ Σ∗, such that x = uvw, and\n\n(1) |uv| ≤ n\n\n(2) |v| ≥ 1\n\n(3) for all i ≥ 0: uviw ∈ L\n\nIt states that there exists some n such that this holds true. In the language a^nb^mc^m if I were to replace m with the pumping length, and choose a pumping length, for example 3, such that a^nb^pc^p I could get the string:\n\naabbbccc\n\nIf I chose:\n\nx = aa\n\ny = b\n\nz = bbccc\n\nthen xy or aab <= p and y > 0. Then if I pump up y by 2 for example, I would get the string aabbbbccc. But this is not in the language as there are more b's then c's. The problem i'm having is if I instead split my string as such:\n\nx = a\n\ny = a\n\nz = bbbccc\n\nthen xy or aa <= p and y > 0. If I pump this string by 2 for example, I would get the string aaabbbccc, which is in the language. So in this case the pumping lemma is passed.\n\nSo my question is, for the pumping lemma does my second example passing mean that the language a^nb^mc^m passes the pumping lemma, or does the fact that I found a case where splitting x and y with this pumping length makes it false, mean that the language does not pass the pumping lemma. And if that is the case then if I had the language:\n\na^nb^m where n >=0 and m >=0 and I choose m to be p, and made the pumping length 3 for example, I could end up with the string:\n\naabbb, where x = a, y = ab, z = bb, which when pumped would produce aababbb which is not in the language, but I know that a^nb^m is a regular language.\n\nHopefully you can see where my confusion lies. I been researching for a while and I just can't find a clear cut answer on this. Any help would be extremely appreciated.\n\nThank you.\n\nApply the pumping lemma on the word $$b^nc^n$$.\n• In order to use the pumping lemma to show that a given language is not regular, you have to find, for each integer $n$, a word of length at least $n$ in the language, such that for each legal decomposition, you can pump it out of the language. I suggest reviewing the statement of the pumping lemma. Jul 13 '19 at 15:19" ]

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[ "# Subtraction 2429 minus 8\n\nWhat is the subtraction of 2429 minus 8 and How much is 2429 - 8 percent?\n\nSubtraction of 2429 and 8 is 2421 and 2429 minus 8 percent is 2234.68\n\n2429 - 8\n=\n2421\n2429 - 8 percent\n=\n2234.68\n\nStep by Step Calculation for what is 2429 minus 8%\n\n= 2429 - 8% of 2429\n\n= 2429 - 8 * 2429 / 100\n\n= 2429 - 8 * 2429/100\n\n= 2429 - 4858/25\n\n= 55867/25\n\n55867/25 or fraction as a decimal is 2234.68" ]

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[ "# Impact d'une galerie étanche peu profonde sur l'écoulement d'une nappe/Impact of an impervious shallow gallery on groundwater flow\n\nAbstract : While many studies have been achieved on the interactions between groundwater and deep tunnels, in order to identify the evolution of pore pressure around the structure and to characterize the flow to its leaky parts, few studies have dealt with the impact of the carrying out of an impervious gallery in a shallow aquifer. The induced change in the piezometric level of the aquifer and the one in the hydraulic gradient of the flow however can, in this case, have significant consequences, in particular when the linear structure is located in an urban environment. This paper investigates, in steady state, the case of a straight tunnel having a horizontal axis perpendicular to the direction of the regional groundwater flow and a circular or square cross section. The aim is to determine the additional lost head Δh s due to the tunnel (i.e. additional to that resulting from the regional flow, supposed to be uniform with a hydraulic gradient i 0). In the context of a horizontal confined aquifer having a thickness 2B and of a tunnel of radius R located in the middle part of the aquifer, an analogy can be established with the flow above a hydraulic threshold resulting from a local rise of the elevation of the base of an aquifer, having a thickness B, on a width 2R and with a vertical maximum amplitude R. When neglecting the vertical component of the hydraulic gradient compared to its horizontal component, analytical solutions are developed for various hydraulic threshold shapes (rectangular, triangular and circular), based on the equivalence with a local change in the transmissivity of an aquifer keeping a constant thickness. The corresponding formulas take the form: ${\\frac{{\\Updelta h_{s} - \\Updelta h_{0} }}{{\\Updelta h_{0} }}} = f(a)$, with $a = {\\frac{R}{B}}$ and Δh 0 = 2Ri 0. The use of these formulas shows that the additional lost head Δh s due to the hydraulic threshold is proportional to i 0 and that, for values of the ratio a < 0.5, the change in the piezometric surface is small. These conclusions are therefore limited by the fact that the vertical conductivity is supposed to be very large. In order to remove this hypothesis, numerical simulations are achieved using the MODFLOW code. It is considered a confined aquifer of length 2L = 110 m and thickness B = 10 m, a ratio $a = {\\frac{R}{B}} = 0.25$ and a horizontal hydraulic conductivity $K_{H} = 10^{ - 5} \\,{\\text{m}}\\,{\\text{s}}^{ - 1}$. In the case of an isotropic medium ($\\alpha = {\\frac{{K_{H} }}{{K_{V} }}} = 1$), the simulations allow to check the linearity of the relationship between Δh s and i 0, with therefore a homogeneous variation in the proportionality coefficient compared to analytical solutions. Simulations also reveal that, in the case considered, the width of influence upstream and downstream L i , corresponding to a value of the vertical component of the hydraulic gradient <1% of i 0, is below 5.5R for the three hydraulic threshold shapes, and that it was few influenced by the hydraulic gradient i 0. In the case of an anisotropy of the horizontal and vertical hydraulic conductivities, simulations reveal the significant importance of the anisotropy ratio $\\alpha = {\\frac{{K_{H} }}{{K_{V} }}}$ when it is more than 1, the most common case, and indicate that the proposed analytical solutions give an asymptotic value of ${\\frac{{\\Updelta h_{s} }}{{\\Updelta h_{0} }}}$ for the isotropic case and for the values of the component α < 1. In the context of an unconfined aquifer, the hydraulic threshold model is not directly applicable. The model studied, using the Dupuit-Forchheimer assumption, is the one of a water table aquifer with a sloped base (slope value: p 0). The simulations focus on an aquifer of length 2L = 85 m, with a tunnel of circular cross section having a diameter 2R = 5 m, bottom of which is located 5 m above the base of the aquifer, the isotropic hydraulic conductivity being equal to $K = 10^{ - 5} \\,{\\text{m}}\\,{\\text{s}}^{ - 1}$. The definitions of water heights d 0 and d between the water table and the top of the tunnel are given in Fig. 7. The water table can be located above (fully submerged tunnel) or below (partially emerged tunnel) the top of the tunnel. The difference d 0 − d represents the half of the additional lost head Δh s due to the tunnel. Simulations are performed for various values of p 0 and d 0. They provide the values of i 0, d and Δh s . In the case of a fully submerged tunnel (d > 0), a significant rise of the water table upstream of the tunnel is obtained only for high values of the hydraulic gradient (5 and 10%), but, even in this case, it remains less than the tenth of the wetted height of the aquifer h m . It is also highlighted that the ratio ${\\frac{{\\Updelta h_{s} }}{{i_{0} }}}$ varies as a linear function of (R + d) and that, in the studied case, there is no influence of the tunnel for d ≥ 4R. In the case of a partially emerged tunnel (d < 0), the aquifer is locally confined under the tunnel. It is suggested that an equivalence is possible with the case of a confined aquifer having a thickness equal to the wetted height in the unconfined aquifer. This is verified with one of the simulations. In the case of a partially emerged tunnel, the change in the water table due to the tunnel remains low.\nKeywords :\nDocument type :\nJournal articles\nComplete list of metadata\n\nhttps://hal-mines-paristech.archives-ouvertes.fr/hal-00460619\nContributor : Pascale Nalon Connect in order to contact the contributor\nSubmitted on : Monday, March 1, 2010 - 6:01:18 PM\nLast modification on : Wednesday, November 17, 2021 - 12:31:17 PM\n\n### Citation\n\nMichel Deveughèle, Pierre Zokimila, Roger Cojean. Impact d'une galerie étanche peu profonde sur l'écoulement d'une nappe/Impact of an impervious shallow gallery on groundwater flow. Bulletin of Engineering Geology and the Environment, Springer Verlag, 2009, 69 (1), pp.143-152. ⟨10.1007/s10064-009-0226-x⟩. ⟨hal-00460619⟩\n\nRecord views" ]

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[ "", null, "", null, "Under the auspices of the Computational Complexity Foundation (CCF)", null, "", null, "", null, "", null, "", null, "REPORTS > DETAIL:\n\n### Paper:\n\nTR00-035 | 6th June 2000 00:00\n\n#### Independent minimum length programs to translate between given strings", null, "TR00-035\nAuthors: Nikolay Vereshchagin, Mikhail V. Vyugin\nPublication: 6th June 2000 18:21\nKeywords:\n\nAbstract:\n\nA string $p$ is called a program to compute $y$ given $x$\nif $U(p,x)=y$, where $U$ denotes universal programming language.\nKolmogorov complexity $K(y|x)$ of $y$ relative to $x$\nis defined as minimum length of\na program to compute $y$ given $x$.\nLet $K(x)$ denote $K(x|\\text{empty string})$\n(Kolmogorov complexity of $x$) and\nlet $I(x:y)=K(x)+K(y)-K(\\pair{x,y})$\n(the amount of mutual information in $x,y$).\nIn the present paper we answer in negative the following\nquestion posed in~\\cite{id}:\nIs it true that for any\nstrings $x,y$ there are\nindependent minimum length programs $p,q$ to translate between $x,y$,\nthat is, is it true that\nfor any $x,y$ there are $p,q$ such that $U(p,x)=y$, $U(q,y)=x$,\nthe length of $p$ is $K(y|x)$,\nthe length of $q$ is $K(x|y)$, and\n$I(p:q)=0$ (where the last three equalities hold up\nto an additive $O(\\log(K(x|y)+K(y|x)))$ term)?\n\nISSN 1433-8092 | Imprint" ]

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[ "", null, "", null, "# If 1111=T; S=2113; N=1233; EZ=40010112; O=1401; F=6220 then 11114001010112123=??\n\n+1 vote\n577 views\n\nIf\n\n``````1111=T\nS=2113\nN=1233\nEZ=40010112\nO=1401\nF=6220\n``````\n\nthen\n\n``````11114001010112123=??\n``````", null, "posted Jul 26, 2015\nLooking for solution? Promote on:\n\nSimilar Puzzles\n–1 vote\n\nIf symbols of alphabets (without any change in their sense and meaning) are\nL M V A K B N C P I D Y O S E Q R F T H G J U W Z X\nthen\nHow we will write \"WORLDCUP\".\n\nwhat is the missing letter?\n? T T F F\nS S E N T\nE T T F F\nS S E N T\nOPTIONS:\n1) E\n2) R\n3) O\n4) P" ]

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[ "More than a year ago a friend of mine wanted to learn a bit more about Rust by trying out a project. He had a nice project in mind which suits Rust quite well I think. For fun I joined his effort and created an implementation at the same time as he did, discussing and comparing along the way. In this post I’ll tell you about the project specifics, but the point of the post is more an encouragement. If you’ve read about Rust before but haven’t tried it yet, find a small project like the one below, and learn Rust in a fun and hands-on way yourself. It’s a great programming language, I highly recommend it.\n\n# The project: consistent overhead byte stuffing\n\nComputer networking can be messy business. Depending on what layer of the network your software is operating in, you need to worry about different kinds of errors. If you’re receiving raw bytes, you might run into the issue of corrupted messages. You could throw away such a message entirely… Or you could try to chop it into chunks with a clear boundary and recover at the next boundary.\n\nByte stuffing is the process of stuffing bytes into a smaller range of values than the full byte, so you can use the unused values for something special like the boundaries of messages. The usual terminology is splitting your bytes of data into frames and using sentinel values to delimit the frames.\n\nThe problem that byte stuffing solves then, is what to do with values in your data that are the sentinel value you picked for delimiting frames. These should be turned into something else that can be reliably decoded again. The project of this post is to implement an algorithm for Consistent Overhead Byte Stuffing, or COBS. This algorithm has an overhead of at least one byte and at most one byte in 254 rounded up. If I’d been presented with this problem myself before I’d heard of COBS, I would have probably done something like pick two byte values, one delimiter and one “escape character”. The worst case for that is something like twice the size of the message. So this algorithm is pretty cool. I’ll explain it in my own words, but the Wikipedia article is very nice too, so browse that if you don’t follow everything here.\n\n# COBS in short\n\nLet’s choose zero as our sentinel values, our frame marker. If we need to recover from some error in the middle of the stream of frames, we just look for the next zero, that’s the end of a frame. Then the first byte is our consistent minimum overhead byte, which starts the COBS encoding. This byte tells us the offset to where the next zero should have been in the message. Until that offset is reached, the bytes should be the original message. If in those bytes you find a zero, the message is definitely corrupted and you should skip to the next frame. Once you reach the offset, instead of a zero you should find another number, which is the offset from there to the next zero in the original message. So each zero is turned into a higher number of where the next zero is. The last zero points to the place where the zero of the end of the frame should be1.\n\nLet’s call these offsets to a zero zero markers. The first zero marker is fake, since it doesn’t mark a zero at its place. We need it to point out the first actual zero. There can be more fake zero markers in the message, and this is where the worst case overhead comes from: what if two zeroes are further apart than the size of a byte? Assuming we’re speaking of octets, which is usually the case these days, we have 0-255 as the normal value range. We’re changing that range to 1-255. So if 255 is the maximum value for our zero marker, we cannot have more than 254 consecutive non-zero bytes. To fix this we say that zero markers with value 255 signify that the next zero marker is fake. Again, a fake zero marker does not signify a zero, but just how many bytes to read until we reach the next zero marker.\n\n# COBS in Rust\n\nNow that we’ve seen a prose description of COBS in Rust, let’s implement an encode and a decode function for COBS. In this case I’ll present an implementation that is uses a sentinel value of zero, adds that zero as part of the encode procedure, and expects it during decode. Note that this is not going to be the most beautiful implementation possible. We’re not using traits from Rust’s standard library, such as Read and Write even though these work with bytes. We’re hard-coding the sentinel value to 0.\n\n## Tests\n\nTo start things off, let’s define the types for encode and decode, and then write some tests to make our understanding of the algorithm executable. We start out with some unit tests, and some property based tests using quickcheck. Property based tests use a function from some input to boolean and given that input check if a property holds.\n\nWith the property based tests we check that encoding and decoding a given vector of bytes comes to the same thing. The quickcheck framework then generates some random vectors of bytes and checks if our property holds. If it doesn’t, the framework shrinks the counter example with some heuristics. It’s pretty cool stuff. I recommend using this form of testing whenever you can.\n\n//! Consistent overhead byte stuffing\n//! =================================\n//!\n//! This encoding allows for packet loss in a stream of bytes by dividing data into frames.\n//!\n//! 0 = framemarker, the thing you search for when you recover in the middle of a stream. It marks\n//! the end of a frame.\n//! Zeromarkers both mark a zero and have a value of where the next zeromarker is. There are also\n//! special zeromarkers, which say the next zeromarker is fake. Fake zeromarkers don't mark zeroes,\n//! they only tell where the next zeromarker is.\n//! The first byte of a frame is a fake zeromarker.\n//! Other bytes are normal bytes.\n//! When normally (in the original data) the next zero occurs, this is another zeromarker. The\n//! first (fake) zeromarker will have the offset after which the next zeromarker occurs.\n//! The special zeromarker 255 predicts that the next zeromarker is fake. The reason for calling it\n//! fake is to support data where the are more than 255 bytes between zeroes.\n\n#[cfg(test)]\n#[macro_use]\nextern crate quickcheck;\n\nuse std::iter;\n\npub fn encode(data: &[u8], encoded_data_buffer: &mut Vec<u8>) -> usize {\nunimplemented!()\n}\n\npub fn decode(encoded_data: &[u8], decoded_data_buffer: &mut Vec<u8>) -> Result<usize, usize> {\nunimplemented!()\n}\n\npub fn max_encoded_size(input_size: usize) -> usize {\ninput_size * (u8::max_value() as f32 / (u8::max_value() - 1) as f32).ceil() as usize + 1\n}\n\npub fn max_decoded_size(encoded_size: usize) -> usize {\nencoded_size - 2\n}\n\n#[cfg(test)]\nmod tests {\nmacro_rules! unit_test_set {\n($assert:path) => { #[test] fn empty() {$assert(&[], &[1, 0]);\n}\n\n#[test]\nfn zero() {\n$assert(&, &[1, 1, 0]); } #[test] fn one() {$assert(&, &[2, 1, 0]);\n}\n\n#[test]\nfn byte_max() {\n$assert(&, &[2, 255, 0]); } #[test] fn five_zeroes() {$assert(&[0; 5], &[1, 1, 1, 1, 1, 1, 0]);\n}\n\n#[test]\nfn five_ones() {\n$assert(&[1; 5], &[6, 1, 1, 1, 1, 1, 0]); } #[test] fn byte_max_zeroes() { let mut output = Vec::with_capacity(257); output.extend_from_slice(&[1; 256]); output.push(0);$assert(&[0; 255], output.as_slice());\n}\n\n#[test]\nfn byte_max_ones() {\nlet mut output = Vec::with_capacity(258);\noutput.push(255);\noutput.extend_from_slice(&[1;254]);\noutput.extend_from_slice(&[2, 1, 0]);\n$assert(&[1; 255], output.as_slice()); } } } mod encode { use super::super::encode; fn check_encoded_vs_given(data: &[u8], encoded: &[u8]) { let mut buffer = Vec::new(); let _ = encode(data, &mut buffer); assert_eq!(encoded, buffer.as_slice()); } unit_test_set!(check_encoded_vs_given); } mod decode { use super::super::decode; fn check_decoded_vs_given(data: &[u8], encoded: &[u8]) { let mut buffer = Vec::new(); let _ = decode(encoded, &mut buffer); assert_eq!(data, buffer.as_slice()); } unit_test_set!(check_decoded_vs_given); } use super::{encode, decode}; quickcheck! { fn encode_decode_identity(data: Vec<u8>) -> bool { let mut b1 = Vec::new(); let mut b2 = Vec::new(); let _ = encode(data.as_slice(), &mut b1); let _ = decode(b1.as_slice(), &mut b2); data == b2 } } } ## Constants I really dislike magic values in my code. Here are some constants we’ll use: const MAXu8: u8 = u8::max_value(); const MAX: usize = MAXu8 as usize; const MAX_CONSECUTIVE: usize = MAX - 1; The MAX* values are just shorter without using 255 literally. I did use them in the tests because there are related numbers off by 2 or 3. It made more sense in the unit tests, which encode specific examples. The code is abstract, should work for any case. ## Decode So let’s decode some bytes. For every byte we can either interpret it as a zero marker if it’s at the right offset, or we can copy the byte to the output verbatim. If the copied byte is zero, that’s an error. Unless it’s a predicted zero marker, in which case we’re successfully finished. If it’s a zero marker that’s not fake, we push a zero instead of the byte value of the marker. We also need to compute the index of the next zero marker with the relative offset. This boils down to the following code: pub fn decode(encoded_data: &[u8], decoded_data_buffer: &mut Vec<u8>) -> Result<usize, usize> { let mut zero_marker_index: usize = 0; let mut fake = true; let mut written = 0; for (index, &byte) in encoded_data.iter().enumerate() { if index == zero_marker_index { if byte == 0 { return Ok(written); // framemarker, we're done } if !fake { decoded_data_buffer.push(0); written += 1; } zero_marker_index = index + byte as usize; fake = byte == MAXu8; } else { if byte == 0 { break; // fail } decoded_data_buffer.push(byte); written += 1; } } Err(written) // fail } The zero_marker_index is the absolute offset from the start of the encoded_data slice. The boolean fake is for remembering if that zero marker will be fake or not. We also track how many bytes we’ve written in the buffer of decoded data. We break from the loop to fail, because we may also find no 0 at all and run out of encoded data, which is also a corner case where we should fail. ## Encode (buffer input) There are two different ways you can implement COBS. One is to look ahead in the data for the zero. This means to need a buffer of 254 bytes at most, but you can sequentially output the encoded bytes. Is to sequentially read the input byte without buffering them, instead buffering the output so you can go back and fill in the space you reserved for the zero marker once you’ve seen the next zero. Let’s first look at the look-ahead version: pub fn encode_lookahead(data: &[u8], encoded_data_buffer: &mut Vec<u8>) -> usize { let start_length_out_buffer = encoded_data_buffer.len(); let mut data_iter = data.iter().chain(iter::once(&0)).peekable(); let mut buf = [0_u8; MAX_CONSECUTIVE]; let mut buf_index: usize; while let Some(_) = data_iter.peek() { buf_index = 0; // Find the next zero, copy bytes seen into buffer for &byte in data_iter.by_ref().take_while(|&&b| b != 0) { buf[buf_index] = byte; buf_index += 1; debug_assert!(buf_index <= buf.len()); if buf_index == buf.len() { break; } } // Write where next zero is, then write the data from the buffer // Note the +1, since buf_index starts at zero and the next zero is always at least one away encoded_data_buffer.push(buf_index as u8 + 1); encoded_data_buffer.extend_from_slice(&buf[0..buf_index]); } encoded_data_buffer.push(0); encoded_data_buffer.len() - start_length_out_buffer } Given that we take in some Vec<u8> that we append to, we should save the length and return the difference in length as the amount of bytes written. Then we create an iterator over the data, followed by an extra zero, which should be peekable. This means we can look ahead without consuming to see if the iterator is done yet. While it isn’t done, we set keep an index of the used part of the buffer. For the bytes in the iterator, we just add them to the buffer unless it’s a zero or the buffer is full. Then we write the zero marker first using the buffer index to see how many non-zeroes we found ahead. And we copy over the buffer. The extra zero on the data iterator compensates for the extra zero-marker at the front of the message. The extra zero pushed at the end ends the message. ## Encode (buffer output) If we buffer the output, we can just write a bogus value for the zero marker, remember its index, and overwrite it later. Again only 255 bytes need to be buffered at a maximum, although this isn’t visible in our implementation. pub fn encode(data: &[u8], encoded_data_buffer: &mut Vec<u8>) -> usize { let start_length_out_buffer = encoded_data_buffer.len(); // Note that we always start from 1, so we count MAX_CONSEQUTIVE bytes of non-zero data let mut non_zero_count = 1_usize; let mut zero_marker_index; macro_rules! next_zero_marker { () => { encoded_data_buffer[zero_marker_index] = non_zero_count as u8; non_zero_count = 1_usize; zero_marker_index = encoded_data_buffer.len(); encoded_data_buffer.push(0); }}; zero_marker_index = encoded_data_buffer.len(); encoded_data_buffer.push(0); // NOTE: the extra zero at the end will become the framemarker for &byte in data.iter().chain(iter::once(&0)) { if byte == 0 { next_zero_marker!(); } else { encoded_data_buffer.push(byte); non_zero_count += 1; debug_assert!(non_zero_count <= MAX); if non_zero_count == MAX { next_zero_marker!(); } } } encoded_data_buffer.len() - start_length_out_buffer } In the end, I don’t think the memory requirements and timing behaviour of the two different options should be very different. But to put that to the test, I’ve written a little benchmark: macro_rules! gen_benches { ($prefix:ident) => {\nmod $prefix { use test::Bencher; use cobs::$prefix;\nuse super::LOREM_IPSUM_RAW;\n\n#[bench]\nfn encode_r1(b: &mut Bencher) {\nb.iter(|| {\nlet mut lorem_ipsum_encoded = Vec::new();\nlet _ = \\$prefix(LOREM_IPSUM_RAW, &mut lorem_ipsum_encoded);\n});\n}\n}\n}\n}\n\ngen_benches!(encode);\n\n\nThis uses the first 4 paragraphs or so from Lorem Ipsum, and on my machine the input buffering version is always faster:\n\n name encode:: ns/iter encode_lookahead:: ns/iter diff ns/iter diff % speedup\nencode_r1 4,243 3,745 -498 -11.74% x 1.13\n\n\n## Faster decode\n\nSay we wanted to seriously speed up our decoding. We could do so by dropping the check of an unexpected zero. Why? Well if you drop that check, a zero marker will tell you exactly how many bytes you can copy over verbatim before the next marker. Which you can do in Rust with the extend_from_slice function, which is probably a bit faster than a manual loop. Let’s try that out:\n\npub fn decode(encoded_data: &[u8], decoded_data_buffer: &mut Vec<u8>) -> Result<usize, usize> {\nif encoded_data.len() == 0 {\nreturn Err(0)\n}\n\nlet start_length_out_buffer = decoded_data_buffer.len();\nlet mut index = 0;\nlet mut zero_marker = encoded_data[index] as usize;\n\nloop {\nlet next_index = index + zero_marker;\nif zero_marker == 0 {\nreturn Ok(decoded_data_buffer.len() - start_length_out_buffer);\n}\nif next_index >= encoded_data.len() {\nreturn Err(decoded_data_buffer.len() - start_length_out_buffer)\n}\n\ndecoded_data_buffer.extend_from_slice(&encoded_data[index+1..next_index]);\n\nif zero_marker != u8::max_value() as usize && encoded_data[next_index] != 0 {\ndecoded_data_buffer.push(0);\n}\n\nzero_marker = encoded_data[next_index] as usize;\nindex = next_index;\n}\n}\n\n\nWe don’t use iterators any more in this code. This is not very idiomatic for Rust, since Rust can more easily eliminate bounds checks for loops over iterators. However, we do get the extend_from_slice which is hopefully more efficient. So what we do is keep the index into the data around, and look up the zero marker. If the zero marker is zero, we’re done, end of message. If the index is out of bounds, that’s an error. Otherwise we extend from the index+1 up to (not including) the next index. The +1 is because index always points to a zeromarker. Should the zero marker be 255 or the next zero marker be 0, then we don’t need to add a zero after the copied data. Then we update the zero marker and index.\n\nTo test this we run another benchmark, this time decoding the encoded lorem ipsum text. The results are quite promising:\n\n name naive_decode:: ns/iter decode:: ns/iter diff ns/iter diff % speedup\ndecode_r1 3,962 361 -3,601 -90.89% x 10.98\n\n\nNaturally, this faster decode is too permissive. So a quickcheck test such as the following will fail most of the time by finding an incorrect COBS encoded message with an unexpected zero.\n\n quickcheck! {\nfn naive_decode_eq_decode(data: Vec<u8>) -> bool {\nlet mut b1 = Vec::new();\nlet mut b2 = Vec::new();\nlet r1 = decode(data.as_slice(), &mut b1);\nlet r2 = naive_decode(data.as_slice(), &mut b2);\nif let (Err(_), Err(_)) = (r1,r2) {\ntrue\n} else {\nr1 == r2 && b1 == b2\n}\n}\n}\n\n\n## Faster encode\n\nPerhaps we can also improve our look-ahead encoding, by not explicitly buffering anything. If instead we just find the position of the next zero, we can use some index juggling:\n\npub fn encode_itertools(data: &[u8], encoded_data_buffer: &mut Vec<u8>) -> usize {\nuse itertools::Itertools;\n\nlet start_length_out_buffer = encoded_data_buffer.len();\nlet mut index = 0_usize;\nfor z_index in data.iter().chain(iter::once(&0)).positions(|&b| b == 0) {\ndebug_assert!(z_index >= index);\n// index is always still-unvisited, so when z_index == index, we need to write a 1\nmacro_rules! offset_between_zeroes { () => {z_index - index + 1}}\nwhile offset_between_zeroes!() >= MAX {\nencoded_data_buffer.push(MAXu8);\nencoded_data_buffer.extend_from_slice(&data[index..index + MAX_CONSECUTIVE]);\nindex += MAX_CONSECUTIVE as usize;\n}\nencoded_data_buffer.push(offset_between_zeroes!() as u8);\nencoded_data_buffer.extend_from_slice(&data[index..z_index]);\nindex = z_index + 1;\n}\nencoded_data_buffer.push(0);\n\nencoded_data_buffer.len() - start_length_out_buffer\n}\n\n\nWe use itertools here for the positions iterator method. We go over the whole data looking for the indices of the zero bytes. We’ve encoded up to index, so if the offset between there and the zero is too far, we need to add the special zero markers and the MAX_CONSECUTIVE number of data bytes. In the end we always write the offset as zero marker, then the data up to the zero, then set the index to after the zero.\n\nThis is again a bit faster because we don’t copy each byte into a buffer only to copy the buffer again. The positions iterator method now does the look-ahead for us.\n\n name encode_lookahead:: ns/iter encode_itertools:: ns/iter diff ns/iter diff % speedup\nencode_r1 3,750 2,756 -994 -26.51% x 1.36\n\n\n# Conclusion\n\nSo we’ve seen some Rust code today that was hopefully readable to you. Tests are easy because they’re built-in. Property based tests are just a crate import away. Benchmarks require the nightly compiler, but only those do, so you can just use cargo +nightly bench to run them. The comparison tables are generated with cargo-benchcmp.\n\nI’ve shown some of my implementations for COBS in Rust, but this was only a learning exercise. I hope this inspires you to find a project of your own to get more experience with Rust. A real implementation of COBS in Rust can be found in the cobs crate, which allows you to use whichever sentinel value you want, can decode in-place, and doesn’t even use vectors so you can use the crate without the standard library.\n\n# Footnotes\n\n1. If you count the zero at the end of a frame as part of the COBS algorithm, it has a minimum offset of 2. But apparently people usually count that as a separate “packetize” step, or so it says on Wikipedia." ]

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[ "# What is the angle between <5,7,1> and <5,1,7> ?\n\nJan 18, 2017\n\nThe angle is $58.7$º\n\n#### Explanation:\n\nThe angle between $\\vec{A}$ and $\\vec{B}$ is given by the dot product definition.\n\nvecA.vecC=∥vecA∥*∥vecB∥costheta\n\nWhere $\\theta$ is the angle between $\\vec{A}$ and $\\vec{B}$\n\nThe dot product is\n\nvecA.vecB=〈5,7,1〉.〈5,1,7〉=25+7+7=39\n\nThe modulus of $\\vec{A}$= ∥〈5,7,1〉∥=sqrt(25+49+1)=sqrt75\n\nThe modulus of $\\vec{C}$= ∥〈5,1,7〉∥=sqrt(15+1+49)=sqrt75\n\nSo,\n\ncostheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=39/(sqrt75*sqrt75)=39/75=0.52\n\n$\\theta = 58.7$º" ]

[ null ]

[ "Grams To Kilograms\n\n# 296 g to kg296 Grams to Kilograms\n\ng\n=\nkg\n\n## How to convert 296 grams to kilograms?\n\n 296 g * 0.001 kg = 0.296 kg 1 g\nA common question is How many gram in 296 kilogram? And the answer is 296000.0 g in 296 kg. Likewise the question how many kilogram in 296 gram has the answer of 0.296 kg in 296 g.\n\n## How much are 296 grams in kilograms?\n\n296 grams equal 0.296 kilograms (296g = 0.296kg). Converting 296 g to kg is easy. Simply use our calculator above, or apply the formula to change the length 296 g to kg.\n\n## Convert 296 g to common mass\n\nUnitMass\nMicrogram296000000.0 µg\nMilligram296000.0 mg\nGram296.0 g\nOunce10.4410927371 oz\nPound0.6525682961 lbs\nKilogram0.296 kg\nStone0.0466120211 st\nUS ton0.0003262841 ton\nTonne0.000296 t\nImperial ton0.0002913251 Long tons\n\n## What is 296 grams in kg?\n\nTo convert 296 g to kg multiply the mass in grams by 0.001. The 296 g in kg formula is [kg] = 296 * 0.001. Thus, for 296 grams in kilogram we get 0.296 kg.\n\n## 296 Gram Conversion Table", null, "## Alternative spelling\n\n296 Grams to Kilograms, 296 Grams in Kilograms, 296 g to kg, 296 g in kg, 296 Gram to Kilograms, 296 Gram in Kilograms, 296 Grams to Kilogram, 296 Grams in Kilogram, 296 Gram to kg, 296 Gram in kg, 296 g to Kilogram, 296 g in Kilogram, 296 g to Kilograms, 296 g in Kilograms" ]

[ null, "https://grams-to-kilograms.appspot.com/image/296.png", null ]

[ "## Description\n\nThis node creates an index based falloff that associates input Start Value to all objects with index less than some input index and the input End Value to all objects with index more than some input index, while the objects at indices in between are associated with values in between the Start Value and End Value evaluated at the input interpolation.\n\n## Illustration", null, "The examples above shows the fade falloff node in action. We offset the vertices of the line by one unit in the z-axis and used the Fade Falloff node as a factor for that offset. The Start Index is set to `2` so all vertices with index less than or equal `2` are offset `2` units in the z-axis because the original offset multiplied by the Start Value is `1x2=2`. The End Index is set to `6` so all vertices with index larger than or equal `6` are not offset because the original offset multiplied by the End Value is `1x0=0`. Indices in between however (from `2` to `6`) are offset with amounts linearly changing between `0` and `2`. Had I used a non-linear interpolation, values will no longer linearly change as in the following example:", null, "Notice how they are changing exponentially and how the start value affected the offset.\n\n## Options\n\nOptions are only different on how the start and end index are defined.\n\n• Start / End - User define start and end indices directly.\n• Start / Amount - User define start index and the amount of indices in the interval.\n• End / Amount - Use define the end index and the amount of indices in the interval." ]

[ null, "https://docs.animation-nodes.com/documentation/nodes/falloff/fade_falloff/fade_falloff_node_illustration.png", null, "https://docs.animation-nodes.com/documentation/nodes/falloff/fade_falloff/fade_falloff_node_illustration2.png", null ]

[ "### Work word problems\n\nSometimes you will encounter a word problem that asks you to determine how long it would take two people working together to finish a job.  Solving this type of problem requires a few steps of logic.  Let’s jump straight to an example.\n\nExample:  Jennifer can mop a warehouse in 8.3 hours.  Heather can mop the same warehouse in 11.2 hours.  Find how long it would take them if they worked together.\n\nSolution:  We set up an equation to model Jen’s work.  We know that Jen can mop a warehouse in 8.3 hours, which means\n\n$$\\Large \\frac{{1{\\text{Warehouse Mopped}}}}{{8.3{\\text{ hours}}}} = 0.12{\\text{Warehouse Mopped in }}1{\\text{ hour}}$$\n\nThat is, Jen can mop 12 percent of the warehouse in one hour.  We set up a similar equation for Heather.  We know that Heather can mop the same warehouse in 11.2 hours, which means\n\n$$\\Large \\frac{{1{\\text{Warehouse Mopped}}}}{{11.2{\\text{ hours}}}} = 0.09{\\text{Warehouse Mopped in }}1{\\text{ hour}}$$\n\nThat is, Heather can mop about 9 percent of the warehouse in one hour.  Now we can find out how much of the warehouse they can mop together in one hour.  We have\n\n$$0.12\\left( {for Jen} \\right) + 0.09\\left( {for Heather} \\right) = 0.21$$\n\nThat is, together they can mop 21 percent of the warehouse in 1 hour.  Let’s set up our final equation to model this word problem.  We use a simple ratio:\n\n$$\\Large \\frac{{1{\\text{Warehouse Mopped}}}}{{x{\\text{ hours}}}} = \\Large \\frac{{0.21{\\text{Warehouse Mopped}}}}{{1{\\text{ hour}}}}$$\n\nCross multiplying gives\n\n$$x = \\Large \\frac{1}{{0.21}} = 4.76{\\text{ hours}}$$\n\nAnother Example:  Molly can clean an attic in 10.6 hours.  Jasmine can clean the same attic in 15 hours.  If they worked together how long would it take them?\n\n$$\\Large \\frac{{1 Attic}}{{10.6 hours}} = 0.09 in one hour$$\n\nFor Jasmine, we have\n\n$$\\Large \\frac{{1 Attic}}{{15 hours}} = 0.07 in one hour$$\n\nTogether, their labor yields\n\n$$0.09 + 0.07 = 0.16 together in one hour$$\n\nThen we use another ratio to solve the problem\n\n$$\\Large \\frac{{1 Attic Cleaned}}{{x hours}} = \\Large \\frac{{0.16 Cleaned}}{{1 hour}}$$\n\nThen, by cross multiplying,\n\n$$x = \\Large \\frac{1}{{0.16}} = 6.25 hours$$", null, "4496 x\n\nThis free worksheet contains 10 assignments each with 24 questions with answers.\n\nExample of one question:", null, "Watch below how to solve this example:", null, "5726 x\n\nThis free worksheet contains 10 assignments each with 24 questions with answers.\n\nExample of one question:", null, "Watch below how to solve this example:", null, "7264 x\n\nThis free worksheet contains 10 assignments each with 24 questions with answers.\n\nExample of one question:", null, "Watch below how to solve this example:\n\n### Geometry\n\nCircles\nCongruent Triangles\nConstructions\nParallel Lines and the Coordinate Plane\nProperties of Triangles\n\n### Algebra and Pre-Algebra\n\nBeginning Algebra\nBeginning Trigonometry\nEquations\nExponents\nFactoring\nLinear Equations and Inequalities\nPercents\nPolynomials" ]

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[ "## Filters\n\nQ&A - Ask Doubts and Get Answers\nQ\n\n# Solve it, The ratio of charge to potential of a body is known as\n\nThe ratio of charge to potential of a body is known as\n\n• Option 1)\n\nCapacitance\n\n• Option 2)\n\nConductance\n\n• Option 3)\n\nInductance\n\n• Option 4)\n\nResistance\n\n99 Views\n\nAs we have learned\n\nCapacitance of Conductor -\n\n- wherein\n\nC - Capacity or capacitance of conductor\n\nV - Potential.\n\ndefinition of capacitance .\n\nOption 1)\n\nCapacitance\n\nOption 2)\n\nConductance\n\nOption 3)\n\nInductance\n\nOption 4)\n\nResistance\n\nExams\nArticles\nQuestions" ]

[ null ]

[ "", null, "##", null, "mill power draw calculation\n\nClick To See More", null, "### mill power draw\n\ncalculate sag mill power draw . CALCULATION OF THE POWER DRAW OF DRY For all model based methods a reliable method to calculate mill power draw for a given mill is required for the calculation of power draw Morrell (1996) proposed a mathematical model for autogenous semi-autogenous and ball mills which is based on the motion of grinding charge inside the mill Battery Bank CalculatorPlease remember that this calculator works out the 'minimum' battery bank size for a given power consumption. When using an inverter, the current draw on the battery side can be extremely high, so you may need a battery bank that is larger than the minimum. For example, 1200W drawn at 240V is only 5A, whereas at 12V this current increases to 100A. We … calculate sag mill power draw - pochiraju.co.inSAG Mill power draw models are used in mill design and grinding circuit Get Price. Polycorp - Mining - ServicesMore details: /price-list.php ball mill power calculation pdf ball mill power calculation pdf, power draw calculation Get Price. DJB Consultants Inc. Millpower 2022 - Twin Metal. Amperage Calculator - Deelat Industrial USACalculate the phase current I in amps (A) by dividing the power P in watts (W) by the square root of 3 times the power factor PF times the line to line RMS voltage VL-L in volts (V): (The power factor is the ratio of the real power flowing to the load to the apparent power in the circuit. Power factor values can range from 0 to 1. Pellet mill designChapter 3: Pellet Mill Design Pellet mill design BY MR. RON TURNER, FEED PELLETING CONSULTANT REVIEWED AND EDITED BY ADAM FAHRENHOLZ, CASSANDRA JONES, AND CHARLES STARK W hen considering at pellet mill design, there are some important aspects to study to achieve maximum efficiency of the machine. Maximum efficiency is defined …\n\n### Calculation of Power Draw of Tumbling Mills | Mill\n\nMorrell, S. 1996. Power draw of wet tumbling mills and its relationship to charge dynamics - Part 2: an empirical approach to modelling of mill power draw. Trans Inst Min Metall, Section C, Vol 105, C54-62. Moys M.H., 1990. A model for mill power as affected by mill speed, load volume and liner design. Mill's Moral and Political Philosophy (Stanford09.10.2022 · Mill and Sidgwick thought that our knowledge of others and our causal powers to do good were limited to those near and dear and other associates with whom we have regular contact, with the result that as individuals we do better overall by focusing our energies and actions on associates of one kind or another, rather than the world at large (U II 19; Sidgwick, … Power draw estimations in experimental 01.02.2022· 1. Introduction. The power draw of a tumbling mill is known to be an important measure in determining its efficiency. Many models have been derived to predict the power draw as a function of characteristics related to charge motion (Harris and Schnock, 1985, Morell, 1992).While these models have been shown to calculate good approximations of mill power, they have been observed to be How To Calculate Power Output Of Wind | Windpower EngineeringJan 26, 2022· The formula for how to calculate power is: Where: P = Power output, kilowatts. Cp = Maximum power coefficient, ranging from 0.25 to 0.45, dimension less (theoretical maximum = 0.59) ρ = Air density, lb/ft3. A = Rotor swept area, ft2 or π D2/4 (D is the rotor diameter in ft, π = 3.1416) Steam Calculators: Steam Turbine Calculator17.03.2022 · Calculation Details Step 1: Determine Inlet Properties Using the Steam Property Calculator, properties are determined using Inlet Pressure and the selected second parameter (Temperature, Specific Enthalpy, Specific Entropy, or Quality). The Specific Enthalpy is then multiplied by the Mass Flow to get the Energy Flow: Inlet Energy Flow = Specific Enthalpy * …\n\n### ball mill circulating load calculations\n\nCALCULATION OF THE POWER DRAW OF DRY · Calculation of the power draw of dry multi-compartment ball mills 225 The mill load that is the volume of charge in the mill is the principal determinant of power draw. Estimation of the ball load that is mixed with the cement charge is difficult and can be highly erroneous. Wind Power Calculator: Energy vs Turbine Size vs SpeedThis simple online wind power calculator lets you find average output wattage of your generator. Here is how it works. First you need to determine the wind class of your site from this online US wind resource atlas (look down their list and open the map for your state). Then find from the above table the average wind density in W/sq.m corresponding to your class at rated height (pick 10 m or Rolling processes - الرئيسيةPower (in Kw) = 5.12 Where F is in newtons, L is in meters, and N is the revolutions per minute of the roll. In traditional English units, the total power can be expressed as Power (in hp) = 5.13 Where F is in pounds and L is in feet. Example: A 300-mm-wide strip 25-mm thick is fed through a rolling mill with two powered Wind PowerWind Power Fundamentals24.01.2022 · – P t Mill I t d d i N th EPost Mill Introduced in Northern Europe – Horizontal-Axis Wind-Mill: sails connected to a horizontal shaft on a tower encasing gears and axles for translating horizontal into rotational motionfor translating horizontal into rotational motion Wind in 19th century US – Wind-rose horizontal-axis wate r-pumping wind-mills found throughout rural … Ohms Law Calculator - RapidTables.comVoltage divider calculator Ohm's law calculation formula. The voltage V in volts (V) is equal to the current I in amps (A) times the resistance R in ohms (Ω): V (V) = I (A) × R (Ω) The power P in watts (W) is equal to the voltage V in volts (V) times the current I in amps (A): P (W) = V (V) × I (A) AC Ohm's law calculator\n\n### formula for grinding mill power draw\n\nCALCULATION OF THE POWER DRAW OF DRY . 2022-10-13 given mill is required for the calculation of power draw. Morrell (1996) proposed a mathematical model for autogenous, semi-autogenous and ball mills which is based on the motion of grinding charge inside the mill. He also verified his approach with various plant data (Napier-Munn et. al., 1996). MILLING CONTROL & OPTIMISATIONFor the most productive milling operation it is often best to operate close to the maximum mill power draw. The power-load relationship is highly non-linear and shifts around as the ore and steel load/liner changes. Traditional control and modeling techniques can therefore not be used. Mintek has developed a Power Optimiser that: • Continuously Power Drawbar: Easy Milling Machine Add-On [Bridgeport26.01.2022 · Easy DIY Power Drawbar for Milling Machines (Bridgeport / R8 / Kurt Style) Here's the power drawbar perched atop my Industrial Hobbies Mill… Click this image for a MOVie file of the power drawbar in operation: Requires QuickTime! I don't know how I got by without one, now that I have it. Ball Mill Power Draw Calculation - holidayhit.plBall Mill Power Draw Calculation . FOB Reference Price: Get Latest Price Vertical Mill Power Calculation Tgrachthuys. Calculation of the power draw of dry multi-compartment ball mills 225 the mill load that is the volume of charge in the mill is the principal determinant of power drawstimation of the ball load that is mixed with the cement charge is difficult and can be highly Wind turbine power coefficientIt's just an example of how to do the calculation! Using the wind power formula shown above, with an assumed standard air density at sea level of 1.225 kilograms per cubic meter (kg/m^3), and a blade diameter of 101 meters (about 331 feet - pretty big), I can calculate that the wind power into the turbine, at 12 meters per second (m/s), is 8,480 kW. Therefore we can easily …", null, "" ]

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[ "### LeetCode", null, "", null, "• ㊗️\n• 大家\n• offer\n• 多多！\n\n## Problem\n\nGiven an array of integers nums and an integer k, return the number of unique k-diff pairs in the array.\n\nA k-diff pair is an integer pair (nums[i], nums[j]), where the following are true:\n\n• 0 <= i, j < nums.length\n• i != j\n• nums[i] - nums[j] == k\n\nNotice that |val| denotes the absolute value of val.\n\nExample 1:\n\nInput: nums = [3,1,4,1,5], k = 2\nOutput: 2\nExplanation: There are two 2-diff pairs in the array, (1, 3) and (3, 5).\nAlthough we have two 1s in the input, we should only return the number of unique pairs.\n\n\nExample 2:\n\nInput: nums = [1,2,3,4,5], k = 1\nOutput: 4\nExplanation: There are four 1-diff pairs in the array, (1, 2), (2, 3), (3, 4) and (4, 5).\n\n\nExample 3:\n\nInput: nums = [1,3,1,5,4], k = 0\nOutput: 1\nExplanation: There is one 0-diff pair in the array, (1, 1).\n\n\nConstraints:\n\n• 1 <= nums.length <= 10^4\n• -10^7 <= nums[i] <= 10^7\n• 0 <= k <= 10^7\n\n## Code\n\nclass Solution {\npublic int findPairs(int[] nums, int k) {\nHashMap<Integer, Integer> map = new HashMap<>();\n\nfor (int num : nums) {\nmap.put(num, map.getOrDefault(num, 0) + 1);\n}\n\nint res = 0;\nfor (int key : map.keySet()) {\nint num1 = key - k;\nint num2 = key + k;\n\nif (k == 0) {\nif (map.get(key) >= 2) {\nres += 2;\n}\n} else {\nif (map.containsKey(num1)) res++;\nif (map.containsKey(num2)) res++;\n}\n}\n\nreturn res / 2;\n}\n}" ]

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[ "It looks like you're using Internet Explorer 11 or older. This website works best with modern browsers such as the latest versions of Chrome, Firefox, Safari, and Edge. If you continue with this browser, you may see unexpected results.\n\n# FINANCE GUIDE: COMPOUND INTEREST\n\nThis guide will cover important topics like simple interest, compound interest, consumer price index (CPI), and the tax system in America. Gaining an understanding in these topics can help you manage your finances more efficiently.\n\n## INTRODUCTION\n\nIn a world full of finances, any person can benefit from learning what interest is, how it applies in certain backgrounds, and how to calculate it. For example, taking out a loan to cover expenses could produce a lot or a little interest depending on the interest rate and how the person calculates the amount. Understanding simple interest vs. compound interest could save you money or earn you more money.\n\n## WHAT IS COMPOUNDED INTEREST?\n\nCompound Interest is the total amount of interest and principal amount reinvested over multiple time periods.\n\nIn other words, at the end of a time period your interest is added to your principal amount and then that amount becomes reinvested over and over again based on the time period in years.\n\nThe chart below shows how much more money is created when using compounded interest compared to simple interest.", null, "## COMPOUNDED INTEREST AS A FORMULA", null, "", null, "", null, "## TWO SCENARIOS\n\nHere are two scenarios that can apply to compounded interest.\n\n1. Borrow/Get a Loan\n\nBorrowing money with a compounded interest rate is bad! Very Bad! Over time, you must pay a lot more back to the bank since the amount exponentially increases.\n\n2. Invest\n\nInvesting money with a compound interest rate is amazing! Very Amazing! Take the opposite of what was said above. Your money will exponentially grow and you will be rewarded with more money than a simple interest rate.", null, "## HOW DO WE CALCULATE COMPOUNDED INTEREST?\n\nRemember: The compound interest formula does not yield interest because it is the principal amount AND the interest.\n\nHere’s a quick example without any context:", null, "" ]

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[ "%A Lin Zhi-Hua, Xu Jiang-He, Liu Shu-Shen, Zheng Xu-Xi, Li Zhi-Liang %T Study On Quantitative Structure-Property Relationship of Chain Hydrocarbons,Aldehydes and Alkanones by Molecular Distance-edge Vector %0 Journal Article %D 2000 %J Acta Phys. -Chim. Sin. %R 10.3866/PKU.WHXB20000211 %P 153-161 %V 16 %N 02 %U {http://www.whxb.pku.edu.cn/CN/abstract/article_22993.shtml} %8 2000-02-15 %X\n\nAn extented molecular distance-edge (MDE,μ) vector of chain hydrocarbons including alkanes, alkens, alkynes, dienes, and alkenynes are defined and calculated. Quantitative structure property relationship (M1) on boiling point of chain hydrocarbons is proposed by multiple linear regression method with good results: root mean square error RMS=4.69K and correlation coefficient R=0.9976. Random sampling modeling and cross validation (CV) procedure are also performed and satisfactory results have been obtained with average correlation coefficient being R=0.9975, 0.9971 and average RMS=4.72K,4.83K respectively. In order to generalize MDE for other organic compounds ,a novel molecular distance edge (MDE) vector, which can be applied to discribe the molecular structure of both fatty aldehydes and fatty alkanones, is developed by using dying factors to express the carboxyl ( > CO) based on a molecular distance edge (MDE) vector of alkanes. and the QSPR model between the new MDE vector and boiling point (Tb) is also developed for 72 aldehyde and alkanone compounds with good results: R=0.9989 and RMS=3.93K.Random modeling procedure is also performed with average correlation coefficient being R=0.9990 and average root mean square error RMS=4.12K; and quantitative prediction is done with R=0.9980 and RMS=5.52K. All these results show that the MDE vector is a useful vector in QSPR/QSAR studies." ]

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[ "Require Import prosa.classic.util.all.\nRequire Import prosa.classic.model.schedule.global.jitter.job prosa.classic.model.schedule.global.jitter.schedule\nprosa.classic.model.schedule.global.jitter.interference prosa.classic.model.schedule.global.jitter.platform.\nFrom mathcomp Require Import ssreflect ssrbool ssrfun eqtype ssrnat seq fintype bigop.\n\nSection Lemmas.\n\nContext {Job: eqType}.\nVariable job_arrival: Job time.\nVariable job_cost: Job time.\nVariable job_jitter: Job time.\n\n(* Consider any job arrival sequence ... *)\nVariable arr_seq: arrival_sequence Job.\n\n(* ... and any schedule of this arrival sequence. *)\nContext {num_cpus: nat}.\nVariable sched: schedule Job num_cpus.\nHypothesis H_jobs_come_from_arrival_sequence:\njobs_come_from_arrival_sequence sched arr_seq.\n\n(* For simplicity, let's define some local names. *)\nLet job_is_pending := pending job_arrival job_cost job_jitter sched.\nLet job_is_backlogged := backlogged job_arrival job_cost job_jitter sched.\n\n(* Next, assume all jobs have valid parameters. *)\nHypothesis H_valid_job_parameters:\nj,\narrives_in arr_seq j\n\n(* In this section we prove the absence of multiple jobs of the same\n\nSection NoMultipleJobs.\n\n(* Assume any work-conserving priority-based scheduler. *)\nVariable higher_eq_priority: JLDP_policy Job.\nHypothesis H_work_conserving:\nwork_conserving job_arrival job_cost job_jitter arr_seq sched.\nHypothesis H_respects_JLDP_policy:\nrespects_JLDP_policy job_arrival job_cost job_jitter arr_seq sched higher_eq_priority.\n\n(* Consider task set ts. *)\n\n(* Assume that all jobs come from the taskset. *)\nj,\narrives_in arr_seq j\n\n(* Suppose that jobs are sequential, ...*)\nHypothesis H_sequential_jobs: sequential_jobs sched.\n(* ... jobs only execute after the jitter, ... *)\nHypothesis H_jobs_execute_after_jitter:\njobs_execute_after_jitter job_arrival job_jitter sched.\n(* ... and jobs do not execute after completion. *)\nHypothesis H_completed_jobs_dont_execute:\ncompleted_jobs_dont_execute job_cost sched.\n\n(* Consider a valid task tsk, ...*)\n\n(*... whose job j ... *)\nVariable j: Job.\nHypothesis H_j_arrives: arrives_in arr_seq j.\nHypothesis H_job_of_tsk: job_task j = tsk.\n\n(*... is backlogged at time t. *)\nVariable t: time.\nHypothesis H_j_backlogged: job_is_backlogged j t.\n\n(* Assume that any previous jobs of tsk have completed by the period. *)\nHypothesis H_all_previous_jobs_completed :\nj_other tsk_other,\narrives_in arr_seq j_other\njob_arrival j_other + task_period tsk_other t\n\n(* Then, there can be at most one pending job of each task at time t. *)\nj1 j2,\narrives_in arr_seq j1\narrives_in arr_seq j2\njob_is_pending j1 t\njob_is_pending j2 t\nj1 = j2.\nProof.\nintros j1 j2 ARRin1 ARRin2 PENDING1 PENDING2 SAMEtsk.\napply/eqP; rewrite -[_ == _]negbK; apply/negP; red; move ⇒ /eqP DIFF.\nmove: PENDING1 PENDING2 ⇒ /andP [ARRIVED1 /negP NOTCOMP1] /andP [ARRIVED2 /negP NOTCOMP2].\nunfold jitter_has_passed, actual_arrival in ×.\ndestruct (leqP (job_arrival j1) (job_arrival j2)) as [BEFORE1 | BEFORE2].\n{\nspecialize (SPO j1 j2 DIFF ARRin1 ARRin2 SAMEtsk BEFORE1).\n{\napply leq_trans with (n := job_arrival j2); first by done.\nby apply leq_trans with (n := job_arrival j2 + job_jitter j2); first by apply leq_addr.\n}\nexploit (PREV j1 (job_task j1)); try (by done).\nintros COMP1; apply NOTCOMP1.\ntry ( by apply completion_monotonic with (t0 := job_arrival j1 + task_period (job_task j1)) ) ||\nby apply completion_monotonic with (t := job_arrival j1 + task_period (job_task j1)).\n}\n{\napply ltnW in BEFORE2.\nexploit (SPO j2 j1); try (by done); [by red; ins; subst | intro SPO'].\n{\napply leq_trans with (n := job_arrival j1); first by done.\nby apply leq_trans with (n := job_arrival j1 + job_jitter j1); first by apply leq_addr.\n}\nexploit (PREV j2 (job_task j2)); try (by done).\nintros COMP2; apply NOTCOMP2.\ntry ( by apply completion_monotonic with (t0 := job_arrival j2 + task_period (job_task j2)) ) ||\nby apply completion_monotonic with (t := job_arrival j2 + task_period (job_task j2)).\n}\nQed.\n\n(* Therefore, all processors are busy with tasks other than tsk. *)\ncount (scheduled_task_other_than tsk) ts = num_cpus.\nProof.\nrename H_all_jobs_from_taskset into FROMTS, H_sequential_jobs into SEQUENTIAL,\nH_work_conserving into WORK, H_jobs_come_from_arrival_sequence into FROMarr,\nH_respects_JLDP_policy into PRIO, H_j_backlogged into BACK,\nH_job_of_tsk into JOBtsk, H_valid_job_parameters into JOBPARAMS,\nH_completed_jobs_dont_execute into COMP, H_jobs_execute_after_jitter into JITTER.\napply work_conserving_eq_work_conserving_count in WORK.\nrespects_JLDP_policy, completed_jobs_dont_execute,\nsequential_jobs in ×.\napply/eqP; rewrite eqn_leq; apply/andP; split.\n{\napply leq_trans with (n := count (fun xtask_is_scheduled job_task sched x t) ts);\nfirst by apply sub_count; first by red; movex /andP [SCHED _].\napply count_exists; first by destruct ts.\n{\nintros cpu x1 x2 SCHED1 SCHED2.\ndestruct (sched cpu t); last by done.\nmove: SCHED1 SCHED2 ⇒ /eqP SCHED1 /eqP SCHED2.\nby rewrite -SCHED1 -SCHED2.\n}\n}\n{\nrewrite -(WORK j t) // -count_predT.\napply leq_trans with (n := count\n(jobs_scheduled_at sched t));\nlast first.\n{\nrewrite -count_map.\napply count_sub_uniqr; last first.\n{\nmovetsk' /mapP [j' INj' JOBtsk']; subst; apply FROMTS.\nby rewrite mem_scheduled_jobs_eq_scheduled in INj'; eauto 2.\n}\nrewrite map_inj_in_uniq; first by apply scheduled_jobs_uniq.\nred; intros j1 j2 SCHED1 SCHED2 SAMEtsk.\nrewrite 2!mem_scheduled_jobs_eq_scheduled in SCHED1 SCHED2.\nhave ARRin1: arrives_in arr_seq j1 by apply (FROMarr j1 t).\nhave ARRin2: arrives_in arr_seq j2 by apply (FROMarr j2 t).\nby apply UNIQ; try (by done); apply scheduled_implies_pending.\n}\n{\napply sub_in_count; intros j' SCHED' _.\nrewrite mem_scheduled_jobs_eq_scheduled in SCHED'.\n{\nmove: SCHED' ⇒ /existsP [cpu /eqP SCHED'].\nby apply/existsP; cpu; rewrite /task_scheduled_on SCHED' eq_refl.\n}\n{\napply/eqP; red; intro SAMEtsk; symmetry in SAMEtsk.\nmove: BACK ⇒ /andP [PENDING NOTSCHED].\ngeneralize SCHED'; intro PENDING'.\nhave ARRin': arrives_in arr_seq j' by apply (FROMarr j' t).\nexploit (UNIQ j j'); try (by done);\n[by apply scheduled_implies_pending | by rewrite -SAMEtsk |].\nby intro EQjob; subst; rewrite SCHED' in NOTSCHED.\n}\n}\n}\nQed.\n\nEnd NoMultipleJobs.\n\n(* In this section we also prove the absence of multiple jobs of the same\nof fixed-priority scheduling.  *)\n\nSection NoMultipleJobsFP.\n\n(* Assume any work-conserving priority-based scheduler. *)\nHypothesis H_work_conserving: work_conserving job_arrival job_cost job_jitter arr_seq sched.\nHypothesis H_respects_JLDP_policy:\nrespects_FP_policy job_arrival job_cost job_task job_jitter arr_seq sched higher_eq_priority.\n\n(* Consider any task set ts. *)\n\n(* Assume that all jobs come from the taskset. *)\nj,\narrives_in arr_seq j\n\n(* Suppose that jobs are sequential, ...*)\nHypothesis H_sequential_jobs: sequential_jobs sched.\n(* ... jobs only execute after the jitter, ... *)\nHypothesis H_jobs_execute_after_jitter:\njobs_execute_after_jitter job_arrival job_jitter sched.\n(* ... and jobs do not execute after completion. *)\nHypothesis H_completed_jobs_dont_execute:\ncompleted_jobs_dont_execute job_cost sched.\n\n(* Consider a valid task tsk, ...*)\n\n(*... whose job j ... *)\nVariable j: Job.\nHypothesis H_j_arrives: arrives_in arr_seq j.\nHypothesis H_job_of_tsk: job_task j = tsk.\n\n(*... is backlogged at time t <= job_arrival j + task_period tsk. *)\nVariable t: time.\nHypothesis H_j_backlogged: job_is_backlogged j t.\nHypothesis H_t_before_period: t < job_arrival j + task_period tsk.\n\n(* Recall the definition of a higher-priority task (with respect to tsk). *)\n\n(* Assume that any jobs of higher-priority tasks complete by their period. *)\nHypothesis H_all_previous_jobs_completed :\nj_other tsk_other,\narrives_in arr_seq j_other\ncompleted job_cost sched j_other (job_arrival j_other + task_period tsk_other).\n\n(* Assume that any jobs of tsk prior to j complete by their period. *)\nHypothesis H_all_previous_jobs_of_tsk_completed :\nj0,\narrives_in arr_seq j0\njob_arrival j0 < job_arrival j\ncompleted job_cost sched j0 (job_arrival j0 + task_period tsk).\n\n(* Then, there can be at most one pending job of higher-priority tasks at time t. *)\nj1 j2,\narrives_in arr_seq j1\narrives_in arr_seq j2\njob_is_pending j1 t\njob_is_pending j2 t\nj1 = j2.\nProof.\nH_all_previous_jobs_completed into PREV.\nintros j1 j2 ARRin1 ARRin2 PENDING1 PENDING2 SAMEtsk INTERF.\napply/eqP; rewrite -[_ == _]negbK; apply/negP; red; move ⇒ /eqP DIFF.\nmove: PENDING1 PENDING2 ⇒ /andP [ARRIVED1 /negP NOTCOMP1] /andP [ARRIVED2 /negP NOTCOMP2].\ndestruct (leqP (job_arrival j1) (job_arrival j2)) as [BEFORE1 | BEFORE2].\n{\nspecialize (SPO j1 j2 DIFF ARRin1 ARRin2 SAMEtsk BEFORE1).\n{\napply leq_trans with (n := job_arrival j2); first by done.\nby apply leq_trans with (n := job_arrival j2 + job_jitter j2); first by apply leq_addr.\n}\nexploit (PREV j1 (job_task j1) ARRin1); [by done | by apply INTERF | intros COMP1].\napply NOTCOMP1.\ntry ( by apply completion_monotonic with (t0 := job_arrival j1 + task_period (job_task j1)) ) ||\nby apply completion_monotonic with (t := job_arrival j1 + task_period (job_task j1)).\n}\n{\napply ltnW in BEFORE2.\nfeed (SPO j2 j1); first by red; ins; subst j2.\nfeed (SPO ARRin2 ARRin1); first by rewrite SAMEtsk.\nfeed SPO; first by done.\n{\napply leq_trans with (n := job_arrival j1); first by done.\nby apply leq_trans with (n := job_arrival j1 + job_jitter j1); first by apply leq_addr.\n}\nexploit (PREV j2 (job_task j2) ARRin2);\n[by done | by rewrite -SAMEtsk | intro COMP2 ].\napply NOTCOMP2.\ntry ( by apply completion_monotonic with (t0 := job_arrival j2 + task_period (job_task j2)) ) ||\nby apply completion_monotonic with (t := job_arrival j2 + task_period (job_task j2)).\n}\nQed.\n\n(* Also, there can be at most one pending job of tsk at time t. *)\nLemma platform_fp_no_multiple_jobs_of_tsk :\nj',\narrives_in arr_seq j'\njob_is_pending j' t\nj' = j.\nProof.\nH_all_previous_jobs_of_tsk_completed into PREVtsk,\nH_all_previous_jobs_completed into PREV,\nH_j_backlogged into BACK, H_job_of_tsk into JOBtsk.\nintros j' ARRin' PENDING' SAMEtsk.\napply/eqP; rewrite -[_ == _]negbK; apply/negP; red; move ⇒ /eqP DIFF.\nmove: BACK PENDING' ⇒ /andP [/andP [ARRIVED /negP NOTCOMP] NOTSCHED]\n/andP [ARRIVED' /negP NOTCOMP'].\ndestruct (leqP (job_arrival j') (job_arrival j)) as [BEFORE | BEFORE'].\n{\nexploit (SPO j' j DIFF ARRin' H_j_arrives); [by rewrite JOBtsk | by done | intro SPO'].\nassert (LEt: job_arrival j' + task_period tsk t).\n{\napply leq_trans with (n := job_arrival j); first by rewrite -SAMEtsk.\nby apply leq_trans with (n := job_arrival j + job_jitter j); first by apply leq_addr.\n}\napply NOTCOMP'.\ntry ( apply completion_monotonic with (t0 := job_arrival j' + task_period tsk); try (by done) ) ||\napply completion_monotonic with (t := job_arrival j' + task_period tsk); try (by done).\napply PREVtsk; try (by done).\napply leq_trans with (n := job_arrival j' + task_period tsk); last by rewrite -SAMEtsk.\n}\n{\nunfold jitter_has_passed, actual_arrival in ×.\nrewrite leqNgt in ARRIVED'; move: ARRIVED' ⇒ /negP BUG; apply BUG.\napply leq_trans with (n := job_arrival j'); last by apply leq_addr.\napply leq_trans with (n := job_arrival j + task_period tsk); first by done.\nby rewrite -JOBtsk; apply SPO; try (by done);\n[by red; ins; subst j' | by rewrite SAMEtsk | by apply ltnW].\n}\nQed.\n\n(* Therefore, all processors are busy with tasks other than tsk. *)\nProof.\nhave UNIQ' := platform_fp_no_multiple_jobs_of_tsk.\nrename H_all_jobs_from_taskset into FROMTS, H_sequential_jobs into SEQUENTIAL,\nH_work_conserving into WORK, H_respects_JLDP_policy into PRIO,\nH_j_backlogged into BACK, H_job_of_tsk into JOBtsk,\nH_completed_jobs_dont_execute into COMP,\nH_jobs_come_from_arrival_sequence into FROMarr,\nH_all_previous_jobs_of_tsk_completed into PREVtsk,\nH_jobs_execute_after_jitter into JITTER.\napply work_conserving_eq_work_conserving_count in WORK.\nrespects_FP_policy, respects_JLDP_policy, FP_to_JLDP,\ncompleted_jobs_dont_execute, sequential_jobs,\napply/eqP; rewrite eqn_leq; apply/andP; split.\n{\napply leq_trans with (n := count (fun xtask_is_scheduled job_task sched x t) ts);\nfirst by apply sub_count; red; movex /andP [SCHED _].\napply count_exists; first by destruct ts.\n{\nintros cpu x1 x2 SCHED1 SCHED2.\ndestruct (sched cpu t); last by done.\nmove: SCHED1 SCHED2 ⇒ /eqP SCHED1 /eqP SCHED2.\nby rewrite -SCHED1 -SCHED2.\n}\n}\n{\nrewrite -(WORK j t) // -count_predT.\napply leq_trans with (n := count (fun j\nlast first.\n{\nrewrite -count_map.\napply leq_trans with (n := count predT\n[seq x <- (map (fun jjob_task j) (jobs_scheduled_at sched t))\nfirst by rewrite count_filter; apply sub_count; red; ins.\napply leq_trans with (n := count predT\n[seq x <- ts | scheduled_task_with_higher_eq_priority x]);\nlast by rewrite count_predT size_filter.\napply count_sub_uniqr; last first.\n{\nred; intros tsk' IN'.\nrewrite mem_filter in IN'; move: IN' ⇒ /andP [SCHED IN'].\nrewrite mem_filter; apply/andP; split; first by done.\nmove: IN' ⇒ /mapP [j' IN'] ->; apply FROMTS.\nby rewrite mem_scheduled_jobs_eq_scheduled in IN'; eauto 2.\n}\n{\nrewrite filter_map.\nrewrite map_inj_in_uniq; first by apply filter_uniq, scheduled_jobs_uniq.\nred; intros j1 j2 SCHED1 SCHED2 SAMEtsk.\nrewrite 2!mem_filter in SCHED1 SCHED2.\nmove: SCHED1 SCHED2 ⇒ /andP [/andP [_ HP1] SCHED1] /andP [/andP [_ HP2] SCHED2].\nrewrite 2!mem_scheduled_jobs_eq_scheduled in SCHED1 SCHED2.\nhave ARRin1: arrives_in arr_seq j1 by apply (FROMarr j1 t).\nhave ARRIn2: arrives_in arr_seq j2 by apply (FROMarr j2 t).\nby apply UNIQ; try (by done); apply scheduled_implies_pending.\n}\n}\n{\napply sub_in_count; intros j' SCHED' _.\nrewrite mem_scheduled_jobs_eq_scheduled in SCHED'.\napply/andP; split.\n{\nmove: SCHED' ⇒ /existsP [cpu /eqP SCHED'].\nby apply/existsP; cpu; rewrite /task_scheduled_on SCHED' eq_refl.\n}\napply/andP; split; first by rewrite -JOBtsk; apply PRIO with (t := t).\n{\napply/eqP; red; intro SAMEtsk.\ngeneralize SCHED'; intro PENDING'.\nhave ARRin': arrives_in arr_seq j' by apply (FROMarr j' t).\ntry ( apply scheduled_implies_pending with (job_arrival0 := job_arrival)\n(job_cost0 := job_cost) (job_jitter0 := job_jitter) in PENDING'; try (by done) ) ||\napply scheduled_implies_pending with (job_arrival := job_arrival)\n(job_cost := job_cost) (job_jitter := job_jitter) in PENDING'; try (by done).\nspecialize (UNIQ' j' ARRin' PENDING' SAMEtsk); subst j'.\nby move: BACK ⇒ /andP [_ NOTSCHED]; rewrite SCHED' in NOTSCHED.\n}\n}\n}\nQed.\n\nEnd NoMultipleJobsFP.\n\nEnd Lemmas." ]

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[ "# Which one has the bigger parameter?", null, "(iii) Both of them are equal\nBoth of them are equal as:\nSquare-2+2+2+2=8 cm\n​  Triangle-2+2+4=8 cm\nHOPE THAT HELPS :) :) :) :) :) :)\n\n• 12\nboth are equal\n• 10\nBOTH ARE EQUAL\n\n• 10\nOPTION 3 AS SUM OF SIDES OF SQUARE = 8 CM\nSUM OF SIDES OF TRIANGLE= 8CM\nHOPE THAT HELPS.....\nTHUMBS UP PLZPLZ\n• 8\nWhat are you looking for?" ]

[ null, "https://s3mn.mnimgs.com/img/shared/ck-files/ck_55e83ef93364a.jpg", null ]

[ "## Fundamentals of Fractals 4: The Sierpinski carpet", null, "In the last tutorial, we plotted a Sierpinski gasket using dots. I figured that I wanted to introduce the Sierpinski Carpet as well, before moving into more heavy fractals.\n\nThe Sierpinski Carpet is very simple as well, but we are approaching this problem from another perspective. In this example, we are going to draw a grid of white quads, where each quad is evaluated towards a function that decides if a given quad is in the Sierpinski carpet or not. If yes, we draw the quad, if not, we don,t.\n\nFirst of all, we need to set up our grid. We are doing this by rendering 200×200 squares, each square representing a “pixel” in our image. If we render the whole thing, we will get a big white square:", null, "Now, this white square is made up of 200×200 small squares. Removing a square will make a green spot where the square was located, making this a drawing area with two colors, white and green.\n\nTo render our white square, we iterate through all the quads and render them one by one. Each quad is also scaled by 0,5 to make them fit with no space between each quad, or any overlap with neighboring quads.\n\nfor(int x = 0; x < 200; x++)\n{\nfor(int y = 0; y < 200; y++)\n{\ng_World1 =  XMMatrixScaling( 0.5f, 0.5f, 1.0f ) * XMMatrixTranslation( -100+x, -100+y, 1.0f );\n\ncb1.mWorld = XMMatrixTranspose( g_World1 );\ncb1.mView = XMMatrixTranspose( g_View );\ncb1.mProjection = XMMatrixTranspose( g_Projection );\n\n// Render a triangle\nbool isInside = isSierpinskiCarpetPixelFilled(x,y,200,200);\ng_pImmediateContext->Draw( 6, 0 );\n}\n}\n\nNow, we will work on the function that will decide if we want to draw a quad or not. If we don’t want to draw a given quad, we must skip the code (marked in red above) that renders the quad.\n\nThis is where our evaluation function comes into play. This function will return true or false if a given quad is inside or outside the Sierpinski carpet.", null, "Next, we do the same for each of the solid quads in 2, remove the middle quad from each.(3). Then we repeat the same step to get to 4 and forward.\n\nLet’s take a look at the code:\n\nbool isInsideSierpinskiCarpet(int x, int y, int w, int h)\n{\nif (x <= 1)\n{\nreturn true;\n}\n\n// This is where we split the quad into 9 parts\n// Now, the pixel/quad we are processing, what\n// part of the 9 quads does this belong?\nint x2 = x * 3 / w;\nint y2 = y * 3 / h;\n\n// Is the quad we are inside of the center?\n// if yes, this quad must not be rendered\nif (x2 == 1 && y2 == 1)\nreturn false;\n\n// Not center? prepare us for a recursice call in\n// order to split the current cube into even\n// smaller cubes\nx -= x2 * w / 3;\ny -= y2 * h / 3;\n\n// ..and go!\nreturn isInsideSierpinskiCarpet(x, y, (w / 3) + 1, (h / 3) + 1);\n}\nThis code is based on the code found on this subject at Wikipedia\n\nIn here, we see that we are splitting the quad being processed into 9 smaller quads, and find out what quad the point x,y is inside:\n\nint x2 = x * 3 / w;\nint y2 = y * 3 / h;\n\nThen we check if the the new x is in the middle-quad, and if so, return from the function and indicate that this quad should not be rendered.\n\nif (x2 == 1 && y2 == 1)\nreturn false;\n\nIf not, we move forward and prepare to split the smaller quad into an additional 9 quads and use the same function again on the subset. The resolution is divided by 3, so the next function is only working on that subset of the Sierpinski carpet.\n\nx -= x2 * w / 3;\ny -= y2 * h / 3;\n\nreturn isInsideSierpinskiCarpet(x, y, (w / 3) + 1, (h / 3) + 1);\n\nNow that we got out function to evaluate if a quad should be drawn or not, we can use this to finish our fractal. In the render code, use this function to evaluate if quad x,y should be drawn:\n\nfor(int x = 0; x < 200; x++)\n{\nfor(int y = 0; y < 200; y++)\n{\nif(isInsideSierpinskiCarpet(x,y,200,200))\n{\ng_World1 =  XMMatrixScaling( 0.5f, 0.5f, 1.0f ) * XMMatrixTranslation( -100+x, -100+y, 1.0f );\n\ncb1.mWorld = XMMatrixTranspose( g_World1 );\ncb1.mView = XMMatrixTranspose( g_View );\ncb1.mProjection = XMMatrixTranspose( g_Projection );\n\n// Render a triangle\n\ng_pImmediateContext->Draw( 6, 0 );\n}\n}\n}\n\nThis is done by using the return value from our function on a given quad, and render it based on the result.\n\nThe result should be something like this:", null, "" ]

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[ "## 51014\n\n51,014 (fifty-one thousand fourteen) is an even five-digits composite number following 51013 and preceding 51015. In scientific notation, it is written as 5.1014 × 104. The sum of its digits is 11. It has a total of 3 prime factors and 8 positive divisors. There are 24,376 positive integers (up to 51014) that are relatively prime to 51014.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Even\n• Number length 5\n• Sum of Digits 11\n• Digital Root 2\n\n## Name\n\nShort name 51 thousand 14 fifty-one thousand fourteen\n\n## Notation\n\nScientific notation 5.1014 × 104 51.014 × 103\n\n## Prime Factorization of 51014\n\nPrime Factorization 2 × 23 × 1109\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 51014 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 51,014 is 2 × 23 × 1109. Since it has a total of 3 prime factors, 51,014 is a composite number.\n\n## Divisors of 51014\n\n1, 2, 23, 46, 1109, 2218, 25507, 51014\n\n8 divisors\n\n Even divisors 4 4 2 2\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 79920 Sum of all the positive divisors of n s(n) 28906 Sum of the proper positive divisors of n A(n) 9990 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 225.863 Returns the nth root of the product of n divisors H(n) 5.10651 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 51,014 can be divided by 8 positive divisors (out of which 4 are even, and 4 are odd). The sum of these divisors (counting 51,014) is 79,920, the average is 9,990.\n\n## Other Arithmetic Functions (n = 51014)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 24376 Total number of positive integers not greater than n that are coprime to n λ(n) 12188 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 5224 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares\n\nThere are 24,376 positive integers (less than 51,014) that are coprime with 51,014. And there are approximately 5,224 prime numbers less than or equal to 51,014.\n\n## Divisibility of 51014\n\n m n mod m 2 3 4 5 6 7 8 9 0 2 2 4 2 5 6 2\n\nThe number 51,014 is divisible by 2.\n\n• Arithmetic\n• Deficient\n\n• Polite\n\n• Square Free\n\n• Sphenic\n\n## Base conversion (51014)\n\nBase System Value\n2 Binary 1100011101000110\n3 Ternary 2120222102\n4 Quaternary 30131012\n5 Quinary 3113024\n6 Senary 1032102\n8 Octal 143506\n10 Decimal 51014\n12 Duodecimal 25632\n20 Vigesimal 67ae\n36 Base36 13d2\n\n## Basic calculations (n = 51014)\n\n### Multiplication\n\nn×i\n n×2 102028 153042 204056 255070\n\n### Division\n\nni\n n⁄2 25507 17004.7 12753.5 10202.8\n\n### Exponentiation\n\nni\n n2 2602428196 132760271990744 6772632515335814416 345499075137341236617824\n\n### Nth Root\n\ni√n\n 2√n 225.863 37.0877 15.0287 8.74053\n\n## 51014 as geometric shapes\n\n### Circle\n\n Diameter 102028 320530 8.17577e+09\n\n### Sphere\n\n Volume 5.56105e+14 3.27031e+10 320530\n\n### Square\n\nLength = n\n Perimeter 204056 2.60243e+09 72144.7\n\n### Cube\n\nLength = n\n Surface area 1.56146e+10 1.3276e+14 88358.8\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 153042 1.12688e+09 44179.4\n\n### Triangular Pyramid\n\nLength = n\n Surface area 4.50754e+09 1.56459e+13 41652.8\n\n## Cryptographic Hash Functions\n\nmd5 84bdc10b5cc3b036ce04a562b0e54d61 e33bfb7b8b6feafd1d705ce49e92e81e63434c77 b5f5fd3ea636510152e486caf9cdf557e7c575bdf61f85672a73a95bbe2d65f5 29db97c7ff98dd4940a4ab77ebc7f551958b7a16b02d2da0a99aaf994478d4a16ade59e69fea50020e629b1fd66c3b584b2429a23c9430a6a710f8a738faf9ca 13bef04a352a8d6b23cdc3ebb6680e9154c83e27" ]

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[ "# On a coordinate grid AB has an end point B at (24,16), the midpoint of AB is P(4,-3), what is the Y-coordinate of point A?\n\nLet's take the $x$ and $y$ co-ordinates separately\nThe $x$ and $y$ of the midpoint are the mean of those of the end points.\nIf $P$ is the midpoint then:\n${x}_{P} = \\frac{{x}_{A} + {x}_{B}}{2} \\to 4 = \\frac{{x}_{A} + 24}{2} \\to {x}_{A} = - 16$\n${y}_{P} = \\frac{{y}_{A} + {y}_{B}}{2} \\to - 3 = \\frac{{y}_{A} + 16}{2} \\to {y}_{A} = - 22$" ]

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[ "# Double Multiplication Worksheets Inspirational Multiplication Triple Digit X Double Digit 10 Printable Worksheets Year 3 4 5 6 Grade 3 4 5 6 Multiplication Practice Multiply by\n\nHomeDouble Multiplication Worksheets ➟ Double Multiplication Worksheets Inspirational Multiplication Triple Digit X Double Digit 10 Printable Worksheets Year 3 4 5 6 Grade 3 4 5 6 Multiplication Practice Multiply by" ]

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[ "Courses\n\n# Test: Compressibility & Consolidation - 1\n\n## 10 Questions MCQ Test Soil Mechanics | Test: Compressibility & Consolidation - 1\n\nDescription\nThis mock test of Test: Compressibility & Consolidation - 1 for Civil Engineering (CE) helps you for every Civil Engineering (CE) entrance exam. This contains 10 Multiple Choice Questions for Civil Engineering (CE) Test: Compressibility & Consolidation - 1 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Compressibility & Consolidation - 1 quiz give you a good mix of easy questions and tough questions. Civil Engineering (CE) students definitely take this Test: Compressibility & Consolidation - 1 exercise for a better result in the exam. You can find other Test: Compressibility & Consolidation - 1 extra questions, long questions & short questions for Civil Engineering (CE) on EduRev as well by searching above.\nQUESTION: 1\n\n### Consider the following statements: 1. Coefficient of consolidation normally increases with decreasing liquid limit of clay. 2. The larger the value of coefficient of consolidation, the longer it takes for full consolidation to occur. Which of these statements is/are correct?\n\nSolution:\n\nFrom Taylor’s formula, we have,", null, "For full consolidation i.e. U = 100%. Tv will more or less be a constant.", null, "Thus, larger the value of coefficient of consolidation, smaller it takes for full consolidation to occur.\n\nQUESTION: 2\n\n### The virgin compression curve for a soil is shown in figure given below. What is the compression index of the soil?", null, "Solution:\n\nCompression index is given by the equation,", null, "QUESTION: 3\n\n### The e-p curve for a soil is shown in the figure below. The coefficient of compressibility (in m2kN) of the soil is", null, "Solution:\n\nCoefficient of compressibility,", null, "QUESTION: 4\n\nThe figure given shows the state of a sample of clay before and after consolidation. Based on these figures, the settlement of a clay layer of initial thickness H will be", null, "Solution:", null, "QUESTION: 5\n\nThe settlement analysis for a clay layer draining from top and bottom shows a settlement of 2.5 cm in 4 years and an ultimate settlement of 10 cm. However detailed subsurface investigation reveals that there is no drainage at the bottom. The ultimate settlement in this condition will be\n\nSolution:\n\nUltimate settlement does not depend upon drainage condition.\n\nQUESTION: 6\n\nMatch List-I with List-II and select the correct answer using the codes given below the lists (notations have their usual meaning):\n\nList-I\nA. Coefficient of compressibility\nB. Com pression index\nC. Time factor\nD. Coefficient of volume compressibility\n\nList-ll\n1. mv\n2. Cvt / H2\n3. av\n4. Cc\n\nCodes:\nA B C D\n(a) 3 2 4 1\n(b) 1 2 4 3\n(c) 1 4 2 3\n(d) 3 4 2 1\n\nSolution:\nQUESTION: 7\n\nThe liquid limit of a saturated normally consolidated soil is 59%. The compression index of this soil for the virgin compression curve will be\n\nSolution:\n\nCc = 0.009 (LL - 10)\n= 0.009 (59 - 10)\n= 0.441\n\nQUESTION: 8\n\nCoefficient of consolidation is used to calculate\n\nSolution:\n\nThe average degree of consolidation depends upon the non-dimensional time factor Tv. The time factor Tv depends upon the coefficient of consolidation Cv, time t and drainage path d.\n\nQUESTION: 9\n\nThe value of a compression index for remoulded sample, whose liquid limit is 50%, is\n\nSolution:\n\nFor remoulded sample,\nCc = 0.007 (LL - 10)\n= 0.007 (50 - 10) = 0.28\n\nQUESTION: 10\n\nRate of consolidation\n\nSolution:\n\nWith increase in temperature viscosity of water decreases and coefficient of permeability increases.\nAs coefficient of consolidation", null, "∴ Cv  increases and rate of consolidation increase." ]

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[ "# Calculate the sustainable growth based on the following information: D= 30% ROE = 25% Answer\n\nCalculate the sustainable growth based on the following information: D= 30% ROE = 25% Answer" ]

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[ "# A Table of Logs\n\nGot nostalgic thinking about how we ran calculations before scientific calculators, and broke out one of my slide rules and calculated the volume of the observable universe. That’s not as daunting at it seems: You just take the furthest point of the observable universe, now known to be 13.8 x 109 light-years, as the radius, and use 4πr3 ÷ 3, the formula for calculating the volume of a sphere. Using scientific notation and 3.141 6 for π, you can calculate it on scratch paper. But when you’re a kid in junior high who’s still impressed by lots of zeros, and who’s discovered slide rules, it’s something big. I still remember running that calculation, my mind blown by the speed of the slide rule. And until the advent of pocket scientific calculators, slide rules would remain the fastest way.\n\nSlide rules are usually good for three digits, which isn’t much, but close enough for most applications. Even when doing this by hand, restricting the number of significant digits was a compromise of speed vs accuracy. Calculating this stuff with pencil and paper gets tedious.\n\nIt was possible to get a few more numbers with less effort than longhand by using logarithmic tables. Science textbooks often had short logarithmic tables with about the same precision as a slide rule in the back, but were slower to use. For work such as astronomy, it was common to use larger tables of logarithms. This level of precision was rarely needed outside of the sciences, engineering, navigation, and surveying, so a book of logarithmic tables wasn’t a common item. I had owned a scientific calculator for several years before I saw a book of logarithmic tables. Even through the scientific calculator was a better solution, a book of logarithms fascinated me for the simple reason they didn’t need batteries. This was at the height of the Cold War, and there was a distinct appeal in an alternative in calculation techniques that didn’t need batteries, just in case.\n\nThis book was at the local library, but thanks to the Internet, it’s possible to find PDFs of public domain works free for downloading. You can even find Charles Babbage’s tables, the one he generated using his famed Difference Engine. Being spoiled by pocket calculators, I wanted something with greater precision, and set out making one in a spreadsheet.\n\nIt didn’t take long to abandon that project. It’s easy enough, but the greater the number of digits, the larger the table. A logarithmic table of 16 digit numbers would weigh in at about 1.785 7 x 1013 pages. Printed up in 1,000 page books, it would make a 1.785 7 x 1010 volume set. If each book was 2 inches thick, it would stretch over 563 thousand miles. That’s well over twice the distance to the moon. So much for that idea.\n\nSo I downloaded Baron Von Vegas’ Logarithmic Tables of Numbers and Trigonometrical Functions, which has seven place logarithms. With logarithm tables, the greater the number of places, or digits, the greater the precision.\n\nThe calculations that follow is how it used to be done. For sentimental reasons, we’ll calculate the volume of the observable universe.\n\nFirst we look up the logarithm of the numbers we need: 13.8 x 109 light-years for the radius of the observable universe; 4; 3; and, of course, 3.141 6 for π.\n\nLogarithmic tables are usually laid out with the number, N, in a column on the far left, with numbers from 0 to 9 across the top. In each row under the header is the logarithm. To find the value of 1.38, we go down under N until we find 1380, and across to the 0 column. There we find 139 8791. That’s our mantissa. Since we’re dealing with 13.8 x 109, and since this is the same as 1.38 x 1010, our characteristic is 10. So the logarithm for 13.8 x 109 is 10.139 8791.\n\nOn to the number 3. Here we look up 3000 under the N column, and again go over to the 0 column. There we find 477 1213. Since 3 is less than 10, it has a characteristic of 0, so the logarithm of 3 is 0.477 1213.\n\nI had hoped to find the logarithm of π as a handy constant, but since it doesn’t seem to have it, we’ll have to look it up. Using the value of 3.141 6, we go down to 3141 and across to the 6 column. Here we find 497 1509. As with the logarithm of 3, the characteristic is 0, so our logarithm of π is 0.497 1509.\n\nThe logarithm of 4 is easily found by looking up 4000 and going over to the 0 column: 0.602 0600.\n\nSince logarithms are exponents of a base number, in this case 10, to multiply and divide, all we have to do is add and subtract. And since raising a number to the 3rd power (cubing the number) is multiplying it by itself 3 times, we multiply the logarithm of 13.8 x 109 by 3\n\nOn to our calculations!\n\n13.8 x 109 to the 3rd power: 10.139 8791 x 3 = 30.419 6373\n\nTo multiply this by 4: 30.419 6373 + 0.602 0600 = 31.021 6973\n\nTo multiply by π: 31.021 6973 + 0.497 1509 = 31.518 8482\n\nTo divide by 3: 31.518 8482 – 0.477 1213 = 31.041 7269\n\nTo find the number of the logarithm, we look in the table until we find 0417 269. This is between 11008 = 041 7084 and 11009 = 041 7479. Now what?\n\nFirst we find the proportional part. The proportional part is how much increase is between the two logarithms. 11008 = 041 7084 and 11009 = 041 7479. As our heading number increased by one, the logarithm increased by 395. Using this, we can interpolate and find our number. 041 7269 – 041 7084 = 185. 185÷395 = 0.468 3. We round this to 0.5 , and add this to 11008 to get 11008.5. Here the decimal point acts as a convenient place holder; without the characteristic, the logarithm for 0.041 7479 is for a number less than 10. This is why we call our number 1.10085. Since these logarithms are base 10, the characteristic of 31 means 1031. So our answer is 1.10085 x 1031\n\nIn this book of logarithms the interpolation is already done for us. On the same page as 11008 we find small, two-column, tables of numbers in the far right column. These are the proportional parts and their interpolated values. First we find the mini-table with the heading 395, and go down the right column until we find the closest match to 185. Here it’s 197.5, and this number corresponds to 5. We now write this to the left of 11008, giving us 110085. Since our characteristic is 31, we can now write our final value: 1.100 85 x 1031. Working this out by pocket calculator, using 3.1416 as π, and rounding to five significant digits, we get the answer of 1.100 85 x 1031.\n\nWe can also interpolate in the other direction. If we had decided to use 3.141 59 for π, we would have found the proportional part between 31415 and 31416 (497 1509 – 497 1371 = 138), and either multiply that by 0.9 or look in the mini-table in the right column under the proportional part to find the number we would have to add to 497 1371. Both methods give 124.2, so 497 1371 + 124.2 = 497 1495.2. Since here the decimal is a convenient place holder, we can either round the logarithm to 497 1495 or call it 497 14952. But if we call it 497 14952, we have to remember to align the mantissa from the right of the characteristic, so, if we had used this value of π, when it came time to add it to 31.021 6973, we would have done so like this:\n\n31.021 6973\n+0.497 14952\n31.518 84682\n\nWe interpolated a lot, especially in trigonometry. Yes, the same technique works for other types of mathematical tables. And the tables in the back of our textbooks didn’t have mini-tables of proportional parts to simplify things for us.\n\nThat was how it was done in the day. Not quite like walking to school uphill both ways, but not very speedy.\n\nHow is this an improvement over doing it longhand? When the calculations were more involved, especially those involving trigonometric functions, it increased speed and accuracy. Let’s say we wanted to calculate the volume of the universe in cubic inches. All we need is the following information:\n\n1 light-year = 5,878,499,810,000 miles (call it 5.878 5 x 1012 miles) = 12.769 2665.\n\n1 mile = 5,280 feet (5.280 x 103 feet) = 3.722 6339.\n\n1 foot = 12 inches = 1.079 1812.\n\nWe simply add the logarithms to find the number of inches in a light-year:\n\n12.769 2665 + 3.722 6339 + 1.079 1812 = 17.571 0816.\n\nTo get the number of cubic inches in a cubic light-year, we multiply this logarithm by 3:\n\n17.571 0816 x 3 = 52.713 2448.\n\nNow we add this to the logarithm of the volume of the universe in cubic light-year:\n\n52.713 2448 + 31.041 7269 = 83.754 9717.\n\nLooking this up, we come up with 5.688 16 x 1083 cubic inches. Comparing this with a scientific calculator, using the same values and rounding to 5 significant digits, we have 5.688 16 x 1083.\n\nThis is more efficient than calculating it longhand. You can even use a mechanical adding machine for the job. It’s still labor intensive, which was why universities often hired people to do the work. They were called computers. I kiddest thee not. So it was that in the early years of the 20th Century, a Harvard computer named Henrietta Swan Leavitt was doing the grunt work of cataloging variable stars and noticed a pattern. Published in 1908, her work enabled astronomers to measure stellar distances that were previously impossible to gauge.\n\nThese days, when we need to run calculations, we just reach for our scientific calculators and have the answer quicker and more accurately than going through conversions to logarithms, processing, then converting them back. And while I have a twinge of nostalgia every now and then, I don’t really miss those days. Running through calculations every now and then with a table of logarithms can be fun, but using logarithmic and trigonometric tables all day long? Shudder.\n\nGiven the choice between daily use of a slide rule, a book of logarithmic and trigonometric tables, and a scientific calculator, I’ll take the calculator. Every time." ]

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[ "December 4, 2020\n\n# AND function in Google Sheet\n\nAND function is used when you want to check multiple conditions in both Excel & Google Sheet. It returns TRUE if all the conditions evaluate to TRUE, else it returns FALSE (Note: it returns FALSE even if one condition is FALSE).\n\n`=AND(logical1, [logical2],…)`\n• logical1 – This is the first condition, you want to evaluate for TRUE or FALSE.\n• [logical2] – Second and rest conditions are optional, that you want to evaluate for TRUE or FALSE.\n\nIn Below Example Price has been mentioned as 15 in cell A2. We have used AND function in cell B2 with 2 conditions.\n\nThe first condition returns TRUE as 15>10.\n\nThe second condition also returns TRUE as 15<30.\n\nHence the result of AND function is TRUE.", null, "See below we have changed the price as 5.\n\nNow the first condition returns FALSE as 5 is not greater than 10.\n\nThe second condition also returns TRUE as 5<30.\n\nHence the result of AND function is FALSE.", null, "#### Notes :\n\n• AND function can be used with other formulas like, in an IF Function, we can test a condition and then specify a value when it’s TRUE and a value when it is FALSE. Using AND function within IF enables users to check multiple conditions at one go.\n• For example, if you have to test whether A1 is greater than 20 and less than 200, here is how you can do it in an IF function:\n=IF(AND(A1>20,A1<200),”Approve”,”Reject”)\n• The arguments must either evaluate to logical values (TRUE/FALSE), or must be arrays/references of logical values.\n• Maximum of 255 conditions can be tested in a single AND function.\n• If the specified range contains no logical value, the AND function returns #VALUE! error.\n• Text & empty cells are ignored.", null, "#### abhraisraja\n\nView all posts by abhraisraja →" ]

[ null, "https://cdn.shortpixel.ai/client/q_glossy,ret_img,w_409,h_119/http://solvergeek.com/wp-content/uploads/2020/03/Untitled105.png", null, "https://cdn.shortpixel.ai/client/q_glossy,ret_img,w_409,h_121/http://solvergeek.com/wp-content/uploads/2020/03/Untitled106.png", null, "https://secure.gravatar.com/avatar/e912ab4f48cc631ddccdd35a3fc99b5e", null ]

[ "# System of four equations of four variables including second powers.\n\nI've been tasked with solving the following system of equations and it seems like I am stuck:\n\n$$a-x^2=y$$$$a-y^2=z$$$$a-z^2=t$$$$a-t^2=x,$$where $a$ is a real number, for which $0\\leq a\\leq 1$. I thought the best way would be to subtract some equations from each other and the exploit $x^2-y^2=(x+y)(x-y)$. Even some estimates could be useful, since we have an estimate for $a$. However, I put this system of equations to WolframAlfa and the solutions (depending on $a$) looked very uneasy. In general, I have got very little experience in solving quadratic systems of equations like this one, could somebody please point me in the right direction? Thanks a lot!\n\n• are $$x,y,z,t$$ the variables? Nov 23 '16 at 16:47\n• Yes, $x,y,z,t$ are the variables. Nov 23 '16 at 16:51\n• Is this in fact a contest problem? If so, one would expect a cute solution, but nobody has found one yet. Nov 23 '16 at 18:45\n• Yes it is, I used tag \"contest-math\". Nov 23 '16 at 20:17\n• This system was considered by Ramanujan. I'm sure he would have won that contest blind-folded. See answer below. Nov 25 '16 at 12:29\n\nfrom the first equation we get $$a-(a-x^2)^2=z$$ then we obtain $$a-(a-(a-x^2)^2)^2=t$$ $$a-(a^2+(a-x^2)^4-2a(a-x^2)^2)=t$$ and finally we obtain $$a-(a^2+(a-x^2)^4-2a(a-x^2)^2)^2=x$$ and now you can try to get $x$ and this is what i'm get: $$a^8+a^7 \\left(-8 x^2-4\\right)+a^6 \\left(28 x^4+24 x^2+6\\right)+a^5 \\left(-56 x^6-60 x^4-24 x^2-6\\right)+a^4 \\left(70 x^8+80 x^6+36 x^4+16 x^2+5\\right)+a^3 \\left(-56 x^{10}-60 x^8-24 x^6-16 x^4-8 x^2-2\\right)+a^2 \\left(28 x^{12}+24 x^{10}+6 x^8+8 x^6+4 x^4+1\\right)+a \\left(-8 x^{14}-4 x^{12}-2 x^8-1\\right)=x \\left(-x^{15}-1\\right)$$\n\n• This doesn't seem like something from which I could get $x$ easily. Nov 23 '16 at 17:09\n• you have asked how to solve the system, i think there is only a numerical way for a value of $a$ Nov 23 '16 at 17:12\n• Yes, but there is that condition for $a$, it must simplify the solution somehow, like to prove that if $0\\leq a\\leq1$, then all other solutions but the trivial ones obtained from taking $x=y=z=t$ would not be real or something. Nov 24 '16 at 13:03\n\nBecause of the symmetry, it is natural to assume $x=y=z=t$, which gives you an easily solvable quadratic. The solutions are $\\frac 12(-1 \\pm \\sqrt{4a+1})$, which are real when $a \\ge -\\frac 14$, covering your region of interest. Another approach is to assume $x=z, y=t$, which gives $a-(a-x^2)^2=x$ and the additional two solutions $x=\\frac 12(1\\pm \\sqrt {4a-3}), y=\\frac 12(1\\mp \\sqrt {4a-3})$, which are real when $a \\ge \\frac 34$. If you don't assume the equality, you get the sixteenth degree polynomial of Dr. Sonnhard Graubner. These will reduce the degree to $12$, but there is still a long way to go.\n\n• Thanks a lot, this is something I was looking for. Honestly it haven't cross my mind I could just take $x=y=z=t$. Nov 23 '16 at 17:33\n• May be worth noting that symmetry arguments are somewhat speculative. For example the following also has the assymetric solutions $(0,1)$ and $(1,0)$: $$\\begin{cases} 1 - x^2 = y \\\\ 1 - y^2 = x \\end{cases}$$\n– dxiv\nNov 23 '16 at 17:49\n• @dxiv: absolutely, but symmetric solutions are often easier to find. They are roots of the full polynomial, so can be divided out. Sometimes one can prove that the symmetric solutions are the only ones, sometimes not. Here they seem not to be. Nov 23 '16 at 18:38\n• @pie314: note that these are some solutions, not all solutions. With great luck, they may be the ones you want. Nov 23 '16 at 18:46\n• Yes, of course. I wrote that the solutions provided by Wolfram Alpha were uneasy looking, there indeed are some other very ugly solutions. I was thinking that because of condition for $a$, these might be the onle ones. Nov 23 '16 at 19:08\n\nGiven the system, $$a-x^2=y\\\\a-y^2=z\\\\a-z^2=t\\\\a-t^2=x$$ Let $(x,\\,y,\\,z,\\,t) = (-x_1,\\,-x_2,\\,-x_3,\\,-x_4)$ and we get, $$x_1^2 = x_2+a\\\\ x_2^2 = x_3+a\\\\ x_3^2 = x_4+a\\\\ x_4^2 = x_1+a$$ This system was considered by Ramanujan, in particular for the case $a=5$, yielding the infinitely nested radical,\n\n$$x_1=\\small+\\sqrt{5+\\sqrt{5+\\sqrt{5\\color{red}-\\sqrt{5+\\sqrt{5+\\sqrt{5+\\sqrt{5\\color{red}-\\cdots}}}}}}} = \\frac{2+\\sqrt 5 +\\sqrt{15-6\\sqrt 5}}{2}=2.7472\\dots$$ and similar expressions for the three other $x_i$. However it can be solved in radicals for any $a$. See this post for more details." ]

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[ "### Journal Menu >>", null, "Energy and Power Engineering, 2010, 39-45 doi:10.4236/epe.2010.21007 Published Online February 2010 (http://www.scirp.org/journal/epe) Copyright © 2010 SciRes EPE Simulation of Electric Fields in Small Size Divertor Tokamak Plasma Edge A. H. BEKHEIT Plasma & Nuclear Fusion Department, Nuclear Research Centre, Atomic Energy Authority, Cairo, Egypt Email: amrbekheitga@yahoo.com Abstract: The fluid simulation of Small Size Divertor Tokamak (SSDT) plasma edge by the B2-SOLPS5.0 2D transport code gives the following results: First, in the vicinity of separatrix the radial electric field result is not close to the neoclassical electric f ield. Second, the shear of radial electric f ield is independent on plasma parameters. Third, switching on poloidal dr ifts (E×B and diamagnetic drifts) leads to asymmetric par-allel and poloidal fluxes from outer to inner plates and upper part of SOL for normal direction of toroidal magnetic field. Fourth, for the normal direction of toroidal magnetic, the radial electric field of SSDT is af-fected by the variation in temperature heating of plasma. Fifth, the parallel flux is directed from inner to outer plate in case of discharge without neutral beam injection (NBI). Keywords: electric field, transport codes, divertor tokamak 1. Introduction A regime of improved confinement is extremely impor-tant for the operation of a thermonuclear reactor. A tran-sition from the low confinement (L-mode) to high con-finement regime (H-mode) was discovered , and since then it has been observed on many tokamaks and stel-larators. The L-H transition may be caused by a strong radial electric field at the edge plasma and suppression of the fluctuation level by strong poloidal rotation in the EB fields [3,4]. As a result, the transport coefficients are strongly reduced in the H-mode and transport barriers with steep density and temperature gradients were formed near the separatrix or close flux surface. The key element in the transition physics is the origin of the strong radial electric field in the edge plasma. If the ra-dial electric field is sufficiently strong, the poloidal EB flow acquires a large shear, which is considered to be necessary for suppression of edge turbulence. The radial electric field in the separatrix vicinity is simulated by using the B2-SOLPS5.0 two dimensional fluid code , in which most complete system of transport equations is solved including all the important perpendicular current and EB drifts for SSDT. This code differs from the similar fluid codes (e.g. codes [5,6]), since it included detail account of parallel viscosity and perpendicular current. The equation system provides a transition to the neoclassical equation when the anomalous transport co-efficients are replaced by classical value. The simulation is performed for SSDT, where the plasma parameters in the separatrix vicinity and the Scrap Off Layer (SOL) correspond to Pfirch-Schlueter regime , thus justifying the applicability of the fluid equations. On the basis of simulation for different power of additional heating, plasma densities, toroidal rotation velocities and mag-netic field directions, it is demonstrated that the radial electric field in the separatrix vicinity is not of the order of the neoclassical field. In this paper the shear of the poloidal EB drifts is calculated. It is shown that the shear of the poloidal rotation is not function of plasma parameters. 2. Simultion Results The computation region for simulation is based on Single Null (SN) magnetic divertor and covers the SOL, core and private regions as shown in Figure 1. In computation region the coordinate which vary in the direction along flux surfaces (x-coordinate or poloidal coordinate) and the coordinate which vary in the direction across flux surfaces (y-coordinate or radial coordinate). The compu-tation mesh is divided into 24×96 units (where-1 x  96, -1  y 24) and the separatrix was at y=12. The simula-tions were performed for L-regimes of SSDT (minor radius a=0.1m, major radius R=0.3 m, I=50kA, BT=1.7 T, electron density at equatorial midplane ne =ni =n =2  1019 m-3, ion temperature Ti=31-93 eV). The anomalous values of diffusion and heat conduc-tivity coefficients were chosen: D=0.5 m2s-1, e,i = 0.7 m2s-1. The perpendicular viscosity was taken in the form =nmiD. The inner boundary flux surface, was located in the core few cm from the separetrix, the boundary conditions for the density of plasma, the av-erage toroidal momentum flux, the electron and ion", null, "A. H. BEKHEIT Copyright © 2010 SciRes EPE 40 Figure 1. Coordinate system and simulation mesh 0510 15 20 25-2-10123456Separatrix Er ( K V / m2 )y (cm ) C o de Ne o c la s s ic a l Figure 2. Radial electric field at edge of SSDT for discharge without neutral beam injection (NBI) at Ti = 31 e v, ni = 41019m-3 heat fluxes were specified . The first result of simula-tions for the radial electric field Er was compared with the neoclassic al el ect ric fiel d E (NEO) which is given by :  dxgdxBVgbdyTdhkdyndheTExiyTyiNEO ||)( )ln1ln1( (1) where bx= Bx/B (Bx is poloidal magnetic field and 22xzBBB where Bz is toroidal magnetic fields), xyzghhh is the metric coefficients, (hx = 1/ x, hy = 1/ y, hz = 1/ z) V is the parallel (toroidal) ve-locity (the coefficient kT = 2.7 corresponds to the Pfirsch-Schlueter regime).Typical radial electric field is shown in Figure 2. The comparison is showed that in the vicinity of separatrix the radial electric field is not order of the neoclassical electric field for both discharges without neutral beam injection (NBI). This fact means that, the radial transport of toroidal (parallel) momentu m is larger than parallel viscosity in parallel momentum balance equation . It is worth to mention that the av-eraged parallel velocity in Equation (1) is determined by the radial transport of parallel (toroidal) momentum i.e. by anomalous values of the diffusion and perpen-dicular viscosity coefficients. The radial profiles of parallel ve-locity for different values of averag e velocity at the inn er boundary are shown in Figure 3. Even in the Ohmic case the contribution from the last term in Equation (1) is not negligible and should take into account. The second re-sult, the radial electric field calculate by the code is negative in core, vicinity of separatrix and is positive in SOL, Figures 2. For normal direction of toroidal mag-netic field in SOL one can see that the Er  BT drifts are directed from outer plate to inner plate in SOL, and from inner plate to outer plate in the private region as shown 0510 15 20 25-50000-40000-30000-20000-1000001000020000300004000050000Core SOLseparatrixVll ( m /sec)y b(N o N BI) b(Co-injection ) b(Contra-injection) Figure 3. Parallel velocity for discharges with co and contra- injection neutral beam (NBI), for parameter of SSDT Figure 4. The arrows shows the direction of E × B drifts in the edge plasma of small size divertor tokamak Inner plate SOL Core Outer plate Private re gio n Separatrix x y", null, "A. H. BEKHEIT Copyright © 2010 SciRes EPE 41 in Figure 4. Plasma rotates in the core in the direction opposite to that in the SOL, thus creating a shear near separatrix. The third result it has been found that radial electric field is affected by the variation in temperature of plasma heating (temperature heating of plasma given by Theating = 2×Ap×, where Ap surface plasma area = 1.53 for SSDT and  is constant given by code. For ex-ample, for  = 98.1 25, Theating = 2×1.53×98.125 = 300.27 eV) for SSDT as shown in Figure 5.Therefore, the in-creasing of the temperature heating of plasma causes a change in the structure of the edge radial electric field of SSDT (the electric field is more negative in the core). The fourth result is in the simulations, the parametric independence of radial electric field and its shear s defined by Equation, s= d VE×B / dy = (RBx / B) d (Ey / RBx)/ hy dy, on plasma parameters has been studied. The independence of shear of the electric field on plasma parameters was obtained, Figures (6–8). 0510 1520 25-6-4-202468Core SOLSeparatrix Er ( K V / m )y Theat ing = 300.27 eV Theat ing = 258.19 eV Figure 5. The radial electric field at different temperatures heating of plasma for SSDT 0510 15 20 2501x1062x1063x1064x106Core SOLSeparatrixs (s-1)y ni= 2 x 1019 ni= 2.25 x 1019 ni= 2.50 x 1019 Figure 6. E  B shear at different plasma density", null, "A. H. BEKHEIT Copyright © 2010 SciRes EPE 42 0510 15 20 2501x1062x1063x1064x106CoreSOLSeparat rix s (s -1) y T i = 78.125 ev T i = 96 ev T i = 98 ev Figure 7. E  B shear at different ion temperature 051015 20 2501x1062x1063x1064x106CoreSOLSep ara trix s (s-1) y < V ll >= 0 < V ll> = 5 K m/s < V ll> = -5 K m/s Figure 8. E  B shear at different average parallel velocity To obtain a scaling for the L-H transition threshold it is necessary to specify the critical shear when the transi-tion starts. For the critical shear we chose the value of s independently of the regime due to limited knowledge of the turbulent processes. This value must be gives best fitting to the experiment. To reach the chosen critical shear it is necessary to increase the heating power pro-portionally to the local density and the toroidal mag- netic field. This result is explained by the neocla- ssical nature of the simulated radial electric field. Indeed, the linear dependence of the threshold heating power on the local density corresponds to the constant critical value of the ion temperature, which determines the criti-cal shear. In the vicinity of separatrix the radial electric field of SSDT, is not of order of neoclassical field. Therefore, for SSDT it’s can’t reach to critical shear to start L-H transition. The deviation of the electric field from the neoclassical value is relatively pronounced", null, "A. H. BEKHEIT Copyright © 2010 SciRes EPE 43 near the separatrix in the outer midplane, as shown in Figure 2. The reason for this difference is connected with the contribution of anomalous radial transport of the toroidal (parallel) momentum to the parallel mo-mentum balance equation . As result, for normal di-rection of toroidal magnetic field, the parallel fluxes inside and outside the separatrix are coupled Figure (9). The fifth result of simulation is studding the influence of drifts (E × B and diamagnetic drifts) on the fluxes in SOL. This could be understood from the analysis of parallel and poloidal fluxes in SOL. The structure of the parallel fluxes in the SOL is governed by the combina-tion of three factors. The first is the Pfirch – Schlueter (PS) parallel fluxes close the vertical ion  B dr ift, and their direction depends on the toroidal magnetic field. The second is the contribution from the radial electric field to the PS fluxes has the same sign as the contribu-tion from  B drift in the SOL since the radial electric 0 102030405060708090100-80000-70000-60000-50000-40000-30000-20000-100000100002000030000400005000060000700008000090000100000110000outer p lateInner plate Vll ( m / sec )x Vll in Co r e Vll in SOL Figure 9. Parallel velocity inside and outside the separatrix 0 102030405060708090-3000-2000-1000010002000300040005000Stagnation pointsTopouter plateinner plateVp ( m / se c )X without drifts w ith d rift s Figure 10. Poloidal velocity in SOL with and without drifts", null, "A. H. BEKHEIT Copyright © 2010 SciRes EPE 44 0 102030405060708090100-160000-140000-120000-100000-80000-60000-40000-20000020000400006000080000100000120000140000160000Topstagnation pointstagnation pointouter plateinner plateVll ( m / sec )x w ith d r ifts w ith o u t d r ifts Figure 11. Parallel velocity with and without drifts 051015 2025-20000-18000-16000-14000-12000-10000-8000-6000-4000-2000Core SOLSeparatrixVll ( m / sec )y ( cm ) Figure 12. Parallel flux in the edge of SSDT field is positive in the SOL as shown in Figure 2 (inside separatrix, where the radial electric field is negative, the contribution to PS fluxes  B drifts and EB drift compensate each other in accordance with neoclassical theory). The third contribution arises from the poloidal fluxes that are responsible for the particle transport to the plates. Those poloidal fluxes are closed the radial diffu-sive particle fluxes. In the absence of the poloidal EB and  B drifts these fluxes coincide with the projection of the parallel fluxes. On the one hand, the outer plate is larger than the inner plate and the integral poloidal flux to this plate should be larger. On the other hand, since the plasma in the vicinity of the inner plate is colder and denser, the particle flux per square meter to this plate that is proportional to n (T/mi)1/2 is larger than particle flux to outer plate (the pressure is almost the same at the plates). Switching on EB drift leads to asymmetric in parallel and poloidal fluxes from outer to inner plates and upper part of SOL as shown as in Fig ures (10,11). Also switch-ing on the poloidal dr ifts leads to decrease of the parallel velocity, because the poloidal projection of the parallel velocity not compensates poloid al EB drifts, so that the poloidal rotation changes. Moreover, the position stagna-tion point for the poloidal did not change much which is", null, "A. H. BEKHEIT Copyright © 2010 SciRes EPE 45 located somewhere near the upper part of the SOL, as shown in Figure 11. In contrast, the position of the stag-nation point for parallel flux may be significantly differ-ent Figure 11. For normal B the stagnation point is strongly shifted toward s the outer plate. This fact is con-sistent with the observations with simulations . Fi-nally the parallel flow pattern is the result of all three factors. For normal B parallel flux in SOL is negative as sh own in Fig ure 12. Therefo re, for normal B parallel flux is directed from outer to inner plate. 3. Conclusions In conclusion, the performed simulations for SSDT demonstrate the following results: (First) In the vicinity of separatrix the radial electric field was not close to the neoclassical electric field. (Second) the independence of the shear of radial electric field on plasma parameters. (Third) For normal direction of torodial magnetic field, the radial electric field of SSDT was affected by the variation in temperature heating of plasma. (Fourth) Switching on poloidal d rifts leads to asymmetric parallel and poloidal fluxes from outer to inner plates and upper part of SOL for normal direction of toroidal magnetic field. Also switching on the poloidal drifts leads to de-crease of the parallel velocity, because the poloidal pro-jection of the parallel velocity not compensates poloidal EB drifts, so that the poloidal rotation changes. (Fifth) The parallel flux is directed from outer to inner plate. (Sixth) The E×B drift are directed from outer plate to inner plate in SOL, and from inner plate to outer plate in the private region. REFERENCES R. Schneider, V. Rozhansky, and P. Xantopoulos, Contri- bution Plasma Physics, No. 40, pp. 4213, 2004. F. Wagner, Physical Review Letters, No. 49, pp. 1408, 1982. H. Biglari, P. H. Diamond, and P. W. Terry, Physics of Fluids, No. B21, pp. 1, 1990. K. H. Burrell, Physics of Plasma, No. 4, pp. 1499, 1997. G. J. Radford, Contribution Plasma Physics, No. 36, pp. 187, 1996. T. D. Rognlien, D. D. Ryutov, N. Mattor, and G. D. Porter, Physics of Plasmas, No. 6, pp. 1851, 1999. S. P. Hirshman and D. J. Sigmar, Nuclear Fusion, No. 21, pp. 1079, 1981. A. H. Bekheit, Egyptian Journal of Fusion energy, Vol. 28, No. 4, pp. 338–345, 2008. T. S. Hahm and K. H. Burrell, Phyics of Plasma, No. 2, pp. 1648, 1994. S. K. Erents, G. Corrigan, and J. Spence, Journal of Nu-clear Materials, No. 290–293, pp. 518, 2001." ]

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[ "i ≤ {\\displaystyle K} above. N total draws) from a population of size k! Now, using Equation (1), A random variable distributed hypergeometrically with parameters Spiegel, M. R. Theory and Problems of Probability and Statistics. total draws. n [K1] is the expected value [K2] the number of crashes expected to occur in a week. Substituting the values obtained in ( ∗ ∗) and ( ∗ ∗ ∗) for the terms in the formula ( ∗) for the expectation of X, we obtain. In probability theory and statistics, the hypergeometric distribution is a discrete probability distribution that describes the probability of k successes (random draws for which the object drawn has a specified feature) in n draws, without replacement, from a finite population of size N that contains exactly K objects with that feature, wherein each draw is either a success or a failure. / {\\displaystyle i^{\\text{th}}} 00 1 nn xx aNa xnx fx N n == ⎛⎞⎛ ⎞− ⎜⎟⎜ ⎟ ⎝⎠⎝ ⎠− == ⎛⎞ ⎜⎟ ⎝⎠ ∑∑. and {\\displaystyle n} {\\displaystyle X\\sim \\operatorname {Hypergeometric} (N,K,n)} The player would like to know the probability of one of the next 2 cards to be shown being a club to complete the flush. (Note that the probability calculated in this example assumes no information is known about the cards in the other players' hands; however, experienced poker players may consider how the other players place their bets (check, call, raise, or fold) in considering the probability for each scenario. Hypergeometric Distribution Examples: For the same experiment (without replacement and totally 52 cards), if we let X = the number of ’s in the rst20draws, then X is still a hypergeometric random variable, but with n = 20, M = 13 and N = 52. Hypergeometric {\\displaystyle n} . k , K Reciprocally, the p-value of a two-sided Fisher's exact test can be calculated as the sum of two appropriate hypergeometric tests (for more information see). {\\displaystyle p=K/N} balls and colouring them red first. Properties of the Hypergeometric Distribution There are several important values that give information about a particular probability distribution. CRC Standard Mathematical Tables, 28th ed. {\\displaystyle n} Hypergeometric Distribution The hypergeometric distribution is a discrete probability distribution that describes the number of successes in a sequence of n draws from a finite population without replacement. This has the same relationship to the multinomial distribution that the hypergeometric distribution has to the binomial distribution—the multinomial distribution is the \"with-replacement\" distribution and the multivariate hypergeometric is the \"without-replacement\" distribution. = , In the first round, 2 The Binomial Distribution as a Limit of Hypergeometric Distributions The connection between hypergeometric and binomial distributions is to the level of the distribution itself, not only their moments. ) k N . 2 The total number of green balls in the sample is X = X 1 + + X n. The X i’s are identically distributed, but dependent. N The mean or expected value of the hypergeometric random variable is given by 1 00 nn x xx N aNa xxfx x nxnx μ − == ⎛⎞ ⎛⎞⎛ ⎞− == =⎜⎟ ⎜⎟⎜ ⎟ ⎝⎠ ⎝⎠⎝ ⎠− ∑∑. K N The th selection has an equal likelihood of 1. N ( n - 1 k - 1). = N K 41-45, 1968. the hypergeometric distribution should be applied. ∼ The three discrete distributions we discuss in this article are the binomial distribution, hypergeometric distribution, and poisson distribution. and − K selection and ways for a \"bad\" Collection of teaching and learning tools built by Wolfram education experts: dynamic textbook, lesson plans, widgets, interactive Demonstrations, and more. also follows from the symmetry of the problem. The probability that one of the next two cards turned is a club can be calculated using hypergeometric with − − The sampling rates are usually defined by law, not statistical design, so for a legally defined sample size n, what is the probability of missing a problem which is present in K precincts, such as a hack or bug? th N {\\displaystyle k=2,n=2,K=9} The test is often used to identify which sub-populations are over- or under-represented in a sample. The Hypergeometric Distribution Proposition If X is the number of S’s in a completely random sample of size n drawn from a population consisting of M S’s and (N –M) F’s, then the probability distribution of X, called the hypergeometric distribution, is given by for x, an integer, satisfying max (0, n –N + M) x min (n, M). The exponential distribution is the continuous analogue of the geometric distribution. 2 https://mathworld.wolfram.com/HypergeometricDistribution.html. 2 c ( {\\displaystyle n} True . 3.5 Expected value of hypergeometric distribution Let p = K=N be the fraction of balls in the urn that are green. Cumulative distribution function (CDF) of the hypergeometric distribution in Excel =IF (k>=expected,1-HYPGEOM.DIST (k-1,s,M,N,TRUE),HYPGEOM.DIST (k,s,M,N,TRUE)) For example, if a problem is present in 5 of 100 precincts, a 3% sample has 86% probability that k = 0 so the problem would not be noticed, and only 14% probability of the problem appearing in the sample (positive k): The sample would need 45 precincts in order to have probability under 5% that k = 0 in the sample, and thus have probability over 95% of finding the problem: In hold'em poker players make the best hand they can combining the two cards in their hand with the 5 cards (community cards) eventually turned up on the table. However, for of these, so there Strictly speaking, the approach to calculating success probabilities outlined here is accurate in a scenario where there is just one player at the table; in a multiplayer game this probability might be adjusted somewhat based on the betting play of the opponents.). = The following conditions characterize the hypergeometric distribution: A random variable and its expected value (mean), variance and standard deviation are, = E(Y) = nr N, ˙2 = V(Y) = n r N N −r N N −n N − 1 , ˙ = p V(Y). In a test for over-representation of successes in the sample, the hypergeometric p-value is calculated as the probability of randomly drawing 1, 3rd ed. Explore anything with the first computational knowledge engine. Bugs are often obscure, and a hacker can minimize detection by affecting only a few precincts, which will still affect close elections, so a plausible scenario is for K to be on the order of 5% of N. Audits typically cover 1% to 10% of precincts (often 3%), so they have a high chance of missing a problem. N X successes (random draws for which the object drawn has a specified feature) in some random draws for the object drawn that has some specified feature) in n no of draws, without any replacement, from a given population size N which includes accurately K objects having that feature, where the draw may succeed or may fail. K Then the colored marbles are put back. If the variable N describes the number of all marbles in the urn (see contingency table below) and K describes the number of green marbles, then N âˆ’ K corresponds to the number of red marbles. b In order The test based on the hypergeometric distribution (hypergeometric test) is identical to the corresponding one-tailed version of Fisher's exact test. The distribution \\eqref{*} is called a negative hypergeometric distribution by analogy with the negative binomial distribution, which arises in the same way for sampling with replacement. {\\textstyle X\\sim \\operatorname {Hypergeometric} (N,K,n)} − N p Define drawing a green marble as a success and drawing a red marble as a failure (analogous to the binomial distribution). + expression. − , draws, without replacement, from a finite population of size The symmetry in This is the probability that k = 0. , Intuitively we would expect it to be even more unlikely that all 5 green marbles will be among the 10 drawn. Hints help you try the next step on your own. 0 From MathWorld--A Wolfram Web Resource. − < ) This is a little digression from Chapter 5 of Using R for Introductory Statistics that led me to the hypergeometric distribution. K Hypergeometric For a population of N objects containing m defective components, it follows the remaining N − m components are non-defective. Beyer, W. H. CRC Standard Mathematical Tables, 28th ed. N = The classical application of the hypergeometric distribution is sampling without replacement. n {\\displaystyle \\left. and {\\displaystyle k} of obtaining correct balls are Approximation to a Hypergeometric Random Variable. Now we can start with the definition of the expected value: E[X]= n ∑ x=0 x(K x) ( M−K n−x) (M n). summation over . The hypergeometric distribution is implemented in the Wolfram Language as HypergeometricDistribution [ N , n, m + n ]. D min If the variable N describes the number of all marbles in the urn (see contingency table below) and K describes the number of green marbles, then N − K corresponds to the number of red marbles. N {\\displaystyle K} k or more successes from the population in , {\\displaystyle X} The actual points you gain from the game is lower than the expected value. 47 EXAMPLE 3 Using the Hypergeometric Probability Distribution Problem: The hypergeometric probability distribution is used in acceptance sam-pling. The random variable X = the number of items from the group of interest. n These are the conditions of a hypergeometric distribution. 9 {\\displaystyle N} Φ 1, 3rd ed. {\\displaystyle N=\\sum _{i=1}^{c}K_{i}} Join the initiative for modernizing math education. for is. Take samples and let equal 1 if selection Think of an urn with two colors of marbles, red and green. K i 9 What is the probability that exactly 4 of the 10 are green? 1 b. will always be one of the values x can take on, although it may not be the highest probability value for the random variable. In the second round, 0 Male or Female ? N = {\\displaystyle k=0,n=2,K=9} , Practice online or make a printable study sheet. ) (about 3.33%), The probability that neither of the next two cards turned are clubs can be calculated using hypergeometric with k {\\displaystyle K} ⁡ n The hypergeometric distribution, the probability of y successes when sampling without15replacement n items from a population with r successes and N − r fail- ures, is p(y) = P (Y = y) = r y N −r n− y N n , 0 ≤ y ≤ r, 0 ≤ n− y ≤ N − r, and its expected value (mean), variance and standard deviation are, µ = E(Y) = nr N, σ2= V(Y) = n r N N −r N N −n N − 1 , σ = p V(Y). is the total number of marbles. is then. The hypergeometric test uses the hypergeometric distribution to measure the statistical significance of having drawn a sample consisting of a specific number of 1 In probability theory and statistics, the hypergeometric distribution is a discrete probability distribution that describes the probability of and the kurtosis excess is given by a complicated is written neutral marbles are drawn from an urn without replacement and coloured green. ( k - 1)! Suppose that a machine shop orders 500 bolts from a supplier.To determine whether to accept the shipment of bolts,the manager of … N = N . N For i = 1,..., n, let X i = 1 if the ith ball is green; 0 otherwise. ∑ If there are Ki marbles of color i in the urn and you take n marbles at random without replacement, then the number of marbles of each color in the sample (k1, k2,..., kc) has the multivariate hypergeometric distribution. . 2 where Since through the = ) K There are 12 crashes in 30 days, so the number of crashes per day is 12/30=0.4. Let are a total of terms Make the change of variable j = k − 1. In contrast, the binomial distribution describes the probability of This is an ex ante probability—that is, it is based on not knowing the results of the previous draws. The problem of finding the probability of such a picking problem is sometimes called the \"urn problem,\" since it asks for the probability that out of balls drawn are n K Let x be a random variable whose value is the number of successes in the sample. k By the lemma, or otherwise, ( ∗ ∗ ∗) k ( r k) = r ( r − 1 k − 1). . {\\displaystyle \\max(0,n+K-N)\\leq k\\leq \\min(K,n)} follows the hypergeometric distribution if its probability mass function (pmf) is given by. n Feller, W. \"The Hypergeometric Series.\" But since and are random Bernoulli variables (each 0 or 1), their product n {\\displaystyle p=K/N} is also a Bernoulli variable. The properties of this distribution are given in the adjacent table, where c is the number of different colors and The mean of a probability distribution is called its expected value. ( n {\\displaystyle n} We find P(x) = (4C3)(48C10) 52C13 ≈ 0.0412 . 2 K because green marbles are bigger/easier to grasp than red marbles) then, This page was last edited on 2 December 2020, at 05:06. ) To improve this 'Hypergeometric distribution Calculator', please fill in questionnaire. Knowledge-based programming for everyone. N The mean of a binomial distribution … ( N ( K N k 6 which essentially follows from Vandermonde's identity from combinatorics. The Hypergeometric Distribution Basic Theory Dichotomous Populations. The following table describes four distributions related to the number of successes in a sequence of draws: The model of an urn with green and red marbles can be extended to the case where there are more than two colors of marbles. The pmf is positive when n n X A hypergeometric distribution is a probability distribution. Then for − {\\displaystyle N} a Let Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more. ( − K N [ For example, a marketing group could use the test to understand their customer base by testing a set of known customers for over-representation of various demographic subgroups (e.g., women, people under 30). stems from the fact that the two rounds are independent, and one could have started by drawing n K Hypergeometric: televisions. In this example, X is the random variable whose outcome is k, the number of green marbles actually drawn in the experiment. n has a geometric distribution taking values in the set {0, 1, 2,...}, with expected value r / (1 − r). ⋅ Note that although we are looking at success/failure, the data are not accurately modeled by the binomial distribution, because the probability of success on each trial is not the same, as the size of the remaining population changes as we remove each marble. 113-114, Exercise 3.7 (The Hypergeometric Probability Distribution) 1. Then, the number of marbles with both colors on them (that is, the number of marbles that have been drawn twice) has the hypergeometric distribution. p {\\displaystyle X\\sim \\operatorname {Hypergeometric} (K,N,n)} = There are 5 cards showing (2 in the hand and 3 on the table) so there are ) {\\displaystyle 0" ]

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[ "# #cd5460 Color Information\n\nIn a RGB color space, hex #cd5460 is composed of 80.4% red, 32.9% green and 37.6% blue. Whereas in a CMYK color space, it is composed of 0% cyan, 59% magenta, 53.2% yellow and 19.6% black. It has a hue angle of 354 degrees, a saturation of 54.8% and a lightness of 56.7%. #cd5460 color hex could be obtained by blending #ffa8c0 with #9b0000. Closest websafe color is: #cc6666.\n\n• R 80\n• G 33\n• B 38\nRGB color chart\n• C 0\n• M 59\n• Y 53\n• K 20\nCMYK color chart\n\n#cd5460 color description : Moderate red.\n\n# #cd5460 Color Conversion\n\nThe hexadecimal color #cd5460 has RGB values of R:205, G:84, B:96 and CMYK values of C:0, M:0.59, Y:0.53, K:0.2. Its decimal value is 13456480.\n\nHex triplet RGB Decimal cd5460 `#cd5460` 205, 84, 96 `rgb(205,84,96)` 80.4, 32.9, 37.6 `rgb(80.4%,32.9%,37.6%)` 0, 59, 53, 20 354°, 54.8, 56.7 `hsl(354,54.8%,56.7%)` 354°, 59, 80.4 cc6666 `#cc6666`\nCIE-LAB 52.026, 48.947, 17.917 30.459, 20.167, 13.354 0.476, 0.315, 20.167 52.026, 52.123, 20.105 52.026, 87.095, 12.331 44.908, 42.481, 13.804 11001101, 01010100, 01100000\n\n# Color Schemes with #cd5460\n\n• #cd5460\n``#cd5460` `rgb(205,84,96)``\n• #54cdc1\n``#54cdc1` `rgb(84,205,193)``\nComplementary Color\n• #cd549d\n``#cd549d` `rgb(205,84,157)``\n• #cd5460\n``#cd5460` `rgb(205,84,96)``\n• #cd8554\n``#cd8554` `rgb(205,133,84)``\nAnalogous Color\n• #549dcd\n``#549dcd` `rgb(84,157,205)``\n• #cd5460\n``#cd5460` `rgb(205,84,96)``\n• #54cd85\n``#54cd85` `rgb(84,205,133)``\nSplit Complementary Color\n• #5460cd\n``#5460cd` `rgb(84,96,205)``\n• #cd5460\n``#cd5460` `rgb(205,84,96)``\n• #60cd54\n``#60cd54` `rgb(96,205,84)``\n• #c154cd\n``#c154cd` `rgb(193,84,205)``\n• #cd5460\n``#cd5460` `rgb(205,84,96)``\n• #60cd54\n``#60cd54` `rgb(96,205,84)``\n• #54cdc1\n``#54cdc1` `rgb(84,205,193)``\n• #a4303c\n``#a4303c` `rgb(164,48,60)``\n• #b83643\n``#b83643` `rgb(184,54,67)``\n• #c7404e\n``#c7404e` `rgb(199,64,78)``\n• #cd5460\n``#cd5460` `rgb(205,84,96)``\n• #d36872\n``#d36872` `rgb(211,104,114)``\n• #d97b85\n``#d97b85` `rgb(217,123,133)``\n• #de8f97\n``#de8f97` `rgb(222,143,151)``\nMonochromatic Color\n\n# Alternatives to #cd5460\n\nBelow, you can see some colors close to #cd5460. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #cd547e\n``#cd547e` `rgb(205,84,126)``\n• #cd5474\n``#cd5474` `rgb(205,84,116)``\n• #cd546a\n``#cd546a` `rgb(205,84,106)``\n• #cd5460\n``#cd5460` `rgb(205,84,96)``\n• #cd5456\n``#cd5456` `rgb(205,84,86)``\n• #cd5c54\n``#cd5c54` `rgb(205,92,84)``\n• #cd6654\n``#cd6654` `rgb(205,102,84)``\nSimilar Colors\n\n# #cd5460 Preview\n\nThis text has a font color of #cd5460.\n\n``<span style=\"color:#cd5460;\">Text here</span>``\n#cd5460 background color\n\nThis paragraph has a background color of #cd5460.\n\n``<p style=\"background-color:#cd5460;\">Content here</p>``\n#cd5460 border color\n\nThis element has a border color of #cd5460.\n\n``<div style=\"border:1px solid #cd5460;\">Content here</div>``\nCSS codes\n``.text {color:#cd5460;}``\n``.background {background-color:#cd5460;}``\n``.border {border:1px solid #cd5460;}``\n\n# Shades and Tints of #cd5460\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, #0b0304 is the darkest color, while #fefbfb is the lightest one.\n\n• #0b0304\n``#0b0304` `rgb(11,3,4)``\n• #1a080a\n``#1a080a` `rgb(26,8,10)``\n• #290c0f\n``#290c0f` `rgb(41,12,15)``\n• #391115\n``#391115` `rgb(57,17,21)``\n• #48151a\n``#48151a` `rgb(72,21,26)``\n• #571920\n``#571920` `rgb(87,25,32)``\n• #661e25\n``#661e25` `rgb(102,30,37)``\n• #75222b\n``#75222b` `rgb(117,34,43)``\n• #852730\n``#852730` `rgb(133,39,48)``\n• #942b36\n``#942b36` `rgb(148,43,54)``\n• #a3303b\n``#a3303b` `rgb(163,48,59)``\n• #b23441\n``#b23441` `rgb(178,52,65)``\n• #c13946\n``#c13946` `rgb(193,57,70)``\n• #c94552\n``#c94552` `rgb(201,69,82)``\n• #cd5460\n``#cd5460` `rgb(205,84,96)``\n• #d1636e\n``#d1636e` `rgb(209,99,110)``\n• #d6727c\n``#d6727c` `rgb(214,114,124)``\n• #da828a\n``#da828a` `rgb(218,130,138)``\n• #df9198\n``#df9198` `rgb(223,145,152)``\n• #e3a0a7\n``#e3a0a7` `rgb(227,160,167)``\n• #e8afb5\n``#e8afb5` `rgb(232,175,181)``\n• #ecbec3\n``#ecbec3` `rgb(236,190,195)``\n• #f1cdd1\n``#f1cdd1` `rgb(241,205,209)``\n• #f5dddf\n``#f5dddf` `rgb(245,221,223)``\n• #f9eced\n``#f9eced` `rgb(249,236,237)``\n• #fefbfb\n``#fefbfb` `rgb(254,251,251)``\nTint Color Variation\n\n# Tones of #cd5460\n\nA tone is produced by adding gray to any pure hue. In this case, #929090 is the less saturated color, while #f82a3e is the most saturated one.\n\n• #929090\n``#929090` `rgb(146,144,144)``\n• #9a8789\n``#9a8789` `rgb(154,135,137)``\n• #a37f82\n``#a37f82` `rgb(163,127,130)``\n• #ab767b\n``#ab767b` `rgb(171,118,123)``\n• #b46e74\n``#b46e74` `rgb(180,110,116)``\n• #bc656e\n``#bc656e` `rgb(188,101,110)``\n• #c55d67\n``#c55d67` `rgb(197,93,103)``\n• #cd5460\n``#cd5460` `rgb(205,84,96)``\n• #d64c59\n``#d64c59` `rgb(214,76,89)``\n• #de4352\n``#de4352` `rgb(222,67,82)``\n• #e73b4c\n``#e73b4c` `rgb(231,59,76)``\n• #ef3245\n``#ef3245` `rgb(239,50,69)``\n• #f82a3e\n``#f82a3e` `rgb(248,42,62)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #cd5460 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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[ "# 2.5. Glossary\n\nAccuracy (error rate)\n\nThe rate of correct (incorrect) predictions made by the model over a data set (cf. coverage). Accuracy is usually estimated by using an independent test set that was not used at any time during the learning process. More complex accuracy estimation techniques, such as cross-validation and the bootstrap, are commonly used, especially with data sets containing a small number of instances.\n\nAssociation learning\n\nTechniques that find conjunctive implication rules of the form \"$$X$$ and $$Y$$ implies $$A$$ and $$B$$ \" (associations) that satisfy given criteria. The conventional association algorithms are sound and complete methods for finding all associations that satisfy criteria for minimum support (at least a specified fraction of the instances must satisfy both sides of the rule) and minimum confidence (at least a specified fraction of instances satisfying the left hand side, or antecedent, must satisfy the right hand side, or consequent).\n\nAttribute (field, variable, feature)\n\nA quantity describing an instance. An attribute has a domain defined by the attribute type, which denotes the values that can be taken by an attribute. The following domain types are common:\n\nCategorical\n\nA finite number of discrete values. The type nominal denotes that there is no ordering between the values, such as last names and colors. The type ordinal denotes that there is an ordering, such as in an attribute taking on the values low, medium, or high.\n\nContinuous (quantitative)\n\nCommonly, subset of real numbers, where there is a measurable difference between the possible values. Integers are usually treated as continuous in practical problems.\n\nA feature is the specification of an attribute and its value. For example, color is an attribute. \"Color is blue\" is a feature of an example. Many transformations to the attribute set leave the feature set unchanged (for example, regrouping attribute values or transforming multi-valued attributes to binary attributes). Some authors use feature as a synonym for attribute (e.g., in feature-subset selection).\n\nClassification\n\nProcess related to categorization, the process in which ideas and objects are recognized, differentiated, and understood.\n\nClassifier\n\nA mapping from unlabeled instances to (discrete) classes. Classifiers have a form (e.g., decision tree) plus an interpretation procedure (including how to handle unknowns, etc.). Some classifiers also provide probability estimates (scores), which can be threshold to yield a discrete class decision thereby taking into account a utility function.\n\nCluster\n\nGroup of loosely coupled objects that belongs to the same category\n\nConfusion matrix\n\nA matrix showing the predicted and actual classifications. A confusion matrix is of size $$LxL$$ , where L is the number of different label values. The following confusion matrix is for $$L=2$$ :\n\nactual / predicted\n\nnegative\n\npositive\n\nNegative\n\na\n\nb\n\nPositive\n\nc\n\nd\n\nThe following terms are defined for a two by two confusion matrix:\n\nAccuracy\n\n$$(a+d) / (a+b+c+d)$$\n\nTrue positive rate (Recall, Sensitivity)\n\n$$d / (c+d)$$\n\nTrue negative rate (Specificity)\n\n$$a / (a+b)$$\n\nPrecision\n\n$$d / (b+d)$$\n\nFalse positive rate\n\n$$b / (a+b)$$\n\nFalse negative rate\n\n$$c / (c+d)$$\n\nCoverage\n\nThe proportion of a data set for which a classifier makes a prediction. If a classifier does not classify all the instances, it may be important to know its performance on the set of cases for which it is \"confident\" enough to make a prediction.\n\nCost (utility/loss/payoff)\n\nA measurement of the cost to the performance task (and/or benefit) of making a prediction Y' when the actual label is y. The use of accuracy to evaluate a model assumes uniform costs of errors and uniform benefits of correct classifications.\n\nCross-validation\n\nA method for estimating the accuracy (or error) of an inducer by dividing the data into k mutually exclusive subsets (the \"folds\") of approximately equal size. The inducer is trained and tested $$k$$ times. Each time it is trained on the data set minus a fold and tested on that fold. The accuracy estimate is the average accuracy for the k folds.\n\nData cleaning/cleansing\n\nThe process of improving the quality of the data by modifying its form or content, for example by removing or correcting data values that are incorrect. This step usually precedes the machine learning step, although the knowledge discovery process may indicate that further cleaning is desired and may suggest ways to improve the quality of the data. For example, learning that the pattern Wife implies Female from the census sample at UCI has a few exceptions may indicate a quality problem.\n\nData mining\n\nThe term data mining is somewhat overloaded. It sometimes refers to the whole process of knowledge discovery and sometimes to the specific machine learning phase.\n\nData set\n\nA schema and a set of instances matching the schema. Generally, no ordering on instances is assumed. Most machine learning work uses a single fixed-format table.\n\nDecision Boundary\n\nIn a statistical-classification problem with two classes, a decision boundary or decision surface is a hyper-surface that partitions the underlying vector space into two sets, one for each class. The classifier will classify all the points on one side of the decision boundary as belonging to one class and all those on the other side as belonging to the other class.\n\nA decision boundary is the region of a problem space in which the output label of a classifier is ambiguous.\n\nDimension\n\nAn attribute or several attributes that together describe a property. For example, a geographical dimension might consist of three attributes: country, state, city. A time dimension might include 5 attributes: year, month, day, hour, minute.\n\nDiscriminative model\n\nClass of models used in machine learning for modeling the dependence of unobserved (target) variables $$y$$ on observed variables $$x$$. Within a probabilistic framework, this is done by modeling the conditional probability distribution $$P(y|x)$$, which can be used for predicting $$y$$ from $$x$$.\n\nDiscriminative models, as opposed to generative models, do not allow one to generate samples from the joint distribution of observed and target variables. However, for tasks such as classification and regression that do not require the joint distribution, discriminative models can yield superior performance (in part because they have fewer variables to compute). On the other hand, generative models are typically more flexible than discriminative models in expressing dependencies in complex learning tasks. In addition, most discriminative models are inherently supervised and cannot easily support unsupervised learning. Application-specific details ultimately dictate the suitability of selecting a discriminative versus generative model.\n\nError rate\n\nSee Accuracy.\n\nExample\n\nSee Instance.\n\nFeature\n\nSee Attribute.\n\nFeature vector (record, tuple)\n\nA list of features describing an instance.\n\nField\n\nSee Attribute.\n\nGenerative Model\n\nIn statistical classification, including machine learning, two main approaches are called the generative approach and the discriminative approach. These compute classifiers by different approaches, differing in the degree of statistical modelling. Terminology is inconsistent,[a] but three major types can be distinguished, following (Jebara 2004):\n\n• Given an observable variable $$X$$ and a target variable $$Y$$, a generative model is a statistical model of the joint probability distribution on $$X × Y$$, $$P(X,Y)$$,\n\n• A discriminative model is a model of the conditional probability of the target $$Y$$, given an observation $$x$$, symbolically, $$P(Y|X=x)$$,\n\n• Classifiers computed without using a probability model are also referred to loosely as \"discriminative\".\n\ni.i.d. sample\n\nA set of independent and identically distributed instances.\n\nInducer / induction algorithm\n\nAn algorithm that takes as input specific instances and produces a model that generalizes beyond these instances.\n\nInstance (example, case, record)\n\nA single object of the world from which a model will be learned, or on which a model will be used (e.g., for prediction). In most machine learning work, instances are described by feature vectors; some work uses more complex representations (e.g., containing relations between instances or between parts of instances).\n\nKnowledge discovery\n\nThe non-trivial process of identifying valid, novel, potentially useful, and ultimately understandable patterns in data. This is the definition used in \"Advances in Knowledge Discovery and Data Mining\", 1996, by Fayyad, Piatetsky-Shapiro, and Smyth.\n\nLearning Algorithm\n\nProcedure that creates classifiers. Finds patterns in training data.\n\nLoss\n\nSee Cost.\n\nMachine learning\n\nIn Knowledge Discovery, machine learning is most commonly used to mean the application of induction algorithms, which is one step in the knowledge discovery process. This is similar to the definition of empirical learning or inductive learning in Readings in Machine Learning by Shavlik and Dietterich. Note that in their definition, training examples are \"externally supplied\", whereas here they are assumed to be supplied by a previous stage of the knowledge discovery process. Machine Learning is the field of scientific study that concentrates on induction algorithms and on other algorithms that can be said to \"learn\".\n\nMissing value\n\nThe value for an attribute is not known or does not exist. There are several possible reasons for a value to be missing, such as: it was not measured; there was an instrument malfunction; the attribute does not apply, or the attribute's value cannot be known. Some algorithms have problems dealing with missing values.\n\nModel\nEstimator\n\nA structure and corresponding interpretation that summarizes or partially summarizes a set of data, for description or prediction. Most inductive algorithms generate models that can then be used as classifiers, as regressors, as patterns for human consumption, and/or as input to subsequent stages of the KDD process.\n\nModel deployment\n\nThe use of a learned model. Model deployment usually denotes applying the model to real data.\n\nObservation\n\nOne row in features and labels table. For example Iris dataset has 150 observations.\n\nOut-of-sample data\n\nData that is not in Observation. In most cases that would be the data to predict.\n\nOLAP (MOLAP, ROLAP)\n\nOn-Line Analytical Processing. Usually synonymous with MOLAP (multi-dimensional OLAP). OLAP engines facilitate the exploration of data along several (predetermined) dimensions. OLAP commonly uses intermediate data structures to store pre-calculated results on multidimensional data, allowing fast computations. ROLAP (relational OLAP) refers to performing OLAP using relational databases.\n\nOverfitting\n\nModels that overfit learns to recognize noise from the signal, than the data.", null, "Figure 2.27. Black line represents the decision boundary and represents the signal. Green line represents overfitted model which learned the noise.\nPreprocessing\n\nIs the module used to do some cleaning/scaling of data prior to machine learning.\n\nRecord\n\nSee Feature vector.\n\nRegression\n\nIs a form of supervised machine learning, which is where the scientist teaches the machine by showing it features and then showing it was the correct answer is, over and over, to teach the machine. Once the machine is taught, the scientist will usually \"test\" the machine on some unseen data, where the scientist still knows what the correct answer is, but the machine doesn't. The machine's answers are compared to the known answers, and the machine's accuracy can be measured. If the accuracy is high enough, the scientist may consider actually employing the algorithm in the real world.\n\nRegressor\n\nA mapping from unlabeled instances to a value within a predefined metric space (e.g., a continuous range).\n\nResubstitution accuracy (error/loss)\n\nThe accuracy (error/loss) made by the model on the training data.\n\nSchema\n\nA description of a data set's attributes and their properties.\n\nSensitivity\n\nTrue positive rate (see Confusion matrix).\n\nSpecificity\n\nTrue negative rate (see Confusion matrix).\n\nSupervised learning\n\nTechniques used to learn the relationship between independent attributes and a designated dependent attribute (the label). Most induction algorithms fall into the supervised learning category.\n\nTuple\n\nSee Feature vector.\n\nUnsupervised learning\n\nLearning techniques that group instances without a pre-specified dependent attribute. Clustering algorithms are usually unsupervised.\n\nUtility\n\nSee Cost." ]

[ null, "https://db.astrotech.io/_images/glossary-overfitting.png", null ]

[ "## Introduction\n\nCavity optomechanical systems, coupling light field with mechanical oscillators via radiation pressure force, have generated considerable interest in recent years, especially for systems in the quantum regime. They have found applications in precise metrology1,2,3,4,5,6,7,8, nonreciprocal coupling9,10, preparation of mechanical quantum states11, entanglement between mechanical and optical systems12, and may also be used to probe the fundamental physics of macroscopic quantum mechanics13,14 and test models of quantum gravity15,16. A prerequisite to operate in the regime of quantum optomechanics is a high mechanical quality factor17, and thus a low mechanical decoherence rate, such that the coherent optomechanical coupling rate can be larger than both mechanical and optical decoherence rates18. Some devices have indeed reached the motional ground state with the assistance of cryogenics and laser cooling19,20,21. Newer platforms are being designed to minimize the mechanical clamping of the resonator to exhibit high mechanical quality factors that can be used to explore quantum phenomena even at room temperature22.\n\nThe main channel by which environmental thermal noise enters the system is via mechanical supports. By forsaking any form of extrinsic clamping, mechanical oscillators can sustain coherent oscillations for extended times and would therefore pose as a better candidate for quantum optomechanics. Optical levitation23,24,25,26 has been identified as a promising route in this direction. In addition to the benefits of environmental isolation, these schemes allow us to fully manipulate the quantum state and mechanical frequency of the levitated mirror via the optical fields. In particular with the proposals of a scattering-free tripod27 or sandwich28 of optical cavities, a milligram-scale mirror is levitated by coherently interacting with the radiation pressure forces provided by optical resonators. Due to the scattering-free feature of these two schemes, the mechanical quality factor of the levitating mirror can be expected to be on the order of 101027 and the quantum cooperativity can be estimated as 10329, making them excellent candidates to explore quantum and nonlinear effects in the macroscopic regime.\n\nHere we set up a reduced version of the optical tripod, where we consider only one vertical optical cavity as a testbed for levitation30. This simplified setup enables us to investigate the system’s dynamics and build a theoretical reference to better understand the underlying physics. Supported by a good agreement between experiment and numerical simulations, we believe that our models will be useful for any free-standing optomechanical system under high power. In particular, under the high power required for levitation, the optical intensity can reach 3 MWcm−2, which is even larger than that of the Laser Interferometer Gravitational-wave Observatory (LIGO)1. The response of our optomechanical system is dominated by different types of nonlinear interactions. The most prominent arises from the photothermal expansion of the mirror coating, which leads to optical bistability31,32 and a hysteretic asymmetry in the cavity response as a function of detuning. Another consequence of photo-absorption is the excitation of the acoustic modes of the mirror, which perturbs the cavity field. The fluctuation of the cavity field is subsequently propagated to other degrees of freedom via optical back action. In particular, we show that the absorbed energy results in effective anti-damping and parametric amplification33,34,35. Finally, the mirror also interacts with the intracavity field via the radiation pressure force, inducing a displacement when the optical push is stronger than the gravitational force. The interplay between all of these interactions results in complex behavior and rich dynamics. Higher-order optical sidebands36,37,38 and a quasi-continuous spectrum of the optical output are observed, which are suggestive of chaotic behavior39,40,41. To assess the stability of the cavity under external control, we implement active feedback42,43 to suppress the excitation of the acoustic vibration. With the analysis and modeling drawn from these investigations, we show a route towards the realization of stable and coherent optical levitation.\n\n## Results\n\n### Experimental setup\n\nOur experimental setup [Fig. 1(a)] consists of an optical resonator in a vertical configuration. The top mirror of the cavity, shown in Fig. 1(b) and hereon also referred to as the levitation mirror (or simply the mirror), is free-standing on a top of the Invar mount. The hole is specifically designed to have three symmetric contact points to minimize Van der Waals interactions [Fig. 1(c)]. When the radiation pressure force is sufficiently strong to balance the gravitational weight of the mirror, the torque exerted heaves a side off one of the contact points. As the mirror lifts, the effective cavity length increases, and as a consequence the laser driving the intracavity field becomes blue-detuned compared to the cavity’s resonance. In this regime the optical spring effect provides a restoring force, which is the origin of the optical trap expected for the full tripod levitation setup27.\n\nTo detect the mechanical displacement directly we introduce a weak optical beam at an angle that is reflected on the backside of the mirror and collected onto a quadrant photodetector. The intensity readout between different quadrants is then subtracted to obtain a relative measurement of the position. We also use the reflected and transmitted outputs of the main cavity field to monitor the evolution of the intracavity optical field. In Fig. 2 we show an example of the dynamics in the system to give an idea of the interactions involved. Note that all of the measurements presented in this and following figures are single time traces but are highly reproducible (see Supplementary Note 2 for details). We will unravel the different elements involved over the next few sections.", null, "Fig. 2: System dynamics at an input power of 4.5 W.\n\nThe most noticeable effect is the onset of mechanical oscillations (Fig. 2) resulting in multiple crossings of the cavity resonance. When the average detuning is scanned linearly from higher to lower frequencies [Fig. 2(a)-(b), piezoelectric actuator moving up shortening the length of the cavity], the resonance is quickly crossed, and the oscillation amplitude starts to decay overtime immediately after the brief excitation of the cavity field. Scanning in the opposite direction [Fig. 2(c)-(d), actuator moving down extending the length of the cavity], one can instead observe a slow build-up of the oscillations even before the first full resonance crossing. Moreover, the oscillations become self-sustained, and the cavity enters a passive feedback loop that is broken only when the scan moves the average detuning too far. We label these two opposite responses as “anti-locking” and “self-locking” respectively, and we will see that they stem from a competition between photothermal and radiation pressure interactions40,41. The observed oscillations are identified as excitations of the mirror’s acoustic modes by the intracavity field. We also note two interesting aspects from the wavelet transform of the transmission output of the cavity during the self-locking scan in Fig. 2(e). The first involves the generation of high-order sidebands of the acoustic modes induced by the nonlinearity of the system44,45. The second relates to the region enclosed by the green box, which shows how the intracavity field modifies the natural frequency of the acoustic mode. This is a consequence of the optical spring effect46,47,48,49,50.\n\n### Equations of motion\n\nBefore entering any in-detail analysis, it is important to develop a simple and effective model of the system to help understand the different phenomena. In our model we separate the position degree of freedom of the mirror into three different entities, each subject to a different type of interaction with the intracavity field: xth accounts for displacements of the reflective coating due to photothermal expansion, xac corresponds to a different position of the mirror’s surface following the vibrations of acoustic mode, and xlev represents a full shift of the center of mass because of radiation pressure force. All three interact with the optical degree of freedom, a, describing the amplitude of the optical field inside the cavity. Our model is far from being comprehensive: other phenomena may be present that are not accounted for, among which we mention for example, the bolometric interaction directly coupling photothermal absorption to the acoustic mode, the dependence of cavity decay rate on either acoustic or photothermal displacements, the Brownian motion and other stochastic dynamics of the three position degrees of freedom, the shot noise and phase fluctuations of the laser, or even the black body radiation of the levitating mirror. We will show, however, that a simpler model based simply on the mutual interaction of the cavity field with the three position degrees of freedom identified before is sufficient to reconstruct a faithful picture of the whole system, in good agreement with the experimental results.\n\nWe therefore consider the following equations to characterize the system in the classical limit:\n\n$${\\dot{x}}_{{\\rm{th}}}=-{\\gamma }_{{\\rm{th}}}[{x}_{{\\rm{th}}}+\\beta {P}_{{\\rm{opt}}}(a)],$$\n(1)\n$${\\ddot{x}}_{{\\rm{ac}}}=-{\\gamma }_{{\\rm{ac}}}{\\dot{x}}_{{\\rm{ac}}}-{\\omega }_{{\\rm{ac}}}^{2}{x}_{{\\rm{ac}}}+{F}_{{\\rm{opt}}}(a)/{m}_{{\\rm{ac}}},$$\n(2)\n$${\\ddot{x}}_{{\\rm{lev}}}=\\left\\{\\begin{array}{ll}-{\\gamma }_{{\\rm{lev}}}{\\dot{x}}_{{\\rm{lev}}}&{F}_{{\\rm{opt}}}\\le {F}_{{\\rm{g}}},\\\\ -{\\gamma }_{{\\rm{lev}}}{\\dot{x}}_{{\\rm{lev}}}+{F}_{{\\rm{opt}}}(a)-{F}_{{\\rm{g}}}&{F}_{{\\rm{opt}}}\\,> \\, {F}_{{\\rm{g}}},\\end{array}\\right.$$\n(3)\n$$\\dot{a}=-[\\kappa /2-i(\\Delta +G({x}_{{\\rm{th}}}+{x}_{{\\rm{ac}}}+{x}_{{\\rm{lev}}}))]a+\\sqrt{{\\kappa }_{{\\rm{in}}}}{a}_{{\\rm{in}}}.$$\n(4)\n\nThe dynamics of the optical field a given by Eq. (4) correspond to the typical evolution of the intracavity field under the additional back-action from the photothermal, acoustic, and center-of-mass modes. The field is in the frame rotating at the laser frequency ωl, initially detuned from the cavity resonance frequency ωopt by Δ = ωl − ωopt. The cavity is driven by a field of amplitude ain coupling through the input mirror at a rate κin. The coefficient G = ωopt/L, with L being the length of the cavity, is the optomechanical coupling between the mirror and the intracavity field converting a position shift into a change in detuning. The constant β is the photothermal response coefficient. The radiation pressure force Fopt is related to the optical power within the cavity Popt as Fopt = Ga2 = 2Popt/c, while Fg = mg represents the gravitational weight of the mirror. The quantities g and c respectively indicate the free-fall gravitational acceleration and the speed of light. The mirror’s total inertial mass is m, while mac is the effective mass of the acoustic mode of frequency ωac. The dissipation mechanisms in the system are described by κ for the intracavity field, γth for photothermal expansion, γac for the acoustic mode, and γlev for the center-of-mass motion. We note that there is no direct coupling between the three displacements and that their interaction is enabled only through the optical field.\n\nFirst we describe the photothermal expansion in Eq. (1). Even though the mirror is highly reflective, the intensity of the field circulating inside the cavity is high enough that even a small fraction of power absorbed in the coating causes it to expand noticeably. Considering that the beam size near the coating is around 100 μm and that the intracavity power is as high as a few kilowatts, the optical intensity can reach 3 MW cm−2, which is even larger than that of LIGO1. Thanks to the ion beam sputter coating, this intensity is still below the laser damage threshold. Nevertheless, the extremely high optical intensity still causes a local rise in temperature, which leads to expansion and therefore a change in cavity length. We model the photothermal displacement under the empirical assumption of an exponential relation with the intracavity power51, governed by the photothermal coefficient β and dissipation rate γth. The heating of the mirror results in a decrease in cavity length, indicating a positive value for β. Note that β can be negative if the photothermal heating were to lengthen the cavity52. This could happen, for example, if the photorefractive effect were dominant, or if the substrate had a negative expansion coefficient.\n\nThe excitation of the acoustic modes is described by Eq. (2). The steady impact of radiation pressure drives the transverse vibrations of the mirror. Given the high aspect ratio of the substrate disk, these vibrations have a significant impact on the overall dynamics of the cavity. Finite-element analysis of the natural mechanical frequencies of the levitating mirror returns a primary acoustic mode of interest around 30 kHz. The direct displacement measurement shown in Fig. 2(a) gives a natural frequency of 28.(6) kHz and a damping rate of 30 Hz. The discrepancy between the theoretical and experimental values of mechanical frequency may be due to the imprecision of simulated coating layers and mirror geometry. Other modes are also observed in the displacement spectrum, but their magnitude is much smaller, and for all practical purposes they can be neglected in the following analysis. We note that the resonant frequencies of the acoustic modes are sensitive to the constraints set by the supporting contacts of the mirror and other imperfections at the time of fabrication.\n\nFinally, in Eq. (3) we examine the displacement of the center of mass by radiation pressure. This is the degree of freedom linked to optical levitation. When the optical push is weaker than the gravitational weight, the radiation pressure force is fully balanced by the constraint of the mechanical support, and the mirror rests on the stage. Above this threshold the net difference between optical and gravitational forces lifts the mirror to a new equilibrium position, tipping it away from one of the contact points of the supporting structure and modifying the relative detuning of the cavity. The tipping angle is generally very small, with the center of mass displacement being in the nanometer scale as opposed to the size of the mirror of a few millimeters. It is, therefore, reasonable to consider linear forces and displacement (rather than torque and angle) in the equation. Note that the threshold force is not generally equivalent to the gravitational weight of the mirror but higher, Fth Fg, as it is also necessary to account for Van der Waals interaction and other static forces. Also, the damping coefficient γlev is considered to differ from that of the acoustic mode, γac, since the former is mostly affected by air viscosity and not internal friction.\n\nThe steady-state solutions of the system dynamics are obtained by setting the derivative terms in Eqs. (1)–(4) to zero:\n\n$${x}_{{\\rm{th}}}^{0}=-\\alpha | {a}_{0}{| }^{2},$$\n(5)\n$${x}_{{\\rm{ac}}}^{0}=\\frac{\\hslash G| {a}_{0}{| }^{2}}{{m}_{{\\rm{ac}}}{\\omega }_{{\\rm{ac}}}^{2}},$$\n(6)\n$$\\left\\{\\begin{array}{ll}{x}_{{\\rm{lev}}}^{0}=0&{F}_{{\\rm{opt}}}\\le {F}_{{\\rm{mg}}}\\\\ \\hslash G| {a}_{0}{| }^{2}=mg&{F}_{{\\rm{opt}}}\\, > \\, {F}_{{\\rm{mg}}}\\end{array}\\right.\\ ,$$\n(7)\n$${a}_{0}=\\frac{\\sqrt{{\\kappa }_{{\\rm{in}}}}{a}_{{\\rm{in}}}}{\\kappa /2-i[\\Delta +G({x}_{{\\rm{th}}}^{0}+{x}_{{\\rm{ac}}}^{0}+{x}_{{\\rm{lev}}}^{0})]},$$\n(8)\n\nwhere α = βωc/τc. It is convenient to define an effective detuning of the cavity: $${\\Delta }_{{\\rm{eff}}}=\\Delta +G({x}_{{\\rm{th}}}^{0}+{x}_{{\\rm{ac}}}^{0}+{x}_{{\\rm{lev}}}^{0})$$. From these equations we can calculate the minimum input power required to optically lift the mirror. Assuming the cavity to be on resonance and static forces on the mirror to be negligible (i.e., Fth = Fg), the input power threshold is given by\n\n$$\\begin{array}{r}{P}_{{\\rm{th}}}=\\frac{mg{\\kappa }^{2}L}{4{\\kappa }_{{\\rm{in}}}}.\\end{array}$$\n(9)\n\nWhen the input power is above threshold, the system reacts by reaching a new equilibrium point where the intracavity power is the same but the effective detuning is different. Thus, above threshold the optical power circulating within the cavity is purely determined by the mass of the mirror and is independent of input power, as suggested by the second line of Eq. (7). A similar argument applies when Fth Fg.\n\nThe parameter values extrapolated directly from the experiment or inferred from the following analysis are provided in Methods. The threshold input power for levitation is expected to be around 4 W. In the following sections we will unravel the combined dynamics of the system, starting from an input power well below threshold and proceeding at increasingly higher power until the anticipated threshold is exceeded.\n\n### Optical bistability\n\nBelow threshold the effective detuning at equilibrium is determined only by the photothermal expansion and acoustic modes, $${\\Delta }_{{\\rm{eff}}}=G({x}_{{\\rm{th}}}^{0}+{x}_{{\\rm{ac}}}^{0})$$. Eqs. (5)-(6) show that both of these are proportional to the average intracavity power, with thermal expansion being negative and corresponding to a decrease in cavity length while the displacement of the acoustic mode is positive and contributes in the opposite direction. Combining the solutions for these modes into Eq. (8) we obtain a cubic equation for the cavity photon number n = a02. Depending on the system’s parameters three possible solutions are possible, with two being stable and one unstable. This bistability phenomenon is well known and has been observed in analogous systems also driven by radiation pressure force53,54 and photothermal expansion31,53.\n\nIn our system, the input laser power is the primary free parameter used to trigger bistability. With sufficiently high power, the two stable solutions can overlap and the system exhibits hysteresis as a function of cavity detuning. Intuitively, we can imagine a scenario where the bottom piezoelectric actuator is set to scan downwards to change the detuning. As the cavity approaches resonance, the power builds up enough thermal expansion in the levitation mirror to compensate for the downward travel of the bottom mirror, resulting in a self-locking response.\n\nIn practice, with two effects competing in opposite directions, the course of bistability is determined by the displacement that is most reactive to laser power. We define the ratio of photothermal displacement to acoustic displacement as\n\n$$\\zeta =\\left|\\frac{{x}_{{\\rm{th}}}^{0}}{{x}_{{\\rm{ac}}}^{0}}\\right|=\\frac{{m}_{{\\rm{ac}}}{\\omega }_{{\\rm{ac}}}^{2}\\beta c}{2},$$\n(10)\n\nfor a quantitative evaluation of the dominant process. For ζ > 1, the photothermal displacement is dominant over the acoustic displacement, and the cavity resonance shifts towards the blue-detuned regime (Δ > 0). For ζ < 1 resonance shifts instead towards the red-detuned regime (Δ < 0) and bistability is phenomenologically the same as expected in the case of optical lifting. In our experiment we find ζ = 16 and photothermal expansion to be dominant, as evidenced by the observation of self-locking when blue-detuning the cavity.\n\nIn our system, the appearance of bistability occurs from a minimum input power of about 180 mW, as shown in Fig. 3. We expose the hysteretic behavior by moving the mirror on the piezoelectric actuator upwards (blue trace, from blue to red detunings) and downwards (red trace, from red to blue detunings). Compared to the typical Lorentzian profile (purple trace, obtained at 8 mW) the resonance appears broadened as the cavity tends to self-lock in the red-detuning regime, and narrowed when encountering the unstable state first by approaching from the opposite direction. This response is easily simulated by numerically solving Eq. (4)–(1) for a detuning varying linearly with time. Thanks to an excellent agreement between the experimental data and theory, we use this data to calibrate the free parameters reported at the start of this section.\n\n### Excitation of acoustic modes\n\nOptical bistability becomes more evident at higher input power, when self-locking causes the resonance to become increasingly broad at the same scan speed. At about 500 mW the cavity starts exhibiting a new effect in the form of an oscillatory process, as demonstrated in Fig. 4. These oscillations are linked to the excitation of the acoustic modes of the levitation mirror as they get parametrically amplified by the photothermal effect. Similarly to the radiation-pressure induced optical spring effect, the positive photothermal stiffness experienced during self-locking by the system is paralleled by a negative damping coefficient34,35. The amount by which the natural damping of the acoustic mode is modified by the photothermal interaction can be estimated by the eigenvalues of the Jacobian matrix30,55,56,57,58.", null, "Fig. 4: Excitation of the mirror’s acoustic mode at different scan speeds of the piezoelectric actuator, at the input power of 500 mW.\n\nThe excitations are easier to observe when the scan speed is slow enough to enable the full accretion of the oscillations, as seen for example in the 1.0 and 1.5 μm s−1 cases in Fig. 4(a). Parametric amplification (corresponding to a negative effective damping coefficient) ensues in the red-detuned regime. When the average detuning imposed by the external scan falls on the opposite side of the resonance, the effective damping of the oscillations turns back positive, and the acoustic mode turns quiescent again. Similar oscillations are also excited when scanning in the opposite direction, in Fig. 4(b). In this case the red-detuning regime is on the right-hand side of the trace, and since the cavity jumps too quickly to the next stable state, there is not sufficient time for them to develop significantly. A more complete physical picture is given in Supplementary Note 3, where we break down the relative contribution of each degree of freedom during the scan.\n\nWe use finite-element analysis to verify that the oscillation frequency (estimated at 28.(6) kHz by direct measurement), corresponds to a specific vibrational eigenmode of the mirror, shown in Fig. 4(e). Other vibrational modes were also found. Their participation factor, however, turned out to be at least two orders of magnitude smaller and therefore negligible. This result is in agreement with the full spectrum obtained by the direct measurement of the displacement in Fig. 2.\n\n### High-order sidebands\n\nThe optical cavity field can generate nonlinear amplitude modulations at high intracavity powers. As a result, higher-order stokes and anti-sidebands can be produced in the spectrum of the cavity output field38,59.\n\nWe tune the input power up to 1.9 W and present the experimental result in Fig. 5(a). It is shown that a long self-locking envelope carries multi-frequency oscillations. The evolution of the oscillatory frequency can be analyzed by performing the wavelet transform of the time-domain reflection data, as shown in Fig. 5(c). The mechanical oscillation of 28.(6) kHz is excited and parametrically amplified at around 0.2 ms. Second-order sidebands start to appear from 0.5 ms, and higher-order sidebands are generated when the average detuning approaches the cavity resonance. The blue and green dashed lines refer to the natural frequency of the acoustic mode and its second-order sideband frequency, respectively.", null, "Fig. 5: Self-locking cavity response at the input power of 1.9 W.\n\n### Optical spring\n\nAnother feature that we observe in Fig. 5 is how the natural frequency of the acoustic mode varies as the mirror interacts with the cavity. This is particularly evident before entering the regime of self-sustained oscillations, where the change in detuning is on average still proportional to the applied linear scan. We attribute this modification to the optical spring effect, which causes a reduction of the effective frequency in the red-detuned regime when the parametric amplification begins, and an increase in the blue-detuned regime when the cavity trails out of resonance60. We note that photothermal effects can also contribute to the optical modification of the natural mechanical frequency35. The pure optical spring is produced by the back-action of the intracavity optical field via the passive feedback loop L1 between the cavity field and the acoustic mode. The interaction between the photothermal effect and the intracavity field adds an additional feedback loop L2 to the system. The presence of the photothermal displacement works as a feedback path influencing the optical path length of the cavity and then the intracavity field. The variation of the optical field induced by loop L2 further modifies the mechanical susceptibility via loop L1.\n\nThe optical spring effect is more prominent in the theoretical plot than in the experimental result, as shown in Fig. 5 (c) and (d). This discrepancy might be the result of the following assumptions in our model: firstly, we neglect the direct interaction between acoustic mode and photothermal effects; secondly, we assume that the photothermal displacement is linear to the intracavity power. We can, however, see a fair agreement between the theory and experiment.\n\n### Optical lift\n\nAbove the threshold input power of 4 W the mirror should experience optical lift. If static electric forces are sufficiently small, the input power of 4.5 W used for the plots in Fig. 2 is expected to satisfy this requirement and to successfully detach the mirror from one of the contact points.\n\nThe experiment does not allow direct observation of this phenomenon. Any analysis of the mirror’s position is performed through a measurement of the reflective coating. Neither the direct measurement by means of the quadrant detector or the indirect deduction from the cavity response will yield the absolute position of the center of mass, especially taking into account the more consequential dynamics examined so far. As a matter of fact, at this power, we do not identify a qualitative difference compared to when the system operates below the lift-off threshold. We remember that, at equilibrium, radiation pressure shifts the detuning point of the cavity so as to achieve the same intracavity power that would be circulating on resonance at the threshold power. Calculating the induced detuning as $$| {\\Delta }_{{\\rm{RP}}}| =(\\kappa /2)\\sqrt{{P}_{{\\rm{in}}}/{P}_{{\\rm{th}}}-1}$$, we can estimate the center-of-mass displacement xlev to be on the order of ΔRP/G, i.e., about 40 pm. For comparison, the distance traveled by the reflective coating due to the acoustic vibrations is proportional to a few linewidths, i.e., a multiple of κ/G, which is equivalent to a few nanometers.\n\nThese projections are corroborated by the numerical simulations, which allow us to investigate the photothermal expansion, acoustic vibrations, and mechanical displacement individually. At an input power of 4.5 W (used in Fig. 2), it is expected that the scale of both acoustic and photothermal displacements is on the order of a few nanometers while the mirror is only lifted off the stage by a few picometers. We also compare the simulated system dynamics in the presence and absence of the optical lifting interaction given by Eq. (3), finding no apparent difference. We conclude that, at this power, the system dynamics are scarcely affected by the optical lift, and it is not possible to identify a visible signature of this effect.\n\nIn general, it may be tempting to increase the power well above the threshold to ensure a noticeable detuning, detectable for example, by monitoring the Pound–Drever–Hall signal of the cavity. We note, however, that the scale of the instabilities makes this task hardly implementable, as the oscillations are such as to alter the instantaneous detuning to a much larger extent.\n\n### Feedback cooling of system instability\n\nThe excited eigenmodes of the mirror destabilize the cavity and make it impossible to evince the possible suspension of the mirror on the optical field. Moreover, as the vibrations cover the whole surface of the mirror they would inevitably propagate to the other independent cavities in the final tripod configuration. We apply active feedback42,43 to the input power in order to suppress such instability and avoid cascading disruptions. We feed the direct displacement measurement of the quadrant detector to an acousto-optic modulator, providing a modulation of the input power and therefore adding an effective drive to the mirror on top of the radiation pressure force. Accounting for a delay τ within the feedback line, the equation for a single acoustic mode is\n\n$${\\ddot{x}}_{{\\rm{ac}}}+{\\gamma }_{{\\rm{ac}}}{\\dot{x}}_{{\\rm{ac}}}+{\\omega }_{{\\rm{ac}}}^{2}{x}_{{\\rm{ac}}}=[{F}_{{\\rm{opt}}}+{F}_{{\\rm{fb}}}(t-\\tau )]/{m}_{{\\rm{ac}}},$$\n(11)\n\nwith Ffb(t − τ) = gx(t − τ) being proportional to the total displacement measured by a factor g, representing the feedback gain. The delay τ can be tuned via a phase shifter or a differentiator for optimal results. The effective mechanical susceptibility is obtained by solving Eq. (11) in the frequency domain:\n\n$${\\chi }_{{\\rm{eff}}}(\\omega )=\\frac{1}{{m}_{{\\rm{ac}}}({\\omega }_{{\\rm{ac}}}^{2}-{\\omega }^{2})-g\\cos (\\omega \\tau )+i[{m}_{{\\rm{ac}}}{\\gamma }_{{\\rm{ac}}}\\omega +g\\sin (\\omega \\tau )]}.$$\n(12)\n\nThe feedback force modifies the natural frequency and damping rate of the acoustic mode. At ωτ = π/2 the frequency remains unchanged but the damping can be modified to become positive and stabilize the mirror.\n\nTo implement feedback control we use a thinner levitation mirror, with a thickness of approximately 30 μm and a mass of 0.966 mg. This mirror displays two major acoustic resonances, one around 30 kHz and the other at 100 kHz. The system dynamics for this mirror exhibit a further nonlinear phenomenon in the form of a continuous spectrum characteristic of chaotic systems, as we report in Supplementary Note 1. The experimental setup for the feedback control is shown in Fig. 1(d). The modulation of the input power is achieved using an acousto-optic modulator, driven by a 80 MHz harmonic oscillation. The signal detected by the quadrant detector is filtered using a bandpass with its bandwidth from 20 kHz to 100 kHz, giving a clean signal containing the information on the mirror displacement. This signal is then phase-shifted, amplified, and fed to the modulator to vary the amplitude of the optical input field. This process forms a closed feedback loop. We process the mechanical displacement on a spectrum analyzer and disclose the obtained results in Fig. 6. Tuning the feedback phase accordingly, we can achieve essentially complete suppression of the acoustic excitations (more than 10 dB for both modes) at powers much lower than the levitation threshold for this particular mirror.\n\nThe technique demonstrated is unfortunately not easily extended to different regimes, as it is most effective for single-mode applications 42. At high power, the instabilities arising from the photothermal interaction prevail and the increasing complexity of the dynamics requires exceedingly stringent feedback parameters. At full operating power for optical levitation, active feedback may altogether be an inadequate choice for stabilization, in which case it may be preferable to consider passive techniques acting directly on the damping properties of the system and therefore addressing the problems closer to the source.\n\n## Discussion\n\nIn this paper, we explored the nonlinear dynamics of a vertical optical cavity where the top reflector consists of a millimeter-scale mirror to be optically levitated. We input up to 4.5 W of laser power onto an ion beam sputtered mirror weighing ≈1 mg. The spot size of our laser beam is 100 μm in diameter. This size corresponds to an optical intensity of 3 MWcm−2, which is significantly larger than that of LIGO1. We found that the intracavity field bridges the interaction of different degrees of freedom linked to photothermal expansion, acoustic modes, and position of the center of mass. The result is a remarkably complex assortment of dynamics that we proceeded to identify by observing the system’s response at different power regimes. We observed optical bistability, parametric amplification, high-order sideband generation, and optical spring corrections to the natural eigenmodes.\n\nThe ideal course of action to reduce the system’s instability would be to reduce the optical absorption of the mirror coating significantly. By doing so, radiation pressure would be the major source of interaction between the optical field and the mirror, the combined system would be less elaborate, and any stabilizing effort such as feedback control can be engaged to more specific aspects. Reducing absorption beyond a certain level, however, may be technically challenging. Alternatively, one may attempt to switch the sign of photothermal interaction so that it collaborates with radiation pressure towards both static and dynamic stability35,61. In our system, the photothermal coefficient is β = 8.6 pm W−1, which is positive. We can flip the sign of this parameter by modifying the thickness of the first coating layer of the levitating mirror61 or by introducing another photothermal effect with opposite interaction. In the first case, the different thicknesses of the coating would change the mass and the frequency of the acoustic model of the levitating mirror, as well as the photothermal coefficients. In the other case, adding a new photothermal degree of freedom may require additional components within the optical resonator, modifying the cavity decay rate and finesse together with the effective photothermal response. With this approach, our estimation suggests that the photothermal coefficient can not only be decreased by as little as a factor of two, but also be swapped in sign and changed by as much as two orders of magnitude.\n\nWe also expect further explorations of the optomechanical nonlinearity to lead to the observation of interesting stochastic phenomena in the system. Many relevant nonlinear effects such as stochastic resonance have been investigated in optically levitated nanoparticles62,63,64,65. A milligram-scale mirror driven by an ultra-intense laser opens up a very different parameter regime, where it might be intriguing to extend the concepts and experiments observed at nano-scale towards a new territory and even identify exclusive nonlinear effects.\n\nDespite being focused on a specific optical levitation system, our investigations may offer methods to understand the physics of high-power optomechanical systems.\n\n## Methods\n\nThe top mirror of the levitation cavity presents a high-reflectivity coating of 99.992% on the curved side of a fused silica spherical cap with a radius of curvature of 25 mm, diameter of 3 mm and thickness of approximately 50 μm. The mirror’s coating is obtained by ion beam sputter deposition to reduce loss and absorption. This coating process produces substantial stress on the substrate, as a result it cannot be applied directly to a thin substrate (of few micrometers  in thickness) without risk of damage. The coating is therefore finished on a thick substrate with a thickness of 3 mm. The mirror is then lapped down to 50 μm from the uncoated side. The lapping technique leaves a relatively rough surface on the backside of the mirror, but at least it helps to dissipate the residual stress baked-in from the coating run. The bottom mirror of the cavity is a conventional 1-inch concave mirror, also with high-reflectivity coating on the concave side of the fused silica substrate. This mirror is attached to a piezoelectric actuator to allow scanning of the cavity length, Lc, and therefore change the detuning. The actuator is pre-loaded via mechanical clamping, improving its stability and performance. The whole cavity is 80 mm long and is enclosed by a monolithic Invar block to reduce thermal fluctuations and isolate the cavity from airflow. A 1050-nm Nd:YAG laser is used to drive the cavity with up to about 15 W of input power. The mechanical noises produced by the actuator and the cavity mount are too small to be clearly seen in the experimental results presented in this paper.\n\nFrom our experimental results we infer the parameters of our system to be L = 80 mm, m = 1.116 mg, mac = 0.38 mg, ωac = 2π × 28.6 kHz, γac = 2π × 30 Hz, γth = 2π × 560 Hz, γlev = 2π × 50 Hz, κ = 2π × 730 kHz, κin = 2π × 180 kHz, G = 2π × 3.6 MHz nm−1, β = 8.6 pm W−1." ]

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[ "# Loss rate calculation in a Poisson process\n\nI have a bit a naive question about Poisson process.\n\nSuppose I have a station that receives a number of packets $n(i)$ each time slot $i$. If the number of received messages is less than $M$ then the $n(i)$ packages will all be sent.\n\nIf $n(i) > M$ then $n(i)-M$ packets will be lost and $M$ packets will be sent. Let assume packets arrive at a Poisson rate $\\lambda$. What is the mean of loss rate (total lost packages / total number of received packets)?\n\nI formulated the problem in this way:\n\n• Let $N(t)$ be the total number of packets at time $t$, $N(t) = \\sum_{i=0}^{t} n(i)$\n• The total number of lost packets is $e(t) = \\sum_{i=0}^{t} (n(i)-M) I_{(n(i) >M)}$\n\nThe loss rate will then be $\\eta(t) = \\frac{e(t)}{N(t)}$.\n\nThe mean loss rate will be:\n\n$$\\lim_{t \\to \\infty} E[\\eta(t)]= E\\left[\\frac{\\sum_{i=0}^{t} (n(i)-M) I_{(n(i) >M)}}{\\sum_{i=0}^{t} n(i)}\\right]$$\n\nBut I don't know how to calculate it.\n\nThe expected number of lost packets can be re-expressed as the expected number of arriving packets minus the expected number of sent packets:\n\n$\\lambda - \\left(\\sum_{n=0}^M np(n;\\lambda) + M(1-P(M;\\lambda))\\right)$\n\nwhere $p(n;\\lambda)$ is the probability of seeing $n$ packets in a time slot given an arrival rate $\\lambda$ and $P(M;\\lambda)$ is the probability of seeing $M$ or fewer packets ... The first term captures the events where $n \\leq M$, when we send all $n$ packets, and the second where $n > M$, when we send $M$ packets.\n\nWriting out the Poisson distribution function in the second term and dropping the term where $n=0$ (as it equals zero, so contributes nothing to the sum) gives:\n\n$\\sum_{n=0}^M np(n;\\lambda) = e^{-\\lambda}\\sum_{n=1}^M n\\left(\\frac{\\lambda^n}{n!}\\right)$\n\nwhich can be rearranged as:\n\n$\\lambda e^{-\\lambda}\\sum_{n=0}^{M-1}\\frac{\\lambda^n}{n!}$\n\nwhich in turn equals $\\lambda P(M-1;\\lambda)$. Substituting this into the first expression results in:\n\n$\\mathbb{E}[\\text{number of lost packets}] = \\lambda[1-P(M-1;\\lambda)] - M[1-P(M;\\lambda)]$\n\nNoting that $P(M;\\lambda) = P(M-1;\\lambda) + p(M;\\lambda)$ allows us some more, albeit slight, simplification:\n\n$\\mathbb{E}[\\text{number of lost packets}] = (\\lambda - M)[1-P(M-1;\\lambda)] + Mp(M;\\lambda)$\n\nChecking our algebra by comparing a simulation with our result gives:\n\n> M <- 3\n> lambda <- 2\n>\n> x <- rpois(1000000,lambda)\n> xM <- pmin(x,M)\n>\n> cat(\" Simulation result: \",lambda-mean(xM), \"\\n\",\n+ \"Mathematical result: \",(lambda-M)*(1-ppois(M-1,lambda)) + M*dpois(M,lambda))\nSimulation result: 0.218311\nMathematical result: 0.2180175\n\n• Hi jbowman, I just want to say waw, thanks a million. I like the way you reformulated the problem, it is more handy. I suspected from the beginning that my problem basically was in the formulation. – sirus Dec 22 '13 at 14:56" ]

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[ "# Lesson:Newton Gets Me Moving\n\n### Quick Look\n\nTime Required: 15 minutes\n\nLesson Dependency: None\n\nSubject Areas: Earth and Space, Physical Science, Science and Technology\n\n### Summary\n\nStudents explore motion, rockets and rocket motion while assisting Spacewoman Tess, Spaceman Rohan and Maya in their explorations. First they learn some basic facts about vehicles, rockets and why we use them. Then, they discover that the motion of all objects—including the flight of a rocket and movement of a canoe—can be described by Newton's three laws of motion.\nThis engineering curriculum meets Next Generation Science Standards (NGSS).\n\n### Engineering Connection\n\nWhenever engineers work on something that moves, they use Newton's laws of motion to help describe how it is going to move. This includes cars, trains, boats, bicycles, skateboards, roller coasters, airplanes and rockets. Really, Newton's laws of motion explain the movement of anything that is—simply—in motion. Knowing how a vehicle will move is very important when designing a successful vehicle. And, similarly, knowing how a rocket will move is very important to designing a successful rocket. Understanding Newton's laws helps engineers figure out how much fuel is needed, how big the rocket must be, how much the rocket can weigh, how long the rocket must burn, and even how fast the rocket will go.\n\n### Learning Objectives\n\nAfter this lesson, students should be able to:\n\n• Describe the characteristics and function of rockets.\n• Identify and explain Newton's three laws of motion.\n• Describe how Newton's laws relate to engineering, rockets and paddling.\n\n### Educational Standards Each TeachEngineering lesson or activity is correlated to one or more K-12 science, technology, engineering or math (STEM) educational standards. All 100,000+ K-12 STEM standards covered in TeachEngineering are collected, maintained and packaged by the Achievement Standards Network (ASN), a project of D2L (www.achievementstandards.org). In the ASN, standards are hierarchically structured: first by source; e.g., by state; within source by type; e.g., science or mathematics; within type by subtype, then by grade, etc.\n\n###### NGSS: Next Generation Science Standards - Science\nNGSS Performance Expectation\n\n3-PS2-1. Plan and conduct an investigation to provide evidence of the effects of balanced and unbalanced forces on the motion of an object. (Grade 3)\n\nDo you agree with this alignment?\n\nClick to view other curriculum aligned to this Performance Expectation\nThis lesson focuses on the following Three Dimensional Learning aspects of NGSS:\nScience & Engineering Practices Disciplinary Core Ideas Crosscutting Concepts\nPlan and conduct an investigation collaboratively to produce data to serve as the basis for evidence, using fair tests in which variables are controlled and the number of trials considered.\n\nAlignment agreement:\n\nScience investigations use a variety of methods, tools, and techniques.\n\nAlignment agreement:\n\nEach force acts on one particular object and has both strength and a direction. An object at rest typically has multiple forces acting on it, but they add to give zero net force on the object. Forces that do not sum to zero can cause changes in the object's speed or direction of motion. (Boundary: Qualitative and conceptual, but not quantitative addition of forces are used at this level.)\n\nAlignment agreement:\n\nObjects in contact exert forces on each other.\n\nAlignment agreement:\n\nCause and effect relationships are routinely identified.\n\nAlignment agreement:\n\n###### Common Core State Standards - Math\n• Multiply or divide to solve word problems involving multiplicative comparison, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem, distinguishing multiplicative comparison from additive comparison. (Grade 4) More Details\n\nDo you agree with this alignment?\n\n• Fluently multiply multi-digit whole numbers using the standard algorithm. (Grade 5) More Details\n\nDo you agree with this alignment?\n\n###### International Technology and Engineering Educators Association - Technology\n• The use of transportation allows people and goods to be moved from place to place. (Grades 3 - 5) More Details\n\nDo you agree with this alignment?\n\n• A transportation system may lose efficiency or fail if one part is missing or malfunctioning or if a subsystem is not working. (Grades 3 - 5) More Details\n\nDo you agree with this alignment?\n\n###### State Standards\n• Develop a scientific explanation regarding relationships of the components of the solar system (Grade 4) More Details\n\nDo you agree with this alignment?\n\nSuggest an alignment not listed above\n\n### More Curriculum Like This\n\nGet Me Off This Planet\n\nThe purpose of this lesson is to teach students how a spacecraft gets from the surface of the Earth to Mars. Students first investigate rockets and how they are able to get us into space. Finally, the nature of an orbit is discussed as well as how orbits enable us to get from planet to planet — spec...\n\nUsing Thrust, Weight & Control: Rocket Me into Space\n\nThrough the continuing storyline of the Rockets unit, this lesson looks more closely at Spaceman Rohan, Spacewoman Tess, their daughter Maya, and their challenges with getting to space, setting up satellites, and exploring uncharted waters via a canoe. Students are introduced to the ideas of thrust,...\n\nMay the Force Be with You: Thrust\n\nStudents study how propellers and jet turbines generate thrust. This lesson focuses on Isaac Newton's third law of motion for every action there is an equal and opposite reaction.\n\nNewton Rocket Car\n\nThrough the use of small wooden cars, this activity demonstrates Newton's third law of motion—which states that every action has an equal and opposite reaction. The \"Newton rocket cars\" that students put together show how action/reaction works and how the mass of a moving object affects the accelera...\n\n### Introduction/Motivation\n\nWhat are vehicles and why do we need them? (With the students, discuss their concept of a vehicle. Help them get to the conclusion that the basic definition of a vehicle is a device that enables something to move from one place to another quicker than if no vehicle existed.) Now, what is motion? What are some different ways a person can get from one place to another? (List students' answers on the classroom board. Possible answers: walk, run, bicycle, skateboard, drive/ ride in a car, train, boat, airplane or rocket.) How do these objects move? You're right! Everything that moves in one way or another involves a push or a pull, which engineers call a force. For example, an engine in a car causes the wheels to turn, which push against the ground, while a sailboat is pulled along by the wind. Every single motion is caused by a force. If tno pushes or pulls existed, objects—or in this case, vehicles—would not go anywhere. This is an example of Newton's first law, which states that an object at rest tends to stay at rest and an object in motion tends to stay in motion, unless a force acts upon that object.\n\nLet's look at our list on the board again. Which of these objects move fast and which move slowly? Now that we have separated the list into fast and slow groups, let's think about the forces (pushes and pulls) acting on the objects. Are the forces acting on the faster objects more or less than the forces acting on the slower objects? (Guide students to realize that the faster objects are faster because larger forces acts on them.) This is an example of Newton's second law, which states that the force of an object is equal to its mass times its acceleration. Larger mass equals larger force.\n\nWhich vehicles will Maya and her family be using in their explorations? Tess and Rohan will need a rocket to carry their communications satellite into space. Maya has a canoe that she will paddle to explore uncharted waters. A rocket is large and will take a large force to get it moving. A canoe is smaller and only requires a smaller force to get it moving.\n\nSo how do we create a force to move an object? Let's think about Maya in her canoe. How will she move it? (Answer: She will only move if a force acts upon her canoe.) Maya needs a force to move and that force could be present in many different forms: Maya could use the movement of the water as a force to move her canoe if the water is going in the direction she wants; she could use a paddle to move or push her canoe; or she could have a friend push her in the canoe. If Maya was holding a bowling ball in her canoe and threw it overboard, would she move? The answer is yes, throwing the bowling ball in one direction would cause Maya and her canoe to move in the opposite direction. Can you see that for every movement, some responding action happens in the opposite direction? This is an example of Newton's third law, which states that every action has an equal and opposite reaction. (optional: Demonstrate this using a skateboard or a rolling chair.)\n\nSo, what about Tess' rocket? What makes a rocket a rocket? A rocket is a device that burns fuel causing extremely hot gasses to be ejected from the rocket out the nozzle (the tailend). The action of all this hot gas moving in one direction causes the rocket to move in the opposite direction. Rockets usually burn either liquid or solid fuel. It takes a lot of engineers to build a modern rocket since they are so complicated. Figure 1 shows a diagram of a liquid fuel rocket.\n\nIn a liquid fuel rocket, the fuel and oxidizer are pumped into a combustion chamber where the fuel and oxidizer burn to create super hot gas that is forced to escape through the nozzle. The rocket works on the same principle as Maya throwing the bowling ball while sitting in her canoe (do you remember: that for every action, there is an opposite reaction), but instead of throwing bowling balls the rocket is throwing hot gas. The rocket throws the hot gas down towards the Earth, which causes the rocket to move upward, away from the Earth. This does not seem like it would push the rocket very far, but the rocket is throwing so much hot gas at such a high speed that it can move very quickly. Other types of rockests use solid fuels. They are simpler rockets since no pump or oxidizer is required, but they cannot be turned on and off. Typically, solid fuel rockets are not as efficient as liquid fuel rockets. Examples of liquid fuel rockets include the space shuttle's main engine as well as the Atlas, Titan and Delta rockets that are used to put satellites into space. Examples of solid fuel rockets include the solid rocket boosters on the space shuttle, rocket powered cars and bottle rockets. Figure 2 shows the liquid and solid fuel rockets on the space shuttle.\n\nFor what purpose do engineers design rockets? We have already talked about designing rockets to go fast, but we have other reasons to design rockets, too. You may have heard or read that we often use rockets on spacecraft and satellites. That is because right now rockets are the only efficient way we have to move in space. Jets or propellers cannot be used to travel in space because they require the presence of air to work, and no air is exists in space. And, we cannot use a canoe to get around space because no water exists in which to paddle. We could get around in space if we had a bunch of bowling balls to throw! By throwing the bowling balls in one direction, we would be able to move in the opposite direction; however, throwing bowling balls is not the best way to move around in space, so we will continue to using rockets for now. Today, we are going to learn more about motion, engineering and about how a man named Isaac Newton formed three laws that describe for us why objects—including rockets and canoes—move. Following the lesson students can conduct the associated activiy Newton Rocket Car to learn about Newton's laws of motion as they build small vehicles that move by launching a mass backward.\n\n### Lesson Background and Concepts for Teachers\n\nNewton's Laws of Motion\n\nThe basic motion of any object is described by Isaac Newton's three laws of motion. His simple laws explain how objects move and, more specifically, how rockets move in the atmosphere and in space or how canoes paddled in the water move. (Note: For more reading on Sir Isaac Newton, see the accompanying reading material.)\n\nNewton's First Law of Motion\n\nNewton's first law states that an object at rest tends to stay at rest and an object in motion tends to stay in motion unless a force acts upon that object. This means that for an object to speed up or slow down, a force must be present to push or pull on the object. Sometimes a force acting on an object causes that object to stay at rest or in motion. This can happen when another force cancels out the first force. For example, a person just standing on the ground has a force acting on him or her. This force is called gravity, but even though gravity is acting on this person, s/he is not moving. How can this be? The reason is because the ground is pushing up on the person with the same force as the gravity is pulling down. This upwards force cancels out gravity, resulting in no change in motion. Engineers call two forces that cancel each other out balanced forces. If the floor was not present, gravity would no longer be canceled out by upward force, and the person would start to move (fall).\n\nAn object at rest stays at rest if the forces acting on that object are balanced or no forces are acting on it. This is obvious for something that is not moving; but it also applies to moving objects in a vacuum. An object in motion stays in motion if balanced forces or no forces act on it. If a spaceship floating through deep space is moving at a constant velocity and has no forces acting on it (for example, gravity), then there is no change in motion, and the spaceship keeps moving in a straight line—forever! Continuous motion is not seen on Earth due to friction and other forces that slow things down.\n\nNewton's Second Law of Motion\n\nIf a bowling ball and a soccer ball were both dropped at the same time from the roof of a tall building, which would hit the ground with greater force? Common sense picks the bowling ball because it is heavier. Might we believe this to be true because we naturally assume that the bowling ball will fall faster? This statement is actually NOT true. Gravity accelerates all objects at the same rate; therefore, the balls would hit the ground at the same time and with the same velocity. However, the bowling ball would hit with greater force because it has a greater mass. Newton stated this relationship in his second law: the force of an object is equal to its mass times its acceleration.\n\nThis law of motion can be expressed as a simple mathematical equation. The three parts of the equation are mass (m), acceleration (a) and force (F). Using the letters to symbolize each part, the equation can be written as follows:\n\nF = m x a\n\nTo explain this law, consider a cannon as an example: when a cannon is fired, an explosion propels a cannonball out the open end of the barrel (top end of the cannon). It is propelled a kilometer or two to its target. At the same time, the cannon itself is pushed backward a meter or two. This is action and reaction at work (Newton's third law, which we will discuss shortly). Figure 3 shows a cannon and how Newton's laws of motion explains the movement of both the cannon ball and the cannon. The force acting on the cannon and the ball is the same force. What happens to the cannon and the ball is determined by the relative masses, according to the following equations:\n\nForce on the cannon = mass (of cannon) x acceleration (of cannon)\n\nForce on the ball = mass (of ball) x acceleration (of ball)\n\nThe first equation refers to the cannon and the second to the cannon ball. In the first equation, the mass is the cannon itself and the acceleration is the movement of the cannon. In the second equation, the mass is the cannon ball and the acceleration is its movement. Because the force (exploding gun powder) is the same for the two equations, the equations can be set equal to each other and rewritten as:\n\nmass (of cannon) x acceleration (of cannon) = mass (of ball) x acceleration (of ball)\n\nIn order to keep the two sides of the equations equal, the accelerations must balance the masses. In other words, since the cannon's mass is large and the cannon ball's mass is small, the only way the equation balances is if the cannon ball has a much larger acceleration than the cannon. This is why the cannon only rolls back a few feet and the cannon ball flies a long distance.\n\nNow, apply this principle to a rocket. Replace the mass of the cannon ball with the mass of the gases (fuel) being ejected out of the rocket engine nozzle. Replace the mass of the cannon with the mass of the rocket moving in the other direction. The force is the pressure created by the controlled explosion taking place inside the rocket's engines (similar to the gun powder explosion inside the cannon). That pressure accelerates the fuel gases one way out the nozzle, which causes the rocket to move the other way.\n\nNewton's Third Law of Motion\n\nThis law states that every action has an equal and opposite reaction. If you have ever run into anything in surprise or by accident, you have a experienced the essence of this law.\n\nThink about Maya in her canoe. Maya pushes the water back using a paddle, which creates a counterforce of similar size that propels the canoe forward. When Maya wants to move forward in the canoe, she paddles in a backward motion; when she wants to move backwards in the canoe (perhaps to avoid rocks, trees or animals), she moves the paddle in a forward motion. Figure 4 illustrates this idea.\n\nAs another example, imagine Spaceman Rohan alone at home with a skateboard. He and his skateboard are in a state of rest (not moving). Spaceman Rohan jumps off the skateboard. In Newton's third law, the act of jumping is called an action. The skateboard responds to that action by traveling some distance in the opposite direction. The skateboard's opposite motion is called a reaction. When the distance traveled by the rider and the skateboard are compared, it would appear that the skateboard has been affected by a much greater force than the rider, but this is not actually the case. The reason the skateboard has traveled farther is that it has less mass than the rider (see Newton's second law).\n\nWith rockets, the action is the expelling of gas out of the engine. The reaction is the movement of the rocket in the opposite direction. To enable a rocket to lift off from the launch pad, the action (or thrust) from the engine must be greater than the downward acceleration of gravity on the mass of the rocket. In space, when the downward acceleration of gravity is balanced, even tiny thrusts cause the rocket to change direction.\n\nRockets work better in space than they do in the air. The surrounding air impedes the action-reaction. In the atmosphere, both the nose of the rocket and the exhaust gases leaving the rocket engine must push away the surrounding air; this uses up some of the energy of the rocket. In space, the exhaust gases can escape freely (action) and no air friction exists to slow the rocket's reaction forward.\n\n### Associated Activities\n\n• Newton Rocket Car - Students learn about Newton's laws of motion as they build small vehicles that move by launching a mass backward.\n\n### Lesson Closure\n\nLet's look around the room and find examples of balanced and unbalanced forces. (Possible classroom examples: an air duct, water faucet, clock, and the students themselves.) Any change in motion indicates unbalanced forces. Anytime something encounters friction, that is a force acting upon that object. Every time there is a force, there is an equal and opposite force. A fan blade hitting an air molecule pushes it away (one force), but the air molecule also applies a reactive force to the fan and slows it down slightly (equal and opposite force). Can you think of examples actions and reactions from everyday life? Do you think rockets would work without the natural behavior described in Newton's third law? (Answer: No way!) It is important to understand that a force is required for an object to start or stop moving. How fast an object speeds up (accelerates) is dependent on the mass of the object and the size of the force acting on it. Lastly, for every action there is always an equal and opposite reaction.\n\nAs we end this lesson, consider the fact that Spacewoman Tess and Spaceman Rohan need a rocket to put a communication satellite or two up in orbit in order to keep in contact with Maya as she goes on her journey. Since you all now understand the laws explaining motion, you are capable of becoming engineers who will be in charge of helping to build such a rocket.\n\n### Vocabulary/Definitions\n\nacceleration: How quickly the speed of an object is changing.\n\nforce: A push or pull that causes motion or change.\n\nIsaac Newton: An English mathematician and physicist who defined three important laws of motion. Born 1642; died 1727.\n\nmass: A measure of the amount of matter in an object.\n\nNewton's first law: No forces = no change in motion. An object at rest tends to stay at rest, and an object in motion tends to stay in motion unless a force acts on the object.\n\nNewton's second law: force = mass X acceleration\n\nNewton's third law: For every action, there is an equal and opposite reaction.\n\nrocket: A vehicle that moves by ejecting mass.\n\nrocket: A vehicle that moves by ejecting mass.\n\n### Assessment\n\nPre-Lesson Assessment\n\nConcept Quiz: Administer the What Is a Rocket Quiz. Have students work together in groups of four. Encourage them to share ideas. Do not give them any answers yet!\n\nPost Introduction Assessment\n\nVehicle Detectives: Organize the students up into teams of three to four. Give the groups each a specific vehicle, such as a skateboard, toy car, toy train, bicycle, pogo stick, etc., but preferably vehicles from the list generated earlier in the lesson, and ask students to describe the vehicle's motion using Newton's laws. Questions to ask:\n\n• What sort of action is used to move the vehicle?\n• What is the reaction to that action?\n• Does the vehicle experience more or less friction depending on where it is used? Why?\n• Is a vehicle that is already in motion more inclined to continue to be in motion? Why? Can you think of an example of one that is?\n• What types of fuel are used to move the vehicle?\n• Does the vehicle move fast or move slow? Why?\n\nLesson Summary Assessment\n\nConcept Quiz: Have students redo their What Is a Rocket? Quiz. Discuss the answers and have students correct each other's papers.\n\nInformal Discussion: Solicit, integrate and summarize student responses.\n\n• Ask students to explain how rocket motion is different from car, airplane or canoe motion and reference Newton's third law (for every action there is an equal and opposite reaction).\n• Ask the students to explain Newton's second law (force = mass x acceleration). Expect students to understand that many different combinations of mass and acceleration can give you the same final force using F=m x a:\n\n12 = 1 x 12\n\n12 = 2 x 6\n\n12 = 3 x 4\n\nUsing the Equation: Have students use the force equations presented in the Lesson Background to calculate the force on the cannon and on the ball, given the following information:\n\n• mass of cannon = 1000 kg\n• mass of ball = 20 kg\n• acceleration of cannon = 2 m/s2\n• acceleration of ball = 100 m/s2\n\n(Answers: Force on cannon = 1000 kg x 2 m/s2 = 2000 kg*m/s2. Force on the ball = 20 kg x 100 m/s2 = 2000 kg*m/s2)\n\nHuman Matching: On 10 pieces of paper, write either the term or the definition of the vocabulary words. Ask for volunteers from the class to come up to the front of the room, and give each person one of the pieces of paper. One at a time, have each volunteer read what is written on his/her paper. Have the remainder of the class match term to definition by voting. Have student \"terms\" stand by their \"definitions.\" At the end, give a brief explanation of the concepts.\n\n### Lesson Extension Activities\n\nTake the class to the gym and have some students sit on scooters or skateboards with their feet off the ground while throwing heavier balls. Note the resulting movement. Is this an efficient way to travel? Why or why not? Have students discuss among themselves. (Answer: Probably not.)\n\n### Contributors\n\nJeff White; Brian Argrow; Geoffrey Hill; Jay Shah; Malinda Schaefer Zarske; Janet Yowell\n\n### Supporting Program\n\nIntegrated Teaching and Learning Program, College of Engineering, University of Colorado Boulder\n\n### Acknowledgements\n\nThe contents of this digital library curriculum were developed under grants from the Fund for the Improvement of Postsecondary Education (FIPSE), U.S. Department of Education and National Science Foundation (GK-12 grant no. 0338326). However, these contents do not necessarily represent the policies of the DOE or NSF, and you should not assume endorsement by the federal government." ]

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[ "# Decimal.Ceiling() Method in C#\n\nThis method is used to round the decimal to the closest integer toward positive infinity.\n\nSyntax: public static decimal Ceiling (decimal d);\nHere d is the decimal whose ceiling value is to be calculated.\n\nReturn Value: It returns the smallest integral value that is greater than or equal to the d parameter. Note that this method returns a Decimal instead of an integral type.\n\nBelow programs illustrate the use of Decimal.Ceiling(Decimal) Method:\n\nExample 1:\n\n `// C# program to demonstrate the ` `// Decimal.Ceiling(Decimal) Method ` `using` `System; ` `using` `System.Globalization; ` ` `  `class` `GFG { ` ` `  `    ``// Main Method ` `    ``public` `static` `void` `Main() ` `    ``{ ` ` `  `        ``// Declaring the decimal variable ` `        ``Decimal a = 4.01m; ` ` `  `        ``// finding the ceiling of  ` `        ``// the Decimal value ` `        ``// using ceiling() method; ` `        ``Decimal value = Decimal.Ceiling(a); ` ` `  `        ``// Display the ceiling ` `        ``Console.WriteLine(``\"Ceiling Value is : {0}\"``, ` `                                            ``value); ` `    ``} ` `} `\n\nOutput:\n\n```Ceiling Value is : 5\n```\n\nExample 2:\n\n `// C# program to demonstrate the ` `// Decimal.Ceiling(Decimal) Method ` `using` `System; ` `using` `System.Globalization; ` ` `  `class` `GFG { ` ` `  `    ``// Main Method ` `    ``public` `static` `void` `Main() ` `    ``{ ` ` `  `        ``// Declaring the decimal variable ` `        ``Decimal a = -5.03m; ` ` `  `        ``// finding the Ceiling of ` `        ``// the Decimal value ` `        ``// using Ceiling() method; ` `        ``Decimal value = Decimal.Ceiling(a); ` ` `  `        ``// Display the Ceiling ` `        ``Console.WriteLine(``\"Ceiling Value is : {0}\"``, ` `                                            ``value); ` `    ``} ` `} `\n\nOutput:\n\n```Ceiling Value is : -5\n```\n\nExample 3:\n\n `// C# program to demonstrate the ` `// Decimal.Ceiling(Decimal) Method ` `using` `System; ` `using` `System.Globalization; ` ` `  `class` `GFG { ` ` `  `    ``// Main Method ` `    ``public` `static` `void` `Main() ` `    ``{ ` ` `  `        ``// Declaring the decimal variable ` `        ``Decimal a = 2.00m; ` ` `  `        ``// finding the Ceiling of ` `        ``// the Decimal value ` `        ``// using Ceiling() method; ` `        ``Decimal value = Decimal.Ceiling(a); ` ` `  `        ``// Display the Ceiling ` `        ``Console.WriteLine(``\"Ceiling Value is : {0}\"``, ` `                                            ``value); ` `    ``} ` `} `\n\nOutput:\n\n```Ceiling Value is : 2\n```\n\nMy Personal Notes arrow_drop_up", null, "Check out this Author's contributed articles.\n\nIf you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.\n\nPlease Improve this article if you find anything incorrect by clicking on the \"Improve Article\" button below.\n\nArticle Tags :\n\n1\n\nPlease write to us at contribute@geeksforgeeks.org to report any issue with the above content." ]

[ null, "https://media.geeksforgeeks.org/auth/profile/yswvujbzywpvc981vsfq", null ]

[ "## 558314\n\n558,314 (five hundred fifty-eight thousand three hundred fourteen) is an even six-digits composite number following 558313 and preceding 558315. In scientific notation, it is written as 5.58314 × 105. The sum of its digits is 26. It has a total of 3 prime factors and 8 positive divisors. There are 262,720 positive integers (up to 558314) that are relatively prime to 558314.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Even\n• Number length 6\n• Sum of Digits 26\n• Digital Root 8\n\n## Name\n\nShort name 558 thousand 314 five hundred fifty-eight thousand three hundred fourteen\n\n## Notation\n\nScientific notation 5.58314 × 105 558.314 × 103\n\n## Prime Factorization of 558314\n\nPrime Factorization 2 × 17 × 16421\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 558314 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 558,314 is 2 × 17 × 16421. Since it has a total of 3 prime factors, 558,314 is a composite number.\n\n## Divisors of 558314\n\n8 divisors\n\n Even divisors 4 4 4 0\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 886788 Sum of all the positive divisors of n s(n) 328474 Sum of the proper positive divisors of n A(n) 110848 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 747.204 Returns the nth root of the product of n divisors H(n) 5.03673 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 558,314 can be divided by 8 positive divisors (out of which 4 are even, and 4 are odd). The sum of these divisors (counting 558,314) is 886,788, the average is 11,084,8.5.\n\n## Other Arithmetic Functions (n = 558314)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 262720 Total number of positive integers not greater than n that are coprime to n λ(n) 65680 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 45839 Total number of primes less than or equal to n r2(n) 16 The number of ways n can be represented as the sum of 2 squares\n\nThere are 262,720 positive integers (less than 558,314) that are coprime with 558,314. And there are approximately 45,839 prime numbers less than or equal to 558,314.\n\n## Divisibility of 558314\n\n m n mod m 2 3 4 5 6 7 8 9 0 2 2 4 2 1 2 8\n\nThe number 558,314 is divisible by 2.\n\n• Deficient\n\n• Polite\n\n• Square Free\n\n• Sphenic\n\n## Base conversion (558314)\n\nBase System Value\n2 Binary 10001000010011101010\n3 Ternary 1001100212022\n4 Quaternary 2020103222\n5 Quinary 120331224\n6 Senary 15544442\n8 Octal 2102352\n10 Decimal 558314\n12 Duodecimal 22b122\n16 Hexadecimal 884ea\n20 Vigesimal 39ffe\n36 Base36 bysq\n\n## Basic calculations (n = 558314)\n\n### Multiplication\n\nn×i\n n×2 1116628 1674942 2233256 2791570\n\n### Division\n\nni\n n⁄2 279157 186105 139578 111663\n\n### Exponentiation\n\nni\n n2 311714522596 174034581968663144 97165943597252194579216 54249106633556261764300401824\n\n### Nth Root\n\ni√n\n 2√n 747.204 82.3429 27.335 14.1051\n\n## 558314 as geometric shapes\n\n### Circle\n\nRadius = n\n Diameter 1.11663e+06 3.50799e+06 9.7928e+11\n\n### Sphere\n\nRadius = n\n Volume 7.28994e+17 3.91712e+12 3.50799e+06\n\n### Square\n\nLength = n\n Perimeter 2.23326e+06 3.11715e+11 789575\n\n### Cube\n\nLength = n\n Surface area 1.87029e+12 1.74035e+17 967028\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 1.67494e+06 1.34976e+11 483514\n\n### Triangular Pyramid\n\nLength = n\n Surface area 5.39905e+11 2.05102e+16 455861\n\n## Cryptographic Hash Functions\n\nmd5 c23803c5e2902b720024b77e0d113504 750823d62fe018ecada26d3c2751fe1abd42bc9f fad69103ad0d6ad0ed466cddabbdb6cce72f114d202bf7f41134639970e77ee5 98e2a55b0704d028cb03115fdc535cf3251a7d9367ef0035892cb48344d276a9272a5ea48f7cf821b890edbbc09505ee1db2ef64f985c9a41681d2889ea7ee00 114a370b645409139b142b207ff72e8f7183882d" ]

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[ "1\nJEE Advanced 2013 Paper 1 Offline\nMCQ (More than One Correct Answer)\n+4\n-1\nLet $$f\\left( x \\right) = x\\sin \\,\\pi x,\\,x > 0.$$ Then for all natural numbers $$n,\\,f'\\left( x \\right)$$ vanishes at\nA\nA unique point in the interval $$\\left( {n,\\,n + {1 \\over 2}} \\right)$$\nB\nA unique point in the interval $$\\left( {n + {1 \\over 2},n + 1} \\right)$$\nC\nA unique point in the interval $$\\left( {n,\\,n + 1} \\right)$$\nD\nTwo points in the interval $$\\left( {n,\\,n + 1} \\right)$$\n2\nIIT-JEE 2012 Paper 1 Offline\nMCQ (More than One Correct Answer)\n+4\n-1\nLet $$\\theta ,\\,\\varphi \\, \\in \\,\\left[ {0,2\\pi } \\right]$$ be such that\n$$2\\cos \\theta \\left( {1 - \\sin \\,\\varphi } \\right) = {\\sin ^2}\\theta \\,\\,\\left( {\\tan {\\theta \\over 2} + \\cot {\\theta \\over 2}} \\right)\\cos \\varphi - 1,\\,\\tan \\left( {2\\pi - \\theta } \\right) > 0$$ and $$- 1 < \\sin \\theta \\, < - {{\\sqrt 3 } \\over 2},$$\n\nthen $$\\varphi$$ cannot satisfy\n\nA\n$$0 < \\varphi < {\\pi \\over 2}$$\nB\n$${\\pi \\over 2} < \\varphi < {{4\\pi } \\over 3}$$\nC\n$${{4\\pi } \\over 3} < \\varphi < {{3\\pi } \\over 2}$$\nD\n$${{3\\pi } \\over 2} < \\varphi < 2\\pi$$\n3\nIIT-JEE 2009 Paper 2 Offline\nMCQ (More than One Correct Answer)\n+4\n-2\nFor $$0 < \\theta < {\\pi \\over 2},$$ the solution (s) of $$\\sum\\limits_{m = 1}^6 {\\cos ec\\,\\left( {\\theta + {{\\left( {m - 1} \\right)\\pi } \\over 4}} \\right)\\,\\cos ec\\,\\left( {\\theta + {{m\\pi } \\over 4}} \\right) = 4\\sqrt 2 }$$\\$ is (are)\nA\n$$\\,{\\pi \\over 4}$$\nB\n$$\\,{\\pi \\over 6 }$$\nC\n$$\\,{\\pi \\over 12}$$\nD\n$$\\,{5\\pi \\over 12}$$\n4\nIIT-JEE 2009 Paper 1 Offline\nMCQ (More than One Correct Answer)\n+4\n-2\nIf $${{{{\\sin }^4}x} \\over 2} + {{{{\\cos }^4}x} \\over 3} = {1 \\over 5},$$ then\nA\n$${\\tan ^2}x = {2 \\over 3}$$\nB\n$${{{{\\sin }^8}x} \\over 8} + {{{{\\cos }^8}x} \\over {27}} = {1 \\over {125}}$$\nC\n$${\\tan ^2}x = {1 \\over 3}$$\nD\n$${{{{\\sin }^8}x} \\over 8} + {{{{\\cos }^8}x} \\over {27}} = {2 \\over {125}}$$\nPhysics\nMechanics\nElectricity\nOptics\nModern Physics\nChemistry\nPhysical Chemistry\nInorganic Chemistry\nOrganic Chemistry\nMathematics\nAlgebra\nTrigonometry\nCoordinate Geometry\nCalculus\nEXAM MAP\nJoint Entrance Examination" ]

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[ "# #0207e9 Color Information\n\nIn a RGB color space, hex #0207e9 is composed of 0.8% red, 2.7% green and 91.4% blue. Whereas in a CMYK color space, it is composed of 99.1% cyan, 97% magenta, 0% yellow and 8.6% black. It has a hue angle of 238.7 degrees, a saturation of 98.3% and a lightness of 46.1%. #0207e9 color hex could be obtained by blending #040eff with #0000d3. Closest websafe color is: #0000ff.\n\n• R 1\n• G 3\n• B 91\nRGB color chart\n• C 99\n• M 97\n• Y 0\n• K 9\nCMYK color chart\n\n#0207e9 color description : Vivid blue.\n\n# #0207e9 Color Conversion\n\nThe hexadecimal color #0207e9 has RGB values of R:2, G:7, B:233 and CMYK values of C:0.99, M:0.97, Y:0, K:0.09. Its decimal value is 133097.\n\nHex triplet RGB Decimal 0207e9 `#0207e9` 2, 7, 233 `rgb(2,7,233)` 0.8, 2.7, 91.4 `rgb(0.8%,2.7%,91.4%)` 99, 97, 0, 9 238.7°, 98.3, 46.1 `hsl(238.7,98.3%,46.1%)` 238.7°, 99.1, 91.4 0000ff `#0000ff`\nCIE-LAB 29.53, 72.78, -100.049 14.806, 6.047, 77.473 0.151, 0.061, 6.047 29.53, 123.721, 306.034 29.53, -8.67, -117.968 24.59, 64.443, -169.582 00000010, 00000111, 11101001\n\n# Color Schemes with #0207e9\n\n• #0207e9\n``#0207e9` `rgb(2,7,233)``\n• #e9e402\n``#e9e402` `rgb(233,228,2)``\nComplementary Color\n• #027be9\n``#027be9` `rgb(2,123,233)``\n• #0207e9\n``#0207e9` `rgb(2,7,233)``\n• #7102e9\n``#7102e9` `rgb(113,2,233)``\nAnalogous Color\n• #7be902\n``#7be902` `rgb(123,233,2)``\n• #0207e9\n``#0207e9` `rgb(2,7,233)``\n• #e97102\n``#e97102` `rgb(233,113,2)``\nSplit Complementary Color\n• #07e902\n``#07e902` `rgb(7,233,2)``\n• #0207e9\n``#0207e9` `rgb(2,7,233)``\n• #e90207\n``#e90207` `rgb(233,2,7)``\n• #02e9e4\n``#02e9e4` `rgb(2,233,228)``\n• #0207e9\n``#0207e9` `rgb(2,7,233)``\n• #e90207\n``#e90207` `rgb(233,2,7)``\n• #e9e402\n``#e9e402` `rgb(233,228,2)``\n• #01059d\n``#01059d` `rgb(1,5,157)``\n• #0205b6\n``#0205b6` `rgb(2,5,182)``\n• #0206d0\n``#0206d0` `rgb(2,6,208)``\n• #0207e9\n``#0207e9` `rgb(2,7,233)``\n• #080dfd\n``#080dfd` `rgb(8,13,253)``\n• #2126fd\n``#2126fd` `rgb(33,38,253)``\n• #3a3efd\n``#3a3efd` `rgb(58,62,253)``\nMonochromatic Color\n\n# Alternatives to #0207e9\n\nBelow, you can see some colors close to #0207e9. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #0241e9\n``#0241e9` `rgb(2,65,233)``\n• #022ee9\n``#022ee9` `rgb(2,46,233)``\n• #021ae9\n``#021ae9` `rgb(2,26,233)``\n• #0207e9\n``#0207e9` `rgb(2,7,233)``\n• #1002e9\n``#1002e9` `rgb(16,2,233)``\n• #2402e9\n``#2402e9` `rgb(36,2,233)``\n• #3702e9\n``#3702e9` `rgb(55,2,233)``\nSimilar Colors\n\n# #0207e9 Preview\n\nThis text has a font color of #0207e9.\n\n``<span style=\"color:#0207e9;\">Text here</span>``\n#0207e9 background color\n\nThis paragraph has a background color of #0207e9.\n\n``<p style=\"background-color:#0207e9;\">Content here</p>``\n#0207e9 border color\n\nThis element has a border color of #0207e9.\n\n``<div style=\"border:1px solid #0207e9;\">Content here</div>``\nCSS codes\n``.text {color:#0207e9;}``\n``.background {background-color:#0207e9;}``\n``.border {border:1px solid #0207e9;}``\n\n# Shades and Tints of #0207e9\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, #000113 is the darkest color, while #ffffff is the lightest one.\n\n• #000113\n``#000113` `rgb(0,1,19)``\n• #000127\n``#000127` `rgb(0,1,39)``\n• #00023a\n``#00023a` `rgb(0,2,58)``\n• #01024d\n``#01024d` `rgb(1,2,77)``\n• #010361\n``#010361` `rgb(1,3,97)``\n• #010374\n``#010374` `rgb(1,3,116)``\n• #010488\n``#010488` `rgb(1,4,136)``\n• #01059b\n``#01059b` `rgb(1,5,155)``\n• #0105af\n``#0105af` `rgb(1,5,175)``\n• #0206c2\n``#0206c2` `rgb(2,6,194)``\n• #0206d6\n``#0206d6` `rgb(2,6,214)``\n• #0207e9\n``#0207e9` `rgb(2,7,233)``\n• #0208fc\n``#0208fc` `rgb(2,8,252)``\n• #151afd\n``#151afd` `rgb(21,26,253)``\n• #292dfd\n``#292dfd` `rgb(41,45,253)``\n• #3c40fd\n``#3c40fd` `rgb(60,64,253)``\n• #5053fd\n``#5053fd` `rgb(80,83,253)``\n• #6366fe\n``#6366fe` `rgb(99,102,254)``\n• #7679fe\n``#7679fe` `rgb(118,121,254)``\n• #8a8cfe\n``#8a8cfe` `rgb(138,140,254)``\n• #9d9ffe\n``#9d9ffe` `rgb(157,159,254)``\n• #b1b3fe\n``#b1b3fe` `rgb(177,179,254)``\n• #c4c6fe\n``#c4c6fe` `rgb(196,198,254)``\n• #d8d9ff\n``#d8d9ff` `rgb(216,217,255)``\n• #ebecff\n``#ebecff` `rgb(235,236,255)``\n• #ffffff\n``#ffffff` `rgb(255,255,255)``\nTint Color Variation\n\n# Tones of #0207e9\n\nA tone is produced by adding gray to any pure hue. In this case, #6e6f7d is the less saturated color, while #0207e9 is the most saturated one.\n\n• #6e6f7d\n``#6e6f7d` `rgb(110,111,125)``\n• #656686\n``#656686` `rgb(101,102,134)``\n• #5c5d8f\n``#5c5d8f` `rgb(92,93,143)``\n• #535598\n``#535598` `rgb(83,85,152)``\n• #4a4ca1\n``#4a4ca1` `rgb(74,76,161)``\n• #4144aa\n``#4144aa` `rgb(65,68,170)``\n• #383bb3\n``#383bb3` `rgb(56,59,179)``\n• #2f32bc\n``#2f32bc` `rgb(47,50,188)``\n• #262ac5\n``#262ac5` `rgb(38,42,197)``\n• #1d21ce\n``#1d21ce` `rgb(29,33,206)``\n• #1418d7\n``#1418d7` `rgb(20,24,215)``\n• #0b10e0\n``#0b10e0` `rgb(11,16,224)``\n• #0207e9\n``#0207e9` `rgb(2,7,233)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #0207e9 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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[ "Community Profile", null, "Rodney Tan\n\nUCSI University\n\nLast seen: 27 jours ago Active since 2015\n\nMy research area are Renewable Energy, Energy Storage, Power Electronics and Power System\n\nStatistics\n\nAll\n•", null, "•", null, "•", null, "•", null, "•", null, "•", null, "•", null, "•", null, "•", null, "•", null, "•", null, "•", null, "Content Feed\n\nView by\n\nSolved\n\nSide of an equilateral triangle\nIf an equilateral triangle has area A, then what is the length of each of its sides, x? <<http://upload.wikimedia.org/wikipe...\n\nenviron 5 ans ago\n\nSolved\n\nFind the sum of the elements in the \"second\" diagonal\nFind the sum of the elements in the diagonal that starts at the top-right corner and ends at the bottom-left corner.\n\nenviron 5 ans ago\n\nSolved\n\nArea of an equilateral triangle\nCalculate the area of an equilateral triangle of side x. <<http://upload.wikimedia.org/wikipedia/commons/e/e0/Equilateral-tr...\n\nenviron 5 ans ago\n\nSolved\n\nSorted highest to lowest?\nReturn 1 if the input is sorted from highest to lowest, 0 if not. Example: 1:7 -> 0 [7 5 2] -> 1\n\nenviron 5 ans ago\n\nSolved\n\nFind all elements less than 0 or greater than 10 and replace them with NaN\nGiven an input vector x, find all elements of x less than 0 or greater than 10 and replace them with NaN. Example: Input ...\n\nenviron 5 ans ago\n\nSolved\n\nFind x in provided equation!\nx^2-2*x+1=0 This polynomial can be expressed by using each term's coefficients, such as [1 -2 1]. Using the polynomial ...\n\nenviron 5 ans ago\n\nSolved\n\nBack to Basics - New Data Type in R2016b - convert a char to a string\nConvert an char array into a string.\n\nenviron 5 ans ago\n\nSolved\n\nInitialize a Natural Number matrix.\nGiven length of matrix initialize a matrix consisting of natural numbers from 1 to n: n = 10; x = [ 1 2 3 4 5 6 7 8 9 10]; ...\n\nenviron 5 ans ago\n\nSolved\n\nCalculate cross product\nMake function for cross product a=[1 3 2]; b=[2 4 1]; y=function(a,b) y=[-5 3 -2]\n\nenviron 5 ans ago\n\nSolved\n\nConvert decimal to hex as shown in test cases\nConvert decimal to hex as shown in test cases.\n\nenviron 5 ans ago\n\nSolved\n\nGenerate vector according to sign of vector\nGenerate vector according to sign of vector Example: If A=[-2 0 5] then output must be[-1 0 1] meaning that for negative n...\n\nenviron 5 ans ago\n\nSolved\n\nComplex transpose\nCalculate the transpose of a matrix having complex numbers as its elements without changing the signs of the imaginary part. ...\n\nenviron 5 ans ago\n\nSolved\n\nSort the vector with the given index\nGiven x = [1 2 4 8 17] and t = [1 3 2 5 4] then y = [1 4 2 17 8].\n\nenviron 5 ans ago\n\nSolved\n\n03 - Matrix Variables 2\nMake the following variable: <<http://samle.dk/STTBDP/Assignment1_3b.png>> A 9x9 matrix of zeros, but with the following v...\n\nenviron 5 ans ago\n\nSolved\n\nBack to basics 7 - Equal NaNs\nCovering some basic topics I haven't seen elsewhere on Cody. Given 2 input variables, output true if they are equal, false ot...\n\nenviron 5 ans ago\n\nSolved\n\nFind the largest value in the 3D matrix\nGiven a 3D matrix, A, find the largest value. E.g. >> A = 1:9; >> A=reshape(A,[3 1 3]); >> islargest(A) ans = 9\n\nenviron 5 ans ago\n\nSolved\n\nColumn Removal\nRemove the nth column from input matrix A and return the resulting matrix in output B. So if A = [1 2 3; 4 5 6]; and ...\n\nenviron 5 ans ago\n\nSolved\n\nConvert array of decimal numbers into hexadecimal numbers array.\nConvert array of decimal numbers into hexadecimal numbers array. Example x =[ 32 33 34 35 36 37 38 ...\n\nenviron 5 ans ago\n\nSolved\n\nSlope of the line passing through the point [x1 y1] and [x2 y2]\nDetermin the slope of Line passing through the points a=[x1 y1] and b=[x2 y2]\n\nenviron 5 ans ago\n\nSolved\n\nFind out DC Gain of transfer function\nFind out DC Gain of transfer function Example, G(s)=5/ (s+2), then DC gain is 2.5 (put s=0)\n\nenviron 5 ans ago\n\nSolved\n\nHow to permute given 3d matrix?\nA(:,:,1)=[1 3] A(:,:,2)=[2 2] A(:,:,3)=[4 3] Change rows to columns and columns to rows, similar to transpose. Resul...\n\nenviron 5 ans ago\n\nSolved\n\nConduct inner product using given matrix\na=[1 2 3]; b=[3 4 5]; y=function(a,b) Output y should be 26\n\nenviron 5 ans ago\n\nSolved\n\nSuper Basic - 3 + true equals ??\nSolve the title\n\nenviron 5 ans ago\n\nSolved\n\nMake random permutation\nMake random permutation that consist of random number from 1 to n.\n\nenviron 5 ans ago\n\nSolved\n\nhow to round off a given number containing decimals?\nhow to round off a given number containing decimals?\n\nenviron 5 ans ago\n\nSolved\n\nAreas\nGiven certain dimensions determine the area of that shape. If given only one value assume its the radius. Use round(x) to round ...\n\nenviron 5 ans ago\n\nSolved\n\nLog of a number\nWrite a script that will give the log of x as output.\n\nenviron 5 ans ago\n\nSolved\n\nMATLAB Basic: rounding IV\n\nenviron 5 ans ago\n\nSolved\n\nMATLAB Basic: rounding III" ]

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[ "## Visual Basic for Access Report- form load no unique values in text boxes\n\nHello-\nNew to coding and having an issue. I'm using Visual Basic in Access reports to calculate a value, but when I form load it's calculating one value and posting that in every text box. How do I get it to run the calculation for each box? I'm pulling from several tables (total freight and LTL freight) and basing the calculation on the carrier ID. See code below:\n\nPrivate Sub Report_Click()\n\nIf [Carrier Id] = 100001 Then\nText13 = ([Total Freight] * 2) + [LTL Freight]\nElseIf [Carrier Id] = 100009 Then Text13 = ([Total Freight] * 2) + [LTL Freight]\nElseIf [Carrier Id] = 106932 Then Text13 = ([Total Freight] * 2) + [LTL Freight]\nElseIf [Carrier Id] = 107078 Then Text13 = ([Total Freight] * 2) + [LTL Freight]\nElseIf [Carrier Id] = 106973 Then Text13 = ([Total Freight] * 2) + [LTL Freight]\nElseIf [Carrier Id] = 106974 Then Text13 = ([Total Freight] * 2) + [LTL Freight]\nElseIf [Carrier Id] = 106975 Then Text13 = ([Total Freight] * 2) + [LTL Freight]\nElseIf [Carrier Id] = 106976 Then Text13 = ([Total Freight] * 2) + [LTL Freight]\nElseIf [Carrier Id] = 107121 Then Text13 = ([Total Freight] * 2) + [LTL Freight]\nElseIf [Carrier Id] = 106977 Then Text13 = ([Total Freight] * 2) + [LTL Freight]\nElseIf [Carrier Id] = 106840 Then Text13 = ([Total Freight] * 2) + [LTL Freight]\nElseIf [Carrier Id] = 100008 Then Text13 = ([Total Freight] * 2) + [LTL Freight]\nElseIf [Carrier Id] = 106895 Then Text13 = ([Total Freight] * 2) + [LTL Freight]\nElseIf [Carrier Id] = 107145 Then Text13 = ([Total Freight] * 2) + [LTL Freight]\nElseIf [Carrier Id] = 100006 Then Text13 = ([Total Freight] * 2) + [LTL Freight]\nElseIf [Carrier Id] = 100003 Then Text13 = ([Total Freight] * 2) + [LTL Freight]\nElseIf [Carrier Id] = 100002 Then Text13 = ([Total Freight] * 2) + [LTL Freight]\nElseIf [Carrier Id] = 100012 Then Text13 = ([Total Freight] * 2) + [LTL Freight]\nElseIf [Carrier Id] = 100010 Then Text13 = ([Total Freight] * 2) + [LTL Freight]\nElseIf [Carrier Id] = 100013 Then Text13 = ([Total Freight] * 2) + [LTL Freight]\nElseIf [Carrier Id] = 100015 Then Text13 = ([Total Freight] * 2) + [LTL Freight]\nElseIf [Carrier Id] = 106982 Then Text13 = ([Total Freight] * 2) + [LTL Freight]\nElseIf [Carrier Id] = 106983 Then Text13 = ([Total Freight] * 2) + [LTL Freight]\nElseIf [Carrier Id] = 106984 Then Text13 = ([Total Freight] * 2) + [LTL Freight]\nElseIf [Carrier Id] = 106985 Then Text13 = ([Total Freight] * 2) + [LTL Freight]\nElseIf [Carrier Id] = 106986 Then Text13 = ([Total Freight] * 2) + [LTL Freight]\nElseIf [Carrier Id] = 106987 Then Text13 = ([Total Freight] * 2) + [LTL Freight]\nElseIf [Carrier Id] = 106990 Then Text13 = ([Total Freight] * 2) + [LTL Freight]\nElseIf [Carrier Id] = 106991 Then Text13 = ([Total Freight] * 2) + [LTL Freight]\nElseIf [Carrier Id] = 106992 Then Text13 = ([Total Freight] * 2) + [LTL Freight]\nElseIf [Carrier Id] = 107688 Then Text13 = ([Total Freight] * 2) + [LTL Freight]\nElseIf [Carrier Id] = 106994 Then Text13 = ([Total Freight] * 2) + [LTL Freight]\nElseIf [Carrier Id] = 100014 Then Text13 = ([Total Freight] * 2) + [LTL Freight]\nElse:\nEnd If\nEnd Sub\n\nWhat am I doing wrong? Any help would be greatly appreciated.\n\nThank you," ]

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